Ergebnisse der Mathematik und ihrer Grenzgebiete 3. Folge
A Series of Modern Surveys in Mathematics
Editorial Board M...

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Ergebnisse der Mathematik und ihrer Grenzgebiete 3. Folge

A Series of Modern Surveys in Mathematics

Editorial Board M. Gromov, Bures-sur-Yvette J. Jost, Leipzig J. Kollár, Princeton H. W. Lenstra, Jr., Leiden J. Tits, Paris D. B. Zagier, Bonn G. Ziegler, Berlin Managing Editor R. Remmert, Münster

Volume 11

Michael D. Fried • Moshe Jarden

Field Arithmetic Third Edition Revised by Moshe Jarden

Michael D. Fried Department of Mathematics Montana State University – Billings Billings MT 59101 USA [email protected]

ISBN 978-3-540-77269-9

Moshe Jarden School of Mathematics Tel Aviv University Ramat Aviv, Tel Aviv 69978 Israel [email protected]

e-ISBN 978-3-540-77270-5

DOI 10.1007/978-3-540-77270-5 Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics ISSN 0071-1136 Library of Congress Control Number: 2008924174 Mathematics Subject Classiﬁcation (2000): 12E30 c 2008 Springer-Verlag Berlin Heidelberg This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microﬁlm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover design: WMXDesign GmbH, Heidelberg Printed on acid-free paper 9 8 7 6 5 4 3 2 1 springer.com

To those precious colleagues who can appreciate the goals of and connections to other areas. To those who acknowledge the depth of what we already know from the absorbed contribution of previous generations before we address our papers. To those who can transcend the hubris of today’s mathematical community.

Table of Contents Chapter 1. Infinite Galois Theory and Profinite Groups 1.1 Inverse Limits . . . . . . . . . . . . . . . 1.2 Profinite Groups . . . . . . . . . . . . . . 1.3 Infinite Galois Theory . . . . . . . . . . . . 1.4 The p-adic Integers and the Pr¨ ufer Group . . . 1.5 The Absolute Galois Group of a Finite Field . . Exercises . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . .

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Chapter 2. Valuations and Linear Disjointness . 2.1 Valuations, Places, and Valuation Rings . . 2.2 Discrete Valuations . . . . . . . . . . . 2.3 Extensions of Valuations and Places. . . . 2.4 Integral Extensions and Dedekind Domains 2.5 Linear Disjointness of Fields . . . . . . . 2.6 Separable, Regular, and Primary Extensions 2.7 The Imperfect Degree of a Field . . . . . 2.8 Derivatives . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .

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19 19 21 24 30 34 38 44 48 50 51

Chapter 3. Algebraic Function Fields of One Variable . . 3.1 Function Fields of One Variable . . . . . . . . . 3.2 The Riemann-Roch Theorem . . . . . . . . . . 3.3 Holomorphy Rings . . . . . . . . . . . . . . . 3.4 Extensions of Function Fields . . . . . . . . . . 3.5 Completions . . . . . . . . . . . . . . . . . . 3.6 The Different . . . . . . . . . . . . . . . . . 3.7 Hyperelliptic Fields . . . . . . . . . . . . . . . 3.8 Hyperelliptic Fields with a Rational quadratic Subfield Exercises . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . .

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Chapter 4. The Riemann Hypothesis for Function Fields . 4.1 Class Numbers . . . . . . . . . . . . . . . . . 4.2 Zeta Functions . . . . . . . . . . . . . . . . . 4.3 Zeta Functions under Constant Field Extensions . . 4.4 The Functional Equation . . . . . . . . . . . . 4.5 The Riemann Hypothesis and Degree 1 Prime Divisors 4.6 Reduction Steps . . . . . . . . . . . . . . . . 4.7 An Upper Bound . . . . . . . . . . . . . . . . 4.8 A Lower Bound . . . . . . . . . . . . . . . .

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77 77 79 81 82 84 86 87 89

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Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 5. Plane Curves . . . . . . . 5.1 Affine and Projective Plane Curves 5.2 Points and prime divisors . . . . 5.3 The Genus of a Plane Curve . . . 5.4 Points on a Curve over a Finite Field Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .

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Chapter 6. The Chebotarev Density Theorem 6.1 Decomposition Groups . . . . . . . 6.2 The Artin Symbol over Global Fields . 6.3 Dirichlet Density . . . . . . . . . . 6.4 Function Fields . . . . . . . . . . 6.5 Number Fields . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . .

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107 107 111 113 115 121 129 130

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132 132 134 135 137 138 139 141 145 147 147 148

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149 149 152 154 156 157 159 160 161 162

Chapter 7. Ultraproducts . . . . . . . 7.1 First Order Predicate Calculus . . 7.2 Structures . . . . . . . . . . . 7.3 Models . . . . . . . . . . . . 7.4 Elementary Substructures . . . . 7.5 Ultrafilters . . . . . . . . . . 7.6 Regular Ultrafilters . . . . . . . 7.7 Ultraproducts . . . . . . . . . 7.8 Regular Ultraproducts . . . . . 7.9 Nonprincipal Ultraproducts of Finite Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .

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91 93

Chapter 8. Decision Procedures . . . . . . . . 8.1 Deduction Theory . . . . . . . . . . . 8.2 G¨odel’s Completeness Theorem . . . . . 8.3 Primitive Recursive Functions . . . . . . 8.4 Primitive Recursive Relations . . . . . . 8.5 Recursive Functions . . . . . . . . . . 8.6 Recursive and Primitive Recursive Procedures 8.7 A Reduction Step in Decidability Procedures Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .

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Chapter 9. Algebraically Closed Fields . 9.1 Elimination of Quantifiers . . . . 9.2 A Quantifiers Elimination Procedure 9.3 Effectiveness . . . . . . . . . . 9.4 Applications . . . . . . . . . . Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .

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163 163 165 168 169 170 170

Chapter 10. Elements of Algebraic Geometry . . 10.1 Algebraic Sets . . . . . . . . . . . . 10.2 Varieties . . . . . . . . . . . . . . . 10.3 Substitutions in Irreducible Polynomials . 10.4 Rational Maps . . . . . . . . . . . . 10.5 Hyperplane Sections . . . . . . . . . . 10.6 Descent . . . . . . . . . . . . . . . 10.7 Projective Varieties . . . . . . . . . . 10.8 About the Language of Algebraic Geometry Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .

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172 172 175 176 178 180 182 185 187 190 191

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192 192 193 199 201 203 207 211 217 218

Chapter 11. Pseudo Algebraically Closed Fields . . 11.1 PAC Fields . . . . . . . . . . . . . . . 11.2 Reduction to Plane Curves . . . . . . . . 11.3 The PAC Property is an Elementary Statement 11.4 PAC Fields of Positive Characteristic . . . 11.5 PAC Fields with Valuations . . . . . . . . 11.6 The Absolute Galois Group of a PAC Field . . . . . 11.7 A non-PAC Field K with Kins PAC Exercises . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . .

Chapter 12. Hilbertian Fields . . . . . . . . . . . . 12.1 Hilbert Sets and Reduction Lemmas . . . . . . 12.2 Hilbert Sets under Separable Algebraic Extensions 12.3 Purely Inseparable Extensions . . . . . . . . 12.4 Imperfect fields . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . .

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219 219 223 224 228 229 230

Chapter 13. The Classical Hilbertian Fields 13.1 Further Reduction . . . . . . . . 13.2 Function Fields over Infinite Fields 13.3 Global Fields . . . . . . . . . . 13.4 Hilbertian Rings . . . . . . . . 13.5 Hilbertianity via Coverings . . . .

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231 231 236 237 241 244

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13.6 Non-Hilbertian g-Hilbertian Fields 13.7 Twisted Wreath Products . . . 13.8 The Diamond Theorem . . . . 13.9 Weissauer’s Theorem . . . . . Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .

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248 252 258 262 264 266

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267 267 268 270 272 274 275 276

Chapter 15. Nonstandard Approach to Hilbert’s Irreducibility Theorem 15.1 Criteria for Hilbertianity . . . . . . . . . . . . 15.2 Arithmetical Primes Versus Functional Primes . . 15.3 Fields with the Product Formula . . . . . . . . 15.4 Generalized Krull Domains . . . . . . . . . . . 15.5 Examples . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . .

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277 277 279 281 283 286 289 290

Chapter 16. Galois Groups over Hilbertian Fields . . . . . . 16.1 Galois Groups of Polynomials . . . . . . . . . . . . 16.2 Stable Polynomials . . . . . . . . . . . . . . . . 16.3 Regular Realization of Finite Abelian Groups . . . . . 16.4 Split Embedding Problems with Abelian Kernels . . . 16.5 Embedding Quadratic Extensions in Z/2n Z-extensions . 16.6 Zp -Extensions of Hilbertian Fields . . . . . . . . . . 16.7 Symmetric and Alternating Groups over Hilbertian Fields 16.8 GAR-Realizations . . . . . . . . . . . . . . . . . 16.9 Embedding Problems over Hilbertian Fields . . . . . 16.10 Finitely Generated Profinite Groups . . . . . . . . 16.11 Abelian Extensions of Hilbertian Fields . . . . . . . 16.12 Regularity of Finite Groups over Complete Discrete Valued Fields . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .

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291 291 294 298 302 306 308 315 321 325 328 332

Chapter 14. Nonstandard Structures . 14.1 Higher Order Predicate Calculus 14.2 Enlargements . . . . . . . . 14.3 Concurrent Relations . . . . 14.4 The Existence of Enlargements 14.5 Examples . . . . . . . . . . Exercises . . . . . . . . . . . . Notes . . . . . . . . . . . . .

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Chapter 17. Free Profinite Groups . . . . . . . . . . . . . . . 338 17.1 The Rank of a Profinite Group . . . . . . . . . . . . . 338

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17.2 Profinite Completions of Groups . . . 17.3 Formations of Finite Groups . . . . . 17.4 Free pro-C Groups . . . . . . . . . . 17.5 Subgroups of Free Discrete Groups . . 17.6 Open Subgroups of Free Profinite Groups 17.7 An Embedding Property . . . . . . . Exercises . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . .

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340 344 346 350 358 360 361 362

Chapter 18. The Haar Measure . . . . . . . . . . 18.1 The Haar Measure of a Profinite Group . . . 18.2 Existence of the Haar Measure . . . . . . . 18.3 Independence . . . . . . . . . . . . . . . 18.4 Cartesian Product of Haar Measures . . . . . 18.5 The Haar Measure of the Absolute Galois Group 18.6 The PAC Nullstellensatz . . . . . . . . . . 18.7 The Bottom Theorem . . . . . . . . . . . 18.8 PAC Fields over Uncountable Hilbertian Fields 18.9 On the Stability of Fields . . . . . . . . . . 18.10 PAC Galois Extensions of Hilbertian Fields . 18.11 Algebraic Groups . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . .

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363 363 366 370 376 378 380 382 386 390 394 397 400 401

Chapter 19. Effective Field Theory and Algebraic Geometry . 19.1 Presented Rings and Fields . . . . . . . . . . . . . 19.2 Extensions of Presented Fields . . . . . . . . . . . 19.3 Galois Extensions of Presented Fields . . . . . . . . 19.4 The Algebraic and Separable Closures of Presented Fields 19.5 Constructive Algebraic Geometry . . . . . . . . . . 19.6 Presented Rings and Constructible Sets . . . . . . . 19.7 Basic Normal Stratification . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .

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403 403 406 411 412 413 422 425 427 428

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Chapter 20. The Elementary Theory of e-Free PAC Fields 20.1 ℵ1 -Saturated PAC Fields . . . . . . . . . . . . 20.2 The Elementary Equivalence Theorem of ℵ1 -Saturated PAC Fields . . . 20.3 Elementary Equivalence of PAC Fields . . . . . . 20.4 On e-Free PAC Fields . . . . . . . . . . . . . 20.5 The Elementary Theory of Perfect e-Free PAC Fields 20.6 The Probable Truth of a Sentence . . . . . . . . 20.7 Change of Base Field . . . . . . . . . . . . . . . . . . . . . . . . 20.8 The Fields Ks (σ1 , . . . , σe )

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430 433 436 438 440 442 444

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20.9 The Transfer Theorem . . . . . . . . 20.10 The Elementary Theory of Finite Fields Exercises . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . .

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446 448 451 453

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454 454 455 460 462 463 467 472 479 489 493 495

Chapter 22. Projective Groups and Frattini Covers 22.1 The Frattini Groups of a Profinite Group . 22.2 Cartesian Squares . . . . . . . . . . . 22.3 On C-Projective Groups . . . . . . . . 22.4 Projective Groups . . . . . . . . . . . 22.5 Frattini Covers . . . . . . . . . . . . 22.6 The Universal Frattini Cover . . . . . . 22.7 Projective Pro-p-Groups . . . . . . . . 22.8 Supernatural Numbers . . . . . . . . . 22.9 The Sylow Theorems . . . . . . . . . 22.10 On Complements of Normal Subgroups . 22.11 The Universal Frattini p-Cover . . . . . 22.12 Examples of Universal Frattini p-Covers . . . 22.13 The Special Linear Group SL(2, Zp ) 22.14 The General Linear Group GL(2, Zp ) . . Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .

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497 497 499 502 506 508 513 515 520 522 524 528 532 534 537 539 542

Chapter 23. PAC Fields and Projective Absolute Galois Groups 23.1 Projective Groups as Absolute Galois Groups . . . . . 23.2 Countably Generated Projective Groups . . . . . . . 23.3 Perfect PAC Fields of Bounded Corank . . . . . . . 23.4 Basic Elementary Statements . . . . . . . . . . . . 23.5 Reduction Steps . . . . . . . . . . . . . . . . . . 23.6 Application of Ultraproducts . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .

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544 544 546 549 550 554 558 561 561

Chapter 21. Problems of Arithmetical Geometry 21.1 The Decomposition-Intersection Procedure 21.2 Ci -Fields and Weakly Ci -Fields . . . . . . . 21.3 Perfect PAC Fields which are Ci 21.4 The Existential Theory of PAC Fields . 21.5 Kronecker Classes of Number Fields . . 21.6 Davenport’s Problem . . . . . . . . 21.7 On permutation Groups . . . . . . . 21.8 Schur’s Conjecture . . . . . . . . . . 21.9 Generalized Carlitz’s Conjecture . . . . Exercises . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . .

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Chapter 24. Frobenius Fields . . . . . . . . . . . . . . 24.1 The Field Crossing Argument . . . . . . . . . . . 24.2 The Beckmann-Black Problem . . . . . . . . . . 24.3 The Embedding Property and Maximal Frattini Covers 24.4 The Smallest Embedding Cover of a Profinite Group . 24.5 A Decision Procedure . . . . . . . . . . . . . . 24.6 Examples . . . . . . . . . . . . . . . . . . . . 24.7 Non-projective Smallest Embedding Cover . . . . . 24.8 A Theorem of Iwasawa . . . . . . . . . . . . . . 24.9 Free Profinite Groups of at most Countable Rank . . 24.10 Application of the Nielsen-Schreier Formula . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . .

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Chapter 25. Free Profinite Groups of Infinite Rank . . 25.1 Characterization of Free Profinite Groups by Embedding Problems 25.2 Applications of Theorem 25.1.7 . . . . . . . . 25.3 The Pro-C Completion of a Free Discrete Group . 25.4 The Group Theoretic Diamond Theorem . . . . 25.5 The Melnikov Group of a Profinite Group . . . 25.6 Homogeneous Pro-C Groups . . . . . . . . . 25.7 The S-rank of Closed Normal Subgroups . . . . 25.8 Closed Normal Subgroups with a Basis Element . 25.9 Accessible Subgroups . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . .

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595 601 604 606 613 615 620 623 625 633

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635 635 640 642

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Chapter 26. Random Elements in Free Profinite Groups . 26.1 Random Elements in a Free Profinite Group . . . 26.2 Random Elements in Free pro-p Groups . . . . . ˆn . . . . . . . . . . . . . 26.3 Random e-tuples in Z 26.4 On the Index of Normal Subgroups Generated by Random Elements . 26.5 Freeness of Normal Subgroups Generated by Random Elements Notes . . . . . . . . . . . . . . . . . . . . . . Chapter 27. Omega-Free PAC Fields . . . . . . . . . 27.1 Model Companions . . . . . . . . . . . . . 27.2 The Model Companion in an Augmented Theory of 27.3 New Non-Classical Hilbertian Fields . . . . . . 27.4 An abundance of ω-Free PAC Fields . . . . . . Notes . . . . . . . . . . . . . . . . . . . . .

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Chapter 28. Undecidability . . . . . . . . . . . . . . 28.1 Turing Machines . . . . . . . . . . . . . . . 28.2 Computation of Functions by Turing Machines . . 28.3 Recursive Inseparability of Sets of Turing Machines 28.4 The Predicate Calculus . . . . . . . . . . . . 28.5 Undecidability in the Theory of Graphs . . . . . 28.6 Assigning Graphs to Profinite Groups . . . . . . 28.7 The Graph Conditions . . . . . . . . . . . . . 28.8 Assigning Profinite Groups to Graphs . . . . . . 28.9 Assigning Fields to Graphs . . . . . . . . . . . 28.10 Interpretation of the Theory of Graphs in the Theory of Fields Exercises . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . .

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671 671 672 676 679 682 687 688 690 694

. . . 694 . . . . 697 . . . . 697

Chapter 29. Algebraically Closed Fields with Distinguished Automorphisms 29.1 The Base Field K . . . . . . . . . . . . . . . . . 29.2 Coding in PAC Fields with Monadic Quantifiers . . . . ˜ σ1 , . . . , σe i’s . . . . . . 29.3 The Theory of Almost all hK, 29.4 The Probability of Truth Sentences . . . . . . . . .

. . . . .

. . . . .

. 698 698 700 704 706

Chapter 30. Galois Stratification . . . . 30.1 The Artin Symbol . . . . . . . . 30.2 Conjugacy Domains under Projection 30.3 Normal Stratification . . . . . . 30.4 Elimination of One Variable . . . 30.5 The Complete Elimination Procedure 30.6 Model-Theoretic Applications . . . 30.7 A Limit of Theories . . . . . . . Exercises . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

708 708 710 715 717 720 722 725 726 729

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . .

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. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

Chapter 31. Galois Stratification over Finite Fields . 31.1 The Elementary Theory of Frobenius Fields . 31.2 The Elementary Theory of Finite Fields . . . 31.3 Near Rationality of the Zeta Function of a Galois Exercises . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . Formula . . . . . . . .

. . 730 . . 730 . . 735 . 739 . . 748 . . 750

Chapter 32. Problems of Field Arithmetic . . . . 32.1 Open Problems of the First Edition . . . . 32.2 Open Problems of the Second Edition . . . 32.3 Open problems . . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

751 751 754 758

Table of Contents

References Index

xv

. . . . . . . . . . . . . . . . . . . . . . . . . 761

. . . . . . . . . . . . . . . . . . . . . . . . . . . 780

Introduction to the Third Edition The third edition of “Field Arithmetic” improves the second edition in two ways. First it removes many typos and mathematical inaccuracies that occur in the second edition. In particular, it fills out a big gap in the References of the second edition, where unfortunately all references between “Gilmore and Robinson” and “Kantor and Lubotzky” are missing. Secondly, the third edition reports on five open problems of the second edition that were solved since that edition appeared in 2005. J´anos Koll´ar solved Problem 2 by proving that if each projective plane curve defined over a field K has a K-rational point, then K is PAC. J´anos Koll´ar also solved Problem 3 and proved that if K is a PAC field, ˜ and V is a variety defined over K, then V (K) is w is a valuation of K, ˜ w-dense in V (K). J´anos Koll´ar partially settled Problem 21. He proved that every PAC field of characteristic 0 is C1 . Problem 31 was affirmatively solved by Lior Bary-Soroker by establishing an analog of the diamond theorem for the finitely generated non-Abelian free profinite groups. Finally, Eric Rosen suggested to reorganize Corollary 28.5.3 of the second edition that led to an affirmative solution of Problem 33. Unfortunately, a full account of the first four solutions is out of the scope of the present volume. Much of the improvment made in the present edition is due to Arno Fehm and Dan Haran. I am really indebted to them for their contribution. Tel Aviv, Autumn 2007

Moshe Jarden

Introduction to the Second Edition The first edition of “Field Arithmetic” appeared in 1986. At the end of that edition we gave a list of twenty-two open problems. It is remarkable that since then fifteen of them were partially or fully solved. Parallel to this, Field Arithmetic has developed in many directions establishing itself as an independent branch of Algebra and Number Theory. Some of these developments have been documented in books. We mention here “Groups as Galois groups” [V¨ olklein] on consequences of the Riemann existence theorem, “Inverse Galois Groups” [Malle-Matzat] with a comprehensive report on finite Galois groups over number fields, “Profinite groups” [Ribes-Zalesskii] including the cohomology of profinite groups, “Analytic prop Groups” [Dixon-du.Sautoy-Mann-Segal] on closed subgroups of GL(n, Zp ), “Subgroup Growth” [Lubotzky-Segal] on counting the number of subgroups of finitely generating groups, and “Multi-Valued Fields” [Ershov7] on the model theory of fields with several valuations. This led to an official recognition of Field Arithmetic by the Mathematical Reviews in the form of MSC number 12E30.

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The extent which Field Arithmetic has reached makes it impossible for us to report in one extended volume about all exciting results which have been achieved. We have therefore made several choices which best suit the spirit of this book but do not extend beyond the scope of one volume. The new results and additional topics have made it necessary to reorganize and to enlarge the sections dealing with background material. Of course, we took the opportunity afforded by editing a second edition to correct flaws and mistakes which occurred in the first edition and to add more details to proofs wherever it seemed useful. We list the major changes and additions we made in the book: Chapter 2 has been reorganized. Sections 2.5–2.9 of the first edition, which survey the theory of algebraic function fields of one variable, were moved to Chapter 3. Sections 2.5–2.8 dealing with linear disjointness, regular extensions, and separability appeared in the first edition as sections 9.1– 9.3. A nice application of linear disjointness is Leptin’s construction (which preceded that of Warehouse) of a Galois group isomorphic to a given profinite group (Proposition 2.6.12). In addition to the introductory material about the theory of algebraic function fields of one variable, Chapter 3 now includes a proof of the RiemannHurwitz formula and a discussion of hyperelliptic curves. The proof of Theorem 4.9 of the first edition, estimating the number of zeros of an absolutely irreducible polynomial over a finite field, had a flaw. This has been fixed in the proof of Theorem 5.4.1. Likewise, the inequality given by [Fried-Jarden3, Prop. 5.16] is inaccurate. This inaccuracy is fixed in Proposition 6.4.8. We find it more convenient to use the language of algebraic sets as introduced in [Weil5] for model theoretic applications. Section 10.8 translates the basic concepts of that language to the now more commonly used language of schemes. Theorem 10.14 of the first edition (due to Frey-Prestel) says that the Henselian hull of a PAC field K is Ks . Proposition 11.5.3 (due to Prestel) ˜ for every valuation strengthens this theorem. It says that K is w-dense in K ˜ w of K. What we called “a separably Hilbertian field” in the first edition, is now called “a Hilbertian field” (Section 12.1). This agrees with the common usage and seems more appropriate for applications. Section 13.5 gives an alternative definition of Hilbertianity via coverings leading to the notion of “g-Hilbertianity”. This sets the stage for a generalization of a theorem of Zannier: Every global field has an infinite normal extension N which is g-Hilbertian but not Hilbertian (Theorem 13.6.2). Moreover, there is a unique factorization subring R of N with infinitely many irreducible elements (Example 15.5.8). This answers negatively Problems 14.20 and 14.21 of the first edition. Chapter 13 includes now one of the major results of Field Arithmetic which we call “Haran’s diamond theorem”: Let M1 and M2 be Galois ex-

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tensions of a Hilbertian field K and M a field between K and M1 M2 not contained in M1 nor in M2 . Then M is Hilbertian (Theorem 13.8.3). In particular, if N is a Galois extension of K, then N is not the compositum of two Galois extensions of K neither of which is contained in the other. This settles Problems 12.18 and 12.19 of the first edition. The immediate goal of Hilbert’s irreducibility theorem was to realize the groups Sn and An as Galois groups over Q. Chapter 16 is dedicated to realizations of Galois groups over arbitrary Hilbertian fields. One of the most important of these results is due to Harbater (Proposition 16.12.1): Let K be a complete valued field, t an indeterminate, and G a finite group. Then G is regular over K, that is, K(t) has a finite Galois extension F , regular over K, with Gal(F/K(t)) ∼ = G. Unfortunately, none of the three proofs of this theorem fits into the scope of this book. Section 16.6 proves a theorem of Whaples: Let K be a field and p a prime number. Suppose Z/pZ (resp. Z/4Z if p = 2) occurs as a Galois group over K. Then Zp is realizable over K. Section 16.7 generalizes a theorem of Hilbert: Let K be a field and n ≥ 2 an integer with char(K) - (n − 1)n. Then An is regular over K. One of the most far-reaching attempts to realize arbitrary finite Galois groups over Hilbertian fields uses Matzat’s notion of GAR realization of simple finite groups: Let K be a Hilbertian field and α: G → Gal(L/K) a finite embedding problem over K. Suppose every composition factor of Ker(α) has a GAR realization over K. Then the embedding problem is solvable. This leads in particular to the realization of many finite groups over Q (Remark 16.9.5). Chapter 17 deals mainly with Melnikov’s formations C (i.e. sets consisting of all finite groups whose composition factors belong to a given set of finite simple groups). We prove that every free abstract group F is residuallyC. Thus, if the free pro-C group with a given rank m exists, then the canonical injection of F into Fˆm (C) is injective (Proposition 17.5.11 – Ribes-Zalesskii). Konrad Neumann improved former results of Fried-Geyer-Jarden and proved that every field is stable (Theorem 18.9.3). This allows the construction of PAC Hilbertian Galois extensions of arbitrary countable Hilbertian fields (Theorems 18.10.2 and 18.10.3). We survey Neumann’s proof in Section 18.9. The full proof unfortunately falls outside the scope of this book. It seemed to be well known that the concept of absolute irreducibility of a variety is elementary. Unfortunately, we could find no solid proof for it in the literature. Proposition 19.5.9 fills in the gap by proving that result. Section 21.2 includes now the classical results about Ci -fields and not only the corresponding results about weakly Ci -fields as was the case in Section 19.2 of the first edition. Sections 21.8 gives a complete proof of Schur’s Conjecture: If f (X) is a polynomial with coefficients in a global field K with char(K) - deg(f ) and f permutes OK /p for infinitely many primes p of K, then each composition factor of f is linearly related over K to a Dickson polynomial of a prime

Introduction to the Second Edition

xix

degree. Section 21.7 proves all lemmas about permutation groups which are used in the proof of Schur’s Conjecture (Theorem 21.8.13). This includes the classification of subgroups of AGL(1, Fl ) (Lemma 21.7.2), and the theorems of Schur (Proposition 21.7.7) and Burnside (Proposition 21.7.8) about doubly transitive permutation groups. Section 21.9 contains the Fried-Cohen version of Lenstra’s proof of the generalized Carlitz’s Conjecture: Let p be a prime number, q a power of p, and f ∈ Fq [X] a polynomial of degree n > 1 which is not a power of p. Suppose f permutes infinitely many finite extensions of Fq . Then gcd(n, q − 1) = 1. The universal Frattini p-cover of a finite group plays a central role in Fried’s theory of modular towers. Section 22.11 introduces the former concept and proves its basic properties. Corollary 22.13.4 shows then that PSL(2, Zp ) is a p-Frattini cover of PSL(2, Fp ) although it is not the universal p-Frattini cover. Chapter 23 puts together material on PAC fields which appeared in Section 20.5 and Chapter 21 of the first edition. The Beckmann-Black Problem is a refinement of the inverse problem of Galois Theory. D´ebes proved that the problem has an affirmative solution over PAC fields (Theorem 24.2.2). Chapter 25 substantially extends the study of free profinite groups F of infinite rank which appeared in Section 24.4 of the first edition. Most of the material goes back to Melnikov. We characterize closed normal subgroups of F by their S-ranks, and prove that a closed subgroup of F is accessible if and only if it is homogeneous. The first part of Chapter 25 reproduces the group theoretic version of Haran’s diamond theorem. Chapter 26 is completely new. It describes the properties of the closed subgroup hxi and the closed normal subgroup [x] generated by a random etuple x = (x1 , . . . , xe ) of elements of a finitely generated free profinite group F of finite rank n ≥ 2. For example, with probability 1, hxi ∼ = Fˆe (Proposition 26.1.7). This solves Problem 16.16 of the first edition. In addition, with a positive probability, [x] has infinite rank and is isomorphic to Fˆω (Theorem 26.4.5 and Corollary 26.5.7). The latter result is based on the Golod-Shafarevich Inequality. Chapter 28 considers an infinite field K which is finitely generated over its base field. It proves that for e ≥ 2 the theory of all sentences θ which hold ˜ σ1 , . . . , σe i with (σ1 , . . . , σe ) ∈ Gal(K)e is undein almost all structures hK, ˜ σ 1 , . . . , σe i cidable. Moreover, the probability that a sentence θ hold in hK, is in general a nonrational number. Perhaps the most significant achievement of Field Arithmetic since the first edition appeared is the solution of Problem 24.41 of that edition: The absolute Galois group of a countable PAC Hilbertian field is free of rank ℵ0 . It was originally proved in characteristic 0 with complex analysis by Fried-V¨olklein. Then it was proved in the general case by Pop using rigid geometry and by Haran-Jarden-V¨ olklein using “algebraic patching”. The two

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Introduction to the First Edition

latter methods also lead to the proof that Gal(C(t)) is a free profinite group if C is an arbitrary algebraically closed field (Harbater, Pop, Haran-Jarden). The method of Fried-V¨ olklein led to the theory of modular towers of Fried. A remote goal in Galois theory is the classification of absolute Galois groups among all profinite groups. In this framework, one tries to construct new absolute Galois groups out of existing ones. For example, for all fields K1 , . . . , Kn there exists a field K with Gal(K) isomorphic to the free product of Gal(K1 ), . . . , Gal(Kn ) (Pop, Melnikov, Ershov, Koenigsmann). Generalization of this result to infinite families of closed subgroups generalize the concepts “projective groups” and “PRC fields” or “PpC fields” to “relatively projective groups” and “pseudo closed fields” (Haran-Jarden-Pop). They generalize the classification of projective groups as those profinite groups appearing as absolute Galois groups of PAC fields. All of the exciting material mentioned in the preceding two paragraphs lie unfortunately outside the scope of this volume. It is my pleasure to thank colleagues and friends who critically read parts of the manuscript of the present edition of “Field Arithmetic”: Michael Bensimhoun, David Brink, Gregory Cherlin, Michael Fried, Wulf-Dieter Geyer, Peter M¨ uller, Dan Haran, Wolfgang Herfort, Alexander Lubotzky, Nikolay Nikolov, Dan Segal, Aharon Razon, and Irene Zimmermann. Tel Aviv, Spring 2004

Moshe Jarden

Introduction to the First Edition Our topic is the use of algebraic tools — coming mainly from algebraic geometry, number theory, and the theory of profinite groups — in the study of the elementary properties of classes of fields, and related algorithmic problems. (We take the precise definition of “elementary” from first order logic.) This subject has its more distant roots in Tarski’s observation that, as a consequence of elimination theory, the full elementary theory of the class of all algebraically closed fields is decidable; this relies on the Euclid algorithm of finding the greatest common divisor of two polynomials in one variable over a field. In its first phase this line of thought led to similar results on real closed fields and p-adic fields. The subject took a new turn with the work of James Ax [Ax2] on the elementary theory of the class of finite fields, which represents a radical departure in terms of the algebraic methods used. The analysis is based entirely on three properties of a finite field K: (1a) K is perfect. (1b) K has a unique extension of each degree. (1c) There is an explicitly computable function q(d, m) such that any absolutely irreducible variety V defined over K will have a K-rational point if |K| > q(dim(V ), deg(V )).

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The validity of the third condition for finite fields is a consequence of Riemann’s hypothesis for curves over finite fields. Methods of logic, specifically ultraproducts, led Ax to consider this condition for infinite fields as well, in which case the lower bound afforded by the function q is vacuous, and the condition becomes: (2) Every absolutely irreducible variety over K has a K-rational point. Fields satisfying (2) are said to be pseudo algebraically closed, or PAC. The second condition may be interpreted as a description of the absolute Galois group Gal(K) as a profinite group: Gal(K) is the free profinite group on one generator. In Ax’ approach it was convenient to have an Abelian absolute Galois group, but a strong trend in later work has been the systematic analysis of situations involving progressively more general Galois groups. One of our central goals here is the presentation of the general theory of PAC fields in its modern form, and its connections with other branches of algebra. From what we have said so far, some connections with algebraic geometry and profinite groups are visible; a number theoretic connection will appear shortly. One important feature of PAC fields is that they occur in profusion in nature and are in fact typical in the following sense. Since the absolute Galois group Gal(Q) of the rationals is a compact topological group, it carries a canonical invariant probability measure, the Haar measure. We can therefore ˜ ask for the probability that the fixed field Q(σ) of a sequence σ = (σ1 , ..., σe ) ˜ will be PAC; and we find that this occurs with probof automorphisms of Q ˜ ability 1. In addition, the absolute Galois group of Q(σ) is free on the e generators σ1 , ..., σe , again with probability 1. These facts are consequences of Hilbert’s irreducibility theorem for Q (Chapter 13), at least in the context of countable fields. We will develop other connections between the PAC property and Hilbertianity. There are also remarkable connections with number theory via the Chebotarev density theorem (Chapters 6, 13, 16, 20, 21, 31). For example, the ˜ probability that a given elementary statement ψ holds for the field Q(σ) coincides with the Dirichlet density of the set of primes for which it holds for the field Fp , and this density is rational. Thus, the “probability 1” theory of ˜ the fixed fields Q(σ) coincides with the theory of “all sufficiently large” finite fields, which by Ax’ work is an algorithmically decidable theory. ˜ Ax’ results extend to the “probability 1” theory of the fields Q(σ) for σ of length e > 1, by somewhat different methods (Chapter 20), although the connection with finite fields is lost. The elementary theories of such fields are largely determined by three properties: PAC, characteristic zero, and having an absolute Galois group which is free on e generators. To determine the full elementary theory of one such field K, it is also necessary to describe the ˜ intersection K ∩ Q. Although the absolute Galois group of a PAC field need not be free, it can be shown to be projective in a natural sense, and conversely any

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projective profinite group occurs as the Galois group of some PAC field. In extending the theory from PAC fields with free Galois group to the general (projective) case, certain obstacles arise: for example, the algorithmic results do not extend. There is nonetheless a quite general theory, which enables us to identify some broad classes of projective profinite groups for which the associated classes of profinite groups behave well, and also to pinpoint unruly behavior in other case. One approach to the algorithmic problems associated with PAC fields leads to the study of profinite groups G with the embedding property (the terminology reflects a preoccupation with the corresponding fields): for each pair of continuous epimorphisms ϕ: G → A, α: B → A, where B is a finite quotient of G, we require that ϕ should factor through α. A perfect PAC field whose absolute Galois group is a group with the embedding property is called a Frobenius field. The elementary theory of all Frobenius fields can be computed quite explicitly. The algorithm has some relationship with elimination theory as used by Tarski. We associate to each elementary statement in the language of PAC fields a stratification of affine space into basic normal locally closed algebraic sets, each equipped with a Galois extension of its function field, and the given statement is reinterpreted as a statement about conjugacy classes of subgroups of the specified Galois groups. When the initial statement has no quantifiers this is a fairly trivial procedure, but addition of quantifiers corresponds to a special kind of “projection” of these Galois stratifications. This procedure has not yet been closely examined from the point of view of computational complexity. Like most procedures which operate by tracing through a series of projections, it is effective but hopelessly inefficient in its present form. It is not yet clear whether it is substantially less efficient than Tarski’s procedure for algebraically closed fields, nor whether, like that procedure, it can significantly reorganized and sped up. The Galois stratification algorithm relies on techniques of effective algebraic geometry, and also involves substantial algorithmic problems of a new type connected with the theory of profinite groups. Specifically, it is necessary to determine, given two collections A1 , ..., Am and B1 , ..., Bn of finite groups, whether or not there is a projective group with the embedding property which has each Ai as (continuous) image, but none of the groups Bj . The solution to this problem depends on recent work on projective covers (Chapter 22) and embedding covers (Chapter 24). Ultimately our decision problem reduces to the determination of the finite quotients of the projective cover of the embedding cover of a single finite group. The theory of projective covers leads also to the undecidability results alluded to earlier. A fairly natural encoding of graphs into profinite groups is lifted by this theory into the class of projective profinite groups, and then by looking at the corresponding PAC fields we see that their elementary theories encode algorithmically undecidable problems (the analogous results for graphs are well known).

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In the final chapter we return to our point of departure, the theory of finite fields. The zeta function of a Galois formula over a finite field is defined, and using a result of Dwork and Bombieri we show that some integral power of each such function is a product of an exponential and a rational function over Q. One of the goals of this book is to serve as a bridge between algebraists and logicians. For the algebraist there is a self contained introduction to the logic and model theory background for PAC fields (Chapter 7). Chapter 14 gives the “nonstandard” framework that suffices for Weissauer’s proof of Hilbert’s irreducibility theorem (Chapter 15), and Chapters 8 and 28 include basic recursion theory. On the other hand, for logicians with basic algebraic background (e.g. Lang’s book “Algebra”) Chapter 4 has the Stepanov-Bombieri elementary proof of the Riemann hypotheses for curves, and Chapter 6 gives an elementary proof of the Chebotarev density theorem. Both groups of readers may find the extensive treatment of profinite groups (Chapters 1, 17, 18, 22, 24, 25 and 26) and of Hilbertian fields (Chapters 12, 13, 15, and 16) valuable. Although PAC fields arise over arithmetically rich fields, they themselves lack properties that we associate with the arithmetic, say, of the rationals. For example, a PAC field F admits no orderings and all Henselizations of F are separably closed (Section 11.5). Many PAC field results generalize to pseudo real closed (PRC) fields. A field F is PRC if each absolutely irreducible variety defined over F has an F -rational point provided it has a nonsingular Fˆ -rational point in each real closure Fˆ of F . Thus, a PRC field without orderings is PAC. This, and the development of the theory of pseudo p-adically closed PpC fields are outside the scope of this book. We refer to [Prestel1], to [Jarden12], [Haran-Jarden2], and to [Haran-Jarden3] for literature about PRC fields and to [Haran-Jarden4] for PpC fields. Similarly, we give no account of the theories of real closed fields and p-adically closed fields that preceded the development of the theory of PAC fields. In particular, for Hilbert’s 17th problem and the Ax-Kochen-Ershov p-adic theory, we refer the reader to [Prestel2], [Ax-Kochen1, Ax-Kochen2, and Ax-Kochen3], and [Prestel-Roquette]. Acknowledgement: We are indebted to several colleagues who corrected errors in the process of critically reading the manuscript. In particular, WulfDieter Geyer, Gregory L. Cherlin, and Dan Haran made crucial contributions. Michael D. Fried, Gainesville, Florida Moshe Jarden, Tel Aviv, Israel Summer 1986

Notation and Convention Z = the ring of rational integers. Zp = the ring of p-adic integers. Q = the field of rational numbers. R = the field of real numbers. C = the field of complex numbers. Fq = the field with q elements. Ks = the separable closure of a field K. Kins = the maximal purely inseparable extension of a field K. ˜ = the algebraic closure of a field K. K Gal(L/K) = the Galois group of a Galois extension L/K. ˜ We call a polynomial f ∈ K[X] separable if f has no multiple root in K. Gal(f, K) = the Galois group of a separable polynomial f ∈ K[X] over a field K viewed as a permutation group of the roots of f . Gal(K) = Gal(Ks /K) = the absolute Galois group of a field K. irr(x, K) = the monic irreducible polynomial of an algebraic element x over a field K. Whenever we form the compositum EF of field extensions of a field K we tacitly assume that E and F are contained in a common field. |A| = #A = the cardinality of a set A. R× = the group of invertible elements of a ring R. Quot(R) = the quotient field of an integral domain R. A ⊂ B means “the set A is properly contained in the set B”. ax = x−1 ax, for elements a and x of a group G. H x = {hx | h ∈ H}, for a subgroup H of G. Given subgroups A, B of a group G, we use “A ≤ B” for “A is a subgroup of B” and “A < B” for “A is a proper subgroup of B”. Given an Abelian (additive) group A and a positive integer n, we write An for the Ssubgroup {a ∈ A | na = 0}. For a prime number p we let ∞ Ap∞ = i=1 Api . For a group B that acts on a group A from the right, we use B n A to denote the semidirect product of A and B. Bold face letters stand for n-tuples, e.g. x = (x1 , . . . , xn ). ord(x) is the order of an element x in a group G. For a positive integer n and an integer a with gcd(a, n) = 1, we use ordn a to denote the order of a modulo n. Thus, ordn a is the minimal positive integer d with ad ≡ 1 mod n. In the context of groups, Sn (resp. An ) stands for the full permutation group (resp. alternative group) of {1, . . . , n}. In the context of groups, Cn stands for the cyclic multiplicative group of order n. Likewise we use Z/nZ for the additive multiplicative group of order n. In S the context of fields, ζn stands for a primitive root of unity of order n. · i∈I Bi is the disjoint union of sets Bi , i ∈ I.

Chapter 1. Infinite Galois Theory and Profinite Groups The usual Galois correspondence between subgroups of Galois groups of finite Galois extensions and intermediate fields is not valid for infinite Galois extensions. The Krull topology restores this correspondence for closed subgroups (Proposition 1.3.1). Since Galois groups are inverse limits of finite groups, they are profinite. Conversely, we define profinite groups, independently of Galois theoretic properties. Each profinite group actually appears as a Galois group ˆ and (Corollary 1.3.4). In particular, we study the procyclic groups Zp and Z prove that every finite field has the latter as its absolute Galois group.

1.1 Inverse Limits Our interest in inverse limits comes from infinite Galois theory: Infinite Galois groups are inverse limits of finite Galois groups. As a preparation to the study of “profinite groups” we define in this section inverse limits of topological spaces and characterize inverse limits of finite topological spaces. Let I be a set with a partial ordering ≤; that is, ≤ is a binary relation which is reflexive, transitive, and a ≤ b and b ≤ a imply a = b. We call (I, ≤) a directed partially ordered set if in addition (1) for all i, j ∈ I there exists k ∈ I with i ≤ k and j ≤ k. An inverse system (also called a projective system) over a directed partially ordered set (I, ≤) is a data (Si , πji )i,j∈I where Si is a set and πji : Sj → Si is a map for all i, j ∈ I with i ≤ j satisfying the following rules: (2a) πii the identity map for each i ∈ I. (2b) πki = πji ◦ πkj if i ≤ j ≤ k. Q Let S be the subset of the cartesian product i∈I Si consisting of all elements Q s = (si )i∈I with πji (sj ) = si for all i ≤ j. Note: S may be empty. Let pri : j∈I Sj → Si be the projection on the ith coordinate. Denote the restriction of pri to S by πi . Then πi = πji ◦ πj for every i ≤ j. We say (S, πi )i∈I is the inverse (or projective) limit of the family (Si )i∈I with respect to the maps πji . Denote S by lim Si . ←− 0 )i,j∈I be another inverse system over I. Suppose for each Let (Si0 , πji 0 ◦ θj = θi ◦ πji for all i ≤ j. (We i ∈ I we are given a map θi : Si → Si0 with πji say that the maps θi , i ∈ I, are compatible.) Then there exists a unique map θ: lim Si → lim Si0 satisfying πi0 ◦ θ = θi ◦ πi for each i ∈ I: θ maps ←− ←− s = (si ) ∈ lim Si onto θ(s) with θ(s)i = θi (si ). Denote θ by lim θi . ←− ←− Similarly, let X be a set and for each i ∈ I let θi : X → Si be a map satisfying πji ◦ θj = θi whenever i ≤ j (Again, we say that the maps θi ,

2

Chapter 1. Infinite Galois Theory and Profinite Groups

i ∈ I, are compatible.) Then there exists a unique map θ: X → lim Si with ←− πi ◦ θ = θi for each i ∈ I. When Si are topological spaces, we assume πji are continuous. Then we equip lim Si with the topology induced from the product topology of ←− Q Q i∈I Si . Recall Q that the product topology on i∈I Si has a basis consisting of the sets i∈I Ui , with Ui open in Si for each i ∈ I, and Ui = Si for all but finitely many i ∈ I. Since pri is continuous, so is πi , i ∈ I. If θi : Si → Si0 are continuous, then θ: lim Si → lim Si0 is also continuous. ←− ←− Lemma 1.1.1: The collection of all subsets of S = lim Si of the form πi−1 (Ui ) ←− with Ui open in Si is a basis for the topology of S. Proof: Let s ∈ S. each basic open neighborhood of s has the Q Q By definition, V × form V = S ∩ j∈J j i∈I r J Sj , where J is a finite subset of I and subset of S , j ∈ J. Take k ∈ I with k ≥ j for all j ∈ J. Vj is an open j T −1 (Vj ) is an open subset of Sk and πk−1 (Uk ) is an open Then Uk = j∈J πkj neighborhood of s in V . Therefore, the collection πi−1 (Ui ), i ∈ I, is a basis for the topology of S. Lemma 1.1.2: In the notation above, if each Si , i ∈ I, is a Hausdorff space, Q then lim Si is a closed subset of Si . ←− Q Proof: Suppose s = (si ) ∈ Si does not belong to lim Si . Then there are ←− Ui i, j ∈ I with i ≤ j and πji (sj ) 6= si . Take open disjoint neighborhoods Q −1 0 and Ui0 of si and πji (sj ), respectively. Then U × π (U ) × S is an i k i ji k6 = i,j Q open neighborhood of s in Si that does not intersect lim Si . ←− Q

If, in addition, each Si is compact, then Tychonoff’s theorem implies Si is also compact. Thus, lim Si with the induced topology is compact. ←−

Lemma 1.1.3: The inverse limit S of an inverse system of nonempty compact Hausdorff spaces Si , i ∈ I, is a nonempty compact Hausdorff space. T Proof: We only need Q to prove that S is nonempty. Indeed, S = k≥j Q Rkj , Si → where Rkj = {s ∈ Si | πkj (sk ) = sj }. The natural map prk × prj : Sk × Sj is continuous. The Hausdorff property of Sj implies T = {(sk , sj ) ∈ Sk × Sj | πkj (sk ) = sj } is a closed Q subset of QSk × Sj . Hence, Rkj = (prk × prj )−1 (T ) is a closed subset of Si . Since Si is compact, we only need to show that the intersection of finitely many of the Rkj is nonempty. Indeed, let j1 ≤ k1 , . . . , jn ≤ kn be n pairs in I. Choose l ∈ I with ki ≤ l, i = 1, . . . , n, and choose sl ∈ Sl . Define sji = πl,ji (sl ) and ski = πl,ki (sl ), for i = 1, . . . , n. For each r ∈ I r 1 , . . . , jn , k1 , . . . , kn } let sr an arbitrary element of Sr . T{j n Then s = (si ) ∈ i=1 Rki ,ji .

1.1 Inverse Limits

3

Corollary 1.1.4: The inverse limit of an inverse system of nonempty finite sets is nonempty. Proof: Equip each of the finite sets with the discrete topology.

0 )i,j∈I be inverse systems of Corollary 1.1.5: Let (Si , πji )i,j∈I and (Si0 , πji 0 compact Hausdorff spaces. Let θi : Si → Si be a compatible system of surjective continuous maps. Put S = lim Si , S 0 = lim Si0 , and θ = lim θi . Then ←− ←− ←− θ: S → S 0 is surjective.

Proof: Let s0 = (s0i ) be an element of S 0 . Then (θi−1 (s0i ), πji )i,j∈I is an inverse system of nonempty compact Hausdorff spaces. By Lemma 1.1.3, the inverse limit of θi−1 (s0i ) is nonempty. Each element in the inverse limit is mapped by θ onto s0 . Corollary 1.1.6: Let X be a compact space, (Si , πji )i,j∈I an inverse system of Hausdorff spaces, and θi : X → Si a compatible system of continuous surjective maps. Put θ = lim θi . Then θ: X → lim Si is surjective. ←− ←− Consider s = (si )i∈I ∈ lim Si . Then θi−1 (si ) is a closed nonempty ←− subset of X. For i1 , . . . , in ∈ I there is a j ∈ I with i1 , . . . , in ≤ j. Then (si1 ) ∩ · · · ∩ θi−1 (sin ), so θi−1 (si1 ) ∩ · · · ∩ θi−1 (sin ) is nonempty. θj−1 (sj ) ⊆ θi−1 1 n 1 n Since X is compact, there is an x ∈ X which belongs to θi−1 (si ) for every i ∈ I. It satisfies, θ(x) = s. Thus, θ is surjective. Proof:

A profinite space is an inverse limit of an inverse system of finite discrete spaces. Lemma 1.1.7: A compact Hausdorff space S is profinite if and only if its topology has a basis consisting of open-closed sets. Proof: Suppose first S = lim Si is an inverse limit of finite discrete spaces ←− Si . Let πi : S → Si be the projection on the ith coordinate, i ∈ I. By Lemma 1.1.1, the sets πi−1 (Ui ), where i ∈ I and Ui is a subset of Si , form a basis of the topology of S. Since Si is discrete, Ui is open-closed. Hence, so is πi−1 (Ui ). Now suppose the topology of S has a basis consisting of open-closed sets. An open-closed partition of S is a finite set A of nonempty openclosed disjoint subsets of S whose union is S. Denote the set of all openclosed partitions of S by A. Let A, A0 , B ∈ A. Write A ≤ B if for each B ∈ B there is an A ∈ A with B ⊆ A. Then A is unique. Next note that C = {A ∩ A0 | A ∈ A, A0 ∈ A0 } belongs to A and satisfies A, A0 ≤ C. Thus, (A, ≤) is a directed partially ordered set. When A ≤ B, define a map πB,A from B to A by πB,A (B) = A, where A is the unique element of A containing B. Equip each A ∈ A with the discrete topology. Then (A, πB,A )A,B∈A is an inverse system of finite discrete spaces.

4

Chapter 1. Infinite Galois Theory and Profinite Groups

Its limit S 0 = lim A is a profinite space. We construct a homeomorphism of ←− S onto S 0 . Let s ∈ S and A ∈ A. Define θA (s) to be the unique A in A which contains s. Then θA : S → A is a continuous surjective map. If A ≤ B, then πB,A ◦ θB = θA . Hence, by Corollary 1.1.6, there is a continuous surjective map θ: S → S 0 satisfying πA ◦ θ = θA for each A ∈ A. Suppose s, s0 are distinct elements of S. Since S is Hausdorff, there are disjoint open-closed subsets A and A0 of S with s ∈ A and s0 ∈ A0 . Put A00 = S r(A ∪ A0 ). Then A = {A, A0 , A00 } is an open-closed partition of S, θA (s) = A, and θA (s0 ) = A0 . Thus, θA (s) 6= θA (s0 ), so θ(s) 6= θ(s0 ). Therefore, θ is bijective. Since S is compact and S 0 is Hausdorff, θ is a homeomorphism. Consequently, S is a profinite space. Remark 1.1.8: Totally disconnected spaces. Let S be a compact Hausdorff space. Suppose S has a basis for its topology consisting of open-closed sets. It is not difficult to see that S is totally disconnected; that is each s in S is its own connected component. Conversely, if S is totally disconnected, then the topology of S has a basis consisting of open-closed subsets [Ribes-Zalesski, Thm. 1.1.12].

1.2 Profinite Groups We survey here the basic properties of compact groups. In particular, this will apply to profinite groups. Topological Groups. A topological group is a group G equipped with a topology in which the product (x, y) 7→ xy and the inverse map x 7→ x−1 are continuous. It follows that for each a ∈ G the maps x 7→ ax, x 7→ xa, and x 7→ x−1 are homeomorphisms. We always assume {1} is a closed subset of G. Consequently, {a} is a closed subset of G for each a ∈ G. It follows that G is a Hausdorff space. Indeed, let a, b be distinct elements of G. The identity 1 · 1−1 = 1 and the continuity of the group operations give open neighborhoods U and W of 1 with U W −1 ⊆ G r{a−1 b}. Thus, aU and bW are disjoint open neighborhoods of a and b, respectively, as needed. In addition, each closed subgroup H of GSof finite index is open. Indeed, there are ai ∈ G, i ∈ I, with I finite and G = · i∈I ai H. Each of theSsets ai H is closed and there is a j ∈ I with aj H = H. Therefore, H = G r · i6=j ai H is open. Conversely, if G is compact, then every open subgroup H of G is of finite index. Otherwise, G would be a disjoint union of infinitely many cosets of H, each of which is open. Let N be a closed normal subgroup of G and π: G → G/N the quotient ¯ of G/N map. Equip G/N with the quotient topology. Thus, a subset U −1 ¯ is open if and only if π (U ) is open. It follows that the group operations of G/N are continuous. In addition, a subset C¯ of G/N is closed if and

1.2 Profinite Groups

5

¯ is closed. In particular, since N = π −1 (1) is closed, {1} only if π −1 (C) is a closed subset of G/N . Consequently, G/N is a topological group and π: G → G/N is a continuous map. Moreover, if U is an open subset of G, S then π −1 (π(U )) = n∈N nU , so π(U ) is open in G/N . Therefore, π is an open map. Suppose again G is compact. Let θ be a continuous homomorphism of G into a topological group H. Since H is Hausdorff, θ is a closed map. Suppose in addition, θ is surjective. Put N = Ker(θ). Let π: G → G/N ¯ G/N → H the map induced by θ. Then θ¯ is be the quotient map and θ: an isomorphism of abstract groups. In addition, θ¯ is a continuous bijective map of the compact group G/N onto the Hausdorff group H. Hence, θ¯ is an isomorphism of topological groups and θ is an open map. This is the first isomorphism theorem for compact groups. Let H be a closed subgroup and N a closed normal subgroup of G. Then HN = {hn | h ∈ H, n ∈ N } is the image of the compact group H × N under the continuous map (h, n) 7→ hn. Hence, HN is a closed subgroup of G. As in the preceding paragraph, the group theoretic isomorphism θ: HN/N → H/H ∩ N defined by θ(hN ) = h(H ∩ N ), h ∈ H, is a homeomorphism, so θ is an isomorphism of topological groups. Similarly, if M and N are closed normal subgroups of G with N ≤ M , then M/N is a closed normal subgroup of G/N and the map G/M → (G/N )/(M/N ) given by gM 7→ (gN )M/N , g ∈ G, is an isomorphism of topological groups. Here is one way to equip an abstract group G with a topology. Let N be a family of normal subgroups of G, closed under finite intersections, such that the intersection of all N ∈ N is 1. Take N to be a basis for the open neighborhoods of 1. A basis for the open neighborhoods of a ∈ G is the family Na = {aN | N ∈ N }. The union of the Na is then a basis for a group topology of G. For example, the identity xN · yN = xyN for normal subgroups N implies multiplication is continuous. Let (Gi , πji )i,j∈I be an inverse system of topological groups and continuous homomorphisms πji : Gj → Gi , for each i, j ∈ I with j ≥ i. Then G = lim Gi is a topological group and the projections πi : G → Gi are con←− 0 ii,j∈I be another system of topological tinuous homomorphisms. Let hG0i , πji 0 0 groups with G = lim Gi . Suppose θi : Gi → G0i , i ∈ I, is a compatible system ←− of continuous homomorphisms. Then the corresponding map θ: G → G0 is a continuous homomorphism. Profinite Groups. In this book, we primarily consider an inverse system of finite groups (Gi , πji )i,j∈I , each equipped with the discrete topology. We call the inverse limit G = lim Gi a profinite group. By Lemma 1.1.3, G is ←− a compact group. By Lemma 1.1.1, the open-closed sets πi−1 (gi ), gi ∈ Gi , i ∈ I, form a basis for the topology on G. In particular, the open normal subgroups of G form a basis for the open neighborhoods of 1.

6

Chapter 1. Infinite Galois Theory and Profinite Groups

Remark 1.2.1: Basic rules. Here are some basic rules for profinite groups G and H which we use in this book without explicit reference: (a) A subgroup H of G is open if and only if H is closed of a finite index. The intersection of all open normal subgroups of G is 1. Every open subset of G is a union of cosets gi Ni with Ni open normal and gi ∈ G. (b) Every profinite group is compact, Hausdorff, and has a basis for its topology consisting of open-closed sets (Lemmas 1.1.3 and 1.1.7). (c) A subset C of a profinite group is closed if and only if C is compact (use (b)). (d) A subset B of a profinite group is open-closed if it is a union of finitely many cosets gi N with N open normal and gi ∈ G (use (a) and the compactness of B). (e) Every homomorphism ϕ: G → H is tacitly assumed to be continuous. In particular, ϕ maps compact subsets of G onto compact subsets of H. Hence, ϕ maps closed subsets of G onto closed subsets of H (use (c)). (f) By the first isomorphism theorem for compact groups, every epimorphism ϕ: G → H of profinite groups is an open map. In particular, ϕ maps open subgroups of G onto open subgroups of H. We list below special properties of profinite groups not shared by all compact groups: Lemma 1.2.2: Let {Hi | i ∈ I} be a directed family of closed subset of a profinite group G; that is: T (1) for every finite subset J of I there is an i ∈ I with Hi ≤ j∈J Hj . T Put H = i∈I Hi . Then: (a) For every open subgroup U of G containing H, there is an i ∈ I with H Ti ≤ U . T (b) i∈I KHi = KH and i∈I Hi K = HK for T every closedTsubset K of G. ¯ be an epimorphism. Then ϕ( (c) Let ϕ: G → G i∈I Hi ) = i∈I ϕ(Hi ). (d) Let K and K 0 be closed subgroups of G. Then the set S = {σ ∈ G | K σ = K 0 } is closed. (e) Let K and K 0 be closed subgroups of G which contain H. Suppose each Hi is normal and KHi and K 0 Hi are conjugate. Then K and K 0 are conjugate. (f) Let G = lim Gi be an inverse limit of finite groups, πi : G → Gi the ←− quotient maps, and K, K 0 closed subgroups of G. Suppose πi (K) and πi (K 0 ) are conjugate in Gi for each i. Then K and K 0 are conjugates. T Proof of (a): The set Hi ∩ (G r U ) is closed for every i ∈ I and i∈I Hi ∩ r (G r U ) = H ∩ (G T U ) = ∅. Since G is compact, there exists a finite subset J of ITsuch that j∈J Hj ∩ (G r U ) = ∅. By (1) there exists i ∈ I such that Hi ≤ j∈J Hj . Hence, Hi ≤ U . T Proof of (b): Let g ∈ i∈I KHi . Then, for each i ∈ I there are ki ∈ K and hi ∈ Hi with g = ki hi . Hence, the closed subset Hi ∩ g −1 K of G is nonempty.

1.2 Profinite Groups

7

−1 By (1), the intersection of finitely many T of the sets Hi ∩ g K is nonempty. Since G is compact, there exists h ∈ i∈I Hi ∩ g −1 K. It satisfies h ∈ H and h = g −1 k for some k ∈ K. Therefore, g = kh−1 ∈ KH.

Proof of (c): Let K = Ker(ϕ). Then ϕ induces a bijection between the set of closed subgroups lying between K and G and the set of closed subgroups ¯ Hence, by (b), ϕ(H) = ϕ(KH) = ϕ(T KHi ) = T ϕ(KHi ) = of G. i∈I i∈I T i∈I ϕ(Hi ). Proof of (d): Denote the set of all open normal subgroups of G by N . For each N ∈ N , let SN be the inverse image under the quotient map G → G/N of the finite set {s ∈ G/N | (KN/N )s = K 0 N/N }. If s ∈ SN for each N ∈ T N , then, by (b), K s = Then, SN is closed. T T s 0 0 N ∈N K N = N ∈N K N = K . Hence, S = N ∈N SN . Therefore, S is closed. Proof of (e): For each i ∈ I, Si = {σ ∈ G | K σ Hi = K 0 Hi } is a nonempty T subset of G. By (d), Si is closed. In theTnotation of (1) we have Si ⊆ j∈J Sj . Hence, by compactness, there is a σ ∈ i∈I Si . Thus, K σ Hi = K 0 Hi for each i ∈ I. We conclude from (b) that K σ = K 0 . Proof of (f): Apply (e) to Hi = Ker(πi ).

Lemma 1.2.3: Each closed subgroup H of a profinite group G is the intersection of open subgroups. Proof: Let T N be the set of all open normal subgroups of G. By Lemma 1.2.2(b), N ∈N N H = H. The following result gives a sufficient condition for a compact group to be profinite. We use it below to prove that the category of profinite groups is closed under various natural operations. Lemma 1.2.4: Let G be a compact group and {Ni | i ∈ I} a directed family of closed T normal subgroups of G of finite index satisfying (2) i∈I Ni = 1. Then G = lim G/Ni and G is a profinite group. ←− Proof: By definition, lim G/Ni is a profinite group. We only need to prove ←− the isomorphism. The quotient maps G → G/Ni define a continuous embedding θ of G into lim G/Ni . If (gi Ni )i∈I is an element of the latter group, ←− then the closed subsets gi Ni ofT G have the finite intersection property. Since G is compact, there exists g ∈ i∈I gi Ni . Then, θ(g) = (gi Ni )i∈I . Thus, θ is bijective. Compactness of G and the Hausdorff property of lim G/Ni imply ←− that θ is an isomorphism of topological groups.

8

Chapter 1. Infinite Galois Theory and Profinite Groups

Lemma 1.2.5: The following statements hold for each closed subgroup H of a profinite group G: (a) The group H is profinite. Moreover, for each open normal subgroup M of H there exists an open normal subgroup N of G with H ∩ N ≤ M . If M is normal in G, then N can be chosen such that H ∩ N = M . (b) Every open subgroup H0 of H is the intersection of an open subgroup of G with H. (c) Every homomorphism ϕ0 of H into a finite group A extends to a homomorphism ϕ: H 0 → A, where H 0 is an open subgroup of G containing H. If H is normal in G, then H 0 can be chosen to be normal. Proof of (a): If N is an open normal subgroup of G, then H ∩ N is an open normal subgroup of H. The family of all groups H ∩ N is directed and its intersection is trivial. Hence, H = lim H/H ∩ N is a profinite group. ←− Let M be an open normal subgroup of H. By Lemma 1.2.2(a), H ∩ N ≤ M for some open normal subgroup N of G. If M / G, then M N / G and H ∩ MN = M. Proof of (b): The intersection of all conjugates of H0 in H is an open normal subgroup of H. Hence, by the first statement of the lemma, there is an open normal subgroup N of G such that M = H ∩ N ≤ H0 . Then G0 = H0 N is an open subgroup of G and H ∩ G0 = H0 . Proof of (c): By (a), G has an open normal subgroup N with H ∩ N ≤ Ker(ϕ0 ). Put H 0 = HN and define ϕ: H 0 → A by ϕ(hn) = ϕ0 (h). Then ϕ is a well defined homomorphism which extends ϕ0 . Lemma 1.2.6: These statements hold: (a) If N is a closed normal subgroup of a profinite group G, then G/N is profinite. Q (b) The cartesian product, G = i∈I Gi , of profinite groups is profinite. (c) Every inverse limit, G = lim Gi , of profinite groups is profinite. ←− Proof of (a): A closed normal group N is an intersection of open subgroups (Lemma 1.2.3) and therefore of open normal subgroups Ni , i ∈ I, of G. Now apply Lemma 1.2.4 to G/N and {Ni /N | i ∈ I}. Proof of (b): Consider a finite subset J of Q I. For each Q j ∈ J let Nj be an open normal subgroup of Gj . Then N = j∈J Nj × i∈I r J Gi is an open normal subgroup of G. The family of all these normal subgroups is directed and its intersection is 1. By Lemma 1.2.4, G = lim G/N is a profinite group. ←− Proof of (c): Observe that G is a closed subgroup of the cartesian product Q i∈I Gi (Lemma 1.1.2). By (b) this product is profinite. Now apply Lemma 1.2.5.

1.3 Infinite Galois Theory

9

Lemma 1.2.7: Each epimorphism ϕ: G → A of profinite groups has a continuous set theoretic section ϕ0 : A → G. That is, ϕ0 is a continuous map satisfying ϕ ◦ ϕ0 = idA . Proof: Let K = Ker(ϕ). We split the proof into two parts. Part A: K is finite. Lemma 1.2.5 gives an open subgroup H of G with K ∩ H = 1. Thus, ϕ maps H bijectively onto B = ϕ(H). As both H and B are Hausdorff and compact, ϕ|H is a homeomorphism. Set β = (ϕ|H )−1 . Next S n let (A : B) = n, choose a1 , . . . , an ∈ A and g1 , . . . , gn ∈ G with A = · i=1 ai B and ϕ(gi ) = ai , i = 1, . . . , n. Define ϕ0 : A → G by ϕ0 (ai b) = gi β(b), b ∈ B. Then ϕ0 is a continuous set theoretic section of ϕ. Part B: K is arbitrary. Denote the set of all closed normal subgroups of G which are contained in K by L. For each L ∈ L, let ϕL : G/L → A be the epimorphism induced by ϕ. For each L0 ∈ L with L0 ≤ L let ϕL0 ,L : G/L0 → G/L be the quotient map. Let Φ0 be the set of all pairs (L, ϕ0L ) where L ∈ L and ϕ0L : A → G/L is a set theoretic section of ϕL . Since ϕK : G/K → A is an isomorphism, it has an inverse ϕ0K . Hence, (K, ϕ0K ) is in Φ0 . Define a partial ordering on Φ0 as follows: (L0 , ϕ0L0 ) ≤ (L, ϕ0L ) if L0 ≤ L and ϕL0 ,L ◦ ϕ0L0 = ϕ0L . Suppose Φ00 = {(Li , ϕ0i ) | i ∈ I} is a descending chain in Φ0 ; that is, every two elements of Φ00 are comparable. Put L00 = T 00 lim G/Li . Thus, i∈I Li . Then, by Lemmas 1.2.6(a) and 1.2.4, G/L = ←− 0 0 the compatible maps ϕi : A → G/Li give a section ϕL00 : A → G/L00 to ϕL00 . Thus, (L00 , ϕ0L00 ) is a lower bound for Φ00 . Zorn’s lemma gives a minimal element (L, ϕ0L ) of Φ0 . Assume L 6= 1. Then L has a proper open subgroup L0 which is normal in G. Consider the epimorphism ϕL0 ,L : G/L0 → G/L with the finite kernel L/L0 . Part A gives a set theoretic section ϕ0L,L0 to ϕL0 ,L . Set ϕ0L0 = ϕ0L,L0 ◦ ϕ0L . Then (L0 , ϕ0L0 ) is an element of Φ0 which is smaller than (L, ϕ0L ). This contradiction to the minimality of (L, ϕ0L ) proves that L = 1. Put ϕ0 = ϕ0L . Then ϕ0 is a continuous set theoretic section of ϕ.

1.3 Infinite Galois Theory Let N be a Galois extension of a field K. The Galois group Gal(N/K) associated with N/K consists of all automorphisms of N that fix each element of K. If N/K is a finite extension and H1 , H2 are subgroups of Gal(N/K) with the same fixed fields in N , then H1 = H2 . This is not the case any more if N/K is infinite. Consider S∞ for example the case where K = Fp for some prime number p and N = i=1 Fpi . Let ϕ be the Frobenius automorphism of N/Fp . It is defined by the rule ϕx = xp for each x ∈ N . Let G0 be the discrete subgroup of Gal(N/Fp ) generated by ϕ. It is a countable group and Fp is its fixed field in N . On the other hand, each element of Gal(Fp2i /Fp ) has exactly

10

Chapter 1. Infinite Galois Theory and Profinite Groups

two extensions to Fp2i+1 . Hence, there are 2ℵ0 sequences (σ1 , σ2 , σ3 , . . .) with σi ∈ Gal(Fp2i /Fp ) such that the restriction of σi+1 to Fp2i is σi for i = 1, 2, 3, . . . . Each such sequence defines a unique σ ∈ Gal(N/Fp ) whose restriction to Fp2i is σi , i = 1, 2, 3, . . . . It follows that the cardinality of Gal(N/Fp ) is 2ℵ0 . In particular, Gal(N/Fp ) is different from G0 but has the same fixed field, namely Fp . The Galois correspondence is restored for closed subgroups of Gal(N/K) in the “Krull topology” which we now introduce. Denote the set of all intermediate fields K ⊆ L ⊆ N , with L/K finite and Galois, by L. To each L ∈ L associate the (finite) Galois group Gal(L/K). If L0 ∈ L and L ⊆ L0 , then resL : Gal(L0 /K) → Gal(L/K) is an epimorphism. Consider the inverse limit lim Gal(L/K), with L ranging over L. Every σ ∈ ←− Gal(N/K) defines a unique element (resL σ)L∈L of lim Gal(L/K). Con←− versely, every (σL )L∈L ∈ lim Gal(L/K) defines a unique σ ∈ Gal(N/K) ←− with resL σ = σL for each L ∈ L. Thus, σ 7→ (resL σ)L∈L is an isomorphism Gal(N/K) ∼ = lim Gal(L/K). This isomorphism induces a topology ←− on Gal(N/K) through the topology on lim Gal(L/K): the Krull topology. ←− Thus, under the Krull topology, Gal(N/K) becomes a profinite group and the family N = {Gal(N/L) | L ∈ L} is a basis for the open neighborhoods of 1. If N/K is a finite extension, then the Krull topology is discrete. Suppose L is a finite extension of K contained in N . Then its Galois ˆ is the smallest Galois extension of K that contains L. It is finite closure L over K and is contained in N . Write Gal(N/L) as a union of right cosets of ˆ to see that Gal(N/L) is an open closed subgroup of Gal(N/K). Gal(N/L) Suppose L is an arbitrary extension of K in N . Then L is the union of a T family {Li | i ∈ I} of finite extensions of K. Hence, Gal(N/L) = i∈I Gal(N/Li ). Therefore, Gal(N/L) is a closed subgroup of Gal(N/K). If S is a set of automorphisms of N , then N (S) = {x ∈ N | σx = x for every σ ∈ S} is the fixed field of S in N . N (S) is also the fixed field in N of the closed subgroup hSi of Gal(N/K) generated by S. If S = {σ1 , . . . , σe } is a finite set, replace N (S) by N (σ1 , . . . , σe ). Let M be a Galois extension of K in N . Denote by resM (or res) the homomorphism from Gal(N/K) into Gal(M/K) that maps σ ∈ Gal(N/K) onto its restriction resM σ to M . It is a continuous surjective map. Proposition 1.3.1: Let N be a Galois extension of a field K. Then L 7→ Gal(N/L) is a bijection from the family of fields L lying between K and N onto the family of closed subgroups of G = Gal(N/K). The inverse map is H 7→ N (H). Proof: Consider a field extension L of K in N . Put H = Gal(N/L). Then L ⊆ N (H). Each x ∈ N (H) is contained in a finite Galois extension M ⊆ N

1.3 Infinite Galois Theory

11

of L. Since the map res: Gal(N/L) → Gal(M/L) is surjective, σx = x for every σ ∈ Gal(M/L). By finite Galois theory, x ∈ L. Hence, N (Gal(N/L)) = L. Conversely, let H be a closed subgroup of G and put L = N (H). Then H ≤ Gal(N/L). Consider σ ∈ Gal(N/L). In order to prove σ ∈ H it suffices to show that σ is in the closure of H. Indeed, let M ⊆ N be a finite Galois extension of K. Then M ∩ L = M (resM H). Finite Galois theory shows resM σ ∈ Gal(M/M ∩ L) = resM H. Hence, H ∩ σGal(N/M ) is nonempty. It follows that Gal(N/N (H)) = H. As in finite Galois theory [Lang7, pp. 192–199], Proposition 1.3.1 gives the following rules for the Galois correspondence: (1a) L1 ⊆ L2 ⇐⇒ Gal(N/L2 ) ≤ Gal(N/L1 ). (1b) H1 ≤ H2 ⇐⇒ N (H2 ) ⊆ N (H1 ). (1c) N (H1 ) ∩ N (H2 ) = N (hH1 , H2 i), where hH1 , H2 i is the closed subgroup of G generated by the closed subgroups H1 and H2 . (1d) Gal(N/L1 ∩ L2 ) = hGal(N/L1 ), Gal(N/L2 )i. (1e) Gal(N/L1 L2 ) = Gal(N/L1 ) ∩ Gal(N/L2 ). (1f) N (H1 ∩ H2 ) = N (H1 )N (H2 ). Since Galois groups are compact, their images under restriction are closed. As in the finite case this produces other theorems of infinite Galois theory. (2a) N (σHσ −1 ) = σN (H) and (2b) Gal(N/σL) = σGal(N/L)σ −1 , for every σ ∈ G. (2c) A closed subgroup H of G is normal if and only if L = N (H) is a Galois extension of K. (2d) If L is a Galois extension of K and L ⊆ N , then res: Gal(N/K) → Gal(L/K) is a continuous open epimorphism with kernel Gal(N/L) and Gal(L/K) ∼ = Gal(N/K)/Gal(N/L). (2e) res: Gal(LM/M ) → Gal(L/L ∩ M ) is an isomorphism for every Galois extension L of K and every extension M of K. (2f) If, in (2e), M is also a Galois extension of K, then σ 7→ (resL σ, resM σ) is an isomorphism Gal(LM/L ∩ M ) ∼ = Gal(L/L ∩ M ) × Gal(M/L ∩ M ) and Gal(LM/K) ∼ = {(σ, τ ) ∈ Gal(L/K) × Gal(M/K) | resL∩M σ=resL∩M τ }. In both cases we use the product topology on products of groups. The first isomorphism of (2f) is a special case of the second one. To prove the second isomorphism, note first that the map σ 7→ (resL σ, resM σ) is a continuous injective map of the left hand side onto the right hand side. Hence, it suffices to prove surjectivity. Thus, consider ρ ∈ Gal(L/K) and τ ∈ Gal(M/K) with resL∩M ρ = resL∩M τ . Extend ρ to an automorphism ρ1

12

Chapter 1. Infinite Galois Theory and Profinite Groups

of LM and let ρ0 = resM ρ1 . Then ρ−1 0 τ ∈ Gal(M/L ∩ M ). By (2e) there is τ . The element σ = ρ1 λ of Gal(LM/K) a λ ∈ Gal(LM/L) with resM λ = ρ−1 0 satisfies resL σ = ρ and resM σ = τ , as desired. These statements are useful when N is the separable closure Ks of K. Denote Gal(Ks /K) by Gal(K), the absolute Galois group of K. Next we show that profinite groups are Galois groups. Lemma 1.3.2: Suppose a profinite group G acts faithfully as automorphisms of a field F . Suppose for each x ∈ F , the stabilizer S(x) = {σ ∈ G | σx = x}, is an open subgroup of G. Then F is a Galois extension of the fixed field K = F (G) and G = Gal(F/K). Proof: If G is a finite group, this is a result of Artin [Lang7, p. 264]. In general the group H = S(x1 ) ∩ · · · ∩ S(xn ) is open in G for every x1 , . . . , xn ∈ F . Therefore, so is the intersection N of all the conjugates of H (Exercise 4). The finite quotient group G/N acts faithfully on the field L = K(Gx1 , . . . , Gxn ). It has K as its fixed field. Thus, L is a finite Galois extension of K and G/N ∼ = Gal(L/K). The field F is the union of all above L and 1 is the intersection of all the above N . Hence, F is a Galois extension of K and Gal(F/K) ∼ = lim Gal(L/K) ∼ = lim G/N ∼ = G. ←− ←− Proposition 1.3.3: Let L/K be a Galois extension and α: G → Gal(L/K) an epimorphism of profinite groups. Then there is a Galois extension F/E and an isomorphism ϕ: Gal(F/E) → G such that F is a purely transcendental extension of L, L ∩ E = K, and α ◦ ϕ = resL . Proof: Let X be the disjoint union of all quotient groups G/N , where N ranges over all open normal subgroups of G. Consider the elements of X as independent over L and set F = L(X). Define an action of each σ of G on F by σ(τ N ) = στ N , for τ N ∈ X, and σ(a) = α(σ)(a) for a ∈ L. This action of G is faithful. We have S(τ N ) = N and S(a) = α−1 (Gal(L/K(a))). Any u ∈ F is a rational function with integral coefficients in a1 , . . . , am ∈ L and x1 , . . . , xn ∈ X. The stabilizer S(u) of u contains the open subgroup S(a1 ) ∩ · · · ∩ S(am ) ∩ S(x1 ) ∩ · · · ∩ S(xn ). Hence, S(u) is an open subgroup. Let E be the fixed field of G in F . By Lemma 1.3.2, G = Gal(F/E) and the conclusion of the proposition follows from the definitions. Corollary 1.3.4 (Leptin): Every profinite group is isomorphic to a Galois group of some Galois extension.

1.4 The p-adic Integers and the Pr¨ ufer Group The first examples of profinite groups are the group Zp of p-adic numbers ˆ = lim Z/nZ. and the Pr¨ ufer group Z ←−

1.4 The p-adic Integers and the Pr¨ ufer Group

13

The p-adic group Zp . Let p be a rational prime. Consider the quotient rings Z/pi Z with their canonical homomorphisms Z/pj Z → Z/pi Z given for j ≥ i by x + pj Z 7→ x + pi Z. The inverse limit Zp = lim Z/pi Z is the ring ←− of p-adic integers. It is a profinite ring. Each x ∈ Zp is a sequence (xi + pi Z)i∈N where xi ∈ Z and xj ≡ xi mod pi Z for j ≥ i. Each integer m ≥ 0 corresponds to a basic neighborhood of x consisting of all elements y = (yi + pi Z)i∈N with ym ≡ xm mod pm Z. The map a 7→ (a + pi Z)i∈N is an embedding of Z into Zp . Identify Z with its image in Zp . The sequence (xi )i∈N converges to x = (xi + pi Z)i∈N in the p-adic topology. Hence, Z is dense in Zp . Yet, Z is not equal to Zp . For Pn−1 example, if p 6= 2, then ( i=0 pi + pn Z)n∈N belongs to Zp but not to Z. For Pn−1 p = 2, ( i=0 4i + 2n Z)n∈N belongs to Z2 but not to Z (see also Exercise 15). Lemma 1.4.1: The ring Zp has the following properties: (a) An element x = (xi + pi Z)i∈N is invertible if and only if p - x1 . (b) Zp is an integral domain. Proof of (a): Suppose x0 = (x0i + pi Z)i∈N is an inverse of x. Then x01 x1 ≡ 1 mod p, hence p - x1 . Conversely, suppose p - x1 . Then for each i, p - xi . Hence, there exists x0i ∈ Z which is unique modulo pi with x0i xi ≡ 1 mod pi . Thus, x0 = (x0i + pi Z)i∈N is in Zp and x0 x = 1. Proof of (b): Let x = (xi + pi )i∈N and y = (yi + pi )i∈N be nonzero elements of Zp . Then there are m, n ∈ N with xm 6≡ 0 mod pm and yn 6≡ 0 mod pn . Hence, xm+n ym+n 6≡ 0 mod pm+n . Therefore, xy 6= 0. Consequently, Zp is an integral domain. Lemma 1.4.2: (a) For each i, pi Zp is the kernel of the projection πi : Zp → Z/pi Z. Thus, pi Zp is an open subgroup of Zp of index pi . (b) If H is a subgroup of Zp of a finite index, then H = pi Zp for some i ∈ N. (c) 0 is the only closed subgroup of Zp of infinite index. (d) pZp is the unique closed maximal subgroup of Zp . (e) All nonzero closed subgroups of Zp are isomorphic to Zp . Proof of (a): Suppose x = (xj + pj Z)j∈N belongs to Ker(πi ). Then xi ≡ 0 mod pi . Hence, xj ≡ 0 mod pi for each j ≥ i. Write xj = pi yj for j ≥ i and yj = yi for j < i. Let z = (yj+i + pj Z)j∈N . Then z ∈ Zp and pi z = x. Indeed, pi zj = pi yj+i = xj+i ≡ xj mod pj for every positive integer j. Proof of (b): Conversely, let H be a subgroup of Zp of index n. Suppose n = kpi where p - k. By Lemma 1.4.1(a), k is invertible in the ring Zp , so nZp = pi Zp . Thus, pi Zp = nZp ≤ H. It follows, pi = (Zp : pi Zp ) ≥ (Zp : H) = kpi . Therefore, k = 1 and H = pi Zp . Proof of (c): A closed subgroup J of Zp of infinite index is the intersection of infinitely many open groups (Lemma 1.2.3). Hence, by (a), all subgroups pi Zp contain J. Consequently J = 0.

14

Chapter 1. Infinite Galois Theory and Profinite Groups

Proof of (d): By (a), (b), and (c), pZp is the unique closed maximal subgroup of Zp . Proof of (e): By Lemma 1.4.1(b), the map x 7→ pi x is an isomorphism of Zp onto pi Zp . Every element xP= (xi + pi Z)i∈N has a unique representation as a ∞ i formal power series i=0 ai p , with 0 ≤ ai < p for all i. Indeed, xn ≡ Pn−1 i n i=0 ai p mod p , for every n ∈ N. Lemma 1.4.3: Let α: Zp → Z/pn Z be an epimorphism with n ≥ 1 and H a closed subgroup of Zp . Suppose α(H) = Z/pn Z. Then H = Zp . Proof: By Lemma 1.4.2(a), Ker(α) = pn Zp . Thus, by assumption, H + pn Zp = Zp . Assume H 6= Zp . Then, by Lemma 1.4.2(d), H ≤ pZp . Therefore, Zp = H + pn Zp ≤ pZp < Zp . It follows from this contradiction that H = Zp . In the terminology of Section 22.5, Lemma 1.4.3 says that α: Zp → Z/pn Z is a Frattini cover. ¨ fer group. For each n ∈ N consider the quotient group Z/nZ The Pru and the canonical homomorphisms Z/nZ → Z/mZ defined for m|n by x + ˆ = lim Z/nZ is the Pr¨ nZ 7→ x + mZ. The inverse limit Z ufer group. Like ←− ˆ by x 7→ (x + nZ)n∈N . Thus, Z ˆ with Zp , embed Z as a dense subgroup of Z ˆ is the closure of the subgroup generated by 1. Write Z = h1i and say that 1 ˆ Also, the subgroups nZ of Z form a basis for the neighborhoods generates Z. of 0 in the induced topology. ˆ is an open subgroup of Z ˆ of index n and Lemma 1.4.4: For each n ∈ N, nZ ∼ ˆ ˆ ˆ ˆ nZ = Z. If H is a subgroup of Z of index n, then H = nZ. Proof of: Suppose x = (xk + kZ)k∈Z lies in the kernel Zn of the projection ˆ → Z/nZ. Then xn ≡ 0 mod n. Hence, for each r ∈ N we have xrn ≡ Z xn ≡ 0 mod n, so xrn = nyrn for some yrn ∈ Z. Let z = (yrn + rZ)r∈Z . If r0 is a multiple of r, then nyr0 n ≡ nyrn mod rn, so yr0 n ≡ yrn mod r. ˆ Moreover, xr ≡ xrn ≡ nyrn = nzr mod r. Hence, x = nz. Therefore, z ∈ Z. ˆ is an open subgroup of Z ˆ of index n. Consequently, nZ ˆ onto nZ. ˆ Indeed, Next note that the map x 7→ nx is an isomorphism of Z if nx = 0, then nxrn ≡ 0 mod rn, so xr ≡ xrn ≡ 0 mod r for each r ∈ N. Hence, x = 0. ˆ of index n, then nZ ˆ is contained in H Finally, if H is a subgroup of Z ˆ and has the same index. Therefore, H = nZ. ˆ to the groups Zp . We conclude by relating Z

1.5 The Absolute Galois Group of a Finite Field

15

ˆ is topologically isomorphic to the cartesian prodLemma1.4.5: The group Z Q uct Zp where p ranges over all primes numbers. Q Proof: Let n = pkp be the decomposition of a positive integer n into a product of prime Q powers. The Chinese remainder theorem gives a canonthe projection ical isomorphism Z/pkp Z → Z/nZ. Combine this with Q Q Q kp Zp → Z/nZ. Zp → Z/p Z to obtain a continuous epimorphism fn : The maps Q fn form a compatible system, so they give a continuous homomorˆ Since Q Zp is compact and Z ˆ is Hausdorff, Im(f ) is a phism f : Zp → Z. ˆ Moreover, Z embeds diagonally in Q Zp and f (m) = m closed subgroup of Z. ˆ ˆ for each m ∈ Z. Thus, Z ⊆ Im(f ). Since T Z is dense in Z, we have Im(f ) = Z, so f is surjective. The kernel of f is Ker(fn ) = 0. Hence, f is also injective. The compactness and Hausdorff properties imply that f is a topological isomorphism. As a consequence of Lemma 1.4.5, Zp is both a closed subgroup and a ˆ for each prime p. quotient of Z,

1.5 The Absolute Galois Group of a Finite Field For every prime power q there exists a field Fq (unique up to isomorphism) with q elements. It is characterized within its algebraic closure F˜q by Fq = {x ∈ F˜q | xq = x}. The field Fq has, for each n ∈ N, exactly one extension, Fqn , of degree n. It is Galois with a cyclic group generated by the Frobenius automorphism a is πq,n defined by πq,n (x) = xq for x ∈ Fqn . The map a + nZ 7→ πq,n an isomorphism of Z/nZ onto Gal(Fqn /Fq ). If m|n, there is a canonical commutative diagram / Z/mZ

Z/nZ Gal(Fqn /Fq )

res

/ Gal(Fqm /Fq ).

ˆ∼ Take the inverse limits to obtain an isomorphism Z = Gal(Fq ) mapping the ˆ to the Frobenius automorphism πq , defined on all identity element 1 of Z ˜ q by πq (x) = xq . of F S∞ (l) Let l be a prime number and Fq = i=1 Fqli . Then the projection ˆ 7→ Zl corresponds to res: Gal(Fq ) → Gal(F(l) Z q /Fq ). By Lemma 1.4.5, Q (l) ∼ Gal(Fq ) = Gal(Fq /Fq ). On the other hand, let Nl be the fixed field of Zl (l) ˜ q . Then Gal(Nl ) = Zl , F(l) ˜ in F q Nl = Fq , and Fq ∩ Nl = Fq . It follows, Y ∼ Zl0 × Zl . Gal(Fq ) = Gal(F(l) q ) × Gal(Nl ) = l0 6=l

16

Chapter 1. Infinite Galois Theory and Profinite Groups

Exercises 1. Let (Si , πji ) be an inverse system of finite sets with all πji surjective and let S = lim Si . Use Lemma 1.1.3 to prove that all maps πi : S → Si ←− determined by the πji ’s are surjective. 2.

Let H1 , . . . , Hr be closed subgroups of a profinite group G. Prove that \ H1 ∩ · · · ∩ Hr = (H1 N ∩ · · · ∩ Hr N ), N

where N ranges over all open normal subgroups of G. Hint: Use Lemma 1.2.2(b). 3. Suppose H is a closed subgroup of a profinite group G. Prove: If HN/N = G/N for every open normal subgroup N of G, then H = G. 4. Let H be an open subgroup of index n of a profinite group G. Denote the intersection of all conjugates of H in G by N . Note: Multiplication of G on the left cosets of H induces a homomorphism of G into the symmetric group Sn with kernel N . Conclude that G/N is isomorphic to a subgroup of Sn and (G : N ) ≤ n!. 5. Let G be a compact group and H an open subgroup. Suppose H is profinite. Prove: G is profinite. Hint: Use Lemma 1.2.4. 6. Let S be a set of rational primes. Consider the profinite group ZS = lim Z/nZ, with n running over all positive integers with prime factors in S. ←− (a) Prove: The finite homomorphic images of ZS are exactly the groups Z/nZ, where the prime factors of n belong to S. (b) Embed Z in ZS and determine the topology on Z induced by that of ZS . (c) Prove that ZS is procyclic (i.e. ZS is the closure of a group generated by one element). Q (d) Follow the proof of Lemma 1.4.5 to prove that ZS ∼ = p∈S Zp . 7. Let G be a procyclic group (Exercise 6). Use that G is a homomorphic ˆ to show there exists a set S of primes with G = Q image of Z p∈S Gp , where ip ∼ for each p ∈ S either Gp ∼ Z for some i ∈ N or G . Z/p Z = p p = p In particular, Q Z . if G is torsion free, then G ∼ = p∈S p ˆ Prove that every finite quotient of G is 8. Let G be a closed subgroup of Z. a cyclic group. Conclude that G is procyclic and therefore that there exists Q a set S of prime numbers with G ∼ = l∈S Zl . 9. Let G be a profinite group. Prove that each of the following statements ˆ is equivalent to G ∼ = Z. (a) G has exactly one open subgroup of each index n. ˆ (b) G is procyclic and there is an epimorphism π: G → Z.

Exercises

17

10. Let G be a procyclic group. Use Exercise 8 to prove that each epimorˆ is an isomorphism. phism π: G → Z 11. Let G be a profinite group with at most one open subgroup of every index n. (a) Prove that every open subgroup is normal. ¯ of G. Con(b) Observe that (a) holds for every finite homomorphic image G ¯ clude that G is nilpotent. (c) Let P be a finite p-group with the above properties. Prove that every element x of P of maximal order generates P . (d) Conclude that G is a procyclic group. ˆ in the following 12. Define powers in a profinite group G with exponents in Z ˆ way: Let g ∈ G and ν ∈ Z. Then there exists a sequence {ν1 , ν2 , ν3 , . . .} of elements of Z that converges to ν. By compactness there is a subsequence of {g ν1 , g ν2 , g ν3 , . . .} that converges to an element h of G. (a) Prove h does not depend on the sequence {ν1 , ν2 , ν3 , . . .}. So we may denote h by g ν . Hint: If N is a normal subgroup of G of index n, then xn ∈ N for every x ∈ G. (b) Prove the usual rules for the power operations. For example, g µ g ν = g µ+ν ,

(g µ )ν = g µν ,

and

g ν hν = (gh)ν if gh = hg.

ˆ into G is continuous. (c) Prove that the map (g, ν) 7→ g ν of G × Z 13. Multiplication in the groups Z/pi Z is compatible with the canonical maps Z/pi+1 Z → Z/pi Z. Therefore, it defines a multiplication in the additive group Zp = lim Z/pi Z. ←− (a) Prove: Zp is an integral domain (the quotient field of which, Qp , is the field of p-adic numbers). (b) Show: Every closed subgroup of Zp is an ideal of Zp . (c) Show: pZp is the unique maximal ideal of Zp ; observe that Zp /pZp ∼ = Fp . (d) Deduce: α ∈ Zp is a unit (i.e. invertible in this ring) if and only if α is congruent modulo pZp to one of the numbers 1, 2, . . . , p − 1. Hint: Form the inverse of 1 + βp, β ∈ Zp using the geometric series for 1/(1 + x). ˆ in a manner analogous to 14. Define multiplication in the additive group Z ˆ a commutative the definition of multiplication in Zp . Prove that this makes Z topological ring with zero divisors. Q ˆ∼ (a) Prove that the isomorphism of additive groups Z = Zp established in Lemma 1.4.5 is an isomorphism of rings. ˆ is also an ideal. (b) Prove that every closed subgroup of Z 15. Use the power series representation of the elements of Zp (Section 1.4) to show |Zp | = 2ℵ0 . Conclude that Qp has elements that are transcendental over Q.

18

Chapter 1. Infinite Galois Theory and Profinite Groups

√ 16. For a prime number p, let K be Q(ζ ) if p = 6 2 and Q( −1) if p = 2. p p S∞ Also, let Lp = i=1 Q(ζpi ). (a) Prove that if K 0 is a field such that Kp ⊆ K 0 ⊂ Lp , then Gal(Lp /K 0 ) ∼ = Zp and [K 0 : Q] < ∞. (b) Prove that Gal(Lp /Q) is isomorphic to Zp × Z/(p − 1)Z if p 6= 2 and to Z2 × Z/2Z if p = 2.

Notes More about topological groups can be found in [Pontryagin]. A detailed exposition on Galois theory of finite extensions appears in [Lang7, Chapter VI, Section 1]. For finite fields see [Langl, Chapter V, Section 5]. Leptin’s proof of Corollary 1.3.4 uses linear disjointness of fields [Leptin]. We reproduce it in Proposition 2.6.12. The proofs of Lemma 1.3.2 and Proposition 1.3.3 appear in [Waterhouse]. This chapter overlaps with [Ribes, Chapter 1].

Chapter 2. Valuations and Linear Disjointness Sections 2.1–2.4 introduce the basic elements of the theory of valuations, especially discrete valuations, and of Dedekind domains. These sections are primarily a survey. We prove that an overring of a Dedekind domain is again a Dedekind domain (Proposition 2.4.7). The rest of the chapter centers around the notion of linear disjointness of fields. We use this notion to define separable, regular, and primary extensions of fields. In particular, we prove that an extension F/K with a K-rational place is regular. Section 2.8 gives a useful criterion for separability with derivatives.

2.1 Valuations, Places, and Valuation Rings The literature treats arithmetic theory of fields through three intimately connected classes of objects: valuations, places, and valuation rings. We briefly review the basic definitions. Call an Abelian (additive) group Γ with a binary relation < an ordered group if the following statements hold for all α, β, γ ∈ Γ. (1a) Either α < β, or α = β, or β < α. (1b) If α < β and β < γ, then α < γ. (1c) If α < β, then α + γ < β + γ. Some examples of ordered groups are the additive groups Z, R, and Z⊕Z with the order (m, n) < (m0 , n0 ) if either m < m0 or m = m0 and n < n0 (the lexicographic order). A valuation v of a field F is a map of F into a set Γ ∪ {∞}, where Γ is an ordered group, with these properties: (2a) v(ab) = v(a) + v(b). (2b) v(a + b) ≥ min v(a), v(b) . (2c) v(a) = ∞ if and only if a = 0. (2d) There exists a ∈ F × with v(a) 6= 0. By definition the symbol ∞ satisfies these rules: (3a) ∞ + ∞ = α + ∞ = ∞ + α = ∞; and (3b) α < ∞ for each α ∈ Γ. Condition (2) implies several more properties of v: (4a) v(1) = 0, v(−a) = v(a). (4b) If v(a) < v(b), then v(a + b) = v(a) (Use the identity a = (a + b) − b and P(2b)); n (4c) If i=1 ai = 0, then there exist i 6= j such that v(ai ) = v(aj ) and v(ai ) = min(v(a1 ), . . . , v(an )) (Use (2b) and (4b)). We refer to the pair (F, v) as a valued field.

20

Chapter 2. Valuations and Linear Disjointness

The subgroup Γv = v(F × ) of Γ is the value group of v. The set Ov = {a ∈ F | v(a) ≥ 0} is the valuation ring of v. It has a unique maximal ideal mv = {a ∈ F | v(a) > 0}. Refer to the residue field F¯v = Ov /mv as the residue field of F at v. Likewise, whenever there is no ambiguity, we ¯ and call it the residue of a at v. denote the coset a + mv by a Two valuations v1 , v2 of a field F with value groups Γ1 , Γ2 are equivalent if there exists an isomorphism f : Γ1 → Γ2 with v2 = f ◦ v1 . Starting from Section 2.2, we abuse our language and say that v1 and v2 are distinct if they are inequivalent. A place of a field F is a map ϕ of F into a set M ∪ {∞}, where M is a field, with these properties: (5a) ϕ(a + b) = ϕ(a) + ϕ(b). (5b) ϕ(ab) = ϕ(a)ϕ(b). (5c) There exist a, b ∈ F with ϕ(a) = ∞ and ϕ(b) 6= 0, ∞. By definition the symbol ∞ satisfies the following rules: (6a) x + ∞ = ∞ + x = ∞ for each x ∈ M . (6b) x · ∞ = ∞ · x = ∞ · ∞ = ∞ for each x ∈ M × . (6c) Neither ∞ + ∞, nor 0 · ∞ are defined. It is understood that (5a) and (5b) hold whenever the right hand side is defined. These conditions imply that ϕ(1) = 1, ϕ(0) = 0 and ϕ(x−1 ) = ϕ(x)−1 . In particular, if x 6= 0, then ϕ(x) = 0 if and only if ϕ(x−1 ) = ∞. We call an element x ∈ F with ϕ(x) 6= ∞ finite at ϕ, and say that ϕ is finite at x. The subring of all elements finite at ϕ, Oϕ = {a ∈ F | ϕ(a) 6= ∞}, is the valuation ring of ϕ. It has a unique maximal ideal mϕ = {a ∈ F | ϕ(a) = 0}. The quotient ring Oϕ /mϕ is a field which is canonically isomorphic to the residue field F¯ϕ = {ϕ(a) | a ∈ Oϕ } of F at ϕ. The latter is a subfield of M . Call ϕ a K-place if K is a subfield of F and ϕ(a) = a for each a ∈ K. Two places ϕ1 and ϕ2 of a field F with residue fields M1 and M2 are equivalent if there exists an isomorphism λ: M1 → M2 with ϕ2 = λ ◦ ϕ1 . A valuation ring of a field F is a proper subring O of F such that if / O} is x ∈ F × , then x ∈ O or x−1 ∈ O. The subset m = {x ∈ O | x−1 ∈ the unique maximal ideal of O (Exercise 1). The map ϕ: F → O/m ∪ {∞} which maps x ∈ O onto its residue class modulo m and maps x ∈ F r O onto ∞ is a place of F with valuation ring O. Denote the units of O by U = {x ∈ O | x−1 ∈ O}. Then F × /U is a multiplicative group ordered by the rule xU ≤ yU ⇐⇒ yx−1 ∈ O. The map x 7→ xU defines a valuation of F with O being its valuation ring. These definitions easily give a bijective correspondence between the valuation classes, the place classes and the valuation rings of a field F . An isomorphism σ: F → F 0 of fields induces a bijective map of the valuations and places of F onto those of F 0 according to the following rule: If v is a valuation of F , then σ(v) is defined by σ(v)(x) = v(σ −1 x) for every x ∈ F 0 . If ϕ is a place of F , then σ(ϕ)(x) = ϕ(σ −1 x). In particular, σ

2.2 Discrete Valuations

21

0 induces an isomorphism F¯ϕ ∼ of residue fields. It is also clear that if = F¯σ(ϕ) ϕ corresponds to v, then σ(ϕ) corresponds to σ(v). A valuation v of a field F is real (or of rank 1) if Γv is isomorphic to a subgroup of R. Real valuations satisfy the so called weak approximation theorem, a generalization of the Chinese remainder theorem [CasselsFr¨ohlich, p. 48]:

Proposition 2.1.1: Consider the following objects: inequivalent real valuations v1 , . . . , vn of a field F , elements x1 , . . . , xn of F , and real numbers γ1 , . . . , γn . Then there exists x ∈ F with vi (x − xi ) ≥ γi , i = 1, . . . , n.

2.2 Discrete Valuations ∼ Z. In this case we normalize A valuation v of a field F is discrete if v(F × ) = v by replacing it with an equivalent valuation such that v(F × ) = Z. Each element π ∈ F with v(π) = 1 is a prime element of Ov . Prime elements of a unique factorization domain R produce discrete valuations of F = Quot(R). If p is a prime element of R, then every element x of F × has a unique representation as x = upm , where u is relatively prime to p and m ∈ Z. Define vp (x) to be m. Then vp is a discrete valuation of F . Suppose p0 is another prime element of R. Then vp0 is equivalent to vp if and only if p0 R = pR, that is if p0 = up with u ∈ R× . Example 2.2.1: Basic examples of discrete valuations. (a) The ring of integers Z is a unique factorization domain. For each prime number p the residue field of Q at vp is Fp . When p ranges over all prime numbers, vp ranges over all valuations of Q (Exercise 3). (b) Let R = K[t] be the ring of polynomials in an indeterminate t over a field K. Then R is a unique factorization domain. Then prime elements of R are the irreducible polynomials p over K. Units of R are the elements u of K × , so vp (u) = 0 and we say vp is trivial on K. The residue field of K(t) at vp is isomorphic to the field K(a), where a is a root of p. There is one additional valuation, v∞ , of K(t) which is trivial on K. It is defined for a quotient fg of elements of K[t] by the formula v∞ ( fg ) = deg(g) − deg(f ). The set of vp ’s and v∞ give all valuation of K(t) trivial on K. Thus, all valuations of K(t) which are trivial on K are discrete (Exercise 4). ˜ An arbitrary irreducible polynomial p may have several roots a ∈ K. ˜ ∪{∞} by ϕa (t) = a and ϕa (c) = c Each of them defines a place ϕa : K(t) → K for each c ∈ K. These places are equivalent. If p(t) = t − a, then ϕa is the unique place of K(t) corresponding to vp . Similarly, there is a unique place ϕ∞ corresponding to v∞ . It is defined by ϕ∞ (t) = ∞. We may view each f (t) ∈ K(t) as a function from K ∪ {∞} into itg(t) with g, h ∈ K[X] and self: f (a) = ϕa (f (t)). Explicitly, write f (t) = h(t) gcd(g, h) = 1. Let a ∈ K. Then f (a) =

g(a) h(a)

if h(a) 6= 0 and f (a) = ∞

22

Chapter 2. Valuations and Linear Disjointness

if h(a) = 0. To compute f (∞) let u = t−1 and write f (t) =

g1 (u) h1 (u)

with

g1 , h1 ∈ K[X] and gcd(g1 , h1 ) = 1. Then f (∞) = hg11(0) (0) if h1 (0) 6= 0 and f (∞) = ∞ if h1 (0) = 0. Suppose for example f (t) ∈ K[t] and f 6= 0. Then f maps K into itself and f (∞) = ∞. Now suppose f (t) = at+b ct+d with ad − bc 6= 0 and c 6= 0, then f (∞) = ac . When K is algebraically closed, each irreducible polynomial is linear. Hence, each valuation of K(t) which is trivial over K is either vt−a for some a ∈ K or v∞ . More examples of discrete valuations arise through extensions of the basic examples (Section 2.3). Lemma 2.2.2: Every discrete valuation ring R is a principal ideal domain. Proof: Let v be the valuation of K = Quot(R) with Ov = R and v(K × ) = Z. Choose a prime element π of R. Now consider a nonzero ideal a of R. Then the minimal integer m with π m ∈ a is positive. It satisfies, a = π m R. As a consequence of Lemma 2.2.2, finitely generated modules over R have a simple structure. Proposition 2.2.3: Let R be a discrete valuation ring, p a prime element ¯ = R/pR. of R, K = Quot(R), and M a finitely generated R-module. Put K Let r = dimK M ⊗R K, n = dimK¯ M/pM , and m = n − r. Then there is a unique m-tuple of positive integers (k1 , k2 , . . . , km ) with k1 ≤ k2 ≤ · · · ≤ km and M ∼ = R/pkm R ⊕ · · · ⊕ R/pk1 R ⊕ Rr . Moreover, r is the maximal number of elements of M which are linearly independent over R and n is the minimal number of generators of M . Proof: By Lemma 2.2.2, R is a principal ideal domain, so M = Mtor ⊕ N , where Mtor = {m ∈ M | rm = 0 for some r ∈ R, r 6= 0} and N is a free R-module [Lang7, p. 147, Thm. 7.3]. Both Mtor and N are finitely generated [Lang7, p. 147, Cor. 7.2]. In particular, N ∼ = Rs for some integer s ≥ 0. Suppose m ∈ Mtor and am = 0 with a ∈ R, a 6= 0. Then, m⊗1 = am⊗ a1 = 0. Hence, Mtor ⊗R K = 0 and M ⊗R K ∼ = K s . Therefore, s = r. By [Lang7, p. 151, Thm. 7.7], Mtor ∼ = R/qm0 R ⊕ · · · ⊕ R/q1 R where q1 , . . . , qm0 are elements of R which are neither zero nor units and qi |qi+1 , i = 1, . . . , m0 − 1. Multiplying each qi by a unit, we may assume qi = pki with ki an integer and 1 ≤ k1 ≤ k2 ≤ · · · ≤ km0 . Moreover, the above cited theorem assures Rq1 , . . . , Rqm0 are uniquely determined by the above conditions. Hence, k1 , . . . , km0 are also uniquely determined. Combining the first two paragraphs gives: M∼ = R/pkm0 R ⊕ · · · ⊕ R/pk1 R ⊕ Rr . 0 ¯ m0 +r , so n = m0 + r and m0 = m. Hence, M/pM = (R/pR)m +r ∼ =K

2.2 Discrete Valuations

23

Now recall Ps that elements v1 , . . . , vs of M are linearly independent over R if i=1 ai vi = 0 with a1 , . . . , as ∈ R implies a1 = · · · = as = 0. Alternatively, v1 ⊗ 1, . . . , vs ⊗ 1 are linearly independent over K. Thus, r is the maximal number of R-linearly independent elements of M . Finally, by Nakayama’s lemma [Lang7, p. 425, Lemma 4.3], n is the minimal number of generators of M . Definition 2.2.4: Let R be an integral domain with quotient field F . An overring of R is a ring R ⊆ R0 ⊂ F . It is said to be proper if R 6= R0 . Lemma 2.2.5: A discrete valuation ring O has no proper overrings. Proof: Let R be an overring of O. Assume there exists x ∈ R r O. Then x−1 is a nonunit of O. Choose a prime element π for O. Then x = uπ −m for some u ∈ O× and a positive integer m. Hence, π −1 = u−1 π m−1 x ∈ R. Therefore, u0 π k ∈ R for all u0 ∈ O× and k ∈ Z. We conclude that R = Quot(O). Composita of places attached to discrete valuations of rational function fields of one variable give rise to useful places of rational function fields of several variables. Construction 2.2.6: Composition of places. Suppose ψ is a place of a field K with residue field L and ϕ is a place of L with residue field M . Then ψ −1 (Oϕ ) is a valuation ring of K with maximal ideal ψ −1 (mϕ ) and residue field ψ −1 (Oϕ )/ψ −1 (mϕ ) ∼ = Oϕ /mϕ ∼ = M . Define a map ϕ ◦ ψ: K → M ∪ {∞} as follows: ϕ ◦ ψ(x) = ϕ(ψ(x)) if ψ(x) 6= ∞ and ϕ ◦ ψ(x) = ∞ if ψ(x) = ∞. Then ϕ ◦ ψ is a homomorphism on ψ −1 (Oϕ ) and {x ∈ K | ϕ ◦ ψ(x) = ∞} = K r ψ −1 (Oϕ ). Therefore, ϕ ◦ ψ is a place of K, called the compositum of ψ and ϕ, Oϕ◦ψ = ψ −1 (Oϕ ), and mϕ◦ψ = ψ −1 (mϕ ). K Oψ Oϕ◦ψ mϕ◦ψ mψ

/ L ∪ {∞}

ϕ

/ M ∪ {∞}

ψ

/L

ϕ

/ M ∪ {∞}

ψ

/ Oϕ

ϕ

/M

ψ

/ mϕ

ϕ

/0

ψ

/0

ϕ

/0

ψ

¯ ψ and M = L ¯ϕ = K ¯ ϕ◦ψ . In addition, L = K

˜ t1 , . . . , tr indeLemma 2.2.7: Let K be a field, a1 , . . . , ar elements of K, terminates, and L a finite extension of K. Then there exists a K-place ϕ: K(t) → K(a) ∪ {∞} such that ϕ(ti ) = ai , i = 1, . . . , r. Moreover, every extension of ϕ to an L-place of L(t) maps L(t) onto L(a) ∪ {∞}. Proof: For each i there is a K(a1 , . . . , ai−1 , ti+1 , . . . , tr )-place ϕi : K(a1 , . . . , ai−1 , ti , ti+1 . . . , tr ) → K(a1 , . . . , ai−1 , ai , ti+1 , . . . , tr )

24

Chapter 2. Valuations and Linear Disjointness

with ϕi (ti ) = ai (Example 2.2.1). The compositum ϕ = ϕr ◦ · · · ◦ ϕ1 is a K-place of K(t1 , . . . , tr ) with residue field K(a1 , . . . , ar ) and ϕ(ti ) = ai , i = 1, . . . , r. Let now ϕ be an extension of ϕ to an L-place of L(t). Choose a basis also a basis for L(t)/K(t). Hence, b1 , . . . , bn for L/K. Then b1 , . . . , bn is P n each f ∈ L(t) has a presentation f = i=1 bi fi with fi ∈ K(t). Assume Pn without loss that ff1i is finite under ϕ for i = 1, . . . , n. Then f = f1 i=1 ff1i bi and ϕ(f ) ∈ L(a1 , . . . , ar ) ∪ {∞}. Thus, ϕ(L(t)) = L(a) ∪ {∞}.

2.3 Extensions of Valuations and Places The examples of Section 2.2 and the following extension results give a handle on describing valuations of function fields in one variable. Proposition 2.3.1 (Chevalley [Lang4, p. 8, Thm. 1]): Let ϕ0 be a homomorphism of an integral domain R into an algebraically closed field M and let F be a field containing R. Then ϕ0 extends either to an embedding ϕ of F into M or to a place ϕ of F into M ∪ {∞}. When F is algebraic over R, the proposition has a more precise form: Let f ∈ R[X] be an irreducible polynomial over E = Quot(R) and f¯ ∈ M [X] the result of applying ϕ0 to the coefficients of f . Suppose f¯ is ˜ and M , not identically zero. Assume x and x ¯ are roots of f and f¯ in E respectively. Then ϕ0 extends to a place ϕ of E(x) into M ∪ {∞} with ϕ(x) = x ¯ [Lang4, p. 10, Thm. 2]. Moreover, if ϕ0 is injective, so is ϕ [Lang4, p. 8, Prop. 2]. In particular, suppose v is a valuation of a field E and F is an extension of E. Then v extends to a valuation w0 of F . Each valuation w of F which is equivalent to w0 lies over v. Thus, w lies over v if and only if Ov ⊆ Ow and mv = mw ∩ Ov . The number ew/v = (w(F × ) : w(E × )) is the ramification index of w over v (and also over E). The field degree [F : E] ¯v embeds in F¯w to give the inequality bounds ew/v (Exercise 5). Similarly, E ¯v ] ≤ [F : E] (Exercise 7). Both the ramification index and fw/v = [F¯w : E the residue field degree are multiplicative. Thus, if (F 0 , w0 ) is an extension of (F, w), then ew0 /v = ew0 /w ew/v and fw0 /v = fw0 /w fw/v . If [F : E] < ∞, then the number of valuations of F that lie over v is finite (a consequence of Proposition 2.3.2). Proposition 2.3.2: Let F/E be a finite extension of fields and v a valuation of E. Let w1 , . . . , wg be all inequivalent extensions of v to F . Then

(1)

g X

ewi /v fwi /v ≤ [F : E]

i=1

[Bourbaki2, p. 420, Thm. 1]. If, in addition, v is discrete and F/E is separa-

2.3 Extensions of Valuations and Places

25

ble, then each wi is discrete and (see [Bourbaki2, p. 425, Cor. 1]) g X

(2)

ewi /v fwi /v = [F : E].

i=1

Suppose (F, w)/(E, v) is an extension of discrete valued fields. In particular, w(a) = v(a) for each a ∈ E. By definition, ew/v = (w(F × ) : v(E × )). However, as in Section 2.2, it is customary to replace v and w by equivalent valuations with v(E × ) = w(F × ) = Z. The new valuations satisfy w(a) = ew/v v(a)

for each a ∈ E.

Whenever we speak about an extension of discrete valuations, we mean they are normalized and satisfy the latter relation. Suppose F is a finite Galois extension of E with a Galois group G. Let w be a discrete valuation of F and let σ ∈ G. Then, σ(w) is a valuation of F (Section 2.1), both w and σ(w) lie over the same valuation v of E, and ew/v = eσ(w)/v

and fw/v = fσ(w)/v .

Conversely, suppose w and w0 are two discrete valuations of F over the same valuation v of E. Then there exists σ ∈ G such that σ(w) = w0 (Exercise 9). Thus, if w1 , . . . , wg are all distinct valuations of F that lie over v, then they all have the same residue degree f and ramification index e over v. In this case formula (2) simplifies to (3)

ef g = [F : E]. The subgroups Dw = Dw/v = {σ ∈ G | σOw = Ow } Iw = Iw/v = {σ ∈ G | w(x − σx) > 0 for all x ∈ Ow }

are the decomposition group and the inertia group, respectively, of w ¯v is separable, then [Serre3, p. 33] over E. Obviously Iw / Dw . If F¯w /E (4)

|Iw | = ew/v

and |Dw | = ew/v fw/v .

Section 2.6 generalizes the notion of separable algebraic extension of fields to arbitrary extensions of fields. In particular, purely transcendental extensions of fields are separable. We use this notion in the following definition. Suppose (F, w)/(E, v) is an arbitrary extension of valued fields. We say ¯v w is unramified (resp. tamely ramified) over v (or also over E) if F¯w /E ¯ is a separable extension and ew/v = 1 (resp. char(Ev ) - ew/v ). We say v is unramified (resp. tamely ramified) in F if each extension of v to F is unramified (resp. tamely ramified) over v.

26

Chapter 2. Valuations and Linear Disjointness

Example 2.3.3: Purely transcendental extensions. Let (E, v) be a valued field. Consider a transcendental element t over E. Extend v to a valuation v 0 of E(t) as follows. First define v 0 on E[t] by the following rule: v0

(5)

m X

ai ti = min v(a0 ), . . . , v(am )

i=0

for a0 , . . . , am ∈ E. The same argument used to prove Gauss’ Lemma proves 0 (g) for all f, g ∈ E[t].P that v 0 (f g) = v 0 (f ) + vP m n Indeed, let f (t) = i=0 ai ti and g(t) = j=0 bj tj . Let r be the minimal integer with v(ar ) = min v(a0 ), . . . , v(am ) and let s be the minimal integer with v(bs ) = min v(b0 ), . . . , v(bn ) . If i + j = r + s and (i, j) 6= (r, s), then either i < r or j < s. In both cases v(ar ) + v(bs ) < v(ai ) + v(bj ). Hence v0

m X

n X ai ti + v 0 bj tj = v(ar ) + v(bs )

i=0

j=0

X

= min

v(ai bj ) | k = 0, . . . , m + n

i+j=k

= v0

m X

ai ti ·

i=0

n X

b j tj ,

j=0

as claimed. We extend v 0 to E(t) by the rule v 0 ( fg ) = v 0 (f ) − v 0 (g). Then we prove v 0 (u1 + u2 ) ≥ min v 0 (u1 ), v 0 (u2 ) first for u1 , u2 ∈ E[t] and then for u1 , u2 ∈ Note that the residue of t at v 0 is E(t). Thus, v 0 is a valuation of E(t). P n ¯ ¯i t¯i = 0 for some a0 , . . . , an ∈ transcendental over Ev . Indeed, suppose i=0 a P n 0 i > 0. Hence, a ¯i = 0, Ov . Then min v(a0 ), . . . , v(an ) = v i=0 ai t i = 0, . . . , n. ¯v (t¯) is a rational function field over E ¯v . By It follows that, E(t)v0 = E 0 definition, Γv0 = Γv . In particular, if v is discrete, then so is v and ev0 /v = 1. Suppose v 00 is another extension of v to E(t) with the residue of t at v 00 ¯v . We show that v 00 = v 0 . Indeed, for a0 , . . . , an ∈ E, transcendental over E not all zero, choose j between 0 and n with v(aj ) = min v(a0 ), . . . , v(an ) . Pn Then i=0 ai /aj t¯i 6= 0. Therefore, v

00

n X i=0

ai t

i

= v(aj ) + v

00

n X

(ai /aj )ti

i=0 n X = min v(a0 ), . . . , v(an ) = v 0 ai ti , i=0

as claimed.

2.3 Extensions of Valuations and Places

27

Lemma 2.3.4: Let v be a discrete valuation of a field E, h ∈ Ov [X] a monic ˜ and F = E(x). irreducible polynomial of degree n, x a root of h(X) in E, ¯ Suppose the residue polynomial h(X) is separable. Then v is unramified in F. Qr ¯ ¯v [X] are distinct Proof: By assumption, h(X) = i=1 hi (X), where hi ∈ E monic irreducible polynomials. For each i between 1 and r choose a root ai ¯v )s . Use Proposition 2.3.1 to extend the residue map Ov → E ¯v of hi (X) in (E to a place ϕi of F with ϕi (x) = ai . Denote the corresponding valuation by ¯v (ai ) ⊆ F¯w . Since hi (X) and hj (X) have no common root for wi . Then E i i 6= j, the valuations w1 , . . . , wr are mutually inequivalent extensions of v. Label any further extensions of v to valuations of F as wr+1 , . . . , wg . By (1) n=

r X

deg(hi ) =

i=1

r X

¯v (ai ) : E ¯v ] ≤ [E

i=1

g X

ewi /v fwi /v ≤ n.

i=1

¯v (ai ) = F¯w for i = 1, . . . , r. Moreover, w1 , . . . , wr are Hence, ewi /v = 1 and E i all extensions of v to F and each of them is unramified over E. Therefore, v is unramified in F . ¯v to be infinite. The converse of Lemma 2.3.4 requires E Lemma 2.3.5: Let v be a discrete valuation of a field E. Let F be a separable ¯v is an infinite extension of E of degree n. Suppose v is unramified in F and E field. Then F/E has a primitive element x with irr(x, E) ∈ Ov [X] and the residue of irr(x, E) at v is a separable polynomial. Proof: Let w1 , . . . , wg be all extensions of v to F . By (2), [F : E] = P g ¯ ¯ ¯ ¯ i=1 [Fwi : Ev ]. Moreover, for each i the extension Fwi /Ev is finite and separable. Hence, we may choose ci in F with wi (ci ) = 0 and the residue ¯v . Let hi = irr(¯ ¯v ). Since ci , E c¯i of ci at wi is a primitive element of F¯wi /E ¯v is infinite, we may choose c1 , . . . , cg such that c¯1 , . . . , c¯g are mutually E ¯v . Thus, h1 , . . . , hg are relatively prime. nonconjugate over E Use Proposition 2.1.1 to find x ∈ F with wi (x − ci ) > 0, i = 1, . . . , g. Then, wi (x) = 0, i = 1, . . . , g. Extend each wi to the Galois closure of F/E. Then all E-conjugates of x have nonnegative values under each extended valuation. Hence, the elementary symmetric polynomials in the E-conjugates of x belong to Ov . Therefore, f (X) = irr(x, E) ∈ Ov [X]. Let f¯ be the residue of f at v. By construction, Q f¯(¯ ci ) = 0, therefore g ¯ hi |f , i = 1, . . . , g. Since h1 , . . . , hg are relatively prime, i=1 hi |f¯. Hence, [F : E] =

g X i=1

¯v ] = [F¯wi : E

g X

deg(hi )

i=1

≤ deg(f¯) = deg(f ) = [E(x) : E] ≤ [F : E]. Consequently, E(x) = F , as desired.

28

Chapter 2. Valuations and Linear Disjointness

¯v to be infinite is necessary Example 3.5.4 shows the assumption on E for Lemma 2.3.5 to hold. The next lemma says that arbitrary change of the base field preserves unramified discrete valuations. Lemma 2.3.6: Let (E, v) be a discrete valued field. Consider a separable algebraic extension F of E and a discrete valued field (E1 , v1 ) which extends (E, v). Suppose v is unramified in F . Then v1 is unramified in F E1 . Proof: Suppose without loss that [F : E] < ∞. Let F1 = F E1 . Suppose ¯v is infinite. Choose x as in Lemma 2.3.5 and let f (X) = irr(x, E). first that E Then F = E(x) and f¯(X) is separable. Hence, F1 = E1 (x) and f¯(X) is still separable. By Lemma 2.3.4, v1 is unramified in F1 . In the general case we consider an extension w1 of v1 to a valuation of F1 . Denote the restriction of w1 to F by w. Let t be transcendental over F1 . Example 2.3.3 extends v (resp. w, v1 , w1 ) in a canonical way to a discrete valuation v 0 (resp. w0 , v10 , w10 ) of E(t) (resp. F (t), E1 (t), F1 (t)). Further, ¯v (t¯) ev0 /v = 1 (resp. ew0 /w = 1, ev10 /v1 = 1, ew10 /w1 = 1) and E(t)v0 = E ¯1,v (t¯), F1 (t) 0 = F¯1,w (t¯)), where t¯ is (resp. F (t)w0 = F¯w (t¯), E1 (t)v0 = E 1 1 w1 1 transcendental over F¯1,w1 . Moreover, w10 extends w0 and v10 extends v 0 giving this diagram: (F (t), w0 ) p p ppp

(F1 (t), w10 ) n n nnn (F1 , w1 )

(E(t), v 0 ) p p ppp

(E1 (t), v10 ) n n nnn

(F, w)

(E, v)

(E1 , v1 )

¯v (t¯) is a separable We claim v 0 is unramified in F (t). Indeed, F (t)w0 = F¯w · E extension of E(t)v0 . Also, ew0 /v0 = ew0 /v0 ev0 /v = ew0 /v = ew0 /w ew/v = 1. Hence, w0 is unramified over v 0 . If u∗ is an arbitrary extension of v 0 to F (t) and u is its restriction to F , then the residue of t at u∗ is t¯, which is transcendental over F¯u . Thus, by uniqueness of the construction in Example 2.3.3, u∗ = u0 , where u0 is the canonical extension of u to F (t). By the above, u∗ is unramified over v 0 . Since E(t)v0 is infinite, the first paragraph of the proof implies v10 is ¯1,v (t¯) is a separable extension and unramified in F1 (t). Thus, F¯1,w1 (t¯)/E 1 ¯1,v is a separable extension and ew10 /v10 = 1. Therefore, F¯1,w1 /E 1 ew1 /v1 = ew1 /v1 ew10 /w1 = ew10 /v1 = ew10 /v10 ev10 /v1 = 1. Consequently, v1 is unramified in F1 .

Combine the multiplicativity of the ramification index and the residue field degree with Lemma 2.3.6 to prove:

2.3 Extensions of Valuations and Places

29

Corollary 2.3.7: Let (E, v) ⊆ (E 0 , v 0 ) ⊆ (E 00 , v 00 ) be a tower of discrete valued fields. The following hold: (a) v 00 /v is unramified if and only if v 00 /v 0 and v 0 /v are unramified. (b) v is unramified in E 00 if and only if v is unramified in E 0 and each extension of v to E 0 is unramified in E 00 . (c) Let F1 and F2 be field extensions of E which are contained in a common field. Suppose F1 /E is separable algebraic and v is unramified in F1 and in F2 . Then v is unramified in F1 F2 . Example 2.3.8: Radical extensions. Let (E, v) be a discrete valued field and ¯v ) - n. Consider an extension F = E(x) of n a positive integer with char(E degree n of E where xn = a is in E. Let w be an extension of v to a valuation of F and let e = ew/v . Assume both v and w are normalized. Then (6)

nw(x) = ev(a)

and e ≤ n.

There are three cases to consider: Case A: gcd n, v(a) = 1. By (6), n|e, so n = e. By (2), w is the unique extension of v to F . Therefore, v totally ramifies in F . Case B: n - v(a). By (6), e 6= 1. Hence, w ramifies over E. Case C: n|v(a). Choose π ∈ E with v(π) = 1. Write a = bπ kn with k ∈ Z and b ∈ E such that v(b) = 0. Then y = xπ −k satisfies y n = b ¯v )s into distinct linear and F = E(y). Moreover, Y n − ¯b decomposes over (E factors. Therefore, by Lemma 2.3.4, v is unramified in F . Example 2.3.9: Artin-Schreier Extensions. Let (E, v) be a discrete valued field of positive characteristic p. An Artin-Schreier extension F of degree p has the form E(x) where xp − x = a with a ∈ E. We consider two cases: Case A: v(a) < 0 and p - v(a). Let w be an extension of v to F . Then w(x) must be negative and w(xp ) < w(x). Hence, pw(x) = ev(a), where e = ew/v . Hence, p = e and w(x) = v(a). Thus, v totally ramifies in F . Case B: v(a) ≥ 0. Then X p − X − a ¯ is a separable polynomial. By Lemma 2.3.4, v is unramified in F . Qp−1 ¯v . Hence, by In particular, if v(a) > 0, then X p − X = i=0 (X − i) in E Proposition 2.3.2, v has exactly p extensions to F . Label them v0 , . . . , vp−1 with vi (x − i) > 0, i = 0, . . . , p − 1. Since vi (x − i) < vi (x − i)p , we conclude from (x − i)p − (x − i) = a that vi (x − i) = v(a). Lemma 2.3.10 (Eisenstein’s Criterion): Let R be a unique factorization domain, p a prime element of R, and f (X) = an X n + an−1 X n−1 + · · · + a0 a polynomial with coefficients ai ∈ R. Then each of the following conditions suffices for f to be irreducible over Quot(R): (a) p - an , p divides a0 , . . . , an−1 , and p2 - a0 .

30

Chapter 2. Valuations and Linear Disjointness

(b) p - a0 , p divides a1 , . . . , an , and p2 - an . Proof of (a): See [Lang7, p. 183]. Proof of (b): By (a), the polynomial X n f (X −1 ) = an + an−1 X + · · · + a0 X n is irreducible over K. Therefore, f (X) is irreducible. Example 2.3.11: Ramification at infinity. Let K be a field, t an indeterminate, and f (X) = an X n + · · · + a0 ∈ K[X] with an 6= 0. By Eisenstein crite] ˜ rion, f (X) − t is irreducible over K(t). Choose a root x of f (X) = t in K(t). Let v = v∞ be the valuation of K(t) with v(t) = −1 which is trivial on K and let w be a valuation of K(x) lying over v. The relation an xn + · · · + a0 = t implies w(x) < 0. Hence, −ew/v = w(t) = w(f (x)) = nw(x). Since ew/v ≤ [K(x) : K(t)] ≤ n, this implies ew/v = [K(x) : K(t)] = n and w(x) = −1. Hence, v is totally ramified in K(x). In particular, w is the unique valuation of K(x) lying over K(t).

2.4 Integral Extensions and Dedekind Domains Integral extensions of Z in number fields are Dedekind domains. Although they are in general not unique factorization domain, their ideals uniquely factor as products of prime ideals. In this section we survey the concepts of integral extensions of rings and of Dedekind domains and prove that every overring of a Dedekind domain is again a Dedekind domain. Let F be a field containing an integral domain R. An element x ∈ F is integral over R if it satisfies an equation of the form xn + an−1 xn−1 + · · · + a0 = 0 with a1 , . . . , an ∈ R. The set of all elements of F which are integral over R form a ring (e.g. by Proposition 2.4.1 below), the integral closure of R in F . Call R integrally closed if R coincides with its integral closure in Quot(R). For example, every valuation ring O of F is integrally closed. Indeed, assume x ∈ F r O and x is integral over O. Then xn + an−1 xn−1 + · · · + a0 = 0 for some a0 , . . . , an−1 ∈ O. Then x−1 is in the maximal ideal m of O and 1 + an−1 x−1 + · · · + a0 x−n = 0. Thus, 1 ∈ m, a contradiction. Proposition 2.4.1 ([Lang4, p. 12]): An element x of F is integral over R if and only if every place of F finite on R is finite at x. Thus, the integral closure of R in F is the intersection of all valuation rings of F which contain R. In particular, every valuation ring of F is integrally closed. Suppose ϕ is a place of a field F and K is a subfield of F . We say that ϕ is trivial on K, or also that ϕ is a place of F/K, if ϕ(x) 6= ∞ for all x ∈ K. Then ϕ(y) 6= 0 for all y ∈ K × . Thus, ϕ maps K isomorphically onto ϕ(K). Lemma 2.4.2: Let K ⊆ L ⊆ F be a tower of fields and ϕ a place of F . Suppose ϕ is trivial on K and L is algebraic over K. Then ϕ is trivial on L. Proof: Each x ∈ L is integral over K, so by Proposition 2.4.1, ϕ(x) 6= ∞. Thus, ϕ is also trivial on L.

2.4 Integral Extensions and Dedekind Domains

31

Let S be a subring of F containing R. Call S integral over R if every element of S is integral over R. If S = R[x1 , . . . , xm ] and S is integral over R, then S is a finitely generated R-module. Indeed, every element of S is a linear αm 1 α2 combination with coefficients in R of the set of monomials xα 1 x2 · · · xm , where 0 ≤ αi < deg irr(xi , Quot(R)) . Propositions 2.3.1 and 2.4.1 give the following: Proposition 2.4.3: Let R ⊆ S be integral domains with S finitely generated as an R-algebra. Suppose S is integral over R. Then the following hold: (a) S is finitely generated as an R-module. (b) Let ϕ: R → M be a homomorphism into an algebraically closed field M . Then the set of all homomorphisms ψ: S → M that extend ϕ is finite and nonempty. Suppose R1 ⊆ R2 ⊆ R3 are integral domains. Proposition 2.4.1 implies that R3 is integral over R1 if and only if R2 is integral over R1 and R3 is integral over R2 . Call an integral domain R Noetherian if every ideal of R is finitely generated. For example, since a discrete valuation ring O is a principal ideal domain, it is integrally closed and Noetherian. If R is an integral domain and p is a prime ideal of R, then Rp =

a | a∈R b

and

b ∈ Rrp

is the local ring of R at p. It has a unique maximal ideal, pRp . If R is a Noetherian domain, then Rp is also Noetherian. If R is integrally closed, then so is Rp . T Lemma 2.4.4: Suppose R is an integral domain. Then R = T Rm , where m ranges over all maximal ideals of R. More generally, a = aRm for each ideal a of R. Proof: Suppose x belongs to each aRm . For each m, x = am /bm , with am ∈ a and bm ∈ R r m. Denote the ideal generated by all the bm ’s by b. If b 6= R, then b is contained in a maximal ideal Pm. Hence, bm ∈ m, a contradiction. Hence, b = R. In particular, 1 = m∈M bm cm where M is a P finite set of maximal P ideals, and cm ∈ R for each m ∈ M . Therefore x = m∈M xbm cm = m∈M am cm ∈ a. Let R be an integral domain with the quotient field F . A nonzero Rsubmodule a of F is said to be a fractional ideal of R if there exists a nonzero x ∈ R with xa ⊆ R. In particular, every ideal of R is a fractional ideal. Define the product, ab, of two fractional ideals a and b to be the R-submodule generated by the products ab, with a ∈ a and b ∈ b. Define the inverse of a fractional ideal a as a−1 = {x ∈ F | xa ⊆ R}. If a ∈ a, then aa−1 ⊆ R. Therefore, both ab and a−1 are fractional ideals.

32

Chapter 2. Valuations and Linear Disjointness

Proposition 2.4.5 ([Cassels-Fr¨ ohlich, p. 6]): The following conditions on an integral domain R are equivalent: (a) R is Noetherian, integrally closed, and its nonzero prime ideals are maximal. (b) R is Noetherian and the local ring, Rp , of every nonzero prime ideal p is a discrete valuation ring. (c) Every fractional ideal a is invertible (i.e. aa−1 = R). When these conditions hold, R is called a Dedekind domain. By Proposition 2.4.5, the set of all fractional ideals of a Dedekind domain R forms an Abelian group, with R being the unit. One proves that this group is free and the maximal ideals of R are free generators of this group. mr 1 m2 Thus, every ideal a of R has a unique presentation a = pm 1 p2 · · · pr , as the product of powers of maximal ideals with positive exponents [CasselsFr¨ohlich, p. 8]. Every principal ideal domain is a Dedekind domain. Thus, Z and K[x], where x is a transcendental element over a field K, are Dedekind domains. By the same reason, every discrete valuation ring is a Dedekind domain. In the notation of Proposition 2.4.5(b), Rp is the valuation ring of a discrete valuation vp of K = Quot(R). The corresponding place ϕp is finite on R. Conversely, if ϕ is such a place, then p = {x ∈ R | ϕ(x) = 0} is a nonzero prime ideal of R. Since Rp ⊆ Oϕ , Lemma 2.2.5 implies that Rp = Oϕ . This establishes a bijection between the nonzero prime ideals of R and the equivalence classes of places of K finite on R. Proposition 2.4.6 ([Cassels-Fr¨ ohlich, p. 13]): Let S be the integral closure of a Dedekind domain R in a finite algebraic extension of Quot(R). Then S is also a Dedekind domain. Let p be a prime ideal of R. Then pS = Pe11 Pe22 · · · Perr , where P1 , P2 , . . . , Pr are the distinct prime ideals of S that lie over p; that is, Pi ∩ R = p, i = 1, . . . , r. For each i we have pSPi = Pei i SPi . Hence, ei is the ramification index of vPi over vp . We say Pi is unramified over K if vPi /vp is unramified; that is, ei = 1 and S/Pi is a separable extension of R/p. The prime ideal p is unramified in L if each Pi is unramified over K. By Proposition 2.4.6, the integral closure of Z in a finite extension L of Q is a Dedekind domain, OL , called the ring of integers of L. Proposition 2.4.7 (Noether-Grell): Every overring R0 of a Dedekind domain R is a Dedekind domain. Proof: We show that R0 satisfies Condition (b) of Proposition 2.4.5. Part A: An injective map. If p0 is a nonzero prime ideal of R0 , then p = R ∩ p0 is a nonzero prime ideal of R. Indeed, for 0 6= x ∈ p0 , write x = ab , where a, b ∈ R. Thus, 0 6= a = bx ∈ R ∩ p0 = p. Since Rp ⊆ Rp0 0 and Rp is a discrete valuation ring, Lemma 2.2.5 implies that Rp = Rp0 0 . Hence, (1)

pRp = p0 Rp0 0 .

2.4 Integral Extensions and Dedekind Domains

33

In addition, p0 Rp0 0 ∩ R0 = p0 . Therefore, the map p0 7→ R ∩ p0 from the set of nonzero prime ideals of R0 into the set of nonzero prime ideals of R is injective. Part B: A finiteness condition. Let x be a nonzero element of R0 , p0 a prime ideal of R0 which contains x, and p = R ∩ p0 . Then Rp = Rp0 0 . Hence, vp (x) > 0, where vp is the valuation of Quot(R) corresponding to p. But this relation holds only for the finitely many prime ideals of R that appear with positive exponents in the factorization of the fractional ideal xR. Hence, by Part A, x belongs to only finitely many prime ideals of R0 . Part C: The ring R0 is Noetherian. Let a be a nonzero ideal of R0 . Choose a nonzero element x ∈ a and denote the finite set of prime ideals of R0 that contain x by P . For each p ∈ P the local ring Rp0 is a discrete valuation domain. Hence, there exists ap ∈ a such that aRp0 = ap Rp0 . Denote the ideal of R0 generated by x and by all ap , for p ∈ P , by a0 . It is contained in a. To show that a is finitely generated, we need only prove that a ⊆ a0 . Indeed, consider a prime ideal q of R0 not in P . Then x ∈ / q,Tso a0 6⊆ q. Rq0 . It followsT from Lemma T2.4.4 that a0 = p∈P a0 Rp0 . Hence, a0 Rq0 = T Therefore, a ⊆ p∈P aRp0 = p∈P ap Rp0 ⊆ p∈P a0 Rp0 = a0 , as desired. Lemma 2.4.8: Let (E, v) be a discrete valued field, F1 , F2 , F finite separable extensions of E with F = F1 F2 , and w an extension of v to F . Suppose v is unramified in F1 . Then the residue fields with respect to w satisfy F¯ = F¯1 F¯2 . Proof: Choose a finite Galois extension N of E which contains F and an extension w0 of w to N . Denote the decomposition groups of w0 over E, F1 , F2 , F by DE , DF1 , DF2 , DF , respectively. Let E 0 , F10 , F20 , F 0 be the fixed fields in N of DE , DF1 , DF2 , DF , respectively. Let v 0 = w0 |E 0 . Since all valuations of N lying over v 0 are conjugate over E 0 , the definition of E 0 as the fixed field of DE implies that w0 is the unique extension of v 0 to N . Also, DF1 = Gal(N/F1 ) ∩ DE , so F1 E 0 = F10 . By Lemma 2.3.6, v 0 is unramified in F10 . Finally, by [Serre3, p. 32, Prop. 21(c)], the residue fields of E, F1 , F2 , F at w coincide with the residue fields of E 0 , F10 , F20 , F 0 at w0 , respectively. We may therefore replace E, F1 , F2 , F , respectively, by E 0 , F10 , F20 , F 0 , if necessary, to assume that w|F1 is the unique extension of v to F1 . Now put wi = w|Fi , i = 1, 2. By Proposition 2.4.1, Ow1 is the integral closure of Ov in ¯ F1 . Since v is unramified in F1 , Proposition 2.3.2 implies [F1 : E] = [F¯1 : E], where the bar denotes reduction modulo w. Choose x ∈ Ow1 such that x ¯ is a primitive element for the separable ¯ Let f = irr(x, E) and p = irr(¯ ¯ Then f ∈ Ov [X] and extension F¯1 /E. x, E). f (x) = 0. Hence, f¯(¯ x) = 0 and p|f¯. Therefore, ¯ = [F1 : E]. [F1 : E] ≥ deg(f ) ≥ deg(¯ p) = [F¯1 : E] ¯ x). Consequently, p = f¯, F1 = E(x), and F¯1 = E(¯

34

Chapter 2. Valuations and Linear Disjointness

By Lemma 2.3.6, w2 is unramified in F . Thus, we may apply the result x). Conseof the preceding paragraph to F/F2 and conclude that F¯ = F¯2 (¯ ¯ x)F¯2 = F¯2 (¯ x) = F¯ . quently, F¯1 F¯2 = E(¯

2.5 Linear Disjointness of Fields Central to field theory is the concept “linear disjointness of fields”, an analog of linear independence of vectors. We repeat the convention made in “Notation and Convention” that whenever we form the compositum of fields, we tacitly assume they are contained in a common field. Lemma 2.5.1: Let E and F be extensions of a field K. The following conditions are equivalent: (a) Each m-tuple (x1 , . . . , xm ) of elements of E which is linearly independent over K is also linearly independent over F . (b) Each n-tuple (y1 , . . . , yn ) of elements of F which is linearly independent over K is also linearly independent over E. Proof: It suffices to prove that (a) implies (b). Let y1 , . . . , yn be elements of F for which there exist a1 , . . . , an ∈ E with a1 y1 + · · · +Pan yn = 0. Let {xj | j ∈ J} be a linear basis for E over K and write ai = j∈J aij xj with aij elements of K, only finitely many different from 0. Then n X X j∈J

aij yi xj = 0.

i=1

P aij yi = 0 for By (a), {xj | j ∈ J} is linearly independent over F . Hence, every j. If y1 , . . . , ym are linearly independent over K, then aij = 0 for every i and j, so ai = 0, i = 1, . . . , m. Thus, y1 , . . . , ym are linearly independent over E. This proves (b). Definition: With E and F field extensions of a field K, refer to E and F as linearly disjoint over K if (a) (or (b)) of Lemma 2.5.1 holds. Corollary 2.5.2: Let E and F be extensions of a field K such that [E : K] < ∞. Then E and F are linearly disjoint over K if and only if [E : K] = [EF : F ]. If in addition [F : K] < ∞, then this is equivalent to [EF : K] = [E : K][F : K]. Proof: If E and F are linearly disjoint over K and w1 , . . . , wn is a basis for E/K, then w1 , . . . , wn is also a basis for EF over F . Hence, [EF : F ] = n = [E : K]. Conversely, suppose [E : K] = [EF : F ] and let x1 , . . . , xm ∈ E be linearly independent over K. Extend {x1 , . . . , xm } to a basis {x1 , . . . , xn } of E/K. Since {x1 , . . . , xn } generates EF over F and n = [EF : F ], {x1 , . . . , xn } is a basis of EF/F . In particular, x1 , . . . , xm are linearly independent over F .

2.5 Linear Disjointness of Fields

35

Let E/K be a finite Galois extension. If E ∩ F = K, then, by Corollary 2.5.2, E and F are linearly disjoint over K. The condition, E ∩ F = K is equivalent to “res: Gal(EF/F ) → Gal(E/K) is an isomorphism” and also to “res: Gal(F ) → Gal(E/K) is surjective.” For arbitrary extensions this condition is clearly necessary, but not sufficient. Let L be a degree n > 1 extension of K for which L0 is conjugate to L over K and L0 ∩ L = K. Then 2.5.2, L and L0 are not [LL0 : K] ≤ n(n − 1). Thus, according to Corollary √ 3 linearly √ disjoint over K. For example, Q( 2) is not linearly disjoint from Q(ζ3 3 2) over Q although their intersection is Q. Lemma 2.5.3 (Tower Property): Let K ⊆ E and K ⊆ L ⊆ F be four fields. Then E is linearly disjoint from F over K if and only if E is linearly disjoint from L over K and EL is linearly disjoint from F over L. Proof: The only nontrivial part is to show that if E and F are linearly disjoint over K, then EL and F are linearly disjoint over L. Apply Lemma 2.5.1. Suppose that y1 , . . . , ym are elements of F which are linearly independent over L, but a1 , . . . , am are elements of EL such Pm that Pi=1 ai yi = 0. Clear denominators to assume that ai ∈ L[E], so that xj with aij ∈ L, where {xj | j ∈ J} is a linear basis for E over K. ai = PaijP yi )xj = 0. By assumption, the xj are linearly independent Then j ( i aijP over F . Hence, j aij yi = 0, so aij = 0 for all i and j. Consequently, ai = 0, i = 1, . . . , m. Lemma 2.5.4: Let L be a separable algebraic extension of a field K and let M be a purely inseparable extension of K. Then L and M are linearly disjoint over K. ˆ be the Galois closure of L/K. Then L ˆ ∩ M = K. Hence, L ˆ Proof: Let L and M are linearly disjoint over K. Therefore, by Lemma 2.5.3, L and M are linearly disjoint over K. Let E1 , . . . , En be n extensions of a field K. We say that E1 , . . . , En are linearly disjoint over K if E1 · · · Em−1 and Em are linearly disjoint over K for m = 2, . . . , n. Induction on n shows that this is the case if of and only if the following condition holds: If wi,ji , ji ∈ Ji , are elements Qn Ei which are linearly independent over K, i = 1, . . . , n, then i=1 wi,ji , (j1 , . . . , jn ) ∈ J1 × · · · × Jn , are linearly independent over K. It follows that E1 , . . . , En are linearly disjoint over K if and only if the canonical homomorphism of E1 ⊗K · · · ⊗K En into E1 · · · En that maps x1 ⊗ · · · ⊗ xn onto x1 · · · xn is injective. It also follows that if E1 , . . . , En are linearly disjoint over K, then Eπ(1) , . . . , Eπ(n) are linearly disjoint over K for every permutation π of {1, . . . , n}. The application of tensor products makes the following lemma an easy observation. Lemma 2.5.5: Let E1 , . . . , En (resp. F1 , . . . , Fn ) be linearly disjoint field extensions of K (resp. L). For each i between 1 and n let ϕi : Ei → Fi ∪ {∞},

36

Chapter 2. Valuations and Linear Disjointness

be either a place or an embedding. Suppose ϕ1 , . . . , ϕn coincide on K and ϕi (K) = L, i = 1, . . . , n. Let E = E1 · · · En and F = F1 · · · Fn . Then there exists a place ϕ: E → F˜ ∪ {∞} that extends each of the ϕi ’s. If each ϕi is an isomorphism of Ei onto Fi , then ϕ is an isomorphism of E onto F . Proof: Let Oi be the valuation ring of ϕi if ϕi is a place and Ei if ϕi is an isomorphism. By assumption, the map x1 · · · xn → x1 ⊗ · · · ⊗ xn is an isomorphism O1 · · · On ∼ = O1 ⊗K · · · ⊗K On of rings. Hence, there exists a ring homomorphism ϕ0 : O1 · · · On → F such that ϕ0 (x) = ϕi (x) for each x ∈ Oi , i = 1, . . . , n. Extend ϕ0 to a place ϕ: E → F˜ ∪ {∞} (Proposition 2.3.1). If x ∈ Ei r Oi , then ϕ(x−1 ) = ϕi (x−1 ) = 0, so ϕ(x) = ϕi (x) = ∞. We conclude that ϕ coincides with ϕi on Ei . Finally, define a family {Ei | i ∈ I} of field extensions of K to be linearly disjoint over K if every finite subfamily is linearly disjoint over K. It follows from the discussion preceding Lemma 2.5.5 that a sequence (E1 , E2 , E3 , . . .) of fields extensions of K is linearly disjoint over K if En is linearly disjoint from E1 · · · En−1 for 2, 3, 4, . . . . Then, Eπ(1) , Eπ(2) , Eπ(3) , . . . are linearly disjoint for every permutation π of N. disjoint family of Galois extenLemma 2.5.6: Let {Li | i ∈ I} Q be a linearlyQ sions of a field K. Then Gal( i∈I Li /K) ∼ = i∈I Gal(Li /K). Q Q Proof: Since i∈I Gal(Li /K) ∼ = lim i∈I0 Gal(Li /K), we may assume I ←− Q Q is finite. In this case, the embedding Gal( i∈I Li /K) → i∈I Gal(Li /K) given by σ 7→ (σ|Li )i∈I is surjective (Lemma 2.5.5). Therefore, it is an isomorphism. Lemma 2.5.7: Let K be a field, K1 , K2 , K3 , . . . a linearly disjoint sequence of extensions of K, and L a finite separable extension of K. Then there exists a positive integer n such that L, Kn , Kn+1 , Kn+2 , . . . are linearly disjoint over K. Proof: Replace L by its Galois closure over K, if necessary, to assume L is Galois over K. Assume for each positive integer n the field L is not linearly disjoint from Kn Kn+1 Kn+2 · · · over K. Then Ln = L ∩ Kn Kn+1 Kn+2 · · · is a proper extension of K. Since L has only finitely many extensions that contain K and since Ln ⊇ Ln+1 ⊇ Ln+2 ⊇ · · ·, there is an m such that Ln = Lm for all n ≥ m. Since Lm is a finite extension of K, there is an n > m with Lm ⊆ Km · · · Kn−1 . Similarly, there exists r > n with Lm ⊆ Kn · · · Kr−1 . By assumption, Km · · · Kn−1 and Kn · · · Kr−1 are linearly disjoint over K. In particular, their intersection is K. Therefore, Lm = K. This contradiction proves there exists n such that L, Kn , Kn+1 , Kn+2 , . . . are linearly disjoint over K. Lemma 2.5.8: Let v be a discrete valuation of a field K and L, M finite extensions of K. Suppose v is unramified in L but totally ramified in M . Then L and M are linearly disjoint over K.

2.5 Linear Disjointness of Fields

37

Proof: Let L0 be the maximal separable extension of K in L and v0 an extension of v to L0 . Then L/L0 is purely inseparable. Hence, v0 is ramified in L. Therefore, L = L0 and L/K is separable. Since v is unramified in each of the conjugates of L over K, it is unramified in their compositum (Corollary 2.3.7). We may therefore replace L by the Galois closure of L/K, if necessary, to assume L/K is Galois. Let m = [L ∩ M : K]. Choose an extension w of v to L ∩ M . Then e(w/v) = 1 on one hand and e(w/v) = m on the other hand. Thus, L ∩ M = K. Therefore, L is linearly disjoint from M over K. Example 2.5.9: Roots of unity. For each n consider the Galois extension Q(ζn ) of Q obtained by adjoining a primitive root of unity of order n. It is well known that ϕ(n) = [Q(ζn ) : Q] is the number of integers between 1 and n which are relatively prime to n [Lang7, p. 278, Thm. 3.1]. If m is relatively prime to n, then ϕ(mn) = ϕ(m)ϕ(n) [LeVeque, p. 28, Thm. 3-7]. In addition, Q(ζm , ζn ) = Q(ζmn ). Hence, [Q(ζm , ζn ) : Q] = [Q(ζmn ) : Q] = ϕ(mn) = ϕ(m)ϕ(n) = [Q(ζm ) : Q][Q(ζn ) : Q]. It follows from Corollary 2.5.2 that Q(ζm ) and Q(ζn ) are linearly disjoint over Q. Here is an application of linear disjointness to integral closures of domains. Lemma 2.5.10: Let K be a field, L a separable algebraic extension of K, and R an integrally closed integral domain containing K. Let E = Quot(R), F = EL, and S the integral closure of R in F . Suppose E and L are linearly disjoint over K. Then S = RL ∼ = R ⊗K L. Proof: Assume without loss L/K is finite. Choose a basis w1 , . . . , wn for L/K. Let σ1 , . . . , σn be the distinct K-embeddings of L into Ks . Then det(σi wj ) 6= 0. Each element of L is integral over K, hence over R, so RL ⊆ S. Conversely, let xP ∈ S. By the linear disjointness, w1 , . . . , wn form a basis for F/E. n Hence, x = j=1 ej wj with ej ∈ E, j = 1, . . . , n. Also, each σi extends to Pn an E-embedding of F into Es (Lemma 2.5.5). Thus, σi x = j=1 ej σi wj , i = 1, . . . , n. Apply Kramer’s law to present each ek as a polynomial in σi x, σi wj , with i, j = 1, . . . , n, divided by det(σi wj ). Thus, ek is an element of E which is integral over R. Since R is integrally closed, ek ∈ R, k = 1, . . . , n. Consequently, x ∈ RL, as needed. We generalize the tower property to families of field extensions: Lemma 2.5.11: Let K be a field and I a set. For each i ∈ I let Fi /Ei be a field extension with K ⊆ Ei . Suppose {Fi | i ∈ I} is linearly disjoint over K. Denote the compositum of all Ei ’s by E. Then the set {Fi E | i ∈ I} is linearly disjoint over E. Moreover, for each i ∈ I, the field Fi is linearly disjoint from E over Ei . Proof: It suffices to consider the case where I = {1, 2, . . . , n}. By induction suppose Fi E1 · · · En−1 , i = 1, . . . , n − 1, are linearly disjoint over E1 · · · En .

38

Chapter 2. Valuations and Linear Disjointness

By assumption, F1 · · · Fn−1 is linearly disjoint from Fn over K. Hence, by the tower property, F1 · · · Fn−1 is linearly disjoint from E over E1 , . . . , En−1 , so Fi E, i = 1, . . . , n − 1, are linearly disjoint over E. F1 · · · Fn−1

F1 · · · Fn−1 E

Fi E1 · · · En−1

Fi E

E1 · · · En−1

E

EFn

K

En

Fn

Moreover, F1 · · · Fn−1 E is linearly disjoint from EFn over E. Consequently, E is linearly disjoint from Fn over En and Fi E, i = 1, . . . , n are linearly disjoint over E, as claimed.

2.6 Separable, Regular, and Primary Extensions Based on the notion of linear disjointness we define here three type of field extensions. We say that a field extension F/K is separable (resp. regular, ˜ Ks ) over K. primary) if F is linearly disjoint from Kins (resp. K, Separable Extensions. We generalize the notion of “separable algebraic extension” to arbitrary field extensions. Let K be a field of positive characteristic p. The field generated over K by the pth roots of all elements of K is denoted K 1/p . We denote the ∞ maximal purely inseparable extension of K by Kins (or K 1/p ). Let F be a finitely generated extension of K. A collection t1 , . . . , tr ∈ F of elements algebraically independent over K is a separating transcendence basis if F/K(t1 , . . . , tr ) is a finite separable extension. Lemma 2.6.1: An extension F of a field K is separable if it satisfies one of the following equivalent conditions: (a) F is linearly disjoint from Kins over K. (b) F is linearly disjoint from K 1/p over K. (c) Every finitely generated extension E of K which is contained in F has a separating transcendence basis. Moreover, a separating transcendence basis can be selected from a given set of generators for F/K. Proof: The implications “(a) => (b)” and “(c) => (a)” are immediate consequences of the tower property (Lemma 2.5.3). For “(b) => (c)” see [Lang 4, p. 54]. Lemma 19.2.4 gives a constructive proof. In particular, every separable algebraic extension satisfies conditions (a), (b), and (c) of Lemma 2.6.1. Now apply the rules of linear disjointness.

2.6 Separable, Regular, and Primary Extensions

39

Corollary 2.6.2: (a) If E/K and F/E are separable extensions, then F/K is also separable. (b) If F/K is a separable extension, then E/K is separable for every field K ⊆ E ⊆ F. (c) Every extension of a perfect field is separable. (d) If E/K is a purely inseparable extension and F/K is a separable extension, then E and F are linearly disjoint over K. Example 2.6.3: A separable tower does not imply separable steps. Consider the tower of fields Fp ⊂ Fp (tp ) ⊂ Fp (t), where t is transcendental over Fp . The extension Fp (t)/Fp is separable, but Fp (t)/Fp (tp ) is not. Regular Extensions. Finitely generated regular extensions characterize absolutely irreducible varieties (Section 10.2) Lemma 2.6.4: A field extension F/K is regular if it satisfies one of the following equivalent conditions: (a) F/K is separable and K is algebraically closed in F . ˜ over K. (b) F is linearly disjoint from K Proof: The implication “(b) => (a)” is immediate. To prove “(a) => (b)”, it suffices to assume that F/K is finitely generated. Then F/K has a separating transcendence basis, t1 , . . . , tr , which is also a separating transcendence basis for the extension F Ks /Ks . Since ˜ over ˜ = (Ks )ins , Lemma 2.6.1 implies that F Ks is linearly disjoint from K K Ks . Also, Ks /K is a Galois extension and F ∩ Ks = K. Hence, F is linearly disjoint from Ks over K. Therefore, by Lemma 2.5.3, F is linearly disjoint ˜ over K. from K Corollary 2.6.5: (a) If E/K and F/E are regular extensions, then F/K is regular. (b) If F/K is a regular extension, then E/K is regular for every field E lying between K and F . (c) Every extension of an algebraically closed field is regular. (d) Let m be a cardinal number andSKα , α ≤ m, an ascending transfinite sequence of fields such that Kγ = α (a).” The implication “(a) => (b)” holds since F ∩ Ks = K and Ks /K is a Galois extension. Corollary 2.6.14: (a) If E/K and F/E are primary extensions, then so is F/K. (b) If F/K is a primary extension, then E/K is primary, for every field K ⊆ E ⊆ F. (c) Every extension of a separably closed field is primary. (d) An extension F/K is regular if and only if it is separable and primary. Lemma 2.6.15: (a) Let E be a primary extension of a field K which is algebraically independent from an extension F of K. Then EF is a primary extension of F. (b) If two primary extensions E and F of K are algebraically independent, then EF/K is primary. Proof: Assertion (b) follows from (a) and from Corollary 2.6.14(a). To prove (a), choose a transcendence base T for E/K and let M be the maximal separable extension of K(T ) in E. Then M is a separable and primary extension of K. Hence, by Lemma 2.6.14(d), it is regular. Also, M is algebraically independent from Fs over K. By Lemma 2.6.7, M F is linearly disjoint from Fs over F . Since EF is a purely inseparable extension of M F , it is linearly disjoint from M Fs . It follows that EF is linearly disjoint from Fs over F ; that is, EF is a primary extension of F .

2.7 The Imperfect Degree of a Field We classify fields of positive characteristic by their imperfect degree and characterize those fields for which every finite extension has a primitive element as fields of imperfect degree 1. Let F be a field of positive characteristic p. Consider a subfield F0 of F that contains the field F p of all pth powers in F . Observe that for x1 , . . . , xn ∈ F , the set of monomials (1)

xi11 · · · xinn ,

0 ≤ i1 , . . . , in ≤ p − 1,

generates F0 (x) over F0 . Hence, [F0 (x) : F0 ] ≤ pn . If [F0 (x) : F0 ] = pn , then x1 , . . . , xn are said to be p-independent over F0 . Equivalently, each of the fields F0 (x1 ), . . . , F0 (xn ) has degree p over F0 and they are linearly disjoint over F0 . This means that the set of monomials (1) is linearly independent over F0 . A subset B of F is p-independent over F0 , if every finite subset of B is p-independent over F0 . If in addition F0 (B) = F , then B is said to be a p-basis for F over F0 . As in the theory of vector spaces, each maximal p-independent subset of F over F0 is a p-basis for F over F0 .

2.7 The Imperfect Degree of a Field

45

If x1 , . . . , xn ∈ F are p-independent over F p , we call them p-independent elements of F . The p-power pn = [F : F p ] is the imperfect degree of F and n is the imperfect exponent of F . We say that F is n-imperfect. Thus, a perfect field has imperfect exponent 0. Both quantities are infinite if [F : F p ] = ∞. In this case F is ∞-imperfect. Lemma 2.7.1 (Exchange Principle): Let F0 be a subfield of F which contains F p. (a) Let x1 , . . . , xm , y1 , . . . , yn ∈ F be such that x1 , . . . , xm are p-independent over F0 and x1 , . . . , xm ∈ F0 (y1 , . . . , yn ). Then m ≤ n, and there is a reordering of y1 , . . . , yn so that y1 , . . . , ym ∈ F0 (x1 , . . . , xm , ym+1 , . . . , yn ). (b) Every subset of F which is p-independent over F0 extends to a p-basis for F over F0 . Proof: We use induction on m. Assume the lemma is true for m = k. Thus, for m = k + 1 we may assume that xk+1 ∈ F0 (x1 , . . . , xk , yk+1 , . . . , yn ) = F1 . Then [F1 : F0 ] ≤ pn and there exists l between k + 1 and n such that yl ∈ F0 (x1 , . . . , xk+1 , yk+1 , . . . , yl−1 ), since otherwise [F1 : F0 ] ≥ pn+1 , a contradiction. Thus, yl can be exchanged for xk+1 . This proves the first part of the lemma for m = k + 1. For the last part start from a subset A of K which is p-independent over F0 . Use Zorn’s lemma to prove the existence of a maximal subset B of F which contains A and which is p-independent over F0 . Then B is a p-basis of F over F0 . Lemma 2.7.2: Suppose F is a finitely generated extension of transcendence degree n of a perfect field K of positive characteristic p. Then the imperfect exponent of F is n. Proof: Choose a separating transcendence basis t1 , . . . , tn for F/K. Then K(t)p = K(tp ) and t1 , . . . , tn is a p-basis for K(t)/K(tp ); that is, [K(t) : K(tp )] = pn . Since K(t) is a purely inseparable extension of K(tp ) and F p is a separable extension of K(tp ), these extensions of K(tp ) are linearly disjoint. Also, F is both a separable extension and a purely inseparable extension of K(t)F p . Hence, F = K(t)F p . Consequently, [F : F p ] = [K(t) : K(tp )] = pn , as claimed. Lemma 2.7.3: Let B a subset of F which is p-independent over F p and F 0 a separable extension of F . Then B is p-independent over (F 0 )p . If, in addition, F 0 is separable algebraic over F , then the imperfect degree of F 0 is equal to that of F . Proof: Assume without loss that B consists of n elements. Then [(F 0 )p (B) : (F 0 )p ] = [F p (B) : F p ] = pn . Hence, B is p-independent over (F 0 )p .

46

Chapter 2. Valuations and Linear Disjointness

Suppose now F 0 /F is separably algebraic. Then F 0 is both separably and purely inseparable over F (F 0 )p , so, F 0 = F (F 0 )p . Hence, [F 0 : (F 0 )p ] = [F : F p ]. Therefore, the imperfect degree of F 0 is equal to that of F . Lemma 2.7.4: Let K be a field of positive characteristic p, let a, b1 , . . . , bm be p-independent elements of K, and let x1 , . . . , xm be algebraically independent over K. Suppose y1 , . . . , ym satisfy (2)

axpi + bi yip = 1,

i = 1, . . . , m.

Then K is algebraically closed in K(x, y) = Km . Proof: We use induction on m. Part A: m = 1. Let x = x1 , y = y1 , and b = b1 and assume that u is a nonzero element of K1 which is algebraic over K. Then u is also algebraic over K(a1/p , b1/p ). But K(x, y, a1/p , b1/p ) = K(x, a1/p , b1/p ) is a purely transcendental extension of K(a1/p , b1/p ). Hence, u ∈ K(a1/p , b1/p ) and therefore up ∈ K. Write (3)

u=

h0 (x) h1 (x) hk (x) k + y + ··· + y h(x) h(x) h(x)

with k ≤ p − 1, h(x), h0 (x), . . . , hk (x) ∈ K[x] and h(x), hk (x) 6= 0. With no loss we may assume that x does not divide the greatest common divisor of h(x), h0 (x), . . . , hk (x). Raise (3) to the pth power, multiply it by h(x)p and substitute y p = (1 − axp )b−1 to obtain: (4) (h(x)u)p = h0 (x)p + h1 (x)p (1 − axp )b−1 + · · · + hk (x)p (1 − axp )k b−k . If h(0) = 0, then the substitution x = 0 in (4) gives 0 = h0 (0)p + h1 (0)p b−1 + · · · + hk (0)p b−k , Therefore, h0 (0) = h1 (0) = · · · = hk (0) = 0, contrary to assumption. Thus, we may assume h(0) 6= 0. Then the substitution x = 0 in (4) shows that u ∈ K(b1/p ). Similarly, u ∈ K(a1/p ). Since a and b are p-independent in K, u ∈ K(a1/p ) ∩ K(b1/p ) = K. Thus, K is algebraically closed in K(x, y). Part B: Induction. Assume the Lemma is true for m − 1. Then K is algebraically closed in Km−1 = K(x1 , . . . , xm−1 , y1 , . . . , ym−1 ). If we prove that a and bm are p-independent in Km−1 , then with Km−1 replacing K in Part A, Km−1 is algebraically closed in Km , so K is algebraically closed in Km . Since x1 , . . . , xm are algebraically independent over K, the field 1/p 1/p 1/p K(a , b1 , . . . , bm ) is linearly disjoint from Em−1 = K(x1 , . . . , xm−1 ) over K. Thus, (5)

1/p

m+1 . [Em−1 (a1/p , b1 , . . . , b1/p m ) : Em−1 ] = p

2.7 The Imperfect Degree of a Field

47

Also, from (2) Km−1 = Em−1 (y1 , . . . , ym−1 )

and 1/p

1/p Km−1 (a1/p , b1/p , b1 , . . . , b1/p m ) = Em−1 (a m ).

Thus, (6)

[Km−1 : Em−1 ] ≤ pm−1

and

2 [Km−1 (a1/p , b1/p m ) : Km−1 ] ≤ p .

Combine (5) and (6) to conclude that (6) consists of equalities. In particular, a and bm are p-independent in Km−1 . Lemma 2.7.5: The following conditions on a field K of positive characteristic p are equivalent: (a) The imperfect exponent of K is at most 1. (b) Every finite extension of K has a primitive element. (c) If K is algebraically closed in a field extension F , then F is regular over K. Proof: If K is perfect, then (a), (b), and (c) are true. Therefore, we may assume char(K) = p > 0 and K is imperfect. Proof of “(a) =⇒ (b)”: By assumption, [K 1/p : K] = [K : K p ] = p. Hence, K1 = K 1/p is the unique purely inseparable extension of K of degree p. n Moreover, K1 = K(a1/p ) for some a ∈ K, so Kn = K(a1/p ) is a purely inseparable extension of K of degree pn . Assume that for each m ≤ n, Km is the unique purely inseparable extension of K of degree pm . Let L be a purely inseparable extension of K of degree pn+1 . If we prove that L = Kn+1 , then we may conclude by induction that each finite purely inseparable extension of K has a primitive element. To this end choose x ∈ L r Kn . Let m be the smallest positive integer m with xp ∈ K. Then K(x) is a purely inseparable extension of K of degree pm . If m ≤ n, then by the induction hypothesis K(x) = Km ⊆ Kn , so x ∈ Kn . This contradiction proves that m = n + 1 and L = K(x). The same argument implies that xp ∈ Kn . Hence, with q = pn , we have Pq−1 n p x = i=0 ci ai/p for some c0 , . . . , cq−1 ∈ K. Therefore, x=

q−1 X

1/p

ci ai/p

n+1

∈ K1 (a1/p

n+1

) = Kn+1 .

i=0

It follows that L ⊆ Kn+1 . As both fields have degree pn+1 over K, they coincide, as desired. Now let E be a finite extension of K. Denote the maximal separable extension of K in E by E0 . By the primitive element theorem, E0 = K(x). Since E0 is both separable and purely inseparable over KE0p we have E0 = KE0p . Therefore [E0 : E0p ] = [K : K p ] = p. Apply the first part of the proof

48

Chapter 2. Valuations and Linear Disjointness

to E0 and conclude that E = E0 (y), for some element y. Thus, E = K(x, y) with x separable over K. By [Waerden3, §6.10], E/K has a primitive element Proof of “(b) =⇒ (c)”: Let K(x) be a finite extension of K and let f = irr(x, K). If K is algebraically closed in F , then f remains irreducible over F . Otherwise, its factors would have coefficients algebraic over K and in F , and therefore in K. Thus, F is linearly disjoint from K(x) over K. Hence, (b) implies that F is regular over K. Proof of “(c) =⇒ (a)”: Assume a and b are p-independent elements of K. Then [K(a1/p , b1/p ) : K] = p2 . Let x and y be transcendental elements over K with axp + by p = 1. Put F = K(x, y). By Lemma 2.7.4, K is algebraically closed in F . Hence, by (c), F is regular over K. Therefore, [F (a1/p , b1/p ) : F ] = [K(a1/p , b1/p ) : K] = p2 . On the other hand, , F (a1/p ) = F (b1/p ), so [F (a1/p , b1/p ) : F ] ≤ p. This contradiction proves that the imperfect exponent of K is at most 1. Remark 2.7.6: Relative algebraic closedness does not imply regularity. Let K be a field of positive characteristic p. Suppose K has p-independent elements a, b (e.g. K = Fp (t, u) where t, u are algebraically independent over Fp ). Let x, y be transcendental elements over K with axp + by p = 1. Put F = K(x, y). The proof of “(c) =⇒ (a)” then shows that K is algebraically closed in F but F is not linearly disjoint from K 1/p over K. Thus, F is not a separable extension of K. A fortiori, F/K is not regular.

2.8 Derivatives We develop a criterion for a finitely generated field extension of positive characteristic p to be separable in terms of derivatives.. Definition 2.8.1: A map D: F → F is called a derivation of the field F if D(x + y) = D(x) + D(y) and D(xy) = D(x)y + xD(y) for all x, y ∈ F . If D vanishes on a subfield K of F , then D is a derivation of F over K (or a K-derivation). Let F (x) be a field extension of F and f ∈ F [X]. Suppose D extends to F (x). Then D satisfies the classical chain rule: (1)

D(f (x)) = f D (x) + f 0 (x)D(x),

where f D is the polynomial obtained by applying D to the coefficients of f and f 0 is the usual derivative of f . There are three cases: Case 1: x is separably algebraic over F . Then, with f = irr(x, F ), f 0 (x) 6= 0. By (1), 0 = f D (x) + f 0 (x)D(x). Thus, D extends uniquely to F (x). Case 2: x is transcendental. Then D extends to F (x) by rule (1) and D(x) may be chosen arbitrarily.

2.8 Derivatives

49 m

Case 3: x satisfies xp = a ∈ F , for some m. Then D extends to F (x) if and only if D(a) = 0. In this case D(x) may be chosen arbitrarily. Lemma 2.8.2: A necessary and sufficient condition for a finitely generated extension F/K to be separably algebraic is that 0 is the only K-derivation of F . Proof: Necessity follows from Case 1. Now suppose F/K is not separably algebraic. Then we may write F = K(x1 , . . . , xn ) such that xi is transcendental over K(x1 , . . . , xi−1 ) for i = 1, . . . , k, xi is separably algebraic over K(x1 , . . . , xi−1 ) for i = k+1, . . . , l, and xi is purely inseparable over K(x1 , . . . , xi−1 ) for i = l + 1, . . . , n. Moreover, either n > l or n = l and k > 0. If n > l, then Case 1 allows us to extend the zero derivation of K(x1 , . . . , xn−1 ) to a nonzero derivation of F . If n = l and k > 0, then by Case 2, the zero derivation of K(x1 , . . . , xk−1 ) extends to a nonzero derivation D of K(x1 , . . . , xk ). Applying Case 3 several times, we may then extend D to a derivation of F . Lemma 2.8.3: Let F/K be a finitely generated extension of positive characteristic p and transcendence degree n. Then F/K is separable if and only if [F : KF p ] = pn . In this case t1 , . . . , tn form a p-basis for F over KF p if and only they form a separating transcendence basis for F/K. Proof: Suppose first [F : KF p ] = pn . Let t1 , . . . , tn be a p-basis for F/KF p . Every derivation D of F vanishes on F p . If D vanishes on K(t), it vanishes on F = K(t) · F p . By Lemma 2.8.2, F/K(t) is separably algebraic and t1 , . . . , tn is a separating transcendence basis for F/K. Conversely, suppose F/K is separable. Let t1 , . . . , tn be a separating transcendence basis for F/K. The extension F/K(t) · KF p is both separable and purely inseparable. Hence, F = K(t) · KF p . Since F p /K(t)p is separably algebraic and since K(tp )F p = KF p , we conclude that KF p /K(tp ) is separably algebraic. K(t) F

K(tp )

KF p

K(t)p

Fp

Therefore, KF p is linearly disjoint from K(t) over K(tp ), and [F : KF p ] = [K(t) : K(tp )] = pn . Moreover, t is a p-basis for F/KF p . Corollary 2.8.4: Let F/K be a finitely generated separable extension of positive characteristic p and let t ∈ F . (a) If there exists a derivation D of F/K such that D(t) 6= 0, then F is a separable extension of K(t).

50

Chapter 2. Valuations and Linear Disjointness

(b) If t is transcendental over K and F/K(t) is separable, then there exists a derivation D of F/K such that D(t) 6= 0. Proof of (a): By assumption, t ∈ / KF p . Let n = trans.deg(F/K). By p n Lemma 2.8.3, [F : KF ] = p . Hence, t can be extended to a p-basis t, t2 , . . . , tn for F/KF p . Again, by Lemma 2.8.3, t, t2 , . . . , tn is a separating transcendence basis for F/K. Therefore, F is a separable extension of K(t). Proof of (b): Let t2 , . . . , tn be a separating transcendence basis for F/K(t). By Case 2, there exists a derivation D0 of K(t, t2 , . . . , tn )/K such that D0 (t) = 1, D0 (t2 ) = 0, . . . , D0 (tn ) = 0. By Case 1, D0 extends to a derivation D of F/K.

Exercises 1. Let O be a valuation ring of a field F and consider the subset m = {x ∈ / O}. Show that if x ∈ m and a ∈ O, then ax ∈ m. Prove that m is O | x−1 ∈ closed under addition. Hint: Use the identity x + y = (1 + xy −1 )y for y 6= 0. Show that m is the unique maximal ideal of O. 2.

Use Exercise 1 to prove that every valuation ring is integrally closed.

3. Let v be a valuation of Q. Observe that v(n) ≥ v(1) = 0, for each n ∈ N. Hence, there exists a smallest p ∈ N such that v(p) > 0. Prove that p is a prime element of Ov and v is equivalent to vp . Hint: If a positive integer m is relatively prime to p, then there exist x, y ∈ Z such that xp + ym = 1. 4. Let v be a valuation of the rational function field F = K(t) which is trivial on K. Suppose there exists p ∈ K[t] with v(p) > 0. Now suppose p has smallest degree with this property. Show that v is equivalent to vp . Otherwise, there exists f ∈ K[t] such that v(f (t)) < 0. Conclude that v(t) < 0, and that v is equivalent to v∞ . 5. Let F/E be a field extension, w a valuation of F , and x1 , . . . , xe elements of F such that w(x1 ), . . . , w(xe ) represent distinct classes of w(F × ) modulo w(E × ). Show that x1 , . . . , xe are linearly independent over E. Thus, (w(F × ) : w(E × )) ≤ [F : E]. Hint: Use (4b) of Section 2.1. 6. Let ∆ be an ordered group containing Z as a subgroup of index e. Show there exists no positive element δ ∈ ∆ such that eδ < 1. Conclude that ∆ contains a smallest positive element and hence that ∆ ∼ = Z. Combine this with Exercise 5 to prove that if the restriction of w to E is discrete, then w is discrete. 7. In the notation of Exercise 5, let v be the restriction of w to E. Let y1 , . . . , yf be elements of F with w(y1 ), . . . , w(yf ) ≥ 0 with residue classes ¯v . Show that y1 , . . . , yf are linearly y¯1 , . . . , y¯f linearly independent over E ¯v ] ≤ [F : E]. Hint: If a1 , . . . , af ∈ ¯ independent over E. Conclude that [Fw : E

Notes

F are not all zero, then there exists j, 1 ≤ j ≤ f such that v 0.

51 a1 aj

,...,v

af aj

≥

8. Let v be a discrete valuation of a field K and let w be an extension of v to a finite Galois extension L of K. Assume that w0 is also an extension of v to L such that w0 6= σ(w) for all σ ∈ Gal(L/K). Combine Exercise 7 with Proposition 2.1.1 to produce x ∈ L such that w0 (x) > 0 and w(σx−1) > 0 for all σ ∈ Gal(L/K). With y = NL/K (x), conclude that the former condition gives v(y) > 0, while the latter implies v(y − 1) > 0. Use this contradiction to prove that Gal(L/K) acts transitively on the extensions of v to L. 9. Let L, K1 , . . . , Kn be extensions of a field K. Let Li = Ki L, i = 1, . . . , n. Suppose Ki is linearly disjoint from L over K for i = 1, . . . , n and L1 , . . . , Ln are linearly disjoint over L. Prove that K1 , . . . , Kn are linearly disjoint over K. 10. Let v be a discrete valuation of a field K and let L and M be two finite extensions of K such that v is unramified in L and totally ramified in M . Prove that L and M are linearly disjoint over K. Hint: Consider the Galois ˆ of L/K. hull L 11. Let E be a regular extension of a perfect field K and let F be a purely inseparable extension of E. Prove that F/K is a regular extension. 12. Let K be a field algebraically closed in an extension F . Prove that K(x) ˜ Hint: Check the irreducibility of is linearly disjoint from F for every x ∈ K. irr(x, K) over F . ˜ = 13. Prove that a field extension F/K is primary if and only if F Kins ∩ K Kins . Use this criterion to give another proof to Lemma 2.6.14(a). 14. Let F/K be a finitely generated field extension of characteristic p > 0 n and of transcendence degree 1. Prove that for each positive integer n, KF p is the unique subfield E of F which contains K such that F/E is a purely inseparable extension of degree pn . 15. (Geyer) The following example shows that Lemma 2.4.8 is false for arbitrary real valuations. Consider the field Q2 of 2-adic numbers. Show that √ the field K = Q2 ( n 2 | n ∈ N) is a totally ramified extension of Q2 with value group Q. Hence, of K is unramified. Prove that the residue √ each extension √ 3) and K( −1) is F2 . However, their compositum contains field of both K( √ K( −3) and therefore has F4 as its residue field.

Notes The terminology “algebraic independence” for field extensions replaces “freeness” which we used in [Fried-Jarden3]. Corollary 4 of [Lang4, p. 61] proves Lemma 2.6.15(a) only under the condition (our notation) that E is a separable extension of K.

Chapter 3. Algebraic Function Fields of One Variable Sections 3.1–3.4 survey the theory of functions of one variable; the RiemannRoch Theorem; properties of holomorphy rings of function fields; and extensions of the field of constants. Sections 3.5–3.6 include a proof of the Riemann-Hurwitz formula. The rest of the chapter applies these concepts and results to hyperelliptic curves.

3.1 Function Fields of One Variable Call a field extension F/K an algebraic function field of one variable (briefly a function field) if these conditions hold: (1a) The transcendence degree of F/K is 1. (1b) F/K is finitely generated and regular. In this case there exists t ∈ F , transcendental over K, with F/K(t) a finite separable extension. All valuations of K(t) trivial on K are discrete (Example 2.2.1), so their extensions to F are also discrete (Proposition 2.3.2). Also, since the residue fields of the valuations of K(t) are finite extensions of K, so are the residue fields of the valuations of F . We define a prime divisor of F/K as an equivalence class of K-places of F . For p a prime divisor of F/K, choose a place ϕp in p. Then ϕp fixes ˜ ∪ {∞}. Denote its residue field by F¯p . the elements of K and maps F into K ¯ As mentioned above, Fp is a finite extension of K of degree deg(p) = [F¯p : K] which we call the degree of p. Also, choose a valuation vp corresponding to p and normalize it so that vp (F × ) = Z. Each element π of F with vp (π) = 1 is a local parameter of F at p. Denote the free Abelian group that the prime divisors of F/K generate by Div(F/K). P Each element a of Div(F/K) is a divisor of F/K. It has the form a = αp p, where p runs over the prime divisors of F/K, the αp are integers and all but finitely many of them are zero. Define a homomorphism vp : Div(F/K) → Z by vp (a) = αp . The symbol vp appears for two distinct functions. We show these uses are compatibleP as follows. Introduce the divisor of a nonzero element x of F as div(x) = vp (x)p. Since vp (x) = 0 for all but finitely many p the right hand side is well defined. If x is a constant (i.e. x ∈ K), then div(x) = 0. If x ∈ F r K, then the place of K(x) taking x to 0 (resp. to ∞) has finitely many extensions to the field F (Proposition 2.3.1). Equivalence classes of these extensions are the zeros (resp. poles) of x. Thus, div(x) is not zero if x is not a constant. Define the divisor of zeros and the divisor of poles

3.2 The Riemann-Roch Theorem

53

of x as follows: div0 (x) =

X

vp (x)p,

vp (x)>0

div∞ (x) = −

X

vp (x)p.

vp (x) 2g − 2 implies δ(a) = 0 and dim(a) = deg(a) − g + 1 . Proof of (a): L(0) = K because each x ∈ F r K has a pole. Similarly, L(div(x)) = Kx−1 . Proof of (b): Take a = 0 and then a = w in (3). Proof of (c): Use that deg(div(x)) = 0. Proof of (d): By (b) deg(w − a) < 0. Hence, by (c), δ(a) = dim(w − a) = 0. Thus, (3) simplifies to dim(a) = deg(a) + 1 − g. Here is our first application of Lemma 3.2.2(d): Lemma 3.2.3: Let F/K be a function field of genus g, p a prime divisor of F/K, and n a positive integer satisfying (n − 1) deg(p) > 2g − 2. Then there exists x ∈ F × with div∞ (x) = np. Proof: By Lemma 3.2.2(d), dim((n − 1)p) = (n − 1) deg(p) + 1 − g and dim(np) = n deg(p)+1−g. Hence, dim((n−1)p) < dim(np), so L((n−1)p) ⊂ L(np). Every element x ∈ L(np) r L((n − 1)p) will satisfy div∞ (x) = np.

56

Chapter 3. Algebraic Function Fields of One Variable

Example 3.2.4: Rational function field. Let F = K(t), where t is a transcendental element over K. Denote the prime divisor corresponding to the valuation v∞ (Section 2.2) by p∞ . We determine the linear space L(np∞ ), where n is a positive integer. If u ∈ L(np∞ ), then vp (u) ≥ 0 for every prime divisor p 6= p∞ . This means u ∈ K[t]. Also vp∞ (u) ≥ −n, so the degree of u as a polynomial in t is bounded by n. Thus, L(np∞ ) = {u ∈ K[t] | deg(u) ≤ n} and dim(np∞ ) = n + 1. For n > 2g − 2, Lemma 3.2.2(d) gives n + 1 = n − g + 1. Hence, g = 0. Conversely, let F/K be a function field of one variable of genus 0. Assume F/K has a prime divisor p of degree 1 (e.g. if K is algebraically closed). Lemma 3.2.3 gives t ∈ F × with div∞ (t) = p. By Section 3.1, [F : K(t)] = deg(div∞ (t)) = 1. Consequently, F = K(t). Pn Consider now a polynomial f (t) = i=0 ai ti with ai ∈ K and an 6= 0. For each i < n we have v∞ (an tn ) < v∞ (ai ti ), so v∞ (f (t)) = −n. For p 6= p∞ we have vp (t) ≥ 0, hence vp (f (t)) ≥ 0. Thus, div∞ (f (t)) = deg(f )p∞ . (t) Suppose g(t) ∈ K[t] is relatively prime to f (t). Put u = fg(t) . Then t is a root of the polynomial h(u, T ) = u · g(T ) − f (T ). Since h(u, T ) is linear in u and gcd(f, g) = 1, h(u, T ) is irreducible over K(u). In addition, degT (h) = max(deg(f ), deg(g)). Therefore,

(4)

[K(t) : K(u)] = max deg(f ), deg(g) .

In particular, suppose K(u) = K(t). Then, f (t) and g(t) are linear and relatively prime. This means, u = at+b ct+d with a, b, c, d ∈ K and ad − bc 6= 0.

3.3 Holomorphy Rings Let F/K be a function field of genus g. Denote the set of prime divisors of F/K by R. For each p ∈ R let Op = {x ∈ F | vp (x) ≥ 0} be the corresponding valuation ring. To every subset S of R we attach the holomorphy ring T OS = p∈S Op . By definition, K ⊆ OS . If S is empty, then, by definition, OS = F . If S = R, then the elements of OS have no poles. They are therefore constants. Thus, OR = K. The case where S is a nonempty proper subset of R requires a strengthening of the weak approximation theorem (Proposition 2.1.1). Proposition 3.3.1 (Strong Approximation Theorem): Let S be a finite subset of R. Consider q ∈ R r S and let S 0 = S ∪ {q}. Suppose for each p ∈ S we have an xp ∈ F and a positive integer mp . Then there exists x ∈ F with (1)

vp (x − xp ) = mp for each p ∈ S and vp (x) ≥ 0 for each p ∈ R r S 0 .

3.3 Holomorphy Rings

57

Moreover, if m is an integer with (2)

m · deg(q) > 2g − 2 +

X

(mp + 1) deg(p),

p∈S

then x can be chosen such that, in addition to (1), it satisfies vq (x) ≥ −m. Proof: Let P m be a positive integer satisfying (2). Consider the divisor a = mq − p∈S mp p. Then deg(a) > 2g − 2. By Lemma 3.2.2(d), δ(a) = 0, so A = F +Λ(a). Define ξ ∈ A by ξp = xp for p ∈ S and ξp = 0 for p ∈ R r S. Then there exists y ∈ F such that y − ξ ∈ Λ(a): (3) vp (y − xp ) ≥ mp for p ∈ S, vq (y) ≥ −m, and vp (y) ≥ 0 for p ∈ R r S 0 . P Now consider the divisor b = mq − p∈S (mp + 1)p. For each p ∈ S we have deg(b + p) > deg(b) > 2g − 2. By Lemma 3.2.2(d), dim(b + p) = deg(b + p) − g + 1 > deg(b) − g + 1 = dim(b). Hence, L(b) ⊂ L(b + p). Choose zp ∈ L(b + p) r L(b). Then vp (zp ) = mp . Also vp0 (zp ) ≥ mp0 + 1 if p0 ∈ S r{p}, vq (zp ) ≥ −m, and vp0 (zp ) ≥ 0 for p0 ∈ R r S 0 . r P Let P = {p ∈ S | vp (y − xp ) > mp } and let Q = S P . Then z = p∈P zp has the following property: (4) vp (z) = mp if p ∈ P , vp (z) ≥ mp + 1 if p ∈ Q, vq (z) ≥ −m, and vp (z) ≥ 0 for p ∈ R r S 0 . Combine (3) and (4) to see that x = z + y satisfies vp (x − xp ) = mp , for p ∈ S, vq (x) ≥ −m, and vp (x) ≥ 0 for p ∈ R r S 0 . If p belongs to a subset S of R, then OS ⊆ Op . Also, P = {x ∈ OS | vp (x) > 0} is a prime ideal of OS , the center of p at OS . Denote the local ring of OS at P by OS,P . Proposition 3.3.2 (Holomorphy Ring Theorem): Let S be a nonempty proper subset of R. Then S has these properties: (a) Quot(OS ) = F . (b) If p ∈ S and P is the center of p at OS , then Op = OS,P . (c) If q ∈ R r S, then OS 6⊆ Oq . (d) Every nonzero prime ideal of OS is the center of a prime p ∈ S. (e) Distinct primes in S have distinct centers at OS , and the center of each p ∈ S is a maximal ideal of OS . (f) OS is a Dedekind domain. Proof of (a): Consider z ∈ F r K. Since S ⊂ R, there is a q ∈ R r S. There are only finitely many p ∈ S with vp (z) < 0. Hence, Proposition 3.3.1 gives y ∈ F such that the following holds for each p ∈ S: If vp (z) < 0, then vp (y − z −1 ) = vp (z −1 ) + 1, so vp (y) = −vp (z); while if vp (z) ≥ 0, then

58

Chapter 3. Algebraic Function Fields of One Variable

vp (y) ≥ 0. Let x = yz. Then both x and y belong to OS . If z ∈ OS , then / OS , then there is a p ∈ S with vp (z) < 0. Hence, z ∈ Quot(OS ). If z ∈ vp (y) = vp (z −1 ) 6= ∞, so y 6= 0. Therefore, z = xy −1 ∈ Quot(OS ). Proof of (b): Let z ∈ Op r K. As in the proof of (a), there exists y ∈ F such that vp (y) = 0, vp0 (y) = −vp0 (z) if p0 ∈ S r{p} and vp0 (z) < 0, while vp0 (y) ≥ 0 if p0 ∈ S r{p} and vp0 (z) ≥ 0. Therefore, x = yz is in OS and z ∈ OS,P . Since the inclusion OS,P ⊆ Op is clear, Op = OS,P . Proof of (c): If S ∪ {q} = R and OS ⊆ Oq , then OS = OR = K, a contradiction to (a). Therefore, assume that S ∪ {q} is a proper subset of R. By Proposition 3.3.1, there exists x ∈ F such that vq (x) = −1 and vp (x) ≥ 0 for each p ∈ S. This element belongs to OS but not to Oq . Proof of (d): Let P be a nonzero prime ideal of OS . Proposition 2.3.1 extends the quotient map OS → OS /P to a place ϕ of F trivial on K. Let p be the prime divisor of F/K which is defined by ϕ. Then OS ⊆ Op and P = {x ∈ OS | vp (x) > 0}. By (c), p ∈ S. Proof of (e): Let p and p0 be two distinct prime divisors in S. By the strong approximation theorem (Proposition 3.3.1), there exists x ∈ OS with vp (x) > 0 and vp0 (x) = 0. This means the center P of p is not contained in the center of p0 . The maximality of P now follows from (d). Proof of (f): Let q ∈ R r S, let S 0 = R r{q} and choose p0 ∈ S 0 . Since OS is an overring of OS 0 it suffices (Proposition 2.4.7) to prove that OS 0 is a Dedekind domain. The strong approximation theorem gives x ∈ OS 0 with vp0 (x) > 0. Since x must have a pole, it must be q. Thus,T K[x] ⊆ Op if and only if p ∈ S 0 . Therefore, by Proposition 2.4.1, OS 0 = p∈S 0 Op is the integral closure of K[x] in F . Since K[x] is a Dedekind domain, OS 0 is also a Dedekind domain (Proposition 2.4.6). Corollary 3.3.3: The following conditions are equivalent for a nonempty subset S of R: (a) S = R; (b) OS = K; and (c) OS is a field. Proof: Every nonconstant element of F has a pole. Therefore, (a) implies (b). The implication “(b) => (c)” is trivial. To prove that (c) implies (a), note that OS 6= F , since S is nonempty. If S 6= R, then, by Proposition 3.3.2(a), the quotient field of OS is F . Hence, OS is not a field. The following converse to Proposition 3.3.2 is useful: Proposition 3.3.4: Let F/K be a function field and R a proper subring of F containing K. Suppose R is integrally closed and Quot(R) = F . Then there is a nonempty subset S of R with R = OS . Thus, R is a Dedekind domain. Proof: Let S be the set of all prime divisors of F/K which are finite on R. By Proposition 2.4.1, OS = R. As R ⊂ F , S is nonempty. By Corollary 3.3.3, S 6= R. Hence, by Proposition 3.3.2(f), R is a Dedekind domain.

3.4 Extensions of Function Fields

59

3.4 Extensions of Function Fields Let E/K and F/L be algebraic function fields of one variable. We say that F/L is an extension of E/K if E ⊆ F , K ⊆ L, and L ∩ E = K. We call F/L a constant field extension of E/K if E is linearly disjoint from L over K and F = EL. Thus, in this case [F : E] = [L : K]. Recall that E/K is a regular extension ((1b) of Section 3.1). Hence, the linear disjointness of E and L over K is automatic if L is an algebraic extension of K. Likewise, E is linearly disjoint from L over K if L = K(x1 , . . . , xn ) and x1 , . . . , xn are algebraically independent over E (Lemma 2.6.7). Let F/L be an extension of E/K, p be a prime divisor of E/K, and P be a prime divisor of F/L that lies over p, that is vP lies over vp . Denote the ramification index of vP over vp by eP/p . Since both vP and vp are discrete and normalized, vP (x) = eP/p vp (x) for each x ∈ E. Refer to P as unramified over E if vP is unramified over E. If F is separable algebraic over E, then only finitely many prime divisors of F/L are ramified over E [Deuring3, p. 111]. Over every prime divisor of E/K there lie only finitely many prime divisors of F/L [Deuring3, p. 96]. Use this result to embed the group of divisors Div(E/K) of E/K into Div(F/L) as follows: For p a prime divisor of E/K and P1 , . . . , Pd the prime divisors of F/L lying over p, map p to the Pd divisor i=1 ePi /p Pi of F/L. Extend this map to Div(E/K) by linearity. The principal divisor of x (in Div(E/K)) maps to the principal divisor of x in Div(F/L), so there is no ambiguity using div(x) for that divisor. In particular, for every divisor a of E/K we have LE (a) = E ∩ LF (a). Suppose [F : E] < ∞. If F/E is separable, apply (2) of Section 2.3 to conclude: (1)

d X

¯p ] = [F : E]. ePi /p [F¯Pi : E

i=1

Even if F/E is not separable, the following argument shows (1) still holds: By Lemma 3.2.3, there are an integer m and x ∈ E with div∞(x) = mp. Apply the rules degF div∞ (x) = [F : L(x)] and degE div∞ (x) = [E : K(x)] to conclude that (1) is true in general [Deuring3, p. 97]. Next consider a function field F/K and let E be a proper extension of ˜ over K, so is E. In K contained in F . Since F is linearly disjoint from K particular, there is a transcendental element x in E. Then [F : K(x)] = degF div∞ (x) < ∞. Thus, [E : K(x)] < ∞, E/K is a function field of one variable, and [F : E] < ∞. Lemma 3.4.1: Let E/K and F/K be algebraic function fields of one variable with E ⊆ F . Then, degF (a) = [F : E] degE (a) for each a ∈ Div(E/K). Proof: Assume by linearity that a = p is a prime divisor. Let P1 , . . . , Pd ¯p ⊆ F¯P , so be the prime divisors of F/K which lie over p. Then K ⊆ E i

60

Chapter 3. Algebraic Function Fields of One Variable

degF Pi = fPi /p degE (p), i = 1, . . . , d. By (1),

degF (p) =

d X

ePi /p degF (Pi )

i=1

=

d X

ePi /p fPi /p degE (p) = [F : E] degE (p).

i=1

Proposition 3.4.2: Let F/L be a constant field extension of an algebraic function field E/K of one variable with L/K separable. Then: (a) For a divisor a of E/K, degE (a) = degF (a), dimE (a) = dimF (a), and LE (a)L = LF (a). (b) genus(E/K) = genus(F/L). (c) If p and P are respective prime divisors of E/K and F/L, with P lying ¯p and P is unramified over p. over p, then F¯P = LE (d) Let R be an integrally closed subring of E containing K. Then RL is the integral closure of R in F . (e) Let p be a prime divisor of E/K of degree 1. Then there is a unique prime divisor P of F/L lying over p and deg(P) = 1. (f) Let x1 , . . . , xm be elements of E. Denote the integral closure of K[x] in E by S. Then SL is the integral closure of L[x] in F . Proof of (a): [Deuring3, p. 126] shows degE (a) = degF (a). We show dimE (a) = dimF (a). Let B be a basis of LE (a) over K. Then B is contained in LF (a) and is linearly independent over L. Therefore, dimE (a) ≤ dimF (a). Conversely, let B 0 be a basis of LF (a). Then B 0 is contained in EL0 for some finitely generated extension L0 /K of L/K. By [Deuring3, p. 132], dimEL0 (a) = dimE (a). Hence, dimF (a) = |B 0 | ≤ dimEL0 (a) = dimE (a). Therefore, dimE (a) = dimF (a). It follows that LE (a)L ⊆ LF (a) have the same dimension over L, so LE (a)L = LF (a). Proof of (b): Let gE = genus(E/K) and gF = genus(F/L). Choose a divisor a of E/K with degE (a) > max(2gE − 2, 2gF − 2). By Lemma 3.2.2(d), dimE (a) = degE (a) + 1 − gE and dimF (a) = degF (a) + 1 − gF . Therefore, by (a), gE = gF . Proof of (c): The general case reduces to the case where L/K is finitely ¯p . generated. In this case [Deuring3, p. 128] proves that F¯P = LE Now choose a primitive element c of L/K and let f = irr(c, K). By ¯ The reduction of f modulo P is f the preceding paragraph, F¯P = E(c). itself. Since L/K is separable, f is separable. Hence, by Lemma 2.3.4, P/p is unramified. Proof of (d): Assume without loss that [L : K] < ∞. Choose a basis w1 , . . . , wn of L/K. Let σ1 , . . . , σn be the K-embeddings of L into Ks . Since

3.5 Completions

61

E is linearly disjoint from L over K each σi extends uniquely to an Eembedding of F into Es . Pn Now consider z ∈ F which is integral over R. Write z = j=1 aj wj with Pn a1 , . . . , an ∈ E. Then σi z = j=1 aj σi wj , i = 1, . . . , n. By Cramer’s rule, b

b ∆

aj = ∆j = ∆j 2 , where bj is in the ring generated over R by the ak ’s and the σk wl , and ∆ = det(σk wl )1≤k,l≤n ∈ K × . By assumption, each bj is integral over R. Since L/K is separable, ∆2 ∈ K × [Lang7, p. 286, Cor. 5.4]. Hence, each aj is integral over R. Since R is integrally closed, aj ∈ R, j = 1, . . . , n. Consequently, z ∈ RL. Proof of (e): To prove (e), it suffices to consider two cases. In one case, L = K(u) with u purely transcendental over K. [Deuring3, pp. 128–129] handles this case. The other case is when L/K is separable and finite. Let in this case P1 , . . . , Pd be the prime divisors of F/L lying over p. By (c), ¯p = K. Hence, by ePi /p = 1, i = 1, . . . , m. By assumption, deg(p) = 1, so E (c), F¯Pi = L, i = 1, . . . , d. Since F/L is a constant field extension of E/K, we have [F : E] = [L : K]. Therefore, by (1), d = 1 and F¯P1 = L. Thus, deg(P1 ) = 1. S∞ Pm Proof of (f): Put S∞(xi ). By Lemma 3.1.1, S = k=1 LE (ka). S∞ a = i=1 div∞,E By (a), SL = k=1 LE (ka)L = k=1 LF (ka). Hence, by Lemma 3.1.1, SL is the integral closure of L[x] in F .

3.5 Completions The completion of a function field F/K at a prime divisor p gives a powerful tool to investigate the behavior of F at p. For example, it allows us to determine the decomposition of p to prime divisors in finite extensions of F (Proposition 3.5.3). We also use completions to define the ‘different’ of an extension. This notion plays a central role in the Riemann-Hurwitz genus formula, to be introduced in the next section. Let v be a rank-1 valuation of a field F . Then v induces a topology on F . Two elements x, y of F are ‘close’ in this topology if v(x − y) is ‘large’. A sequence {xi }∞ i=1 of elements of F is a Cauchy sequence if for every integer m there exists k such that i, j ≥ k implies v(xi − xj ) ≥ m. If every Cauchy sequence converges, F is complete. Each F embeds as a dense subfield in a complete field Fˆv with a valuation v extending the valuation of F [BorevichShafarevich, Chap. 1, Sec. 4.1]. In particular, Fˆv has the same residue field and value group at v as F . We call (Fˆv , v) (or also just Fˆv ) the completion of (F, v). We also say that Fˆv is the completion of F at v. The completion (Fˆv , v) of (F, v) is unique up to an F -automorphism. Example 3.5.1: Qp and K((t)). The completion of Q at the p-adic valuation vp (Section 2.2) is the field Qp of p-adic numbers. Every element x of Q× p has a unique presentation as a convergent (in the vp -topology) power series

62

Chapter 3. Algebraic Function Fields of One Variable

P∞

n n=m an p with m = vp (x), an ∈ Z, 0 ≤ an ≤ p − 1, and am 6= 0. The valuation ring Zp of Qp consists of all x ∈ Qp with m ≥ 0. Next consider a field K and a transcendental element t over K. Let v be the unique valuation of K(t) with v(t) = 1 (Section 2.2). The completion of K(t) at v is the field K((t)) of formal power series in t with coefficients in K. Each nonzero element f of K((t)) has a unique presentation f = P∞ n n=m an t with m = v(f ) and an ∈ K. The valuation ring of K((t)) is the ring K[[t]] of formal power series in t with coefficients in K. The residue field of K((t)) under v is K. By Lemma 2.6.9(b), K((t)) is a regular extension of K. Consider now a finite extension L of K with a basis wP 1 , . . . , wd . Then ∞ w1 , . . . , wd are linearly independent over K((t)). Let x = n=m an tn with Pd an ∈ L be an element of L((t)). For each n write an = i=1 ani wi with Pd P∞ ani ∈ K. Then, x = i=1 ( n=m ani tn )wi . Therefore, L((t)) = K((t))L and w1 , . . . , wd form a basis for L((t))/K((t)).

The next result contains various versions of Hensel’s Lemma: Proposition 3.5.2: Let (F, v) be a complete discrete valued field. (a) Let f ∈ Ov [X] and a ∈ Ov with v(f (a)) > 2v(f 0 (a)). Then there is a unique x ∈ F with f (x) = 0 and v(x − a) ≥ v(f (a)) − v(f 0 (a)) [CasselsFr¨ohlich, p. 83]. (b) Let f ∈ Ov [X] be a monic polynomial. Denote reduction at v by a bar. Suppose f¯(X) = ζ(X)η(X) with ζ, η ∈ F¯v [X] monic and relatively prime. Then there are monic polynomials g, h ∈ Ov [X] with g¯ = ζ, ¯ = η, and f (X) = g(X)h(X) [Zariski-Samuel2, p. 279, Thm. 17]. h (c) Let F 0 be a finite algebraic extension of F . Then v has a unique extension v 0 to F 0 and F 0 is complete under v 0 [Cassels-Fr¨ohlich, p. 56, Thm.]. Here is a global application of completions: Proposition 3.5.3: Let (E, v) be a discrete valued field. Denote its comˆ vˆ). Consider a finite separable extension F of E. Let z be a pletion by (E, primitive element for F/E which is integral over Ov . Put h = irr(z, E). Let h = h1 · · · hr be the decomposition of h into a product of irreducible polyˆs . Denote the unique ˆ For each i let zi be a root of hi in E nomials over E. ˆ ˆ extension of vˆ to Fi = E(zi ) by vˆi . The following holds: (a) The map z 7→ zi extends to a E-embedding of F into Fˆi . (b) Fˆi is the completion of F at the restriction vi of vˆi to F . (c) The valuations v1 , . . . , vr are mutually nonequivalent. Every extension of v to F coincides with one of the vi ’s. ¯ i = he with hi0 ∈ E[X] ¯ (d) evˆi /ˆv = evi /v , fvˆi /ˆv = fvi /v , and h irreducible i0 and e ∈ N. ˆ ˆ (e) The map g(z) 7→ g(z1 ), . . . , g(zr )) for g ∈ E[X] is an E-isomorphism of L r ˆ ˆ E ⊗E F onto i=1 Fi .

3.5 Completions

(f) Each z in F satisfies traceF/E z = Qr ˆ zi . i=1 normFˆi /E

63

Pr

i=1

traceFˆi /Eˆ zi and normF/E z =

Proof: See [Cassels-Fr¨ ohlich, p. 57, Thm.]. The last statement of (d) follows from Proposition 3.5.2(b). Example 3.5.4: Dedekind. We apply Proposition 3.5.3 to show the necessity ¯v is infinite” in Lemma 2.3.5. of the assumption “E Consider the polynomial f (X) = X 3 − X 2 − 2X − 8. Observe that f (X) has no root modulo 3, hence no root in Q. It is therefore irreducible. Let z ˜ and F = Q(z). Then [F : Q] = 3. Next observe that be a root of f (X) in Q 0 v2 (f (0)) = 3, v2 (f (0)) = 1, v2 (f (1)) = 1, v2 (f 0 (1)) = 0, v2 (f (2)) = 3, and v2 (f 0 (2)) = 1. Hence, by Proposition 3.5.2(a), f has three roots x1 , x2 , x3 in Q2 with v2 (x1 ) ≥ 2, v2 (x2 − 1) ≥ 1, and v2 (x3 − 2) ≥ 2. Thus, f (X) = (X − x1 )(X − x2 )(X − x3 ) is a decomposition of f (X) over Q2 into three distinct irreducible polynomials. By Proposition 3.5.3, v2 has three distinct extensions to F : w, w0 , and w00 . By Proposition 2.3.2, F¯w = F2 . Now assume F/Q has a primitive element z with g = irr(z, Q) ∈ Ov2 [X] such that g¯ is separable. Then deg(g) = 3. By the preceding paragraph, F¯w = F2 . Hence, all three distinct roots of g¯ belong to F2 . But F2 has only two elements, so z does not exist. Lemma 3.5.5: Let (F, v) be a discrete valued field and I a nonzero Ov submodule of F which is not F . Then I is a fractional ideal of Ov and there exists m ∈ Z with I = m−m v . If F is a finite extension of a field E, and Ov is the unique valuation ring of F lying over Ov ∩ E (e.g. (E, v) is complete), then traceF/E I is a fractional ideal of Ov ∩ E. Proof: Choose π ∈ F with v(π) = 1. Let x be a nonzero element of I. If 0 x0 ∈ F and v(x0 ) ≥ v(x), then x0 = xx x ∈ I. Since I 6= F , this implies v(I) is bounded from below. Hence, −m = inf(v(x) | x ∈ I) is an integer. Therefore, I = π −m Ov . Now suppose Ov is the unique valuation ring of F over O = Ov ∩ E. By Proposition 2.4.1, Ov is the integral closure of O in F . Let a be an element of A with v(a) sufficiently large. By the preceding paragraph, aI ⊆ Ov . Then a · traceF/E (I) = traceF/E (aI) ⊆ traceF/E (Ov ) ⊆ A. Thus, traceF/E (I) is a fractional ideal of A. Using completions, we generalize the notions of repartition and differential (Section 3.2) of a function field. Let p be a prime divisor of F/K. Suppose the residue field F¯p of F at p is separable over K. By Hensel’s Lemma, F¯p embeds into Fˆp . Indeed, a). Put f = irr(¯ a, K). Then vp (f (a)) > 0 and choose a ∈ F with F¯p = K(¯ vp (f 0 (a)) = 0. By Proposition 3.5.2(a), there is an x0 ∈ Fˆp with f (x0 ) = 0. The map a ¯ 7→ x0 extends to a K-embedding of F¯p into Fˆp . Let π be an element of Fˆp with vp (π) = 1. Then the completion Fˆp is isomorphic to the field F¯p ((π)) of formal power series in π over F¯p . Every

64

Chapter 3. Algebraic Function Fields of One Variable

P∞ element of this field has the form α = i=m ai π i , where m is an integer and ai ∈ F¯p [Chevalley2, p. 46]. If am 6= 0, then vp (α) = m. Q Consider the cartesian product Fˆp with p ranging over all prime diviQ sors of F/K. An α ∈ Fˆp is an adele if vp (αp ) ≥ 0 for all but finitely many p. In particular, each repartition of F/K is an adele. Denote the set of adeles ˆ ˆ by Q ˆA (or by AF , if F is not clear from the context). It is an F -subalgebra of Fp which contains the algebra A of repartitions. For each prime divisor p ˆ as follows. Identify x ∈ Fˆp with the adele ξ having of F/K embed Fˆp in A 0 ξp = x and ξp = 0 if p0 6= p. For each a ∈ Div(F/K) consider the K-vector space ˆ | vp (α) + vp (a) ≥ 0 for each p}, ˆ Λ(a) = {α ∈ A ˆ = Λ(a) and A + where we have abbreviated vp (αp ) by vp (α). Then A ∩ Λ(a) ˆ Indeed, let α ˆ For each p choose αp ∈ F with vp (αp − α ˆ Λ(a) = A. ˆ ∈ A. ˆp) ≥ ˆ ˆ is in Λ(a), and α ˆ = α − (α − α ˆ) ∈ −vp (a). Then α = (αp ) belongs to A, α − α ˆ A + Λ(a). Therefore (1)

ˆ ˆ ˆ A ∩ (Λ(a) + F ) = Λ(a) + F and A + (Λ(a) + F ) = A.

∼ A/ Λ(a) + F . Thus, in the notation of Section 3.2, ˆ Λ(a) ˆ Hence, A/ +F = (2)

ˆ Λ(a) ˆ + F ) = δ(a) = dim(w − a), dimK A/(

where w is a canonical divisor of F/K. Recall that a differential of F/K is a K-linear map, ω: A → K, which vanishes on a subspace of the form Λ(a)+F . By (1) we can extend ω uniquely ˆ → K which vanishes on Λ(a) ˆ + F . So, from now on, a to a K-linear map ω ˆ: A ˆ ˆ differential of F/K is a K-linear map ω ˆ : A → K which vanishes on Λ(a)+F for some a ∈ Div(F/K). The restriction of ω to A is a differential in the old sense. The divisor of ω ˆ is the maximum of all a ∈ Div(F/K) such that ω ˆ ˆ vanishes on Λ(a). Denote it by div(ˆ ω ). By the above, div(ˆ ω ) = div(ω). ˆ Denote the K-vector space of all differentials that vanish on Λ(a) +F ˆ ˆ by Ω(a). The natural map ω 7→ ω ˆ is an isomorphism of Ω(a) onto Ω(a). In particular, these spaces have the same dimension δ(a). Suppose F/E is a finite separable extension of function fields of one ˆ E . Given a prime divisor P of F/K, denote its variable over K. Let α ∈ A ˆ E as a subalgebra of restriction to E by p and let αP = αp . This identifies A L ˆ F . The isomorphisms E ˆp ⊗E F ∼ A = P|p FˆP of Proposition 3.5.3(e) combine to an isomorphism (3)

ˆ E ⊗E F ∼ ˆF A =A

3.5 Completions

65

[Cassels-Fr¨ohlich, p. 64 or Artin3, p. 244, Thm. 2]. In addition, define a trace ˆF → A ˆE: function traceF/E : A traceF/E (β)p =

X

traceFˆP /Eˆp (βP ).

P|p

Proposition 3.5.6: Let F/E be a finite separable extension of function ˆF : fields of one variable over a field K. The following holds for all α, β ∈ A (a) traceF/E (α + β) = traceF/E (α) + traceF/E (β). ˆE. (b) traceF/E (αβ) = α · traceF/E (β) if α ∈ A (c) The trace of an element x of F coincides with the trace of x as an adele. (d) Let P be a prime divisor of F/K and p = P|E . Then, the trace of an ˆp coincides with the trace of x as an adele. element x of FˆP to E ˆ F such that traceF/E (α) 6= 0. (e) There exists α ∈ A Proof: Statements (a) and (b) follow from the corresponding properties of the trace function on fields. Statement (c) follows from Proposition 3.5.3(f). Statement (d) follows from the definition. Finally, use (c) and the corresponding fact for the trace of fields [Lang7, p. 286, Thm. 5.2] to prove (e). Remark 3.5.7: Complementary modules. (a) Let (E, v) be a complete discrete valued field. Denote the valuation ring of (E, v) by OE . Let F be a finite separable extension of E. Denote the unique extension of v to F by v. Choose a generator πF of the maximal ideal of OF . The complementary module of OF over E is 0 OF/E = {x ∈ F | traceF/E (xOF ) ⊆ OE }.

It is a fractional ideal of OF which contains OF [Lang5, p. 58, Cor.]. By −d

0 Lemma 3.5.5, OF/E = πF F /E OF for some nonnegative integer dF/E . Call dF/E the different exponent of F/E. It is known [Lang5, p. 62, Prop. 8] that dF/E > 0 if and only if F/E is ramified. Moreover, dF/E ≥ eF/E − 1. If the residue field extension is separable, then equality holds if and only if F/E is tamely ramified [Serre4, p. 67, Prop. 13]. (b) Let F/E be a finite separable extension of function fields of one variable over a field K. Consider prime divisors p and P of E/K and F/K, ˆp and FˆP with respectively, with P lying over p. Choose completions E ˆp and FˆP , by O ˆ p and O ˆ P , reˆp ⊆ FˆP . Denote the valuation rings of E E spectively. Occasionally write dFˆP /Eˆp as dP/p or dP/E and call dP/p the different exponent of P over p (or over E). By (a), dP/p > 0 if and only if P/p is ramified. This happens for only finitely many P’s.

66

Chapter 3. Algebraic Function Fields of One Variable

Lemma 3.5.8: Let E ⊆ F ⊆ F 0 be function fields of one variable over a field K with F 0 /E separable. Consider prime divisors p, P, P0 of E/K, F/K, F 0 /K, respectively, with P lying over p and P0 lying over P. Then dP0 /p = eP0 /P dP/p + dP0 /P . In particular, if P0 /P is unramified, then dP0 /p = dP/p . If P/p is unramified, then dP0 /p = dP0 /P . Proof: Suppose the formula holds. If P0 /P is unramified, then dP0 /P = 0 and eP0 /P = 1. Hence, dP0 /p = dP/p . If P/p is unramified, then dP/p = 0, so dP0 /p = dP0 /P . To prove the formula, assume without loss, E, F , and F 0 are complete with respective valuation rings OE , OF , and OF 0 . Choose prime elements πE , πF , and πF 0 for the respective maximal ideals. Put d = dF/E , d0 = dF 0 /F , and e0 = eF 0 /F . We have to prove that e0 d + d0 = dF 0 /E . 0

First note that πFe 0 OF 0 = πF OF 0 . Hence, 0

0

0

traceF 0 /E (πF−e0 d−d OF 0 ) = traceF/E (traceF 0 /F (πF−d πF−d0 OF 0 )) 0

= traceF/E (πF−d traceF 0 /F (πF−d0 OF 0 )) ⊆ traceF/E (πF−d OF ) ⊆ OE . Thus, 0

0

πF−e0 d−d OF 0 ⊆ OF0 0 /E .

(4)

0 By definition, πF−d−1 ∈ / OF/E . Hence, traceF/E (πF−d−1 OF ) 6⊆ OE . By the

second part of Lemma 3.5.5, traceF/E (πF−d−1 OF ) is a fractional ideal of OE . −m Hence, by the first part of Lemma 3.5.5, traceF/E (πF−d−1 OF ) = πE OE for −1 −d−1 some positive integer m. Therefore, πE OE ⊆ traceF/E (πF OF ). Simi0

larly, πF−1 OF ⊆ traceF 0 /F (πF−d0 −1 OF 0 ). 0

0

0

0

Assume πF−e0 d−d OF 0 6= OF0 0 /E . Then, by (4), πF−e0 d−d −1 OF 0 ⊆ OF0 0 /E . Therefore, by the preceding paragraph, 0

0

OE ⊇ traceF 0 /E (πF−e0 d−d −1 OF 0 ) 0

= traceF/E (πF−d traceF 0 /F (πF−d0 −1 OF 0 )) −1 ⊇ traceF/E (πF−d−1 OF ) ⊇ πE OE , 0

0

−d

which is a contradiction. Therefore, πF−e0 d−d OF 0 = OF0 0 /E = πF 0 F Consequently, e0 d + d0 = dF 0 /E , as claimed.

0 /E

OF 0 .

3.6 The Different

67

3.6 The Different The Riemann-Hurwitz genus formula enables us to compute the genus of a function field of one variable F over a field K from the genus of a function subfield E in terms of the ‘different’ of the extension F/E. This formula is in particular useful when the genus of E is known, e.g. when E is the field of rational functions over K. Let F/E be a finite separable extension of function fields of one variable over a field K. The different of F/E is a divisor of F/K: Diff(F/E) =

X

dP/E P,

where P ranges over all prime divisors of F/K and dP/E is the different exponents of P/E. By Remark 3.5.7(a), dP/E ≥ 0 for all P. Moreover, dP/E > 0 if and only if P is ramified over E. Since only finitely many P ramify over E, Diff(F/E) is well defined, Diff(F/E) ≥ 0, and deg(Diff(F/E)) ≥ 0. Theorem 3.6.1 (Riemann-Hurwitz Genus Formula): Let F/E be a finite separable extension of function fields of one variable over a field K. Put gE = genus(E/K) and gF = genus(F/K). Then (1)

2gF − 2 = [F : E](2gE − 2) + deg(Diff(F/E)).

Proof: Let pi , i ∈ I, be the prime divisors of E/K. Choose a nonzero P differential ω of E/K. Write div(ω) = i∈I ki pi with ki ∈ Z and ki = 0 for all but finitely many i P ∈ I. By definition, P ˆ ˆE( (2) ω vanishes on Λ i∈I ki pi ) but not on ΛE ((kr + 1)pr + i6=r ki pi ) for any r ∈ I. ˆ F → K: Define a map Ω: A Ω(α) = ω(traceF/E (α)). It is K-linear and vanishes P on F (Proposition 3.5.6). For each i ∈ I let pi = j∈Ji eij pij , where the pij are the distinct prime divisors of F/K lying over pi and the eij are the corresponding ramification ˆp , indices. We simplify notation and for all i ∈ I and j ∈ Ji let Ei = E i ˆ ˆ ˆ Oi = Opi , vi = vpi , Fij = Fpij , Oij = Opij , vij = vpij , and (3)

0 dij = vij (Diff(F/E)) = −vij (Oij ),

0 is the complementary module of Oij with respect to traceF/E where Oij ˆ E and β ∈ A ˆ F let αi = αp and βij = βp . (Remark 3.5.7). Also, for all α ∈ A i ij We prove that ˆ F (P (eij ki + dij )pij ) and (4a) Ω vanishes on Λ i,j (4b) for each r ∈ I and each sP ∈ Jr the differential Ω does not vanish on ˆ F (ers kr + drs + 1)prs + Λ (e k + d )p ij ij . (i,j)6=(r,s) ij i

68

Chapter 3. Algebraic Function Fields of One Variable

This will imply that divF (Ω) =

X

(eij ki + dij )pij = divE (ω) + Diff(F/E).

i,j

The formulas deg(divF (Ω)) = 2gF − 2 and deg(divE (ω)) = [F : E](2gE − 2) (Lemmas 3.2.2(b) and 3.4.1) will give (1). For all i, j choose πi ∈ E and πij ∈ F with vi (πi ) = 1 and vij (πij ) = 1. ˆ F (P (eij ki + dij )pij ). Then vij (α) ≥ Proof of (4a): Consider α ∈ Λ ij −eij ki − dij .

Hence, vij (πiki α) ≥ −dij .

0 By (3), (πiki α)ij ∈ Oij , so

traceFij /Ei (πiki α)ij ∈ Oi . Thus, (traceF/E (πiki α))i =

X

traceFij /Ei (πiki α) ∈ Oi ,

j∈Ji

hence vi (traceF/E (α)) + ki ≥ 0 for each i ∈ I. Therefore, traceF/E (α) ∈ ˆ E (P ki pi ). Consequently, by (2), Ω(α) = ω(traceF/E (α)) = 0. Λ i∈I

Proof of (4b): Assume there exist r ∈ I and s ∈ Jr such that Ω vanishes on X ˆ F (ers kr + drs + 1)prs + (eij ki + dij )pij . V =Λ (i,j)6=(r,s)

By Lemma 3.5.5, there is an m ∈ Z with −drs −1 Ors ) = πr−m Or Irs = traceFrs /Er (πr−kr πrs

We distinguish between two cases: ˆ E (kr + 1)pr + P Consider β ∈ Λ i6=r ki pi . Then ˆ E (P ki pi ), so ω(β) = vr (β) ≥ −kr − 1. If vr (β) ≥ −kr , then β ∈ Λ i∈I 0 (by (2)). Otherwise, vr (β) = −kr − 1. In this case write β = α + γ, ˆ E satisfy αr = (1 − πr )βr , γr = πr βr , αi = 0, and γi = βi where α, γ ∈ A for i 6= r. Then, αr ∈ πr−kr −1 Or ⊆ πr−m Or = Irs , so there is a δrs ∈ −drs −1 Ors with traceFrs /Er δrs = αr . For (i, j) 6= (r, s) let δij = 0. πr−kr πrs Then δ ∈ V , α = traceF/E δ, and ω(α) = ω(trace(δ)) = Ω(δ) = 0. Also, ˆ E (P ki pi ), so ω(γ) = 0 (by (2)). It follows that ω(β) = 0. Thus, ω γ∈Λ i∈I ˆ E (kr + 1)pr + P vanishes on Λ ki pi , in contradiction to (2). Case A: m ≥ kr + 1.

i6=r

Case B: m ≤ kr .

Then,

−drs −1 Ors ) = πr−m Or ⊆ πr−kr Or . traceFrs /Er (πr−kr πrs −drs −1 −drs −1 0 Hence, traceFrs /Er (πrs Ors ) ⊆ Or . Therefore, πrs Ors ⊆ Ors , so 0 vrs (Ors ) ≤ −drs − 1. This contradiction to (3) completes the proof of Case B and the proof of the whole theorem.

3.6 The Different

69

Remark 3.6.2: Applications of the Riemann-Hurwitz formula. Let F/E be a finite separable extension of function fields of one variable of a field K. We say that F/E is unramified (resp. tamely ramified) if each prime divisor of F/K is unramified (resp. tamely ramified) over E. Let gE = genus(E/K) and gF = genus(F/K). Suppose [F : E] ≥ 2. (a) Comparison of genera: We have mentioned at the beginning of this section that deg(Div(F/E)) ≥ 0. Hence, by (1), gF ≥ gE . Both gE and gF have the same value if and only if gE = 1 and F/E is unramified, or gE = 0 and deg(Diff(F/E)) = 2 [F : E] − 1 . In particular, if F = K(t), then gF = 0 (Example 3.2.4). Hence, gE = 0. Each prime divisor of F/K of degree 1 induces a prime divisor of E/K of degree 1. We conclude from Example 3.2.4 that E = K(u) is also a rational function field. This is L¨ uroth’s theorem. This theorem actually holds for arbitrary algebraic extension K(t)/E and not only for separable extensions and may be proved by elementary arguments on polynomials [Waerden1, p. 218]. (b) An analog of a theorem of Minkowski: Suppose F/E is unramified. In this case the Riemann-Hurwitz formula simplifies to gF − 1 = [F : E](gE − 1). Hence, gE > 0. In other words, a function field E/K of genus 0 has no proper finite unramified extension F which is regular over K. In particular, K(t) has no finite proper separable unramified extension F which is regular over K. This is an analog of a theorem of Minkowski saying that Q has no proper unramified extensions [Janusz, p. 57, Cor. 11.11]. (c) The Hurwitz-Riemann formula for tamely ramified extensions: Suppose F/E is tamely ramified. By Remark 3.5.7, the Riemann-Hurwitz formula simplifies to XX eP/p − 1 deg(P). 2gF − 2 = [F : E](2gE − 2) + p

P|p

(d) An analog of Minkowski’s theorem in the tamely ramified case: Suppose K is algebraically closed and gE = 0. Then E = K(t) and the degree of each prime divisor is 1. Suppose F/E is a proper tamely ramified extension. Then E has at least two prime divisors that ramify in F . Indeed, assume E has only one prime divisor p that ramifies in F . Let P1 , . . . , Pr be the prime divisors of p in F . Then −2 ≤ 2gF − 2 = −2[F : E] +

r X

ePi /p − 1

i=1

= −2[F : E] + [F : E] − r = −[F : E] − r Hence, 3 ≤ [F : E] + r ≤ 2, a contradiction.

70

Chapter 3. Algebraic Function Fields of One Variable

In particular, if K is algebraically closed and char(K) = 0, then every proper extension of K(t) is ramified over at least two prime divisors. (e) Generation of Galois groups by inertia groups: Let K be an algebraically closed field, E = K(t), and F a finite Galois extension of E. Denote the prime divisors of F/K which ramify over E by P1 , . . . , Pr . For each i let Di be the decomposition group of Pi over E. Since K is algebraically closed, Di is also the inertia group of Pi over E. Let Ei be the fixed field of Di in F . Then Pi |Ei is unramified over E. Hence, E0 = E1 ∩ · · · ∩ Er is unramified over E. By (b), E0 = E. In other words, the inertia groups of the prime divisors of F/K which ramify over E generate Gal(F/E). (f) Quasi-p groups: Let K be an algebraically closed field of positive characteristic p. Consider the rational field E = K(t), a prime divisor p of E/K, and a finite Galois extension F of E which is ramified only over p. Denote the fixed field in F of all p-Sylow subgroups of Gal(F/E) by Ep . Then Ep /E is a Galois extension of degree relatively prime to p. Hence, Ep /E is tamely ramified. The only prime divisor of E/K which is possibly ramified in Ep is p. It follows from (d) that Ep = E. In other words, Gal(F/E) is generated by its p-Sylow subgroups. One says that Gal(F/E) is quasi-p. (g) Abhyankar’s conjecture: Let K be an algebraically closed field of positive characteristic p. Let E = K(t) and p a prime divisor of E/K. In 1957, Abhyankar conjectured that for each finite quasi-p group G there exists a finite Galois extension F of E which is unramified outside p and Gal(F/E) ∼ = G [Abhyankar]. Serre proved the conjecture for solvable G in 1990 [Serre9]. Raynaud treated all other cases in 1991 [Raynaud]. (h) The generalized Abhyankar’s conjecture: Let again K be an algebraically closed field of positive characteristic p. Put E = K(t). Consider a set S = {p1 , . . . , pr } of prime divisors of E/K and a finite Galois extension F of E which is unramified over E outside S. Let G = Gal(F/E). Denote the subgroup of G generated by all p-Sylow subgroups of G by G(p). Let Ep be the fixed field of G(p) in F . Then Ep is a Galois extension of E which is tamely ramified over E and unramified outside S. By [Grothendieck, XIII, Cor. 2.12], Gal(Ep /E) is generated by r − 1 elements. This led Abhyankar to conjecture that for every finite group G such that G/G(p) is generated by r − 1 elements there is a Galois extension F of E which is unramified outside S and Gal(F/E) ∼ = G. Harbater proved this conjecture in [Harbater2] by reducing it to the special case r = 1 proved by Raynaud.

3.7 Hyperelliptic Fields We demonstrate the concepts and results of this chapter to a study of a special kind of algebraic function fields of one variable which we now introduce. A function field F/K is hyperelliptic if its genus is at least 2 and if it is a quadratic extension of a function field E/K of genus 0. We then say E is a quadratic subfield of F . It turns out that E is uniquely determined.

3.7 Hyperelliptic Fields

71

Moreover, we will be able to identify E from the arithmetic of F . Lemma 3.7.1: Let F/K be a function field of genus g. Let a, b be nonnegative divisors of F/K and let w be a canonical divisor of F/K. (a) If L(b − a) = L(b), and dim(b) ≥ 1, then dim(a) = 1. (b) If g ≥ 1 and a > 0, then dim(w − a) < dim(w). (c) If g ≥ 1 and x1 , . . . , xg is a basis for L(w), then −w = min div(x1 ), . . . , div(xg ) . Proof of (a): Since a ≥ 0, we have K ⊆ L(a). Conversely, let x ∈ L(a). Then div(x) + a ≥ 0. Consider y ∈ L(b). By assumption, y ∈ L(b − a), hence div(y) + b − a ≥ 0. Therefore, div(xy) + b ≥ 0, so xy ∈ L(b). Apply this Pn result to a basis y1 , . . . , yn of L(b). Find aij ∈ K such that xyi = j=1 aij yj , i = 1, . . . , n. Hence, det(xI − A) = 0, where A = (aij )1≤i,j≤n . Therefore, x satisfies a monic equation with coefficients in K. Since K is algebraically closed in F , we have x ∈ K. Consequently, dim(a) = 1. Proof of (b): Assume dim(w − a) = dim(w). Then L(w − a) = L(w). By Lemma 3.2.2(b), dim(w − a) = dim(w) = g. So, by (a), dim(a) = 1. By Theorem 3.2.1, dim(a) = deg(a) + 1 − g + dim(w − a). Hence, deg(a) = 0, which is a contradiction. Therefore, dim(w − a) < dim(w). Proof of (c): Denote min div(x1 ), . . . , div(xg ) by m. Since div(xi ) + w ≥ 0 we have m ≥ −w. If m > −w, then there exists a prime divisor p of F/K with m − p ≥ −w. Hence, div(xi ) + w − p ≥ 0, so xi ∈ L(w − p) for i = 1, . . . , g. Therefore, L(w − p) = L(w), contradicting (b). Consequently, m = −w. Proposition 3.7.2: Let F/K be a function field of genus g ≥ 2. Consider a canonical divisor w of F/K. Let x1 , . . . , xg be a basis of L(w). If E = x K xx12 , . . . , xg1 is a proper subfield of F , then genus(E/K) = 0 and [F : E] = 2. Thus, F/K is a hyperelliptic field. Proof: Let w0 = div(x1 ) + w. Then w0 is also a canonical divisor and x 1, xx21 , . . . , xg1 is a basis of L(w0 ). Replace w by w0 , if necessary, to assume that x1 = 1 and E = K(x2 , . . . , xg ). Since g ≥ 2, the element x2 is transcendental over K. Hence, E is also an algebraic function field of one variable over K and d = [F : E] < ∞ (Section divisor of E/K. 3.4). Denote genus(E/K) by gE and let wE be a canonical By Lemma 3.7.1(c), w = − min 0, div(x2 ), . . . , div(xg ) . In particular, w ≥ 0. By Lemma 3.2.2(b), deg(w) = 2g − 2 ≥ 2. Therefore, (1)

w > 0.

Observe that div(xi ) ∈ Div(E/K), i = 2, . . . , g, so w ∈ Div(E/K). We may therefore apply Riemann-Roch to E/K and w: (2) dim LE (w) = degE (w) + 1 − gE + dim LE (wE − w) .

72

Chapter 3. Algebraic Function Fields of One Variable

Since 1, x1 , . . . , xg are in E and generateLF (w), we have LE (w) = LF (w). Hence, by Lemma 3.2.2(b), dim LE (w) = g. Applying Lemma 3.2.2(b) again and Lemma 3.4.1, we have 2g−2 = degF (w) = d·degE (w). Substituting this in (2) gives (3)

g=

2g − 2 + 1 − gE + dim LE (wE − w) , d

which may be rewritten as (4)

d dim LE (wE − w) − gE ) = (g − 1)(d − 2)

By assumption, both g and d are at least 2. Hence, the right hand side of (4) is at least 0. On the other hand, by Lemma 3.2.2(b) and by (1), (5) dim LE (wE − w) ≤ dim LE (wE ) = gE . Hence, both sides of (4) are 0. Since g ≥ 2, this gives d = 2. Finally, by (3), gE = dim(LE (wE − w)). Hence, by (5), dim(LE (wE − w)) = dim(LE (wE )). We conclude from Lemma 3.7.1(b) that gE = 0. Lemma 3.7.3: Let F/K be a function field of genus g ≥ 1, w a canonical divisor of F/K, and x1P , . . . , xg a basis of LF (w). Let E/K be a subfield of g genus 0 of F/K. Then i=1 Exi 6= F . ˆ E (0)xi ⊆ Λ ˆ F (w). Hence, Proof: For each i we have Λ g X

(6)

ˆ F (w) + F. ˆ E (0) + E xi ⊆ Λ Λ

i=1

Choose a canonical divisor wE of E/K. By (2) of Section 3.5, ˆ E /(Λ ˆ E (0) + E) = dim(wE ) = genus(E/K) = 0. dimK A Hence, ˆE = Λ ˆ E (0) + E. A

(7) On the other hand,

ˆ F /(Λ ˆ F (w) + F ) = dim(0) = 1. dimK A ˆ F , so by (6) and (7) ˆ F (w) + F ⊂ A Hence, Λ (8)

g X

ˆ E xi ⊂ A ˆF . A

i=1

Pg Pg ˆ ˆ If i=1 Exi = F , then Pg by (3) of Section 3.5, AF = i=1 AE xi , contradicting (8). Consequently, i=1 Exi 6= F .

3.8 Hyperelliptic Fields with a Rational Quadratic Subfield

73

Proposition 3.7.4: Let F/K be a hyperelliptic function field, w a canonical x divisor of F/K, and x1 , . . . , xg a basis of LF (w). Then E = K xx21 , . . . , xg1 is the only quadratic subfield of F . Proof: By definition, F has subfields of genus 0. Let E 0 be one Pquadratic g 0 of them. By Lemma 3.7.3, i=1 E xi 6= F . Since [F : E 0 ] = 2, this implies xi ∈ E 0 , i = 1, . . . , g. It follows that E ⊆ E 0 . In particular, E ⊂ F . By Proposition 3.7.2, [F : E] = 2. Comparing degrees implies E 0 = E.

3.8 Hyperelliptic Fields with a Rational Quadratic Subfield A hyperelliptic field with a rational quadratic subfield is generated by two generators which satisfy an equation of a special type. This situation arises, for example, when the hyperelliptic field has a prime divisor of degree 1 (Example 3.2.4). Proposition 3.8.1: Let F/K be a hyperelliptic field of genus g. Suppose the quadratic subfield of F is K(x) with x indeterminate. Then F = K(x, y), where y satisfies a relation y 2 + h1 (x)y + h2 (x) = 0 with h1 , h2 ∈ K[X], deg(h1 ) ≤ g + 1, and deg(h2 ) ≤ 2g + 2. If char(K) 6= 2, we may choose y with y 2 = f (x), where f ∈ K[X] is a polynomial with no multiple root and deg(f (x)) ≤ 2g + 2. Proof: Let p∞ be the pole of x in K(x)/K. By Lemma 3.4.1, degF (np∞ ) = 2n for each positive integer n. If n ≥ g, then 2n > 2g − 2. By Lemma 3.2.2(d), dim(LF (np∞ )) = 2n + 1 − g. In particular, for n = g the elements 1, x, . . . , xg form a basis for LF (gp∞ ). For n = g + 1 we have dim LF ((g + 1)p∞ ) = g + 3. Hence, two more elements are needed to complete 1, x, . . . , xg to a basis of LF (g + 1)p∞ . We take one of them as xg+1 and denote the other one by y. By RiemannRoch, dim(LK(x) ((g + 1)p∞ )) = g + 2. So, 1, x, . . . , xg+1 form a basis of LK(x) (g + 1)p∞ . If y is in K(x), then y is also in LK(x) (g + 1)p∞ . This implies that 1, x, . . . , xg+1 , y are linearly dependent over K. We conclude from this contradiction that y ∈ / K(x), so F = K(x, y). Next note that dim LF ((2g + 2)p∞ ) = 3g + 5. All 3g + 6 elements 1, x, . . . , x2g+2 , y, yx, . . . , yxg+1 , y 2 belong to LF (2g + 2)p∞ . Therefore, P2g+2 Pg+1 there are ai , bj , c ∈ K, not all zero, with i=0 ai xi + j=0 bj yxj + cy 2 = 0. Pg+1 −1 bj X j , h2 (X) = Since y ∈ / K(x), we have c 6= 0. Let h1 (X) = j=0 c P2g+2 −1 ai X i . Then i=0 c (1)

y 2 + h1 (x)y + h2 (x) = 0.

If char(K) 6= 2, replace y by 2y + h1 (x), if necessary, to assume (1) has the form y 2 = f (x). Here f ∈ K[X] has degree at most 2g + 2. Finally, if f has

74

Chapter 3. Algebraic Function Fields of One Variable

multiple roots, then f = g 2 h with h having no multiple roots. Replace y by yg(x)−1 to assume f has no multiple roots. Our first task is to compute the genus of the hyperelliptic field in characteristic 6= 2 from deg(f ). Proposition 3.8.2: Let K be a field of characteristic 6= 2 and f ∈ K[X] a polynomial of degree d ≥ 1 with no multiple roots. Let F = K(x, y), with x transcendental over K and y 2 = f (x). Then F/K is an algebraic function d−2 field of one variable of genus d−1 2 if d is odd and of genus 2 if d is even. In particular, if d ≥ 5, then F/K is hyperelliptic. ˜ Proof: Since f (x) is not a square in K(x), ˜ ˜ [K(x, y) : K(x)] = [K(x, y) : K(x)] = 2. ˜ Hence, K(x, y) is linearly disjoint from K(x) over K(x). Since x is transcen˜ over K. Hence, by the tower dental over K, K(x) is linearly disjoint form K ˜ property of linear disjointness (Lemma 2.5.3), F is linearly disjoint from K over K. It follows that F/K is an algebraic function field of one variable. Since char(K) 6= 2, F/K(x) is tamely ramified. Since the genus of K(x) is 0, the Riemann-Hurwitz formula reduces to XX (eP/p − 1) deg(P) (2) 2g − 2 = −4 + p

P|p

(Remark 3.6.2(c)). If p is a prime divisor of K(x) which ramifies in F , then p has only one extension P to F , the ramification index of P is 2, and its residue degree is 1. Hence, (2) simplifies to (3)

2g = −2 +

X

deg(p),

where p ranges over all prime divisors of K(x)/K which ramify in F . Let f (x) = p1 (x) · · · pr (x) be the decomposition of f (x) into a product of distinct irreducible polynomials in K[x]. To each pi there corresponds a prime divisor pi of K(x)/K of degree deg(pi ) and a valuation vi such that vi (f (x)) = vi (pi (x)) = 1 (Example 2.2.1(b)). By Example 2.3.8, each pi ramifies in F . In addition, let v∞ be the valuation of K(x)/K with v∞ (x) = −1 and let p∞ be the corresponding prime divisor; its degree is 1. Since v∞ (f (x)) = −d, the prime divisor p∞ is ramified in F if d is odd and unramified if d is even (Example 2.3.8). All other prime divisors of K(x) are unramified in F . The Pr sum of the degrees of the ramified prime divisors is δ + i=1 deg(pi ) = δ + d, where δ = 1 if d is odd and δ = 0 if d is even. It follows from (3) that g = d−1 2 if d is odd and g = d−2 2 if d is even. In characteristic 2 we compute the genus of only a special type of a hyperelliptic field.

Exercises

75

Lemma 3.8.3: Let (K, v) be a valued field of characteristic 2 and t an element of K with v(t) = 1. Consider an Artin-Schreier extension L = K(x) with x2 + x = 1t . Denote the unique extension of v to L by w. Let y = tx. Then 0 = y −2 Ow . w(y) = 1, Ow = Ov [y] is the integral closure of Ov in L, and Ow Proof: By Example 2.3.11, v has a unique extension w to L. It totally ramifies over K and satisfies w(x) = −1, w(t) = 2, and w(y) = 1. In particular, Ow is the integral closure of Ov in L (Proposition 2.4.1). Since ¯ w , each z ∈ Ow can be written as z = a + by with a ∈ Ov and b ∈ Ow . ¯v = L K Hence, Ow = Ov [y] [Lang5, p. 26, Prop. 23]. 0 To compute Ow = {z ∈ L | traceL/K (zOw ) ⊆ Ov } observe that g(X) = 2 0 = t−1 Ow = y −2 Ow irr(y, K) = Y + tY + t and g 0 (X) = t. Therefore, Ow 0 [Lang5, p. 59, Cor.]. Of course, one may also compute Ow directly. Proposition 3.8.4: Let K be a field of characteristic 2 and a1 , . . . , ad distinct elements of K. Let F = K(x, y) with x transcendental over K, h ∈ K[X] with deg(h) ≤ d, and y 2 + y = (x−a1h(x) )···(x−ad ) . Then F/K is an algebraic function field of one variable of genus d − 1. In particular, if d ≥ 3, then F/K is hyperelliptic. Proof: For each i between 1 and d let vi be the discrete normalized valuation of K(x)/K with vi (x − ai ) = 1. Let t = (x − a1 ) · · · (x − ad ). Then vi (t) = 1. ˜ ˜ If y is in K(x), ˜ K. then 2˜ vi (y) = −1, Extend vi to a valuation v˜i of K(x)/ ˜ which is a contradiction. Hence, y is not in K(x), so [K(x, y) : K(x)] = ˜ ˜ ˜ over K, so [K(x, y) : K(x)] = 2. It follows that F is linearly disjoint from K F/K is an algebraic function field of one variable. By Example 2.3.9, v1 , . . . , vd are the only valuations of K(x)/K that ramify in F . Hence, only v1 , . . . , vd contribute to the different of F/K(x). [ (y). By Let wi be the unique extension of vi to F . Then Fˆwi = K(x) vi Lemma 3.8.3, the contribution of wi to deg Diff(F/K(x)) is 2. Hence, deg(Diff(F/K(x))) = 2d. Let g = genus(F/K). By Riemann-Hurwitz (Theorem 3.6.1), 2g − 2 = −4 + 2d. Hence, g = d − 1, as claimed.

Exercises In the following exercises F is a function field of one variable over a field K and t is a transcendental element over K. 1. Let a be a divisor of F/K. Note: If a ≥ 0 and deg(a) = 0, then a = 0. (a) Prove that if deg(a) = 0 and a is not a principal divisor, then dim(a) = 0. (b) Let g be the genus of F . Suppose a is a noncanonical divisor with deg(a) = 2g − 2. Show that dim(a) = g − 1. Hint: Use Riemann-Roch and (a). 2. Let a and b be divisors of F/K with b ≥ 0. Use Riemann-Roch to prove that (a) dim(a) ≤ dim(a + b) ≤ dim(a) + deg(b); and

76

Chapter 3. Algebraic Function Fields of One Variable

(b) dim(a) ≤ max(0, deg(a) + 1). Hint: Write a as the difference of its “positive” and “negative” parts. 3. Let p1 , . . . , pn be distinct Pnprime divisors of F/K and let m1 , . . . , mn be If K is infinite, prove positive integers such that i=1 mi deg(pi ) > 2g − P1. n × such that div (x) = that there exists x ∈ F ∞ i=1 Pn mi pi . In particular, Pn [F : K(x)] = i=1 mi deg(pi ). Hint: Consider a = i=1 mi pi and aj = a−pj for j = 1, . . . , n. For each j, 1 ≤ j ≤ n, distinguish between the cases deg(aj ) ≤ 2g − 2 and deg(aj ) > 2g − 2, and prove that L(aj ) ⊂ L(a). 4. Suppose F = K(t). Prove, in the notation of Example 3.2.4, that −2p∞ is a canonical divisor of F/K. 5. Suppose the genus of F/K is 0. Prove that every divisor a with deg(a) = 0 is principal. Hint: Compute dim(a) and apply Exercise 1(a). kr ∈ N, and 6. In the notation of Proposition 3.3.2 let p1 , . . . , pr ∈ S, k1 , . . . , Q n 1, . . . , r. Consider the ideal A = i=1 Piki of Pi be the center of pi at OS , i =P r OS . Prove that the divisor a = i=1 ki pi of F/K satisfies deg(a) = (OS : A). Qr Hint: Prove that OS /A ∼ = i=1 OS /Piki and that OS /Pi ∼ = Pik /Pik+1 , as groups, for each nonnegative integer k. 7. Prove that every algebraic function field F/K of genus 2 is hyperelliptic with a rational quadratic subfield. Hint: Choose a positive canonical divisor w of F/K. Then choose a nonconstant x in L(w). 8. Let K be a field of characteristic that does not divide n. Consider the function field F = K(x, y) over K with x, y satisfying xn +y n = 1. Prove that . Hint: Use Example 2.3.8 and the Riemann-Hurwitz genus(F ) = (n−1)(n−2) 2 formula (Remark 3.6.2(c)).

Notes In addition to [Chevalley2] one may find a proof of the Riemann-Roch theorem in [Lang4, Chapter 10, Section 2], [Deuring3, Section 15], and [Stichtenoth, Section I.5]. The content of Section 3.7 is borrowed from [Artin3]. One may also find it in [Stichtenoth, Section VI.2] along with computations of genera of various function fields using the Riemann-Hurwitz genus formula. However, in contrast to our exposition, [Stichtenoth] assumes the fields of constants to be perfect.

Chapter 4. The Riemann Hypothesis for Function Fields In this chapter K is a finite field of characteristic p with q elements. Let F be an algebraic function field of one variable over K and g the genus of F/K. Denote the group of divisors and the group of divisor classes of F/K by D and C, respectively. P ∞ The series ζ(s) = n=1 n−s defines the classical Riemann zeta function of a complex variable. It converges absolutely for Re(s) > 1. Hence, ζ(s) is an analytic function in this domain. The series diverges for s = 1. The function, however, can be analytically continued to a function meromorphic on the whole s-plane. The production of this requires two stages: an analytic 1 to the half plane Re(s) > 0 by a rearrangement of continuation of ζ(s) + 1−s the series via Abel summation; and then, (the difficult part) a demonstration ζ(s) is the product of the classical Gamma function and an elementary that ζ(1−s) function in the domain 0 < Re(s) < 1. The expression that results from the last stage is called the functional equation for ζ(s). The resulting analytic continuation of ζ(s) yields a function with a simple pole at s = 1 with residue 1 and zeros at the points −2, −4, −6, −8, . . . . There are no other zeros in the domains Re(s) ≥ 1 and Re(s) ≤ 0 [Titchmarsh, p. 30]. The classical Riemann hypothesis is still unproven. It states that the only zeros of ζ(s) in the strip 0 < Re(s) < 1 lie on the line Re(s) = 12 ; its applications are legion. There is an analog for F/K of the Riemann zeta function (Section 4.2). It satisfies a functional equation (Proposition 4.4.1). Our main goal is the proof of an analog to the Riemann hypothesis (Theorem 4.5.1). In Chapter 5 we extract from this an explicit estimate for the number of points on any curve over a finite field.

4.1 Class Numbers The assumption that K is finite results in the finiteness of other sets connected to F/K. For example, F/K has only finitely many ideal classes of degree 0 and only finitely many nonnegative divisors of a given degree n. The main result of this section (Lemma 4.1.4) computes the latter number in terms of the former one. Let p be a prime divisor of F/K. Its residue field F¯p is a finite extension of K of degree deg(p). Thus, F¯p is a finite field whose order we indicate by N p = q deg(p) , the norm of p. Extend the definition of the norm to arbitrary divisors by the formula N a = q deg(a) . Then N (a + b) = N a · N b. Definition 4.1.1: Denote the number of divisor classes of F/K of degree zero by h; the class number of F/K. If g = 0, then every divisor of degree 0 of F/K is principal (Exercise 5

78

Chapter 4. The Riemann Hypothesis for Function Fields

of Chapter 6). Therefore, h = 1 if F is K(x) or any genus 0 function field. Lemma 4.1.2: Only finitely many nonnegative divisors of F/K have degree equal to a given integer m. In addition, the class number of F/K is finite. Proof: Let x ∈ F be transcendental element over K and let E = K(x). Denote the collection of all prime divisors of E/K of degree ≤ m by S0 . Each element of S0 , except possibly p∞ , corresponds to a monic irreducible polynomial in K[x] of degree ≤ m. Hence, S0 is a finite set. Only finitely many prime divisors of F/K lie over a given prime divisor p0 of E/K. Each of them has degree at least as large as deg(p0 ). Thus, there are only finitely many prime divisors of F/K of degree ≤ m. Therefore, the set Am of all nonnegative divisors of F/K of degree m is finite. For the second part of the lemma choose a nonnegative divisor m of degree m ≥ g. Denote the set of divisor classes of degree m by Cm . For each b ∈ Cm Riemann-Roch (Theorem 3.2.1) implies that dim(b) ≥ m − g + 1 ≥ 1. Hence, there is an x ∈ F × with div(x) + b ≥ 0, so the class of b contains a nonnegative divisor. It follows that the map Am → Cm mapping each a ∈ Am onto each class is surjective. Therefore, by the preceding paragraph, Cm is finite. Finally, the map a 7→ a + m induces a bijective map of C0 onto Cm . Consequently, C0 is a finite group. Lemma 4.1.3: The number of nonnegative divisors in a given class of divisors C of F/K is

q dim(C) −1 . q−1

Proof: If a ≥ 0, then div(a) + a ≥ 0 for each a ∈ K × , so K ⊆ L(a). Hence, dim(a) ≥ 1. This gives the formula if dim(C) = 0. Suppose dim(C) = n > 0. Let c be a divisor in C. The number of nonnegative divisors in C is equal to . , xn be a the number of principal divisors div(x) with x ∈ L(c). Let x1 , . .P n basis for L(c) over K. The number of elements of F × of the form i=1 ai xi , with a1 , . . . , an ∈ K is equal to q n − 1. Since div(x) = div(x0 ) if and only if there exists an a ∈ K × such that x0 = ax, the formula follows. Denote the greatest common divisor of the degrees of the divisors of F/K by δ. Eventually we prove that δ = 1. In the meantime notice that a positive integer n is a multiple of δ if and only if there exists a divisor of degree n. Indeed, there are divisors a1 , . . . , ar with δ = gcd(deg(a Pr 1 ), . . . , deg(ar )). For , . . . , a ∈ Z with each multiple n of δ there are a 1 r i=1 ai deg(ai ) = n. The Pr divisor a = i=1 ai ai satisfies deg(a) = n. In particular, since the degree of the canonical divisor is 2g − 2 (Lemma 3.2.2(b)) δ divides 2g − 2. Lemma 4.1.4: Let An be the number of nonnegative divisors of F/K of degree n. If n ≥ 0 is a multiple of δ larger than 2g − 2, then An = h q

n−g+1

q−1

−1

.

Proof: Let n be a multiple of δ with n > 2g − 2. Choose a divisor c of degree n. By Riemann-Roch, dim(c) = n − g + 1. Moreover, the map a 7→ a + c defines a bijection of the set of divisor classes of degree 0 with the set of

4.2 Zeta Functions

79

divisor classes of degree m. Hence, there are h divisor classes of degree n. It n−g+1 follows that, An = h q q−1 −1 .

4.2 Zeta Functions The zeta function of a function field over K is a rational function with coefficients in Q and with simple poles at several points including 1. We define the zeta function of the function field F/K to be the Dirichlet series X (N a)−s (1) ζ(s) = ζF/K (s) = a≥0

where a runs over the nonnegative divisors of F/K. Check the domain of convergence of the series (1) using the substitution t = q −s and the identity N a = q deg(a) . We obtain a power series for ζ(s) in terms of t: (2)

Z(t) =

X

tdeg(a) =

∞ X

An tn ,

n=0

a≥0

where An is the number of nonnegative divisors of degree n. By Lemma 4.1.4 Z(t) =

2g−2 X

An tn + h

n=0

∞ X q mδ−g+1 − 1 mδ t q−1

m=d

where d = 2g−2+δ . The right hand side converges for |t| < q −1 ; (i.e. for δ Re(s) > 1) and (3)

Z(t) = Φ(t) +

hq g−1+δ t2g−2+δ t2g−2+δ h · · − , q−1 1 − (qt)δ q − 1 1 − tδ

where Φ(t) =

2g−2 X

An tn

n=0

is a polynomial of degree ≤ 2g − 2. We summarize: Proposition 4.2.1: The power series Z(t) in (2) converges in the circle |t| < q −1 . Formula (3) continues Z(t) to a meromorphic function on the whole plane. The only poles of Z(t) occur for values of t with tδ = 1 or tδ = q −δ , and they are simple. The Dirichlet series for ζ(s) in (1) converges in the right half plane Re(s) > 1. The substitution t = q −s in (3) continues ζ(s) to a meromorphic function in the whole plane. Like the Riemann zeta function, ζF/K (s) has a multiplicative presentation. Let a1 , a2 , a3 , . . . be a sequence of complex numbers of absolute value

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Chapter 4. The Riemann Hypothesis for Function Fields

less than 1. We say that the infinite product solutely converges) if the limit n Y

1 n→∞ 1 − ai i=1 lim

Q∞

1 i=1 1−ai

converges (resp. ab-

n Y

1 n→∞ 1 − |ai | i=1

resp. lim

converges to a nonzero complex number. Proposition 4.2.2: If Re(s) > 1 and |t| < q −1 , then Y Y 1 1 = , (4) ζ(s) = −s deg(p) 1 − (N p) 1 − t p p where p runs over the prime divisors of F/K. The product converges absolutely. Therefore, it is independent of the order of the factors. In particular, if Re(s) > 1, then ζ(s) 6= 0. Proof: The prime divisors are free generators of the group of divisors. Thus, for every positive integer m, if Re(s) > 1, then Y N p≤m

∞ Y X X X0 1 = (N p)−sk = (N a)−s + (N a)−s , −s 1 − (N p) a≥0 a≥0 N p≤m k=0

N a>m

N a≤m

where the prime in the second sum means that a runs over all nonnegative divisors with norm exceeding m whose prime divisors p satisfy N p ≤ m. It follows that X Y 1 |(N a)−s | ≤ ζ(s) − 1 − (N p)−s a≥0 N p≤m

N a>m

and the right hand side converges to zero as m tends to infinity. Now we prove that ζ(s) 6= 0 when Re(s) > 1. Indeed, consider a positive integer m. Then, Y Y 1 1 − (N p)−s = ζ(s) 1 − (N p)−s N p>m N p≤m X X00 (N a)−s ≥ 1 − (N a)Re(s) > 0, = 1 + a>0 N a>m

(where the double primes mean summation over all positive divisors a with only prime divisors p satisfying N p > m) if m is large enough. Therefore, ζ(s) 6= 0. Q In particular, p 1−(N p)1 −Re(s) converges. Since, |N p−s | = N p−Re(s) , this Q means that p 1−(N1p)−s absolutely converges. By Exercise 3, the value of the product is independent of the order of the factors.

4.3 Zeta Functions under Constant Field Extensions

81

4.3 Zeta Functions under Constant Field Extensions The analytic properties of the zeta function of a function field F/K proved in Proposition 4.2.2 result in the conclusion that δ = 1 (Corollary 4.3.3). Denote the unique extension of K of degree r by Kr . Then Fr = F Kr is a function field of one variable over Kr . Use r as a subscript to denote the “extension” of objects of F to Fr . For example, if a is a divisor of F/K, then degr a denotes the degree of a as a divisor of Fr /Kr . We have already noted that degr a = deg(a), dimr a = dim(a) and gr = g (Proposition 3.4.2). Lemma 4.3.1: Let p be a prime divisor of F/K. Then p decomposes in Fr as p = P1 + P2 + · · · + Pd , where P1 , P2 , . . . , Pd are distinct prime divisors of Fr /Kr , deg(Pi ) = r−1 · lcm(r, deg(p)) and d = gcd(r, deg(p)). Proof:

Put m = deg(p). Since p is unramified in Fr (Proposition 3.4.2), P1 , . . . , Pd are distinct. Moreover, (Fr )Pi = Kr F¯p . Hence, [(Fr )Pi : K] =

lcm(r, m), and thus deg(Pi ) = [(Fr )Pi : Kr ] = r−1 · lcm(r, m), i = 1, . . . , d. Also, by Propositions 2.6 and 3.4.2, r = [Kr : K] = [Fr : F ] = d · [F¯r,Pi : F¯p ] = d · lcm(r, m)m−1 . Therefore, d = gcd(r, m). Proposition 4.3.2: For every complex number t, Y (1) Zr (tr ) = Z(ξt), ξ r =1

where ξ runs over the rth roots of unity. Proof: Since both sides of (1) are rational functions of t (by (3) of Section 4.2), it suffices to prove (1) for |t| < q −r . First apply the product formula (4) of Section 4.2, then (Lemma 4.3.1): YY Y (1 − tr·deg(P) ) = (1 − tlcm(r,deg(p)) )gcd(r,deg(p)) (2) Zr (tr )−1 = p P|p

(3)

Y ξ r =1

Z(ξt)−1 =

p

Y Y p

(1 − (ξt)deg(p) ).

ξ r =1

Thus, (1) follows if we show equality of the corresponding factors on the right hand sides of (2) and (3). Indeed, for a fixed p let m = deg(p) and d = gcd(r, m). We must show that Y (1 − (ξt)m ). (4) (1 − trm/d )d = ξ r =1

Substitute tm = x−1 in (4) to rewrite it as Y (x − ξ m ) . (5) (xr/d − 1)d = ξ r =1

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Chapter 4. The Riemann Hypothesis for Function Fields

Both monic polynomials in (5) have each (r/d)th root of unity as a zero of multiplicity d. Indeed, if ζr is a primitive root of unity of order r, then ζrm is a primitive root of unity of order r/d and each power of ζrm appears d times (r−1)m among 1, ζrm , ζr2m , . . . , ζr . Therefore, the polynomials are equal. Corollary 4.3.3 (F. K. Schmidt): δ = 1. P∞ Proof: By (2) of Section 4.2, Z(t) = m=0 Amδ tmδ . Hence, if ξ δ = 1, then Z(ξt) = Z(t). From (1), Zδ (tδ ) = Z(t)δ . However, by (3) of Section 4.2 (applied to Zδ (tδ ) instead of to Z(t)), Zδ (tδ ) has a simple pole at t = 1, while Z(t)δ has a pole of order δ at t = 1. Consequently, δ = 1. Corollary 4.3.4: For every integer n there are exactly h divisor classes of F/K of degree n. Proof: By Corollary 4.3.3 there is a divisor c of degree n. The map a → a+c induces a bijection of the set of divisor classes of F/K of degree 0 onto the set of divisor classes of degree m. Hence, the number of elements in the latter set equals the number of elements in the former set, namely h. Example 4.3.5: By definition, A1 is the number of prime divisors of F/K of degree 1. If g = 0, then h = 1 and A1 = q +1 (Lemma 4.1.4). Thus, F/K has prime divisors of degree 1. By Example 3.2.4, F = K(x) is a rational function field. If δ = 1 (Corollary 4.3.3) and g = 1, then A1 = h (Lemma 4.1.4); in other words the class number is equal to the number of prime divisors of degree 1. Since this is the order of a group (Definition 4.1.1 ), it is again 1 . positive. By Lemma 4.1.4 and by (2) of Section 4.1.2, Z(t) = (1−t)(1−qt)

4.4 The Functional Equation Like the Riemann zeta functions, Z(t) satisfies a functional equation relating 1 . The main tool in the proof is the Riemann-Roch its values in t and qt theorem. Proposition 4.4.1: Z(t) satisfies the functional equation 1 √ √ ( qt)1−g Z(t) = ( qt)g−1 Z . qt Proof:

If g = 0, the result follows from the explicit presentation Z(t) = in Example 4.3.5. Therefore, assume g > 0. The basic idea is to split Z(t) into the sum of a polynomial P (t) and an infinite series Q(t), each of which satisfies the same functional equation in the statement of the proposition. Apply Lemmas 4.1.3 and 4.1.4 with δ = 1 (Corollary 4.3.3) to obtain: 1 (1−t)(1−qt)

Z(t) =

X a≥0

tdeg(a) =

2g−2 X deg(C)=0

X a≥0 a∈C

tdeg(a) +

∞ X n=2g−1

An tn

4.4 The Functional Equation

=

2g−2 X deg(C)=0

=

h 1 q−1

83

∞ X q dim(C) − 1 deg(C) q n−g+1 − 1 n t t + h q−1 q−1 n=2g−1 2g−2 X

q dim(C) tdeg(C)

i

deg(C)=0

+

∞ ∞ h h X h X ni q n−g+1 tn − t q − 1 n=2g−1 q − 1 n=0

= P (t) + Q(t), where P (t) (resp. Q(t)) is the expression in the first (resp. second) brackets. We have also used that F/K has exactly h divisor classes of each degree (Corollary 4.3.4). First we analyze P (t). The Riemann-Roch theorem relates dim(C) to dim(W − C), where W is the canonical class. Recall that deg(W ) = 2g − 2 (Lemma 3.2.2(b)). Hence 1 1 deg(C) = deg(C) + 1 − g + dim(W − C) 2 2 1 = dim(W − C) − deg(W − C). 2 As C varies over all divisor classes of degree between 0 and 2g − 2 so does W − C. Hence √ 2g−2 ( qt)2−2g X √ 2−2g ( qt) P (t) = q dim(C) tdeg(C) q−1 dim(C) −

deg(C)=0

=

=

=

1 q−1 1 q−1 1 q−1

2g−2 X

1 √ q dim(C)− 2 deg(C) ( qt)2−2g+deg(C)

deg(C)=0 2g−2 X

1 √ q dim(W −C)− 2 deg(W −C) ( qt)− deg(W −C)

deg(C)=0 2g−2 X

0

0

q dim(C ) (qt)− deg(C ) = P

deg(C 0 )=0

1 . qt

Now evaluate the geometric series involved in the expression for Q(t): ∞ ∞ h h q g t2g−1 h h X n−g+1 n X n i 1 i Q(t) = − q t − t = q − 1 n=2g−1 q − 1 1 − qt 1−t n=0 A direct computation shows that √

( qt)

2−2g

1 . Q(t) = Q qt

84

Chapter 4. The Riemann Hypothesis for Function Fields

This completes the proof of the proposition.

4.5 The Riemann Hypothesis and Degree 1 Prime Divisors We reformulate here the Riemann hypothesis for a function field F of one variable over a finite field and draw an estimate for the number of prime divisors of F of degree 1. Rewrite formula (3) of Section 4.2 with δ = 1 (Corollary 4.3.3) as Z(t) = Φ(t) +

t2g−1 h t2g−1 hq g · − · q − 1 1 − qt q − 1 1 − t

where Φ(t) a polynomial of degree ≤ 2g − 2. Hence Z(t) =

L(t) (1 − t)(1 − qt)

with L(t) = a0 + a1 t + · · · + a2g t2g a polynomial with rational coefficients. We determine some of these coefficients: First: a0 = L(0) = Z(0) = A0 = 1, since the zero divisor is the only nonnegative divisor of degree 0. Second: A1 is equal to the number of prime divisors of F/K of degree 1. Write N = A1 , so that L(t) = (1 − t)(1 − qt)

∞ X

An tn ≡ 1 + (N − (q + 1))t mod t2 .

n=0

Therefore, a1 = N − (q + 1)

(1)

Now let x ∈ F be transcendental over K and write F0 = K(x). The Zeta 1 (Example 4.3.5). By Proposition function of F0 /K is Z0 (t) = (1−t)(1−qt) 4.4.1 1 q g−1 t2g−2 Z qt Z(t) 1 g 2g (2) L(t) = = −1 −2 . t L = q 1 Z0 (t) qt q t Z0 qt This functional equation for L(t), written explicitly, has the form 2g X i=0

ai ti =

2g X

a2g−i q i−g ti .

i=0

Equivalently, ai = q i−g a2g−i . In particular, deduce for i = 0 and i = 1 that (3) a2g = q g and a2g−1 = q g−1 N − (q + 1) .

4.5 The Riemann Hypothesis and Degree 1 Prime Divisors

85

These formulas imply that deg(L(t)) = 2g. Decompose L(t) over C as (4)

L(t) =

2g Y

(1 − ωi t)

i=1

where the ωi−1 ’s are the zeros of L(t). Formulas (1), (3), and (4) give (5)

qg =

2g Y

ωi

and N − (q + 1) = −

i=1

2g X

ωi .

i=1

Moreover, the functional equation (2) for L(t) implies that 1 ωi = 0. L = 0 if and only if L ωi q √ √ Rename the roots ω1 , . . . , ω2g as ω1 , ω10 , . . . , ωf , ωf0 , q, . . . , q, √ √ − q, . . . , − q with f ≤ g such that ωi ωi0 = q, i = 1, . . . , f , and q √ (resp. − q) appear k (resp. l) times. Then 2f + k + l = 2g shows that if k is odd, then l is odd. In this case (5) gives q g = q f q k/2 (−1)l q l/2 = −q g , a contradiction. Hence, both k and l are even and we may take f = g. Thus, √

L(t) =

g Y

(1 − ωi t)(1 − ωi0 t),

i=1

with ωi ωi0 = q for i = 1, . . . , g. Here is a reformulation of the Riemann hypothesis for the function field F/K. Sections 4.6 - 4.8 complete the proof. Theorem 4.5.1: (a) The zeros of the function ζF/K (s) lie on the line Re(s) = 12 . 1

(b) The zeros of the function ZF/K (t) lie on the circle |t| = q − 2 . (c) If the ωi are the inverses of the zeros of the polynomial LF/K (t), then |ωi | =

√ q,

i = 1, 2, . . . , 2g.

Note that (c) is equivalent to (b), since the poles of Z(t) are t = 1 and t = q −1 . Theorem 4.5.1 with (5) provides an estimate on the number of prime divisors of degree 1: Theorem 4.5.2: Let F be a function field of one variable over a finite field K of q elements. Denote the genus of F/K by g and let N be the number of √ prime divisors of F/K of degree 1. Then |N − (q + 1)| ≤ 2g q.

86

Chapter 4. The Riemann Hypothesis for Function Fields

4.6 Reduction Steps Theorem 4.5.2 is a consequence of the Riemann hypothesis. As a first step, this section shows that an appropriate version of Theorem 4.5.2 implies the Riemann hypothesis. As in Section 4.3, denote the unique extension of K of degree r by Kr . Lemma 4.6.1: The Riemann hypothesis holds for the function field F/K if and only if it holds for the function field Fr /Kr . Proof: Use (2) and (4) of Section 4.5 to express L(t). Then apply Proposition 4.3.2 to compute Lr (tr ): Lr (tr ) =

Y Z(ξt) Y Zr (tr ) = = L(ξt) r Zr,0 (t ) Z0 (ξt) r r ξ =1

ξ =1

=

2g Y Y ξ r =1

(1 − ωi ξt) =

i=1

2g Y

(1 − ωir tr ).

i=1

Q2g Hence, Lr (t) = i=1 (1 − ωir t). Thus, r (1) ω1r , . . . , ω2g are the inverses of the zeros of Lr . √ √ Since |ωi | = q if and only if |ωir | = q r , the lemma follows.

Denote the number of prime divisors of Fr /Kr of degree 1 by Nr . Lemma 4.6.2: Let F be a function field of one variable over a field K of q elements. If there exists a constant c such that |Nr − (q r + 1)| ≤ cq r/2 for every positive integer r, then the Riemann hypothesis holds for F/K. log Proof: Apply the differential operator D = −t ddt to both sides of the Q2g formula L(t) = i=1 (1 − ωi t):

(2)

D(L(t)) =

2g X i=1

2g ∞ X X ωi t = ωir tr 1 − ωi t n=1 r=1

P2g Combine (1) with (5) of Section 4.5 to obtain − i=1 ωir = Nr − (q r + 1). P2g r The hypothesis of the lemma thus implies | i=1 ωi | ≤ cq r/2 . Therefore 1 the radius of convergence R of the right hand side of (2) satisfies R ≥ q − 2 . But (2) implies that the ωi−1 are the only singularities of D(L(t)). Hence, √ R = min1≤i≤2g |ωi−1 |. Therefore, |ωi | ≤ q for i = 1, . . . , 2g. This together Q2g √ with the equality q g = i=1 ωi ((5) of Section 4.5), implies that |ωi | = q for i = 1, . . . , 2g.

4.7 An Upper Bound

87

4.7 An Upper Bound Assume, by extension of constants if necessary, that K and F satisfy these conditions: (1a) q = a2 is a square; (1b) q > (g + 1)4 ; and (1c) F has a prime divisor o of degree 1. By Lemma 4.6.1, a proof of the Riemann hypothesis under these conditions suffices for the general case. We prove a result that has, as a special case, the inequality √ N − (q + 1) < (2g + 1) q.

(2)

Let σ be an automorphism of F over K. It induces a permutation of the prime divisors of F/K. If p is a prime divisor of F/K, the pσ is the prime divisor corresponding to the place ϕσp (x) = ϕp (σx) for x ∈ F (where, as in Section 2.1, ϕσp = σ −1 ϕp ). Also, recall that the map x 7→ xq is an automorphism F˜ over K. Hence, the formula ϕqp (x) = ϕp (x)q defines a place ϕqp of F/K. Although ϕqp is equivalent to ϕp , it is convenient to use ϕqp , because ϕp = ϕqp if and only if deg(p) = 1. The remainder of this section investigates the expression N (σ) =

X

deg(p)

q ϕσ p =ϕp

in order to show that (3)

√ N (σ) − (q + 1) < (2g + 1) q.

If σ is the identity automorphism, then N (σ) = N and (3) becomes (2). With the notation m = a − 1, n = a + 2g, and r = m + an, rewrite (3) as (4)

N (σ) − 1 ≤ r.

With o as in (1c), consider the ascending sequence of K-vector spaces L(o) ⊆ L(2o) ⊆ L(3o) ⊆ · · ·. By Exercise 2 of Chapter 3, (5)

dim(L(io)) − dim(L((i − 1))o) ≤ 1.

88

Chapter 4. The Riemann Hypothesis for Function Fields

With k a positive integer, let Ik be the set of i, 1 ≤ i ≤ k for which equality holds in (5). For each i ∈ Ik , choose ui ∈ L(io)−L((i−1)o). Then div∞ (ui ) = io and the system {ui | i ∈ Ik } is a basis for L(ko). In particular, this holds for k = m. Since a2 = q, a is a power of char(K). Thus, the set L(no)a = {y a | y ∈ L(no)}, consisting of elements in the field F a is a Kvector space with basis {uaj | j ∈ In } and the same dimension as L(no). Therefore, o nX ui yia | yi ∈ L(no) L= i∈Im

is a K-vector space generated by the set U = {ui uaj | i ∈ Im and j ∈ In }. Lemma 4.7.1: The set U is linearly independent over K. Proof: It suffices to prove that {ui | i ∈ Im } is linearly independent over a . Indeed, assume that there exist yi ∈ F , i ∈ Im , not all zero, the field FP a such that i∈Im ui yi = 0. Then there exist distinct i, j ∈ Im such that a a vo (ui yi ) = vo (uj yj ), yi 6= 0, yj 6= 0 ((2) of Section 2.1). Thus, −i+avo (yi ) = −j + avo (yj ), so i ∼ = j mod a. Since 1 ≤ i, j ≤ m < a, this is a contradiction. By Lemma 4.7.1, dim(L) = dim(L(mo)) · dim(L(no)). Apply RiemannRoch to the right hand side terms: (6)

dim(L) ≥ (m − g + 1)(n − g + 1) = q +

√ q − g(g + 1).

Now consider the K-vector space L0 =

nX

o (σ −1 ui )a yi | yi ∈ L(no) .

i∈Im

Check that L0 ⊆ L(maoσ + no) and deg(maoσ + no) = q + 2g > 2g − 2. By Riemann-Roch, (7)

dim(L0 ) ≤ dim(maoσ + no) = deg(maoσ + no) − g + 1 = q + g + 1.

√ By (1b), q − g(g + 1) > g + 1. Thus, the right side of (6) is greater than the right side of (7). Therefore, dim(L0 ) < dim(L).

(8)

Finally, define an additive map σ ∗ from L into L0 by σ∗

X i∈Im

X ui yia = (σ −1 ui )a yi i∈Im

4.8 A Lower Bound

89

By (8), the kernel of σ ∗ is nontrivial. Hence, there exist yi ∈ L(no), i ∈ Im , not all zero, with X (9) (σ −1 ui )a yi = 0. i∈Im

In particular, u = i∈Im ui yia is a nonzero element of L(ro). If p is a prime divisor of F/K and p 6= o, then ϕp (yi ) 6= ∞ and ϕp (ui ) 6= ∞. If in addition ϕσp = ϕqp , then (9) implies X X ϕp (u) = ϕp (ui )ϕp (yi )a = ϕp (σ −1 ui )q ϕp (yi )a P

i∈Im

= ϕp

i∈Im

X

(σ −1 ui )a yi

a

=0

i∈Im

Hence, p occurs in the divisor of zeros of u. In other words, X p ≤ div0 (u). p6=o pσ =pq

Thus, N (σ) − 1 ≤ deg(div0 (u)) = deg(div∞ (u)) ≤ r and the proof of (4) is complete.

4.8 A Lower Bound In the notation of Section 4.7, we establish a lower bound inequality of the √ form N (σ) − (q + 1) > c0 q where c0 is an explicit constant depending only on F . This will complete the construction of Lemma 4.6.2 toward the proof of the Riemann hypothesis for F/K. Lemma 4.8.1: Let F be a function field of one variable over a field K with q elements. Let F 0 be a finite Galois extension of F with a Galois group G such that K algebraically closed in F 0 and let σ ∈ G. Then N (σ) (F ) = Pis also (στ 0 −1 ) (F 0 ). [F : F ] τ ∈G N Proof: Let p0 be a prime divisor of F 0 /K and let p be its restriction to F . q Then ϕσp = ϕqp if and only if there exists τ ∈ G such that ϕστ p0 = ϕp0 . The 0 number of such τ is the ramification index ep of p over p (Section 2.3). Put fp0 /p = [F¯p0 0 : F¯p ] and denote the number of prime divisors of F 0 lying over p by gp0 /p . Then: X X X X X N (στ ) (F 0 ) = deg(p0 ) = ep0 /p deg(p0 ) q τ ∈G ϕστ 0 =ϕ 0

τ ∈G

p

=

X

p

q 0 ϕσ p =ϕp p |p

ep0 /p fp0 /p gp0 /p deg(p)

q ϕσ p =ϕp

= [F 0 : F ]

X q ϕσ p =ϕp

deg(p) = [F 0 : F ]N (σ) (F ).

90

Chapter 4. The Riemann Hypothesis for Function Fields

Let F be a function field of one variable over a field K of q elements and let σ be an automorphism of F over K of finite order. Denote the fixed field of σ in F by E. Then F is a finite Galois extension of E. As a finite field, K is perfect. Hence, there exists x ∈ E, transcendental over K, such that E is a finite separable extension of K(x). Let Fˆ be the Galois closure of F/K(x) ˆ be the algebraic closure of K in Fˆ . Then Fˆ , as well as F K, ˆ are and let K ˆ and σ extends to an automorphism of function fields of one variable over K ˆ ˆ (if necessary), assume Fˆ over K(x). After an additional finite extension of K ˆ and therefore for F K/ ˆ K ˆ that Condition (1) of Section 4.7 holds for Fˆ /K, (use Proposition 3.4.2). Starting with a given F we have extended the field of constants so as to assume these conditions: (1a) F/K has a separating transcendence element x; the field F has a finite extension Fˆ , which is Galois over K(x), and K is algebraically closed in Fˆ ; ˆ and (1b) q is a square larger than (ˆ g + 1)4 where gˆ is the genus of Fˆ /K; ˆ has a prime divisor of degree 1. (1c) Fˆ /K Lemma 4.8.2: Under these conditions N (σ) (F ) − (q + 1) ≥ −

(2)

n−m √ (2ˆ g + 1) q, m

where m = [Fˆ : F ] and n = [Fˆ : K(x)]. Proof: Let H = Gal(Fˆ /F ) and G = Gal(Fˆ /K(x)). From Lemma 4.8.1 (3)

N (σ) (F ) =

1 X (στ ) ˆ 1 X (θ) ˆ N (F ) and q + 1 = N (K(x)) = N (F ). m n τ ∈H

θ∈G

Apply inequality (3) of Section 4.7: X

N (θ) (Fˆ ) =

X

N (στ ) (Fˆ ) +

τ ∈H

θ∈G

≤

X

=

N (θ) (Fˆ )

θ∈G r σH

N (στ ) (Fˆ ) +

τ ∈H

X

X X

√ (q + 1 + (2ˆ g + 1) q)

θ∈G r σH

N

(στ )

√ (Fˆ ) + (n − m)(q + 1 + (2ˆ g + 1) q).

τ ∈H

From the second half of (3), X τ ∈H

√ N (στ ) (Fˆ ) ≥ n(q + 1) − (n − m)(q + 1 + (2ˆ g + 1) q) √ = m(q + 1) − (n − m)(2ˆ g + 1) q.

Exercises

91

Thus, the first half of (3) implies that N (σ) (F ) ≥ (q + 1) −

n−m √ (2ˆ g + 1) q. m

Note that the condition (1), as well as the numbers gˆ, m, n, are independent of extension of the field of constants. We may therefore combine Lemma 4.8.2 with the results of Section 4.7 to conclude: Proposition 4.8.3: Let F be a function field of one variable over a finite field K and let σ be an automorphism of F over K of finite order. Then K has a finite extension K 0 with q 0 elements and there exists a positive constant c such that for every positive integer r we have |N (σ) (Fr0 ) − ((q 0 )r − 1)| ≤ c(q 0 )r/2 , where Kr0 is the unique extension of K 0 of degree r, Fr0 = F 0 Kr0 , and σ extends to an automorphism, also denoted by σ, of Fr0 over Kr0 . In particular, Proposition 4.8.3 is valid in the case σ = 1. By Lemma 4.6.2, the Riemann hypothesis is true for the function field F 0 /K 0 . It follows from Lemma 4.6.1 that it is also true for F/K.

Exercises 1.

For real valued functions f, g write f (x) = O(g(x)) as x → a

if there exists a positive constant c such that |f (x)| ≤ c|g(x)| for all values of x in a neighborhood of a. Let F be a function field of one variable over a field K of q elements. Denote the set of prime divisors of F/K of degree r by Pr (F/K). Follow these instructions to prove that |Pr (F/K)| =

1 r q + O(q r/2 ) r

(a) Use Theorem 4.5.2 to prove that |P1 (Fr /Kr )| = q r + O(q r/2 ). (b) Observe that if P ∈ P1 (Fr /Kr ) and if p is the prime divisor of F/K that lies below P, then deg(p)|r. (c) Deduce from Lemma 4.3.1 that if d|r, then over each p ∈ Pd (F/K) there lie exactly d elements of P1 (Fr /Kr ). Thus, X |P1 (Fr /Kr )| = d|Pd (F/K)|. d|r

(d) Use the estimates |Pd (F/K)| = O(q d ) and d ≤ 2r for proper divisors d of r r r and the inequality q + q 2 + · · · + q 2 ≤ 2q 2 to complete the proof.

92

Chapter 4. The Riemann Hypothesis for Function Fields

2. Let F be a function field of one variable over a field K of q elements and genus g. Let σ be an automorphism of F over K of finite order and let N (σ) √ be as in Section 4.7. Prove that |N (σ) − (q + 1)| ≤ 2g q. Hint: Extend σ to ˜ K ˜ by σ ˜ Then let E be an automorphism σ ˜ of F K/ ˜ x = xq for each x ∈ K. ˜ the fixed field of σ ˜ in F K. Show that E is a function field of one variable ˜ = F K. ˜ over K and that E K 3. This exercise establishes basic facts about infinite products which lie behind the multiplicative presentation of the zeta functions. Consider a se∞ quence {zi }Q i=1 of nonzero complex numbers. If the sequence of the partial n to a nonzero complex number z, we say that products i=1 zi converges Q∞ z converges and the infinite product i=1 i Q∞that z is its value. We say that Qn (1 + z ) absolutely converges if i i=1 i=1 (1 + |zi |) converges. The logarithm function makes a connection between the theory of infinite products and the theory of infinite series: Proposition ([Knopp, p. 434]): Q∞ Let ai 6= −1, i = 1, 2, 3, . . . be complex numbers. Then the product i=1 (1 + ai ) converges if and only if, the series P∞ i=1 log(1+ai ) whose terms are the Q∞principal values of log(1+ai ) converges. If l is the sum of this series, then i=1 (1 + ai ) = el . (a) Prove that if |z| ≤ 12 , then |z| ≤ 2| log(1 + z)| ≤ 2|z|. (b) Suppose 0 < |ai | ≤ 12 for i = 1, 2, 3, . . . and consider the following series: ∞ ∞ ∞ X X X | log(1 + ai )|, |ai |, log(1 + |ai |). i=1

i=1

i=1

Use (a) to prove that the convergence of each these series implies the convergence of the two others. Q∞ − |ai |)−1 con(c) Suppose 0 < |ai | ≤ 12 for i = 1, 2, 3, . . . and i=1 Q(1 ∞ verges. Then for each permutation π of N, the product i=1 (1 − aπ(i) )−1 converges and its value is independent of π. 4. Let F be a function field of one variable over a field K with q elements. Consider σ ∈ Aut(F/K). As in Section 4.7, define N (σ) to be the sum of all deg(p), where p ranges over all prime divisors of F/K for which ϕσp = ϕqp . ˜ by the formula σ ˜ Extend σ to an automorphism of F K ˜ x = xq for each x ∈ K. Use Lemma 4.3.1 for r a multiple of deg(p) to prove that N (σ) is also the ˜ K ˜ for which ϕσ˜ = ϕq . number of prime divisors P of F K/ P P 5. Let F be a function field of one variable over a field with q elements and of genus g. Let σ be an element of Aut(F/K) of order n. Define N (σ) as in Exercise 4. Prove that √ |N (σ) − (q + 1)| ≤ 2g q. ˜ by F 0 . Prove Hint: Let σ ˜ be as in Exercise 4. Denote the fixed field of σ ˜ in F K 0 ˜ in F Kr for each multiple r of n. Deduce that F is also the fixed field of σ

Notes

93

˜ = F 0 K. ˜ Use that F 0 is a function field of one variable over K such that F K Exercise 4 to prove that N (σ) is the number of prime divisors of F 0 /K of degree 1. Then apply Theorem 4.5.2. 6. Let q be a power of an odd prime. Prove that ax2 + by 2 = c has a solution (x, y) ∈ F2q if a, b, c ∈ F× q . Let (x0 , y0 ) be one of these solutions. Use the substitution x = x0 + tz, y = y0 + t and solve for t in terms of z to show that the function field of aX 2 + bY 2 − c = 0 is isomorphic to Fq (z). Conclude that the function field has exactly q + 1 degree 1 places. Hint: Assume a = 1 and observe that c − by 2 takes on (q + 1)/2 values in Fq . On the other hand there are q+1 2 squares in Fq . 7. Let q be a prime power ≡ 1 mod 3 and let α be a generator of F× q . Follow these instructions to prove that αX 3 + α2 Y 3 = 1 has a solution in Fq : Let Ai = {αi x3 | x ∈ F× q }, i = 0, 1, 2 and note that αA0 = A1 , αA1 = A2 and αA2 = A0 . Let Ai + Aj = {x + y | x ∈ Ai ∧ y ∈ Aj }. Now assume that A1 + A2 ⊆ A1 ∪ A2 to obtain a contradiction. For this use the notation A1 + A2 = m1 A1 ∪ m2 A2 to indicate that for each ai ∈ Ai there are mi pairs (x, y) ∈ A1 × A2 such that x + y = ai (independent of ai ∈ Ai ), i = 1, 2. Multiply this “equality” by α (resp. α2 ) to compute A2 + A0 (resp. A0 + A1 ). For a0 ∈ A0 , use this to compute the number of triples (x, y, z) ∈ A0 ×A1 ×A2 such that x + y + z = a0 in two ways: first compute (A0 + A1 ) + A2 and then A0 + (A1 + A2 ). The resulting expressions for m1 and m2 will lead to the contradiction m1 = m2 = 0. The attractiveness of the Riemann hypothesis for curves over finite fields has resulted in extensive lists of problems in a number of books treating the combinatorics of finite fields. For the sake of completeness we reference two such problem sources: [Ireland-Rosen, Chap. 8, pp. 105–107; Chap. 11, pp. 169–171] [Lidl-Niederreiter, see notes, Chap. 6, pp. 339–346]. In those chapters that use the Riemann hypothesis, our material and problems will tend to concentrate on the connections between the Riemann hypothesis and arithmetic properties of fields that are special to this book (e.g. an explicit form of Hilbert’s irreducibility theorem for global fields that follows from the Riemann hypothesis - Theorem 13.3.4).

Notes An extensive survey of the literature giving estimates on the number of points on an affine variety V appears in [Lidl-Niederreiter, pp. 317–339]. Although considerable literature on the Riemann hypothesis for curves over finite fields (Theorem 4.5.1) existed long before the two proofs given by Weil [Weil2], subsequent concerns included two sophisticated — and interrelated — developments. Both of Weil’s proofs employed elements of the theory of algebraic geometry outside the domain of algebraic curves. Indeed, Weil’s far reaching generalization of Theorem 4.5.1 was suggested to him

94

Chapter 4. The Riemann Hypothesis for Function Fields

by the latter proof. This generalization, now called the Riemann hypothesis for nonsingular projective varieties over finite fields (proved by Deligne in [Deligne]). Since, however, Weil’s theorem had so many applications to apparently elementary results about finite fields, many practitioners were anxious for a more accessible proof. Stepanov [Stepanov1] was the first to make serious progress on an elementary proof (e.g. not applying the theory of algebraic surfaces) of Weil’s result. He introduced elements of diophantine approximation to the problem in the case of hyperelliptic curves, in a style suggested by the Thue-SiegelRoth theorem. To do this he constructed an auxiliary nonvanishing function on the hyperelliptic curve with a “lot” of prescribed zeros of “high” multiplicity. Eventually [Stepanov2] realized Theorem 4.5.1 for all hyperelliptic curves over finite fields. Continuing with the purely diophantine approximation approach, Stepanov [Stepanov3], for prime fields (and some extra conditions on the equations) and W. M. Schmidt [W.M.Schmidt1 and W.M.Schmidt2] were able to prove Theorem 4.5.1 for all curves. Indeed, Schmidt’s method were even applicable to prove results like those of Deligne even for some complete intersections. Our proof, however, of Theorem 4.5.1 follows [Bombieri1]. In this proof the Riemann hypotheses replaces diophantine approximation to give the construction of the auxiliary functions that appear in Stepanov’s proof (the function u at the end of Section 4.7). Voloch [Voloch] has an elementary proof of the Riemann hypothesis for function fields that sometimes gives a better bound than Weil’s estimate (see also [St¨ohr-Voloch]).

Chapter 5. Plane Curves The estimate on the number of prime divisors of degree 1 of a function field F over Fq (Theorem 4.5.2) leads in this chapter to an estimate on the number N of K-rational zeros of an absolutely irreducible polynomial f ∈ Fq [X, Y ]. √ We prove (Theorem 5.4.1) that |N + (q + 1)| ≤ (d − 1)(d − 2) q + d, where d = deg(f ).

5.1 Affine and Projective Plane Curves Let K be a field and let Ω be an algebraically closed extension of K of infinite transcendence degree over K. We denote by A2 the affine plane: all pairs (x, y) ∈ Ω2 . We denote by P2 the projective plane: all nonzero triples x = (x0 , x1 , x2 ) ∈ Ω3 modulo the equivalence relation x ∼ x0 if and only if x0 = cx for some c ∈ Ω. We denote the equivalence class of (x0 , x1 , x2 ) by (x0 :x1 :x2 ). Embed the affine plane A2 in P2 by the map (x, y) → (1:x:y). With this understood, the points of A2 are then referred to as the finite points on P2 , whereas the points of the form (0:x1 :x2 ) are the points at infinity on P2 . An affine plane curve defined over K is a set (1)

Γ = {(x, y) ∈ A2 | f (x, y) = 0},

where f ∈ K[X, Y ] is a nonconstant absolutely irreducible polynomial. Write it in the form (2)

f (X, Y ) = fd (X, Y ) + fd−1 (X, Y ) + · · · + f0 (X, Y )

where fk (X, Y ) is a homogeneous polynomial of degree k, for k = 0, . . . , d, and fd (X, Y ) 6= 0. Then d is the degree of Γ. Attach to f the homogeneous polynomial f ∗ (X0 , X1 , X2 ) of degree d: (3) f ∗ (X0 , X1 , X2 ) = fd (X1 , X2 ) + X0 fd−1 (X1 , X2 ) + · · · + X0d f0 (X1 , X2 ) and let Γ∗ = {(x0 :x1 :x2 ) ∈ P2 | f ∗ (x0 , x1 , x2 ) = 0} Then Γ∗ is the projective plane curve corresponding to Γ. It is also called the projective completion of Γ. We have Γ = Γ∗ ∩ A2 , and Γ∗ r Γ is a finite set corresponding to the points of fd (X1 , X2 ) = 0 in P1 . The infinite points of Γ∗ are sometimes referred to as the points at infinity on Γ. If f (X, Y ) = a+bX+cY with a, b, c ∈ K and b 6= 0 or c 6= 0 (i.e. deg(f ) = 1), then Γ is a line. The points on the corresponding projective line satisfy aX0 + bX1 + cX2 = 0. The line at infinity is given by X0 = 0.

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Call a point (x, y) of an affine curve Γ generic if trans.degK K(x, y) = 1. Because f is absolutely irreducible, the field F = K(x, y) is a regular extension of K. Thus, F is a function field of one variable which we call the function field of Γ over K. Up to K-isomorphism it is independent of the choice of the generic point. The map (X, Y ) 7→ (x, y) extends to a K-epimorphism of rings K[X, Y ] → K[x, y] with f (X, Y )K[X, Y ] as the kernel (Gauss’ lemma). Thus, a polynomial g ∈ K[X, Y ] vanishes on Γ if and only if g(x, y) = 0, or, equivalently, g is a multiple of f . Define the genus of Γ (and of Γ∗ ) to be the genus of F . The coordinate ring of Γ over K is R = K[x, y]. There is a bijective correspondence between the set of all maximal ideals m of R with quotient field R/m = K and the set Γ(K) of all K-rational points of Γ. If (a, b) ∈ Γ(K), then the corresponding maximal ideal of R is m = {g(x, y) ∈ R | g(a, b) = 0}. If p is a prime ideal of R and p ⊆ m, then the transcendence degree of the quotient field of R/p is either 0 or 1. In the latter case p = 0, and in the former case R/p is already a field so that p = m. The local ring of Γ at a = (a, b) over K is the local ring of R at m: OΓ,(a,b),K = Rm =

o n g(x, y) g(x, y), h(x, y) ∈ R and h(a, b) 6= 0 . h(x, y)

The unique nonzero prime ideal of Rm is generated by the elements of m. As a local ring of a Noetherian domain, Rm is itself a Noetherian domain. Similarly, a point (x0 :x1 :x2 ) of Γ∗ with x0 6= 0 is said to be generic over K, if xx10 , xx20 is a generic point of Γ. Define the local ring of a point a = (a0 :a1 :a2 ) of Γ∗ over K as OΓ∗ ,a,K =

n g(x , x , x ) g, h are homogeneous polynomials of o 0 1 2 h(x0 , x1 , x2 ) the same degree and h(a0 , a1 , a2 ) 6= 0

If a0 6= 0, then this ring coincides with the local ring of the corresponding point aa10 , aa20 of Γ. If a1 6= 0, then regard a as a point on the affine curve Γ1 , defined by the equation f ∗ (X0 , 1, X2 ) = 0. Then the projective completion of Γ1 is also Γ∗ , but the line X1 = 0 is taken as “the line at infinity”. This shows that the local ring of a projective plane curve at each point is equal to the local ring of some affine curve at a point. It is therefore a Noetherian domain. In any case, call a K-rational point of Γ (or of Γ∗ ) K-normal if its local ring over K is integrally closed. Two affine plane curves Γ1 and Γ2 are K- isomorphic if their coordinate rings are K-isomorphic. In this case generators of the coordinate ring of Γ2 (resp. Γ1 ) can be expressed as polynomials in the generators of the coordinate ring of Γ1 (resp. Γ2 ), and the composition of these two polynomial maps is the identity when applied to generators of the coordinate ring of Γ1 (resp. Γ2 ).

5.2 Points and Prime Divisors

97

Two projective plane curves Γ∗1 and Γ∗2 are K-isomorphic if (4a) for each x ∈ Γ∗1 there exist homogeneous polynomials g0 , g1 , g2 in K[X0 , X1 , X2 ] of the same degree such that yi = gi (x) 6= 0 for at least one i, 0 ≤ i ≤ 2, y ∈ Γ∗2 , and there exist homogeneous polynomials h0 , h1 , h2 in K[X0 , X1 , X2 ] of the same degree such that hi (y) = xi , i = 0, 1, 2; and (4b) the same condition with the roles of Γ∗1 and Γ∗2 exchanged. For example, if g0 , g1 , g2 are linear polynomials with a nonsingular coefficient matrix, then (g0 , g1 , g2 ) is called nonsingular homogeneous linear transformation. The function fields of two isomorphic plane curves are K-isomorphic. So are the local rings of corresponding points. It follows that the genera of isomorphic plane curves are the same; and if p1 and p2 are corresponding points of the curves, then p1 is K-normal if and only if p2 is K-normal. In particular, both curves have the same number of nonnormal points. If Γ is a plane curve defined over a field K and L is an algebraic extension of K, then Γ is also defined over L. The function field FL of Γ over L is the extension of F by the field of constants L. By Proposition 3.4.2, the genus of Γ remains unchanged if L is separable over K.

5.2 Points and Prime Divisors Let Γ be an affine plane curve of degree d defined by an absolutely irreducible equation f (X, Y ) = 0 over a field K. We establish a bijective correspondence between the K-rational points of f and the prime divisors of the function field of f of degree 1 (Lemma 5.2.2). We prove that a K-rational of Γ is normal if and only if it is simple (Lemma 5.2.3). Let (x, y) be a generic point of Γ over K. Denote the coordinate ring and the function field, respectively, of Γ over K by R = K[x, y] and F = K(x, y). Consider a K-rational point (a, b) of Γ. Then the map (x, y) 7→ (a, b) uniquely extends to a K-homomorphism ϕ of the local ring O(a,b) into K. The ˜ homomorphism ϕ extends further (not necessarily uniquely) to a K-valued 0 place ϕ of F . Call (a, b) the center of the corresponding prime divisor p of F/K. Then p lies over the unique prime divisor p0 of K(x)/K determined by the map x 7→ a. Since [F : K(x)] ≤ d, there exist at most d prime divisors p of F/K with the point (a, b) as a center on Γ. Since each point of Γ∗ is a finite point of some affine representative of Γ∗ (Section 5.1) this holds for each K-rational point of the projective completion Γ∗ of Γ. If a point p of Γ∗ (K) is K-normal, then its local ring Op is a Noetherian integrally closed domain. By Section 5.1 , each nonzero prime ideal of Op is maximal. Hence, by Proposition 2.4.5, Op is a discrete valuation domain. Therefore, there exists a unique prime divisor p of F/K with p as a center on Γ. The degree of p is 1. Conversely, consider a prime divisor p of F/K of degree 1. Suppose that ϕp is finite on R. Then (a, b) = ϕp (x, y) is a K-rational point of Γ and it is

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the center of p on Γ. The following result shows that the number of prime divisors p of F/K which are not finite on R does not exceed d. Proposition 5.2.1 (Noether’s Normalization Theorem): Let K[x1 , . . . , xn ] be a finitely generated integral domain over a field K with a quotient field F . If the transcendence degree of F over K is r, then there exist elements t1 , . . . , tr in K[x] such that K[x] = K[t1 , . . . , tr , x1 , . . . , xn−r ] (after rearranging the xi ’s) and K[x] is integral over K[t] [Lang4, p. 22]. If K is an infinite field, then t1 , . . . , tr can be chosen to be linear combinations of x1 , . . . , xr with coefficients in K [Zariski-Samuel1, p. 266]. Lemma 19.5.1 gives a constructive version of Noether’s normalization theorem. Return now to the plane curve Γ. Noether’s normalization theorem, in its linear form, allows us to replace x by a linear combination t of x and y such that R is integral over K[t] and [F : K(t)] ≤ d. If K is finite, replace K by a suitable finite extension to achieve the linear dependence of t on x and y. If a place ϕp of F is not finite on R, then it is infinite at t (Lemma 2.4.1). Hence, p lies over the infinite prime divisor p∞ of K(t)/K. There are at most d prime divisors of F/K that lie over p∞ . This proves our contention. Now we summarize. Lemma 5.2.2: Let Γ be an affine plane curve of degree d defined over a field K. Denote the coordinate ring and the function field, respectively, of Γ over K by R and F . Then: (a) For each K-normal point p ∈ Γ(K) there exists exactly one prime divisor p of F/K with center p on Γ; the degree of p is 1. (b) There are at most d prime divisors of F/K whose centers on Γ are a given point p. (c) If a prime divisor p of F/K is of degree 1 and if ϕp is finite on R, then its center is in Γ(K). (d) There are at most d prime divisors of F/K which are not finite on R. Finally, we point out that the K-normal points and simple points on ∂f (a, b) 6= 0 or Γ are the same. Here, a point (a, b) on Γ is simple if ∂X ∂f ∂Y (a, b) 6= 0. Lemma 5.2.3: A K-rational point (a, b) of Γ is normal if and only if it is ∂f (a, b) 6= 0, then F has a discrete normalized valuation simple. Moreover, if ∂X v with v(y − b) = 1. Proof: Suppose first that (a, b) is normal. Then, its local ring R = O(a,b) over K is a discrete valuation domain. Hence, R(x − a) ⊆ R(y − b) or R(y − b) ⊆ R(x − a). Suppose for example that R(x − a) ⊆ R(y − b). Then there exist g, h, q ∈ K[X, Y ] such that h(a, b) 6= 0 and h(X, Y )(X − a) = g(X, Y )(Y − b) + q(X, Y )f (X, Y ).

5.3 The Genus of a Plane Curve

Apply

∂ ∂X

99

to both sides and substitute (a, b) for (X, Y ): h(a, b) = q(a, b)

∂f (a, b). ∂X

∂f (a, b) 6= 0 and therefore (a, b) is simple. Hence, ∂X ∂f (a, b) 6= 0. We show that y − b generates Conversely, suppose that ∂X the maximal ideal m of R. Indeed,

f (X, Y ) =

∂f ∂f (a, b)(X − a) + (a, b)(Y − b) + higher terms, ∂X ∂Y

and 0 = f (x, y) =

∂f ∂f (a, b)(x − a) + (a, b)(y − b) + (x − a)u + (y − b)v, ∂X ∂Y

with u, v ∈ m. Hence, 0=

∂f (a, b) + u (x − a) + (a, b) + v (y − b). ∂X ∂Y

∂f

∂f (a, b) is a nonzero constant and u ∈ m, the coefficient of x − a But, since ∂X is a unit of R. Therefore, x − a ∈ R(y − b),Tso m = R(x − a, y − b) = R(y − b). ∞ Since R is Noetherian, the ideal a = n=1 mn is a finitely generated Rmodule. Since ma = a, Nakayama’s Lemma [Lang4, p. 195, Prop. 1] implies that a = 0 (alternatively, use Krull’s intersection theorem [Eisenbud, p. 152]). Hence, each z ∈ R has a unique representation z = w(y − b)n , with w a unit of R and n ≥ 0. Thus, R is a discrete valuation domain. Therefore, R is integrally closed (Exercise 3 of Chapter 2). Let v be the normalized discrete valuation of F corresponding to R. Since y − b generates m, we have v(y − b) = 1.

For a projective curve Γ∗ defined by f ∗ (X0 , X1 , X2 ) = 0 as in (3) of ∂f (a0 , a1 , a2 ) 6= 0, for some i, Section 5.1, the point (a0 :a1 :a2 ) is simple if ∂X i i = 0, 1 or 2. Otherwise, (a0 :a1 :a2 ) is singular.

5.3 The Genus of a Plane Curve Here we bound the genus and the number of nonnormal points of a plane curve by a function of its degree. We first prove a finiteness result for the coordinate ring of a plane curve. Lemma 5.3.1: Let R = K[x, y] be an integral domain with quotient field F of transcendence degree 1 over K. Let S be the integral closure of R in F . Then S/R is a finitely generated K-vector space. Proof: It is well known that S is a finitely generated R-module ([Lang4, p. 120] or [Zariski-Samuel1, p. 267]). Hence, there exists a nonzero element

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Chapter 5. Plane Curves

z ∈ R such that zS ⊆ R. Thus, S/R ∼ = zS/zR ⊆ R/zR. It suffices to prove that dimK R/zR < ∞. By Noether’s normalization theorem (Proposition 5.2.1), we may assume without loss that R is integral over K[x]. The element z satisfies an equation of the form z m + a(x)z m−1 + · · · + g(x) = 0, with 0 6= g(x) ∈ K[x] of degree, say, k. Since g(x) belongs to zR, every power of x is a linear combination modulo zR of 1, x, . . . , xk−1 with coefficients in K. If d is the degree of a monic equation for y over K[x], then each element of R can be written as h(x, y), with h ∈ K[X, Y ] is of degree at most d − 1 in Y . Our lemma follows. In the notation of the proof of Lemma 5.3.1, S is the integral closure in F of the Dedekind domain K[x]. Hence, S is a Dedekind domain and therefore every nonzero prime ideal of S is maximal. By Section 5.2, every nonzero prime ideal of R is maximal. Consider a nonzero prime ideal p of R. Let Rp be the local ring of R at p. Then the integral closure of Rp in F is Sp = sb | s ∈ S and b ∈ R r p . Lemma 5.3.2: P (a) The quotient S/R is isomorphic to the direct sum Sp /Rp , where p runs over the nonzero prime ideals of R. (b) Let p1 , . . . , pk be the prime ideals of S that lie over a prime ideal p of R. Then dimK Sp /Rp ≥ k − 1. Proof of (a): Denote the conductor of S over R by c = {c ∈ R | cS ⊆ R}. It is a nonzero ideal of both R and S. The local ring Rp is integrally closed, and therefore equal to Sp , if and only if c 6⊆ p [Zariski-Samuel1, p. 269]. Let Qk Ql(i) e c = i=1 j=1 piji,j be a factorization of c into the product of prime ideal powers of S where pi1 , . . . , pi,l(i) all lie over the same prime ideal pi of R, i = 1, . . . , k. Then Spij ⊆ Rpi , i = 1, . . . , k. If p is a nonzero prime ideal of R not in the set {p1 , . . . , pk }, then Rp = Sp . Hence, with Ri T = Rpi and Si = Spi , i = 1, . . . , k, Lemma 2.4.4 gives S ∩ R1 ∩ · · · ∩ Rk = Rp = R, where p runs over the nonzero prime ideals of R. Therefore, the map Pk s+R 7→ (s+R1 , . . . , s+Rk ) for s ∈ S, is an injective homomorphism into i=1 Si /Ri . The proof of the lemma is complete if we show that the map is surjective. / pi . Let s1 , . . . , sk be elements of S and let a1 , . . . , ak ∈ R such that ai ∈ Ql(i) ei,j The ideals qi = j=1 pij of S are pairwise relatively prime, i = 1, . . . , k. Hence, by the weak approximation theorem (Proposition 2.1.1), there exist s ∈ S, qi ∈ qi , and a0i ∈ R r pi such that s−

qi si = 0 ai ai

i = 1, . . . , k.

Q For each i choose bi ∈ (R r pi ) ∩ j6=i qj . Then ci = bi qi ∈ c ⊆ R and s = asii + bicai 0 . Thus, s + Ri = asii + Ri for i = 1, . . . , k. i

Proof of (b): Now let p1 , . . . , pk be the prime ideals of S that lie over p. For i = 1, . . . , k − 1 use the weak approximation theorem to find si ∈ S

5.3 The Genus of a Plane Curve

101

such that si ≡ 1 mod pi and si ≡ 0 mod pj , for j = 1, . . . , k and j 6= i. We need only show that s1 , . . . , sk−1 are K-linearly independent modulo Rp . Pk−1 Indeed, if r = i=1 ai si ∈ Rp with ai ∈ K (and si ∈ pk ), then, r belongs to Pk−1 pRp . Since aj ≡ i=1 ai si ≡ 0 mod pk , this gives aj = 0, j = 1, . . . , k − 1. Consequently, dimK Sp /Rp ≥ k − 1. Lemma 5.3.3: Let Γ be an affine plane curve defined over a field K with a generic point x = (x1 , x2 ) and let a ∈ Γ(K). Consider a separable algebraic 0 ) be the local ring of Γ at a over extension L of K and let Oa,K (resp. Oa,K K (resp. its integral closure). Then (a) Oa,L ∩ K(x) = Oa,K ; and 0 0 (b) dimK Oa,K /Oa,K ≤ dimL Oa,L /Oa,L . Proof: Without loss assume that L/K is finite. Let w1 , . . . , wn be a linear basis for L/K. Since Γ is defined by an absolutely irreducible polynomial, K(x)/K is a regular extension of K. Hence, K(x) and L are linearly disjoint over K. Therefore, w1 , . . . , wn is also a linear basis for L(x)/K(x). Consider u ∈ Oa,L ∩ K(x). Then P f (x), g(x) ∈ P u = f (x)/g(x) with = gi (x)wi , with L[x] and g(a) 6= 0. Write f (x) = fP i (x)wi and g(x) P ugi (x)wi = fi (x)wi deduce that fi (x), gi (x) ∈ K[x], i = 1, . . . , n. From ugi (x) = fi (x), i = 1, . . . , n. Since there is an i with gi (a) 6= 0, we have u ∈ Oa,K . This proves (a). 0 To prove (b), consider u1 , . . . , um ∈ Oa,K linearly dependent over L modulo Oa,L . We show that they are also linearly dependent over K modulo that Oa,K . Indeed, there existPb1 , . . . , bm ∈ L not all zero and v ∈ Oa,L Psuch P n m = j=1 aij wj with aij ∈ K and let vj = i=1 aij ui . bi ui = v. Write bi P Then vj ∈ K(x) and vj wj = v. For each σ of L(x) into P K(x)-embedding the algebraic closure of K(x) we have vj wjσ = v σ . Since L(x)/K(x) is ) 6= 0 [Lang7, p. 286, Cor. 5.4]. Apply Cramer’s rule to separable, det(wjσP solve the system vj wjσ = v σ , all σ, and write vj as a linear combination σ ˆ of L/K. Since a is of the v ’s with coefficients in the Galois closure L K-rational, each v σ is in Oa,Lˆ . Thus, vj ∈ Oa,Lˆ ∩ K(x). Hence, by (a), vj ∈ Oa,K . Therefore, the ui ’s are linearly dependent over K modulo Oa,K . Proposition 5.3.4: Let K be an algebraically closed field and Γ∗ a projective plane curve of degree d and genus g defined over K. Then (1)

g=

X 1 (d − 1)(d − 2) − dimK Op0 /Op , 2

where p runs over the points of Γ∗ (K), Op denotes the local ring of Γ∗ at p and Op0 is the integral closure of Op in the function field of Γ∗ over K. Proof: The five parts of the proof draw information from the form of Γ∗ after a change of variables that puts d distinct points at infinity.

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Chapter 5. Plane Curves

Part A: A linear transformation. Choose a line L over K passing through no singular point of Γ∗ and is not tangent to Γ∗ . Then L cuts Γ∗ in d distinct simple points p1 , . . . , pd of Γ∗ [Seidenberg, p. 37]. By Lemma 5.2.3, they are normal. After a suitable linear homogeneous transformation [Seidenberg, Chapter 5] we may assume that L is the line at infinity and that the infinite points on the X-axis and the Y -axis, (0:1:0) and (0:0:1) do not belong to the set {p1 , . . . , pd }. Such a transformation does not change the genus or the degree of the curve. Moreover, corresponding points have the same local rings. View Γ∗ as the projective completion of an affine curve Γ defined by an equation f (X, Y ) = 0, as given by (1) of Section 5.1. Let (x, y) be a generic point of Γ over K, let R = K[x, y] be the coordinate ring of Γ, and let F be its function field. Part B: The divisor at ∞. By Part A, pi = (0:ai :1), with ai ∈ K × for i = 1, . . . , d, and a1 , . . . , ad distinct. In the notation of (2) of Section 5.1, obtain the factorization (2)

fd (X, Y ) = c

d Y

(X − ai Y ), with 0 6= c ∈ K.

i=1

Therefore, f (x, Y ) is an irreducible polynomial of degree d over K[x]. Thus, R is integral over K[x] and each element of R can be uniquely expressed as a polynomial h(x, y) with coefficients in K[x] such that degY (h) ≤ d − 1. Similarly R is integral over K[y]. Over the infinite prime divisor p∞ of K(x) there lie exactly d distinct prime divisors p1 , . . . , pd of F/K, with pi being the unique prime divisor with center pi . In particular, p∞ is unramified in F (Proposition 2.3.2) and we may therefore normalize vpi such that vpi(x) = −1. Since R is integral over K[x], this implies vpi (y) < 0, so vpi xy ≥ 0. by (2) of Section 5.1, Qd c i=1 xy − ai + fd−1 xy , 1 y −1 + · · · + f0 xy , 1 y −d = 0. Hence, xy has residue ai at pi , so vpi (y) = −1. Let o = p1 + · · · + pd . Then the pole divisors div∞ (x) and div∞ (y) of x and y in F are both o. Part C: The integral closure of R in F . Let S be the integral closure of R S∞ in F . By Lemma 3.1.1, S = n=1 Sn , where Sn = L(no). Let Rn = R ∩ Sn , n = 1, 2, . . . . By Lemma 5.3.1, dimK S/R < ∞. Hence, for n sufficiently large R + Sn = R + Sn+1 = R + Sn+2 = . . . . Therefore, S = R + Sn , so Sn /Rn ∼ = S/R. For n > 2g − 2, Lemma 3.2.2(d) implies that dimK Sn = nd − g + 1. It follows that (3) Part D: (4)

nd − g + 1 = dimK Rn + dimK S/R. For n > d, we have Rn = {h(x, y) ∈ R | deg(h) ≤ n and degY (h) ≤ d − 1}.

5.3 The Genus of a Plane Curve

103

Indeed, let k ≥ 0 be an integer and hk ∈ K[X, Y ] a homogeneous polynomial of degree k with degY (hk ) ≤ d − 1. If i + j = k, then div∞ (xi y j ) = ko (Part B). Hence, hk (x, y) ∈ Sk . / Sk−1 : Since K is algebraically closed, we may We prove that hk (x, y) ∈ factor hk (x, y) as hk (x, y) =

d−1 Y

(x − bj y) · axk−d+1

j=1

with bj ∈ K and a ∈ K × . Then vpi (hk (x, y)) =

d−1 X

vpi (x − bj y) − (k − d + 1).

j=1

Pd−1 If hk (x, y) ∈ Sk−1 , then for each i we have j=1 vpi (x − bj y) − (k − d + 1) + k − 1 ≥ 0. Hence, there exists 1 ≤ j ≤ d − 1 with vpi (x − bj y) ≥ 0. Therefore, bj = ai . But this implies that two of the ai ’s are equal, a contradiction. To complete Pm Part D, write each h(x, y) in K[x, y] with degY (h) ≤ d − 1 as h(x, y) = k=1 hk (x, y) with hk as above and hm (x, y) 6= 0. If m ≤ n, then hk (x, y) ∈ Sk ⊆ Sn , k = 0, . . . , m, and therefore h(x, y) ∈ Rn . If m > n, / Rn . then h(x, y) ∈ Sm r Sm−1 and therefore h(x, y) ∈ Sd−1 Part E: Computation of the genus. By Part D, the set j=0 {xi y j | i = 0, . . . , n − j} is a basis of Rn . Hence dimK Rn =

d−1 X

1 (n − j + 1) = nd + 1 − (d − 1)(d − 2). 2 j=0

Substitute this in (3) to conclude that 1 (d − 1)(d − 2) − dimK S/R. 2 L By Lemma 5.3.2(a), S/R = SP /RP , where P ranges over all nonzero prime ideals of R. Since K is algebraically closed, there is a bijective correspondence between the nonzero prime ideals P ofL R and the finite points p of Γ∗ and 0 we have RP = Op . Hence, S/R = p∈Γ∗ (K) Op /Op . Substituting this expression for S/R in (5), we get (1). (5)

g=

Corollary 5.3.5: Let Γ∗ be a projective plane curve of degree d defined over a perfect field K. Then genus(Γ∗ ) ≤ 12 (d − 1)(d − 2). Proof: The genus of Γ∗ does not change by going from K to its algebraic closure (Proposition 3.4.2). Now apply Lemma 5.3.3 and Proposition 5.3.4.

104

Chapter 5. Plane Curves

Corollary 5.3.6: Let Γ∗ be a projective plane curve of degree d defined over an algebraically closed field K. Then genus(Γ∗ ) = 12 (d − 1)(d − 2) if and only if Γ∗ is smooth; that is, all K-rational points of Γ∗ are simple. Proof: The condition “Γ∗ is smooth” is equivalent to Op = Op0 for all K rational points p of Γ∗ . Now apply Proposition 5.3.4.

5.4 Points on a Curve over a Finite Field Using the formula for the genus of an absolutely irreducible curve Γ over Fq (Proposition 5.3.4), we translate the estimate of the number of prime divisors of degree 1 of the function field of Γ to an estimate on |Γ(Fq )| which involves only q and the degree of Γ. Theorem 5.4.1: Let f ∈ Fq [X, Y ] be an absolutely irreducible polynomial of degree d. Denote the affine curve defined by the equation f (X, Y ) = 0 by Γ. Then √ √ q + 1 − (d − 1)(d − 2) q − d ≤ |Γ(Fq )| ≤ q + 1 + (d − 1)(d − 2) q. Proof: Put K = Fq . Denote the function field of Γ over K by F , and denote the number of prime divisors of degree 1 of F/K by N . By Theorem 4.5.2, √ √ (1) q + 1 − 2g q ≤ N ≤ q + 1 + 2g q , where g = genus(F/K). Let Γ∗ be the projective completion of Γ. For each p ∈ Γ∗ (K) let k(p) (resp. k1 (p)) be the number of prime divisors (resp. prime divisors of degree 1) of F/K with center at p. Then X X (2) N − |Γ∗ (K)| = (k1 (p) − 1) ≤ (k(p) − 1). p∈Γ∗ (K)

p∈Γ∗ (K)

Let (3)

δ=

X

0 dimK Op,K /Op,K and δ˜ =

p∈Γ∗ (K)

X

0 dimK˜ Op, ˜. ˜ /Op,K K

p∈Γ∗ (K)

0 /Op,K = 0 and Consider p ∈ Γ∗ (K). If p is K-normal, then dimK Op,K k1 (p) = k(p) = 1 (Lemma 5.2.2(a)). If p is not K-normal, then 0 dimK Op,K /Op,K ≥ 1. Hence, by (3), and by Lemma 5.3.3,

X

˜ (1 − k1 (p)) ≤ δ ≤ δ.

p∈Γ∗ (K)

We conclude from (2), (1), and Proposition 5.3.4 that X √ |Γ(K)| ≤ |Γ∗ (K)| = N + (1 − k1 (p)) ≤ q + 1 + 2g q + δ˜ p∈Γ∗ (K)

√ √ √ = q + 1 + (d − 1)(d − 2) q − 2δ˜ q + δ˜ ≤ q + 1 + (d − 1)(d − 2) q.

Exercises

105

Next we give a lower bound for |Γ(K)|. Note first that Γ∗ has at most d points at infinity [Seidenberg, p. 37]. Hence, |Γ(K)| ≥ |Γ∗ (K)|−d. Therefore, by (2), Lemma 5.3.2(b), (1), and Proposition 5.3.4 |Γ(K)| ≥ |Γ∗ (K)| − d ≥ N −

X

(k(p) − 1) − d

p∈Γ∗ (K)

≥N−

X

√ 0 dimK OK,p /OK,p − d ≥ q + 1 − 2g q − δ˜ − d

p∈Γ∗ (K)

√ √ = q + 1 − (d − 1)(d − 2) q + 2δ˜ q − δ˜ − d √ ≥ q + 1 − (d − 1)(d − 2) q − d. This completes the proof of the theorem.

Corollary 5.4.2: The curve Γ of Theorem 5.4.1 satisfies the following conditions: (a) For each m there exists q0 such that |Γ(Fq )| ≥ m for all q ≥ q0 . (b) If q > (d − 1)4 , then Γ(K) is not empty. √ Proof of (b): By Theorem 5.4.1, |Γ(Fq )| ≥ q + 1 − (d − 1)(d − 2) q − d = √ √ √ q q − (d − 1)(d − 2) − (d − 1) > q − (d − 1) > 0.

Exercises 1. Use Proposition 5.3.4 to prove that any projective plane curve of degree 2 is smooth. 2. Let Γ∗ be a projective plane curve of degree 3. Use Proposition 5.3.4 to show that either Γ∗ is smooth, in which case it has genus 1, or Γ∗ has exactly one singular point, in which case it has genus 0 and therefore its function field is rational (Example 3.2.4). 2 −2X 2 −X 3 = 3. Consider the affine plane curve Γ defined by the equation Y√ 0 over a field K of characteristic 6= 2 that does not contain 2. Note that (0, 0) is a singular point of Γ. Let (x, y) be a generic point of Γ over K. Show that the map (x, y) → (0, 0) does not extend to a K-rational (rather than ˜ K-rational) place of K(x, y) (i.e. the singular point of Γ is not the center of a K-rational place of the function field of Γ). Hint: Consider the element xy of K(x, y).

4. Prove directly that the function field of the curve Y 2 − 2X 2 − X 3 = 0 is rational. 5. Consider the projective plane curve Γ∗ defined for d ≥ 2 by X0 X1d−1 − X0 X2d−1 − X2d = 0 over an algebraically closed field K. (a) Use Lemma 5.2.3 to show that the only nonnormal point of Γ∗ is (1:0:0).

106

Chapter 5. Plane Curves

(b) Let (x, y) be a generic point of the affine part Γ of Γ∗ defined by X d−1 − Y d−1 − Y d = 0 and let z = xy . Use Lemma 2.4.4 to conclude that K[y, z] is the integral closure of K[x, y] in K(x, y). Conclude that the genus of Γ is 0. (y,z) | f ∈ K[Y, Z], g ∈ K[X, Y ], g(0, 0) 6= 0} is the (c) Prove that S0 = { fg(x,y) integral closure of the local ring R0 of Γ at (0, 0). (d) Use Proposition 5.3.4 to conclude that the set {xi z j | j = 1, . . . , d−2; i = 0, . . . , j − 1} gives a basis for S0 /R0 over K. 6. Count the number of points on the projective plane curve X03 +X13 +X23 = 0 over the finite field Fq , where q is a prime power such that gcd(q − 1, 3) = 1. Hint: The map x 7→ x3 from F× q into itself is bijective. (X) 7. Let f ∈ Fq [X] be a polynomial of degree d such that g(X, Y ) = f (YY)−f −X is absolutely irreducible. Suppose that either q is not a power of 2 or d ≥ 3. Use Theorem 5.4.1 to prove that if q > (d − 1)4 , then there exist distinct x, y ∈ Fq such that f (y) = f (x). Hint: Observe that g(X, X), the derivative of f (X), has at most d − 1 zeros.

Notes The proof of Proposition 5.3.4 is an elaboration on [Samuel, p. 52]. Denote the maximum number of Fq -points on a curve of genus g which is defined over Fq by Nq (g). For fixed q, put A(q) = lim sup g1 Nq (g). Weil’s g→∞ √ estimate Nq (g) ≤ q + 1 + 2g q (Theorem 4.5.2 and the proof of Theorem 1 5.4.1) implies that A(q) ≤ 2q 2 . [Serre6] improves Weil’s estimate (via interpretation of the Frobenius as an endomorphism on Jacobians) to give the √ bound Nq (g) ≤ q + 1 + g[2 q] (where [x] is the greatest integer not exceeding √ adut] obtains the much improved x). Thus, A(q) ≤ [2 q]. But [Drinfeld-Vlˆ √ estimate A(q) ≤ q − 1. When q is a square [Ihara] and [Tsfasman-VlˆadutZink] have shown this bound to be exact. As for a lower estimate, [Serre6] 8 ). The proves the existence of c > 0 such that A(q) ≥ c log(q) (e.g. A(2) ≥ 39 exact lower and upper bound for A(q) for general q have yet to be found. The case q = 2 has application to coding theory as first noted by [Goppa] (or p. 530 of [Lidl-Niederreiter] for a survey of work in this direction).

Chapter 6. The Chebotarev Density Theorem The major connection between the theory of finite fields and the arithmetic of number fields and function fields is the Chebotarev density theorem. Explicit decision procedures and transfer principles of Chapters 20 and 31 depend on the theorem or some analogs. In the function field case our proof, using the Riemann hypothesis for curves, is complete and elementary. In particular, we make no use of the theory of analytic functions. The number field case, however, uses an asymptotic formula for the number of ideals in an ideal class, and only simple properties of analytic functions. In particular, we do not use Artin’s reciprocity law (or any equivalent formulation of class field theory). This proof is close to Chebotarev’s original field crossing argument, which gave a proof of a piece of Artin’s reciprocity law for cyclotomic extensions.

6.1 Decomposition Groups Let R be an integrally closed domain with quotient field K. Consider a finite Galois extension L of K with Galois group G, and denote the integral closure of R in L by S. Suppose p is a prime ideal of R. By Chevalley’s theorem (Proposition 2.3.1), there exists a prime ideal P of S lying over p (i.e. p = R ∩ P). Denote the quotient fields of R/p and S/P, respectively, ¯ and by L. ¯ The set of all σ ∈ G satisfying σ(P) = P is a group DP by K called the decomposition group of P over K. Its fixed field in L is the decomposition field of P over K. For x ∈ S denote the equivalence class of x modulo P by x ¯. Each ¯ over K ¯ satisfying σ ¯ of L ¯x ¯ = σx σ ∈ DP induces a unique automorphism σ ¯ K). ¯ for each x ∈ S. The map σ 7→ σ ¯ is a homomorphism of DP into Aut(L/ Its kernel is the inertia group IP of P over K, IP = {σ ∈ G | σx ∈ x + P for every x ∈ S}. The fixed field of IP in L is the inertia field of P over K. If σ ∈ G, then σS = S and σP is another prime ideal of S that lies over p. In this case DσP = σ ·DP ·σ −1 and IσP = σ ·IP ·σ −1 . Conversely, any two prime ideals of S lying over the same prime ideal of R are conjugate over K [Lang7, p. 340]. In the proof of Lemma 6.1.1 we use the expression “to localize R and S at p”. This means that we replace R and Rp , p by pRp , S by Sp = { as | s ∈ S and a ∈ R r p}, and P by PSp . The local ring Rp is integrally ¯ = Rp /pRp . In addition, Sp is the closed, pRp is its maximal ideal, and K integral closure of Rp in L [Lang7, p. 338, Prop. 1.8 and 1.9], and PSp is a prime ideal that lies over pRp . Thus, Sp /PSp is a domain which is integral ¯ is a field [Lang7, p. 339, Prop. 1.11] ¯ Hence, Sp /PSp = L over the field K. and PSp is a maximal ideal. Moreover, DPSp = DP and IPSp = IP .

108

Chapter 6. The Chebotarev Density Theorem

Lemma 6.1.1: ¯ K ¯ is normal, and the map σ 7→ σ (a) The field extension L/ ¯ from DP into ¯ K) ¯ is surjective. Aut(L/ ¯ K ¯ is separable and [L ¯ : K] ¯ = [L : K]. Then L/ ¯ K ¯ is Galois, (b) Suppose L/ ¯ is an isomorphism of Gal(L/K) DP = Gal(L/K), and the map σ 7→ σ ¯ K). ¯ onto Gal(L/ Proof of (a): Denote the decomposition field of P by L0 , let S0 = S ∩ Lo , and let P0 = P ∩ L0 . We prove S0 /P0 = R/p. Suppose x ∈ S0 . We need only to find a ∈ R such that x ≡ a mod P0 . For each σ ∈ G r DP we have σ −1 P 6= P. If σ −1 P ∩ L0 = P0 , then there exists τ ∈ Gal(L/L0 ) = DP such that τ σ −1 P = P. Therefore, σ ∈ DP , a contradiction. Thus, σ −1 P ∩ L0 6= P0 . Localize at p, to assume that p is a maximal ideal of R. Then P0 and σ −1 P∩L0 are maximal ideals of S0 . Hence, P0 + σ −1 P0 ∩ L0 = S0 . By the Chinese remainder theorem [Lang7, p. 94], there exists y ∈ S0 with y ≡ x mod P0 and y ≡ 1 mod σ −1 P ∩ L0 for every σ ∈ G r DP . Thus, y ≡ x mod P and σy ≡ 1 mod P for every σ ∈ G r DP . Since L0 /K is a separable extension, the element a = normL0 /K (y) of R is a product of y and elements σy with σ running over nonidentity coset representatives of DP in G. Consequently a ≡ x mod P0 , as desired. ¯ = S0 /P0 and L ¯ = S/P. To continue localize S0 and S at P0 to assume K ¯ Write each element of L as x ¯, with x ∈ S, and let f = irr(x, L0 ). Then Qm f (X) = i=1 (X − xi ), where xi ∈ S, i = 1, . . . , m, are the conjugates of x over L0 . The coefficients of f (X) belong to L0 ∩ S =QS0 . Hence, m ¯ xi ) f¯(X) ∈ K[X]. In addition, x ¯ is a root of the polynomial f¯(X) = i=1 (X−¯ ¯ ¯ ¯ ¯ x) : with roots in L. Hence, L is a normal extension of K. Moreover, [K(¯ ¯ ≤ deg(f ) ≤ [L : L0 ]. K] Separable extensions have primitive generators [Lang7, p. 243]. Hence, ¯ which is separable over K ¯ satisfies [E : K] ¯ ≤ the maximal subfield E of L ¯ x) for some x ∈ S. In the above notation, if [L : L0 ]. Thus, E = K(¯ ¯ K) ¯ ∼ ¯ we have τ x τ ∈ Aut(L/ ¯=x ¯j for some j, 1 ≤ j ≤ m. The = Gal(E/K), ¯ = τ. map x 7→ xj extends to a field automorphism σ ∈ Gal(L/L0 ) with σ ¯ K) ¯ is surjective. Consequently, The map σ 7→ σ ¯ from DP into Aut(L/ ¯ : K] ¯ = |Gal(L/ ¯ K)| ¯ ≤ |DP | ≤ [L : Proof of (b): By assumption, [L : K] = [L ¯ K)|. ¯ K]. Hence, |DP | = |Gal(L/K)| = |Gal(L/ Therefore, DP = Gal(L/K) ¯ K) ¯ is an isomorphism. and the map Gal(L/K) → Gal(L/ The discriminant of f ∈ R[X], Qn disc(f ), gives information about ramification. Assume f is monic and i=1 (X − xi ) is the factorization of f into linear factors. Then (1)

disc(f ) = (−1)

n(n−1) 2

n Y n(n−1) Y 2 (xi − xj ) = (−1) f 0 (xj ). i6=j

j=1

This is an element of R, and disc(f ) 6= 0 if and only if the xi ’s are distinct.

6.1 Decomposition Groups

109

Assume f is irreducible. Then (1) implies disc(f ) = (−1)

n(n−1) 2

normK(x1 )/K (f 0 (x1 )).

We call normK(x1 )/K (f 0 (x1 )) the discriminant of x1 over K. Lemma 6.1.2: Let R be an integrally closed domain with quotient field K. Let S be the integral closure of R in a finite separable extension L of K. Assume L = K(z) with z integral over R and let f = irr(z, K). Suppose d = disc(f ) is a unit of R. Then S = R[z]. Proof: Let n = [L : K]. Assume y = a0 + a1 z + · · · + an−1 z n−1 , with ai ∈ K, is element of S. We must prove ai ∈ R, i = 0, . . . , n − 1. To this ˜ end, let 1 , . . . , σn be the isomorphisms of L over K into K. Then each of Pσn−1 σi y = j=0 aj σi z j , i = 1, . . . , n, is integral over R. b

∆b

Solve for the aj ’s by Cramer’s rule. This gives aj = ∆j = ∆2j , j = 0, . . . , n − 1, where ∆ = det(σi z j ) with bj integral over R, j = 0, . . . , n − 1. But ∆ is a Vandermonde determinant: Y ∆2 = ± (σi z − σj z) = ±normL/K f 0 (z) = ±d. i6=j

Since d is a unit of R, aj is integral over R, j = 0, . . . , n − 1. Since R is integrally closed, all aj are in R. Definition 6.1.3: Ring covers. As in the preceding lemmas, consider two integrally closed integral domains R ⊆ S with K ⊆ L their respective quotient fields such that L/K is finite and separable. Suppose S = R[z], where z is integral over R and the discriminant of z over K is a unit of R. In this set up we say S/R is a ring cover and L/K is the corresponding field cover. In this case, Lemma 6.1.2 implies that S is the integral closure of R in L. Call the element z a primitive element for the cover. If in addition L/K is Galois, call S/R is a Galois ring cover. We summarize the previous results for ring covers: Lemma 6.1.4: Let S/R be a Galois ring cover with L/K the corresponding field cover. Then for every prime ideal p of R and for every prime ideal P ¯ of S/P is a of S lying over p the following holds. The quotient field, L, ¯ Galois extension of the quotient field, K, of R/p. The map σ 7→ σ ¯ of DP ¯ K) ¯ given by σ into Gal(L/ ¯x ¯ = σx for x ∈ S is an isomorphism. Remark 6.1.5: Creating ring covers. Let K = K0 (x1 , . . . , xn ) be a finitely generated extension of a field K0 . The subring R = K0 [x1 , . . . , xn ] of K is not necessarily integrally closed. But, there exists a nonzero xn+1 ∈ K with R0 = K0 [x1 , . . . , xn+1 ] integrally closed ([Lang4, p. 120]; a constructive proof of this fact appears in Section 19.7). Suppose z is a primitive generator for L/K, f ∈ R[Z] is irreducible polynomial over K, and f (z) = 0. Multiply xn+1 by the inverse of the product of the leading coefficient and the discriminant of f . Then S 0 = R0 [z] is a ring cover of R0 with z a primitive element.

110

Chapter 6. The Chebotarev Density Theorem

Remark 6.1.6: Decomposition groups of places. Suppose L/K is a finite ¯ = ϕ(O) Galois extension and ϕ is a place of L with a valuation ring O. Then L is the residue field of L under ϕ. Also, R = O ∩ K is the valuation ring of the ¯ = ϕ(R) is its residue field. By Proposition 2.4.1, restriction of ϕ to K and K O contains the integral closure S of R in L. Let m be the maximal ideal of O, P = S ∩ m, and p = P ∩ R. Then O = SP [Lang4, p. 18, Thm. 4] and ¯ = S/P and R/p ∼ ¯ P is maximal [Lang7, p. 339, Prop 1.11]. Hence, L = K. We call Dϕ = DP and Iϕ = IP the decomposition group and inertia group, respectively, of ϕ over K. The fixed fields of Dϕ and Iϕ in L are the decomposition field and the inertia field, respectively, of ϕ over K. By ¯ K ¯ is a normal extension and the map σ 7→ σ lemma 6.1.1, L/ ¯ from DP into ¯ K) ¯ is surjective. Aut(L/ ¯ K ¯ is separable and Iϕ = 1 (This holds if S/R is a ringSuppose now L/ ¯ K) ¯ ¯ K ¯ is Galois and Dϕ is isomorphic to Gal(L/ cover.) By Lemma 6.1.1, L/ under the map σ 7→ σ ¯ . Denote the decomposition field of ϕ over K by L0 . ¯ For each x ∈ O ∩ L0 and each σ ∈ Dϕ we have σ ¯x ¯ = σx = x ¯. Hence, x ¯ ∈ K. ¯ Therefore, ϕ(L0 ) = K ∪ {∞}. Let P0 = P ∩ L0 . Then each prime ideal of S that lies over P0 is conjugate to P by an element of DP . Hence, P is the only prime ideal of ¯ :L ¯ 0 ] ≤ [L : L0 ]. Since S lying over P0 . By Proposition 2.3.2, e(P/P0 )[L ¯ ¯ ¯ ¯ [L : L0 ] = [L : K] = [L : L0 ], we have e(P/P0 ) = 1. In particular, if O is discrete, p is unramified in L. Remark 6.1.7: Ring covers under change of base ring. Consider an integrally closed domain R with quotient field K. Let L be a finite separable extension of K, S the integral closure of R in L, z anQelement of S with L = K(z), and f = irr(z, K). Then normL/K (f 0 (z)) = f 0 (z)σ , where σ ranges over all K-embeddings of L into Ks . Each f 0 (z)σ is integral over R. Hence, normL/K (f 0 (z)) is a unit of R if and only if f 0 (z) is a unit of S. Suppose f 0 (z) is a unit of S. By Lemma 6.1.2, S = R[z] and z is a primitive element of the ring-cover S/R. Let ϕ be a homomorphism of R ¯ Put K ¯ = Quot(R). ¯ Then ϕ extends to into an integrally closed domain R. ¯ (Proposition 2.4.1). a homomorphism ψ of S into the algebraic closure of K ¯ z ], L ¯ = Quot(S), ¯ f¯ = ϕ(f ), and g = irr(¯ ¯ Then Put z¯ = ψ(z), S¯ = R[¯ z , K). ¯ z ], and there is a monic polynomial h ∈ K[X] ¯ z ) is a unit of R[¯ f¯(¯ z ) = 0, f¯0 (¯ with f¯(X) = g(X)h(X). The coefficients of h are polynomials in the roots ¯ we have h ∈ R[X]. ¯ of f¯. Since the latter are integral over R, Deduce from 0 0 0 ¯ ¯ ¯ R ¯ z )h(¯ z ) = f (¯ z ) that g (¯ z ) is a unit of R[¯ z ]. Hence, by Lemma 6.1.2, S/ g (¯ ¯ is a ring-cover with z¯ as a primitive element. In particular, S is the integral ¯ in L. ¯ Thus, if R ¯ = K, ¯ then S¯ = L. ¯ closure of R As an example, let L/K be a finite separable extension, z a primitive element for L/K and f = irr(z, K). Then f 0 (z) 6= 0. Hence, L = K[z] is a ring-cover of K. Therefore, R[z]/R is a ring-cover whenever R is an integrally closed ring containing K. For another example suppose in the notation of the first two paragraphs

6.2 The Artin Symbol over Global Fields

111

¯=K ¯ that R is a valuation ring and Ker(ϕ) is the maximal ideal of R. Then R ¯ z ] is an integral extension of K. ¯ Hence, S¯ = L ¯ is a field. is a field and S¯ = K[¯ In addition, the local ring of S at Ker(ψ) is a valuation ring lying over R [Lang4, p. 18, Thm. 4.7]. Hence, ϕ and ψ extend uniquely to places of K ¯ and L, ¯ respectively. and L with residue fields K The next result is another consequence of Lemma 6.1.2 which may be applied to covers. Lemma 6.1.8: (a) Let R be an integral domain with quotient field K, L a finite Galois extension of K, and S the integral closure of R in L. Consider a monic polynomial f ∈ R[X] having all of its roots in L, a prime ideal p of R, and a prime ideal P of S lying over p. Assume disc(f ) ∈ / p and denote reduction modulo P by a bar. Also let σ ∈ DP and F a field containing ¯ such that L ¯ ∩ F = L(¯ ¯ σ ). Then the number of the roots of f in L(σ) K is equal to the number of the roots f¯ in F . (b) Suppose L = K(z) with z integral over R, f = irr(z, K), and disc(f ) ∈ / p. ¯ K ¯ is separable, IP = 1, and p is unramified in L. Then L/ Proof of (a): The roots of f¯ are distinct, because disc(f¯) 6= 0. In addition deg(f ) = deg(f¯). Thus, x 7→ x ¯ maps the roots of f bijectively onto the roots of f¯. For x a root of f , σx = x if and only if σ ¯x ¯ = x ¯. Moreover, since ¯ we have x ¯ σ ) if and only if x all of the roots of f¯ belong to L, ¯ ∈ L(¯ ¯ ∈ F. Consequently, the number of the roots of f in L(σ) is equal to the number of the roots of f¯ in in F . Proof of (b): Replace R by Rp , if necessary, to assume R is a local ring and p is its maximal ideal. By Lemma 6.1.2, S = R[z] is the integral closure of ¯ = K(¯ ¯ z ). From (a), L ¯ is a separable R in L. Under the assumptions of (b), L ¯ ¯ z¯ = z¯. Hence, σz = z, extension of K. Also, if σ ¯ = 1 for some σ ∈ DP , then σ so σ = 1. Hence, IP = 1. By (4) of Section 2.3, p is unramified in L.

6.2 The Artin Symbol over Global Fields The Artin symbol over number fields is a generalization of the Legendre symbol for quadratic residues. Here we define the Artin symbol over global fields and state some of its basic properties. Let R be a Dedekind domain with quotient field K. Consider a finite separable extension L of K. Let S be the integral closure of R in L. Take z ∈ S with L = K(z). If f = irr(z, K), then d = disc(f ) ∈ R and d 6= 0. Consider R1 = R[d−1 ] and S1 = S[d−1 ]. Then d is a unit in R1 and S1 = R1 [z] (Lemma 6.1.2). Thus, adjoining d−1 gives a ring cover S1 /R1 for L/K. Maximal ideals P for which PS1 = S1 are exactly those containing d. For z ]. all others S/P ∼ = S1 /PS1 and S/P = (R/p)[¯ If in addition L/K is a Galois extension, then extending P to S1 leaves the decomposition group and the inertia group unchanged. Hence, if P does

112

Chapter 6. The Chebotarev Density Theorem

not contain d, then PS1 , and therefore also P, is unramified over K (Lemma 6.1.8). Thus, a prime ideal p of R not containing d, does not ramify in L. Since only finitely many prime ideals of R contain d, only finitely many prime ideals ramify in L. Denote the greatest common divisor of all the principal ideals f 0 (z)S with z ∈ S by Diff(S/R) and call it the different of S over R. Then (Diff(S/R))−1 = {x ∈ L | traceL/K (xS) ⊆ R}. A prime ideal P of S ramifies over K if and only if it divides Diff(S/R) [Lang5, p. 62]. Hence, the prime divisors of the discriminant DS/R = normL/K Diff(S/R) of S over R (an ideal of R), are exactly those primes that ramify in L. Call K a global field if K is either a finite extension of Q (K is a number field) or K is a function field of one variable over a finite field. In the number field case denote the integral closure of Z in K by OK . In the function field case K is a finite separable extension of Fp (t), where p = char(K) and t is transcendental over Fp . Denote the integral closure of Fp [t] in K by OK . With the understanding it depends on t, call OK the ring of integers of K. The local ring of OK at a prime ideal p is a valuation ring. Denote its ¯ p is a finite field. We call N p = |K ¯ p| ¯ p . Note that K residue class field by K the absolute norm of p. Let L be a finite Galois extension of K. Suppose p is unramified in L. If P is a prime ideal of OL over p, then reduction modulo P gives a ¯ P /Kp) ¯ canonical isomorphism of the decomposition group DP and Gal(L (Lemma 6.1.4). The latter group is cyclic. It contains a canonical generator ¯ P by this rule: Frob, the Frobenius automorphism. It acts on L (1)

Frob(x) = xN p

¯P. for x ∈ L

Call the element of DP that corresponds to Frob the Frobenius automor phism at P and denote it by L/K P . It is uniquely determined in Gal(L/K) by the condition L/K x ≡ xN p mod P for all x ∈ OL . (2) P Let Fqn be the algebraic closure of Fq in L and let N p = q k . By (2), the restriction of L/K to Fqn is Frobkq . P If K ⊆ K 0 ⊆ L and K 0 /K is a Galois extension, this immediately implies 0 K /K L/K (3) resK 0 = . P P ∩ K0 −1 If σ ∈ Gal(L/K), then L/K . Therefore, as P ranges over = σ L/K σP P σ prime ideals of OL lying over p, the Frobenius automorphism ranges over

6.3 Dirichlet Density

113

some conjugacy class in Gal(L/K) that depends on p. This conjugacy class . It is tacit in this symbol that p is unramified is the Artin symbol, L/K p in L. If L/K is Abelian, then L/K is one element, L/K p P . In this case write L/K for L/K p P . In defining the Frobenius automorphism and the Artin symbol, replacing OK by the local ring OK,p does not change these objects. If K is a function field over Fq , the local rings OK,p bijectively correspond to prime divisors p0 of K/Fq finite at t. We use L/K as a substitute for L/K . Since each p0 p prime divisor p0 of K/Fq is either finite at t or at t−1 , the symbol L/K is p0 0 well defined if p is unramified in L. Example 6.2.1: Quadratic extensions of Q. Let a be a nonsquare integer √ and p an odd prime number not dividing a. Put L = Q( a). Then p is unramified in L (Example 2.3.8). Let p be a prime divisor of L lying over p. By elementary number theory [LeVeque, p. 46],

√ p p−1 √ L/Q √ a √ 2 a ≡ ( a) = a a≡ a mod p. p p L/Q p

Thus, the Frobenius symbol

acts

√

a as the Legendre symbol

a p

.

6.3 Dirichlet Density For K a global field denote the set of all prime ideals of OK by P (K). If A is a subset of P (K), then the Dirichlet density, δ(A), of A is the limit P δ(A) = lim+ P s→1

p∈A (N p)

−s

p∈P (K) (N p)

−s

,

if it exists. The Dirichlet density is a quantitative measure on subsets of P (K). We apply it to test if specific subsets are infinite. Clearly δ(A) is a real number between 0 and 1. For example, δ(P (K)) = 1. If K is a number field, then X (N p)−s = ∞ lim s→1+

p∈P (K)

[Lang5, p. 162]. Relation (19) of Section 6.4 implies that P the same holds if K is a function field. Hence, in both cases, δ(A) = 0 if p∈A (N p)−1 is finite. In particular, δ(A) = 0 if A is finite. If A and B are disjoint subsets of P (K) having a density, then δ(A ∪ B) = δ(A) + δ(B). Here is the main result:

114

Chapter 6. The Chebotarev Density Theorem

Theorem 6.3.1 (Chebotarev Density Theorem): Let L/K be a finite Galois extension of global fields and let C be a conjugacy class in Gal(L/K). Then |C| . = C exists and is equal to [L:K] the Dirichlet density of p ∈ P (K) | L/K p Section 6.4 proves Theorem 6.3.1 for function fields and Section 6.5 proves the theorem for number fields. A non obvious special case is Dirichlet’s theorem showing the arithmetic progression {a, a + n, a + 2n, . . .} has infinitely many primes when gcd(a, n) = 1. Corollary 6.3.2 (Dirichlet): Suppose a and n are relatively prime positive 1 , integers. Then the Dirichlet density of {p ∈ P (Q) | p ≡ a mod n} is ϕ(n) where ϕ(n) is the Euler totient function. Proof: Denote a primitive nth root of unity by ζn and let L = Q(ζn ). Then Gal(L/K) is isomorphic to (Z/nZ)× . If σ ∈ Gal(L/K) and σζn = ζna , then this isomorphism maps σ to a mod n. Also, for a, b relatively prime to n, we have ζna ≡ ζnb mod p if and only if ζna = ζnb . Thus, for p - n, p ≡ a mod n if a and only if L/Q p (ζn ) = ζn . Now apply Theorem 6.3.1. Example 6.3.3:

Let f ∈ Z[X] be a monic polynomial. Write f (X) =

r Y

fi (X)

i=1

with f1 (X), . . . , fr (X) monic and irreducible and let d be the product of the discriminants of f1 , . . . , fr ((1) of Section 6.1). Consider the following hypotheses: (1) f (X) ≡ 0 mod p has a solution for all but finitely many primes p. (2) f (X) ≡ 0 mod p has a solution for all primes p - d. Let L be the splitting field of f over Q. According to Theorem 6.3.1, each for infinitely many prime ideals p of element of Gal(L/Q) has the form L/Q p OL . Therefore, Lemma 6.1.8(a) implies each of (1) and (2) is equivalent to the following. (3) Each σ ∈ Gal(L/Q) fixes a root of f (X). In particular, (1) and (2) are equivalent. Example 6.3.4: Let K be a number field and B the set of all P prime ideals 1 of OK whose absolute norm is not a prime number. Then p∈B N p ≤ P [K:Q] p p2 < ∞. Hence, δ(B) = 0. Suppose L is a finite Galois extension of K and C a conjugacy class in Gal(L/K). Then, in view of the preceding paragraph, the Chebotarev density = C and theorem gives infinitely many prime ideals p of OK such that L/K p N p is a prime number.

6.4 Function Fields

115

6.4 Function Fields This section contains the proof of the Chebotarev density theorem in the function field case. Apart from elementary algebraic manipulations it depends only on the Riemann hypothesis for curves. To fix notation, let q be a power of a prime number. Consider a function field K over Fq , a finite Galois extension L of K, and a conjugacy class C of Gal(L/K) with c elements. Let Fqn be the algebraic closure of Fq in L and fix a separating transcendence element t for K/Fq . Denote the Frobenius element of Fqn /Fq by Frobq . As in Section 6.2, let OK be the integral closure of Fq [t] in L and let P (K) be the set of all nonzero prime ideals of OK . Denote the set of prime divisors of K/Fq by P(K). Identify P (K) with a cofinite subset of P(K). Thus, C = {p ∈ P(K) | L/K = C} and {p ∈ P (K) | L/K = C} differ by p p c finitely many elements. It suffices therefore to prove that δ(C) = [L:K] . n In addition to n = [Fq : Fq ], two more degrees enter the proof: d = [K : Fq (t)] and m = [L : KFqn ] as in the following diagram. K

n

K · Fq n

n

Fqn (t)

m

L

d

Fq (t)

Fix the following notation: P0 (K) = {p ∈ P(K) | p is unramified over Fq (t) and in L} Pk (K) = {p ∈ P0 (K) | deg(p) = k} P0k (K) = {p ∈ P0 (K) | deg(p) = k} Ck (L/K, C) = {p ∈ P0k (K) | L/K = C} p Dk (L/K, τ ) = {P ∈ P(L) | P ∩ K ∈ Pk (K) and L/K = τ }, P for τ ∈ Gal(L/K) S∞ C 0 = k=1 Ck (L/K, C) gK = the genus of K = the Frobenius automorphism of Gal(Fq ) and also of Frobq Gal(Fqn /Fk ) for each k. The sets C 0 and C differ by only finitely many elements. Hence, they have the same Dirichlet density. To compute this density, we compute the cardinality of each finite set Ck (L/K, C). This is also of independent interest, especially when k = 1. Lemma 6.4.1: Let k be a positive integer, p ∈ Ck (L/K, C), and τ ∈ C. (a) There are exactly [L : K]/ord(τ ) primes of P(L) over p. (b) If Ck is a subset of Ck (L/K, C) and Dk (τ ) = {P ∈ Dk (L/K, τ ) | P ∩ K ∈ Ck },

116

Chapter 6. The Chebotarev Density Theorem

then |Ck | = |C| · ord(τ ) · |Dk (τ )| · [L : K]−1 . Proof of (a): Suppose P ∈ P(L) lies over p. Then, by (2) of Section 2.3, [L : K] = eP/p fP/p gP/p where fP/p is the order of the decomposition group h L/K P i. In our case eP/p = 1 and τ is conjugate to L/K P . Thus, fP/p = ord(τ ) and (a) holds. 0 Proof of (b): For each σ ∈ Gal(L/K), Dk (στ σ −1 ) = S σDk (τ ). If τ ∈ C 0 and τ 6= τ , then Dk (τ ) and Dk (τ ) are disjoint. Thus, · τ ∈C Dk (τ ) is the set of primes of P(L) lying over the elements of Ck . By (a),

|Ck | · [L : K] X = |Dk (τ )| = |C| · |Dk (τ )|, ord(τ ) τ ∈C

and the formula follows.

Lemma 6.4.2: Let K ⊆ K 0 ⊆ L and τ ∈ Gal(L/K 0 ). Denote the algebraic closure of Fq in K 0 by Fqr . Suppose r|k. Then Dk (L/K, τ ) = Dk/r (L/K 0 , τ ) ∩ {P ∈ P(L) | deg(P ∩ K) = k}. Proof: Let P ∈ P(L). Suppose p = P∩K ∈ Pk (K) and p0 = P∩K 0 ∈ P(K 0 ). ¯ 0 0 = q rl , where l = deg(p0 ) = [K ¯ 0 0 : K 0 ]. ¯ p = q k and N p0 = K Then, N p = K p p By (2) of Section 6.2, (1)

k L/K = τ ⇐⇒ τ x ≡ xq mod P for every x ∈ OL ; and P

(2)

rl L/K 0 = τ ⇐⇒ τ x ≡ xq mod P for every x ∈ OL . P

Thus, it suffices to show

L/K P

= τ implies rl = k. k

Since τ ∈ Gal(L/K 0 ), (1) implies x ≡ xq mod P for every x ∈ OK 0 . ¯ p = Fqk . Therefore, Hence, K 0 p0 ⊆ Fqk . On the other hand, K 0 p0 ⊇ K 0 Fqrl = K p0 = Fqk . Consequently, rl = k. Corollary 6.4.3: With the hypotheses of Lemma 6.4.2, let C and C 0 be the respective conjugacy classes of τ in Gal(L/K) and in Gal(L/K 0 ) and 0 = Ck/r (L/K 0 , C 0 ) r{p0 ∈ P (K 0 ) | deg(p0 ∩ K) ≤ k2 }. Then Ck/r |Ck (L/K, C)| =

0 | |C||Ck/r

|C 0 |[K 0 : K]

.

6.4 Function Fields

117

Proof: Let l = kr . Then Dk0 (τ ) = Dl (L/K 0 , τ ) ∩ {P ∈ P(L) | deg(P ∩ K) = k} is the set of primes in Dl (L/K 0 , τ ) lying over Ck00 = Cl (L/K 0 , C 0 ) ∩ {p0 ∈ P(K 0 ) | deg(p0 ∩ K) = k}. Dk0 (τ ) is also the set of primes in Dl (L/K, τ ) over Ck (L/K, C). By Lemma 6.4.2, Dk0 (τ ) = Dk (L/K, τ ). Applying Lemma 6.4.1 twice gives a chain of equalities: [L : K] [L : K 0 ] |Ck (L/K, C)| = |Dk (L/K, τ )| = |Dk0 (τ )| = 0 |C 00 |. |C| · ord(τ ) |C | · ord(τ ) k Thus, it suffices to show that Ck00 = Cl0 . Indeed, if p0 ∈ P(K 0 ) is of degree l ¯ p ⊆ K 0 p0 = Fqrl = Fqk . Hence, deg(p)|k. Thus, and p = p0 ∩ K, then Fq ⊆ K either deg(p) = k or deg(p) ≤ k2 . Therefore, Ck00 = Cl0 . Lemma 6.4.4: Let k be a positive integer such that (3)

resFqn (τ ) = resFqn (Frobkq )

for every τ ∈ C. Let n0 be a multiple of n and L0 = LFqn0 . Then L0 /K is a Galois extension, Fqn0 is the algebraic closure of Fq in L0 and gL = gL0 (Proposition 3.4.2). Moreover, for each τ ∈ C there exists a unique τ 0 ∈ Gal(L0 /K) with resL τ 0 = τ and resFqn0 (τ 0 ) = resFqn0 (Frobkq ). Furthermore: (a) ord(τ 0 ) = lcm(ord(τ ), [Fqn0 : Fqn0 ∩ Fqk ]); (b) C 0 = {τ 0 | τ ∈ C} is a conjugacy class in Gal(L0 /K); and (c) Ck (L0 /K, C) = Ck (L/K, C 0 ). Proof: Given τ ∈ C, the existence of τ 0 follows from (3), because KFqn0 ∩L = KFqn . Uniqueness follows from L0 = Fqn0 L. To prove (a), note that ord(τ 0 ) = lcm ord(resL τ 0 ), ord(resFqn0 τ 0 ) = lcm ord(τ ), [Fqn0 : Fqn0 ∩ Fqk ] . Since Fqn0 /Fq is an Abelian extension, assertion (b) follows from the uniqueness of τ 0 . To prove (c), we show that P ∈ P(L0 ) and p = P ∩ K ∈ Pk (K) imply L/K L0 /K L/K k = τ 0 . Indeed, if P∩L = τ , then τ x ≡ xq mod P ∩ L P∩L = τ ⇐⇒ P k

for each x ∈ OL . By definition, Frobkq x = xq for each x ∈ Fqn0 . Since k

OL0 = Fqn OL (Proposition 3.4.2(d)), τ 0 x ≡ xq mod P for each x ∈ OL0 . 0 Hence, L P/K = τ 0 . The converse is a special case of (3) of Section 6.2.

118

Chapter 6. The Chebotarev Density Theorem

Corollary 6.4.5: If L = Fqn K and τ ∈ Gal(L/K) satisfies (3), then Ck (K/K, 1) = Ck (L/K, {τ }). Now we give estimates for the key sets. Lemma 6.4.6: Suppose L = KFqn , C = {τ }, and τ |Fqn = Frobq . Then √ (4) |#C1 (L/Fq , C) − q| < 2(gL q + gL + d). Proof: First note that C1 (L/Fq , C) = P01 (K) and each p ∈ P(Fq ) is unramified in L. Thus, P1 (Fq ) r C1 (L/K, C) consists exactly of all prime divisors of Diff(K/Fq (t)). By Riemann-Hurwitz (Theorem 3.6.1), deg(Diff(K/Fq (t))) = 2(gK + d − 1). By Theorem 4.5.2, √ |#P1 (K) − (q + 1)| ≤ 2gK q. √ Hence, |#C1 (L/K, C) − q| ≤ 2gK q + 1 + 2(gK + d − 1). Since gK = gL , this proves (4). Denote the situation when a divides b and a < b by a pd b. Lemma 6.4.7: Let K 0 be a degree km extension of K containing Fqk . Then (5)

#{q ∈ P(K 0 ) | deg(q ∩ K) pd k} ≤ 2m(q k/2 + (2gK + 1)q k/4 ).

Proof: If j|k and p ∈ Pj (K), then Fqj ⊆ Fqk . By Lemma 4.3.1, p decomposes in KFqj into j prime divisors of degree 1. Each has exactly one extension to KFqk . The latter decomposes in K 0 into at most m prime divisors. Hence, X {q ∈ P (K 0 ) | deg(q ∩ K) = j} #{q ∈ P (K 0 ) | deg(q ∩ K) pd k} = j|k j≤k/2

(6)

≤m

X

{q ∈ P (KFqk ) | deg(q ∩ K) = j}

j|k j≤k/2

=m

X

{q ∈ P (KFqj ) | deg(q ∩ K) = j}

j|k j≤k/2

≤m

X

|P1 (KFqj )|

j|k j≤k/2

By Theorem 4.5.2, |P1 (KFqj )| ≤ q j +2gK q j/2 +1. Now verify the inequalities P j k/2 and k2 ≤ 2q k/4 and use them to deduce (5) from (6). j≤k/2 q ≤ 2q We now come to the key result of this Section, Proposition 6.4.8. It is the main result from which the Chebotarev density theorem follows. It is also the main ingredient in an arithmetic proof of Hilbert irreducibility theorem (Lemma 13.3.3). A variant of it (Lemma 31.2.1) is crucial to the Galois stratification procedure for the elementary theory of finite fields. Recall: Fqn is the algebraic closure of Fq in L.

6.4 Function Fields

119

Proposition 6.4.8: Let a be a positive integer with resFqn τ = resFqn Frobaq for each τ ∈ C. Let k be a positive integer. If k 6≡ a mod n, then Ck (L/K, C) is empty. If k ≡ a mod n, then c k q #Ck (L/K, C) − km 2c (m + gL )q k/2 + m(2gK + 1)q k/4 + gL + dm . < km Proof: Suppose P ∈ P(L) lies over p ∈ Ck (L/K, C). Then, L/K ∈ C, so P

(7)

resFqn (Frobaq )

L/K = resFqn (Frobkq ). P

= resFqn

Hence, k ≡ a mod n. Conversely, suppose k ≡ a mod n. Let τ ∈ C and n0 = nk · ord(τ ). Extend L to L0 = LFqn0 . Then [L0 : KFqn0 ] = [L : KFqn ] = m. Since k ≡ a mod n, there exists τ 0 ∈ Gal(L0 /K) with τ 0 |L = τ and τ 0 |Fqn0 = Frobkq . Thus, ord(τ 0 ) = lcm(ord(τ ), ord(Frobkq )) = lcm(ord(τ ), [Fqn0 : Fqk ]) = lcm(ord(τ ), n · ord(τ )) = n · ord(τ ). Denote the conjugacy class of τ 0 in Gal(L0 /K) by C 0 . By Lemma 6.4.4(c), Ck (L0 /K, C 0 ) = Ck (L/K, C). ˜ q = Fqk Denote the fixed field of τ 0 in L0 by K 0 . Then K 0 ∩ Fqn0 = K 0 ∩ F 0 0 and K Fqn0 = L . K0 m

K d

KFqk

ord(τ 0 )

L0 m

KFqn0

d

Fq (t)

Fqk (t)

Fqn0 (t)

Fq

Fq k

Fqn0

120

Chapter 6. The Chebotarev Density Theorem

Thus, [K 0 : KFqk ] = [L0 : KFqn0 ] = [L : KFqn ] = m. Hence, [K 0 : Fqk (t)] = dm. Applying Lemma 6.4.3 with L0 , K, C 0 , {τ 0 }, k replacing F , E, C, C 0 , r we conclude that c |C1 (L0 /K 0 , {τ 0 }) r{q ∈ P(K 0 ) | deg(q ∩ K) pd k}|. |Ck (L0 /K, C 0 )| = [K 0 : K] Since [K 0 : K] = km, Lemma 6.4.7 implies #Ck (L0 /K, C 0 )− c #C1 (L0 /K 0 , {τ 0 }) (8) km c #{q ∈ P(K 0 ) | deg(q ∩ K) pd k} ≤ km c · 2m(q k/2 + (2gK + 1)q k/4 ) ≤ km By Lemma 6.4.6, now with K 0 , L0 , n0 , τ 0 , q k replacing K, L, n, τ , q, |#C1 (L0 /K 0 , {τ 0 }) − q k | < 2(gL0 q k/2 + gL0 + dm).

(9)

Now we combine (8) and (9) using the equalities gL = gL0 and Ck (L/K, C) = Ck (L0 /K, C 0 ) to prove (7): c k c k q | = |#Ck (L0 /K, C 0 ) − q | |#Ck (L/K, C) − km km c c #C1 (L0 /K 0 , {τ 0 }) + #C1 (L0 /K 0 , {τ 0 }) − q k ≤ #Ck (L0 /K, C 0 ) − km km 2c c k/2 k/4 k/2 · 2m q gL q + gL + dm + (2gK + 1) + ≤ km km 2c k/2 k/4 (m + gL )q + (2gK + 1)q + gL + dm . = km We deduce the function field case of Theorem 6.3.1 by summing over k in the conclusion of Proposition 6.4.8. We use the big O notation, i.e. for real valued functions f (x) and g(x) write f (x) = O(g(x)),

x→a

to mean there exists a constant c with |f (x)| ≤ c|g(x)| for all x close to a. In particular, if g(x) = 1, then f (x) is bounded near a. Lemma 6.4.9: Let a and n be positive integers. Then ∞ X xa+jn 1 = − log(1 − x) + O(1), a + jn n j=0

x → 1− .

Proof: If ζ 6= 1 is an nth root of unity, then 1 + ζ + · · · + ζ n−1 = 0. Hence −

n−1 ∞ n−1 1X 1 X xk X i(k−a) log(1 − ζ i x)ζ −ia = ζ = n i=0 n k i=0 k=1

X k≡a mod n

xk . k

Since, for 1 ≤ i ≤ n − 1, log(1 − ζ i x) is bounded in the neighborhood of 1, the result follows.

6.5 Number Fields

121

Lemma 6.4.10: Suppose 0 < a ≤ n is an integer with resFqn τ = resFqn Frobaq for each τ ∈ C. Then, (10)

X

(N p)−s = −

p∈C

c log(1 − q 1−s ) + O(1), [L : K]

s → 1+ .

S∞ Proof: The set C 0 = k=1 Ck (L/K, C) differs from C by only finitely many elements. We apply Proposition 6.4.8 and Lemma 6.4.9 for x = q 1−s to compute: X

(N p)−s =

∞ X

X

(N p)−s

j=0 p∈Ca+jn (L/K,C)

p∈C

=

∞ X j=0

=

1 c q a+jn + O(q 2 (a+jn) ) q −(a+jn)s m(a + jn)

∞ ∞ X 1 1 c X q (1−s)(a+jn) + O q ( 2 −s)a q ( 2 −s)jn m j=0 a + jn j=0 1

q ( 2 −s)a c log(1 − q 1−s ) + O(1) + O 1 mn 1 − q ( 2 −s)n c log(1 − q 1−s ) + O(1), s → 1+ . =− [L : K] =−

When L = K, Lemma 6.4.10 simplifies to (19)

X

N p−s = − log(1 − q 1−s ) + O(1),

s → 1+ .

p∈P (E)

Dividing (18) by (19) and taking the limit as s → 1+ gives the Dirichlet density of C: P −s c p∈C (N p) δ(C) = lim P = −s + [L : K] s→1 p∈P (E) (N p) This concludes the Chebotarev density theorem for function fields.

6.5 Number Fields Let L/K be a finite Galois extension of number fields. The proof of the Chebotarev density theorem for L/K splits into eight parts. It uses the asymptotic formula (2) for counting ideals with bounded norm in a given class (which we quote without proof). We say L/K is cyclotomic if L ⊆ K(ζ) with ζ a root of 1. The case that L/K is cyclotomic produces the general case from an easy reduction to L/K cyclic.

122

Chapter 6. The Chebotarev Density Theorem

Part A: Ideals with a bounded norm in a given class. Let c be a nonzero ideal of OK . Denote the group of all fractional ideals of K relatively prime to c by J(c). Let P (c) be the subgroup of all principal fractional ideals xOK , where x satisfies the following conditions. (1a) If p is a prime ideal of OK that divides c, then x lies in the local ring Op of OK at p and x ≡ 1 mod cOp . (1b) x is totally positive: σx is positive for each embedding σ: K → R. The factor group G(c) = J(c)/P (c) is finite [Lang5, p. 127]. Denote the order of G(c) by hc . Extend the absolute norm N p of prime ideals multiplicatively to all fractional ideals. Consider a class K of J(c) modulo P (c). Denote the number of ideals a ∈ K (of OK ) with N a ≤ n by j(K, n) The key asymptotic formula is: (2)

−1

j(K, n) = ρc n + O(n1−[K:Q] ),

n → ∞,

where ρc is a positive constant dependent on c and K but not on K [Lang5, p. 132]. Part B: Abelian characters. A character of a finite Abelian group G is a homomorphism χ: G → C× . Define multiplication of characters by ˆ of characters of G forms a group iso(χ1 χ2 )(σ) = χ1 (σ)χ2 (σ). The set G morphic P to G. Here are the standard character formulas. (3a) Pσ∈G χ1 (σ −1 )χ2 (σ) = |G| if χ1 = χ2 and 0 otherwise. −1 )χ(τ ) = |G| if σ = τ and 0 otherwise. (3b) ˆ χ(σ Q χ∈G (3c) χ∈Gˆ (1 − χ(σ)X) = (1 − X f )|G|/f if f = ord(σ). Formulas (3a) and (3b) are known as the orthogonality relations [Goldstein, p. 113]. Formula (3c) follows in the case where G = hσi by observing that the zeros of both sides are the roots of 1 of order f and from ˆ = |G|. The general case follows from the cyclic case and the the relation |G| ⊥ c = ∼ G/hσi ˆ ˆ | χ(σ) = 1} canonical isomorphism hσi , where hσi⊥ = {χ ∈ G [Goldstein, p. 112]. Part C: L-series. For a given ideal c of OK and a character χ of G(c) consider the Dirichlet series X χ(a) (4) Lc (s, χ) = , Re(s) > 1, (N a)s gcd(a,c)=1

where a ranges over all ideals of OK relatively prime to c, and χ(a) = χ(K) if K is the class of a. Call Lc (s, χ) an L-series. The function χ(a) is multiplicative on J(c). Therefore, Lc (s, χ) satisfies the Euler identity Y χ(p) −1 (5) Lc (s, χ) = for Re(s) > 1 1− (N p)s p-c

We quote the following result of complex analysis:

6.5 Number Fields

123

Lemma 6.5.1 ([Lang5, p. 158]): Let {ai }∞ i=1 be a sequence of complex numbers, for which there is a 0 ≤ σ < 1 and a complex number ρ with n X

ai = ρn + O(nσ )

as

n → ∞.

i=1

P∞ Then f (s) = n=1 an n−s for Re(s) > 1 analytically continues to Re(s) > σ, except for a simple pole with residue ρ at s = 1. Lemma 6.5.2: The function Lc (s, χ) has an analytic continuation to the half 1 . If χ = 1, then it has a simple pole at s = 1 with plane Re(s) > 1 − [K:Q] residue hc ρc . If χ 6= 1, then Lc (s, χ) is analytic in the entire half plane. Proof: We use (2) to substitute for j(K, t) and use the orthogonality relation (3a) for the finite group G(c), to conclude that X

χ(a) =

(a,c)=1 N a≤n

X

X

K∈G(c)

a∈K N a≤n

=

χ(a) =

X

−1

χ(K) ρc n + O(n1−[K:Q] )

K∈G(c) −1

hc ρc n + O(n1−[K:Q] ) −1 O(n1−[K:Q] )

if χ = 1 if χ = 6 1,

Thus, our Lemma is a special case of Lemma 6.5.1.

n → ∞.

Part D: Special case of Artin’s reciprocity law. This law is the central result of class field theory. Consider a finite Abelian extension L/K of number fields. Let c be an ideal of OK divisible by all prime ideals ramifying in L (we say that c is admissible). If a prime ideal p does not divide c, then L/K defines a unique element of Gal(L/K). The map p 7→ L/K extends p p to a homomorphism ωc : J(c) → Gal(L/K) called the reciprocity map. When referring to the extension L/K we will denote ωc by ωL/K,c . Let L0 be any Abelian extension of K containing L. Suppose each prime ideal of OK ramified in L0 divides c. Then, resL (ωL0 /K,c (p)) = ωL/K,c (p) for each prime p ∈ J(c) ((3) of Section 6.2). So, (6)

resL (ωL0 /K,c (a)) = ωL/K,c (a)

for each a ∈ J(c)

Class field theory proves that ωc is surjective [Lang5, p. 199, Thm. 1]. In order to describe its kernel, let Norm = NormL/K be the norm map of fractional ideals of OL onto fractional ideals of OK . If A is an ideal of OL , then Norm(A) is the ideal of OK generated by normL/K (a) for all a ∈ A. If B is another ideal, then Norm(AB) = Norm(A)Norm(B), so Norm extends multiplicatively to the group of all fractional ideals of OL . If P is a prime ideal of OL , p = P ∩ OK , and f = fP/p , then Norm(P) = pf . Finally, NormL/Q and the absolute norm of ideals relate to each other by the formula NormL/Q (A) = N (A)Z. [Janusz, pp. 35–37].

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Chapter 6. The Chebotarev Density Theorem

Artin reciprocity law gives an admissible ideal c of OK such that ωc : J(c) → Gal(L/K) is surjective and Ker(ωc ) = Norm(c)P (c) [Lang5, p. 205, Thm. 3]. Here we prove one part of Artin reciprocity law for cyclotomic extensions of K. Lemma 6.5.3: Let ζ be a primitive mth root of 1, L a subfield of K(ζ) containing K, and c an ideal of OK divisible by m. Then P (c) ⊆ Ker(ωc ). Proof: Each prime ideal of OK which ramifies in K(ζ) divides m [Goldstein, p. 98]. This defines ωc = ωL/K,c Use (6) to assume that L = K(ζ). Consider the natural embedding i: Gal(L/K) → (Z/mZ)× determined by σ(ζ) = ζ i(σ) , σ ∈ Gal(L/K). If p ∈ J(c) is prime and P is a prime ideal of OL lying over p, then ωc (p)(ζ) ≡ ζ N p mod P. Since reduction modulo P is injective on {1, ζ, . . . , ζ m−1 } [Goldstein, p. 97, Prop. 6-2-2], ωc (p)(ζ) = ζ N p . Hence, i(ωc (p)) ≡ N p mod m. Therefore, i(ωc (a)) ≡ N a mod m for each a ∈ J(c).

(7)

Now let xOK ∈ P (c), with x ∈ K × . Since x is totally positive (by (1b)), NK/Q (x) is a positive rational number congruent by (1a) to 1 mod m. It generates the same fractional Z-ideal as N (xOK ) [Janusz, p. 37]. Since N (xOK ) is also a positive rational number, N (xOK ) = NK/Q (x) ≡ 1 mod m. We conclude from (7) that ωc (xOK ) = 1. Part E: Cyclotomic L-series and the Dedekind zeta function. Let L and c be as in Lemma 6.5.3. Then ωc induces a homomorphism ω ¯ c from G(c) = J(c)/P (c) onto a subgroup G of Gal(L/K). (In Corollary 6.5.5(c), we prove ˆ χ◦ω that G = Gal(L/K).) Thus, for each χ ∈ G, ¯ c is a character of G(c). In the notation of Part C, the L-series Lc (s, 1) of the trivial character is the Dedekind zeta function of K with respect to c. X Y 1 −1 1 1 − = , Re(s) > 1 ζc (s, K) = (N a)s (N p)s gcd(a,c)=1

p-c

Lemma 6.5.4: In the above notation let C = cOL and let n = (Gal(L/K) : G). Then Y Lc (s, χ ◦ ω ¯ c )n . (8) ζC (s, L) = ˆ χ∈G

Proof: Let p be a prime ideal of OK with p - c. Suppose p factors in L into a product of g prime ideals P, each of degree f . Since p is unramified in L, we have f g = [L : K] = n|G| and ωc (p) = L/K is of order f . By (3c) and p the relation N P = N pf , Y χ◦ω ¯ c (p) n 1 1 n|G|/f Y 1− 1 − . = 1 − = (N p)s (N p)sf (N P)s ˆ χ∈G

Applying the product over all p - c, we conclude (8).

P|p

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125

Corollary 6.5.5: Let χ1 be a nontrivial character of G. Then: ¯ c ) 6= 0; (a) Lc (1, χ1 ◦ ω (b) log ζc (s, K) = − log(s − 1) + O(1), s → 1+ ; and (c) G = Gal(L/K). ¯ c is a nontrivial character Proof: Since ω ¯ c : G(c) → G is surjective, χ ◦ ω of G(c) for each nontrivial character of G. By Lemma 6.5.2, ζc (s, K) has a simple pole at s = 1. All other factors of the right hand side of (8) are ¯ c ) has a zero at s = 1, then the zero of regular at s = 1. Assume Lc (s, χ1 ◦ ω ¯ c )n in (8) at s = 1 cancels the pole of ζc (s, K)n . Hence, ζC (s, L) is Lc (s, χ ◦ ω analytic at s = 1, contradicting Lemma 6.5.1. This proves (a). Formula (b) follows from Lemma 6.5.2. Finally, the left side of (8) has a pole of order 1 at s = 1. By (a), the right side has pole of order n. Therefore, n = 1 and G = Gal(L/K). Lemma 6.5.6: If χ is a character of G, then ¯c) = log Lc (s, χ ◦ ω

Xχ◦ω ¯ c (p) p-c

(N p)s

+ O(1),

s → 1+ .

Proof: We apply the Euler identity (5) for Re(s) > 1 to obtain: log Lc (s, χ ◦ ω ¯c) = −

X

∞ ¯ c (p)k χ◦ω ¯ c (p) X X χ ◦ ω log 1 − . = s (N p) k · (N p)ks p-c k=1

p-c

Next let σ = Re(s). Then ∞ ∞ XX |χ ◦ ω ¯ c (p)|k X X X 1 ≤ f kσ ks p/p k(N p) p p p-c k=2

p|p k=2

≤ [K : Q]

∞ XX X 1 1 1 = [K : Q] pkσ p2σ 1 − p−σ p p k=2

X 1 1 with P (N p)−s ε 1 P p∈Cσ (13) − , > −s [L : K] [L : K] p∈P (K) (N p)

1 < s < s0 .

As σ ranges on Gal(L/K), the sum of the left hand sided of (13) is 1. Hence, by (13) P (N p)−s [L : K] − 1 1 P p∈Cσ (14) +ε , 1 < s < s0 . < −s (N p) [L : K] [L : K] p∈P (K) It follows from (13) and (14) that each Cσ has Dirichlet density [L : K]−1 . This proves the Chebotarev density theorem for Abelian extensions. Part H: Reduction to the cyclic case. We now reduce the Chebotarev density theorem for an arbitrary finite Galois extension L/K of number fields to the case that L/K is cyclic. Let C be a conjugacy class in Gal(L/K) = G and let L/K =C . C = p ∈ P (K) | p Choose τ ∈ C and let K 0 = L(τ ) be its fixed field. Denote the set of primes q ∈ P (K 0 ) which are unramified over K, have a relative degree 1 over K, and 0 satisfy L/K = {τ } by D0 . If q ∈ D0 and p = q ∩ K, then N p = N q. Hence, q L/K 0 = P = τ , so p ∈ C. Moreover, since if P ∈ P (L) lies over q, then L/K P ord(τ ) = [L : K 0 ], P is the unique element of P (L) over q. Conversely, if p ∈ C, then there exists a prime P ∈ P (L) that lies over p with L/K = τ. P Then q = P ∩ K 0 belongs to D0 and lies over p. Let CG (τ ) be the centralizer of τ in G. The map σ 7→ στ σ −1 from G onto C has fibers whose order is |CG (τ )|, so |G = |C| · |CG (τ )|. All primes P0 ∈ P (L) over p that satisfy L/K = τ are conjugate to P by some element P0 |G| G (τ )| σ ∈ CG (τ ). Hence, their number is |C|D = [L:K 0 ]·|C| . P| 0 This is therefore the number of q ∈ D that lie over p. Hence, for s > 1

(15)

X p∈C

1 1 |C|[L : K 0 ] X = . s (N p) [L : K] (N q)s 0 q∈D

0 = {τ } by only primes of The set D0 differs from D = q ∈ P (K 0 ) | L/K q relative degree at least 2 over K. For these primes we have X deg(q)≥2

X 1 1 0 ≤ [K : Q] < ∞. (N q)s q2 q

Exercises

129

(Note: This elimination of the set of absolute degree 2 primes won’t work in the function field case, for there are only finitely many primes of degree 1.) Hence, by Part G for cyclic extensions X

(N q)−s =

q∈D 0

X

(N q)−s + O(1) = −

q∈D

1 log(s − 1) + O(1), [L : K 0 ]

s → 1+ .

Combining this with (15) gives X

(N p)−s = −

p∈C

Finally, by (9), δ(C) =

|C| log(s − 1) + O(1), [L : K]

|C| [L:K] ,

as stated.

s → 1+ .

Exercises 1. Consider the ring R = Z[ζp ][x], where p is a prime number and x is an indeterminate. Let K be the quotient field of R and L = K(x1/p ). Give an ¯ K ¯ is not example, in Lemma 6.1, where the residue class field extension L/ Galois. 2. Let S/R be a ring cover. Consider prime ideals p of R and P of S with P ∩ R = p. Prove that pSP = PSP (This, together with the separability of SP /PSP over Rp /pRp means that S/R is an unramified extension of rings). Hint: Assume without loss that R and S are local rings with maximal ideals p and P, respectively. Let z be a primitive element for S/R and f = irr(z, Quot(R)). Denote the reduction modulo pS with a bar. Observe that ¯ Prove that S¯ ∼ ¯ ¯ S¯ is a local ring with maximal ideal P. f¯(X)R[X] and = R[X]/ the right hand side is a direct sum of fields corresponding to the irreducible factors of f¯(X). Conclude that there is only one such factor. 3. Let L1 , L2 be finite Galois extensions of a global field K and let L = L1 L2 . Consider a prime ideal P of OL and put Pi = P ∩ Li , i = 1, 2. Prove that L/K i /K is the unique element of Gal(L/K) whose restriction to Li is LP , P i i = 1, 2. 4. Show that if f (X) ∈ Z[X] is irreducible, then there exists infinitely many primes p for which f (X) ≡ 0 mod p has no solution. Hint: Use the equivalence of (1) and (3) of Section 6.3 and Lemma 13.3.2. 5. Let L/K be a finite Galois extension of global fields. Denote the set of prime ideals p of OK that split completely in L (pOL = P1 · · · Pn , where n = [L : K]) by Splt(L/K). (a) Show that a prime p ∈ P (K), unramified in L, belongs to Splt(L/K) = 1. if and only if L/K p

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Chapter 6. The Chebotarev Density Theorem

(b) (Bauer) Suppose L and L0 are finite Galois extensions of a global field K such that Splt(L/K) and Splt(L0 /K) differ by a finite set., Prove that L = L0 . Hint: Apply the Chebotarev density theorem to the field LL0 . 6. Let K and L be number fields with equal zeta functions (Section 6.5, Part E). For each positive integer n prove that the number of ideals of OK with absolute norm n is the number of ideals of OL with absolute norm n. Apply this to any a prime p unramified in KL. Prove that p splits completely in K if and only if p splits completely in L. Conclude from Exercise 5 that if K and L are Galois over Q, then K = L 7. Let K be a global field. Denote the set of all prime ideals p of OK whose absolute degree is at least 2 by P 0 (K). That is, N p = pd where d ≥ 2. When K is a number field prove that the Dirichlet density as well as the natural density of P 0 (K) is 0. If, however K is a function field show that almost all primes of P (K) belong to P 0 (K). Thus, δ(P 0 (K)) = 1.

Notes Frobenius [Frobenius] conjectured what we now call the Chebotarev density theorem for finite Galois extension L/K of number fields. His result replaced the conjugacy class appearing in the conjecture by the union of all conjugates of σ i , where σ is a given element of Gal(L/K) and i ranges over all integers relatively prime to ord(σ). A fine account of the Frobenius density theorem appears in [Janusz]. Chebotarev [Tschebotarev] used cyclotomic fields to prove the Frobenius conjecture via a more difficult version of the field crossing argument of Part G of Section 6.5. Artin [Artin1] introduced his L-series; then he proved his reciprocity law and applied it to reprove the conjecture [Artin2]. Our proof is a mixture of both methods, with the addition of Deuring’s reduction to the cyclic case [Deuring1]. It was elaborated for this book by Haran. Deuring’s reduction was reproduced in [MacCluer]. For the function field case note that Reichardt proved Proposition 6.19 when a = k = 1 and K is algebraically closed in F , (i.e. m = 1) [Reichardt]. The restriction m = 1 does not appear explicitly in [Reichardt]. Without it, however, the result as well as its proof would be false. It is also interesting to note that [Reichardt] appeared before Weil proved the Riemann hypothesis for curves. Thus, Reichardt’s (analytic) proof uses only that the maximum of the real parts of the zeros of the Zeta function is less than 1. Serre [Serre2] gives a unified approach to the number field and function field case. He considers a scheme X of finite type over Z, takes an ´etale Galois covering of X and attaches an L-series L(X, χ; s) to the cover. He says that an induction on dim(X) shows if χ 6= 1, then L(X, χ; s) is holomorphic and 6= 0 at the point s = dim(X). This implies the Chebotarev density theorem by the classical Dirichlet argument (e.g. as in Part F of Section 6.5). Serre’s program for the function field case appears in [Fried9]. An early version of our proof for the function field case appears in [Jarden10].

Notes

131

The proof of [Fried-Jarden3, Prop. 5.16] applies [Fried-Jarden3, Lemma 5.14] in a faulty way. Indeed, d on [Fried-Jarden3, p. 63, line -3] should be replaced by md. This version of Field Arithmetic follows [Geyer-Jarden4, Appendix] and corrects this mistake also improving the estimate of [FriedJarden3, Prop. 5.16]. There are several effective versions of the Chebotarev density theorem. One of the most valuable for the problems in this book is [LagariasMontegomery-Odlyzko, p. 416, Theorem]. This isolates the contribution of the absolute discriminant dL of a number field L over Q to the error term. It proves that there is an effectively computable constant A with the following property: For each Galois extension L/K of Gal(L/K) of number fields and each conjugacy class C of Gal(L/K) there is a prime p of K, unramified in L = C, p = NL/Q p is a rational prime and p ≤ 2dA with L/K L . This result is p independent of the generalized Riemann hypothesis.

Chapter 7. Ultraproducts We develop the basic concepts of logic and model theory required for applications to field theory. These include the Skolem-L¨owenheim theorem, Loˇs theorem and an ℵ1 -saturation property for ultraproducts. Finally, we apply regular ultraproducts of families of models to the theory of finite fields.

7.1 First Order Predicate Calculus There is no general test to decide whether a given polynomial f (X1 , . . . , Xn ) with integral coefficients has a zero in Zn ; this is the negative solution to Hilbert’s 10th problem. A partial decision test must satisfy two criteria: it must be conclusive for a significant body of polynomials; and it must be effective in concrete situations. The simplest, and most famous, such test is the congruence test, whereby we test the congruence f (X1 , . . . , Xn ) ≡ 0 mod p for solutions for all primes p. Regard the coefficients of f as elements of Fp to see that the above congruence is equivalent to solving the equation f (X1 , . . . , Xn ) = 0 in Fp . If f (X1 , . . . , Xn ) = 0 has no solution in Fp for one p, then f (X1 , . . . , Xn ) = 0 has no solution in Z. This chapter develops language and technique for the formulation of analogs of the diophantine problem and of the corresponding congruence test over general rings and fields. We start with the introduction of a first order language, the concept of a theory in the first order language, and a model for this theory. A language (more precisely, first order language) consists of letters, rules for combining letters into meaningful words, and, finally, an interpretation of the meaningful words. The language we now describe depends on functions µ and ν from sets I and J to N and on a set K. It is denoted L(µ, ν, K). Here are its letters: (1a) Countably many variable symbols: X1 , X2 , X3 , . . . ; (1b) constant symbols ck , one for each k ∈ K; (1c) a µ(i)-ary relation symbol, Ri , one for each i ∈ I; (1d) the equality symbol =; (1e) a ν(i)-ary function symbol, Fj , one for each j ∈ J; (1f) the negation symbol ¬, and the disjunction symbol ∨; (1g) the existential symbol ∃; and (1h) parentheses ( ) and brackets [ ]. A finite sequence of letters of L(µ, ν, K) is a string. Among the strings of L(µ, ν, K) the (meaningful) words include, ”terms”, ”formulas” and ”sentences”; which we now define. The collection of terms of L(µ, ν, K) is the smallest collection of strings that contains all of the following: (2a) all the variable symbols Xi ;

7.1 First Order Predicate Calculus

133

(2b) all the constant symbols ck ; and (2c) all the strings Fj (t1 , . . . , tν(j) ) where j ∈ J and (t1 , . . . , tν(j) ) is a ν(j)tuple of previously defined terms. We list the atomic formulas: (3a) t = t0 for each pair of terms (t, t0 ); and (3b) Ri (t1 , . . . , tµ(i) ), for all i ∈ I and all µ(i)-tuples (t1 , . . . , tµ(i) ) of terms. The set of formulas is the smallest collection of strings containing all atomic formulas and satisfying the following: (4a) ¬[ϕ] is a formula, if ϕ is a formula; (4b) ϕ1 ∨ ϕ2 is a formula if ϕ1 and ϕ2 are formulas; and (4c) (∃Xl )[ϕ] is a formula, if ϕ is a formula and l ∈ N. This definition allows us to prove a property of formulas or to make definitions depending on formulas by an induction on structure. We first prove the property for atomic formulas. Then, assuming its validity for ϕ, ϕ1 and ϕ2 we prove it for ¬[ϕ], ϕ1 ∨ ϕ2 and (∃Xl )[ϕ]. As a first example we define the notion of free occurrence of a variable in a formula by an induction on structure: Any occurrence of X in an atomic formula ϕ is free. If an occurrence of X in a formula ϕ is free, and ψ is an arbitrary formula, then this occurrence is free in ¬ϕ, ϕ ∨ ψ and (∃Y )[ϕ], for Y distinct from X. Any occurrence of X which is not free is bounded. Any variable X which has a free occurrence in a formula ϕ is said to be a free variable of ϕ. Frequently we write ϕ(X1 , . . . , Xn ) (or t(X1 , . . . , Xn )) to indicate that X1 , . . . , Xn include all the free variables of ϕ (or t). Example: In the formula 1 2 3 (∃X) X = Y ∨ (∃X)[X = c] ∨ ¬R(X , Y ) occurrences 1 and 2 are bounded, while occurrence 3 is free; both occurrences of Y are free. Hence, X and Y are free variables of the formula. A formula without free variables is a sentence. Some abbreviations simplify this language: (5a) ϕ ∧ ψ for ¬[¬ϕ ∨ ¬ψ] (∧ is the conjunction symbol); (5b) ϕ → ψ for ¬ϕ ∨ ψ (→ is the implication symbol); (5c) ϕ ↔ ψ for [ϕ → ψ] ∧ [ψ → ϕ] (↔ is the double implication symbol); (5d) (∀Xl )[ϕ] for ¬(∃Xl )[¬ϕ] (∀ is the universal quantifier); Vn (5e) i=1 ϕi for ϕ1 ∧ ϕ2 ∧ · · · ∧ ϕn ; and Wn (5f) i=1 ϕi for ϕ1 ∨ ϕ2 ∨ · · · ∨ ϕn .

134

Chapter 7. Ultraproducts

7.2 Structures The sentences of a first order language are interpreted in “structures” of this language. In each of these structures, they are either true or false. A structure for the language L(µ, ν, K) is a system ¯ i , F¯j , c¯k ii∈I, j∈J, k∈K A = hA; R ¯ i is a µ(i)-ary relation where A is a nonvoid set, called the domain of A, R µ(i) ν(j) ¯ → A is a ν(j)-ary function on A, and of A (i.e. a subset of A ), Fj : A c¯k is an element of A, called a constant. Sometimes we use the same letter for the logical symbol and its interpretation in the structure. Also, for well known binary relations and binary functions we write the relation and function symbols as usual, between the argument (e.g. “a ≤ b” for “a less than or equal to b”). Occasionally we add to L = L(µ, ν, K) a new constant symbol a ˜ for each a | a ∈ A} . a ∈ A. This gives an extended structure L(A) = L µ, ν, K ∪· {˜ A substitution into A is a function, f (Xi ) = xi , from the set of variables into A. The following recursive rules extend this uniquely to a function from the set of terms into A: (1a) f (ck ) = c¯k , and (1b) f (Fj (t1 , . . . , tν(j) )) = F¯j (f (t1 ), . . . , f (tν(j) )), where t1 , . . . , tν(j) , are terms for which f has already been defined. Define the truth value of a formula ϕ under a substitution f (either “true” or “false”) by induction on structure: (2a) t = t0 is true if f (t) = f (t0 ); and ¯i. (2b) Ri (t1 , . . . , tµ(i) ) is true if (f (t1 ), . . . , f (tµ(i) )) ∈ R Continue by assuming that the truth values of ϕ, ϕ1 , and ϕ2 have been defined for all possible substitutions. Then (3a) ¬ϕ is true if ϕ is false; (3b) ϕ1 ∨ ϕ2 is true if ϕ1 is true or if ϕ2 is true (so if both ϕ1 and ϕ2 are true, then ϕ1 ∨ ϕ2 is also true); and (3c) (∃Xl )[ϕ] is true if there exists an x in A such that ϕ is true under the substitution g defined by: g(Xl ) = x and g(Xm ) = f (Xm ) if m 6= l. The truth values of the additional logical symbols introduced above are as follows: (4a) ϕ1 ∧ ϕ2 is true if both ϕ1 and ϕ2 are true; (4b) ϕ → ψ is true if the truth of ϕ implies the truth of ψ (i.e. either ϕ is false, or both ϕ and ψ are true); (4c) ϕ ↔ ψ is true if both ϕ and ψ are true or both ϕ and ψ are false; and (4d) (∀Xl )[ϕ] is true if for each x in A, ϕ is true under the substitution g defined by g(Xl ) = x and g(Xm ) = f (Xm ) if m 6= l. By an easy induction on structure one observes that the truth value of a formula ϕ(X1 , . . . , Xn ) under a substitution f depends only on f (X1 ) = x1 , . . . , f (Xn ) = xn . If ϕ is true under f , write A |= ϕ(x1 , . . . , xn ). In

7.3 Models

135

particular, for ϕ a sentence, the truth value of ϕ is independent of f . It is either true in A, or false in A. In the former case write A |= ϕ, and in the latter, A 6|= ϕ

7.3 Models Models of a first order language L are generalization of groups, rings, fields, and ordered sets. They are structures where a given set of sentences, called axioms, is true. A theory in a first order language L = L(µ, ν, K) is a set of sentences T of L. A structure A for L is called a model of T if A |= θ for every θ ∈ T . In this case write A |= T . If T 0 is another theory in L for which every model of T 0 is also a model of T , write T 0 |= T . If Π is a theory of L and T is the theory of all sentences θ of L such that Π |= θ, then Π is said to be a set of axioms for T . Denote the class of all models of a theory T by Mod(T ). If A is a structure for a language L, then Th(A, L) is the set of all sentences of L which are true in A. Example 7.3.1: The theory of fields. Denote the first order language that contains the two binary functions symbols + (addition) and · (multiplication), and two constant symbols 0 and 1 by L(ring). For each integral domain R let L(ring, R) be the language L(ring) extended by all elements of R as constant symbols. Denote the usual axioms for the theory of fields by Π: (∀X)(∀Y )(∀Z)[(X + Y ) + Z = X + (Y + Z)]; (∀X)(∀Y )[X + Y = Y + X]; (∀X)[X + 0 = X]; (∀X)(∃Y )[X + Y = 0]; (∀X)(∀Y )(∀Z)[(XY )Z = X(Y Z)]; (∀X)(∀Y )[XY = Y X]; (∀X)[1 · X = X]; (∀X)[X 6= 0 → (∃Y )[XY = 1]]; 1 6= 0; and (∀X)(∀Y )(∀Z)[X(Y + Z) = XY + XZ]. Every model of Π is a field. Extend Π by all equalities — the positive diagram of R — (1)

a1 + b1 = c1 and a2 b2 = c2 , for ai , bi , ci ∈ R

that are true in R. Denote the set obtained by Π(R). A model of Π(R) is ¯ = {¯ a field that contains a subset R a | a ∈ R} whose elements satisfy the equalities ¯2¯b2 = c¯2 a ¯1 + ¯b1 = c¯1 and a

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Chapter 7. Ultraproducts

¯ is a whenever the corresponding equalities of (1) are true in R. That is, R homomorphic image of R. ¯ is an isomorphic copy of K. Thus, a model If R = K is a field, then K of Π(K) is, (up to an isomorphism) a field containing K. Example 7.3.2: Irreducible Polynomials. Let R be an integral domain. An elementary statement about models of Π(R) is a mathematical statement that applies to each member of Mod(Π(R)) and for which there exists a sentence θ of L(ring, R) which is true in any given model F if and only if the statement is true. Consider, for example, a polynomial f (X1 , . . . , Xn ) of degree d with coefficients in R. Then “f (X) is irreducible” is an elementary statement about models of Π(R). Indeed, it is equivalent to the conjunction of the statements “there exist no polynomials g, h of degree d1 and d2 respectively such that f (X) = g(X)h(X),” where (d1 , d2 ) runs over all pairs of positive integers with d1 + d2 = d. Rewrite the phrase “there exists no polynomial g(X1 , . . . , Xn ) of degree d1 ”, as “¬(∃u1 ) · · · (∃uk )” where u1 , . . . , uk are variables for the coefficients of g(X). A system of equalities between corresponding coefficients on both sides of “=” replaces “f (X) = g(X)h(X).” Similarly, we may consider a polynomial f (u, X1 , . . . , Xn ) =

X

ui X1i1 · · · Xnin

with intermediate coefficients ui . The same argument as above gives a formula ϕ(u) in L(ring) such that for each field K and all tuples a with entries in K, the polynomial f (a, X) is irreducible in K[X] if and only if ϕ(a) is true in K. Two structures A = hA, Ri , Fj , ck i and B = hB, Si , Gj , dk i of the language L = L(µ, ν, K) are isomorphic, if there exists a bijective function f : A → B such that (2a) (a1 , . . . , aµ(i) ) ∈ Ri ⇐⇒ (f (a1 ), . . . , f (aµ(i) )) ∈ Si , for each i ∈ I; (2b) f (Fj (a1 , . . . , aν(j) )) = Gj (f (a1 ), . . . , f (aν(j) )), for each j ∈ J; and (2c) f (ck ) = dk , for each k ∈ K. In this case write A ∼ = B. The structures A and B are elementarily equivalent if A |= θ ⇐⇒ B |= θ for every sentence θ of L. If this is the case, we write A ≡ B. Clearly, if A ∼ = B, then A ≡ B. But we will have many examples that show the converse is false. Two fields L and L0 that contain a field K are isomorphic as models of Π(K) if and only if there exists a field isomorphism of L onto L0 that fixes every element of K: write L ∼ =K L0 . If, however, they are elementarily equivalent as models of Π(K), we write L ≡K L0 . Call A a substructure of B, (A ⊆ B, and B is an extension of A) if A ⊆ B, Ri = Aµ(i) ∩ Si for each i ∈ I, Fj (a1 , . . . , aν(j) ) = Gj (a1 , . . . , aν(j) ) for each j ∈ J and all a1 , . . . , aν(j) ∈ A; and ck = dk for each k ∈ K.

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137

More generally, an embedding of A into B is an injective map f : A → B that satisfies Condition (2). Note that an arbitrary map f : A → B is an embedding of A into B if and only if for each quantifier free formula ϕ(X1 , . . . , Xn ) of L and for all a1 , . . . , an ∈ A, the condition A |= ϕ(a1 , . . . , an ) implies B |= ϕ(f (a1 ), . . . , f (an )). Indeed, an application of the latter condition to the formula X1 6= X2 implies that f is injective. Suppose now that A ⊆ B. We say A is existentially closed in B if for each quantifier free formula ϕ(X1 , . . . , Xn ) and for all b1 , . . . , bn ∈ B with B |= ϕ(b1 , . . . , bn ) there exist a1 , . . . , an ∈ A with A |= ϕ(a1 , . . . , an ). Call A an elementary substructure of B and B an elementary extension of A (in symbols A ≺ B) if A ⊆ B and if for each formula ϕ(X1 , . . . , Xn ) of L and for every a1 , . . . , an in A, the truth of ϕ(a1 , . . . , an ) in A is equivalent to its truth in B. It follows, in particular, that a sentence θ of L is true in A if and only if it is true in B, (i.e. A ≡ B). The converse is false (Example 7.3.3). If, however, A ⊆ B then “A ≡ B as models of L(A)” is equivalent to “A ≺ B as models of L”. Transitivity of elementarily equivalence follows immediately: A ≺ B, B ≺ C implies A ≺ C. In addition, A ⊆ B ≺ C and A ≺ C imply A ≺ B. Example 7.3.3: Elementary subfields. If a field K is an elementary subfield of a field F , then F is a regular extension of K. In other words, K is algebraically closed in F and F/K is separable (Lemma 2.6.4). ˜ ∩ F and f = irr(x, K). Then the sentence First of all let x ∈ K (∃X)[f (X) = 0] holds in F , so also in K. Therefore, deg(f ) = 1, hence ˜ ∩ F = K. x ∈ K. Consequently, K Pn 1/p bn ∈ F with i=1 bi ui = 0. Then, Pn Letp now u1 , . . . , un ∈ K and b1 , . . . ,P n p i=1 bi ui = 0. Hence, (∃X1 ) · · · (∃Xn ) i=1 Xi ui is true in F , so also in K. Pn 1/p In other words, there exist a1 , . . . , an ∈ K with i=1 ai ui = 0. Therefore, 1/p over K. Consequently, F/K is separable. F is linearly disjoint from K For example, let x be an indeterminate. Then, Q(x2 ) ∼ = Q(x). Hence, 2 Q(x ) ≡ Q(x). But Q(x) is a proper algebraic extension of Q(x2 ). Therefore, Q(x2 ) is not an elementary subfield of Q(x).

7.4 Elementary Substructures We develop criteria for one structure to be an elementary substructure of another. Let m be a cardinal number. Consider a transfinite sequence {Aα | α < m} of structures for a language L = L(µ, ν, K) with Aα = hAα , RS αi , Fαj , cαk i. Suppose Aα ⊆ Aβ for each α ≤ β < m and define the union α<m Aα to S = hA , R , F , c i with A = A be the structure A m m mi mj mk m α , Rmi = α<m S α<m Rmi , Fmj (x1 , . . . , xν(j) ) = Fαj (x1 , . . . , xν(j) ) if x1 , . . . , xν(j) ∈ Aα , and cmk = c0k . Then Aα ⊆ Am for each α < m.

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Lemma 7.4.1: (a) If Aα ≺ Aβ for each α ≤ β < m, then Aα ≺ Am for each α < m. (b) If B is another structure of L such that Aα ≺ B for each α < m, then Am ≺ B. Proof of (a): Use an induction on structure to prove that for each formula ϕ(X1 , . . . , Xn ) of L, for each α < m, and for every x1 , . . . , xn ∈ Aα Aα |= ϕ(x) ⇐⇒ Am |= ϕ(x).

(1)

Proof of (b): Let ϕ(X1 , . . . , Xn ) be a formula of L and let x1 , . . . , xn ∈ Am . Then there exists α < m such that x1 , . . . , xn ∈ Aα . It follows from (1) that Am |= ϕ(x) ⇐⇒ Aα |= ϕ(x) ⇐⇒ B |= ϕ(x).

Proposition 7.4.2 (Skolem-L¨ owenheim): Let L = L(µ, ν, K) be a countable language, let B = hB, Si , Gj , dk i be a structure of L and let A0 be a countable subset of B. Then B has a countable elementary substructure A = hA, Ri , Fj , ck i such that A0 ⊆ A. Proof: We construct an ascending chain of countable sets A0 ⊆ A1 ⊆ A2 ⊆ · · · ⊆ B. Suppose that An has already been constructed, then An+1 consists of all dk ’s with k ∈ K; the Gj (y1 , . . . , yν(j) ) for all j ∈ J and ν(j)-tuple (y1 , . . . , yν(j) ) of elements of An ; and an element xm ∈ B such that B |= ϕ(x1 , . . . , xm ) for each formula ϕ(X1 , . . . , Xm ) and for every x1 , . . . , xm−1 ∈ An satisfying B |= (∃Xm )[ϕ(x1S, . . . , xm−1 , Xm )]. ∞ Define A as follows: A = n=1 An , Ri = Aµ(i) ∩ Si , Fj (y1 , . . . , yµ(j) ) = Gj (y1 , . . . , yµ(j) ) for y1 , . . . , yµ(j) ∈ A, and ck = dk . By the choice of the xn ’s above, the function Fj is well defined. Hence, A is a countable substructure of B. Now use an induction on structure to prove that for each formula ϕ(X1 , . . . , Xn ) of L and for each x ∈ An , A |= ϕ(x) if and only if B |= ϕ(x). This proves that A ≺ B

7.5 Ultrafilters A filter on a set S is a nonempty family D of subsets of S t which are ”big” in a sense made precise by the following condition: (1a) ∅ ∈ / D. (1b) If A, B ∈ D, then A ∩ B ∈ D. (1c) If A ∈ D and A ⊆ B ⊆ S, then B ∈ D. If, in addition, (2) For each A ⊆ S either A ∈ D or S r A ∈ D, then D is an ultrafilter. In this case D also satisfies (3) A ∪ B ∈ D implies A ∈ D or B ∈ D.

7.6 Regular Ultrafilters

139

Example 7.5.1: (a) The family of all cofinite subsets of S (i.e. those subsets whose complements are finite) is a filter of S. (b) The family Da of all subsets of S that contain a given element a of S is an ultrafilter on S, called a principal ultrafilter. From (3), an ultrafilter D is principal if and only if it contains a finite set. A family D0 of subsets of S satisfies the finite intersection property if A1 , . . . , An ∈ D0 implies A1 ∩ · · · ∩ An 6= ∅. If one adds to D0 all the sets B ⊆ S that contain finite intersections A1 ∩ · · · ∩ An of elements of D0 , then one obtains a filter D1 . By Zorn’s Lemma there exists a maximal filter D of S that contains D1 . Our next Lemma says that D is an ultrafilter. Lemma 7.5.2: A filter D on a set S is an ultrafilter if and only if it is maximal. Proof: Suppose D is maximal and let A ⊆ S. Assume that S r A 6∈ D. Then D ∪ {A} has the finite intersection property. Indeed, for D1 , . . . , Dn ∈ D let D = D1 ∩ · · · ∩ Dn . If D ∩ A = ∅, then D ⊆ S r A. Hence, S r A ∈ D, a contradiction. By the comment above there exists a filter D0 on S containing D ∪ {A}. By the maximality of D, A ∈ D. Thus, D is an ultrafilter. The converse is clear. Corollary 7.5.3: Every family D0 of subsets of S that satisfies the finite intersection property is contained in an ultrafilter. A somewhat stronger assumption implies the existence of nonprincipal ultrafilters. Lemma 7.5.4: Let D0 be a family of subsets of a set S that has the following property: If A1 , . . . , An ∈ D0 , then A1 ∩· · ·∩An is an infinite set. Then there exists a nonprincipal ultrafilter D on S that contains D0 . Proof: The family D1 that consists of all subsets of S that contain a set in D0 and all cofinite subsets of S has the finite intersection property. Choose an ultrafilter D that contains D1 ; it is a nonprincipal ultrafilter.

7.6 Regular Ultrafilters The family F of all finite subsets of an infinite set S has properties dual to those of a filter: (1a) S 6∈ F; (1b) A, B ∈ F implies A ∪ B ∈ F; and (1c) B ∈ F and A ⊆ B imply A ∈ F. Later we will work with another family that satisfies the same conditions the family of zero sets of a measure space S. Therefore we give both cases a unified treatment. Let F be a nonempty family of subsets of S that satisfies (1). Call the elements of F small sets.

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Define Boolean polynomials in Z1 , . . . , Zm recursively: The variables Z1 , . . . , Zm are Boolean polynomials, and if U, U1 , U2 are Boolean polynomials, then U 0 , U1 ∪ U2 , and U1 ∩ U2 are Boolean polynomials. Evaluate a Boolean polynomial P (Z1 , . . . , Zm ) at subsets A1 , . . . , Am of S by interpreting the symbols ∪, ∩, and 0 as union, intersection, and taking the complement, respectively. With addition and multiplication given by A + B = (A r B) ∪ (B r A),

A · B = A ∩ B,

0 = ∅,

1 = S,

the family of all subsets of S becomes an algebra over the field F2 , in which each element except 1 is a zero divisor. The family F of small sets is an ideal of this algebra. Two subsets A and B of S are congruent modulo F in this algebra if and only if they differ from each other by a small set, (i.e. (A r B)∪(B r A) ∈ F). We then say that A and B are almost equal, and write A ≈ B. Clearly, if A1 ≈ B1 and A2 ≈ B2 , then S r A1 ≈ S r B1 , A1 ∪ A2 ≈ B1 ∪ B2 , and A1 ∩ A2 ≈ B1 ∩ B2 . Thus, in general, if P (Z1 , . . . , Zm ) is a Boolean polynomial in Z1 , . . . , Zm and Ai ≈ Bi , i = 1, . . . , m, are subsets of S, then P (A1 , . . . , Am ) ≈ P (B1 , . . . , Bm ). If the difference A r B of two subsets A, B of S is a small set, then we say that A is almost contained in B. A family of subsets of S which is closed under unions, intersections and taking complements is called a Boolean algebra of sets. The Boolean algebra generated by a family A0 of subsets of S is the intersection of all Boolean algebras of families of S that contain S. It consists of all expressions P (A1 , . . . , Am ), where P (Z1 , . . . , Zm ) is a Boolean polynomial and A1 , . . . , Am belong to A0 . Denote this family by A. The family A0 of all subsets of S which are almost equal to a set in A is a Boolean algebra that contains both A and F. A0 is the Boolean algebra generated by A0 and F. Call an ultrafilter D on S regular (with respect to F) if it contains no small set. In particular, if A ∈ D and B ≈ A, then B ∈ D. For example, nonprincipal ultrafilters on S are regular with respect to the family of finite subsets of S. Lemma 7.6.1: Let S be a set and let F be a family of small subsets of S. Suppose that a family D0 of subsets of S satisfies (2) A1 , . . . , An ∈ D0 implies A1 ∩ · · · ∩ An is not a small set. Then there exists a regular ultrafilter D on S that contains D0 . Proof: Repeat the proof of Lemma 7.5.4.

We will apply the next result to model theoretic results for families of fields.

7.7 Ultraproducts

141

Proposition 7.6.2 ([Ax2, p. 265]): Let S be a set, F a family of small subsets of S, A a Boolean algebra of subsets of S that contains F, and C a subset of S. Suppose C ∈ / A. Then there exist two regular ultrafilters D and / D0 . D0 such that D ∩ A = D0 ∩ A but C ∈ D and C ∈ Proof: Denote the collection of all A ∈ A that almost contain C or almost contain S r C by A0 . Suppose Ai ∈ A0 almost contains C for i = 1, . . . , m, and Bj ∈ A0 almost contains S r C, j = 1, . . . , n, and A1 ∩ · · · ∩ Am ∩ B1 ∩ · · · ∩ Bn ≈ ∅. Take complements to deduce that (S r A1 ) ∪ · · · ∪ (S r Am ) ∪ (S r B1 ) ∪ · · · ∪ (S r Bn ) ≈ S Each S r Ai is almost contained in S r C and each S r Bi is almost contained in C. Hence C ≈ (S r B1 ) ∪ · · · ∪ (S r Bn ), and therefore C ∈ A, a contradiction. Thus, A0 satisfies (2). By Zorn’s Lemma, A has a maximal subcollection A1 that contains A0 and has property (2). In particular, A1 is closed under finite intersections. Claim: The family A1 ∪ {C} satisfies (2). Indeed, if A1 , . . . , Am ∈ A1 and A1 ∩ · · · ∩ Am ∩ C ≈ ∅, then C is almost contained in S r A, where A = A1 ∩· · ·∩Am ∈ A. Hence, S r A ∈ A0 ⊆ A1 . But this is a contradiction, because A ∈ A1 . By Lemma 7.6.1 there exists a regular ultrafilter D on S that contains A1 ∪ {C}. Obviously D ∩ A contains A1 and satisfies (2). The maximality of A1 implies that D ∩ A = A1 . Similarly, there exists a regular ultrafilter D0 on S that contains A1 ∪ {S r C}. It also satisfies D0 ∩ A = A1 . Hence, D0 ∩ A = D ∩ A.

7.7 Ultraproducts From a given family of structures, ultraproducts allow us to create new structures which retain, sometimes in a particularly useful form, those elementary properties that hold for almost all structures in the family. Furthermore, those elementary properties that hold only for a small subfamily no longer hold in the new structures. In many cases we are able to establish simple criteria under which two such new models are elementarily equivalent (e.g. Lemma 20.3.3). This has been a successful route to the investigation of the elementary theory of many algebraic structures. To be more explicit, consider a language L = L(µ, ν, K) and a set S together with an ultrafilter D on S. Suppose that for each s ∈ S we are given a structure As = hAs , Ris , Fjs , cks i for Q L. We construct the ultraproduct As /D = A = hA, Ri , Fj , ck i, in the of As , s ∈ S, modulo D, denoted following way. Q Define an equivalence relation on the cartesian product s∈S As by a ∼ b ⇐⇒ {s ∈ S | as = bs } ∈ D.

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Q The domain A of A is the elements of As modulo this relation. For simplicity use representatives of the equivalence classes instead of the classes themselves. With this convention define Ri by (a1 , . . . , aµ(i) ) ∈ Ri ⇐⇒ {s ∈ S| (a1s , . . . , aµ(i),s ) ∈ Ris } ∈ D. Define Fj by the rule: Fj (b1 , . . . , bν(j) ) is the equivalence class of the function s 7→ Fjs (b1s , . . . , bν(j),s ). Similarly ck is the equivalence class of the function s 7→ cks . Check that Ri , Fj , and ck are well defined. An induction on the structure of terms shows that if t(X1 , . . . , Xn ) is a term of L and x1 , . . . , xn ∈ A, then (1)

{s ∈ S | t(x1 , . . . , xn )s = t(x1s , . . . , xns )} ∈ D.

This gives rise to the fundamental property of ultraproducts: Proposition 7.7.1 (Loˇs): Let D be an ultrafilter on S. Suppose for each for a language L(µ, ν, K). s ∈ S that As = hAs , Ris , Gjs , cks i is a structure Q Consider the ultraproduct A = hA, Ri , Gj , ck i = As /D. Then, for every formula ϕ(X1 , . . . , Xn ) and for every x1 , . . . , xn ∈ A: (2)

A |= ϕ(x1 , . . . , xn ) ⇐⇒ {s ∈ S | As |= ϕ(x1s , . . . , xns )} ∈ D

Proof: Do an induction on structure: Statement (2) follows easily for atomic formulas from (1) and the induction steps present no difficulty. We demonstrate the induction step for the existential quantifier. Suppose that (2) is true for ϕ and for each n-tuple of elements of A. Let x1 , . . . , xn−1 ∈ A and assume that S0 = {s ∈ S | As |= (∃Xn )[ϕ(x1s , . . . , xn−1,s , Xn )]} ∈ D. For each s ∈ S0 there exists xns ∈ As such that As |= ϕ(x1s , . . . , xns ). If s 6∈ S0 , let xns ∈ As be arbitrary. Let xn be the equivalence class defined by the xns ’s. Then the set {s ∈ S | As |= ϕ(x1s , . . . , xns )} contains S0 , hence belongs to D. By the induction assumption, A |= ϕ(x1 , . . . , xn ). Therefore, A |= (∃Xn )[ϕ(x1 , . . . , xn−1 , Xn )]. Corollary 7.7.2: If θ is a sentence of L, then A |= θ ⇐⇒ {s ∈ S | As |= θ} ∈ D. Example 7.7.3: Ultraproducts of fields. If the structures As is a field for each s ∈ S, then so is A. If each of them is algebraically closed, then so is A. But even if each of them is algebraic over a given field K, A may be not. For example, let S = N and let D be a nonprincipal ultrafilter on N. ˜ For each n ∈ N choose an element Q xn ∈ Q of degree greater than n over Q. Then let Kn = Q(xn ), K = Kn /D, and x be the equivalence class of

7.7 Ultraproducts

143

(x1 , x2 , x3 , . . .). For each n, almost all xi satisfy no equation of degree n over Q. Hence, by Corollary 7.7.2, so does x. Consequently, x is transcendental over Q. Ultraproducts also satisfy a saturation property. Let A be a structure with a domain A for a language L. Extend L to a language L(A) by adding a new constant symbol for each element of A. We say that A is ℵ1 -saturated if the following holds. If r(1) < r(2) < r(3) < · · · is an increasing sequence of positive integers and for each n ∈ N , ϕn (X1 , . . . , Xr(n) ) is a formula of L(A) such that (3)

A |= (∃X1 ) · · · (∃Xr(n) )

n ^

ϕt (X1 , . . . , Xr(t) ),

t=1

then there exist x1 , x2 , x3 , . . . in A such that A |= ϕn (x1 , . . . , xr(n) ) for each n ∈ N. Lemma 7.7.4: Let D be a nonprincipal ultrafilter on N. Suppose for each n ∈ N that An is aQstructure with a domain An for a language L. Then the ultraproduct A = An /D is ℵ1 -saturated. Proof: To simplify notation assume that rn = n in the preceding definition. Suppose (3) holds for each n ∈ N. Then Dn = {s ∈ N | As |= (∃X1 ) · · · (∃Xn )

n ^

ϕt (X1 , . . . , Xt )} ∈ D

t=1

for each n ∈ N. Clearly D1 ⊇ D2 ⊇ D3 ⊇ · · ·. Since D is nonprincipal, Dn0 = Dn r{1, 2, . . . , n} ∈ D. Also D10 , D20 , D30 , . . . is a decreasing sequence with empty intersection. 0 Now define x1 , x2 , x3 , . . . in A asV follows. If s ∈ Dn0 r Dn−1 , choose n x1s , . . . , xns in As such that As |= t=1 ϕt (x1s , . . . , xts ). Thus, for each n ∈ N, xns is well defined for all s ∈ Dn0 . For s ∈ N r Dn0 choose xns ∈ As arbitrarily. From this definition, for each S∞ n ∈ N, 0the set {s0 ∈ N | As |= ϕn (x1s , . . . , xns )} contains the union p=n (Dp0 r Dp−1 ) = Dn . Therefore Proposition 7.7.1 gives A |= ϕn (x1 , . . . , xn ). If all structures As are the same, say As = A, then the ultraproduct As /D, denoted by AS /D, is called the ultrapower of A to S modulo D. Denote the domain of A by A. Consider the diagonal embedding of A into AS . That is, map a ∈ A onto the constant function as = a. This gives a canonical injective map of A into AS /D. Indeed, if the images of two elements a and b of A are equal, then the set {s ∈ S | as = bs } belongs to D and is therefore nonempty. It follows that a = b. We identify A with its image to conclude from Proposition 7.7.1 the following result: Q

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Proposition 7.7.5: If D is an ultrafilter on a set S and A is a structure for a language L, then A is an elementary substructure of AS /D. The compactness theorem of model theory is now an easy corollary. Proposition 7.7.6 (The Compactness Theorem): Let T be a set of sentences in a first order language L. If each finite subset of T has a model, then T has a model. Proof: Denote the collection of all finite subsets of T by F . For each Φ ∈ F let DΦ = {Φ0 ∈ F | Φ ⊆ Φ0 }. Then, DΦ ∩ DΦ0 = DΦ∪Φ0 . Hence, the family D0 = {DΦ | Φ ∈ F } has the finite intersection property. By Corollary 7.5.3, there exists an ultrafilter DQon F that contains D0 . Choose a model, MΦ , for each Φ ∈ F . Then, M = MΦ /D is a model of T , because if θ ∈ T , then D{θ} ∈ D A similar construction characterizes existential closedness by ultraproducts: Proposition 7.7.7: Let A ⊆ B be structures of a language L with domains A ⊆ B. Then A is existentially closed in B if and only if there exists an ultrapower A∗ of A and there exists an embedding β: B → A∗ whose restriction to A is the canonical embedding A → A∗ . Proof: Suppose first that there exist A∗ and β as above. Let ϕ(X1 , . . . , Xn ) be a quantifier free formula in L(A) and let b1 , . . . , bn be elements of B such that B |= ϕ(b1 , . . . , bn ). Then, A∗ |= ϕ(β(b1 ), . . . , β(bn )). Hence, A∗ |= (∃X1 ) · · · (∃Xn )[ϕ(X1 , . . . , Xn )]. We conclude from Proposition 7.7.5 that there exist a1 , . . . , an ∈ A with A |= ϕ(a1 , . . . , an ). Conversely, suppose that A is existentially closed in B. Denote the constant symbol of L(B) associated with an element b ∈ B by ˜b. Let T be the set of sentences ϕ(˜b1 , . . . , ˜bn ) of L(B) such that ϕ(X1 , . . . , Xn ) is a quantifier free formula of L(A) and b1 , . . . , bn are elements of B with B |= ϕ(b1 , . . . , bn ). Denote the collection of all finite subsets of T by F . For each Φ ∈ F let DΦ = {Φ0 ∈ F | Φ ⊆ Φ0 }. Then DΦ ∩ DΦ0 = DΦ∪Φ0 , hence the family D0 = {DΦ | Φ ∈ F } has the finite intersection property. By Corollary 7.5.3 there exists an ultrafilter D on F that contains D0 . Consider an element Φ in F . List the elements of Φ as ϕi (˜bi1 , . . . , ˜bin ), where ϕi (X1 , . . . , Xn ) is a quantifier free formula of L(A) and bi1 , . . . , bin are elements of B such that B |= ϕi (bi1 , . . . , bin ), i = 1, . . . , m. For each i choose new variable symbols Xi1 , . . . , Xin and consider the quantifier free formula m ^

ϕi (Xi1 , . . . , Xin ) ∧

^

[Xij = Xkl ]

i=1

of L(A), where in the second Vm conjunct, i, j, k, lVrange over all indices such that bij = bkl . Since B |= i=1 ϕi (bi1 , . . . , bin ) ∧ [bij = bkl ] there exist bΦ ij ∈ A V Φ Vm Φ Φ Φ such that A |= i=1 ϕi (bi1 , . . . , bin ) ∧ [bij = bkl ]. Extend A to a structure

7.8 Regular Ultraproducts

145

˜ AΦ of L(B) by choosing bΦ ij for the constant symbol bij , for all i and j. If b is an element of B which is different from all bij , choose an arbitrary a ∈ A for ˜b. Then AΦ is well defined and each of the sentences ϕi (˜bi1 , . . . , ˜bin ) is valid in AΦ . Q Finally, consider the ultraproduct AΦ /D with domain A∗ = AF /D. ∗ ∗ For each b ∈ B let b be the element of A that corresponds to the constant symbol ˜b. Put A∗ = AF /D. Then the map β: B → A∗ that maps b onto b∗ is an embedding of B into A∗ whose restriction to A is the canonical map A → A∗ . Indeed, consider a quantifier free formula ϕ(X1 , . . . , Xn ) of L(A) and elements b1 , . . . , bn ∈ B with B |= ϕ(b1 , . . . , bn ). Then AΦ |= Φ Φ each Φ in F that contains ϕ. Hence, by Corollary 7.7.2, ϕ(b Q 1 , . . . , bn ) for AΦ /D |= ϕ(b∗1 , . . . , b∗n ). Consequently, A∗ |= ϕ(b∗1 , . . . , b∗n ). A variation of the proof of Proposition 7.7.7 proves the following characterization of elementary embedding: Proposition 7.7.8 (Scott): An embedding α: A → B of structures is elementary if and only if there exists an ultrapower A∗ of A and an elementary embedding β: B → A∗ such that β ◦ α: A → A∗ is the canonical embedding. Likewise a lemma of Frane [Bell-Slomson, p. 161] says that A ≡ B if and only if B is elementarily embeddable in an ultrapower of A. A theorem of Shelah supersedes Frane’s lemma: A ≡ B if and only if there exists a set I and an ultrafilter D on I such that AI /D ∼ = BI /D [Shelah].

7.8 Regular Ultraproducts Regular ultraproducts generalize non-principal ultraproducts. They appear ˜ in Chapter 20 as ultraproducts of fields Q(s) with sets of measure 0 replacing finite sets of σ’s. The general model theoretic notions that we develop in this section will be applied in Chapter 20 to investigate the model theory of the ˜ Q(σ)’s. Let S be a set equipped with a family F of small subsets. Suppose that for each s ∈ S, As is a structure for a fixed language L. The truth set of a sentence θ of L is defined to be the following subset of S: A(θ) = {s ∈ S | As |= θ}. The map θ 7→ A(θ) preserves the Boolean operations: A(θ1 ∨ θ2 ) = A(θ1 ) ∪ A(θ2 ), A(θ1 ∧ θ2 ) = A(θ1 ) ∩ A(θ2 ), A(¬θ) = S r A(θ). More generally, if P (Z1 , . . . , Zm ) is a Boolean polynomial, then (1)

A(P (θ1 , . . . , θm )) = P (A(θ1 ), . . . , A(θm )),

where P (θ1 , . . . , θm ) is obtained from P (Z1 , . . . , Zm ) by first replacing ∪, ∩, and 0 by ∨, ∧, and ¬, respectively, and then substituting θ1 , . . . , θm for the Z1 , . . . , Zm .

146

Chapter 7. Ultraproducts

Let T be the theory of all sentences θ of L that are true in As for almost all s ∈ S (i.e. for all s ∈ S excluding a small subset). Q If D is a regular ultrafilter, we say that As /D is a regular ultraproduct. Proposition 7.8.1: (a) A sentence θ of L is in T if and only if it is true in every regular ultraproduct of the As ’. (b) Every model of T is elementarily equivalent to a regular ultraproduct of the As ’. Proof of (a): If θ belongs to T , then, by Corollary 7.7.2, θ is true in every regular ultraproduct of the As ’. Conversely, if θ ∈ / T , then A(¬θ) is not small. Hence, by Lemma 7.6.1, there exists a regular ultrafilter Q D on S which contains A(¬θ). Therefore, by Corollary 7.7.2, θ is false in As /D. Proof of (b): Let A be a model of T . Then, by (1), D0 = {A(θ) | A |= θ} is closed under finite intersections. Observe that no A(θ) ∈ D0 is small, since this would mean then ¬θ ∈ T , so A |= ¬θ. By Lemma 7.6.1, there existsQa regular ultrafilter D on S which contains D0 . By Corollary 7.7.2, A ≡ As /D. In concrete situations we often seek, in addition to the above data, a special set Λ0 of sentences of L with this property: (2) If A and A0 are models of T , then A ≡ A0 if and only if A and A0 satisfy the same sentences of Λ0 . Call Λ0 a set of basic test sentences. Every Boolean combination of basic test sentences is called a test sentence. Denote the Boolean algebra generated by {A(λ) | λ ∈ Λ0 } ∪ F by Λ. From (2), if D and D0 are two regular ultrafilters on S, then (3)

Y

As /D ≡

Y

As /D0 ⇐⇒ D ∩ Λ = D0 ∩ Λ.

Proposition 7.8.2: Suppose that Λ0 is a set of basic test sentences. Then for each sentence θ of L there exists a test sentence λ such that (a) A(θ) ≈ A(λ), and (b) θ ↔ λ belongs to T . Proof: We have only to prove that A(θ) ∈ Λ. Assume A(θ) does not belong to Λ. By Proposition 7.6.2, there exist regular ultrafilters D and D0 of S such 0 ∩ Λ, but A(θ) ∈ D and A(θ) ∈ / D0 . This contradicts (3), that D Q∩ Λ = D Q 0 since As /D ≡ As /D .

Exercises

147

7.9 Nonprincipal Ultraproducts of Finite Fields We return to the starting point and the motivation of this book, the theory of finite fields. Let S be a countable set. Define small sets as the finite subsets of S. For each s ∈ S let Fs be a finite field. Suppose (1) for each n ∈ N there are only finitely many s ∈ S such that |Fs | ≤ n. A concrete example arises by letting S be the collection of prime divisors of a global field K. The finite fields are the corresponding residue fields. Every finite field is perfect. That is, for every prime p, every Fs satisfies the following sentence of L(ring): (2)

p = 0 → (∀X)(∃Y )[Y p = X].

By Section 1.5: (3)

ˆ Gal(Fs ) ∼ =Z

for each s ∈ S. Also, by Theorem 5.4.2(a), and (1), given positive integers d and n, the following statement is true for almost all s ∈ S: (4) For every absolutely irreducible polynomial f ∈ Fs [X, Y ] of degree d there are n distinct points (xi , yi ) ∈ Fs × Fs with f (xi , yi ) = 0, i = 1, . . . , n. In Proposition 20.4.4 we will display a sequence Π1 of sentences of ˆ L(ring) such that a field F is a model of Π1 , if and only if Gal(F ) ∼ = Z. Also, there are sentences πd,n such that F |= πd,n if and only if F satisfies (4) (Proposition 11.3.2). We will be able then to prove the following result: Q Proposition 7.9.1: Suppose S satisfies (1) and F = Fs /D is a nonprinˆ and for every cipal ultraproduct. Then F is a perfect field, Gal(F ) ∼ = Z nonconstant absolutely irreducible polynomial f ∈ F (X, Y ) there exist infinitely many points (x, y) ∈ F × F with f (x, y) = 0.

Exercises 1. Consider the language L which has only one relation symbol, 0. Finally, certain long summations and long multiplications are primitive recursive. If Pg(x1 , . . . , xn , y) is a primitive Q recursive function, then both of the functions y γ. It follows that w(x − σx) > γ. Since γ is arbitrary, x = σx. Consequently, x ∈ Kv . ˜ Proof of (d): Assume without loss that char(K) = p > 0. Let x ∈ K × ˜ and let γ ∈ w(K ). Then there are a power q of p and an a ∈ Ks with b ∈ Ks× with v(b) > max(qγ − w(x), v(a)). xq = a. Use (a) to choose Qq q Then Pq X − bX − a =q i=1 (X − xi ) with x1 , . . . , xq ∈ Ks . It follows that i=1 w(x−xi ) = w(x −bx−a) = w(bx) > qγ. Hence, there exists i between ˜ 1 and q with w(x − xi ) > γ. Consequently, Ks is w-dense in K. Proposition 11.5.3 (Prestel): Let K be a PAC field and let w be a valua˜ Then K is w-dense in K. ˜ tion of K. ˜ is Proof: By Lemma 11.5.2(b) and Corollary 11.2.5 the w-closure of K in K ˜ and prove that a PAC field. Thus, we may assume that K is w-closed in K ˜ K = K. To this end, let f ∈QK[X] be a separable irreducible polynomial of degree n ˜ Consider n ≥ 1 and let f (X) = i=1 (X − xi ) be its factorization in K[X]. × × ˜ γ ∈ Γ = v(K ) and choose c ∈ K such that w(c) ≥ nγ (Lemma 11.5.2(a)). ˜ ] to a linear Apply Lemma 2.3.10(b) (Eisenstein’s criterion) over the ring K[Y 2 factor of f (Y ) to deduce that f (X)f (Y ) − c is absolutely irreducible. Since K is PAC, there exist x, y ∈ K such that f (x)f (y) = c2 . It follows from w(f (x)) + w(f (y)) = 2w(c) that w(f (x)) ≥ nγ or Pnw(f (y)) ≥ nγ. Suppose for example that the first possibility occurs. Then i=1 w(x − xi ) ≥ nγ. Hence, there is an i with w(x − xi ) ≥ γ. Since {x1 , . . . , xn } is a finite set, the latter conclusion implies that there exists i such that for each γ ∈ Γ there exists an x ∈ K with w(x − xi ) ≥ γ. Thus, xi belongs to the w-closure of K, which is, by assumption, K itself.

11.5 PAC Fields with Valuations

205

Since f is irreducible, n = 1. Therefore, K = Ks . By Lemma 11.5.2(d), K ˜ Consequently, K = K. ˜ is w-dense in K. Remark 11.5.4: Density. Exploiting Prestel’s trick in the proof of Proposition 11.5.3, J´anos Koll´ ar proves in [Koll´ ar2, Thm. 2] that if K is a PAC field, ˜ and V is a variety over K, Then V (K) is w-dense in w is a valuation of K, ˜ V (K). Corollary 11.5.5 (Frey-Prestel): The Henselian closure Kv of a PAC field ¯ v is separably closed K with respect to a valuation v is Ks . Consequently, K × and v(K ) is a divisible group. Proof: Let w be an extension of v to Ks . By Proposition 11.5.3, K is wdense in Ks . By Lemma 11.5.2(c), the w-closure of K in Ks is contained in Kv . Hence, Kv = Ks . Now recall that K and Kv have the same residue field and the same ¯ v [X] of positive value group. Consider a monic separable polynomial f¯ ∈ K ¯ degree. Lift f to a monic polynomial f ∈ Ov [X]. Then f is separable, so has ¯v. a root in Kv . The residue of this root is a root of f¯ in K × Next consider a ∈ K and a positive integer n which is not divisible by char(K). Choose x ∈ Kv with xn = a. Then n1 v(a) ∈ v((Kv )× ) = v(K × ). Finally, suppose char(K) = p > 0 and v(a) < 0. Then there is an x ∈ Kv with xp − x − a = 0. Hence, p1 v(a) = v(x) ∈ v(K × ). Consequently, v(K × ) is divisible. Corollary 11.5.6: Let K be a PAC field which is not separably closed. Then K is Henselian with respect to no valuation. Denote the maximal Abelian, nilpotent, and solvable extensions of Q by Qab , Qnil and Qsolv , respectively. Corollary 11.5.7 ([Frey]): The fields Qab and Qnil are not PAC fields. Proof: By Corollary 11.2.5 we have only to prove the statement for Qnil . For ˜ ∩ Qp is Henselian. Assume Qnil is PAC. Then Qnil Qp,alg is each p, Qp,alg = Q ˜ PAC (Corollary 11.2.5) and Henselian. By Corollary 11.5.5, Qnil Qp,alg = Q. ∼ Hence, Gal(Qp,alg ) = Gal(Qnil ∩ Qnil Qp,alg ) is pronilpotent. That is, the Galois group of every irreducible separable polynomial over Qp,alg is nilpotent. In particular, this is true for p = 5. However, X 3 + 5 is irreducible over Q5 (e.g. by Eisenstein’s criterion) and its discriminant −27 · 52 [Lang7, p. 270] is not a square in Q5,alg (e.g. −27 is a quadratic non-residue modulo 5). Hence, X 3 + 5 is irreducible over Q5,alg and its discriminant is not a square in Q5,alg . Therefore, Gal(X 3 + 5, Q5,alg ) is S3 which is not nilpotent. We conclude from this contradiction to the preceding paragraph that Qnil is not a PAC field. ˜ for each prime In contrast to Qab and Qnil , we have Qsolv Qp,alg = Q number p.

206

Chapter 11. Pseudo Algebraically Closed Fields

˜ Proposition 11.5.8: The Henselian closure of each valuation of Qsolv is Q. Proof: Let v be a valuation of Qsolv . Replacing v by an equivalent valuation, we may assume that v|Q = vp for some prime number p. Thus, Qsolv can be ˜ p such that v is the restriction to Qsolv of the unique extension embedded in Q ˜ ˜ of vp to Qp . Let Qp,alg = Q∩Q p and M = Qsolv Qp,alg . Then M is a Henselian ˜ closure of M at v. Assume M 6= Q. Since Gal(Qp ) is prosolvable [Cassels-Fr¨ohlich, p. 31, Cor. 1], so is Gal(M ). Since √ M contains all roots of unity, there exist b ∈ M and n ∈ N / M . If a ∈ M is sufficiently vp -close to M , then by Krasner, such that n b ∈ √ √ n M ( b) ⊆ M ( n a) [Jarden14, Lemma Pm 12.1]. Choose w1 , . . . , wm ∈ Qsolv and b1 , . . . , bm ∈ Qp,alg such that b = i=1 bi wi . Since Q is vp -dense P in Qp , we m may choose a1 , . . . , ap which are vp -close to b1 , . . . , bp . Then a = i=1 ai wi √ √ n is v-close to b and lies in Qsolv . It follows that n a ∈ Qsolv ⊆ M , so b ∈ M . ˜ We conclude from this contradiction that M = Q. Thus, Corollary 11.5.5 fails to solve the following problem. Problem 11.5.9: (a) Is Qsolv a PAC field? (b) Does there exist an infinite non-PAC field K of a finite transcendence degree over its prime field such that K is not formally real and all of its Henselian closures are separably closed? Remark 11.5.10: On Problem 11.5.9(a). Problem 11.5.9(a) leads to other problems with respectable classical connections. Here is one example: By Theorem 11.2.3 we have only to check if each absolutely irreducible curve, f (X, Y ) = 0, with f ∈ Q[X, Y ] has a point with coordinates in Qsolv . Let E = Q(x, y) be the function field of this curve. Suppose there exists t ∈ E such ˆ is solvable over Q(t). that E/Q(t) is an extension whose Galois closure, E, Then, each specialization t → t0 such that t0 ∈ Q and x and y are integral over the corresponding local ring extends to a specialization (x, y) → (x0 , y0 ) with x0 , y0 ∈ Qsolv . Since there are infinitely many such specializations, f (X, Y ) = 0 certainly has infinitely many Qsolv -rational points. This idea fails, however, if the function field F = C(x, y) over C for a “general” curve, f (X, Y ) = 0, has no subfield C(t) with Gal(Fˆ /C(t)) solvable. The proper subfields of a “general” curve of genus g > 1 are of genus 0 [Fried7, p. 26-27]. Therefore, with no loss, assume that there are no proper fields between C(t) and F ; that is, Gal(Fˆ /C(t)) is a primitive solvable group. A theorem of Galois implies that [F : C(t)] = pr for some prime p [Burnside2, p. 202]. Combining [Fried7, p. 26] and [Ritt] one may prove that for F “general” of genus suitably large (> 6) that if r = 1 then it is impossible for Gal(Fˆ /C(t)) to be solvable. But higher values of r have not yet been excluded. Remark 11.5.11: On Problem 11.5.9(b). Theorem D of [Geyer-Jarden5] gives for each characteristic p (including p = 0) an example of an infinite

11.6 The Absolute Galois Group of a PAC Field

207

non-PAC field K of characteristic p which is not formally real and all of its Henselian closures of K are separably closed. This gives an affirmative answer to Problem 11.5.1(b) of [Fried-Jarden3]. The proof is based on results and ideas of [Efrat]. It uses Galois cohomology, valuations of higher rank, and the Jacobian varieties of curves. Problem 11.5.9(b) is a reformulation of the older problem.

11.6 The Absolute Galois Group of a PAC Field We show that the absolute Galois groups of PAC fields are “projective” (Theorem 11.6.2). All decidability and undecidability results on PAC fields depend on this result. Lemma 11.6.1: Let L/K be a finite Galois extension, B a finite group, and α: B → Gal(L/K) an epimorphism. Then there exists a finite Galois extension F/E with Gal(F/E) = B such that E is a regular finitely generated extension of K, F is a purely transcendental extension of L with trans.deg(F/L) = |B|, and α = resF/L . Proof: Let {y β | β ∈ B} be a set of indeterminates of cardinality |B|. Put 0 0 F = L(y β | β ∈ B). Define an action of B on F by (y β )β = y ββ and β0 α(β 0 ) 0 for β, β ∈ B and a ∈ L. Then let E be the fixed field of B a = a in F . By Galois theory, F is a Galois extension of E with Galois group B [Lang7, p. 264, Thm. 1.8] and α(β) = resF/L (β) for each β ∈ B. Moreover, F is a finitely generated extension of K. Hence, by Lemma 10.5.1, E is a finitely generated extension of K. By construction, E ∩ L = K and F is a ˜ over L. purely transcendental extension of L, so F is linearly disjoint from K ˜ Hence, EL is linearly disjoint from K over L. It follows from Lemma 2.5.3, ˜ over K; that is E is a regular extension of that E is linearly disjoint from K K. Let E be a finitely generated extension of a field K and F a finite Galois extension of E. By Remark 6.1.5, F/E has a Galois ring cover S/R. Thus, R = K[x1 , . . . , xm ] is an integrally closed domain with quotient field E; S is the integral closure of R in F , and S = R[z] where, if f = irr(z, E), then f 0 (z) is a unit of S (Definition 6.1.3). Every homomorphism ϕ0 of R onto ¯ extends to a homomorphism ϕ of S onto a Galois extension F¯ of a field E ¯ → Gal(F/E) such that ¯ The map ϕ induces an embedding ϕ∗ : Gal(F¯ /E) E. ¯ and x ∈ S (Lemma 6.1.4). ϕ ϕ∗ (σ)(x) = σ(ϕ(x)) for each σ ∈ Gal(F¯ /E) Theorem 11.6.2 ([Ax2, p. 269]): Let K be a PAC field, A and B finite groups, and ρ: Gal(K) → A and α: B → A epimorphisms. Then there exists a homomorphism γ: Gal(K) → B such that ρ = α ◦ γ (i.e. Gal(K) is projective). Proof (Haran): Denote the fixed field of Ker(ρ) in Ks by L. Then L is a finite Galois extension of K and ρ defines an isomorphism Gal(L/K) → A. Thus, we may identify A with Gal(L/K) and ρ with the restriction map.

208

Chapter 11. Pseudo Algebraically Closed Fields

Let E and F be as in Lemma 11.6.1. Let S/R be a Galois ring cover for F/E. Since K is a PAC field, there exists a K-homomorphism ϕ0 : R → K (Proposition 11.1.3). Let ϕ be an extension of ϕ0 to S which is the identity on L. Then M = ϕ(S) is a Galois extension of K which contains L and ϕ induces an embedding ϕ∗ : Gal(M/K) → Gal(F/E) such that resF/L ◦ ϕ∗ = resM/L . Compose ϕ∗ with the map resM : Gal(K) → Gal(M/K) to obtain the desired homomorphism γ: Gal(K) → B with ρ = α ◦ γ. There are non PAC fields K with Gal(K) projective (e.g. K is finite or K = C(t)). On the other hand, if G is a projective group, then there exists some PAC field K such that G ∼ = Gal(K) (Corollary 23.1.2). The projectivity of the absolute Galois group of a field K is closely related to the vanishing of the Brauer group Br(K) of K, although it is not equivalent to it. We survey the concept of the Brauer group and prove that Br(K) = 0 if K is PAC. A central simple K-algebra is a K-algebra A whose center is K and which has no two sided ideals except 0 and A. In particular, if D is a division ring with center K, then the ring Mn (D) of all n × n matrices with entries in D is a central simple K-algebra for each positive integer n [Huppert, p. 472]. Conversely, if A is a finite dimensional central simple K-algebra, then, by a theorem of Wedderburn, there exists a unique division ring D with center K and a positive integer n such that A ∼ =K Mn (D) [Huppert, p. 472]. Suppose A0 is another finite dimensional central simple K-algebra. Then A0 is equivalent to A if there exists a positive integer n0 such that A0 ∼ = Mn0 (D). In particular, A is equivalent to D. We denote the equivalence class of A by [A] and let Br(K) be the set of all equivalence classes of finite dimensional central simple K-algebras. The tensor product of two finite dimensional central simple K-algebras is again a finite dimensional central simple K-algebra [Weil6, p. 166]. Moreover, the tensor product respects the equivalence relation between finite dimensional central simple K-algebras. Hence, [A] · [B] = [A ⊗K B] is an associative multiplication rule on Br(K). Since A ⊗K B ∼ = B ⊗K A, multiplication in Br(K) is commutative. Further, the equivalence class [K] is a unit in Br(K), because A ⊗K K ∼ = A. Finally, let Ao be the opposite algebra of A. It consists of all elements ao , with a ∈ A. Addition and multiplication are defined by the rules ao + bo = (a + b)o and ao bo = (ba)o . One proves that A ⊗K Ao ∼ = Mn (K), where n = dimK (A). Thus, [Ao ] = [A]−1 . Therefore, Br(K) is an Abelian group. For each field extension L of K the map A 7→ A ⊗K L induces a homomorphism resL/K : Br(K) → Br(L). The kernel of resL/K consists of all [A] such that A ⊗K L ∼ =L Mn (L) for some positive integer n. If A satisfies the latter relation, then A is said to split over L. If L0 contains L, then A also splits over L0 . It is known that each A splits over Ks [Weil6, p. 167]. Thus, Br(Ks ) is trivial.

11.6 The Absolute Galois Group of a PAC Field

209

Construction 11.6.3: The reduced norm of a central simple algebra. Let A be a finite dimensional central simple algebra A over a field K. Choose a Ks -isomorphism α: A ⊗K Ks → Mn (Ks ) for some positive integer n. In particular, dimK A = dimKs (Mn (Ks )) = n2 . Let {eij | 1 ≤ i, j ≤ n} be a basis of A over K. Then eij = α(eij ⊗ 1), 1 ≤ i, j ≤ n, form Pna basis of Mn (Ks ) over Ks . Each a ∈ A has a unique presentation as a = i,j=1 aij eij with aij ∈ K. The matrix a = (aij )1≤i,j≤n satisfies α(a ⊗ 1) =

n X

aij eij = λkl (a)

1≤k,l≤n

,

i,j 2 where λkl are linear forms over s in the n variables Xij . Indeed, if eij = PK n (εij,kl )1≤k,l≤n , then λkl (X) = ij=1 εij,kl Xij , where X = (Xij )1≤i,j≤n . The reduced norm of a is defined by

(1)

red.norm(a) = det(α(a ⊗ 1)).

If α0 : A ⊗K Ks → Mn (Ks ) is another Ks -isomorphism, then α0 ◦ α−1 is a Ks isomorphism of Mn (Ks ). By Skolem-Noether [Weil5, p. 166, Prop. 4], α0 ◦α−1 is a conjugation by an invertible matrix of Mn (Ks ). Hence, det(α0 (a ⊗ 1)) = det(α(a⊗1)). Thus, red.norm(a) is independent of the particular choice of α. Moreover, each σ ∈ Gal(K) fixes red.norm(a). Therefore, red.norm(a) ∈ K. Indeed, σ induces an isomorphism 1 ⊗ σ −1 of A ⊗K Ks and an isomorphism σn of Mn (Ks ). Then α0 = σn ◦ α ◦ 1 ⊗ σ −1 : A ⊗K Ks → Mn (Ks ) is a Ks -isomorphism satisfying α0 (a ⊗ 1) = σn α(a ⊗ 1) for each a ∈ K. It follows that σ red.norm(a) = σ det(α(a ⊗ 1)) = det σn (α(a ⊗ 1)) = det α0 (a ⊗ 1) = red.norm(a), as claimed. Now let p(X) = det λkl (X) . It is a homogeneous polynomial of degree n over Ks such that p(a) ∈ K for each a ∈ Mn (K). It follows that the coefficients of p belong to K (Exercise 9). Next observe that the linear forms λkl are linearly independent over Ks because the n2 ×n2 matrix (eij )1≤i,j≤n is nonsingular. We may therefore form a change of variables Ykl = λkl (X). It maps p(X) onto det(Y), which is an absolutely irreducible polynomial. Hence, p(X) is also absolutely irreducible. We call p(X) the reduced form of A. Theorem 11.6.4: Let K be a PAC field. Then its Brauer group Br(K) is trivial. Proof ([Ax2, p. 269]): Assume Br(K) is nontrivial. Then there exists a division ring D with center K such that dimK (D) = n2 and n > 1. Let p(X) be the associated reduced form. Since p(X) is an absolutely irreducible polynomial (Construction 11.6.3), p(X) has a nontrivial zero a ∈ Mn (K) (Proposition Pn 11.1.1). In the notation of Construction 11.6.3 (with A = D), let a = i,j=1 aij eij . By (1), red.norm(a) = p(a) = 0. On the other hand, a

210

Chapter 11. Pseudo Algebraically Closed Fields

is a nonzero element of D, hence invertible. Therefore, α(a ⊗ 1) is a regular matrix, so red.norm(a) = det(α(a ⊗ 1)) 6= 0. This contradiction proves that Br(K) is trivial. Remark 11.6.5: Varieties of Severi-Brauer. An alternative proof of Theorem 11.6.4 uses varieties of Severi-Brauer. They are varieties V which are defined over a field K and are isomorphic over Ks to Pn for some positive integer n. There is a bijective correspondence between K-isomorphism classes of varieties V of Severi-Brauer and equivalence classes of finite dimensional central simple K-algebras A. If V has a K-rational point, then A splits over K [Jacobson, p. 113]. In particular, if K is PAC, this implies that Br(K) = 0. The connection between the projectivity of the absolute Galois group of a field K and its Brauer group is based on the canonical isomorphism (2)

H 2 (Gal(K), Ks× ) ∼ = Br(K)

[Deuring2, p. 56, Satz 1 or Serre4, §X5]. Here we assume that the reader is familiar with Galois cohomology, e.g. as presented in [Ribes1] or in [Serre9]. In particular, it follows from (2) that (3) every element of Br(K) has a finite order [Ribes1, p. 138, Cor. 6.7]. For each prime number p and a profinite group G the notation cdp (G) stands for the pth cohomological dimension of G. It is the maximal positive integer n such that H n (G, A)p∞ = 0 for each torsion G-module A. Finally, cd(G) = supp (cdp (G)) is the cohomological dimension of G. Proposition 11.6.6: The following conditions on a field K are equivalent: (a) Gal(K) is projective. (b) cd(Gal(K)) ≤ 1. (c) For each prime number p 6= char(K) and for each finite separable extension L of K, Br(L)p∞ is trivial. Proof of “(a) ⇐⇒ (b)”: Let p be a prime number. By [Ribes, p. 211, Prop. 3.1], cdp (Gal(K)) ≤ 1 if and only if for every finite Galois extension L of K and for every short exact sequence α

0 −→ (Z/pZ)m −→ B −→ Gal(L/K) −→ 1 there exists a homomorphism β: Gal(K) → B such that α ◦ β = resKs /L . By a theorem of Gruenberg (Corollary 22.4.3), the latter condition holds for all p if and only if Gal(K) is projective. Consequently, (a) and (b) are equivalent. Proof of “(b) ⇐⇒ (c)”: Let p be again a prime number. First suppose that p 6= char(K). Then cdp (Gal(K)) ≤ 1 if and only if Br(L)p∞ is trivial for every finite separable extension L of K [Ribes, p. 261, Cor. 3.7]. If p = char(K), then cdp (Gal(L)) ≤ 1 for every field L of characteristic p [Ribes, p. 256, Thm. 3.3]. Since the Brauer group of each field is torsion (by (3)), this establishes the equivalence of (b) and (c).

11.7 A non-PAC Field K with Kins PAC

211

Proposition 11.6.7 ([Ribes, p. 264, Prop. 3.10]): The following conditions on a field K are equivalent. (a) Br(L) is trivial for every finite separable extension L of K. (b) The norm map, norm: N × → L× , is surjective for every finite separable extension L of K and for every finite Galois extension N of L. We summarize consequences of the previous results for PAC fields: Corollary 11.6.8: The following statements hold for every PAC field K: (a) Gal(K) is projective. (b) Br(K) is trivial. (c) cd(Gal(K)) ≤ 1. (d) The map norm: N × → K × is surjective for each finite Galois extension N of K. Proof: Let L be a finite separable extension of K. By Corollary 11.2.5, L is PAC. Hence, by Theorem 11.6.4, Br(L) is trivial. Therefore, cd(Gal(K)) ≤ 1 (Proposition 11.6.6) and the norm map N × → K × is surjective for each finite Galois extension N of K. Example 11.6.9 (Geyer): We construct an example of a field K with Gal(K) projective, a finite Galois extension K 0 of K, and an element u of K which is not a norm of an element of K 0 . By Proposition 11.6.7, K has a finite separable extension L such that Br(L) 6= 0. This will show that it is impossible to omit the condition “p 6= char(K)” in Condition (c) of Proposition 11.6.6. We start from a transcendental element u over F2 and let K0 = F2 (u)s . Then choose an transcendental element t over K0 and let K = K0 (t). By Tsen’s Theorem, Gal(K) is projective [Ribes2, p. 276 or Jarden17, Thm. 1.1]. Consider the Artin-Schreier extension K 0 = K(x) of K with x2 + x + t = 0. Each element y of K 0 has the form y = v + wx with v, w ∈ K and (t) normK 0 /K (y) = (v + wx)(v + w(1 + x)) = v 2 + vw + w2 t. Write v = fh(t) and w=

g(t) h(t) ,

where f, g, h ∈ K0 [t] and h 6= 0. Let a (resp. b, c) be the leading

coefficient of f (resp. g, h). If normK 0 /K y = u, then f (t)2 +f (t)g(t)+g(t)2 t = h(t)2 u. Compare the leading coefficients of both sides of this equality. If deg(f ) > deg(g), then a2 = c2 u. If deg(f ) ≤ deg(g), then b2 = c2 u. In both cases we find that u is a square in F2 (u)s , which is not the case. This contradiction proves that u is not a norm of an element of K 0 .

11.7 A non-PAC Field K with Kins PAC Let L/K be a purely inseparable extension of fields. If K is PAC, then so is L (Corollary 11.2.5). Problem 12.4 of [Geyer-Jarden3] asks whether the converse is true. An example of Hrushovski shows that this is not the case. The main ingredient of this example is the analog of Mordell conjecture for function fields:

212

Chapter 11. Pseudo Algebraically Closed Fields

Proposition 11.7.1 (Grauert-Manin [Samuel, pp. 107 and 118]): Let K be a finitely generated regular extension of a field K0 and C a nonconstant ˜ is at least 2. Then C(K) curve over K/K0 . Suppose the genus of C over K is a finite set. Here we say that C is a nonconstant curve over K/K0 if C is defined ˜ to a curve C0 defined over K and if C is not birationally equivalent over K ˜ over K0 . Lemma 11.7.2: Let F = K(x1 , . . . , xn ) be a finitely generated extension of a field K of positive characteristic p. Suppose K is algebraically closed in F . Then ∞ \

(1)

k

k

K(xp1 , . . . , xpn ) = K.

k=1

Proof: Denote the left hand side of (1) by F0 . First suppose that K is perfect. Thus, K p = K, so F0 = F0p is also perfect. In addition F0 , as a subfield of F , is finitely generated over K (Lemma 10.5.1). Assume F0 is transcendental over K. Choose a transcendental basis t1 , . . . , tr with r ≥ 1. Then F0 has a finite degree over E = K(t1 , . . . , tr ). On the other hand, since 1/pm F0 is perfect, E(t1 ) is contained in F0 and has degree pm over E for each positive integer m. This contradiction proves that F0 is algebraic over K. Since K is algebraically closed in F , we conclude that F0 = K. In the general case Kins is a perfect field. Hence, by the preceding paragraph, F0 ⊆ F ∩

∞ \

k

k

Kins (xp1 , . . . , xpn ) = F ∩ Kins = K.

k=1

Therefore, F0 = K.

Lemma 11.7.3: Let K be a finitely generated regular transcendental extension of a field K0 of positive characteristic p. Let C be a curve which is ˜ 0 is at least 2. Let F be a finitely defined over K and whose genus over K K generated regular extension of K. Suppose C is a nonconstant curve over F/K0 . Then K has a finitely generated extension E which is contained in F such that F/E is a finite purely inseparable extension and C(K) = C(E). k

k

Proof: Let F = K(x1 , . . . , xn ), and for each k write Fk = K(xp1 , . . . , xpn ). ˜ 0 /K K ˜ 0 is a By Lemma 11.7.2, the intersection of all Fk is K. Since F K ˜ regular extension, the genus of C over F K0 is the same as the genus of C ˜ 0 (Proposition 3.4.2(b)), so at least 2. By Proposition 11.7.1, C(F ) is over K K a finite set. Hence, there exists a positive integer k such that C(Fk ) = C(K), so E = Fk satisfies the assertion of the Lemma.

11.7 A non-PAC Field K with Kins PAC

213

Remark 11.7.4: On M¨ obius transformations. Let K be a field and x an in obius determinate. To each matrix A = ac db in GL(2, K) we associate a M¨ transformation τA (also called a linear fractional transformation). It is the K-isomorphism of K(x) into K(x) defined by the following rule: (3)

τA (x) =

ax + b . cx + d

If B is another matrix in GL(2, K), then τA ◦ τB = τBA . If I is the unit matrix, then τI is the identity map of K(x). In particular, τA−1 ◦ τA = id, so τA is an automorphism of K(x)/K. If k ∈ K × , then τkI = id. Conversely, if τA = id, then, by (3), c(x0 )2 + (d − a)(x0 ) − b = 0 for all x0 ∈ K. Hence, c = b = 0 and d = a, so A = aI. Therefore, the kernel of the map A 7→ τA consists of the group of scalar matrices. If τ is arbitrary element of Aut K(x)/K , then K(x) = K(τ (x)). Hence, by Example 3.2.4, there exists A ∈ GL(2, K) such that τ (x) = τA (x). Thus, the map A 7→ τA defines an isomorphism (4)

PGL(2, K) ∼ = Aut K(x)/K .

Substituting elements of K ∪ {∞} in (3), we may also view τA as a a bijective map of K ∪ {∞} onto itself. For example, if c 6= 0, then a·∞+b c·∞+d = c . Note that since the pairs (a, b) and (c, d) are linearly independent over K, no x0 ∈ K satisfies both ax0 +b = 0 and cx0 +d = 0. Hence, the value of τA (x0 ) is well defined. The arithmetic with ∞ becomes clearer if we substitute x = xx10 in (3) and view τA as a bijective map of P1 (K) onto itself: (5)

τA (x0 :x1 ) = (cx1 + dx0 : ax1 + bx0 ).

The map x0 7→ τA (x0 ) of K ∪ {∞} onto itself defines determines τA . This is one of the consequences of Lemma 11.7.5: Lemma 11.7.5: Let K/K0 be an extension of fields. Let (x1 , x2 , x3 ) and (y1 , y2 , y3 ) be triples of distinct elements of K ∪ {∞}. Then there is a unique M¨obius transformation τ over K such that τ (xi ) = yi , i = 1, 2, 3. Moreover, τ can be presented as τA , where the entries of A belong to the field K0 (x1 , x2 , x3 , y1 , y2 , y3 ). Proof of uniqueness: If τi (xj ) = yj for i = 1, 2 and j = 1, 2, 3, then τ = τ1−1 τ2 satisfies τ (xj ) = xj for j = 1, 2, 3. Suppose first that none of the xj is ∞ and τ = τA with A = ac db . Then cx2j + (d − a)xj − b = 0 for j = 1, 2, 3. Hence, b = c = 0 and a = d. Now assume that x1 = ∞; that is, x1 is (0:1) in homogeneous coordinates. By (5), c = 0. Then we may assume that d = 1 and conclude from axj + b = xj , j = 2, 3, that a = 1 and b = 0. In both cases τ = id and τ1 = τ2 .

214

Chapter 11. Pseudo Algebraically Closed Fields

1 Proof of existence: The M¨ obius transformation τ (x) = x−x 0 maps the ele1 0 ment x of K onto ∞. Likewise, τ (x) = x exchanges 0 and ∞. Hence, we may assume that x1 = ∞ and y1 = ∞. Since x2 6= x3 ,

τ (x) =

y 2 − y3 x2 y3 − x3 y2 x+ x2 − x3 x2 − x3

maps xi onto yi , i = 1, 2, 3, as desired.

Remark 11.7.6: Conservation of branch points. Let K be an algebraically closed field, x an indeterminate, and F a finite separable extension of K(x). Then F/K is a function field of one variable. For each a ∈ K ∪ {∞} let ϕa : K(x) → K ∪ {∞} be the K-place of K(x) with ϕa (x) = a. Denote the corresponding prime divisor of K(x)/K by pa . We say that a is a branch point of F/K(x) (with respect to x) if pa ramifies in F . There are only finitely many prime divisors of K(x)/K which ramify in F (Section 3.4), so F/K(x) has only finitely many branch points. If τ is a M¨obius transformation of K(x) and τ (x) = x0 , then τ maps the set of branch points of F/K(x) with respect to x onto the set of branch points of F/K(x) with respect to x0 . By Lemma 11.7.5, there exists a M¨obius transformation τ of K(x) mapping ∞ onto a finite nonbranch point of F/K(x). Replacing x by τ (x), if necessary, we may assume that ∞ is not a branch point. Let S = K[y1 , . . . , yn ] be the integral closure of K[x] in F . Then S is a Dedekind domain (Proposition 2.4.6). Denote the curve generated in An over K by y by C. The local ring OC,b of C at each b ∈ C(K) is the local ring of S at the kernel of the K-homomorphism mapping y onto b. This kernel is a maximal ideal of S. Hence, OC,b is a valuation ring. The corresponding prime divisor qb of F/K is uniquely determined by b. Conversely, each prime divisor q of F/K with ϕq finite at x is also finite on S and ϕq (y) ∈ C(K). Thus, the map b 7→ qb is a bijective correspondence between C(K) and the set of prime divisors of F/K which are finite at x. Choose a polynomial f ∈ K[Y1 , . . . , Yn ] such that x = f (y). Then f defines a morphism π: C → A1 by π(b) = f (b) for each b ∈ C(K). The prime divisor qb of F/K lies over pa if and only if π(b) = a. Now let p be a prime divisor of K(x)/K which is finite at x. Denote the prime divisors of F/K lying over p by q1 , . . . , qr and the corresponding Pr ramification indices by e1 , . . . , er . Since K is algebraically closed, i=1 ei = [F : K(x)] (Proposition 2.3.2). Hence, p is ramified in F if and only if r < [F : K(x)]. Therefore, by the preceding two paragraphs, an element a of K is a branch point of F/K(x) if and only if there are less than [F : K(x)] points of C(K) lying over a. Next consider an algebraically closed subfield K0 of K. Suppose there is a function field F0 over K0 which contains x, y1 , . . . , yn such that F0 K = F and f has coefficients in K0 . Then [F0 : K0 (x)] = [F : K(x)] and C is already defined over K0 . By the preceding paragraph, “a is a branch point

11.7 A non-PAC Field K with Kins PAC

215

with respect to x” is an elementary statement in the language L(ring, K0 ). Let a1 , . . . , am be all branch points of F0 /K0 (x). Then “a is a branch point with respect to x if and only if a = ai for some i between 1 and m” is an elementary statement which holds over K0 . Since K0 is an elementary subfield of K (Corollary 9.3.2), the same statement holds over K. It follows that each branch point of F/K(x) with respect to x belongs to K0 . Remark 11.7.7: Construction of nonconstant curves. Let K0 be a field. Choose five distinct elements t1 , . . . , t5 in some regular extension of K0 such ˜ 0 (t1 , t2 , t3 ). Consider a regular field extension K of K0 containing /K that t4 ∈ Q5 t1 , t2 , t3 , t4 , t5 . Put f (X) = i=1 (X − ti ) and define a curve C over K by 1 if char(K) = 2. the equation Y 2 = f (X) if char(K) 6= 2 and Y 2 + Y = f (X) By Propositions 3.8.2 and 3.8.4, C is a hyperelliptic curve. More precisely, the genus g of C is 2 if char(K) 6= 2 and 4 if char(K) = 2. Claim: C is a nonconstant curve over K/K0 . ˜ Proof: Choose a generic point (x, y) for C over K and let F = K(x, y). Then ˜ K(x) is a quadratic subfield of F . Assume that C is a constant curve over ˜ 0 such that K/K0 . Then there exists a function field of one variable F0 over K ˜ 0 is a regular extension, F0 is linearly disjoint from ˜ = F . Since F0 /K F0 K ˜ 0 ) = genus(F/K) ˜ ˜ over K ˜ 0 (Lemma 2.6.7). In particular, g = genus(F0 /K K ˜ 0 by w0 and de(Proposition 3.4.2). Denote the canonical divisor of F0 /K ˜ by w. By Proposition 3.4.2 and Lemma note the image of w0 in Div(F/K) 3.2.2(b), deg(w) = deg(w0 ) and dim(w) = dim(w0 ). Hence, w is the canoni˜ (Exercise 1(b) of Chapter 3). Choose a basis z1 , z2 , . . . , zg cal divisor of F/K ˜ for L(w0 ) over K0 . Then, by the linear disjointness, z1 , z2 , . . . , zg form a basis ˜ By Proposition 3.7.4, K( ˜ z2 , . . . , zg ) is the unique quadratic for L(w) over K. z1 z1 ˜ z2 , . . . , zg ) = K(x). ˜ ˜ z2 , . . . , zg ) is a funcsubfield of F , so K( In particular, K( z1

z1

z1

z1

z

˜ Hence, K ˜ 0 ( z2 , . . . , g ) is a function field of genus tion field of genus 0 over K. z1 z1 ˜ By Example 3.2.4, there is an x0 with K ˜ 0 ( z2 , . . . , zg ) = K ˜ 0 (x0 ). 0 over K. z1

z1

˜ 0 ) = K(x). ˜ It satisfies K(x By Lemma 11.7.5, there exists a M¨obius transfor˜ mation τ over K such that τ (x) = x0 . ˜ ˜ ∪ {∞} be the K-place ˜ →K with For each i between 1 and 5 let ϕi : K(x) ˜ with ϕi (x) = ti . By Examples 2.3.8 and 2.3.9, ti is a branch point of F/K(x) respect to x. Put ai = τ (ti ). Then ϕi (x0 ) = ϕi (τ (x)) = τ (ϕi (x)) = τ (ti ) = ˜ 0 ) with respect to x0 . It follows from ai . Thus, ai is a branch point of F/K(x ˜ Remark 11.7.6 that ai ∈ K0 . ˜ 0 (t1 , t2 , t3 ). Hence, t4 = By Lemma 11.7.5, τ is already defined over K −1 ˜ τ (a4 ) ∈ K0 (t1 , t2 , t3 ). This contradiction to the assumption we made above proves that C is a nonconstant curve over K/K0 . Theorem 11.7.8 ([Hrushovski, Cor. 5]): For each prime number p there exists a countable non-PAC field E of characteristic p such that Eins is PAC. Proof:

Choose algebraically independent elements t1 , t2 , t3 , t4 , t5 over Fp .

216

Chapter 11. Pseudo Algebraically Closed Fields

Let C be the curve defined over K = Fp (t1 , t2 , t3 , t4 , t5 ) in Remark 11.7.7. Then, C is a nonconstant curve over F/Fp for every regular field extension F of Fp that contains K. By Proposition 11.7.1, C(K) is a finite set. By induction we construct two ascending towers of fields K = E1 ⊆ E2 ⊆ E3 ⊆ · · · and F1 ⊆ F2 ⊆ F3 ⊆ · · · and for each positive integer m we enumerate the absolutely irreducible varieties which are defined over Em in a sequence, Vm1 , Vm2 , Vm3 , . . . such that (6a) Em and Fm are finitely generated regular extensions of K, (6b) Fm is a finite purely inseparable extension of Em , (6c) C(Em ) = C(K), and (6d) Vij (Fm ) 6= ∅ for i, j = 1, . . . , m − 1. Indeed, suppose E1 , . . . , Em−1 , F1 , . . . , Fm−1 , and Vij for i < m and all j have been defined such that they satisfy (6). Let V be the direct product of Vij for i, j = 1, . . . , m − 1. It is an absolutely irreducible variety defined 0 over Em−1 . Let x be a generic point of V over Em−1 . Then Em = Em−1 (x) is a finitely generated regular extension of Em−1 and therefore also of K. 0 (instead K, F ) and construct an extension Apply Lemma 11.7.3 to Em−1 , Em 0 0 such that Em /Em is a finite purely Em of Em−1 which is contained in Em inseparable extension and C(Em ) = C(Em−1 ). By (6c) for m − 1 we have 0 is a regular extension of Em−1 , it is linearly C(Em ) = C(K). Since Em 0 ˜ disjoint from Fm−1 K over Em−1 . Hence, Fm = Em Fm−1 is linearly disjoint ˜ from Fm−1 K over Fm−1 . 0 Em

Fm

Em Em−1 K

Fm−1

˜ Fm−1 K ˜ K

˜ over K, Fm is linearly disjoint from K ˜ Since Fm−1 is linearly disjoint from K over K. Thus, Fm is a regular extension of K. By construction, Fm is a finite Vij (Fm ) 6= ∅ for i, j = 1 . . . , m − 1. purely inseparable S∞ extension of EmSand ∞ Let E = m=1 Em and F = m=1 Fm . Then E and F are countable regular extensions of K and F is purely inseparable over E. Hence, in order to prove that Eins is PAC it suffices, by Theorem 11.2.3, to prove that each absolutely irreducible variety V defined over E has an F -rational point. Indeed, if V is such a variety, then V = Vij for some i and j. Let m = max{i, j} + 1. By (6d), V has an Fm -rational point, which is, of course, an F -rational point. Finally, each point of C(E) belongs to C(Em ) for some m and therefore, by (6c), to C(K). Thus, C(E) = C(K) is a finite set. By Proposition 11.1.1, E is not PAC.

Exercises

217

Problem 11.7.9: Does there exists a finitely generated field extension E of Fp such that E is not PAC but Eins is PAC?

Exercises 1. Let V be an absolutely irreducible variety which is defined over a PAC field K. Suppose that A and B are Zariski K-closed subset of V such that V (K) = A(K) ∪ B(K). Use Proposition 11.1.1 to prove that V = A or V = B. Conclude that V (K) is connected in the Zariski-topology. 2. By the Artin-Schreier theorem, the only torsion that occurs in the absolute Galois group of a field K comes from real closed fields. Prove that if K is PAC, then Gal(K) is torsion free. 3. A more elementary characterization of PAC fields, just as useful as Theorem 11.2.3, can be derived from Lemma 10.4.1. Prove that a field K is PAC if and only if for each nonconstant absolutely irreducible polynomial f ∈ K[T1 , . . . , Tr , X], separable in X, and for every nonzero polynomial g ∈ K[T1 , . . . , Tr ], there exist a1 , . . . , ar , b ∈ K such that f (a, b) = 0 and g(a) 6= 0. 4. Prove the following assertion without using descent: If K is a PAC field of characteristic p > 0, then Kins is also a PAC field. 5. Let Γ be the plane curve defined over F3 by the equation X04 + X14 − X24 − X02 X22 + X13 X2 = 0. Prove that Γ is absolutely irreducible and has no F3 -rational point. Hint: Show that Γ is smooth. 6. (a) Let f ∈ L[X1 , . . . , Xn ] be a nonconstant homogeneous polynomial with at least one absolutely irreducible factor of multiplicity 1. Prove that if a ∈ L× , then f (X1 , . . . , Xn ) − a is absolutely irreducible. (b) Let M/L be a finite Galois extension with basis w1 , . . . , wn . Describe the norm map normM/L : M → L as a homogeneous polynomial of degree n in X1 , . . . , Xn . (c) Combine (a) and (b) to show that if L is a finite separable extension of a PAC field K, then normM/L is surjective. Thus, show directly property (b) of Lemma 11.6.7. 7. This exercise suggest an alternative way to achieve some consequences of Corollary 11.5.5. (a) Use Eisenstein’s criterion to prove that the polynomial f (X, Y ) = (X q −X)(Y q −Y )+1 defined over the field Fq (q a prime power) is absolutely irreducible and has no Fq -rational zeros. Deduce that a PAC field is infinite. (b) (Ax) Let K be a field with a valuation v whose residue class field has q elements. Let π ∈ K with v(π) > 0 and show that (X q − X − 1)(Y q − Y − 1) − π = 0 defines an absolutely irreducible variety with no K-rational point. Thus, show directly that K is not a PAC field. Conclude, in particular, that fields finitely generated over their prime fields are not PAC fields.

218

Chapter 11. Pseudo Algebraically Closed Fields

8. Use the idea included in the proof of Corollary 11.5.7 to show that Gal(Qp ) is pronilpotent for no prime p. 9. Let K be a field, L an infinite extension of K, and f ∈ L[X1 , . . . , Xn ]. Prove that if f (a) ∈ K for each a ∈ Ln , then the coefficients of f belong to K. Hint: For n = 1 use Cramer’s rule to compute the coefficients of f . Then continue by induction on n.

Notes Apparently PAC fields appear for the first time in J. Ax’s papers [Ax1] and [Ax2]. Ax observes that the Riemann Hypothesis for curves implies that nonprincipal ultraproducts of finite fields are PAC fields. From this he establishes a recursive decision procedure for the theory of finite fields. Following a suggestion of the second author, Frey introduces in [Frey] the name PAC for fields over which each variety has rational points. He also proves a place theoretic predecessor to Lemma 11.2.1. Remark 11.2.1, due to J´anos Koll´ar, settles Problem 11.2.10 of the second edition. Our proof of Proposition 11.4.1 appears to be simpler than Tamagawa’s. In [Frey] Frey uses the theory of homogeneous spaces attached to elliptic curves and Tate’s theory of bad reduction in order to prove Corollary 11.5.5 for real valuations. Prestel showed the authors the elementary direct proof of the Corollary for arbitrary valuations (private communication). A variation of that proof serves also for Proposition 11.5.3.

Chapter 12. Hilbertian Fields David Hilbert proved his celebrated irreducibility theorem during his attempt to solve a central problem of Galois theory: Is every finite group realizable over Q? He proved that a general specialization of the coefficients of the general polynomial of degree n to elements of Q gives a polynomial whose Galois group is Sn . Further, if f ∈ Q[T1 , . . . , Tr , X] is an irreducible polynomial, then there exist a1 , . . . , ar ∈ Q such that f (a, X) remains irreducible. This result is now known as Hilbert’s irreducibility theorem. Since then, many more finite groups have been realized over Q. Most of those have been realized via Hilbert’s theorem. This has brought the theorem to the center of the theory of fields. Various alternative proofs of the irreducibility theorem apply to other fields (including all infinite finitely generated fields). We call them Hilbertian fields. We give several reductions of the irreducibility theorem, and one especially valuable result (Corollary 12.2.3): Let K be a Hilbertian field, L a finite separable extension of K, and f ∈ L[T1 , . . . , Tr , X] an irreducible polynomial. Then there exist a1 , . . . , ar ∈ K such that f (a, X) is irreducible in L[X]. Chapters 13 and 15 give a proof of the Hilbertian property for most known Hilbertian fields. This lays the foundation to a subject central to the book, the model theory of PAC fields.

12.1 Hilbert Sets and Reduction Lemmas Consider a field K and two sets T1 , . . . , Tr and X1 , . . . , Xn of variables. Let f1 (T, X), . . . , fm (T, X) be polynomials in X1 , . . . , Xn with coefficients in K(T). Assume these are irreducible in the ring K(T)[X]. For g ∈ K[T] a nonzero polynomial, denote the set of all r-tuples (a1 , . . . , ar ) ∈ K r with g(a) 6= 0 and f1 (a, X), . . . , fm (a, X) defined and irreducible in K[X] by HK (f1 , . . . , fm ; g). Call HK (f1 , . . . , fm ; g) a Hilbert subset of K r . If in addition n = 1 and each fi is separable in X, call HK (f1 , . . . , fm ; g) a separable Hilbert subset of K r . A Hilbert set (resp. separable Hilbert set) of K is a Hilbert subset (resp. separable Hilbert subset) of K r for some positive integer r. The intersection of finitely many Hilbert subsets of K r is again a Hilbert subset of K r . Hence, if each Hilbert subset of K r is nonempty, then each Hilbert subset of K r is Zariski K-dense in K r . By Lemma 10.2.5, each Hilbert subset of K r is Zariski dense in K r . The same holds for separable Hilbert sets. We say that K is Hilbertian if each separable Hilbert set of K is nonempty. In particular, a Hilbertian field must be infinite.

220

Chapter 12. Hilbertian Fields

The next lemmas reduce the infiniteness of arbitrary Hilbert sets (resp. separable Hilbert sets) to the infiniteness of special Hilbert sets (resp. separable Hilbert sets). Lemma 12.1.1: Each Hilbert subset (resp. separable Hilbert subset) HK (f1 , . . . , fm ; g) of K r contains a Hilbert subset (resp. separable Hilbert 0 subset) HK (f10 , . . . , fm ; g 0 ) of K r with fi0 irreducible in K[T, X] and 0 / K[T], i = 1, . . . , m. fi ∈ Proof: By assumption, fi is irreducible in K(T)[X]. Hence, at least one Xj occurs in fi . Multiply fi (T, X) by a polynomial, gi (T), to ensure that its coefficients lie in K[T]. Then divide the resulting polynomial by the greatest common divisor, di (T), of its coefficients. Since K[T, X] has unique factorization, we obtain an irreducible polynomial fi0 ∈ K[T, X]. Now put g 0 = 0 ; g 0 ) ⊆ HK (f1 , . . . , fm ; g). g · g1 d1 . . . gm dm and conclude that HK (f10 , . . . , fm Finally, suppose n = 1 and each fi (T, X) is separable in X. Then so is each fi0 (T, X). Lemma 12.1.2: Suppose every Hilbert set of K of the form HK (f1 , . . . , fm ; g) with fi irreducible in K[T, X1 , . . . , Xn ], i = 1, . . . , m, is nonempty. Then every Hilbert set of K is nonempty. Proof: Start with irreducible polynomials fi ∈ K[T1 , . . . , Tr , X1 , . . . , Xn ], i = 1, . . . , m, and a nonzero polynomial g ∈ K[T1 , . . . , Tr ]. By assumption, there exists a1 ∈ K with fi (a1 , T2 , . . . , Tr , X) irreducible, i = 1, . . . , m, and g(a1 , T2 , . . . , Tr ) 6= 0. Repeat this procedure r times to find a1 , . . . , ar ∈ K with fi (a, X) irreducible and g(a) 6= 0, i = 1, . . . , m. By Lemma 12.1.1, every Hilbert set of K is nonempty. Lemma 12.1.3: Every Hilbert subset of K contains a Hilbert set of the form H(f1 , . . . , fm ; g), where fi is an irreducible polynomial in K[T, X] with degX (fi ) ≥ 1, i = 1, . . . , m. Proof: Let f ∈ K[T, X1 , . . . , Xn ] be an irreducible polynomial with f ∈ / K[T ] and 0 6= g0 ∈ K[T ]. By Lemma 12.1.1, it suffices to find irreducible polynomials h1 , . . . , hr ∈ K[T, Y ] r K[T ] and 0 6= g ∈ K[T ] with HK (h1 , . . . , hr ; g) ⊆ HK (f ; g0 ). Indeed, let d > max1≤j≤n degXj (f ). Apply the Kronecker substitution Sd : SK[T ] (n, d) → SK[T ] (1, dn ) on f (Section 11.3). Consider the factorization of Sd (f ) into irreducible factors of K[T, Y ]: Y hi (T, Y ). (1) Sd (f )(T, Y ) = i∈I

The polynomials Sd (f )(T, Y ) and f (T, X1 , . . . , Xn ) have the same coefficients in K[T ]. Since f is irreducible in K[T, X1 , . . . , Xn ], the greatest common divisor of its coefficients in K[T ] is 1, so none of the hi is in K[T ]. Let I = J ∪· J 0 be a nontrivialQpartition of I. The Q exponent to which Y appears in each of the polynomials i∈J hi (T, Y ) and i∈J 0 hi (T, Y ) does not

12.1 Hilbert Sets and Reduction Lemmas

221

exceed dn − 1. Since Sd is bijective on SK[T ] (n, d) (Section 11.3), there exist polynomials pJ , pJ 0 ∈ K[T, X] with degXj (pJ ), degXj (pJ 0 ) < d, j = 1, . . . , n, and Y Y hi (T, Y ) and Sd (pJ 0 )(T, Y ) = hi (T, Y ). (2) Sd (pJ )(T, Y ) = i∈J 0

i∈J

Note: The product pJ (T, X)pJ 0 (T, X) contains a monomial of the form gJ (T )X1ν1 · · · Xnνn in which at least one of the νj exceeds d − 1. Otherwise, the relation Sd (f ) = Sd (pJ ) · Sd (pJ 0 ) = Sd (pJ pJ 0 ) would imply a nontrivial factorization of the irreducible polynomial f into pJ pJ 0 . Let g be the product of g0 with all gJ (one for each nontrivial partition). Let a be an element of K with each hi (a, Y ) irreducible in K[Y ] and g(a) 6= 0. We show f (a, X) is also irreducible. Indeed, assume f (a, X) = q(X)q 0 (X) is a nontrivial factorization of f (a, X) in K[X], then (1) implies Y

hi (a, Y ) = Sd (f )(a, Y ) = Sd (q)(Y ) · Sd (q 0 )(Y ).

i∈I

Hence, (2) implies a nontrivial partition I = J ∪· J 0 of I with Sd (q)(Y ) =

Y

hi (a, Y ) = Sd (pJ )(a, Y )

and

i∈J

Sd (q 0 )(Y ) =

Y

hi (a, Y ) = Sd (pJ 0 )(a, Y ).

i∈J 0

Hence, q(X) = pJ (a, X) and q 0 (X) = pJ 0 (a, X). Thus, pJ (a, X)pJ 0 (a, X) = f (a, X). The left hand side contains the monomial gJ (a)X1ν1 · · · Xnνn in which at least one νj exceeds d − 1, while f (a, X) contains no such monomial. Therefore, f (a, X) is irreducible. Lemma 12.1.4: Let H = HK (g1 , . . . , gm ; h) be a Hilbert subset of K r with gi ∈ K[T1 , . . . , Tr , X] irreducible and degX (gi ) ≥ 1, i = 1, . . . , m. Then H contains a Hilbert set of the form HK (f1 , . . . , fm ) in which fi is a monic irreducible polynomial in K[T, X] of degree at least 2 in X, i = 1, . . . , m. If g1 , . . . , gm are separable in X, then so are f1 , . . . , fm . Moreover, suppose a ∈ K r and none of the polynomials fi (a, X) has a root in K. Then none of the polynomials gi (a, X) with degX (gi ) ≥ 2 has a root in K. Proof: Let ci be the leading coefficient of gi as a polynomial in X, ni = degX (gi ), and q = hc1 · · · cm . Choose a prime number l 6= char(K) with q not an lth power in K[T]. By assumption, ni ≥ 1. If m = 0 or ni = 1, let fi (T, X) = X l − q(T). If ni ≥ 2, let fi (T, X) = q(T)ni ci (T)−1 gi (T, q(T)−1 X).

222

Chapter 12. Hilbertian Fields

Then fi (T, X) = X ni + bi,ni −1 (T)X ni −1 + q(T)

nX i −2

bij (T)X j .

j=0

with bij ∈ K[T]. In each case fi (T, X) is monic in X and irreducible in K[T, X] (see [Lang7, p. 297, Lemma 9.1] for the case m = 0 or ni = 1). We prove that HK (f1 , . . . , fm ) ⊆ H. Let a be in K r with f1 (a, X), . . . , fm (a, X) irreducible in K[X]. Consider an i between 1 and m. Suppose first ni = 1. Then X l − q(a) is irreducible. Hence, q(a) 6= 0. Therefore, h(a) 6= 0, ci (a) 6= 0, and gi (a, X) = ci (a)X + bi for some bi ∈ K. Thus, gi (a, X) is irreducible. Now suppose ni ≥ 2. Then q(a) 6= 0 (otherwise fi (a, X) = X ni + bi,ni −1 (a)X ni is reducible) and gi (a, q(a)−1 X) = q(a)−ni ci (a)fi (a, X) is irreducible. Hence, so is gi (a, X). Similarly, if ni ≥ 2 and fi (a, X) has no root in K, neither has gi (a, X). Corollary 12.1.5: Suppose each Hilbert set of the form HK (f1 , . . . , fm ) with fi ∈ K[T, X] monic and of degree at least 2 in X is nonempty. Then every Hilbert set of K is nonempty. Proof: By Lemma 12.1.4 for r = 1, each Hilbert subset of K of the form HK (g1 , . . . , gm ; h) with gi ∈ K[T, X] irreducible and degX (gi ) ≥ 1, i = 1, . . . , m, is nonempty. Hence, by Lemma 12.1.3, each Hilbert subset of K is nonempty. Consequently, by Lemma 12.1.2, every Hilbert set of K is nonempty. The corresponding result for separable Hilbert sets is also true. We postpone its proof to Section 13.2. Here we show that separable Hilbert subsets of K r contain especially simple separable Hilbert subsets of K r . Lemma 12.1.6: Let H be a separable Hilbert subset of K r . Then there exists an irreducible polynomial f ∈ K[T1 , . . . , Tr , X], separable, monic, and of degree at least 2 in X with HK (f ) ⊆ H. Proof: Use Lemma 12.1.1 to assume that H = HK (f1 , . . . , fm ; g) with f1 , . . . , fm ∈ K[T, X] r K[T] irreducible and separable in X and 0 6= g ∈ K[T]. For each i, 1 ≤ i ≤ m, let xi be a root of fi (T, X) in K(T)s . Then the finite separable extension K(T, x1 , . . . , xm ) of K(T) is contained in a separable extension K(T, y) of K(T) of degree at least 2. Assume without loss, y is integral over K[T]. Let h1 = irr(y, K(T)). Then h1 (T, X) is separable, monic, and of degree at least 2 in X. Rewrite xi as ppi 0(T,y) (T) with p1 , . . . , pm ∈ K[T, X] and 0 6= p0 ∈ K[T]. Let ui (T, xi ) be the leading coefficient of irr(y, K(T, xi )). Write normK(T,xi )/K(T) (ui (T, xi )) as qq0i (T) (T) with q0 , q1 , . . . , qm ∈ K[T]. Finally, let ri (T) be the leading coefficient of fi (T, X),

12.2 Hilbert Sets under Separable Algebraic Extensions

223

i = 1, . . . , m. Define g1 to be the product g · p0 · q0 · · · qm · r1 · · · rm . We show that HK (h1 ; g1 ) ⊆ HK (f1 , . . . , fm ; g). Indeed, if a ∈ HK (h1 ; g1 ) and c is a root of h1 (a, X), then (3) Moreover, with bi =

[K(c) : K] = [K(T, y) : K(T)]. pi (a,c) p0 (a) ,

the nonvanishing of g1 (a) implies

[K(bi ) : K] ≤ [K(T, xi ) : K(T)] and [K(c) : K(bi )] ≤ [K(T, y) : K(T, xi )]. Multiply the terms in these two inequalities and apply (3) to conclude that they are equalities. Therefore, f1 (a, X), . . . , fm (a, X) are irreducible over K. By Lemma 12.1.4, there exists an irreducible polynomial f ∈ K[T, X], separable, monic, and of degree at least 2 in X, with HK (f ) ⊆ HK (h1 ; g1 ).

12.2 Hilbert Sets under Separable Algebraic Extensions Let L/K be a finite separable extension. We prove every Hilbert subset of Lr contains a Hilbert subset of K r . Lemma 12.2.1: Let L be a separable extension of degree d of an infinite field K and σ0 , . . . , σd−1 distinct representatives of the cosets of Gal(L) in Gal(K). Suppose f ∈ L[X1 , . . . , Xn ] is a nonconstant polynomial. Then there exist c1 , . . . , cn ∈ L with f (X1 + c1 , . . . , Xn + cn )σi , i = 0, . . . , d − 1, ˜ 1 , . . . , Xn ]. pairwise relatively prime in K[X Proof: Let θ be a primitive element for L/K. Choose algebraically independent elements tik , i = 0, . . . , d − 1 and k = 1, . . . , n, over K. For every i and k consider d−1 X (θσi )j tjk . uik = j=0

Write ui = (ui1 , . . . , uin ) and u = (u0 , . . . , ud−1 Q ). The linear transformation t → u has as determinant the nth power of i<j (θσj − θσi ) 6= 0. Thus, the ˜ so the uik are tik are linear combinations of the uik with coefficients in K; algebraically independent over K. ˜ factors. Since fµ is Write f = f1 · · · fm , a product of K-irreducible σ nonconstant and the uik are algebraically independent, fµσi (ui ) 6= fν j (uj ) for i 6= j and each µ and ν, 1 ≤ µ, ν ≤ m. Therefore YY (fµσi (ui ) − fνσj (uj )) 6= 0, h(t) = i<j µ,ν

so there exist aik ∈ K with h(a) 6= 0. Let ck =

d−1 X j=0

ajk θj ,

k = 1, . . . , n.

224

Chapter 12. Hilbertian Fields

˜ The K-specialization t → a maps uik to cσk i . Hence fµ (c)σi 6= fν (c)σj for i 6= j. Therefore, fµ (X + c)σi 6= fν (X + c)σj for all i, j, µ, ν with 0 ≤ i < j ≤ d − 1 and 1 ≤ µ, ν ≤ m. Since f1 (X + c)σi , . . . , fm (X + c)σi are exactly the ˜ f (X + c)σ0 , . . . , f (X + c)σd−1 are irreducible factors of f (X + c)σi in K[X], relatively prime in pairs. Lemma 12.2.2: Let L be a finite separable extension of a field K and f ∈ L(T1 , . . . , Tr )[X1 , . . . , Xn ] irreducible. Then, there exists an irreducible p ∈ K(T)[X] with HK (p) ⊆ HL (f ). If n = 1 and f is separable in X, then p is separable in X. Proof: Let S be a set of representatives of the left cosets of Gal(L) in Gal(K). Denote the algebraic closure of K(T) by F . First consider the case Q where the f σ , with σ ∈ S, are pairwise relatively prime in F [X]. Let p = σ∈S f σ . Since Gal(K) fixes the coefficients of p, p ∈ K(T)[X]. Moreover, if q is an irreducible factor of p in K(T)[X], then one of the f σ ’s (and therefore all) ˜ divides q in K(T)[X]. Since the f σ are pairwise relatively prime, p divides q. Hence, p is irreducible in K(T)[X]. In addition, if n = 1 and f is separable in X, then p is separable in X. Let a1 , . . . , ar be elements of K such that p(a, X) is defined and irreducible in K[X]. If f (a, X) Q were a nontrivial decomposition Q = g(X)h(X) in L[X], then p(a, X) = ( σ∈S g σ (X))( σ∈S hσ (X)) would be a nontrivial decomposition in K[X], which is a contradiction. Therefore, f (a, X) is irreducible in L[X] and HK (p) ⊆ HL (f ). In the general case apply Lemma 12.2.1 to find c1 , . . . , cr ∈ L(T) with f (T, X + c)σ pairwise relatively prime in F [X], σ running over S. Let g(T, X) = f (T, X + c). Irreducibility of g(a, X) is equivalent to that of f (a, X) for each a ∈ K r such that both polynomials are defined. Thus, the lemma follows from the first part of the proof. Corollary 12.2.3: Let L be a finite separable extension of a field K. Then every Hilbert subset (resp. separable Hilbert subset) of Lr contains a Hilbert subset (resp. separable Hilbert subset) of K r . In particular, if K is Hilbertian, so is L. Remark 12.2.4: The converse of Corollary 12.2.3 is false: Example 13.9.5 gives a field K with an empty Hilbert set and a finite separable extension L whose Hilbert sets are all nonempty.

12.3 Purely Inseparable Extensions Corollary 12.2.3 is false if L/K is inseparable. For example, let K = Fp (t), with t transcendental over Fp , and let L = K(t1/p ). Then X p −T is irreducible over L. Yet every a ∈ K is a pth power in L, so X p − a is reducible over L. Hence, HL (X p − T ) contains no elements of K. Still, X p − t1/p is irreducible over L, so HL (X p − T ) is nonempty. The following results generalize this

12.3 Purely Inseparable Extensions

225

observation. They assert: if every Hilbert set of K is nonempty, then every Hilbert set of L is nonempty. First we list simple properties of purely inseparable extensions. Lemma 12.3.1: Let L/K be a purely inseparable extension of fields of characteristic p. (a) For each irreducible g ∈ K[X] there exist an irreducible f ∈ L[X] and e e ≥ 0 with g = f p . e e (b) Consider f ∈ L[X]. Suppose Lp ⊆ K and g = f p is irreducible in K[X]. Then f is irreducible in L[X]. e (c) Let f ∈ L[X] be a non pth power with g = f p irreducible in K[X]. Then f is irreducible in L[X]. (d) If f ∈ L[X] is irreducible and e ≥ 0 is the least nonnegative integer with e g = f p ∈ K[X], then g is irreducible in K[X]. Proof of (a):

Let f1 · · · fm be a factorization of g into irreducible factors k

in L[X]. Choose k ≥ 0 such that fip ∈ K[X], i = 1, . . . , m. The relation k

k

k

k

p g p = f1p · · · fm and the unique factorization in K[X] imply that f1p = g r for some positive integer r. Unique factorization over L[X] then gives a k

positive integer s with g = f1s . Then f1p = f1rs implies that s = pe , so e g = f1p for some positive integer e. Proof of (b): Let f1 · · · fm be a factorization of f into irreducible factors in e pe is a factorization of g in K[X]. Since g is irreducible L[X]. Then f1p · · · fm over K, we have m = 1. Therefore, f is irreducible over L. Proof of (c): e0

e0

By (a), g = hp e

for some irreducible h ∈ L[X] and e0 ≥ 0. e0 −e

Hence, hp = f p , so e0 ≥ e. Therefore, hp = f . The assumption on f implies that e0 = e and f = h. Thus, f is irreducible in L[X]. Proof of (d): Let g1 · · · gm be a factorization of g into irreducible factors ei in K[X]. By (a), gi = fip for some irreducible polynomial fi in L[X] and e e 1 pem ei ≥ 0. Hence, f p = f1p · · · fm . Therefore, fi = f and pe1 +· · ·+pem = pe . pei = gi ∈ K[X], so by assumption, ei ≥ e. It follows that In addition, f m = 1 and g is irreducible in K[X]. Lemma 12.3.2: Let K be a field with char(K) = p > 0, L a purely inseparable extension of K, and f an irreducible polynomial in L[T1 , . . . , Tr , X] with degX f ≥ 1. (a) If f is separable in X, then HL (f ) contains a separable Hilbert subset of Kr. (b) If all occurrences of T1 , . . . , Tr , X in f are powers of p, then HL (f ) contains a Hilbert subset of K r . e

Proof: Let e be the least nonnegative integer with g = f p ∈ K[T, X]. By Lemma 12.3.1(d), g is irreducible in K[T, X]. Distinguish between two cases to find a Hilbert subset of K contained in HL (f ).

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Chapter 12. Hilbertian Fields

Case A: f is separable with respect to X. Consider a ∈ K with g(a, X) irreducible in K[X] and f (a, X) separable. Then, f (a, X) is not a pth power in L[X]. By Lemma 12.3.1(c), f (a, X) is irreducible in L[X]. Case B: All occurrences of T1 , . . . , Tr , X in f are powers of p. By assumption, f is not a pth power in L[T, X], so at least one coefficient c(T) of f (T, X) is not a pth power in L[T]. In particular, L is an infinite field. Hence, K is also infinite. such that c(a) ∈ / Claim: There is a nonempty Zariski L-open subset U of Ar P ci T1i1 p · · · Trir p Lp for all a ∈ U (L). To prove the claim, write c(T) = with i = (i1 , . . . , ir ), 0 ≤ i1 ≤ d1 , . . . , 0 ≤ ir ≤ dr , and ci ∈ Lr . Let d = (d1 + 1) · · · (dr + 1). Assume there are d r-tuples (bk1 , . . . , bkr )

(1)

with 0 ≤ kj ≤ d1 and bkj 6= bkj0 if j 6= j 0 and 1 ≤ j, j 0 ≤ r such that c(bk1 , . . . , bkr ) = λp with λ ∈ L. Thus, there is a λk1 ,...,kr ∈ L with (2)

X

ci bki11p · · · bkirrp = λpk1 ,...,kr

i

Consider (2) as a system of d linear equations in the ci ’s. The coefficient matrix of (2) is a Kronecker product of r Vandermonde matrices. The determinant of this matrix is D =det(bki11p · · · bkirrp ) =

r Y

i

0

det(bkjj )dj p

j=1

=

r Y Y

0

(bkj0 − bkj )dj p

j=1 j<j 0

where d0j = (dj +1)−1 (d1 +1) · · · (dr +1), j = 1, . . . , r (See [Bourbki, Algebra, p. 534] for the determinant of the Kronecker product of two matrices, from which the determinant of the Kronecker product of r matrices can be derived.) Thus, D is a pth power of a nonzero element of L. Applying Cramer’s rule to the system (2), we conclude that ci ∈ Lp for each i. This contradiction to the assumption we made on c(T) proves that there are at most d − 1 r-tuples (1). Let B be the set of all these r-tuples. Define, / B and xj 6= xj 0 if j 6= j 0 }. U = {(x1 , . . . , xr ) ∈ Ar | (x1 , . . . , xr ) ∈ Then U is a Zariski L-open subset of Ar satisfying our claim. Now we may use the claim to choose a ∈ K r with c(a) 6∈ Lp (therefore f (a, X) is not a pth power in L[X]) and g(a, X) irreducible in K[X]. By Lemma 12.3.1(c), f (a, X) is irreducible in L[X].

12.3 Purely Inseparable Extensions

227

Proposition 12.3.3: Let L/K be an algebraic extension of fields of a finite separable degree. Then each separable Hilbert subset of Lr contains a separable Hilbert subset of K r . Thus, if K is Hilbertian, then so is L. Proof: By assumption, L is a purely inseparable extension of a finite separable extension of K. Apply Corollary 12.2.3 to assume L/K is purely inseparable. Let H be a separable Hilbert subset of Lr . By Lemma 12.1.6, H contains HL (f ) with irreducible f ∈ L[T, X] separable in X. From Lemma 12.3.2, HL (f ) contains a separable Hilbert subset of K r . Lemma 12.3.4: Let K be a field of positive characteristic p, L an extension r of K with Lp ⊆ K for some r ≥ 0, and f1 , . . . , fm irreducible polynomials in L[T, X]. Suppose degX fi ≥ 1 and fi is separable in T . Suppose in addition that each Hilbert subset of K is nonempty. Then there exists γ ∈ L such that HL (fi (T + γ, X)) contains a Hilbert subset of K, i = 1, . . . , m. Proof: Let i be between 1 and m. At least one of the coefficients ci (T ) of fi (T, X) has nonzero derivative c0i (T ). Choose a nonzero coefficient γi of e−1 e c0i (T ). Let e be the least nonnegative integer with Lp ⊆ K. Then Lp 6⊆ K, Sm −pe−1 pe−1 pe−1 pe−1 6⊆ γi K, i = 1, . . . , m. Since K is infinite, L 6⊆ i=1 γi K. so L pe−1 ∈ / K. This implies Hence, there exists λ ∈ L independent of i with (γi λ) e−1 6∈ K[T ]. (c0i (T )λ)p The Taylor expansion ci (T + λZ) = ci (T ) + c0i (T ) · λZ + · · · shows that e−1 e ci (T + λZ)p 6∈ K[T, Z]. Hence, hi (T, Z) = ci (T + λZ)p is not a pth power of a polynomial belonging to K[T, Z]. Therefore, X p − hi (T, Z) is irreducible over K. By assumption, there exists b ∈ K, independent of i, with X p − hi (T, b) irreducible over K. Thus, e is the least nonnegative integer for e which ci (T + λb)p is in K[T ]. Hence, e is the least nonnegative integer for e which gi (T, X) = fi (T + λb, X)p ∈ K[T, X]. By Lemma 12.3.1(d), gi (T, X) is irreducible in K[T, X]. If for a ∈ K, the polynomial gi (a, X) is irreducible in K[X], then from Lemma 12.3.1(b), fi (a + λb, X) is irreducible in L[X]. Thus, HK (gi ) ⊆ HL (fi (T + λb, X)). Proposition 12.3.5: Let K be a field with all Hilbert subsets nonempty and L an algebraic extension of K. Then in each of the following cases all Hilbert sets of L are nonempty. r (a) char(K) = p > 0 and Lp ⊆ K for some r ≥ 0. (b) L is a finite extension of K. Proof of (a): Consider a Hilbert subset HL (f1 , . . . , fn ) of L with fi ∈ L[T, X] irreducible and degX fi ≥ 2, i = 1, . . . , n. Reorder the fi ’s, if necessary, to assume the following: For i = 1, . . . , m, fi is separable in X or inseparable in both T and X. For i = m + 1, . . . , n, fi is separable in T . Lemma 12.3.4 gives γ in L such that HL (fi (T + γ, X)) contains a Hilbert subset Hi of K.

228

Chapter 12. Hilbertian Fields

For each i between 1 and m, the polynomial fi (T + γ, X) is either separable in X or inseparable in both T and X. By Lemma 12.3.2, HL (fi (T + γ, X)) contains a Hilbert subset Hi of K, so H = H1 ∩ · · · Hm is a Hilbert subset of K which is contained in HL (f1 , . . . , fn ). By Corollary 12.1.5, every Hilbert set of L is nonempty. Proof of (b): Combine (a) with Lemma 12.2.2.

12.4 Imperfect fields Suppose char(K) = p 6= 0 and every Hilbert set of K is nonempty. Then there exists a ∈ K with X p − a is irreducible. Thus, K is imperfect. We show that, conversely, every Hilbert set of an imperfect Hilbertian field is nonempty. Lemma 12.4.1: Let char(K) = p > 0 and let f ∈ K[X] r K p [X] be irrem ducible and monic. Then f (X p ) is irreducible in K[X] for each m ≥ 0. Proof: By induction it suffices to prove the Lemma for m = 1. If x is a root of f (X p ), then y = xp is a root of f (X). We show [K(x) : K(y)] = p. If not, then K(x) = K(y) and irr(x, K) = X n + bn−1 X n−1 + · · · + b0 , where n = [K(x) : K] = [K(y) : K] = deg(f ). Therefore, y is a root of g(X) = X n + bpn−1 X n−1 + · · · + bp0 . Hence, f = g and f ∈ K p [X], contrary to our hypotheses. It follows, [K(x) : K] = p · deg(f ) = deg(f (X p )). Consequently, f (X p ) is irreducible. Lemma 12.4.2: Let K be an imperfect field of characteristic p. Suppose h ∈ K[T ] is not a pth power. Then: (a) The additive group K has infinitely many congruence classes modulo K p . (b) All a ∈ K with h(T + a) ∈ K p [T ] are congruent modulo K p . (c) For each b ∈ K with h(T + b) 6∈ K p [T ] only finitely many c ∈ K satisfy h(cp + b) ∈ K p . Proof of (a): Since K is imperfect, K/K p is a nonzero vector space over the infinite field K p . Hence, K/K p is infinite. Proof of (b): Let a, b ∈ K with h(T + a), h(T + b) ∈ K p [T ]. Since h(T + a) is not a pth power, h(T + a) = g(T p ) + f (T ) where g, f ∈ K p [T ] but p - deg(f ). Then, with c = b − a, we have h(T + b) = g(T p + cp ) + f (T + c). Let f (T ) = dpm T m + dpm−1 T m−1 + · · · + dp0 with dm 6= 0. Then, p - m and the coefficient of T m−1 in f (T +c) is mcdpm +dpm−1 . Since h(T +b) and g(T p +cp ) are in K p [T ], this element belongs to K p . Finally, mp = m 6= 0. Therefore, c is a pth power. Proof of (c): Write h(T + b) 6∈ K p [T ] in the form h(T + b) = f1 (T ) + f2 (T )u2 + · · · + fr (T )ur ,

Notes

229

where r ≥ 2, u2 , . . . , ur ∈ K, 1, u2 , . . . , ur linearly independent over K p , f1 , . . . , fr ∈ K p [T ], and f2 , . . . , fr 6= 0. If c ∈ K and h(cp + b) ∈ K p , then f1 (cp ) + f2 (cp )u2 + · · · + fr (cp )ur ∈ K p . Therefore, f2 (cp ) = · · · = fr (cp ) = 0. The number of such c’s is finite. Proposition 12.4.3 ([Uchida]): Let K be a field satisfying these conditions: (1a) Every separable Hilbert subset of K is nonempty. (1b) If char(K) > 0, then K is imperfect. Then every Hilbert set of K is nonempty. In particular, K is Hilbertian. Proof: If char(K) = 0, then each Hilbert set of K is separable. Corollary 12.1.5 settles this case, so suppose char(K) = p > 0 and K is imperfect. In this case, Corollary 12.1.5 shows it suffices to prove that each Hilbert set of the form HK (f1 , . . . , fm ) with f1 , . . . , fm ∈ K[T, X] irreducible, monic, and of degree at least 2 in X, is nonempty. Assume f1 , . . . , fl are not separable in X and fl+1 , . . . , fm are separable in X. For each i, 1 ≤ i ≤ l there exists gi ∈ K[T, X] irreducible, separable and monic in X and qi , a power of p different from 1, with fi (T, X) = gi (T, X qi ). Since fi (T, X) is irreducible, gi has a coefficient hi ∈ K[T ] which is not a pth power. Choose ai ∈ K with hi (T + ai ) ∈ K p [T ] if there exists any, otherwise let ai = 0. By (1b), K is infinite. Hence, by Lemma 12.4.2(a), Sl / K p [T ], there exists b ∈ K r i=1 (ai + K p ). By Lemma 12.4.2(b), hi (T + b) ∈ p so gi (T +b, X) ∈ K(X)[T ] r K(X) [T ], i = 1, . . . , l. Thus (Lemma 12.4.2(c)), the set C of all elements c ∈ K with hi (cp + b) ∈ K p for some i, 1 ≤ i ≤ l, is Q finite. Let d(T ) = c∈C (T − c) and let gi = fi , i = l + 1, . . . , m. By Lemma 12.4.1, all gi (T p + b, X) are irreducible, monic, and separable in X. By (1a), there exists a ∈ K such that d(a) 6= 0 and gi (ap + b, X) is separable and irreducible in K[X], i = 1, . . . , m. Thus, fi (ap + b, X) is irreducible for i = l + 1, . . . , m. Now consider i between 1 and l. Since / K p . Hence, gi (ap + b, X) ∈ / K p [X]. d(a) 6= 0, we have a ∈ / C, so hi (cp + b) ∈ p p qi By Lemma 12.4.1, fi (a + b, X) = gi (a + b, X ) is irreducible over K. It follows that ap + b ∈ HK (f1 , . . . , fm ).

Exercises 1. Let K be a Hilbertian field and let f ∈ K[T1 , . . . , Tn , X] be a polynomial such that for every t ∈ K n there exists a x ∈ K with f (t, y) = 0. Prove that f (T, X) has a factor of degree 1 in X. 2. Let {Kα | α < m} be a transfinite ascending tower of Hilbertian fields. Suppose Kα+1 S is a regular extension of Kα for each α < m. Prove that the union L = α<m Kα is a Hilbertian field.

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Notes A field K is defined to be Hilbertian in [Fried-Jarden3] if each Hilbert set of K is nonempty. Most applications however use only separable Hilbertian sets. We have therefore decided to follow the convention of other authors and call a field Hilbertian if each separable Hilbert set of K is nonempty. Of course, if char(K) = 0, there is no difference between the two notions. When char(K) > 0 there is a simple relation between them. By Uchida’s theorem (Proposition 12.4.3), Hilbertian fields in the old sense are imperfect Hilbertian fields in the new sense. A great part of Section 12.1 occurs in Hilbert’s original paper [Hilbert]. [Lang3, Chapter VIII] reproduces it, though its version of our Lemma 12.2.2 contains a flaw. Following Inaba [Lang3, p. 151, Proposition 3] claims that if L is a finite separable extension of a field K of degree d and if an irreducible polynomial f ∈ L[X] has d distinct conjugates f σ1 , . . . , f σd over K, then d is irreducible in K[X]. Here is a counter-example. g = f σ1 · · · f σ√ Let γ = 3 2 and ω be a primitive 3rd root of unity. Then the polynomial f (X) = X 2 + γX + γ 2 = (X − γω)(X − γω 2 ) is irreducible over L = Q(γ). It has distinct conjugates over Q: f σ2 (X) = X 2 + γωX + γ 2 ω 2 = (X − γω 2 )(X − γ) and f σ3 (X) = X 2 + γω 2 X + γ 2 ω = (X − γ)(X − γω). Yet the product f (X)f σ2 (X)f σ3 (X) = (X 3 − 2)2 is reducible in Q[X]. Inaba treats purely inseparable extensions, though [Inaba, p. 12] claims that if L is a finite purely inseparable extension of a field K, then every Hilbert subset of L contains a Hilbert subset of K. This contradicts the example appearing in the first paragraph of Section 12.3. Finally, the results of Sections 12.3 and 12.4 are due mainly to [Uchida].

Chapter 13. The Classical Hilbertian Fields Global fields and functions fields of several variables have been known to be Hilbertian for three quarters of a century. These are the “classical Hilbertian fields”. The Hilbert property for rational function fields of one variable K = K0 (t) over infinite fields K0 is basically a combination of the MatsusakaZariski theorem (Proposition 10.5.2) with the Bertini-Noether theorem (Proposition 10.4.2). We show that every separable Hilbert subset H of K contains all elements of the form a+bt with (a, b) in a nonempty Zariski K0 -open subset of A2 (Proposition 13.2.1). Our unified approach to the proof of the irreducibility theorem for both number fields and function fields over finite fields uses Proposition 6.4.8, the main ingredient in the proof of the Chebotarev density theorem for function fields, to show that every separable Hilbert subset of a global field contains an arithmetic progression. Thus, we display an intimate connection between the irreducibility theorem and the Riemann hypothesis for curves. Our proof of the irreducibility theorem for global fields specializes the parameters of the irreducible polynomials to integral elements. This leads to the notion of Hilbertian rings. Thus, finitely generated integral domains over Z and finitely generated transcendental ring extensions of an arbitrary field K0 are Hilbertian (Proposition 13.4.1). Section 13.5 describes the connection between the classical definition and the geometric definition of the Hilbert property due to Colliot-Th´el`ene. This leads to the notion of a g-Hilbertian field. We prove that for each g ≥ 0 each global field has an infinite normal extension which is g-Hilbertian but not Hilbertian (Theorem 13.6.2). Each finite proper separable extension of a Galois extension of a Hilbertian field K is Hilbertian. This is a theorem of Weissauer. Moreover, if L1 and L2 are Galois extensions of K and neither of them contains the other, then L1 L2 is Hilbertian. The diamond theorem says even more: each extension M of K in L1 L2 which is contained in neither L1 nor in L2 is Hilbertian (Theorem 13.8.3). An essential tool in the proof is the twisted wreath product (Section 13.7).

13.1 Further Reduction The main application of Hilbertianity of a field K is to reduce Galois extensions of K(t1 , . . . , tr ) to Galois extensions of K with the same Galois group. Lemma 13.1.1 (Hilbert): Let K be a field, t1 , . . . , tr indeterminates, and Fj = K(t, xj ) a finite separable extension of K(t) with xj 6= 0, j = 1, . . . , m. Then: (a) Ar has a nonempty Zariski K-open subset U such that each every K-place

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˜ ∪ {∞} with a = ϕ(t) ∈ U (K) ˜ and ϕ(K(t)) = K(a) ∪ {∞} ϕ: Fj → K satisfies this: (a1) bj = ϕ(xj ) ∈ Ks× and ϕ(Fj ) = K(a, bj ) ∪ {∞}, j = 1, . . . , m. (a2) For each j with Fj /K(t) Galois, K(a, bj )/K(a) is Galois. Moreover, the map σ 7→ σ ¯ defined by σ ¯ (ϕ(y)) = ϕ(σy) for σ ∈ Dϕ and y ∈ Fj with ϕ(y) 6= ∞ is an isomorphism of the decomposition group Dϕ onto Gal(K(a, bj )/K). (b) K r has a separable Hilbert subset H satisfying this: For each a ∈ H ∩ ˜ ∪ {∞} satisfying (a1) and (a2), U (K) and for each K-place ϕ: Fj → K the map σ 7→ σ ¯ defined in (a2) is an isomorphism Gal(Fj /K(t)) ∼ = Gal(K(bj )/K), j = 1, . . . , m. Proof: For each j between 1 and m let fj ∈ K[T, X] be an irreducible polynomial with fj (t, xj ) = 0. Let gj0 ∈ K[T] be the leading coefficient of fj viewed as a polynomial in X. Let gj1 ∈ K[T] be a nonzero polynomial and g

(t)

k a positive integer such that gj0j1(t)k is the discriminant of fj (t, X). Write Qs g = j=1 gj0 gj1 , R = K[t, g(t)−1 ], and Sj = R[xj ]. By Lemma 6.1.2, Sj /R is a Galois ring-cover with xj a primitive element. Define U as the set of all a ∈ Ar with g(a) 6= 0. ˜ ∪ {∞} be a K-place such that a = ϕ(t) ∈ U (K) ˜ and Let ϕ: Fj → K ϕ(K(t)) = K(a)∪{∞} (By Lemma 2.2.7, there exists ϕ with these properties ˜ for each a ∈ U (K).) Put R = Oϕ ∩ K(t). By Remark 6.1.7, R[xj ]/R is a ring-cover and ϕ(Fj ) = K(a, bj ) ∪ {∞}. This proves (a1). Assertion (a2) then follows from Lemma 6.1.4. The set H = HK (f1 , . . . , fm ) satisfies the requirements of (b). Consider an arbitrary field K. Let hi ∈ K[T, X] be irreducible with degX (hi ) ≥ 2, i = 1, . . . , m, and let 0 6= g ∈ K[T ]. Put 0 HK (h1 , . . . , hm ; g) = {a ∈ K | g(a) 6= 0 and

m Y

hi (a, b) 6= 0 for each b ∈ K}.

i=1 0 (h1 , . . . , hm ; g) is nonempty. The next If K is Hilbertian, then each set HK lemma shows the converse:

Lemma 13.1.2: Let f ∈ K[T1 , . . . , Tr , X] be an irreducible polynomial, monic and separable in X with degX (f ) ≥ 2. Then there exist absolutely irreducible polynomials h1 , . . . , hm ∈ K[T1 , . . . , Tr , X], monic and separable 0 (h1 , . . . , hm ) ⊆ HK (f ). in X, with degX (hi ) ≥ 2, i = 1, . . . , m, such that HK Proof: We prove the lemma in two ways. The first proof uses decomposition groups. The second one uses the beginning of what we later call “the intersection decomposition procedure” (Section 21.1.1). In both proofs we start from algebraically independent elements t1 , . . . , tr over K and write E = K(t). Proof A: Let x be a root of f (t, X) in Es , F = E(x), and Fˆ the splitting field of f (t, X) over E. Choose a primitive element z for Fˆ /E. List all proper

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233

extensions of E in Fˆ which are regular over K as E1 , . . . , Em . For each i between 1 and m choose a primitive element yi for Ei /E which is integral over K[t]. Thus, there is an irreducible polynomial hi ∈ K[T, X] monic and separable in X with hi (t, yi ) = 0 and degX (hi ) = [Ei : E] ≥ 2. By Example 2.6.11, hi is absolutely irreducible. Lemma 13.1.1(a) gives a nonzero polynomial g ∈ K[T] such that for ˜ ∪ {∞} with ϕ(t) = a each a ∈ K r with g(a) 6= 0 each K-place ϕ: Fˆ → K has this property: (2a) ϕ(F ) = K(ϕ(x)) ∪ {∞}. (2b) Let N be the residue field of Fˆ under ϕ. Then the decomposition group of ϕ is isomorphic to Gal(N/K). 0 0 Claim: HK (h1 , . . . , hm ; g) ⊆ HK (f ). Indeed, let a ∈ HK (h1 , . . . , hm ; g). ˜ ∪ {∞} with ϕ(t) = a Use Lemma 2.2.7 to construct a K-place ϕ: Fˆ → K ˜ Denote and ϕ(K(t)) = K ∪ {∞}. Since f (a, X) is monic, b = ϕ(x) is in K. the decomposition field of ϕ in Fˆ by E 0 . The restriction of ϕ to E 0 maps E 0 onto K ∪ {∞} (Remark 6.1.6). Hence, by Lemma 2.6.9(b), E 0 /K is regular. Assume E 0 6= E. Then E 0 = Ei for some i between 1 and m. Hence, c = ϕ(yi ) ∈ K and hi (a, c) = 0. This contradiction to the choice of a proves that E 0 = E. Denote the residue field of Fˆ under ϕ by N . By the preceding paragraph and (2b), [N : K] = [Fˆ : E]. By (2a), the residue field of F under ϕ is K(b). Then [K(b) : K] ≤ [F : E] and [N : K(b)] ≤ [Fˆ : F ]. Hence, [K(b) : K] = [F : E]. Therefore, f (a, X) is irreducible over K, as claimed. Finally, use Lemma 12.1.4 to eliminate g. Qn Proof B: Let f (t, X) = i=1 (X − xi ) be the factorization of f (t, X) in E Qs [X]. Consider a nonempty proper subset I of {1, . . . , n} and let fI (X) = / E[X]. Hence, fI (X) has a coeffii∈I (X − xi ). Since f is irreducible, fI ∈ cient yI 6∈ E. Thus, there is an irreducible polynomial gI ∈ K[T, X], monic and separable in X with degX (gI ) ≥ 2, such that gI (t, X) = irr(yI , E) ∈ K[t, X]. ˜ gI = gI,1 . . . gI,k , where each Suppose gI factors nontrivially over K: ˜ gI,j ∈ K[T, X] is irreducible and k ≥ 2. Since gI (t, X) is monic and separable in X, the factors gI,j are relatively prime. By Hilbert Nullstellensatz (Proposition 9.4.2), V (gI,i ) 6⊆ V (gI,j ) for i 6= j. Thus (Lemma 10.1.2), WI = V (gI,1 , . . . , gI,k ) is a Zariski K-closed subset of Ar+1 of dimension at most r − 1. ˜ by Denote the union of all WI such that gI factors nontrivially over K W . Let A be the K-Zariski closure of the projection of W on the first r coordinates. Then dim(A) ≤ r − 1. Hence, there exists a nonzero polynomial q ∈ K[T] which vanishes on A. Now list all gI which are absolutely irreducible as h1 , . . . , hm . Then

(3)

0 0 (h1 , . . . , hm ; q) ⊆ HK (gI | I 6= ∅, I ⊂ {1, . . . , n}). HK

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Indeed, suppose a is in the left hand side of (3). Then gI (a, b) 6= 0 for each I with gI absolutely irreducible and each b ∈ K. Assume there is an I with ˜ and there is a b ∈ K with gI (a, b) = 0. Then, in the gI reducible over K above notation, there exists j between 1 and k with gI,j (a, b) = 0. As gI is irreducible over K, all gI,i are K-conjugate to gI,j . Hence, gI,i (a, b) = 0 for each i. Thus, (a, b) ∈ W (K) and so a ∈ A(K). Therefore, q(a) = 0, in contradiction to the assumption on a. Next we prove that the right hand side of (3) is contained in HK (f ). Assume for a ∈ K r that f (a, X) = p(X)q(X) factors nontrivially in K[X] with p and q monic. Extend the K-specialization t → a to a K-specialization ˜ (Propositions 2.3.1 and . . , cn ) with c1 , . . . , cn ∈ K (t, x1 , . . . , xn ) → (a, c1 , .Q n 2.3.3). Then, f (a, X) Q = i=1 (X − ci ). For some nonempty proper subset I of {1, . . . , n}, p(X) = i∈I (X − ci ), the polynomial fI (X) maps onto p(X), and yI maps onto a coefficient b of p(X). Then b lies in K and satisfies gI (a, b) = 0. Thus, a does not belong to the right hand side of (3). Finally, use Lemma 12.1.4 to eliminate q. The following result is one ingredient of the proof of Lemma 13.1.4. That lemma will be used in the proof of the diamond theorem (Theorem 13.8.3). Lemma 13.1.3: Let K be an infinite field, t1 , . . . , tr algebraically independent elements over K, E0 = K(t), and E a finite separable extension of E0 . Then there are fields L, F0 , and F with these properties: (a) L/K is a finite Galois extension. ˆ where E ˆ is the Galois closure of E/K. (b) F = EL, (c) F0 L = F and F0 ∩ L = K. (d) Gal(F/E0 ) = Gal(F/L(t)) n Gal(F/F0 ) (Definition 13.7.1). (e) There is an L-place ϕ: F → L ∪ {∞} with ϕ(F0 ) = K ∪ {∞}. (f) Both F0 /K and F/L are regular extensions. F0

F { {{ { {E {{{ L(t) E0 ˆ be the Galois closure of E/E0 . Choose a primitive element Proof: Let E ˆ x for E/E0 . Remark 6.1.5 gives a nonzero polynomial q ∈ K[T] such that K[t, q(t)−1 , x]/K[t, q(t)−1 ] is a Galois ring-cover with x being a primitive element. Choose a ∈ K r with q(a) 6= 0. Extend the specialization t → a to a K-place ψ0 : E0 → K ∪ {∞} (Lemma 2.2.7). Then extend ψ0 to a place ˆ into K ˜ ∪ {∞}. By Remark 6.1.7, ψ(E) ˆ = L ∪ {∞} where L = ψ of E ˆ K(ψ(x)) is a finite Galois extension of K. Let F = EL. By Remark 6.1.7, −1 −1 L[t, q(t) , x]/L[t, q(t) ] is a ring-cover and ψ extends to an L-place ϕ: F → L ∪ {∞}. Hence, by Lemma 2.6.9(b), F/L is regular.

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235

Denote the decomposition field of ψ over E0 by F0 . By Remark 6.1.6, ψ(E0 ) = K∪{∞}. Hence, by Lemma 2.6.9(b), F0 /K is regular. In particular, F0 ∩ L = K. By Remark 6.1.7, L[t, q(t)−1 ]/K[t, q(t)−1 ] is a ring-cover. Hence, L[t, q(t)−1 , x] is the integral closure of K[t, q(t)−1 ] in F . Suppose σ is in the inertia group of ϕ over E0 ; that is, σ ∈ Gal(F/F0 ). Let a ∈ L. Then ψ(σa) = σ ¯ ψ(a) = ψ(a). Since ψ is an isomorphism on L (see discussion after Proposition 2.3.1), σa = a. Thus, σ lies in the inertia group of ψ over L(t). Since L[t, q(t)−1 , x]/L[t, q(t)−1 ] is a ring-cover, the latter is trivial. Therefore, σ = 1. Thus, by Remark 6.1.6, Gal(F/F0 ) ∼ = Gal(L/K) and therefore F0 L = F . Finally, (d) is a reinterpretation of (c). Let f ∈ K[X] be a polynomial. We say f is Galois over K if f is irreducible, separable, and K(x) is the splitting field of f over K for every root x of f . If L0 is an extension of L and f is irreducible over L0 , then L0 is linearly disjoint from L(x) over L. Hence, f is Galois over L0 and Gal(L0 (x)/L0 ) ∼ = Gal(L(x)/L). Lemma 13.1.4: Suppose K is an infinite field. Let f ∈ K[T1 , . . . , Tr , X] be an irreducible polynomial which is monic and separable in X. Then there are a finite Galois extension L of K and an absolutely irreducible polynomial g ∈ K[T, X] which as a polynomial in X is monic, separable, and Galois over L(T) such that K r ∩ HL (g) ⊆ HK (f ). Proof: Let t1 , . . . , tr be algebraically independent elements over K, E0 = K(t), x ∈ E0,s a root of f (t, X), and E = E0 (x). Let L, F0 and F be the fields given by 13.1.3. Choose a primitive element y for F0 /E0 which is integral over K[t]. Then there is an irreducible polynomial g ∈ K[T, X], monic and separable in X such that g(t, y) = 0. By Example 2.6.11, g is absolutely irreducible. Since F = L(t, y), the polynomial g(t, X) is Galois over L(t). Lemma 13.1.1(a) gives q ∈ K[T], q 6= 0 such that for each K-place ˜ ∪ {∞} with a = ϕ(t) 6= ∞, ϕ(K(t)) = K ∪ {∞}, and q(a) 6= 0 ϕ: F → K this holds: (4) Let b = ϕ(x) and c = ϕ(y). Then ϕ(E) = K(b) ∪ {∞}, ϕ(F0 ) = K(c) ∪ {∞}, and ϕ(F ) = L(c) ∪ {∞}. Suppose a ∈ K r ∩ HL (g; q). Then g(a, Y ) is irreducible over L, q(a) 6= 0,, q(a) 6= 0, and g(a, c) = 0. Use Lemma 2.2.7 to construct a K-place ˜ ∪ {∞} with ϕ(t) = a and ϕ(K(t)) = K ∪ {∞}. By Lemma 13.1.3, ϕ: F → K degY (g) = [F0 : E0 ] = [F : L(t)]. Hence, [L(c) : K] = [L(c) : L][L : K] = degY (g)[L : K] = [F : L(t)][L(t : K(t)] = [F : E0 ]. In addition, [K(b) : K] ≤ [E : E0 ] and [L(c) : K(b)] ≤ [F : E]. Hence, [K(b) : K] = [E : E0 ]. Since f (a, b) = 0, the polynomial f (a, X) is irreducible over K.

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Finally, use Lemma 12.1.4 to eliminate q.

Problem 13.1.5 ([D`ebes-Haran, p. 284]): Let K be an infinite field and H a separable Hilbertian subset of K r . Does there exist an absolutely irreducible polynomial f ∈ K[T1 , . . . , Tr , X] which is monic and separable in X such that HK (f ) ⊆ H?

13.2 Function Fields over Infinite Fields Function fields of several variables over infinite fields are Hilbertian. This is a consequence of the following result. Proposition 13.2.1: Let K0 be an infinite field, t an indeterminate, and K = K0 (t). Then every separable Hilbert subset H of K contains a set {a + bt | (a, b) ∈ U (K0 )} for some nonempty Zariski K0 -open subset U of A2 . Thus, every Hilbert set of K is nonempty and K is Hilbertian. Proof: The first statement implies that every separable Hilbert subset of K is nonempty. If char(K) > 0, then K is imperfect. Hence, by Uchida (Proposition 12.4.3), every Hilbert set of K is nonempty and K is Hilbertian. Lemmas 12.1.6 and 13.1.2 reduce the proof of the first statement to the following one: For an absolutely irreducible polynomial f ∈ K[X, Y ], monic and separable in Y , with degY (f ) > 1, there exists a nonempty Zariski K0 -open subset U of A2 such that f (a + bt, Y ) has no zeros in K for each (a, b) ∈ U (K0 ). If necessary, multiply f by a suitable element of K and make a linear change in the variable Y to assume that f (X, Y ) = g(t, X, Y ) ∈ K0 [t, X, Y ] is ∂g 6= 0. Then g(t, Z0 +Z1 t, Y ) is an an absolutely irreducible polynomial and ∂Y ˜ Y ], where L = K0 (Z0 , Z1 ) (Proposition 10.5.4). irreducible polynomial in L[t, By Proposition 9.4.3, there exists a nonzero polynomial h ∈ K0 [Z0 , Z1 ] such ˜ 0 [t, Y ] for every a, b ∈ K0 such that that g(t, a + bt, Y ) is irreducible in K h(a, b) 6= 0. In particular, f (a + bt, Y ) has no zeros in K if h(a, b) 6= 0. Proposition 13.2.1 implies the analog to Corollary 12.1.5 for separable Hilbert sets: Proposition 13.2.2: Let K be a field. Suppose each separable Hilbert subset of K of the form HK (f ) with irreducible f ∈ K[T, X], separable, monic, and of degree at least 2 in X, is nonempty. Then K is Hilbertian. Proof: By Lemma 12.1.6, it suffices to consider a separable irreducible polynomial f ∈ K[T1 , . . . , Tr , X] and to prove that HK (f ) 6= ∅. The case r = 1 is covered by the assumption of the Proposition. Now, suppose r ≥ 2 and the statement holds for r − 1. The assumption of the Proposition implies K is infinite. Hence, by Proposition 13.2.1, f (T1 , . . . , Tr−1 , g(T1 , . . . , Tr−1 ), X) is irreducible and separable in K(T1 , . . . , Tr−1 )[X] for some g ∈ K(T1 , . . . , Tr−1 ). 1 ,...,Tr−1 ) Write g(t1 , . . . , Tr−1 ) = gg10 (T (T1 ,...,Tr−1 ) with g0 , g1 ∈ K[T1 , . . . , Tr−1 ]. Lemma

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12.1.6 gives an irreducible polynomial h ∈ K[T1 , . . . , Tr−1 , X], separable, monic, and of degree at least 2 in X, such that HK (h) ⊆ HK f (T1 , . . . , Tr−1 , g(T1 , . . . , Tr−1 , X)); g0 (T1 , . . . , Tr−1 ) . The induction hypotheses gives a1 , . . . , ar−1 ∈ K such that h(a1 , . . . , ar−1 , X) is irreducible and separable in K[X]. Let ar = g(a1 , . . . , ar−1 ). Then f (a1 , . . . , ar , X) is well defined and irreducible in K[X].

13.3 Global Fields For K a global field it is easy to choose the hi ’s in Lemma 13.1.2 with coefficients in the ring of integers OK . If, in particular, K is a number field, r = 1, and each of the curves hi = 0 has positive genus, a celebrated theorem 2 of Siegel implies that each of the hi ’s has only finitely many zeros (a, b) ∈ OK 0 ([Lang3, p. 121] or [Robinson-Roquette]). In this case HK (h1 , . . . , hm ) is clearly infinite. If, however, the curve hi = 0 is of genus zero, we may use Riemann-Hurwitz to replace hi (T, x) by gi (T, X) = hi (m(T ), X) for some m(T ) ∈ OK [T ], so that gi (T, X) = 0 has positive genus. Thus, Siegel’s theorem gives Hilbert’s theorem effortlessly. Of course, Siegel’s theorem, a deep result in arithmetic, applies only to number fields (more generally, to fields which are finitely generated over Q [Lang3, p. 127, Thm. 4]). Besides, its power masks subtle connections between the irreducibility theorem and other arithmetic results. Our approach to the Hilbert irreducibility theorem for global fields is based on the Chebotarev density theorem for function fields over finite fields. More accurately, we use a special case of Proposition 6.4.8. As a bonus we will prove that each Hilbert set over Q contains arithmetic progressions. We start the proof with a weak consequence of Bauer’s theorem. In keeping with our elementary treatment we use only Euclid’s argument for proving the infinitude of primes: Lemma 13.3.1: Let L/K be a finite separable extension of global fields. ¯ p for every ¯P = K Then there exist infinitely many primes p of K such that L prime P of L lying over p. Proof: Assume, without loss, that K = Q if char(K) = 0 and K = Fp (t) if char(K) = p. In particular, OK is a principal ideal domain with only finitely many units. Replace L by the Galois hull of L/K to assume that L/K is Galois. Consider a primitive element z ∈ OL for the extension L/K with discriminant d ∈ OK (Section 6.1). Suppose p1 , . . . , pk are prime ideals of OK satisfying the conclusion of the lemma. For each i between 1 and k choose a nonzero πi ∈ pi and let π = d · π1 · · · πk . Consider also f (X) = irr(z, K) = X n + cn−1 X n−1 + · · · + c0 . Since c0 6= 0 and OK has only finitely many units, there exists a positive integer m such that n−1 mn m π + cn−1 cn−2 π m(n−1) + · · · + 1 c−1 0 f (c0 π ) = c0 0

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is a nonunit of OK . Therefore, this element has a prime divisor p = pk+1 different from p1 , . . . , pk and relatively prime to d. Let P be a prime ideal of OL lying over pk+1 . Denote the reduction modulo P by a bar. Then f¯(X) ¯ p , namely c0 π m modulo p, which we may assume to be z¯. By has a root in K ¯ p (¯ ¯ p. ¯P = K z) = K Remark 6.1.7, L Lemma S 13.3.2: For H a proper subgroup of a finite group G, the set G r σ∈G σ −1 Hσ is nonempty. For f ∈ K[X] an irreducible separable polynomial with roots x1 , . . . , xn and splitting field N = K(x1 , . . . , xn ), there exists σ ∈ Gal(N/K) such that σxi 6= xi , for each i = 1, . . . , n. Proof: Let H1 , . . . , Hm be the distinct subgroups of G conjugate to H. Assume without loss that m ≥ 2. Then m is equal to the index of the normalizer of H. (G : H). The intersection H1 ∩ · · · ∩ Hm Sm Thus, mP≤ m contains 1, so | i=1 Hi | < i=1 |Hi | = m · |H| ≤ |G|. With Gal(N/K) = G and Gal(N/K(x1 )) = H the conjugates of H are Gal(N/K(x1 )), . . . , Gal(N/K(x m )). The last statement of the lemma follows S by choosing σ ∈ G r τ ∈G τ −1 Hτ . ¯ = Fq , t an indeterminate, and Lemma 13.3.3: Let q be a prime power, K ¯ ¯ ¯ ¯ z) where g¯(t, z) = 0 with E = K(t). Consider a Galois extension F = K(t, ¯ ¯ is algebraically closed in g ∈ K[T, Z] an irreducible polynomial. Suppose K ¯ ¯ g ), and C be a conjugacy class in Gal(F¯ /E). F . Let d = deg(¯ g ), m = degZ (¯ ¯ ¯ ¯ Then the number N of primes of E/K of degree 1, unramified in F , with Artin symbol equal to C satisfies (1)

N − |C| q < 10d2 |C|√q. m

¯ is algebraically closed in F¯ , g¯ is Proof: Write gF¯ = genus(F¯ ). Since K absolutely irreducible (Corollary 10.2.2). Hence, g(T, Z) = 0 defines a curve ¯ Recalling that E ¯ has genus 0, Proposition 6.4.8, with of degree d over K. k = 1, gives the inequality h i N − |C| q < 2|C| (m + gF¯ )q 12 + mq 14 + gF¯ + m . m m Combining this with the inequality gF¯ ≤ 12 (d − 1)(d − 2) for the genus of F¯ (Corollary 5.3.5) we obtain (1). An absolute value of a field K is a map | |: K → R which satisfies the following conditions for all x, y ∈ K: (2a) |x| ≥ 0 and |x| = 0 if and only if x = 0. (2b) |xy| = |x||y|. (2c) |x + y| ≤ |x| + |y|. (2d) There is an a ∈ K with |a| 6= 1.

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If, instead of (2c), | | satisfies the stronger condition (2c’) |x + y| ≤ max(|x|, |y|), it is ultrametric (or non-archimedean), otherwise, it is metric (or archimedean). An absolute value | |0 is equivalent to | | if there exists c > 0 such that |x|0 = |x|c for all x ∈ K. If | | is ultrametric, then O| | = {x ∈ K | |x| ≤ 1} is a valuation ring of K. In particular, |n| ≤ 1 for each positive integer n. Conversely, suppose |n| ≤ 1 for each positive integer n. Let z be an element of K with |z| ≤ 1. Pk Then |1 + z|k = i=0 ki z i ≤ k + 1. Hence, |1 + z| ≤ (k + 1)1/k . Letting k go to infinity, we find that |1 + z| ≤ 1. Next let x, y ∈ K with |x| ≤ |y| and y 6= 0. Then |x + y| = |1 + xy ||y| ≤ |y| = max(|x|, |y|). Consequently, | | is ultrametric. In particular, suppose char(K) = p > 0. Then np−1 = 1 for each positive integer n not divisible by p. Hence, |n| = 1. Therefore, | | is ultrametric. If K is a global field, the map | | → O| | is a bijection between the equivalence classes of ultrametric absolute values of K and the valuation rings of K. In particular, if K is a function field of one variable over a finite field K0 , then the equivalence classes of absolute values of K bijectively correspond to the prime divisors of K/K0 . If K is a number field, then the equivalence classes of ultrametric absolute values bijectively correspond to the prime ideals of OK . In addition, each embedding σ: K → C defines a metric absolute value | |σ of K: |x|σ = |σx|, where | | is here the usual absolute value of C. Each metric absolute value of K is equivalent to some | |σ and there are at most [K : Q] such equivalence classes [Cassels-Fr¨ohlich, p. 57, Thm.]. In each case we call an equivalence class of an absolute value a prime of K. For each prime p of K we choose an absolute value | |p which represents it. The set of all primes of a global field K satisfies the strong approximation theorem: Let p0 , p1 , . . . , pn be distinct primes of K. For each i between 1 and n consider an element ai of K and let ε > 0. Then there exists x ∈ K such that |x − ai |pi < ε, i = 1, . . . , n, and |x|p ≤ 1 for each ohlich, p. 67]. In the function field prime p not in {p0 , p1 , . . . , pn } [Cassels-Fr¨ case, this theorem is also a consequence of the strong approximation theorem (Proposition 3.3.1). Generalizations of the following lemma appear as ingredients in the Galois stratification procedure of Chapter 30. Lemma 13.3.4: Let K be a global field and f ∈ K[T, X] an absolutely irreducible polynomial, monic and separable in X, with degX (f ) > 1. Then K has infinitely many ultrametric primes p for which there is an ap ∈ OK with this property: if a ∈ OK satisfies a ≡ ap mod p, then f (a, b) 6= 0 for every b ∈ K. Proof: Assume without loss that f ∈ OK [T, X]. Let E = K(t), with t an indeterminate, and let x1 , . . . , xn be the roots of f (t, X) in Es . Denote the algebraic closure of K in the Galois extension F = E(x1 , . . . , xn ) of E by L.

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Then F is a regular extension of L; write it as F = L(t, z) with z integral over OL [t]. Thus, there is an absolutely irreducible polynomial g ∈ L[T, X] with g(t, X) = irr(z, L(t)). By Lemma 13.3.1, the set of ultrametric primes ¯ p for each prime P of L over p, is infinite; call it ¯P = K p of K for which L A1 . The rest of the proof derives the result from the case where L replaces K and g replaces f . Part A: Geometrically exceptional primes. Let A2 be the restriction to K of the set of ultrametric primes P of L for which both f mod P and g mod P are defined, separable in X, and absolutely irreducible. By Bertini-Noether (Proposition 9.4.3), A2 is a cofinite set. Express z as a polynomial in x1 , . . . , xn (with coefficients that are ratios of elements of OK [t]) and express each xi as a polynomial in z (with coefficients that are ratios of elements of OL [t]). Denote the cofinite set of ultrametric primes of K which divide none of the denominators of these coefficients by A3 . ∂g (t, z)) ∈ OL [t]. Denote the Finally, consider d(g; t) = normF/LE ( ∂X cofinite set consisting of the restriction to K of the ultrametric primes P of L for which deg(d(g; t)) = deg(d(g; t) mod P) by A4 . The set A = A1 ∩ A2 ∩ A3 ∩ A4 is infinite. Part B: Reduction modulo primes of A. For p ∈ A and P a prime of L ¯P = K ¯ p to a place ϕ of F by choosing over p, extend the residue map OL → L ¯ ¯ t = ϕ(t) as a transcendental element over Kp . Denote ϕ(x) by x ¯ for every ¯ z ) = E(¯ ¯ x1 , . . . , x ¯=K ¯ p (t¯) and F¯ = E(¯ ¯n ), where x ∈ F finite under ϕ. Let E ¯n are the distinct roots of the irreducible polynomial f¯(t¯, X). x ¯1 , . . . , x ¯ Hence, By the choice of p, the polynomial f¯(t¯, X) is irreducible over E. ¯ ¯ ¯i , i = 1, . . . , n. by Lemma 13.3.2, there exists σ ∈ Gal(F /E) such that σ x ¯i 6= x ¯ by Con(σ). Next note that since Denote the conjugacy class of σ in Gal(F¯ /E) ¯ p is algebraically closed g¯ is absolutely irreducible and g¯(t¯, z¯) = 0, the field K ∂ g ¯ ¯ (t, z¯)) with coefficients in F¯ . In addition, the polynomial d(¯ g ; t¯) = NF¯ /E¯ ( ∂X ¯ ¯ ¯ ∈ Kp is not a root of d(¯ g ; t¯), then in Kp has the same degree as d(g; t). If a ¯ ¯ ¯ of E is unramified in the prime pa¯ corresponding to the specialization t → a ¯ which ramify in F¯ F¯ (Lemma 6.1.8). The number of primes of degree 1 of E is therefore bounded by 1 + deg(d(g; t)). ¯p ¯p ∈ K Let A0 be the set of all p ∈ A finite at t for which there exists a ¯ ¯ with d(f¯; a ¯p ) 6= 0 such that the Artin symbol F /E (here a ¯ = a ¯p ) correpa ¯

sponding to the prime pa¯ equals Con(σ). By Lemma 13.3.3, A0 is cofinite in A. For each p ∈ A0 there exists a prime Q of F¯ lying over pa¯ with these ¯ p| properties: for q = |K ¯ (3a) σx ≡ xq mod Q Qfor every x ∈ F integral with respect to Q; xi − x ¯j ), and, since d(f¯; t¯) = i6=j (¯ (3b) x ¯i 6≡ x ¯j mod Q for every i 6= j. Part C: Finding ap . For each i between 1 and n let j be an integer with ¯j . Then i 6= j. Hence, by (3a) and (3b), x ¯i ≡ 6 x ¯qi mod Q. That is, σ¯ xi = x

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241

¯ p. the polynomial f¯(¯ ap , X) has no roots in K ¯p by ϕ. If a ∈ OK Let ap be an element of Ok which is mapped to a satisfies a ≡ ap mod p, then a ¯ = a ¯p , and therefore f (a, b) 6= 0 for each b ∈ OK . Since f (a, X) is monic, this proves f (a, X) has no zero in K. Let K be a global field. An arithmetic progression of OK is a set of the form a + a, where a ∈ OK and a is an ideal of OK . Theorem 13.3.5: Let K be a global field and H a separable Hilbert subset of K. Then: (a) H contains an arithmetic progression of OK . (b) Let q0 , q1 , . . . , qn be distinct primes of K, b1 , . . . , bn elements of K, and ε > 0. Then there exists x ∈ H with |x − bi |qi < ε for i = 1, . . . , n and |x|q ≤ 1 for each prime of K not in {q0 , q1 , . . . , qn }. (c) The intersection of H with each arithmetical progression of OK is nonempty. Proof: By Lemma 13.1.2 there exist absolutely irreducible polynomials h1 , . . . , hm ∈ OK [T, X], monic and separable in X, with degX (hi ) > 1, 0 (h1 , . . . , hm ) ⊆ H. Apply Lemma 13.3.4 to find disi = 1, . . . , m, and HK tinct prime ideals p1 , . . . , pm of OK and elements a1 , . . . , am ∈ OK such 0 that x ∈ HK (hi ) for each x ∈ K with |x − ai |pi < 1, i = 1, . . . , m. The Chinese remainder theorem [Lang7, p. 94] produces an element a ∈ OK with a+pi = ai +pi for i = 1, . . . , m. Thus, with a = p1 . . . pm , we have a+a ⊆ H. This proves (a). For the proof of (b) choose p1 , . . . , pm above not in {q0 , q1 , . . . , qn }. Then use the strong approximation theorem for global fields [Cassels-Fr¨ohlich, p. 67] to find x ∈ K with |x − ai |pi < 1, i = 1, . . . , m, |x − bj |qj < ε, j = 1, . . . , n, and |x|q ≤ 1 for each prime q not in {p1 , . . . , pm , q0 , . . . , qn }. Then x ∈ H and |x|pi ≤ 1 for i = 1, . . . , m, as desired. Each arithmetical progression in OK has the form A = {x ∈ OK | vqi (x− a) ≥ ki , i = 1, . . . , r} for some a ∈ OK , ultrametric primes q1 , . . . , qr which are finite on OK , and positive integers k1 , . . . , kr . If K is a number field, choose q0 to be a metric prime. If K is a function field of one variable over a finite field, choose q0 to be a prime which is not finite on OK . Statement (b) gives x ∈ A with vq (x) ≥ 0 for all ultrametric primes q of K which are finite on OK . Thus, x ∈ OK .

13.4 Hilbertian Rings We call an integral domain R with quotient field K Hilbertian if every separable Hilbert subset of K r contains elements all of whose coordinates are in R. In this case, each overring of R is Hilbertian. In particular, K is Hilbertian.

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Proposition 13.4.1: Let R be an integral domain with quotient field K. Suppose either R is finitely generated over Z or R is finitely generated over a field K0 and K/K0 is transcendental. Then R is Hilbertian. Proof: Let R0 be either Z or K0 . In the former case let K0 = Q. By assumption, R = R0 [u1 , . . . , un ]. Assume without loss u1 , . . . , um with m ≤ n form a transcendental base for K/K0 . By Proposition 12.3.3, every separable Hilbertian subset of K r contains a separable Hilbertian subset of K0 (u1 , . . . , um ). We may therefore assume that u1 , . . . , un are algebraically independent over K0 . Consider an irreducible polynomial f ∈ K[T1 , . . . , Tr , X] which is separable in X. We have to prove that HK (f ) ∩ Rr 6= ∅ (Lemma 12.1.6). There are several cases to consider. Case 1: r = 1 and n = 0. Then R = R0 = Z. By Theorem 13.3.5 there exists a ∈ R such that f (a, X) is irreducible. Case 2: r = 1, n = 1, and R0 = K0 . If K0 is finite, then K is global and R = OK . Thus, Theorem 13.3.5 gives a ∈ R with f (a, X) irreducible. If K0 is infinite, then Proposition 13.2.1 gives a, b ∈ K0 such that f (a + bu1 , X) is irreducible. Case 3: r = 1, n = 1, and R0 = Z. Proposition 13.2.1 gives a nonempty Zariski Q-open subset U of A2 such that f (a + bu1 , X) is irreducible for all (a, b) ∈ U (Q). Since Z is infinite, we may choose (a, b) in U (Z). Case 4: r = 1 and n ≥ 2. Then R0 [u1 , . . . , un−1 ] is an infinite ring. Hence, by Proposition 13.2.1, it has elements a, b such that f (a + bu1 , X) is irreducible. Case 5: r ≥ 2. Consider f as a polynomial in Tr , X with coefficients in the infinite ring R0 [u, T1 , . . . , Tr−1 ]. As such, f is irreducible. Replace R0 and K0 in Cases 2, 3, and 4 by R0 [u, T1 , . . . , Tr−1 ] and K0 (u, T1 , . . . , Tr−1 ) to find g ∈ R0 [u, T1 , . . . , Tr−1 ] such that f (T1 , . . . , Tr−1 , g(T1 , . . . , Tr−1 ), X) is irreducible. Now use an induction hypothesis on r to find a1 , . . . , ar−1 ∈ R with f (a1 , . . . , ar−1 , g(a1 , . . . , ar−1 ), X) irreducible. Then ar = g(a1 , . . . , ar−1 ) is in R and f (a, X) is irreducible. The quotient fields of the rings mentioned in Proposition 13.4.1 are the classical Hilbertian fields. Theorem 13.4.2: Suppose K is a global field or a finitely generated transcendental extension of an arbitrary field K0 . Then K is Hilbertian. Moreover, each Hilbert set of K is nonempty. Proof: As a consequence of Proposition 13.4.1, K is Hilbertian. To prove the second statement, assume char(K) = p > 0. Then K is imperfect. Hence, by Uchida (Proposition 12.4.3), every Hilbertian set of K is nonempty. The proof of Proposition 13.4.1 gives another useful version of Theorem 13.4.2:

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Proposition 13.4.3: Let K be a finitely generated separable extension of a field K0 with m = trans.deg(K/K0 ) > 0. Let r ≤ m and H a separable Hilbert subset of K r . Then H contains a point (u1 , . . . , ur ) with u1 , . . . , ur algebraically independent over K0 . Proof: Choose a separating transcendence base t1 , . . . , tm for K/K0 . Then K/K0 (t) is a finite separable extension. By Corollary 12.2.3, H contains a separable Hilbert subset of K0 (t)r . We may therefore assume K = K0 (t). By Lemma 12.1.6, we may assume H = HK (f ) with an irreducible f ∈ K[T1 , . . . , Tr , X], monic and separable in X. For m = 1 and K0 finite, Theorem 13.3.5 gives u ∈ H transcendental over K0 . For m = 1 and K0 infinite, Proposition 13.2.1 does the job. Assume m ≥ 2. Then K 0 = K0 (t2 , . . . , tm , T2 , . . . , Tr ) is an infinite field, t1 is transcendental over K 0 , and K 0 (t1 ) = K(T2 , . . . , Tr ). Consider f as an irreducible polynomial in T1 , X over K 0 (t1 ). Proposition 13.2.1 gives nonzero g, h ∈ K0 [t2 , . . . , tm , T2 , . . . , Tr ] such that f1 (T2 , . . . , Tr , X) = f g(T2 , . . . , Tr ) + h(T2 , . . . , Tr )t1 , T2 , . . . , Tr , X

is irreducible over K. Let K1 = K0 (t1 ). Apply the induction hypothesis to K1 and f1 instead of to K and f and find u2 , . . . , ur in K algebraically independent over K1 such that f (g(u2 , . . . , ur ) + h(u2 , . . . , ur )t1 , u2 , . . . , ur , X) is irreducible over K. Put u1 = g(u2 , . . . , ur ) + h(u2 , . . . , ur )t1 . Then f (u, X) is irreducible over K and K0 (u1 , u2 , . . . , ur ) = K0 (t1 , u2 , . . . , ur ). Hence, u1 , . . . , ur are algebraically independent over K0 , as desired. Remark 13.4.4: More Hilbertian rings. Exercise 4 proves that each valuation ring of a Hilbertian field is Hilbertian. Geyer extends this result to several valuations of rank 1. He considers a Hilbertian field K, nonequivalent absolute values | |1 , . . . , | |n of K, and a separable Hilbert subset H of K r . Let a1 , . . . , ar be tuples in K r and let ε > 0. Then there exists x ∈ H such that |x − ai |i < ε, i = 1, . . . , n [Geyer1, Lemma 3.4]. In particular, if v1 , . . . , vn are valuations of rank 1, then their holomorphy ring, R = {a ∈ K | vi (a) ≥ 0, i = 1, . . . , n} is Hilbertian. Proposition 19.7 of [Jarden14] generalizes Geyer’s result. Here one considers valuations v1 , . . . , vm and orderings αi , i = 1, . . . , m, and −cj <j x − bj <j cj , j = 1, . . . , n. Moreover, the holomorphy ring of arbitrary finitely many valuations of K is Hilbertian [Jarden14, Proposition 19.6]. In contrast, holomorphy ring of infinitely many valuations of a field K may not be Hilbertian even if K is Hilbertian. For example, if K0 is a finite

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field, the holomorphy ring of all valuations of K0 (t) is K0 . The latter field is not Hilbertian, although K0 (t) is Hilbertian. More interesting examples can be found in Example 15.5.6. The following Lemma improves Exercise 2 of Section 12: Lemma 13.4.5: Let m be a cardinal number and {Kα | α < m} a transfinite sequence of fields. Suppose that for each α < m the field S Kα+1 is a proper finitely generated regular extension of Kα . Then K = α<m Kα is a Hilbertian field. Proof: Let f ∈ K(T1 , . . . , Tr )[X] be a monic separable irreducible polynomial. By Lemma 12.1.6, it suffices to prove that HK (f ) 6= ∅. Indeed, the coefficients of f belong to Kα for some α < m. By Theorem 13.4.2, Kα+1 is Hilbertian. Find a1 , . . . , ar ∈ Kα+1 such that f (a, X) is separable and irreducible in Kα+1 [X]. Choose a zero b of f (a, X) in the separable closure of Kα+1 . By Corollary 2.6.5(d), K is a regular extension of Kα+1 . In particular, K is linearly disjoint from Kα+1 (b) over Kα+1 . Therefore, f (a, X) is irreducible in K[X]. The PAC and the Hilbert properties are in a sense opposing. Nevertheless there are fields that are both PAC and Hilbertian: Proposition 13.4.6 ([Fried3]): Every field K has a regular extension F which is PAC and Hilbertian. Proof: Wellorder the varieties of dimension at least 1 that are defined over K in a transfinite sequence {Vα | α < m}. Use transfinite induction to define, for each Q β < m, a function field Fβ for Vβ which is algebraically independent from α max char(K), [K(x) : K(u, t)] . Put ˜ ˜ y) : K(x)] = l. Hence, K(x, y)/K y = u1/l . Then, [K(x, y) : K(x)] = [K(x, is a regular extension. By Corollary 10.2.2(a), (x, y) generate a variety W0 over K. Let ϕ0 : W0 → V be the projection ϕ0 (x, y) = x. Then ϕ0 is a dominant separable rational map of degree l. By construction, (x0 , 0) ∈ W . If a ∈ V0 (K), then the specialization x0 → a extends to a specialization (x0 , 0) → (a, 0). Hence, (a, 0) ∈ W0 (K) and ϕ0 (a, 0) = a. Thus, ϕ0 (W0 (K)) = V0 (K). (g) Suppose V is a variety over K. For each i between 1 and m let hi ∈ K[X1 , . . . , Xn , Y ] be a polynomial which is monic and of degree at least 2 in Y . Let yi be a root of hi (x, Y ) in K(x)s . Suppose K(x, yi ) is a regular extension of K. Let g ∈ K[X] be a polynomial which does not vanish on V . Generalize the notation of Section 13.1 and define 0 HK,V = {a ∈ V (K) | g(a) 6= 0 and

m Y

hi (a, b) 6= 0 for all b ∈ K}.

i=1 0 6= 0. Indeed, Wi = V ∩ V (hi ) is a variety If V (K) is nonthin, then HK,V n+1 in A with generic point (x, yi ) over K . Let ϕi : Wi → V be the projection on the first n coordinates and let V0 = V (g). Then V (K) 6⊆ V0 (K) ∪ Sm ϕ (W (K)) (because V (K) is nonthin), so there is a a ∈ V (K) r V0 (K)∪ i i i=1 Sm 0 i=1 ϕi (Wi (K)) = HK,V , as claimed. 0 0 6= ∅ for all HK,V as above such that K(x, yi ) Conversely, suppose HK,V is a regular extension of K, i = 1, . . . , m. Giving ϕi and Wi as in (1), we may assume that Wi is absolutely irreducible (by (e)). Let Fi be the function field of Wi over K. Then Fi can be chosen to be a finite separable extension of K(x). Choose a primitive element yi for Fi /K(x) which is integral over K[x]. Let hi ∈ K[X, Y ] a polynomial such that hi (x, Y ) = irr(yi , K(x)). By (a), we may assume that (x, yi ) is a generic point of Wi over K and ϕi : Wi → V

13.5 Hilbertianity via Coverings

247

0 is the projection on the first n coordinates. Let V0 = V (g). Then HK,V 6= ∅ Sm is equivalent to V (K) 6⊆ V0 (K) ∪ i=1 ϕi (Wi (K)). (h) An (K) is nonthin if and only if for all absolutely irreducible polynomials h1 , . . . , hm ∈ K[X1 , . . . , Xn , Y ] which are separable and monic in Y of 0 degree at least 2 in Y and for all nonzero g ∈ K[X1 , . . . , Xn ], the set HK,A n is nonempty. Indeed, a generic point x = (x1 , . . . , xn ) for An over K consists of algebraically independent elements x1 , . . . , xn over K. If y ∈ K(x)s is integral over K and h ∈ K[X, Y ] is a monic polynomial in Y such that h(x, Y ) = irr(y, K(x)), then h is absolutely irreducible if and only if K(x, y) is a regular extension of K (Corollary 10.2.2). Thus, our statement is a special case of (g). (i) Suppose A1 (K) is a thin set. Then so is V (K) for every variety V over K. Indeed, suppose dim(V ) = r > 0. Let x be a generic point of V over K. Then K(x) is a regular extension of K of transcendence degree r. By (e) and (f), there exist varieties W1 , . . . , Wn over KSand dominant separable m maps ϕi : Wi → A1 , i = 1, . . . , m such that K = i=1 ϕi (Wi (K)). Use (a) to assume that Wi is a curve defined by gi (T, Y ) = 0, where gi ∈ K[T, Y ] is absolutely irreducible and is monic and separable in Y . Further, assume ϕi is the projection on the first coordinate. Proposition 13.4.3 gives a transcendental element t ∈ K(x) such that gi (t, Y ) is irreducible over K(x), i = 1, . . . , m. Choose yi ∈ K(t)s with gi (t, yi ) = 0. Then K(x) and K(t, yi ) are linearly disjoint over K(t). Let ψ: V → A1 be the K-rational map defined by ψ(x) = t. Then ψ is defined outside a Zariski K-closed subset V0 of V of dimension less than r. Let Wi0 be the K-variety generated by (x, yi ). Let ϕ0i :SWi0 → V be the projection m ϕ0i (x, yi ) = x. We prove that V (K) = V0 (K) ∪ i=1 ϕ0i (Wi0 (K)).

K(x)

K(x, yi )

V o

ϕ0i

ψi0

ψ

K(t)

K(t, yi )

A1 o

Wi0

ϕi

Wi

Indeed, let b ∈ V (K) r V0 (K). Then a = ψ(b) ∈ K. By assumption, there exist i between 1 and m and ci ∈ Wi (K) with ϕi (ci ) = a. By linear disjointness, K[x, yi ] = K[x] ⊗K[t] K[yi ]. Hence, (x, yi ) → (b, ci ) is a Kspecialization. In other words, (b, ci ) ∈ Wi0 (K) and ϕ0i (b, ci ) = b, as desired. In the language of schemes, Wi0 is the fiber product of ϕ: V → A1 and ϕi : Wi → A1 . All of this gives a new characterization of Hilbertian field. Proposition 13.5.3: The following conditions on a field K are equivalent: (a) K is Hilbertian.

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(b) A1 (K) is nonthin. (c) There is a variety V over K such that V (K) is nonthin. Proof: By Lemma 13.1.2, K is Hilbertian if and only if the following condi0 tion holds: HK,A 1 (h1 , . . . , hm ; g) 6= ∅ for all absolutely irreducible polynomials hi ∈ K[X, Y ] which are separable and of degree at least 2 in Y , and for all nonzero g ∈ K[X]. By Remark 13.5.2(h), the latter condition is equivalent to A1 (K) being nonthin. Thus, (a) is equivalent to (b). Remark 13.5.2(i) asserts that (b) is equivalent to (c). Remark 13.5.4: Examples of thin sets. Even if A1 (K) is nonthin, there may be varieties V over K such that V (K) is thin. For example, each finitely generated extension K of Q is Hilbertian (Theorem 13.4.2). Hence, by Proposition 13.5.3, A1 (K) is nonthin. However, by Faltings, C(K) is finite for each curve C over K of genus at least 2 [Faltings-W¨ ustholz, p. 205]. In particular, C(K) is thin. Moreover, let A be an Abelian variety over K of dimension d. For each integer n ≥ 2, multiplication by n is a dominant rational map of degree n2d of A onto itself [Mumford1, pp. 42 and 64]. By Mordell-Weil, A(K)/nA(K) is a finite group [Lang3, p. 71]. Let a1 , . . . , am be representatives for the cosets of A(K) modulo nA(K). For each i, the map S ϕi (x) = ai + nx gives a morphism m of degree n2d of A onto A. Also, A(K) = i=1 ϕi (A(K)). Therefore, A(K) is thin. For example, let E be an elliptic curve over K with E(K) infinite. Then E(K) is Zariski K-dense in E but E(K) is thin. Thus, the converse of Remark 13.5.2(b) is not true. Proposition 27.3.5 characterizes fields K for which V (K) is nonthin for every variety V over K as ‘ω-free PAC fields’.

13.6 Non-Hilbertian g-Hilbertian Fields In case where A = A1 , one may try to bound the genera of the curves Wi appearing in (1) of Section 13.5. This leads to a weaker version of Hilbertianity. Definition 13.6.1: g-Hilbertian fields. Let K be a field and g a nonnegative integer. We say that K is g-Hilbertian if K is not a finite union of sets ϕ(C(K)) with C a curve over K of genus at most g and ϕ: C → A1 a dominant separable rational map of degree at least 2. By Proposition 13.5.3, K is Hilbertian if and only if K is g-Hilbertian for each g ≥ 0. One may ask whether being g-Hilbertian for large g suffices for K to be Hilbertian. We show here that this is not the case. Theorem 13.6.2 (Zannier): Let K be a global field and g a nonnegative integer. Then K has an infinite normal extension N which is g-Hilbertian but not Hilbertian.

13.6 Non-Hilbertian g -Hilbertian Fields

249

Proof: The construction of N starts with a prime number l 6= char(K) and another prime number q with (2)

q > max(char(K), 2g − 2 + 2l).

Extend K, if necessary, to assume ζl , ζq ∈ K. Then use the Hilbertianity of K (Theorem 13.4.2) to choose u ∈ K which is not a qth power. Thus, √ q u∈ / Kins . By Eisenstein’s criterion (Lemma 2.3.10), f (X, Y ) = Y l − X lq + u is an absolutely irreducible polynomial. Now define an ascending sequence √ of Galois extensions Kn of K inductively: K0 = K and Kn+1 = Kn ( l alq − u | a ∈ Kn ). Then Kn /K is a pro-l-extension. That is, Kn /K is Galois and [L : K] is an l-power for each finite subextension L/K of Kn /K. S∞ Let M (K) S= n=0 Kn . This is an infinite pro-l extension of K. Let N = N (K) = M (K 0 ) with K 0 ranging over all finite purely inseparable extensions of K. Then N is a perfect field which is infinite and normal over K. For each a ∈ N there exists b ∈ N such that bl − alq + u = 0. Hence, N is non-Hilbertian. We prove however that N is g-Hilbertian. Note: We may replace K in the proof by any finite extension L in N . Indeed, then M (L) is contained in M (K 0 ) for some finite purely inseparable extension K 0 of K. Hence, N = N (L). Claim A: Suppose M √ is a finite Galois extension of K in N . Then [M : K] is a power of l and M ( q u) is a Galois extension of M of degree q. Indeed, there is a finite purely inseparable extension K 0 of K with M ⊆ 0 ), so [M : K] = [M K 0 : K 0 ] is√ a power of q. By the choice of u, M (K √ q K( u)/K is Galois of degree 1, so M ( q u)/M is a Galois extension of degree q. Let P be the set of all ultrametric prime divisors p of K with vp (u) = 0, √ q vp (l) = 0, vp (q) = 0, and K( pu/K 6= 1. Claim B: Each p in P is unramified in M (K). Indeed, it suffices to consider a finite extension L of K in M (K) in which p is unramified, to take a prime divisor √ q of L over p and an element a ∈ L, and to prove that q is unramified in L( l alq − u). By Example 2.3.8, it suffices to prove that l divides vq (alq − u). Suppose first vq (a) < 0. Then vq (alq − u) = vq (alq ) = lqvq (a). Now suppose vq (a) ≥ 0. Denote reduction modulo q by a bar. The assumptions √ √ √ ¯ qu ¯) = K( q u). on q imply Op [ q u]/Op is a ring-cover. By Remark 6.1.7, K( √ √ q ¯ = q. Next, let M be the ¯ qu Since K( pu/K 6= 1, this implies [K( ¯) : K] ¯ : K] ¯ is a Galois closure of L/K. Then M ⊆ M (K). Hence, by Claim√A, [M q ¯ ¯ ¯ ¯ ¯) 6⊆ L, so u ¯ is power of l. Therefore, [L : K] is also a power of L. Thus, K( u ¯ In particular, a ¯. Consequently, vq (alq − u) = 0. not a qth power in L. ¯lq 6= u

250

Chapter 13. The Classical Hilbertian Fields

Claim C: Let L be a finite extension of K in M (K). Consider nonconstant polynomials h1 , . . . , hr ∈ L[X]. Suppose hi has an irreducible L-factor of fi which is separable, of multiplicity not divisible by l; suppose further the degree of each irreducible L-factor of hi is less p than q, i = 1, . . . , r. Then / N for all a ∈ A and OK has an arithmetical progression A with l hi (a) ∈ i = 1, . . . , r. Indeed, write hi = fik gi with fi , gi ∈ L[X], fi irreducible, l - k, and gi relatively prime to fi . Choose k 0 , l0 ∈ Z with k 0 ≥ 1 and kk 0 + ll0 = 1. Then p 0 0 0 0 0 / N if and only if hki = fikk gik = fi gik fi−ll . For every a ∈ K, l hi (a) ∈ p 0 0 l k fi (a)gi (a) ∈ / N . Each root of fi in Ks is a simple root of fi gik . Therefore, 0 we may replace hi by fi gik , if necessary, to assume hi has a simple root ai in Ks . Now extend L, if necessary, to assume L is Galois over K with Galois √ group G. By Claim A, Q L( q u)/L Q is a Galois extension of degree q. Let M r be the splitting field of i=1 σ∈G hσi (X) over L. Since the degree of each irreducible L-factor of h1 , . . . , hr is less than q, M is a finite Galois extension of K whose degree is divisible √ only by prime numbers smaller than q. Hence, q u). Denote the set of all ultrametric primes M is a proper subfield of M ( √ p of K which unramify in M ( q u) with vp (u) = 0, vp (l) = 0, vp (q) = 0, and √ √ M ( q u)/K ⊆ Gal(M ( q u)/M ) r{1} by P0 . Then P0 is a subset of P . By p Chebotarev (Theorem 6.3.1), P0 is infinite. Omit finitely many elements from P0 to assume that for each p ∈ P0 all nonzero coefficients of the hi are p-units and ai is a simple root of hi modulo p. Now choose distinct primes p1 , . . . , pr in P0 . Consider i between 1 and r. Let p = pi . Choose a prime q = qi of L over p, a prime q0 of M over q, and denote reduction modulo q0 by a bar. Then ¯ i is a polynomial with coefficients in K ¯ which decomposes into ¯ = K, ¯ so h M ¯ ¯ 0 (¯ ¯ linear factors. In particular, a ¯i ∈ K, hi (¯ ai ) = 0, and h i ai ) 6= 0. Choose bi ∈ K with vq0 (bi − ai ) > 0. Then, vq (hi (bi )) ≥ 1 and vq (h0i (bi )) = 0. If vq (hi (bi )) ≥ 2, choose πi ∈ K with vp (πi ) = 1. Since p is unramified in M , vq (πi ) = 1. Then vq hi (bi + πi ) − hi (bi ) − h0i (bi )πi ≥ 2. Hence, vq (hi (bi + πi )) = 1. Thus, replacing bi by bi + πi , if necessary, we may assume that in any case vq (hi (bi )) = 1. The weak approximation theorem (Proposition 2.1.1) gives b, π ∈ K with vpi (b − bi ) ≥ 2 and vpi (π − πi ) ≥ 2, i = 1, . . . , r. Let A = b + π 2 OK . Then vpi (hi (a)) = 1 for each i and all a ∈ A. Consider an arbitrary finite purely inseparable extension K 0 of K. Denote the unique extension of pi to K 0 by p0i . Then vp0i (hi (a)) = 1 if char(K) = 0 and vp0i (hi (a)) is a power of char(K) if char(K) > 0. In each case, l - vp0i (hi (a)). By Claim B, applied to K 0 instead of to K, p0i is unramified in M (K 0 ). Hence, l - vi (hi (a)) for each extension vi of vp0i to M (K 0 ), so p p l hi (a) ∈ / M (K 0 ). Consequently, l hi (a) ∈ / N , as claimed. Claim D: Let L be a finite extension of K and C a curve of genus at most

13.6 Non-Hilbertian g -Hilbertian Fields

251

Qs g over L. Suppose C is defined by Y l = c j=1 fj (X)mj , where c ∈ K × , f1 , . . . , fs are distinct monic irreducible polynomials in L[X], and l - mj , j = 1, . . . , s. Then deg(fj ) < q, j = 1, . . . , s. Indeed, replace mj by its residue modulo l, if necessary, to assume 1 ≤ − 1. Choose a transcendental element x over L and y ∈ L(x)s with mj ≤ l Q s y l = c j=1 fj (x)mj . Then F = L(x, y) is the function field of C over L. Denote the prime divisor of L(x)/L corresponding to fj by Pj (Section 2.2). Then deg(Pj )Q= deg(fj ), vPj (fj (x)) = 1, and vPj (fi (x)) = 0 for j 6= i. s Hence, vPj (c i=1 fi (x)mi ) = mj is not divisible by l. By Example 2.3.8, Pj totally ramifies in F . Thus, F has a unique prime divisor Qj lying over Pj , eQj /Pj = l, and deg(Qj ) = deg(Pj ) = deg(fj ). Since l 6= char(L), the ramification is tame. Hence, by Riemann-Hurwitz (Remark 3.6.2(c)), 2g − 2 ≥ 2 · genus(F ) − 2 ≥ −2l + (l − 1) deg(fj ). Therefore, by (2), deg(fj ) ≤ 2g − 2 + 2l < q, j = 1, . . . , s, as claimed. Claim E: N is g-Hilbertian. Sm Indeed, assume N = i=1 ϕi (Ci (N )) with Ci a curve over N of genus at most g and ϕi : Ci → A1 a dominant separable rational map over N of degree at least 2. Replace K by a finite extension in N , if necessary, to assume that Ci and ϕi are defined over K and the genus of Ci over K is the same as the genus of Ci over N . Choose a transcendental element x over K and a generic point zi for Ci over K with ϕi (zi ) = x, i = 1, . . . , m. Then Ei = K(zi ) is a finite separable extension of K(x) of degree at least 2. Denote the Galois closure of Ei by Fi . Next, list the cyclic extensions of K(x) of degree l which are contained in one of the fields Ei as D1 , . . . , Dr . For each j between 1 and r choose a primitive element yj for Dj /K(x) such that yjl = hj (x) with hj ∈ K[X]. Then hj (x) is not an lth power in K(x). Since each Ei is a regular extension of K, so is Dj /K. Hence, deg(hj ) ≥ 1. Since N is perfect, we may again replace a finite purely inseparable extension in N to assume that hj = Qsj Kmby hjkjk such that cj ∈ K × , hjk are distinct monic separable irreducible cj k=1 polynomials of positive degrees, and mjk are positive integers. If l|mjk for some j, k, replace yj by yj hjk (x)−mjk /l . Thus, assume without loss, l - mjk for all j and k. By Claim D, deg(hjk ) < q. Hence, by Claim C, there exists an arithmetip l / N for each a ∈ A, j = 1, . . . , r. cal progression Qm A of OK such that hj (a) ∈ Let F = i=1 Fi . Let H be the set of all a ∈ K such that each K-place ˜ ∪ {∞} with ψ(x) = a maps Fi onto F¯i ∪ {∞} and Ei onto ψ: F → K ¯ Ei ∪ {∞} with these properties: (3a) F¯i /K is a Galois extension and Gal(F¯i /K) ∼ = Gal(Fi /K(x)). ¯i = K(ci ) and [K(ci ) : K] = (3b) ψ is finite at zi and ci = ψ(zi ) satisfies E [Ei : K(x)]. By Lemma 13.1.1, H contains a separable Hilbert subset. Theorem 13.3.5(c) gives a ∈ A ∩ H. By assumption, a = ϕi (c) with c ∈ Ci (N ) for some i between 1 and m. Extend the K-specialization (x, zi ) → (a, c) to a

252

Chapter 13. The Classical Hilbertian Fields

˜ ∪ {∞}. Then (3) is true with ci = c. Since Fi is the Galois place ψ: F → K closure of Ei /K(x), F¯i is the Galois closure of K(c)/K). Since K(c) ⊆ N and N/K is normal, F¯i ⊆ N . By Claim A, Gal(F¯i /K) is an l-group. By ¯i ) is a proper subgroup of ¯i : K] = [Ei : K(x)] > 1. Thus, Gal(F¯i /E (3b), [E Gal(F¯i /K). Hence, Gal(F¯i /Ei ) is contained in a normal subgroup of index l of Gal(F¯i /K) [Hall, p. 45, Cor. 4.2.2]. By (3a), Gal(Fi /Ei ) is contained in a normal subgroup of Gal(Fi /K(x)) of index l, so K(x) has a cyclic extension of degree l in Ei . It is Dj for some j between 1 and r. Then b = ψ(yj ) satisfies ¯i ⊆ N . This contradicts the choice of a in A. Thus, our bl = hj (a) and b ∈ E Sr initial assumption N ⊆ i=1 ϕi (Ci (N )) is false, so N is g-Hilbertian.

13.7 Twisted Wreath Products Given finite groups A and G, there are several ways of constructing a new group out of them. We describe here two of them: the semidirect product and the wreath product. Definition 13.7.1: Semidirect products. Let A and G be profinite groups. Suppose G acts on A continuously from the right. That is, there is a continuous map G × A → A mapping (σ, a) onto aσ and satisfying these rules: (ab)σ = aσ bσ , a1 = a, and (aσ )τ = aστ for a, b ∈ A and σ, τ ∈ G. The semidirect product G n A consists of all pairs (σ, a) ∈ G × A with the product rule (σ, a)(τ, b) = (στ, aτ b). This makes G n A a profinite group −1 with unit element (1, 1) and (σ, a)−1 = (σ −1 , a−σ ). Identify each σ ∈ G (resp. a ∈ A) with the pair (σ, 1) (resp. (1, a)). This embeds G and A into G n A such that (1) A is normal, G ∩ A = 1, and GA = G n A. Each element of G n A has a unique presentation as σa with σ ∈ G and a ∈ A. The product rule becomes (σa)(τ b) = (στ )(aτ b) and the action of G on A coincides with conjugation: aσ = σ −1 aσ. The projection σa 7→ σ of G n A on G is an epimorphism with kernel A. Conversely, suppose H is a profinite group and A, G are closed subgroups satisfying (1). Then H is the semidirect product G n A. Likewise, consider a short exact sequence α

1 −→ A −→ H −→ G −→ 1 Suppose the sequence splits. That is, there exists a homomorphism α0 : G → H satisfying α(α0 (g)) = g for each g ∈ G. Then α0 is an embedding which we call a group theoretic section of α. Identifying G with α0 (G), we have H = G n A. Suppose ϕ1 : G → E and ϕ2 : A → E are homomorphisms of profinite groups and ϕ2 (aσ ) = ϕ2 (a)ϕ1 (σ) for all a ∈ A and σ ∈ G. Then ϕ(σa) = ϕ1 (σ)ϕ2 (a) is a homomorphism ϕ: G n A → E. Here is a Galois theoretic interpretation of semidirect products: Let K, L, E, F be fields, with L/K Galois, F/K Galois, EL = F , and E ∩L = K. Then Gal(F/K) = Gal(F/E) n Gal(F/L).

13.7 Twisted Wreath Products

253

Definition 13.7.2: Twisted wreath products. Let A and G be finite groups, G0 a subgroup of G, and Σ is a system of representatives for the right cosets G0 σ, σ ∈ G. Thus, [ [ G = · G0 σ = · σ −1 G0 . σ∈Σ

σ∈Σ

Suppose G0 acts on A from the right. Let σ0 IndG for all σ ∈ G and σ0 ∈ G0 }. G0 (A) = {f : G → A | f (σσ0 ) = f (σ) g Make IndG G0 (A) a group by the rule (f g)(σ) = f (σ)g(σ). Then f (σ) = f (σ)g(σ) , where f g denotes conjugation in IndG G0 (A) and the right hand side is conjugation in A. σ The group G acts on IndG G0 (A) by f (τ ) = f (στ ). This gives rise to the semidirect product G n IndG G0 (A), which we also denote by A wrG0 G. Each element of this group has a unique presentation as a product σf with σ ∈ G and f ∈ IndG G0 (A). The product and the inverse operation in A wrG0 G −1

are given by (σf )(τ g) = στ f τ g and (σf )−1 = σ −1 f −σ . The identification σ = σ · 1 and f = 1 · σ identifies G and IndG G0 (A) as subgroups of (A) is normal, G ∩ IndG A wrG0 G. Under this identification, IndG G0 G0 (A) = 1, and −1 (1 · f )(σ · 1) = 1 · f σ , conjugaG · IndG G0 (A) = A wrG0 G. Since (σ · 1) tion of f by σ in A wrG0 G coincides with the action of σ on f . The map σf 7→ σ is an epimorphism π: A wrG0 G → G with Ker(π) = IndG G0 (A). We call A wrG0 G the twisted wreath product of A and G with respect to G0 . Twisted wreath products are usually non-Abelian: Lemma 13.7.3: Let A be a nontrivial finite group, G a finite group, and G0 a proper subgroup of G. Then A wrG0 G is not commutative. σ Proof: Choose a ∈ A, a 6= 1. Define a function f ∈ IndG G0 (A) by f (σ) = a for σ ∈ G0 and f (σ) = 1 for σ ∈ G r G0 . Choose σ ∈ G r G0 . Then f σ (1) = f (σ) = 1 6= a = f (1). Hence, f σ 6= f . Consequently, A wrG0 G is not commutative.

The next result shows that not only twisted wreath products are in general non-Abelian but their centers are small: Lemma 13.7.4: Let π: A wrG0 G → G be a twisted wreath product of finite groups, H1 / A wrG0 G, and h2 ∈ A wrG0 G. Put I = IndG G0 (A) = Ker(π) and G1 = π(H1 ). Suppose A 6= 1. (a) Suppose π(h2 ) ∈ / G0 and (G1 G0 : G0 ) > 2. Then there is an h1 ∈ H1 ∩ I with h1 h2 6= h2 h1 .

254

Chapter 13. The Classical Hilbertian Fields

(b) Suppose G1 6≤ G0 and π(h2 ) ∈ / G1 G0 . Then there is an h1 ∈ H1 ∩ I with 0 / hh1 ih for all h0 ∈ π −1 (G1 G0 ). In particular, h1 h2 6= h2 h1 . hh1 2 ∈ Proof: Put σ2 = π(h2 ). Consider σ1 ∈ G1 and g ∈ I. By definition, there are f1 , f2 ∈ I with σ1 f1 ∈ H1 and h2 = σ2 f2 . Put h1 = g σ1 f1 g −1 . Then h1 = [σ1 f1 , g −1 ] ∈ [H1 , I] ≤ H1 ∩ I. For each τ ∈ G h1 (τ ) = (g σ1 )f1 (τ )g(τ )−1 = g(σ1 τ )f1 (τ ) g(τ )−1 . Hence, for all τ ∈ G and f 0 ∈ I we have: (2a)

hh1 2 (1) = hσ1 2 f2 (1) = h1 (σ2 )f2 (1) = g(σ1 σ2 )f1 (σ2 )f2 (1) g(σ2 )−f2 (1) , 0

(2b) hτ1 f (1) = h1 (τ )f (2c)

0

(1)

f1 (1)

h1 (1) = g(σ1 )

= g(σ1 τ )f1 (τ )f −1

g(1)

0

(1)

g(τ )−f

0

(1)

, and

.

We apply (2) in the proofs of (a) and (b) to special values σ1 and g. Choose a ∈ A, a 6= 1. Proof of (a): Since (G1 G0 : G0 ) > 2, there is a σ1 ∈ G1 with distinct cosets σ1−1 G0 , σ2 G0 , G0 . Thus, none of the cosets σ1 G0 , σ2 G0 , σ1 σ2 G0 is G0 . Therefore, by definition of I, there is a g ∈ I with g(σ1 ) = g(σ2 ) = g(σ1 σ2 ) = 1 and g(1) = a. By (2a), hh1 2 (1) = 1. By (2c), h1 (1) 6= 1. Consequently, hh1 2 6= h1 , as desired. σ −1

Proof of (b): Since Gσ1 2 = G1 6≤ G0 , we have G1 6≤ G0 2 . Hence, G1 ∩ G0 σ −1

and G1 ∩ G0 2

are proper subgroups of G1 . Their union is a proper subset σ −1

of G1 . Thus, there is an element σ1 ∈ G1 r G0 ∪ G0 2 . It follows that / σ1 σ2 G0 . By assumption, σ2 ∈ / G1 G0 . Therefore, there is a g ∈ I with σ2 ∈ g(G1 G0 ) = 1, g(σ1 σ2 ) = 1, and g(σ2 ) = a−1 . Consider τ ∈ G1 G0 and f 0 ∈ I. By (2a), hh1 2 (1) = af2 (1) 6= 1. By 0

0

(2b), hτ1 f (1) = 1. Hence, (hk1 )τ f (1) = 1 for all integers k. It follows that 0 hh1 2 ∈ / hh1 ih for all h0 ∈ π −1 (G1 G0 ). Remark 13.7.5: Decomposition of IndG G0 (A) into a direct product. a ∈ A associate a function fa : G → A: σ a if σ ∈ G0 fa (σ) = 1 if σ ∈ / G0 .

To each

These functions satisfy the following rules: fa fb = fab and g −1 fa g = fag(1) for all a, b ∈ A and g ∈ IndG G0 (A). Thus, the map a 7→ fa identifies A with r G0 ) = 1} of IndG the normal subgroup {f ∈ IndG G0 (A) | f (G G0 (A). Applying σ ∈ G on A gives the following normal subgroup of IndG G0 (A): r σ −1 G0 ) = 1}. Aσ = {f ∈ IndG G0 (A) | f (G

13.7 Twisted Wreath Products

255

Q An arbitrary element f ∈ IndG G0 (A) has a unique presentation f = σ∈Σ fσ , σ −1 −1 (σ σ0 ) = f (σ −1 σ0 ) with fσ ∈ A . Specifically, fσ (G r σ G0 ) = 1 and fσQ σ for all σ ∈ Σ and σ0 ∈ G0 . It follows that IndG G0 (A) = σ∈Σ A . The latter relation allows us to present A as a quotient of IndG G0 (A) in G various ways: Let N = {f ∈ IndG0 (A) | f (1) = 1}. For each σ ∈ G let N σ = −1 = 1}. Then the map f 7→ f (σ −1 ) {f σ | f ∈ N } = {f ∈ IndG G0 (A) | f (σ gives rise to a short exact sequence 1 → N σ → IndG G0 (A) → A → 1. Like in the preceding paragraph, we find that Y −1 ) = 1} = Aτ . (3) N σ = {f ∈ IndG G0 (A) | f (σ τ ∈Σ G0 τ 6=G0 σ

Note that G0 leaves N = N 1 invariant, so N / IndG G0 (A)G0 . We summarize some of the groups mentioned above in the following diagram: (4)

IndG G0 (A)

IndG G0 (A)G0

N

N G0

1

G0

A wrG0 G

G

Remark 13.7.6: Interpretation of twisted wreath products in Galois theory. (a) Let Fˆ /K be a finite Galois extension with Gal(Fˆ /K) ∼ = A wrG0 G and A 6= 1. View IndG G0 (A) and G under this isomorphism as subgroups of ˆ be the fixed fields in Fˆ , respectively, of the Gal(Fˆ /K). Let F , L, L0 , and K G G subgroups N , IndG0 (A), IndG0 (A)G0 , and G. Galois theory interprets the various relations among the subgroups of A wrG0 G as relations among fields: (5a) K ⊆ L0 ⊆ L ⊂ F ⊆ Fˆ . ˆ = K and LK ˆ = Fˆ . (5b) L ∩ K (5c) L/K, F/L0 , and Fˆ /K are finite Galois extensions. (5d) The Q fields F σ with σ ∈ Σ are linearly disjoint over L. Moreover, Fˆ = σ∈Σ F σ . (5e) There is a field F0 with L ∩ F0 = L0 and F = LF0 . L G0

L0

K

A

F G0

F0

N

Fˆ G0

ˆ L0 K

ˆ K

256

Chapter 13. The Classical Hilbertian Fields

The assertion “F/L0 is Galois” follows from “N / IndG G0 (A)G0 ”. Thus, Conˆ dition (5e) follows from (5a)-(5d) by taking F0 = F ∩ L0 K. ˆ satisfying Conditions (b) Conversely, consider fields K, L0 , L, F, Fˆ , K (5a)-(5d). Let ˆ F0 = F ∩ (L0 K),

ˆ ∼ G = Gal(Fˆ /K) = Gal(L/K),

ˆ ∼ G0 = Gal(Fˆ /L0 K)) = Gal(L/L0 ), = Gal(F/F0 ) ∼ A = Gal(F/L) ∼ = Gal(F0 /L0 ). In particular, Σ is a subset of Gal(Fˆ /K). Suppose also that (5d) holds. Then A / Gal(F/L0 ). Thus, G0 viewed as a subgroup of Gal(F/L0 ) acts on A by conjugation. We construct an isomorphism ϕ: A wrG0 G → Gal(Fˆ /K) which ˆ is the identity on G and maps IndG G0 (A) onto Gal(F /L). Construction of ϕ: By (5d), Fˆ /L is a Galois extension of degree |A||Σ| . ˆ n Gal(Fˆ /L). Hence, by (5b), Gal(Fˆ /K) = Gal(Fˆ /K) G For each σ ∈ G, the group IndG0 (A) acts on F σ by the rule zf =

(6)

zσ

−1

f (σ−1 ) σ

f ∈ IndG G0 (A),

,

z ∈ F σ. 0

This action does not depend on σ. Indeed, assume F σ = F σ with σ, σ 0 ∈ G. Write σ = σ0 τ and σ 0 = σ00 τ 0 with σ0 , σ00 ∈ G0 and τ, τ 0 ∈ Σ. Since F/F0 0 is Galois, F τ = F τ , so by (5a) and (5d), τ = τ 0 . Thus, σ 0 = ρσ with ρ = λ0 λ−1 ∈ G0 . Moreover, ρ−1 f (σ −1 ρ−1 )ρ = f (σ −1 ρ−1 )ρ = f (σ −1 ρ−1 ρ) = f (σ −1 ). Hence,

z (ρσ)

−1

f ((ρσ)−1 ) ρσ

=

zσ

−1

f (σ−1 ) σ

.

−1

f

If z ∈ L, then z = z, because f (σ ) as an element of A fixes L. σ Thus, the action (6) defines a homomorphism ϕσ : IndG G0 (A) → Gal(F /L). −1

−1

−1

−1

If ϕσ (f ) = 1, then (z σ )f (σ ) = z σ for each z ∈ F σ . Hence, z f (σ ) = z for each z ∈ F . Therefore, f (σ −1 ) = 1. It follows that Ker(ϕσ ) = N σ , with N σ as in (3). Using (5d), the ϕσ ’s, with σ ranging on Σ, define a homomorphism ϕ0 : IndG G0 (A) →

Y

Gal(F σ /L) = Gal(Fˆ /L)

σ∈Σ

Q T by ϕ0 (f ) = σ∈Σ ϕσ (f ). The kernel of ϕ0 is σ∈Σ N σ = 1 and |IndG G0 (A)| = |Σ| ˆ |A| = |Gal(F /L)|. Hence, ϕ0 is an isomorphism.

13.7 Twisted Wreath Products

257

Next we show that ϕ0 is compatible with the action of G. To this end −1 −1 let z ∈ F σ and τ ∈ G. Then z τ ∈ F στ and −1 ) τ −1 τ −1 −1 f (τ σ −1 ) (στ f = (z τ )(τ σ ) zτ = (z σ

−1

)f (τ σ

−1

) σ

=

zσ

−1

f τ (σ−1 ) σ

τ

= zf .

Hence, τ −1 ϕ0 (f )τ = ϕ0 (f τ ), as desired. This allows us to combine ϕ0 with the identity map of G and define the isomorphism ϕ: A wrG0 G → Gal(Fˆ /K). ˆ L0 , L, F, Fˆ realize the Having established ϕ, we say that the fields K, K, twisted wreath product A wrG0 G. We say that the fields K, L0 , L, F, Fˆ ˆ such that realize the twisted wreath product if there exists a field K ˆ ˆ , L, F, F realize the twisted wreath product. K, K, L0 ˆ L0 , L, F, Fˆ that realize A wrG G. Suppose E (c) Consider fields K, K, 0 ¯ = Gal(E/L) and with L ⊆ E ⊆ F . Let A is a Galois extension of L 0 σ ˆ =Q ˆ ˆ ˆ ˆ E σ∈Σ E . Then E is a Galois extension of K in F . Let J = E ∩ K. ∼ Gal(Fˆ /K) ∼ ˆK ˆ = Fˆ . Hence, Gal(E/J) ˆ ˆ = G and Gal(E/L ˆ 0 J) = Then E = ˆ ˆ = G0 . Moreover, L ∩ J = K and LJ = E. Gal(Fˆ /L0 K)

L

L0

E

E0

}} }} } }

} }} }}

F ˆ E

Fˆ ww w ww ww

F0

K ˆ realize A¯ wrG G. By (b), K, J, L0 , L, E, E 0

L0 J

J

ˆ L0 K w ww ww

ˆ vK vv v v vv

Remark 13.7.7: Wreath products. Suppose G0 is the trivial subgroup of G. Then the twisted wreath product A wrG0 G simplifies to the (usual) wreath G product A wr G. In this case IndG G0 (A) is just the group A of all functions f : G → A. Multiplication is carried out componentwise. Again, G acts on AG by the formula f σ (τ ) = f (στ ). Thus, A wr G is the semidirect product G n AG . Each element of G n AG has a unique representation as a product σf with σ ∈ G and f ∈ AG . The multiplication rule is σf · τ g = (στ )(f τ g). Identify each a ∈ A with the function Q fa : G → A given by fa (1) = a and σ fa (σ) = 1 for σ 6= 1. Then IndG G0 (A) = σ∈G A .

258

Chapter 13. The Classical Hilbertian Fields

ˆ is a field Now Suppose K ⊆ L ⊆ F ⊆ Fˆ is a tower of fields and K satisfying these conditions: (7a) L/K, F/L, and Fˆ /K are finite Galois extensions. ˆ = K and LK ˆ = Fˆ . Write G = Gal(Fˆ /K). ˆ (7b) L ∩ K Q (7c) The fields F σ with σ ∈ G are linearly disjoint over L and Fˆ = σ∈G F σ . By Remark 13.7.6(b), there is an isomorphism ϕ: A wr GQ→ Gal(Fˆ /K) which is the identity on G, maps AG onto Gal(Fˆ /L), and σ6=1 Aσ onto Gal(Fˆ /F ). By (7b), restriction to L maps G isomorphically onto Gal(L/K). ˆ L, F, Fˆ realize the wreath product A wr G. We say that the fields K, K,

13.8 The Diamond Theorem The diamond theorem proved in this section says all fields ‘captured’ between two Galois extensions of a Hilbertian field are Hilbertian. In particular, this theorem implies that a non-Hilbertian Galois extension N of a Hilbertian field K is not the compositum of two Galois extensions of K which are properly contained in N . For example, Ks is not the compositum of two proper subfields which are Galois over K. Lemma 13.8.1: Let K ⊆ L0 ⊆ L be fields with L/K finite and Galois. Let c1 , . . . , cn be a basis of L0 /K and t1 , . . . , tn algebraically independent elements over K. Let f ∈ L0 [U, X] be an absolutely irreducible polynomial which is monic in X and Galois over L(U ) with degX f ≥ 2. Put G = Gal(L/K) = Gal(L(t)/K(t)), G0 = Gal(L/L0 ) = Gal(L(t)/L0 (t)), and A = Gal(f (U, X), L(U )). Then G0 acts on A and there exist fields F and Fˆ with the following properties: (a) F P is a regular extension of L. Moreover, F = L(t, z) with irr(z, L(t)) = n f ( i=1 ci ti , Z) ∈ L0 [t, Z]. (b) K(t), L0 (t), L(t), F, Fˆ realize A wrG0 G. Proof: Fix x ∈ L(U )s with f (U, x) = 0. Since f is absolutely irreducible, [L(U, x) : L(U )] = [L0 (U, x) : L0 (U )], so L(U ) is linearly disjoint from L0 (U, x) over L0 (U ) and we may identify both Gal(L(U )/L0 (U ) and Gal(L(U, x), L0 (U, x)) with G0 via restriction. Since f (U, X) is Galois over L(U ) with respect to X, the extension L(U, x)/L(U ) is Galois with Galois group A. In addition, A 6= 1 because degX f ≥ 2. The fixed field in L(U, x) of the subgroup of Aut(L(U, x)) generated by A and G0 is L0 (U ). By Artin [Lang7, p. 264, Thm. 1.8], L(U, x)/L0 (U ) is Galois and Gal(L(U, x)/L0 (U )) ∼ = Gal(L(U, x)/L0 (U, x))nGal(L(U, x)/L(U )) ∼ = G0 nA. This defines an action of G0 on A. As in Section 13.7, choose a system of representatives Σ for the right cosets G0 σ of G0 in G. Let S = {G0 σ | σ ∈ Σ}. Choose algebraically independent elements uS , S ∈ S, over K. Write Q = L(uS | S ∈ S). For

13.8 The Diamond Theorem

259

each S ∈ S write cSi = cσi and f S = f σ for some σ ∈ S. Since ci ∈ L0 and f ∈ L0 (U, X), both cSi and f S are independent of σ. Then choose a root z S of f S (uS , Z) in Qs and write F S = Q(z S ). Since f S is absolutely irreducible, LS0 (uS , z S ) is a regular extension of LS0 (Example 2.6.11). Hence, L(uS , z S ) is a regular extension of L. Since L(uS , z S ), S ∈ S, are algebraically independent over L, they are linearly disjoint over L (Lemma 2.6.7). Moreover, their compositum Fˆ = L(uS , z S | S ∈ S) = Q(z S | S ∈ S) is a regular extension of L. By Lemma 2.5.11, in each rectangle of the following diagram, the fields lying in the left upper corner and the right lower corner are linearly disjoint over the field in the left lower corner and their compositum is the field in the right upper corner: 0

0

L(uS , z S | S 0 6= S)

0

0

L(u)(z S | S 0 6= S)

A

0

0

ˆ L(uS , z S | S 0 ∈ S) = F

A

L(u, z S ) = F S

L(uS )

A

L(uS , z S )

S LS 0 (u )

A

S S LS 0 (u , z )

L(uS | S 0 6= S)

L(t) = L(u) = Q

L

LS 0

It follows that, F S /Q is a Galois extension with Galois group isomorphic to A, the F S , S ∈ S, are linearly disjoint over Q, and Fˆ is their compositum. Thus, Fˆ /Q is Galois. The matrix (cSi ) ∈ Mn (L) is invertible [Lang7, p. 286, Cor. 5.4]. Hence, the system of equations (1)

n X

cSi Ti = uS ,

S∈S

i=1

has a unique solution t01 , . . . , t0n . It satisfies L(t01 , . . . , t0n ) = L(uS | S ∈ S) = Q. Since trans.deg(Q/K) = n, the elements t01 , . . . , t0n are algebraically independent over K. We may therefore assume that t0i = ti , i = 1, . . . , n, so Q = L(t). Use the linear disjointness of the F S ’s over Q to extend the action of G on L to an action on Fˆ : (uS )τ = uSτ and (z S )τ = z Sτ . In particular, τ permutes the equations of (1). Hence, (tτ1 , . . . , tτn ) is a solution of (1), so it coincides with (t1 , . . . , tn ). In other words, the action of G on Q is the unique extension of the given action on L that fixes t1 , . . . , tn . In particular, K(t) is the fixed field of G in Q. Now write u = uG0 ·1 , z = z G0 ·1 , andQF = F G0 ·1 . Then F S = F σ for σ ˆ all PnS ∈ S and each σ ∈ S. Hence, F = σ∈Σ F . Moreover, by (1), u = i=1 ci ti ∈ L0 (t), u is transcendental over K, f (u, z) = 0, and F = Q(z).

260

Chapter 13. The Classical Hilbertian Fields

Hence, f (u, Z) ∈ L0 (u)[Z], and f (u, Z) is Galois over L(u). Since Q = L(u) is a purely transcendental extension of L(u), f (u, Z) is irreducible and Galois over Q. Therefore, f (u, Z) = irr(z, Q) (this settles (a)), and F is a Galois extension of L0 (t). Finally, since Fˆ /L(t) and L(t)/K(t) are Galois and each τ ∈ G = Gal(L(t)/K(t)) extends to an automorphism of Fˆ , the extension Fˆ /K(t) ˆ be the fixed field of G in Fˆ . Then K ˆ ∩ L(t) = K(t) is Galois. Let K ˆ = Fˆ . By Remark 13.7.6(b), is the fixed field of G in L(t) and L(t)K ˆ L0 (t), L(t), F, Fˆ realize A wrG G. K(t), K, 0 Proposition 13.8.2 (Realization of Twisted Wreath Products): Let K ⊆ L0 ⊆ L be a tower of fields with K Hilbertian and L/K finite Galois. Consider an absolutely irreducible polynomial f (T, X) ∈ L0 [T, X] which is Galois over L(T ). Let G = Gal(L/K), G0 = Gal(L/L0 ), and A = Gal(f (T, X), L(T )). Then G0 acts on A and there are fields M , N , such that K, L0 , L, M, N realize A wrG0 G. Proof: Let t1 , . . . , tn be algebraically independent elements over K with n = ˆ F , and Fˆ such that K(t), K, ˆ L0 (t), [L0 : K]. Lemma 13.8.1 gives fields K, L(t), F, Fˆ realize A wrG0 G. Thus, Conditions (5a)-(5d) of Section 13.7 hold for K(t), L0 (t), L(t) instead of for K, L0 , L. Since K is Hilbertian, Lemma 13.1.1 gives a ∈ K r such that those conditions hold for the residue fields under each L-place ϕ of Fˆ with ϕ(K(t)) = K ∪ {∞}. Lemma 2.6.6 gives such a place. Moreover, the residue fields of L0 (t) and L(t) under ϕ are L0 ˆ 0 , F 0 , and Fˆ 0 be the residue fields of K, ˆ F , and Fˆ , and L, respectively. Let K respectively, under ϕ. By Remark 13.7.6(b), they realize A wrG0 G. Theorem 13.8.3 (Diamond Theorem [Haran4, Thm. 4.1]): Let K be a Hilbertian field, M1 and M2 Galois extensions of K, and M an intermediate field of M1 M2 /K. Suppose that M 6⊆ M1 and M 6⊆ M2 . Then M is Hilbertian. Proof: Corollary 12.2.3 allows us to assume that [M : K] = ∞. Part A of the proof strengthens this assumption: Part A: We may assume [M : (M1 ∩ M )] = ∞. Otherwise, [M : (M1 ∩ M )] < ∞. Then K has a finite Galois extension M20 such that M ⊆ (M1 ∩ M )M20 . Then M ⊆ M1 M20 and [M : M ∩ M20 ] = ∞. Replace M1 by M20 and M2 by M1 , if necessary, to restore our assumption. Part B: Reduction to an absolutely irreducible Galois polynomial. By Lemma 13.1.4, each separable Hilbert subset of M contains a subset of the form M ∩HM 0 (f ), where M 0 is a finite Galois extension of M and f ∈ M [T, X] is an absolutely irreducible polynomial which is monic and separable in X and f (T, X) is Galois over M 0 (T ). It suffices to find a ∈ M such that f (a, X) is irreducible over M 0 . We may assume M 0 ⊆ M1 M2 . Indeed, K has a finite Galois extension K 0 such that M 0 ⊆ M K 0 . If M 0 6⊆ M2 K 0 , replace M2 by M2 K 0 . If M 0 ⊆ M2 K 0 ,

13.8 The Diamond Theorem

261

replace M1 by K 0 . In the latter case we still have [M : (K 0 ∩ M )] = ∞, because [M : K] = ∞ and [K 0 : K] < ∞. Thus, we may assume that f (T, X) is Galois over M1 M2 . It suffices to produce a ∈ M such that f (a, X) is irreducible over M1 M2 . Part C: Finite Galois extensions. Write M0 = M and N = M1 M2 . Then N/K is Galois. For each finite Galois extension L of K in N let Li = Mi ∩ L, i = 0, 1, 2. Then Li /K is Galois, i = 1, 2. Use the assumptions M0 6⊆ Mi , i = 1, 2 and [M0 : M1 ∩ M0 ] = ∞ (Part A) to choose L large with L0 6⊆ Li for i = 1, 2, [L0 : L1 ∩ L0 ] > 2, and (2) f ∈ L0 [T, X] and f (T, X) is Galois over L(T ). The conditions on the fields Li translate into conditions on the groups Gi = Gal(L/Li ), i = 0, 1, 2: (3a) G1 , G2 6≤ G0 . (3b) (G0 G1 : G0 ) > 2. Part D: Twisted wreath products. Let A0 = Gal(f, L(T )) = Gal(f, Ks (T )). Choose a basis b1 , . . . , bn for L0 /K and algebraically independent elements t1 , . . . , tn over K. By (2) and Lemma 13.8.1, the group G0 acts on A0 . Moreover, there are fields F, Fˆ such that (4a) K(t), L0 (t), L(t), F, Fˆ realize A0 wr PGn0 G and (4b) F = L(t, z) with irr(z, L(t)) = f ( i=1 bi ti , X). Since K is Hilbertian, Lemma 13.1.1 gives c1 , . . . , cn ∈ K such that the specialization t → c extends to an L-place of Fˆ onto a Galois extension Fˆ 0 of K with Galois group isomorphic to Gal(Fˆ /K(t)). Thus, Fˆ 0 has a subfield F 0 with these properties: G and (5a) K, L0 , L, F 0 , Fˆ 0 realize A0 wrG0P n (5b) F 0 = L(z 0 ) with irr(z 0 , L) = f ( i=1 bi ci , X). Pn Let a = i=1 bi ci . Then a is in L0 , hence in M . If we prove N ∩ F 0 = L, it will follow from (5b) that [N (z 0 ) : N ] = [F 0 : L] = deg(f (a, X)). Thus, f (a, X) will be irreducible over N , as desired. Part E: Conclusion of the proof. Let E = N ∩ F 0 and A = Gal(E/L). By ˆ of K such Remark 13.7.6(c), G0 acts on A and there is a Galois extension E that ˆ realize H = A wrG G. (6) K, L0 , L, E, E 0 ˆ ⊂ N. In particular, all conjugates of E over K are in N . Hence, E ˆ ˆ Identify H with Gal(E/K) such that resE/L : Gal( E/K) → Gal(L/K) ˆ coincides with the projection π: H → G. Then π ◦ resN/Eˆ = resN/L . For i = 1, 2 let Hi = resN/Eˆ (Gal(N/Mi )). Then Hi / H and π(Hi ) = resN/L (Gal(N/Mi )) = Gi . Since Gal(N/M1 ) and Gal(N/M2 ) are normal subgroups of Gal(N/K) with a trivial intersection, they commute. Hence, H1 and H2 commute.

262

Chapter 13. The Classical Hilbertian Fields

By (3a) there exists h2 ∈ H2 with π(h2 ) ∈ / G0 . If A were not trivial, then by (3b) and Lemma 13.7.4 there would be h1 ∈ H1 which does not commute with h2 . We conclude from this contradiction that A = 1 and therefore N ∩ F 0 = L, as desired. We conclude this section with an application of the diamond theorem that solves Problems 12.18 and 12.19 of [Fried-Jarden3]: Corollary 13.8.4 ([Haran-Jarden5]): Let K be a Hilbertian field and let N be a Galois extension of K which is not Hilbertian. Then N is not the compositum of two Galois extensions of K neither of which is contained in the other. In particular, this conclusion holds for Ks .

13.9 Weissauer’s Theorem The most useful application of the diamond theorem is part (b) of the following result: Theorem 13.9.1 (Weissauer): Let K be a Hilbertian field. (a) Let M be a separable algebraic extension of K and M 0 a finite proper separable extension of M . Suppose M 0 is not contained in the Galois closure of M/K. Then M 0 is Hilbertian. (b) Let N be a Galois extension of K and N 0 a finite proper separable extension of N . Then N 0 is Hilbertian. (c) Let N be a Galois extension of a Hilbertian field K and L a finite proper separable extension of K. Suppose that N ∩ L = K. Then N L is Hilbertian. Proof: Statement (c) is a special case of Statement (b). Statement (b) is a special case of Statement (a). Statement (a) follows from Corollary 12.2.3 if [M : K] < ∞. Suppose therefore that [M : K] = ∞. Then K has a finite Galois extension L with M 0 ⊆ M L. Let N be the Galois closure of M/K. Then M 0 is contained neither in L nor in N . By Theorem 13.8.3, M 0 is Hilbertian. Proposition 13.9.3 below is a stronger version of Theorem 13.9.1(c) which gives information about the Hilbert sets of N L. That version implies Theorem 13.9.1(a), hence also Theorem 13.9.1(b). The proof of Proposition 13.9.3 depends on the following lemma rather that on the diamond theorem: Lemma 13.9.2: Let L = K(α) be a finite proper separable extension of a field ˆ be the Galois hull of L/K. Consider an absolutely irreducible K and let L polynomial h ∈ L[T, X] with degX (h) > 1 which is separable with respect to X. For u, v algebraically independent elements over K, let E be a Galois extension of K(u, v) such that E ∩ L(u, v) = K(u, v). Then the polynomial ˆ h(u + αv, X) has no root in the field E L. ˆ Then choose σ ∈ Gal(F/E) with Proof: Put t = u + αv and F = E L. 0 0 α = σα 6= α and put t = σt = u + α0 v and h0 = σh. Assume that there

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263

exists x ∈ F with h(t, x) = 0. Then x0 = σx ∈ F and h0 (t0 , x0 ) = 0. Since ˆ t0 ) = L(u, ˆ v). Hence, t, t0 are algebraically independent α 6= α0 , we have L(t, over K. ˆ t0 , x) ˆ x) F L(t, L(t, ˆ L(t)

ˆ t0 ) L(t,

ˆ t0 , x0 ) L(t,

ˆ L

ˆ 0) L(t

ˆ 0 , x0 ) L(t

ˆ Since ˆ x) and L(t ˆ 0 , x0 ) are algebraically independent over L. Therefore, L(t, ˆ they are regular extensions of L they are linearly disjoint (Lemma 2.6.7). By ˆ t0 , x0 ) are linearly disjoint over L(t, ˆ t0 ). In ˆ t0 , x) and L(t, Lemma 2.5.11, L(t, particular, (1)

ˆ t0 , x0 ) = L(t, t0 ). ˆ t0 , x) ∩ L(t, L(t,

ˆ t0 ), we have Gal(F/E0 ) = Gal(F/E)× On the other hand, with E0 = E ∩ L(t, ˆ t0 )). Hence, στ = τ σ for all τ ∈ Gal(F/L(t, ˆ t0 , x)). In particular, Gal(F/L(t, ˆ t0 , x). By (1), L(t, ˆ t0 , x0 ) = τ x0 = τ σx = στ x = σx = x0 . Therefore, x0 ∈ L(t, ˆ t0 ), a contradiction to degX (h0 ) > 1. L(t, Proposition 13.9.3: Let L be a finite proper separable extension of a Hilbertian field K and let M be a Galois extension of K such that M ∩L = K. Then the field N = M L is Hilbertian. Moreover, every separable Hilbert subset of N contains elements of L. Proof: By Lemma 13.1.2, it suffices to consider absolutely irreducible polynomials h1 , . . . , hm ∈ N [T, X], separable monic and of degree at least 2 in X and to find c ∈ L such that hi (c, X) has no root in N , i = 1, . . . , m. Let u and v be algebraically independent over K. Choose a primitive element α for L/K and let L0 be a finite extension of L which is contained in N and contains all coefficients of h1 , . . . , hm . Put K 0 = M ∩ L0 and let F be a finite Galois extension of K 0 (u, v) that contains L0 (u, v) and over which all polynomials h1 (u + αv, X), . . . , hm (u + αv, X) decompose into linear factors. Let g ∈ K[u, v] be the product of the discriminants of h1 (u + αv, X), . . . , hm (u + αv, X) with respect to X. Let B 0 be the set of all (a, b) ∈ (K 0 )2 with g(a, b) 6= 0 satisfying the following condition: (2) The L0 -specialization (u, v) → (a, b) extends to a place ϕ of F which induces an isomorphism ϕ0 : Gal(F/K 0 (u, v)) → Gal(F 0 /K 0 ) (with F 0 being the residue field of F ) such that (ϕ0 σ)(ϕx) = ϕ(σx) for all σ ∈ Gal(F/K 0 (u, v)) and x ∈ F with ϕx 6= ∞. In particular ϕ maps the set of all zeros of hi (u + αv, X) bijectively onto the set of zeros of hi (a + αb, X), i = 1 . . . , m. By Lemma 13.1.1 and Example 2.6.10, B 0 contains a separable Hilbert subset A0 of (K 0 )2 . By Corollary 12.2.3, A0 contains a separable Hilbert subset of K 2 .

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Thus, there are a, b ∈ K satisfying (2). Let c = a + αb. Assume there is an i ¯ in N . Then between 1 and m such that the polynomial hi (c, X) has a root x x ¯ ∈ F 0 ∩ N . Since res: Gal(N/L0 ) → Gal(M/K 0 ) is an isomorphism, there exists a Galois extension E 0 of K 0 contained in M such that E 0 L0 = F 0 ∩ N . Since E 0 ∩ L0 = K 0 = M ∩ L0 , Gal(F 0 /E 0 ) · Gal(F 0 /L0 ) = Gal(F 0 /K 0 ) Gal(F 0 /E 0 ) ∩ Gal(F 0 /L0 ) = Gal(F 0 /F 0 ∩ N ).

(3)

Therefore, K 0 (u, v) has a Galois extension E in F such that ϕ0 (Gal(F/E)) = Gal(F 0 /E 0 ). From (3), E ∩ L0 (u, v) = K 0 (u, v) and ϕ0 (Gal(F/EL0 )) = Gal(F 0 /F 0 ∩ N ). Moreover, the polynomial hi (u + αv, X) has a root x such x) = x ¯= that ϕ(x) = x ¯. For each σ ∈ Gal(F/EL0 ) we have ϕ(σx) = ϕ0 (σ)(¯ ϕ(x). Hence, σx = x. In particular, x ∈ EL0 , a contradiction to Lemma 13.9.2. We deduce Theorem 13.9.1(a) from Proposition 13.9.3: Proposition 13.9.4: Let M be a separable algebraic extension of a Hilbertian field K and M 0 a proper finite separable extension of M . Suppose M 0 is not contained in the Galois closure of M/K. Then M 0 is Hilbertian. ˆ . With no loss replace M Proof: Denote the Galois closure of M/K by M ˆ ∩ M 0 . Choose a primitive element α for M 0 /M . Put L0 = K(α), and by M ˆ ∩ L0 . Then the conditions of Proposition 13.9.3 are satisfied for K0 = M 0 0 ˆ K , L , M replacing K, L, M . Let f ∈ M 0 [T, X] be an irreducible polynomial, monic and separable in X with degX (f ) > 1. By Lemma 13.1.2, there exist h1 , . . . , hm ∈ M 0 [T, X] 0 absolutely irreducible and separable in X such that HM 0 (h1 , . . . , hm ) ⊆ 0 ˆ HM 0 (f ). Since h1 , . . . , hm are irreducible in M [T, X], Proposition 13.9.3, ˆ (α)[X] and theregives a ∈ L0 with h1 (a, X), . . . , hm (a, X) irreducible in M 0 0 fore in M [X]. By Proposition 13.2.2, M is Hilbertian. ˜ whose finite proper extensions Example 13.9.5: Non-Hilbertian subfield of Q are Hilbertian. Denote the compositum of all finite solvable extensions of Q by Qsolv . It is not Hilbertian; there exists no a ∈ Qsolv such that X 2 − a is irreducible over Qsolv . However, Qsolv is a Galois extension of Q. Hence, by Corollary 13.9.1(b), every finite proper extension of Qsolv is Hilbertian. By Corollary 13.8.4, Qsolv is not the compositum of two Galois extensions of Q neither of which is contained in the other.

Exercises 1. [Fried10, §2] This exercise shows that it is not always possible to take m = 1 in Lemma 13.1.2. Consider an irreducible polynomial f ∈ Q[T, X], monic in X, and g1 , g2 ∈ Q[Y ], polynomials of positive degree for which f (gi (Y ), X)

Exercises

265

is reducible, i = 1, 2. In addition, suppose there exists an irreducible h ∈ Q[T, X] with degX (h) > 1 such that HQ0 (h) = {a ∈ Q | h(a, X) has no zero in Q} ⊆ HQ (f ). (a) Use Exercise 1 of Chapter 12 to conclude that h(gi (Y ), X) has a factor of degree 1 in X, say X − mi (Y ), where mi ∈ Q(Y ), i = 1, 2. (b) Use field theory to interpret (a): Let t be an indeterminate, yi a zero of gi (Y ) − t, and xi = mi (yi ), i = 1, 2. Note that x1 and x2 are both zeros of h(t, X), so that they are conjugate over Q(t). Conclude that [Q(y1 ) ∩ Q(y20 ) : Q(t)] > 1 for some conjugate y20 of y2 over Q(t). (c) Consider the case f (T, X) = X 4 + 2X 2 − T and g1 (Y ) = Y 4 + 2Y 2 , g2 (Y ) = −4Y 4 − 4Y 2 − 1. Prove that the splitting field of gi (Y ) − t over Q(t) has the dihedral group of order 8 as its Galois group over Q(t). Observe that since Q(y1 )/Q(t) is a nonnormal extension of degree 4, Q(y12 )/Q(t) is its only quadratic subextension. Prove that the prime of Q(t) corresponding to the specialization t → 0 is unramified in Q(y12 ) but ramified in Q(y22 ). Conclude that there is no irreducible h ∈ Q[T, X] with degX (h) > 1 and HQ0 (h) ⊆ HQ (f ). 2. Let f1 (T, X), . . . , fm (T, X), with degX (fi ) > 1, i = 1, . . . , m, be absolutely irreducible polynomials, separable in X, with coefficients in a global field K. Let t be transcendental over K and denote the splitting fields of f1 (t, X), . . . , fm (t, X), respectively, over E = K(t) by F1 , . . . , Fm . Assume F1 , . . . , Fm are linearly disjoint over E and let F = F1 . . . Fm . Observe that there is a σ ∈ Gal(F/E) that fixes none of the roots of f1 (t, X) . . . fm (t, X) and improve Lemma 13.3.4. Show there exists a prime ideal p of OK and an element ap ∈ OK such that for a ∈ OK , if a ≡ ap mod p, then fi (a, X) has no zeros in K, i = 1, . . . , m. 3.

Consider the three absolutely irreducible polynomials f1 (T, X) = X 2 − T, f2 (T, X) = X 2 − (T + 1), f3 (T, X) = X 2 − T (T + 1)

and let H = HQ (f1 , f2 , f3 ). Choose a prime number p and an integer a such that both a and a + 1 are quadratic nonresidues modulo p. Prove that a + pZ ⊆ H, even though F1 , F2 , and F3 (in the notation of Exercise 2) are not linearly disjoint over E. 4. Let K be a Hilbertian field with valuation v. Prove that every Hilbert subset H of K r is v-dense in K r . Hint: Let H = HK (f1 , . . . , fm ), where f1 , . . . , fm are irreducible polynomials in K(T1 , . . . , Tr )[X1 , . . . , Xn ]. For (a1 , . . . , ar ) ∈ K r , and γ = v(c) an element of the value group, each polynomial in the set {fi (a1 + cT1ε1 , . . . , ar + cTrεr , X) | 1 ≤ i ≤ m and ε1 , . . . , εr ∈ {±1}}, is irreducible in K(T)[X]. Substitute elements t1 , . . . , tr for T1 , . . . , Tr so that these polynomials remain irreducible in K(X). Thus, find (b1 , . . . , br ) ∈ H such that v(bi − ai ) ≥ γ, i = 1, . . . , r.

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5. Suppose K is the quotient field of a Hilbertian ring R. Let M1 , M2 be Galois extensions of K and M an extension of K in M1 M2 which is contained in neither M1 nor in M2 . Prove the integral closure of R in M is Hilbertian. Hint: In the proof of Theorem 13.8.3 choose b1 , . . . , bn integral over R. Then choose c1 , . . . , cn in R.

Notes Hilbert [Hilbert, p. 280] proves Lemma 13.1.1(b) for number fields. Our proof of Proposition 13.2.1 is a version of Inaba’s proof [Inaba]. Theorem 13.3.5(a) for number fields appears in [Eichler]. We follow [Fried5]. Lang [Lang3, p. 152] reproduces Franz’s power series expansion proof [Franz]. Theorem 13.6.2 for g = 0 and K a number field appears in [CorvajaZannier]. Generalization to arbitrary g and arbitrary countable Hilbertian field K of characteristic 0 appears in [Fried-Jarden4]. The proof uses a deep group theoretic result due to Guralnick, Thompson, et al. and the Riemann existence theorem. Thus, one cannot generalize the proof to positive characteristic. An elementary proof of the theorem for arbitrary g but still for K a number field can be found in [Zannier]. Our proof generalizes that of Zannier. We replace the polynomial Y 2 − X 2q + 2 which Zannier uses by the polynomial Y l − X lq + u in order to make the proof works for each global field K. Weissauer proves that every finite proper separable extension of a Galois extension of a Hilbertian field is Hilbertian [Weissauer, Satz 9.7]. Section 13.9 replaces Weissauer’s “nonstandard” proof by a simpler “standard” one still using Weissauer’s auxiliary variable trick as in Lemma 13.9.2, [Fried10, Thm. 1.3]. Haran’s diamond theorem 13.8.3 generalizes Weissauer’s theorem (Theorem 13.9.1) and Corollary 13.8.5 [Haran-Jarden5]. Its proof is an offspring of the proofs of the two earlier results. Proposition 13.4.1 says the ring of integers OK of a number field K is Hilbertian. Thus, OK ∩ HK (f ) is an infinite set for each irreducible polynomial f ∈ K[T, X]. One may further ask when OK r HK (f ) is finite. This is not always the case. For example, OK r HK (X 2 − T ) is the infinite set of uller proves the set Z r HQ (f ) is finite all squares in OK . But, for K = Q, M¨ in each of the following cases: degX (f ) is a prime number and the curve f (T, X) = 0 has positive genus or Gal(f (T, X), Q(T )) ∼ = Sn for some positive integer n 6= 5 [M¨ uller, Thm. 1.2]. The proof uses Siegel’s theorem about integral points on algebraic curves and classical results about finite groups.

Chapter 14. Nonstandard Structures A. Robinson invented “nonstandard” methods in order to supplement the Weierstrass ε, δ formalism of the calculus by a rigorous version of the classical calculus of infinitesimals in the spirit of Leibniz and other formalists. We will apply the nonstandard approach to algebra in Chapter 15 in order to find new Hilbertian fields. Its main virtue, from an algebraic point of view, is that it creates additional algebraic structures to which well known theorems can be applied. Here we present the basics of the nonstandard method: the higher order structure on a set M (Section 14.1); the concept of an enlargement M ∗ of M (Sections 14.2 and 14.3); and the existence of M ∗ (Section 14.4) via ultraproducts.

14.1 Higher Order Predicate Calculus Sentences that speak of arbitrary subsets, functions, relations, etc., are common in mathematics, even though they are usually outside the scope of first order languages. Here we introduce a language which includes such sentences. First the notion of a type (of a higher order object) is inductively defined by the following rules: (1a) The number 0 is a type. (1b) If n is a positive integer and τ (1), . . . , τ (n) are types, then the sequence (τ (1), . . . , τ (n)) is a type. Denote the set of all types by T . Given a set M , attach a set Mτ to each type τ as follows: (2a) M0 = M . (2b) If τ (1), . . . , τ (n) are types and τ = (τ (1), . . . , τ (n)), then Mτ = P(Mτ (1) × · · · × Mτ (n) ) = {all subsets of Mτ (1) × · · · × Mτ (n) }. Elements of Mτ are objects of type τ over M . We call them sets (or relations) of type τ if τ 6= 0. A higher order set is a set of type τ for some τ 6= 0. Call the system M = hMτ | τ ∈ T i the higher order structure over M . For each higher order set A of M introduce a sequence of variables XA1 , XA2 , . . . . Inductively define formulas of the higher order language L∞ (M ) as follows: (3a) XAi = XAj , XAi = a, and a = b are formulas for each higher order set A, all a, b ∈ A, and i, j ∈ N. (3b) If A(1), . . . , A(n) are higher order sets, A(0) is a subset of P(A(1) × · · · × A(n)), and for ν between 0 and n either tν is a variable XA(ν),i(ν) or an element of A(ν), then (t1 , . . . , tn ) ∈ t0 is a formula.

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(3c) Negations and disjunctions (hence also conjunctions and implications) of formulas are formulas. (3d) If ϕ is a formula, then (∃XAi ∈ A)[ϕ] (hence also (∀XAi ∈ A)[ϕ]) is a formula for all higher order set A and i ∈ N. Define free variables of a formula as usual (Section 7.1). In particular, a sentence is a formula without free variables. Likewise, a substitution is a function f that, for each higher order set A of M , replaces each XAi by an element xAi of A and fixes the elements of A. Interpret the truth of a formula ϕ under f by reading “(t1 , . . . , tn ) ∈ t0 ” as “(f (t1 ), · · · , f (tn )) belongs to f (t0 ).” Note that each object t of type τ = (τ (1), . . . , τ (n)) 6= 0 has two roles in the language L∞ (M ). First, it is a constant: t is equal or not equal to another object of the same type. Second, it is a relation between objects of types τ (1), . . . , τ (n). If the set M has a first order structure, embed this in the higher order structure of M in a natural way. For example, if M is a field, then addition corresponds to a subset A of M × M × M , and X1 + b = c in the first order language becomes (XM 1 , b, c) ∈ A. Remark: Let N∗ be a proper elementary extension of N. Then N(0) (resp. (N∗ )(0) ) is the collection of subsets of N (resp. N∗ ). The second order structure N∗ ∪ (N∗ )(0) is not an elementary extension of the second order structure N ∪ N(0) . For example the induction axiom (∀X ∈ N(0) ) [1 ∈ X ∧ (∀x ∈ N)[x ∈ X → x + 1 ∈ X]] → (∀x ∈ N)[x ∈ X] holds in N ∪ N(0) but it fails in N∗ ∪ (N∗ )(0) if we replace N and N(0) by N∗ and (N∗ )(0) . Indeed, the induction axiom does not hold for X = N. In order to restore the elementary extension property for higher order structures of the elementary extension N∗ of N, we must restrict quantification of subsets of N∗ to a proper subcollection of (N∗ )(0) , to be denoted N∗(0) , the “inner” subset of N∗ . We define these in the next section.

14.2 Enlargements We consider a set M together with its higher order structure and define an enlargement of M as a special model of the higher order theory of M satisfying Conditions I and II below and III of Section 14.3. The enlargement will be saturated with respect to all higher order relations. The underlying set M ∗ of this model has the property that to each object a of type τ of M there corresponds an object a∗ of M ∗ of the same type (i.e. a∗ is an element of (M ∗ )τ ). Call such an a∗ standard. In particular, Mτ itself, an element of M(τ ) , corresponds to an element (Mτ )∗ of (M ∗ )(τ ) . This means (Mτ )∗ is a subset of (M ∗ )τ . Call the elements of (Mτ )∗ the internal objects of type τ of M ∗ . All other elements of (M ∗ )τ are external.

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269

Note: For τ = 0 we have (M0 )∗ = (M ∗ )0 = M ∗ . Thus, every element of M ∗ is internal, although (M ∗ )(0) itself may have external subsets. To simplify notation write Mτ∗ for (Mτ )∗ . If τ (1), . . . , τ (n) are types and if R ⊆ Mτ (1) × · · · × Mτ (n) , then R ∈ Mτ for τ = (τ (1), . . . , τ (n)). Hence, R∗ ∈ Mτ∗ , that is, R∗ ⊆ (M ∗ )τ (1) × · · · × (M ∗ )τ (n) . We demand, however, that an enlargement satisfy a stronger condition: I. Internal condition on n-ary relations. If τ (1), . . . , τ (n) are types and if R ⊆ Mτ (1) × · · · × Mτ (n) , then R∗ ⊆ Mτ∗(1) × · · · × Mτ∗(n) . That is, the elements of a standard n-ary relation are n-tuples with internal coordinates. Call them internal n-tuples. Suppose τ is a type and A is a subset of Mτ . By Condition I, A∗ ⊆ Mτ∗ . So, each element of A∗ is an element of Mτ∗ , hence internal. Consequently, each element of a standard set is internal. On the other hand, each subset of A∗ is an element of (M ∗ )(τ ) . It is ∗ internal exactly when it belongs to M(τ ). To interpret a formula of L∞ (M ) in M ∗ consider only internal substitutions f . These satisfy the condition f (XAi ) ∈ A∗ for each higher order set A. Define the truth value of the formula ϕ under f first by placing an asterisk to the right of each constant or relation symbol that appears in ϕ to obtain a formula ϕ∗ . Then interpret the formula ϕ∗ (under f ) as usual. Note: If (∃X ∈ A) is part of ϕ, then (∃X ∈ A∗ ) is a part of ϕ∗ . Since each element of A∗ is internal, this means one quantifies only on internal objects. For example, consider a formula ϕ(X1 , . . . , Xn ) where Xν = XMτ (ν) ,i(ν) , ν = 1, . . . , n. Let R = {(a1 , . . . , an ) ∈ Mτ (1) × · · · × Mτ (n) | M |= ϕ(a1 , . . . , an )}. Then the sentence θ (∀X1 ∈ Mτ (1) ) · · · (∀Xn ∈ Mτ (n) )[(X1 , . . . , Xn ) ∈ R ↔ ϕ(X1 , . . . , Xn )] is true in M . The sentence θ∗ takes the form (∀X1 ∈ Mτ∗(1) ) . . . (∀Xn ∈ Mτ∗(n) )[(X1 , . . . , Xn ) ∈ R∗ ↔ ϕ∗ (X1 , . . . , Xn )]. It is reasonable to ask that θ∗ will be true in M ∗ . Since Condition I implies that R∗ contains only elements with internal coordinates, we may rephrase θ∗ as R∗ = {(a1 , . . . , an ) ∈ Mτ∗(1) × · · · × Mτ∗(n) | M ∗ |= ϕ∗ (a1 , . . . , an )}. Here is the condition guaranteeing this indeed holds:

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II. Elementary extension condition. Let A1 , . . . , An be higher order sets, ϕ(XA1 ,i(1) , . . . , XAn ,i(n) ) a formula of L∞ (M ), and aν ∈ Aν , ν = 1, . . . , n. Then ϕ(a1 , . . . , an ) is true in M if and only if ϕ∗ (a∗1 , . . . , a∗n ) is true in M ∗ . For example, if a ∈ Mτ , then a∗ ∈ Mτ∗ . Thus, standard objects are internal. Also, if a, b ∈ Mτ and a 6= b, then a∗ 6= b∗ : the canonical map a 7→ a∗ of Mτ into Mτ∗ is injective. Therefore, we occasionally regard Mτ as a subset of Mτ∗ . In addition, for R ⊆ A1 × · · · × An , if a ∈ A1 × · · · × An belongs to R∗ , then a ∈ R. Thus, R∗ ∩ (A1 × · · · × An ) = R and the relation R∗ is an extension of R. In particular, if M has a first order structure, then it is an elementary substructure of a natural extension to M ∗ . Now consider the notion of an internal function. Let A and B be two higher order sets. We view a function f : A → B as a subset of A × B that satisfies these two conditions: (1a) (1b)

(∀a ∈ A)(∃b ∈ B)[(a, b) ∈ f ]; (∀a ∈ A)(∀b1 , b2 ∈ B)[(a, b1 ) ∈ f ∧ (a, b2 ) ∈ f → b1 = b2 ].

The subset f ∗ of A∗ × B ∗ satisfies the corresponding conditions: it is a function from A∗ to B ∗ . Call it a standard function. For F , the set of all functions from A to B, the elements of F ∗ are the internal functions from A∗ to B ∗ . Lemma 14.2.1: The image of each internal subset of A∗ under an internal function f : A∗ → B ∗ is an internal subset of B ∗ . Proof: For each f ∈ F the following statement holds in M : (3)

(∀f ∈ F )(∀X ⊆ A)(∃Y ⊆ B) [(∀a ∈ X) f (a) ∈ Y ]

∧ [(∀b ∈ Y )(∃a ∈ X) f (a) = b]

The close “∀X ⊆ A” is not part of the language L∞ (M ). However, it can be reinterpreted as “(∀X ∈ M(τ ) )(∀x ∈ Mτ )[x ∈ X → x ∈ A]”, assuming A ⊆ Mτ and with X = XM(τ ) ,1 and x = XMτ ,1 . The interpretation of this ∗ ∗ ∗ in M ∗ is “(∀X ∈ M(τ ) )(∀x ∈ Mτ )[x ∈ X → X ∈ A ]”. This means “for all ∗ internal subsets X of A ”. Similarly, “∃Y ⊆ B” interprets in M ∗ as “there is an internal subset of B ∗ ”. Thus, our claim is the interpretation of (3) in M ∗.

14.3 Concurrent Relations Suppose a higher order relation R of M is finite with elements a1 , . . . , an . Then M |= (∀X ∈ R)[X = a1 ∨ · · · ∨ X = an ], so that M ∗ |= (∀X ∈ R∗ )[X = a∗1 ∨ · · · ∨ X = a∗n ]. That is, each element of R∗ is standard. We impose a final condition on M ∗ guaranteeing R∗ will be a true “enlargement” of R if R is an infinite relation.

14.3 Concurrent Relations

271

Definition 14.3.1: Let A and B be two higher order sets of M . Call a binary relation R ⊆ A × B concurrent if for all a1 , . . . , an ∈ A there exists b ∈ B with (ai , b) ∈ R, i = 1, . . . , n. III. Compactness (or saturation) condition. If A, B are two higher order sets of M and if R ⊆ A × B is a concurrent relation, then there exists b ∈ B ∗ such that (a, b) ∈ R∗ for each a ∈ A. Corollary 14.3.2: Let A be a higher order set of M . (a) If A = {a1 , . . . , an } is a finite set, then A∗ = {a∗1 , . . . , a∗n } and A is identified with A∗ as higher order sets. (b) If A is infinite, then A is properly contained in A∗ . Wn XA,1 = a1 ] holds in M . Hence, Proof of (a): The sentence (∀XA,1 ∈ A)[ i=1 W n by Condition II of Section 14.2, (∀XA,1 ∈ A∗ )[ i=1 XA,1 = a∗1 ] is true in M ∗ . ∗ ∗ ∗ Thus, A = {a1 , . . . , an }. Proof of (b): For every a1 , . . . , an in A there exists b ∈ A with ai 6= b, i = 1, . . . , n. That is, the inequality relation on A is concurrent. Hence, there exists b ∈ A∗ such that a 6= b for each a ∈ A (although not for each a ∈ A∗ ). Remark 14.3.3: Warning. Let A be a higher order set of M . Suppose A is an element of another higher order set B. In Section 14.2, we have identified B with a subset of B ∗ . Under this identification we have identified the element A of B with the element A∗ of B ∗ . Corollary 14.3.2 shows that the latter identification identifies A with A∗ as higher order sets if and only if A is finite. The case N = M , the natural numbers, provides the first example of an external object. Replacing M with N ∪ M , if necessary, we tacitly assume from now on that N ⊆ M . Lemma 14.3.4: The set N is an external subset of N∗ . Proof: By Corollary 14.3.2, N∗ contains a nonstandard element c. Since there are no elements of N between n and n + 1, the same is true for N∗ . In the extension I0 > I1 > I2 > · · · is a strictly descending sequence. 15. (a) Prove that N∗ r N is not an internal subset of N∗ . (b) Prove that for each a ∈ N the set {x ∈ N∗ | x > a} is internal.

Notes This chapter is in the spirit of Chapter 2 of the [Robinson-Roquette], which contains a nonstandard proof of the Siegel-Mahler theorem.

Chapter 15. Nonstandard Approach to Hilbert’s Irreducibility Theorem We use the nonstandard methods of Chapter 14 to give a new criterion for a field K to be Hilbertian: There exists a nonstandard element t of an enlargement K ∗ of K such that t has only finitely many poles in K(t)s ∩ K ∗ . From this there results a second and uniform proof (Theorem 15.3.4) that the classical Hilbertian fields are indeed Hilbertian. In addition, a formal power series field, K0 ((X1 , . . . , Xn )) of n ≥ 2 variables over an arbitrary field K0 , is also Hilbertian (Example 15.5.2).

15.1 Criteria for Hilbertianity We give two criteria for a field K to be Hilbertian in terms of an enlargement K ∗ of K. The first is a straightforward application of the compactness property of K. Proposition 15.1.1 (Gilmore-Robinson): The field K is Hilbertian if and only if there exists t ∈ K ∗ r K such that K(t)s ∩ K ∗ = K(t). Proof: Suppose K is Hilbertian. If f1 , . . . , fm ∈ K[T, X] irreducible polynomials which are separable in X and g1 , . . . , gm ∈ K[T ] are nonzero polynomials, then there exists a ∈ K with fi (a, X) irreducible K[X] and gi (a) 6= 0, i = 1, . . . , m. The compactness property (Condition III of Section 14.3) gives t ∈ K ∗ such that for each irreducible f ∈ K[T, X] which is separable in X and each 0 6= g ∈ K[T ], the polynomial f (t, X) is irreducible in K ∗ [X] and g(t) 6= 0. The second condition implies t 6∈ K. Consider x ∈ K(t)s ∩ K ∗ . Let f ∈ K[T, X] be an irreducible polynomial which is separable in X with f (t, x) = 0. Since f (t, X) is irreducible over K ∗ , it is linear. Hence, x ∈ K(t). Conversely, suppose t is an element of K ∗ r K with K(t)s ∩ K ∗ = K. Let f1 , . . . , fm ∈ K[T, X] be irreducible polynomials which are separable, monic, and of degree at least 2 in X and let 0 6= g ∈ K[T ]. If for some i between 1 and m, fi (t, X) is reducible over K ∗ , the coefficients of its factors lie in K(t)s ∩ K ∗ = K(t). Thus, fi (t, X) is reducible over K(t). But since t is transcendental over K, this means fi (T, X) is reducible over K, contrary to our assumption. Finally, since K ∗ is an elementary extension of K, there exists a ∈ K such that g(a) 6= 0 and fi (a, X) is irreducible over K, i = 1, . . . , m. By Lemma 12.1.4, K is Hilbertian. One may reapproach the classical Hilbertian fields through Proposition 15.1.1 [Roquette2], but it is easier to apply the following weaker condition of Weissauer. Consider t ∈ K ∗ r K and let Ωt = K(t)s ∩ K ∗ . Then Ωt is a separable algebraic (possibly infinite) extension of K(t). Since Ωt is regular over K

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(Example 14.5.3) consider it as a generalized function field of one variable over K. It is a union of function fields of one variable over K. Call an equivalence class of valuations of Ωt which are trivial on K a prime divisor of Ωt /K. Refer to a prime divisor as a pole of t if t has a negative value. Finally, call t polefinite if t has only finitely many poles in Ωt . In other words, there is an integer m such that the number of poles of t in any function field F with K(t) ⊆ F ⊆ Ωt is at most m. Proposition 15.1.2 ([Weissauer, Folgerung 3.2]): If K ∗ contains a polefinite element, then K is Hilbertian. Proof: The existence of a polefinite element implies K 6= K ∗ . Hence, K is infinite (discussion preceding Corollary 14.3.2). Assume K is not Hilbertian. Then Proposition 13.2.2 gives an irreducible polynomial f ∈ K[T, X], separable, monic, and of degree at least 2 in X, such that f (a, X) is reducible over K for each a ∈ K. The same statement holds in K ∗ . In particular, for each t ∈ K ∗ r K, (1) f (t, X) is irreducible over K(t), but reducible over K ∗ . Consider now an element t of K ∗ r K. The remaining parts of the proof consider properties of the prime divisors of K(t). Part A: Removing ramification over ∞. Let F be the splitting field of ˜ ∪ {∞} let pa f (t, X) over K(t). Suppose degt (f ) = d. For each a ∈ K be the prime divisor of K(t)/K defined by t 7→ a. In particular, p∞ is the unique pole of t in K(t)/K. Since K is infinite, we may choose a ∈ K with pa unramified in F and with residue field F¯ Galois over K. Apply 1 to replace f (T, X) by the K-automorphism of K(t) defined by t 7→ t−a 1 d T g(T, X) = f ( T −a , X), the field F by the splitting field of g(t, X), and pa by p∞ (Section 2.2). Thus assume, along with (1), that (2) p∞ is unramified in F . Each a ∈ K defines a K-automorphism σa of K(t) by t 7→ t + a. It fixes p∞ . Extend σa to an automorphism of K(t)s . Denote σa (F ) by Fa . Thus, p∞ is unramified in Fa and F¯a is Galois over K. By Corollary 2.3.7(c), (3) p∞ is unramified in the compositum, F 0 , of Fa for all a ∈ K. Part B: The finiteness of F 0 over ∞. Let P be a prime divisor of F 0 over p∞ . Denote reduction modulo P by a bar. The residue fields F¯ and F¯a are conjugate over K (Section 2.3). Since both are Galois extensions of K (Lemma 6.1.1), they coincide. The compositum of the residue class fields of unramified extensions of K(t) is the residue class field of their compositum (Lemma 2.4.8). Thus, F¯ 0 = F¯ is a finite extension of K which is independent of P. Part C: The infinitude of [F 0 ∩ K ∗ : K(t)]. By (1), Fa is not linearly disjoint from K ∗ over K(t). Hence, Ea = Fa ∩ K ∗ , a regular extension of K contained in Ωt , is a proper extension of K(t). Let R(Ea ) (resp. R(Fa ))

15.2 Arithmetical Primes Versus Functional Primes

279

be the set of prime divisors of K(t)/K that ramify in Ea (resp. in Fa ). By Remark 3.6.2(b), R(Ea ) is finite and nonempty. Note: pc = pc0 if and only if c0 is conjugate to c over K. If R(F ) = {pc1 , . . . , pcn }, then R(Fa ) = {pc1 −a , . . . , pcn −a }. For L a finite separable extension of K(t) let R(L) be {pd1 , . . . , pdr }. Since K is an infinite field, we may choose a ∈ K with ci − a 6= τ (dj ) for all i and j and for every K(t)-isomorphism τ of L. Hence R(Ea ) ∩ R(L) ⊆ R(Fa ) ∩ R(L) = ∅. Since R(Ea ) 6= ∅, the field Ea is not contained in L. Therefore, the compositum E 0 of Ea for all a ∈ K is an infinite extension of K(t) contained in Ωt . Part D: Conclusion of the proof. Assume t has only m poles in Ωt . Use Part C to choose a finite extension N of K(t) in E 0 with [N : K(t)] > m[F¯ : K]. Let q1 , . . . , qk be all extensions of p∞ to N . Each of them extends to a ¯q ⊆ F¯ . Hence, by Proposition 2.3.2, pole of t in Ωt , so k ≤ m. By Part B, N i [N : K(t)] =

k X

¯q : K] ≤ m[F¯ : K]. [N i

i=1

This contradiction to the choice of N prove that t is not a polefinite element.

15.2 Arithmetical Primes Versus Functional Primes To apply Proposition 15.1.2 we start with a field K that carries arithmetic structure and extend this structure to an enlargement K ∗ of K. For each t ∈ K ∗ r K we consider finite extensions F of K(t) in K ∗ . Then we compare the function field structure of F/K with the arithmetic structure induced on F from K ∗ . The goal is to find conditions on t to be a polefinite element. Let S is a set of primes of K. Thus, S is a set of equivalent classes of absolute values of K (Section 13.3). For each p ∈ S choose a representative | |p . Define a map vp : K → R ∪{∞} by vp (a) = − log(|a|p ). Conditions (3a)– (3d) of Section 13.3 on the valuation translate into the following properties of vp : (1a) vp (a) = ∞ if and only if a = 0. (1b) vp (ab) = vp (a) + vp (b). (1c) vp (a + b) ≥ min(vp (a), vp (b)) − log 2. (1d) There is an a ∈ K × with vp (a) 6= 0. We call vp an additive absolute value of K. If p is ultrametric, then |a + b|p ≤ max(|a|p , |b|p ), hence vp (a + b) ≥ min(vp (a), vp (b)) for all a, b ∈ K. Thus, vp is a valuation of K. The following lemma gives a simple criterion for p to be metric: Lemma 15.2.1: A prime p of a field K is metric if and only if vp (2) < 0. Proof: Suppose first that p is metric. By Section 13.3, there is a positive integer n with |n|p > 1. Hence, char(K) = 0 and the restriction of | |p to Q is

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a metric absolute value. By a theorem of Ostrowski, | |p is equivalent to the ordinary absolute value of Q. Thus, there is a positive real number c such that |x|p = |x|c for each x ∈ Q. In particular, |2|p = |2|c = |nlog 2/ log n |c > 1. Consequently, vp (2) < 0. Conversely, if vp (2) < 0, then |2|p > 1, so p is metric. We assume now that S satisfies the following finiteness condition: (2) If a ∈ K × , then {p ∈ S | vp (a) 6= 0} is a finite set. By Lemma 15.2.1, there are only finitely many archimedean primes in S. Consider again an enlargement of K ∗ of K. Then S extends to a set ∗ S of arithmetical primes of K ∗ . To each p ∈ S ∗ there corresponds a star-additive absolute value vp : (K ∗ )× → R∗ . In particular, each prime in S extends to an element p of S ∗ , a standard prime. The elements of S ∗ r S are the nonstandard primes. Condition (2) becomes: (2)∗ If a 6= a ∈ K ∗ , then {p ∈ S ∗ | vp (a) 6= 0} is starfinite (Example 14.5.1). In particular, if 0 6= a ∈ K, then the finite set {p ∈ S | vp (a) 6= 0} does not grow in the enlargement Thus, vp (a) = 0 for all p ∈ S ∗ r S and a ∈ K × . Therefore, by (2), there are only finitely many archimedean primes in S ∗ ; they are all standard. For an arbitrary prime p ∈ S ∗ consider the convex hull of vp (K × ): Γp = {r ∈ R∗ | ∃x ∈ K × with |r| ≤ vp (x)}. In the notation of Example 14.5.2, (3) Γp = Rfin if p ∈ S and Γp = 0 if p ∈ S ∗ r S. Unlike nonstandard primes, standard primes form valuations that are nontrivial on K. We modify them so that they will “behave” like the nonstandard primes. However, these “modified” primes will not be internal objects. Combining internal and external objects is the key to the nonstandard machinery. ˙ p = R∗ /Γp . By (3), R˙ p = For p ∈ S ∗ , consider the ordered group R ˙ (Example 14.5.2) if p is standard and R ˙ p = R∗ otherwise. Define the R ˙ p by v˙ p (x) = vp (x) + Γp . modified valuation v˙ p : (K ∗ )× → R Suppose p is standard. Then p may be metric or ultrametric. In the second case vp is a valuation, whereas in the first case it is not. In both cases, however, vp satisfies (3). Since log 2 ∈ R ⊆ Γp , v˙ p (a+b) ≥ min{v˙ p (a), v˙ p (b)}. Hence, v˙ p is a valuation. In addition, if 0 6= x ∈ K, then vp (x) ∈ R. Hence, v˙ p (x) = 0. In summary: ˙ p defined Lemma 15.2.2: For each p ∈ S ∗ the function v˙ p : (K ∗ )× → R by v˙ p (x) = vp (x) + Γp is a valuation which is trivial on K. Moreover, if p ∈ S ∗ r S, then v˙ p = vp . Consider an element t ∈ K ∗ r K. Let F be a finite extension of K(t) in K . By Example 14.5.3, F is a regular extension of K, hence a function field ∗

15.3 Fields with the Product Formula

281

over K. Consider p ∈ S ∗ with v˙ p not vanishing on F × . Then, by Lemma 15.2.2, the restriction of v˙ p to F is a valuation which is trivial valuation on K. Thus, it defines a prime P of F/K. We say P is induced from p. Conversely, we may ask if a prime divisor P of F/K (i.e. a functional prime) if it is induced by an arithmetical prime. This question is inspired by the following observation. Proposition 15.2.3: Let S be a set of primes of a field K with {p ∈ S | vp (a) 6= 0} being a finite set for each a ∈ K × . Let t be a nonstandard element of an enlargement K ∗ of K. Suppose S(t) = {p ∈ S ∗ | v˙ p (t) < 0} is a finite set and for each finite separable extension F of K(t) in K ∗ all poles of t in F are induced by arithmetical primes (i.e. elements of S ∗ ). Then t is a polefinite element. Proof: Let P be a pole of t in Ωt . For each finite extension F of K in Ωt the restriction of P to F is also a pole of t. Hence, there is a pF ∈ S ∗ which induces P |F . In particular, v˙ pF (t) < 0, so pF ∈ S(t). Since S(t) is finite, there is a p ∈ S(t) which induces P |F for each F as above (Use compactness of S(t).) Therefore, p determines P . Thus, the number of poles of t in Ωt is at most |S(t)|. Consequently, t is a polefinite element.

15.3 Fields with the Product Formula As in Section 15.2, suppose S is a nonempty set of primes of K. For each p ∈ S choose an absolute value | |p representing p and let vp be the corresponding additive absolute value. We say S satisfies a product formula if the following statement holds: for each p ∈ S there exists a positive real number λp with the following property: Q λ (1) For each a ∈ K × the set {p ∈ S | |a|p 6= 1} is finite and p∈S |a|p p = 1. In this case call K a field with a product formula. Example 15.3.1: Basic examples of product formulas. For K = Q and p a prime number, define | |p by |a|p = p−r for a = xy · pr with r ∈ Z and x, y ∈ Z relatively prime to p. The infinite absolute value | |∞ is the usual absolute value. This set S of primes satisfies the product formula with λp = 1 for each p ∈ S. For K0 (t), a rational function field over an arbitrary field K0 , choose a real number 0 < c < 1. For each irreducible polynomial p ∈ K0 [t] define the absolute value | |p by |u|p = c−r for u = fg ·pr with f, g polynomials relatively prime to p. Let λp be deg(p). For a quotient f /g of polynomials, define the infinite absolute value as |f /g|∞ = cdeg(f )−deg(g) , and let λ∞ = 1. Check that the corresponding set of primes of K0 (t) satisfies the product formula. If a set S of primes of a field K is a field with a the product formula and if S 0 is the set of primes of a finite extension K 0 /K lying over S, then

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Chapter 15. Nonstandard Approach to Hilbert’s Irreducibility Theorem

S 0 satisfies the product formula [Lang3, p. 20]. We conclude from Example 15.3.1 that every global field is a field with a product formula. Again let K be any field with the product formula with respect to a nonempty set of primes S. Rewrite condition (1) additively: P (2) For each a ∈ K × , {p ∈ S | vp (a) 6= 0} is finite and p∈S λp vp (a) = 0. For an enlargement K ∗ of K, Condition (2) has a similar form: ∗ ∗ × ∗ (2) For P each a ∈ (K ) , the set {p ∈∗ S | vp (a) 6= 0} is starfinite and p∈S P∗ λp vp (a) = 0, where λp ∈ R and λp > 0 are independent of p and here is the starfinite summation (Example 14.5.1). Proposition 15.3.2: Let t be a nonstandard element of K ∗ and F a finite separable extension of K(t) in K ∗ . If there exists q ∈ S ∗ with v˙ q (t) < 0, then each functional prime of F is induced by an arithmetical prime of K ∗ . Proof: Denote the set of all functional primes of F which are induced by elements of S ∗ by P. Let D = {x ∈ F | vP (x) ≥ 0 for all P ∈ P} = {x ∈ F | v˙ p (x) ≥ 0 for all p ∈ S ∗ } be the holomorphy ring of P. The assumption v˙ q (t) < 0 implies that t 6∈ D and P 6= ∅, so D 6= F . Assume P does not contain all the functional primes of F . Then F = Quot(D) (Proposition 3.3.2(a)). We now show that D is a field. This will give a contradiction to D 6= F and establish the theorem. We need to prove that if x ∈ D× and x 6= 0, then x−1 ∈ D. Indeed, by (2)∗ the set S(x) = {p ∈ S ∗ | vp (x) < 0} is starfinite. For p ∈ S ∗ r S, Lemma 15.2.2 shows that vp (x) = v˙ p (x) ≥ 0. This means S(x) contains only standard primes. Hence, by Corollary 14.3.4(a), S(x) is a finite set. For p ∈ S(x) we have vp (x) < 0 and v˙ p (x) ≥ 0, so v˙ p (x) = 0. Now use the summation formula (2)∗ and the additivity of starfinite summation (Example 14.5.1): X X (3) λp vp (x) = ˙ λp vp (x) = ˙ 0. p∈S ∗ r S(x)

p∈S ∗

If p0 ∈ S ∗ r S(x) and v˙ p0 (x) 6= 0, then v˙ p0 (x) > 0. This means X ˙ 0, λp vp (x) ≥ λp0 vp0 (x) > p∈S ∗ −S(x)

a contradiction to (3). Thus, v˙ p (x) = 0 for each p ∈ S ∗ r S(x) and therefore for each p ∈ S ∗ . It follows that v˙ p (x−1 ) = 0 for each p ∈ S ∗ . Consequently, x−1 ∈ D, as desired. Theorem 15.3.3 ([Weissauer, Satz 6.2]): Every field K with the product formula is Hilbertian. Proof: Let S be a nonempty set of primes of a field K satisfying the product formula. Choose q ∈ S and a ∈ K with vq (a) < 0. In an enlargement

15.4 Generalized Krull Domains

283

K ∗ of K, pick a nonstandard positive integer ω ∈ N∗ r N. We prove the nonstandard element t = aω of K ∗ is polefinite. Indeed, S(t) = {p ∈ S ∗ | vp (t) < 0} = {p ∈ S ∗ | vp (a) < 0} = {p ∈ S ∗ | v˙ p (t) < 0}. Therefore, S(t) is a starfinite set that contains q. In particular, v˙ q (t) < 0. By Lemma 15.1.2, if p ∈ S ∗ r S, then v˙ p (a) = 0. Thus, S(t) consists only of standard primes, so, by Corollary 14.3.4(b), S(t) is a finite set. Suppose F is a finite separable extension of K(t) in K ∗ . By Propositions 15.3.2, each functional prime P of F/K(t) is induced by some p ∈ S ∗ . If P is a pole of t, then p ∈ S(t). By Proposition 15.2.3, t is polefinite. Consequently, by Proposition 15.1.2, K is Hilbertian. Corollary 15.3.4: Every number field and every function field of several variables over an arbitrary field is Hilbertian.

15.4 Generalized Krull Domains Non-standard methods produce new Hilbertian fields: The quotient field of each ‘generalized Krull domain of dimension at least 2’ is Hilbertian (Theorem 15.4.6). In particular, fields of formal power series of at least two variables over arbitrary fields are Hilbertian (Theorem 15.4.6). Let R be an integral domain with quotient field K. We call R a generalized Krull domain if K has a nonempty set S of primes satisfying the following conditions: (1a) For each p ∈ S, vp is a real valuation. (1b) The valuation ring Op of vp is the local ring of R relative to mp = {a ∈ R | vT p (a) > 0}. (1c) R = p∈S Op . (1d) For each a ∈ K × the set {p ∈ S | vp (a) 6= 0} is finite. The dimension of a ring R is the maximal integer n for which there is a sequence p0 ⊂ p1 ⊂ · · · ⊂ pn of n + 1 distinct prime ideals. Thus, dim(R) ≥ 2 if and only if (1e) R has a maximal ideal M which properly contains a nonzero prime ideal. Thus, R is a generalized Krull ring of dimension exceeding 1 if and only if it satisfies Condition (1). Lemma 15.4.1: Let R be an integral domain satisfying Condition (1). Then: (a) For each nonunit b of R, b 6= 0, there exists p ∈ S with vp (b) > 0. (b) For each p ∈ S, mp is minimal among nonzero prime ideals of R. (c) If p, q are distinct primes in S, then mp 6⊆ mq and mp 6= M . T Proof of (a): Otherwise b−1 ∈ p∈S Op = R. Proof of (2b): Let n ⊆ mp be a nonzero prime ideal. Choose 0 6= a ∈ n. Since vp is real, for each b ∈ mp there is a positive integer n with vp (a) ≤

284

Chapter 15. Nonstandard Approach to Hilbert’s Irreducibility Theorem n

n

nvp (b) = vp (bn ), so ba ∈ Op . By (1b), there are c, d ∈ R with ba = dc and d 6∈ mp , so dbn = ac ∈ n. Since d 6∈ n, we have b ∈ n. Consequently, n = mp . Proof of (c): By (1b), mp 6= mq . By (b), mp 6⊂ mq Hence, mp 6⊆ mq . In addition, by (b) and (1e), m 6= M . Lemma 15.4.2: The local ring RM of R at M is a generalized Krull domain of dimension exceeding 1 with respect to S 0 = {p ∈ S | mp ⊂ M }. Proof: Conditions (1a), (1b), (1d), and (1e) follow from T the basic definitions of the local ring RM . It remains to prove that RM = p∈S 0 Op . This follows if we show, for x ∈ K with vp (x) ≥ 0 for each p ∈ S 0 , that x ∈ RM . By (1d), the set T = {q ∈ S | vq (x) < 0} is finite. If q ∈ T , then q ∈ / S0, r so mq 6⊆ M . Choose aq ∈ mq M . By (1a) there is a positive integer n(q) Q n(q) n(q) and y = ax. Then a ∈ R r M with vq (aq ) > −vq (x). Let a = q∈T aq and vp (y) ≥ 0 for each p ∈ S. Therefore, by (1c), y ∈ R. Consequently, x ∈ RM . For the goal of this section - a proof that K is Hilbertian - we may replace R by RM to assume that (2) R is a local ring and M is its maximal ideal. Consider an enlargement K ∗ of K. Then R∗ is a local ring with maximal ideal M ∗ . Also, S ∗ has these properties: (1a)∗ For p ∈ S ∗ , vp is a valuation of K ∗ with values in R∗ . (1b)∗ The valuation ring Op∗ of vp is the local ring of R∗ relative to m∗p = {a ∈ R∗ | vp (a) > 0}. T ∗ (1c) R∗ = p∈S ∗ Op∗ . (1d)∗ For each a ∈ K ∗ r{0} the set {p ∈ S ∗ | vp (a) 6= 0} is starfinite. (1e)∗ M ∗ properly contains the prime ideals m∗p . As with fields with a product formula, consider the modified valuations v˙ p and their valuation rings O˙ p = {x ∈ K ∗ | v˙ p (x) ≥ 0}. The holomorphy T ring of these valuations, R˙ = p∈S ∗ O˙ p , contains both K (Lemma 15.2.2) and R∗ . Moreover: Lemma 15.4.3: The ring R˙ equals K · R∗ . Proof: We have to show only that each x ∈ R˙ lies in the composite K · R∗ . Indeed, v˙ p (x) ≥ 0 for each p ∈ S ∗ . From (1d)∗ and lemma 15.2.2 the starfinite set S(x) = {p ∈ S ∗ | vp (x) < 0} contains only standard primes. Hence, (Corollary 14.3.5(b)), it is actually finite. For each p ∈ S(x) there exists v˙ p (x) ≥ 0). By (1b) we may find 0 6= ap ∈ R rp ∈ R with vp (x) ≥ rp (because Q with vp (ap ) ≥ −vp (x). Let a = p∈S(x) ap and y = ax. Then vp (y) ≥ 0 for each p ∈ S ∗ . By (1c)∗ , y ∈ R∗ . Consequently, x = a−1 y ∈ K · R∗ . ˙ Suppose for each p ∈ S we Lemma 15.4.4: Let x and y be nonunits of R. ˙ have v˙ p (x) > 0 or v˙ p (y) > 0. Then αx + βy 6= 1 for all α, β ∈ R.

15.4 Generalized Krull Domains

285

Proof: Assume there exist α, β ∈ R˙ with αx + βy = 1. Apply Lemma 15.4.3 to write (3)

ax by + = 1 with a, b ∈ R∗ and c ∈ K × . c c

We show that both summands on the left hand side of (3) belong to M ∗ . This contradiction to (3) will conclude the proof of the lemma. Indeed, let p ∈ S ∗ with vp (c) > 0. Then v˙ p (c) = 0 and p is standard (Lemma 15.2.2). Hence, v˙ p (x) > 0 or v˙ p (y) > 0. If v˙ p (x) > 0, then v˙ p ( ax c )> by ax ˙ 0. So, by (3), vp ( c ) = 0. Similarly, if v˙ p (y) > 0, then 0. Hence, vp ( c ) > by ˙ 0 and vp ( ax vp ( c ) > c ) = 0. by Next consider p ∈ S ∗ with vp (c) = 0. Then vp ( ax c ) ≥ 0 and vp ( c ) ≥ 0. by ∗ It follows from (1c)∗ that both ax c and c belong to R . ˙ there exists p ∈ S ∗ with v˙ p (x) > 0. Hence Since x is a nonunit in R, ax ax ∗ v˙ p ( c ) > 0 and therefore c ∈ M ∗ . Similarly by c ∈M . Our next lemma is “standard”: Lemma 15.4.5: Let {p1 , . . . , pm } and {q1 , . . . , qn } be two disjoint finite subsets of S. Then there exists an element a ∈ R such that (5) vpi (a) = 0, i = 1, . . . , m, and vqj (a) > 0, j = 1, . . . , n. Proof: Proceed by induction on m. Suppose m = 1. By Lemma 15.4.1(c), mqj 6⊆ mp1 , 1 ≤ j ≤ n. Let aj ∈ mqj r mp1 . Then a = a1 · · · an satisfies (5). Suppose m > 1. The induction hypothesis gives a1 ∈ R with vp1 (a1 ) = · · · = vpm−1 (a1 ) = 0 and vpm (a1 ), vq1 (a1 ), . . . , vqn (a1 ) > 0. By the case m = 1, there exists a2 ∈ R with vpm (a2 ) = 0 and vp1 (a2 ), . . . , vpm−1 (a2 ), vq1 (a2 ), . . . , vqn (a2 ) > 0. The element a = a1 + a2 satisfies (5).

We now prove the main theorem of this section. Theorem 15.4.6 (Weissauer): The quotient field of a generalized Krull domain of dimension exceeding 1 is Hilbertian. Proof: As previously, let K, R, S, M satisfy (1)-(2). Let q ∈ S and choose b ∈ M r mq (by (Lemma 15.4.1(c)). By (1d) and Lemma 15.4.1(a), the set T (b) = {p ∈ S | vp (b) > 0} is finite and nonempty. For each finite set T with T (b) ⊆ T ⊆ S there exists aT ∈ R such that vp (aT ) = 0 for p ∈ T (b) and vp (aT ) > 0 for each p ∈ T r T (b) (Lemma 15.4.5). Proposition 14.3.6 gives a ∈ R∗ such that vp (a) = 0 for each p ∈ T (b) and vp (a) > 0 for each p ∈ S r T (b). By Lemma 15.2.2, vp (b) = 0 for each p ∈ S ∗ r S. Thus, (5) T (a) = {p ∈ S ∗ | vp (a) > 0} is disjoint from T (b) = {p ∈ S ∗ | vp (b) > 0}. Choose ω ∈ N∗ r N. Put x = aω and y = bω . We conclude the proof in parts, from Lemma 15.1.2, by showing that t = xy is polefinite.

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Part A: S(t) = {p ∈ S ∗ | vp (t) < 0} is a finite set. Indeed, as q ∈ S r T (b), we have vq (a) > 0. Hence, v˙ q (x) > 0. Next choose q0 ∈ T (b). Then vq0 (b) > 0 ˙ Moreover, if and therefore v˙ q0 (y) > 0. Thus, x and y are nonunits of R. p ∈ T (b), then v˙ p (y) > 0 and if p ∈ S r T (b), then v˙ p (x) > 0. By Lemma 15.4.4, ˙ αx + βy 6= 1 for all α, β ∈ R.

(6)

In particular, αt + β 6= 1 for all α, β ∈ K, so t ∈ / K. For each p ∈ S ∗ , (5) gives (7a) (7b)

vp (x) > 0 =⇒ vp (y) = 0 and vp (t) = vp (x); and vp (y) > 0 =⇒ vp (x) = 0 and vp (t) = −vp (y).

Therefore, S(t) = T (b) is a finite set. Part B: Application of Proposition 15.2.3. To complete the proof it suffices to show that if F/K(t) is a finite separable extension in K ∗ , then the functional primes of F are induced by arithmetical primes. Let P be the set of functional primes of F which are induced by arithT metical primes. Let D = P ∈P OP be the corresponding holomorphy ring in F . Since vq (t) = v˙ q (x) > 0, the set P is nonempty. Assume P excludes a functional prime of F . Then the holomorphy ring theorem (Proposition 3.3.2) says that D is a Dedekind domain. By (7), A = {z = {z B = {z = {z

∈ D| ∈ D| ∈ D| ∈ D|

v˙ p (z) ≥ v˙ p (x) for all p ∈ S ∗ } vP (z) ≥ max(vP (t), 0) for all P ∈ P}; and v˙ p (z) ≥ v˙ p (y) for all p ∈ S ∗ } vP (z) ≥ max(−vP (t), 0) for all P ∈ P}.

Every maximal ideal of D is the center of a prime P ∈ P (Proposition 3.3.2(d)). Thus, A is the ideal of zeros of t and B is the ideal of poles of t. That is, A = P1k1 · · · Prkr and B = Ql11 · · · Qlss , where P1 , . . . , Pr , Q1 , . . . , Qs are distinct maximal ideals of D and k1 , . . . , kr , l1 , . . . , ls are positive integers with vPi (t) = ki , i = 1, . . . , r and vQj (t) = −lj , j = 1, . . . , s. In particular, A and B are relatively prime ideals of D. Hence, A + B = D. Thus, there exist ˙ λ ∈ A and µ ∈ B such that λ + µ = 1. By definition, A ⊆ xR˙ and B ⊆ y R. ˙ a contradiction to (6). Thus, λ = αx and µ = βy with α, β ∈ R,

15.5 Examples Let R be an integral domain with quotient field K. Suppose S is a nonempty set of primes of K which satisfies Conditions (1b)–(1d) of Section 15.4. Suppose in addition vp is a discrete valuation, p ∈ S. Then R is a Krull domain, hence a generalized Krull domain.

15.5 Examples

287

Every Dedekind ring R is a Krull domain with S being the set of primes of K associated with the maximal ideals of R. Since each nonzero prime ideal of R is maximal, dim(R) = 1. Thus, R does not satisfy Condition (1e) of Section 15.4. Example 15.5.1: Polynomial rings over fields. Every unique factorization domain R is a Krull domain. Here S corresponds to the set of nonzero prime ideals of R. For example, the polynomial ring R = K0 [X1 , . . . , Xn ] over an arbitrary field K0 is a unique factorization domain [Zariski-Samuel1, p. 38, Thm. 13]. When n ≥ 2, R/(RX1 + RX2 ) ∼ = K0 [X3 , . . . , Xn ] is an integral domain. Hence, RX1 +RX2 is a prime ideal of R which properly contains each of the prime ideals RX1 and RX2 . Thus, dim(R) ≥ 2. (Indeed, dim(R) = n [Matsumura, p. 117, Thm. 15.4].) By Theorem 15.4.6, K0 (X1 , . . . , Xn ) is Hilbertian. This is a weaker result than Theorem 13.4.2 which says that K0 (X1 , . . . , Xn ) is Hilbertian for each n ≥ 1. Example 15.5.2 [Weissauer p. 203]: Formal power series over a field. Let K0 be a field and n ≥ 2. The ring of formal power series R = K0 [[X1 , . . . , Xn ]] is a local integral domain with the maximal ideal

M=

∞ X

fi | fi ∈ K0 [X1 , . . . , Xn ] is a form of degree i .

i=1

Denote its quotient field by K0 ((X1 , . . . , Xn )): the field of formal power series over K0 in X1 , . . . , Xn . The Weierstrass preparation theorem implies that R is a unique factorization domain [Zariski-Samuel2, p. 148]. For each prime element p of R one of the elements X1 or X2 does not belong to Rp. Thus, Rp is properly contained in M , so dim(R) ≥ 2. (Again, dim(R) = n [Zariski-Samuel2, p. 218 or Matsumura, p. 117, Thm. 15.4].) By Theorem 15.4.6, K is Hilbertian. This settles a problem of [Lang3, p. 142, end of third paragraph]. Example 15.5.3: Formal power series over a ring. Let A be a Krull domain and n ≥ 1. Then the ring of polynomials A[X1 , . . . , Xn ] and the ring of formal power series A[[X1 , . . . , Xn ]] in the variables X1 , . . . , Xn over A are Krull domains (see [Matsumura, p. 89] for the case n = 1; the general case follows by induction). Suppose in addition that A is not a field. Let p0 be a nonzero prime ideal of A. Then P =

∞ X

fi ∈ A[X1 , . . . , Xn ] is a form of degree i

i=1

P0 =

∞ X i=0

fi ∈ A[X1 , . . . , Xn ] is a form of degree i and f0 ∈ p0

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are nonzero prime ideals of R = A[[X1 , . . . , Xn ]] and P0 ⊂ P . Indeed, A/P ∼ = A and A/P ∼ = A/p0 are integral domains. Thus, dim(A[[X1 , . . . , Xn ]]) ≥ 2 (Actually, dim(A) = dim(A) + n [Matsumura, p. 117]). Consequently, by Theorem 15.4.6, the quotient field of A[[X1 , . . . , Xn ]] is a Hilbertian field. For example, the quotient fields of Z[[X1 , . . . , Xn ]] and O[[X1 , . . . , Xn ]], where O is a discrete valuation rings, are Hilbertian. Lemma 15.5.4: No Henselian field is Hilbertian. Proof: Let K be a Henselian field with valuation ring R and maximal ideal m. Choose m ∈ m, m 6= 0 and a prime number p 6= char(K). Consider the irreducible polynomials f (T, X) = X p + mT − 1 and g(T, X) = X p + T −1 − 1 of K(T )[X]. Assume K is Hilbertian. Then there exists a ∈ K × with both f (a, X) and g(a, X) irreducible in K[X]. In particular, none of them has a zero in K. But either a ∈ R or a−1 ∈ m. Suppose first a ∈ R. Then f (a, 1) ≡ ∂f (a, 1) 6≡ 0 mod m. Since K is Henselian, f (a, X) has a zero 0 mod m and ∂X in K. Similarly, if a−1 ∈ m, then g(a, X) has a zero in K. This contradiction to the preceding paragraph proves that K is not Hilbertian. Example 15.5.5: Qp and the formal power series K0 ((X)) are complete discrete valuation fields. Hence, they are Henselian (Proposition 3.5.2). By Lemma 15.5.4, they are not Hilbertian. Thus, the assumption “n ≥ 2” in Example 15.5.2 is necessary. Example 15.5.6 (Geyer): In the notation of Example 15.5.2, the ring R = K0 [[X1 , . . . , Xn ]] of formal power series is not Hilbertian. Indeed, let K = Quot(R) and check forPchar(K0 ) 6= 2 (resp. char(K0 ) 6= 3) that every power series f (X1 , X2 ) = 1+ i+j>0 aij X1i X2j is a square (resp. a cube) in R. Thus, the polynomial Z 2 −(1+X1 T ) (resp. Z 3 −(1+X1 T )) is irreducible in K[T, Z] but Z 2 − (1 + X1 t) (resp. Z 3 − (1 + X1 t)) is reducible for each t ∈ R. Lemma 15.5.7: Let R0 be a unique factorization domain with quotient field K0 . Consider a set P of unequivalent prime elements of R0 . For each p ∈ P let vp be the corresponding discrete valuation of K0 . Let K be an algebraic extension of K0 . Suppose each vp with p ∈ P is unramified in K. For each p ∈ P choose an extension wp of vp to K. Then the holomorphy ring R = {x ∈ K | wp (x) ≥ 0 for each p ∈ P } is a unique factorization domain with quotient field K. Proof: The assumptions imply wp (p) = 1 and wp0 (p) = 0 for all distinct p, p0 ∈ P . Consider x ∈ R. Then wp (x) ≥ 0 for each p ∈ P . Moreover, 0. Indeed, if f = irr(x, K0 ) there are only finitely many p ∈ P with wp (x) >Q and vp (f (0)) = 0, then wp (x) = 0. Thus, u = x p∈P p−wp (x) is an element ofQK and wp (u) = 0 for each p ∈ P . Hence, u is a unit of R and x = u p∈P pwp (x) is the desired decomposition of x. Finally, observe that R contains the integral closure of R0 in K. Therefore, K = Quot(R).

Exercises

289

The following example generalizes [Corvaja-Zannier, Theorem 1 (i) and (ii)]: Example 15.5.8: Unique factorization domain with a non-Hilbertian quotient field. Let R0 be either Z or F [t] for some finite field F . It is a unique factorization domain. Put K0 = Quot(R0 ). The proof of Theorem 13.6.2 (with K0 replacing K) gives prime numbers l and q, an element u ∈ K × , an infinite set P of nonequivalent prime elements of R0 , and a field extension K (in the notation of Theorem 13.6.2 equals to M (K0 )) with these properties: (1a) For every x ∈ K there is a y ∈ K with y l − xlq + u = 0. (1b) For each p ∈ P , the discrete valuation vp is unramified in K. Since Y l − X lq + u is absolutely irreducible, Condition (1a) implies K is not Hilbertian. Let wp , p ∈ P , and R be as in Lemma 15.5.7. Then K = Quot(R), {wp | p ∈ P } is an infinite set of discrete valuations, and R is a unique factorization domain. This answers negatively the questions posed in Problems 14.20 and 14.21 of [Fried-Jarden3]. The following problem asks for a generalization of Example 15.5.3: Problem 15.5.9: Let A be a generalized Krull domain which is not a field. (a) Is A[[X]] a generalized Krull domain? (b) Is Quot(A[[X]]) Hilbertian? By Theorem 15.4.6, an affirmative answer to Problem 15.5.9(a) will give an affirmative answer to Problem 15.5.9(b).

Exercises 1. Let K be a countable Hilbertian field. List the irreducible polynomials in K[T, X] as f1 (T, X), f2 (T, X), . . .. For each i choose ti ∈ K for which f1 (ti , Y ), . . . , fi (ti , Y ) are irreducible in K[Y ]. (a) Observe that the infinite set H = {t1 , t2 , . . .} has the universal Hilbert subset property: H r HK (g1 , . . . , gm ) is finite for every collection {g1 , . . . , gm } of irreducible polynomials in K[T, X]. (b) Let K ∗ be an enlargement of K. Prove that if t ∈ H ∗ r H, then K(t) is algebraically closed in K ∗ . 2. Let A = {1, 22! , 33! , . . .} and considerSan enlargement Q∗ of Q. Prove ∞ for each nonstandard element t ∈ A∗ that n=1 Q(t1/n ) ⊆ Q∗ . 3.

Let f1 , f2 , f3 , . . . be a sequence in Q(Y ). Put gn (Y ) = f1 (. . . (fn−1 (fn (Y ))).

Suppose that if y is transcendental over Q and x = gn (y), then x has at least n distinct poles in Q(y). Let A = {g1 (1), g2 (2), g3 (3), . . .}. Consider a nonstandard element t of A∗ . Prove that the equation t = gn (Y ) is solvable in Q∗ for each n. Conclude that t is not polefinite (Section 15.1).

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Chapter 15. Nonstandard Approach to Hilbert’s Irreducibility Theorem

4. [Weissauer, Satz 2.3] Generalize the Gilmore-Robinson criterion as follows. Let K ∗ be an enlargement of a field K. Then K is Hilbertian if and only if there is a Hilbertian field F with K ⊆ F ⊆ K ∗ and F is separably closed in K ∗ . 5. Consider the set A = {2, 3!, 4!, 5!, . . .} and let t be a nonstandard element of A∗ . Show that the metric absolute values of Q induces via Q∗ (Section 15.2) the infinite prime, t → ∞ of Q(t), while every ultrametric absolute value of Q induces the prime, t → 0, of Q(t). Conclude from Proposition 15.3.2, that all other primes of Q(t) are induced by nonstandard arithmetical primes of Q∗ . 6. Consider the field K = K0 ((X1 , . . . , Xn )) of formal power series in n variables over a basic field K0 . Construct a K0 -place ϕ: K → K0 ∪ {∞} with ϕ(Xi ) = 0, i = 1, . . . , n. Conclude that K/K0 is a regular extension. Hint: The case n = 1 is easy. For arbitrary n embed K in the field of iterated power series L = K0 ((X1 ))((X2 )) · · · ((Xn )).

Notes The nonstandard characterization of Hilbertian fields appears in [GilmoreRobinson]. [Roquette2] has exploited the Gilmore-Robinson criterion through a nonstandard interpretation of the Siegel-Mahler theorem (compare with the remarks at the beginning of Section 13.3). Most of this Chapter is from Weissauer’s thesis [Weissauer]. The proof of 15.3.2 for number field that appear in [Robinson] based on nonstandard interpretation of the “distributions” that appear in Weil’s thesis [Weil1]. The influence of [Robinson-Roquette] in Section 15.2 and parts of Section 15.3 should be obvious. The polefinite property of Section 5.1 and its relation to Hilbert’s irreducibility theorem appears in a standard form based on p-adic analysis in the case K = Q in [Sprind¸zuk1]. Standard simplified proofs of both Sprind¸zuk’s result and Weissauer’s approach to Hilbertianity of fields with a product formula (Section 15.3), featuring their common concepts, appear in [Fried10] which also gives the concept of a universal Hilbert subset (e.g. in Exercise 1) whose existence is a consequence of the Gilmore-Robinson observation (Proposition 15.1). As far as we know [Sprind¸zuk2] is the first to give an explicit universal Hilbert subset (over Q): p 2 H = {[exp( log(log(m))] + m!2m }∞ m=1 . It has the property that H r HQ (g1 , . . . , gt ) is finite for every collection {g1 , . . . , gt } of irreducible polynomials in Q[T, X] [Fried10, Thm. 4.9]. Finally, [Klein] gives “standard” proofs of Weissauer’s results, Theorems 15.3.3 and 15.4.6.

Chapter 16. Galois Groups over Hilbertian Fields Given a field K, one may ask which finite groups occur as Galois groups over K. If K is Hilbertian, then every finite group that occurs over K(t), with t transcendental over K, also occurs over K. Moreover, suppose F/K(t) is Galois with Galois group G and F/K is regular. Then K has a linearly disjoint sequence of Galois extensions, L1 , L2 , L3 , . . ., with Gal(Li /K) ∼ = G, i = 1, 2, 3, . . . (Lemma 16.2.6). We prove that this set up occurs for symmetric groups (Corollary 16.2.7), Abelian groups (Proposition 16.3.5), and when char(K) = 0 for alternating groups (Proposition 16.7.6). If K is PAC (but not necessarily Hilbertian), every finite group is regular over K (Proposition 16.12.2). If K is Hilbertian, then Zp occurs over K (Corollary 16.6.7) but is not necessarily regular over K. For example, Zp is not regular over Q (Corollary 16.6.11). Realization of a finite nonsimple group over K is usually done in steps. First one realizes a quotient of the group and then embeds the solution field in a larger Galois extension with the given Galois group. The latter step is always possible when K is Hilbertian and the kernel is a product of nonAbelian simple groups each of which has a GAR realization over K (Sections 16.8 and 16.9). For example, An with n = 5 or n ≥ 7 is GAR over K when char(K) - (n − 1)n (Corollary 16.9.2). A Zp -extension N of K is an example of a Galois extension with finitely generated Galois groups. If K is Hilbertian, then so is N (Proposition 16.11.1). This is one ingredient of the proof that each Abelian extension of K is Hilbertian (Theorem 16.11.3). Finally, the regularity of Z/pZ over a Hilbertian field K has some implications to the structure of Gal(K). For example, Gal(K) has no closed prosolvable normal subgroup. In particular, the center of Gal(K) is trivial. Chapter 18 will exploit the result about the regularity of Sn .

16.1 Galois Groups of Polynomials We prove two theorems about preservation of Galois groups of polynomials under specializations of parameters. One of them (Proposition 16.1.5) is a polynomial analog of Lemma 13.1.1(b). It assumes the ground field to be Hilbertian. The other one (Proposition 16.1.4) is an application of BertiniNoether. Here the ground field is arbitrary but the polynomial in question is absolutely irreducible and Galois. We start with an analog of Lemma 13.1.1(a) for Galois groups of polynomials: Let f ∈ K[X] be a separable polynomial of degree n. By definition, f

292

Chapter 16. Galois Groups over Hilbertian Fields

has n distinct roots x1 , . . . , xn in Ks . Thus, L = K(x1 , . . . , xn ) is a finite Galois extension. Restriction of elements of Gal(L/K) to {x1 , . . . , xn } gives an embedding of Gal(L/K) into the group of all permutations of {x1 , . . . , xn }, called a permutation representation. The image of Gal(L/K) under this representation is Gal(f, K). ¯ and a separable polynomial f¯ ∈ K[X] ¯ Next consider another field K of ∼ ¯ ¯ degree n. We write Gal(f, K) = Gal(f , K) to indicate that the two groups in question are isomorphic as permutation groups. Thus, there exists an ¯ as abstract groups and there isomorphism σ 7→ σ ¯ of Gal(f, K) and Gal(f¯, K) ¯ ¯n of the roots of f with σxi = σ ¯x ¯i , σ ∈ Gal(f, K), is a listing x ¯1 , . . . , x ¯ into i = 1, . . . , n. Similarly we speak about an embedding of Gal(f¯, K) ∼ Gal(f, K) as permutation groups. Finally, we write Gal(f, K) = G for a finite group G when Gal(f, K) and G are isomorphic as abstract groups. Lemma 16.1.1: Let L/K be a finite Galois extension, f ∈ K[X] a separable polynomial, and ϕ a place of L. Denote the residue field of K (resp. L) under ¯ (resp. L). ¯ Suppose L is the splitting field of f over K and f¯ = ϕ(f ) ϕ by K ¯ is a separable polynomial in K[X] with deg(f¯) = deg(f ). ∗ ¯ → Gal(f, K). (a) Then there is an embedding ϕ : Gal(f¯, K) (b) Suppose in addition, f is irreducible and Galois, and f¯ is irreducible. ¯ is the splitting field of f¯ Then f¯ is Galois, ϕ∗ is an isomorphism, and L ¯ over K. ¯ K ¯ is separable. Then L ¯ is the splitting field of (c) Alternatively, suppose L/ ¯ f¯ over K. Qn Proof of (a): Let a be the leading coefficient of f . Then f (X) = a i=1 (X − xi ) with distinct x1 , . . . , xn ∈ Ks and L = K(x1 , . . . , xn ) is the splitting field of f over K. Extend ϕ to a place of L(X) with the same notation ¯ ×. such that ϕ(X) = X. Since deg(f¯) = deg(f ), we haveQ a ¯ = ϕ(a) ∈ K n × ¯ and f¯(X) = a ¯ i=1 (X − x ¯i ). By Hence, x ¯i = ϕ(xi ), i = 1, . . . , n, are in K s ¯n are distinct. Therefore, the map xi 7→ x ¯i , i = 1, . . . , n, assumption, x ¯1 , . . . , x is bijective. Lemma 6.1.1 gives an epimorphism σ 7→ σ ¯ of the decomposition group ¯ K). ¯ It satisfies σ ¯ ¯ y¯ = σy for each y ∈ K with y¯ = ϕ(y) ∈ L. Dϕ onto Aut(L/ ¯x ¯i = x ¯i . Since σ permutes x1 , . . . , xn , we have Suppose σ ¯ = 1. Then σxi = σ σxi = xi , i = 1, . . . , n. Hence, σ = 1. ¯ K(¯ ¯ x1 , . . . , x ¯n )). Then there is a σ ∈ Dϕ with Moreover, let τ ∈ Aut(L/ ¯ K) ¯ → σ ¯ = τ . By the preceding paragraph, σ ¯ = 1. It follows that res: Aut(L/ ¯ ¯ ¯n )/K) is an isomorphism. Gal(K(¯ x1 , . . . , x Finally, consider each σ ∈ Gal(L/K) as a permutation of {x1 , . . . , xn }. ¯n }. Then σ ¯ 7→ σ is an Likewise, consider σ ¯ as a permutation of {¯ x1 , . . . , x ¯ → Gal(f, K). embedding ϕ∗ : Gal(f¯, K) Proof of (b): Since f is irreducible and Galois, deg(f ) = |Gal(f, K)|. Since

16.1 Galois Groups of Polynomials

293

¯ f¯ is irreducible, deg(f¯) ≤ |Gal(f¯, K)|. By (a), ¯ ¯ = [K(¯ ¯ x1 , . . . , x ¯n ) : K] deg(f ) = deg(f¯) ≤ |Gal(f¯, K)| ¯ : K] ¯ ≤ [L : K] = |Gal(f, K)| = deg(f ). ≤ [L ¯ = |Gal(f, K)| and deg(f¯) = [K(¯ ¯ x1 , . . . , x ¯ Thus, f¯ Hence, |Gal(f¯, K)| ¯n ) : K]. ¯ = K(¯ ¯ x1 , . . . , x ¯n ) is Galois is Galois and ϕ∗ is an isomorphism. In addition, L ¯ over K. ¯ K ¯ is normal and separable, it is Galois. By the proof Proof of (c): Since L/ ¯ = K(¯ ¯ x1 , . . . , x ¯ ¯ ¯n )) = 1. Therefore, L ¯n ). of (a), Gal(L/K(¯ x1 , . . . , x An essential assumption in Lemma 16.1.1 is the irreducibility of f¯. Lemma 16.1.4 below uses Bertini-Noether to satisfy this assumption. Proposition 16.1.5 applies Hilbert irreducibility theorem to achieve the same goal. Lemma 16.1.2: Let V be a variety in An over a field K and x a generic point of V over K. Then V has a nonempty Zariski K-open subset V0 with ˜ there is a K-place of K(x) with the following property: For each a ∈ V0 (K) ϕ(x) = a and with residue field K(a). Proof: By Corollary 10.2.2(a), K(x)/K is a regular extension. Assume without loss that x1 , . . . , xr form a separating transcendence base for K(x)/K. Example 2.6.10 produces a K-place ϕ0 of K(x1 , . . . , xr ) with ϕ(xi ) = ai , i = 1, . . . , r, and with residue field K(a1 , . . . , ar ). Use Remark 6.1.5 to find a nonzero polynomial g ∈ K[X1 , . . . , Xr ] such that K[x1 , . . . , xi+1 , g(x1 , . . . , xr )−1 ]/K[x1 , . . . , xi , g(x1 , . . . , xr )−1 ] is a ring cover, i = r, . . . , n − 1. Let V0 = {a ∈ V | g(a1 , . . . , ar ) 6= 0}. ˜ Let ϕ a place of K(x) which extends ϕ0 with Suppose a ∈ V0 (K). ϕ(x) = a. By Remark 6.1.7, the residue field of K(x) at ϕ is K(a). Remark 16.1.3: Simple points. The conclusion of Lemma 16.1.2 is actually true for each simple point a of V [Jarden-Roquette, Cor. A2]. Proposition 16.1.4: Let V be a variety over a field K0 in Am , u a generic point of V over K0 , K = K0 (u), and h1 , . . . , hk ∈ K0 [U1 , . . . , Um , T1 , . . . , Tr , X] polynomials. Suppose hj (u, T, X) are absolutely irreducible as polynomials in (T, X) over K and Galois as polynomial in X over K(T). Then there exists a nonempty Zariski K0 -open subset V0 of V with the following property: For each a ∈ V0 (K0 ), hj (a, T, X) is absolutely irreducible, Galois over K0 (T), and Gal(hj (a, T, X), K0 (T)) ∼ = Gal(hj (u, T, X), K(T)), j = 1, . . . , k. Proof: Bertini-Noether (Proposition 9.4.3) gives a nonempty Zariski K0 open subset V1 of V such that hj (a, T, X) are absolutely irreducible for each

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˜ 0 ), j = 1, . . . , k. Lemma 16.1.2, applied to V over K0 (T), gives a a ∈ V1 (K nonempty Zariski K0 (T)-open subset V20 of V satisfying this: For each a ∈ ^ V20 (K 0 (T)) there is a K0 (T)-place ϕ of K0 (u, T) with residue field K0 (a, T ). ˜ 0) ⊆ Choose a nonempty Zariski K0 -open subset V2 of V such that V2 (K ˜ 0 ). Finally, there is a nonempty Zariski K0 -open subset V3 of V with V20 (K hj (a, T, X) separable and degX (hj (a, T, X)) = degX (hj (u, T, X)) for each ˜ 0 ), j = 1, . . . , k. a ∈ V3 (K V0 = V1 ∩ V2 ∩ V3 is a nonempty Zariski K0 -open subset of V . For each a ∈ V0 (K) let ϕ be a K0 -place of K(T) with residue field K0 (T). By Lemma 16.1.1, hj (a, T, X) is Galois over K0 (T) and (1) holds. Proposition 16.1.5: Let K be a Hilbertian field and hj ∈ K[T1 , . . . , Tr , X] a separable polynomial in X, j = 1, . . . , k. Then K r has a separable Hilbert subset H with Gal(hj (a, X), K) ∼ = Gal(hj (T, K), K(T)) for each a ∈ H, j = 1, . . . , k. Proof: Denote the splitting field of hj (T, X) over K(T) by Fj . Proposition 13.1.1(a) gives a nonempty Zariski K-open subset U1 of Ar satisfying the following assertion: For each a ∈ U1 (K) there is a K-place ϕj of Fj extending T 7→ a such that the residue field F¯j of Fj under ϕj is Galois over K, j = 1, . . . , k. Proposition 13.1.1(b) gives a separable Hilbert subset H 0 of K r such that for each a ∈ H 0 ∩ U1 (K) and every K-place ϕ of Fj with residue field F¯j we have Gal(F¯j /K) ∼ = Gal(Fj /K(T)). Finally, there is a nonempty Zariski open subset U2 of Ar satisfying this: hj (a, X) is separable and degX (hj (a, X)) = degX (hj (T, X)) for each a ∈ U2 (K), j = 1, . . . , k. Put H = H 0 ∩ U1 (K) ∩ U2 (K). Consider a ∈ H and let ϕj be as above. Then Lemma 16.1.1 gives an isomorphism ϕ∗j : Gal(hj (a, X), K) → Gal(hj (T, X), K(T)), j = 1, . . . , k, as claimed.

16.2 Stable Polynomials Let K be a field and G a profinite group. Suppose K has a Galois extension L with Gal(L/K) ∼ = G. Then, G occurs (or is realizable) over K and L is a G-extension of K. If G belongs to a family G of profinite groups, we also say that L is a G-extension of K. For example, when G is Abelian, we say L is an Abelian extension of K. If G is an inverse limit of finite solvable groups, we say L is a prosolvable extension of K. The main problem of Galois theory is whether every finite group occurs over Q. Even if this holds, it is not clear whether the same holds for every Hilbertian field.

16.2 Stable Polynomials

295

To approach the latter problem, consider algebraically independent elements t1 , . . . , tr over K. If K is Hilbertian, G is finite, and G occurs over K(t), then G occurs over K (Lemma 13.1.1(b)). If we want G to occur more than once over K, we have to assume more, as we now explain. We say G is regular over K if there exist algebraically independent elements t1 , . . . , tr over K such that K(t) has a Galois extension F which is regular over K with Gal(F/K(t)) ∼ = G. This stronger property is inherited by all extensions of K: Lemma 16.2.1: Let K be a field and G a profinite group. Suppose G is regular over K. Then G is regular over every extension L of K. Proof: Let t = (t1 , . . . , tr ) and F be as above. Consider a field extension L of K. Assume without loss that t1 , . . . , tr are algebraically independent over L. Then F is linearly disjoint from L over K (Lemma 2.6.7). Hence, F L is a regular extension of L (Corollary 2.6.8), F is linearly disjoint from L(t) over K(t) (Lemma 2.5.3), and Gal(F L/L(t)) ∼ = G. Consequently, G is regular over L. The regular inverse Galois problem asks whether every finite group is regular over every field. By Lemma 16.2.1, it suffices to check the problem over Q and over each of the fields Fp . In this generality the problem is still wide open. Nevertheless, there are many special cases where groups G are proved to be regular over specific fields. This chapter discusses cases when this happens. Remark 16.2.2: Reinterpretation of ‘regularity’ by polynomials. Consider an irreducible polynomial f ∈ K[T1 , . . . , Tr , X], separable with respect to X. Call f X-stable over K if (1)

Gal(f (T, X), K(T)) ∼ = Gal(f (T, X), L(T))

for every extension L of K. In this case denote the splitting field of f (T, X) over K(T) by F . Then Gal(F/K(T)) ∼ = Gal(F L/L(T)). Hence, F is linearly disjoint from L over K (Lemma 2.5.3). Thus, F is a regular extension of K and G is regular over K. In addition, f is absolutely irreducible (Corollary 10.2.2). Conversely, suppose K(T1 , . . . , Tr ) has a Galois extension F which is regular over K and with Galois group G. Choose a primitive element x for F/K(T) and let f ∈ K[T, X] be an irreducible polynomial with f (T, x) = 0. Then f (T, X) is X-stable over K, Galois with respect to X over K(T), and Gal(f (T, X), K(T)) ∼ = G. Lemma 16.2.3: Let K be a field and f a polynomial in K[T1 , . . . , Tr , X], separable in X. Suppose Gal(f (T, X), K(T)) ∼ = Gal(f (T, X), Ks (T)). Then f is X-stable over K. Proof: Condition (1) holds when L is a purely inseparable or a regular extension of K. Hence, we have to consider only the case where L is a

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Chapter 16. Galois Groups over Hilbertian Fields

separable algebraic extension of K. In this case both maps res: Gal(f (T, X), Ks (T)) → Gal(f (T, X), L(T)) res: Gal(f (T, X), L(T)) → Gal(f (T, X), K(T)) are injective. By assumption, their compositum is bijective. Hence, each of them is bijective. Lemma 16.2.4: Let K be a field and N a Galois extension of K. (a) Suppose f is a polynomial in N [T1 , . . . , Tr , X] separable in X. Then K has a finite Galois extension L in N with f ∈ L[T, X] and Gal(f (T, X), L(T)) ∼ = Gal(f (T, X), N (T)). (b) If f is Galois over N (T), then f is Galois over L(T). (c) If f is X-stable over N , then f is X-stable over L. Proof of (a): Choose a finite extension L0 of K in N which contains the coefficients of f . Denote the splitting field of f (T, X) over L0 (T) by F . Then L1 = F ∩ N is a finitely generated extension of K (Lemma 10.5.1). In addition, L1 /K is a separable algebraic extension. Hence, L1 /K is finite. Also, Gal(F/L1 (T)) ∼ = Gal(F N/N (T)). Finally, let L be the Galois closure of L1 /K. Then L satisfies the requirements of (a). Proof of (b): Suppose f and L satisfy the conclusion of (a) and f is Galois over N (T). Let x1 , . . . , xn be the roots of f (T, X) in K(T)s . Consider σ ∈ Gal(f (T, X), L(T)). Suppose xσ1 = x1 . By (a), σ extends to an element σ in Gal(f (T, X), N (T)). Hence, xσi = xi , i = 1, . . . , n. Therefore, f is Galois over L(T). Proof of (c): By assumption, Gal(f (T, X), N (T)) ∼ = Gal(f (T, X), Ks (T)). Hence, by (a), Gal(f (T, X)), L(T)) ∼ = Gal(f (T, X), Ks (T)). By Lemma 16.2.3, f is X-stable over L. Example 16.2.5: Stable polynomials. (a) The general polynomial of degree n is f (T1 , . . . , Tn , X) = X n + T1 X n−1 + · · · + Tn . It satisfies Gal(f (T, X), K(T)) ∼ = Sn for every positive integer n and for every field K [Lang7, p. 272, Example 4]. Thus, f is X-stable over every field. (b) The polynomial X n −T satisfies Gal(X n −T, K(T )) ∼ = Z/nZ for every field K with char(K) - n that contains ζn . Thus, X n − T is X-stable over K. If however ζn 6∈ K, then Gal(X n − T, K(T )) ∼ = Gal(K(ζn )/K) n Z/nZ. Therefore, X n − T is not X-stable over K. (c) The polynomial X p − X − T satisfies Gal(X p − X − T, K(T )) ∼ = Z/pZ for every field K of characteristic p [Lang7, p. 290, Thm. 6.4]. Therefore, f is X-stable over K.

16.2 Stable Polynomials

297

(d) Suppose Gal(f (T, X), K(T)) is simple. Let F be the splitting field of f (T, X) over K(T). Suppose F 6⊆ Ks (T). Then F ∩ Ks (T) = K(T). Therefore, f is X-stable over K. (e) Suppose f ∈ K[T, X] is absolutely irreducible and Galois over K(T). Then f is X-stable over K. Indeed, let L be an extension of K. Then Gal(f (T, X), L(T)) is a subgroup of Gal(f (T, X), K(T)). On the other hand, f (T, X) is also Galois over L(T). Hence, |Gal(f (T, X), L(T))| = degX (f (T, X)) = |Gal(f (T, X), K(T))|. Thus, Gal(f (T, X), L(T)) ∼ = Gal(f (T, X), K(T)), as claimed.

We use stable polynomial to construct linearly disjoint sequences of Galois extensions over Hilbertian fields with a given Galois groups: Lemma 16.2.6: Let fi (T1 , . . . , Tr , X) be an X-stable polynomial over a Hilbertian field K and Gi = Gal(fi (T, X), K(T)), i = 1, 2, 3, . . . . Given a finite separable extension L0 of K, there is a sequence, L1 , L2 , L3 , . . . of Galois extensions of K with the following properties: (a) Gal(Li /K) ∼ = Gi , and fi has an Li -rational zero ci , i = 1, 2, 3, . . . . (b) c1 , c2 , c3 , . . . are distinct. (c) The sequence L0 , L1 , L2 , . . . is linearly disjoint over K. Proof: Suppose by induction, there are Galois extensions L1 , . . . , Ln of K such that Gal(Li /K) ∼ = Gi , i = 1, . . . , n, and L0 , L1 , . . . , Ln are linearly disjoint over K. In addition suppose there are ai ∈ K r and bi ∈ Li such that fi (ai , bi ) = 0 and a1 , . . . , an are distinct. Then L = L0 L1 · · · Ln is a finite separable extension of K. By assumption, Gal(fn+1 (T, X), L(T)) ∼ = Gal(fn+1 (T, X), K(T)) = Gn+1 . Apply Lemma 16.1.5 and Corollary 12.2.3 to find an+1 ∈ K r with Gal(fn+1 (an+1 , X), L) ∼ = Gal(fn+1 (an+1 , X), K) ∼ = Gn+1 and an+1 6= a1 , . . . , an . Let Ln+1 be the splitting field of f (a, X) over K. Then Gal(Ln+1 /K) ∼ = Gn+1 and Ln+1 is linearly disjoint from L over K. In particular, L0 , . . . , Ln+1 are linearly disjoint over K. Finally, choose a zero bn+1 of fn+1 (an+1 , X). Then bn+1 ∈ Ln+1 . Our next result is an immediate application of this lemma to the polynomials of Example 16.2.5(a),(b),(c): Corollary 16.2.7: Let K be a Hilbertian field and G a finite group. Then K has a linearly disjoint sequence of Galois extensions with G as Galois group in each of the following cases: (a) G = Sn , n ∈ N. (b) ζn ∈ K and G = Z/nZ, n ∈ N, char(K) - n.

298

Chapter 16. Galois Groups over Hilbertian Fields

(c) char(K) = p > 0 and G = Z/pZ. We prove in Section 16.4 that every finite Abelian group A satisfies the conclusion of Corollary 16.2.7. But this simple corollary already has interesting implications to closed normal subgroups of Gal(K) (Section 16.12). The following result shows that in many cases it suffices to take r = 1 in the definition of “G is regular over K”: Proposition 16.2.8: Let K be an infinite field and G a finite group. Suppose G is K-regular. Then: (a) There is an X-stable polynomial h ∈ K[T, X] which is Galois with respect to X and with Gal(h(T, X), K(T )) ∼ = G. (b) If in addition K is Hilbertian, then K has a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions such that Gal(Li /K) ∼ = G for all i. Proof: Statement (b) follows from (a) by Lemma 16.2.6. We prove (a): Remark 16.2.2 gives a polynomial f ∈ K[T1 , . . . , Tr , X] which is Xstable over K with r ≥ 2 and Galois with respect to X over K(T) such that Gal(f (T, X), K(T)) ∼ = G. Assume without loss f is monic with respect to X. Let u1 , u2 be algebraically independent elements over K. Put L = K(u1 , u2 ), T0 = (T1 , . . . , Tr−1 ), and g(T0 , X) = f (T1 , . . . , Tr−1 , u1 + u2 Tr−1 , X). By Proposition 10.5.4, g(T0 , X) is absolutely irreducible. Extend the map T → (T0 , u1 + u2 Tr−1 ) to an L(T0 )-place of L(T) with residue field L(T0 ) (Lemma 2.2.7). By Lemma 16.1.1, G∼ = Gal(f (T, X), L(T)) ∼ = Gal(g(T0 , X), L(T0 )). By assumption, K is infinite. Hence, Proposition 16.1.4 gives b1 , b2 ∈ K with g¯(T0 , X) = f (T1 , . . . , Tr−1 , b1 + b2 Tr−1 , X) absolutely irreducible, Galois over g (T0 , X), K(T0 )) ∼ K(T0 ), and Gal(¯ = Gal(g(T0 , X), L(T0 )) ∼ = G. Finally, use induction on r to find an absolutely irreducible polynomial h ∈ K[T1 , X] with Gal(h(T1 , X), K(T1 )) ∼ = G. Problem 16.2.9:

Does Proposition 16.2.8 hold if K is finite?

16.3 Regular Realization of Finite Abelian Groups The inverse problem of Galois theory has an affirmative solutions for every finite Abelian group A and every Hilbertian field K (Corollary 16.3.6). Moreover, every finite Abelian group is even regular over K (Proposition 16.3.5). The proof of the latter result moves from the case where A is cyclic to the general case. In the cyclic case we have to distinguish between the case where char(K) does not divide the order of the group and where the order of the group is a power of char(K). In the general case, we distinguish between the cases where K is finite and K is infinite.

16.3 Regular Realization of Finite Abelian Groups

299

Lemma 16.3.1: Let K be a field, n a positive integer with char(K) - n, and t an indeterminate. Then K(t) has a cyclic extension F of degree n which is contained in K((t)). Proof: Choose a root of unity ζn of order n in Ks . Let L = K(ζn ) and G = Gal(L/K). Then there is a map χ: G → {1, . . . , n − 1} such that χ(σ) σ(ζn ) = ζn . Then gcd(χ(σ), n) = 1 and (1)

χ(στ ) ≡ χ(σ)χ(τ ) mod n

for all σ, τ ∈ G. By Example 3.5.1, K((t)) is a regular extension of K and L((t)) = K((t))(ζn ). Thus, we may identify G with Gal(L((t))/K((t))). Choose a primitive element c of L/K. Consider the element g(t) =

Y

χ(σ−1 )

1 + σ(c)t

σ∈G

of L[t]. Since char(K) - n, Hensel’s lemma (Proposition 3.5.2) gives an x ∈ Q −1 L[[t]] with xn = 1 + ct. Then y = σ∈G σ(x)χ(σ ) ∈ L[[t]] and y n = Q Q −1 n χ(σ −1 ) = σ∈G (1 + σ(c)t)χ(σ ) = g(t). Since ζn ∈ L, F = σ∈G σ(x ) L(t, y) is a cyclic extension of degree d of L(t), where d|n and y d ∈ L(t) [Lang7, p. 289, Thm. 6.2(ii)]. Since χ(σ −1 ) is relatively prime to n, we must have d = n. The Galois group Gal(F/L(t)) is generated by an element ω satisfying ω(y) = ζn y. By (1) there exist for each τ, ρ ∈ G a positive integer k(τ, ρ) and a polynoQ Q −1 −1 mial fτ (t) ∈ L[t] such that τ (y) = σ∈G τ σ(x)χ(σ ) = ρ∈G ρ(x)χ(ρ τ ) = Q Q χ(ρ−1 )χ(τ )+k(τ,ρ)n = y χ(τ ) ρ∈G (1 + ρ(c)t)k(τ,ρ) = y χ(τ ) fτ (t). It ρ∈G ρ(x) follows that G leaves F invariant. Let E be the fixed field of G in F . K((t)) E K(t) K

L((t)) F = L(t, y) L(t) L= K(ζn )

Denote the subgroup of Aut(F/K(t)) generated by G and Gal(F/L(t)) by H. Then the fixed field of H is K(t), so F/K(t) is a Galois extension with Gal(F/K(t)) = G · Gal(F/L(t)). Moreover, given τ ∈ G, put m = χ(τ ). Then τ ω(y) = τ (ζn y) = ζnm y m fτ (t) = ω(y)m fτ (t) = ω(y m fτ (t)) = ωτ (y). Thus, τ ω = ωτ , so G commutes with Gal(F/L(t)). Therefore, E/K(t) is a Galois extension with Gal(E/K(t)) ∼ = Gal(F/L(t)) ∼ = Z/nZ.

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Lemma 16.3.2: Suppose p = char(K). Let L be a cyclic extension of degree pn , n ≥ 1, of K. Then K has a Z/pn+1 Z-extension L0 which contains L. Proof: Define L0 to be L(x) where x is a zero of X p − X − a with a ∈ L. The three parts of the proof produce a, and then show L0 has the desired properties. Part A: Construction of a. Since L/K is separable, there is a b1 ∈ L with c = traceL/K (b1 ) 6= 0 [Lang7, p. 286, Thm. 5.2]. Put b = bc1 . Then traceL/K (b) = 1 and traceL/K (bp − b) = (traceL/K (b))p − traceL/K (b) = 0. With σ a generator of Gal(L/K), the additive form of Hilbert’s Theorem 90 [Lang7, p. 290, Thm. 6.3] gives a ∈ L with σa − a = bp − b.

(2)

Part B: Irreducibility of X p − X − a. Assume X p − X − a is reducible over L. Then x ∈ L [Lang7, p. 290, Thm. 6.4(b)]. Thus (3) (σx − x)p − (σx − x) − (bp − b) = (σx − x)p − (σx − x) − (σa − a) = (σxp − σx − σa) − (xp − x − a) = 0 Since b is a root of X p − X − (bp − b), there is an i with σx − x = b + i [Lang7, p. 290, Thm. 6.4(b)]. Apply traceL/K to both sides to get 0 on the left and 1 on the right. This contradiction proves X p − X − a is irreducible. Part C: Extension of σ to σ 0 that maps x to x+b. Equality (2) implies x+b is a zero of X p − X − σa. Thus, by Part B, σ extends to an automorphism σ 0 of L0 with σ 0 (x) = x + b. We need only prove that σ 0 has order pn+1 . Induction shows (σ 0 )j (x) = x + b + σb + · · · + σ j−1 b. In particular, n

(σ 0 )p (x) = x + traceL/K (b) = x + 1.

(4) n

Hence, (σ 0 )ip (x) = x + i, i = 1, . . . , p. Therefore, the order of σ 0 is pn+1 , as contended. Remark 16.3.3: Lemma 16.3.2 is a special case of a theorem of Witt. Suppose char(K) = p. Then AS(K) = {xp − x | x ∈ K} is a subgroup of the additive group of K and K/AS(K) is a vector space over Fp of dimension, say, r. Consider an embedding problem G → Gal(L/K) over K with G a finite p-group which is generated by r elements. Witt’s theorem says this problem is solvable (If r = ∞, there is no restriction on the number of generators of G.) The technique of Galois cohomology [Ribes, p. 257, Cor. 3.4] simplifies Witt’s original proof [Witt]. Lemma 16.3.4: Let K be a field, t an indeterminate, and A a finite cyclic group. Then K(t) has a Galois extension F such that Gal(F/K(t)) ∼ = A and F/K is regular. Proof: We put p = char(K) and divide the proof into three parts:

16.3 Regular Realization of Finite Abelian Groups

301

Part A: A ∼ = Z/mZ and p - m. By Lemma 16.3.1, K(t) has a cyclic extension Em of degree m which is contained in K((t)). By Example 3.5.1, K((t)) is a regular extension of K. Hence, so is Em (Corollary 2.6.5(b)). Part B: A ∼ = Z/pk Z. Assume without loss that k ≥ 1. By Eisenstein’s criterion and Gauss’ lemma, the polynomial X p − X − t is irreducible over ˜ K(t). Let x be a root of X p −X −t in K(t)s . Then, by Artin-Schreier, [Lang7, p. 290, Thm. 6.4(b)], K(x) is a cyclic extension of degree p of K(t). Lemma 16.3.2 gives a cyclic extension Epk of K(t) of degree pk which contains K(x). ˜ By the preceding paragraph, K(x) ∩ K(t) = K(t). Since Gal(Epk /K(t)) k is a cyclic group of order p , each subextension of Epk which properly contains ˜ K(t) must contain K(x). Hence, Epk ∩ K(t) = K(t). Thus, Epk is linearly ˜ disjoint from K(t) over K(t). By the tower property (Lemma 2.5.3), Epk is ˜ over K; that is, Epk /K is regular. linearly disjoint from K Part C: A ∼ = Z/nZ, n = mpk , p - m. The compositum En = Em Epk is a ˜ cyclic extension of K(t) of degree n. Moreover, En ∩ K(t) decomposes into a cyclic extension of K(t) of degree which divides m and a cyclic extension of K(t) degree dividing pk . By Parts A and B, both subextensions must be K(t). It follows that En is a regular extension of K. We generalize Lemma 16.3.4 from cyclic groups to arbitrary Abelian groups: Proposition 16.3.5: Let K be a field, t an indeterminate, and A a finite Abelian group. Then K(t) has a Galois extension F such that Gal(F/K(t)) ∼ = A and F/K is regular. Proof: The first two parts of the proof prove the proposition in the case where A = (Z/qZ)n , q = pr , p is a prime number, and n, r are positive integers. Part A: A = (Z/qZ)n is as above and K is infinite. Choose algebraically independent elements t1 , . . . , tn over K. For each i between 1 and n, Lemma 16.3.4 gives a finite cyclic extension Ei of K(ti ) of degree q which is regular over K. Then E1 , . . . , En are algebraically independent over K. Hence, by Corollary 2.6.8, E = E1 · · · En is a regular extension of K. In addition, by Lemma 2.6.7, E1 , . . . , En are linearly disjoint over K. Hence, by Lemma 2.5.11, E1 (t), . . . , En (t) are linearly disjoint over K(t) and each Ei (t) is a cyclic extension of K(t) of degree q. It follows that Gal(E/K(t)) ∼ = (Z/qZ)n . Consequently, by Proposition 16.2.8, K(t) has a cyclic extension F of degree q which is regular over K. Part B: A = (Z/qZ)n is as above and K is finite. Choose a transcendental element u over K(t). Lemma 16.3.4 gives a cyclic extension E of K(t, u) of degree q which is regular over K(t). Choose a polynomial h ∈ K(t)[u, X] which is X-stable with Gal(h(u, X), K(t, u)) ∼ = Z/qZ (e.g. use Proposition

302

Chapter 16. Galois Groups over Hilbertian Fields

16.2.8). Denote the unique extension of K of degree q by Kq . By Theorem 13.4.2, K(t) is Hilbertian. Hence, by Lemma 16.2.6, K(t) has sequence F1 , F2 , F3 , . . . of cyclic extensions of degree q such that Kq (t), F1 , F2 , . . . , Fn are linearly disjoint over K(t). Then F = F1 F2 · · · Fn is a Galois extension of K(t) with Gal(F/K(t)) ∼ = (Z/qZ)n . Moreover, F ∩ Kq (t) = K(t). ˜ The group A = Gal(F ∩ K(t)/K) is cyclic quotient of Gal(F/K(t)), so ˜ A is of exponent q. Hence, [F ∩ K(t) : K(t)] = pj with j ≤ r. Therefore, ˜ ˜ F ∩ K(t) ⊆ Kq (t). It follows from the preceding paragraph that F ∩ K(t) = K(t). Consequently, F/K is regular. ∼ Qm (Z/qi Z)ni , where qi = pri , m, ni , ri are positive integers, Part C: A = i i=1 and p1 , . . . , pr are distinct prime numbers. Parts A and B give for each i a Galois extension Fi of K(t) which is regular over K and with Gal(Fi /K(t)) ∼ = (Z/qi Z)ni . Since q1 , . . . , qm are pairwise relatively prime, F1 , . . . , Fm are linearly disjoint over K(t). Hence, F = F1 · · · Fm is a Galois extension with Gal(F/K(t)) ∼ = A. ˜ Let E = F ∩ K(t). By the preceding paragraph, Fi ∩ E = K(t), so i = 1, . . . , m. Since q1 , . . . , qm are pairwise relatively prime, qini |[F : E] for Qm [F : K(t)] = i=1 q ni divides [F : E]. Hence, E = K(t). Consequently, F is a regular extension of K. Part D: A is an arbitrary finite Abelian group. Then A=

mi m Y Y

r

(Z/pi ij Z)nij

i=1 j=1

where p1 , . . . , pm are distinct prime numbers, m ≥ 0, and mi , nij are positive integers. For each i let ri = max(ri1 , . . . , ri,mi ) and ni = max(ni1 , . . . , ni,mi ). Part C gives a GaloisQextension Fˆ of K(t) which is regular over K and m with Gal(Fˆ /K(t)) ∼ = i=1 (Z/pri i Z)ni . By construction, A is a quotient of Gal(Fˆ /K(t). Hence, K(t) has a Galois extension F in Fˆ with Gal(F/K(t)) ∼ = A. Since Fˆ /K is regular, so is F/K. Corollary 16.3.6: Let K be a Hilbertian field and A a finite Abelian group. Then A is realizable over K. Moreover, K has a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions with Gal(Li /K) ∼ = A for each i. Proof: By Proposition 16.3.5, A is regular over K. Hence, by Proposition 16.2.8, K has a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions with Gal(Li /K) ∼ = A for each i.

16.4 Split Embedding Problems with Abelian Kernels Attempts to realize a finite Abelian group A over a Hilbertian field K usually lead to an extension of K with roots of unity. This gives a “split embedding problem with Abelian kernel”. The main result of this section is that each

16.4 Split Embedding Problems with Abelian Kernels

303

such problem is solvable. Consequently, A occurs over K. Indeed, for A to be regular over K it is not necessary that K is Hilbertian. We prove that it is true for arbitrary field. Definition 16.4.1: Embedding problems. Let L/K be a Galois extension, G a profinite group, and α: G → Gal(L/K) an epimorphism. The embedding problem associated with α consists of embedding L in a Galois extension N of K with an isomorphism β: Gal(N/K) → G satisfying α ◦ β = resN/L . Refer to α as an embedding problem over K, β its solution, and N the solution field. Call the problem finite if G is finite. The problem splits if α has a section, that is, an embedding α0 : Gal(L/K) → G with α ◦ α0 = idGal(L/K) . The latter case occurs when G = Gal(L/K) n Ker(α) and α is the projection of G onto Gal(L/K). Let t1 , . . . , tn be algebraically independent elements over K. Then res: Gal(L(t)/K(t)) → Gal(L/K) is an isomorphism. Hence, α: G → Gal(L/K) gives rise to an embedding problem αt : G → Gal(L(t)/K(t)) over K(t) with resL(t)/L ◦ αt = α. Refer to a solution of αt as a solution of α over K(t). Refer to a solution field F of αt as a regular solution of α if F/L is regular. We say α is regularly solvable if there are t1 , . . . , tn as above and αt has a solution field F which is regular over L. Lemma 16.4.2: Let K be a Hilbertian field, α: H → Gal(L/K) a finite embedding problem, t1 , . . . , tn algebraically independent elements over K, and M a finite separable extension of L. If α is solvable over K(t), then α is also solvable over K. If α is regularly solvable, then α has a solution field N over K which is linearly disjoint from M over L. Proof: Let F be a solution field of α over K(t). Thus, F is a Galois extension of K(t) which contains L and there is an isomorphism θ: Gal(F/K(t)) → H with α ◦ θ = resF/L . Lemma 13.1.1 gives a separable Hilbert subset A of K r having the following property: For each a ∈ A there is a K-place ϕ: F → ˜ ∪ {∞} satisfying these conditions: K (3a) ϕ(t) = a, K(t)ϕ = K, and L(t)ϕ = L. (3b) There is an isomorphism ϕ∗ : Gal(F¯ /K) → Gal(F/K(t)) with resF/L ◦ ϕ∗ = resF¯ /L . Here, F¯ = F¯ϕ . The map β = θ ◦ ϕ∗ solves embedding problem α. Suppose now F/L is regular. Then Gal(F M/M (t)) ∼ = Gal(F/L(t)). Lemma 13.1.1 gives a separable Hilbert subset A0 of M r satisfying this: For each a ∈ A ∩ A0 and each M -place ϕ of F M satisfying (3), Gal(F M/M (t)) ∼ = Gal(F¯ M/M ). By Corollary 12.2.3, there is an a ∈ A ∩ A0 . The preceding paragraph gives ϕ satisfying (3). The corresponding field F¯ is linearly disjoint from M over L. Let A and G be finite groups. Recall that AG is the group of all functions f : G → A and G acts on AG by the rule f τ (σ) = f (τ σ). The semidirect product G n AG is the wreath product A wr G (Remark 13.7.7).

304

Chapter 16. Galois Groups over Hilbertian Fields

Lemma 16.4.3: Let G and A be finite groups with A Abelian and G acting on A. Then there is an epimorphism A wr G → G n A with kernel in AG . Q −1 Proof: Define a map α: AG → A by α(f ) = σ∈G f (σ)σ . Since A is Abelian, the right hand side is a well defined homomorphism. For each a ∈ A define a function fa : G → A by fa (1) = a and fa (σ) = 1 for σ ∈ G, σ 6= 1. Then α(fa ) = a, so α is surjective. Next consider τ ∈ G. Then Y Y Y −1 −1 −1 f τ (σ)σ = f (τ σ)σ = f (ρ)ρ τ = α(f )τ . α(f τ ) = σ∈G

σ∈G

ρ∈G

Thus, α respects the action of G. Therefore, α extends to an epimorphism α: G n AG → G n A satisfying α(σf ) = σα(f ) for all σ ∈ G and f ∈ AG . This gives rise to the following commutative diagram: 1

/ AG

/ G n AG

α

1

/A

By definition, Ker(α) ≤ AG .

/G

/1

/G

/1

α

/ GnA

Proposition 16.4.4: Let L/K be a finite Galois extension of degree n with Galois group G. Suppose G acts on a finite Abelian group A. Let π: GnA → G be the projection map. Then π is regularly solvable. Proof: By Proposition 16.3.5, K(T ) has a Galois extension N such that Gal(N/K(T )) ∼ = A and N/K is regular. By Proposition 16.2.8, there exists an X-stable polynomial h ∈ K[T, X] with Gal(h(T, X), K(T )) ∼ = A. In particular, h(T, X) is absolutely irreducible and Gal(h(T, X), L(T )) ∼ = A. Let π ˆ : G n AG → G be the projection on G. Let t1 , . . . , tn be algebraically independent elements over K with n = [L : K]. Lemma 13.8.1, with G0 trivial, gives a Galois extension Fˆ of K(t) which contains L and an isomorphism ˆ ◦ γ = res. Moreover, Fˆ is a regular γ: Gal(Fˆ /K(t)) → G n AG with π extension of L. Lemma 16.4.3 gives an epimorphism α: G n AG → G n A which is the identity map on G. Thus, π ˆ = π ◦ α. Let F be the fixed field in Fˆ of Ker(α ◦ γ). Then there is an epimorphism β: Gal(F/K(t)) → G n A with α ◦ γ = β ◦ resFˆ /F and π ◦ β = resF/L . Thus, β is a solution of π over K(t) with F being the solution field. By construction, L ⊆ F ⊆ Fˆ . Since Fˆ /L is regular, so is F/L (Corollary 2.6.5(b)). Proposition 16.4.5 ([Ikeda, p. 126]): Let K be a Hilbertian field. Then every finite split embedding problem over K with Abelian kernel is solvable Proof: Combine Propositions 16.4.2 and 16.4.4.

16.4 Split Embedding Problems with Abelian Kernels

305

Remark 16.4.6: Proposition 16.4.5 does not hold for an arbitrary profinite Abelian group. For example, Zp is regular over Q for no p (Proposition 16.6.10), but it is not known if every finite p-group is regular over Q. Nevertheless, every finite p-group occurs over Q. More generally, every finite solvable group occurs over every global field. This is a theorem of Shafarevich. Its proof does not use Hilbert irreducibility theorem but rather class field theory and complicated combinatorial arguments. See [Neukirch-SchmidtWingberg, Section 9.5] for a proof that uses cohomological arguments and for a reference to the original articles. Class field theory is not available over an arbitrary Hilbertian field K. Thus, Shafarevich’s proof does not apply to K. However, when p = char(K) > 0, every finite p-group occurs over K. This follows from a theorem of Witt (Remark 16.3.3), but it is not known if each finite p-group occurs over K when char(K) = 0. Amazingly enough, both realization problems raised in Remark 16.4.6 are easy consequences of Shafarevich’s theorem when char(K) > 0. This is the content of the following result: Theorem 16.4.7: Let G be a finite p-group and K a field of positive characteristic. Then G is regular over K. If in addition K is Hilbertian, then G is realizable over K. Proof: The second statement of the theorem follows from the first one by Hilbert (Lemma 13.1.1). In order to prove that G is regular over every field of positive characteristic, it suffices to prove that G is regular over every finite field K (Lemma 16.2.1). Assume without loss G is nontrivial. Let t be an indeterminate. By Shafarevich, G × G occurs over K(t). Thus, K(t) has linearly disjoint Galois extensions F1 and F2 with Gal(Fi /K(t)) ∼ = G, i = 1, 2. Assume none of them ˜ ∩ Fi is a cyclic extension of is regular over K. Then, for i = 1, 2, the field K K of degree pki with ki ≥ 1. Hence, Fi contains the unique extension Kp of K of degree p. Therefore, both F1 and F2 contain Kp (t) which is a proper extension of K(t). This contradiction to the linear disjointness of F1 and F2 over K(t) proves that one of them is regular over K. Corollary 16.4.8: Let K be a field, G a finite group, and A a finite Abelian group. Suppose G is regular over K and G acts on A. Then G n A is regular over K. Proof: There exists t1 , . . . , tn and a finite Galois extension E of K(t) such that Gal(E/K(t)) ∼ = G and E/K is regular. Choose an indeterminate u. Proposition 16.4.4 gives a Galois extension F of K(t, u) such that Gal(F/K(t, u)) ∼ = G n A and F is a regular extension of E. By Corollary 6.2.5, F is a regular extension of K. Thus, G n A is regular over K. Remark 16.4.9: Realization of p-groups of low order. The results of this section imply that each p-group of order at most p4 is regular over every field K.

306

Chapter 16. Galois Groups over Hilbertian Fields

To this end let A be the smallest family of all finite groups satisfying this: (4a) Every Abelian group belongs to A. (4b) Suppose H = G · A with G ∈ A and A Abelian and normal. Then H ∈ A. Note that the group H in (4b) is a quotient of G n A. Hence, if G is regular over K, so is H (Proposition 16.4.8). It follows by induction on the order of the group that each G ∈ A is regular over K. Thus, if K is Hilbertian, every G ∈ A occurs over K (Lemma 13.1.1(b)). The family A contains each finite group G which satisfies one of the following conditions: (5a) G has nilpotence class at most 2, i.e. [G, G] ≤ Z(G) (Thompson [MalleMatzat, p. 277, Prop. 2.9(a)]). (5b) G is solvable and every Sylow subgroup of G is Abelian (Thompson [Malle-Matzat, p. 277, Prop. 2.9(b)]). (5c) G is a p-group of order at most p4 (Dentzer [Malle-Matzat, p. 278, Cor. 2.10]. (5d) G is a 2-group of order 25 (Dentzer [Malle-Matzat, p. 278, Cor. 2.10]). However, there are groups of order p5 (for p 6= 2) and 26 which do not belong to G [Malle-Matzat, p. 278].

16.5 Embedding Quadratic Extensions in Z/2n Z-Extensions Nonsplit embedding problems with Abelian kernel √ over Hilbertian fields need not be solvable. For example Z/4Z → Gal(Q( −1)/Q) √ is not solvable. Otherwise, Q has a Galois extension N containing Q( −1) with Gal(N/Q) ∼ = √ Z/4Z. The only subfields of N are Q, Q( −1), and N . Hence, N ∩ R = Q. Hence, [N R : R] = 4, which is a contradiction. Here is a general criterion for a quadratic extension to be embeddable in a Z/4Z-extension. Proposition 16.5.1: Let K be a field with char(K) 6=√2 and a a nonsquare in K. Then the embedding problem Z/4Z → Gal(K( a)/K) is solvable if and only if there are x, y ∈ K with a = x2 + y 2 . √ Proof: By assumption, L = K( a) is a quadratic extension of K. Suppose K has a Galois extension N √containing L with Gal(N/K) ∼ = Z/4Z. Then u). Let σ be a generator of Gal(N/K). Then there is a u ∈ L with N = L( √ √ √ √ u). By √ Kummer theory, σ u = v u with v ∈ L. Hence, L(√σu) = L( √ of K σ 2 u = v · σv u = b u with b = v · σv. Thus, b is an element √ 2 generates Gal(L( u)/L), so which is a norm from L. On the other had, σ √ √ √ √ σ 2 u = − u. Therefore, b =√−1, v · σv u = − u, and v · σv = −1. Finally, write √ suppose d = 0. Then √ v =√c + d a with c, d ∈ K. First v ∈ K and v 2 u = − u. Hence, v 2 = −1 and −1 ∈ K. Therefore, the

16.5 Embedding Quadratic Extensions in Z/2n Z-Extensions

307

2 2 √ identity a = −1(1 − a4 ) + 1 + a4 yields a representation of a as a sum of two squares in K. Now suppose d 6= 0. Then −1 = v · σv = c2 − ad2 . 2 2 Hence, a = dc + d1 , as desired. Conversely, suppose a = x2 + y 2 with x, y ∈ K. Then x,√ y 6= 0 and −1 = c2 − ad2 , where c = xy and d = y1 . Put v = c + d a. Then normL/K v = c2 − ad2 = −1. Hence, normL/K v 2 = 1. Let σ be an element of Gal(K) whose restriction to L generates Gal(L/K). Hilbert’s Theorem 90 gives u ∈ L with v 2 = σu u [Lang7, p. 288, Thm. 6.1]. Thus, √ √ σ√ u2 σ√ u 2 and u = ±v. Replacing v by −v, if necessary, we may assume v = u √ √ √ √ √ √ √ √ √ u = − u, σ 3 u = −v u,√and σ 4 u = u. σ u =√v u,√σ 2 u√= v · σv √ Thus, u, σ u, σ 2 u, √ σ 3 u are distinct conjugates of u over K. All of of degree 4 them belong to N = L( u). Therefore, N is a cyclic extension √ of K which contains L. Consequently, Z/4Z → Gal(K( a)/K) is solvable. Remark 16.5.2: Embedding in cyclic extensions of higher order. It is possible to slightly generalize Proposition 16.5.1: Suppose K is a field and a is √ a nonsquare in K. Then K( a) can be embedded in a Z/8Z-extension of K if and only if a is a sum of two squares in K and there are x1 , x2 , x3 , x4 in K, not all zero, with x21 + 2x22 + ax23 − 2ax24 = 0. This result is proved in [Kiming, Thm. 3] in a slightly different form. The formulation we give appears in [Geyer-Jensen2, 20◦ ]. Let√K be a number field and a ∈ K. Suppose −1 is a nonsquare √ in K and K( −1) is embeddable in a Z/16Z-extension of K. Then K( −1) is embeddable in a Z/2n Z-extension of K for each positive integer n [GeyerJensen2, Thm. 4]. The proof requires class field theory. √ √ As an example, let K = Q( −14). Then K( −1) is embeddable in a Z/8Z-extension of K [Arason-Fein-Schacher-Sonn, p. 846, Cor. 4 or Geyer√ Jensen2, Remark 19◦ ]. But K( −1) is not embeddable in a Z/16Z-extension of K [Geyer-Jensen2, Example to Proposition 3]. Let OK be the ring of integers of K. Consider a nonzero prime ideal p of OK such that ˆ p. (1) −1 a nonsquare but −1 is the sum of two squares in the completion K Since X 2 +Y 2 +1 is absolutely irreducible. Hence, −1 is a sum of two squares ˆp in Fp for all large p. By Hensel’s lemma, −1 is a sum of two squares in K for all but finitely many p. Hence, by Chebotarev, Condition (1) holds with √ ˆ p ( −1) is embeddable in a Z/2n Z-extension of K ˆ p for all density 12 . Then K n [Geyer-Jensen2, Prop. 3]. Thus, there could be no criterion for embedding a quadratic extension of a field in a Z/16Z-extension similar to the one we gave above for embedding in a Z/8Z-extension. √ √ Consider now the field K = Q( −17). Then K( −1) is embeddable in a Z/2n Z-extension for each positive integer n but is not embeddable in a Z2 -extension [Geyer-Jensen1, p. 371].

308

Chapter 16. Galois Groups over Hilbertian Fields

16.6 Zp -Extensions of Hilbertian Fields Lemma 16.3.2 implies that if K is a field of characteristic p, then every Z/pZ-extension of K can be embedding in a Zp -extension. This result does not generalize to the general case. Remark 16.5.2 gives an example of a quadratic extension L/K (with char(K) 6= 2) which can not be embedded in a Z2 -extension. So, we have to settle for less. We fix a field K, a prime number p, and let q = p if p = 2 and q = 4 if p = 2. We ask when Zp occurs over K. We prove this is the case when Z/qZ occurs over K. This condition is satisfied when K is Hilbertian. Lemma 16.6.1: Suppose p 6= char(K). Let L be a cyclic extension of K of degree pn with n ≥ 1. (a) Suppose ζpn+1 ∈ K. Then K has a Z/pn+1 Z-extension which contains L. (b) Suppose ζpk ∈ K for all k. Then K has a Zp -extension that contains L. Proof: Statement (b) follows from (a) by induction and taking inverse limit. It remains to prove Statement (a). n The theory of cyclic extensions gives a ∈ K with L = K(a1/p ) [Lang7, n+1 / K. The field L0 = K(a1/p ) p. 289, Thm. 6.2(i)]. In particular, a1/p ∈ contains L. By [Lang7, p. 289, Thm. 6.2(ii)], L0 is a cyclic extension of n+1−k ∈ K. If k ≤ n, then a1/p = degree pk for some k ≤ n + 1 and a1/p n+1−k n−k (a1/p )p ∈ K. We conclude from this contradiction that k = n + 1, as desired. Remark 16.6.2: Abelian pro-p groups as Zp -modules. Let A be a finite Abelian additive p-group of exponent pm . Then Z/pm Z acts on A through the rule (k + pm Z)a = ka, a ∈ A. Since Z/pm Z is a quotient of Zp , this defines a continuous action of Zp on A with the discrete topology. This action commutes with homomorphisms of finite Abelian p-groups. Hence, it defines a continuous action of Zp on projective limits of Abelian p-groups (also called Abelian pro-p groups which commutes with homomorphisms. r Consider elements a1 , . . . , ar of A. Then the map Pr Pr Zp → A given by compact, (z1 , . . . , zr ) 7→ i=1 zi ai is continuous. Its image i=1 Zp ai is P r hence closed. On the P other hand, one can approximate each sum i=1 zi ai n with zi ∈ Zp by sums i=1 ki ai with ki ∈ Z. Therefore, the closed subgroup of A generated by a1 , . . . , ar coincides with the Zp -submodule of A generated by a1 , . . . , ar . Lemma 16.6.3: Let 0 → B → A → Z/pZ → 0 be an exact sequence of ∼ Zp . Then either A ∼ Abelian pro-p groups with B = = Zp or A = B ⊕ A0 with A0 ∼ = Z/pZ. Proof: Consider A as a Zp -module (Remark 16.6.2). The exact sequence yields pA ≤ Zp . By Lemma 1.4.2(e), pA is generated by one element. On the other hand, A is generated by two elements, one which generates B and the other with image in Z/pZ generating that module. Assume A ∼ 6 Zp = and A ∼ 6 Zp ⊕ Z/pZ. Then, by Proposition 2.2.3, A ∼ = = Zp ⊕ Z/pk Z with k ≥ 2

16.6 Zp -Extensions of Hilbertian Fields

309

or A ∼ = Zp ⊕ Zp . In each cases pA ∼ = Z/pZ ⊕ Z/pZ, so pA is not generated by one element, contradicting the preceding paragraph. Therefore, A ∼ = Zp or A ∼ Z ⊕ Z/pZ. In the second case A has a subgroup A isomorphic to = p 0 Z/pZ. Since Zp contain no nontrivial closed subgroups of finite order (Lemma 1.4.2(d)), A0 is not contained in B. Consequently, A = B ⊕ A0 . Lemma 16.6.4: Suppose p 6= char(K). Let L = K(ζpn | n = 1, 2, 3, . . .). Then L/K is an Abelian extension and there is a field K∞ satisfying the following conditions: (a) K∞ (ζq ) = L and K∞ ∩ K(ζq ) = L. (b) Gal(L/K) = Gal(L/K(ζq )) × Gal(L/K∞ ). (c) Gal(L/K∞ ) ∼ = Gal(K(ζq )/K) and Gal(K(ζq )/K) is isomorphic to a subgroup of Z/(p − 1)Z (resp. Z/2Z) if p 6= 2 (resp. p = 2). (d) If L 6= K(ζq ), then Gal(K∞ /K) ∼ = Gal(L/K(ζq )) ∼ = Zp . Proof: Let n be a positive integer. Then K(ζpn ) is the splitting field of the n separable polynomial X p − 1 over K. Hence, K(ζpn )/K is Galois. Embed Gal(K(ζpn )/K) into (Z/pn Z)× by mapping each σ ∈ Gal(K(ζpn )/K) onto s(σ)

the element s(σ) of (Z/pn Z)× satisfying ζpσn = ζpn . Thus, K(ζpn )/K is Abelian. Hence, L/K is also Abelian. In particular, Gal(K(ζq )/K) is isomorphic to a subgroup of Z/(p−1)Z if p 6= 2 and of Z/2Z if p = 2. If K(ζq ) = L, then K∞ = K satisfies Conditions (a)-(c). Assume from now on K(ζq ) 6= L. Let m be the maximal positive integer with ζpm ∈ K(ζq ). Then ζpm 6= xp for all x ∈ K(ζq ). If p = 2, then m ≥ 2 and −1 = ζ42 is a square in K(ζq ), so ζpm 6= −4y 4 for all y ∈ K(ζ4 ). Let 1/pn−m

n−m

) and X p n ≥ m. Then K(ζpn ) = K(ζq )(ζpm K(ζq ) [Lang7, p. 297, Thm. 9.1]. So, (1)

− ζpm is irreducible over

[K(ζpn ) : K(ζq )] = pn−m .

Divide the rest of the proof into two parts. Part A: Suppose p 6= 2 and q = p. Then (Z/pn Z)× ∼ = Z/(p − 1)Z × s(σ) n−1 Z/p Z. Let σ ∈ Gal(K(ζpn )/K(ζp )). Then ζp = ζpσ = ζp . Hence, n−1

s(σ) ≡ 1 mod p. Therefore, s(σ)p ≡ 1 mod pn [LeVeque, p. 50, Thm. 45]. Thus, Gal(K(ζpn )/K(ζp )) is isomorphic to a subgroup of Z/pn−1 Z. It follows from (1) that Gal(K(ζpn )/K(ζp )) ∼ = Z/pn−m Z. By the beginning of the preceding paragraph, Gal(K(ζpn )/K) is cyclic of order dpn−m with d = [K(ζp ) : K]. Let τ be a generator of this group. n−m n−m Then ord(τ p ) = d. Denote the fixed field of τ p in K(ζpn ) by Kn . Then Kn ∩ K(ζp ) = K and Kn (ζp ) = K(ζpn ). Moreover, Kn is the unique field with these properties. Hence, Kn ⊆ Kn0 for n0 ≥ m. Now let K∞ = S ∞ n=m Kn . Then K∞ ∩ K(ζp ) = K and K∞ (ζp ) = L. It follows from the preceding paragraph that Gal(L/K∞ ) ∼ = Gal(K(ζp )/K) and Gal(K∞ /K) ∼ = ∼ Gal(L/K(ζp )) = Zp .

310

Chapter 16. Galois Groups over Hilbertian Fields n−3

Part B: Suppose p = 2 and q = 4. Let n ≥ max(3, m). Then 52 ≡1+ 2n−1 mod 2n and ord2n 5 = 2n−2 [LeVeque, p. 54]. In addition, −1+2n Z does not belong to the subgroup of (Z/2n Z)× generated by 5 + 2n Z. Otherwise, 5k ≡ −1 mod 2n for some 1 ≤ k < 2n−2 . Raising both sides to an odd l+1 power, we may assume k = 2l with 1 ≤ l ≤ n − 3. Then 52 ≡ 1 mod 2n . n−3 2 n ≡ −1 mod 2 . It follows that Hence, n − 2 ≤ l + 1, so l = n − 3 and 5 −1 ≡ 1 + 2n−1 mod 2n , which is a contradiction. Consequently, (Z/2n Z)× ∼ = Z/2Z × Z/2n−2 Z with −1 generating the first factor and 5 generating the second factor. s(σ) Consider σ ∈ Gal(K(ζ2n )/K(ζ4 )). Then ζ4 = ζ4σ = ζ4 . Hence, s(σ) ≡ 1 mod 4. By the preceding paragraph, s(σ) ≡ (−1)i 5j mod 2n with 0 ≤ i ≤ 1 and 0 ≤ j ≤ n−2. Then (−1)i ≡ 1 mod 4, so i = 0 and s(σ) ≡ 5j mod 2n . Thus, Gal(K(ζ2n )/K(ζ4 )) is isomorphic to a subgroup of Z/2n−2 Z. By (1), Gal(K(ζ2n /K(ζ4 ))) ∼ = Z/2n−m Z. Taking inverse limit, we get Gal(L/K(ζ4 )) ∼ = Z2 . This gives a short exact sequence of Abelian pro-2 groups 0 → Z2 → Gal(L/K) → Gal(K(ζ4 )/K) → 1 with Gal(K(ζ4 )/K) trivial or isomorphic to Z/2Z. By Lemma 16.6.3, Gal(L/K) = Gal(L/K(ζ4 )) × A0 with A0 trivial or of order 2. Denote the fixed field of A0 in L by K∞ . It satisfies Conditions (a)–(d). The case where p 6= char(K), and 1 < [L : K] < ∞ is the most complicated. We need some concepts and facts from group theory. Let G be a profinite group and x, y, z ∈ G. Define the commutator of x, y by [x, y] = x−1 y −1 xy. It satisfies the following identities: (2)

[x, y]−1 = [y, x], [x, y]z = [xz , y z ], [x, yz] = [x, z][x, y]z , [xy, z] = [x, z]y [y, z]

The commutator subgroup of G is [G, G] = h[x, y] | x, y ∈ Gi. Suppose N is a closed subgroup of G which contains [G, G]. For each n ∈ N and g ∈ G we have ng = n[n, g]. Hence, N / G and G/N is Abelian. Conversely, if N is a closed normal subgroup of G and G/N is Abelian, then [G, G] ≤ N . Consider now a profinite group C and an epimorphism g: G → C. Suppose A = Ker(g) is Abelian. Define an action of C on A in the following way: γ for a ∈ A. For each γ ∈ C choose γ˜ ∈ G with g(˜ γ ) = γ. Then let aγ = γ˜ −1 a˜ Since A is Abelian, this action is independent of γ˜ . P Suppose C is finite. The group ring Z[C] consists of all formal sums γ∈C kγ γ with kγ ∈ Z. Addition is defined componentwise. Multiplication in Z[C] is a linear extension of multiplication in C. Thus, X X X X kγ γ · lδ δ = kγ lδ ε. γ∈C

δ∈C

ε∈C

γδ=ε

The action of C on A naturally extends to an action of Z[C] on A: P Y akγ )γ . a γ∈C kγ γ = γ∈C

16.6 Zp -Extensions of Hilbertian Fields

311

Let γ be an element of C with γ m = 1. Put c = (1 − γ)c = 1 − γ m = 0. Hence, (3)

(1 − γ k )c =

k−1 X

γ i (1 − γ)c = 0,

Pm−1 i=0

γ i . Then

k = 0, . . . , m − 1

i=0

Lemma 16.6.5: Let p be a prime number, C a finite cyclic group, G a profinite group, and g: G → C an epimorphism. Suppose A = Ker(g) is an Abelian pro-p group and f0 : A → Zp is an epimorphism. Let q 0 be a power of p and let π: Zp → Z/q 0 Z an epimorphism. Put α = π ◦ f0 . Suppose one of the following conditions holds: (a) q 0 6= 1 and p - |C|. (b) p = |C| = 2 and q 0 ≥ 4. (c) p = q 0 = |C| = 2 and α extends to an epimorphism β: G → Z/4Z. Then there exists an epimorphism f : A → Zp with Ker(f ) / G such that G/Ker(f ) is Abelian. Proof: The first two parts of the proof are common to all cases. The rest of the proof handles each case separately. Part A: The commutator of G. As A is Abelian, C acts on A by lifting and conjugating. Extend this to an action of Z[C] on A. Since C is Abelian, [G, G] ≤ A. Moreover, [G, G] = ha1−γ | a ∈ A, γ ∈ Ci.

(4)

Indeed, choose a generator γ0 of C and an element γ˜0 in G with g(˜ γ0 ) = γ0 . ˜ = 1 for all γ, δ ∈ C and For each i let γei = γ˜ i . Then [˜ γ , δ] 0

(5)

0

a1−γ = [˜ γ , a]

for each a ∈ A.

Thus, the right hand side of (4) (which we denote by G1 ) is contained in [G, G]. To prove the other inclusion, it suffices to prove [u, v] ∈ G1 for all u, v ∈ G. To this end, write u = a˜ γ and v = bδ˜ with a, b ∈ A and γ, δ ∈ C. Now use (2), (5) and the hypothesis that A is Abelian: ˜ = [a˜ ˜ γ , b]δ [u, v] = [a˜ γ , bδ] γ , δ][a˜ ˜ γ [˜ ˜ b]γδ [˜ ˜ γ , b δ ] ∈ G1 = [a, δ] γ , δ][a, γ , b]δ = [aγ , δ][˜ Pm−1 Part B: Twist of A. Let m = |C| and put c = i=0 γ0i . Let µ: A → A be the homomorphism given by µ(a) = ac . Since γ0m = 1, (3) implies (1−γ)c = 0 for each γ ∈ C. Hence, (a1−γ )c = a(1−γ)c = 1 for each a ∈ A. Therefore, by (4), µ(w) = wc = 1 for each w ∈ [G, G]. γ , a]) = 0. Hence, α(aγ ) = α(a) for all a ∈ A and By (5), α(a1−γ ) = α([˜ Pm−1 i γ ∈ C. Consequently, α(µ(a)) = i=0 α(aγ0 ) = mα(a).

312

Chapter 16. Galois Groups over Hilbertian Fields

Part C: Suppose q 0 > 1 and p - |C|. Then m is an invertible element of Zp and Z/q 0 Z. Let m−1 : Zp → Zp and m−1 : Z/q 0 Z → Z/q 0 Z be multiplications by m−1 . Define a homomorphism f : A → Zp by f = m−1 ◦ f0 ◦ µ. By Part B, [G, G] ≤ Ker(µ) ≤ Ker(f ). Thus, Ker(f ) / G and G/Ker(f ) is Abelian. By Part B, this establishes the following commutative diagram: f µ

A α

Z/q 0 Z

f0 /A / Zp DD DD DDα DD π D! m / Z/q 0 Z

'/

m−1

Zp π

−1

m

/ Z/q 0 Z

In particular, α = π ◦ f . Thus, π(f (A)) = α(A) = Z/q 0 Z. It follows from Lemma 1.4.3 that f is surjective. Part D: Suppose p = |C| = 2 and q 0 ≥ 4. In this case Part B gives the following commutative diagram: A α

Z/q 0 Z

µ

f0 /A / Z2 DD DD α DD DD π " 2 / Z/q 0 Z

Let H = f0 (µ(A)). By Remark 1.2.1(e), H is a closed subgroup of Z2 . It satisfies, π(H) = 2(Z/q 0 Z) = π(2Z2 ). Hence, H + q 0 Z2 = 2Z2 + q 0 Z2 = 2Z2 . By Lemma 1.4.2, H is trivial or H = 2n Z2 for some positive integer n. It follows from q 0 ≥ 4 that H = 2Z2 . Part E: Suppose p = q 0 = 2 = |C| and α extends to an epimorphism β: G → Z/4Z. Then (G : A) = |C| = 2. Hence, in the notation of Part B, a = γ˜02 ∈ A. So, aγ0 = a. Hence, µ(a) = aaγ0 = a2 . The existence of β gives a commutative diagram 1

0

/A

/G

α

β

/ Z/2Z

/ Z/4Z

g

/C

/1

β¯

π ¯

/ Z/2Z

/1

¯ γ0 )) = 1 + 2Z. Hence, with both rows exact. In particular, π ¯ (β(˜ γ0 )) = β(g(˜ 2 γ0 ) = 1 + 2Z. Therefore, by β(˜ γ0 ) = ±1 + 4Z. So, π(f0 (a)) = α(a) = β(˜ Lemma 1.4.3, hf0 (a)i = Z2 . Hence, hf0 (µ(a))i = hf0 (a2 )i = 2Z2 . It follows that f0 (µ(A)) = Z2 or f0 (µ(A)) = 2Z2 .

16.6 Zp -Extensions of Hilbertian Fields

313

Suppose first f0 (µ(A)) = Z2 . Then f = f0 ◦ µ maps A onto Z2 . For each w ∈ [G, G], Part B implies f (w) = f0 (µ(w)) = 1. Therefore, [G, G] ≤ Ker(f ), Ker(f ) / G, and G/Ker(f ) is Abelian. The case f0 (µ(A)) = 2Z2 (which we henceforth assume), will be handled in Part F. Part F: Conclusion of the proof in cases (b) and (c). Multiplication by 2 gives an isomorphism of Z2 onto 2Z2 . Hence, by Parts D and E, there is an epimorphism f : A → Z2 with 2f = f0 ◦ µ. For each w ∈ [G, G], Part B implies 2f (w) = f0 (µ(w)) = 0. Hence, f (w) = 0. Therefore, [G, G] ≤ Ker(f ), Ker(f ) / G, and G/Ker(f ) is Abelian. Theorem 16.6.6 ([Whaples, Thm. 2]): Let K be a field and p a prime number. Put q = p if p 6= 2 and q = 4 if p = 2. Suppose Z/qZ occurs over K. Then Zp occurs over K. Proof: Let K 0 be a Z/qZ-extension of K. Suppose first p = char(K). Lemma 16.3.2 embeds K 0 in a Zp -extension of K. Assume from now on, p 6= char(K). Let L = K(ζpn | n = 1, 2, 3, . . . ). When L = K, Lemma 16.6.1 gives a Zp -extension of K. When L 6= K(ζq ), Lemma 16.6.4 gives a Zp -extension of K. Assume from now on, L = K(ζq ) and L 6= K. Let C = Gal(L/K). Then C is a cyclic group of order which divides p − 1 if p 6= 2 and of order 2 of p = 2. Put L0 = LK 0 . Then L0 is a finite Abelian extension of K. Moreover, 0 L /L is a cyclic extension of degree q 0 which is a power of p. We distinguish between three cases: Case A: p 6= 2. Then K 0 /K is a cyclic extension of degree p and [L : K]|p − 1. Hence, q 0 = [L0 : L] = p and p - |C|. Case B: p = 2 and L ∩ K 0 = K. 4, |C| = 2, and q 0 = [L0 : L] = 4.

Then K 0 /K is a cyclic extension of degree

Case C: p = 2 and L ∩ K 0 6⊆ K. Then, L ⊆ K 0 , K 0 = L0 , and q 0 = |C| = 2. In each case Lemma 16.6.1 gives a Zp -extension F of L which contains L0 . Denote the compositum of all finite Abelian extensions of L of a ppower order by N . In particular, F ⊆ N . Since L/K is Galois, so is N/K. Put G = Gal(N/K) and A = Gal(N/L). Let f0 be resN/F : Gal(N/L) → Gal(F/L) and let π be resF/L0 : Gal(F/L) → Gal(L0 /L). Then α = π ◦ f0 = resN/L0 : Gal(N/L) → Gal(L0 /L). In Case C, Gal(K 0 /K) ∼ = Z/4Z and β = resN/L0 : Gal(N/K) → Gal(L0 /K) extends α. In each case, Lemma 16.6.5 gives an extension F 0 of L in N with F 0 /K Abelian and Gal(F 0 /L) ∼ = Zp . Suppose first p 6= 2. Then [L : K]|p − 1. So, Gal(F 0 /K) has a unique subgroup of order p − 1. This is a special case of a profinite version of the Schur-Zassenhaus theorem (Lemma 22.10.1). Its fixed field E in F 0 satisfies Gal(E/K) ∼ = Zp . Now suppose p = 2. Lemma

314

Chapter 16. Galois Groups over Hilbertian Fields

16.6.3 gives a Z2 -extension E of K in F 0 . Consequently, in each case K has a Zp -extension. Corollary 16.6.7: Let K be a Hilbertian field and p a prime number. Then Zp -occurs over K. Proof: By Corollary 16.3.6, Z/pZ and Z/p2 Z occur over K. Hence, by Theorem 16.6.6, Zp occurs over K. Remark 16.6.8: The assumption in Theorem 16.6.6 that Z/4Z rather than Z/2Z occurs over K is necessary for the theorem to hold. Indeed, Gal(C/R) ∼ = Z/2Z but Z2 does not occur over R. Proposition 16.6.10 below shows it is impossible to conclude that Zp is regular over K in Corollary 16.6.7. Lemma 16.6.9: Let E be a field, p a prime number, v a discrete valuation of ¯v ) and E ¯v (ζp , ζp2 , ζp3 , . . .) E, and F a Zp -extension of E. Suppose p - char(E ¯ is an infinite extension of Ev . Then v is unramified in F . Proof: Let w be a valuation of F extending v. Denote reduction at w by a bar. Assume w is ramified over E. Then its inertia group Iw/v is nontrivial. By Lemma 1.4.2, Iw/v is an open subgroup of Gal(F/E). Replace E by the fixed field of Iw/v in F , if necessary, to assume Iw/v = Gal(F/E). Let n be a positive integer. Denote the unique extension of E in F of degree pn by En . Let vn be a normalized valuation of En over v. It is totally ˆ of E under v. Then E ˆn = ramified over E. Consider the completion E ˆn /E) ˆ ∼ ˆ is the completion of En under vn (Proposition 3.5.3) and Gal(E En E = n ∼ Gal(En /E) = Z/p Z [Cassels-Fr¨ ohlich, p. 41, Prop. 3]. ˆn and x ∈ E ˆ with vn (y) = 1 and v(x) = 1. Then vn (x) = Choose y ∈ E n n ¯n = E, ¯ pn = vn (y p ). Hence, y p = ux with u ∈ En and vn (u) = 0. Since E 0 −1 0 there is an a ∈ E with a ¯=u ¯. Let u = ua and x1 = ax. Then vn (u − 1) > ¯v ), Hensel’s lemma (Proposition 3.5.2(a)) 0, and x1 ∈ E. Since p - char(E n pn gives u1 ∈ En with up1 = u0 . Put y1 = u−1 1 y. Then y1 = x1 . Since En /E is ¯v . cyclic of degree n, this implies ζpn ∈ En . Taking residues, we find ζpn ∈ E ¯ ¯ Thus, Ev (ζp , ζp2 , ζp3 , . . .) = Ev , contrary to the assumption of the lemma. Consequently, v is unramified in F . Proposition 16.6.10: Let K be a field and p a prime number. Suppose p 6= char(K) and K(ζp , ζp2 , ζp3 , . . .) is an infinite extension of K. Then Zp is not regular over K. Proof: Assume Zp is regular over K. Then there are algebraically independent elements t1 , . . . , tr over K and there is a Zp -extension F of E = K(t1 , . . . , tr ). Induction on r proves it suffices to consider the case where r = 1. Put t = t1 . Let E1 be the unique extension of E in F of degree p. Then E1 /K is regular. Remark 3.6.2(b) gives a prime divisor p of E/K which is ramified in E1 . The valuation vp of E associated with p is discrete (Section 3.1). Its

16.7 Symmetric and Alternating Groups over Hilbertian Fields

315

¯p is a finite extension of K. Hence, E ¯p (ζp , ζp2 , ζp3 , . . .) is an residue field E ¯v ) = char(K) - p. Hence, by infinite extension of K. In addition, char(E Lemma 16.6.9, p is unramified in F . In particular, p is unramified in E1 . It follows from this contradiction that Zp is not regular over K. Corollary 16.6.11: Let K be a field and p a prime number. Suppose p 6= char(K) and K is a finitely generated over its prime field. Then Zp is not regular over K.

16.7 Symmetric and Alternating Groups over Hilbertian Fields The Galois group of the general polynomial of degree n is Sn (Example 16.2.5(a)). There is a standard strategy to construct special polynomials with Galois groups Sn . Lemma 16.7.1: Let K be a field, t1 , . . . , tr algebraically independent elements over K, and F a separable extension of K(t) of degree n. Denote ∼ ˜ K(t)) ˜ the Galois closure of F/K(t) by Fˆ . Suppose Gal(Fˆ K/ = Sn . Then, ∼ ˆ Gal(F /K(t)) = Sn . Proof: Since [F : K(t)] = n, the group G = Gal(Fˆ /K(t)) is isomorphic to a ˜ : K(t)] ˜ subgroup of Sn , On the other hand, |Sn | = [Fˆ K ≤ [Fˆ : K(t)] = |G|. ∼ Consequently, G = Sn . Consider for example the case when r = 1. Put t = t1 . Let x be a primitive element of F/K(t), and f ∈ K[T, X] an absolutely irreducible polynomial which separable and of degree n in X with f (t, x) = 0. Then ˜ = Gal(f (t, X), K(t)) ˜ G acts transitively on the n distinct roots of f (t, X) in ] ˜ K(t) ˜ ˜ one K(t). Inspecting inertia groups of prime divisors of F K/ in Fˆ K, ˜ may prove that G contains cycles which generate Sn . Then, by Lemma 16.7.1, Gal(Fˆ /K(t)) ∼ = Sn . When the Galois group is Sn , one may consider the fixed field E of An in Fˆ . It is a quadratic extension of K(t). If we prove that E is rational over K, then An becomes regular over K. We are able to do it, for example, if char(K) 6= 2 and K(t)/K has at most three prime divisors which ramify in F , each of degree 1 (Lemma 16.7.5). We start with a valuation theoretic condition that yields e-cycles for the Galois group of a polynomial. By an e-cycle of Sn we mean a cycle of length e. Lemma 16.7.2: Let (E, v) be a discrete valued field, f ∈ Ov [X] a monic separable irreducible polynomial over E, and x a root of f in Es . Denote ¯ is separably closed, f¯(X) = (X −¯ reduction at v by a bar. Suppose E a)e η(X) ¯ ¯ with a ∈ Ov , e ≥ 1, char(E) - e, and η ∈ E[X] monic separable polynomial with η(¯ a) 6= 0. Suppose v extends to a valuation w of E(x) with w(x − a) > 0 and ew/v = e. Then Gal(f, E) contains an e-cycle.

316

Chapter 16. Galois Groups over Hilbertian Fields

ˆ vˆ). Embed Es into E ˆs and Proof: Denote the completion of (E, v) by (E, ˆ be the valuation ring of E. ˆ By ˆs . Let O extend vˆ to a valuation vˆ of E assumption, η(X) is relatively prime to (X − a ¯)e . Hence, Hensel’s lemma (Proposition 3.5.2(b)) gives a factorization f (X) = g(X)h(X) with g, h ∈ ¯ ˆ = η(X). O[X] monic, g¯(X) = (X − a ¯)e and h(X) Let g(X) = g1 (X) · · · gk (X) with g1 , . . . , gk distinct monic irreducible ˆ polynomials in E[X]. For each i choose a root yi of gi in Es . Then there is a σi ∈ Gal(E) with σi x = yi . The formula wi (z) = vˆ(σi z) for z ∈ E(x) ˆ i ) as completion defines a valuation wi of E(x) which extends v and with E(y (Proposition 3.5.3(a),(b)). Similarly, for each root y of h(X) there is a σ ∈ Gal(E) with σx = y and v 0 (z) = vˆ(σz) for z ∈ E(x) defines an extension of v. By assumption, ¯ y ) = 0 and h(¯ ¯ a) 6= 0. Hence, v 0 (x − a) = vˆ(y − a) = 0. Since w(x − a) > 0, h(¯ this implies, w 6= v 0 . Thus, w is one of the wi ’s, say w = w1 (Proposition 3.5.3(c)). Therefore, e = ew/v ≤

k X i=1

ewi /v ≤

k X i=1

ˆ i ) : E] ˆ = [E(y

k X

deg(gi ) = deg(g) = e.

i=1

ˆ List the roots of g in Es as It follows that k = 1 and g is irreducible over E. x1 , . . . , xe . ˆ ˆ = deg(g) = e. Since char(E) ¯ - e, By the preceding paragraph, [E(x) : E] 1/e ˆ ˆ ˆ ˆ ˆ E(x)/E is tamely ramified. Hence, E(x) = E(π ) for some π ∈ E [Lang5, ˆ ¯ is separably closed, it contains ζe . By Hensel’s lemma, ζe ∈ E. p. 52]. Since E ˆ ˆ ˆ Hence, E(x)/E is a cyclic group of order e. In other words, Gal(g, E) is a cyclic group of order e. It acts transitively on x1 , . . . , xe . Therefore, it is generated by an e-cycle, say (x1 x2 . . . xe ). ¯ Finally, by assumption, h(X) is separable. Hence, it is a product of ¯ distinct monic linear factors in E[X]. By Hensel’s lemma, h(X) Qn decomposes ˆ into distinct monic linear factors in E[X], say, h(X) = i=e+1 (X − xi ). ˆ is ˆ ˆ acts trivially on xe+1 , . . . , xn . Thus, Gal(f, E) Therefore, Gal(E(x)/ E) ˆ ≤ Gal(f, E), the group generated by the cycle (x1 x2 . . . xe ). Since Gal(f, E) Gal(f, E) contains an e-cycle Lemma 16.7.3: Let K be a field, q ∈ K[X] with deg(q) = n, and x a transcendental element over K. Suppose char(K) - n. Put t = q(x) and f (T, X) = q(X) − T . Then: ˜ (a) f (t, X) is irreducible over K(t), separable in X, and [K(x) : K(t)] = n. (b) Denote the derivative of q by q 0 . Let ϕ be a K-place of K(t). Denote the corresponding prime divisor of K(t)/K by p. Suppose ϕ(t) 6= ∞ and ϕ(t) 6= q(b) for each root b of q 0 (X). Then p is unramified in K(x). ˜ (c) Gal(f (t, X), K(t)) contains an e-cycle. ˜ such that q(X) − q(b) = (X − b)e η(X) with (d) Suppose there is a b ∈ K ˜ ˜ e ≥ 2, char(K) - e, η(X) ∈ K[X], and η(b) 6= 0. Then Gal(f (t, X), K) contains an e-cycle.

16.7 Symmetric and Alternating Groups over Hilbertian Fields

317

˜ λ ∈ K[X], ˜ (e) Suppose q 0 (X) = (X − b)λ(X) with b ∈ K, and λ(b) 6= 0. ˜ Then Gal(f (t, X), K(t)) contains a 2-cycle. Proof: Let c be the coefficient of X n in q. Replace x by cx and t by cn−1 t to assume q is monic. Proof of (a): Since f (T, X) is linear in T , it is absolutely irreducible. Hence, ˜ f (t, X) is irreducible over K(t) and over K(t). Therefore, [K(x) : K(t)] = n. The separability of f (t, X) follows from the assumption char(K) - n. Proof of (b): Extend ϕ to a place ψ of K(x). Then ϕ(t) = ψ(q(x)) = q(ψ(x)). Hence, by assumption, ψ(q 0 (x)) = q 0 (ψ(x)) 6= 0. Therefore, the value of ϕ at the product of all K(t)-conjugates of q 0 (x) is not 0. In other words, ϕ(NK(x)/K(t) q 0 (x)) 6= 0. By (a), f (t, X) = irr(x, K(t)). In addition, ∂f (t, x). It follows from Lemma 6.1.8 that p is unramified in K(x). q 0 (x) = ∂X Proof of (c): Let q(X) = X n + cn−1 X n−1 + · · · + c0 . Then xn + cn−1 xn−1 + ˜ ˜ with v∞ (t) = −1 and let K · · · + c0 = t. Let v∞ be the valuation of K(t)/ ˜ w be an extension of v∞ to K(x). Put e = ew/v∞ . By Example 2.3.11, w(x) = −1 and e = n. Now make a change of variables: u = t−1 and y = t−1 x. Put g(U, Y ) = ˜ ˜ ˜ ˜ = K(t), K(y) = K(x), Y n + cn−1 U Y n−1 + · · · + c0 U n − U n−1 . Then K(u) ˜ w(u) = n, w(y) = n − 1, g(u, Y ) is irreducible over K(u), separable in ˜ Y , g(u, y) = 0, and g(0, Y ) = Y n . By Lemma 16.7.2, Gal(g(u, Y ), K(u)) ˜ contains an n-cycle. Therefore, Gal(f (t, X), K(t)) which is isomorphic to ˜ Gal(g(u, X), K(u)) has an n-cycle. ˜ ˜ with va (t − Proof of (d): Put q(b) = a. Let va be the valuation of K(t)/ K e ˜ a) = 1. By assumption, f (a, X) = (X − b) η(X) with η ∈ K[X], η(b) 6= 0. ˆ (It is K((t ˜ ˆ be the ˜ − a)).) Let O Denote the completion of (K(t), va ) by E ˆ valuation ring of E. By Hensel’s lemma, f (t, X) = g(X)h(X) with g, h monic ˆ polynomials in O[X] whose residue at va are (X − b)e , η(X), respectively. ˆ ¯ = b. Let w be the corresponding extension Embed K(x) in Es such that x ˆ ˆ ≤ deg(g) = e. : E] of va to K(x). Then ew/va ≤ [E(x) By the preceding paragraph, (x − b)e η(x) = q(x) − a = t − a. Since η(b) 6= 0, we have w(η(x)) = 0. Hence, ew(x − b) = ew/va ≤ e. Thus, w(x − b) = 1 and ew/va = e. ˜ We conclude from Lemma 16.7.2 that Gal(f (t, X), K(t)) has an e-cycle. e ˜ Proof of (e): Write q(X) − q(b) = (X − b) η(X) with e ≥ 1, η ∈ K[X], and η(b) 6= 0. Then (1) q 0 (X) = (X − b)e−1 eη(X) + (X − b)η 0 (X) If e = char(K), then q 0 (X) = (X − b)e η 0 (X). Hence, e = 1, which is impossible. Therefore, e 6= char(K), so the second factor on the right hand side of (1) does not vanish in b. On the other hand, q 0 (X) = (X − b)λ(X) ˜ with λ(b) 6= 0. Hence, e = 2. Consequently, by (d), Gal(f (t, X), K(t)) has a 2-cycle.

318

Chapter 16. Galois Groups over Hilbertian Fields

Lemma 16.7.4: Let K be a field, n ≥ 2 an integer, x an indeterminate, and t = xn −xn−1 . Suppose char(K) - (n−1)n. Then X n −X n−1 −t is separable, ˜ irreducible over both K(t) and K(t), and ˜ = Sn . Gal(X n − X n−1 − t, K(t)) = Gal(X n − X n−1 − t, K(t)) Moreover, K(t) has at most three prime divisors which ramify in K(x). Each of them has degree 1. Proof: The polynomial q(X) = X n − X n−1 factors as q(X) = X n−1 (X − 1). ˜ By Lemma 16.7.3(d) (with b = 0), Gal(X n − X n−1 − t, K(t)) contains an 0 n−2 (X − n−1 (n − 1)-cycle. The derivative of q(X) is q (X) = nX n ). By n−1 n n−1 ˜ Lemma 16.7.3(e) (with b = n ), Gal(X − X − t, K(t)) contains a 2˜ cycle. Hence, by [Waerden1, p. 199], Gal(X n − X n−1 − t, K(t)) = Sn . It ˜ = Sn . follows from Lemma 16.7.1 that Gal(X n − X n−1 − t, K(t)) n−1 The derivative q 0 (X) has exactly two roots, 0 and n−1 =a n . Put q n and observe that q(0) = 0. Let ϕ be a K-place of K(t) and p the prime divisor of K(t)/K corresponding to ϕ. Suppose first that ϕ(t) 6= a, 0, ∞. By Lemma 16.7.3(b), p is unramified in K(x). Thus, p ramifies in K(x) in at most three cases, when ϕ(t) is a, or 0, or ∞. In each of these cases deg(p) = 1. Lemma 16.7.5: Let K be a field, t an indeterminate, and E a quadratic extension of K(t). Suppose char(K) 6= 2, E/K is regular, and K(t)/K has at most three prime divisors which ramify in E, each of degree 1. Then E = K(u) with u transcendental over K. Proof: Denote the genus of E by g. Let p1 , . . . , pk be the number of prime divisors of K(t)/K which ramify in E. By assumption, k ≤ 3 and deg(pi ) = 1, i = 1, . . . , k. Since [E : K(t)] = 2, each pi extends uniquely to a prime divisor qi of E/K of degree 1. Since char(K) 6= 2, the ramification of pi in E is tame. By Riemann-Hurwitz (Remark 3.6.2(c)), 2g−2 = −4+k ≤ −1. Hence, g ≤ 12 . This implies, g = 0 and k = 2. It follows from Example 3.2.4 that E = K(u) with u transcendental over K. Proposition 16.7.6: Let K be a field and an integer n ≥ 2 with char(K) (n − 1)n. Then An is regular over K. Specifically, there is a tower of fields K(t) ⊆ K(u) ⊆ F satisfying: F/K is regular, F/K(t) is Galois, Gal(F/K(t)) ∼ = Sn , Gal(F/K(u)) ∼ = An , and K(t)/K has at most three prime divisors which ramify in F , each of degree 1. Proof: Let t and x be transcendental elements over K with xn − xn−1 = t. By Proposition 16.7.4, K(x)/K(t) is separable. Denote the Galois closure of ∼ ˜ K(t)) ˜ K(x)/K(t) by F . By Proposition 16.7.4, Gal(F/K(t)) ∼ = Gal(F K/ = Sn . In particular, F/K is regular. In addition, K(t)/K has at most three prime divisors p1 , p2 , p3 which may ramify in K(x), each of degree 1. Denote the fixed field of An in F by E. By Corollary 2.3.7(c), p1 , p2 , p3 are the

16.7 Symmetric and Alternating Groups over Hilbertian Fields

319

only prime divisors of K(t)/K which may ramify in F . Hence, they are the only prime divisors of K(t)/K which may ramify in E. By Lemma 16.7.5, E = K(u) with u transcendental over K and Gal(F/K(u)) ∼ = An , as desired. Next we use the polynomial X n − X n−1 − t to solve embedding problems √ of the form Sn → Gal(K( a)/K) when K is Hilbertian. Remark 16.7.7: On the discriminant of a polynomial. Let K be aQfield and n f ∈ K[X] a monic separable polynomial of degree n. Write f (X) = i=1 (X− xi ) with xi ∈ Ks . Put N = K(x1 , . . . , xn ) and embed Gal(N/K) into Sn by σ(i) = j if σ(xi ) = xj . Formula (1) of Section 6.1 for the discriminant disc(f ) of f is (2)

disc(f ) = (−1)

n(n−1) 2

n Y n(n−1) Y 2 (xi − xj ) = (−1) f 0 (xj ), j=1

i6=j

with f 0 the derivative of f . It can be rewritten as disc(f ) = Thus, Y p disc(f ) = (xi − xj ).

Q

i<j (xi

− xj )2 .

i<j

For each σ ∈ Gal(N/K) we have Y p p (xσ(i) − xσ(j) ) = (−1)sgn(σ) disc(f ). σ( disc(f )) = i<j

where sgn(σ) is the number of pairs (i, j) with i < j and σ(i) > σ(j). If char(K) 6= 2, then (−1)sgn(σ) 6= 1 for odd σ and (−1)sgn(σ) = 1 for even σ. In this case, p Gal N/K( disc(f ) ) = An ∩ Gal(N/K). p In particular, if Gal(N/K) = Sn , then K( disc(f )) is the unique quadratic p extension of K in N and Gal(N/K( disc(f ))) = An . Qn−1 Suppose now char(K) - n. Then f 0 (X) = n j=1 (X − yj ) with yj ∈ ˜ Substituting in (2) and changing the order of multiplication gives an K. alternative formula for the discriminant: (3)

disc(f ) = (−1)

n(n−1) 2

nn

n−1 Y

f (yj ).

j=1

The proof of part (a) of the following result gives an explicit polynomial over K(t) with Galois group An if char(K) - (n − 1)n.

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Chapter 16. Galois Groups over Hilbertian Fields

Proposition 16.7.8 (Brink): Let K be a Hilbertian field, a a nonsquare in K, and n ≥ 2 an integer with char(K) - (n − 1)n. Then (a) An is regular over K and √ (b) the embedding problem Sn → Gal(K( a)/K) is solvable. Proof: Let f (t, X) = X n − X n−1 − t and F be the splitting field of f (t, X) over K(t). By Proposition 16.7.4, Gal(F/K(t)) ∼ = Sn . Since char(K) - n, we n−1 0 n−2 have f (t, X) = n X − n X . Hence, by (3), n − 1 f (0)n−2 n i h n − 1 n n − 1 n−1 n(n−1) = (−1) 2 nn (−t)n−2 − −t n n

disc(f ) = (−1)

= (−1)

n(n−1) 2

nn f

(n−1)(n+2) 2

[(n − 1)n−1 tn−2 + nn tn−1 ].

p By Remark 16.7.7, K( disc(f )) is a quadratic extension of K(t) in F and p Gal F/K( disc(f ) ) ∼ = An . Let ( u=

disc(f )t1−n = (−1)

(n−1)(n+2) 2

2−n

(n−1)(n+2) 2

disc(f )t

= (−1)

[(n − 1)n−1 t−1 + nn ] n−1

[(n − 1)

n

+ n t]

if n is odd if n is even

√ and u0 = u. Then K(u) = K(t), so both u and u0 are transcendental over K. pMoreover, u is a product of disc(f ) with a square of K(t), so K(u0 ) = K( disc(f ) ). Thus, Gal(F/K(u0 )) ∼ = An . This proves (a). Next, express t in terms of u and substitute in f (t, X) to get the irreducible polynomial for the roots of f (t, X) in terms of u:

g(u, X) =

X n − X n−1 −

X n − X n−1 −

(n−1)n−1 (−1)(n−1)(n−2)/2 u−nn

if n is odd

(−1)(n−1)(n−2)/2 u−(n−1)n−1 nn

if n is even.

Then Gal(g(u, X), K(u)) = Sn and Gal(g((u0 )2 , X), K(u0 )) = An . √ Let v = ua and v 0 = u0 a. The map u 7→ v extends to an automorphism of K(u) over K. This automorphism further extends to an isomorphism of K(u0 ) onto K(v 0 ) and further to an isomorphism of F onto a field F 0 . Thus, Gal(F 0 /K(u)) ∼ = Sn and Gal(F 0 /K(v 0 )) ∼ = An . Since a is not a square in K, there is no c ∈ K with v = c2 u. Hence, K(u0 ) and K(v 0 ) are distinct quadratic extensions of K(u). Therefore, K(u0 ) 6⊆ F 0 . Put F 00 = K(u0 )F 0 . Then 00 0 0 /K(u0 )) ∼ Gal(F√ = Sn and Gal(F 00 /K(u0 , v 0 )) ∼ = An . In addition, K(u √, v ) = 0 K(u , a). By Lemma 16.4.2, the embedding problem Sn → Gal(K( a)/K) is solvable.

16.8 GAR-Realizations

321

16.8 GAR-Realizations Let K be a field, L a finite Galois extension of K, and (1)

α

1 −→ C −→ G −→ Gal(L/K) −→ 1

an embedding problem over K with kernel C having a trivial center. We give a sufficient condition for the embedding problem to have a regular solution. As usual, Aut(C) denotes the group of all automorphisms of C. For each c ∈ C let ι(c) be the inner automorphism of C induced by conjugation with c. The map c 7→ ι(c) identifies C with the group Inn(C) of all inner automorphisms of C. Let F/K be an extension of fields. We say F is rational over K (or F is K-rational) if F = K(T ) with T being a set of algebraically independent elements over K. Definition 16.8.1: GAR-Realizations. Let C be a finite group with a trivial center. We say C is GA over K if there are algebraically independent elements t1 , . . . , tr over K satisfying: (2) K(t) has a finite extension F which is regular over K such that Aut(C) acts on F with K(t) being the fixed field of C. Denote the fixed field of Aut(C) in F by E. We say C is GAR over K if in addition to (2) this holds: (3) Every extension E 0 of E satisfying E 0 Ks = Ks (t) is a purely transcen dental extension of E 0 ∩ Ks . Remark 16.8.2: (a) The “G” in GAR abbreviates “Galois”, “A” abbreviates “Automorphisms”, and “R” abbreviates “Rational”. (b) The GAR-condition implies that the fixed field E of Aut(C) is regular over K. But, it does not require E to be rational over K. This is however the case when r = 1 (L¨ uroth’s theorem, Remark 3.6.2(a)). Lemma Qr 16.8.3: Let C1 , . . . , Cr be finite non-Abelian simple groups. Put C = i=1 Ci . Then: (a) Each normal simple subgroup N of C coincides with Ci for some i between 1 and r. ∼ (b) Suppose Qr αi : C1 → Ci is an isomorphism, i = 1, . . . , r. Then Aut(C) = Sr n i=1 Aut(Ci ). Proof of (a): Choose n ∈ N , n 6= 1. Write n = c1 · · · cr with ci ∈ Ci . Assume, without loss, c1 6= 1. Since C1 is simple, its center is trivial. Hence, there is a c01 ∈ C1 with c1 c01 6= c01 c1 . For each i ≥ 2 we have c01 ci = ci c01 . Therefore, nc01 6= c01 n and (c01 )−1 n−1 c01 n1 6= 1. It follows that, N ∩ C1 6= 1. Consequently, N = C1 . Proof of (b): To simplify notation identify Ci with C1 via αi . Thus, each element of C is an r-tuple c = (c1 , . . . , cr ) with ci ∈ C1 . Each Ci is then the

322

Chapter 16. Galois Groups over Hilbertian Fields

group of all c with cj = 1 for j 6= i. We embed Aut(C1 )r and Sr in Aut(C) by the rules: cγ = (cγ11 , . . . , cγr r )

and

(c1 , . . . , cr )σ = (c1σ−1 , . . . , crσ−1 )

This gives: (4)

σ −1 γσ = (γ1σ−1 , . . . , γrσ−1 ).

Thus, Sr normalizes Aut(C1 )r . In addition observe that Aut(C1 )r ∩ Sr = 1. Finally, consider γ ∈ Aut(C). By (a), there is a unique σ ∈ Sr with Ciγ = Ciσ , i = 1, . . . , r. So, there are γ1 , . . . , γr in Aut(C1 ) with γ = (γ1 , . . . , γr )σ −1 . Thus, Aut(C) is the semidirect product of Aut(C1 )r with Sr , where the action of Sr on Aut(C1 )r given by (4). Remark 16.8.4: Minimal normal subgroups. Let G be a finite group and C a normal subgroup. Suppose C is a minimal normal subgroup of G. Thus, C 6= 1, C / G, and G has no normal subgroup in C other than 1 and C. Let C1 be a simple normal subgroup of C. The conjugates C1 , C2 , . . . , Cr of C1 in G generate a nontrivial normal subgroup of G which is contained in C. Hence, they generate C. Suppose in addition, C1 is non-Abelian. Let k be Qkan integer between 0 and r − 1. Suppose by induction hC1 , . . . , Ck i = i=1 Ci . By Lemma Qk Qk+1 16.8.3(a), Qr Ck+1 6≤ i=1 Ci . Hence, hC1 , . . . , Ck+1 i = i=1 Ci . In particular, C = i=1 Ci . By construction, G acts transitively on the set {C1 , . . . , Cr } by conjugation. Lemma 16.8.5 (Semilinear Rationality Criterion): Let K, L, E, F be fields with L/K Galois, K ⊆ E, E ∩ L = K, and EL = F . Let V be a L-subspace of F . Consider the K-subspace U = {v ∈ V | v σ = v for all σ ∈ Gal(F/E)} of E. (a) Suppose V is Gal(F/E)-invariant. Then each K-basis of U is an L-basis of V . Thus, V ∼ = U ⊗K L. (b) Suppose, in addition to the assumption of (a), F = L(V ). Then E = K(U ). (c) Suppose, in addition to the assumptions of (a) and (b), dim(V ) is finite and equal trans.deg(F/L). Then F is L-rational and E is K-rational. Proof of (a): By assumption, E is linearly disjoint from L over K. Thus, each K-basis of U is linearly independent over L. Hence, it suffices to prove that U spans V over L. To this end consider v ∈ V . Choose a finite Galois extension F0 of E in F containing v. Let L0 = F0 ∩ L. By assumption, F0 L = F and res: Gal(F/E) → Gal(L/K) is an isomorphism. Hence, res(Gal(F/F0 )) = Gal(L/L0 ) and res: Gal(F0 /E) → Gal(L0 /K) is an isomorphism. Put m = [F0 : E] = [L0 : K] and G = Gal(F0 /E).

16.8 GAR-Realizations

323

P Choose a basis c1 , . . . , cm for L0 /K. For each i let ui = σ∈G cσi v σ . σ Then ui ∈ U . Since det(ci ) 6= 0 [Lang7, p. 286, Cor. 5.4] and |G| = m, each v σ is a linear combination of the ui with coefficients which are rational functions in cτi , τ ∈ G, i = 1, . . . , m. In particular, v is in the L-vector space spanned by U , as desired. Proof of (b): By the tower property (Lemma 2.5.3), E is linearly disjoint from L(U ) over K(U ). By (a), F = L(V ) = L(U ). Therefore, E = K(U ). Proof of (c): Let u1 , . . . , un be a K-basis of U . By (a), u1 , . . . , un is an L-basis of V . By (b), F = L(u1 , . . . , un ). Since n = trans.deg(F/L), the elements u1 , . . . , un are algebraically independent over L. Therefore, F is L-rational. By (b), E = K(u1 , . . . , un ). Since F/E and L/K are algebraic, trans.deg(E/K) = trans.deg(F/L) = n. Consequently, E is K-rational.

Proposition 16.8.6 (Matzat): Consider an embedding problem (1), where Qr C a minimal normal subgroup of G and C = i=1 Ci with Ci simple nonAbelian and conjugate to C1 in G, i = 1, . . . , r. Suppose C1 is GAR over K. Then embedding problem (1) is regularly solvable over K. Proof: We break the proof into several parts. Part A: Group theory. Gal(L/K) with (5)

The map g 7→ (ι(g), α(g)) embeds G into Aut(C)×

G ∩ Aut(C) = C,

α = pr2 ,

For each i let Di = NG (Ci ). Put D = (5), (6)

pr2 (G) = Gal(L/K).

Tr

i=1

Di . Then C ≤ D ≤ Di ≤ G. By

Di ∩ Aut(Ci ) = Ci .

For each σ ∈ G, Remark 16.8.4 gives a unique π ∈ Sr with Ciσ = Ciπ , i = 1, . . . , r. Thus, Diσ = Diπ , i = 1, . . . , r. Therefore, Qr D is normal in G. Finally, as in Lemma 16.8.3(b), we embed i=1 Aut(Ci ) in Aut(C) by the rule: (c1 , . . . , cr )(γ1 ,...,γr ) = (cγ11 , . . . , cγr r ). Part B: GA-realization of C. The assumption on C1 gives algebraically independent elements t1 , . . . , tm over K and fields E1 , F1 satisfying this: (7a) K ⊆ E1 ⊆ K(t1 , . . . , tm ) ⊆ F1 , F1 /K regular, F1 /E1 Galois, (7b) Gal(F1 /E1 ) = Aut(C1 ), and Gal(F1 /K(t1 , . . . , tm )) = C1 . (7c) Suppose M is an extension of E1 with M Ks = Ks (t1 , . . . , tm ). Then M is a field of rational functions over M ∩ Ks .

324

Chapter 16. Galois Groups over Hilbertian Fields

Choose algebraically independent elements tij , i = 1, . . . , r, j = 1, . . . , m over K with t1j = tj , j = 1, . . . , m. For each i let ti = (ti1 , . . . , tim ). i = tij , j = Also, let αi : K(t1 ) → K(ti ) be the K-isomorphism with tα j 1, . . . , m. Put Ei = E1αi and extend αi to an isomorphism of F1 on a field Fi . The latter induces an isomorphism of Gal(F1 /K(t1 )) and Gal(F1 /E1 ) onto Gal(Fi /K(ti )) and Gal(Fi /Ei ), respectively. They are given by γ 7→ αi−1 γαi . Identify Gal(Fi /K(ti )) and Gal(Fi /Ei ) through this isomorphism with Ci and Aut(Ci ), respectively. Hence, Ei , K(ti ), Fi satisfy (7) with i replacing 1. Also, F1 , . . . , Fr are algebraically independent over K and regular over K (by (7a)). Therefore, they are linearly disjoint over K (Lemma 2.6.7). Next let E = E1 · · · Er , t = (t1 , · · · , tr ), and F = F1 · · · Fr . Then (8a) K ⊆ E ⊆ K(t) ⊆ F , F/K is regular, F/E is Galois, Qr Qr (8b) Gal(F/E) = i=1 Aut(Ci ), and Gal(F/K(t)) = i=1 Ci . The group Sr acts on F by permuting the triples (Ei , ti , Fi ), i = 1, . . . , r. −1 Namely, xπ = xαi αj for π ∈ Sr with iπ = j and for x ∈ Fi . This induces an action on Gal(F/E) which coincides with the action given by (4). (Recall: αi with the To obtain (4) we identify Ci with C1 via αi and then identify Qr identity map.) So, by Lemma 16.8.3(b), the action of i=1 Aut(Ci ) on F extends to an action of Aut(C) on F over K. Denote the fixed field of Aut(C) in F by E0 . Part C: The field crossing argument. By 7(a), F is linearly disjoint from L over K. Put Q = F L. Then Gal(Q/E0 ) ∼ = Aut(C) × Gal(L/K) with pr2 = resQ/L . Also, Q/L is regular. Identify Gal(Q/E0 L) with Aut(C) and Gal(Q/L(t)) with C. Part A identifies G with a subgroup of Gal(Q/E0 ). Denote the fixed field of G in Q by P . By (5), Gal(Q/P ) = G and restriction of G into Gal(L/K) is surjective. (9)

P L = L(t)

and P ∩ L = K.

Thus, all that remains to be proved is the rationality of P over K. Part D: New transcendence basis for L(t)/L. The group D1 acts on L because L/K is Galois. Denote the fixed field by M0 . Each σ ∈ D1 satisfies C1σ = C1 . Hence, the permutation of {1, . . . , r} corresponding to σ fixes 1, so F1σ = F1 . Thus, D1 acts on F1 L. Denote the fixed field by M . By (6), D1 ∩Aut(C1 ) = C1 . The fixed field of C1 in F1 L is L(t1 ) and that of Aut(C1 ) in F1 L is E1 L. Therefore, M E1 L = L(t1 ). Next observe that the restriction of D1 to F1 maps D1 into Aut(C1 ), hence into Gal(F1 /E1 ). Hence, E1 ⊆ M . Thus, M L = L(t1 ), so M Ks = Ks (t1 ). By construction, M ∩ L = M0 . It follows from (7c) that there are algebraically independent elements v1 , . . . , vm over K with M = M0 (v1 , . . . , vm ). They satisfy (10)

L(v1 , . . . , vm ) = L(t1 ).

16.9 Embedding Problems over Hilbertian Fields

325

Here is a partial diagram of the fields involved: F1

F1 L

K(t1 )

E1

M

K

M0

L(t1 ) y y yy yy y y

L

For each i between 1 and r choose σi ∈ G with C1σi = Ci (Remark 16.8.4). Then D1σi = Di . Put vij = vjσi , j = 1, . . . , m, and vi = (vi1 , . . . , vim ). Then let v = (v1 , . . . , vr ). By (10), L(vi ) = L(ti ), i = 1, . . . , r, so L(v) = L(t). Hence, the vij are algebraically independent over K. Since vj ∈ M , we have vjδ = vj for all δ ∈ D1 and j = 1, . . . , m. If (σ δσ −1 )σi

δ = vj i i 1 ≤ i ≤ r and δ ∈ Di , then σi δσi−1 ∈ D1 , hence vij Tr δ = vij for all i and j. It follows for δ = D = i=1 Di that vij

σ

= vj j = vij .

Part E: Rationality of P . Denote the fixed field of D in Q by N . By Part δ A, N/P is Galois. Put N0 = N ∩ L. By Part D, vij = vij for each δ ∈ D and all i, j. Thus, N0 (v) ⊆ N . As [L(v) : N0 (v)] = [L : N0 ] = [L(v) : N ], we have N = N0 (v). By (9), P N0 = N . P

N

L(t)

K

N0

L

Q

Let V be the vector space spanned by the vij ’s over N0 . For each σ ∈ G σ −1 σi0

and each i there exists i0 with Ciσ = Ci0 = Ci i −1 (σσi−1 0 σi )σi σi0

. Hence, σσi−1 0 σi ∈ Di , so

σi−1 σi0

σ vij = vij = vij = vi0 ,j . Thus, Gal(N/P ) leaves V invariant. By Part D, dim(V ) = rm = trans.deg(P/K). It follows from Lemma 16.8.5 that P is K-rational.

16.9 Embedding Problems over Hilbertian Fields An affirmative solution to the inverse problem of Galois theory, the realization of all finite groups over Q, seems at present to be out of reach. Matzat’s method of GAR-realization gives an effective tool for a partial solution of

326

Chapter 16. Galois Groups over Hilbertian Fields

the problem, namely for the of realization of finite groups with non-Abelian composition factors. Every finite group G has a sequence N0 , . . . , Nn of subgroups with N0 = G, Nn = 1, Ni / Ni−1 , and Ni−1 /Ni simple, i = 1, . . . , n. For each finite simple group C, the number of i with C ∼ = Ni−1 /Ni depends only on G and not on the sequence [Huppert, p. 64, Satz 11.7]. If this number is positive, C is a composition factor of G. Proposition 16.9.1 (Matzat): Let α: G → Gal(L/K) be a finite embedding problem over a field K. Suppose every finite embedding problem for G with an Abelian kernel is solvable. (a) If K is Hilbertian and each composition factor of Ker(α) is GAR over K, then α is solvable. (b) If each composition factor of Ker(α) is GAR over every extension of K, then α is regularly solvable. Proof: Assume without loss that Ker(α) is nontrivial. Choose a minimal normal subgroup C of G in Ker(α). Then α induces an isomorphism α0 : G/C → Gal(L/K). The rest of the proof splits into two parts. Part A: Proof of (a). An induction hypothesis gives a Galois extension L0 of K containing L and an isomorphism β 0 : Gal(L0 /K) → G/C with α0 ◦ β 0 = resL0 /L . 0 −1 Let π: G → G/C be the quotient map. Qr The kernel of (β ) ◦ π: G → 0 ∼ Gal(L /K) is C. By Remark 16.8.4, C = i=1 Ci with Ci simple non-Abelian and Ci conjugate to C1 in G, i = 1, . . . , r. In particular, C1 is a composition factor of G. By assumption, α is solvable if C1 is Abelian. Suppose C1 is nonAbelian. Then C1 is GAR over K. Hence, by Proposition 16.8.6, (β 0 )−1 ◦ π is regularly solvable. Since K is Hilbertian, (β 0 )−1 ◦ π is solvable (Lemma 16.4.2). In other words, K has a Galois extension N containing L0 and there is an isomorphism β: Gal(N/K) → G with (β 0 )−1 ◦ π ◦ β = resN/L0 . Then, α ◦ β = resN/L and β solves embedding problem α. Gal(N/K) res β 0 Gal(L /K) β 0 res α / Gal(L/K) G π α0 / Gal(L/K) G/C

16.9 Embedding Problems over Hilbertian Fields

327

Part B: Proof of (b). An induction hypothesis gives algebraically independent elements t1 , . . . , tm over K, a Galois extension E of K(t), and an epimorphism β 0 : Gal(E/K(t)) → G/C satisfying: E is a regular extension of L and α0 ◦ β 0 = resE/L . As above, let π: G → G/C be the quotient map. By assumption, each composition factor of C is regular over K(t). Proposition 16.8.6 gives algebraically independent elements u1 , . . . , un over K(t), a Galois extension F of K(t, u), and an isomorphism β: Gal(F/K(t, u)) → G satisfying: F is a regular extension of E and (β 0 )−1 ◦ π ◦ β = resF/E . Then F is a regular extension of L and β is a regular solution of embedding problem α. Proposition 16.9.2: Let K be a field and n a positive integer with char(K) - (n − 1)n and n 6= 2, 3, 6. Then An is GAR over K. Proof: Use the notation of Proposition 16.7.6. The assumption n 6= 2 implies that the center of Sn is trivial. The assumption n 6= 2, 3, 6 implies Sn = Aut(An ) [Suzuki, p. 299, Statement 2.17]. Hence, F/K(u) is a GArealization of An . We must still prove Condition (3) of Section 16.8. Consider an extension E 0 of K(t) with E 0 Ks = Ks (u). Let L = E 0 ∩ Ks . We have to prove E 0 is L-rational. To begin, note that E 0 ∩Ks (t) = L(t) and [E 0 : L(t)] = [Ks (u) : Ks (t)] = [K(u) : K(t)] = 2. In particular, E 0 is a function field of one variable over L. Since the genus of Ks (u) is 0, so is the genus of E 0 (Proposition 3.4.2(b)). Suppose p0 is a prime divisor of L(t)/L which ramifies in E 0 . Then deg(p0 ) = 1. Hence, there are exactly one prime divisor ps of Ks (t)/Ks over p0 (Proposition 3.4.2(e)) and ps ramifies in Ks (u) (Proposition 3.4.2(c)). Let p be the common restriction of ps and of p0 to K(t). Then p ramifies in K(u) (Proposition 3.4.2(c)). But there are at most three such p and each of them is of degree 1. Hence, there are at most three possibilities for p0 (Proposition 3.4.2(e)). It follows from Lemma 16.7.5 that E 0 is L-rational. The combination of Propositions 16.9.1 and 16.9.2 gives concrete solvable embedding problems with non-Abelian kernels. Proposition 16.9.3: Let K be a field and α: G → Gal(L/K) a finite embedding problem over K. Suppose each composition factor of Ker(α) is An with char(K) - (n − 1)n and n 6= 6. Then α is regularly solvable over K. If, in addition, K is Hilbertian, then α is solvable over K. Take L = K in Proposition 16.9.3: Corollary 16.9.4: Let K be a field and G a finite group. Suppose each composition factor of G is An with char(K) - (n − 1)n and n 6= 6. Then G is regular over K. If, in addition, K is Hilbertian, then G is realizable over K. Remark 16.9.5: More GAR-realizations. There is a long list of finite nonAbelian simple groups which are known to be GAR over Q. Beside An (with n 6= 6), this list includes PSL2 (Fp ) with p odd and p 6≡ ±1 mod 24 and

328

Chapter 16. Galois Groups over Hilbertian Fields

all sporadic groups with the possible exception of M23 . See [Malle-Matzat, Thm. IV.4.3]. The list becomes longer over the maximal Abelian extension Qab of Q. In addition to the groups that are GAR over Q it contains A6 , PSL2 (Fp ) with p odd, and M23 . See [Malle-Matzat, Thm. IV.4.6]. It is still unknown whether every finite non-Abelian simple group is GAR over Qab . If this is the case, each finite embedding problem over Qab would be solvable, as we will see in Example 24.8.5. It is even unknown if every finite non-Abelian simple group is GAR over ˜ An affirmative answer to this question would each field K containing Q. ˜ 1 , t2 )) which is enable us to solve embedding problems over the field Q((t Hilbertian by Example 15.5.2. The lists over Q and Qab have been established in large part by using the Riemann Existence Theorem. This partially explains the lack of knowledge of GAR realizations in characteristic p. An exception to our lack of knowledge of GAR realization is the family An of alternative groups. Theorem 15 of [Brink] improves Proposition 16.9.2 and proves that An is GAR over every field K if n 6= 2, 6 and char(K) 6= 2. The case char(K) = 2 is left open.

16.10 Finitely Generated Profinite Groups Let S be a subset of a profinite group G. Denote the closed subgroup generated by S by hSi. We say S generates G if hSi = G. In this case each map ϕ0 of S into a profinite group H has at most one extension to a (continuous) homomorphism ϕ: G → H. A profinite group G is finitely generated if it has a finite set of generators. In this case, the minimal number of generators of G is the rank of G. Note that the rank of a quotient of G does not exceed the rank of G. Example 16.10.1: Consider the group (Z/pZ)n as a vector space over Fp of dimension n. Each group theoretic set of generators of (Z/pZ)n generates (Z/pZ)n as a vector space over Fp . Hence, rank(Z/pZ)n = n. Since (Z/pZ)n is a quotient of Znp , the rank of the latter group is at least n. For each i between 1 and n consider the element ei = (0, . . . , 1, . . . , 0) with 1 in the ith coordinate and 0 elsewhere. Then e1 , . . . , en generates Znp . It follows that rank(Znp ) = n. Lemma 16.10.2: A finitely generated profinite group G has, for each positive integer n, only finitely many open subgroups of index at most n. Proof: Each open subgroup M of G of index ≤ n contains an open normal subgroup N with G/N isomorphic to a subgroup of Sn . Indeed, suppose m = (G : M ) ≤ n. Then, B = {gM | g ∈ G} is a set of order m. Multiplication from the left with an element x of G induces a permutation π(x) of B. Specifically, π(x)(gM ) = xgM . Thus, π is a

16.10 Finitely Generated Profinite Groups

329

T homomorphism of G into Sn with Ker(π) = x∈G M x . Therefore, Ker(π) is an open normal subgroup of G which is contained in M . Moreover, G/Ker(π) is isomorphic to a subgroup of Sm , hence to a subgroup of Sn . Thus, it suffices to prove G has only finitely many open normal subgroups N with G/N isomorphic to a subgroup of Sn . The map α 7→ Ker(α) maps the set of all homomorphisms α: G → Sn onto the set of all open normal subgroups N of G such that G/N is isomorphic to a subgroup of Sn . Hence, the number ν of those N ’s does not exceed the number of the α’s. Let S be a finite set of generators of G. Then every homomorphism α: G → Sn is determined by its values on S. Therefore, ν ≤ (n!)|S| . We call a profinite group G small if for each positive integer n the group G has only finitely many open subgroups of index n. By Lemma 16.10.2, every finitely generated profinite group is small. Thus, each of the results we prove in this section for small profinite groups holds for finitely generated profinite groups. Remark 16.10.3: Small profinite groups and open subgroups. Let G be a profinite group. (a) Denote the intersection of all open subgroups of G of index at most n by Gn . Then Gn is a closed normal subgroup of G. Moreover, G is small if and only if Gn is open in G for all n. (b) Let H be an open subgroup of index m of G. Consider open subgroups G0 and H 0 of G and H, respectively. Then (G : H 0 ) = (G : H)(H : H 0 ) and (H : H ∩ G0 ) ≤ (G : G0 ). Thus, Gmn ≤ Hn ≤ Gn . By (a), G is small if and only if H is small. (c) Let G and H be as in (b). Suppose H is finitely generated, say by h1 , . . . , hd . Let g1 , . . . , gn be representatives of G/H. Then g1 , . . . , gn , h1 , . . . , hd generate G. Conversely, if G is finitely generated, then H is finitely generated (Exercise 7). We prove a qualitative version of this result in Section 17.6. (d) Let α: G → H be an epimorphism of profinite groups. If g1 , . . . , gn are generators of G, then α(g1 ), . . . , α(gn ) are generators of H. If G is small, so is H. Indeed, let n be a positive integer and H0 an open subgroup of H of index n. Then G0 = α−1 (H0 ) is an open subgroup of G of index n. The map H0 7→ α−1 (H0 ) is injective. Since there are only finitely many G0 ’s, there are only finitely many H0 ’s. Example 16.10.4: A small profinite group which is not finitely generated. Q p Let A = Zp with p ranging over all prime numbers. Consider an open Q subgroup N of A of index n with n = p≤m pkp . Then N contains the Q Q open subgroup p≤m pkp Zpp × p>m Zpp . Hence, there are only finitely many possibilities for N . Thus, A is small. On the other hand, rank(A) ≥ rank(Zpp ) = p for each p (Example 16.10.1). Therefore, A is not finitely generated.

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Remark 16.10.5: Characteristic subgroups. A closed subgroup N of a profinite group G is characteristic if it is invariant under every automorphism of G. In particular, N is normal in G. Suppose G is small. Then, Gn is a characteristic open subgroup. The decreasing sequence G ≥ G2 ≥ G3 ≥ · · · intersects in 1. Thus, it consists of a basis of open neighborhoods of 1 in G. Let N be an open characteristic subgroup of G. Then Aut(G/N ) is a finite group. For each α ∈ Aut(G) define αN ∈ Aut(G/N ) by αN (gN ) = α(g)N . The map α 7→ αN gives a homomorphism ΦN : Aut(G) → Aut(G/N ). By the preceding paragraph, the intersection of all these N is the trivial group. Hence, the ΦN combine to an embedding Aut(G) → lim Aut(G/N ) ←− which is actually surjective. Thus, Aut(G) is a profinite group. The most useful properties of small profinite groups are embodied in the following result: Proposition 16.10.6: Let G be a small profinite group. Then: (a) Every epimorphism of G onto itself is an automorphism. (b) Let α: G → H and β: H → G be epimorphisms. Then both α and β are isomorphisms. Proof of (a): Let θ: G → G be an epimorphism. Let Gn be the finite set of all open subgroups of G of index at most n. The map H 7→ θ−1 (H) maps Gn injectively into itself. Hence, it maps Gn onto itself. Therefore, in the notation of Remark 16.10.3 \ \ \ Gn = H= θ−1 (H) = θ−1 ( H) = θ−1 (Gn ). H∈Gn

If θ(g) = 1, then g ∈

H∈Gn

T∞

n=1

θ−1 (Gn ) =

H∈Gn

T∞

n=1

Gn = 1. Therefore, θ is injective.

Proof of (b): By (a), β ◦ α is an isomorphism. Hence, both α and β are injective. For a profinite group G, we denote the set of all finite quotients (up to an isomorphism) of G by Im(G). Proposition 16.10.7: Let G and H be profinite groups with G small. (a) If Im(H) ⊆ Im(G), then H is a quotient of G. (b) If Im(H) = Im(G), then H is isomorphic to G. Proof of (a): First we prove H is small. Indeed, let n be a positive number and B1 , . . . , Br distinct open subgroups of H of indexTat most n. Choose an r open normal subgroup N of H which is contained in i=1 Bi . Then H/N ∈ Im(H). By assumption, G has an open normal subgroup M with G/M ∼ = H/N . Hence, G has r open subgroups of index at most n. Consequently, r is bounded. In the notation of Remark 16.10.3(a), the finite group H/Hn belongs to Im(H), and therefore to Im(G). Thus, G has an open normal subgroup K

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331

with G/K ∼ = H/Hn . In particular, K is an intersection of open subgroups of index ≤ n. Hence, Gn ≤ K. Therefore, there is an epimorphism from G/Gn to H/Hn . Denote the finite nonempty set of all epimorphisms of G/Gn → H/Hn by Φn . Let ϕ: G/Gn+1 → H/Hn+1 be an epimorphism. It maps the set of all subgroups of G/Gn of index at most n onto the set of all subgroups of H/Hn of index at most n. Hence, ϕ(Gn /Gn+1 ) ≤ Hn /Hn+1 . Therefore, ϕ induces an epimorphism ϕ: ¯ G/Gn → H/Hn . This defines a map Φn+1 → Φn . By Corollary 1.1.4, lim Φn is nonempty. Each element in lim Φn gives ←− ←− a compatible system of epimorphisms βn : G/Gn → H/Hn . It defines an epimorphism β: G → H, as desired. Proof of (b): Statement (a) gives epimorphisms ϕ: G → H and ψ: H → G. Thus, ψ ◦ ϕ: G → G and ϕ ◦ ψ: H → H are epimorphisms. By (a), H is small. Hence, by Proposition 16.10.6, both ψ◦ϕ and ϕ◦ϕ are automorphisms. Consequently, both ϕ and ψ are isomorphisms. Corollary 16.10.8: Let α: G → H be an epimorphism of profinite groups. Suppose G is small and Im(G) ⊆ Im(H). Then α is an isomorphism. Proof: Since G is small, so is H (remark 16.10.3(d)). By Proposition 16.10.7(a), G is a quotient of H. Therefore, by Proposition 16.10.6(b), α is an isomorphism. Example 16.10.9: Small Galois groups. ˜ (Section 1.5). Thus, Gal(K) is (a) For each finite field K, Gal(K) ∼ =Z generated by one element. (b) Let p be a prime number. The local compactness of Qp and Krasner’s lemma imply that Gal(Qp ) is small [Lang5, p. 54, Prop. 14]. Deeper arguments show Gal(Qp ) is generated by 4 elements [Jannsen, Satz 3.6]. (c) Let K be a number field, OK its ring of integers, and S a finite number of prime ideals of OK . Denote the maximal algebraic extension of K unramified outside S by KS . It is a Galois extension of K (Corollary 2.3.7(c)). We prove Gal(KS /K) is a small profinite group. Let T be the set of all prime numbers which ramify in K or lie under a prime ideal belonging to S. Then T is finite and KS ⊆ QT . Suppose we already know that Gal(QT /Q) is small. Then, by Remark 16.10.3(b),(d), Gal(QT /K) is small. Hence, Gal(KS /K) is also small. We may therefore assume K = Q and S is a finite set of prime numbers. Suppose L is a finite extension of Q in QS of degree at most n. By [Serre5, p. 130, Proposition 6], X log p + n|S| log n. log discriminant(L/Q) ≤ (n − 1) p∈S

Thus, discriminant(L/Q) is bounded. By Hermite-Minkowski [Lang5, p. 121, Thm. 5], there are only finitely many extensions of Q a given discriminant. Consequently, there are only finitely many possibilities for L.

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Alternatively, one may follow [Serre9, p. 107] and first observe that d = discriminant(L/Q) is divisible only by p ∈ S. For each such p, [LQp : Qp ] ≤ n. By (b), there are only finitely many possibilities for LQp , hence for the pth part of the different of L/Q. Therefore, there are only finitely many possibilities for d. It is, however, not clear whether Gal(KS /K) is finitely generated [Shafarevich1, §3]. (d) There are small Q absolute Galois groups which are not finitely generated. The group A = Zpp of Example 16.10.4 is one example. To construct a field with absolute Galois group A, we start from a field K of characteristic 0 that contains all roots of unity. Then Gal(K((t))) ∼ = ˆ [Geyer-Jarden2, Cor. 4.2]. Thus, for each p there is an algebraic Gal(K) × Z extension of K((t)) with absolute Galois group Gal(K) × Zp . In particular, taking K to be algebraically closed, we find a field K 0 with Gal(K 0 ) ∼ = Zp . all prime numbers and Induction gives fields Kp,i with p ranging overQ i = 1, . . . , p satisfying these conditions: Gal(Kp,i ) = l

A Series of Modern Surveys in Mathematics

Editorial Board M. Gromov, Bures-sur-Yvette J. Jost, Leipzig J. Kollár, Princeton H. W. Lenstra, Jr., Leiden J. Tits, Paris D. B. Zagier, Bonn G. Ziegler, Berlin Managing Editor R. Remmert, Münster

Volume 11

Michael D. Fried • Moshe Jarden

Field Arithmetic Third Edition Revised by Moshe Jarden

Michael D. Fried Department of Mathematics Montana State University – Billings Billings MT 59101 USA [email protected]

ISBN 978-3-540-77269-9

Moshe Jarden School of Mathematics Tel Aviv University Ramat Aviv, Tel Aviv 69978 Israel [email protected]

e-ISBN 978-3-540-77270-5

DOI 10.1007/978-3-540-77270-5 Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics ISSN 0071-1136 Library of Congress Control Number: 2008924174 Mathematics Subject Classiﬁcation (2000): 12E30 c 2008 Springer-Verlag Berlin Heidelberg This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microﬁlm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover design: WMXDesign GmbH, Heidelberg Printed on acid-free paper 9 8 7 6 5 4 3 2 1 springer.com

To those precious colleagues who can appreciate the goals of and connections to other areas. To those who acknowledge the depth of what we already know from the absorbed contribution of previous generations before we address our papers. To those who can transcend the hubris of today’s mathematical community.

Table of Contents Chapter 1. Infinite Galois Theory and Profinite Groups 1.1 Inverse Limits . . . . . . . . . . . . . . . 1.2 Profinite Groups . . . . . . . . . . . . . . 1.3 Infinite Galois Theory . . . . . . . . . . . . 1.4 The p-adic Integers and the Pr¨ ufer Group . . . 1.5 The Absolute Galois Group of a Finite Field . . Exercises . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . .

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Chapter 2. Valuations and Linear Disjointness . 2.1 Valuations, Places, and Valuation Rings . . 2.2 Discrete Valuations . . . . . . . . . . . 2.3 Extensions of Valuations and Places. . . . 2.4 Integral Extensions and Dedekind Domains 2.5 Linear Disjointness of Fields . . . . . . . 2.6 Separable, Regular, and Primary Extensions 2.7 The Imperfect Degree of a Field . . . . . 2.8 Derivatives . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .

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19 19 21 24 30 34 38 44 48 50 51

Chapter 3. Algebraic Function Fields of One Variable . . 3.1 Function Fields of One Variable . . . . . . . . . 3.2 The Riemann-Roch Theorem . . . . . . . . . . 3.3 Holomorphy Rings . . . . . . . . . . . . . . . 3.4 Extensions of Function Fields . . . . . . . . . . 3.5 Completions . . . . . . . . . . . . . . . . . . 3.6 The Different . . . . . . . . . . . . . . . . . 3.7 Hyperelliptic Fields . . . . . . . . . . . . . . . 3.8 Hyperelliptic Fields with a Rational quadratic Subfield Exercises . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . .

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52 52 54 56 59 61 67 70 73 75 76

Chapter 4. The Riemann Hypothesis for Function Fields . 4.1 Class Numbers . . . . . . . . . . . . . . . . . 4.2 Zeta Functions . . . . . . . . . . . . . . . . . 4.3 Zeta Functions under Constant Field Extensions . . 4.4 The Functional Equation . . . . . . . . . . . . 4.5 The Riemann Hypothesis and Degree 1 Prime Divisors 4.6 Reduction Steps . . . . . . . . . . . . . . . . 4.7 An Upper Bound . . . . . . . . . . . . . . . . 4.8 A Lower Bound . . . . . . . . . . . . . . . .

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Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 5. Plane Curves . . . . . . . 5.1 Affine and Projective Plane Curves 5.2 Points and prime divisors . . . . 5.3 The Genus of a Plane Curve . . . 5.4 Points on a Curve over a Finite Field Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .

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. 95 . 95 . 97 . 99 . 104 . 105 . 106

Chapter 6. The Chebotarev Density Theorem 6.1 Decomposition Groups . . . . . . . 6.2 The Artin Symbol over Global Fields . 6.3 Dirichlet Density . . . . . . . . . . 6.4 Function Fields . . . . . . . . . . 6.5 Number Fields . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . .

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107 107 111 113 115 121 129 130

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132 132 134 135 137 138 139 141 145 147 147 148

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149 149 152 154 156 157 159 160 161 162

Chapter 7. Ultraproducts . . . . . . . 7.1 First Order Predicate Calculus . . 7.2 Structures . . . . . . . . . . . 7.3 Models . . . . . . . . . . . . 7.4 Elementary Substructures . . . . 7.5 Ultrafilters . . . . . . . . . . 7.6 Regular Ultrafilters . . . . . . . 7.7 Ultraproducts . . . . . . . . . 7.8 Regular Ultraproducts . . . . . 7.9 Nonprincipal Ultraproducts of Finite Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .

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Chapter 8. Decision Procedures . . . . . . . . 8.1 Deduction Theory . . . . . . . . . . . 8.2 G¨odel’s Completeness Theorem . . . . . 8.3 Primitive Recursive Functions . . . . . . 8.4 Primitive Recursive Relations . . . . . . 8.5 Recursive Functions . . . . . . . . . . 8.6 Recursive and Primitive Recursive Procedures 8.7 A Reduction Step in Decidability Procedures Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .

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Chapter 9. Algebraically Closed Fields . 9.1 Elimination of Quantifiers . . . . 9.2 A Quantifiers Elimination Procedure 9.3 Effectiveness . . . . . . . . . . 9.4 Applications . . . . . . . . . . Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .

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163 163 165 168 169 170 170

Chapter 10. Elements of Algebraic Geometry . . 10.1 Algebraic Sets . . . . . . . . . . . . 10.2 Varieties . . . . . . . . . . . . . . . 10.3 Substitutions in Irreducible Polynomials . 10.4 Rational Maps . . . . . . . . . . . . 10.5 Hyperplane Sections . . . . . . . . . . 10.6 Descent . . . . . . . . . . . . . . . 10.7 Projective Varieties . . . . . . . . . . 10.8 About the Language of Algebraic Geometry Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .

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172 172 175 176 178 180 182 185 187 190 191

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192 192 193 199 201 203 207 211 217 218

Chapter 11. Pseudo Algebraically Closed Fields . . 11.1 PAC Fields . . . . . . . . . . . . . . . 11.2 Reduction to Plane Curves . . . . . . . . 11.3 The PAC Property is an Elementary Statement 11.4 PAC Fields of Positive Characteristic . . . 11.5 PAC Fields with Valuations . . . . . . . . 11.6 The Absolute Galois Group of a PAC Field . . . . . 11.7 A non-PAC Field K with Kins PAC Exercises . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . .

Chapter 12. Hilbertian Fields . . . . . . . . . . . . 12.1 Hilbert Sets and Reduction Lemmas . . . . . . 12.2 Hilbert Sets under Separable Algebraic Extensions 12.3 Purely Inseparable Extensions . . . . . . . . 12.4 Imperfect fields . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . .

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219 219 223 224 228 229 230

Chapter 13. The Classical Hilbertian Fields 13.1 Further Reduction . . . . . . . . 13.2 Function Fields over Infinite Fields 13.3 Global Fields . . . . . . . . . . 13.4 Hilbertian Rings . . . . . . . . 13.5 Hilbertianity via Coverings . . . .

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231 231 236 237 241 244

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13.6 Non-Hilbertian g-Hilbertian Fields 13.7 Twisted Wreath Products . . . 13.8 The Diamond Theorem . . . . 13.9 Weissauer’s Theorem . . . . . Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .

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248 252 258 262 264 266

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267 267 268 270 272 274 275 276

Chapter 15. Nonstandard Approach to Hilbert’s Irreducibility Theorem 15.1 Criteria for Hilbertianity . . . . . . . . . . . . 15.2 Arithmetical Primes Versus Functional Primes . . 15.3 Fields with the Product Formula . . . . . . . . 15.4 Generalized Krull Domains . . . . . . . . . . . 15.5 Examples . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . .

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277 277 279 281 283 286 289 290

Chapter 16. Galois Groups over Hilbertian Fields . . . . . . 16.1 Galois Groups of Polynomials . . . . . . . . . . . . 16.2 Stable Polynomials . . . . . . . . . . . . . . . . 16.3 Regular Realization of Finite Abelian Groups . . . . . 16.4 Split Embedding Problems with Abelian Kernels . . . 16.5 Embedding Quadratic Extensions in Z/2n Z-extensions . 16.6 Zp -Extensions of Hilbertian Fields . . . . . . . . . . 16.7 Symmetric and Alternating Groups over Hilbertian Fields 16.8 GAR-Realizations . . . . . . . . . . . . . . . . . 16.9 Embedding Problems over Hilbertian Fields . . . . . 16.10 Finitely Generated Profinite Groups . . . . . . . . 16.11 Abelian Extensions of Hilbertian Fields . . . . . . . 16.12 Regularity of Finite Groups over Complete Discrete Valued Fields . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .

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291 291 294 298 302 306 308 315 321 325 328 332

Chapter 14. Nonstandard Structures . 14.1 Higher Order Predicate Calculus 14.2 Enlargements . . . . . . . . 14.3 Concurrent Relations . . . . 14.4 The Existence of Enlargements 14.5 Examples . . . . . . . . . . Exercises . . . . . . . . . . . . Notes . . . . . . . . . . . . .

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Chapter 17. Free Profinite Groups . . . . . . . . . . . . . . . 338 17.1 The Rank of a Profinite Group . . . . . . . . . . . . . 338

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17.2 Profinite Completions of Groups . . . 17.3 Formations of Finite Groups . . . . . 17.4 Free pro-C Groups . . . . . . . . . . 17.5 Subgroups of Free Discrete Groups . . 17.6 Open Subgroups of Free Profinite Groups 17.7 An Embedding Property . . . . . . . Exercises . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . .

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340 344 346 350 358 360 361 362

Chapter 18. The Haar Measure . . . . . . . . . . 18.1 The Haar Measure of a Profinite Group . . . 18.2 Existence of the Haar Measure . . . . . . . 18.3 Independence . . . . . . . . . . . . . . . 18.4 Cartesian Product of Haar Measures . . . . . 18.5 The Haar Measure of the Absolute Galois Group 18.6 The PAC Nullstellensatz . . . . . . . . . . 18.7 The Bottom Theorem . . . . . . . . . . . 18.8 PAC Fields over Uncountable Hilbertian Fields 18.9 On the Stability of Fields . . . . . . . . . . 18.10 PAC Galois Extensions of Hilbertian Fields . 18.11 Algebraic Groups . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . .

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363 363 366 370 376 378 380 382 386 390 394 397 400 401

Chapter 19. Effective Field Theory and Algebraic Geometry . 19.1 Presented Rings and Fields . . . . . . . . . . . . . 19.2 Extensions of Presented Fields . . . . . . . . . . . 19.3 Galois Extensions of Presented Fields . . . . . . . . 19.4 The Algebraic and Separable Closures of Presented Fields 19.5 Constructive Algebraic Geometry . . . . . . . . . . 19.6 Presented Rings and Constructible Sets . . . . . . . 19.7 Basic Normal Stratification . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

. . . . . . . . . .

403 403 406 411 412 413 422 425 427 428

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Chapter 20. The Elementary Theory of e-Free PAC Fields 20.1 ℵ1 -Saturated PAC Fields . . . . . . . . . . . . 20.2 The Elementary Equivalence Theorem of ℵ1 -Saturated PAC Fields . . . 20.3 Elementary Equivalence of PAC Fields . . . . . . 20.4 On e-Free PAC Fields . . . . . . . . . . . . . 20.5 The Elementary Theory of Perfect e-Free PAC Fields 20.6 The Probable Truth of a Sentence . . . . . . . . 20.7 Change of Base Field . . . . . . . . . . . . . . . . . . . . . . . . 20.8 The Fields Ks (σ1 , . . . , σe )

. . . . .

. . . . 429 . . . . 429 . . . . . . . . . . . . . . . . . . . . . . . . . . .

430 433 436 438 440 442 444

xii

20.9 The Transfer Theorem . . . . . . . . 20.10 The Elementary Theory of Finite Fields Exercises . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . .

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446 448 451 453

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454 454 455 460 462 463 467 472 479 489 493 495

Chapter 22. Projective Groups and Frattini Covers 22.1 The Frattini Groups of a Profinite Group . 22.2 Cartesian Squares . . . . . . . . . . . 22.3 On C-Projective Groups . . . . . . . . 22.4 Projective Groups . . . . . . . . . . . 22.5 Frattini Covers . . . . . . . . . . . . 22.6 The Universal Frattini Cover . . . . . . 22.7 Projective Pro-p-Groups . . . . . . . . 22.8 Supernatural Numbers . . . . . . . . . 22.9 The Sylow Theorems . . . . . . . . . 22.10 On Complements of Normal Subgroups . 22.11 The Universal Frattini p-Cover . . . . . 22.12 Examples of Universal Frattini p-Covers . . . 22.13 The Special Linear Group SL(2, Zp ) 22.14 The General Linear Group GL(2, Zp ) . . Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .

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497 497 499 502 506 508 513 515 520 522 524 528 532 534 537 539 542

Chapter 23. PAC Fields and Projective Absolute Galois Groups 23.1 Projective Groups as Absolute Galois Groups . . . . . 23.2 Countably Generated Projective Groups . . . . . . . 23.3 Perfect PAC Fields of Bounded Corank . . . . . . . 23.4 Basic Elementary Statements . . . . . . . . . . . . 23.5 Reduction Steps . . . . . . . . . . . . . . . . . . 23.6 Application of Ultraproducts . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .

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544 544 546 549 550 554 558 561 561

Chapter 21. Problems of Arithmetical Geometry 21.1 The Decomposition-Intersection Procedure 21.2 Ci -Fields and Weakly Ci -Fields . . . . . . . 21.3 Perfect PAC Fields which are Ci 21.4 The Existential Theory of PAC Fields . 21.5 Kronecker Classes of Number Fields . . 21.6 Davenport’s Problem . . . . . . . . 21.7 On permutation Groups . . . . . . . 21.8 Schur’s Conjecture . . . . . . . . . . 21.9 Generalized Carlitz’s Conjecture . . . . Exercises . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . .

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Chapter 24. Frobenius Fields . . . . . . . . . . . . . . 24.1 The Field Crossing Argument . . . . . . . . . . . 24.2 The Beckmann-Black Problem . . . . . . . . . . 24.3 The Embedding Property and Maximal Frattini Covers 24.4 The Smallest Embedding Cover of a Profinite Group . 24.5 A Decision Procedure . . . . . . . . . . . . . . 24.6 Examples . . . . . . . . . . . . . . . . . . . . 24.7 Non-projective Smallest Embedding Cover . . . . . 24.8 A Theorem of Iwasawa . . . . . . . . . . . . . . 24.9 Free Profinite Groups of at most Countable Rank . . 24.10 Application of the Nielsen-Schreier Formula . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . .

. . . 562 . . . 562 . . . 565 . . 567 . . . 569 . . . 574 . . . 576 . . . 579 . . . 581 . . . 583 . . . 586 . . . 591 . . . 592

Chapter 25. Free Profinite Groups of Infinite Rank . . 25.1 Characterization of Free Profinite Groups by Embedding Problems 25.2 Applications of Theorem 25.1.7 . . . . . . . . 25.3 The Pro-C Completion of a Free Discrete Group . 25.4 The Group Theoretic Diamond Theorem . . . . 25.5 The Melnikov Group of a Profinite Group . . . 25.6 Homogeneous Pro-C Groups . . . . . . . . . 25.7 The S-rank of Closed Normal Subgroups . . . . 25.8 Closed Normal Subgroups with a Basis Element . 25.9 Accessible Subgroups . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . .

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595 601 604 606 613 615 620 623 625 633

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635 635 640 642

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Chapter 26. Random Elements in Free Profinite Groups . 26.1 Random Elements in a Free Profinite Group . . . 26.2 Random Elements in Free pro-p Groups . . . . . ˆn . . . . . . . . . . . . . 26.3 Random e-tuples in Z 26.4 On the Index of Normal Subgroups Generated by Random Elements . 26.5 Freeness of Normal Subgroups Generated by Random Elements Notes . . . . . . . . . . . . . . . . . . . . . . Chapter 27. Omega-Free PAC Fields . . . . . . . . . 27.1 Model Companions . . . . . . . . . . . . . 27.2 The Model Companion in an Augmented Theory of 27.3 New Non-Classical Hilbertian Fields . . . . . . 27.4 An abundance of ω-Free PAC Fields . . . . . . Notes . . . . . . . . . . . . . . . . . . . . .

. . . . 646 . . . . 651 . . . . 654

. . . . . . Fields . . . . . . . . .

. . 655 . . 655 . 659 . . 664 . . 667 . . 670

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Chapter 28. Undecidability . . . . . . . . . . . . . . 28.1 Turing Machines . . . . . . . . . . . . . . . 28.2 Computation of Functions by Turing Machines . . 28.3 Recursive Inseparability of Sets of Turing Machines 28.4 The Predicate Calculus . . . . . . . . . . . . 28.5 Undecidability in the Theory of Graphs . . . . . 28.6 Assigning Graphs to Profinite Groups . . . . . . 28.7 The Graph Conditions . . . . . . . . . . . . . 28.8 Assigning Profinite Groups to Graphs . . . . . . 28.9 Assigning Fields to Graphs . . . . . . . . . . . 28.10 Interpretation of the Theory of Graphs in the Theory of Fields Exercises . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . .

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671 671 672 676 679 682 687 688 690 694

. . . 694 . . . . 697 . . . . 697

Chapter 29. Algebraically Closed Fields with Distinguished Automorphisms 29.1 The Base Field K . . . . . . . . . . . . . . . . . 29.2 Coding in PAC Fields with Monadic Quantifiers . . . . ˜ σ1 , . . . , σe i’s . . . . . . 29.3 The Theory of Almost all hK, 29.4 The Probability of Truth Sentences . . . . . . . . .

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. 698 698 700 704 706

Chapter 30. Galois Stratification . . . . 30.1 The Artin Symbol . . . . . . . . 30.2 Conjugacy Domains under Projection 30.3 Normal Stratification . . . . . . 30.4 Elimination of One Variable . . . 30.5 The Complete Elimination Procedure 30.6 Model-Theoretic Applications . . . 30.7 A Limit of Theories . . . . . . . Exercises . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . .

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708 708 710 715 717 720 722 725 726 729

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Chapter 31. Galois Stratification over Finite Fields . 31.1 The Elementary Theory of Frobenius Fields . 31.2 The Elementary Theory of Finite Fields . . . 31.3 Near Rationality of the Zeta Function of a Galois Exercises . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . .

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. . 730 . . 730 . . 735 . 739 . . 748 . . 750

Chapter 32. Problems of Field Arithmetic . . . . 32.1 Open Problems of the First Edition . . . . 32.2 Open Problems of the Second Edition . . . 32.3 Open problems . . . . . . . . . . . . .

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751 751 754 758

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References Index

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Introduction to the Third Edition The third edition of “Field Arithmetic” improves the second edition in two ways. First it removes many typos and mathematical inaccuracies that occur in the second edition. In particular, it fills out a big gap in the References of the second edition, where unfortunately all references between “Gilmore and Robinson” and “Kantor and Lubotzky” are missing. Secondly, the third edition reports on five open problems of the second edition that were solved since that edition appeared in 2005. J´anos Koll´ar solved Problem 2 by proving that if each projective plane curve defined over a field K has a K-rational point, then K is PAC. J´anos Koll´ar also solved Problem 3 and proved that if K is a PAC field, ˜ and V is a variety defined over K, then V (K) is w is a valuation of K, ˜ w-dense in V (K). J´anos Koll´ar partially settled Problem 21. He proved that every PAC field of characteristic 0 is C1 . Problem 31 was affirmatively solved by Lior Bary-Soroker by establishing an analog of the diamond theorem for the finitely generated non-Abelian free profinite groups. Finally, Eric Rosen suggested to reorganize Corollary 28.5.3 of the second edition that led to an affirmative solution of Problem 33. Unfortunately, a full account of the first four solutions is out of the scope of the present volume. Much of the improvment made in the present edition is due to Arno Fehm and Dan Haran. I am really indebted to them for their contribution. Tel Aviv, Autumn 2007

Moshe Jarden

Introduction to the Second Edition The first edition of “Field Arithmetic” appeared in 1986. At the end of that edition we gave a list of twenty-two open problems. It is remarkable that since then fifteen of them were partially or fully solved. Parallel to this, Field Arithmetic has developed in many directions establishing itself as an independent branch of Algebra and Number Theory. Some of these developments have been documented in books. We mention here “Groups as Galois groups” [V¨ olklein] on consequences of the Riemann existence theorem, “Inverse Galois Groups” [Malle-Matzat] with a comprehensive report on finite Galois groups over number fields, “Profinite groups” [Ribes-Zalesskii] including the cohomology of profinite groups, “Analytic prop Groups” [Dixon-du.Sautoy-Mann-Segal] on closed subgroups of GL(n, Zp ), “Subgroup Growth” [Lubotzky-Segal] on counting the number of subgroups of finitely generating groups, and “Multi-Valued Fields” [Ershov7] on the model theory of fields with several valuations. This led to an official recognition of Field Arithmetic by the Mathematical Reviews in the form of MSC number 12E30.

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The extent which Field Arithmetic has reached makes it impossible for us to report in one extended volume about all exciting results which have been achieved. We have therefore made several choices which best suit the spirit of this book but do not extend beyond the scope of one volume. The new results and additional topics have made it necessary to reorganize and to enlarge the sections dealing with background material. Of course, we took the opportunity afforded by editing a second edition to correct flaws and mistakes which occurred in the first edition and to add more details to proofs wherever it seemed useful. We list the major changes and additions we made in the book: Chapter 2 has been reorganized. Sections 2.5–2.9 of the first edition, which survey the theory of algebraic function fields of one variable, were moved to Chapter 3. Sections 2.5–2.8 dealing with linear disjointness, regular extensions, and separability appeared in the first edition as sections 9.1– 9.3. A nice application of linear disjointness is Leptin’s construction (which preceded that of Warehouse) of a Galois group isomorphic to a given profinite group (Proposition 2.6.12). In addition to the introductory material about the theory of algebraic function fields of one variable, Chapter 3 now includes a proof of the RiemannHurwitz formula and a discussion of hyperelliptic curves. The proof of Theorem 4.9 of the first edition, estimating the number of zeros of an absolutely irreducible polynomial over a finite field, had a flaw. This has been fixed in the proof of Theorem 5.4.1. Likewise, the inequality given by [Fried-Jarden3, Prop. 5.16] is inaccurate. This inaccuracy is fixed in Proposition 6.4.8. We find it more convenient to use the language of algebraic sets as introduced in [Weil5] for model theoretic applications. Section 10.8 translates the basic concepts of that language to the now more commonly used language of schemes. Theorem 10.14 of the first edition (due to Frey-Prestel) says that the Henselian hull of a PAC field K is Ks . Proposition 11.5.3 (due to Prestel) ˜ for every valuation strengthens this theorem. It says that K is w-dense in K ˜ w of K. What we called “a separably Hilbertian field” in the first edition, is now called “a Hilbertian field” (Section 12.1). This agrees with the common usage and seems more appropriate for applications. Section 13.5 gives an alternative definition of Hilbertianity via coverings leading to the notion of “g-Hilbertianity”. This sets the stage for a generalization of a theorem of Zannier: Every global field has an infinite normal extension N which is g-Hilbertian but not Hilbertian (Theorem 13.6.2). Moreover, there is a unique factorization subring R of N with infinitely many irreducible elements (Example 15.5.8). This answers negatively Problems 14.20 and 14.21 of the first edition. Chapter 13 includes now one of the major results of Field Arithmetic which we call “Haran’s diamond theorem”: Let M1 and M2 be Galois ex-

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Introduction to the Second Edition

tensions of a Hilbertian field K and M a field between K and M1 M2 not contained in M1 nor in M2 . Then M is Hilbertian (Theorem 13.8.3). In particular, if N is a Galois extension of K, then N is not the compositum of two Galois extensions of K neither of which is contained in the other. This settles Problems 12.18 and 12.19 of the first edition. The immediate goal of Hilbert’s irreducibility theorem was to realize the groups Sn and An as Galois groups over Q. Chapter 16 is dedicated to realizations of Galois groups over arbitrary Hilbertian fields. One of the most important of these results is due to Harbater (Proposition 16.12.1): Let K be a complete valued field, t an indeterminate, and G a finite group. Then G is regular over K, that is, K(t) has a finite Galois extension F , regular over K, with Gal(F/K(t)) ∼ = G. Unfortunately, none of the three proofs of this theorem fits into the scope of this book. Section 16.6 proves a theorem of Whaples: Let K be a field and p a prime number. Suppose Z/pZ (resp. Z/4Z if p = 2) occurs as a Galois group over K. Then Zp is realizable over K. Section 16.7 generalizes a theorem of Hilbert: Let K be a field and n ≥ 2 an integer with char(K) - (n − 1)n. Then An is regular over K. One of the most far-reaching attempts to realize arbitrary finite Galois groups over Hilbertian fields uses Matzat’s notion of GAR realization of simple finite groups: Let K be a Hilbertian field and α: G → Gal(L/K) a finite embedding problem over K. Suppose every composition factor of Ker(α) has a GAR realization over K. Then the embedding problem is solvable. This leads in particular to the realization of many finite groups over Q (Remark 16.9.5). Chapter 17 deals mainly with Melnikov’s formations C (i.e. sets consisting of all finite groups whose composition factors belong to a given set of finite simple groups). We prove that every free abstract group F is residuallyC. Thus, if the free pro-C group with a given rank m exists, then the canonical injection of F into Fˆm (C) is injective (Proposition 17.5.11 – Ribes-Zalesskii). Konrad Neumann improved former results of Fried-Geyer-Jarden and proved that every field is stable (Theorem 18.9.3). This allows the construction of PAC Hilbertian Galois extensions of arbitrary countable Hilbertian fields (Theorems 18.10.2 and 18.10.3). We survey Neumann’s proof in Section 18.9. The full proof unfortunately falls outside the scope of this book. It seemed to be well known that the concept of absolute irreducibility of a variety is elementary. Unfortunately, we could find no solid proof for it in the literature. Proposition 19.5.9 fills in the gap by proving that result. Section 21.2 includes now the classical results about Ci -fields and not only the corresponding results about weakly Ci -fields as was the case in Section 19.2 of the first edition. Sections 21.8 gives a complete proof of Schur’s Conjecture: If f (X) is a polynomial with coefficients in a global field K with char(K) - deg(f ) and f permutes OK /p for infinitely many primes p of K, then each composition factor of f is linearly related over K to a Dickson polynomial of a prime

Introduction to the Second Edition

xix

degree. Section 21.7 proves all lemmas about permutation groups which are used in the proof of Schur’s Conjecture (Theorem 21.8.13). This includes the classification of subgroups of AGL(1, Fl ) (Lemma 21.7.2), and the theorems of Schur (Proposition 21.7.7) and Burnside (Proposition 21.7.8) about doubly transitive permutation groups. Section 21.9 contains the Fried-Cohen version of Lenstra’s proof of the generalized Carlitz’s Conjecture: Let p be a prime number, q a power of p, and f ∈ Fq [X] a polynomial of degree n > 1 which is not a power of p. Suppose f permutes infinitely many finite extensions of Fq . Then gcd(n, q − 1) = 1. The universal Frattini p-cover of a finite group plays a central role in Fried’s theory of modular towers. Section 22.11 introduces the former concept and proves its basic properties. Corollary 22.13.4 shows then that PSL(2, Zp ) is a p-Frattini cover of PSL(2, Fp ) although it is not the universal p-Frattini cover. Chapter 23 puts together material on PAC fields which appeared in Section 20.5 and Chapter 21 of the first edition. The Beckmann-Black Problem is a refinement of the inverse problem of Galois Theory. D´ebes proved that the problem has an affirmative solution over PAC fields (Theorem 24.2.2). Chapter 25 substantially extends the study of free profinite groups F of infinite rank which appeared in Section 24.4 of the first edition. Most of the material goes back to Melnikov. We characterize closed normal subgroups of F by their S-ranks, and prove that a closed subgroup of F is accessible if and only if it is homogeneous. The first part of Chapter 25 reproduces the group theoretic version of Haran’s diamond theorem. Chapter 26 is completely new. It describes the properties of the closed subgroup hxi and the closed normal subgroup [x] generated by a random etuple x = (x1 , . . . , xe ) of elements of a finitely generated free profinite group F of finite rank n ≥ 2. For example, with probability 1, hxi ∼ = Fˆe (Proposition 26.1.7). This solves Problem 16.16 of the first edition. In addition, with a positive probability, [x] has infinite rank and is isomorphic to Fˆω (Theorem 26.4.5 and Corollary 26.5.7). The latter result is based on the Golod-Shafarevich Inequality. Chapter 28 considers an infinite field K which is finitely generated over its base field. It proves that for e ≥ 2 the theory of all sentences θ which hold ˜ σ1 , . . . , σe i with (σ1 , . . . , σe ) ∈ Gal(K)e is undein almost all structures hK, ˜ σ 1 , . . . , σe i cidable. Moreover, the probability that a sentence θ hold in hK, is in general a nonrational number. Perhaps the most significant achievement of Field Arithmetic since the first edition appeared is the solution of Problem 24.41 of that edition: The absolute Galois group of a countable PAC Hilbertian field is free of rank ℵ0 . It was originally proved in characteristic 0 with complex analysis by Fried-V¨olklein. Then it was proved in the general case by Pop using rigid geometry and by Haran-Jarden-V¨ olklein using “algebraic patching”. The two

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Introduction to the First Edition

latter methods also lead to the proof that Gal(C(t)) is a free profinite group if C is an arbitrary algebraically closed field (Harbater, Pop, Haran-Jarden). The method of Fried-V¨ olklein led to the theory of modular towers of Fried. A remote goal in Galois theory is the classification of absolute Galois groups among all profinite groups. In this framework, one tries to construct new absolute Galois groups out of existing ones. For example, for all fields K1 , . . . , Kn there exists a field K with Gal(K) isomorphic to the free product of Gal(K1 ), . . . , Gal(Kn ) (Pop, Melnikov, Ershov, Koenigsmann). Generalization of this result to infinite families of closed subgroups generalize the concepts “projective groups” and “PRC fields” or “PpC fields” to “relatively projective groups” and “pseudo closed fields” (Haran-Jarden-Pop). They generalize the classification of projective groups as those profinite groups appearing as absolute Galois groups of PAC fields. All of the exciting material mentioned in the preceding two paragraphs lie unfortunately outside the scope of this volume. It is my pleasure to thank colleagues and friends who critically read parts of the manuscript of the present edition of “Field Arithmetic”: Michael Bensimhoun, David Brink, Gregory Cherlin, Michael Fried, Wulf-Dieter Geyer, Peter M¨ uller, Dan Haran, Wolfgang Herfort, Alexander Lubotzky, Nikolay Nikolov, Dan Segal, Aharon Razon, and Irene Zimmermann. Tel Aviv, Spring 2004

Moshe Jarden

Introduction to the First Edition Our topic is the use of algebraic tools — coming mainly from algebraic geometry, number theory, and the theory of profinite groups — in the study of the elementary properties of classes of fields, and related algorithmic problems. (We take the precise definition of “elementary” from first order logic.) This subject has its more distant roots in Tarski’s observation that, as a consequence of elimination theory, the full elementary theory of the class of all algebraically closed fields is decidable; this relies on the Euclid algorithm of finding the greatest common divisor of two polynomials in one variable over a field. In its first phase this line of thought led to similar results on real closed fields and p-adic fields. The subject took a new turn with the work of James Ax [Ax2] on the elementary theory of the class of finite fields, which represents a radical departure in terms of the algebraic methods used. The analysis is based entirely on three properties of a finite field K: (1a) K is perfect. (1b) K has a unique extension of each degree. (1c) There is an explicitly computable function q(d, m) such that any absolutely irreducible variety V defined over K will have a K-rational point if |K| > q(dim(V ), deg(V )).

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The validity of the third condition for finite fields is a consequence of Riemann’s hypothesis for curves over finite fields. Methods of logic, specifically ultraproducts, led Ax to consider this condition for infinite fields as well, in which case the lower bound afforded by the function q is vacuous, and the condition becomes: (2) Every absolutely irreducible variety over K has a K-rational point. Fields satisfying (2) are said to be pseudo algebraically closed, or PAC. The second condition may be interpreted as a description of the absolute Galois group Gal(K) as a profinite group: Gal(K) is the free profinite group on one generator. In Ax’ approach it was convenient to have an Abelian absolute Galois group, but a strong trend in later work has been the systematic analysis of situations involving progressively more general Galois groups. One of our central goals here is the presentation of the general theory of PAC fields in its modern form, and its connections with other branches of algebra. From what we have said so far, some connections with algebraic geometry and profinite groups are visible; a number theoretic connection will appear shortly. One important feature of PAC fields is that they occur in profusion in nature and are in fact typical in the following sense. Since the absolute Galois group Gal(Q) of the rationals is a compact topological group, it carries a canonical invariant probability measure, the Haar measure. We can therefore ˜ ask for the probability that the fixed field Q(σ) of a sequence σ = (σ1 , ..., σe ) ˜ will be PAC; and we find that this occurs with probof automorphisms of Q ˜ ability 1. In addition, the absolute Galois group of Q(σ) is free on the e generators σ1 , ..., σe , again with probability 1. These facts are consequences of Hilbert’s irreducibility theorem for Q (Chapter 13), at least in the context of countable fields. We will develop other connections between the PAC property and Hilbertianity. There are also remarkable connections with number theory via the Chebotarev density theorem (Chapters 6, 13, 16, 20, 21, 31). For example, the ˜ probability that a given elementary statement ψ holds for the field Q(σ) coincides with the Dirichlet density of the set of primes for which it holds for the field Fp , and this density is rational. Thus, the “probability 1” theory of ˜ the fixed fields Q(σ) coincides with the theory of “all sufficiently large” finite fields, which by Ax’ work is an algorithmically decidable theory. ˜ Ax’ results extend to the “probability 1” theory of the fields Q(σ) for σ of length e > 1, by somewhat different methods (Chapter 20), although the connection with finite fields is lost. The elementary theories of such fields are largely determined by three properties: PAC, characteristic zero, and having an absolute Galois group which is free on e generators. To determine the full elementary theory of one such field K, it is also necessary to describe the ˜ intersection K ∩ Q. Although the absolute Galois group of a PAC field need not be free, it can be shown to be projective in a natural sense, and conversely any

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projective profinite group occurs as the Galois group of some PAC field. In extending the theory from PAC fields with free Galois group to the general (projective) case, certain obstacles arise: for example, the algorithmic results do not extend. There is nonetheless a quite general theory, which enables us to identify some broad classes of projective profinite groups for which the associated classes of profinite groups behave well, and also to pinpoint unruly behavior in other case. One approach to the algorithmic problems associated with PAC fields leads to the study of profinite groups G with the embedding property (the terminology reflects a preoccupation with the corresponding fields): for each pair of continuous epimorphisms ϕ: G → A, α: B → A, where B is a finite quotient of G, we require that ϕ should factor through α. A perfect PAC field whose absolute Galois group is a group with the embedding property is called a Frobenius field. The elementary theory of all Frobenius fields can be computed quite explicitly. The algorithm has some relationship with elimination theory as used by Tarski. We associate to each elementary statement in the language of PAC fields a stratification of affine space into basic normal locally closed algebraic sets, each equipped with a Galois extension of its function field, and the given statement is reinterpreted as a statement about conjugacy classes of subgroups of the specified Galois groups. When the initial statement has no quantifiers this is a fairly trivial procedure, but addition of quantifiers corresponds to a special kind of “projection” of these Galois stratifications. This procedure has not yet been closely examined from the point of view of computational complexity. Like most procedures which operate by tracing through a series of projections, it is effective but hopelessly inefficient in its present form. It is not yet clear whether it is substantially less efficient than Tarski’s procedure for algebraically closed fields, nor whether, like that procedure, it can significantly reorganized and sped up. The Galois stratification algorithm relies on techniques of effective algebraic geometry, and also involves substantial algorithmic problems of a new type connected with the theory of profinite groups. Specifically, it is necessary to determine, given two collections A1 , ..., Am and B1 , ..., Bn of finite groups, whether or not there is a projective group with the embedding property which has each Ai as (continuous) image, but none of the groups Bj . The solution to this problem depends on recent work on projective covers (Chapter 22) and embedding covers (Chapter 24). Ultimately our decision problem reduces to the determination of the finite quotients of the projective cover of the embedding cover of a single finite group. The theory of projective covers leads also to the undecidability results alluded to earlier. A fairly natural encoding of graphs into profinite groups is lifted by this theory into the class of projective profinite groups, and then by looking at the corresponding PAC fields we see that their elementary theories encode algorithmically undecidable problems (the analogous results for graphs are well known).

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In the final chapter we return to our point of departure, the theory of finite fields. The zeta function of a Galois formula over a finite field is defined, and using a result of Dwork and Bombieri we show that some integral power of each such function is a product of an exponential and a rational function over Q. One of the goals of this book is to serve as a bridge between algebraists and logicians. For the algebraist there is a self contained introduction to the logic and model theory background for PAC fields (Chapter 7). Chapter 14 gives the “nonstandard” framework that suffices for Weissauer’s proof of Hilbert’s irreducibility theorem (Chapter 15), and Chapters 8 and 28 include basic recursion theory. On the other hand, for logicians with basic algebraic background (e.g. Lang’s book “Algebra”) Chapter 4 has the Stepanov-Bombieri elementary proof of the Riemann hypotheses for curves, and Chapter 6 gives an elementary proof of the Chebotarev density theorem. Both groups of readers may find the extensive treatment of profinite groups (Chapters 1, 17, 18, 22, 24, 25 and 26) and of Hilbertian fields (Chapters 12, 13, 15, and 16) valuable. Although PAC fields arise over arithmetically rich fields, they themselves lack properties that we associate with the arithmetic, say, of the rationals. For example, a PAC field F admits no orderings and all Henselizations of F are separably closed (Section 11.5). Many PAC field results generalize to pseudo real closed (PRC) fields. A field F is PRC if each absolutely irreducible variety defined over F has an F -rational point provided it has a nonsingular Fˆ -rational point in each real closure Fˆ of F . Thus, a PRC field without orderings is PAC. This, and the development of the theory of pseudo p-adically closed PpC fields are outside the scope of this book. We refer to [Prestel1], to [Jarden12], [Haran-Jarden2], and to [Haran-Jarden3] for literature about PRC fields and to [Haran-Jarden4] for PpC fields. Similarly, we give no account of the theories of real closed fields and p-adically closed fields that preceded the development of the theory of PAC fields. In particular, for Hilbert’s 17th problem and the Ax-Kochen-Ershov p-adic theory, we refer the reader to [Prestel2], [Ax-Kochen1, Ax-Kochen2, and Ax-Kochen3], and [Prestel-Roquette]. Acknowledgement: We are indebted to several colleagues who corrected errors in the process of critically reading the manuscript. In particular, WulfDieter Geyer, Gregory L. Cherlin, and Dan Haran made crucial contributions. Michael D. Fried, Gainesville, Florida Moshe Jarden, Tel Aviv, Israel Summer 1986

Notation and Convention Z = the ring of rational integers. Zp = the ring of p-adic integers. Q = the field of rational numbers. R = the field of real numbers. C = the field of complex numbers. Fq = the field with q elements. Ks = the separable closure of a field K. Kins = the maximal purely inseparable extension of a field K. ˜ = the algebraic closure of a field K. K Gal(L/K) = the Galois group of a Galois extension L/K. ˜ We call a polynomial f ∈ K[X] separable if f has no multiple root in K. Gal(f, K) = the Galois group of a separable polynomial f ∈ K[X] over a field K viewed as a permutation group of the roots of f . Gal(K) = Gal(Ks /K) = the absolute Galois group of a field K. irr(x, K) = the monic irreducible polynomial of an algebraic element x over a field K. Whenever we form the compositum EF of field extensions of a field K we tacitly assume that E and F are contained in a common field. |A| = #A = the cardinality of a set A. R× = the group of invertible elements of a ring R. Quot(R) = the quotient field of an integral domain R. A ⊂ B means “the set A is properly contained in the set B”. ax = x−1 ax, for elements a and x of a group G. H x = {hx | h ∈ H}, for a subgroup H of G. Given subgroups A, B of a group G, we use “A ≤ B” for “A is a subgroup of B” and “A < B” for “A is a proper subgroup of B”. Given an Abelian (additive) group A and a positive integer n, we write An for the Ssubgroup {a ∈ A | na = 0}. For a prime number p we let ∞ Ap∞ = i=1 Api . For a group B that acts on a group A from the right, we use B n A to denote the semidirect product of A and B. Bold face letters stand for n-tuples, e.g. x = (x1 , . . . , xn ). ord(x) is the order of an element x in a group G. For a positive integer n and an integer a with gcd(a, n) = 1, we use ordn a to denote the order of a modulo n. Thus, ordn a is the minimal positive integer d with ad ≡ 1 mod n. In the context of groups, Sn (resp. An ) stands for the full permutation group (resp. alternative group) of {1, . . . , n}. In the context of groups, Cn stands for the cyclic multiplicative group of order n. Likewise we use Z/nZ for the additive multiplicative group of order n. In S the context of fields, ζn stands for a primitive root of unity of order n. · i∈I Bi is the disjoint union of sets Bi , i ∈ I.

Chapter 1. Infinite Galois Theory and Profinite Groups The usual Galois correspondence between subgroups of Galois groups of finite Galois extensions and intermediate fields is not valid for infinite Galois extensions. The Krull topology restores this correspondence for closed subgroups (Proposition 1.3.1). Since Galois groups are inverse limits of finite groups, they are profinite. Conversely, we define profinite groups, independently of Galois theoretic properties. Each profinite group actually appears as a Galois group ˆ and (Corollary 1.3.4). In particular, we study the procyclic groups Zp and Z prove that every finite field has the latter as its absolute Galois group.

1.1 Inverse Limits Our interest in inverse limits comes from infinite Galois theory: Infinite Galois groups are inverse limits of finite Galois groups. As a preparation to the study of “profinite groups” we define in this section inverse limits of topological spaces and characterize inverse limits of finite topological spaces. Let I be a set with a partial ordering ≤; that is, ≤ is a binary relation which is reflexive, transitive, and a ≤ b and b ≤ a imply a = b. We call (I, ≤) a directed partially ordered set if in addition (1) for all i, j ∈ I there exists k ∈ I with i ≤ k and j ≤ k. An inverse system (also called a projective system) over a directed partially ordered set (I, ≤) is a data (Si , πji )i,j∈I where Si is a set and πji : Sj → Si is a map for all i, j ∈ I with i ≤ j satisfying the following rules: (2a) πii the identity map for each i ∈ I. (2b) πki = πji ◦ πkj if i ≤ j ≤ k. Q Let S be the subset of the cartesian product i∈I Si consisting of all elements Q s = (si )i∈I with πji (sj ) = si for all i ≤ j. Note: S may be empty. Let pri : j∈I Sj → Si be the projection on the ith coordinate. Denote the restriction of pri to S by πi . Then πi = πji ◦ πj for every i ≤ j. We say (S, πi )i∈I is the inverse (or projective) limit of the family (Si )i∈I with respect to the maps πji . Denote S by lim Si . ←− 0 )i,j∈I be another inverse system over I. Suppose for each Let (Si0 , πji 0 ◦ θj = θi ◦ πji for all i ≤ j. (We i ∈ I we are given a map θi : Si → Si0 with πji say that the maps θi , i ∈ I, are compatible.) Then there exists a unique map θ: lim Si → lim Si0 satisfying πi0 ◦ θ = θi ◦ πi for each i ∈ I: θ maps ←− ←− s = (si ) ∈ lim Si onto θ(s) with θ(s)i = θi (si ). Denote θ by lim θi . ←− ←− Similarly, let X be a set and for each i ∈ I let θi : X → Si be a map satisfying πji ◦ θj = θi whenever i ≤ j (Again, we say that the maps θi ,

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Chapter 1. Infinite Galois Theory and Profinite Groups

i ∈ I, are compatible.) Then there exists a unique map θ: X → lim Si with ←− πi ◦ θ = θi for each i ∈ I. When Si are topological spaces, we assume πji are continuous. Then we equip lim Si with the topology induced from the product topology of ←− Q Q i∈I Si . Recall Q that the product topology on i∈I Si has a basis consisting of the sets i∈I Ui , with Ui open in Si for each i ∈ I, and Ui = Si for all but finitely many i ∈ I. Since pri is continuous, so is πi , i ∈ I. If θi : Si → Si0 are continuous, then θ: lim Si → lim Si0 is also continuous. ←− ←− Lemma 1.1.1: The collection of all subsets of S = lim Si of the form πi−1 (Ui ) ←− with Ui open in Si is a basis for the topology of S. Proof: Let s ∈ S. each basic open neighborhood of s has the Q Q By definition, V × form V = S ∩ j∈J j i∈I r J Sj , where J is a finite subset of I and subset of S , j ∈ J. Take k ∈ I with k ≥ j for all j ∈ J. Vj is an open j T −1 (Vj ) is an open subset of Sk and πk−1 (Uk ) is an open Then Uk = j∈J πkj neighborhood of s in V . Therefore, the collection πi−1 (Ui ), i ∈ I, is a basis for the topology of S. Lemma 1.1.2: In the notation above, if each Si , i ∈ I, is a Hausdorff space, Q then lim Si is a closed subset of Si . ←− Q Proof: Suppose s = (si ) ∈ Si does not belong to lim Si . Then there are ←− Ui i, j ∈ I with i ≤ j and πji (sj ) 6= si . Take open disjoint neighborhoods Q −1 0 and Ui0 of si and πji (sj ), respectively. Then U × π (U ) × S is an i k i ji k6 = i,j Q open neighborhood of s in Si that does not intersect lim Si . ←− Q

If, in addition, each Si is compact, then Tychonoff’s theorem implies Si is also compact. Thus, lim Si with the induced topology is compact. ←−

Lemma 1.1.3: The inverse limit S of an inverse system of nonempty compact Hausdorff spaces Si , i ∈ I, is a nonempty compact Hausdorff space. T Proof: We only need Q to prove that S is nonempty. Indeed, S = k≥j Q Rkj , Si → where Rkj = {s ∈ Si | πkj (sk ) = sj }. The natural map prk × prj : Sk × Sj is continuous. The Hausdorff property of Sj implies T = {(sk , sj ) ∈ Sk × Sj | πkj (sk ) = sj } is a closed Q subset of QSk × Sj . Hence, Rkj = (prk × prj )−1 (T ) is a closed subset of Si . Since Si is compact, we only need to show that the intersection of finitely many of the Rkj is nonempty. Indeed, let j1 ≤ k1 , . . . , jn ≤ kn be n pairs in I. Choose l ∈ I with ki ≤ l, i = 1, . . . , n, and choose sl ∈ Sl . Define sji = πl,ji (sl ) and ski = πl,ki (sl ), for i = 1, . . . , n. For each r ∈ I r 1 , . . . , jn , k1 , . . . , kn } let sr an arbitrary element of Sr . T{j n Then s = (si ) ∈ i=1 Rki ,ji .

1.1 Inverse Limits

3

Corollary 1.1.4: The inverse limit of an inverse system of nonempty finite sets is nonempty. Proof: Equip each of the finite sets with the discrete topology.

0 )i,j∈I be inverse systems of Corollary 1.1.5: Let (Si , πji )i,j∈I and (Si0 , πji 0 compact Hausdorff spaces. Let θi : Si → Si be a compatible system of surjective continuous maps. Put S = lim Si , S 0 = lim Si0 , and θ = lim θi . Then ←− ←− ←− θ: S → S 0 is surjective.

Proof: Let s0 = (s0i ) be an element of S 0 . Then (θi−1 (s0i ), πji )i,j∈I is an inverse system of nonempty compact Hausdorff spaces. By Lemma 1.1.3, the inverse limit of θi−1 (s0i ) is nonempty. Each element in the inverse limit is mapped by θ onto s0 . Corollary 1.1.6: Let X be a compact space, (Si , πji )i,j∈I an inverse system of Hausdorff spaces, and θi : X → Si a compatible system of continuous surjective maps. Put θ = lim θi . Then θ: X → lim Si is surjective. ←− ←− Consider s = (si )i∈I ∈ lim Si . Then θi−1 (si ) is a closed nonempty ←− subset of X. For i1 , . . . , in ∈ I there is a j ∈ I with i1 , . . . , in ≤ j. Then (si1 ) ∩ · · · ∩ θi−1 (sin ), so θi−1 (si1 ) ∩ · · · ∩ θi−1 (sin ) is nonempty. θj−1 (sj ) ⊆ θi−1 1 n 1 n Since X is compact, there is an x ∈ X which belongs to θi−1 (si ) for every i ∈ I. It satisfies, θ(x) = s. Thus, θ is surjective. Proof:

A profinite space is an inverse limit of an inverse system of finite discrete spaces. Lemma 1.1.7: A compact Hausdorff space S is profinite if and only if its topology has a basis consisting of open-closed sets. Proof: Suppose first S = lim Si is an inverse limit of finite discrete spaces ←− Si . Let πi : S → Si be the projection on the ith coordinate, i ∈ I. By Lemma 1.1.1, the sets πi−1 (Ui ), where i ∈ I and Ui is a subset of Si , form a basis of the topology of S. Since Si is discrete, Ui is open-closed. Hence, so is πi−1 (Ui ). Now suppose the topology of S has a basis consisting of open-closed sets. An open-closed partition of S is a finite set A of nonempty openclosed disjoint subsets of S whose union is S. Denote the set of all openclosed partitions of S by A. Let A, A0 , B ∈ A. Write A ≤ B if for each B ∈ B there is an A ∈ A with B ⊆ A. Then A is unique. Next note that C = {A ∩ A0 | A ∈ A, A0 ∈ A0 } belongs to A and satisfies A, A0 ≤ C. Thus, (A, ≤) is a directed partially ordered set. When A ≤ B, define a map πB,A from B to A by πB,A (B) = A, where A is the unique element of A containing B. Equip each A ∈ A with the discrete topology. Then (A, πB,A )A,B∈A is an inverse system of finite discrete spaces.

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Chapter 1. Infinite Galois Theory and Profinite Groups

Its limit S 0 = lim A is a profinite space. We construct a homeomorphism of ←− S onto S 0 . Let s ∈ S and A ∈ A. Define θA (s) to be the unique A in A which contains s. Then θA : S → A is a continuous surjective map. If A ≤ B, then πB,A ◦ θB = θA . Hence, by Corollary 1.1.6, there is a continuous surjective map θ: S → S 0 satisfying πA ◦ θ = θA for each A ∈ A. Suppose s, s0 are distinct elements of S. Since S is Hausdorff, there are disjoint open-closed subsets A and A0 of S with s ∈ A and s0 ∈ A0 . Put A00 = S r(A ∪ A0 ). Then A = {A, A0 , A00 } is an open-closed partition of S, θA (s) = A, and θA (s0 ) = A0 . Thus, θA (s) 6= θA (s0 ), so θ(s) 6= θ(s0 ). Therefore, θ is bijective. Since S is compact and S 0 is Hausdorff, θ is a homeomorphism. Consequently, S is a profinite space. Remark 1.1.8: Totally disconnected spaces. Let S be a compact Hausdorff space. Suppose S has a basis for its topology consisting of open-closed sets. It is not difficult to see that S is totally disconnected; that is each s in S is its own connected component. Conversely, if S is totally disconnected, then the topology of S has a basis consisting of open-closed subsets [Ribes-Zalesski, Thm. 1.1.12].

1.2 Profinite Groups We survey here the basic properties of compact groups. In particular, this will apply to profinite groups. Topological Groups. A topological group is a group G equipped with a topology in which the product (x, y) 7→ xy and the inverse map x 7→ x−1 are continuous. It follows that for each a ∈ G the maps x 7→ ax, x 7→ xa, and x 7→ x−1 are homeomorphisms. We always assume {1} is a closed subset of G. Consequently, {a} is a closed subset of G for each a ∈ G. It follows that G is a Hausdorff space. Indeed, let a, b be distinct elements of G. The identity 1 · 1−1 = 1 and the continuity of the group operations give open neighborhoods U and W of 1 with U W −1 ⊆ G r{a−1 b}. Thus, aU and bW are disjoint open neighborhoods of a and b, respectively, as needed. In addition, each closed subgroup H of GSof finite index is open. Indeed, there are ai ∈ G, i ∈ I, with I finite and G = · i∈I ai H. Each of theSsets ai H is closed and there is a j ∈ I with aj H = H. Therefore, H = G r · i6=j ai H is open. Conversely, if G is compact, then every open subgroup H of G is of finite index. Otherwise, G would be a disjoint union of infinitely many cosets of H, each of which is open. Let N be a closed normal subgroup of G and π: G → G/N the quotient ¯ of G/N map. Equip G/N with the quotient topology. Thus, a subset U −1 ¯ is open if and only if π (U ) is open. It follows that the group operations of G/N are continuous. In addition, a subset C¯ of G/N is closed if and

1.2 Profinite Groups

5

¯ is closed. In particular, since N = π −1 (1) is closed, {1} only if π −1 (C) is a closed subset of G/N . Consequently, G/N is a topological group and π: G → G/N is a continuous map. Moreover, if U is an open subset of G, S then π −1 (π(U )) = n∈N nU , so π(U ) is open in G/N . Therefore, π is an open map. Suppose again G is compact. Let θ be a continuous homomorphism of G into a topological group H. Since H is Hausdorff, θ is a closed map. Suppose in addition, θ is surjective. Put N = Ker(θ). Let π: G → G/N ¯ G/N → H the map induced by θ. Then θ¯ is be the quotient map and θ: an isomorphism of abstract groups. In addition, θ¯ is a continuous bijective map of the compact group G/N onto the Hausdorff group H. Hence, θ¯ is an isomorphism of topological groups and θ is an open map. This is the first isomorphism theorem for compact groups. Let H be a closed subgroup and N a closed normal subgroup of G. Then HN = {hn | h ∈ H, n ∈ N } is the image of the compact group H × N under the continuous map (h, n) 7→ hn. Hence, HN is a closed subgroup of G. As in the preceding paragraph, the group theoretic isomorphism θ: HN/N → H/H ∩ N defined by θ(hN ) = h(H ∩ N ), h ∈ H, is a homeomorphism, so θ is an isomorphism of topological groups. Similarly, if M and N are closed normal subgroups of G with N ≤ M , then M/N is a closed normal subgroup of G/N and the map G/M → (G/N )/(M/N ) given by gM 7→ (gN )M/N , g ∈ G, is an isomorphism of topological groups. Here is one way to equip an abstract group G with a topology. Let N be a family of normal subgroups of G, closed under finite intersections, such that the intersection of all N ∈ N is 1. Take N to be a basis for the open neighborhoods of 1. A basis for the open neighborhoods of a ∈ G is the family Na = {aN | N ∈ N }. The union of the Na is then a basis for a group topology of G. For example, the identity xN · yN = xyN for normal subgroups N implies multiplication is continuous. Let (Gi , πji )i,j∈I be an inverse system of topological groups and continuous homomorphisms πji : Gj → Gi , for each i, j ∈ I with j ≥ i. Then G = lim Gi is a topological group and the projections πi : G → Gi are con←− 0 ii,j∈I be another system of topological tinuous homomorphisms. Let hG0i , πji 0 0 groups with G = lim Gi . Suppose θi : Gi → G0i , i ∈ I, is a compatible system ←− of continuous homomorphisms. Then the corresponding map θ: G → G0 is a continuous homomorphism. Profinite Groups. In this book, we primarily consider an inverse system of finite groups (Gi , πji )i,j∈I , each equipped with the discrete topology. We call the inverse limit G = lim Gi a profinite group. By Lemma 1.1.3, G is ←− a compact group. By Lemma 1.1.1, the open-closed sets πi−1 (gi ), gi ∈ Gi , i ∈ I, form a basis for the topology on G. In particular, the open normal subgroups of G form a basis for the open neighborhoods of 1.

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Chapter 1. Infinite Galois Theory and Profinite Groups

Remark 1.2.1: Basic rules. Here are some basic rules for profinite groups G and H which we use in this book without explicit reference: (a) A subgroup H of G is open if and only if H is closed of a finite index. The intersection of all open normal subgroups of G is 1. Every open subset of G is a union of cosets gi Ni with Ni open normal and gi ∈ G. (b) Every profinite group is compact, Hausdorff, and has a basis for its topology consisting of open-closed sets (Lemmas 1.1.3 and 1.1.7). (c) A subset C of a profinite group is closed if and only if C is compact (use (b)). (d) A subset B of a profinite group is open-closed if it is a union of finitely many cosets gi N with N open normal and gi ∈ G (use (a) and the compactness of B). (e) Every homomorphism ϕ: G → H is tacitly assumed to be continuous. In particular, ϕ maps compact subsets of G onto compact subsets of H. Hence, ϕ maps closed subsets of G onto closed subsets of H (use (c)). (f) By the first isomorphism theorem for compact groups, every epimorphism ϕ: G → H of profinite groups is an open map. In particular, ϕ maps open subgroups of G onto open subgroups of H. We list below special properties of profinite groups not shared by all compact groups: Lemma 1.2.2: Let {Hi | i ∈ I} be a directed family of closed subset of a profinite group G; that is: T (1) for every finite subset J of I there is an i ∈ I with Hi ≤ j∈J Hj . T Put H = i∈I Hi . Then: (a) For every open subgroup U of G containing H, there is an i ∈ I with H Ti ≤ U . T (b) i∈I KHi = KH and i∈I Hi K = HK for T every closedTsubset K of G. ¯ be an epimorphism. Then ϕ( (c) Let ϕ: G → G i∈I Hi ) = i∈I ϕ(Hi ). (d) Let K and K 0 be closed subgroups of G. Then the set S = {σ ∈ G | K σ = K 0 } is closed. (e) Let K and K 0 be closed subgroups of G which contain H. Suppose each Hi is normal and KHi and K 0 Hi are conjugate. Then K and K 0 are conjugate. (f) Let G = lim Gi be an inverse limit of finite groups, πi : G → Gi the ←− quotient maps, and K, K 0 closed subgroups of G. Suppose πi (K) and πi (K 0 ) are conjugate in Gi for each i. Then K and K 0 are conjugates. T Proof of (a): The set Hi ∩ (G r U ) is closed for every i ∈ I and i∈I Hi ∩ r (G r U ) = H ∩ (G T U ) = ∅. Since G is compact, there exists a finite subset J of ITsuch that j∈J Hj ∩ (G r U ) = ∅. By (1) there exists i ∈ I such that Hi ≤ j∈J Hj . Hence, Hi ≤ U . T Proof of (b): Let g ∈ i∈I KHi . Then, for each i ∈ I there are ki ∈ K and hi ∈ Hi with g = ki hi . Hence, the closed subset Hi ∩ g −1 K of G is nonempty.

1.2 Profinite Groups

7

−1 By (1), the intersection of finitely many T of the sets Hi ∩ g K is nonempty. Since G is compact, there exists h ∈ i∈I Hi ∩ g −1 K. It satisfies h ∈ H and h = g −1 k for some k ∈ K. Therefore, g = kh−1 ∈ KH.

Proof of (c): Let K = Ker(ϕ). Then ϕ induces a bijection between the set of closed subgroups lying between K and G and the set of closed subgroups ¯ Hence, by (b), ϕ(H) = ϕ(KH) = ϕ(T KHi ) = T ϕ(KHi ) = of G. i∈I i∈I T i∈I ϕ(Hi ). Proof of (d): Denote the set of all open normal subgroups of G by N . For each N ∈ N , let SN be the inverse image under the quotient map G → G/N of the finite set {s ∈ G/N | (KN/N )s = K 0 N/N }. If s ∈ SN for each N ∈ T N , then, by (b), K s = Then, SN is closed. T T s 0 0 N ∈N K N = N ∈N K N = K . Hence, S = N ∈N SN . Therefore, S is closed. Proof of (e): For each i ∈ I, Si = {σ ∈ G | K σ Hi = K 0 Hi } is a nonempty T subset of G. By (d), Si is closed. In theTnotation of (1) we have Si ⊆ j∈J Sj . Hence, by compactness, there is a σ ∈ i∈I Si . Thus, K σ Hi = K 0 Hi for each i ∈ I. We conclude from (b) that K σ = K 0 . Proof of (f): Apply (e) to Hi = Ker(πi ).

Lemma 1.2.3: Each closed subgroup H of a profinite group G is the intersection of open subgroups. Proof: Let T N be the set of all open normal subgroups of G. By Lemma 1.2.2(b), N ∈N N H = H. The following result gives a sufficient condition for a compact group to be profinite. We use it below to prove that the category of profinite groups is closed under various natural operations. Lemma 1.2.4: Let G be a compact group and {Ni | i ∈ I} a directed family of closed T normal subgroups of G of finite index satisfying (2) i∈I Ni = 1. Then G = lim G/Ni and G is a profinite group. ←− Proof: By definition, lim G/Ni is a profinite group. We only need to prove ←− the isomorphism. The quotient maps G → G/Ni define a continuous embedding θ of G into lim G/Ni . If (gi Ni )i∈I is an element of the latter group, ←− then the closed subsets gi Ni ofT G have the finite intersection property. Since G is compact, there exists g ∈ i∈I gi Ni . Then, θ(g) = (gi Ni )i∈I . Thus, θ is bijective. Compactness of G and the Hausdorff property of lim G/Ni imply ←− that θ is an isomorphism of topological groups.

8

Chapter 1. Infinite Galois Theory and Profinite Groups

Lemma 1.2.5: The following statements hold for each closed subgroup H of a profinite group G: (a) The group H is profinite. Moreover, for each open normal subgroup M of H there exists an open normal subgroup N of G with H ∩ N ≤ M . If M is normal in G, then N can be chosen such that H ∩ N = M . (b) Every open subgroup H0 of H is the intersection of an open subgroup of G with H. (c) Every homomorphism ϕ0 of H into a finite group A extends to a homomorphism ϕ: H 0 → A, where H 0 is an open subgroup of G containing H. If H is normal in G, then H 0 can be chosen to be normal. Proof of (a): If N is an open normal subgroup of G, then H ∩ N is an open normal subgroup of H. The family of all groups H ∩ N is directed and its intersection is trivial. Hence, H = lim H/H ∩ N is a profinite group. ←− Let M be an open normal subgroup of H. By Lemma 1.2.2(a), H ∩ N ≤ M for some open normal subgroup N of G. If M / G, then M N / G and H ∩ MN = M. Proof of (b): The intersection of all conjugates of H0 in H is an open normal subgroup of H. Hence, by the first statement of the lemma, there is an open normal subgroup N of G such that M = H ∩ N ≤ H0 . Then G0 = H0 N is an open subgroup of G and H ∩ G0 = H0 . Proof of (c): By (a), G has an open normal subgroup N with H ∩ N ≤ Ker(ϕ0 ). Put H 0 = HN and define ϕ: H 0 → A by ϕ(hn) = ϕ0 (h). Then ϕ is a well defined homomorphism which extends ϕ0 . Lemma 1.2.6: These statements hold: (a) If N is a closed normal subgroup of a profinite group G, then G/N is profinite. Q (b) The cartesian product, G = i∈I Gi , of profinite groups is profinite. (c) Every inverse limit, G = lim Gi , of profinite groups is profinite. ←− Proof of (a): A closed normal group N is an intersection of open subgroups (Lemma 1.2.3) and therefore of open normal subgroups Ni , i ∈ I, of G. Now apply Lemma 1.2.4 to G/N and {Ni /N | i ∈ I}. Proof of (b): Consider a finite subset J of Q I. For each Q j ∈ J let Nj be an open normal subgroup of Gj . Then N = j∈J Nj × i∈I r J Gi is an open normal subgroup of G. The family of all these normal subgroups is directed and its intersection is 1. By Lemma 1.2.4, G = lim G/N is a profinite group. ←− Proof of (c): Observe that G is a closed subgroup of the cartesian product Q i∈I Gi (Lemma 1.1.2). By (b) this product is profinite. Now apply Lemma 1.2.5.

1.3 Infinite Galois Theory

9

Lemma 1.2.7: Each epimorphism ϕ: G → A of profinite groups has a continuous set theoretic section ϕ0 : A → G. That is, ϕ0 is a continuous map satisfying ϕ ◦ ϕ0 = idA . Proof: Let K = Ker(ϕ). We split the proof into two parts. Part A: K is finite. Lemma 1.2.5 gives an open subgroup H of G with K ∩ H = 1. Thus, ϕ maps H bijectively onto B = ϕ(H). As both H and B are Hausdorff and compact, ϕ|H is a homeomorphism. Set β = (ϕ|H )−1 . Next S n let (A : B) = n, choose a1 , . . . , an ∈ A and g1 , . . . , gn ∈ G with A = · i=1 ai B and ϕ(gi ) = ai , i = 1, . . . , n. Define ϕ0 : A → G by ϕ0 (ai b) = gi β(b), b ∈ B. Then ϕ0 is a continuous set theoretic section of ϕ. Part B: K is arbitrary. Denote the set of all closed normal subgroups of G which are contained in K by L. For each L ∈ L, let ϕL : G/L → A be the epimorphism induced by ϕ. For each L0 ∈ L with L0 ≤ L let ϕL0 ,L : G/L0 → G/L be the quotient map. Let Φ0 be the set of all pairs (L, ϕ0L ) where L ∈ L and ϕ0L : A → G/L is a set theoretic section of ϕL . Since ϕK : G/K → A is an isomorphism, it has an inverse ϕ0K . Hence, (K, ϕ0K ) is in Φ0 . Define a partial ordering on Φ0 as follows: (L0 , ϕ0L0 ) ≤ (L, ϕ0L ) if L0 ≤ L and ϕL0 ,L ◦ ϕ0L0 = ϕ0L . Suppose Φ00 = {(Li , ϕ0i ) | i ∈ I} is a descending chain in Φ0 ; that is, every two elements of Φ00 are comparable. Put L00 = T 00 lim G/Li . Thus, i∈I Li . Then, by Lemmas 1.2.6(a) and 1.2.4, G/L = ←− 0 0 the compatible maps ϕi : A → G/Li give a section ϕL00 : A → G/L00 to ϕL00 . Thus, (L00 , ϕ0L00 ) is a lower bound for Φ00 . Zorn’s lemma gives a minimal element (L, ϕ0L ) of Φ0 . Assume L 6= 1. Then L has a proper open subgroup L0 which is normal in G. Consider the epimorphism ϕL0 ,L : G/L0 → G/L with the finite kernel L/L0 . Part A gives a set theoretic section ϕ0L,L0 to ϕL0 ,L . Set ϕ0L0 = ϕ0L,L0 ◦ ϕ0L . Then (L0 , ϕ0L0 ) is an element of Φ0 which is smaller than (L, ϕ0L ). This contradiction to the minimality of (L, ϕ0L ) proves that L = 1. Put ϕ0 = ϕ0L . Then ϕ0 is a continuous set theoretic section of ϕ.

1.3 Infinite Galois Theory Let N be a Galois extension of a field K. The Galois group Gal(N/K) associated with N/K consists of all automorphisms of N that fix each element of K. If N/K is a finite extension and H1 , H2 are subgroups of Gal(N/K) with the same fixed fields in N , then H1 = H2 . This is not the case any more if N/K is infinite. Consider S∞ for example the case where K = Fp for some prime number p and N = i=1 Fpi . Let ϕ be the Frobenius automorphism of N/Fp . It is defined by the rule ϕx = xp for each x ∈ N . Let G0 be the discrete subgroup of Gal(N/Fp ) generated by ϕ. It is a countable group and Fp is its fixed field in N . On the other hand, each element of Gal(Fp2i /Fp ) has exactly

10

Chapter 1. Infinite Galois Theory and Profinite Groups

two extensions to Fp2i+1 . Hence, there are 2ℵ0 sequences (σ1 , σ2 , σ3 , . . .) with σi ∈ Gal(Fp2i /Fp ) such that the restriction of σi+1 to Fp2i is σi for i = 1, 2, 3, . . . . Each such sequence defines a unique σ ∈ Gal(N/Fp ) whose restriction to Fp2i is σi , i = 1, 2, 3, . . . . It follows that the cardinality of Gal(N/Fp ) is 2ℵ0 . In particular, Gal(N/Fp ) is different from G0 but has the same fixed field, namely Fp . The Galois correspondence is restored for closed subgroups of Gal(N/K) in the “Krull topology” which we now introduce. Denote the set of all intermediate fields K ⊆ L ⊆ N , with L/K finite and Galois, by L. To each L ∈ L associate the (finite) Galois group Gal(L/K). If L0 ∈ L and L ⊆ L0 , then resL : Gal(L0 /K) → Gal(L/K) is an epimorphism. Consider the inverse limit lim Gal(L/K), with L ranging over L. Every σ ∈ ←− Gal(N/K) defines a unique element (resL σ)L∈L of lim Gal(L/K). Con←− versely, every (σL )L∈L ∈ lim Gal(L/K) defines a unique σ ∈ Gal(N/K) ←− with resL σ = σL for each L ∈ L. Thus, σ 7→ (resL σ)L∈L is an isomorphism Gal(N/K) ∼ = lim Gal(L/K). This isomorphism induces a topology ←− on Gal(N/K) through the topology on lim Gal(L/K): the Krull topology. ←− Thus, under the Krull topology, Gal(N/K) becomes a profinite group and the family N = {Gal(N/L) | L ∈ L} is a basis for the open neighborhoods of 1. If N/K is a finite extension, then the Krull topology is discrete. Suppose L is a finite extension of K contained in N . Then its Galois ˆ is the smallest Galois extension of K that contains L. It is finite closure L over K and is contained in N . Write Gal(N/L) as a union of right cosets of ˆ to see that Gal(N/L) is an open closed subgroup of Gal(N/K). Gal(N/L) Suppose L is an arbitrary extension of K in N . Then L is the union of a T family {Li | i ∈ I} of finite extensions of K. Hence, Gal(N/L) = i∈I Gal(N/Li ). Therefore, Gal(N/L) is a closed subgroup of Gal(N/K). If S is a set of automorphisms of N , then N (S) = {x ∈ N | σx = x for every σ ∈ S} is the fixed field of S in N . N (S) is also the fixed field in N of the closed subgroup hSi of Gal(N/K) generated by S. If S = {σ1 , . . . , σe } is a finite set, replace N (S) by N (σ1 , . . . , σe ). Let M be a Galois extension of K in N . Denote by resM (or res) the homomorphism from Gal(N/K) into Gal(M/K) that maps σ ∈ Gal(N/K) onto its restriction resM σ to M . It is a continuous surjective map. Proposition 1.3.1: Let N be a Galois extension of a field K. Then L 7→ Gal(N/L) is a bijection from the family of fields L lying between K and N onto the family of closed subgroups of G = Gal(N/K). The inverse map is H 7→ N (H). Proof: Consider a field extension L of K in N . Put H = Gal(N/L). Then L ⊆ N (H). Each x ∈ N (H) is contained in a finite Galois extension M ⊆ N

1.3 Infinite Galois Theory

11

of L. Since the map res: Gal(N/L) → Gal(M/L) is surjective, σx = x for every σ ∈ Gal(M/L). By finite Galois theory, x ∈ L. Hence, N (Gal(N/L)) = L. Conversely, let H be a closed subgroup of G and put L = N (H). Then H ≤ Gal(N/L). Consider σ ∈ Gal(N/L). In order to prove σ ∈ H it suffices to show that σ is in the closure of H. Indeed, let M ⊆ N be a finite Galois extension of K. Then M ∩ L = M (resM H). Finite Galois theory shows resM σ ∈ Gal(M/M ∩ L) = resM H. Hence, H ∩ σGal(N/M ) is nonempty. It follows that Gal(N/N (H)) = H. As in finite Galois theory [Lang7, pp. 192–199], Proposition 1.3.1 gives the following rules for the Galois correspondence: (1a) L1 ⊆ L2 ⇐⇒ Gal(N/L2 ) ≤ Gal(N/L1 ). (1b) H1 ≤ H2 ⇐⇒ N (H2 ) ⊆ N (H1 ). (1c) N (H1 ) ∩ N (H2 ) = N (hH1 , H2 i), where hH1 , H2 i is the closed subgroup of G generated by the closed subgroups H1 and H2 . (1d) Gal(N/L1 ∩ L2 ) = hGal(N/L1 ), Gal(N/L2 )i. (1e) Gal(N/L1 L2 ) = Gal(N/L1 ) ∩ Gal(N/L2 ). (1f) N (H1 ∩ H2 ) = N (H1 )N (H2 ). Since Galois groups are compact, their images under restriction are closed. As in the finite case this produces other theorems of infinite Galois theory. (2a) N (σHσ −1 ) = σN (H) and (2b) Gal(N/σL) = σGal(N/L)σ −1 , for every σ ∈ G. (2c) A closed subgroup H of G is normal if and only if L = N (H) is a Galois extension of K. (2d) If L is a Galois extension of K and L ⊆ N , then res: Gal(N/K) → Gal(L/K) is a continuous open epimorphism with kernel Gal(N/L) and Gal(L/K) ∼ = Gal(N/K)/Gal(N/L). (2e) res: Gal(LM/M ) → Gal(L/L ∩ M ) is an isomorphism for every Galois extension L of K and every extension M of K. (2f) If, in (2e), M is also a Galois extension of K, then σ 7→ (resL σ, resM σ) is an isomorphism Gal(LM/L ∩ M ) ∼ = Gal(L/L ∩ M ) × Gal(M/L ∩ M ) and Gal(LM/K) ∼ = {(σ, τ ) ∈ Gal(L/K) × Gal(M/K) | resL∩M σ=resL∩M τ }. In both cases we use the product topology on products of groups. The first isomorphism of (2f) is a special case of the second one. To prove the second isomorphism, note first that the map σ 7→ (resL σ, resM σ) is a continuous injective map of the left hand side onto the right hand side. Hence, it suffices to prove surjectivity. Thus, consider ρ ∈ Gal(L/K) and τ ∈ Gal(M/K) with resL∩M ρ = resL∩M τ . Extend ρ to an automorphism ρ1

12

Chapter 1. Infinite Galois Theory and Profinite Groups

of LM and let ρ0 = resM ρ1 . Then ρ−1 0 τ ∈ Gal(M/L ∩ M ). By (2e) there is τ . The element σ = ρ1 λ of Gal(LM/K) a λ ∈ Gal(LM/L) with resM λ = ρ−1 0 satisfies resL σ = ρ and resM σ = τ , as desired. These statements are useful when N is the separable closure Ks of K. Denote Gal(Ks /K) by Gal(K), the absolute Galois group of K. Next we show that profinite groups are Galois groups. Lemma 1.3.2: Suppose a profinite group G acts faithfully as automorphisms of a field F . Suppose for each x ∈ F , the stabilizer S(x) = {σ ∈ G | σx = x}, is an open subgroup of G. Then F is a Galois extension of the fixed field K = F (G) and G = Gal(F/K). Proof: If G is a finite group, this is a result of Artin [Lang7, p. 264]. In general the group H = S(x1 ) ∩ · · · ∩ S(xn ) is open in G for every x1 , . . . , xn ∈ F . Therefore, so is the intersection N of all the conjugates of H (Exercise 4). The finite quotient group G/N acts faithfully on the field L = K(Gx1 , . . . , Gxn ). It has K as its fixed field. Thus, L is a finite Galois extension of K and G/N ∼ = Gal(L/K). The field F is the union of all above L and 1 is the intersection of all the above N . Hence, F is a Galois extension of K and Gal(F/K) ∼ = lim Gal(L/K) ∼ = lim G/N ∼ = G. ←− ←− Proposition 1.3.3: Let L/K be a Galois extension and α: G → Gal(L/K) an epimorphism of profinite groups. Then there is a Galois extension F/E and an isomorphism ϕ: Gal(F/E) → G such that F is a purely transcendental extension of L, L ∩ E = K, and α ◦ ϕ = resL . Proof: Let X be the disjoint union of all quotient groups G/N , where N ranges over all open normal subgroups of G. Consider the elements of X as independent over L and set F = L(X). Define an action of each σ of G on F by σ(τ N ) = στ N , for τ N ∈ X, and σ(a) = α(σ)(a) for a ∈ L. This action of G is faithful. We have S(τ N ) = N and S(a) = α−1 (Gal(L/K(a))). Any u ∈ F is a rational function with integral coefficients in a1 , . . . , am ∈ L and x1 , . . . , xn ∈ X. The stabilizer S(u) of u contains the open subgroup S(a1 ) ∩ · · · ∩ S(am ) ∩ S(x1 ) ∩ · · · ∩ S(xn ). Hence, S(u) is an open subgroup. Let E be the fixed field of G in F . By Lemma 1.3.2, G = Gal(F/E) and the conclusion of the proposition follows from the definitions. Corollary 1.3.4 (Leptin): Every profinite group is isomorphic to a Galois group of some Galois extension.

1.4 The p-adic Integers and the Pr¨ ufer Group The first examples of profinite groups are the group Zp of p-adic numbers ˆ = lim Z/nZ. and the Pr¨ ufer group Z ←−

1.4 The p-adic Integers and the Pr¨ ufer Group

13

The p-adic group Zp . Let p be a rational prime. Consider the quotient rings Z/pi Z with their canonical homomorphisms Z/pj Z → Z/pi Z given for j ≥ i by x + pj Z 7→ x + pi Z. The inverse limit Zp = lim Z/pi Z is the ring ←− of p-adic integers. It is a profinite ring. Each x ∈ Zp is a sequence (xi + pi Z)i∈N where xi ∈ Z and xj ≡ xi mod pi Z for j ≥ i. Each integer m ≥ 0 corresponds to a basic neighborhood of x consisting of all elements y = (yi + pi Z)i∈N with ym ≡ xm mod pm Z. The map a 7→ (a + pi Z)i∈N is an embedding of Z into Zp . Identify Z with its image in Zp . The sequence (xi )i∈N converges to x = (xi + pi Z)i∈N in the p-adic topology. Hence, Z is dense in Zp . Yet, Z is not equal to Zp . For Pn−1 example, if p 6= 2, then ( i=0 pi + pn Z)n∈N belongs to Zp but not to Z. For Pn−1 p = 2, ( i=0 4i + 2n Z)n∈N belongs to Z2 but not to Z (see also Exercise 15). Lemma 1.4.1: The ring Zp has the following properties: (a) An element x = (xi + pi Z)i∈N is invertible if and only if p - x1 . (b) Zp is an integral domain. Proof of (a): Suppose x0 = (x0i + pi Z)i∈N is an inverse of x. Then x01 x1 ≡ 1 mod p, hence p - x1 . Conversely, suppose p - x1 . Then for each i, p - xi . Hence, there exists x0i ∈ Z which is unique modulo pi with x0i xi ≡ 1 mod pi . Thus, x0 = (x0i + pi Z)i∈N is in Zp and x0 x = 1. Proof of (b): Let x = (xi + pi )i∈N and y = (yi + pi )i∈N be nonzero elements of Zp . Then there are m, n ∈ N with xm 6≡ 0 mod pm and yn 6≡ 0 mod pn . Hence, xm+n ym+n 6≡ 0 mod pm+n . Therefore, xy 6= 0. Consequently, Zp is an integral domain. Lemma 1.4.2: (a) For each i, pi Zp is the kernel of the projection πi : Zp → Z/pi Z. Thus, pi Zp is an open subgroup of Zp of index pi . (b) If H is a subgroup of Zp of a finite index, then H = pi Zp for some i ∈ N. (c) 0 is the only closed subgroup of Zp of infinite index. (d) pZp is the unique closed maximal subgroup of Zp . (e) All nonzero closed subgroups of Zp are isomorphic to Zp . Proof of (a): Suppose x = (xj + pj Z)j∈N belongs to Ker(πi ). Then xi ≡ 0 mod pi . Hence, xj ≡ 0 mod pi for each j ≥ i. Write xj = pi yj for j ≥ i and yj = yi for j < i. Let z = (yj+i + pj Z)j∈N . Then z ∈ Zp and pi z = x. Indeed, pi zj = pi yj+i = xj+i ≡ xj mod pj for every positive integer j. Proof of (b): Conversely, let H be a subgroup of Zp of index n. Suppose n = kpi where p - k. By Lemma 1.4.1(a), k is invertible in the ring Zp , so nZp = pi Zp . Thus, pi Zp = nZp ≤ H. It follows, pi = (Zp : pi Zp ) ≥ (Zp : H) = kpi . Therefore, k = 1 and H = pi Zp . Proof of (c): A closed subgroup J of Zp of infinite index is the intersection of infinitely many open groups (Lemma 1.2.3). Hence, by (a), all subgroups pi Zp contain J. Consequently J = 0.

14

Chapter 1. Infinite Galois Theory and Profinite Groups

Proof of (d): By (a), (b), and (c), pZp is the unique closed maximal subgroup of Zp . Proof of (e): By Lemma 1.4.1(b), the map x 7→ pi x is an isomorphism of Zp onto pi Zp . Every element xP= (xi + pi Z)i∈N has a unique representation as a ∞ i formal power series i=0 ai p , with 0 ≤ ai < p for all i. Indeed, xn ≡ Pn−1 i n i=0 ai p mod p , for every n ∈ N. Lemma 1.4.3: Let α: Zp → Z/pn Z be an epimorphism with n ≥ 1 and H a closed subgroup of Zp . Suppose α(H) = Z/pn Z. Then H = Zp . Proof: By Lemma 1.4.2(a), Ker(α) = pn Zp . Thus, by assumption, H + pn Zp = Zp . Assume H 6= Zp . Then, by Lemma 1.4.2(d), H ≤ pZp . Therefore, Zp = H + pn Zp ≤ pZp < Zp . It follows from this contradiction that H = Zp . In the terminology of Section 22.5, Lemma 1.4.3 says that α: Zp → Z/pn Z is a Frattini cover. ¨ fer group. For each n ∈ N consider the quotient group Z/nZ The Pru and the canonical homomorphisms Z/nZ → Z/mZ defined for m|n by x + ˆ = lim Z/nZ is the Pr¨ nZ 7→ x + mZ. The inverse limit Z ufer group. Like ←− ˆ by x 7→ (x + nZ)n∈N . Thus, Z ˆ with Zp , embed Z as a dense subgroup of Z ˆ is the closure of the subgroup generated by 1. Write Z = h1i and say that 1 ˆ Also, the subgroups nZ of Z form a basis for the neighborhoods generates Z. of 0 in the induced topology. ˆ is an open subgroup of Z ˆ of index n and Lemma 1.4.4: For each n ∈ N, nZ ∼ ˆ ˆ ˆ ˆ nZ = Z. If H is a subgroup of Z of index n, then H = nZ. Proof of: Suppose x = (xk + kZ)k∈Z lies in the kernel Zn of the projection ˆ → Z/nZ. Then xn ≡ 0 mod n. Hence, for each r ∈ N we have xrn ≡ Z xn ≡ 0 mod n, so xrn = nyrn for some yrn ∈ Z. Let z = (yrn + rZ)r∈Z . If r0 is a multiple of r, then nyr0 n ≡ nyrn mod rn, so yr0 n ≡ yrn mod r. ˆ Moreover, xr ≡ xrn ≡ nyrn = nzr mod r. Hence, x = nz. Therefore, z ∈ Z. ˆ is an open subgroup of Z ˆ of index n. Consequently, nZ ˆ onto nZ. ˆ Indeed, Next note that the map x 7→ nx is an isomorphism of Z if nx = 0, then nxrn ≡ 0 mod rn, so xr ≡ xrn ≡ 0 mod r for each r ∈ N. Hence, x = 0. ˆ of index n, then nZ ˆ is contained in H Finally, if H is a subgroup of Z ˆ and has the same index. Therefore, H = nZ. ˆ to the groups Zp . We conclude by relating Z

1.5 The Absolute Galois Group of a Finite Field

15

ˆ is topologically isomorphic to the cartesian prodLemma1.4.5: The group Z Q uct Zp where p ranges over all primes numbers. Q Proof: Let n = pkp be the decomposition of a positive integer n into a product of prime Q powers. The Chinese remainder theorem gives a canonthe projection ical isomorphism Z/pkp Z → Z/nZ. Combine this with Q Q Q kp Zp → Z/nZ. Zp → Z/p Z to obtain a continuous epimorphism fn : The maps Q fn form a compatible system, so they give a continuous homomorˆ Since Q Zp is compact and Z ˆ is Hausdorff, Im(f ) is a phism f : Zp → Z. ˆ Moreover, Z embeds diagonally in Q Zp and f (m) = m closed subgroup of Z. ˆ ˆ for each m ∈ Z. Thus, Z ⊆ Im(f ). Since T Z is dense in Z, we have Im(f ) = Z, so f is surjective. The kernel of f is Ker(fn ) = 0. Hence, f is also injective. The compactness and Hausdorff properties imply that f is a topological isomorphism. As a consequence of Lemma 1.4.5, Zp is both a closed subgroup and a ˆ for each prime p. quotient of Z,

1.5 The Absolute Galois Group of a Finite Field For every prime power q there exists a field Fq (unique up to isomorphism) with q elements. It is characterized within its algebraic closure F˜q by Fq = {x ∈ F˜q | xq = x}. The field Fq has, for each n ∈ N, exactly one extension, Fqn , of degree n. It is Galois with a cyclic group generated by the Frobenius automorphism a is πq,n defined by πq,n (x) = xq for x ∈ Fqn . The map a + nZ 7→ πq,n an isomorphism of Z/nZ onto Gal(Fqn /Fq ). If m|n, there is a canonical commutative diagram / Z/mZ

Z/nZ Gal(Fqn /Fq )

res

/ Gal(Fqm /Fq ).

ˆ∼ Take the inverse limits to obtain an isomorphism Z = Gal(Fq ) mapping the ˆ to the Frobenius automorphism πq , defined on all identity element 1 of Z ˜ q by πq (x) = xq . of F S∞ (l) Let l be a prime number and Fq = i=1 Fqli . Then the projection ˆ 7→ Zl corresponds to res: Gal(Fq ) → Gal(F(l) Z q /Fq ). By Lemma 1.4.5, Q (l) ∼ Gal(Fq ) = Gal(Fq /Fq ). On the other hand, let Nl be the fixed field of Zl (l) ˜ q . Then Gal(Nl ) = Zl , F(l) ˜ in F q Nl = Fq , and Fq ∩ Nl = Fq . It follows, Y ∼ Zl0 × Zl . Gal(Fq ) = Gal(F(l) q ) × Gal(Nl ) = l0 6=l

16

Chapter 1. Infinite Galois Theory and Profinite Groups

Exercises 1. Let (Si , πji ) be an inverse system of finite sets with all πji surjective and let S = lim Si . Use Lemma 1.1.3 to prove that all maps πi : S → Si ←− determined by the πji ’s are surjective. 2.

Let H1 , . . . , Hr be closed subgroups of a profinite group G. Prove that \ H1 ∩ · · · ∩ Hr = (H1 N ∩ · · · ∩ Hr N ), N

where N ranges over all open normal subgroups of G. Hint: Use Lemma 1.2.2(b). 3. Suppose H is a closed subgroup of a profinite group G. Prove: If HN/N = G/N for every open normal subgroup N of G, then H = G. 4. Let H be an open subgroup of index n of a profinite group G. Denote the intersection of all conjugates of H in G by N . Note: Multiplication of G on the left cosets of H induces a homomorphism of G into the symmetric group Sn with kernel N . Conclude that G/N is isomorphic to a subgroup of Sn and (G : N ) ≤ n!. 5. Let G be a compact group and H an open subgroup. Suppose H is profinite. Prove: G is profinite. Hint: Use Lemma 1.2.4. 6. Let S be a set of rational primes. Consider the profinite group ZS = lim Z/nZ, with n running over all positive integers with prime factors in S. ←− (a) Prove: The finite homomorphic images of ZS are exactly the groups Z/nZ, where the prime factors of n belong to S. (b) Embed Z in ZS and determine the topology on Z induced by that of ZS . (c) Prove that ZS is procyclic (i.e. ZS is the closure of a group generated by one element). Q (d) Follow the proof of Lemma 1.4.5 to prove that ZS ∼ = p∈S Zp . 7. Let G be a procyclic group (Exercise 6). Use that G is a homomorphic ˆ to show there exists a set S of primes with G = Q image of Z p∈S Gp , where ip ∼ for each p ∈ S either Gp ∼ Z for some i ∈ N or G . Z/p Z = p p = p In particular, Q Z . if G is torsion free, then G ∼ = p∈S p ˆ Prove that every finite quotient of G is 8. Let G be a closed subgroup of Z. a cyclic group. Conclude that G is procyclic and therefore that there exists Q a set S of prime numbers with G ∼ = l∈S Zl . 9. Let G be a profinite group. Prove that each of the following statements ˆ is equivalent to G ∼ = Z. (a) G has exactly one open subgroup of each index n. ˆ (b) G is procyclic and there is an epimorphism π: G → Z.

Exercises

17

10. Let G be a procyclic group. Use Exercise 8 to prove that each epimorˆ is an isomorphism. phism π: G → Z 11. Let G be a profinite group with at most one open subgroup of every index n. (a) Prove that every open subgroup is normal. ¯ of G. Con(b) Observe that (a) holds for every finite homomorphic image G ¯ clude that G is nilpotent. (c) Let P be a finite p-group with the above properties. Prove that every element x of P of maximal order generates P . (d) Conclude that G is a procyclic group. ˆ in the following 12. Define powers in a profinite group G with exponents in Z ˆ way: Let g ∈ G and ν ∈ Z. Then there exists a sequence {ν1 , ν2 , ν3 , . . .} of elements of Z that converges to ν. By compactness there is a subsequence of {g ν1 , g ν2 , g ν3 , . . .} that converges to an element h of G. (a) Prove h does not depend on the sequence {ν1 , ν2 , ν3 , . . .}. So we may denote h by g ν . Hint: If N is a normal subgroup of G of index n, then xn ∈ N for every x ∈ G. (b) Prove the usual rules for the power operations. For example, g µ g ν = g µ+ν ,

(g µ )ν = g µν ,

and

g ν hν = (gh)ν if gh = hg.

ˆ into G is continuous. (c) Prove that the map (g, ν) 7→ g ν of G × Z 13. Multiplication in the groups Z/pi Z is compatible with the canonical maps Z/pi+1 Z → Z/pi Z. Therefore, it defines a multiplication in the additive group Zp = lim Z/pi Z. ←− (a) Prove: Zp is an integral domain (the quotient field of which, Qp , is the field of p-adic numbers). (b) Show: Every closed subgroup of Zp is an ideal of Zp . (c) Show: pZp is the unique maximal ideal of Zp ; observe that Zp /pZp ∼ = Fp . (d) Deduce: α ∈ Zp is a unit (i.e. invertible in this ring) if and only if α is congruent modulo pZp to one of the numbers 1, 2, . . . , p − 1. Hint: Form the inverse of 1 + βp, β ∈ Zp using the geometric series for 1/(1 + x). ˆ in a manner analogous to 14. Define multiplication in the additive group Z ˆ a commutative the definition of multiplication in Zp . Prove that this makes Z topological ring with zero divisors. Q ˆ∼ (a) Prove that the isomorphism of additive groups Z = Zp established in Lemma 1.4.5 is an isomorphism of rings. ˆ is also an ideal. (b) Prove that every closed subgroup of Z 15. Use the power series representation of the elements of Zp (Section 1.4) to show |Zp | = 2ℵ0 . Conclude that Qp has elements that are transcendental over Q.

18

Chapter 1. Infinite Galois Theory and Profinite Groups

√ 16. For a prime number p, let K be Q(ζ ) if p = 6 2 and Q( −1) if p = 2. p p S∞ Also, let Lp = i=1 Q(ζpi ). (a) Prove that if K 0 is a field such that Kp ⊆ K 0 ⊂ Lp , then Gal(Lp /K 0 ) ∼ = Zp and [K 0 : Q] < ∞. (b) Prove that Gal(Lp /Q) is isomorphic to Zp × Z/(p − 1)Z if p 6= 2 and to Z2 × Z/2Z if p = 2.

Notes More about topological groups can be found in [Pontryagin]. A detailed exposition on Galois theory of finite extensions appears in [Lang7, Chapter VI, Section 1]. For finite fields see [Langl, Chapter V, Section 5]. Leptin’s proof of Corollary 1.3.4 uses linear disjointness of fields [Leptin]. We reproduce it in Proposition 2.6.12. The proofs of Lemma 1.3.2 and Proposition 1.3.3 appear in [Waterhouse]. This chapter overlaps with [Ribes, Chapter 1].

Chapter 2. Valuations and Linear Disjointness Sections 2.1–2.4 introduce the basic elements of the theory of valuations, especially discrete valuations, and of Dedekind domains. These sections are primarily a survey. We prove that an overring of a Dedekind domain is again a Dedekind domain (Proposition 2.4.7). The rest of the chapter centers around the notion of linear disjointness of fields. We use this notion to define separable, regular, and primary extensions of fields. In particular, we prove that an extension F/K with a K-rational place is regular. Section 2.8 gives a useful criterion for separability with derivatives.

2.1 Valuations, Places, and Valuation Rings The literature treats arithmetic theory of fields through three intimately connected classes of objects: valuations, places, and valuation rings. We briefly review the basic definitions. Call an Abelian (additive) group Γ with a binary relation < an ordered group if the following statements hold for all α, β, γ ∈ Γ. (1a) Either α < β, or α = β, or β < α. (1b) If α < β and β < γ, then α < γ. (1c) If α < β, then α + γ < β + γ. Some examples of ordered groups are the additive groups Z, R, and Z⊕Z with the order (m, n) < (m0 , n0 ) if either m < m0 or m = m0 and n < n0 (the lexicographic order). A valuation v of a field F is a map of F into a set Γ ∪ {∞}, where Γ is an ordered group, with these properties: (2a) v(ab) = v(a) + v(b). (2b) v(a + b) ≥ min v(a), v(b) . (2c) v(a) = ∞ if and only if a = 0. (2d) There exists a ∈ F × with v(a) 6= 0. By definition the symbol ∞ satisfies these rules: (3a) ∞ + ∞ = α + ∞ = ∞ + α = ∞; and (3b) α < ∞ for each α ∈ Γ. Condition (2) implies several more properties of v: (4a) v(1) = 0, v(−a) = v(a). (4b) If v(a) < v(b), then v(a + b) = v(a) (Use the identity a = (a + b) − b and P(2b)); n (4c) If i=1 ai = 0, then there exist i 6= j such that v(ai ) = v(aj ) and v(ai ) = min(v(a1 ), . . . , v(an )) (Use (2b) and (4b)). We refer to the pair (F, v) as a valued field.

20

Chapter 2. Valuations and Linear Disjointness

The subgroup Γv = v(F × ) of Γ is the value group of v. The set Ov = {a ∈ F | v(a) ≥ 0} is the valuation ring of v. It has a unique maximal ideal mv = {a ∈ F | v(a) > 0}. Refer to the residue field F¯v = Ov /mv as the residue field of F at v. Likewise, whenever there is no ambiguity, we ¯ and call it the residue of a at v. denote the coset a + mv by a Two valuations v1 , v2 of a field F with value groups Γ1 , Γ2 are equivalent if there exists an isomorphism f : Γ1 → Γ2 with v2 = f ◦ v1 . Starting from Section 2.2, we abuse our language and say that v1 and v2 are distinct if they are inequivalent. A place of a field F is a map ϕ of F into a set M ∪ {∞}, where M is a field, with these properties: (5a) ϕ(a + b) = ϕ(a) + ϕ(b). (5b) ϕ(ab) = ϕ(a)ϕ(b). (5c) There exist a, b ∈ F with ϕ(a) = ∞ and ϕ(b) 6= 0, ∞. By definition the symbol ∞ satisfies the following rules: (6a) x + ∞ = ∞ + x = ∞ for each x ∈ M . (6b) x · ∞ = ∞ · x = ∞ · ∞ = ∞ for each x ∈ M × . (6c) Neither ∞ + ∞, nor 0 · ∞ are defined. It is understood that (5a) and (5b) hold whenever the right hand side is defined. These conditions imply that ϕ(1) = 1, ϕ(0) = 0 and ϕ(x−1 ) = ϕ(x)−1 . In particular, if x 6= 0, then ϕ(x) = 0 if and only if ϕ(x−1 ) = ∞. We call an element x ∈ F with ϕ(x) 6= ∞ finite at ϕ, and say that ϕ is finite at x. The subring of all elements finite at ϕ, Oϕ = {a ∈ F | ϕ(a) 6= ∞}, is the valuation ring of ϕ. It has a unique maximal ideal mϕ = {a ∈ F | ϕ(a) = 0}. The quotient ring Oϕ /mϕ is a field which is canonically isomorphic to the residue field F¯ϕ = {ϕ(a) | a ∈ Oϕ } of F at ϕ. The latter is a subfield of M . Call ϕ a K-place if K is a subfield of F and ϕ(a) = a for each a ∈ K. Two places ϕ1 and ϕ2 of a field F with residue fields M1 and M2 are equivalent if there exists an isomorphism λ: M1 → M2 with ϕ2 = λ ◦ ϕ1 . A valuation ring of a field F is a proper subring O of F such that if / O} is x ∈ F × , then x ∈ O or x−1 ∈ O. The subset m = {x ∈ O | x−1 ∈ the unique maximal ideal of O (Exercise 1). The map ϕ: F → O/m ∪ {∞} which maps x ∈ O onto its residue class modulo m and maps x ∈ F r O onto ∞ is a place of F with valuation ring O. Denote the units of O by U = {x ∈ O | x−1 ∈ O}. Then F × /U is a multiplicative group ordered by the rule xU ≤ yU ⇐⇒ yx−1 ∈ O. The map x 7→ xU defines a valuation of F with O being its valuation ring. These definitions easily give a bijective correspondence between the valuation classes, the place classes and the valuation rings of a field F . An isomorphism σ: F → F 0 of fields induces a bijective map of the valuations and places of F onto those of F 0 according to the following rule: If v is a valuation of F , then σ(v) is defined by σ(v)(x) = v(σ −1 x) for every x ∈ F 0 . If ϕ is a place of F , then σ(ϕ)(x) = ϕ(σ −1 x). In particular, σ

2.2 Discrete Valuations

21

0 induces an isomorphism F¯ϕ ∼ of residue fields. It is also clear that if = F¯σ(ϕ) ϕ corresponds to v, then σ(ϕ) corresponds to σ(v). A valuation v of a field F is real (or of rank 1) if Γv is isomorphic to a subgroup of R. Real valuations satisfy the so called weak approximation theorem, a generalization of the Chinese remainder theorem [CasselsFr¨ohlich, p. 48]:

Proposition 2.1.1: Consider the following objects: inequivalent real valuations v1 , . . . , vn of a field F , elements x1 , . . . , xn of F , and real numbers γ1 , . . . , γn . Then there exists x ∈ F with vi (x − xi ) ≥ γi , i = 1, . . . , n.

2.2 Discrete Valuations ∼ Z. In this case we normalize A valuation v of a field F is discrete if v(F × ) = v by replacing it with an equivalent valuation such that v(F × ) = Z. Each element π ∈ F with v(π) = 1 is a prime element of Ov . Prime elements of a unique factorization domain R produce discrete valuations of F = Quot(R). If p is a prime element of R, then every element x of F × has a unique representation as x = upm , where u is relatively prime to p and m ∈ Z. Define vp (x) to be m. Then vp is a discrete valuation of F . Suppose p0 is another prime element of R. Then vp0 is equivalent to vp if and only if p0 R = pR, that is if p0 = up with u ∈ R× . Example 2.2.1: Basic examples of discrete valuations. (a) The ring of integers Z is a unique factorization domain. For each prime number p the residue field of Q at vp is Fp . When p ranges over all prime numbers, vp ranges over all valuations of Q (Exercise 3). (b) Let R = K[t] be the ring of polynomials in an indeterminate t over a field K. Then R is a unique factorization domain. Then prime elements of R are the irreducible polynomials p over K. Units of R are the elements u of K × , so vp (u) = 0 and we say vp is trivial on K. The residue field of K(t) at vp is isomorphic to the field K(a), where a is a root of p. There is one additional valuation, v∞ , of K(t) which is trivial on K. It is defined for a quotient fg of elements of K[t] by the formula v∞ ( fg ) = deg(g) − deg(f ). The set of vp ’s and v∞ give all valuation of K(t) trivial on K. Thus, all valuations of K(t) which are trivial on K are discrete (Exercise 4). ˜ An arbitrary irreducible polynomial p may have several roots a ∈ K. ˜ ∪{∞} by ϕa (t) = a and ϕa (c) = c Each of them defines a place ϕa : K(t) → K for each c ∈ K. These places are equivalent. If p(t) = t − a, then ϕa is the unique place of K(t) corresponding to vp . Similarly, there is a unique place ϕ∞ corresponding to v∞ . It is defined by ϕ∞ (t) = ∞. We may view each f (t) ∈ K(t) as a function from K ∪ {∞} into itg(t) with g, h ∈ K[X] and self: f (a) = ϕa (f (t)). Explicitly, write f (t) = h(t) gcd(g, h) = 1. Let a ∈ K. Then f (a) =

g(a) h(a)

if h(a) 6= 0 and f (a) = ∞

22

Chapter 2. Valuations and Linear Disjointness

if h(a) = 0. To compute f (∞) let u = t−1 and write f (t) =

g1 (u) h1 (u)

with

g1 , h1 ∈ K[X] and gcd(g1 , h1 ) = 1. Then f (∞) = hg11(0) (0) if h1 (0) 6= 0 and f (∞) = ∞ if h1 (0) = 0. Suppose for example f (t) ∈ K[t] and f 6= 0. Then f maps K into itself and f (∞) = ∞. Now suppose f (t) = at+b ct+d with ad − bc 6= 0 and c 6= 0, then f (∞) = ac . When K is algebraically closed, each irreducible polynomial is linear. Hence, each valuation of K(t) which is trivial over K is either vt−a for some a ∈ K or v∞ . More examples of discrete valuations arise through extensions of the basic examples (Section 2.3). Lemma 2.2.2: Every discrete valuation ring R is a principal ideal domain. Proof: Let v be the valuation of K = Quot(R) with Ov = R and v(K × ) = Z. Choose a prime element π of R. Now consider a nonzero ideal a of R. Then the minimal integer m with π m ∈ a is positive. It satisfies, a = π m R. As a consequence of Lemma 2.2.2, finitely generated modules over R have a simple structure. Proposition 2.2.3: Let R be a discrete valuation ring, p a prime element ¯ = R/pR. of R, K = Quot(R), and M a finitely generated R-module. Put K Let r = dimK M ⊗R K, n = dimK¯ M/pM , and m = n − r. Then there is a unique m-tuple of positive integers (k1 , k2 , . . . , km ) with k1 ≤ k2 ≤ · · · ≤ km and M ∼ = R/pkm R ⊕ · · · ⊕ R/pk1 R ⊕ Rr . Moreover, r is the maximal number of elements of M which are linearly independent over R and n is the minimal number of generators of M . Proof: By Lemma 2.2.2, R is a principal ideal domain, so M = Mtor ⊕ N , where Mtor = {m ∈ M | rm = 0 for some r ∈ R, r 6= 0} and N is a free R-module [Lang7, p. 147, Thm. 7.3]. Both Mtor and N are finitely generated [Lang7, p. 147, Cor. 7.2]. In particular, N ∼ = Rs for some integer s ≥ 0. Suppose m ∈ Mtor and am = 0 with a ∈ R, a 6= 0. Then, m⊗1 = am⊗ a1 = 0. Hence, Mtor ⊗R K = 0 and M ⊗R K ∼ = K s . Therefore, s = r. By [Lang7, p. 151, Thm. 7.7], Mtor ∼ = R/qm0 R ⊕ · · · ⊕ R/q1 R where q1 , . . . , qm0 are elements of R which are neither zero nor units and qi |qi+1 , i = 1, . . . , m0 − 1. Multiplying each qi by a unit, we may assume qi = pki with ki an integer and 1 ≤ k1 ≤ k2 ≤ · · · ≤ km0 . Moreover, the above cited theorem assures Rq1 , . . . , Rqm0 are uniquely determined by the above conditions. Hence, k1 , . . . , km0 are also uniquely determined. Combining the first two paragraphs gives: M∼ = R/pkm0 R ⊕ · · · ⊕ R/pk1 R ⊕ Rr . 0 ¯ m0 +r , so n = m0 + r and m0 = m. Hence, M/pM = (R/pR)m +r ∼ =K

2.2 Discrete Valuations

23

Now recall Ps that elements v1 , . . . , vs of M are linearly independent over R if i=1 ai vi = 0 with a1 , . . . , as ∈ R implies a1 = · · · = as = 0. Alternatively, v1 ⊗ 1, . . . , vs ⊗ 1 are linearly independent over K. Thus, r is the maximal number of R-linearly independent elements of M . Finally, by Nakayama’s lemma [Lang7, p. 425, Lemma 4.3], n is the minimal number of generators of M . Definition 2.2.4: Let R be an integral domain with quotient field F . An overring of R is a ring R ⊆ R0 ⊂ F . It is said to be proper if R 6= R0 . Lemma 2.2.5: A discrete valuation ring O has no proper overrings. Proof: Let R be an overring of O. Assume there exists x ∈ R r O. Then x−1 is a nonunit of O. Choose a prime element π for O. Then x = uπ −m for some u ∈ O× and a positive integer m. Hence, π −1 = u−1 π m−1 x ∈ R. Therefore, u0 π k ∈ R for all u0 ∈ O× and k ∈ Z. We conclude that R = Quot(O). Composita of places attached to discrete valuations of rational function fields of one variable give rise to useful places of rational function fields of several variables. Construction 2.2.6: Composition of places. Suppose ψ is a place of a field K with residue field L and ϕ is a place of L with residue field M . Then ψ −1 (Oϕ ) is a valuation ring of K with maximal ideal ψ −1 (mϕ ) and residue field ψ −1 (Oϕ )/ψ −1 (mϕ ) ∼ = Oϕ /mϕ ∼ = M . Define a map ϕ ◦ ψ: K → M ∪ {∞} as follows: ϕ ◦ ψ(x) = ϕ(ψ(x)) if ψ(x) 6= ∞ and ϕ ◦ ψ(x) = ∞ if ψ(x) = ∞. Then ϕ ◦ ψ is a homomorphism on ψ −1 (Oϕ ) and {x ∈ K | ϕ ◦ ψ(x) = ∞} = K r ψ −1 (Oϕ ). Therefore, ϕ ◦ ψ is a place of K, called the compositum of ψ and ϕ, Oϕ◦ψ = ψ −1 (Oϕ ), and mϕ◦ψ = ψ −1 (mϕ ). K Oψ Oϕ◦ψ mϕ◦ψ mψ

/ L ∪ {∞}

ϕ

/ M ∪ {∞}

ψ

/L

ϕ

/ M ∪ {∞}

ψ

/ Oϕ

ϕ

/M

ψ

/ mϕ

ϕ

/0

ψ

/0

ϕ

/0

ψ

¯ ψ and M = L ¯ϕ = K ¯ ϕ◦ψ . In addition, L = K

˜ t1 , . . . , tr indeLemma 2.2.7: Let K be a field, a1 , . . . , ar elements of K, terminates, and L a finite extension of K. Then there exists a K-place ϕ: K(t) → K(a) ∪ {∞} such that ϕ(ti ) = ai , i = 1, . . . , r. Moreover, every extension of ϕ to an L-place of L(t) maps L(t) onto L(a) ∪ {∞}. Proof: For each i there is a K(a1 , . . . , ai−1 , ti+1 , . . . , tr )-place ϕi : K(a1 , . . . , ai−1 , ti , ti+1 . . . , tr ) → K(a1 , . . . , ai−1 , ai , ti+1 , . . . , tr )

24

Chapter 2. Valuations and Linear Disjointness

with ϕi (ti ) = ai (Example 2.2.1). The compositum ϕ = ϕr ◦ · · · ◦ ϕ1 is a K-place of K(t1 , . . . , tr ) with residue field K(a1 , . . . , ar ) and ϕ(ti ) = ai , i = 1, . . . , r. Let now ϕ be an extension of ϕ to an L-place of L(t). Choose a basis also a basis for L(t)/K(t). Hence, b1 , . . . , bn for L/K. Then b1 , . . . , bn is P n each f ∈ L(t) has a presentation f = i=1 bi fi with fi ∈ K(t). Assume Pn without loss that ff1i is finite under ϕ for i = 1, . . . , n. Then f = f1 i=1 ff1i bi and ϕ(f ) ∈ L(a1 , . . . , ar ) ∪ {∞}. Thus, ϕ(L(t)) = L(a) ∪ {∞}.

2.3 Extensions of Valuations and Places The examples of Section 2.2 and the following extension results give a handle on describing valuations of function fields in one variable. Proposition 2.3.1 (Chevalley [Lang4, p. 8, Thm. 1]): Let ϕ0 be a homomorphism of an integral domain R into an algebraically closed field M and let F be a field containing R. Then ϕ0 extends either to an embedding ϕ of F into M or to a place ϕ of F into M ∪ {∞}. When F is algebraic over R, the proposition has a more precise form: Let f ∈ R[X] be an irreducible polynomial over E = Quot(R) and f¯ ∈ M [X] the result of applying ϕ0 to the coefficients of f . Suppose f¯ is ˜ and M , not identically zero. Assume x and x ¯ are roots of f and f¯ in E respectively. Then ϕ0 extends to a place ϕ of E(x) into M ∪ {∞} with ϕ(x) = x ¯ [Lang4, p. 10, Thm. 2]. Moreover, if ϕ0 is injective, so is ϕ [Lang4, p. 8, Prop. 2]. In particular, suppose v is a valuation of a field E and F is an extension of E. Then v extends to a valuation w0 of F . Each valuation w of F which is equivalent to w0 lies over v. Thus, w lies over v if and only if Ov ⊆ Ow and mv = mw ∩ Ov . The number ew/v = (w(F × ) : w(E × )) is the ramification index of w over v (and also over E). The field degree [F : E] ¯v embeds in F¯w to give the inequality bounds ew/v (Exercise 5). Similarly, E ¯v ] ≤ [F : E] (Exercise 7). Both the ramification index and fw/v = [F¯w : E the residue field degree are multiplicative. Thus, if (F 0 , w0 ) is an extension of (F, w), then ew0 /v = ew0 /w ew/v and fw0 /v = fw0 /w fw/v . If [F : E] < ∞, then the number of valuations of F that lie over v is finite (a consequence of Proposition 2.3.2). Proposition 2.3.2: Let F/E be a finite extension of fields and v a valuation of E. Let w1 , . . . , wg be all inequivalent extensions of v to F . Then

(1)

g X

ewi /v fwi /v ≤ [F : E]

i=1

[Bourbaki2, p. 420, Thm. 1]. If, in addition, v is discrete and F/E is separa-

2.3 Extensions of Valuations and Places

25

ble, then each wi is discrete and (see [Bourbaki2, p. 425, Cor. 1]) g X

(2)

ewi /v fwi /v = [F : E].

i=1

Suppose (F, w)/(E, v) is an extension of discrete valued fields. In particular, w(a) = v(a) for each a ∈ E. By definition, ew/v = (w(F × ) : v(E × )). However, as in Section 2.2, it is customary to replace v and w by equivalent valuations with v(E × ) = w(F × ) = Z. The new valuations satisfy w(a) = ew/v v(a)

for each a ∈ E.

Whenever we speak about an extension of discrete valuations, we mean they are normalized and satisfy the latter relation. Suppose F is a finite Galois extension of E with a Galois group G. Let w be a discrete valuation of F and let σ ∈ G. Then, σ(w) is a valuation of F (Section 2.1), both w and σ(w) lie over the same valuation v of E, and ew/v = eσ(w)/v

and fw/v = fσ(w)/v .

Conversely, suppose w and w0 are two discrete valuations of F over the same valuation v of E. Then there exists σ ∈ G such that σ(w) = w0 (Exercise 9). Thus, if w1 , . . . , wg are all distinct valuations of F that lie over v, then they all have the same residue degree f and ramification index e over v. In this case formula (2) simplifies to (3)

ef g = [F : E]. The subgroups Dw = Dw/v = {σ ∈ G | σOw = Ow } Iw = Iw/v = {σ ∈ G | w(x − σx) > 0 for all x ∈ Ow }

are the decomposition group and the inertia group, respectively, of w ¯v is separable, then [Serre3, p. 33] over E. Obviously Iw / Dw . If F¯w /E (4)

|Iw | = ew/v

and |Dw | = ew/v fw/v .

Section 2.6 generalizes the notion of separable algebraic extension of fields to arbitrary extensions of fields. In particular, purely transcendental extensions of fields are separable. We use this notion in the following definition. Suppose (F, w)/(E, v) is an arbitrary extension of valued fields. We say ¯v w is unramified (resp. tamely ramified) over v (or also over E) if F¯w /E ¯ is a separable extension and ew/v = 1 (resp. char(Ev ) - ew/v ). We say v is unramified (resp. tamely ramified) in F if each extension of v to F is unramified (resp. tamely ramified) over v.

26

Chapter 2. Valuations and Linear Disjointness

Example 2.3.3: Purely transcendental extensions. Let (E, v) be a valued field. Consider a transcendental element t over E. Extend v to a valuation v 0 of E(t) as follows. First define v 0 on E[t] by the following rule: v0

(5)

m X

ai ti = min v(a0 ), . . . , v(am )

i=0

for a0 , . . . , am ∈ E. The same argument used to prove Gauss’ Lemma proves 0 (g) for all f, g ∈ E[t].P that v 0 (f g) = v 0 (f ) + vP m n Indeed, let f (t) = i=0 ai ti and g(t) = j=0 bj tj . Let r be the minimal integer with v(ar ) = min v(a0 ), . . . , v(am ) and let s be the minimal integer with v(bs ) = min v(b0 ), . . . , v(bn ) . If i + j = r + s and (i, j) 6= (r, s), then either i < r or j < s. In both cases v(ar ) + v(bs ) < v(ai ) + v(bj ). Hence v0

m X

n X ai ti + v 0 bj tj = v(ar ) + v(bs )

i=0

j=0

X

= min

v(ai bj ) | k = 0, . . . , m + n

i+j=k

= v0

m X

ai ti ·

i=0

n X

b j tj ,

j=0

as claimed. We extend v 0 to E(t) by the rule v 0 ( fg ) = v 0 (f ) − v 0 (g). Then we prove v 0 (u1 + u2 ) ≥ min v 0 (u1 ), v 0 (u2 ) first for u1 , u2 ∈ E[t] and then for u1 , u2 ∈ Note that the residue of t at v 0 is E(t). Thus, v 0 is a valuation of E(t). P n ¯ ¯i t¯i = 0 for some a0 , . . . , an ∈ transcendental over Ev . Indeed, suppose i=0 a P n 0 i > 0. Hence, a ¯i = 0, Ov . Then min v(a0 ), . . . , v(an ) = v i=0 ai t i = 0, . . . , n. ¯v (t¯) is a rational function field over E ¯v . By It follows that, E(t)v0 = E 0 definition, Γv0 = Γv . In particular, if v is discrete, then so is v and ev0 /v = 1. Suppose v 00 is another extension of v to E(t) with the residue of t at v 00 ¯v . We show that v 00 = v 0 . Indeed, for a0 , . . . , an ∈ E, transcendental over E not all zero, choose j between 0 and n with v(aj ) = min v(a0 ), . . . , v(an ) . Pn Then i=0 ai /aj t¯i 6= 0. Therefore, v

00

n X i=0

ai t

i

= v(aj ) + v

00

n X

(ai /aj )ti

i=0 n X = min v(a0 ), . . . , v(an ) = v 0 ai ti , i=0

as claimed.

2.3 Extensions of Valuations and Places

27

Lemma 2.3.4: Let v be a discrete valuation of a field E, h ∈ Ov [X] a monic ˜ and F = E(x). irreducible polynomial of degree n, x a root of h(X) in E, ¯ Suppose the residue polynomial h(X) is separable. Then v is unramified in F. Qr ¯ ¯v [X] are distinct Proof: By assumption, h(X) = i=1 hi (X), where hi ∈ E monic irreducible polynomials. For each i between 1 and r choose a root ai ¯v )s . Use Proposition 2.3.1 to extend the residue map Ov → E ¯v of hi (X) in (E to a place ϕi of F with ϕi (x) = ai . Denote the corresponding valuation by ¯v (ai ) ⊆ F¯w . Since hi (X) and hj (X) have no common root for wi . Then E i i 6= j, the valuations w1 , . . . , wr are mutually inequivalent extensions of v. Label any further extensions of v to valuations of F as wr+1 , . . . , wg . By (1) n=

r X

deg(hi ) =

i=1

r X

¯v (ai ) : E ¯v ] ≤ [E

i=1

g X

ewi /v fwi /v ≤ n.

i=1

¯v (ai ) = F¯w for i = 1, . . . , r. Moreover, w1 , . . . , wr are Hence, ewi /v = 1 and E i all extensions of v to F and each of them is unramified over E. Therefore, v is unramified in F . ¯v to be infinite. The converse of Lemma 2.3.4 requires E Lemma 2.3.5: Let v be a discrete valuation of a field E. Let F be a separable ¯v is an infinite extension of E of degree n. Suppose v is unramified in F and E field. Then F/E has a primitive element x with irr(x, E) ∈ Ov [X] and the residue of irr(x, E) at v is a separable polynomial. Proof: Let w1 , . . . , wg be all extensions of v to F . By (2), [F : E] = P g ¯ ¯ ¯ ¯ i=1 [Fwi : Ev ]. Moreover, for each i the extension Fwi /Ev is finite and separable. Hence, we may choose ci in F with wi (ci ) = 0 and the residue ¯v . Let hi = irr(¯ ¯v ). Since ci , E c¯i of ci at wi is a primitive element of F¯wi /E ¯v is infinite, we may choose c1 , . . . , cg such that c¯1 , . . . , c¯g are mutually E ¯v . Thus, h1 , . . . , hg are relatively prime. nonconjugate over E Use Proposition 2.1.1 to find x ∈ F with wi (x − ci ) > 0, i = 1, . . . , g. Then, wi (x) = 0, i = 1, . . . , g. Extend each wi to the Galois closure of F/E. Then all E-conjugates of x have nonnegative values under each extended valuation. Hence, the elementary symmetric polynomials in the E-conjugates of x belong to Ov . Therefore, f (X) = irr(x, E) ∈ Ov [X]. Let f¯ be the residue of f at v. By construction, Q f¯(¯ ci ) = 0, therefore g ¯ hi |f , i = 1, . . . , g. Since h1 , . . . , hg are relatively prime, i=1 hi |f¯. Hence, [F : E] =

g X i=1

¯v ] = [F¯wi : E

g X

deg(hi )

i=1

≤ deg(f¯) = deg(f ) = [E(x) : E] ≤ [F : E]. Consequently, E(x) = F , as desired.

28

Chapter 2. Valuations and Linear Disjointness

¯v to be infinite is necessary Example 3.5.4 shows the assumption on E for Lemma 2.3.5 to hold. The next lemma says that arbitrary change of the base field preserves unramified discrete valuations. Lemma 2.3.6: Let (E, v) be a discrete valued field. Consider a separable algebraic extension F of E and a discrete valued field (E1 , v1 ) which extends (E, v). Suppose v is unramified in F . Then v1 is unramified in F E1 . Proof: Suppose without loss that [F : E] < ∞. Let F1 = F E1 . Suppose ¯v is infinite. Choose x as in Lemma 2.3.5 and let f (X) = irr(x, E). first that E Then F = E(x) and f¯(X) is separable. Hence, F1 = E1 (x) and f¯(X) is still separable. By Lemma 2.3.4, v1 is unramified in F1 . In the general case we consider an extension w1 of v1 to a valuation of F1 . Denote the restriction of w1 to F by w. Let t be transcendental over F1 . Example 2.3.3 extends v (resp. w, v1 , w1 ) in a canonical way to a discrete valuation v 0 (resp. w0 , v10 , w10 ) of E(t) (resp. F (t), E1 (t), F1 (t)). Further, ¯v (t¯) ev0 /v = 1 (resp. ew0 /w = 1, ev10 /v1 = 1, ew10 /w1 = 1) and E(t)v0 = E ¯1,v (t¯), F1 (t) 0 = F¯1,w (t¯)), where t¯ is (resp. F (t)w0 = F¯w (t¯), E1 (t)v0 = E 1 1 w1 1 transcendental over F¯1,w1 . Moreover, w10 extends w0 and v10 extends v 0 giving this diagram: (F (t), w0 ) p p ppp

(F1 (t), w10 ) n n nnn (F1 , w1 )

(E(t), v 0 ) p p ppp

(E1 (t), v10 ) n n nnn

(F, w)

(E, v)

(E1 , v1 )

¯v (t¯) is a separable We claim v 0 is unramified in F (t). Indeed, F (t)w0 = F¯w · E extension of E(t)v0 . Also, ew0 /v0 = ew0 /v0 ev0 /v = ew0 /v = ew0 /w ew/v = 1. Hence, w0 is unramified over v 0 . If u∗ is an arbitrary extension of v 0 to F (t) and u is its restriction to F , then the residue of t at u∗ is t¯, which is transcendental over F¯u . Thus, by uniqueness of the construction in Example 2.3.3, u∗ = u0 , where u0 is the canonical extension of u to F (t). By the above, u∗ is unramified over v 0 . Since E(t)v0 is infinite, the first paragraph of the proof implies v10 is ¯1,v (t¯) is a separable extension and unramified in F1 (t). Thus, F¯1,w1 (t¯)/E 1 ¯1,v is a separable extension and ew10 /v10 = 1. Therefore, F¯1,w1 /E 1 ew1 /v1 = ew1 /v1 ew10 /w1 = ew10 /v1 = ew10 /v10 ev10 /v1 = 1. Consequently, v1 is unramified in F1 .

Combine the multiplicativity of the ramification index and the residue field degree with Lemma 2.3.6 to prove:

2.3 Extensions of Valuations and Places

29

Corollary 2.3.7: Let (E, v) ⊆ (E 0 , v 0 ) ⊆ (E 00 , v 00 ) be a tower of discrete valued fields. The following hold: (a) v 00 /v is unramified if and only if v 00 /v 0 and v 0 /v are unramified. (b) v is unramified in E 00 if and only if v is unramified in E 0 and each extension of v to E 0 is unramified in E 00 . (c) Let F1 and F2 be field extensions of E which are contained in a common field. Suppose F1 /E is separable algebraic and v is unramified in F1 and in F2 . Then v is unramified in F1 F2 . Example 2.3.8: Radical extensions. Let (E, v) be a discrete valued field and ¯v ) - n. Consider an extension F = E(x) of n a positive integer with char(E degree n of E where xn = a is in E. Let w be an extension of v to a valuation of F and let e = ew/v . Assume both v and w are normalized. Then (6)

nw(x) = ev(a)

and e ≤ n.

There are three cases to consider: Case A: gcd n, v(a) = 1. By (6), n|e, so n = e. By (2), w is the unique extension of v to F . Therefore, v totally ramifies in F . Case B: n - v(a). By (6), e 6= 1. Hence, w ramifies over E. Case C: n|v(a). Choose π ∈ E with v(π) = 1. Write a = bπ kn with k ∈ Z and b ∈ E such that v(b) = 0. Then y = xπ −k satisfies y n = b ¯v )s into distinct linear and F = E(y). Moreover, Y n − ¯b decomposes over (E factors. Therefore, by Lemma 2.3.4, v is unramified in F . Example 2.3.9: Artin-Schreier Extensions. Let (E, v) be a discrete valued field of positive characteristic p. An Artin-Schreier extension F of degree p has the form E(x) where xp − x = a with a ∈ E. We consider two cases: Case A: v(a) < 0 and p - v(a). Let w be an extension of v to F . Then w(x) must be negative and w(xp ) < w(x). Hence, pw(x) = ev(a), where e = ew/v . Hence, p = e and w(x) = v(a). Thus, v totally ramifies in F . Case B: v(a) ≥ 0. Then X p − X − a ¯ is a separable polynomial. By Lemma 2.3.4, v is unramified in F . Qp−1 ¯v . Hence, by In particular, if v(a) > 0, then X p − X = i=0 (X − i) in E Proposition 2.3.2, v has exactly p extensions to F . Label them v0 , . . . , vp−1 with vi (x − i) > 0, i = 0, . . . , p − 1. Since vi (x − i) < vi (x − i)p , we conclude from (x − i)p − (x − i) = a that vi (x − i) = v(a). Lemma 2.3.10 (Eisenstein’s Criterion): Let R be a unique factorization domain, p a prime element of R, and f (X) = an X n + an−1 X n−1 + · · · + a0 a polynomial with coefficients ai ∈ R. Then each of the following conditions suffices for f to be irreducible over Quot(R): (a) p - an , p divides a0 , . . . , an−1 , and p2 - a0 .

30

Chapter 2. Valuations and Linear Disjointness

(b) p - a0 , p divides a1 , . . . , an , and p2 - an . Proof of (a): See [Lang7, p. 183]. Proof of (b): By (a), the polynomial X n f (X −1 ) = an + an−1 X + · · · + a0 X n is irreducible over K. Therefore, f (X) is irreducible. Example 2.3.11: Ramification at infinity. Let K be a field, t an indeterminate, and f (X) = an X n + · · · + a0 ∈ K[X] with an 6= 0. By Eisenstein crite] ˜ rion, f (X) − t is irreducible over K(t). Choose a root x of f (X) = t in K(t). Let v = v∞ be the valuation of K(t) with v(t) = −1 which is trivial on K and let w be a valuation of K(x) lying over v. The relation an xn + · · · + a0 = t implies w(x) < 0. Hence, −ew/v = w(t) = w(f (x)) = nw(x). Since ew/v ≤ [K(x) : K(t)] ≤ n, this implies ew/v = [K(x) : K(t)] = n and w(x) = −1. Hence, v is totally ramified in K(x). In particular, w is the unique valuation of K(x) lying over K(t).

2.4 Integral Extensions and Dedekind Domains Integral extensions of Z in number fields are Dedekind domains. Although they are in general not unique factorization domain, their ideals uniquely factor as products of prime ideals. In this section we survey the concepts of integral extensions of rings and of Dedekind domains and prove that every overring of a Dedekind domain is again a Dedekind domain. Let F be a field containing an integral domain R. An element x ∈ F is integral over R if it satisfies an equation of the form xn + an−1 xn−1 + · · · + a0 = 0 with a1 , . . . , an ∈ R. The set of all elements of F which are integral over R form a ring (e.g. by Proposition 2.4.1 below), the integral closure of R in F . Call R integrally closed if R coincides with its integral closure in Quot(R). For example, every valuation ring O of F is integrally closed. Indeed, assume x ∈ F r O and x is integral over O. Then xn + an−1 xn−1 + · · · + a0 = 0 for some a0 , . . . , an−1 ∈ O. Then x−1 is in the maximal ideal m of O and 1 + an−1 x−1 + · · · + a0 x−n = 0. Thus, 1 ∈ m, a contradiction. Proposition 2.4.1 ([Lang4, p. 12]): An element x of F is integral over R if and only if every place of F finite on R is finite at x. Thus, the integral closure of R in F is the intersection of all valuation rings of F which contain R. In particular, every valuation ring of F is integrally closed. Suppose ϕ is a place of a field F and K is a subfield of F . We say that ϕ is trivial on K, or also that ϕ is a place of F/K, if ϕ(x) 6= ∞ for all x ∈ K. Then ϕ(y) 6= 0 for all y ∈ K × . Thus, ϕ maps K isomorphically onto ϕ(K). Lemma 2.4.2: Let K ⊆ L ⊆ F be a tower of fields and ϕ a place of F . Suppose ϕ is trivial on K and L is algebraic over K. Then ϕ is trivial on L. Proof: Each x ∈ L is integral over K, so by Proposition 2.4.1, ϕ(x) 6= ∞. Thus, ϕ is also trivial on L.

2.4 Integral Extensions and Dedekind Domains

31

Let S be a subring of F containing R. Call S integral over R if every element of S is integral over R. If S = R[x1 , . . . , xm ] and S is integral over R, then S is a finitely generated R-module. Indeed, every element of S is a linear αm 1 α2 combination with coefficients in R of the set of monomials xα 1 x2 · · · xm , where 0 ≤ αi < deg irr(xi , Quot(R)) . Propositions 2.3.1 and 2.4.1 give the following: Proposition 2.4.3: Let R ⊆ S be integral domains with S finitely generated as an R-algebra. Suppose S is integral over R. Then the following hold: (a) S is finitely generated as an R-module. (b) Let ϕ: R → M be a homomorphism into an algebraically closed field M . Then the set of all homomorphisms ψ: S → M that extend ϕ is finite and nonempty. Suppose R1 ⊆ R2 ⊆ R3 are integral domains. Proposition 2.4.1 implies that R3 is integral over R1 if and only if R2 is integral over R1 and R3 is integral over R2 . Call an integral domain R Noetherian if every ideal of R is finitely generated. For example, since a discrete valuation ring O is a principal ideal domain, it is integrally closed and Noetherian. If R is an integral domain and p is a prime ideal of R, then Rp =

a | a∈R b

and

b ∈ Rrp

is the local ring of R at p. It has a unique maximal ideal, pRp . If R is a Noetherian domain, then Rp is also Noetherian. If R is integrally closed, then so is Rp . T Lemma 2.4.4: Suppose R is an integral domain. Then R = T Rm , where m ranges over all maximal ideals of R. More generally, a = aRm for each ideal a of R. Proof: Suppose x belongs to each aRm . For each m, x = am /bm , with am ∈ a and bm ∈ R r m. Denote the ideal generated by all the bm ’s by b. If b 6= R, then b is contained in a maximal ideal Pm. Hence, bm ∈ m, a contradiction. Hence, b = R. In particular, 1 = m∈M bm cm where M is a P finite set of maximal P ideals, and cm ∈ R for each m ∈ M . Therefore x = m∈M xbm cm = m∈M am cm ∈ a. Let R be an integral domain with the quotient field F . A nonzero Rsubmodule a of F is said to be a fractional ideal of R if there exists a nonzero x ∈ R with xa ⊆ R. In particular, every ideal of R is a fractional ideal. Define the product, ab, of two fractional ideals a and b to be the R-submodule generated by the products ab, with a ∈ a and b ∈ b. Define the inverse of a fractional ideal a as a−1 = {x ∈ F | xa ⊆ R}. If a ∈ a, then aa−1 ⊆ R. Therefore, both ab and a−1 are fractional ideals.

32

Chapter 2. Valuations and Linear Disjointness

Proposition 2.4.5 ([Cassels-Fr¨ ohlich, p. 6]): The following conditions on an integral domain R are equivalent: (a) R is Noetherian, integrally closed, and its nonzero prime ideals are maximal. (b) R is Noetherian and the local ring, Rp , of every nonzero prime ideal p is a discrete valuation ring. (c) Every fractional ideal a is invertible (i.e. aa−1 = R). When these conditions hold, R is called a Dedekind domain. By Proposition 2.4.5, the set of all fractional ideals of a Dedekind domain R forms an Abelian group, with R being the unit. One proves that this group is free and the maximal ideals of R are free generators of this group. mr 1 m2 Thus, every ideal a of R has a unique presentation a = pm 1 p2 · · · pr , as the product of powers of maximal ideals with positive exponents [CasselsFr¨ohlich, p. 8]. Every principal ideal domain is a Dedekind domain. Thus, Z and K[x], where x is a transcendental element over a field K, are Dedekind domains. By the same reason, every discrete valuation ring is a Dedekind domain. In the notation of Proposition 2.4.5(b), Rp is the valuation ring of a discrete valuation vp of K = Quot(R). The corresponding place ϕp is finite on R. Conversely, if ϕ is such a place, then p = {x ∈ R | ϕ(x) = 0} is a nonzero prime ideal of R. Since Rp ⊆ Oϕ , Lemma 2.2.5 implies that Rp = Oϕ . This establishes a bijection between the nonzero prime ideals of R and the equivalence classes of places of K finite on R. Proposition 2.4.6 ([Cassels-Fr¨ ohlich, p. 13]): Let S be the integral closure of a Dedekind domain R in a finite algebraic extension of Quot(R). Then S is also a Dedekind domain. Let p be a prime ideal of R. Then pS = Pe11 Pe22 · · · Perr , where P1 , P2 , . . . , Pr are the distinct prime ideals of S that lie over p; that is, Pi ∩ R = p, i = 1, . . . , r. For each i we have pSPi = Pei i SPi . Hence, ei is the ramification index of vPi over vp . We say Pi is unramified over K if vPi /vp is unramified; that is, ei = 1 and S/Pi is a separable extension of R/p. The prime ideal p is unramified in L if each Pi is unramified over K. By Proposition 2.4.6, the integral closure of Z in a finite extension L of Q is a Dedekind domain, OL , called the ring of integers of L. Proposition 2.4.7 (Noether-Grell): Every overring R0 of a Dedekind domain R is a Dedekind domain. Proof: We show that R0 satisfies Condition (b) of Proposition 2.4.5. Part A: An injective map. If p0 is a nonzero prime ideal of R0 , then p = R ∩ p0 is a nonzero prime ideal of R. Indeed, for 0 6= x ∈ p0 , write x = ab , where a, b ∈ R. Thus, 0 6= a = bx ∈ R ∩ p0 = p. Since Rp ⊆ Rp0 0 and Rp is a discrete valuation ring, Lemma 2.2.5 implies that Rp = Rp0 0 . Hence, (1)

pRp = p0 Rp0 0 .

2.4 Integral Extensions and Dedekind Domains

33

In addition, p0 Rp0 0 ∩ R0 = p0 . Therefore, the map p0 7→ R ∩ p0 from the set of nonzero prime ideals of R0 into the set of nonzero prime ideals of R is injective. Part B: A finiteness condition. Let x be a nonzero element of R0 , p0 a prime ideal of R0 which contains x, and p = R ∩ p0 . Then Rp = Rp0 0 . Hence, vp (x) > 0, where vp is the valuation of Quot(R) corresponding to p. But this relation holds only for the finitely many prime ideals of R that appear with positive exponents in the factorization of the fractional ideal xR. Hence, by Part A, x belongs to only finitely many prime ideals of R0 . Part C: The ring R0 is Noetherian. Let a be a nonzero ideal of R0 . Choose a nonzero element x ∈ a and denote the finite set of prime ideals of R0 that contain x by P . For each p ∈ P the local ring Rp0 is a discrete valuation domain. Hence, there exists ap ∈ a such that aRp0 = ap Rp0 . Denote the ideal of R0 generated by x and by all ap , for p ∈ P , by a0 . It is contained in a. To show that a is finitely generated, we need only prove that a ⊆ a0 . Indeed, consider a prime ideal q of R0 not in P . Then x ∈ / q,Tso a0 6⊆ q. Rq0 . It followsT from Lemma T2.4.4 that a0 = p∈P a0 Rp0 . Hence, a0 Rq0 = T Therefore, a ⊆ p∈P aRp0 = p∈P ap Rp0 ⊆ p∈P a0 Rp0 = a0 , as desired. Lemma 2.4.8: Let (E, v) be a discrete valued field, F1 , F2 , F finite separable extensions of E with F = F1 F2 , and w an extension of v to F . Suppose v is unramified in F1 . Then the residue fields with respect to w satisfy F¯ = F¯1 F¯2 . Proof: Choose a finite Galois extension N of E which contains F and an extension w0 of w to N . Denote the decomposition groups of w0 over E, F1 , F2 , F by DE , DF1 , DF2 , DF , respectively. Let E 0 , F10 , F20 , F 0 be the fixed fields in N of DE , DF1 , DF2 , DF , respectively. Let v 0 = w0 |E 0 . Since all valuations of N lying over v 0 are conjugate over E 0 , the definition of E 0 as the fixed field of DE implies that w0 is the unique extension of v 0 to N . Also, DF1 = Gal(N/F1 ) ∩ DE , so F1 E 0 = F10 . By Lemma 2.3.6, v 0 is unramified in F10 . Finally, by [Serre3, p. 32, Prop. 21(c)], the residue fields of E, F1 , F2 , F at w coincide with the residue fields of E 0 , F10 , F20 , F 0 at w0 , respectively. We may therefore replace E, F1 , F2 , F , respectively, by E 0 , F10 , F20 , F 0 , if necessary, to assume that w|F1 is the unique extension of v to F1 . Now put wi = w|Fi , i = 1, 2. By Proposition 2.4.1, Ow1 is the integral closure of Ov in ¯ F1 . Since v is unramified in F1 , Proposition 2.3.2 implies [F1 : E] = [F¯1 : E], where the bar denotes reduction modulo w. Choose x ∈ Ow1 such that x ¯ is a primitive element for the separable ¯ Let f = irr(x, E) and p = irr(¯ ¯ Then f ∈ Ov [X] and extension F¯1 /E. x, E). f (x) = 0. Hence, f¯(¯ x) = 0 and p|f¯. Therefore, ¯ = [F1 : E]. [F1 : E] ≥ deg(f ) ≥ deg(¯ p) = [F¯1 : E] ¯ x). Consequently, p = f¯, F1 = E(x), and F¯1 = E(¯

34

Chapter 2. Valuations and Linear Disjointness

By Lemma 2.3.6, w2 is unramified in F . Thus, we may apply the result x). Conseof the preceding paragraph to F/F2 and conclude that F¯ = F¯2 (¯ ¯ x)F¯2 = F¯2 (¯ x) = F¯ . quently, F¯1 F¯2 = E(¯

2.5 Linear Disjointness of Fields Central to field theory is the concept “linear disjointness of fields”, an analog of linear independence of vectors. We repeat the convention made in “Notation and Convention” that whenever we form the compositum of fields, we tacitly assume they are contained in a common field. Lemma 2.5.1: Let E and F be extensions of a field K. The following conditions are equivalent: (a) Each m-tuple (x1 , . . . , xm ) of elements of E which is linearly independent over K is also linearly independent over F . (b) Each n-tuple (y1 , . . . , yn ) of elements of F which is linearly independent over K is also linearly independent over E. Proof: It suffices to prove that (a) implies (b). Let y1 , . . . , yn be elements of F for which there exist a1 , . . . , an ∈ E with a1 y1 + · · · +Pan yn = 0. Let {xj | j ∈ J} be a linear basis for E over K and write ai = j∈J aij xj with aij elements of K, only finitely many different from 0. Then n X X j∈J

aij yi xj = 0.

i=1

P aij yi = 0 for By (a), {xj | j ∈ J} is linearly independent over F . Hence, every j. If y1 , . . . , ym are linearly independent over K, then aij = 0 for every i and j, so ai = 0, i = 1, . . . , m. Thus, y1 , . . . , ym are linearly independent over E. This proves (b). Definition: With E and F field extensions of a field K, refer to E and F as linearly disjoint over K if (a) (or (b)) of Lemma 2.5.1 holds. Corollary 2.5.2: Let E and F be extensions of a field K such that [E : K] < ∞. Then E and F are linearly disjoint over K if and only if [E : K] = [EF : F ]. If in addition [F : K] < ∞, then this is equivalent to [EF : K] = [E : K][F : K]. Proof: If E and F are linearly disjoint over K and w1 , . . . , wn is a basis for E/K, then w1 , . . . , wn is also a basis for EF over F . Hence, [EF : F ] = n = [E : K]. Conversely, suppose [E : K] = [EF : F ] and let x1 , . . . , xm ∈ E be linearly independent over K. Extend {x1 , . . . , xm } to a basis {x1 , . . . , xn } of E/K. Since {x1 , . . . , xn } generates EF over F and n = [EF : F ], {x1 , . . . , xn } is a basis of EF/F . In particular, x1 , . . . , xm are linearly independent over F .

2.5 Linear Disjointness of Fields

35

Let E/K be a finite Galois extension. If E ∩ F = K, then, by Corollary 2.5.2, E and F are linearly disjoint over K. The condition, E ∩ F = K is equivalent to “res: Gal(EF/F ) → Gal(E/K) is an isomorphism” and also to “res: Gal(F ) → Gal(E/K) is surjective.” For arbitrary extensions this condition is clearly necessary, but not sufficient. Let L be a degree n > 1 extension of K for which L0 is conjugate to L over K and L0 ∩ L = K. Then 2.5.2, L and L0 are not [LL0 : K] ≤ n(n − 1). Thus, according to Corollary √ 3 linearly √ disjoint over K. For example, Q( 2) is not linearly disjoint from Q(ζ3 3 2) over Q although their intersection is Q. Lemma 2.5.3 (Tower Property): Let K ⊆ E and K ⊆ L ⊆ F be four fields. Then E is linearly disjoint from F over K if and only if E is linearly disjoint from L over K and EL is linearly disjoint from F over L. Proof: The only nontrivial part is to show that if E and F are linearly disjoint over K, then EL and F are linearly disjoint over L. Apply Lemma 2.5.1. Suppose that y1 , . . . , ym are elements of F which are linearly independent over L, but a1 , . . . , am are elements of EL such Pm that Pi=1 ai yi = 0. Clear denominators to assume that ai ∈ L[E], so that xj with aij ∈ L, where {xj | j ∈ J} is a linear basis for E over K. ai = PaijP yi )xj = 0. By assumption, the xj are linearly independent Then j ( i aijP over F . Hence, j aij yi = 0, so aij = 0 for all i and j. Consequently, ai = 0, i = 1, . . . , m. Lemma 2.5.4: Let L be a separable algebraic extension of a field K and let M be a purely inseparable extension of K. Then L and M are linearly disjoint over K. ˆ be the Galois closure of L/K. Then L ˆ ∩ M = K. Hence, L ˆ Proof: Let L and M are linearly disjoint over K. Therefore, by Lemma 2.5.3, L and M are linearly disjoint over K. Let E1 , . . . , En be n extensions of a field K. We say that E1 , . . . , En are linearly disjoint over K if E1 · · · Em−1 and Em are linearly disjoint over K for m = 2, . . . , n. Induction on n shows that this is the case if of and only if the following condition holds: If wi,ji , ji ∈ Ji , are elements Qn Ei which are linearly independent over K, i = 1, . . . , n, then i=1 wi,ji , (j1 , . . . , jn ) ∈ J1 × · · · × Jn , are linearly independent over K. It follows that E1 , . . . , En are linearly disjoint over K if and only if the canonical homomorphism of E1 ⊗K · · · ⊗K En into E1 · · · En that maps x1 ⊗ · · · ⊗ xn onto x1 · · · xn is injective. It also follows that if E1 , . . . , En are linearly disjoint over K, then Eπ(1) , . . . , Eπ(n) are linearly disjoint over K for every permutation π of {1, . . . , n}. The application of tensor products makes the following lemma an easy observation. Lemma 2.5.5: Let E1 , . . . , En (resp. F1 , . . . , Fn ) be linearly disjoint field extensions of K (resp. L). For each i between 1 and n let ϕi : Ei → Fi ∪ {∞},

36

Chapter 2. Valuations and Linear Disjointness

be either a place or an embedding. Suppose ϕ1 , . . . , ϕn coincide on K and ϕi (K) = L, i = 1, . . . , n. Let E = E1 · · · En and F = F1 · · · Fn . Then there exists a place ϕ: E → F˜ ∪ {∞} that extends each of the ϕi ’s. If each ϕi is an isomorphism of Ei onto Fi , then ϕ is an isomorphism of E onto F . Proof: Let Oi be the valuation ring of ϕi if ϕi is a place and Ei if ϕi is an isomorphism. By assumption, the map x1 · · · xn → x1 ⊗ · · · ⊗ xn is an isomorphism O1 · · · On ∼ = O1 ⊗K · · · ⊗K On of rings. Hence, there exists a ring homomorphism ϕ0 : O1 · · · On → F such that ϕ0 (x) = ϕi (x) for each x ∈ Oi , i = 1, . . . , n. Extend ϕ0 to a place ϕ: E → F˜ ∪ {∞} (Proposition 2.3.1). If x ∈ Ei r Oi , then ϕ(x−1 ) = ϕi (x−1 ) = 0, so ϕ(x) = ϕi (x) = ∞. We conclude that ϕ coincides with ϕi on Ei . Finally, define a family {Ei | i ∈ I} of field extensions of K to be linearly disjoint over K if every finite subfamily is linearly disjoint over K. It follows from the discussion preceding Lemma 2.5.5 that a sequence (E1 , E2 , E3 , . . .) of fields extensions of K is linearly disjoint over K if En is linearly disjoint from E1 · · · En−1 for 2, 3, 4, . . . . Then, Eπ(1) , Eπ(2) , Eπ(3) , . . . are linearly disjoint for every permutation π of N. disjoint family of Galois extenLemma 2.5.6: Let {Li | i ∈ I} Q be a linearlyQ sions of a field K. Then Gal( i∈I Li /K) ∼ = i∈I Gal(Li /K). Q Q Proof: Since i∈I Gal(Li /K) ∼ = lim i∈I0 Gal(Li /K), we may assume I ←− Q Q is finite. In this case, the embedding Gal( i∈I Li /K) → i∈I Gal(Li /K) given by σ 7→ (σ|Li )i∈I is surjective (Lemma 2.5.5). Therefore, it is an isomorphism. Lemma 2.5.7: Let K be a field, K1 , K2 , K3 , . . . a linearly disjoint sequence of extensions of K, and L a finite separable extension of K. Then there exists a positive integer n such that L, Kn , Kn+1 , Kn+2 , . . . are linearly disjoint over K. Proof: Replace L by its Galois closure over K, if necessary, to assume L is Galois over K. Assume for each positive integer n the field L is not linearly disjoint from Kn Kn+1 Kn+2 · · · over K. Then Ln = L ∩ Kn Kn+1 Kn+2 · · · is a proper extension of K. Since L has only finitely many extensions that contain K and since Ln ⊇ Ln+1 ⊇ Ln+2 ⊇ · · ·, there is an m such that Ln = Lm for all n ≥ m. Since Lm is a finite extension of K, there is an n > m with Lm ⊆ Km · · · Kn−1 . Similarly, there exists r > n with Lm ⊆ Kn · · · Kr−1 . By assumption, Km · · · Kn−1 and Kn · · · Kr−1 are linearly disjoint over K. In particular, their intersection is K. Therefore, Lm = K. This contradiction proves there exists n such that L, Kn , Kn+1 , Kn+2 , . . . are linearly disjoint over K. Lemma 2.5.8: Let v be a discrete valuation of a field K and L, M finite extensions of K. Suppose v is unramified in L but totally ramified in M . Then L and M are linearly disjoint over K.

2.5 Linear Disjointness of Fields

37

Proof: Let L0 be the maximal separable extension of K in L and v0 an extension of v to L0 . Then L/L0 is purely inseparable. Hence, v0 is ramified in L. Therefore, L = L0 and L/K is separable. Since v is unramified in each of the conjugates of L over K, it is unramified in their compositum (Corollary 2.3.7). We may therefore replace L by the Galois closure of L/K, if necessary, to assume L/K is Galois. Let m = [L ∩ M : K]. Choose an extension w of v to L ∩ M . Then e(w/v) = 1 on one hand and e(w/v) = m on the other hand. Thus, L ∩ M = K. Therefore, L is linearly disjoint from M over K. Example 2.5.9: Roots of unity. For each n consider the Galois extension Q(ζn ) of Q obtained by adjoining a primitive root of unity of order n. It is well known that ϕ(n) = [Q(ζn ) : Q] is the number of integers between 1 and n which are relatively prime to n [Lang7, p. 278, Thm. 3.1]. If m is relatively prime to n, then ϕ(mn) = ϕ(m)ϕ(n) [LeVeque, p. 28, Thm. 3-7]. In addition, Q(ζm , ζn ) = Q(ζmn ). Hence, [Q(ζm , ζn ) : Q] = [Q(ζmn ) : Q] = ϕ(mn) = ϕ(m)ϕ(n) = [Q(ζm ) : Q][Q(ζn ) : Q]. It follows from Corollary 2.5.2 that Q(ζm ) and Q(ζn ) are linearly disjoint over Q. Here is an application of linear disjointness to integral closures of domains. Lemma 2.5.10: Let K be a field, L a separable algebraic extension of K, and R an integrally closed integral domain containing K. Let E = Quot(R), F = EL, and S the integral closure of R in F . Suppose E and L are linearly disjoint over K. Then S = RL ∼ = R ⊗K L. Proof: Assume without loss L/K is finite. Choose a basis w1 , . . . , wn for L/K. Let σ1 , . . . , σn be the distinct K-embeddings of L into Ks . Then det(σi wj ) 6= 0. Each element of L is integral over K, hence over R, so RL ⊆ S. Conversely, let xP ∈ S. By the linear disjointness, w1 , . . . , wn form a basis for F/E. n Hence, x = j=1 ej wj with ej ∈ E, j = 1, . . . , n. Also, each σi extends to Pn an E-embedding of F into Es (Lemma 2.5.5). Thus, σi x = j=1 ej σi wj , i = 1, . . . , n. Apply Kramer’s law to present each ek as a polynomial in σi x, σi wj , with i, j = 1, . . . , n, divided by det(σi wj ). Thus, ek is an element of E which is integral over R. Since R is integrally closed, ek ∈ R, k = 1, . . . , n. Consequently, x ∈ RL, as needed. We generalize the tower property to families of field extensions: Lemma 2.5.11: Let K be a field and I a set. For each i ∈ I let Fi /Ei be a field extension with K ⊆ Ei . Suppose {Fi | i ∈ I} is linearly disjoint over K. Denote the compositum of all Ei ’s by E. Then the set {Fi E | i ∈ I} is linearly disjoint over E. Moreover, for each i ∈ I, the field Fi is linearly disjoint from E over Ei . Proof: It suffices to consider the case where I = {1, 2, . . . , n}. By induction suppose Fi E1 · · · En−1 , i = 1, . . . , n − 1, are linearly disjoint over E1 · · · En .

38

Chapter 2. Valuations and Linear Disjointness

By assumption, F1 · · · Fn−1 is linearly disjoint from Fn over K. Hence, by the tower property, F1 · · · Fn−1 is linearly disjoint from E over E1 , . . . , En−1 , so Fi E, i = 1, . . . , n − 1, are linearly disjoint over E. F1 · · · Fn−1

F1 · · · Fn−1 E

Fi E1 · · · En−1

Fi E

E1 · · · En−1

E

EFn

K

En

Fn

Moreover, F1 · · · Fn−1 E is linearly disjoint from EFn over E. Consequently, E is linearly disjoint from Fn over En and Fi E, i = 1, . . . , n are linearly disjoint over E, as claimed.

2.6 Separable, Regular, and Primary Extensions Based on the notion of linear disjointness we define here three type of field extensions. We say that a field extension F/K is separable (resp. regular, ˜ Ks ) over K. primary) if F is linearly disjoint from Kins (resp. K, Separable Extensions. We generalize the notion of “separable algebraic extension” to arbitrary field extensions. Let K be a field of positive characteristic p. The field generated over K by the pth roots of all elements of K is denoted K 1/p . We denote the ∞ maximal purely inseparable extension of K by Kins (or K 1/p ). Let F be a finitely generated extension of K. A collection t1 , . . . , tr ∈ F of elements algebraically independent over K is a separating transcendence basis if F/K(t1 , . . . , tr ) is a finite separable extension. Lemma 2.6.1: An extension F of a field K is separable if it satisfies one of the following equivalent conditions: (a) F is linearly disjoint from Kins over K. (b) F is linearly disjoint from K 1/p over K. (c) Every finitely generated extension E of K which is contained in F has a separating transcendence basis. Moreover, a separating transcendence basis can be selected from a given set of generators for F/K. Proof: The implications “(a) => (b)” and “(c) => (a)” are immediate consequences of the tower property (Lemma 2.5.3). For “(b) => (c)” see [Lang 4, p. 54]. Lemma 19.2.4 gives a constructive proof. In particular, every separable algebraic extension satisfies conditions (a), (b), and (c) of Lemma 2.6.1. Now apply the rules of linear disjointness.

2.6 Separable, Regular, and Primary Extensions

39

Corollary 2.6.2: (a) If E/K and F/E are separable extensions, then F/K is also separable. (b) If F/K is a separable extension, then E/K is separable for every field K ⊆ E ⊆ F. (c) Every extension of a perfect field is separable. (d) If E/K is a purely inseparable extension and F/K is a separable extension, then E and F are linearly disjoint over K. Example 2.6.3: A separable tower does not imply separable steps. Consider the tower of fields Fp ⊂ Fp (tp ) ⊂ Fp (t), where t is transcendental over Fp . The extension Fp (t)/Fp is separable, but Fp (t)/Fp (tp ) is not. Regular Extensions. Finitely generated regular extensions characterize absolutely irreducible varieties (Section 10.2) Lemma 2.6.4: A field extension F/K is regular if it satisfies one of the following equivalent conditions: (a) F/K is separable and K is algebraically closed in F . ˜ over K. (b) F is linearly disjoint from K Proof: The implication “(b) => (a)” is immediate. To prove “(a) => (b)”, it suffices to assume that F/K is finitely generated. Then F/K has a separating transcendence basis, t1 , . . . , tr , which is also a separating transcendence basis for the extension F Ks /Ks . Since ˜ over ˜ = (Ks )ins , Lemma 2.6.1 implies that F Ks is linearly disjoint from K K Ks . Also, Ks /K is a Galois extension and F ∩ Ks = K. Hence, F is linearly disjoint from Ks over K. Therefore, by Lemma 2.5.3, F is linearly disjoint ˜ over K. from K Corollary 2.6.5: (a) If E/K and F/E are regular extensions, then F/K is regular. (b) If F/K is a regular extension, then E/K is regular for every field E lying between K and F . (c) Every extension of an algebraically closed field is regular. (d) Let m be a cardinal number andSKα , α ≤ m, an ascending transfinite sequence of fields such that Kγ = α (a).” The implication “(a) => (b)” holds since F ∩ Ks = K and Ks /K is a Galois extension. Corollary 2.6.14: (a) If E/K and F/E are primary extensions, then so is F/K. (b) If F/K is a primary extension, then E/K is primary, for every field K ⊆ E ⊆ F. (c) Every extension of a separably closed field is primary. (d) An extension F/K is regular if and only if it is separable and primary. Lemma 2.6.15: (a) Let E be a primary extension of a field K which is algebraically independent from an extension F of K. Then EF is a primary extension of F. (b) If two primary extensions E and F of K are algebraically independent, then EF/K is primary. Proof: Assertion (b) follows from (a) and from Corollary 2.6.14(a). To prove (a), choose a transcendence base T for E/K and let M be the maximal separable extension of K(T ) in E. Then M is a separable and primary extension of K. Hence, by Lemma 2.6.14(d), it is regular. Also, M is algebraically independent from Fs over K. By Lemma 2.6.7, M F is linearly disjoint from Fs over F . Since EF is a purely inseparable extension of M F , it is linearly disjoint from M Fs . It follows that EF is linearly disjoint from Fs over F ; that is, EF is a primary extension of F .

2.7 The Imperfect Degree of a Field We classify fields of positive characteristic by their imperfect degree and characterize those fields for which every finite extension has a primitive element as fields of imperfect degree 1. Let F be a field of positive characteristic p. Consider a subfield F0 of F that contains the field F p of all pth powers in F . Observe that for x1 , . . . , xn ∈ F , the set of monomials (1)

xi11 · · · xinn ,

0 ≤ i1 , . . . , in ≤ p − 1,

generates F0 (x) over F0 . Hence, [F0 (x) : F0 ] ≤ pn . If [F0 (x) : F0 ] = pn , then x1 , . . . , xn are said to be p-independent over F0 . Equivalently, each of the fields F0 (x1 ), . . . , F0 (xn ) has degree p over F0 and they are linearly disjoint over F0 . This means that the set of monomials (1) is linearly independent over F0 . A subset B of F is p-independent over F0 , if every finite subset of B is p-independent over F0 . If in addition F0 (B) = F , then B is said to be a p-basis for F over F0 . As in the theory of vector spaces, each maximal p-independent subset of F over F0 is a p-basis for F over F0 .

2.7 The Imperfect Degree of a Field

45

If x1 , . . . , xn ∈ F are p-independent over F p , we call them p-independent elements of F . The p-power pn = [F : F p ] is the imperfect degree of F and n is the imperfect exponent of F . We say that F is n-imperfect. Thus, a perfect field has imperfect exponent 0. Both quantities are infinite if [F : F p ] = ∞. In this case F is ∞-imperfect. Lemma 2.7.1 (Exchange Principle): Let F0 be a subfield of F which contains F p. (a) Let x1 , . . . , xm , y1 , . . . , yn ∈ F be such that x1 , . . . , xm are p-independent over F0 and x1 , . . . , xm ∈ F0 (y1 , . . . , yn ). Then m ≤ n, and there is a reordering of y1 , . . . , yn so that y1 , . . . , ym ∈ F0 (x1 , . . . , xm , ym+1 , . . . , yn ). (b) Every subset of F which is p-independent over F0 extends to a p-basis for F over F0 . Proof: We use induction on m. Assume the lemma is true for m = k. Thus, for m = k + 1 we may assume that xk+1 ∈ F0 (x1 , . . . , xk , yk+1 , . . . , yn ) = F1 . Then [F1 : F0 ] ≤ pn and there exists l between k + 1 and n such that yl ∈ F0 (x1 , . . . , xk+1 , yk+1 , . . . , yl−1 ), since otherwise [F1 : F0 ] ≥ pn+1 , a contradiction. Thus, yl can be exchanged for xk+1 . This proves the first part of the lemma for m = k + 1. For the last part start from a subset A of K which is p-independent over F0 . Use Zorn’s lemma to prove the existence of a maximal subset B of F which contains A and which is p-independent over F0 . Then B is a p-basis of F over F0 . Lemma 2.7.2: Suppose F is a finitely generated extension of transcendence degree n of a perfect field K of positive characteristic p. Then the imperfect exponent of F is n. Proof: Choose a separating transcendence basis t1 , . . . , tn for F/K. Then K(t)p = K(tp ) and t1 , . . . , tn is a p-basis for K(t)/K(tp ); that is, [K(t) : K(tp )] = pn . Since K(t) is a purely inseparable extension of K(tp ) and F p is a separable extension of K(tp ), these extensions of K(tp ) are linearly disjoint. Also, F is both a separable extension and a purely inseparable extension of K(t)F p . Hence, F = K(t)F p . Consequently, [F : F p ] = [K(t) : K(tp )] = pn , as claimed. Lemma 2.7.3: Let B a subset of F which is p-independent over F p and F 0 a separable extension of F . Then B is p-independent over (F 0 )p . If, in addition, F 0 is separable algebraic over F , then the imperfect degree of F 0 is equal to that of F . Proof: Assume without loss that B consists of n elements. Then [(F 0 )p (B) : (F 0 )p ] = [F p (B) : F p ] = pn . Hence, B is p-independent over (F 0 )p .

46

Chapter 2. Valuations and Linear Disjointness

Suppose now F 0 /F is separably algebraic. Then F 0 is both separably and purely inseparable over F (F 0 )p , so, F 0 = F (F 0 )p . Hence, [F 0 : (F 0 )p ] = [F : F p ]. Therefore, the imperfect degree of F 0 is equal to that of F . Lemma 2.7.4: Let K be a field of positive characteristic p, let a, b1 , . . . , bm be p-independent elements of K, and let x1 , . . . , xm be algebraically independent over K. Suppose y1 , . . . , ym satisfy (2)

axpi + bi yip = 1,

i = 1, . . . , m.

Then K is algebraically closed in K(x, y) = Km . Proof: We use induction on m. Part A: m = 1. Let x = x1 , y = y1 , and b = b1 and assume that u is a nonzero element of K1 which is algebraic over K. Then u is also algebraic over K(a1/p , b1/p ). But K(x, y, a1/p , b1/p ) = K(x, a1/p , b1/p ) is a purely transcendental extension of K(a1/p , b1/p ). Hence, u ∈ K(a1/p , b1/p ) and therefore up ∈ K. Write (3)

u=

h0 (x) h1 (x) hk (x) k + y + ··· + y h(x) h(x) h(x)

with k ≤ p − 1, h(x), h0 (x), . . . , hk (x) ∈ K[x] and h(x), hk (x) 6= 0. With no loss we may assume that x does not divide the greatest common divisor of h(x), h0 (x), . . . , hk (x). Raise (3) to the pth power, multiply it by h(x)p and substitute y p = (1 − axp )b−1 to obtain: (4) (h(x)u)p = h0 (x)p + h1 (x)p (1 − axp )b−1 + · · · + hk (x)p (1 − axp )k b−k . If h(0) = 0, then the substitution x = 0 in (4) gives 0 = h0 (0)p + h1 (0)p b−1 + · · · + hk (0)p b−k , Therefore, h0 (0) = h1 (0) = · · · = hk (0) = 0, contrary to assumption. Thus, we may assume h(0) 6= 0. Then the substitution x = 0 in (4) shows that u ∈ K(b1/p ). Similarly, u ∈ K(a1/p ). Since a and b are p-independent in K, u ∈ K(a1/p ) ∩ K(b1/p ) = K. Thus, K is algebraically closed in K(x, y). Part B: Induction. Assume the Lemma is true for m − 1. Then K is algebraically closed in Km−1 = K(x1 , . . . , xm−1 , y1 , . . . , ym−1 ). If we prove that a and bm are p-independent in Km−1 , then with Km−1 replacing K in Part A, Km−1 is algebraically closed in Km , so K is algebraically closed in Km . Since x1 , . . . , xm are algebraically independent over K, the field 1/p 1/p 1/p K(a , b1 , . . . , bm ) is linearly disjoint from Em−1 = K(x1 , . . . , xm−1 ) over K. Thus, (5)

1/p

m+1 . [Em−1 (a1/p , b1 , . . . , b1/p m ) : Em−1 ] = p

2.7 The Imperfect Degree of a Field

47

Also, from (2) Km−1 = Em−1 (y1 , . . . , ym−1 )

and 1/p

1/p Km−1 (a1/p , b1/p , b1 , . . . , b1/p m ) = Em−1 (a m ).

Thus, (6)

[Km−1 : Em−1 ] ≤ pm−1

and

2 [Km−1 (a1/p , b1/p m ) : Km−1 ] ≤ p .

Combine (5) and (6) to conclude that (6) consists of equalities. In particular, a and bm are p-independent in Km−1 . Lemma 2.7.5: The following conditions on a field K of positive characteristic p are equivalent: (a) The imperfect exponent of K is at most 1. (b) Every finite extension of K has a primitive element. (c) If K is algebraically closed in a field extension F , then F is regular over K. Proof: If K is perfect, then (a), (b), and (c) are true. Therefore, we may assume char(K) = p > 0 and K is imperfect. Proof of “(a) =⇒ (b)”: By assumption, [K 1/p : K] = [K : K p ] = p. Hence, K1 = K 1/p is the unique purely inseparable extension of K of degree p. n Moreover, K1 = K(a1/p ) for some a ∈ K, so Kn = K(a1/p ) is a purely inseparable extension of K of degree pn . Assume that for each m ≤ n, Km is the unique purely inseparable extension of K of degree pm . Let L be a purely inseparable extension of K of degree pn+1 . If we prove that L = Kn+1 , then we may conclude by induction that each finite purely inseparable extension of K has a primitive element. To this end choose x ∈ L r Kn . Let m be the smallest positive integer m with xp ∈ K. Then K(x) is a purely inseparable extension of K of degree pm . If m ≤ n, then by the induction hypothesis K(x) = Km ⊆ Kn , so x ∈ Kn . This contradiction proves that m = n + 1 and L = K(x). The same argument implies that xp ∈ Kn . Hence, with q = pn , we have Pq−1 n p x = i=0 ci ai/p for some c0 , . . . , cq−1 ∈ K. Therefore, x=

q−1 X

1/p

ci ai/p

n+1

∈ K1 (a1/p

n+1

) = Kn+1 .

i=0

It follows that L ⊆ Kn+1 . As both fields have degree pn+1 over K, they coincide, as desired. Now let E be a finite extension of K. Denote the maximal separable extension of K in E by E0 . By the primitive element theorem, E0 = K(x). Since E0 is both separable and purely inseparable over KE0p we have E0 = KE0p . Therefore [E0 : E0p ] = [K : K p ] = p. Apply the first part of the proof

48

Chapter 2. Valuations and Linear Disjointness

to E0 and conclude that E = E0 (y), for some element y. Thus, E = K(x, y) with x separable over K. By [Waerden3, §6.10], E/K has a primitive element Proof of “(b) =⇒ (c)”: Let K(x) be a finite extension of K and let f = irr(x, K). If K is algebraically closed in F , then f remains irreducible over F . Otherwise, its factors would have coefficients algebraic over K and in F , and therefore in K. Thus, F is linearly disjoint from K(x) over K. Hence, (b) implies that F is regular over K. Proof of “(c) =⇒ (a)”: Assume a and b are p-independent elements of K. Then [K(a1/p , b1/p ) : K] = p2 . Let x and y be transcendental elements over K with axp + by p = 1. Put F = K(x, y). By Lemma 2.7.4, K is algebraically closed in F . Hence, by (c), F is regular over K. Therefore, [F (a1/p , b1/p ) : F ] = [K(a1/p , b1/p ) : K] = p2 . On the other hand, , F (a1/p ) = F (b1/p ), so [F (a1/p , b1/p ) : F ] ≤ p. This contradiction proves that the imperfect exponent of K is at most 1. Remark 2.7.6: Relative algebraic closedness does not imply regularity. Let K be a field of positive characteristic p. Suppose K has p-independent elements a, b (e.g. K = Fp (t, u) where t, u are algebraically independent over Fp ). Let x, y be transcendental elements over K with axp + by p = 1. Put F = K(x, y). The proof of “(c) =⇒ (a)” then shows that K is algebraically closed in F but F is not linearly disjoint from K 1/p over K. Thus, F is not a separable extension of K. A fortiori, F/K is not regular.

2.8 Derivatives We develop a criterion for a finitely generated field extension of positive characteristic p to be separable in terms of derivatives.. Definition 2.8.1: A map D: F → F is called a derivation of the field F if D(x + y) = D(x) + D(y) and D(xy) = D(x)y + xD(y) for all x, y ∈ F . If D vanishes on a subfield K of F , then D is a derivation of F over K (or a K-derivation). Let F (x) be a field extension of F and f ∈ F [X]. Suppose D extends to F (x). Then D satisfies the classical chain rule: (1)

D(f (x)) = f D (x) + f 0 (x)D(x),

where f D is the polynomial obtained by applying D to the coefficients of f and f 0 is the usual derivative of f . There are three cases: Case 1: x is separably algebraic over F . Then, with f = irr(x, F ), f 0 (x) 6= 0. By (1), 0 = f D (x) + f 0 (x)D(x). Thus, D extends uniquely to F (x). Case 2: x is transcendental. Then D extends to F (x) by rule (1) and D(x) may be chosen arbitrarily.

2.8 Derivatives

49 m

Case 3: x satisfies xp = a ∈ F , for some m. Then D extends to F (x) if and only if D(a) = 0. In this case D(x) may be chosen arbitrarily. Lemma 2.8.2: A necessary and sufficient condition for a finitely generated extension F/K to be separably algebraic is that 0 is the only K-derivation of F . Proof: Necessity follows from Case 1. Now suppose F/K is not separably algebraic. Then we may write F = K(x1 , . . . , xn ) such that xi is transcendental over K(x1 , . . . , xi−1 ) for i = 1, . . . , k, xi is separably algebraic over K(x1 , . . . , xi−1 ) for i = k+1, . . . , l, and xi is purely inseparable over K(x1 , . . . , xi−1 ) for i = l + 1, . . . , n. Moreover, either n > l or n = l and k > 0. If n > l, then Case 1 allows us to extend the zero derivation of K(x1 , . . . , xn−1 ) to a nonzero derivation of F . If n = l and k > 0, then by Case 2, the zero derivation of K(x1 , . . . , xk−1 ) extends to a nonzero derivation D of K(x1 , . . . , xk ). Applying Case 3 several times, we may then extend D to a derivation of F . Lemma 2.8.3: Let F/K be a finitely generated extension of positive characteristic p and transcendence degree n. Then F/K is separable if and only if [F : KF p ] = pn . In this case t1 , . . . , tn form a p-basis for F over KF p if and only they form a separating transcendence basis for F/K. Proof: Suppose first [F : KF p ] = pn . Let t1 , . . . , tn be a p-basis for F/KF p . Every derivation D of F vanishes on F p . If D vanishes on K(t), it vanishes on F = K(t) · F p . By Lemma 2.8.2, F/K(t) is separably algebraic and t1 , . . . , tn is a separating transcendence basis for F/K. Conversely, suppose F/K is separable. Let t1 , . . . , tn be a separating transcendence basis for F/K. The extension F/K(t) · KF p is both separable and purely inseparable. Hence, F = K(t) · KF p . Since F p /K(t)p is separably algebraic and since K(tp )F p = KF p , we conclude that KF p /K(tp ) is separably algebraic. K(t) F

K(tp )

KF p

K(t)p

Fp

Therefore, KF p is linearly disjoint from K(t) over K(tp ), and [F : KF p ] = [K(t) : K(tp )] = pn . Moreover, t is a p-basis for F/KF p . Corollary 2.8.4: Let F/K be a finitely generated separable extension of positive characteristic p and let t ∈ F . (a) If there exists a derivation D of F/K such that D(t) 6= 0, then F is a separable extension of K(t).

50

Chapter 2. Valuations and Linear Disjointness

(b) If t is transcendental over K and F/K(t) is separable, then there exists a derivation D of F/K such that D(t) 6= 0. Proof of (a): By assumption, t ∈ / KF p . Let n = trans.deg(F/K). By p n Lemma 2.8.3, [F : KF ] = p . Hence, t can be extended to a p-basis t, t2 , . . . , tn for F/KF p . Again, by Lemma 2.8.3, t, t2 , . . . , tn is a separating transcendence basis for F/K. Therefore, F is a separable extension of K(t). Proof of (b): Let t2 , . . . , tn be a separating transcendence basis for F/K(t). By Case 2, there exists a derivation D0 of K(t, t2 , . . . , tn )/K such that D0 (t) = 1, D0 (t2 ) = 0, . . . , D0 (tn ) = 0. By Case 1, D0 extends to a derivation D of F/K.

Exercises 1. Let O be a valuation ring of a field F and consider the subset m = {x ∈ / O}. Show that if x ∈ m and a ∈ O, then ax ∈ m. Prove that m is O | x−1 ∈ closed under addition. Hint: Use the identity x + y = (1 + xy −1 )y for y 6= 0. Show that m is the unique maximal ideal of O. 2.

Use Exercise 1 to prove that every valuation ring is integrally closed.

3. Let v be a valuation of Q. Observe that v(n) ≥ v(1) = 0, for each n ∈ N. Hence, there exists a smallest p ∈ N such that v(p) > 0. Prove that p is a prime element of Ov and v is equivalent to vp . Hint: If a positive integer m is relatively prime to p, then there exist x, y ∈ Z such that xp + ym = 1. 4. Let v be a valuation of the rational function field F = K(t) which is trivial on K. Suppose there exists p ∈ K[t] with v(p) > 0. Now suppose p has smallest degree with this property. Show that v is equivalent to vp . Otherwise, there exists f ∈ K[t] such that v(f (t)) < 0. Conclude that v(t) < 0, and that v is equivalent to v∞ . 5. Let F/E be a field extension, w a valuation of F , and x1 , . . . , xe elements of F such that w(x1 ), . . . , w(xe ) represent distinct classes of w(F × ) modulo w(E × ). Show that x1 , . . . , xe are linearly independent over E. Thus, (w(F × ) : w(E × )) ≤ [F : E]. Hint: Use (4b) of Section 2.1. 6. Let ∆ be an ordered group containing Z as a subgroup of index e. Show there exists no positive element δ ∈ ∆ such that eδ < 1. Conclude that ∆ contains a smallest positive element and hence that ∆ ∼ = Z. Combine this with Exercise 5 to prove that if the restriction of w to E is discrete, then w is discrete. 7. In the notation of Exercise 5, let v be the restriction of w to E. Let y1 , . . . , yf be elements of F with w(y1 ), . . . , w(yf ) ≥ 0 with residue classes ¯v . Show that y1 , . . . , yf are linearly y¯1 , . . . , y¯f linearly independent over E ¯v ] ≤ [F : E]. Hint: If a1 , . . . , af ∈ ¯ independent over E. Conclude that [Fw : E

Notes

F are not all zero, then there exists j, 1 ≤ j ≤ f such that v 0.

51 a1 aj

,...,v

af aj

≥

8. Let v be a discrete valuation of a field K and let w be an extension of v to a finite Galois extension L of K. Assume that w0 is also an extension of v to L such that w0 6= σ(w) for all σ ∈ Gal(L/K). Combine Exercise 7 with Proposition 2.1.1 to produce x ∈ L such that w0 (x) > 0 and w(σx−1) > 0 for all σ ∈ Gal(L/K). With y = NL/K (x), conclude that the former condition gives v(y) > 0, while the latter implies v(y − 1) > 0. Use this contradiction to prove that Gal(L/K) acts transitively on the extensions of v to L. 9. Let L, K1 , . . . , Kn be extensions of a field K. Let Li = Ki L, i = 1, . . . , n. Suppose Ki is linearly disjoint from L over K for i = 1, . . . , n and L1 , . . . , Ln are linearly disjoint over L. Prove that K1 , . . . , Kn are linearly disjoint over K. 10. Let v be a discrete valuation of a field K and let L and M be two finite extensions of K such that v is unramified in L and totally ramified in M . Prove that L and M are linearly disjoint over K. Hint: Consider the Galois ˆ of L/K. hull L 11. Let E be a regular extension of a perfect field K and let F be a purely inseparable extension of E. Prove that F/K is a regular extension. 12. Let K be a field algebraically closed in an extension F . Prove that K(x) ˜ Hint: Check the irreducibility of is linearly disjoint from F for every x ∈ K. irr(x, K) over F . ˜ = 13. Prove that a field extension F/K is primary if and only if F Kins ∩ K Kins . Use this criterion to give another proof to Lemma 2.6.14(a). 14. Let F/K be a finitely generated field extension of characteristic p > 0 n and of transcendence degree 1. Prove that for each positive integer n, KF p is the unique subfield E of F which contains K such that F/E is a purely inseparable extension of degree pn . 15. (Geyer) The following example shows that Lemma 2.4.8 is false for arbitrary real valuations. Consider the field Q2 of 2-adic numbers. Show that √ the field K = Q2 ( n 2 | n ∈ N) is a totally ramified extension of Q2 with value group Q. Hence, of K is unramified. Prove that the residue √ each extension √ 3) and K( −1) is F2 . However, their compositum contains field of both K( √ K( −3) and therefore has F4 as its residue field.

Notes The terminology “algebraic independence” for field extensions replaces “freeness” which we used in [Fried-Jarden3]. Corollary 4 of [Lang4, p. 61] proves Lemma 2.6.15(a) only under the condition (our notation) that E is a separable extension of K.

Chapter 3. Algebraic Function Fields of One Variable Sections 3.1–3.4 survey the theory of functions of one variable; the RiemannRoch Theorem; properties of holomorphy rings of function fields; and extensions of the field of constants. Sections 3.5–3.6 include a proof of the Riemann-Hurwitz formula. The rest of the chapter applies these concepts and results to hyperelliptic curves.

3.1 Function Fields of One Variable Call a field extension F/K an algebraic function field of one variable (briefly a function field) if these conditions hold: (1a) The transcendence degree of F/K is 1. (1b) F/K is finitely generated and regular. In this case there exists t ∈ F , transcendental over K, with F/K(t) a finite separable extension. All valuations of K(t) trivial on K are discrete (Example 2.2.1), so their extensions to F are also discrete (Proposition 2.3.2). Also, since the residue fields of the valuations of K(t) are finite extensions of K, so are the residue fields of the valuations of F . We define a prime divisor of F/K as an equivalence class of K-places of F . For p a prime divisor of F/K, choose a place ϕp in p. Then ϕp fixes ˜ ∪ {∞}. Denote its residue field by F¯p . the elements of K and maps F into K ¯ As mentioned above, Fp is a finite extension of K of degree deg(p) = [F¯p : K] which we call the degree of p. Also, choose a valuation vp corresponding to p and normalize it so that vp (F × ) = Z. Each element π of F with vp (π) = 1 is a local parameter of F at p. Denote the free Abelian group that the prime divisors of F/K generate by Div(F/K). P Each element a of Div(F/K) is a divisor of F/K. It has the form a = αp p, where p runs over the prime divisors of F/K, the αp are integers and all but finitely many of them are zero. Define a homomorphism vp : Div(F/K) → Z by vp (a) = αp . The symbol vp appears for two distinct functions. We show these uses are compatibleP as follows. Introduce the divisor of a nonzero element x of F as div(x) = vp (x)p. Since vp (x) = 0 for all but finitely many p the right hand side is well defined. If x is a constant (i.e. x ∈ K), then div(x) = 0. If x ∈ F r K, then the place of K(x) taking x to 0 (resp. to ∞) has finitely many extensions to the field F (Proposition 2.3.1). Equivalence classes of these extensions are the zeros (resp. poles) of x. Thus, div(x) is not zero if x is not a constant. Define the divisor of zeros and the divisor of poles

3.2 The Riemann-Roch Theorem

53

of x as follows: div0 (x) =

X

vp (x)p,

vp (x)>0

div∞ (x) = −

X

vp (x)p.

vp (x) 2g − 2 implies δ(a) = 0 and dim(a) = deg(a) − g + 1 . Proof of (a): L(0) = K because each x ∈ F r K has a pole. Similarly, L(div(x)) = Kx−1 . Proof of (b): Take a = 0 and then a = w in (3). Proof of (c): Use that deg(div(x)) = 0. Proof of (d): By (b) deg(w − a) < 0. Hence, by (c), δ(a) = dim(w − a) = 0. Thus, (3) simplifies to dim(a) = deg(a) + 1 − g. Here is our first application of Lemma 3.2.2(d): Lemma 3.2.3: Let F/K be a function field of genus g, p a prime divisor of F/K, and n a positive integer satisfying (n − 1) deg(p) > 2g − 2. Then there exists x ∈ F × with div∞ (x) = np. Proof: By Lemma 3.2.2(d), dim((n − 1)p) = (n − 1) deg(p) + 1 − g and dim(np) = n deg(p)+1−g. Hence, dim((n−1)p) < dim(np), so L((n−1)p) ⊂ L(np). Every element x ∈ L(np) r L((n − 1)p) will satisfy div∞ (x) = np.

56

Chapter 3. Algebraic Function Fields of One Variable

Example 3.2.4: Rational function field. Let F = K(t), where t is a transcendental element over K. Denote the prime divisor corresponding to the valuation v∞ (Section 2.2) by p∞ . We determine the linear space L(np∞ ), where n is a positive integer. If u ∈ L(np∞ ), then vp (u) ≥ 0 for every prime divisor p 6= p∞ . This means u ∈ K[t]. Also vp∞ (u) ≥ −n, so the degree of u as a polynomial in t is bounded by n. Thus, L(np∞ ) = {u ∈ K[t] | deg(u) ≤ n} and dim(np∞ ) = n + 1. For n > 2g − 2, Lemma 3.2.2(d) gives n + 1 = n − g + 1. Hence, g = 0. Conversely, let F/K be a function field of one variable of genus 0. Assume F/K has a prime divisor p of degree 1 (e.g. if K is algebraically closed). Lemma 3.2.3 gives t ∈ F × with div∞ (t) = p. By Section 3.1, [F : K(t)] = deg(div∞ (t)) = 1. Consequently, F = K(t). Pn Consider now a polynomial f (t) = i=0 ai ti with ai ∈ K and an 6= 0. For each i < n we have v∞ (an tn ) < v∞ (ai ti ), so v∞ (f (t)) = −n. For p 6= p∞ we have vp (t) ≥ 0, hence vp (f (t)) ≥ 0. Thus, div∞ (f (t)) = deg(f )p∞ . (t) Suppose g(t) ∈ K[t] is relatively prime to f (t). Put u = fg(t) . Then t is a root of the polynomial h(u, T ) = u · g(T ) − f (T ). Since h(u, T ) is linear in u and gcd(f, g) = 1, h(u, T ) is irreducible over K(u). In addition, degT (h) = max(deg(f ), deg(g)). Therefore,

(4)

[K(t) : K(u)] = max deg(f ), deg(g) .

In particular, suppose K(u) = K(t). Then, f (t) and g(t) are linear and relatively prime. This means, u = at+b ct+d with a, b, c, d ∈ K and ad − bc 6= 0.

3.3 Holomorphy Rings Let F/K be a function field of genus g. Denote the set of prime divisors of F/K by R. For each p ∈ R let Op = {x ∈ F | vp (x) ≥ 0} be the corresponding valuation ring. To every subset S of R we attach the holomorphy ring T OS = p∈S Op . By definition, K ⊆ OS . If S is empty, then, by definition, OS = F . If S = R, then the elements of OS have no poles. They are therefore constants. Thus, OR = K. The case where S is a nonempty proper subset of R requires a strengthening of the weak approximation theorem (Proposition 2.1.1). Proposition 3.3.1 (Strong Approximation Theorem): Let S be a finite subset of R. Consider q ∈ R r S and let S 0 = S ∪ {q}. Suppose for each p ∈ S we have an xp ∈ F and a positive integer mp . Then there exists x ∈ F with (1)

vp (x − xp ) = mp for each p ∈ S and vp (x) ≥ 0 for each p ∈ R r S 0 .

3.3 Holomorphy Rings

57

Moreover, if m is an integer with (2)

m · deg(q) > 2g − 2 +

X

(mp + 1) deg(p),

p∈S

then x can be chosen such that, in addition to (1), it satisfies vq (x) ≥ −m. Proof: Let P m be a positive integer satisfying (2). Consider the divisor a = mq − p∈S mp p. Then deg(a) > 2g − 2. By Lemma 3.2.2(d), δ(a) = 0, so A = F +Λ(a). Define ξ ∈ A by ξp = xp for p ∈ S and ξp = 0 for p ∈ R r S. Then there exists y ∈ F such that y − ξ ∈ Λ(a): (3) vp (y − xp ) ≥ mp for p ∈ S, vq (y) ≥ −m, and vp (y) ≥ 0 for p ∈ R r S 0 . P Now consider the divisor b = mq − p∈S (mp + 1)p. For each p ∈ S we have deg(b + p) > deg(b) > 2g − 2. By Lemma 3.2.2(d), dim(b + p) = deg(b + p) − g + 1 > deg(b) − g + 1 = dim(b). Hence, L(b) ⊂ L(b + p). Choose zp ∈ L(b + p) r L(b). Then vp (zp ) = mp . Also vp0 (zp ) ≥ mp0 + 1 if p0 ∈ S r{p}, vq (zp ) ≥ −m, and vp0 (zp ) ≥ 0 for p0 ∈ R r S 0 . r P Let P = {p ∈ S | vp (y − xp ) > mp } and let Q = S P . Then z = p∈P zp has the following property: (4) vp (z) = mp if p ∈ P , vp (z) ≥ mp + 1 if p ∈ Q, vq (z) ≥ −m, and vp (z) ≥ 0 for p ∈ R r S 0 . Combine (3) and (4) to see that x = z + y satisfies vp (x − xp ) = mp , for p ∈ S, vq (x) ≥ −m, and vp (x) ≥ 0 for p ∈ R r S 0 . If p belongs to a subset S of R, then OS ⊆ Op . Also, P = {x ∈ OS | vp (x) > 0} is a prime ideal of OS , the center of p at OS . Denote the local ring of OS at P by OS,P . Proposition 3.3.2 (Holomorphy Ring Theorem): Let S be a nonempty proper subset of R. Then S has these properties: (a) Quot(OS ) = F . (b) If p ∈ S and P is the center of p at OS , then Op = OS,P . (c) If q ∈ R r S, then OS 6⊆ Oq . (d) Every nonzero prime ideal of OS is the center of a prime p ∈ S. (e) Distinct primes in S have distinct centers at OS , and the center of each p ∈ S is a maximal ideal of OS . (f) OS is a Dedekind domain. Proof of (a): Consider z ∈ F r K. Since S ⊂ R, there is a q ∈ R r S. There are only finitely many p ∈ S with vp (z) < 0. Hence, Proposition 3.3.1 gives y ∈ F such that the following holds for each p ∈ S: If vp (z) < 0, then vp (y − z −1 ) = vp (z −1 ) + 1, so vp (y) = −vp (z); while if vp (z) ≥ 0, then

58

Chapter 3. Algebraic Function Fields of One Variable

vp (y) ≥ 0. Let x = yz. Then both x and y belong to OS . If z ∈ OS , then / OS , then there is a p ∈ S with vp (z) < 0. Hence, z ∈ Quot(OS ). If z ∈ vp (y) = vp (z −1 ) 6= ∞, so y 6= 0. Therefore, z = xy −1 ∈ Quot(OS ). Proof of (b): Let z ∈ Op r K. As in the proof of (a), there exists y ∈ F such that vp (y) = 0, vp0 (y) = −vp0 (z) if p0 ∈ S r{p} and vp0 (z) < 0, while vp0 (y) ≥ 0 if p0 ∈ S r{p} and vp0 (z) ≥ 0. Therefore, x = yz is in OS and z ∈ OS,P . Since the inclusion OS,P ⊆ Op is clear, Op = OS,P . Proof of (c): If S ∪ {q} = R and OS ⊆ Oq , then OS = OR = K, a contradiction to (a). Therefore, assume that S ∪ {q} is a proper subset of R. By Proposition 3.3.1, there exists x ∈ F such that vq (x) = −1 and vp (x) ≥ 0 for each p ∈ S. This element belongs to OS but not to Oq . Proof of (d): Let P be a nonzero prime ideal of OS . Proposition 2.3.1 extends the quotient map OS → OS /P to a place ϕ of F trivial on K. Let p be the prime divisor of F/K which is defined by ϕ. Then OS ⊆ Op and P = {x ∈ OS | vp (x) > 0}. By (c), p ∈ S. Proof of (e): Let p and p0 be two distinct prime divisors in S. By the strong approximation theorem (Proposition 3.3.1), there exists x ∈ OS with vp (x) > 0 and vp0 (x) = 0. This means the center P of p is not contained in the center of p0 . The maximality of P now follows from (d). Proof of (f): Let q ∈ R r S, let S 0 = R r{q} and choose p0 ∈ S 0 . Since OS is an overring of OS 0 it suffices (Proposition 2.4.7) to prove that OS 0 is a Dedekind domain. The strong approximation theorem gives x ∈ OS 0 with vp0 (x) > 0. Since x must have a pole, it must be q. Thus,T K[x] ⊆ Op if and only if p ∈ S 0 . Therefore, by Proposition 2.4.1, OS 0 = p∈S 0 Op is the integral closure of K[x] in F . Since K[x] is a Dedekind domain, OS 0 is also a Dedekind domain (Proposition 2.4.6). Corollary 3.3.3: The following conditions are equivalent for a nonempty subset S of R: (a) S = R; (b) OS = K; and (c) OS is a field. Proof: Every nonconstant element of F has a pole. Therefore, (a) implies (b). The implication “(b) => (c)” is trivial. To prove that (c) implies (a), note that OS 6= F , since S is nonempty. If S 6= R, then, by Proposition 3.3.2(a), the quotient field of OS is F . Hence, OS is not a field. The following converse to Proposition 3.3.2 is useful: Proposition 3.3.4: Let F/K be a function field and R a proper subring of F containing K. Suppose R is integrally closed and Quot(R) = F . Then there is a nonempty subset S of R with R = OS . Thus, R is a Dedekind domain. Proof: Let S be the set of all prime divisors of F/K which are finite on R. By Proposition 2.4.1, OS = R. As R ⊂ F , S is nonempty. By Corollary 3.3.3, S 6= R. Hence, by Proposition 3.3.2(f), R is a Dedekind domain.

3.4 Extensions of Function Fields

59

3.4 Extensions of Function Fields Let E/K and F/L be algebraic function fields of one variable. We say that F/L is an extension of E/K if E ⊆ F , K ⊆ L, and L ∩ E = K. We call F/L a constant field extension of E/K if E is linearly disjoint from L over K and F = EL. Thus, in this case [F : E] = [L : K]. Recall that E/K is a regular extension ((1b) of Section 3.1). Hence, the linear disjointness of E and L over K is automatic if L is an algebraic extension of K. Likewise, E is linearly disjoint from L over K if L = K(x1 , . . . , xn ) and x1 , . . . , xn are algebraically independent over E (Lemma 2.6.7). Let F/L be an extension of E/K, p be a prime divisor of E/K, and P be a prime divisor of F/L that lies over p, that is vP lies over vp . Denote the ramification index of vP over vp by eP/p . Since both vP and vp are discrete and normalized, vP (x) = eP/p vp (x) for each x ∈ E. Refer to P as unramified over E if vP is unramified over E. If F is separable algebraic over E, then only finitely many prime divisors of F/L are ramified over E [Deuring3, p. 111]. Over every prime divisor of E/K there lie only finitely many prime divisors of F/L [Deuring3, p. 96]. Use this result to embed the group of divisors Div(E/K) of E/K into Div(F/L) as follows: For p a prime divisor of E/K and P1 , . . . , Pd the prime divisors of F/L lying over p, map p to the Pd divisor i=1 ePi /p Pi of F/L. Extend this map to Div(E/K) by linearity. The principal divisor of x (in Div(E/K)) maps to the principal divisor of x in Div(F/L), so there is no ambiguity using div(x) for that divisor. In particular, for every divisor a of E/K we have LE (a) = E ∩ LF (a). Suppose [F : E] < ∞. If F/E is separable, apply (2) of Section 2.3 to conclude: (1)

d X

¯p ] = [F : E]. ePi /p [F¯Pi : E

i=1

Even if F/E is not separable, the following argument shows (1) still holds: By Lemma 3.2.3, there are an integer m and x ∈ E with div∞(x) = mp. Apply the rules degF div∞ (x) = [F : L(x)] and degE div∞ (x) = [E : K(x)] to conclude that (1) is true in general [Deuring3, p. 97]. Next consider a function field F/K and let E be a proper extension of ˜ over K, so is E. In K contained in F . Since F is linearly disjoint from K particular, there is a transcendental element x in E. Then [F : K(x)] = degF div∞ (x) < ∞. Thus, [E : K(x)] < ∞, E/K is a function field of one variable, and [F : E] < ∞. Lemma 3.4.1: Let E/K and F/K be algebraic function fields of one variable with E ⊆ F . Then, degF (a) = [F : E] degE (a) for each a ∈ Div(E/K). Proof: Assume by linearity that a = p is a prime divisor. Let P1 , . . . , Pd ¯p ⊆ F¯P , so be the prime divisors of F/K which lie over p. Then K ⊆ E i

60

Chapter 3. Algebraic Function Fields of One Variable

degF Pi = fPi /p degE (p), i = 1, . . . , d. By (1),

degF (p) =

d X

ePi /p degF (Pi )

i=1

=

d X

ePi /p fPi /p degE (p) = [F : E] degE (p).

i=1

Proposition 3.4.2: Let F/L be a constant field extension of an algebraic function field E/K of one variable with L/K separable. Then: (a) For a divisor a of E/K, degE (a) = degF (a), dimE (a) = dimF (a), and LE (a)L = LF (a). (b) genus(E/K) = genus(F/L). (c) If p and P are respective prime divisors of E/K and F/L, with P lying ¯p and P is unramified over p. over p, then F¯P = LE (d) Let R be an integrally closed subring of E containing K. Then RL is the integral closure of R in F . (e) Let p be a prime divisor of E/K of degree 1. Then there is a unique prime divisor P of F/L lying over p and deg(P) = 1. (f) Let x1 , . . . , xm be elements of E. Denote the integral closure of K[x] in E by S. Then SL is the integral closure of L[x] in F . Proof of (a): [Deuring3, p. 126] shows degE (a) = degF (a). We show dimE (a) = dimF (a). Let B be a basis of LE (a) over K. Then B is contained in LF (a) and is linearly independent over L. Therefore, dimE (a) ≤ dimF (a). Conversely, let B 0 be a basis of LF (a). Then B 0 is contained in EL0 for some finitely generated extension L0 /K of L/K. By [Deuring3, p. 132], dimEL0 (a) = dimE (a). Hence, dimF (a) = |B 0 | ≤ dimEL0 (a) = dimE (a). Therefore, dimE (a) = dimF (a). It follows that LE (a)L ⊆ LF (a) have the same dimension over L, so LE (a)L = LF (a). Proof of (b): Let gE = genus(E/K) and gF = genus(F/L). Choose a divisor a of E/K with degE (a) > max(2gE − 2, 2gF − 2). By Lemma 3.2.2(d), dimE (a) = degE (a) + 1 − gE and dimF (a) = degF (a) + 1 − gF . Therefore, by (a), gE = gF . Proof of (c): The general case reduces to the case where L/K is finitely ¯p . generated. In this case [Deuring3, p. 128] proves that F¯P = LE Now choose a primitive element c of L/K and let f = irr(c, K). By ¯ The reduction of f modulo P is f the preceding paragraph, F¯P = E(c). itself. Since L/K is separable, f is separable. Hence, by Lemma 2.3.4, P/p is unramified. Proof of (d): Assume without loss that [L : K] < ∞. Choose a basis w1 , . . . , wn of L/K. Let σ1 , . . . , σn be the K-embeddings of L into Ks . Since

3.5 Completions

61

E is linearly disjoint from L over K each σi extends uniquely to an Eembedding of F into Es . Pn Now consider z ∈ F which is integral over R. Write z = j=1 aj wj with Pn a1 , . . . , an ∈ E. Then σi z = j=1 aj σi wj , i = 1, . . . , n. By Cramer’s rule, b

b ∆

aj = ∆j = ∆j 2 , where bj is in the ring generated over R by the ak ’s and the σk wl , and ∆ = det(σk wl )1≤k,l≤n ∈ K × . By assumption, each bj is integral over R. Since L/K is separable, ∆2 ∈ K × [Lang7, p. 286, Cor. 5.4]. Hence, each aj is integral over R. Since R is integrally closed, aj ∈ R, j = 1, . . . , n. Consequently, z ∈ RL. Proof of (e): To prove (e), it suffices to consider two cases. In one case, L = K(u) with u purely transcendental over K. [Deuring3, pp. 128–129] handles this case. The other case is when L/K is separable and finite. Let in this case P1 , . . . , Pd be the prime divisors of F/L lying over p. By (c), ¯p = K. Hence, by ePi /p = 1, i = 1, . . . , m. By assumption, deg(p) = 1, so E (c), F¯Pi = L, i = 1, . . . , d. Since F/L is a constant field extension of E/K, we have [F : E] = [L : K]. Therefore, by (1), d = 1 and F¯P1 = L. Thus, deg(P1 ) = 1. S∞ Pm Proof of (f): Put S∞(xi ). By Lemma 3.1.1, S = k=1 LE (ka). S∞ a = i=1 div∞,E By (a), SL = k=1 LE (ka)L = k=1 LF (ka). Hence, by Lemma 3.1.1, SL is the integral closure of L[x] in F .

3.5 Completions The completion of a function field F/K at a prime divisor p gives a powerful tool to investigate the behavior of F at p. For example, it allows us to determine the decomposition of p to prime divisors in finite extensions of F (Proposition 3.5.3). We also use completions to define the ‘different’ of an extension. This notion plays a central role in the Riemann-Hurwitz genus formula, to be introduced in the next section. Let v be a rank-1 valuation of a field F . Then v induces a topology on F . Two elements x, y of F are ‘close’ in this topology if v(x − y) is ‘large’. A sequence {xi }∞ i=1 of elements of F is a Cauchy sequence if for every integer m there exists k such that i, j ≥ k implies v(xi − xj ) ≥ m. If every Cauchy sequence converges, F is complete. Each F embeds as a dense subfield in a complete field Fˆv with a valuation v extending the valuation of F [BorevichShafarevich, Chap. 1, Sec. 4.1]. In particular, Fˆv has the same residue field and value group at v as F . We call (Fˆv , v) (or also just Fˆv ) the completion of (F, v). We also say that Fˆv is the completion of F at v. The completion (Fˆv , v) of (F, v) is unique up to an F -automorphism. Example 3.5.1: Qp and K((t)). The completion of Q at the p-adic valuation vp (Section 2.2) is the field Qp of p-adic numbers. Every element x of Q× p has a unique presentation as a convergent (in the vp -topology) power series

62

Chapter 3. Algebraic Function Fields of One Variable

P∞

n n=m an p with m = vp (x), an ∈ Z, 0 ≤ an ≤ p − 1, and am 6= 0. The valuation ring Zp of Qp consists of all x ∈ Qp with m ≥ 0. Next consider a field K and a transcendental element t over K. Let v be the unique valuation of K(t) with v(t) = 1 (Section 2.2). The completion of K(t) at v is the field K((t)) of formal power series in t with coefficients in K. Each nonzero element f of K((t)) has a unique presentation f = P∞ n n=m an t with m = v(f ) and an ∈ K. The valuation ring of K((t)) is the ring K[[t]] of formal power series in t with coefficients in K. The residue field of K((t)) under v is K. By Lemma 2.6.9(b), K((t)) is a regular extension of K. Consider now a finite extension L of K with a basis wP 1 , . . . , wd . Then ∞ w1 , . . . , wd are linearly independent over K((t)). Let x = n=m an tn with Pd an ∈ L be an element of L((t)). For each n write an = i=1 ani wi with Pd P∞ ani ∈ K. Then, x = i=1 ( n=m ani tn )wi . Therefore, L((t)) = K((t))L and w1 , . . . , wd form a basis for L((t))/K((t)).

The next result contains various versions of Hensel’s Lemma: Proposition 3.5.2: Let (F, v) be a complete discrete valued field. (a) Let f ∈ Ov [X] and a ∈ Ov with v(f (a)) > 2v(f 0 (a)). Then there is a unique x ∈ F with f (x) = 0 and v(x − a) ≥ v(f (a)) − v(f 0 (a)) [CasselsFr¨ohlich, p. 83]. (b) Let f ∈ Ov [X] be a monic polynomial. Denote reduction at v by a bar. Suppose f¯(X) = ζ(X)η(X) with ζ, η ∈ F¯v [X] monic and relatively prime. Then there are monic polynomials g, h ∈ Ov [X] with g¯ = ζ, ¯ = η, and f (X) = g(X)h(X) [Zariski-Samuel2, p. 279, Thm. 17]. h (c) Let F 0 be a finite algebraic extension of F . Then v has a unique extension v 0 to F 0 and F 0 is complete under v 0 [Cassels-Fr¨ohlich, p. 56, Thm.]. Here is a global application of completions: Proposition 3.5.3: Let (E, v) be a discrete valued field. Denote its comˆ vˆ). Consider a finite separable extension F of E. Let z be a pletion by (E, primitive element for F/E which is integral over Ov . Put h = irr(z, E). Let h = h1 · · · hr be the decomposition of h into a product of irreducible polyˆs . Denote the unique ˆ For each i let zi be a root of hi in E nomials over E. ˆ ˆ extension of vˆ to Fi = E(zi ) by vˆi . The following holds: (a) The map z 7→ zi extends to a E-embedding of F into Fˆi . (b) Fˆi is the completion of F at the restriction vi of vˆi to F . (c) The valuations v1 , . . . , vr are mutually nonequivalent. Every extension of v to F coincides with one of the vi ’s. ¯ i = he with hi0 ∈ E[X] ¯ (d) evˆi /ˆv = evi /v , fvˆi /ˆv = fvi /v , and h irreducible i0 and e ∈ N. ˆ ˆ (e) The map g(z) 7→ g(z1 ), . . . , g(zr )) for g ∈ E[X] is an E-isomorphism of L r ˆ ˆ E ⊗E F onto i=1 Fi .

3.5 Completions

(f) Each z in F satisfies traceF/E z = Qr ˆ zi . i=1 normFˆi /E

63

Pr

i=1

traceFˆi /Eˆ zi and normF/E z =

Proof: See [Cassels-Fr¨ ohlich, p. 57, Thm.]. The last statement of (d) follows from Proposition 3.5.2(b). Example 3.5.4: Dedekind. We apply Proposition 3.5.3 to show the necessity ¯v is infinite” in Lemma 2.3.5. of the assumption “E Consider the polynomial f (X) = X 3 − X 2 − 2X − 8. Observe that f (X) has no root modulo 3, hence no root in Q. It is therefore irreducible. Let z ˜ and F = Q(z). Then [F : Q] = 3. Next observe that be a root of f (X) in Q 0 v2 (f (0)) = 3, v2 (f (0)) = 1, v2 (f (1)) = 1, v2 (f 0 (1)) = 0, v2 (f (2)) = 3, and v2 (f 0 (2)) = 1. Hence, by Proposition 3.5.2(a), f has three roots x1 , x2 , x3 in Q2 with v2 (x1 ) ≥ 2, v2 (x2 − 1) ≥ 1, and v2 (x3 − 2) ≥ 2. Thus, f (X) = (X − x1 )(X − x2 )(X − x3 ) is a decomposition of f (X) over Q2 into three distinct irreducible polynomials. By Proposition 3.5.3, v2 has three distinct extensions to F : w, w0 , and w00 . By Proposition 2.3.2, F¯w = F2 . Now assume F/Q has a primitive element z with g = irr(z, Q) ∈ Ov2 [X] such that g¯ is separable. Then deg(g) = 3. By the preceding paragraph, F¯w = F2 . Hence, all three distinct roots of g¯ belong to F2 . But F2 has only two elements, so z does not exist. Lemma 3.5.5: Let (F, v) be a discrete valued field and I a nonzero Ov submodule of F which is not F . Then I is a fractional ideal of Ov and there exists m ∈ Z with I = m−m v . If F is a finite extension of a field E, and Ov is the unique valuation ring of F lying over Ov ∩ E (e.g. (E, v) is complete), then traceF/E I is a fractional ideal of Ov ∩ E. Proof: Choose π ∈ F with v(π) = 1. Let x be a nonzero element of I. If 0 x0 ∈ F and v(x0 ) ≥ v(x), then x0 = xx x ∈ I. Since I 6= F , this implies v(I) is bounded from below. Hence, −m = inf(v(x) | x ∈ I) is an integer. Therefore, I = π −m Ov . Now suppose Ov is the unique valuation ring of F over O = Ov ∩ E. By Proposition 2.4.1, Ov is the integral closure of O in F . Let a be an element of A with v(a) sufficiently large. By the preceding paragraph, aI ⊆ Ov . Then a · traceF/E (I) = traceF/E (aI) ⊆ traceF/E (Ov ) ⊆ A. Thus, traceF/E (I) is a fractional ideal of A. Using completions, we generalize the notions of repartition and differential (Section 3.2) of a function field. Let p be a prime divisor of F/K. Suppose the residue field F¯p of F at p is separable over K. By Hensel’s Lemma, F¯p embeds into Fˆp . Indeed, a). Put f = irr(¯ a, K). Then vp (f (a)) > 0 and choose a ∈ F with F¯p = K(¯ vp (f 0 (a)) = 0. By Proposition 3.5.2(a), there is an x0 ∈ Fˆp with f (x0 ) = 0. The map a ¯ 7→ x0 extends to a K-embedding of F¯p into Fˆp . Let π be an element of Fˆp with vp (π) = 1. Then the completion Fˆp is isomorphic to the field F¯p ((π)) of formal power series in π over F¯p . Every

64

Chapter 3. Algebraic Function Fields of One Variable

P∞ element of this field has the form α = i=m ai π i , where m is an integer and ai ∈ F¯p [Chevalley2, p. 46]. If am 6= 0, then vp (α) = m. Q Consider the cartesian product Fˆp with p ranging over all prime diviQ sors of F/K. An α ∈ Fˆp is an adele if vp (αp ) ≥ 0 for all but finitely many p. In particular, each repartition of F/K is an adele. Denote the set of adeles ˆ ˆ by Q ˆA (or by AF , if F is not clear from the context). It is an F -subalgebra of Fp which contains the algebra A of repartitions. For each prime divisor p ˆ as follows. Identify x ∈ Fˆp with the adele ξ having of F/K embed Fˆp in A 0 ξp = x and ξp = 0 if p0 6= p. For each a ∈ Div(F/K) consider the K-vector space ˆ | vp (α) + vp (a) ≥ 0 for each p}, ˆ Λ(a) = {α ∈ A ˆ = Λ(a) and A + where we have abbreviated vp (αp ) by vp (α). Then A ∩ Λ(a) ˆ Indeed, let α ˆ For each p choose αp ∈ F with vp (αp − α ˆ Λ(a) = A. ˆ ∈ A. ˆp) ≥ ˆ ˆ is in Λ(a), and α ˆ = α − (α − α ˆ) ∈ −vp (a). Then α = (αp ) belongs to A, α − α ˆ A + Λ(a). Therefore (1)

ˆ ˆ ˆ A ∩ (Λ(a) + F ) = Λ(a) + F and A + (Λ(a) + F ) = A.

∼ A/ Λ(a) + F . Thus, in the notation of Section 3.2, ˆ Λ(a) ˆ Hence, A/ +F = (2)

ˆ Λ(a) ˆ + F ) = δ(a) = dim(w − a), dimK A/(

where w is a canonical divisor of F/K. Recall that a differential of F/K is a K-linear map, ω: A → K, which vanishes on a subspace of the form Λ(a)+F . By (1) we can extend ω uniquely ˆ → K which vanishes on Λ(a) ˆ + F . So, from now on, a to a K-linear map ω ˆ: A ˆ ˆ differential of F/K is a K-linear map ω ˆ : A → K which vanishes on Λ(a)+F for some a ∈ Div(F/K). The restriction of ω to A is a differential in the old sense. The divisor of ω ˆ is the maximum of all a ∈ Div(F/K) such that ω ˆ ˆ vanishes on Λ(a). Denote it by div(ˆ ω ). By the above, div(ˆ ω ) = div(ω). ˆ Denote the K-vector space of all differentials that vanish on Λ(a) +F ˆ ˆ by Ω(a). The natural map ω 7→ ω ˆ is an isomorphism of Ω(a) onto Ω(a). In particular, these spaces have the same dimension δ(a). Suppose F/E is a finite separable extension of function fields of one ˆ E . Given a prime divisor P of F/K, denote its variable over K. Let α ∈ A ˆ E as a subalgebra of restriction to E by p and let αP = αp . This identifies A L ˆ F . The isomorphisms E ˆp ⊗E F ∼ A = P|p FˆP of Proposition 3.5.3(e) combine to an isomorphism (3)

ˆ E ⊗E F ∼ ˆF A =A

3.5 Completions

65

[Cassels-Fr¨ohlich, p. 64 or Artin3, p. 244, Thm. 2]. In addition, define a trace ˆF → A ˆE: function traceF/E : A traceF/E (β)p =

X

traceFˆP /Eˆp (βP ).

P|p

Proposition 3.5.6: Let F/E be a finite separable extension of function ˆF : fields of one variable over a field K. The following holds for all α, β ∈ A (a) traceF/E (α + β) = traceF/E (α) + traceF/E (β). ˆE. (b) traceF/E (αβ) = α · traceF/E (β) if α ∈ A (c) The trace of an element x of F coincides with the trace of x as an adele. (d) Let P be a prime divisor of F/K and p = P|E . Then, the trace of an ˆp coincides with the trace of x as an adele. element x of FˆP to E ˆ F such that traceF/E (α) 6= 0. (e) There exists α ∈ A Proof: Statements (a) and (b) follow from the corresponding properties of the trace function on fields. Statement (c) follows from Proposition 3.5.3(f). Statement (d) follows from the definition. Finally, use (c) and the corresponding fact for the trace of fields [Lang7, p. 286, Thm. 5.2] to prove (e). Remark 3.5.7: Complementary modules. (a) Let (E, v) be a complete discrete valued field. Denote the valuation ring of (E, v) by OE . Let F be a finite separable extension of E. Denote the unique extension of v to F by v. Choose a generator πF of the maximal ideal of OF . The complementary module of OF over E is 0 OF/E = {x ∈ F | traceF/E (xOF ) ⊆ OE }.

It is a fractional ideal of OF which contains OF [Lang5, p. 58, Cor.]. By −d

0 Lemma 3.5.5, OF/E = πF F /E OF for some nonnegative integer dF/E . Call dF/E the different exponent of F/E. It is known [Lang5, p. 62, Prop. 8] that dF/E > 0 if and only if F/E is ramified. Moreover, dF/E ≥ eF/E − 1. If the residue field extension is separable, then equality holds if and only if F/E is tamely ramified [Serre4, p. 67, Prop. 13]. (b) Let F/E be a finite separable extension of function fields of one variable over a field K. Consider prime divisors p and P of E/K and F/K, ˆp and FˆP with respectively, with P lying over p. Choose completions E ˆp and FˆP , by O ˆ p and O ˆ P , reˆp ⊆ FˆP . Denote the valuation rings of E E spectively. Occasionally write dFˆP /Eˆp as dP/p or dP/E and call dP/p the different exponent of P over p (or over E). By (a), dP/p > 0 if and only if P/p is ramified. This happens for only finitely many P’s.

66

Chapter 3. Algebraic Function Fields of One Variable

Lemma 3.5.8: Let E ⊆ F ⊆ F 0 be function fields of one variable over a field K with F 0 /E separable. Consider prime divisors p, P, P0 of E/K, F/K, F 0 /K, respectively, with P lying over p and P0 lying over P. Then dP0 /p = eP0 /P dP/p + dP0 /P . In particular, if P0 /P is unramified, then dP0 /p = dP/p . If P/p is unramified, then dP0 /p = dP0 /P . Proof: Suppose the formula holds. If P0 /P is unramified, then dP0 /P = 0 and eP0 /P = 1. Hence, dP0 /p = dP/p . If P/p is unramified, then dP/p = 0, so dP0 /p = dP0 /P . To prove the formula, assume without loss, E, F , and F 0 are complete with respective valuation rings OE , OF , and OF 0 . Choose prime elements πE , πF , and πF 0 for the respective maximal ideals. Put d = dF/E , d0 = dF 0 /F , and e0 = eF 0 /F . We have to prove that e0 d + d0 = dF 0 /E . 0

First note that πFe 0 OF 0 = πF OF 0 . Hence, 0

0

0

traceF 0 /E (πF−e0 d−d OF 0 ) = traceF/E (traceF 0 /F (πF−d πF−d0 OF 0 )) 0

= traceF/E (πF−d traceF 0 /F (πF−d0 OF 0 )) ⊆ traceF/E (πF−d OF ) ⊆ OE . Thus, 0

0

πF−e0 d−d OF 0 ⊆ OF0 0 /E .

(4)

0 By definition, πF−d−1 ∈ / OF/E . Hence, traceF/E (πF−d−1 OF ) 6⊆ OE . By the

second part of Lemma 3.5.5, traceF/E (πF−d−1 OF ) is a fractional ideal of OE . −m Hence, by the first part of Lemma 3.5.5, traceF/E (πF−d−1 OF ) = πE OE for −1 −d−1 some positive integer m. Therefore, πE OE ⊆ traceF/E (πF OF ). Simi0

larly, πF−1 OF ⊆ traceF 0 /F (πF−d0 −1 OF 0 ). 0

0

0

0

Assume πF−e0 d−d OF 0 6= OF0 0 /E . Then, by (4), πF−e0 d−d −1 OF 0 ⊆ OF0 0 /E . Therefore, by the preceding paragraph, 0

0

OE ⊇ traceF 0 /E (πF−e0 d−d −1 OF 0 ) 0

= traceF/E (πF−d traceF 0 /F (πF−d0 −1 OF 0 )) −1 ⊇ traceF/E (πF−d−1 OF ) ⊇ πE OE , 0

0

−d

which is a contradiction. Therefore, πF−e0 d−d OF 0 = OF0 0 /E = πF 0 F Consequently, e0 d + d0 = dF 0 /E , as claimed.

0 /E

OF 0 .

3.6 The Different

67

3.6 The Different The Riemann-Hurwitz genus formula enables us to compute the genus of a function field of one variable F over a field K from the genus of a function subfield E in terms of the ‘different’ of the extension F/E. This formula is in particular useful when the genus of E is known, e.g. when E is the field of rational functions over K. Let F/E be a finite separable extension of function fields of one variable over a field K. The different of F/E is a divisor of F/K: Diff(F/E) =

X

dP/E P,

where P ranges over all prime divisors of F/K and dP/E is the different exponents of P/E. By Remark 3.5.7(a), dP/E ≥ 0 for all P. Moreover, dP/E > 0 if and only if P is ramified over E. Since only finitely many P ramify over E, Diff(F/E) is well defined, Diff(F/E) ≥ 0, and deg(Diff(F/E)) ≥ 0. Theorem 3.6.1 (Riemann-Hurwitz Genus Formula): Let F/E be a finite separable extension of function fields of one variable over a field K. Put gE = genus(E/K) and gF = genus(F/K). Then (1)

2gF − 2 = [F : E](2gE − 2) + deg(Diff(F/E)).

Proof: Let pi , i ∈ I, be the prime divisors of E/K. Choose a nonzero P differential ω of E/K. Write div(ω) = i∈I ki pi with ki ∈ Z and ki = 0 for all but finitely many i P ∈ I. By definition, P ˆ ˆE( (2) ω vanishes on Λ i∈I ki pi ) but not on ΛE ((kr + 1)pr + i6=r ki pi ) for any r ∈ I. ˆ F → K: Define a map Ω: A Ω(α) = ω(traceF/E (α)). It is K-linear and vanishes P on F (Proposition 3.5.6). For each i ∈ I let pi = j∈Ji eij pij , where the pij are the distinct prime divisors of F/K lying over pi and the eij are the corresponding ramification ˆp , indices. We simplify notation and for all i ∈ I and j ∈ Ji let Ei = E i ˆ ˆ ˆ Oi = Opi , vi = vpi , Fij = Fpij , Oij = Opij , vij = vpij , and (3)

0 dij = vij (Diff(F/E)) = −vij (Oij ),

0 is the complementary module of Oij with respect to traceF/E where Oij ˆ E and β ∈ A ˆ F let αi = αp and βij = βp . (Remark 3.5.7). Also, for all α ∈ A i ij We prove that ˆ F (P (eij ki + dij )pij ) and (4a) Ω vanishes on Λ i,j (4b) for each r ∈ I and each sP ∈ Jr the differential Ω does not vanish on ˆ F (ers kr + drs + 1)prs + Λ (e k + d )p ij ij . (i,j)6=(r,s) ij i

68

Chapter 3. Algebraic Function Fields of One Variable

This will imply that divF (Ω) =

X

(eij ki + dij )pij = divE (ω) + Diff(F/E).

i,j

The formulas deg(divF (Ω)) = 2gF − 2 and deg(divE (ω)) = [F : E](2gE − 2) (Lemmas 3.2.2(b) and 3.4.1) will give (1). For all i, j choose πi ∈ E and πij ∈ F with vi (πi ) = 1 and vij (πij ) = 1. ˆ F (P (eij ki + dij )pij ). Then vij (α) ≥ Proof of (4a): Consider α ∈ Λ ij −eij ki − dij .

Hence, vij (πiki α) ≥ −dij .

0 By (3), (πiki α)ij ∈ Oij , so

traceFij /Ei (πiki α)ij ∈ Oi . Thus, (traceF/E (πiki α))i =

X

traceFij /Ei (πiki α) ∈ Oi ,

j∈Ji

hence vi (traceF/E (α)) + ki ≥ 0 for each i ∈ I. Therefore, traceF/E (α) ∈ ˆ E (P ki pi ). Consequently, by (2), Ω(α) = ω(traceF/E (α)) = 0. Λ i∈I

Proof of (4b): Assume there exist r ∈ I and s ∈ Jr such that Ω vanishes on X ˆ F (ers kr + drs + 1)prs + (eij ki + dij )pij . V =Λ (i,j)6=(r,s)

By Lemma 3.5.5, there is an m ∈ Z with −drs −1 Ors ) = πr−m Or Irs = traceFrs /Er (πr−kr πrs

We distinguish between two cases: ˆ E (kr + 1)pr + P Consider β ∈ Λ i6=r ki pi . Then ˆ E (P ki pi ), so ω(β) = vr (β) ≥ −kr − 1. If vr (β) ≥ −kr , then β ∈ Λ i∈I 0 (by (2)). Otherwise, vr (β) = −kr − 1. In this case write β = α + γ, ˆ E satisfy αr = (1 − πr )βr , γr = πr βr , αi = 0, and γi = βi where α, γ ∈ A for i 6= r. Then, αr ∈ πr−kr −1 Or ⊆ πr−m Or = Irs , so there is a δrs ∈ −drs −1 Ors with traceFrs /Er δrs = αr . For (i, j) 6= (r, s) let δij = 0. πr−kr πrs Then δ ∈ V , α = traceF/E δ, and ω(α) = ω(trace(δ)) = Ω(δ) = 0. Also, ˆ E (P ki pi ), so ω(γ) = 0 (by (2)). It follows that ω(β) = 0. Thus, ω γ∈Λ i∈I ˆ E (kr + 1)pr + P vanishes on Λ ki pi , in contradiction to (2). Case A: m ≥ kr + 1.

i6=r

Case B: m ≤ kr .

Then,

−drs −1 Ors ) = πr−m Or ⊆ πr−kr Or . traceFrs /Er (πr−kr πrs −drs −1 −drs −1 0 Hence, traceFrs /Er (πrs Ors ) ⊆ Or . Therefore, πrs Ors ⊆ Ors , so 0 vrs (Ors ) ≤ −drs − 1. This contradiction to (3) completes the proof of Case B and the proof of the whole theorem.

3.6 The Different

69

Remark 3.6.2: Applications of the Riemann-Hurwitz formula. Let F/E be a finite separable extension of function fields of one variable of a field K. We say that F/E is unramified (resp. tamely ramified) if each prime divisor of F/K is unramified (resp. tamely ramified) over E. Let gE = genus(E/K) and gF = genus(F/K). Suppose [F : E] ≥ 2. (a) Comparison of genera: We have mentioned at the beginning of this section that deg(Div(F/E)) ≥ 0. Hence, by (1), gF ≥ gE . Both gE and gF have the same value if and only if gE = 1 and F/E is unramified, or gE = 0 and deg(Diff(F/E)) = 2 [F : E] − 1 . In particular, if F = K(t), then gF = 0 (Example 3.2.4). Hence, gE = 0. Each prime divisor of F/K of degree 1 induces a prime divisor of E/K of degree 1. We conclude from Example 3.2.4 that E = K(u) is also a rational function field. This is L¨ uroth’s theorem. This theorem actually holds for arbitrary algebraic extension K(t)/E and not only for separable extensions and may be proved by elementary arguments on polynomials [Waerden1, p. 218]. (b) An analog of a theorem of Minkowski: Suppose F/E is unramified. In this case the Riemann-Hurwitz formula simplifies to gF − 1 = [F : E](gE − 1). Hence, gE > 0. In other words, a function field E/K of genus 0 has no proper finite unramified extension F which is regular over K. In particular, K(t) has no finite proper separable unramified extension F which is regular over K. This is an analog of a theorem of Minkowski saying that Q has no proper unramified extensions [Janusz, p. 57, Cor. 11.11]. (c) The Hurwitz-Riemann formula for tamely ramified extensions: Suppose F/E is tamely ramified. By Remark 3.5.7, the Riemann-Hurwitz formula simplifies to XX eP/p − 1 deg(P). 2gF − 2 = [F : E](2gE − 2) + p

P|p

(d) An analog of Minkowski’s theorem in the tamely ramified case: Suppose K is algebraically closed and gE = 0. Then E = K(t) and the degree of each prime divisor is 1. Suppose F/E is a proper tamely ramified extension. Then E has at least two prime divisors that ramify in F . Indeed, assume E has only one prime divisor p that ramifies in F . Let P1 , . . . , Pr be the prime divisors of p in F . Then −2 ≤ 2gF − 2 = −2[F : E] +

r X

ePi /p − 1

i=1

= −2[F : E] + [F : E] − r = −[F : E] − r Hence, 3 ≤ [F : E] + r ≤ 2, a contradiction.

70

Chapter 3. Algebraic Function Fields of One Variable

In particular, if K is algebraically closed and char(K) = 0, then every proper extension of K(t) is ramified over at least two prime divisors. (e) Generation of Galois groups by inertia groups: Let K be an algebraically closed field, E = K(t), and F a finite Galois extension of E. Denote the prime divisors of F/K which ramify over E by P1 , . . . , Pr . For each i let Di be the decomposition group of Pi over E. Since K is algebraically closed, Di is also the inertia group of Pi over E. Let Ei be the fixed field of Di in F . Then Pi |Ei is unramified over E. Hence, E0 = E1 ∩ · · · ∩ Er is unramified over E. By (b), E0 = E. In other words, the inertia groups of the prime divisors of F/K which ramify over E generate Gal(F/E). (f) Quasi-p groups: Let K be an algebraically closed field of positive characteristic p. Consider the rational field E = K(t), a prime divisor p of E/K, and a finite Galois extension F of E which is ramified only over p. Denote the fixed field in F of all p-Sylow subgroups of Gal(F/E) by Ep . Then Ep /E is a Galois extension of degree relatively prime to p. Hence, Ep /E is tamely ramified. The only prime divisor of E/K which is possibly ramified in Ep is p. It follows from (d) that Ep = E. In other words, Gal(F/E) is generated by its p-Sylow subgroups. One says that Gal(F/E) is quasi-p. (g) Abhyankar’s conjecture: Let K be an algebraically closed field of positive characteristic p. Let E = K(t) and p a prime divisor of E/K. In 1957, Abhyankar conjectured that for each finite quasi-p group G there exists a finite Galois extension F of E which is unramified outside p and Gal(F/E) ∼ = G [Abhyankar]. Serre proved the conjecture for solvable G in 1990 [Serre9]. Raynaud treated all other cases in 1991 [Raynaud]. (h) The generalized Abhyankar’s conjecture: Let again K be an algebraically closed field of positive characteristic p. Put E = K(t). Consider a set S = {p1 , . . . , pr } of prime divisors of E/K and a finite Galois extension F of E which is unramified over E outside S. Let G = Gal(F/E). Denote the subgroup of G generated by all p-Sylow subgroups of G by G(p). Let Ep be the fixed field of G(p) in F . Then Ep is a Galois extension of E which is tamely ramified over E and unramified outside S. By [Grothendieck, XIII, Cor. 2.12], Gal(Ep /E) is generated by r − 1 elements. This led Abhyankar to conjecture that for every finite group G such that G/G(p) is generated by r − 1 elements there is a Galois extension F of E which is unramified outside S and Gal(F/E) ∼ = G. Harbater proved this conjecture in [Harbater2] by reducing it to the special case r = 1 proved by Raynaud.

3.7 Hyperelliptic Fields We demonstrate the concepts and results of this chapter to a study of a special kind of algebraic function fields of one variable which we now introduce. A function field F/K is hyperelliptic if its genus is at least 2 and if it is a quadratic extension of a function field E/K of genus 0. We then say E is a quadratic subfield of F . It turns out that E is uniquely determined.

3.7 Hyperelliptic Fields

71

Moreover, we will be able to identify E from the arithmetic of F . Lemma 3.7.1: Let F/K be a function field of genus g. Let a, b be nonnegative divisors of F/K and let w be a canonical divisor of F/K. (a) If L(b − a) = L(b), and dim(b) ≥ 1, then dim(a) = 1. (b) If g ≥ 1 and a > 0, then dim(w − a) < dim(w). (c) If g ≥ 1 and x1 , . . . , xg is a basis for L(w), then −w = min div(x1 ), . . . , div(xg ) . Proof of (a): Since a ≥ 0, we have K ⊆ L(a). Conversely, let x ∈ L(a). Then div(x) + a ≥ 0. Consider y ∈ L(b). By assumption, y ∈ L(b − a), hence div(y) + b − a ≥ 0. Therefore, div(xy) + b ≥ 0, so xy ∈ L(b). Apply this Pn result to a basis y1 , . . . , yn of L(b). Find aij ∈ K such that xyi = j=1 aij yj , i = 1, . . . , n. Hence, det(xI − A) = 0, where A = (aij )1≤i,j≤n . Therefore, x satisfies a monic equation with coefficients in K. Since K is algebraically closed in F , we have x ∈ K. Consequently, dim(a) = 1. Proof of (b): Assume dim(w − a) = dim(w). Then L(w − a) = L(w). By Lemma 3.2.2(b), dim(w − a) = dim(w) = g. So, by (a), dim(a) = 1. By Theorem 3.2.1, dim(a) = deg(a) + 1 − g + dim(w − a). Hence, deg(a) = 0, which is a contradiction. Therefore, dim(w − a) < dim(w). Proof of (c): Denote min div(x1 ), . . . , div(xg ) by m. Since div(xi ) + w ≥ 0 we have m ≥ −w. If m > −w, then there exists a prime divisor p of F/K with m − p ≥ −w. Hence, div(xi ) + w − p ≥ 0, so xi ∈ L(w − p) for i = 1, . . . , g. Therefore, L(w − p) = L(w), contradicting (b). Consequently, m = −w. Proposition 3.7.2: Let F/K be a function field of genus g ≥ 2. Consider a canonical divisor w of F/K. Let x1 , . . . , xg be a basis of L(w). If E = x K xx12 , . . . , xg1 is a proper subfield of F , then genus(E/K) = 0 and [F : E] = 2. Thus, F/K is a hyperelliptic field. Proof: Let w0 = div(x1 ) + w. Then w0 is also a canonical divisor and x 1, xx21 , . . . , xg1 is a basis of L(w0 ). Replace w by w0 , if necessary, to assume that x1 = 1 and E = K(x2 , . . . , xg ). Since g ≥ 2, the element x2 is transcendental over K. Hence, E is also an algebraic function field of one variable over K and d = [F : E] < ∞ (Section divisor of E/K. 3.4). Denote genus(E/K) by gE and let wE be a canonical By Lemma 3.7.1(c), w = − min 0, div(x2 ), . . . , div(xg ) . In particular, w ≥ 0. By Lemma 3.2.2(b), deg(w) = 2g − 2 ≥ 2. Therefore, (1)

w > 0.

Observe that div(xi ) ∈ Div(E/K), i = 2, . . . , g, so w ∈ Div(E/K). We may therefore apply Riemann-Roch to E/K and w: (2) dim LE (w) = degE (w) + 1 − gE + dim LE (wE − w) .

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Chapter 3. Algebraic Function Fields of One Variable

Since 1, x1 , . . . , xg are in E and generateLF (w), we have LE (w) = LF (w). Hence, by Lemma 3.2.2(b), dim LE (w) = g. Applying Lemma 3.2.2(b) again and Lemma 3.4.1, we have 2g−2 = degF (w) = d·degE (w). Substituting this in (2) gives (3)

g=

2g − 2 + 1 − gE + dim LE (wE − w) , d

which may be rewritten as (4)

d dim LE (wE − w) − gE ) = (g − 1)(d − 2)

By assumption, both g and d are at least 2. Hence, the right hand side of (4) is at least 0. On the other hand, by Lemma 3.2.2(b) and by (1), (5) dim LE (wE − w) ≤ dim LE (wE ) = gE . Hence, both sides of (4) are 0. Since g ≥ 2, this gives d = 2. Finally, by (3), gE = dim(LE (wE − w)). Hence, by (5), dim(LE (wE − w)) = dim(LE (wE )). We conclude from Lemma 3.7.1(b) that gE = 0. Lemma 3.7.3: Let F/K be a function field of genus g ≥ 1, w a canonical divisor of F/K, and x1P , . . . , xg a basis of LF (w). Let E/K be a subfield of g genus 0 of F/K. Then i=1 Exi 6= F . ˆ E (0)xi ⊆ Λ ˆ F (w). Hence, Proof: For each i we have Λ g X

(6)

ˆ F (w) + F. ˆ E (0) + E xi ⊆ Λ Λ

i=1

Choose a canonical divisor wE of E/K. By (2) of Section 3.5, ˆ E /(Λ ˆ E (0) + E) = dim(wE ) = genus(E/K) = 0. dimK A Hence, ˆE = Λ ˆ E (0) + E. A

(7) On the other hand,

ˆ F /(Λ ˆ F (w) + F ) = dim(0) = 1. dimK A ˆ F , so by (6) and (7) ˆ F (w) + F ⊂ A Hence, Λ (8)

g X

ˆ E xi ⊂ A ˆF . A

i=1

Pg Pg ˆ ˆ If i=1 Exi = F , then Pg by (3) of Section 3.5, AF = i=1 AE xi , contradicting (8). Consequently, i=1 Exi 6= F .

3.8 Hyperelliptic Fields with a Rational Quadratic Subfield

73

Proposition 3.7.4: Let F/K be a hyperelliptic function field, w a canonical x divisor of F/K, and x1 , . . . , xg a basis of LF (w). Then E = K xx21 , . . . , xg1 is the only quadratic subfield of F . Proof: By definition, F has subfields of genus 0. Let E 0 be one Pquadratic g 0 of them. By Lemma 3.7.3, i=1 E xi 6= F . Since [F : E 0 ] = 2, this implies xi ∈ E 0 , i = 1, . . . , g. It follows that E ⊆ E 0 . In particular, E ⊂ F . By Proposition 3.7.2, [F : E] = 2. Comparing degrees implies E 0 = E.

3.8 Hyperelliptic Fields with a Rational Quadratic Subfield A hyperelliptic field with a rational quadratic subfield is generated by two generators which satisfy an equation of a special type. This situation arises, for example, when the hyperelliptic field has a prime divisor of degree 1 (Example 3.2.4). Proposition 3.8.1: Let F/K be a hyperelliptic field of genus g. Suppose the quadratic subfield of F is K(x) with x indeterminate. Then F = K(x, y), where y satisfies a relation y 2 + h1 (x)y + h2 (x) = 0 with h1 , h2 ∈ K[X], deg(h1 ) ≤ g + 1, and deg(h2 ) ≤ 2g + 2. If char(K) 6= 2, we may choose y with y 2 = f (x), where f ∈ K[X] is a polynomial with no multiple root and deg(f (x)) ≤ 2g + 2. Proof: Let p∞ be the pole of x in K(x)/K. By Lemma 3.4.1, degF (np∞ ) = 2n for each positive integer n. If n ≥ g, then 2n > 2g − 2. By Lemma 3.2.2(d), dim(LF (np∞ )) = 2n + 1 − g. In particular, for n = g the elements 1, x, . . . , xg form a basis for LF (gp∞ ). For n = g + 1 we have dim LF ((g + 1)p∞ ) = g + 3. Hence, two more elements are needed to complete 1, x, . . . , xg to a basis of LF (g + 1)p∞ . We take one of them as xg+1 and denote the other one by y. By RiemannRoch, dim(LK(x) ((g + 1)p∞ )) = g + 2. So, 1, x, . . . , xg+1 form a basis of LK(x) (g + 1)p∞ . If y is in K(x), then y is also in LK(x) (g + 1)p∞ . This implies that 1, x, . . . , xg+1 , y are linearly dependent over K. We conclude from this contradiction that y ∈ / K(x), so F = K(x, y). Next note that dim LF ((2g + 2)p∞ ) = 3g + 5. All 3g + 6 elements 1, x, . . . , x2g+2 , y, yx, . . . , yxg+1 , y 2 belong to LF (2g + 2)p∞ . Therefore, P2g+2 Pg+1 there are ai , bj , c ∈ K, not all zero, with i=0 ai xi + j=0 bj yxj + cy 2 = 0. Pg+1 −1 bj X j , h2 (X) = Since y ∈ / K(x), we have c 6= 0. Let h1 (X) = j=0 c P2g+2 −1 ai X i . Then i=0 c (1)

y 2 + h1 (x)y + h2 (x) = 0.

If char(K) 6= 2, replace y by 2y + h1 (x), if necessary, to assume (1) has the form y 2 = f (x). Here f ∈ K[X] has degree at most 2g + 2. Finally, if f has

74

Chapter 3. Algebraic Function Fields of One Variable

multiple roots, then f = g 2 h with h having no multiple roots. Replace y by yg(x)−1 to assume f has no multiple roots. Our first task is to compute the genus of the hyperelliptic field in characteristic 6= 2 from deg(f ). Proposition 3.8.2: Let K be a field of characteristic 6= 2 and f ∈ K[X] a polynomial of degree d ≥ 1 with no multiple roots. Let F = K(x, y), with x transcendental over K and y 2 = f (x). Then F/K is an algebraic function d−2 field of one variable of genus d−1 2 if d is odd and of genus 2 if d is even. In particular, if d ≥ 5, then F/K is hyperelliptic. ˜ Proof: Since f (x) is not a square in K(x), ˜ ˜ [K(x, y) : K(x)] = [K(x, y) : K(x)] = 2. ˜ Hence, K(x, y) is linearly disjoint from K(x) over K(x). Since x is transcen˜ over K. Hence, by the tower dental over K, K(x) is linearly disjoint form K ˜ property of linear disjointness (Lemma 2.5.3), F is linearly disjoint from K over K. It follows that F/K is an algebraic function field of one variable. Since char(K) 6= 2, F/K(x) is tamely ramified. Since the genus of K(x) is 0, the Riemann-Hurwitz formula reduces to XX (eP/p − 1) deg(P) (2) 2g − 2 = −4 + p

P|p

(Remark 3.6.2(c)). If p is a prime divisor of K(x) which ramifies in F , then p has only one extension P to F , the ramification index of P is 2, and its residue degree is 1. Hence, (2) simplifies to (3)

2g = −2 +

X

deg(p),

where p ranges over all prime divisors of K(x)/K which ramify in F . Let f (x) = p1 (x) · · · pr (x) be the decomposition of f (x) into a product of distinct irreducible polynomials in K[x]. To each pi there corresponds a prime divisor pi of K(x)/K of degree deg(pi ) and a valuation vi such that vi (f (x)) = vi (pi (x)) = 1 (Example 2.2.1(b)). By Example 2.3.8, each pi ramifies in F . In addition, let v∞ be the valuation of K(x)/K with v∞ (x) = −1 and let p∞ be the corresponding prime divisor; its degree is 1. Since v∞ (f (x)) = −d, the prime divisor p∞ is ramified in F if d is odd and unramified if d is even (Example 2.3.8). All other prime divisors of K(x) are unramified in F . The Pr sum of the degrees of the ramified prime divisors is δ + i=1 deg(pi ) = δ + d, where δ = 1 if d is odd and δ = 0 if d is even. It follows from (3) that g = d−1 2 if d is odd and g = d−2 2 if d is even. In characteristic 2 we compute the genus of only a special type of a hyperelliptic field.

Exercises

75

Lemma 3.8.3: Let (K, v) be a valued field of characteristic 2 and t an element of K with v(t) = 1. Consider an Artin-Schreier extension L = K(x) with x2 + x = 1t . Denote the unique extension of v to L by w. Let y = tx. Then 0 = y −2 Ow . w(y) = 1, Ow = Ov [y] is the integral closure of Ov in L, and Ow Proof: By Example 2.3.11, v has a unique extension w to L. It totally ramifies over K and satisfies w(x) = −1, w(t) = 2, and w(y) = 1. In particular, Ow is the integral closure of Ov in L (Proposition 2.4.1). Since ¯ w , each z ∈ Ow can be written as z = a + by with a ∈ Ov and b ∈ Ow . ¯v = L K Hence, Ow = Ov [y] [Lang5, p. 26, Prop. 23]. 0 To compute Ow = {z ∈ L | traceL/K (zOw ) ⊆ Ov } observe that g(X) = 2 0 = t−1 Ow = y −2 Ow irr(y, K) = Y + tY + t and g 0 (X) = t. Therefore, Ow 0 [Lang5, p. 59, Cor.]. Of course, one may also compute Ow directly. Proposition 3.8.4: Let K be a field of characteristic 2 and a1 , . . . , ad distinct elements of K. Let F = K(x, y) with x transcendental over K, h ∈ K[X] with deg(h) ≤ d, and y 2 + y = (x−a1h(x) )···(x−ad ) . Then F/K is an algebraic function field of one variable of genus d − 1. In particular, if d ≥ 3, then F/K is hyperelliptic. Proof: For each i between 1 and d let vi be the discrete normalized valuation of K(x)/K with vi (x − ai ) = 1. Let t = (x − a1 ) · · · (x − ad ). Then vi (t) = 1. ˜ ˜ If y is in K(x), ˜ K. then 2˜ vi (y) = −1, Extend vi to a valuation v˜i of K(x)/ ˜ which is a contradiction. Hence, y is not in K(x), so [K(x, y) : K(x)] = ˜ ˜ ˜ over K, so [K(x, y) : K(x)] = 2. It follows that F is linearly disjoint from K F/K is an algebraic function field of one variable. By Example 2.3.9, v1 , . . . , vd are the only valuations of K(x)/K that ramify in F . Hence, only v1 , . . . , vd contribute to the different of F/K(x). [ (y). By Let wi be the unique extension of vi to F . Then Fˆwi = K(x) vi Lemma 3.8.3, the contribution of wi to deg Diff(F/K(x)) is 2. Hence, deg(Diff(F/K(x))) = 2d. Let g = genus(F/K). By Riemann-Hurwitz (Theorem 3.6.1), 2g − 2 = −4 + 2d. Hence, g = d − 1, as claimed.

Exercises In the following exercises F is a function field of one variable over a field K and t is a transcendental element over K. 1. Let a be a divisor of F/K. Note: If a ≥ 0 and deg(a) = 0, then a = 0. (a) Prove that if deg(a) = 0 and a is not a principal divisor, then dim(a) = 0. (b) Let g be the genus of F . Suppose a is a noncanonical divisor with deg(a) = 2g − 2. Show that dim(a) = g − 1. Hint: Use Riemann-Roch and (a). 2. Let a and b be divisors of F/K with b ≥ 0. Use Riemann-Roch to prove that (a) dim(a) ≤ dim(a + b) ≤ dim(a) + deg(b); and

76

Chapter 3. Algebraic Function Fields of One Variable

(b) dim(a) ≤ max(0, deg(a) + 1). Hint: Write a as the difference of its “positive” and “negative” parts. 3. Let p1 , . . . , pn be distinct Pnprime divisors of F/K and let m1 , . . . , mn be If K is infinite, prove positive integers such that i=1 mi deg(pi ) > 2g − P1. n × such that div (x) = that there exists x ∈ F ∞ i=1 Pn mi pi . In particular, Pn [F : K(x)] = i=1 mi deg(pi ). Hint: Consider a = i=1 mi pi and aj = a−pj for j = 1, . . . , n. For each j, 1 ≤ j ≤ n, distinguish between the cases deg(aj ) ≤ 2g − 2 and deg(aj ) > 2g − 2, and prove that L(aj ) ⊂ L(a). 4. Suppose F = K(t). Prove, in the notation of Example 3.2.4, that −2p∞ is a canonical divisor of F/K. 5. Suppose the genus of F/K is 0. Prove that every divisor a with deg(a) = 0 is principal. Hint: Compute dim(a) and apply Exercise 1(a). kr ∈ N, and 6. In the notation of Proposition 3.3.2 let p1 , . . . , pr ∈ S, k1 , . . . , Q n 1, . . . , r. Consider the ideal A = i=1 Piki of Pi be the center of pi at OS , i =P r OS . Prove that the divisor a = i=1 ki pi of F/K satisfies deg(a) = (OS : A). Qr Hint: Prove that OS /A ∼ = i=1 OS /Piki and that OS /Pi ∼ = Pik /Pik+1 , as groups, for each nonnegative integer k. 7. Prove that every algebraic function field F/K of genus 2 is hyperelliptic with a rational quadratic subfield. Hint: Choose a positive canonical divisor w of F/K. Then choose a nonconstant x in L(w). 8. Let K be a field of characteristic that does not divide n. Consider the function field F = K(x, y) over K with x, y satisfying xn +y n = 1. Prove that . Hint: Use Example 2.3.8 and the Riemann-Hurwitz genus(F ) = (n−1)(n−2) 2 formula (Remark 3.6.2(c)).

Notes In addition to [Chevalley2] one may find a proof of the Riemann-Roch theorem in [Lang4, Chapter 10, Section 2], [Deuring3, Section 15], and [Stichtenoth, Section I.5]. The content of Section 3.7 is borrowed from [Artin3]. One may also find it in [Stichtenoth, Section VI.2] along with computations of genera of various function fields using the Riemann-Hurwitz genus formula. However, in contrast to our exposition, [Stichtenoth] assumes the fields of constants to be perfect.

Chapter 4. The Riemann Hypothesis for Function Fields In this chapter K is a finite field of characteristic p with q elements. Let F be an algebraic function field of one variable over K and g the genus of F/K. Denote the group of divisors and the group of divisor classes of F/K by D and C, respectively. P ∞ The series ζ(s) = n=1 n−s defines the classical Riemann zeta function of a complex variable. It converges absolutely for Re(s) > 1. Hence, ζ(s) is an analytic function in this domain. The series diverges for s = 1. The function, however, can be analytically continued to a function meromorphic on the whole s-plane. The production of this requires two stages: an analytic 1 to the half plane Re(s) > 0 by a rearrangement of continuation of ζ(s) + 1−s the series via Abel summation; and then, (the difficult part) a demonstration ζ(s) is the product of the classical Gamma function and an elementary that ζ(1−s) function in the domain 0 < Re(s) < 1. The expression that results from the last stage is called the functional equation for ζ(s). The resulting analytic continuation of ζ(s) yields a function with a simple pole at s = 1 with residue 1 and zeros at the points −2, −4, −6, −8, . . . . There are no other zeros in the domains Re(s) ≥ 1 and Re(s) ≤ 0 [Titchmarsh, p. 30]. The classical Riemann hypothesis is still unproven. It states that the only zeros of ζ(s) in the strip 0 < Re(s) < 1 lie on the line Re(s) = 12 ; its applications are legion. There is an analog for F/K of the Riemann zeta function (Section 4.2). It satisfies a functional equation (Proposition 4.4.1). Our main goal is the proof of an analog to the Riemann hypothesis (Theorem 4.5.1). In Chapter 5 we extract from this an explicit estimate for the number of points on any curve over a finite field.

4.1 Class Numbers The assumption that K is finite results in the finiteness of other sets connected to F/K. For example, F/K has only finitely many ideal classes of degree 0 and only finitely many nonnegative divisors of a given degree n. The main result of this section (Lemma 4.1.4) computes the latter number in terms of the former one. Let p be a prime divisor of F/K. Its residue field F¯p is a finite extension of K of degree deg(p). Thus, F¯p is a finite field whose order we indicate by N p = q deg(p) , the norm of p. Extend the definition of the norm to arbitrary divisors by the formula N a = q deg(a) . Then N (a + b) = N a · N b. Definition 4.1.1: Denote the number of divisor classes of F/K of degree zero by h; the class number of F/K. If g = 0, then every divisor of degree 0 of F/K is principal (Exercise 5

78

Chapter 4. The Riemann Hypothesis for Function Fields

of Chapter 6). Therefore, h = 1 if F is K(x) or any genus 0 function field. Lemma 4.1.2: Only finitely many nonnegative divisors of F/K have degree equal to a given integer m. In addition, the class number of F/K is finite. Proof: Let x ∈ F be transcendental element over K and let E = K(x). Denote the collection of all prime divisors of E/K of degree ≤ m by S0 . Each element of S0 , except possibly p∞ , corresponds to a monic irreducible polynomial in K[x] of degree ≤ m. Hence, S0 is a finite set. Only finitely many prime divisors of F/K lie over a given prime divisor p0 of E/K. Each of them has degree at least as large as deg(p0 ). Thus, there are only finitely many prime divisors of F/K of degree ≤ m. Therefore, the set Am of all nonnegative divisors of F/K of degree m is finite. For the second part of the lemma choose a nonnegative divisor m of degree m ≥ g. Denote the set of divisor classes of degree m by Cm . For each b ∈ Cm Riemann-Roch (Theorem 3.2.1) implies that dim(b) ≥ m − g + 1 ≥ 1. Hence, there is an x ∈ F × with div(x) + b ≥ 0, so the class of b contains a nonnegative divisor. It follows that the map Am → Cm mapping each a ∈ Am onto each class is surjective. Therefore, by the preceding paragraph, Cm is finite. Finally, the map a 7→ a + m induces a bijective map of C0 onto Cm . Consequently, C0 is a finite group. Lemma 4.1.3: The number of nonnegative divisors in a given class of divisors C of F/K is

q dim(C) −1 . q−1

Proof: If a ≥ 0, then div(a) + a ≥ 0 for each a ∈ K × , so K ⊆ L(a). Hence, dim(a) ≥ 1. This gives the formula if dim(C) = 0. Suppose dim(C) = n > 0. Let c be a divisor in C. The number of nonnegative divisors in C is equal to . , xn be a the number of principal divisors div(x) with x ∈ L(c). Let x1 , . .P n basis for L(c) over K. The number of elements of F × of the form i=1 ai xi , with a1 , . . . , an ∈ K is equal to q n − 1. Since div(x) = div(x0 ) if and only if there exists an a ∈ K × such that x0 = ax, the formula follows. Denote the greatest common divisor of the degrees of the divisors of F/K by δ. Eventually we prove that δ = 1. In the meantime notice that a positive integer n is a multiple of δ if and only if there exists a divisor of degree n. Indeed, there are divisors a1 , . . . , ar with δ = gcd(deg(a Pr 1 ), . . . , deg(ar )). For , . . . , a ∈ Z with each multiple n of δ there are a 1 r i=1 ai deg(ai ) = n. The Pr divisor a = i=1 ai ai satisfies deg(a) = n. In particular, since the degree of the canonical divisor is 2g − 2 (Lemma 3.2.2(b)) δ divides 2g − 2. Lemma 4.1.4: Let An be the number of nonnegative divisors of F/K of degree n. If n ≥ 0 is a multiple of δ larger than 2g − 2, then An = h q

n−g+1

q−1

−1

.

Proof: Let n be a multiple of δ with n > 2g − 2. Choose a divisor c of degree n. By Riemann-Roch, dim(c) = n − g + 1. Moreover, the map a 7→ a + c defines a bijection of the set of divisor classes of degree 0 with the set of

4.2 Zeta Functions

79

divisor classes of degree m. Hence, there are h divisor classes of degree n. It n−g+1 follows that, An = h q q−1 −1 .

4.2 Zeta Functions The zeta function of a function field over K is a rational function with coefficients in Q and with simple poles at several points including 1. We define the zeta function of the function field F/K to be the Dirichlet series X (N a)−s (1) ζ(s) = ζF/K (s) = a≥0

where a runs over the nonnegative divisors of F/K. Check the domain of convergence of the series (1) using the substitution t = q −s and the identity N a = q deg(a) . We obtain a power series for ζ(s) in terms of t: (2)

Z(t) =

X

tdeg(a) =

∞ X

An tn ,

n=0

a≥0

where An is the number of nonnegative divisors of degree n. By Lemma 4.1.4 Z(t) =

2g−2 X

An tn + h

n=0

∞ X q mδ−g+1 − 1 mδ t q−1

m=d

where d = 2g−2+δ . The right hand side converges for |t| < q −1 ; (i.e. for δ Re(s) > 1) and (3)

Z(t) = Φ(t) +

hq g−1+δ t2g−2+δ t2g−2+δ h · · − , q−1 1 − (qt)δ q − 1 1 − tδ

where Φ(t) =

2g−2 X

An tn

n=0

is a polynomial of degree ≤ 2g − 2. We summarize: Proposition 4.2.1: The power series Z(t) in (2) converges in the circle |t| < q −1 . Formula (3) continues Z(t) to a meromorphic function on the whole plane. The only poles of Z(t) occur for values of t with tδ = 1 or tδ = q −δ , and they are simple. The Dirichlet series for ζ(s) in (1) converges in the right half plane Re(s) > 1. The substitution t = q −s in (3) continues ζ(s) to a meromorphic function in the whole plane. Like the Riemann zeta function, ζF/K (s) has a multiplicative presentation. Let a1 , a2 , a3 , . . . be a sequence of complex numbers of absolute value

80

Chapter 4. The Riemann Hypothesis for Function Fields

less than 1. We say that the infinite product solutely converges) if the limit n Y

1 n→∞ 1 − ai i=1 lim

Q∞

1 i=1 1−ai

converges (resp. ab-

n Y

1 n→∞ 1 − |ai | i=1

resp. lim

converges to a nonzero complex number. Proposition 4.2.2: If Re(s) > 1 and |t| < q −1 , then Y Y 1 1 = , (4) ζ(s) = −s deg(p) 1 − (N p) 1 − t p p where p runs over the prime divisors of F/K. The product converges absolutely. Therefore, it is independent of the order of the factors. In particular, if Re(s) > 1, then ζ(s) 6= 0. Proof: The prime divisors are free generators of the group of divisors. Thus, for every positive integer m, if Re(s) > 1, then Y N p≤m

∞ Y X X X0 1 = (N p)−sk = (N a)−s + (N a)−s , −s 1 − (N p) a≥0 a≥0 N p≤m k=0

N a>m

N a≤m

where the prime in the second sum means that a runs over all nonnegative divisors with norm exceeding m whose prime divisors p satisfy N p ≤ m. It follows that X Y 1 |(N a)−s | ≤ ζ(s) − 1 − (N p)−s a≥0 N p≤m

N a>m

and the right hand side converges to zero as m tends to infinity. Now we prove that ζ(s) 6= 0 when Re(s) > 1. Indeed, consider a positive integer m. Then, Y Y 1 1 − (N p)−s = ζ(s) 1 − (N p)−s N p>m N p≤m X X00 (N a)−s ≥ 1 − (N a)Re(s) > 0, = 1 + a>0 N a>m

(where the double primes mean summation over all positive divisors a with only prime divisors p satisfying N p > m) if m is large enough. Therefore, ζ(s) 6= 0. Q In particular, p 1−(N p)1 −Re(s) converges. Since, |N p−s | = N p−Re(s) , this Q means that p 1−(N1p)−s absolutely converges. By Exercise 3, the value of the product is independent of the order of the factors.

4.3 Zeta Functions under Constant Field Extensions

81

4.3 Zeta Functions under Constant Field Extensions The analytic properties of the zeta function of a function field F/K proved in Proposition 4.2.2 result in the conclusion that δ = 1 (Corollary 4.3.3). Denote the unique extension of K of degree r by Kr . Then Fr = F Kr is a function field of one variable over Kr . Use r as a subscript to denote the “extension” of objects of F to Fr . For example, if a is a divisor of F/K, then degr a denotes the degree of a as a divisor of Fr /Kr . We have already noted that degr a = deg(a), dimr a = dim(a) and gr = g (Proposition 3.4.2). Lemma 4.3.1: Let p be a prime divisor of F/K. Then p decomposes in Fr as p = P1 + P2 + · · · + Pd , where P1 , P2 , . . . , Pd are distinct prime divisors of Fr /Kr , deg(Pi ) = r−1 · lcm(r, deg(p)) and d = gcd(r, deg(p)). Proof:

Put m = deg(p). Since p is unramified in Fr (Proposition 3.4.2), P1 , . . . , Pd are distinct. Moreover, (Fr )Pi = Kr F¯p . Hence, [(Fr )Pi : K] =

lcm(r, m), and thus deg(Pi ) = [(Fr )Pi : Kr ] = r−1 · lcm(r, m), i = 1, . . . , d. Also, by Propositions 2.6 and 3.4.2, r = [Kr : K] = [Fr : F ] = d · [F¯r,Pi : F¯p ] = d · lcm(r, m)m−1 . Therefore, d = gcd(r, m). Proposition 4.3.2: For every complex number t, Y (1) Zr (tr ) = Z(ξt), ξ r =1

where ξ runs over the rth roots of unity. Proof: Since both sides of (1) are rational functions of t (by (3) of Section 4.2), it suffices to prove (1) for |t| < q −r . First apply the product formula (4) of Section 4.2, then (Lemma 4.3.1): YY Y (1 − tr·deg(P) ) = (1 − tlcm(r,deg(p)) )gcd(r,deg(p)) (2) Zr (tr )−1 = p P|p

(3)

Y ξ r =1

Z(ξt)−1 =

p

Y Y p

(1 − (ξt)deg(p) ).

ξ r =1

Thus, (1) follows if we show equality of the corresponding factors on the right hand sides of (2) and (3). Indeed, for a fixed p let m = deg(p) and d = gcd(r, m). We must show that Y (1 − (ξt)m ). (4) (1 − trm/d )d = ξ r =1

Substitute tm = x−1 in (4) to rewrite it as Y (x − ξ m ) . (5) (xr/d − 1)d = ξ r =1

82

Chapter 4. The Riemann Hypothesis for Function Fields

Both monic polynomials in (5) have each (r/d)th root of unity as a zero of multiplicity d. Indeed, if ζr is a primitive root of unity of order r, then ζrm is a primitive root of unity of order r/d and each power of ζrm appears d times (r−1)m among 1, ζrm , ζr2m , . . . , ζr . Therefore, the polynomials are equal. Corollary 4.3.3 (F. K. Schmidt): δ = 1. P∞ Proof: By (2) of Section 4.2, Z(t) = m=0 Amδ tmδ . Hence, if ξ δ = 1, then Z(ξt) = Z(t). From (1), Zδ (tδ ) = Z(t)δ . However, by (3) of Section 4.2 (applied to Zδ (tδ ) instead of to Z(t)), Zδ (tδ ) has a simple pole at t = 1, while Z(t)δ has a pole of order δ at t = 1. Consequently, δ = 1. Corollary 4.3.4: For every integer n there are exactly h divisor classes of F/K of degree n. Proof: By Corollary 4.3.3 there is a divisor c of degree n. The map a → a+c induces a bijection of the set of divisor classes of F/K of degree 0 onto the set of divisor classes of degree m. Hence, the number of elements in the latter set equals the number of elements in the former set, namely h. Example 4.3.5: By definition, A1 is the number of prime divisors of F/K of degree 1. If g = 0, then h = 1 and A1 = q +1 (Lemma 4.1.4). Thus, F/K has prime divisors of degree 1. By Example 3.2.4, F = K(x) is a rational function field. If δ = 1 (Corollary 4.3.3) and g = 1, then A1 = h (Lemma 4.1.4); in other words the class number is equal to the number of prime divisors of degree 1. Since this is the order of a group (Definition 4.1.1 ), it is again 1 . positive. By Lemma 4.1.4 and by (2) of Section 4.1.2, Z(t) = (1−t)(1−qt)

4.4 The Functional Equation Like the Riemann zeta functions, Z(t) satisfies a functional equation relating 1 . The main tool in the proof is the Riemann-Roch its values in t and qt theorem. Proposition 4.4.1: Z(t) satisfies the functional equation 1 √ √ ( qt)1−g Z(t) = ( qt)g−1 Z . qt Proof:

If g = 0, the result follows from the explicit presentation Z(t) = in Example 4.3.5. Therefore, assume g > 0. The basic idea is to split Z(t) into the sum of a polynomial P (t) and an infinite series Q(t), each of which satisfies the same functional equation in the statement of the proposition. Apply Lemmas 4.1.3 and 4.1.4 with δ = 1 (Corollary 4.3.3) to obtain: 1 (1−t)(1−qt)

Z(t) =

X a≥0

tdeg(a) =

2g−2 X deg(C)=0

X a≥0 a∈C

tdeg(a) +

∞ X n=2g−1

An tn

4.4 The Functional Equation

=

2g−2 X deg(C)=0

=

h 1 q−1

83

∞ X q dim(C) − 1 deg(C) q n−g+1 − 1 n t t + h q−1 q−1 n=2g−1 2g−2 X

q dim(C) tdeg(C)

i

deg(C)=0

+

∞ ∞ h h X h X ni q n−g+1 tn − t q − 1 n=2g−1 q − 1 n=0

= P (t) + Q(t), where P (t) (resp. Q(t)) is the expression in the first (resp. second) brackets. We have also used that F/K has exactly h divisor classes of each degree (Corollary 4.3.4). First we analyze P (t). The Riemann-Roch theorem relates dim(C) to dim(W − C), where W is the canonical class. Recall that deg(W ) = 2g − 2 (Lemma 3.2.2(b)). Hence 1 1 deg(C) = deg(C) + 1 − g + dim(W − C) 2 2 1 = dim(W − C) − deg(W − C). 2 As C varies over all divisor classes of degree between 0 and 2g − 2 so does W − C. Hence √ 2g−2 ( qt)2−2g X √ 2−2g ( qt) P (t) = q dim(C) tdeg(C) q−1 dim(C) −

deg(C)=0

=

=

=

1 q−1 1 q−1 1 q−1

2g−2 X

1 √ q dim(C)− 2 deg(C) ( qt)2−2g+deg(C)

deg(C)=0 2g−2 X

1 √ q dim(W −C)− 2 deg(W −C) ( qt)− deg(W −C)

deg(C)=0 2g−2 X

0

0

q dim(C ) (qt)− deg(C ) = P

deg(C 0 )=0

1 . qt

Now evaluate the geometric series involved in the expression for Q(t): ∞ ∞ h h q g t2g−1 h h X n−g+1 n X n i 1 i Q(t) = − q t − t = q − 1 n=2g−1 q − 1 1 − qt 1−t n=0 A direct computation shows that √

( qt)

2−2g

1 . Q(t) = Q qt

84

Chapter 4. The Riemann Hypothesis for Function Fields

This completes the proof of the proposition.

4.5 The Riemann Hypothesis and Degree 1 Prime Divisors We reformulate here the Riemann hypothesis for a function field F of one variable over a finite field and draw an estimate for the number of prime divisors of F of degree 1. Rewrite formula (3) of Section 4.2 with δ = 1 (Corollary 4.3.3) as Z(t) = Φ(t) +

t2g−1 h t2g−1 hq g · − · q − 1 1 − qt q − 1 1 − t

where Φ(t) a polynomial of degree ≤ 2g − 2. Hence Z(t) =

L(t) (1 − t)(1 − qt)

with L(t) = a0 + a1 t + · · · + a2g t2g a polynomial with rational coefficients. We determine some of these coefficients: First: a0 = L(0) = Z(0) = A0 = 1, since the zero divisor is the only nonnegative divisor of degree 0. Second: A1 is equal to the number of prime divisors of F/K of degree 1. Write N = A1 , so that L(t) = (1 − t)(1 − qt)

∞ X

An tn ≡ 1 + (N − (q + 1))t mod t2 .

n=0

Therefore, a1 = N − (q + 1)

(1)

Now let x ∈ F be transcendental over K and write F0 = K(x). The Zeta 1 (Example 4.3.5). By Proposition function of F0 /K is Z0 (t) = (1−t)(1−qt) 4.4.1 1 q g−1 t2g−2 Z qt Z(t) 1 g 2g (2) L(t) = = −1 −2 . t L = q 1 Z0 (t) qt q t Z0 qt This functional equation for L(t), written explicitly, has the form 2g X i=0

ai ti =

2g X

a2g−i q i−g ti .

i=0

Equivalently, ai = q i−g a2g−i . In particular, deduce for i = 0 and i = 1 that (3) a2g = q g and a2g−1 = q g−1 N − (q + 1) .

4.5 The Riemann Hypothesis and Degree 1 Prime Divisors

85

These formulas imply that deg(L(t)) = 2g. Decompose L(t) over C as (4)

L(t) =

2g Y

(1 − ωi t)

i=1

where the ωi−1 ’s are the zeros of L(t). Formulas (1), (3), and (4) give (5)

qg =

2g Y

ωi

and N − (q + 1) = −

i=1

2g X

ωi .

i=1

Moreover, the functional equation (2) for L(t) implies that 1 ωi = 0. L = 0 if and only if L ωi q √ √ Rename the roots ω1 , . . . , ω2g as ω1 , ω10 , . . . , ωf , ωf0 , q, . . . , q, √ √ − q, . . . , − q with f ≤ g such that ωi ωi0 = q, i = 1, . . . , f , and q √ (resp. − q) appear k (resp. l) times. Then 2f + k + l = 2g shows that if k is odd, then l is odd. In this case (5) gives q g = q f q k/2 (−1)l q l/2 = −q g , a contradiction. Hence, both k and l are even and we may take f = g. Thus, √

L(t) =

g Y

(1 − ωi t)(1 − ωi0 t),

i=1

with ωi ωi0 = q for i = 1, . . . , g. Here is a reformulation of the Riemann hypothesis for the function field F/K. Sections 4.6 - 4.8 complete the proof. Theorem 4.5.1: (a) The zeros of the function ζF/K (s) lie on the line Re(s) = 12 . 1

(b) The zeros of the function ZF/K (t) lie on the circle |t| = q − 2 . (c) If the ωi are the inverses of the zeros of the polynomial LF/K (t), then |ωi | =

√ q,

i = 1, 2, . . . , 2g.

Note that (c) is equivalent to (b), since the poles of Z(t) are t = 1 and t = q −1 . Theorem 4.5.1 with (5) provides an estimate on the number of prime divisors of degree 1: Theorem 4.5.2: Let F be a function field of one variable over a finite field K of q elements. Denote the genus of F/K by g and let N be the number of √ prime divisors of F/K of degree 1. Then |N − (q + 1)| ≤ 2g q.

86

Chapter 4. The Riemann Hypothesis for Function Fields

4.6 Reduction Steps Theorem 4.5.2 is a consequence of the Riemann hypothesis. As a first step, this section shows that an appropriate version of Theorem 4.5.2 implies the Riemann hypothesis. As in Section 4.3, denote the unique extension of K of degree r by Kr . Lemma 4.6.1: The Riemann hypothesis holds for the function field F/K if and only if it holds for the function field Fr /Kr . Proof: Use (2) and (4) of Section 4.5 to express L(t). Then apply Proposition 4.3.2 to compute Lr (tr ): Lr (tr ) =

Y Z(ξt) Y Zr (tr ) = = L(ξt) r Zr,0 (t ) Z0 (ξt) r r ξ =1

ξ =1

=

2g Y Y ξ r =1

(1 − ωi ξt) =

i=1

2g Y

(1 − ωir tr ).

i=1

Q2g Hence, Lr (t) = i=1 (1 − ωir t). Thus, r (1) ω1r , . . . , ω2g are the inverses of the zeros of Lr . √ √ Since |ωi | = q if and only if |ωir | = q r , the lemma follows.

Denote the number of prime divisors of Fr /Kr of degree 1 by Nr . Lemma 4.6.2: Let F be a function field of one variable over a field K of q elements. If there exists a constant c such that |Nr − (q r + 1)| ≤ cq r/2 for every positive integer r, then the Riemann hypothesis holds for F/K. log Proof: Apply the differential operator D = −t ddt to both sides of the Q2g formula L(t) = i=1 (1 − ωi t):

(2)

D(L(t)) =

2g X i=1

2g ∞ X X ωi t = ωir tr 1 − ωi t n=1 r=1

P2g Combine (1) with (5) of Section 4.5 to obtain − i=1 ωir = Nr − (q r + 1). P2g r The hypothesis of the lemma thus implies | i=1 ωi | ≤ cq r/2 . Therefore 1 the radius of convergence R of the right hand side of (2) satisfies R ≥ q − 2 . But (2) implies that the ωi−1 are the only singularities of D(L(t)). Hence, √ R = min1≤i≤2g |ωi−1 |. Therefore, |ωi | ≤ q for i = 1, . . . , 2g. This together Q2g √ with the equality q g = i=1 ωi ((5) of Section 4.5), implies that |ωi | = q for i = 1, . . . , 2g.

4.7 An Upper Bound

87

4.7 An Upper Bound Assume, by extension of constants if necessary, that K and F satisfy these conditions: (1a) q = a2 is a square; (1b) q > (g + 1)4 ; and (1c) F has a prime divisor o of degree 1. By Lemma 4.6.1, a proof of the Riemann hypothesis under these conditions suffices for the general case. We prove a result that has, as a special case, the inequality √ N − (q + 1) < (2g + 1) q.

(2)

Let σ be an automorphism of F over K. It induces a permutation of the prime divisors of F/K. If p is a prime divisor of F/K, the pσ is the prime divisor corresponding to the place ϕσp (x) = ϕp (σx) for x ∈ F (where, as in Section 2.1, ϕσp = σ −1 ϕp ). Also, recall that the map x 7→ xq is an automorphism F˜ over K. Hence, the formula ϕqp (x) = ϕp (x)q defines a place ϕqp of F/K. Although ϕqp is equivalent to ϕp , it is convenient to use ϕqp , because ϕp = ϕqp if and only if deg(p) = 1. The remainder of this section investigates the expression N (σ) =

X

deg(p)

q ϕσ p =ϕp

in order to show that (3)

√ N (σ) − (q + 1) < (2g + 1) q.

If σ is the identity automorphism, then N (σ) = N and (3) becomes (2). With the notation m = a − 1, n = a + 2g, and r = m + an, rewrite (3) as (4)

N (σ) − 1 ≤ r.

With o as in (1c), consider the ascending sequence of K-vector spaces L(o) ⊆ L(2o) ⊆ L(3o) ⊆ · · ·. By Exercise 2 of Chapter 3, (5)

dim(L(io)) − dim(L((i − 1))o) ≤ 1.

88

Chapter 4. The Riemann Hypothesis for Function Fields

With k a positive integer, let Ik be the set of i, 1 ≤ i ≤ k for which equality holds in (5). For each i ∈ Ik , choose ui ∈ L(io)−L((i−1)o). Then div∞ (ui ) = io and the system {ui | i ∈ Ik } is a basis for L(ko). In particular, this holds for k = m. Since a2 = q, a is a power of char(K). Thus, the set L(no)a = {y a | y ∈ L(no)}, consisting of elements in the field F a is a Kvector space with basis {uaj | j ∈ In } and the same dimension as L(no). Therefore, o nX ui yia | yi ∈ L(no) L= i∈Im

is a K-vector space generated by the set U = {ui uaj | i ∈ Im and j ∈ In }. Lemma 4.7.1: The set U is linearly independent over K. Proof: It suffices to prove that {ui | i ∈ Im } is linearly independent over a . Indeed, assume that there exist yi ∈ F , i ∈ Im , not all zero, the field FP a such that i∈Im ui yi = 0. Then there exist distinct i, j ∈ Im such that a a vo (ui yi ) = vo (uj yj ), yi 6= 0, yj 6= 0 ((2) of Section 2.1). Thus, −i+avo (yi ) = −j + avo (yj ), so i ∼ = j mod a. Since 1 ≤ i, j ≤ m < a, this is a contradiction. By Lemma 4.7.1, dim(L) = dim(L(mo)) · dim(L(no)). Apply RiemannRoch to the right hand side terms: (6)

dim(L) ≥ (m − g + 1)(n − g + 1) = q +

√ q − g(g + 1).

Now consider the K-vector space L0 =

nX

o (σ −1 ui )a yi | yi ∈ L(no) .

i∈Im

Check that L0 ⊆ L(maoσ + no) and deg(maoσ + no) = q + 2g > 2g − 2. By Riemann-Roch, (7)

dim(L0 ) ≤ dim(maoσ + no) = deg(maoσ + no) − g + 1 = q + g + 1.

√ By (1b), q − g(g + 1) > g + 1. Thus, the right side of (6) is greater than the right side of (7). Therefore, dim(L0 ) < dim(L).

(8)

Finally, define an additive map σ ∗ from L into L0 by σ∗

X i∈Im

X ui yia = (σ −1 ui )a yi i∈Im

4.8 A Lower Bound

89

By (8), the kernel of σ ∗ is nontrivial. Hence, there exist yi ∈ L(no), i ∈ Im , not all zero, with X (9) (σ −1 ui )a yi = 0. i∈Im

In particular, u = i∈Im ui yia is a nonzero element of L(ro). If p is a prime divisor of F/K and p 6= o, then ϕp (yi ) 6= ∞ and ϕp (ui ) 6= ∞. If in addition ϕσp = ϕqp , then (9) implies X X ϕp (u) = ϕp (ui )ϕp (yi )a = ϕp (σ −1 ui )q ϕp (yi )a P

i∈Im

= ϕp

i∈Im

X

(σ −1 ui )a yi

a

=0

i∈Im

Hence, p occurs in the divisor of zeros of u. In other words, X p ≤ div0 (u). p6=o pσ =pq

Thus, N (σ) − 1 ≤ deg(div0 (u)) = deg(div∞ (u)) ≤ r and the proof of (4) is complete.

4.8 A Lower Bound In the notation of Section 4.7, we establish a lower bound inequality of the √ form N (σ) − (q + 1) > c0 q where c0 is an explicit constant depending only on F . This will complete the construction of Lemma 4.6.2 toward the proof of the Riemann hypothesis for F/K. Lemma 4.8.1: Let F be a function field of one variable over a field K with q elements. Let F 0 be a finite Galois extension of F with a Galois group G such that K algebraically closed in F 0 and let σ ∈ G. Then N (σ) (F ) = Pis also (στ 0 −1 ) (F 0 ). [F : F ] τ ∈G N Proof: Let p0 be a prime divisor of F 0 /K and let p be its restriction to F . q Then ϕσp = ϕqp if and only if there exists τ ∈ G such that ϕστ p0 = ϕp0 . The 0 number of such τ is the ramification index ep of p over p (Section 2.3). Put fp0 /p = [F¯p0 0 : F¯p ] and denote the number of prime divisors of F 0 lying over p by gp0 /p . Then: X X X X X N (στ ) (F 0 ) = deg(p0 ) = ep0 /p deg(p0 ) q τ ∈G ϕστ 0 =ϕ 0

τ ∈G

p

=

X

p

q 0 ϕσ p =ϕp p |p

ep0 /p fp0 /p gp0 /p deg(p)

q ϕσ p =ϕp

= [F 0 : F ]

X q ϕσ p =ϕp

deg(p) = [F 0 : F ]N (σ) (F ).

90

Chapter 4. The Riemann Hypothesis for Function Fields

Let F be a function field of one variable over a field K of q elements and let σ be an automorphism of F over K of finite order. Denote the fixed field of σ in F by E. Then F is a finite Galois extension of E. As a finite field, K is perfect. Hence, there exists x ∈ E, transcendental over K, such that E is a finite separable extension of K(x). Let Fˆ be the Galois closure of F/K(x) ˆ be the algebraic closure of K in Fˆ . Then Fˆ , as well as F K, ˆ are and let K ˆ and σ extends to an automorphism of function fields of one variable over K ˆ ˆ (if necessary), assume Fˆ over K(x). After an additional finite extension of K ˆ and therefore for F K/ ˆ K ˆ that Condition (1) of Section 4.7 holds for Fˆ /K, (use Proposition 3.4.2). Starting with a given F we have extended the field of constants so as to assume these conditions: (1a) F/K has a separating transcendence element x; the field F has a finite extension Fˆ , which is Galois over K(x), and K is algebraically closed in Fˆ ; ˆ and (1b) q is a square larger than (ˆ g + 1)4 where gˆ is the genus of Fˆ /K; ˆ has a prime divisor of degree 1. (1c) Fˆ /K Lemma 4.8.2: Under these conditions N (σ) (F ) − (q + 1) ≥ −

(2)

n−m √ (2ˆ g + 1) q, m

where m = [Fˆ : F ] and n = [Fˆ : K(x)]. Proof: Let H = Gal(Fˆ /F ) and G = Gal(Fˆ /K(x)). From Lemma 4.8.1 (3)

N (σ) (F ) =

1 X (στ ) ˆ 1 X (θ) ˆ N (F ) and q + 1 = N (K(x)) = N (F ). m n τ ∈H

θ∈G

Apply inequality (3) of Section 4.7: X

N (θ) (Fˆ ) =

X

N (στ ) (Fˆ ) +

τ ∈H

θ∈G

≤

X

=

N (θ) (Fˆ )

θ∈G r σH

N (στ ) (Fˆ ) +

τ ∈H

X

X X

√ (q + 1 + (2ˆ g + 1) q)

θ∈G r σH

N

(στ )

√ (Fˆ ) + (n − m)(q + 1 + (2ˆ g + 1) q).

τ ∈H

From the second half of (3), X τ ∈H

√ N (στ ) (Fˆ ) ≥ n(q + 1) − (n − m)(q + 1 + (2ˆ g + 1) q) √ = m(q + 1) − (n − m)(2ˆ g + 1) q.

Exercises

91

Thus, the first half of (3) implies that N (σ) (F ) ≥ (q + 1) −

n−m √ (2ˆ g + 1) q. m

Note that the condition (1), as well as the numbers gˆ, m, n, are independent of extension of the field of constants. We may therefore combine Lemma 4.8.2 with the results of Section 4.7 to conclude: Proposition 4.8.3: Let F be a function field of one variable over a finite field K and let σ be an automorphism of F over K of finite order. Then K has a finite extension K 0 with q 0 elements and there exists a positive constant c such that for every positive integer r we have |N (σ) (Fr0 ) − ((q 0 )r − 1)| ≤ c(q 0 )r/2 , where Kr0 is the unique extension of K 0 of degree r, Fr0 = F 0 Kr0 , and σ extends to an automorphism, also denoted by σ, of Fr0 over Kr0 . In particular, Proposition 4.8.3 is valid in the case σ = 1. By Lemma 4.6.2, the Riemann hypothesis is true for the function field F 0 /K 0 . It follows from Lemma 4.6.1 that it is also true for F/K.

Exercises 1.

For real valued functions f, g write f (x) = O(g(x)) as x → a

if there exists a positive constant c such that |f (x)| ≤ c|g(x)| for all values of x in a neighborhood of a. Let F be a function field of one variable over a field K of q elements. Denote the set of prime divisors of F/K of degree r by Pr (F/K). Follow these instructions to prove that |Pr (F/K)| =

1 r q + O(q r/2 ) r

(a) Use Theorem 4.5.2 to prove that |P1 (Fr /Kr )| = q r + O(q r/2 ). (b) Observe that if P ∈ P1 (Fr /Kr ) and if p is the prime divisor of F/K that lies below P, then deg(p)|r. (c) Deduce from Lemma 4.3.1 that if d|r, then over each p ∈ Pd (F/K) there lie exactly d elements of P1 (Fr /Kr ). Thus, X |P1 (Fr /Kr )| = d|Pd (F/K)|. d|r

(d) Use the estimates |Pd (F/K)| = O(q d ) and d ≤ 2r for proper divisors d of r r r and the inequality q + q 2 + · · · + q 2 ≤ 2q 2 to complete the proof.

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Chapter 4. The Riemann Hypothesis for Function Fields

2. Let F be a function field of one variable over a field K of q elements and genus g. Let σ be an automorphism of F over K of finite order and let N (σ) √ be as in Section 4.7. Prove that |N (σ) − (q + 1)| ≤ 2g q. Hint: Extend σ to ˜ K ˜ by σ ˜ Then let E be an automorphism σ ˜ of F K/ ˜ x = xq for each x ∈ K. ˜ the fixed field of σ ˜ in F K. Show that E is a function field of one variable ˜ = F K. ˜ over K and that E K 3. This exercise establishes basic facts about infinite products which lie behind the multiplicative presentation of the zeta functions. Consider a se∞ quence {zi }Q i=1 of nonzero complex numbers. If the sequence of the partial n to a nonzero complex number z, we say that products i=1 zi converges Q∞ z converges and the infinite product i=1 i Q∞that z is its value. We say that Qn (1 + z ) absolutely converges if i i=1 i=1 (1 + |zi |) converges. The logarithm function makes a connection between the theory of infinite products and the theory of infinite series: Proposition ([Knopp, p. 434]): Q∞ Let ai 6= −1, i = 1, 2, 3, . . . be complex numbers. Then the product i=1 (1 + ai ) converges if and only if, the series P∞ i=1 log(1+ai ) whose terms are the Q∞principal values of log(1+ai ) converges. If l is the sum of this series, then i=1 (1 + ai ) = el . (a) Prove that if |z| ≤ 12 , then |z| ≤ 2| log(1 + z)| ≤ 2|z|. (b) Suppose 0 < |ai | ≤ 12 for i = 1, 2, 3, . . . and consider the following series: ∞ ∞ ∞ X X X | log(1 + ai )|, |ai |, log(1 + |ai |). i=1

i=1

i=1

Use (a) to prove that the convergence of each these series implies the convergence of the two others. Q∞ − |ai |)−1 con(c) Suppose 0 < |ai | ≤ 12 for i = 1, 2, 3, . . . and i=1 Q(1 ∞ verges. Then for each permutation π of N, the product i=1 (1 − aπ(i) )−1 converges and its value is independent of π. 4. Let F be a function field of one variable over a field K with q elements. Consider σ ∈ Aut(F/K). As in Section 4.7, define N (σ) to be the sum of all deg(p), where p ranges over all prime divisors of F/K for which ϕσp = ϕqp . ˜ by the formula σ ˜ Extend σ to an automorphism of F K ˜ x = xq for each x ∈ K. Use Lemma 4.3.1 for r a multiple of deg(p) to prove that N (σ) is also the ˜ K ˜ for which ϕσ˜ = ϕq . number of prime divisors P of F K/ P P 5. Let F be a function field of one variable over a field with q elements and of genus g. Let σ be an element of Aut(F/K) of order n. Define N (σ) as in Exercise 4. Prove that √ |N (σ) − (q + 1)| ≤ 2g q. ˜ by F 0 . Prove Hint: Let σ ˜ be as in Exercise 4. Denote the fixed field of σ ˜ in F K 0 ˜ in F Kr for each multiple r of n. Deduce that F is also the fixed field of σ

Notes

93

˜ = F 0 K. ˜ Use that F 0 is a function field of one variable over K such that F K Exercise 4 to prove that N (σ) is the number of prime divisors of F 0 /K of degree 1. Then apply Theorem 4.5.2. 6. Let q be a power of an odd prime. Prove that ax2 + by 2 = c has a solution (x, y) ∈ F2q if a, b, c ∈ F× q . Let (x0 , y0 ) be one of these solutions. Use the substitution x = x0 + tz, y = y0 + t and solve for t in terms of z to show that the function field of aX 2 + bY 2 − c = 0 is isomorphic to Fq (z). Conclude that the function field has exactly q + 1 degree 1 places. Hint: Assume a = 1 and observe that c − by 2 takes on (q + 1)/2 values in Fq . On the other hand there are q+1 2 squares in Fq . 7. Let q be a prime power ≡ 1 mod 3 and let α be a generator of F× q . Follow these instructions to prove that αX 3 + α2 Y 3 = 1 has a solution in Fq : Let Ai = {αi x3 | x ∈ F× q }, i = 0, 1, 2 and note that αA0 = A1 , αA1 = A2 and αA2 = A0 . Let Ai + Aj = {x + y | x ∈ Ai ∧ y ∈ Aj }. Now assume that A1 + A2 ⊆ A1 ∪ A2 to obtain a contradiction. For this use the notation A1 + A2 = m1 A1 ∪ m2 A2 to indicate that for each ai ∈ Ai there are mi pairs (x, y) ∈ A1 × A2 such that x + y = ai (independent of ai ∈ Ai ), i = 1, 2. Multiply this “equality” by α (resp. α2 ) to compute A2 + A0 (resp. A0 + A1 ). For a0 ∈ A0 , use this to compute the number of triples (x, y, z) ∈ A0 ×A1 ×A2 such that x + y + z = a0 in two ways: first compute (A0 + A1 ) + A2 and then A0 + (A1 + A2 ). The resulting expressions for m1 and m2 will lead to the contradiction m1 = m2 = 0. The attractiveness of the Riemann hypothesis for curves over finite fields has resulted in extensive lists of problems in a number of books treating the combinatorics of finite fields. For the sake of completeness we reference two such problem sources: [Ireland-Rosen, Chap. 8, pp. 105–107; Chap. 11, pp. 169–171] [Lidl-Niederreiter, see notes, Chap. 6, pp. 339–346]. In those chapters that use the Riemann hypothesis, our material and problems will tend to concentrate on the connections between the Riemann hypothesis and arithmetic properties of fields that are special to this book (e.g. an explicit form of Hilbert’s irreducibility theorem for global fields that follows from the Riemann hypothesis - Theorem 13.3.4).

Notes An extensive survey of the literature giving estimates on the number of points on an affine variety V appears in [Lidl-Niederreiter, pp. 317–339]. Although considerable literature on the Riemann hypothesis for curves over finite fields (Theorem 4.5.1) existed long before the two proofs given by Weil [Weil2], subsequent concerns included two sophisticated — and interrelated — developments. Both of Weil’s proofs employed elements of the theory of algebraic geometry outside the domain of algebraic curves. Indeed, Weil’s far reaching generalization of Theorem 4.5.1 was suggested to him

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Chapter 4. The Riemann Hypothesis for Function Fields

by the latter proof. This generalization, now called the Riemann hypothesis for nonsingular projective varieties over finite fields (proved by Deligne in [Deligne]). Since, however, Weil’s theorem had so many applications to apparently elementary results about finite fields, many practitioners were anxious for a more accessible proof. Stepanov [Stepanov1] was the first to make serious progress on an elementary proof (e.g. not applying the theory of algebraic surfaces) of Weil’s result. He introduced elements of diophantine approximation to the problem in the case of hyperelliptic curves, in a style suggested by the Thue-SiegelRoth theorem. To do this he constructed an auxiliary nonvanishing function on the hyperelliptic curve with a “lot” of prescribed zeros of “high” multiplicity. Eventually [Stepanov2] realized Theorem 4.5.1 for all hyperelliptic curves over finite fields. Continuing with the purely diophantine approximation approach, Stepanov [Stepanov3], for prime fields (and some extra conditions on the equations) and W. M. Schmidt [W.M.Schmidt1 and W.M.Schmidt2] were able to prove Theorem 4.5.1 for all curves. Indeed, Schmidt’s method were even applicable to prove results like those of Deligne even for some complete intersections. Our proof, however, of Theorem 4.5.1 follows [Bombieri1]. In this proof the Riemann hypotheses replaces diophantine approximation to give the construction of the auxiliary functions that appear in Stepanov’s proof (the function u at the end of Section 4.7). Voloch [Voloch] has an elementary proof of the Riemann hypothesis for function fields that sometimes gives a better bound than Weil’s estimate (see also [St¨ohr-Voloch]).

Chapter 5. Plane Curves The estimate on the number of prime divisors of degree 1 of a function field F over Fq (Theorem 4.5.2) leads in this chapter to an estimate on the number N of K-rational zeros of an absolutely irreducible polynomial f ∈ Fq [X, Y ]. √ We prove (Theorem 5.4.1) that |N + (q + 1)| ≤ (d − 1)(d − 2) q + d, where d = deg(f ).

5.1 Affine and Projective Plane Curves Let K be a field and let Ω be an algebraically closed extension of K of infinite transcendence degree over K. We denote by A2 the affine plane: all pairs (x, y) ∈ Ω2 . We denote by P2 the projective plane: all nonzero triples x = (x0 , x1 , x2 ) ∈ Ω3 modulo the equivalence relation x ∼ x0 if and only if x0 = cx for some c ∈ Ω. We denote the equivalence class of (x0 , x1 , x2 ) by (x0 :x1 :x2 ). Embed the affine plane A2 in P2 by the map (x, y) → (1:x:y). With this understood, the points of A2 are then referred to as the finite points on P2 , whereas the points of the form (0:x1 :x2 ) are the points at infinity on P2 . An affine plane curve defined over K is a set (1)

Γ = {(x, y) ∈ A2 | f (x, y) = 0},

where f ∈ K[X, Y ] is a nonconstant absolutely irreducible polynomial. Write it in the form (2)

f (X, Y ) = fd (X, Y ) + fd−1 (X, Y ) + · · · + f0 (X, Y )

where fk (X, Y ) is a homogeneous polynomial of degree k, for k = 0, . . . , d, and fd (X, Y ) 6= 0. Then d is the degree of Γ. Attach to f the homogeneous polynomial f ∗ (X0 , X1 , X2 ) of degree d: (3) f ∗ (X0 , X1 , X2 ) = fd (X1 , X2 ) + X0 fd−1 (X1 , X2 ) + · · · + X0d f0 (X1 , X2 ) and let Γ∗ = {(x0 :x1 :x2 ) ∈ P2 | f ∗ (x0 , x1 , x2 ) = 0} Then Γ∗ is the projective plane curve corresponding to Γ. It is also called the projective completion of Γ. We have Γ = Γ∗ ∩ A2 , and Γ∗ r Γ is a finite set corresponding to the points of fd (X1 , X2 ) = 0 in P1 . The infinite points of Γ∗ are sometimes referred to as the points at infinity on Γ. If f (X, Y ) = a+bX+cY with a, b, c ∈ K and b 6= 0 or c 6= 0 (i.e. deg(f ) = 1), then Γ is a line. The points on the corresponding projective line satisfy aX0 + bX1 + cX2 = 0. The line at infinity is given by X0 = 0.

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Chapter 5. Plane Curves

Call a point (x, y) of an affine curve Γ generic if trans.degK K(x, y) = 1. Because f is absolutely irreducible, the field F = K(x, y) is a regular extension of K. Thus, F is a function field of one variable which we call the function field of Γ over K. Up to K-isomorphism it is independent of the choice of the generic point. The map (X, Y ) 7→ (x, y) extends to a K-epimorphism of rings K[X, Y ] → K[x, y] with f (X, Y )K[X, Y ] as the kernel (Gauss’ lemma). Thus, a polynomial g ∈ K[X, Y ] vanishes on Γ if and only if g(x, y) = 0, or, equivalently, g is a multiple of f . Define the genus of Γ (and of Γ∗ ) to be the genus of F . The coordinate ring of Γ over K is R = K[x, y]. There is a bijective correspondence between the set of all maximal ideals m of R with quotient field R/m = K and the set Γ(K) of all K-rational points of Γ. If (a, b) ∈ Γ(K), then the corresponding maximal ideal of R is m = {g(x, y) ∈ R | g(a, b) = 0}. If p is a prime ideal of R and p ⊆ m, then the transcendence degree of the quotient field of R/p is either 0 or 1. In the latter case p = 0, and in the former case R/p is already a field so that p = m. The local ring of Γ at a = (a, b) over K is the local ring of R at m: OΓ,(a,b),K = Rm =

o n g(x, y) g(x, y), h(x, y) ∈ R and h(a, b) 6= 0 . h(x, y)

The unique nonzero prime ideal of Rm is generated by the elements of m. As a local ring of a Noetherian domain, Rm is itself a Noetherian domain. Similarly, a point (x0 :x1 :x2 ) of Γ∗ with x0 6= 0 is said to be generic over K, if xx10 , xx20 is a generic point of Γ. Define the local ring of a point a = (a0 :a1 :a2 ) of Γ∗ over K as OΓ∗ ,a,K =

n g(x , x , x ) g, h are homogeneous polynomials of o 0 1 2 h(x0 , x1 , x2 ) the same degree and h(a0 , a1 , a2 ) 6= 0

If a0 6= 0, then this ring coincides with the local ring of the corresponding point aa10 , aa20 of Γ. If a1 6= 0, then regard a as a point on the affine curve Γ1 , defined by the equation f ∗ (X0 , 1, X2 ) = 0. Then the projective completion of Γ1 is also Γ∗ , but the line X1 = 0 is taken as “the line at infinity”. This shows that the local ring of a projective plane curve at each point is equal to the local ring of some affine curve at a point. It is therefore a Noetherian domain. In any case, call a K-rational point of Γ (or of Γ∗ ) K-normal if its local ring over K is integrally closed. Two affine plane curves Γ1 and Γ2 are K- isomorphic if their coordinate rings are K-isomorphic. In this case generators of the coordinate ring of Γ2 (resp. Γ1 ) can be expressed as polynomials in the generators of the coordinate ring of Γ1 (resp. Γ2 ), and the composition of these two polynomial maps is the identity when applied to generators of the coordinate ring of Γ1 (resp. Γ2 ).

5.2 Points and Prime Divisors

97

Two projective plane curves Γ∗1 and Γ∗2 are K-isomorphic if (4a) for each x ∈ Γ∗1 there exist homogeneous polynomials g0 , g1 , g2 in K[X0 , X1 , X2 ] of the same degree such that yi = gi (x) 6= 0 for at least one i, 0 ≤ i ≤ 2, y ∈ Γ∗2 , and there exist homogeneous polynomials h0 , h1 , h2 in K[X0 , X1 , X2 ] of the same degree such that hi (y) = xi , i = 0, 1, 2; and (4b) the same condition with the roles of Γ∗1 and Γ∗2 exchanged. For example, if g0 , g1 , g2 are linear polynomials with a nonsingular coefficient matrix, then (g0 , g1 , g2 ) is called nonsingular homogeneous linear transformation. The function fields of two isomorphic plane curves are K-isomorphic. So are the local rings of corresponding points. It follows that the genera of isomorphic plane curves are the same; and if p1 and p2 are corresponding points of the curves, then p1 is K-normal if and only if p2 is K-normal. In particular, both curves have the same number of nonnormal points. If Γ is a plane curve defined over a field K and L is an algebraic extension of K, then Γ is also defined over L. The function field FL of Γ over L is the extension of F by the field of constants L. By Proposition 3.4.2, the genus of Γ remains unchanged if L is separable over K.

5.2 Points and Prime Divisors Let Γ be an affine plane curve of degree d defined by an absolutely irreducible equation f (X, Y ) = 0 over a field K. We establish a bijective correspondence between the K-rational points of f and the prime divisors of the function field of f of degree 1 (Lemma 5.2.2). We prove that a K-rational of Γ is normal if and only if it is simple (Lemma 5.2.3). Let (x, y) be a generic point of Γ over K. Denote the coordinate ring and the function field, respectively, of Γ over K by R = K[x, y] and F = K(x, y). Consider a K-rational point (a, b) of Γ. Then the map (x, y) 7→ (a, b) uniquely extends to a K-homomorphism ϕ of the local ring O(a,b) into K. The ˜ homomorphism ϕ extends further (not necessarily uniquely) to a K-valued 0 place ϕ of F . Call (a, b) the center of the corresponding prime divisor p of F/K. Then p lies over the unique prime divisor p0 of K(x)/K determined by the map x 7→ a. Since [F : K(x)] ≤ d, there exist at most d prime divisors p of F/K with the point (a, b) as a center on Γ. Since each point of Γ∗ is a finite point of some affine representative of Γ∗ (Section 5.1) this holds for each K-rational point of the projective completion Γ∗ of Γ. If a point p of Γ∗ (K) is K-normal, then its local ring Op is a Noetherian integrally closed domain. By Section 5.1 , each nonzero prime ideal of Op is maximal. Hence, by Proposition 2.4.5, Op is a discrete valuation domain. Therefore, there exists a unique prime divisor p of F/K with p as a center on Γ. The degree of p is 1. Conversely, consider a prime divisor p of F/K of degree 1. Suppose that ϕp is finite on R. Then (a, b) = ϕp (x, y) is a K-rational point of Γ and it is

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Chapter 5. Plane Curves

the center of p on Γ. The following result shows that the number of prime divisors p of F/K which are not finite on R does not exceed d. Proposition 5.2.1 (Noether’s Normalization Theorem): Let K[x1 , . . . , xn ] be a finitely generated integral domain over a field K with a quotient field F . If the transcendence degree of F over K is r, then there exist elements t1 , . . . , tr in K[x] such that K[x] = K[t1 , . . . , tr , x1 , . . . , xn−r ] (after rearranging the xi ’s) and K[x] is integral over K[t] [Lang4, p. 22]. If K is an infinite field, then t1 , . . . , tr can be chosen to be linear combinations of x1 , . . . , xr with coefficients in K [Zariski-Samuel1, p. 266]. Lemma 19.5.1 gives a constructive version of Noether’s normalization theorem. Return now to the plane curve Γ. Noether’s normalization theorem, in its linear form, allows us to replace x by a linear combination t of x and y such that R is integral over K[t] and [F : K(t)] ≤ d. If K is finite, replace K by a suitable finite extension to achieve the linear dependence of t on x and y. If a place ϕp of F is not finite on R, then it is infinite at t (Lemma 2.4.1). Hence, p lies over the infinite prime divisor p∞ of K(t)/K. There are at most d prime divisors of F/K that lie over p∞ . This proves our contention. Now we summarize. Lemma 5.2.2: Let Γ be an affine plane curve of degree d defined over a field K. Denote the coordinate ring and the function field, respectively, of Γ over K by R and F . Then: (a) For each K-normal point p ∈ Γ(K) there exists exactly one prime divisor p of F/K with center p on Γ; the degree of p is 1. (b) There are at most d prime divisors of F/K whose centers on Γ are a given point p. (c) If a prime divisor p of F/K is of degree 1 and if ϕp is finite on R, then its center is in Γ(K). (d) There are at most d prime divisors of F/K which are not finite on R. Finally, we point out that the K-normal points and simple points on ∂f (a, b) 6= 0 or Γ are the same. Here, a point (a, b) on Γ is simple if ∂X ∂f ∂Y (a, b) 6= 0. Lemma 5.2.3: A K-rational point (a, b) of Γ is normal if and only if it is ∂f (a, b) 6= 0, then F has a discrete normalized valuation simple. Moreover, if ∂X v with v(y − b) = 1. Proof: Suppose first that (a, b) is normal. Then, its local ring R = O(a,b) over K is a discrete valuation domain. Hence, R(x − a) ⊆ R(y − b) or R(y − b) ⊆ R(x − a). Suppose for example that R(x − a) ⊆ R(y − b). Then there exist g, h, q ∈ K[X, Y ] such that h(a, b) 6= 0 and h(X, Y )(X − a) = g(X, Y )(Y − b) + q(X, Y )f (X, Y ).

5.3 The Genus of a Plane Curve

Apply

∂ ∂X

99

to both sides and substitute (a, b) for (X, Y ): h(a, b) = q(a, b)

∂f (a, b). ∂X

∂f (a, b) 6= 0 and therefore (a, b) is simple. Hence, ∂X ∂f (a, b) 6= 0. We show that y − b generates Conversely, suppose that ∂X the maximal ideal m of R. Indeed,

f (X, Y ) =

∂f ∂f (a, b)(X − a) + (a, b)(Y − b) + higher terms, ∂X ∂Y

and 0 = f (x, y) =

∂f ∂f (a, b)(x − a) + (a, b)(y − b) + (x − a)u + (y − b)v, ∂X ∂Y

with u, v ∈ m. Hence, 0=

∂f (a, b) + u (x − a) + (a, b) + v (y − b). ∂X ∂Y

∂f

∂f (a, b) is a nonzero constant and u ∈ m, the coefficient of x − a But, since ∂X is a unit of R. Therefore, x − a ∈ R(y − b),Tso m = R(x − a, y − b) = R(y − b). ∞ Since R is Noetherian, the ideal a = n=1 mn is a finitely generated Rmodule. Since ma = a, Nakayama’s Lemma [Lang4, p. 195, Prop. 1] implies that a = 0 (alternatively, use Krull’s intersection theorem [Eisenbud, p. 152]). Hence, each z ∈ R has a unique representation z = w(y − b)n , with w a unit of R and n ≥ 0. Thus, R is a discrete valuation domain. Therefore, R is integrally closed (Exercise 3 of Chapter 2). Let v be the normalized discrete valuation of F corresponding to R. Since y − b generates m, we have v(y − b) = 1.

For a projective curve Γ∗ defined by f ∗ (X0 , X1 , X2 ) = 0 as in (3) of ∂f (a0 , a1 , a2 ) 6= 0, for some i, Section 5.1, the point (a0 :a1 :a2 ) is simple if ∂X i i = 0, 1 or 2. Otherwise, (a0 :a1 :a2 ) is singular.

5.3 The Genus of a Plane Curve Here we bound the genus and the number of nonnormal points of a plane curve by a function of its degree. We first prove a finiteness result for the coordinate ring of a plane curve. Lemma 5.3.1: Let R = K[x, y] be an integral domain with quotient field F of transcendence degree 1 over K. Let S be the integral closure of R in F . Then S/R is a finitely generated K-vector space. Proof: It is well known that S is a finitely generated R-module ([Lang4, p. 120] or [Zariski-Samuel1, p. 267]). Hence, there exists a nonzero element

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Chapter 5. Plane Curves

z ∈ R such that zS ⊆ R. Thus, S/R ∼ = zS/zR ⊆ R/zR. It suffices to prove that dimK R/zR < ∞. By Noether’s normalization theorem (Proposition 5.2.1), we may assume without loss that R is integral over K[x]. The element z satisfies an equation of the form z m + a(x)z m−1 + · · · + g(x) = 0, with 0 6= g(x) ∈ K[x] of degree, say, k. Since g(x) belongs to zR, every power of x is a linear combination modulo zR of 1, x, . . . , xk−1 with coefficients in K. If d is the degree of a monic equation for y over K[x], then each element of R can be written as h(x, y), with h ∈ K[X, Y ] is of degree at most d − 1 in Y . Our lemma follows. In the notation of the proof of Lemma 5.3.1, S is the integral closure in F of the Dedekind domain K[x]. Hence, S is a Dedekind domain and therefore every nonzero prime ideal of S is maximal. By Section 5.2, every nonzero prime ideal of R is maximal. Consider a nonzero prime ideal p of R. Let Rp be the local ring of R at p. Then the integral closure of Rp in F is Sp = sb | s ∈ S and b ∈ R r p . Lemma 5.3.2: P (a) The quotient S/R is isomorphic to the direct sum Sp /Rp , where p runs over the nonzero prime ideals of R. (b) Let p1 , . . . , pk be the prime ideals of S that lie over a prime ideal p of R. Then dimK Sp /Rp ≥ k − 1. Proof of (a): Denote the conductor of S over R by c = {c ∈ R | cS ⊆ R}. It is a nonzero ideal of both R and S. The local ring Rp is integrally closed, and therefore equal to Sp , if and only if c 6⊆ p [Zariski-Samuel1, p. 269]. Let Qk Ql(i) e c = i=1 j=1 piji,j be a factorization of c into the product of prime ideal powers of S where pi1 , . . . , pi,l(i) all lie over the same prime ideal pi of R, i = 1, . . . , k. Then Spij ⊆ Rpi , i = 1, . . . , k. If p is a nonzero prime ideal of R not in the set {p1 , . . . , pk }, then Rp = Sp . Hence, with Ri T = Rpi and Si = Spi , i = 1, . . . , k, Lemma 2.4.4 gives S ∩ R1 ∩ · · · ∩ Rk = Rp = R, where p runs over the nonzero prime ideals of R. Therefore, the map Pk s+R 7→ (s+R1 , . . . , s+Rk ) for s ∈ S, is an injective homomorphism into i=1 Si /Ri . The proof of the lemma is complete if we show that the map is surjective. / pi . Let s1 , . . . , sk be elements of S and let a1 , . . . , ak ∈ R such that ai ∈ Ql(i) ei,j The ideals qi = j=1 pij of S are pairwise relatively prime, i = 1, . . . , k. Hence, by the weak approximation theorem (Proposition 2.1.1), there exist s ∈ S, qi ∈ qi , and a0i ∈ R r pi such that s−

qi si = 0 ai ai

i = 1, . . . , k.

Q For each i choose bi ∈ (R r pi ) ∩ j6=i qj . Then ci = bi qi ∈ c ⊆ R and s = asii + bicai 0 . Thus, s + Ri = asii + Ri for i = 1, . . . , k. i

Proof of (b): Now let p1 , . . . , pk be the prime ideals of S that lie over p. For i = 1, . . . , k − 1 use the weak approximation theorem to find si ∈ S

5.3 The Genus of a Plane Curve

101

such that si ≡ 1 mod pi and si ≡ 0 mod pj , for j = 1, . . . , k and j 6= i. We need only show that s1 , . . . , sk−1 are K-linearly independent modulo Rp . Pk−1 Indeed, if r = i=1 ai si ∈ Rp with ai ∈ K (and si ∈ pk ), then, r belongs to Pk−1 pRp . Since aj ≡ i=1 ai si ≡ 0 mod pk , this gives aj = 0, j = 1, . . . , k − 1. Consequently, dimK Sp /Rp ≥ k − 1. Lemma 5.3.3: Let Γ be an affine plane curve defined over a field K with a generic point x = (x1 , x2 ) and let a ∈ Γ(K). Consider a separable algebraic 0 ) be the local ring of Γ at a over extension L of K and let Oa,K (resp. Oa,K K (resp. its integral closure). Then (a) Oa,L ∩ K(x) = Oa,K ; and 0 0 (b) dimK Oa,K /Oa,K ≤ dimL Oa,L /Oa,L . Proof: Without loss assume that L/K is finite. Let w1 , . . . , wn be a linear basis for L/K. Since Γ is defined by an absolutely irreducible polynomial, K(x)/K is a regular extension of K. Hence, K(x) and L are linearly disjoint over K. Therefore, w1 , . . . , wn is also a linear basis for L(x)/K(x). Consider u ∈ Oa,L ∩ K(x). Then P f (x), g(x) ∈ P u = f (x)/g(x) with = gi (x)wi , with L[x] and g(a) 6= 0. Write f (x) = fP i (x)wi and g(x) P ugi (x)wi = fi (x)wi deduce that fi (x), gi (x) ∈ K[x], i = 1, . . . , n. From ugi (x) = fi (x), i = 1, . . . , n. Since there is an i with gi (a) 6= 0, we have u ∈ Oa,K . This proves (a). 0 To prove (b), consider u1 , . . . , um ∈ Oa,K linearly dependent over L modulo Oa,L . We show that they are also linearly dependent over K modulo that Oa,K . Indeed, there existPb1 , . . . , bm ∈ L not all zero and v ∈ Oa,L Psuch P n m = j=1 aij wj with aij ∈ K and let vj = i=1 aij ui . bi ui = v. Write bi P Then vj ∈ K(x) and vj wj = v. For each σ of L(x) into P K(x)-embedding the algebraic closure of K(x) we have vj wjσ = v σ . Since L(x)/K(x) is ) 6= 0 [Lang7, p. 286, Cor. 5.4]. Apply Cramer’s rule to separable, det(wjσP solve the system vj wjσ = v σ , all σ, and write vj as a linear combination σ ˆ of L/K. Since a is of the v ’s with coefficients in the Galois closure L K-rational, each v σ is in Oa,Lˆ . Thus, vj ∈ Oa,Lˆ ∩ K(x). Hence, by (a), vj ∈ Oa,K . Therefore, the ui ’s are linearly dependent over K modulo Oa,K . Proposition 5.3.4: Let K be an algebraically closed field and Γ∗ a projective plane curve of degree d and genus g defined over K. Then (1)

g=

X 1 (d − 1)(d − 2) − dimK Op0 /Op , 2

where p runs over the points of Γ∗ (K), Op denotes the local ring of Γ∗ at p and Op0 is the integral closure of Op in the function field of Γ∗ over K. Proof: The five parts of the proof draw information from the form of Γ∗ after a change of variables that puts d distinct points at infinity.

102

Chapter 5. Plane Curves

Part A: A linear transformation. Choose a line L over K passing through no singular point of Γ∗ and is not tangent to Γ∗ . Then L cuts Γ∗ in d distinct simple points p1 , . . . , pd of Γ∗ [Seidenberg, p. 37]. By Lemma 5.2.3, they are normal. After a suitable linear homogeneous transformation [Seidenberg, Chapter 5] we may assume that L is the line at infinity and that the infinite points on the X-axis and the Y -axis, (0:1:0) and (0:0:1) do not belong to the set {p1 , . . . , pd }. Such a transformation does not change the genus or the degree of the curve. Moreover, corresponding points have the same local rings. View Γ∗ as the projective completion of an affine curve Γ defined by an equation f (X, Y ) = 0, as given by (1) of Section 5.1. Let (x, y) be a generic point of Γ over K, let R = K[x, y] be the coordinate ring of Γ, and let F be its function field. Part B: The divisor at ∞. By Part A, pi = (0:ai :1), with ai ∈ K × for i = 1, . . . , d, and a1 , . . . , ad distinct. In the notation of (2) of Section 5.1, obtain the factorization (2)

fd (X, Y ) = c

d Y

(X − ai Y ), with 0 6= c ∈ K.

i=1

Therefore, f (x, Y ) is an irreducible polynomial of degree d over K[x]. Thus, R is integral over K[x] and each element of R can be uniquely expressed as a polynomial h(x, y) with coefficients in K[x] such that degY (h) ≤ d − 1. Similarly R is integral over K[y]. Over the infinite prime divisor p∞ of K(x) there lie exactly d distinct prime divisors p1 , . . . , pd of F/K, with pi being the unique prime divisor with center pi . In particular, p∞ is unramified in F (Proposition 2.3.2) and we may therefore normalize vpi such that vpi(x) = −1. Since R is integral over K[x], this implies vpi (y) < 0, so vpi xy ≥ 0. by (2) of Section 5.1, Qd c i=1 xy − ai + fd−1 xy , 1 y −1 + · · · + f0 xy , 1 y −d = 0. Hence, xy has residue ai at pi , so vpi (y) = −1. Let o = p1 + · · · + pd . Then the pole divisors div∞ (x) and div∞ (y) of x and y in F are both o. Part C: The integral closure of R in F . Let S be the integral closure of R S∞ in F . By Lemma 3.1.1, S = n=1 Sn , where Sn = L(no). Let Rn = R ∩ Sn , n = 1, 2, . . . . By Lemma 5.3.1, dimK S/R < ∞. Hence, for n sufficiently large R + Sn = R + Sn+1 = R + Sn+2 = . . . . Therefore, S = R + Sn , so Sn /Rn ∼ = S/R. For n > 2g − 2, Lemma 3.2.2(d) implies that dimK Sn = nd − g + 1. It follows that (3) Part D: (4)

nd − g + 1 = dimK Rn + dimK S/R. For n > d, we have Rn = {h(x, y) ∈ R | deg(h) ≤ n and degY (h) ≤ d − 1}.

5.3 The Genus of a Plane Curve

103

Indeed, let k ≥ 0 be an integer and hk ∈ K[X, Y ] a homogeneous polynomial of degree k with degY (hk ) ≤ d − 1. If i + j = k, then div∞ (xi y j ) = ko (Part B). Hence, hk (x, y) ∈ Sk . / Sk−1 : Since K is algebraically closed, we may We prove that hk (x, y) ∈ factor hk (x, y) as hk (x, y) =

d−1 Y

(x − bj y) · axk−d+1

j=1

with bj ∈ K and a ∈ K × . Then vpi (hk (x, y)) =

d−1 X

vpi (x − bj y) − (k − d + 1).

j=1

Pd−1 If hk (x, y) ∈ Sk−1 , then for each i we have j=1 vpi (x − bj y) − (k − d + 1) + k − 1 ≥ 0. Hence, there exists 1 ≤ j ≤ d − 1 with vpi (x − bj y) ≥ 0. Therefore, bj = ai . But this implies that two of the ai ’s are equal, a contradiction. To complete Pm Part D, write each h(x, y) in K[x, y] with degY (h) ≤ d − 1 as h(x, y) = k=1 hk (x, y) with hk as above and hm (x, y) 6= 0. If m ≤ n, then hk (x, y) ∈ Sk ⊆ Sn , k = 0, . . . , m, and therefore h(x, y) ∈ Rn . If m > n, / Rn . then h(x, y) ∈ Sm r Sm−1 and therefore h(x, y) ∈ Sd−1 Part E: Computation of the genus. By Part D, the set j=0 {xi y j | i = 0, . . . , n − j} is a basis of Rn . Hence dimK Rn =

d−1 X

1 (n − j + 1) = nd + 1 − (d − 1)(d − 2). 2 j=0

Substitute this in (3) to conclude that 1 (d − 1)(d − 2) − dimK S/R. 2 L By Lemma 5.3.2(a), S/R = SP /RP , where P ranges over all nonzero prime ideals of R. Since K is algebraically closed, there is a bijective correspondence between the nonzero prime ideals P ofL R and the finite points p of Γ∗ and 0 we have RP = Op . Hence, S/R = p∈Γ∗ (K) Op /Op . Substituting this expression for S/R in (5), we get (1). (5)

g=

Corollary 5.3.5: Let Γ∗ be a projective plane curve of degree d defined over a perfect field K. Then genus(Γ∗ ) ≤ 12 (d − 1)(d − 2). Proof: The genus of Γ∗ does not change by going from K to its algebraic closure (Proposition 3.4.2). Now apply Lemma 5.3.3 and Proposition 5.3.4.

104

Chapter 5. Plane Curves

Corollary 5.3.6: Let Γ∗ be a projective plane curve of degree d defined over an algebraically closed field K. Then genus(Γ∗ ) = 12 (d − 1)(d − 2) if and only if Γ∗ is smooth; that is, all K-rational points of Γ∗ are simple. Proof: The condition “Γ∗ is smooth” is equivalent to Op = Op0 for all K rational points p of Γ∗ . Now apply Proposition 5.3.4.

5.4 Points on a Curve over a Finite Field Using the formula for the genus of an absolutely irreducible curve Γ over Fq (Proposition 5.3.4), we translate the estimate of the number of prime divisors of degree 1 of the function field of Γ to an estimate on |Γ(Fq )| which involves only q and the degree of Γ. Theorem 5.4.1: Let f ∈ Fq [X, Y ] be an absolutely irreducible polynomial of degree d. Denote the affine curve defined by the equation f (X, Y ) = 0 by Γ. Then √ √ q + 1 − (d − 1)(d − 2) q − d ≤ |Γ(Fq )| ≤ q + 1 + (d − 1)(d − 2) q. Proof: Put K = Fq . Denote the function field of Γ over K by F , and denote the number of prime divisors of degree 1 of F/K by N . By Theorem 4.5.2, √ √ (1) q + 1 − 2g q ≤ N ≤ q + 1 + 2g q , where g = genus(F/K). Let Γ∗ be the projective completion of Γ. For each p ∈ Γ∗ (K) let k(p) (resp. k1 (p)) be the number of prime divisors (resp. prime divisors of degree 1) of F/K with center at p. Then X X (2) N − |Γ∗ (K)| = (k1 (p) − 1) ≤ (k(p) − 1). p∈Γ∗ (K)

p∈Γ∗ (K)

Let (3)

δ=

X

0 dimK Op,K /Op,K and δ˜ =

p∈Γ∗ (K)

X

0 dimK˜ Op, ˜. ˜ /Op,K K

p∈Γ∗ (K)

0 /Op,K = 0 and Consider p ∈ Γ∗ (K). If p is K-normal, then dimK Op,K k1 (p) = k(p) = 1 (Lemma 5.2.2(a)). If p is not K-normal, then 0 dimK Op,K /Op,K ≥ 1. Hence, by (3), and by Lemma 5.3.3,

X

˜ (1 − k1 (p)) ≤ δ ≤ δ.

p∈Γ∗ (K)

We conclude from (2), (1), and Proposition 5.3.4 that X √ |Γ(K)| ≤ |Γ∗ (K)| = N + (1 − k1 (p)) ≤ q + 1 + 2g q + δ˜ p∈Γ∗ (K)

√ √ √ = q + 1 + (d − 1)(d − 2) q − 2δ˜ q + δ˜ ≤ q + 1 + (d − 1)(d − 2) q.

Exercises

105

Next we give a lower bound for |Γ(K)|. Note first that Γ∗ has at most d points at infinity [Seidenberg, p. 37]. Hence, |Γ(K)| ≥ |Γ∗ (K)|−d. Therefore, by (2), Lemma 5.3.2(b), (1), and Proposition 5.3.4 |Γ(K)| ≥ |Γ∗ (K)| − d ≥ N −

X

(k(p) − 1) − d

p∈Γ∗ (K)

≥N−

X

√ 0 dimK OK,p /OK,p − d ≥ q + 1 − 2g q − δ˜ − d

p∈Γ∗ (K)

√ √ = q + 1 − (d − 1)(d − 2) q + 2δ˜ q − δ˜ − d √ ≥ q + 1 − (d − 1)(d − 2) q − d. This completes the proof of the theorem.

Corollary 5.4.2: The curve Γ of Theorem 5.4.1 satisfies the following conditions: (a) For each m there exists q0 such that |Γ(Fq )| ≥ m for all q ≥ q0 . (b) If q > (d − 1)4 , then Γ(K) is not empty. √ Proof of (b): By Theorem 5.4.1, |Γ(Fq )| ≥ q + 1 − (d − 1)(d − 2) q − d = √ √ √ q q − (d − 1)(d − 2) − (d − 1) > q − (d − 1) > 0.

Exercises 1. Use Proposition 5.3.4 to prove that any projective plane curve of degree 2 is smooth. 2. Let Γ∗ be a projective plane curve of degree 3. Use Proposition 5.3.4 to show that either Γ∗ is smooth, in which case it has genus 1, or Γ∗ has exactly one singular point, in which case it has genus 0 and therefore its function field is rational (Example 3.2.4). 2 −2X 2 −X 3 = 3. Consider the affine plane curve Γ defined by the equation Y√ 0 over a field K of characteristic 6= 2 that does not contain 2. Note that (0, 0) is a singular point of Γ. Let (x, y) be a generic point of Γ over K. Show that the map (x, y) → (0, 0) does not extend to a K-rational (rather than ˜ K-rational) place of K(x, y) (i.e. the singular point of Γ is not the center of a K-rational place of the function field of Γ). Hint: Consider the element xy of K(x, y).

4. Prove directly that the function field of the curve Y 2 − 2X 2 − X 3 = 0 is rational. 5. Consider the projective plane curve Γ∗ defined for d ≥ 2 by X0 X1d−1 − X0 X2d−1 − X2d = 0 over an algebraically closed field K. (a) Use Lemma 5.2.3 to show that the only nonnormal point of Γ∗ is (1:0:0).

106

Chapter 5. Plane Curves

(b) Let (x, y) be a generic point of the affine part Γ of Γ∗ defined by X d−1 − Y d−1 − Y d = 0 and let z = xy . Use Lemma 2.4.4 to conclude that K[y, z] is the integral closure of K[x, y] in K(x, y). Conclude that the genus of Γ is 0. (y,z) | f ∈ K[Y, Z], g ∈ K[X, Y ], g(0, 0) 6= 0} is the (c) Prove that S0 = { fg(x,y) integral closure of the local ring R0 of Γ at (0, 0). (d) Use Proposition 5.3.4 to conclude that the set {xi z j | j = 1, . . . , d−2; i = 0, . . . , j − 1} gives a basis for S0 /R0 over K. 6. Count the number of points on the projective plane curve X03 +X13 +X23 = 0 over the finite field Fq , where q is a prime power such that gcd(q − 1, 3) = 1. Hint: The map x 7→ x3 from F× q into itself is bijective. (X) 7. Let f ∈ Fq [X] be a polynomial of degree d such that g(X, Y ) = f (YY)−f −X is absolutely irreducible. Suppose that either q is not a power of 2 or d ≥ 3. Use Theorem 5.4.1 to prove that if q > (d − 1)4 , then there exist distinct x, y ∈ Fq such that f (y) = f (x). Hint: Observe that g(X, X), the derivative of f (X), has at most d − 1 zeros.

Notes The proof of Proposition 5.3.4 is an elaboration on [Samuel, p. 52]. Denote the maximum number of Fq -points on a curve of genus g which is defined over Fq by Nq (g). For fixed q, put A(q) = lim sup g1 Nq (g). Weil’s g→∞ √ estimate Nq (g) ≤ q + 1 + 2g q (Theorem 4.5.2 and the proof of Theorem 1 5.4.1) implies that A(q) ≤ 2q 2 . [Serre6] improves Weil’s estimate (via interpretation of the Frobenius as an endomorphism on Jacobians) to give the √ bound Nq (g) ≤ q + 1 + g[2 q] (where [x] is the greatest integer not exceeding √ adut] obtains the much improved x). Thus, A(q) ≤ [2 q]. But [Drinfeld-Vlˆ √ estimate A(q) ≤ q − 1. When q is a square [Ihara] and [Tsfasman-VlˆadutZink] have shown this bound to be exact. As for a lower estimate, [Serre6] 8 ). The proves the existence of c > 0 such that A(q) ≥ c log(q) (e.g. A(2) ≥ 39 exact lower and upper bound for A(q) for general q have yet to be found. The case q = 2 has application to coding theory as first noted by [Goppa] (or p. 530 of [Lidl-Niederreiter] for a survey of work in this direction).

Chapter 6. The Chebotarev Density Theorem The major connection between the theory of finite fields and the arithmetic of number fields and function fields is the Chebotarev density theorem. Explicit decision procedures and transfer principles of Chapters 20 and 31 depend on the theorem or some analogs. In the function field case our proof, using the Riemann hypothesis for curves, is complete and elementary. In particular, we make no use of the theory of analytic functions. The number field case, however, uses an asymptotic formula for the number of ideals in an ideal class, and only simple properties of analytic functions. In particular, we do not use Artin’s reciprocity law (or any equivalent formulation of class field theory). This proof is close to Chebotarev’s original field crossing argument, which gave a proof of a piece of Artin’s reciprocity law for cyclotomic extensions.

6.1 Decomposition Groups Let R be an integrally closed domain with quotient field K. Consider a finite Galois extension L of K with Galois group G, and denote the integral closure of R in L by S. Suppose p is a prime ideal of R. By Chevalley’s theorem (Proposition 2.3.1), there exists a prime ideal P of S lying over p (i.e. p = R ∩ P). Denote the quotient fields of R/p and S/P, respectively, ¯ and by L. ¯ The set of all σ ∈ G satisfying σ(P) = P is a group DP by K called the decomposition group of P over K. Its fixed field in L is the decomposition field of P over K. For x ∈ S denote the equivalence class of x modulo P by x ¯. Each ¯ over K ¯ satisfying σ ¯ of L ¯x ¯ = σx σ ∈ DP induces a unique automorphism σ ¯ K). ¯ for each x ∈ S. The map σ 7→ σ ¯ is a homomorphism of DP into Aut(L/ Its kernel is the inertia group IP of P over K, IP = {σ ∈ G | σx ∈ x + P for every x ∈ S}. The fixed field of IP in L is the inertia field of P over K. If σ ∈ G, then σS = S and σP is another prime ideal of S that lies over p. In this case DσP = σ ·DP ·σ −1 and IσP = σ ·IP ·σ −1 . Conversely, any two prime ideals of S lying over the same prime ideal of R are conjugate over K [Lang7, p. 340]. In the proof of Lemma 6.1.1 we use the expression “to localize R and S at p”. This means that we replace R and Rp , p by pRp , S by Sp = { as | s ∈ S and a ∈ R r p}, and P by PSp . The local ring Rp is integrally ¯ = Rp /pRp . In addition, Sp is the closed, pRp is its maximal ideal, and K integral closure of Rp in L [Lang7, p. 338, Prop. 1.8 and 1.9], and PSp is a prime ideal that lies over pRp . Thus, Sp /PSp is a domain which is integral ¯ is a field [Lang7, p. 339, Prop. 1.11] ¯ Hence, Sp /PSp = L over the field K. and PSp is a maximal ideal. Moreover, DPSp = DP and IPSp = IP .

108

Chapter 6. The Chebotarev Density Theorem

Lemma 6.1.1: ¯ K ¯ is normal, and the map σ 7→ σ (a) The field extension L/ ¯ from DP into ¯ K) ¯ is surjective. Aut(L/ ¯ K ¯ is separable and [L ¯ : K] ¯ = [L : K]. Then L/ ¯ K ¯ is Galois, (b) Suppose L/ ¯ is an isomorphism of Gal(L/K) DP = Gal(L/K), and the map σ 7→ σ ¯ K). ¯ onto Gal(L/ Proof of (a): Denote the decomposition field of P by L0 , let S0 = S ∩ Lo , and let P0 = P ∩ L0 . We prove S0 /P0 = R/p. Suppose x ∈ S0 . We need only to find a ∈ R such that x ≡ a mod P0 . For each σ ∈ G r DP we have σ −1 P 6= P. If σ −1 P ∩ L0 = P0 , then there exists τ ∈ Gal(L/L0 ) = DP such that τ σ −1 P = P. Therefore, σ ∈ DP , a contradiction. Thus, σ −1 P ∩ L0 6= P0 . Localize at p, to assume that p is a maximal ideal of R. Then P0 and σ −1 P∩L0 are maximal ideals of S0 . Hence, P0 + σ −1 P0 ∩ L0 = S0 . By the Chinese remainder theorem [Lang7, p. 94], there exists y ∈ S0 with y ≡ x mod P0 and y ≡ 1 mod σ −1 P ∩ L0 for every σ ∈ G r DP . Thus, y ≡ x mod P and σy ≡ 1 mod P for every σ ∈ G r DP . Since L0 /K is a separable extension, the element a = normL0 /K (y) of R is a product of y and elements σy with σ running over nonidentity coset representatives of DP in G. Consequently a ≡ x mod P0 , as desired. ¯ = S0 /P0 and L ¯ = S/P. To continue localize S0 and S at P0 to assume K ¯ Write each element of L as x ¯, with x ∈ S, and let f = irr(x, L0 ). Then Qm f (X) = i=1 (X − xi ), where xi ∈ S, i = 1, . . . , m, are the conjugates of x over L0 . The coefficients of f (X) belong to L0 ∩ S =QS0 . Hence, m ¯ xi ) f¯(X) ∈ K[X]. In addition, x ¯ is a root of the polynomial f¯(X) = i=1 (X−¯ ¯ ¯ ¯ ¯ x) : with roots in L. Hence, L is a normal extension of K. Moreover, [K(¯ ¯ ≤ deg(f ) ≤ [L : L0 ]. K] Separable extensions have primitive generators [Lang7, p. 243]. Hence, ¯ which is separable over K ¯ satisfies [E : K] ¯ ≤ the maximal subfield E of L ¯ x) for some x ∈ S. In the above notation, if [L : L0 ]. Thus, E = K(¯ ¯ K) ¯ ∼ ¯ we have τ x τ ∈ Aut(L/ ¯=x ¯j for some j, 1 ≤ j ≤ m. The = Gal(E/K), ¯ = τ. map x 7→ xj extends to a field automorphism σ ∈ Gal(L/L0 ) with σ ¯ K) ¯ is surjective. Consequently, The map σ 7→ σ ¯ from DP into Aut(L/ ¯ : K] ¯ = |Gal(L/ ¯ K)| ¯ ≤ |DP | ≤ [L : Proof of (b): By assumption, [L : K] = [L ¯ K)|. ¯ K]. Hence, |DP | = |Gal(L/K)| = |Gal(L/ Therefore, DP = Gal(L/K) ¯ K) ¯ is an isomorphism. and the map Gal(L/K) → Gal(L/ The discriminant of f ∈ R[X], Qn disc(f ), gives information about ramification. Assume f is monic and i=1 (X − xi ) is the factorization of f into linear factors. Then (1)

disc(f ) = (−1)

n(n−1) 2

n Y n(n−1) Y 2 (xi − xj ) = (−1) f 0 (xj ). i6=j

j=1

This is an element of R, and disc(f ) 6= 0 if and only if the xi ’s are distinct.

6.1 Decomposition Groups

109

Assume f is irreducible. Then (1) implies disc(f ) = (−1)

n(n−1) 2

normK(x1 )/K (f 0 (x1 )).

We call normK(x1 )/K (f 0 (x1 )) the discriminant of x1 over K. Lemma 6.1.2: Let R be an integrally closed domain with quotient field K. Let S be the integral closure of R in a finite separable extension L of K. Assume L = K(z) with z integral over R and let f = irr(z, K). Suppose d = disc(f ) is a unit of R. Then S = R[z]. Proof: Let n = [L : K]. Assume y = a0 + a1 z + · · · + an−1 z n−1 , with ai ∈ K, is element of S. We must prove ai ∈ R, i = 0, . . . , n − 1. To this ˜ end, let 1 , . . . , σn be the isomorphisms of L over K into K. Then each of Pσn−1 σi y = j=0 aj σi z j , i = 1, . . . , n, is integral over R. b

∆b

Solve for the aj ’s by Cramer’s rule. This gives aj = ∆j = ∆2j , j = 0, . . . , n − 1, where ∆ = det(σi z j ) with bj integral over R, j = 0, . . . , n − 1. But ∆ is a Vandermonde determinant: Y ∆2 = ± (σi z − σj z) = ±normL/K f 0 (z) = ±d. i6=j

Since d is a unit of R, aj is integral over R, j = 0, . . . , n − 1. Since R is integrally closed, all aj are in R. Definition 6.1.3: Ring covers. As in the preceding lemmas, consider two integrally closed integral domains R ⊆ S with K ⊆ L their respective quotient fields such that L/K is finite and separable. Suppose S = R[z], where z is integral over R and the discriminant of z over K is a unit of R. In this set up we say S/R is a ring cover and L/K is the corresponding field cover. In this case, Lemma 6.1.2 implies that S is the integral closure of R in L. Call the element z a primitive element for the cover. If in addition L/K is Galois, call S/R is a Galois ring cover. We summarize the previous results for ring covers: Lemma 6.1.4: Let S/R be a Galois ring cover with L/K the corresponding field cover. Then for every prime ideal p of R and for every prime ideal P ¯ of S/P is a of S lying over p the following holds. The quotient field, L, ¯ Galois extension of the quotient field, K, of R/p. The map σ 7→ σ ¯ of DP ¯ K) ¯ given by σ into Gal(L/ ¯x ¯ = σx for x ∈ S is an isomorphism. Remark 6.1.5: Creating ring covers. Let K = K0 (x1 , . . . , xn ) be a finitely generated extension of a field K0 . The subring R = K0 [x1 , . . . , xn ] of K is not necessarily integrally closed. But, there exists a nonzero xn+1 ∈ K with R0 = K0 [x1 , . . . , xn+1 ] integrally closed ([Lang4, p. 120]; a constructive proof of this fact appears in Section 19.7). Suppose z is a primitive generator for L/K, f ∈ R[Z] is irreducible polynomial over K, and f (z) = 0. Multiply xn+1 by the inverse of the product of the leading coefficient and the discriminant of f . Then S 0 = R0 [z] is a ring cover of R0 with z a primitive element.

110

Chapter 6. The Chebotarev Density Theorem

Remark 6.1.6: Decomposition groups of places. Suppose L/K is a finite ¯ = ϕ(O) Galois extension and ϕ is a place of L with a valuation ring O. Then L is the residue field of L under ϕ. Also, R = O ∩ K is the valuation ring of the ¯ = ϕ(R) is its residue field. By Proposition 2.4.1, restriction of ϕ to K and K O contains the integral closure S of R in L. Let m be the maximal ideal of O, P = S ∩ m, and p = P ∩ R. Then O = SP [Lang4, p. 18, Thm. 4] and ¯ = S/P and R/p ∼ ¯ P is maximal [Lang7, p. 339, Prop 1.11]. Hence, L = K. We call Dϕ = DP and Iϕ = IP the decomposition group and inertia group, respectively, of ϕ over K. The fixed fields of Dϕ and Iϕ in L are the decomposition field and the inertia field, respectively, of ϕ over K. By ¯ K ¯ is a normal extension and the map σ 7→ σ lemma 6.1.1, L/ ¯ from DP into ¯ K) ¯ is surjective. Aut(L/ ¯ K ¯ is separable and Iϕ = 1 (This holds if S/R is a ringSuppose now L/ ¯ K) ¯ ¯ K ¯ is Galois and Dϕ is isomorphic to Gal(L/ cover.) By Lemma 6.1.1, L/ under the map σ 7→ σ ¯ . Denote the decomposition field of ϕ over K by L0 . ¯ For each x ∈ O ∩ L0 and each σ ∈ Dϕ we have σ ¯x ¯ = σx = x ¯. Hence, x ¯ ∈ K. ¯ Therefore, ϕ(L0 ) = K ∪ {∞}. Let P0 = P ∩ L0 . Then each prime ideal of S that lies over P0 is conjugate to P by an element of DP . Hence, P is the only prime ideal of ¯ :L ¯ 0 ] ≤ [L : L0 ]. Since S lying over P0 . By Proposition 2.3.2, e(P/P0 )[L ¯ ¯ ¯ ¯ [L : L0 ] = [L : K] = [L : L0 ], we have e(P/P0 ) = 1. In particular, if O is discrete, p is unramified in L. Remark 6.1.7: Ring covers under change of base ring. Consider an integrally closed domain R with quotient field K. Let L be a finite separable extension of K, S the integral closure of R in L, z anQelement of S with L = K(z), and f = irr(z, K). Then normL/K (f 0 (z)) = f 0 (z)σ , where σ ranges over all K-embeddings of L into Ks . Each f 0 (z)σ is integral over R. Hence, normL/K (f 0 (z)) is a unit of R if and only if f 0 (z) is a unit of S. Suppose f 0 (z) is a unit of S. By Lemma 6.1.2, S = R[z] and z is a primitive element of the ring-cover S/R. Let ϕ be a homomorphism of R ¯ Put K ¯ = Quot(R). ¯ Then ϕ extends to into an integrally closed domain R. ¯ (Proposition 2.4.1). a homomorphism ψ of S into the algebraic closure of K ¯ z ], L ¯ = Quot(S), ¯ f¯ = ϕ(f ), and g = irr(¯ ¯ Then Put z¯ = ψ(z), S¯ = R[¯ z , K). ¯ z ], and there is a monic polynomial h ∈ K[X] ¯ z ) is a unit of R[¯ f¯(¯ z ) = 0, f¯0 (¯ with f¯(X) = g(X)h(X). The coefficients of h are polynomials in the roots ¯ we have h ∈ R[X]. ¯ of f¯. Since the latter are integral over R, Deduce from 0 0 0 ¯ ¯ ¯ R ¯ z )h(¯ z ) = f (¯ z ) that g (¯ z ) is a unit of R[¯ z ]. Hence, by Lemma 6.1.2, S/ g (¯ ¯ is a ring-cover with z¯ as a primitive element. In particular, S is the integral ¯ in L. ¯ Thus, if R ¯ = K, ¯ then S¯ = L. ¯ closure of R As an example, let L/K be a finite separable extension, z a primitive element for L/K and f = irr(z, K). Then f 0 (z) 6= 0. Hence, L = K[z] is a ring-cover of K. Therefore, R[z]/R is a ring-cover whenever R is an integrally closed ring containing K. For another example suppose in the notation of the first two paragraphs

6.2 The Artin Symbol over Global Fields

111

¯=K ¯ that R is a valuation ring and Ker(ϕ) is the maximal ideal of R. Then R ¯ z ] is an integral extension of K. ¯ Hence, S¯ = L ¯ is a field. is a field and S¯ = K[¯ In addition, the local ring of S at Ker(ψ) is a valuation ring lying over R [Lang4, p. 18, Thm. 4.7]. Hence, ϕ and ψ extend uniquely to places of K ¯ and L, ¯ respectively. and L with residue fields K The next result is another consequence of Lemma 6.1.2 which may be applied to covers. Lemma 6.1.8: (a) Let R be an integral domain with quotient field K, L a finite Galois extension of K, and S the integral closure of R in L. Consider a monic polynomial f ∈ R[X] having all of its roots in L, a prime ideal p of R, and a prime ideal P of S lying over p. Assume disc(f ) ∈ / p and denote reduction modulo P by a bar. Also let σ ∈ DP and F a field containing ¯ such that L ¯ ∩ F = L(¯ ¯ σ ). Then the number of the roots of f in L(σ) K is equal to the number of the roots f¯ in F . (b) Suppose L = K(z) with z integral over R, f = irr(z, K), and disc(f ) ∈ / p. ¯ K ¯ is separable, IP = 1, and p is unramified in L. Then L/ Proof of (a): The roots of f¯ are distinct, because disc(f¯) 6= 0. In addition deg(f ) = deg(f¯). Thus, x 7→ x ¯ maps the roots of f bijectively onto the roots of f¯. For x a root of f , σx = x if and only if σ ¯x ¯ = x ¯. Moreover, since ¯ we have x ¯ σ ) if and only if x all of the roots of f¯ belong to L, ¯ ∈ L(¯ ¯ ∈ F. Consequently, the number of the roots of f in L(σ) is equal to the number of the roots of f¯ in in F . Proof of (b): Replace R by Rp , if necessary, to assume R is a local ring and p is its maximal ideal. By Lemma 6.1.2, S = R[z] is the integral closure of ¯ = K(¯ ¯ z ). From (a), L ¯ is a separable R in L. Under the assumptions of (b), L ¯ ¯ z¯ = z¯. Hence, σz = z, extension of K. Also, if σ ¯ = 1 for some σ ∈ DP , then σ so σ = 1. Hence, IP = 1. By (4) of Section 2.3, p is unramified in L.

6.2 The Artin Symbol over Global Fields The Artin symbol over number fields is a generalization of the Legendre symbol for quadratic residues. Here we define the Artin symbol over global fields and state some of its basic properties. Let R be a Dedekind domain with quotient field K. Consider a finite separable extension L of K. Let S be the integral closure of R in L. Take z ∈ S with L = K(z). If f = irr(z, K), then d = disc(f ) ∈ R and d 6= 0. Consider R1 = R[d−1 ] and S1 = S[d−1 ]. Then d is a unit in R1 and S1 = R1 [z] (Lemma 6.1.2). Thus, adjoining d−1 gives a ring cover S1 /R1 for L/K. Maximal ideals P for which PS1 = S1 are exactly those containing d. For z ]. all others S/P ∼ = S1 /PS1 and S/P = (R/p)[¯ If in addition L/K is a Galois extension, then extending P to S1 leaves the decomposition group and the inertia group unchanged. Hence, if P does

112

Chapter 6. The Chebotarev Density Theorem

not contain d, then PS1 , and therefore also P, is unramified over K (Lemma 6.1.8). Thus, a prime ideal p of R not containing d, does not ramify in L. Since only finitely many prime ideals of R contain d, only finitely many prime ideals ramify in L. Denote the greatest common divisor of all the principal ideals f 0 (z)S with z ∈ S by Diff(S/R) and call it the different of S over R. Then (Diff(S/R))−1 = {x ∈ L | traceL/K (xS) ⊆ R}. A prime ideal P of S ramifies over K if and only if it divides Diff(S/R) [Lang5, p. 62]. Hence, the prime divisors of the discriminant DS/R = normL/K Diff(S/R) of S over R (an ideal of R), are exactly those primes that ramify in L. Call K a global field if K is either a finite extension of Q (K is a number field) or K is a function field of one variable over a finite field. In the number field case denote the integral closure of Z in K by OK . In the function field case K is a finite separable extension of Fp (t), where p = char(K) and t is transcendental over Fp . Denote the integral closure of Fp [t] in K by OK . With the understanding it depends on t, call OK the ring of integers of K. The local ring of OK at a prime ideal p is a valuation ring. Denote its ¯ p is a finite field. We call N p = |K ¯ p| ¯ p . Note that K residue class field by K the absolute norm of p. Let L be a finite Galois extension of K. Suppose p is unramified in L. If P is a prime ideal of OL over p, then reduction modulo P gives a ¯ P /Kp) ¯ canonical isomorphism of the decomposition group DP and Gal(L (Lemma 6.1.4). The latter group is cyclic. It contains a canonical generator ¯ P by this rule: Frob, the Frobenius automorphism. It acts on L (1)

Frob(x) = xN p

¯P. for x ∈ L

Call the element of DP that corresponds to Frob the Frobenius automor phism at P and denote it by L/K P . It is uniquely determined in Gal(L/K) by the condition L/K x ≡ xN p mod P for all x ∈ OL . (2) P Let Fqn be the algebraic closure of Fq in L and let N p = q k . By (2), the restriction of L/K to Fqn is Frobkq . P If K ⊆ K 0 ⊆ L and K 0 /K is a Galois extension, this immediately implies 0 K /K L/K (3) resK 0 = . P P ∩ K0 −1 If σ ∈ Gal(L/K), then L/K . Therefore, as P ranges over = σ L/K σP P σ prime ideals of OL lying over p, the Frobenius automorphism ranges over

6.3 Dirichlet Density

113

some conjugacy class in Gal(L/K) that depends on p. This conjugacy class . It is tacit in this symbol that p is unramified is the Artin symbol, L/K p in L. If L/K is Abelian, then L/K is one element, L/K p P . In this case write L/K for L/K p P . In defining the Frobenius automorphism and the Artin symbol, replacing OK by the local ring OK,p does not change these objects. If K is a function field over Fq , the local rings OK,p bijectively correspond to prime divisors p0 of K/Fq finite at t. We use L/K as a substitute for L/K . Since each p0 p prime divisor p0 of K/Fq is either finite at t or at t−1 , the symbol L/K is p0 0 well defined if p is unramified in L. Example 6.2.1: Quadratic extensions of Q. Let a be a nonsquare integer √ and p an odd prime number not dividing a. Put L = Q( a). Then p is unramified in L (Example 2.3.8). Let p be a prime divisor of L lying over p. By elementary number theory [LeVeque, p. 46],

√ p p−1 √ L/Q √ a √ 2 a ≡ ( a) = a a≡ a mod p. p p L/Q p

Thus, the Frobenius symbol

acts

√

a as the Legendre symbol

a p

.

6.3 Dirichlet Density For K a global field denote the set of all prime ideals of OK by P (K). If A is a subset of P (K), then the Dirichlet density, δ(A), of A is the limit P δ(A) = lim+ P s→1

p∈A (N p)

−s

p∈P (K) (N p)

−s

,

if it exists. The Dirichlet density is a quantitative measure on subsets of P (K). We apply it to test if specific subsets are infinite. Clearly δ(A) is a real number between 0 and 1. For example, δ(P (K)) = 1. If K is a number field, then X (N p)−s = ∞ lim s→1+

p∈P (K)

[Lang5, p. 162]. Relation (19) of Section 6.4 implies that P the same holds if K is a function field. Hence, in both cases, δ(A) = 0 if p∈A (N p)−1 is finite. In particular, δ(A) = 0 if A is finite. If A and B are disjoint subsets of P (K) having a density, then δ(A ∪ B) = δ(A) + δ(B). Here is the main result:

114

Chapter 6. The Chebotarev Density Theorem

Theorem 6.3.1 (Chebotarev Density Theorem): Let L/K be a finite Galois extension of global fields and let C be a conjugacy class in Gal(L/K). Then |C| . = C exists and is equal to [L:K] the Dirichlet density of p ∈ P (K) | L/K p Section 6.4 proves Theorem 6.3.1 for function fields and Section 6.5 proves the theorem for number fields. A non obvious special case is Dirichlet’s theorem showing the arithmetic progression {a, a + n, a + 2n, . . .} has infinitely many primes when gcd(a, n) = 1. Corollary 6.3.2 (Dirichlet): Suppose a and n are relatively prime positive 1 , integers. Then the Dirichlet density of {p ∈ P (Q) | p ≡ a mod n} is ϕ(n) where ϕ(n) is the Euler totient function. Proof: Denote a primitive nth root of unity by ζn and let L = Q(ζn ). Then Gal(L/K) is isomorphic to (Z/nZ)× . If σ ∈ Gal(L/K) and σζn = ζna , then this isomorphism maps σ to a mod n. Also, for a, b relatively prime to n, we have ζna ≡ ζnb mod p if and only if ζna = ζnb . Thus, for p - n, p ≡ a mod n if a and only if L/Q p (ζn ) = ζn . Now apply Theorem 6.3.1. Example 6.3.3:

Let f ∈ Z[X] be a monic polynomial. Write f (X) =

r Y

fi (X)

i=1

with f1 (X), . . . , fr (X) monic and irreducible and let d be the product of the discriminants of f1 , . . . , fr ((1) of Section 6.1). Consider the following hypotheses: (1) f (X) ≡ 0 mod p has a solution for all but finitely many primes p. (2) f (X) ≡ 0 mod p has a solution for all primes p - d. Let L be the splitting field of f over Q. According to Theorem 6.3.1, each for infinitely many prime ideals p of element of Gal(L/Q) has the form L/Q p OL . Therefore, Lemma 6.1.8(a) implies each of (1) and (2) is equivalent to the following. (3) Each σ ∈ Gal(L/Q) fixes a root of f (X). In particular, (1) and (2) are equivalent. Example 6.3.4: Let K be a number field and B the set of all P prime ideals 1 of OK whose absolute norm is not a prime number. Then p∈B N p ≤ P [K:Q] p p2 < ∞. Hence, δ(B) = 0. Suppose L is a finite Galois extension of K and C a conjugacy class in Gal(L/K). Then, in view of the preceding paragraph, the Chebotarev density = C and theorem gives infinitely many prime ideals p of OK such that L/K p N p is a prime number.

6.4 Function Fields

115

6.4 Function Fields This section contains the proof of the Chebotarev density theorem in the function field case. Apart from elementary algebraic manipulations it depends only on the Riemann hypothesis for curves. To fix notation, let q be a power of a prime number. Consider a function field K over Fq , a finite Galois extension L of K, and a conjugacy class C of Gal(L/K) with c elements. Let Fqn be the algebraic closure of Fq in L and fix a separating transcendence element t for K/Fq . Denote the Frobenius element of Fqn /Fq by Frobq . As in Section 6.2, let OK be the integral closure of Fq [t] in L and let P (K) be the set of all nonzero prime ideals of OK . Denote the set of prime divisors of K/Fq by P(K). Identify P (K) with a cofinite subset of P(K). Thus, C = {p ∈ P(K) | L/K = C} and {p ∈ P (K) | L/K = C} differ by p p c finitely many elements. It suffices therefore to prove that δ(C) = [L:K] . n In addition to n = [Fq : Fq ], two more degrees enter the proof: d = [K : Fq (t)] and m = [L : KFqn ] as in the following diagram. K

n

K · Fq n

n

Fqn (t)

m

L

d

Fq (t)

Fix the following notation: P0 (K) = {p ∈ P(K) | p is unramified over Fq (t) and in L} Pk (K) = {p ∈ P0 (K) | deg(p) = k} P0k (K) = {p ∈ P0 (K) | deg(p) = k} Ck (L/K, C) = {p ∈ P0k (K) | L/K = C} p Dk (L/K, τ ) = {P ∈ P(L) | P ∩ K ∈ Pk (K) and L/K = τ }, P for τ ∈ Gal(L/K) S∞ C 0 = k=1 Ck (L/K, C) gK = the genus of K = the Frobenius automorphism of Gal(Fq ) and also of Frobq Gal(Fqn /Fk ) for each k. The sets C 0 and C differ by only finitely many elements. Hence, they have the same Dirichlet density. To compute this density, we compute the cardinality of each finite set Ck (L/K, C). This is also of independent interest, especially when k = 1. Lemma 6.4.1: Let k be a positive integer, p ∈ Ck (L/K, C), and τ ∈ C. (a) There are exactly [L : K]/ord(τ ) primes of P(L) over p. (b) If Ck is a subset of Ck (L/K, C) and Dk (τ ) = {P ∈ Dk (L/K, τ ) | P ∩ K ∈ Ck },

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Chapter 6. The Chebotarev Density Theorem

then |Ck | = |C| · ord(τ ) · |Dk (τ )| · [L : K]−1 . Proof of (a): Suppose P ∈ P(L) lies over p. Then, by (2) of Section 2.3, [L : K] = eP/p fP/p gP/p where fP/p is the order of the decomposition group h L/K P i. In our case eP/p = 1 and τ is conjugate to L/K P . Thus, fP/p = ord(τ ) and (a) holds. 0 Proof of (b): For each σ ∈ Gal(L/K), Dk (στ σ −1 ) = S σDk (τ ). If τ ∈ C 0 and τ 6= τ , then Dk (τ ) and Dk (τ ) are disjoint. Thus, · τ ∈C Dk (τ ) is the set of primes of P(L) lying over the elements of Ck . By (a),

|Ck | · [L : K] X = |Dk (τ )| = |C| · |Dk (τ )|, ord(τ ) τ ∈C

and the formula follows.

Lemma 6.4.2: Let K ⊆ K 0 ⊆ L and τ ∈ Gal(L/K 0 ). Denote the algebraic closure of Fq in K 0 by Fqr . Suppose r|k. Then Dk (L/K, τ ) = Dk/r (L/K 0 , τ ) ∩ {P ∈ P(L) | deg(P ∩ K) = k}. Proof: Let P ∈ P(L). Suppose p = P∩K ∈ Pk (K) and p0 = P∩K 0 ∈ P(K 0 ). ¯ 0 0 = q rl , where l = deg(p0 ) = [K ¯ 0 0 : K 0 ]. ¯ p = q k and N p0 = K Then, N p = K p p By (2) of Section 6.2, (1)

k L/K = τ ⇐⇒ τ x ≡ xq mod P for every x ∈ OL ; and P

(2)

rl L/K 0 = τ ⇐⇒ τ x ≡ xq mod P for every x ∈ OL . P

Thus, it suffices to show

L/K P

= τ implies rl = k. k

Since τ ∈ Gal(L/K 0 ), (1) implies x ≡ xq mod P for every x ∈ OK 0 . ¯ p = Fqk . Therefore, Hence, K 0 p0 ⊆ Fqk . On the other hand, K 0 p0 ⊇ K 0 Fqrl = K p0 = Fqk . Consequently, rl = k. Corollary 6.4.3: With the hypotheses of Lemma 6.4.2, let C and C 0 be the respective conjugacy classes of τ in Gal(L/K) and in Gal(L/K 0 ) and 0 = Ck/r (L/K 0 , C 0 ) r{p0 ∈ P (K 0 ) | deg(p0 ∩ K) ≤ k2 }. Then Ck/r |Ck (L/K, C)| =

0 | |C||Ck/r

|C 0 |[K 0 : K]

.

6.4 Function Fields

117

Proof: Let l = kr . Then Dk0 (τ ) = Dl (L/K 0 , τ ) ∩ {P ∈ P(L) | deg(P ∩ K) = k} is the set of primes in Dl (L/K 0 , τ ) lying over Ck00 = Cl (L/K 0 , C 0 ) ∩ {p0 ∈ P(K 0 ) | deg(p0 ∩ K) = k}. Dk0 (τ ) is also the set of primes in Dl (L/K, τ ) over Ck (L/K, C). By Lemma 6.4.2, Dk0 (τ ) = Dk (L/K, τ ). Applying Lemma 6.4.1 twice gives a chain of equalities: [L : K] [L : K 0 ] |Ck (L/K, C)| = |Dk (L/K, τ )| = |Dk0 (τ )| = 0 |C 00 |. |C| · ord(τ ) |C | · ord(τ ) k Thus, it suffices to show that Ck00 = Cl0 . Indeed, if p0 ∈ P(K 0 ) is of degree l ¯ p ⊆ K 0 p0 = Fqrl = Fqk . Hence, deg(p)|k. Thus, and p = p0 ∩ K, then Fq ⊆ K either deg(p) = k or deg(p) ≤ k2 . Therefore, Ck00 = Cl0 . Lemma 6.4.4: Let k be a positive integer such that (3)

resFqn (τ ) = resFqn (Frobkq )

for every τ ∈ C. Let n0 be a multiple of n and L0 = LFqn0 . Then L0 /K is a Galois extension, Fqn0 is the algebraic closure of Fq in L0 and gL = gL0 (Proposition 3.4.2). Moreover, for each τ ∈ C there exists a unique τ 0 ∈ Gal(L0 /K) with resL τ 0 = τ and resFqn0 (τ 0 ) = resFqn0 (Frobkq ). Furthermore: (a) ord(τ 0 ) = lcm(ord(τ ), [Fqn0 : Fqn0 ∩ Fqk ]); (b) C 0 = {τ 0 | τ ∈ C} is a conjugacy class in Gal(L0 /K); and (c) Ck (L0 /K, C) = Ck (L/K, C 0 ). Proof: Given τ ∈ C, the existence of τ 0 follows from (3), because KFqn0 ∩L = KFqn . Uniqueness follows from L0 = Fqn0 L. To prove (a), note that ord(τ 0 ) = lcm ord(resL τ 0 ), ord(resFqn0 τ 0 ) = lcm ord(τ ), [Fqn0 : Fqn0 ∩ Fqk ] . Since Fqn0 /Fq is an Abelian extension, assertion (b) follows from the uniqueness of τ 0 . To prove (c), we show that P ∈ P(L0 ) and p = P ∩ K ∈ Pk (K) imply L/K L0 /K L/K k = τ 0 . Indeed, if P∩L = τ , then τ x ≡ xq mod P ∩ L P∩L = τ ⇐⇒ P k

for each x ∈ OL . By definition, Frobkq x = xq for each x ∈ Fqn0 . Since k

OL0 = Fqn OL (Proposition 3.4.2(d)), τ 0 x ≡ xq mod P for each x ∈ OL0 . 0 Hence, L P/K = τ 0 . The converse is a special case of (3) of Section 6.2.

118

Chapter 6. The Chebotarev Density Theorem

Corollary 6.4.5: If L = Fqn K and τ ∈ Gal(L/K) satisfies (3), then Ck (K/K, 1) = Ck (L/K, {τ }). Now we give estimates for the key sets. Lemma 6.4.6: Suppose L = KFqn , C = {τ }, and τ |Fqn = Frobq . Then √ (4) |#C1 (L/Fq , C) − q| < 2(gL q + gL + d). Proof: First note that C1 (L/Fq , C) = P01 (K) and each p ∈ P(Fq ) is unramified in L. Thus, P1 (Fq ) r C1 (L/K, C) consists exactly of all prime divisors of Diff(K/Fq (t)). By Riemann-Hurwitz (Theorem 3.6.1), deg(Diff(K/Fq (t))) = 2(gK + d − 1). By Theorem 4.5.2, √ |#P1 (K) − (q + 1)| ≤ 2gK q. √ Hence, |#C1 (L/K, C) − q| ≤ 2gK q + 1 + 2(gK + d − 1). Since gK = gL , this proves (4). Denote the situation when a divides b and a < b by a pd b. Lemma 6.4.7: Let K 0 be a degree km extension of K containing Fqk . Then (5)

#{q ∈ P(K 0 ) | deg(q ∩ K) pd k} ≤ 2m(q k/2 + (2gK + 1)q k/4 ).

Proof: If j|k and p ∈ Pj (K), then Fqj ⊆ Fqk . By Lemma 4.3.1, p decomposes in KFqj into j prime divisors of degree 1. Each has exactly one extension to KFqk . The latter decomposes in K 0 into at most m prime divisors. Hence, X {q ∈ P (K 0 ) | deg(q ∩ K) = j} #{q ∈ P (K 0 ) | deg(q ∩ K) pd k} = j|k j≤k/2

(6)

≤m

X

{q ∈ P (KFqk ) | deg(q ∩ K) = j}

j|k j≤k/2

=m

X

{q ∈ P (KFqj ) | deg(q ∩ K) = j}

j|k j≤k/2

≤m

X

|P1 (KFqj )|

j|k j≤k/2

By Theorem 4.5.2, |P1 (KFqj )| ≤ q j +2gK q j/2 +1. Now verify the inequalities P j k/2 and k2 ≤ 2q k/4 and use them to deduce (5) from (6). j≤k/2 q ≤ 2q We now come to the key result of this Section, Proposition 6.4.8. It is the main result from which the Chebotarev density theorem follows. It is also the main ingredient in an arithmetic proof of Hilbert irreducibility theorem (Lemma 13.3.3). A variant of it (Lemma 31.2.1) is crucial to the Galois stratification procedure for the elementary theory of finite fields. Recall: Fqn is the algebraic closure of Fq in L.

6.4 Function Fields

119

Proposition 6.4.8: Let a be a positive integer with resFqn τ = resFqn Frobaq for each τ ∈ C. Let k be a positive integer. If k 6≡ a mod n, then Ck (L/K, C) is empty. If k ≡ a mod n, then c k q #Ck (L/K, C) − km 2c (m + gL )q k/2 + m(2gK + 1)q k/4 + gL + dm . < km Proof: Suppose P ∈ P(L) lies over p ∈ Ck (L/K, C). Then, L/K ∈ C, so P

(7)

resFqn (Frobaq )

L/K = resFqn (Frobkq ). P

= resFqn

Hence, k ≡ a mod n. Conversely, suppose k ≡ a mod n. Let τ ∈ C and n0 = nk · ord(τ ). Extend L to L0 = LFqn0 . Then [L0 : KFqn0 ] = [L : KFqn ] = m. Since k ≡ a mod n, there exists τ 0 ∈ Gal(L0 /K) with τ 0 |L = τ and τ 0 |Fqn0 = Frobkq . Thus, ord(τ 0 ) = lcm(ord(τ ), ord(Frobkq )) = lcm(ord(τ ), [Fqn0 : Fqk ]) = lcm(ord(τ ), n · ord(τ )) = n · ord(τ ). Denote the conjugacy class of τ 0 in Gal(L0 /K) by C 0 . By Lemma 6.4.4(c), Ck (L0 /K, C 0 ) = Ck (L/K, C). ˜ q = Fqk Denote the fixed field of τ 0 in L0 by K 0 . Then K 0 ∩ Fqn0 = K 0 ∩ F 0 0 and K Fqn0 = L . K0 m

K d

KFqk

ord(τ 0 )

L0 m

KFqn0

d

Fq (t)

Fqk (t)

Fqn0 (t)

Fq

Fq k

Fqn0

120

Chapter 6. The Chebotarev Density Theorem

Thus, [K 0 : KFqk ] = [L0 : KFqn0 ] = [L : KFqn ] = m. Hence, [K 0 : Fqk (t)] = dm. Applying Lemma 6.4.3 with L0 , K, C 0 , {τ 0 }, k replacing F , E, C, C 0 , r we conclude that c |C1 (L0 /K 0 , {τ 0 }) r{q ∈ P(K 0 ) | deg(q ∩ K) pd k}|. |Ck (L0 /K, C 0 )| = [K 0 : K] Since [K 0 : K] = km, Lemma 6.4.7 implies #Ck (L0 /K, C 0 )− c #C1 (L0 /K 0 , {τ 0 }) (8) km c #{q ∈ P(K 0 ) | deg(q ∩ K) pd k} ≤ km c · 2m(q k/2 + (2gK + 1)q k/4 ) ≤ km By Lemma 6.4.6, now with K 0 , L0 , n0 , τ 0 , q k replacing K, L, n, τ , q, |#C1 (L0 /K 0 , {τ 0 }) − q k | < 2(gL0 q k/2 + gL0 + dm).

(9)

Now we combine (8) and (9) using the equalities gL = gL0 and Ck (L/K, C) = Ck (L0 /K, C 0 ) to prove (7): c k c k q | = |#Ck (L0 /K, C 0 ) − q | |#Ck (L/K, C) − km km c c #C1 (L0 /K 0 , {τ 0 }) + #C1 (L0 /K 0 , {τ 0 }) − q k ≤ #Ck (L0 /K, C 0 ) − km km 2c c k/2 k/4 k/2 · 2m q gL q + gL + dm + (2gK + 1) + ≤ km km 2c k/2 k/4 (m + gL )q + (2gK + 1)q + gL + dm . = km We deduce the function field case of Theorem 6.3.1 by summing over k in the conclusion of Proposition 6.4.8. We use the big O notation, i.e. for real valued functions f (x) and g(x) write f (x) = O(g(x)),

x→a

to mean there exists a constant c with |f (x)| ≤ c|g(x)| for all x close to a. In particular, if g(x) = 1, then f (x) is bounded near a. Lemma 6.4.9: Let a and n be positive integers. Then ∞ X xa+jn 1 = − log(1 − x) + O(1), a + jn n j=0

x → 1− .

Proof: If ζ 6= 1 is an nth root of unity, then 1 + ζ + · · · + ζ n−1 = 0. Hence −

n−1 ∞ n−1 1X 1 X xk X i(k−a) log(1 − ζ i x)ζ −ia = ζ = n i=0 n k i=0 k=1

X k≡a mod n

xk . k

Since, for 1 ≤ i ≤ n − 1, log(1 − ζ i x) is bounded in the neighborhood of 1, the result follows.

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121

Lemma 6.4.10: Suppose 0 < a ≤ n is an integer with resFqn τ = resFqn Frobaq for each τ ∈ C. Then, (10)

X

(N p)−s = −

p∈C

c log(1 − q 1−s ) + O(1), [L : K]

s → 1+ .

S∞ Proof: The set C 0 = k=1 Ck (L/K, C) differs from C by only finitely many elements. We apply Proposition 6.4.8 and Lemma 6.4.9 for x = q 1−s to compute: X

(N p)−s =

∞ X

X

(N p)−s

j=0 p∈Ca+jn (L/K,C)

p∈C

=

∞ X j=0

=

1 c q a+jn + O(q 2 (a+jn) ) q −(a+jn)s m(a + jn)

∞ ∞ X 1 1 c X q (1−s)(a+jn) + O q ( 2 −s)a q ( 2 −s)jn m j=0 a + jn j=0 1

q ( 2 −s)a c log(1 − q 1−s ) + O(1) + O 1 mn 1 − q ( 2 −s)n c log(1 − q 1−s ) + O(1), s → 1+ . =− [L : K] =−

When L = K, Lemma 6.4.10 simplifies to (19)

X

N p−s = − log(1 − q 1−s ) + O(1),

s → 1+ .

p∈P (E)

Dividing (18) by (19) and taking the limit as s → 1+ gives the Dirichlet density of C: P −s c p∈C (N p) δ(C) = lim P = −s + [L : K] s→1 p∈P (E) (N p) This concludes the Chebotarev density theorem for function fields.

6.5 Number Fields Let L/K be a finite Galois extension of number fields. The proof of the Chebotarev density theorem for L/K splits into eight parts. It uses the asymptotic formula (2) for counting ideals with bounded norm in a given class (which we quote without proof). We say L/K is cyclotomic if L ⊆ K(ζ) with ζ a root of 1. The case that L/K is cyclotomic produces the general case from an easy reduction to L/K cyclic.

122

Chapter 6. The Chebotarev Density Theorem

Part A: Ideals with a bounded norm in a given class. Let c be a nonzero ideal of OK . Denote the group of all fractional ideals of K relatively prime to c by J(c). Let P (c) be the subgroup of all principal fractional ideals xOK , where x satisfies the following conditions. (1a) If p is a prime ideal of OK that divides c, then x lies in the local ring Op of OK at p and x ≡ 1 mod cOp . (1b) x is totally positive: σx is positive for each embedding σ: K → R. The factor group G(c) = J(c)/P (c) is finite [Lang5, p. 127]. Denote the order of G(c) by hc . Extend the absolute norm N p of prime ideals multiplicatively to all fractional ideals. Consider a class K of J(c) modulo P (c). Denote the number of ideals a ∈ K (of OK ) with N a ≤ n by j(K, n) The key asymptotic formula is: (2)

−1

j(K, n) = ρc n + O(n1−[K:Q] ),

n → ∞,

where ρc is a positive constant dependent on c and K but not on K [Lang5, p. 132]. Part B: Abelian characters. A character of a finite Abelian group G is a homomorphism χ: G → C× . Define multiplication of characters by ˆ of characters of G forms a group iso(χ1 χ2 )(σ) = χ1 (σ)χ2 (σ). The set G morphic P to G. Here are the standard character formulas. (3a) Pσ∈G χ1 (σ −1 )χ2 (σ) = |G| if χ1 = χ2 and 0 otherwise. −1 )χ(τ ) = |G| if σ = τ and 0 otherwise. (3b) ˆ χ(σ Q χ∈G (3c) χ∈Gˆ (1 − χ(σ)X) = (1 − X f )|G|/f if f = ord(σ). Formulas (3a) and (3b) are known as the orthogonality relations [Goldstein, p. 113]. Formula (3c) follows in the case where G = hσi by observing that the zeros of both sides are the roots of 1 of order f and from ˆ = |G|. The general case follows from the cyclic case and the the relation |G| ⊥ c = ∼ G/hσi ˆ ˆ | χ(σ) = 1} canonical isomorphism hσi , where hσi⊥ = {χ ∈ G [Goldstein, p. 112]. Part C: L-series. For a given ideal c of OK and a character χ of G(c) consider the Dirichlet series X χ(a) (4) Lc (s, χ) = , Re(s) > 1, (N a)s gcd(a,c)=1

where a ranges over all ideals of OK relatively prime to c, and χ(a) = χ(K) if K is the class of a. Call Lc (s, χ) an L-series. The function χ(a) is multiplicative on J(c). Therefore, Lc (s, χ) satisfies the Euler identity Y χ(p) −1 (5) Lc (s, χ) = for Re(s) > 1 1− (N p)s p-c

We quote the following result of complex analysis:

6.5 Number Fields

123

Lemma 6.5.1 ([Lang5, p. 158]): Let {ai }∞ i=1 be a sequence of complex numbers, for which there is a 0 ≤ σ < 1 and a complex number ρ with n X

ai = ρn + O(nσ )

as

n → ∞.

i=1

P∞ Then f (s) = n=1 an n−s for Re(s) > 1 analytically continues to Re(s) > σ, except for a simple pole with residue ρ at s = 1. Lemma 6.5.2: The function Lc (s, χ) has an analytic continuation to the half 1 . If χ = 1, then it has a simple pole at s = 1 with plane Re(s) > 1 − [K:Q] residue hc ρc . If χ 6= 1, then Lc (s, χ) is analytic in the entire half plane. Proof: We use (2) to substitute for j(K, t) and use the orthogonality relation (3a) for the finite group G(c), to conclude that X

χ(a) =

(a,c)=1 N a≤n

X

X

K∈G(c)

a∈K N a≤n

=

χ(a) =

X

−1

χ(K) ρc n + O(n1−[K:Q] )

K∈G(c) −1

hc ρc n + O(n1−[K:Q] ) −1 O(n1−[K:Q] )

if χ = 1 if χ = 6 1,

Thus, our Lemma is a special case of Lemma 6.5.1.

n → ∞.

Part D: Special case of Artin’s reciprocity law. This law is the central result of class field theory. Consider a finite Abelian extension L/K of number fields. Let c be an ideal of OK divisible by all prime ideals ramifying in L (we say that c is admissible). If a prime ideal p does not divide c, then L/K defines a unique element of Gal(L/K). The map p 7→ L/K extends p p to a homomorphism ωc : J(c) → Gal(L/K) called the reciprocity map. When referring to the extension L/K we will denote ωc by ωL/K,c . Let L0 be any Abelian extension of K containing L. Suppose each prime ideal of OK ramified in L0 divides c. Then, resL (ωL0 /K,c (p)) = ωL/K,c (p) for each prime p ∈ J(c) ((3) of Section 6.2). So, (6)

resL (ωL0 /K,c (a)) = ωL/K,c (a)

for each a ∈ J(c)

Class field theory proves that ωc is surjective [Lang5, p. 199, Thm. 1]. In order to describe its kernel, let Norm = NormL/K be the norm map of fractional ideals of OL onto fractional ideals of OK . If A is an ideal of OL , then Norm(A) is the ideal of OK generated by normL/K (a) for all a ∈ A. If B is another ideal, then Norm(AB) = Norm(A)Norm(B), so Norm extends multiplicatively to the group of all fractional ideals of OL . If P is a prime ideal of OL , p = P ∩ OK , and f = fP/p , then Norm(P) = pf . Finally, NormL/Q and the absolute norm of ideals relate to each other by the formula NormL/Q (A) = N (A)Z. [Janusz, pp. 35–37].

124

Chapter 6. The Chebotarev Density Theorem

Artin reciprocity law gives an admissible ideal c of OK such that ωc : J(c) → Gal(L/K) is surjective and Ker(ωc ) = Norm(c)P (c) [Lang5, p. 205, Thm. 3]. Here we prove one part of Artin reciprocity law for cyclotomic extensions of K. Lemma 6.5.3: Let ζ be a primitive mth root of 1, L a subfield of K(ζ) containing K, and c an ideal of OK divisible by m. Then P (c) ⊆ Ker(ωc ). Proof: Each prime ideal of OK which ramifies in K(ζ) divides m [Goldstein, p. 98]. This defines ωc = ωL/K,c Use (6) to assume that L = K(ζ). Consider the natural embedding i: Gal(L/K) → (Z/mZ)× determined by σ(ζ) = ζ i(σ) , σ ∈ Gal(L/K). If p ∈ J(c) is prime and P is a prime ideal of OL lying over p, then ωc (p)(ζ) ≡ ζ N p mod P. Since reduction modulo P is injective on {1, ζ, . . . , ζ m−1 } [Goldstein, p. 97, Prop. 6-2-2], ωc (p)(ζ) = ζ N p . Hence, i(ωc (p)) ≡ N p mod m. Therefore, i(ωc (a)) ≡ N a mod m for each a ∈ J(c).

(7)

Now let xOK ∈ P (c), with x ∈ K × . Since x is totally positive (by (1b)), NK/Q (x) is a positive rational number congruent by (1a) to 1 mod m. It generates the same fractional Z-ideal as N (xOK ) [Janusz, p. 37]. Since N (xOK ) is also a positive rational number, N (xOK ) = NK/Q (x) ≡ 1 mod m. We conclude from (7) that ωc (xOK ) = 1. Part E: Cyclotomic L-series and the Dedekind zeta function. Let L and c be as in Lemma 6.5.3. Then ωc induces a homomorphism ω ¯ c from G(c) = J(c)/P (c) onto a subgroup G of Gal(L/K). (In Corollary 6.5.5(c), we prove ˆ χ◦ω that G = Gal(L/K).) Thus, for each χ ∈ G, ¯ c is a character of G(c). In the notation of Part C, the L-series Lc (s, 1) of the trivial character is the Dedekind zeta function of K with respect to c. X Y 1 −1 1 1 − = , Re(s) > 1 ζc (s, K) = (N a)s (N p)s gcd(a,c)=1

p-c

Lemma 6.5.4: In the above notation let C = cOL and let n = (Gal(L/K) : G). Then Y Lc (s, χ ◦ ω ¯ c )n . (8) ζC (s, L) = ˆ χ∈G

Proof: Let p be a prime ideal of OK with p - c. Suppose p factors in L into a product of g prime ideals P, each of degree f . Since p is unramified in L, we have f g = [L : K] = n|G| and ωc (p) = L/K is of order f . By (3c) and p the relation N P = N pf , Y χ◦ω ¯ c (p) n 1 1 n|G|/f Y 1− 1 − . = 1 − = (N p)s (N p)sf (N P)s ˆ χ∈G

Applying the product over all p - c, we conclude (8).

P|p

6.5 Number Fields

125

Corollary 6.5.5: Let χ1 be a nontrivial character of G. Then: ¯ c ) 6= 0; (a) Lc (1, χ1 ◦ ω (b) log ζc (s, K) = − log(s − 1) + O(1), s → 1+ ; and (c) G = Gal(L/K). ¯ c is a nontrivial character Proof: Since ω ¯ c : G(c) → G is surjective, χ ◦ ω of G(c) for each nontrivial character of G. By Lemma 6.5.2, ζc (s, K) has a simple pole at s = 1. All other factors of the right hand side of (8) are ¯ c ) has a zero at s = 1, then the zero of regular at s = 1. Assume Lc (s, χ1 ◦ ω ¯ c )n in (8) at s = 1 cancels the pole of ζc (s, K)n . Hence, ζC (s, L) is Lc (s, χ ◦ ω analytic at s = 1, contradicting Lemma 6.5.1. This proves (a). Formula (b) follows from Lemma 6.5.2. Finally, the left side of (8) has a pole of order 1 at s = 1. By (a), the right side has pole of order n. Therefore, n = 1 and G = Gal(L/K). Lemma 6.5.6: If χ is a character of G, then ¯c) = log Lc (s, χ ◦ ω

Xχ◦ω ¯ c (p) p-c

(N p)s

+ O(1),

s → 1+ .

Proof: We apply the Euler identity (5) for Re(s) > 1 to obtain: log Lc (s, χ ◦ ω ¯c) = −

X

∞ ¯ c (p)k χ◦ω ¯ c (p) X X χ ◦ ω log 1 − . = s (N p) k · (N p)ks p-c k=1

p-c

Next let σ = Re(s). Then ∞ ∞ XX |χ ◦ ω ¯ c (p)|k X X X 1 ≤ f kσ ks p/p k(N p) p p p-c k=2

p|p k=2

≤ [K : Q]

∞ XX X 1 1 1 = [K : Q] pkσ p2σ 1 − p−σ p p k=2

X 1 1 with P (N p)−s ε 1 P p∈Cσ (13) − , > −s [L : K] [L : K] p∈P (K) (N p)

1 < s < s0 .

As σ ranges on Gal(L/K), the sum of the left hand sided of (13) is 1. Hence, by (13) P (N p)−s [L : K] − 1 1 P p∈Cσ (14) +ε , 1 < s < s0 . < −s (N p) [L : K] [L : K] p∈P (K) It follows from (13) and (14) that each Cσ has Dirichlet density [L : K]−1 . This proves the Chebotarev density theorem for Abelian extensions. Part H: Reduction to the cyclic case. We now reduce the Chebotarev density theorem for an arbitrary finite Galois extension L/K of number fields to the case that L/K is cyclic. Let C be a conjugacy class in Gal(L/K) = G and let L/K =C . C = p ∈ P (K) | p Choose τ ∈ C and let K 0 = L(τ ) be its fixed field. Denote the set of primes q ∈ P (K 0 ) which are unramified over K, have a relative degree 1 over K, and 0 satisfy L/K = {τ } by D0 . If q ∈ D0 and p = q ∩ K, then N p = N q. Hence, q L/K 0 = P = τ , so p ∈ C. Moreover, since if P ∈ P (L) lies over q, then L/K P ord(τ ) = [L : K 0 ], P is the unique element of P (L) over q. Conversely, if p ∈ C, then there exists a prime P ∈ P (L) that lies over p with L/K = τ. P Then q = P ∩ K 0 belongs to D0 and lies over p. Let CG (τ ) be the centralizer of τ in G. The map σ 7→ στ σ −1 from G onto C has fibers whose order is |CG (τ )|, so |G = |C| · |CG (τ )|. All primes P0 ∈ P (L) over p that satisfy L/K = τ are conjugate to P by some element P0 |G| G (τ )| σ ∈ CG (τ ). Hence, their number is |C|D = [L:K 0 ]·|C| . P| 0 This is therefore the number of q ∈ D that lie over p. Hence, for s > 1

(15)

X p∈C

1 1 |C|[L : K 0 ] X = . s (N p) [L : K] (N q)s 0 q∈D

0 = {τ } by only primes of The set D0 differs from D = q ∈ P (K 0 ) | L/K q relative degree at least 2 over K. For these primes we have X deg(q)≥2

X 1 1 0 ≤ [K : Q] < ∞. (N q)s q2 q

Exercises

129

(Note: This elimination of the set of absolute degree 2 primes won’t work in the function field case, for there are only finitely many primes of degree 1.) Hence, by Part G for cyclic extensions X

(N q)−s =

q∈D 0

X

(N q)−s + O(1) = −

q∈D

1 log(s − 1) + O(1), [L : K 0 ]

s → 1+ .

Combining this with (15) gives X

(N p)−s = −

p∈C

Finally, by (9), δ(C) =

|C| log(s − 1) + O(1), [L : K]

|C| [L:K] ,

as stated.

s → 1+ .

Exercises 1. Consider the ring R = Z[ζp ][x], where p is a prime number and x is an indeterminate. Let K be the quotient field of R and L = K(x1/p ). Give an ¯ K ¯ is not example, in Lemma 6.1, where the residue class field extension L/ Galois. 2. Let S/R be a ring cover. Consider prime ideals p of R and P of S with P ∩ R = p. Prove that pSP = PSP (This, together with the separability of SP /PSP over Rp /pRp means that S/R is an unramified extension of rings). Hint: Assume without loss that R and S are local rings with maximal ideals p and P, respectively. Let z be a primitive element for S/R and f = irr(z, Quot(R)). Denote the reduction modulo pS with a bar. Observe that ¯ Prove that S¯ ∼ ¯ ¯ S¯ is a local ring with maximal ideal P. f¯(X)R[X] and = R[X]/ the right hand side is a direct sum of fields corresponding to the irreducible factors of f¯(X). Conclude that there is only one such factor. 3. Let L1 , L2 be finite Galois extensions of a global field K and let L = L1 L2 . Consider a prime ideal P of OL and put Pi = P ∩ Li , i = 1, 2. Prove that L/K i /K is the unique element of Gal(L/K) whose restriction to Li is LP , P i i = 1, 2. 4. Show that if f (X) ∈ Z[X] is irreducible, then there exists infinitely many primes p for which f (X) ≡ 0 mod p has no solution. Hint: Use the equivalence of (1) and (3) of Section 6.3 and Lemma 13.3.2. 5. Let L/K be a finite Galois extension of global fields. Denote the set of prime ideals p of OK that split completely in L (pOL = P1 · · · Pn , where n = [L : K]) by Splt(L/K). (a) Show that a prime p ∈ P (K), unramified in L, belongs to Splt(L/K) = 1. if and only if L/K p

130

Chapter 6. The Chebotarev Density Theorem

(b) (Bauer) Suppose L and L0 are finite Galois extensions of a global field K such that Splt(L/K) and Splt(L0 /K) differ by a finite set., Prove that L = L0 . Hint: Apply the Chebotarev density theorem to the field LL0 . 6. Let K and L be number fields with equal zeta functions (Section 6.5, Part E). For each positive integer n prove that the number of ideals of OK with absolute norm n is the number of ideals of OL with absolute norm n. Apply this to any a prime p unramified in KL. Prove that p splits completely in K if and only if p splits completely in L. Conclude from Exercise 5 that if K and L are Galois over Q, then K = L 7. Let K be a global field. Denote the set of all prime ideals p of OK whose absolute degree is at least 2 by P 0 (K). That is, N p = pd where d ≥ 2. When K is a number field prove that the Dirichlet density as well as the natural density of P 0 (K) is 0. If, however K is a function field show that almost all primes of P (K) belong to P 0 (K). Thus, δ(P 0 (K)) = 1.

Notes Frobenius [Frobenius] conjectured what we now call the Chebotarev density theorem for finite Galois extension L/K of number fields. His result replaced the conjugacy class appearing in the conjecture by the union of all conjugates of σ i , where σ is a given element of Gal(L/K) and i ranges over all integers relatively prime to ord(σ). A fine account of the Frobenius density theorem appears in [Janusz]. Chebotarev [Tschebotarev] used cyclotomic fields to prove the Frobenius conjecture via a more difficult version of the field crossing argument of Part G of Section 6.5. Artin [Artin1] introduced his L-series; then he proved his reciprocity law and applied it to reprove the conjecture [Artin2]. Our proof is a mixture of both methods, with the addition of Deuring’s reduction to the cyclic case [Deuring1]. It was elaborated for this book by Haran. Deuring’s reduction was reproduced in [MacCluer]. For the function field case note that Reichardt proved Proposition 6.19 when a = k = 1 and K is algebraically closed in F , (i.e. m = 1) [Reichardt]. The restriction m = 1 does not appear explicitly in [Reichardt]. Without it, however, the result as well as its proof would be false. It is also interesting to note that [Reichardt] appeared before Weil proved the Riemann hypothesis for curves. Thus, Reichardt’s (analytic) proof uses only that the maximum of the real parts of the zeros of the Zeta function is less than 1. Serre [Serre2] gives a unified approach to the number field and function field case. He considers a scheme X of finite type over Z, takes an ´etale Galois covering of X and attaches an L-series L(X, χ; s) to the cover. He says that an induction on dim(X) shows if χ 6= 1, then L(X, χ; s) is holomorphic and 6= 0 at the point s = dim(X). This implies the Chebotarev density theorem by the classical Dirichlet argument (e.g. as in Part F of Section 6.5). Serre’s program for the function field case appears in [Fried9]. An early version of our proof for the function field case appears in [Jarden10].

Notes

131

The proof of [Fried-Jarden3, Prop. 5.16] applies [Fried-Jarden3, Lemma 5.14] in a faulty way. Indeed, d on [Fried-Jarden3, p. 63, line -3] should be replaced by md. This version of Field Arithmetic follows [Geyer-Jarden4, Appendix] and corrects this mistake also improving the estimate of [FriedJarden3, Prop. 5.16]. There are several effective versions of the Chebotarev density theorem. One of the most valuable for the problems in this book is [LagariasMontegomery-Odlyzko, p. 416, Theorem]. This isolates the contribution of the absolute discriminant dL of a number field L over Q to the error term. It proves that there is an effectively computable constant A with the following property: For each Galois extension L/K of Gal(L/K) of number fields and each conjugacy class C of Gal(L/K) there is a prime p of K, unramified in L = C, p = NL/Q p is a rational prime and p ≤ 2dA with L/K L . This result is p independent of the generalized Riemann hypothesis.

Chapter 7. Ultraproducts We develop the basic concepts of logic and model theory required for applications to field theory. These include the Skolem-L¨owenheim theorem, Loˇs theorem and an ℵ1 -saturation property for ultraproducts. Finally, we apply regular ultraproducts of families of models to the theory of finite fields.

7.1 First Order Predicate Calculus There is no general test to decide whether a given polynomial f (X1 , . . . , Xn ) with integral coefficients has a zero in Zn ; this is the negative solution to Hilbert’s 10th problem. A partial decision test must satisfy two criteria: it must be conclusive for a significant body of polynomials; and it must be effective in concrete situations. The simplest, and most famous, such test is the congruence test, whereby we test the congruence f (X1 , . . . , Xn ) ≡ 0 mod p for solutions for all primes p. Regard the coefficients of f as elements of Fp to see that the above congruence is equivalent to solving the equation f (X1 , . . . , Xn ) = 0 in Fp . If f (X1 , . . . , Xn ) = 0 has no solution in Fp for one p, then f (X1 , . . . , Xn ) = 0 has no solution in Z. This chapter develops language and technique for the formulation of analogs of the diophantine problem and of the corresponding congruence test over general rings and fields. We start with the introduction of a first order language, the concept of a theory in the first order language, and a model for this theory. A language (more precisely, first order language) consists of letters, rules for combining letters into meaningful words, and, finally, an interpretation of the meaningful words. The language we now describe depends on functions µ and ν from sets I and J to N and on a set K. It is denoted L(µ, ν, K). Here are its letters: (1a) Countably many variable symbols: X1 , X2 , X3 , . . . ; (1b) constant symbols ck , one for each k ∈ K; (1c) a µ(i)-ary relation symbol, Ri , one for each i ∈ I; (1d) the equality symbol =; (1e) a ν(i)-ary function symbol, Fj , one for each j ∈ J; (1f) the negation symbol ¬, and the disjunction symbol ∨; (1g) the existential symbol ∃; and (1h) parentheses ( ) and brackets [ ]. A finite sequence of letters of L(µ, ν, K) is a string. Among the strings of L(µ, ν, K) the (meaningful) words include, ”terms”, ”formulas” and ”sentences”; which we now define. The collection of terms of L(µ, ν, K) is the smallest collection of strings that contains all of the following: (2a) all the variable symbols Xi ;

7.1 First Order Predicate Calculus

133

(2b) all the constant symbols ck ; and (2c) all the strings Fj (t1 , . . . , tν(j) ) where j ∈ J and (t1 , . . . , tν(j) ) is a ν(j)tuple of previously defined terms. We list the atomic formulas: (3a) t = t0 for each pair of terms (t, t0 ); and (3b) Ri (t1 , . . . , tµ(i) ), for all i ∈ I and all µ(i)-tuples (t1 , . . . , tµ(i) ) of terms. The set of formulas is the smallest collection of strings containing all atomic formulas and satisfying the following: (4a) ¬[ϕ] is a formula, if ϕ is a formula; (4b) ϕ1 ∨ ϕ2 is a formula if ϕ1 and ϕ2 are formulas; and (4c) (∃Xl )[ϕ] is a formula, if ϕ is a formula and l ∈ N. This definition allows us to prove a property of formulas or to make definitions depending on formulas by an induction on structure. We first prove the property for atomic formulas. Then, assuming its validity for ϕ, ϕ1 and ϕ2 we prove it for ¬[ϕ], ϕ1 ∨ ϕ2 and (∃Xl )[ϕ]. As a first example we define the notion of free occurrence of a variable in a formula by an induction on structure: Any occurrence of X in an atomic formula ϕ is free. If an occurrence of X in a formula ϕ is free, and ψ is an arbitrary formula, then this occurrence is free in ¬ϕ, ϕ ∨ ψ and (∃Y )[ϕ], for Y distinct from X. Any occurrence of X which is not free is bounded. Any variable X which has a free occurrence in a formula ϕ is said to be a free variable of ϕ. Frequently we write ϕ(X1 , . . . , Xn ) (or t(X1 , . . . , Xn )) to indicate that X1 , . . . , Xn include all the free variables of ϕ (or t). Example: In the formula 1 2 3 (∃X) X = Y ∨ (∃X)[X = c] ∨ ¬R(X , Y ) occurrences 1 and 2 are bounded, while occurrence 3 is free; both occurrences of Y are free. Hence, X and Y are free variables of the formula. A formula without free variables is a sentence. Some abbreviations simplify this language: (5a) ϕ ∧ ψ for ¬[¬ϕ ∨ ¬ψ] (∧ is the conjunction symbol); (5b) ϕ → ψ for ¬ϕ ∨ ψ (→ is the implication symbol); (5c) ϕ ↔ ψ for [ϕ → ψ] ∧ [ψ → ϕ] (↔ is the double implication symbol); (5d) (∀Xl )[ϕ] for ¬(∃Xl )[¬ϕ] (∀ is the universal quantifier); Vn (5e) i=1 ϕi for ϕ1 ∧ ϕ2 ∧ · · · ∧ ϕn ; and Wn (5f) i=1 ϕi for ϕ1 ∨ ϕ2 ∨ · · · ∨ ϕn .

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7.2 Structures The sentences of a first order language are interpreted in “structures” of this language. In each of these structures, they are either true or false. A structure for the language L(µ, ν, K) is a system ¯ i , F¯j , c¯k ii∈I, j∈J, k∈K A = hA; R ¯ i is a µ(i)-ary relation where A is a nonvoid set, called the domain of A, R µ(i) ν(j) ¯ → A is a ν(j)-ary function on A, and of A (i.e. a subset of A ), Fj : A c¯k is an element of A, called a constant. Sometimes we use the same letter for the logical symbol and its interpretation in the structure. Also, for well known binary relations and binary functions we write the relation and function symbols as usual, between the argument (e.g. “a ≤ b” for “a less than or equal to b”). Occasionally we add to L = L(µ, ν, K) a new constant symbol a ˜ for each a | a ∈ A} . a ∈ A. This gives an extended structure L(A) = L µ, ν, K ∪· {˜ A substitution into A is a function, f (Xi ) = xi , from the set of variables into A. The following recursive rules extend this uniquely to a function from the set of terms into A: (1a) f (ck ) = c¯k , and (1b) f (Fj (t1 , . . . , tν(j) )) = F¯j (f (t1 ), . . . , f (tν(j) )), where t1 , . . . , tν(j) , are terms for which f has already been defined. Define the truth value of a formula ϕ under a substitution f (either “true” or “false”) by induction on structure: (2a) t = t0 is true if f (t) = f (t0 ); and ¯i. (2b) Ri (t1 , . . . , tµ(i) ) is true if (f (t1 ), . . . , f (tµ(i) )) ∈ R Continue by assuming that the truth values of ϕ, ϕ1 , and ϕ2 have been defined for all possible substitutions. Then (3a) ¬ϕ is true if ϕ is false; (3b) ϕ1 ∨ ϕ2 is true if ϕ1 is true or if ϕ2 is true (so if both ϕ1 and ϕ2 are true, then ϕ1 ∨ ϕ2 is also true); and (3c) (∃Xl )[ϕ] is true if there exists an x in A such that ϕ is true under the substitution g defined by: g(Xl ) = x and g(Xm ) = f (Xm ) if m 6= l. The truth values of the additional logical symbols introduced above are as follows: (4a) ϕ1 ∧ ϕ2 is true if both ϕ1 and ϕ2 are true; (4b) ϕ → ψ is true if the truth of ϕ implies the truth of ψ (i.e. either ϕ is false, or both ϕ and ψ are true); (4c) ϕ ↔ ψ is true if both ϕ and ψ are true or both ϕ and ψ are false; and (4d) (∀Xl )[ϕ] is true if for each x in A, ϕ is true under the substitution g defined by g(Xl ) = x and g(Xm ) = f (Xm ) if m 6= l. By an easy induction on structure one observes that the truth value of a formula ϕ(X1 , . . . , Xn ) under a substitution f depends only on f (X1 ) = x1 , . . . , f (Xn ) = xn . If ϕ is true under f , write A |= ϕ(x1 , . . . , xn ). In

7.3 Models

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particular, for ϕ a sentence, the truth value of ϕ is independent of f . It is either true in A, or false in A. In the former case write A |= ϕ, and in the latter, A 6|= ϕ

7.3 Models Models of a first order language L are generalization of groups, rings, fields, and ordered sets. They are structures where a given set of sentences, called axioms, is true. A theory in a first order language L = L(µ, ν, K) is a set of sentences T of L. A structure A for L is called a model of T if A |= θ for every θ ∈ T . In this case write A |= T . If T 0 is another theory in L for which every model of T 0 is also a model of T , write T 0 |= T . If Π is a theory of L and T is the theory of all sentences θ of L such that Π |= θ, then Π is said to be a set of axioms for T . Denote the class of all models of a theory T by Mod(T ). If A is a structure for a language L, then Th(A, L) is the set of all sentences of L which are true in A. Example 7.3.1: The theory of fields. Denote the first order language that contains the two binary functions symbols + (addition) and · (multiplication), and two constant symbols 0 and 1 by L(ring). For each integral domain R let L(ring, R) be the language L(ring) extended by all elements of R as constant symbols. Denote the usual axioms for the theory of fields by Π: (∀X)(∀Y )(∀Z)[(X + Y ) + Z = X + (Y + Z)]; (∀X)(∀Y )[X + Y = Y + X]; (∀X)[X + 0 = X]; (∀X)(∃Y )[X + Y = 0]; (∀X)(∀Y )(∀Z)[(XY )Z = X(Y Z)]; (∀X)(∀Y )[XY = Y X]; (∀X)[1 · X = X]; (∀X)[X 6= 0 → (∃Y )[XY = 1]]; 1 6= 0; and (∀X)(∀Y )(∀Z)[X(Y + Z) = XY + XZ]. Every model of Π is a field. Extend Π by all equalities — the positive diagram of R — (1)

a1 + b1 = c1 and a2 b2 = c2 , for ai , bi , ci ∈ R

that are true in R. Denote the set obtained by Π(R). A model of Π(R) is ¯ = {¯ a field that contains a subset R a | a ∈ R} whose elements satisfy the equalities ¯2¯b2 = c¯2 a ¯1 + ¯b1 = c¯1 and a

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¯ is a whenever the corresponding equalities of (1) are true in R. That is, R homomorphic image of R. ¯ is an isomorphic copy of K. Thus, a model If R = K is a field, then K of Π(K) is, (up to an isomorphism) a field containing K. Example 7.3.2: Irreducible Polynomials. Let R be an integral domain. An elementary statement about models of Π(R) is a mathematical statement that applies to each member of Mod(Π(R)) and for which there exists a sentence θ of L(ring, R) which is true in any given model F if and only if the statement is true. Consider, for example, a polynomial f (X1 , . . . , Xn ) of degree d with coefficients in R. Then “f (X) is irreducible” is an elementary statement about models of Π(R). Indeed, it is equivalent to the conjunction of the statements “there exist no polynomials g, h of degree d1 and d2 respectively such that f (X) = g(X)h(X),” where (d1 , d2 ) runs over all pairs of positive integers with d1 + d2 = d. Rewrite the phrase “there exists no polynomial g(X1 , . . . , Xn ) of degree d1 ”, as “¬(∃u1 ) · · · (∃uk )” where u1 , . . . , uk are variables for the coefficients of g(X). A system of equalities between corresponding coefficients on both sides of “=” replaces “f (X) = g(X)h(X).” Similarly, we may consider a polynomial f (u, X1 , . . . , Xn ) =

X

ui X1i1 · · · Xnin

with intermediate coefficients ui . The same argument as above gives a formula ϕ(u) in L(ring) such that for each field K and all tuples a with entries in K, the polynomial f (a, X) is irreducible in K[X] if and only if ϕ(a) is true in K. Two structures A = hA, Ri , Fj , ck i and B = hB, Si , Gj , dk i of the language L = L(µ, ν, K) are isomorphic, if there exists a bijective function f : A → B such that (2a) (a1 , . . . , aµ(i) ) ∈ Ri ⇐⇒ (f (a1 ), . . . , f (aµ(i) )) ∈ Si , for each i ∈ I; (2b) f (Fj (a1 , . . . , aν(j) )) = Gj (f (a1 ), . . . , f (aν(j) )), for each j ∈ J; and (2c) f (ck ) = dk , for each k ∈ K. In this case write A ∼ = B. The structures A and B are elementarily equivalent if A |= θ ⇐⇒ B |= θ for every sentence θ of L. If this is the case, we write A ≡ B. Clearly, if A ∼ = B, then A ≡ B. But we will have many examples that show the converse is false. Two fields L and L0 that contain a field K are isomorphic as models of Π(K) if and only if there exists a field isomorphism of L onto L0 that fixes every element of K: write L ∼ =K L0 . If, however, they are elementarily equivalent as models of Π(K), we write L ≡K L0 . Call A a substructure of B, (A ⊆ B, and B is an extension of A) if A ⊆ B, Ri = Aµ(i) ∩ Si for each i ∈ I, Fj (a1 , . . . , aν(j) ) = Gj (a1 , . . . , aν(j) ) for each j ∈ J and all a1 , . . . , aν(j) ∈ A; and ck = dk for each k ∈ K.

7.4 Elementary Substructures

137

More generally, an embedding of A into B is an injective map f : A → B that satisfies Condition (2). Note that an arbitrary map f : A → B is an embedding of A into B if and only if for each quantifier free formula ϕ(X1 , . . . , Xn ) of L and for all a1 , . . . , an ∈ A, the condition A |= ϕ(a1 , . . . , an ) implies B |= ϕ(f (a1 ), . . . , f (an )). Indeed, an application of the latter condition to the formula X1 6= X2 implies that f is injective. Suppose now that A ⊆ B. We say A is existentially closed in B if for each quantifier free formula ϕ(X1 , . . . , Xn ) and for all b1 , . . . , bn ∈ B with B |= ϕ(b1 , . . . , bn ) there exist a1 , . . . , an ∈ A with A |= ϕ(a1 , . . . , an ). Call A an elementary substructure of B and B an elementary extension of A (in symbols A ≺ B) if A ⊆ B and if for each formula ϕ(X1 , . . . , Xn ) of L and for every a1 , . . . , an in A, the truth of ϕ(a1 , . . . , an ) in A is equivalent to its truth in B. It follows, in particular, that a sentence θ of L is true in A if and only if it is true in B, (i.e. A ≡ B). The converse is false (Example 7.3.3). If, however, A ⊆ B then “A ≡ B as models of L(A)” is equivalent to “A ≺ B as models of L”. Transitivity of elementarily equivalence follows immediately: A ≺ B, B ≺ C implies A ≺ C. In addition, A ⊆ B ≺ C and A ≺ C imply A ≺ B. Example 7.3.3: Elementary subfields. If a field K is an elementary subfield of a field F , then F is a regular extension of K. In other words, K is algebraically closed in F and F/K is separable (Lemma 2.6.4). ˜ ∩ F and f = irr(x, K). Then the sentence First of all let x ∈ K (∃X)[f (X) = 0] holds in F , so also in K. Therefore, deg(f ) = 1, hence ˜ ∩ F = K. x ∈ K. Consequently, K Pn 1/p bn ∈ F with i=1 bi ui = 0. Then, Pn Letp now u1 , . . . , un ∈ K and b1 , . . . ,P n p i=1 bi ui = 0. Hence, (∃X1 ) · · · (∃Xn ) i=1 Xi ui is true in F , so also in K. Pn 1/p In other words, there exist a1 , . . . , an ∈ K with i=1 ai ui = 0. Therefore, 1/p over K. Consequently, F/K is separable. F is linearly disjoint from K For example, let x be an indeterminate. Then, Q(x2 ) ∼ = Q(x). Hence, 2 Q(x ) ≡ Q(x). But Q(x) is a proper algebraic extension of Q(x2 ). Therefore, Q(x2 ) is not an elementary subfield of Q(x).

7.4 Elementary Substructures We develop criteria for one structure to be an elementary substructure of another. Let m be a cardinal number. Consider a transfinite sequence {Aα | α < m} of structures for a language L = L(µ, ν, K) with Aα = hAα , RS αi , Fαj , cαk i. Suppose Aα ⊆ Aβ for each α ≤ β < m and define the union α<m Aα to S = hA , R , F , c i with A = A be the structure A m m mi mj mk m α , Rmi = α<m S α<m Rmi , Fmj (x1 , . . . , xν(j) ) = Fαj (x1 , . . . , xν(j) ) if x1 , . . . , xν(j) ∈ Aα , and cmk = c0k . Then Aα ⊆ Am for each α < m.

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Lemma 7.4.1: (a) If Aα ≺ Aβ for each α ≤ β < m, then Aα ≺ Am for each α < m. (b) If B is another structure of L such that Aα ≺ B for each α < m, then Am ≺ B. Proof of (a): Use an induction on structure to prove that for each formula ϕ(X1 , . . . , Xn ) of L, for each α < m, and for every x1 , . . . , xn ∈ Aα Aα |= ϕ(x) ⇐⇒ Am |= ϕ(x).

(1)

Proof of (b): Let ϕ(X1 , . . . , Xn ) be a formula of L and let x1 , . . . , xn ∈ Am . Then there exists α < m such that x1 , . . . , xn ∈ Aα . It follows from (1) that Am |= ϕ(x) ⇐⇒ Aα |= ϕ(x) ⇐⇒ B |= ϕ(x).

Proposition 7.4.2 (Skolem-L¨ owenheim): Let L = L(µ, ν, K) be a countable language, let B = hB, Si , Gj , dk i be a structure of L and let A0 be a countable subset of B. Then B has a countable elementary substructure A = hA, Ri , Fj , ck i such that A0 ⊆ A. Proof: We construct an ascending chain of countable sets A0 ⊆ A1 ⊆ A2 ⊆ · · · ⊆ B. Suppose that An has already been constructed, then An+1 consists of all dk ’s with k ∈ K; the Gj (y1 , . . . , yν(j) ) for all j ∈ J and ν(j)-tuple (y1 , . . . , yν(j) ) of elements of An ; and an element xm ∈ B such that B |= ϕ(x1 , . . . , xm ) for each formula ϕ(X1 , . . . , Xm ) and for every x1 , . . . , xm−1 ∈ An satisfying B |= (∃Xm )[ϕ(x1S, . . . , xm−1 , Xm )]. ∞ Define A as follows: A = n=1 An , Ri = Aµ(i) ∩ Si , Fj (y1 , . . . , yµ(j) ) = Gj (y1 , . . . , yµ(j) ) for y1 , . . . , yµ(j) ∈ A, and ck = dk . By the choice of the xn ’s above, the function Fj is well defined. Hence, A is a countable substructure of B. Now use an induction on structure to prove that for each formula ϕ(X1 , . . . , Xn ) of L and for each x ∈ An , A |= ϕ(x) if and only if B |= ϕ(x). This proves that A ≺ B

7.5 Ultrafilters A filter on a set S is a nonempty family D of subsets of S t which are ”big” in a sense made precise by the following condition: (1a) ∅ ∈ / D. (1b) If A, B ∈ D, then A ∩ B ∈ D. (1c) If A ∈ D and A ⊆ B ⊆ S, then B ∈ D. If, in addition, (2) For each A ⊆ S either A ∈ D or S r A ∈ D, then D is an ultrafilter. In this case D also satisfies (3) A ∪ B ∈ D implies A ∈ D or B ∈ D.

7.6 Regular Ultrafilters

139

Example 7.5.1: (a) The family of all cofinite subsets of S (i.e. those subsets whose complements are finite) is a filter of S. (b) The family Da of all subsets of S that contain a given element a of S is an ultrafilter on S, called a principal ultrafilter. From (3), an ultrafilter D is principal if and only if it contains a finite set. A family D0 of subsets of S satisfies the finite intersection property if A1 , . . . , An ∈ D0 implies A1 ∩ · · · ∩ An 6= ∅. If one adds to D0 all the sets B ⊆ S that contain finite intersections A1 ∩ · · · ∩ An of elements of D0 , then one obtains a filter D1 . By Zorn’s Lemma there exists a maximal filter D of S that contains D1 . Our next Lemma says that D is an ultrafilter. Lemma 7.5.2: A filter D on a set S is an ultrafilter if and only if it is maximal. Proof: Suppose D is maximal and let A ⊆ S. Assume that S r A 6∈ D. Then D ∪ {A} has the finite intersection property. Indeed, for D1 , . . . , Dn ∈ D let D = D1 ∩ · · · ∩ Dn . If D ∩ A = ∅, then D ⊆ S r A. Hence, S r A ∈ D, a contradiction. By the comment above there exists a filter D0 on S containing D ∪ {A}. By the maximality of D, A ∈ D. Thus, D is an ultrafilter. The converse is clear. Corollary 7.5.3: Every family D0 of subsets of S that satisfies the finite intersection property is contained in an ultrafilter. A somewhat stronger assumption implies the existence of nonprincipal ultrafilters. Lemma 7.5.4: Let D0 be a family of subsets of a set S that has the following property: If A1 , . . . , An ∈ D0 , then A1 ∩· · ·∩An is an infinite set. Then there exists a nonprincipal ultrafilter D on S that contains D0 . Proof: The family D1 that consists of all subsets of S that contain a set in D0 and all cofinite subsets of S has the finite intersection property. Choose an ultrafilter D that contains D1 ; it is a nonprincipal ultrafilter.

7.6 Regular Ultrafilters The family F of all finite subsets of an infinite set S has properties dual to those of a filter: (1a) S 6∈ F; (1b) A, B ∈ F implies A ∪ B ∈ F; and (1c) B ∈ F and A ⊆ B imply A ∈ F. Later we will work with another family that satisfies the same conditions the family of zero sets of a measure space S. Therefore we give both cases a unified treatment. Let F be a nonempty family of subsets of S that satisfies (1). Call the elements of F small sets.

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Define Boolean polynomials in Z1 , . . . , Zm recursively: The variables Z1 , . . . , Zm are Boolean polynomials, and if U, U1 , U2 are Boolean polynomials, then U 0 , U1 ∪ U2 , and U1 ∩ U2 are Boolean polynomials. Evaluate a Boolean polynomial P (Z1 , . . . , Zm ) at subsets A1 , . . . , Am of S by interpreting the symbols ∪, ∩, and 0 as union, intersection, and taking the complement, respectively. With addition and multiplication given by A + B = (A r B) ∪ (B r A),

A · B = A ∩ B,

0 = ∅,

1 = S,

the family of all subsets of S becomes an algebra over the field F2 , in which each element except 1 is a zero divisor. The family F of small sets is an ideal of this algebra. Two subsets A and B of S are congruent modulo F in this algebra if and only if they differ from each other by a small set, (i.e. (A r B)∪(B r A) ∈ F). We then say that A and B are almost equal, and write A ≈ B. Clearly, if A1 ≈ B1 and A2 ≈ B2 , then S r A1 ≈ S r B1 , A1 ∪ A2 ≈ B1 ∪ B2 , and A1 ∩ A2 ≈ B1 ∩ B2 . Thus, in general, if P (Z1 , . . . , Zm ) is a Boolean polynomial in Z1 , . . . , Zm and Ai ≈ Bi , i = 1, . . . , m, are subsets of S, then P (A1 , . . . , Am ) ≈ P (B1 , . . . , Bm ). If the difference A r B of two subsets A, B of S is a small set, then we say that A is almost contained in B. A family of subsets of S which is closed under unions, intersections and taking complements is called a Boolean algebra of sets. The Boolean algebra generated by a family A0 of subsets of S is the intersection of all Boolean algebras of families of S that contain S. It consists of all expressions P (A1 , . . . , Am ), where P (Z1 , . . . , Zm ) is a Boolean polynomial and A1 , . . . , Am belong to A0 . Denote this family by A. The family A0 of all subsets of S which are almost equal to a set in A is a Boolean algebra that contains both A and F. A0 is the Boolean algebra generated by A0 and F. Call an ultrafilter D on S regular (with respect to F) if it contains no small set. In particular, if A ∈ D and B ≈ A, then B ∈ D. For example, nonprincipal ultrafilters on S are regular with respect to the family of finite subsets of S. Lemma 7.6.1: Let S be a set and let F be a family of small subsets of S. Suppose that a family D0 of subsets of S satisfies (2) A1 , . . . , An ∈ D0 implies A1 ∩ · · · ∩ An is not a small set. Then there exists a regular ultrafilter D on S that contains D0 . Proof: Repeat the proof of Lemma 7.5.4.

We will apply the next result to model theoretic results for families of fields.

7.7 Ultraproducts

141

Proposition 7.6.2 ([Ax2, p. 265]): Let S be a set, F a family of small subsets of S, A a Boolean algebra of subsets of S that contains F, and C a subset of S. Suppose C ∈ / A. Then there exist two regular ultrafilters D and / D0 . D0 such that D ∩ A = D0 ∩ A but C ∈ D and C ∈ Proof: Denote the collection of all A ∈ A that almost contain C or almost contain S r C by A0 . Suppose Ai ∈ A0 almost contains C for i = 1, . . . , m, and Bj ∈ A0 almost contains S r C, j = 1, . . . , n, and A1 ∩ · · · ∩ Am ∩ B1 ∩ · · · ∩ Bn ≈ ∅. Take complements to deduce that (S r A1 ) ∪ · · · ∪ (S r Am ) ∪ (S r B1 ) ∪ · · · ∪ (S r Bn ) ≈ S Each S r Ai is almost contained in S r C and each S r Bi is almost contained in C. Hence C ≈ (S r B1 ) ∪ · · · ∪ (S r Bn ), and therefore C ∈ A, a contradiction. Thus, A0 satisfies (2). By Zorn’s Lemma, A has a maximal subcollection A1 that contains A0 and has property (2). In particular, A1 is closed under finite intersections. Claim: The family A1 ∪ {C} satisfies (2). Indeed, if A1 , . . . , Am ∈ A1 and A1 ∩ · · · ∩ Am ∩ C ≈ ∅, then C is almost contained in S r A, where A = A1 ∩· · ·∩Am ∈ A. Hence, S r A ∈ A0 ⊆ A1 . But this is a contradiction, because A ∈ A1 . By Lemma 7.6.1 there exists a regular ultrafilter D on S that contains A1 ∪ {C}. Obviously D ∩ A contains A1 and satisfies (2). The maximality of A1 implies that D ∩ A = A1 . Similarly, there exists a regular ultrafilter D0 on S that contains A1 ∪ {S r C}. It also satisfies D0 ∩ A = A1 . Hence, D0 ∩ A = D ∩ A.

7.7 Ultraproducts From a given family of structures, ultraproducts allow us to create new structures which retain, sometimes in a particularly useful form, those elementary properties that hold for almost all structures in the family. Furthermore, those elementary properties that hold only for a small subfamily no longer hold in the new structures. In many cases we are able to establish simple criteria under which two such new models are elementarily equivalent (e.g. Lemma 20.3.3). This has been a successful route to the investigation of the elementary theory of many algebraic structures. To be more explicit, consider a language L = L(µ, ν, K) and a set S together with an ultrafilter D on S. Suppose that for each s ∈ S we are given a structure As = hAs , Ris , Fjs , cks i for Q L. We construct the ultraproduct As /D = A = hA, Ri , Fj , ck i, in the of As , s ∈ S, modulo D, denoted following way. Q Define an equivalence relation on the cartesian product s∈S As by a ∼ b ⇐⇒ {s ∈ S | as = bs } ∈ D.

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Q The domain A of A is the elements of As modulo this relation. For simplicity use representatives of the equivalence classes instead of the classes themselves. With this convention define Ri by (a1 , . . . , aµ(i) ) ∈ Ri ⇐⇒ {s ∈ S| (a1s , . . . , aµ(i),s ) ∈ Ris } ∈ D. Define Fj by the rule: Fj (b1 , . . . , bν(j) ) is the equivalence class of the function s 7→ Fjs (b1s , . . . , bν(j),s ). Similarly ck is the equivalence class of the function s 7→ cks . Check that Ri , Fj , and ck are well defined. An induction on the structure of terms shows that if t(X1 , . . . , Xn ) is a term of L and x1 , . . . , xn ∈ A, then (1)

{s ∈ S | t(x1 , . . . , xn )s = t(x1s , . . . , xns )} ∈ D.

This gives rise to the fundamental property of ultraproducts: Proposition 7.7.1 (Loˇs): Let D be an ultrafilter on S. Suppose for each for a language L(µ, ν, K). s ∈ S that As = hAs , Ris , Gjs , cks i is a structure Q Consider the ultraproduct A = hA, Ri , Gj , ck i = As /D. Then, for every formula ϕ(X1 , . . . , Xn ) and for every x1 , . . . , xn ∈ A: (2)

A |= ϕ(x1 , . . . , xn ) ⇐⇒ {s ∈ S | As |= ϕ(x1s , . . . , xns )} ∈ D

Proof: Do an induction on structure: Statement (2) follows easily for atomic formulas from (1) and the induction steps present no difficulty. We demonstrate the induction step for the existential quantifier. Suppose that (2) is true for ϕ and for each n-tuple of elements of A. Let x1 , . . . , xn−1 ∈ A and assume that S0 = {s ∈ S | As |= (∃Xn )[ϕ(x1s , . . . , xn−1,s , Xn )]} ∈ D. For each s ∈ S0 there exists xns ∈ As such that As |= ϕ(x1s , . . . , xns ). If s 6∈ S0 , let xns ∈ As be arbitrary. Let xn be the equivalence class defined by the xns ’s. Then the set {s ∈ S | As |= ϕ(x1s , . . . , xns )} contains S0 , hence belongs to D. By the induction assumption, A |= ϕ(x1 , . . . , xn ). Therefore, A |= (∃Xn )[ϕ(x1 , . . . , xn−1 , Xn )]. Corollary 7.7.2: If θ is a sentence of L, then A |= θ ⇐⇒ {s ∈ S | As |= θ} ∈ D. Example 7.7.3: Ultraproducts of fields. If the structures As is a field for each s ∈ S, then so is A. If each of them is algebraically closed, then so is A. But even if each of them is algebraic over a given field K, A may be not. For example, let S = N and let D be a nonprincipal ultrafilter on N. ˜ For each n ∈ N choose an element Q xn ∈ Q of degree greater than n over Q. Then let Kn = Q(xn ), K = Kn /D, and x be the equivalence class of

7.7 Ultraproducts

143

(x1 , x2 , x3 , . . .). For each n, almost all xi satisfy no equation of degree n over Q. Hence, by Corollary 7.7.2, so does x. Consequently, x is transcendental over Q. Ultraproducts also satisfy a saturation property. Let A be a structure with a domain A for a language L. Extend L to a language L(A) by adding a new constant symbol for each element of A. We say that A is ℵ1 -saturated if the following holds. If r(1) < r(2) < r(3) < · · · is an increasing sequence of positive integers and for each n ∈ N , ϕn (X1 , . . . , Xr(n) ) is a formula of L(A) such that (3)

A |= (∃X1 ) · · · (∃Xr(n) )

n ^

ϕt (X1 , . . . , Xr(t) ),

t=1

then there exist x1 , x2 , x3 , . . . in A such that A |= ϕn (x1 , . . . , xr(n) ) for each n ∈ N. Lemma 7.7.4: Let D be a nonprincipal ultrafilter on N. Suppose for each n ∈ N that An is aQstructure with a domain An for a language L. Then the ultraproduct A = An /D is ℵ1 -saturated. Proof: To simplify notation assume that rn = n in the preceding definition. Suppose (3) holds for each n ∈ N. Then Dn = {s ∈ N | As |= (∃X1 ) · · · (∃Xn )

n ^

ϕt (X1 , . . . , Xt )} ∈ D

t=1

for each n ∈ N. Clearly D1 ⊇ D2 ⊇ D3 ⊇ · · ·. Since D is nonprincipal, Dn0 = Dn r{1, 2, . . . , n} ∈ D. Also D10 , D20 , D30 , . . . is a decreasing sequence with empty intersection. 0 Now define x1 , x2 , x3 , . . . in A asV follows. If s ∈ Dn0 r Dn−1 , choose n x1s , . . . , xns in As such that As |= t=1 ϕt (x1s , . . . , xts ). Thus, for each n ∈ N, xns is well defined for all s ∈ Dn0 . For s ∈ N r Dn0 choose xns ∈ As arbitrarily. From this definition, for each S∞ n ∈ N, 0the set {s0 ∈ N | As |= ϕn (x1s , . . . , xns )} contains the union p=n (Dp0 r Dp−1 ) = Dn . Therefore Proposition 7.7.1 gives A |= ϕn (x1 , . . . , xn ). If all structures As are the same, say As = A, then the ultraproduct As /D, denoted by AS /D, is called the ultrapower of A to S modulo D. Denote the domain of A by A. Consider the diagonal embedding of A into AS . That is, map a ∈ A onto the constant function as = a. This gives a canonical injective map of A into AS /D. Indeed, if the images of two elements a and b of A are equal, then the set {s ∈ S | as = bs } belongs to D and is therefore nonempty. It follows that a = b. We identify A with its image to conclude from Proposition 7.7.1 the following result: Q

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Proposition 7.7.5: If D is an ultrafilter on a set S and A is a structure for a language L, then A is an elementary substructure of AS /D. The compactness theorem of model theory is now an easy corollary. Proposition 7.7.6 (The Compactness Theorem): Let T be a set of sentences in a first order language L. If each finite subset of T has a model, then T has a model. Proof: Denote the collection of all finite subsets of T by F . For each Φ ∈ F let DΦ = {Φ0 ∈ F | Φ ⊆ Φ0 }. Then, DΦ ∩ DΦ0 = DΦ∪Φ0 . Hence, the family D0 = {DΦ | Φ ∈ F } has the finite intersection property. By Corollary 7.5.3, there exists an ultrafilter DQon F that contains D0 . Choose a model, MΦ , for each Φ ∈ F . Then, M = MΦ /D is a model of T , because if θ ∈ T , then D{θ} ∈ D A similar construction characterizes existential closedness by ultraproducts: Proposition 7.7.7: Let A ⊆ B be structures of a language L with domains A ⊆ B. Then A is existentially closed in B if and only if there exists an ultrapower A∗ of A and there exists an embedding β: B → A∗ whose restriction to A is the canonical embedding A → A∗ . Proof: Suppose first that there exist A∗ and β as above. Let ϕ(X1 , . . . , Xn ) be a quantifier free formula in L(A) and let b1 , . . . , bn be elements of B such that B |= ϕ(b1 , . . . , bn ). Then, A∗ |= ϕ(β(b1 ), . . . , β(bn )). Hence, A∗ |= (∃X1 ) · · · (∃Xn )[ϕ(X1 , . . . , Xn )]. We conclude from Proposition 7.7.5 that there exist a1 , . . . , an ∈ A with A |= ϕ(a1 , . . . , an ). Conversely, suppose that A is existentially closed in B. Denote the constant symbol of L(B) associated with an element b ∈ B by ˜b. Let T be the set of sentences ϕ(˜b1 , . . . , ˜bn ) of L(B) such that ϕ(X1 , . . . , Xn ) is a quantifier free formula of L(A) and b1 , . . . , bn are elements of B with B |= ϕ(b1 , . . . , bn ). Denote the collection of all finite subsets of T by F . For each Φ ∈ F let DΦ = {Φ0 ∈ F | Φ ⊆ Φ0 }. Then DΦ ∩ DΦ0 = DΦ∪Φ0 , hence the family D0 = {DΦ | Φ ∈ F } has the finite intersection property. By Corollary 7.5.3 there exists an ultrafilter D on F that contains D0 . Consider an element Φ in F . List the elements of Φ as ϕi (˜bi1 , . . . , ˜bin ), where ϕi (X1 , . . . , Xn ) is a quantifier free formula of L(A) and bi1 , . . . , bin are elements of B such that B |= ϕi (bi1 , . . . , bin ), i = 1, . . . , m. For each i choose new variable symbols Xi1 , . . . , Xin and consider the quantifier free formula m ^

ϕi (Xi1 , . . . , Xin ) ∧

^

[Xij = Xkl ]

i=1

of L(A), where in the second Vm conjunct, i, j, k, lVrange over all indices such that bij = bkl . Since B |= i=1 ϕi (bi1 , . . . , bin ) ∧ [bij = bkl ] there exist bΦ ij ∈ A V Φ Vm Φ Φ Φ such that A |= i=1 ϕi (bi1 , . . . , bin ) ∧ [bij = bkl ]. Extend A to a structure

7.8 Regular Ultraproducts

145

˜ AΦ of L(B) by choosing bΦ ij for the constant symbol bij , for all i and j. If b is an element of B which is different from all bij , choose an arbitrary a ∈ A for ˜b. Then AΦ is well defined and each of the sentences ϕi (˜bi1 , . . . , ˜bin ) is valid in AΦ . Q Finally, consider the ultraproduct AΦ /D with domain A∗ = AF /D. ∗ ∗ For each b ∈ B let b be the element of A that corresponds to the constant symbol ˜b. Put A∗ = AF /D. Then the map β: B → A∗ that maps b onto b∗ is an embedding of B into A∗ whose restriction to A is the canonical map A → A∗ . Indeed, consider a quantifier free formula ϕ(X1 , . . . , Xn ) of L(A) and elements b1 , . . . , bn ∈ B with B |= ϕ(b1 , . . . , bn ). Then AΦ |= Φ Φ each Φ in F that contains ϕ. Hence, by Corollary 7.7.2, ϕ(b Q 1 , . . . , bn ) for AΦ /D |= ϕ(b∗1 , . . . , b∗n ). Consequently, A∗ |= ϕ(b∗1 , . . . , b∗n ). A variation of the proof of Proposition 7.7.7 proves the following characterization of elementary embedding: Proposition 7.7.8 (Scott): An embedding α: A → B of structures is elementary if and only if there exists an ultrapower A∗ of A and an elementary embedding β: B → A∗ such that β ◦ α: A → A∗ is the canonical embedding. Likewise a lemma of Frane [Bell-Slomson, p. 161] says that A ≡ B if and only if B is elementarily embeddable in an ultrapower of A. A theorem of Shelah supersedes Frane’s lemma: A ≡ B if and only if there exists a set I and an ultrafilter D on I such that AI /D ∼ = BI /D [Shelah].

7.8 Regular Ultraproducts Regular ultraproducts generalize non-principal ultraproducts. They appear ˜ in Chapter 20 as ultraproducts of fields Q(s) with sets of measure 0 replacing finite sets of σ’s. The general model theoretic notions that we develop in this section will be applied in Chapter 20 to investigate the model theory of the ˜ Q(σ)’s. Let S be a set equipped with a family F of small subsets. Suppose that for each s ∈ S, As is a structure for a fixed language L. The truth set of a sentence θ of L is defined to be the following subset of S: A(θ) = {s ∈ S | As |= θ}. The map θ 7→ A(θ) preserves the Boolean operations: A(θ1 ∨ θ2 ) = A(θ1 ) ∪ A(θ2 ), A(θ1 ∧ θ2 ) = A(θ1 ) ∩ A(θ2 ), A(¬θ) = S r A(θ). More generally, if P (Z1 , . . . , Zm ) is a Boolean polynomial, then (1)

A(P (θ1 , . . . , θm )) = P (A(θ1 ), . . . , A(θm )),

where P (θ1 , . . . , θm ) is obtained from P (Z1 , . . . , Zm ) by first replacing ∪, ∩, and 0 by ∨, ∧, and ¬, respectively, and then substituting θ1 , . . . , θm for the Z1 , . . . , Zm .

146

Chapter 7. Ultraproducts

Let T be the theory of all sentences θ of L that are true in As for almost all s ∈ S (i.e. for all s ∈ S excluding a small subset). Q If D is a regular ultrafilter, we say that As /D is a regular ultraproduct. Proposition 7.8.1: (a) A sentence θ of L is in T if and only if it is true in every regular ultraproduct of the As ’. (b) Every model of T is elementarily equivalent to a regular ultraproduct of the As ’. Proof of (a): If θ belongs to T , then, by Corollary 7.7.2, θ is true in every regular ultraproduct of the As ’. Conversely, if θ ∈ / T , then A(¬θ) is not small. Hence, by Lemma 7.6.1, there exists a regular ultrafilter Q D on S which contains A(¬θ). Therefore, by Corollary 7.7.2, θ is false in As /D. Proof of (b): Let A be a model of T . Then, by (1), D0 = {A(θ) | A |= θ} is closed under finite intersections. Observe that no A(θ) ∈ D0 is small, since this would mean then ¬θ ∈ T , so A |= ¬θ. By Lemma 7.6.1, there existsQa regular ultrafilter D on S which contains D0 . By Corollary 7.7.2, A ≡ As /D. In concrete situations we often seek, in addition to the above data, a special set Λ0 of sentences of L with this property: (2) If A and A0 are models of T , then A ≡ A0 if and only if A and A0 satisfy the same sentences of Λ0 . Call Λ0 a set of basic test sentences. Every Boolean combination of basic test sentences is called a test sentence. Denote the Boolean algebra generated by {A(λ) | λ ∈ Λ0 } ∪ F by Λ. From (2), if D and D0 are two regular ultrafilters on S, then (3)

Y

As /D ≡

Y

As /D0 ⇐⇒ D ∩ Λ = D0 ∩ Λ.

Proposition 7.8.2: Suppose that Λ0 is a set of basic test sentences. Then for each sentence θ of L there exists a test sentence λ such that (a) A(θ) ≈ A(λ), and (b) θ ↔ λ belongs to T . Proof: We have only to prove that A(θ) ∈ Λ. Assume A(θ) does not belong to Λ. By Proposition 7.6.2, there exist regular ultrafilters D and D0 of S such 0 ∩ Λ, but A(θ) ∈ D and A(θ) ∈ / D0 . This contradicts (3), that D Q∩ Λ = D Q 0 since As /D ≡ As /D .

Exercises

147

7.9 Nonprincipal Ultraproducts of Finite Fields We return to the starting point and the motivation of this book, the theory of finite fields. Let S be a countable set. Define small sets as the finite subsets of S. For each s ∈ S let Fs be a finite field. Suppose (1) for each n ∈ N there are only finitely many s ∈ S such that |Fs | ≤ n. A concrete example arises by letting S be the collection of prime divisors of a global field K. The finite fields are the corresponding residue fields. Every finite field is perfect. That is, for every prime p, every Fs satisfies the following sentence of L(ring): (2)

p = 0 → (∀X)(∃Y )[Y p = X].

By Section 1.5: (3)

ˆ Gal(Fs ) ∼ =Z

for each s ∈ S. Also, by Theorem 5.4.2(a), and (1), given positive integers d and n, the following statement is true for almost all s ∈ S: (4) For every absolutely irreducible polynomial f ∈ Fs [X, Y ] of degree d there are n distinct points (xi , yi ) ∈ Fs × Fs with f (xi , yi ) = 0, i = 1, . . . , n. In Proposition 20.4.4 we will display a sequence Π1 of sentences of ˆ L(ring) such that a field F is a model of Π1 , if and only if Gal(F ) ∼ = Z. Also, there are sentences πd,n such that F |= πd,n if and only if F satisfies (4) (Proposition 11.3.2). We will be able then to prove the following result: Q Proposition 7.9.1: Suppose S satisfies (1) and F = Fs /D is a nonprinˆ and for every cipal ultraproduct. Then F is a perfect field, Gal(F ) ∼ = Z nonconstant absolutely irreducible polynomial f ∈ F (X, Y ) there exist infinitely many points (x, y) ∈ F × F with f (x, y) = 0.

Exercises 1. Consider the language L which has only one relation symbol, 0. Finally, certain long summations and long multiplications are primitive recursive. If Pg(x1 , . . . , xn , y) is a primitive Q recursive function, then both of the functions y γ. It follows that w(x − σx) > γ. Since γ is arbitrary, x = σx. Consequently, x ∈ Kv . ˜ Proof of (d): Assume without loss that char(K) = p > 0. Let x ∈ K × ˜ and let γ ∈ w(K ). Then there are a power q of p and an a ∈ Ks with b ∈ Ks× with v(b) > max(qγ − w(x), v(a)). xq = a. Use (a) to choose Qq q Then Pq X − bX − a =q i=1 (X − xi ) with x1 , . . . , xq ∈ Ks . It follows that i=1 w(x−xi ) = w(x −bx−a) = w(bx) > qγ. Hence, there exists i between ˜ 1 and q with w(x − xi ) > γ. Consequently, Ks is w-dense in K. Proposition 11.5.3 (Prestel): Let K be a PAC field and let w be a valua˜ Then K is w-dense in K. ˜ tion of K. ˜ is Proof: By Lemma 11.5.2(b) and Corollary 11.2.5 the w-closure of K in K ˜ and prove that a PAC field. Thus, we may assume that K is w-closed in K ˜ K = K. To this end, let f ∈QK[X] be a separable irreducible polynomial of degree n ˜ Consider n ≥ 1 and let f (X) = i=1 (X − xi ) be its factorization in K[X]. × × ˜ γ ∈ Γ = v(K ) and choose c ∈ K such that w(c) ≥ nγ (Lemma 11.5.2(a)). ˜ ] to a linear Apply Lemma 2.3.10(b) (Eisenstein’s criterion) over the ring K[Y 2 factor of f (Y ) to deduce that f (X)f (Y ) − c is absolutely irreducible. Since K is PAC, there exist x, y ∈ K such that f (x)f (y) = c2 . It follows from w(f (x)) + w(f (y)) = 2w(c) that w(f (x)) ≥ nγ or Pnw(f (y)) ≥ nγ. Suppose for example that the first possibility occurs. Then i=1 w(x − xi ) ≥ nγ. Hence, there is an i with w(x − xi ) ≥ γ. Since {x1 , . . . , xn } is a finite set, the latter conclusion implies that there exists i such that for each γ ∈ Γ there exists an x ∈ K with w(x − xi ) ≥ γ. Thus, xi belongs to the w-closure of K, which is, by assumption, K itself.

11.5 PAC Fields with Valuations

205

Since f is irreducible, n = 1. Therefore, K = Ks . By Lemma 11.5.2(d), K ˜ Consequently, K = K. ˜ is w-dense in K. Remark 11.5.4: Density. Exploiting Prestel’s trick in the proof of Proposition 11.5.3, J´anos Koll´ ar proves in [Koll´ ar2, Thm. 2] that if K is a PAC field, ˜ and V is a variety over K, Then V (K) is w-dense in w is a valuation of K, ˜ V (K). Corollary 11.5.5 (Frey-Prestel): The Henselian closure Kv of a PAC field ¯ v is separably closed K with respect to a valuation v is Ks . Consequently, K × and v(K ) is a divisible group. Proof: Let w be an extension of v to Ks . By Proposition 11.5.3, K is wdense in Ks . By Lemma 11.5.2(c), the w-closure of K in Ks is contained in Kv . Hence, Kv = Ks . Now recall that K and Kv have the same residue field and the same ¯ v [X] of positive value group. Consider a monic separable polynomial f¯ ∈ K ¯ degree. Lift f to a monic polynomial f ∈ Ov [X]. Then f is separable, so has ¯v. a root in Kv . The residue of this root is a root of f¯ in K × Next consider a ∈ K and a positive integer n which is not divisible by char(K). Choose x ∈ Kv with xn = a. Then n1 v(a) ∈ v((Kv )× ) = v(K × ). Finally, suppose char(K) = p > 0 and v(a) < 0. Then there is an x ∈ Kv with xp − x − a = 0. Hence, p1 v(a) = v(x) ∈ v(K × ). Consequently, v(K × ) is divisible. Corollary 11.5.6: Let K be a PAC field which is not separably closed. Then K is Henselian with respect to no valuation. Denote the maximal Abelian, nilpotent, and solvable extensions of Q by Qab , Qnil and Qsolv , respectively. Corollary 11.5.7 ([Frey]): The fields Qab and Qnil are not PAC fields. Proof: By Corollary 11.2.5 we have only to prove the statement for Qnil . For ˜ ∩ Qp is Henselian. Assume Qnil is PAC. Then Qnil Qp,alg is each p, Qp,alg = Q ˜ PAC (Corollary 11.2.5) and Henselian. By Corollary 11.5.5, Qnil Qp,alg = Q. ∼ Hence, Gal(Qp,alg ) = Gal(Qnil ∩ Qnil Qp,alg ) is pronilpotent. That is, the Galois group of every irreducible separable polynomial over Qp,alg is nilpotent. In particular, this is true for p = 5. However, X 3 + 5 is irreducible over Q5 (e.g. by Eisenstein’s criterion) and its discriminant −27 · 52 [Lang7, p. 270] is not a square in Q5,alg (e.g. −27 is a quadratic non-residue modulo 5). Hence, X 3 + 5 is irreducible over Q5,alg and its discriminant is not a square in Q5,alg . Therefore, Gal(X 3 + 5, Q5,alg ) is S3 which is not nilpotent. We conclude from this contradiction to the preceding paragraph that Qnil is not a PAC field. ˜ for each prime In contrast to Qab and Qnil , we have Qsolv Qp,alg = Q number p.

206

Chapter 11. Pseudo Algebraically Closed Fields

˜ Proposition 11.5.8: The Henselian closure of each valuation of Qsolv is Q. Proof: Let v be a valuation of Qsolv . Replacing v by an equivalent valuation, we may assume that v|Q = vp for some prime number p. Thus, Qsolv can be ˜ p such that v is the restriction to Qsolv of the unique extension embedded in Q ˜ ˜ of vp to Qp . Let Qp,alg = Q∩Q p and M = Qsolv Qp,alg . Then M is a Henselian ˜ closure of M at v. Assume M 6= Q. Since Gal(Qp ) is prosolvable [Cassels-Fr¨ohlich, p. 31, Cor. 1], so is Gal(M ). Since √ M contains all roots of unity, there exist b ∈ M and n ∈ N / M . If a ∈ M is sufficiently vp -close to M , then by Krasner, such that n b ∈ √ √ n M ( b) ⊆ M ( n a) [Jarden14, Lemma Pm 12.1]. Choose w1 , . . . , wm ∈ Qsolv and b1 , . . . , bm ∈ Qp,alg such that b = i=1 bi wi . Since Q is vp -dense P in Qp , we m may choose a1 , . . . , ap which are vp -close to b1 , . . . , bp . Then a = i=1 ai wi √ √ n is v-close to b and lies in Qsolv . It follows that n a ∈ Qsolv ⊆ M , so b ∈ M . ˜ We conclude from this contradiction that M = Q. Thus, Corollary 11.5.5 fails to solve the following problem. Problem 11.5.9: (a) Is Qsolv a PAC field? (b) Does there exist an infinite non-PAC field K of a finite transcendence degree over its prime field such that K is not formally real and all of its Henselian closures are separably closed? Remark 11.5.10: On Problem 11.5.9(a). Problem 11.5.9(a) leads to other problems with respectable classical connections. Here is one example: By Theorem 11.2.3 we have only to check if each absolutely irreducible curve, f (X, Y ) = 0, with f ∈ Q[X, Y ] has a point with coordinates in Qsolv . Let E = Q(x, y) be the function field of this curve. Suppose there exists t ∈ E such ˆ is solvable over Q(t). that E/Q(t) is an extension whose Galois closure, E, Then, each specialization t → t0 such that t0 ∈ Q and x and y are integral over the corresponding local ring extends to a specialization (x, y) → (x0 , y0 ) with x0 , y0 ∈ Qsolv . Since there are infinitely many such specializations, f (X, Y ) = 0 certainly has infinitely many Qsolv -rational points. This idea fails, however, if the function field F = C(x, y) over C for a “general” curve, f (X, Y ) = 0, has no subfield C(t) with Gal(Fˆ /C(t)) solvable. The proper subfields of a “general” curve of genus g > 1 are of genus 0 [Fried7, p. 26-27]. Therefore, with no loss, assume that there are no proper fields between C(t) and F ; that is, Gal(Fˆ /C(t)) is a primitive solvable group. A theorem of Galois implies that [F : C(t)] = pr for some prime p [Burnside2, p. 202]. Combining [Fried7, p. 26] and [Ritt] one may prove that for F “general” of genus suitably large (> 6) that if r = 1 then it is impossible for Gal(Fˆ /C(t)) to be solvable. But higher values of r have not yet been excluded. Remark 11.5.11: On Problem 11.5.9(b). Theorem D of [Geyer-Jarden5] gives for each characteristic p (including p = 0) an example of an infinite

11.6 The Absolute Galois Group of a PAC Field

207

non-PAC field K of characteristic p which is not formally real and all of its Henselian closures of K are separably closed. This gives an affirmative answer to Problem 11.5.1(b) of [Fried-Jarden3]. The proof is based on results and ideas of [Efrat]. It uses Galois cohomology, valuations of higher rank, and the Jacobian varieties of curves. Problem 11.5.9(b) is a reformulation of the older problem.

11.6 The Absolute Galois Group of a PAC Field We show that the absolute Galois groups of PAC fields are “projective” (Theorem 11.6.2). All decidability and undecidability results on PAC fields depend on this result. Lemma 11.6.1: Let L/K be a finite Galois extension, B a finite group, and α: B → Gal(L/K) an epimorphism. Then there exists a finite Galois extension F/E with Gal(F/E) = B such that E is a regular finitely generated extension of K, F is a purely transcendental extension of L with trans.deg(F/L) = |B|, and α = resF/L . Proof: Let {y β | β ∈ B} be a set of indeterminates of cardinality |B|. Put 0 0 F = L(y β | β ∈ B). Define an action of B on F by (y β )β = y ββ and β0 α(β 0 ) 0 for β, β ∈ B and a ∈ L. Then let E be the fixed field of B a = a in F . By Galois theory, F is a Galois extension of E with Galois group B [Lang7, p. 264, Thm. 1.8] and α(β) = resF/L (β) for each β ∈ B. Moreover, F is a finitely generated extension of K. Hence, by Lemma 10.5.1, E is a finitely generated extension of K. By construction, E ∩ L = K and F is a ˜ over L. purely transcendental extension of L, so F is linearly disjoint from K ˜ Hence, EL is linearly disjoint from K over L. It follows from Lemma 2.5.3, ˜ over K; that is E is a regular extension of that E is linearly disjoint from K K. Let E be a finitely generated extension of a field K and F a finite Galois extension of E. By Remark 6.1.5, F/E has a Galois ring cover S/R. Thus, R = K[x1 , . . . , xm ] is an integrally closed domain with quotient field E; S is the integral closure of R in F , and S = R[z] where, if f = irr(z, E), then f 0 (z) is a unit of S (Definition 6.1.3). Every homomorphism ϕ0 of R onto ¯ extends to a homomorphism ϕ of S onto a Galois extension F¯ of a field E ¯ → Gal(F/E) such that ¯ The map ϕ induces an embedding ϕ∗ : Gal(F¯ /E) E. ¯ and x ∈ S (Lemma 6.1.4). ϕ ϕ∗ (σ)(x) = σ(ϕ(x)) for each σ ∈ Gal(F¯ /E) Theorem 11.6.2 ([Ax2, p. 269]): Let K be a PAC field, A and B finite groups, and ρ: Gal(K) → A and α: B → A epimorphisms. Then there exists a homomorphism γ: Gal(K) → B such that ρ = α ◦ γ (i.e. Gal(K) is projective). Proof (Haran): Denote the fixed field of Ker(ρ) in Ks by L. Then L is a finite Galois extension of K and ρ defines an isomorphism Gal(L/K) → A. Thus, we may identify A with Gal(L/K) and ρ with the restriction map.

208

Chapter 11. Pseudo Algebraically Closed Fields

Let E and F be as in Lemma 11.6.1. Let S/R be a Galois ring cover for F/E. Since K is a PAC field, there exists a K-homomorphism ϕ0 : R → K (Proposition 11.1.3). Let ϕ be an extension of ϕ0 to S which is the identity on L. Then M = ϕ(S) is a Galois extension of K which contains L and ϕ induces an embedding ϕ∗ : Gal(M/K) → Gal(F/E) such that resF/L ◦ ϕ∗ = resM/L . Compose ϕ∗ with the map resM : Gal(K) → Gal(M/K) to obtain the desired homomorphism γ: Gal(K) → B with ρ = α ◦ γ. There are non PAC fields K with Gal(K) projective (e.g. K is finite or K = C(t)). On the other hand, if G is a projective group, then there exists some PAC field K such that G ∼ = Gal(K) (Corollary 23.1.2). The projectivity of the absolute Galois group of a field K is closely related to the vanishing of the Brauer group Br(K) of K, although it is not equivalent to it. We survey the concept of the Brauer group and prove that Br(K) = 0 if K is PAC. A central simple K-algebra is a K-algebra A whose center is K and which has no two sided ideals except 0 and A. In particular, if D is a division ring with center K, then the ring Mn (D) of all n × n matrices with entries in D is a central simple K-algebra for each positive integer n [Huppert, p. 472]. Conversely, if A is a finite dimensional central simple K-algebra, then, by a theorem of Wedderburn, there exists a unique division ring D with center K and a positive integer n such that A ∼ =K Mn (D) [Huppert, p. 472]. Suppose A0 is another finite dimensional central simple K-algebra. Then A0 is equivalent to A if there exists a positive integer n0 such that A0 ∼ = Mn0 (D). In particular, A is equivalent to D. We denote the equivalence class of A by [A] and let Br(K) be the set of all equivalence classes of finite dimensional central simple K-algebras. The tensor product of two finite dimensional central simple K-algebras is again a finite dimensional central simple K-algebra [Weil6, p. 166]. Moreover, the tensor product respects the equivalence relation between finite dimensional central simple K-algebras. Hence, [A] · [B] = [A ⊗K B] is an associative multiplication rule on Br(K). Since A ⊗K B ∼ = B ⊗K A, multiplication in Br(K) is commutative. Further, the equivalence class [K] is a unit in Br(K), because A ⊗K K ∼ = A. Finally, let Ao be the opposite algebra of A. It consists of all elements ao , with a ∈ A. Addition and multiplication are defined by the rules ao + bo = (a + b)o and ao bo = (ba)o . One proves that A ⊗K Ao ∼ = Mn (K), where n = dimK (A). Thus, [Ao ] = [A]−1 . Therefore, Br(K) is an Abelian group. For each field extension L of K the map A 7→ A ⊗K L induces a homomorphism resL/K : Br(K) → Br(L). The kernel of resL/K consists of all [A] such that A ⊗K L ∼ =L Mn (L) for some positive integer n. If A satisfies the latter relation, then A is said to split over L. If L0 contains L, then A also splits over L0 . It is known that each A splits over Ks [Weil6, p. 167]. Thus, Br(Ks ) is trivial.

11.6 The Absolute Galois Group of a PAC Field

209

Construction 11.6.3: The reduced norm of a central simple algebra. Let A be a finite dimensional central simple algebra A over a field K. Choose a Ks -isomorphism α: A ⊗K Ks → Mn (Ks ) for some positive integer n. In particular, dimK A = dimKs (Mn (Ks )) = n2 . Let {eij | 1 ≤ i, j ≤ n} be a basis of A over K. Then eij = α(eij ⊗ 1), 1 ≤ i, j ≤ n, form Pna basis of Mn (Ks ) over Ks . Each a ∈ A has a unique presentation as a = i,j=1 aij eij with aij ∈ K. The matrix a = (aij )1≤i,j≤n satisfies α(a ⊗ 1) =

n X

aij eij = λkl (a)

1≤k,l≤n

,

i,j 2 where λkl are linear forms over s in the n variables Xij . Indeed, if eij = PK n (εij,kl )1≤k,l≤n , then λkl (X) = ij=1 εij,kl Xij , where X = (Xij )1≤i,j≤n . The reduced norm of a is defined by

(1)

red.norm(a) = det(α(a ⊗ 1)).

If α0 : A ⊗K Ks → Mn (Ks ) is another Ks -isomorphism, then α0 ◦ α−1 is a Ks isomorphism of Mn (Ks ). By Skolem-Noether [Weil5, p. 166, Prop. 4], α0 ◦α−1 is a conjugation by an invertible matrix of Mn (Ks ). Hence, det(α0 (a ⊗ 1)) = det(α(a⊗1)). Thus, red.norm(a) is independent of the particular choice of α. Moreover, each σ ∈ Gal(K) fixes red.norm(a). Therefore, red.norm(a) ∈ K. Indeed, σ induces an isomorphism 1 ⊗ σ −1 of A ⊗K Ks and an isomorphism σn of Mn (Ks ). Then α0 = σn ◦ α ◦ 1 ⊗ σ −1 : A ⊗K Ks → Mn (Ks ) is a Ks -isomorphism satisfying α0 (a ⊗ 1) = σn α(a ⊗ 1) for each a ∈ K. It follows that σ red.norm(a) = σ det(α(a ⊗ 1)) = det σn (α(a ⊗ 1)) = det α0 (a ⊗ 1) = red.norm(a), as claimed. Now let p(X) = det λkl (X) . It is a homogeneous polynomial of degree n over Ks such that p(a) ∈ K for each a ∈ Mn (K). It follows that the coefficients of p belong to K (Exercise 9). Next observe that the linear forms λkl are linearly independent over Ks because the n2 ×n2 matrix (eij )1≤i,j≤n is nonsingular. We may therefore form a change of variables Ykl = λkl (X). It maps p(X) onto det(Y), which is an absolutely irreducible polynomial. Hence, p(X) is also absolutely irreducible. We call p(X) the reduced form of A. Theorem 11.6.4: Let K be a PAC field. Then its Brauer group Br(K) is trivial. Proof ([Ax2, p. 269]): Assume Br(K) is nontrivial. Then there exists a division ring D with center K such that dimK (D) = n2 and n > 1. Let p(X) be the associated reduced form. Since p(X) is an absolutely irreducible polynomial (Construction 11.6.3), p(X) has a nontrivial zero a ∈ Mn (K) (Proposition Pn 11.1.1). In the notation of Construction 11.6.3 (with A = D), let a = i,j=1 aij eij . By (1), red.norm(a) = p(a) = 0. On the other hand, a

210

Chapter 11. Pseudo Algebraically Closed Fields

is a nonzero element of D, hence invertible. Therefore, α(a ⊗ 1) is a regular matrix, so red.norm(a) = det(α(a ⊗ 1)) 6= 0. This contradiction proves that Br(K) is trivial. Remark 11.6.5: Varieties of Severi-Brauer. An alternative proof of Theorem 11.6.4 uses varieties of Severi-Brauer. They are varieties V which are defined over a field K and are isomorphic over Ks to Pn for some positive integer n. There is a bijective correspondence between K-isomorphism classes of varieties V of Severi-Brauer and equivalence classes of finite dimensional central simple K-algebras A. If V has a K-rational point, then A splits over K [Jacobson, p. 113]. In particular, if K is PAC, this implies that Br(K) = 0. The connection between the projectivity of the absolute Galois group of a field K and its Brauer group is based on the canonical isomorphism (2)

H 2 (Gal(K), Ks× ) ∼ = Br(K)

[Deuring2, p. 56, Satz 1 or Serre4, §X5]. Here we assume that the reader is familiar with Galois cohomology, e.g. as presented in [Ribes1] or in [Serre9]. In particular, it follows from (2) that (3) every element of Br(K) has a finite order [Ribes1, p. 138, Cor. 6.7]. For each prime number p and a profinite group G the notation cdp (G) stands for the pth cohomological dimension of G. It is the maximal positive integer n such that H n (G, A)p∞ = 0 for each torsion G-module A. Finally, cd(G) = supp (cdp (G)) is the cohomological dimension of G. Proposition 11.6.6: The following conditions on a field K are equivalent: (a) Gal(K) is projective. (b) cd(Gal(K)) ≤ 1. (c) For each prime number p 6= char(K) and for each finite separable extension L of K, Br(L)p∞ is trivial. Proof of “(a) ⇐⇒ (b)”: Let p be a prime number. By [Ribes, p. 211, Prop. 3.1], cdp (Gal(K)) ≤ 1 if and only if for every finite Galois extension L of K and for every short exact sequence α

0 −→ (Z/pZ)m −→ B −→ Gal(L/K) −→ 1 there exists a homomorphism β: Gal(K) → B such that α ◦ β = resKs /L . By a theorem of Gruenberg (Corollary 22.4.3), the latter condition holds for all p if and only if Gal(K) is projective. Consequently, (a) and (b) are equivalent. Proof of “(b) ⇐⇒ (c)”: Let p be again a prime number. First suppose that p 6= char(K). Then cdp (Gal(K)) ≤ 1 if and only if Br(L)p∞ is trivial for every finite separable extension L of K [Ribes, p. 261, Cor. 3.7]. If p = char(K), then cdp (Gal(L)) ≤ 1 for every field L of characteristic p [Ribes, p. 256, Thm. 3.3]. Since the Brauer group of each field is torsion (by (3)), this establishes the equivalence of (b) and (c).

11.7 A non-PAC Field K with Kins PAC

211

Proposition 11.6.7 ([Ribes, p. 264, Prop. 3.10]): The following conditions on a field K are equivalent. (a) Br(L) is trivial for every finite separable extension L of K. (b) The norm map, norm: N × → L× , is surjective for every finite separable extension L of K and for every finite Galois extension N of L. We summarize consequences of the previous results for PAC fields: Corollary 11.6.8: The following statements hold for every PAC field K: (a) Gal(K) is projective. (b) Br(K) is trivial. (c) cd(Gal(K)) ≤ 1. (d) The map norm: N × → K × is surjective for each finite Galois extension N of K. Proof: Let L be a finite separable extension of K. By Corollary 11.2.5, L is PAC. Hence, by Theorem 11.6.4, Br(L) is trivial. Therefore, cd(Gal(K)) ≤ 1 (Proposition 11.6.6) and the norm map N × → K × is surjective for each finite Galois extension N of K. Example 11.6.9 (Geyer): We construct an example of a field K with Gal(K) projective, a finite Galois extension K 0 of K, and an element u of K which is not a norm of an element of K 0 . By Proposition 11.6.7, K has a finite separable extension L such that Br(L) 6= 0. This will show that it is impossible to omit the condition “p 6= char(K)” in Condition (c) of Proposition 11.6.6. We start from a transcendental element u over F2 and let K0 = F2 (u)s . Then choose an transcendental element t over K0 and let K = K0 (t). By Tsen’s Theorem, Gal(K) is projective [Ribes2, p. 276 or Jarden17, Thm. 1.1]. Consider the Artin-Schreier extension K 0 = K(x) of K with x2 + x + t = 0. Each element y of K 0 has the form y = v + wx with v, w ∈ K and (t) normK 0 /K (y) = (v + wx)(v + w(1 + x)) = v 2 + vw + w2 t. Write v = fh(t) and w=

g(t) h(t) ,

where f, g, h ∈ K0 [t] and h 6= 0. Let a (resp. b, c) be the leading

coefficient of f (resp. g, h). If normK 0 /K y = u, then f (t)2 +f (t)g(t)+g(t)2 t = h(t)2 u. Compare the leading coefficients of both sides of this equality. If deg(f ) > deg(g), then a2 = c2 u. If deg(f ) ≤ deg(g), then b2 = c2 u. In both cases we find that u is a square in F2 (u)s , which is not the case. This contradiction proves that u is not a norm of an element of K 0 .

11.7 A non-PAC Field K with Kins PAC Let L/K be a purely inseparable extension of fields. If K is PAC, then so is L (Corollary 11.2.5). Problem 12.4 of [Geyer-Jarden3] asks whether the converse is true. An example of Hrushovski shows that this is not the case. The main ingredient of this example is the analog of Mordell conjecture for function fields:

212

Chapter 11. Pseudo Algebraically Closed Fields

Proposition 11.7.1 (Grauert-Manin [Samuel, pp. 107 and 118]): Let K be a finitely generated regular extension of a field K0 and C a nonconstant ˜ is at least 2. Then C(K) curve over K/K0 . Suppose the genus of C over K is a finite set. Here we say that C is a nonconstant curve over K/K0 if C is defined ˜ to a curve C0 defined over K and if C is not birationally equivalent over K ˜ over K0 . Lemma 11.7.2: Let F = K(x1 , . . . , xn ) be a finitely generated extension of a field K of positive characteristic p. Suppose K is algebraically closed in F . Then ∞ \

(1)

k

k

K(xp1 , . . . , xpn ) = K.

k=1

Proof: Denote the left hand side of (1) by F0 . First suppose that K is perfect. Thus, K p = K, so F0 = F0p is also perfect. In addition F0 , as a subfield of F , is finitely generated over K (Lemma 10.5.1). Assume F0 is transcendental over K. Choose a transcendental basis t1 , . . . , tr with r ≥ 1. Then F0 has a finite degree over E = K(t1 , . . . , tr ). On the other hand, since 1/pm F0 is perfect, E(t1 ) is contained in F0 and has degree pm over E for each positive integer m. This contradiction proves that F0 is algebraic over K. Since K is algebraically closed in F , we conclude that F0 = K. In the general case Kins is a perfect field. Hence, by the preceding paragraph, F0 ⊆ F ∩

∞ \

k

k

Kins (xp1 , . . . , xpn ) = F ∩ Kins = K.

k=1

Therefore, F0 = K.

Lemma 11.7.3: Let K be a finitely generated regular transcendental extension of a field K0 of positive characteristic p. Let C be a curve which is ˜ 0 is at least 2. Let F be a finitely defined over K and whose genus over K K generated regular extension of K. Suppose C is a nonconstant curve over F/K0 . Then K has a finitely generated extension E which is contained in F such that F/E is a finite purely inseparable extension and C(K) = C(E). k

k

Proof: Let F = K(x1 , . . . , xn ), and for each k write Fk = K(xp1 , . . . , xpn ). ˜ 0 /K K ˜ 0 is a By Lemma 11.7.2, the intersection of all Fk is K. Since F K ˜ regular extension, the genus of C over F K0 is the same as the genus of C ˜ 0 (Proposition 3.4.2(b)), so at least 2. By Proposition 11.7.1, C(F ) is over K K a finite set. Hence, there exists a positive integer k such that C(Fk ) = C(K), so E = Fk satisfies the assertion of the Lemma.

11.7 A non-PAC Field K with Kins PAC

213

Remark 11.7.4: On M¨ obius transformations. Let K be a field and x an in obius determinate. To each matrix A = ac db in GL(2, K) we associate a M¨ transformation τA (also called a linear fractional transformation). It is the K-isomorphism of K(x) into K(x) defined by the following rule: (3)

τA (x) =

ax + b . cx + d

If B is another matrix in GL(2, K), then τA ◦ τB = τBA . If I is the unit matrix, then τI is the identity map of K(x). In particular, τA−1 ◦ τA = id, so τA is an automorphism of K(x)/K. If k ∈ K × , then τkI = id. Conversely, if τA = id, then, by (3), c(x0 )2 + (d − a)(x0 ) − b = 0 for all x0 ∈ K. Hence, c = b = 0 and d = a, so A = aI. Therefore, the kernel of the map A 7→ τA consists of the group of scalar matrices. If τ is arbitrary element of Aut K(x)/K , then K(x) = K(τ (x)). Hence, by Example 3.2.4, there exists A ∈ GL(2, K) such that τ (x) = τA (x). Thus, the map A 7→ τA defines an isomorphism (4)

PGL(2, K) ∼ = Aut K(x)/K .

Substituting elements of K ∪ {∞} in (3), we may also view τA as a a bijective map of K ∪ {∞} onto itself. For example, if c 6= 0, then a·∞+b c·∞+d = c . Note that since the pairs (a, b) and (c, d) are linearly independent over K, no x0 ∈ K satisfies both ax0 +b = 0 and cx0 +d = 0. Hence, the value of τA (x0 ) is well defined. The arithmetic with ∞ becomes clearer if we substitute x = xx10 in (3) and view τA as a bijective map of P1 (K) onto itself: (5)

τA (x0 :x1 ) = (cx1 + dx0 : ax1 + bx0 ).

The map x0 7→ τA (x0 ) of K ∪ {∞} onto itself defines determines τA . This is one of the consequences of Lemma 11.7.5: Lemma 11.7.5: Let K/K0 be an extension of fields. Let (x1 , x2 , x3 ) and (y1 , y2 , y3 ) be triples of distinct elements of K ∪ {∞}. Then there is a unique M¨obius transformation τ over K such that τ (xi ) = yi , i = 1, 2, 3. Moreover, τ can be presented as τA , where the entries of A belong to the field K0 (x1 , x2 , x3 , y1 , y2 , y3 ). Proof of uniqueness: If τi (xj ) = yj for i = 1, 2 and j = 1, 2, 3, then τ = τ1−1 τ2 satisfies τ (xj ) = xj for j = 1, 2, 3. Suppose first that none of the xj is ∞ and τ = τA with A = ac db . Then cx2j + (d − a)xj − b = 0 for j = 1, 2, 3. Hence, b = c = 0 and a = d. Now assume that x1 = ∞; that is, x1 is (0:1) in homogeneous coordinates. By (5), c = 0. Then we may assume that d = 1 and conclude from axj + b = xj , j = 2, 3, that a = 1 and b = 0. In both cases τ = id and τ1 = τ2 .

214

Chapter 11. Pseudo Algebraically Closed Fields

1 Proof of existence: The M¨ obius transformation τ (x) = x−x 0 maps the ele1 0 ment x of K onto ∞. Likewise, τ (x) = x exchanges 0 and ∞. Hence, we may assume that x1 = ∞ and y1 = ∞. Since x2 6= x3 ,

τ (x) =

y 2 − y3 x2 y3 − x3 y2 x+ x2 − x3 x2 − x3

maps xi onto yi , i = 1, 2, 3, as desired.

Remark 11.7.6: Conservation of branch points. Let K be an algebraically closed field, x an indeterminate, and F a finite separable extension of K(x). Then F/K is a function field of one variable. For each a ∈ K ∪ {∞} let ϕa : K(x) → K ∪ {∞} be the K-place of K(x) with ϕa (x) = a. Denote the corresponding prime divisor of K(x)/K by pa . We say that a is a branch point of F/K(x) (with respect to x) if pa ramifies in F . There are only finitely many prime divisors of K(x)/K which ramify in F (Section 3.4), so F/K(x) has only finitely many branch points. If τ is a M¨obius transformation of K(x) and τ (x) = x0 , then τ maps the set of branch points of F/K(x) with respect to x onto the set of branch points of F/K(x) with respect to x0 . By Lemma 11.7.5, there exists a M¨obius transformation τ of K(x) mapping ∞ onto a finite nonbranch point of F/K(x). Replacing x by τ (x), if necessary, we may assume that ∞ is not a branch point. Let S = K[y1 , . . . , yn ] be the integral closure of K[x] in F . Then S is a Dedekind domain (Proposition 2.4.6). Denote the curve generated in An over K by y by C. The local ring OC,b of C at each b ∈ C(K) is the local ring of S at the kernel of the K-homomorphism mapping y onto b. This kernel is a maximal ideal of S. Hence, OC,b is a valuation ring. The corresponding prime divisor qb of F/K is uniquely determined by b. Conversely, each prime divisor q of F/K with ϕq finite at x is also finite on S and ϕq (y) ∈ C(K). Thus, the map b 7→ qb is a bijective correspondence between C(K) and the set of prime divisors of F/K which are finite at x. Choose a polynomial f ∈ K[Y1 , . . . , Yn ] such that x = f (y). Then f defines a morphism π: C → A1 by π(b) = f (b) for each b ∈ C(K). The prime divisor qb of F/K lies over pa if and only if π(b) = a. Now let p be a prime divisor of K(x)/K which is finite at x. Denote the prime divisors of F/K lying over p by q1 , . . . , qr and the corresponding Pr ramification indices by e1 , . . . , er . Since K is algebraically closed, i=1 ei = [F : K(x)] (Proposition 2.3.2). Hence, p is ramified in F if and only if r < [F : K(x)]. Therefore, by the preceding two paragraphs, an element a of K is a branch point of F/K(x) if and only if there are less than [F : K(x)] points of C(K) lying over a. Next consider an algebraically closed subfield K0 of K. Suppose there is a function field F0 over K0 which contains x, y1 , . . . , yn such that F0 K = F and f has coefficients in K0 . Then [F0 : K0 (x)] = [F : K(x)] and C is already defined over K0 . By the preceding paragraph, “a is a branch point

11.7 A non-PAC Field K with Kins PAC

215

with respect to x” is an elementary statement in the language L(ring, K0 ). Let a1 , . . . , am be all branch points of F0 /K0 (x). Then “a is a branch point with respect to x if and only if a = ai for some i between 1 and m” is an elementary statement which holds over K0 . Since K0 is an elementary subfield of K (Corollary 9.3.2), the same statement holds over K. It follows that each branch point of F/K(x) with respect to x belongs to K0 . Remark 11.7.7: Construction of nonconstant curves. Let K0 be a field. Choose five distinct elements t1 , . . . , t5 in some regular extension of K0 such ˜ 0 (t1 , t2 , t3 ). Consider a regular field extension K of K0 containing /K that t4 ∈ Q5 t1 , t2 , t3 , t4 , t5 . Put f (X) = i=1 (X − ti ) and define a curve C over K by 1 if char(K) = 2. the equation Y 2 = f (X) if char(K) 6= 2 and Y 2 + Y = f (X) By Propositions 3.8.2 and 3.8.4, C is a hyperelliptic curve. More precisely, the genus g of C is 2 if char(K) 6= 2 and 4 if char(K) = 2. Claim: C is a nonconstant curve over K/K0 . ˜ Proof: Choose a generic point (x, y) for C over K and let F = K(x, y). Then ˜ K(x) is a quadratic subfield of F . Assume that C is a constant curve over ˜ 0 such that K/K0 . Then there exists a function field of one variable F0 over K ˜ 0 is a regular extension, F0 is linearly disjoint from ˜ = F . Since F0 /K F0 K ˜ 0 ) = genus(F/K) ˜ ˜ over K ˜ 0 (Lemma 2.6.7). In particular, g = genus(F0 /K K ˜ 0 by w0 and de(Proposition 3.4.2). Denote the canonical divisor of F0 /K ˜ by w. By Proposition 3.4.2 and Lemma note the image of w0 in Div(F/K) 3.2.2(b), deg(w) = deg(w0 ) and dim(w) = dim(w0 ). Hence, w is the canoni˜ (Exercise 1(b) of Chapter 3). Choose a basis z1 , z2 , . . . , zg cal divisor of F/K ˜ for L(w0 ) over K0 . Then, by the linear disjointness, z1 , z2 , . . . , zg form a basis ˜ By Proposition 3.7.4, K( ˜ z2 , . . . , zg ) is the unique quadratic for L(w) over K. z1 z1 ˜ z2 , . . . , zg ) = K(x). ˜ ˜ z2 , . . . , zg ) is a funcsubfield of F , so K( In particular, K( z1

z1

z1

z1

z

˜ Hence, K ˜ 0 ( z2 , . . . , g ) is a function field of genus tion field of genus 0 over K. z1 z1 ˜ By Example 3.2.4, there is an x0 with K ˜ 0 ( z2 , . . . , zg ) = K ˜ 0 (x0 ). 0 over K. z1

z1

˜ 0 ) = K(x). ˜ It satisfies K(x By Lemma 11.7.5, there exists a M¨obius transfor˜ mation τ over K such that τ (x) = x0 . ˜ ˜ ∪ {∞} be the K-place ˜ →K with For each i between 1 and 5 let ϕi : K(x) ˜ with ϕi (x) = ti . By Examples 2.3.8 and 2.3.9, ti is a branch point of F/K(x) respect to x. Put ai = τ (ti ). Then ϕi (x0 ) = ϕi (τ (x)) = τ (ϕi (x)) = τ (ti ) = ˜ 0 ) with respect to x0 . It follows from ai . Thus, ai is a branch point of F/K(x ˜ Remark 11.7.6 that ai ∈ K0 . ˜ 0 (t1 , t2 , t3 ). Hence, t4 = By Lemma 11.7.5, τ is already defined over K −1 ˜ τ (a4 ) ∈ K0 (t1 , t2 , t3 ). This contradiction to the assumption we made above proves that C is a nonconstant curve over K/K0 . Theorem 11.7.8 ([Hrushovski, Cor. 5]): For each prime number p there exists a countable non-PAC field E of characteristic p such that Eins is PAC. Proof:

Choose algebraically independent elements t1 , t2 , t3 , t4 , t5 over Fp .

216

Chapter 11. Pseudo Algebraically Closed Fields

Let C be the curve defined over K = Fp (t1 , t2 , t3 , t4 , t5 ) in Remark 11.7.7. Then, C is a nonconstant curve over F/Fp for every regular field extension F of Fp that contains K. By Proposition 11.7.1, C(K) is a finite set. By induction we construct two ascending towers of fields K = E1 ⊆ E2 ⊆ E3 ⊆ · · · and F1 ⊆ F2 ⊆ F3 ⊆ · · · and for each positive integer m we enumerate the absolutely irreducible varieties which are defined over Em in a sequence, Vm1 , Vm2 , Vm3 , . . . such that (6a) Em and Fm are finitely generated regular extensions of K, (6b) Fm is a finite purely inseparable extension of Em , (6c) C(Em ) = C(K), and (6d) Vij (Fm ) 6= ∅ for i, j = 1, . . . , m − 1. Indeed, suppose E1 , . . . , Em−1 , F1 , . . . , Fm−1 , and Vij for i < m and all j have been defined such that they satisfy (6). Let V be the direct product of Vij for i, j = 1, . . . , m − 1. It is an absolutely irreducible variety defined 0 over Em−1 . Let x be a generic point of V over Em−1 . Then Em = Em−1 (x) is a finitely generated regular extension of Em−1 and therefore also of K. 0 (instead K, F ) and construct an extension Apply Lemma 11.7.3 to Em−1 , Em 0 0 such that Em /Em is a finite purely Em of Em−1 which is contained in Em inseparable extension and C(Em ) = C(Em−1 ). By (6c) for m − 1 we have 0 is a regular extension of Em−1 , it is linearly C(Em ) = C(K). Since Em 0 ˜ disjoint from Fm−1 K over Em−1 . Hence, Fm = Em Fm−1 is linearly disjoint ˜ from Fm−1 K over Fm−1 . 0 Em

Fm

Em Em−1 K

Fm−1

˜ Fm−1 K ˜ K

˜ over K, Fm is linearly disjoint from K ˜ Since Fm−1 is linearly disjoint from K over K. Thus, Fm is a regular extension of K. By construction, Fm is a finite Vij (Fm ) 6= ∅ for i, j = 1 . . . , m − 1. purely inseparable S∞ extension of EmSand ∞ Let E = m=1 Em and F = m=1 Fm . Then E and F are countable regular extensions of K and F is purely inseparable over E. Hence, in order to prove that Eins is PAC it suffices, by Theorem 11.2.3, to prove that each absolutely irreducible variety V defined over E has an F -rational point. Indeed, if V is such a variety, then V = Vij for some i and j. Let m = max{i, j} + 1. By (6d), V has an Fm -rational point, which is, of course, an F -rational point. Finally, each point of C(E) belongs to C(Em ) for some m and therefore, by (6c), to C(K). Thus, C(E) = C(K) is a finite set. By Proposition 11.1.1, E is not PAC.

Exercises

217

Problem 11.7.9: Does there exists a finitely generated field extension E of Fp such that E is not PAC but Eins is PAC?

Exercises 1. Let V be an absolutely irreducible variety which is defined over a PAC field K. Suppose that A and B are Zariski K-closed subset of V such that V (K) = A(K) ∪ B(K). Use Proposition 11.1.1 to prove that V = A or V = B. Conclude that V (K) is connected in the Zariski-topology. 2. By the Artin-Schreier theorem, the only torsion that occurs in the absolute Galois group of a field K comes from real closed fields. Prove that if K is PAC, then Gal(K) is torsion free. 3. A more elementary characterization of PAC fields, just as useful as Theorem 11.2.3, can be derived from Lemma 10.4.1. Prove that a field K is PAC if and only if for each nonconstant absolutely irreducible polynomial f ∈ K[T1 , . . . , Tr , X], separable in X, and for every nonzero polynomial g ∈ K[T1 , . . . , Tr ], there exist a1 , . . . , ar , b ∈ K such that f (a, b) = 0 and g(a) 6= 0. 4. Prove the following assertion without using descent: If K is a PAC field of characteristic p > 0, then Kins is also a PAC field. 5. Let Γ be the plane curve defined over F3 by the equation X04 + X14 − X24 − X02 X22 + X13 X2 = 0. Prove that Γ is absolutely irreducible and has no F3 -rational point. Hint: Show that Γ is smooth. 6. (a) Let f ∈ L[X1 , . . . , Xn ] be a nonconstant homogeneous polynomial with at least one absolutely irreducible factor of multiplicity 1. Prove that if a ∈ L× , then f (X1 , . . . , Xn ) − a is absolutely irreducible. (b) Let M/L be a finite Galois extension with basis w1 , . . . , wn . Describe the norm map normM/L : M → L as a homogeneous polynomial of degree n in X1 , . . . , Xn . (c) Combine (a) and (b) to show that if L is a finite separable extension of a PAC field K, then normM/L is surjective. Thus, show directly property (b) of Lemma 11.6.7. 7. This exercise suggest an alternative way to achieve some consequences of Corollary 11.5.5. (a) Use Eisenstein’s criterion to prove that the polynomial f (X, Y ) = (X q −X)(Y q −Y )+1 defined over the field Fq (q a prime power) is absolutely irreducible and has no Fq -rational zeros. Deduce that a PAC field is infinite. (b) (Ax) Let K be a field with a valuation v whose residue class field has q elements. Let π ∈ K with v(π) > 0 and show that (X q − X − 1)(Y q − Y − 1) − π = 0 defines an absolutely irreducible variety with no K-rational point. Thus, show directly that K is not a PAC field. Conclude, in particular, that fields finitely generated over their prime fields are not PAC fields.

218

Chapter 11. Pseudo Algebraically Closed Fields

8. Use the idea included in the proof of Corollary 11.5.7 to show that Gal(Qp ) is pronilpotent for no prime p. 9. Let K be a field, L an infinite extension of K, and f ∈ L[X1 , . . . , Xn ]. Prove that if f (a) ∈ K for each a ∈ Ln , then the coefficients of f belong to K. Hint: For n = 1 use Cramer’s rule to compute the coefficients of f . Then continue by induction on n.

Notes Apparently PAC fields appear for the first time in J. Ax’s papers [Ax1] and [Ax2]. Ax observes that the Riemann Hypothesis for curves implies that nonprincipal ultraproducts of finite fields are PAC fields. From this he establishes a recursive decision procedure for the theory of finite fields. Following a suggestion of the second author, Frey introduces in [Frey] the name PAC for fields over which each variety has rational points. He also proves a place theoretic predecessor to Lemma 11.2.1. Remark 11.2.1, due to J´anos Koll´ar, settles Problem 11.2.10 of the second edition. Our proof of Proposition 11.4.1 appears to be simpler than Tamagawa’s. In [Frey] Frey uses the theory of homogeneous spaces attached to elliptic curves and Tate’s theory of bad reduction in order to prove Corollary 11.5.5 for real valuations. Prestel showed the authors the elementary direct proof of the Corollary for arbitrary valuations (private communication). A variation of that proof serves also for Proposition 11.5.3.

Chapter 12. Hilbertian Fields David Hilbert proved his celebrated irreducibility theorem during his attempt to solve a central problem of Galois theory: Is every finite group realizable over Q? He proved that a general specialization of the coefficients of the general polynomial of degree n to elements of Q gives a polynomial whose Galois group is Sn . Further, if f ∈ Q[T1 , . . . , Tr , X] is an irreducible polynomial, then there exist a1 , . . . , ar ∈ Q such that f (a, X) remains irreducible. This result is now known as Hilbert’s irreducibility theorem. Since then, many more finite groups have been realized over Q. Most of those have been realized via Hilbert’s theorem. This has brought the theorem to the center of the theory of fields. Various alternative proofs of the irreducibility theorem apply to other fields (including all infinite finitely generated fields). We call them Hilbertian fields. We give several reductions of the irreducibility theorem, and one especially valuable result (Corollary 12.2.3): Let K be a Hilbertian field, L a finite separable extension of K, and f ∈ L[T1 , . . . , Tr , X] an irreducible polynomial. Then there exist a1 , . . . , ar ∈ K such that f (a, X) is irreducible in L[X]. Chapters 13 and 15 give a proof of the Hilbertian property for most known Hilbertian fields. This lays the foundation to a subject central to the book, the model theory of PAC fields.

12.1 Hilbert Sets and Reduction Lemmas Consider a field K and two sets T1 , . . . , Tr and X1 , . . . , Xn of variables. Let f1 (T, X), . . . , fm (T, X) be polynomials in X1 , . . . , Xn with coefficients in K(T). Assume these are irreducible in the ring K(T)[X]. For g ∈ K[T] a nonzero polynomial, denote the set of all r-tuples (a1 , . . . , ar ) ∈ K r with g(a) 6= 0 and f1 (a, X), . . . , fm (a, X) defined and irreducible in K[X] by HK (f1 , . . . , fm ; g). Call HK (f1 , . . . , fm ; g) a Hilbert subset of K r . If in addition n = 1 and each fi is separable in X, call HK (f1 , . . . , fm ; g) a separable Hilbert subset of K r . A Hilbert set (resp. separable Hilbert set) of K is a Hilbert subset (resp. separable Hilbert subset) of K r for some positive integer r. The intersection of finitely many Hilbert subsets of K r is again a Hilbert subset of K r . Hence, if each Hilbert subset of K r is nonempty, then each Hilbert subset of K r is Zariski K-dense in K r . By Lemma 10.2.5, each Hilbert subset of K r is Zariski dense in K r . The same holds for separable Hilbert sets. We say that K is Hilbertian if each separable Hilbert set of K is nonempty. In particular, a Hilbertian field must be infinite.

220

Chapter 12. Hilbertian Fields

The next lemmas reduce the infiniteness of arbitrary Hilbert sets (resp. separable Hilbert sets) to the infiniteness of special Hilbert sets (resp. separable Hilbert sets). Lemma 12.1.1: Each Hilbert subset (resp. separable Hilbert subset) HK (f1 , . . . , fm ; g) of K r contains a Hilbert subset (resp. separable Hilbert 0 subset) HK (f10 , . . . , fm ; g 0 ) of K r with fi0 irreducible in K[T, X] and 0 / K[T], i = 1, . . . , m. fi ∈ Proof: By assumption, fi is irreducible in K(T)[X]. Hence, at least one Xj occurs in fi . Multiply fi (T, X) by a polynomial, gi (T), to ensure that its coefficients lie in K[T]. Then divide the resulting polynomial by the greatest common divisor, di (T), of its coefficients. Since K[T, X] has unique factorization, we obtain an irreducible polynomial fi0 ∈ K[T, X]. Now put g 0 = 0 ; g 0 ) ⊆ HK (f1 , . . . , fm ; g). g · g1 d1 . . . gm dm and conclude that HK (f10 , . . . , fm Finally, suppose n = 1 and each fi (T, X) is separable in X. Then so is each fi0 (T, X). Lemma 12.1.2: Suppose every Hilbert set of K of the form HK (f1 , . . . , fm ; g) with fi irreducible in K[T, X1 , . . . , Xn ], i = 1, . . . , m, is nonempty. Then every Hilbert set of K is nonempty. Proof: Start with irreducible polynomials fi ∈ K[T1 , . . . , Tr , X1 , . . . , Xn ], i = 1, . . . , m, and a nonzero polynomial g ∈ K[T1 , . . . , Tr ]. By assumption, there exists a1 ∈ K with fi (a1 , T2 , . . . , Tr , X) irreducible, i = 1, . . . , m, and g(a1 , T2 , . . . , Tr ) 6= 0. Repeat this procedure r times to find a1 , . . . , ar ∈ K with fi (a, X) irreducible and g(a) 6= 0, i = 1, . . . , m. By Lemma 12.1.1, every Hilbert set of K is nonempty. Lemma 12.1.3: Every Hilbert subset of K contains a Hilbert set of the form H(f1 , . . . , fm ; g), where fi is an irreducible polynomial in K[T, X] with degX (fi ) ≥ 1, i = 1, . . . , m. Proof: Let f ∈ K[T, X1 , . . . , Xn ] be an irreducible polynomial with f ∈ / K[T ] and 0 6= g0 ∈ K[T ]. By Lemma 12.1.1, it suffices to find irreducible polynomials h1 , . . . , hr ∈ K[T, Y ] r K[T ] and 0 6= g ∈ K[T ] with HK (h1 , . . . , hr ; g) ⊆ HK (f ; g0 ). Indeed, let d > max1≤j≤n degXj (f ). Apply the Kronecker substitution Sd : SK[T ] (n, d) → SK[T ] (1, dn ) on f (Section 11.3). Consider the factorization of Sd (f ) into irreducible factors of K[T, Y ]: Y hi (T, Y ). (1) Sd (f )(T, Y ) = i∈I

The polynomials Sd (f )(T, Y ) and f (T, X1 , . . . , Xn ) have the same coefficients in K[T ]. Since f is irreducible in K[T, X1 , . . . , Xn ], the greatest common divisor of its coefficients in K[T ] is 1, so none of the hi is in K[T ]. Let I = J ∪· J 0 be a nontrivialQpartition of I. The Q exponent to which Y appears in each of the polynomials i∈J hi (T, Y ) and i∈J 0 hi (T, Y ) does not

12.1 Hilbert Sets and Reduction Lemmas

221

exceed dn − 1. Since Sd is bijective on SK[T ] (n, d) (Section 11.3), there exist polynomials pJ , pJ 0 ∈ K[T, X] with degXj (pJ ), degXj (pJ 0 ) < d, j = 1, . . . , n, and Y Y hi (T, Y ) and Sd (pJ 0 )(T, Y ) = hi (T, Y ). (2) Sd (pJ )(T, Y ) = i∈J 0

i∈J

Note: The product pJ (T, X)pJ 0 (T, X) contains a monomial of the form gJ (T )X1ν1 · · · Xnνn in which at least one of the νj exceeds d − 1. Otherwise, the relation Sd (f ) = Sd (pJ ) · Sd (pJ 0 ) = Sd (pJ pJ 0 ) would imply a nontrivial factorization of the irreducible polynomial f into pJ pJ 0 . Let g be the product of g0 with all gJ (one for each nontrivial partition). Let a be an element of K with each hi (a, Y ) irreducible in K[Y ] and g(a) 6= 0. We show f (a, X) is also irreducible. Indeed, assume f (a, X) = q(X)q 0 (X) is a nontrivial factorization of f (a, X) in K[X], then (1) implies Y

hi (a, Y ) = Sd (f )(a, Y ) = Sd (q)(Y ) · Sd (q 0 )(Y ).

i∈I

Hence, (2) implies a nontrivial partition I = J ∪· J 0 of I with Sd (q)(Y ) =

Y

hi (a, Y ) = Sd (pJ )(a, Y )

and

i∈J

Sd (q 0 )(Y ) =

Y

hi (a, Y ) = Sd (pJ 0 )(a, Y ).

i∈J 0

Hence, q(X) = pJ (a, X) and q 0 (X) = pJ 0 (a, X). Thus, pJ (a, X)pJ 0 (a, X) = f (a, X). The left hand side contains the monomial gJ (a)X1ν1 · · · Xnνn in which at least one νj exceeds d − 1, while f (a, X) contains no such monomial. Therefore, f (a, X) is irreducible. Lemma 12.1.4: Let H = HK (g1 , . . . , gm ; h) be a Hilbert subset of K r with gi ∈ K[T1 , . . . , Tr , X] irreducible and degX (gi ) ≥ 1, i = 1, . . . , m. Then H contains a Hilbert set of the form HK (f1 , . . . , fm ) in which fi is a monic irreducible polynomial in K[T, X] of degree at least 2 in X, i = 1, . . . , m. If g1 , . . . , gm are separable in X, then so are f1 , . . . , fm . Moreover, suppose a ∈ K r and none of the polynomials fi (a, X) has a root in K. Then none of the polynomials gi (a, X) with degX (gi ) ≥ 2 has a root in K. Proof: Let ci be the leading coefficient of gi as a polynomial in X, ni = degX (gi ), and q = hc1 · · · cm . Choose a prime number l 6= char(K) with q not an lth power in K[T]. By assumption, ni ≥ 1. If m = 0 or ni = 1, let fi (T, X) = X l − q(T). If ni ≥ 2, let fi (T, X) = q(T)ni ci (T)−1 gi (T, q(T)−1 X).

222

Chapter 12. Hilbertian Fields

Then fi (T, X) = X ni + bi,ni −1 (T)X ni −1 + q(T)

nX i −2

bij (T)X j .

j=0

with bij ∈ K[T]. In each case fi (T, X) is monic in X and irreducible in K[T, X] (see [Lang7, p. 297, Lemma 9.1] for the case m = 0 or ni = 1). We prove that HK (f1 , . . . , fm ) ⊆ H. Let a be in K r with f1 (a, X), . . . , fm (a, X) irreducible in K[X]. Consider an i between 1 and m. Suppose first ni = 1. Then X l − q(a) is irreducible. Hence, q(a) 6= 0. Therefore, h(a) 6= 0, ci (a) 6= 0, and gi (a, X) = ci (a)X + bi for some bi ∈ K. Thus, gi (a, X) is irreducible. Now suppose ni ≥ 2. Then q(a) 6= 0 (otherwise fi (a, X) = X ni + bi,ni −1 (a)X ni is reducible) and gi (a, q(a)−1 X) = q(a)−ni ci (a)fi (a, X) is irreducible. Hence, so is gi (a, X). Similarly, if ni ≥ 2 and fi (a, X) has no root in K, neither has gi (a, X). Corollary 12.1.5: Suppose each Hilbert set of the form HK (f1 , . . . , fm ) with fi ∈ K[T, X] monic and of degree at least 2 in X is nonempty. Then every Hilbert set of K is nonempty. Proof: By Lemma 12.1.4 for r = 1, each Hilbert subset of K of the form HK (g1 , . . . , gm ; h) with gi ∈ K[T, X] irreducible and degX (gi ) ≥ 1, i = 1, . . . , m, is nonempty. Hence, by Lemma 12.1.3, each Hilbert subset of K is nonempty. Consequently, by Lemma 12.1.2, every Hilbert set of K is nonempty. The corresponding result for separable Hilbert sets is also true. We postpone its proof to Section 13.2. Here we show that separable Hilbert subsets of K r contain especially simple separable Hilbert subsets of K r . Lemma 12.1.6: Let H be a separable Hilbert subset of K r . Then there exists an irreducible polynomial f ∈ K[T1 , . . . , Tr , X], separable, monic, and of degree at least 2 in X with HK (f ) ⊆ H. Proof: Use Lemma 12.1.1 to assume that H = HK (f1 , . . . , fm ; g) with f1 , . . . , fm ∈ K[T, X] r K[T] irreducible and separable in X and 0 6= g ∈ K[T]. For each i, 1 ≤ i ≤ m, let xi be a root of fi (T, X) in K(T)s . Then the finite separable extension K(T, x1 , . . . , xm ) of K(T) is contained in a separable extension K(T, y) of K(T) of degree at least 2. Assume without loss, y is integral over K[T]. Let h1 = irr(y, K(T)). Then h1 (T, X) is separable, monic, and of degree at least 2 in X. Rewrite xi as ppi 0(T,y) (T) with p1 , . . . , pm ∈ K[T, X] and 0 6= p0 ∈ K[T]. Let ui (T, xi ) be the leading coefficient of irr(y, K(T, xi )). Write normK(T,xi )/K(T) (ui (T, xi )) as qq0i (T) (T) with q0 , q1 , . . . , qm ∈ K[T]. Finally, let ri (T) be the leading coefficient of fi (T, X),

12.2 Hilbert Sets under Separable Algebraic Extensions

223

i = 1, . . . , m. Define g1 to be the product g · p0 · q0 · · · qm · r1 · · · rm . We show that HK (h1 ; g1 ) ⊆ HK (f1 , . . . , fm ; g). Indeed, if a ∈ HK (h1 ; g1 ) and c is a root of h1 (a, X), then (3) Moreover, with bi =

[K(c) : K] = [K(T, y) : K(T)]. pi (a,c) p0 (a) ,

the nonvanishing of g1 (a) implies

[K(bi ) : K] ≤ [K(T, xi ) : K(T)] and [K(c) : K(bi )] ≤ [K(T, y) : K(T, xi )]. Multiply the terms in these two inequalities and apply (3) to conclude that they are equalities. Therefore, f1 (a, X), . . . , fm (a, X) are irreducible over K. By Lemma 12.1.4, there exists an irreducible polynomial f ∈ K[T, X], separable, monic, and of degree at least 2 in X, with HK (f ) ⊆ HK (h1 ; g1 ).

12.2 Hilbert Sets under Separable Algebraic Extensions Let L/K be a finite separable extension. We prove every Hilbert subset of Lr contains a Hilbert subset of K r . Lemma 12.2.1: Let L be a separable extension of degree d of an infinite field K and σ0 , . . . , σd−1 distinct representatives of the cosets of Gal(L) in Gal(K). Suppose f ∈ L[X1 , . . . , Xn ] is a nonconstant polynomial. Then there exist c1 , . . . , cn ∈ L with f (X1 + c1 , . . . , Xn + cn )σi , i = 0, . . . , d − 1, ˜ 1 , . . . , Xn ]. pairwise relatively prime in K[X Proof: Let θ be a primitive element for L/K. Choose algebraically independent elements tik , i = 0, . . . , d − 1 and k = 1, . . . , n, over K. For every i and k consider d−1 X (θσi )j tjk . uik = j=0

Write ui = (ui1 , . . . , uin ) and u = (u0 , . . . , ud−1 Q ). The linear transformation t → u has as determinant the nth power of i<j (θσj − θσi ) 6= 0. Thus, the ˜ so the uik are tik are linear combinations of the uik with coefficients in K; algebraically independent over K. ˜ factors. Since fµ is Write f = f1 · · · fm , a product of K-irreducible σ nonconstant and the uik are algebraically independent, fµσi (ui ) 6= fν j (uj ) for i 6= j and each µ and ν, 1 ≤ µ, ν ≤ m. Therefore YY (fµσi (ui ) − fνσj (uj )) 6= 0, h(t) = i<j µ,ν

so there exist aik ∈ K with h(a) 6= 0. Let ck =

d−1 X j=0

ajk θj ,

k = 1, . . . , n.

224

Chapter 12. Hilbertian Fields

˜ The K-specialization t → a maps uik to cσk i . Hence fµ (c)σi 6= fν (c)σj for i 6= j. Therefore, fµ (X + c)σi 6= fν (X + c)σj for all i, j, µ, ν with 0 ≤ i < j ≤ d − 1 and 1 ≤ µ, ν ≤ m. Since f1 (X + c)σi , . . . , fm (X + c)σi are exactly the ˜ f (X + c)σ0 , . . . , f (X + c)σd−1 are irreducible factors of f (X + c)σi in K[X], relatively prime in pairs. Lemma 12.2.2: Let L be a finite separable extension of a field K and f ∈ L(T1 , . . . , Tr )[X1 , . . . , Xn ] irreducible. Then, there exists an irreducible p ∈ K(T)[X] with HK (p) ⊆ HL (f ). If n = 1 and f is separable in X, then p is separable in X. Proof: Let S be a set of representatives of the left cosets of Gal(L) in Gal(K). Denote the algebraic closure of K(T) by F . First consider the case Q where the f σ , with σ ∈ S, are pairwise relatively prime in F [X]. Let p = σ∈S f σ . Since Gal(K) fixes the coefficients of p, p ∈ K(T)[X]. Moreover, if q is an irreducible factor of p in K(T)[X], then one of the f σ ’s (and therefore all) ˜ divides q in K(T)[X]. Since the f σ are pairwise relatively prime, p divides q. Hence, p is irreducible in K(T)[X]. In addition, if n = 1 and f is separable in X, then p is separable in X. Let a1 , . . . , ar be elements of K such that p(a, X) is defined and irreducible in K[X]. If f (a, X) Q were a nontrivial decomposition Q = g(X)h(X) in L[X], then p(a, X) = ( σ∈S g σ (X))( σ∈S hσ (X)) would be a nontrivial decomposition in K[X], which is a contradiction. Therefore, f (a, X) is irreducible in L[X] and HK (p) ⊆ HL (f ). In the general case apply Lemma 12.2.1 to find c1 , . . . , cr ∈ L(T) with f (T, X + c)σ pairwise relatively prime in F [X], σ running over S. Let g(T, X) = f (T, X + c). Irreducibility of g(a, X) is equivalent to that of f (a, X) for each a ∈ K r such that both polynomials are defined. Thus, the lemma follows from the first part of the proof. Corollary 12.2.3: Let L be a finite separable extension of a field K. Then every Hilbert subset (resp. separable Hilbert subset) of Lr contains a Hilbert subset (resp. separable Hilbert subset) of K r . In particular, if K is Hilbertian, so is L. Remark 12.2.4: The converse of Corollary 12.2.3 is false: Example 13.9.5 gives a field K with an empty Hilbert set and a finite separable extension L whose Hilbert sets are all nonempty.

12.3 Purely Inseparable Extensions Corollary 12.2.3 is false if L/K is inseparable. For example, let K = Fp (t), with t transcendental over Fp , and let L = K(t1/p ). Then X p −T is irreducible over L. Yet every a ∈ K is a pth power in L, so X p − a is reducible over L. Hence, HL (X p − T ) contains no elements of K. Still, X p − t1/p is irreducible over L, so HL (X p − T ) is nonempty. The following results generalize this

12.3 Purely Inseparable Extensions

225

observation. They assert: if every Hilbert set of K is nonempty, then every Hilbert set of L is nonempty. First we list simple properties of purely inseparable extensions. Lemma 12.3.1: Let L/K be a purely inseparable extension of fields of characteristic p. (a) For each irreducible g ∈ K[X] there exist an irreducible f ∈ L[X] and e e ≥ 0 with g = f p . e e (b) Consider f ∈ L[X]. Suppose Lp ⊆ K and g = f p is irreducible in K[X]. Then f is irreducible in L[X]. e (c) Let f ∈ L[X] be a non pth power with g = f p irreducible in K[X]. Then f is irreducible in L[X]. (d) If f ∈ L[X] is irreducible and e ≥ 0 is the least nonnegative integer with e g = f p ∈ K[X], then g is irreducible in K[X]. Proof of (a):

Let f1 · · · fm be a factorization of g into irreducible factors k

in L[X]. Choose k ≥ 0 such that fip ∈ K[X], i = 1, . . . , m. The relation k

k

k

k

p g p = f1p · · · fm and the unique factorization in K[X] imply that f1p = g r for some positive integer r. Unique factorization over L[X] then gives a k

positive integer s with g = f1s . Then f1p = f1rs implies that s = pe , so e g = f1p for some positive integer e. Proof of (b): Let f1 · · · fm be a factorization of f into irreducible factors in e pe is a factorization of g in K[X]. Since g is irreducible L[X]. Then f1p · · · fm over K, we have m = 1. Therefore, f is irreducible over L. Proof of (c): e0

e0

By (a), g = hp e

for some irreducible h ∈ L[X] and e0 ≥ 0. e0 −e

Hence, hp = f p , so e0 ≥ e. Therefore, hp = f . The assumption on f implies that e0 = e and f = h. Thus, f is irreducible in L[X]. Proof of (d): Let g1 · · · gm be a factorization of g into irreducible factors ei in K[X]. By (a), gi = fip for some irreducible polynomial fi in L[X] and e e 1 pem ei ≥ 0. Hence, f p = f1p · · · fm . Therefore, fi = f and pe1 +· · ·+pem = pe . pei = gi ∈ K[X], so by assumption, ei ≥ e. It follows that In addition, f m = 1 and g is irreducible in K[X]. Lemma 12.3.2: Let K be a field with char(K) = p > 0, L a purely inseparable extension of K, and f an irreducible polynomial in L[T1 , . . . , Tr , X] with degX f ≥ 1. (a) If f is separable in X, then HL (f ) contains a separable Hilbert subset of Kr. (b) If all occurrences of T1 , . . . , Tr , X in f are powers of p, then HL (f ) contains a Hilbert subset of K r . e

Proof: Let e be the least nonnegative integer with g = f p ∈ K[T, X]. By Lemma 12.3.1(d), g is irreducible in K[T, X]. Distinguish between two cases to find a Hilbert subset of K contained in HL (f ).

226

Chapter 12. Hilbertian Fields

Case A: f is separable with respect to X. Consider a ∈ K with g(a, X) irreducible in K[X] and f (a, X) separable. Then, f (a, X) is not a pth power in L[X]. By Lemma 12.3.1(c), f (a, X) is irreducible in L[X]. Case B: All occurrences of T1 , . . . , Tr , X in f are powers of p. By assumption, f is not a pth power in L[T, X], so at least one coefficient c(T) of f (T, X) is not a pth power in L[T]. In particular, L is an infinite field. Hence, K is also infinite. such that c(a) ∈ / Claim: There is a nonempty Zariski L-open subset U of Ar P ci T1i1 p · · · Trir p Lp for all a ∈ U (L). To prove the claim, write c(T) = with i = (i1 , . . . , ir ), 0 ≤ i1 ≤ d1 , . . . , 0 ≤ ir ≤ dr , and ci ∈ Lr . Let d = (d1 + 1) · · · (dr + 1). Assume there are d r-tuples (bk1 , . . . , bkr )

(1)

with 0 ≤ kj ≤ d1 and bkj 6= bkj0 if j 6= j 0 and 1 ≤ j, j 0 ≤ r such that c(bk1 , . . . , bkr ) = λp with λ ∈ L. Thus, there is a λk1 ,...,kr ∈ L with (2)

X

ci bki11p · · · bkirrp = λpk1 ,...,kr

i

Consider (2) as a system of d linear equations in the ci ’s. The coefficient matrix of (2) is a Kronecker product of r Vandermonde matrices. The determinant of this matrix is D =det(bki11p · · · bkirrp ) =

r Y

i

0

det(bkjj )dj p

j=1

=

r Y Y

0

(bkj0 − bkj )dj p

j=1 j<j 0

where d0j = (dj +1)−1 (d1 +1) · · · (dr +1), j = 1, . . . , r (See [Bourbki, Algebra, p. 534] for the determinant of the Kronecker product of two matrices, from which the determinant of the Kronecker product of r matrices can be derived.) Thus, D is a pth power of a nonzero element of L. Applying Cramer’s rule to the system (2), we conclude that ci ∈ Lp for each i. This contradiction to the assumption we made on c(T) proves that there are at most d − 1 r-tuples (1). Let B be the set of all these r-tuples. Define, / B and xj 6= xj 0 if j 6= j 0 }. U = {(x1 , . . . , xr ) ∈ Ar | (x1 , . . . , xr ) ∈ Then U is a Zariski L-open subset of Ar satisfying our claim. Now we may use the claim to choose a ∈ K r with c(a) 6∈ Lp (therefore f (a, X) is not a pth power in L[X]) and g(a, X) irreducible in K[X]. By Lemma 12.3.1(c), f (a, X) is irreducible in L[X].

12.3 Purely Inseparable Extensions

227

Proposition 12.3.3: Let L/K be an algebraic extension of fields of a finite separable degree. Then each separable Hilbert subset of Lr contains a separable Hilbert subset of K r . Thus, if K is Hilbertian, then so is L. Proof: By assumption, L is a purely inseparable extension of a finite separable extension of K. Apply Corollary 12.2.3 to assume L/K is purely inseparable. Let H be a separable Hilbert subset of Lr . By Lemma 12.1.6, H contains HL (f ) with irreducible f ∈ L[T, X] separable in X. From Lemma 12.3.2, HL (f ) contains a separable Hilbert subset of K r . Lemma 12.3.4: Let K be a field of positive characteristic p, L an extension r of K with Lp ⊆ K for some r ≥ 0, and f1 , . . . , fm irreducible polynomials in L[T, X]. Suppose degX fi ≥ 1 and fi is separable in T . Suppose in addition that each Hilbert subset of K is nonempty. Then there exists γ ∈ L such that HL (fi (T + γ, X)) contains a Hilbert subset of K, i = 1, . . . , m. Proof: Let i be between 1 and m. At least one of the coefficients ci (T ) of fi (T, X) has nonzero derivative c0i (T ). Choose a nonzero coefficient γi of e−1 e c0i (T ). Let e be the least nonnegative integer with Lp ⊆ K. Then Lp 6⊆ K, Sm −pe−1 pe−1 pe−1 pe−1 6⊆ γi K, i = 1, . . . , m. Since K is infinite, L 6⊆ i=1 γi K. so L pe−1 ∈ / K. This implies Hence, there exists λ ∈ L independent of i with (γi λ) e−1 6∈ K[T ]. (c0i (T )λ)p The Taylor expansion ci (T + λZ) = ci (T ) + c0i (T ) · λZ + · · · shows that e−1 e ci (T + λZ)p 6∈ K[T, Z]. Hence, hi (T, Z) = ci (T + λZ)p is not a pth power of a polynomial belonging to K[T, Z]. Therefore, X p − hi (T, Z) is irreducible over K. By assumption, there exists b ∈ K, independent of i, with X p − hi (T, b) irreducible over K. Thus, e is the least nonnegative integer for e which ci (T + λb)p is in K[T ]. Hence, e is the least nonnegative integer for e which gi (T, X) = fi (T + λb, X)p ∈ K[T, X]. By Lemma 12.3.1(d), gi (T, X) is irreducible in K[T, X]. If for a ∈ K, the polynomial gi (a, X) is irreducible in K[X], then from Lemma 12.3.1(b), fi (a + λb, X) is irreducible in L[X]. Thus, HK (gi ) ⊆ HL (fi (T + λb, X)). Proposition 12.3.5: Let K be a field with all Hilbert subsets nonempty and L an algebraic extension of K. Then in each of the following cases all Hilbert sets of L are nonempty. r (a) char(K) = p > 0 and Lp ⊆ K for some r ≥ 0. (b) L is a finite extension of K. Proof of (a): Consider a Hilbert subset HL (f1 , . . . , fn ) of L with fi ∈ L[T, X] irreducible and degX fi ≥ 2, i = 1, . . . , n. Reorder the fi ’s, if necessary, to assume the following: For i = 1, . . . , m, fi is separable in X or inseparable in both T and X. For i = m + 1, . . . , n, fi is separable in T . Lemma 12.3.4 gives γ in L such that HL (fi (T + γ, X)) contains a Hilbert subset Hi of K.

228

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For each i between 1 and m, the polynomial fi (T + γ, X) is either separable in X or inseparable in both T and X. By Lemma 12.3.2, HL (fi (T + γ, X)) contains a Hilbert subset Hi of K, so H = H1 ∩ · · · Hm is a Hilbert subset of K which is contained in HL (f1 , . . . , fn ). By Corollary 12.1.5, every Hilbert set of L is nonempty. Proof of (b): Combine (a) with Lemma 12.2.2.

12.4 Imperfect fields Suppose char(K) = p 6= 0 and every Hilbert set of K is nonempty. Then there exists a ∈ K with X p − a is irreducible. Thus, K is imperfect. We show that, conversely, every Hilbert set of an imperfect Hilbertian field is nonempty. Lemma 12.4.1: Let char(K) = p > 0 and let f ∈ K[X] r K p [X] be irrem ducible and monic. Then f (X p ) is irreducible in K[X] for each m ≥ 0. Proof: By induction it suffices to prove the Lemma for m = 1. If x is a root of f (X p ), then y = xp is a root of f (X). We show [K(x) : K(y)] = p. If not, then K(x) = K(y) and irr(x, K) = X n + bn−1 X n−1 + · · · + b0 , where n = [K(x) : K] = [K(y) : K] = deg(f ). Therefore, y is a root of g(X) = X n + bpn−1 X n−1 + · · · + bp0 . Hence, f = g and f ∈ K p [X], contrary to our hypotheses. It follows, [K(x) : K] = p · deg(f ) = deg(f (X p )). Consequently, f (X p ) is irreducible. Lemma 12.4.2: Let K be an imperfect field of characteristic p. Suppose h ∈ K[T ] is not a pth power. Then: (a) The additive group K has infinitely many congruence classes modulo K p . (b) All a ∈ K with h(T + a) ∈ K p [T ] are congruent modulo K p . (c) For each b ∈ K with h(T + b) 6∈ K p [T ] only finitely many c ∈ K satisfy h(cp + b) ∈ K p . Proof of (a): Since K is imperfect, K/K p is a nonzero vector space over the infinite field K p . Hence, K/K p is infinite. Proof of (b): Let a, b ∈ K with h(T + a), h(T + b) ∈ K p [T ]. Since h(T + a) is not a pth power, h(T + a) = g(T p ) + f (T ) where g, f ∈ K p [T ] but p - deg(f ). Then, with c = b − a, we have h(T + b) = g(T p + cp ) + f (T + c). Let f (T ) = dpm T m + dpm−1 T m−1 + · · · + dp0 with dm 6= 0. Then, p - m and the coefficient of T m−1 in f (T +c) is mcdpm +dpm−1 . Since h(T +b) and g(T p +cp ) are in K p [T ], this element belongs to K p . Finally, mp = m 6= 0. Therefore, c is a pth power. Proof of (c): Write h(T + b) 6∈ K p [T ] in the form h(T + b) = f1 (T ) + f2 (T )u2 + · · · + fr (T )ur ,

Notes

229

where r ≥ 2, u2 , . . . , ur ∈ K, 1, u2 , . . . , ur linearly independent over K p , f1 , . . . , fr ∈ K p [T ], and f2 , . . . , fr 6= 0. If c ∈ K and h(cp + b) ∈ K p , then f1 (cp ) + f2 (cp )u2 + · · · + fr (cp )ur ∈ K p . Therefore, f2 (cp ) = · · · = fr (cp ) = 0. The number of such c’s is finite. Proposition 12.4.3 ([Uchida]): Let K be a field satisfying these conditions: (1a) Every separable Hilbert subset of K is nonempty. (1b) If char(K) > 0, then K is imperfect. Then every Hilbert set of K is nonempty. In particular, K is Hilbertian. Proof: If char(K) = 0, then each Hilbert set of K is separable. Corollary 12.1.5 settles this case, so suppose char(K) = p > 0 and K is imperfect. In this case, Corollary 12.1.5 shows it suffices to prove that each Hilbert set of the form HK (f1 , . . . , fm ) with f1 , . . . , fm ∈ K[T, X] irreducible, monic, and of degree at least 2 in X, is nonempty. Assume f1 , . . . , fl are not separable in X and fl+1 , . . . , fm are separable in X. For each i, 1 ≤ i ≤ l there exists gi ∈ K[T, X] irreducible, separable and monic in X and qi , a power of p different from 1, with fi (T, X) = gi (T, X qi ). Since fi (T, X) is irreducible, gi has a coefficient hi ∈ K[T ] which is not a pth power. Choose ai ∈ K with hi (T + ai ) ∈ K p [T ] if there exists any, otherwise let ai = 0. By (1b), K is infinite. Hence, by Lemma 12.4.2(a), Sl / K p [T ], there exists b ∈ K r i=1 (ai + K p ). By Lemma 12.4.2(b), hi (T + b) ∈ p so gi (T +b, X) ∈ K(X)[T ] r K(X) [T ], i = 1, . . . , l. Thus (Lemma 12.4.2(c)), the set C of all elements c ∈ K with hi (cp + b) ∈ K p for some i, 1 ≤ i ≤ l, is Q finite. Let d(T ) = c∈C (T − c) and let gi = fi , i = l + 1, . . . , m. By Lemma 12.4.1, all gi (T p + b, X) are irreducible, monic, and separable in X. By (1a), there exists a ∈ K such that d(a) 6= 0 and gi (ap + b, X) is separable and irreducible in K[X], i = 1, . . . , m. Thus, fi (ap + b, X) is irreducible for i = l + 1, . . . , m. Now consider i between 1 and l. Since / K p . Hence, gi (ap + b, X) ∈ / K p [X]. d(a) 6= 0, we have a ∈ / C, so hi (cp + b) ∈ p p qi By Lemma 12.4.1, fi (a + b, X) = gi (a + b, X ) is irreducible over K. It follows that ap + b ∈ HK (f1 , . . . , fm ).

Exercises 1. Let K be a Hilbertian field and let f ∈ K[T1 , . . . , Tn , X] be a polynomial such that for every t ∈ K n there exists a x ∈ K with f (t, y) = 0. Prove that f (T, X) has a factor of degree 1 in X. 2. Let {Kα | α < m} be a transfinite ascending tower of Hilbertian fields. Suppose Kα+1 S is a regular extension of Kα for each α < m. Prove that the union L = α<m Kα is a Hilbertian field.

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Chapter 12. Hilbertian Fields

Notes A field K is defined to be Hilbertian in [Fried-Jarden3] if each Hilbert set of K is nonempty. Most applications however use only separable Hilbertian sets. We have therefore decided to follow the convention of other authors and call a field Hilbertian if each separable Hilbert set of K is nonempty. Of course, if char(K) = 0, there is no difference between the two notions. When char(K) > 0 there is a simple relation between them. By Uchida’s theorem (Proposition 12.4.3), Hilbertian fields in the old sense are imperfect Hilbertian fields in the new sense. A great part of Section 12.1 occurs in Hilbert’s original paper [Hilbert]. [Lang3, Chapter VIII] reproduces it, though its version of our Lemma 12.2.2 contains a flaw. Following Inaba [Lang3, p. 151, Proposition 3] claims that if L is a finite separable extension of a field K of degree d and if an irreducible polynomial f ∈ L[X] has d distinct conjugates f σ1 , . . . , f σd over K, then d is irreducible in K[X]. Here is a counter-example. g = f σ1 · · · f σ√ Let γ = 3 2 and ω be a primitive 3rd root of unity. Then the polynomial f (X) = X 2 + γX + γ 2 = (X − γω)(X − γω 2 ) is irreducible over L = Q(γ). It has distinct conjugates over Q: f σ2 (X) = X 2 + γωX + γ 2 ω 2 = (X − γω 2 )(X − γ) and f σ3 (X) = X 2 + γω 2 X + γ 2 ω = (X − γ)(X − γω). Yet the product f (X)f σ2 (X)f σ3 (X) = (X 3 − 2)2 is reducible in Q[X]. Inaba treats purely inseparable extensions, though [Inaba, p. 12] claims that if L is a finite purely inseparable extension of a field K, then every Hilbert subset of L contains a Hilbert subset of K. This contradicts the example appearing in the first paragraph of Section 12.3. Finally, the results of Sections 12.3 and 12.4 are due mainly to [Uchida].

Chapter 13. The Classical Hilbertian Fields Global fields and functions fields of several variables have been known to be Hilbertian for three quarters of a century. These are the “classical Hilbertian fields”. The Hilbert property for rational function fields of one variable K = K0 (t) over infinite fields K0 is basically a combination of the MatsusakaZariski theorem (Proposition 10.5.2) with the Bertini-Noether theorem (Proposition 10.4.2). We show that every separable Hilbert subset H of K contains all elements of the form a+bt with (a, b) in a nonempty Zariski K0 -open subset of A2 (Proposition 13.2.1). Our unified approach to the proof of the irreducibility theorem for both number fields and function fields over finite fields uses Proposition 6.4.8, the main ingredient in the proof of the Chebotarev density theorem for function fields, to show that every separable Hilbert subset of a global field contains an arithmetic progression. Thus, we display an intimate connection between the irreducibility theorem and the Riemann hypothesis for curves. Our proof of the irreducibility theorem for global fields specializes the parameters of the irreducible polynomials to integral elements. This leads to the notion of Hilbertian rings. Thus, finitely generated integral domains over Z and finitely generated transcendental ring extensions of an arbitrary field K0 are Hilbertian (Proposition 13.4.1). Section 13.5 describes the connection between the classical definition and the geometric definition of the Hilbert property due to Colliot-Th´el`ene. This leads to the notion of a g-Hilbertian field. We prove that for each g ≥ 0 each global field has an infinite normal extension which is g-Hilbertian but not Hilbertian (Theorem 13.6.2). Each finite proper separable extension of a Galois extension of a Hilbertian field K is Hilbertian. This is a theorem of Weissauer. Moreover, if L1 and L2 are Galois extensions of K and neither of them contains the other, then L1 L2 is Hilbertian. The diamond theorem says even more: each extension M of K in L1 L2 which is contained in neither L1 nor in L2 is Hilbertian (Theorem 13.8.3). An essential tool in the proof is the twisted wreath product (Section 13.7).

13.1 Further Reduction The main application of Hilbertianity of a field K is to reduce Galois extensions of K(t1 , . . . , tr ) to Galois extensions of K with the same Galois group. Lemma 13.1.1 (Hilbert): Let K be a field, t1 , . . . , tr indeterminates, and Fj = K(t, xj ) a finite separable extension of K(t) with xj 6= 0, j = 1, . . . , m. Then: (a) Ar has a nonempty Zariski K-open subset U such that each every K-place

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Chapter 13. The Classical Hilbertian Fields

˜ ∪ {∞} with a = ϕ(t) ∈ U (K) ˜ and ϕ(K(t)) = K(a) ∪ {∞} ϕ: Fj → K satisfies this: (a1) bj = ϕ(xj ) ∈ Ks× and ϕ(Fj ) = K(a, bj ) ∪ {∞}, j = 1, . . . , m. (a2) For each j with Fj /K(t) Galois, K(a, bj )/K(a) is Galois. Moreover, the map σ 7→ σ ¯ defined by σ ¯ (ϕ(y)) = ϕ(σy) for σ ∈ Dϕ and y ∈ Fj with ϕ(y) 6= ∞ is an isomorphism of the decomposition group Dϕ onto Gal(K(a, bj )/K). (b) K r has a separable Hilbert subset H satisfying this: For each a ∈ H ∩ ˜ ∪ {∞} satisfying (a1) and (a2), U (K) and for each K-place ϕ: Fj → K the map σ 7→ σ ¯ defined in (a2) is an isomorphism Gal(Fj /K(t)) ∼ = Gal(K(bj )/K), j = 1, . . . , m. Proof: For each j between 1 and m let fj ∈ K[T, X] be an irreducible polynomial with fj (t, xj ) = 0. Let gj0 ∈ K[T] be the leading coefficient of fj viewed as a polynomial in X. Let gj1 ∈ K[T] be a nonzero polynomial and g

(t)

k a positive integer such that gj0j1(t)k is the discriminant of fj (t, X). Write Qs g = j=1 gj0 gj1 , R = K[t, g(t)−1 ], and Sj = R[xj ]. By Lemma 6.1.2, Sj /R is a Galois ring-cover with xj a primitive element. Define U as the set of all a ∈ Ar with g(a) 6= 0. ˜ ∪ {∞} be a K-place such that a = ϕ(t) ∈ U (K) ˜ and Let ϕ: Fj → K ϕ(K(t)) = K(a)∪{∞} (By Lemma 2.2.7, there exists ϕ with these properties ˜ for each a ∈ U (K).) Put R = Oϕ ∩ K(t). By Remark 6.1.7, R[xj ]/R is a ring-cover and ϕ(Fj ) = K(a, bj ) ∪ {∞}. This proves (a1). Assertion (a2) then follows from Lemma 6.1.4. The set H = HK (f1 , . . . , fm ) satisfies the requirements of (b). Consider an arbitrary field K. Let hi ∈ K[T, X] be irreducible with degX (hi ) ≥ 2, i = 1, . . . , m, and let 0 6= g ∈ K[T ]. Put 0 HK (h1 , . . . , hm ; g) = {a ∈ K | g(a) 6= 0 and

m Y

hi (a, b) 6= 0 for each b ∈ K}.

i=1 0 (h1 , . . . , hm ; g) is nonempty. The next If K is Hilbertian, then each set HK lemma shows the converse:

Lemma 13.1.2: Let f ∈ K[T1 , . . . , Tr , X] be an irreducible polynomial, monic and separable in X with degX (f ) ≥ 2. Then there exist absolutely irreducible polynomials h1 , . . . , hm ∈ K[T1 , . . . , Tr , X], monic and separable 0 (h1 , . . . , hm ) ⊆ HK (f ). in X, with degX (hi ) ≥ 2, i = 1, . . . , m, such that HK Proof: We prove the lemma in two ways. The first proof uses decomposition groups. The second one uses the beginning of what we later call “the intersection decomposition procedure” (Section 21.1.1). In both proofs we start from algebraically independent elements t1 , . . . , tr over K and write E = K(t). Proof A: Let x be a root of f (t, X) in Es , F = E(x), and Fˆ the splitting field of f (t, X) over E. Choose a primitive element z for Fˆ /E. List all proper

13.1 Further Reduction

233

extensions of E in Fˆ which are regular over K as E1 , . . . , Em . For each i between 1 and m choose a primitive element yi for Ei /E which is integral over K[t]. Thus, there is an irreducible polynomial hi ∈ K[T, X] monic and separable in X with hi (t, yi ) = 0 and degX (hi ) = [Ei : E] ≥ 2. By Example 2.6.11, hi is absolutely irreducible. Lemma 13.1.1(a) gives a nonzero polynomial g ∈ K[T] such that for ˜ ∪ {∞} with ϕ(t) = a each a ∈ K r with g(a) 6= 0 each K-place ϕ: Fˆ → K has this property: (2a) ϕ(F ) = K(ϕ(x)) ∪ {∞}. (2b) Let N be the residue field of Fˆ under ϕ. Then the decomposition group of ϕ is isomorphic to Gal(N/K). 0 0 Claim: HK (h1 , . . . , hm ; g) ⊆ HK (f ). Indeed, let a ∈ HK (h1 , . . . , hm ; g). ˜ ∪ {∞} with ϕ(t) = a Use Lemma 2.2.7 to construct a K-place ϕ: Fˆ → K ˜ Denote and ϕ(K(t)) = K ∪ {∞}. Since f (a, X) is monic, b = ϕ(x) is in K. the decomposition field of ϕ in Fˆ by E 0 . The restriction of ϕ to E 0 maps E 0 onto K ∪ {∞} (Remark 6.1.6). Hence, by Lemma 2.6.9(b), E 0 /K is regular. Assume E 0 6= E. Then E 0 = Ei for some i between 1 and m. Hence, c = ϕ(yi ) ∈ K and hi (a, c) = 0. This contradiction to the choice of a proves that E 0 = E. Denote the residue field of Fˆ under ϕ by N . By the preceding paragraph and (2b), [N : K] = [Fˆ : E]. By (2a), the residue field of F under ϕ is K(b). Then [K(b) : K] ≤ [F : E] and [N : K(b)] ≤ [Fˆ : F ]. Hence, [K(b) : K] = [F : E]. Therefore, f (a, X) is irreducible over K, as claimed. Finally, use Lemma 12.1.4 to eliminate g. Qn Proof B: Let f (t, X) = i=1 (X − xi ) be the factorization of f (t, X) in E Qs [X]. Consider a nonempty proper subset I of {1, . . . , n} and let fI (X) = / E[X]. Hence, fI (X) has a coeffii∈I (X − xi ). Since f is irreducible, fI ∈ cient yI 6∈ E. Thus, there is an irreducible polynomial gI ∈ K[T, X], monic and separable in X with degX (gI ) ≥ 2, such that gI (t, X) = irr(yI , E) ∈ K[t, X]. ˜ gI = gI,1 . . . gI,k , where each Suppose gI factors nontrivially over K: ˜ gI,j ∈ K[T, X] is irreducible and k ≥ 2. Since gI (t, X) is monic and separable in X, the factors gI,j are relatively prime. By Hilbert Nullstellensatz (Proposition 9.4.2), V (gI,i ) 6⊆ V (gI,j ) for i 6= j. Thus (Lemma 10.1.2), WI = V (gI,1 , . . . , gI,k ) is a Zariski K-closed subset of Ar+1 of dimension at most r − 1. ˜ by Denote the union of all WI such that gI factors nontrivially over K W . Let A be the K-Zariski closure of the projection of W on the first r coordinates. Then dim(A) ≤ r − 1. Hence, there exists a nonzero polynomial q ∈ K[T] which vanishes on A. Now list all gI which are absolutely irreducible as h1 , . . . , hm . Then

(3)

0 0 (h1 , . . . , hm ; q) ⊆ HK (gI | I 6= ∅, I ⊂ {1, . . . , n}). HK

234

Chapter 13. The Classical Hilbertian Fields

Indeed, suppose a is in the left hand side of (3). Then gI (a, b) 6= 0 for each I with gI absolutely irreducible and each b ∈ K. Assume there is an I with ˜ and there is a b ∈ K with gI (a, b) = 0. Then, in the gI reducible over K above notation, there exists j between 1 and k with gI,j (a, b) = 0. As gI is irreducible over K, all gI,i are K-conjugate to gI,j . Hence, gI,i (a, b) = 0 for each i. Thus, (a, b) ∈ W (K) and so a ∈ A(K). Therefore, q(a) = 0, in contradiction to the assumption on a. Next we prove that the right hand side of (3) is contained in HK (f ). Assume for a ∈ K r that f (a, X) = p(X)q(X) factors nontrivially in K[X] with p and q monic. Extend the K-specialization t → a to a K-specialization ˜ (Propositions 2.3.1 and . . , cn ) with c1 , . . . , cn ∈ K (t, x1 , . . . , xn ) → (a, c1 , .Q n 2.3.3). Then, f (a, X) Q = i=1 (X − ci ). For some nonempty proper subset I of {1, . . . , n}, p(X) = i∈I (X − ci ), the polynomial fI (X) maps onto p(X), and yI maps onto a coefficient b of p(X). Then b lies in K and satisfies gI (a, b) = 0. Thus, a does not belong to the right hand side of (3). Finally, use Lemma 12.1.4 to eliminate q. The following result is one ingredient of the proof of Lemma 13.1.4. That lemma will be used in the proof of the diamond theorem (Theorem 13.8.3). Lemma 13.1.3: Let K be an infinite field, t1 , . . . , tr algebraically independent elements over K, E0 = K(t), and E a finite separable extension of E0 . Then there are fields L, F0 , and F with these properties: (a) L/K is a finite Galois extension. ˆ where E ˆ is the Galois closure of E/K. (b) F = EL, (c) F0 L = F and F0 ∩ L = K. (d) Gal(F/E0 ) = Gal(F/L(t)) n Gal(F/F0 ) (Definition 13.7.1). (e) There is an L-place ϕ: F → L ∪ {∞} with ϕ(F0 ) = K ∪ {∞}. (f) Both F0 /K and F/L are regular extensions. F0

F { {{ { {E {{{ L(t) E0 ˆ be the Galois closure of E/E0 . Choose a primitive element Proof: Let E ˆ x for E/E0 . Remark 6.1.5 gives a nonzero polynomial q ∈ K[T] such that K[t, q(t)−1 , x]/K[t, q(t)−1 ] is a Galois ring-cover with x being a primitive element. Choose a ∈ K r with q(a) 6= 0. Extend the specialization t → a to a K-place ψ0 : E0 → K ∪ {∞} (Lemma 2.2.7). Then extend ψ0 to a place ˆ into K ˜ ∪ {∞}. By Remark 6.1.7, ψ(E) ˆ = L ∪ {∞} where L = ψ of E ˆ K(ψ(x)) is a finite Galois extension of K. Let F = EL. By Remark 6.1.7, −1 −1 L[t, q(t) , x]/L[t, q(t) ] is a ring-cover and ψ extends to an L-place ϕ: F → L ∪ {∞}. Hence, by Lemma 2.6.9(b), F/L is regular.

13.1 Further Reduction

235

Denote the decomposition field of ψ over E0 by F0 . By Remark 6.1.6, ψ(E0 ) = K∪{∞}. Hence, by Lemma 2.6.9(b), F0 /K is regular. In particular, F0 ∩ L = K. By Remark 6.1.7, L[t, q(t)−1 ]/K[t, q(t)−1 ] is a ring-cover. Hence, L[t, q(t)−1 , x] is the integral closure of K[t, q(t)−1 ] in F . Suppose σ is in the inertia group of ϕ over E0 ; that is, σ ∈ Gal(F/F0 ). Let a ∈ L. Then ψ(σa) = σ ¯ ψ(a) = ψ(a). Since ψ is an isomorphism on L (see discussion after Proposition 2.3.1), σa = a. Thus, σ lies in the inertia group of ψ over L(t). Since L[t, q(t)−1 , x]/L[t, q(t)−1 ] is a ring-cover, the latter is trivial. Therefore, σ = 1. Thus, by Remark 6.1.6, Gal(F/F0 ) ∼ = Gal(L/K) and therefore F0 L = F . Finally, (d) is a reinterpretation of (c). Let f ∈ K[X] be a polynomial. We say f is Galois over K if f is irreducible, separable, and K(x) is the splitting field of f over K for every root x of f . If L0 is an extension of L and f is irreducible over L0 , then L0 is linearly disjoint from L(x) over L. Hence, f is Galois over L0 and Gal(L0 (x)/L0 ) ∼ = Gal(L(x)/L). Lemma 13.1.4: Suppose K is an infinite field. Let f ∈ K[T1 , . . . , Tr , X] be an irreducible polynomial which is monic and separable in X. Then there are a finite Galois extension L of K and an absolutely irreducible polynomial g ∈ K[T, X] which as a polynomial in X is monic, separable, and Galois over L(T) such that K r ∩ HL (g) ⊆ HK (f ). Proof: Let t1 , . . . , tr be algebraically independent elements over K, E0 = K(t), x ∈ E0,s a root of f (t, X), and E = E0 (x). Let L, F0 and F be the fields given by 13.1.3. Choose a primitive element y for F0 /E0 which is integral over K[t]. Then there is an irreducible polynomial g ∈ K[T, X], monic and separable in X such that g(t, y) = 0. By Example 2.6.11, g is absolutely irreducible. Since F = L(t, y), the polynomial g(t, X) is Galois over L(t). Lemma 13.1.1(a) gives q ∈ K[T], q 6= 0 such that for each K-place ˜ ∪ {∞} with a = ϕ(t) 6= ∞, ϕ(K(t)) = K ∪ {∞}, and q(a) 6= 0 ϕ: F → K this holds: (4) Let b = ϕ(x) and c = ϕ(y). Then ϕ(E) = K(b) ∪ {∞}, ϕ(F0 ) = K(c) ∪ {∞}, and ϕ(F ) = L(c) ∪ {∞}. Suppose a ∈ K r ∩ HL (g; q). Then g(a, Y ) is irreducible over L, q(a) 6= 0,, q(a) 6= 0, and g(a, c) = 0. Use Lemma 2.2.7 to construct a K-place ˜ ∪ {∞} with ϕ(t) = a and ϕ(K(t)) = K ∪ {∞}. By Lemma 13.1.3, ϕ: F → K degY (g) = [F0 : E0 ] = [F : L(t)]. Hence, [L(c) : K] = [L(c) : L][L : K] = degY (g)[L : K] = [F : L(t)][L(t : K(t)] = [F : E0 ]. In addition, [K(b) : K] ≤ [E : E0 ] and [L(c) : K(b)] ≤ [F : E]. Hence, [K(b) : K] = [E : E0 ]. Since f (a, b) = 0, the polynomial f (a, X) is irreducible over K.

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Finally, use Lemma 12.1.4 to eliminate q.

Problem 13.1.5 ([D`ebes-Haran, p. 284]): Let K be an infinite field and H a separable Hilbertian subset of K r . Does there exist an absolutely irreducible polynomial f ∈ K[T1 , . . . , Tr , X] which is monic and separable in X such that HK (f ) ⊆ H?

13.2 Function Fields over Infinite Fields Function fields of several variables over infinite fields are Hilbertian. This is a consequence of the following result. Proposition 13.2.1: Let K0 be an infinite field, t an indeterminate, and K = K0 (t). Then every separable Hilbert subset H of K contains a set {a + bt | (a, b) ∈ U (K0 )} for some nonempty Zariski K0 -open subset U of A2 . Thus, every Hilbert set of K is nonempty and K is Hilbertian. Proof: The first statement implies that every separable Hilbert subset of K is nonempty. If char(K) > 0, then K is imperfect. Hence, by Uchida (Proposition 12.4.3), every Hilbert set of K is nonempty and K is Hilbertian. Lemmas 12.1.6 and 13.1.2 reduce the proof of the first statement to the following one: For an absolutely irreducible polynomial f ∈ K[X, Y ], monic and separable in Y , with degY (f ) > 1, there exists a nonempty Zariski K0 -open subset U of A2 such that f (a + bt, Y ) has no zeros in K for each (a, b) ∈ U (K0 ). If necessary, multiply f by a suitable element of K and make a linear change in the variable Y to assume that f (X, Y ) = g(t, X, Y ) ∈ K0 [t, X, Y ] is ∂g 6= 0. Then g(t, Z0 +Z1 t, Y ) is an an absolutely irreducible polynomial and ∂Y ˜ Y ], where L = K0 (Z0 , Z1 ) (Proposition 10.5.4). irreducible polynomial in L[t, By Proposition 9.4.3, there exists a nonzero polynomial h ∈ K0 [Z0 , Z1 ] such ˜ 0 [t, Y ] for every a, b ∈ K0 such that that g(t, a + bt, Y ) is irreducible in K h(a, b) 6= 0. In particular, f (a + bt, Y ) has no zeros in K if h(a, b) 6= 0. Proposition 13.2.1 implies the analog to Corollary 12.1.5 for separable Hilbert sets: Proposition 13.2.2: Let K be a field. Suppose each separable Hilbert subset of K of the form HK (f ) with irreducible f ∈ K[T, X], separable, monic, and of degree at least 2 in X, is nonempty. Then K is Hilbertian. Proof: By Lemma 12.1.6, it suffices to consider a separable irreducible polynomial f ∈ K[T1 , . . . , Tr , X] and to prove that HK (f ) 6= ∅. The case r = 1 is covered by the assumption of the Proposition. Now, suppose r ≥ 2 and the statement holds for r − 1. The assumption of the Proposition implies K is infinite. Hence, by Proposition 13.2.1, f (T1 , . . . , Tr−1 , g(T1 , . . . , Tr−1 ), X) is irreducible and separable in K(T1 , . . . , Tr−1 )[X] for some g ∈ K(T1 , . . . , Tr−1 ). 1 ,...,Tr−1 ) Write g(t1 , . . . , Tr−1 ) = gg10 (T (T1 ,...,Tr−1 ) with g0 , g1 ∈ K[T1 , . . . , Tr−1 ]. Lemma

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237

12.1.6 gives an irreducible polynomial h ∈ K[T1 , . . . , Tr−1 , X], separable, monic, and of degree at least 2 in X, such that HK (h) ⊆ HK f (T1 , . . . , Tr−1 , g(T1 , . . . , Tr−1 , X)); g0 (T1 , . . . , Tr−1 ) . The induction hypotheses gives a1 , . . . , ar−1 ∈ K such that h(a1 , . . . , ar−1 , X) is irreducible and separable in K[X]. Let ar = g(a1 , . . . , ar−1 ). Then f (a1 , . . . , ar , X) is well defined and irreducible in K[X].

13.3 Global Fields For K a global field it is easy to choose the hi ’s in Lemma 13.1.2 with coefficients in the ring of integers OK . If, in particular, K is a number field, r = 1, and each of the curves hi = 0 has positive genus, a celebrated theorem 2 of Siegel implies that each of the hi ’s has only finitely many zeros (a, b) ∈ OK 0 ([Lang3, p. 121] or [Robinson-Roquette]). In this case HK (h1 , . . . , hm ) is clearly infinite. If, however, the curve hi = 0 is of genus zero, we may use Riemann-Hurwitz to replace hi (T, x) by gi (T, X) = hi (m(T ), X) for some m(T ) ∈ OK [T ], so that gi (T, X) = 0 has positive genus. Thus, Siegel’s theorem gives Hilbert’s theorem effortlessly. Of course, Siegel’s theorem, a deep result in arithmetic, applies only to number fields (more generally, to fields which are finitely generated over Q [Lang3, p. 127, Thm. 4]). Besides, its power masks subtle connections between the irreducibility theorem and other arithmetic results. Our approach to the Hilbert irreducibility theorem for global fields is based on the Chebotarev density theorem for function fields over finite fields. More accurately, we use a special case of Proposition 6.4.8. As a bonus we will prove that each Hilbert set over Q contains arithmetic progressions. We start the proof with a weak consequence of Bauer’s theorem. In keeping with our elementary treatment we use only Euclid’s argument for proving the infinitude of primes: Lemma 13.3.1: Let L/K be a finite separable extension of global fields. ¯ p for every ¯P = K Then there exist infinitely many primes p of K such that L prime P of L lying over p. Proof: Assume, without loss, that K = Q if char(K) = 0 and K = Fp (t) if char(K) = p. In particular, OK is a principal ideal domain with only finitely many units. Replace L by the Galois hull of L/K to assume that L/K is Galois. Consider a primitive element z ∈ OL for the extension L/K with discriminant d ∈ OK (Section 6.1). Suppose p1 , . . . , pk are prime ideals of OK satisfying the conclusion of the lemma. For each i between 1 and k choose a nonzero πi ∈ pi and let π = d · π1 · · · πk . Consider also f (X) = irr(z, K) = X n + cn−1 X n−1 + · · · + c0 . Since c0 6= 0 and OK has only finitely many units, there exists a positive integer m such that n−1 mn m π + cn−1 cn−2 π m(n−1) + · · · + 1 c−1 0 f (c0 π ) = c0 0

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is a nonunit of OK . Therefore, this element has a prime divisor p = pk+1 different from p1 , . . . , pk and relatively prime to d. Let P be a prime ideal of OL lying over pk+1 . Denote the reduction modulo P by a bar. Then f¯(X) ¯ p , namely c0 π m modulo p, which we may assume to be z¯. By has a root in K ¯ p (¯ ¯ p. ¯P = K z) = K Remark 6.1.7, L Lemma S 13.3.2: For H a proper subgroup of a finite group G, the set G r σ∈G σ −1 Hσ is nonempty. For f ∈ K[X] an irreducible separable polynomial with roots x1 , . . . , xn and splitting field N = K(x1 , . . . , xn ), there exists σ ∈ Gal(N/K) such that σxi 6= xi , for each i = 1, . . . , n. Proof: Let H1 , . . . , Hm be the distinct subgroups of G conjugate to H. Assume without loss that m ≥ 2. Then m is equal to the index of the normalizer of H. (G : H). The intersection H1 ∩ · · · ∩ Hm Sm Thus, mP≤ m contains 1, so | i=1 Hi | < i=1 |Hi | = m · |H| ≤ |G|. With Gal(N/K) = G and Gal(N/K(x1 )) = H the conjugates of H are Gal(N/K(x1 )), . . . , Gal(N/K(x m )). The last statement of the lemma follows S by choosing σ ∈ G r τ ∈G τ −1 Hτ . ¯ = Fq , t an indeterminate, and Lemma 13.3.3: Let q be a prime power, K ¯ ¯ ¯ ¯ z) where g¯(t, z) = 0 with E = K(t). Consider a Galois extension F = K(t, ¯ ¯ is algebraically closed in g ∈ K[T, Z] an irreducible polynomial. Suppose K ¯ ¯ g ), and C be a conjugacy class in Gal(F¯ /E). F . Let d = deg(¯ g ), m = degZ (¯ ¯ ¯ ¯ Then the number N of primes of E/K of degree 1, unramified in F , with Artin symbol equal to C satisfies (1)

N − |C| q < 10d2 |C|√q. m

¯ is algebraically closed in F¯ , g¯ is Proof: Write gF¯ = genus(F¯ ). Since K absolutely irreducible (Corollary 10.2.2). Hence, g(T, Z) = 0 defines a curve ¯ Recalling that E ¯ has genus 0, Proposition 6.4.8, with of degree d over K. k = 1, gives the inequality h i N − |C| q < 2|C| (m + gF¯ )q 12 + mq 14 + gF¯ + m . m m Combining this with the inequality gF¯ ≤ 12 (d − 1)(d − 2) for the genus of F¯ (Corollary 5.3.5) we obtain (1). An absolute value of a field K is a map | |: K → R which satisfies the following conditions for all x, y ∈ K: (2a) |x| ≥ 0 and |x| = 0 if and only if x = 0. (2b) |xy| = |x||y|. (2c) |x + y| ≤ |x| + |y|. (2d) There is an a ∈ K with |a| 6= 1.

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239

If, instead of (2c), | | satisfies the stronger condition (2c’) |x + y| ≤ max(|x|, |y|), it is ultrametric (or non-archimedean), otherwise, it is metric (or archimedean). An absolute value | |0 is equivalent to | | if there exists c > 0 such that |x|0 = |x|c for all x ∈ K. If | | is ultrametric, then O| | = {x ∈ K | |x| ≤ 1} is a valuation ring of K. In particular, |n| ≤ 1 for each positive integer n. Conversely, suppose |n| ≤ 1 for each positive integer n. Let z be an element of K with |z| ≤ 1. Pk Then |1 + z|k = i=0 ki z i ≤ k + 1. Hence, |1 + z| ≤ (k + 1)1/k . Letting k go to infinity, we find that |1 + z| ≤ 1. Next let x, y ∈ K with |x| ≤ |y| and y 6= 0. Then |x + y| = |1 + xy ||y| ≤ |y| = max(|x|, |y|). Consequently, | | is ultrametric. In particular, suppose char(K) = p > 0. Then np−1 = 1 for each positive integer n not divisible by p. Hence, |n| = 1. Therefore, | | is ultrametric. If K is a global field, the map | | → O| | is a bijection between the equivalence classes of ultrametric absolute values of K and the valuation rings of K. In particular, if K is a function field of one variable over a finite field K0 , then the equivalence classes of absolute values of K bijectively correspond to the prime divisors of K/K0 . If K is a number field, then the equivalence classes of ultrametric absolute values bijectively correspond to the prime ideals of OK . In addition, each embedding σ: K → C defines a metric absolute value | |σ of K: |x|σ = |σx|, where | | is here the usual absolute value of C. Each metric absolute value of K is equivalent to some | |σ and there are at most [K : Q] such equivalence classes [Cassels-Fr¨ohlich, p. 57, Thm.]. In each case we call an equivalence class of an absolute value a prime of K. For each prime p of K we choose an absolute value | |p which represents it. The set of all primes of a global field K satisfies the strong approximation theorem: Let p0 , p1 , . . . , pn be distinct primes of K. For each i between 1 and n consider an element ai of K and let ε > 0. Then there exists x ∈ K such that |x − ai |pi < ε, i = 1, . . . , n, and |x|p ≤ 1 for each ohlich, p. 67]. In the function field prime p not in {p0 , p1 , . . . , pn } [Cassels-Fr¨ case, this theorem is also a consequence of the strong approximation theorem (Proposition 3.3.1). Generalizations of the following lemma appear as ingredients in the Galois stratification procedure of Chapter 30. Lemma 13.3.4: Let K be a global field and f ∈ K[T, X] an absolutely irreducible polynomial, monic and separable in X, with degX (f ) > 1. Then K has infinitely many ultrametric primes p for which there is an ap ∈ OK with this property: if a ∈ OK satisfies a ≡ ap mod p, then f (a, b) 6= 0 for every b ∈ K. Proof: Assume without loss that f ∈ OK [T, X]. Let E = K(t), with t an indeterminate, and let x1 , . . . , xn be the roots of f (t, X) in Es . Denote the algebraic closure of K in the Galois extension F = E(x1 , . . . , xn ) of E by L.

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Chapter 13. The Classical Hilbertian Fields

Then F is a regular extension of L; write it as F = L(t, z) with z integral over OL [t]. Thus, there is an absolutely irreducible polynomial g ∈ L[T, X] with g(t, X) = irr(z, L(t)). By Lemma 13.3.1, the set of ultrametric primes ¯ p for each prime P of L over p, is infinite; call it ¯P = K p of K for which L A1 . The rest of the proof derives the result from the case where L replaces K and g replaces f . Part A: Geometrically exceptional primes. Let A2 be the restriction to K of the set of ultrametric primes P of L for which both f mod P and g mod P are defined, separable in X, and absolutely irreducible. By Bertini-Noether (Proposition 9.4.3), A2 is a cofinite set. Express z as a polynomial in x1 , . . . , xn (with coefficients that are ratios of elements of OK [t]) and express each xi as a polynomial in z (with coefficients that are ratios of elements of OL [t]). Denote the cofinite set of ultrametric primes of K which divide none of the denominators of these coefficients by A3 . ∂g (t, z)) ∈ OL [t]. Denote the Finally, consider d(g; t) = normF/LE ( ∂X cofinite set consisting of the restriction to K of the ultrametric primes P of L for which deg(d(g; t)) = deg(d(g; t) mod P) by A4 . The set A = A1 ∩ A2 ∩ A3 ∩ A4 is infinite. Part B: Reduction modulo primes of A. For p ∈ A and P a prime of L ¯P = K ¯ p to a place ϕ of F by choosing over p, extend the residue map OL → L ¯ ¯ t = ϕ(t) as a transcendental element over Kp . Denote ϕ(x) by x ¯ for every ¯ z ) = E(¯ ¯ x1 , . . . , x ¯=K ¯ p (t¯) and F¯ = E(¯ ¯n ), where x ∈ F finite under ϕ. Let E ¯n are the distinct roots of the irreducible polynomial f¯(t¯, X). x ¯1 , . . . , x ¯ Hence, By the choice of p, the polynomial f¯(t¯, X) is irreducible over E. ¯ ¯ ¯i , i = 1, . . . , n. by Lemma 13.3.2, there exists σ ∈ Gal(F /E) such that σ x ¯i 6= x ¯ by Con(σ). Next note that since Denote the conjugacy class of σ in Gal(F¯ /E) ¯ p is algebraically closed g¯ is absolutely irreducible and g¯(t¯, z¯) = 0, the field K ∂ g ¯ ¯ (t, z¯)) with coefficients in F¯ . In addition, the polynomial d(¯ g ; t¯) = NF¯ /E¯ ( ∂X ¯ ¯ ¯ ∈ Kp is not a root of d(¯ g ; t¯), then in Kp has the same degree as d(g; t). If a ¯ ¯ ¯ of E is unramified in the prime pa¯ corresponding to the specialization t → a ¯ which ramify in F¯ F¯ (Lemma 6.1.8). The number of primes of degree 1 of E is therefore bounded by 1 + deg(d(g; t)). ¯p ¯p ∈ K Let A0 be the set of all p ∈ A finite at t for which there exists a ¯ ¯ with d(f¯; a ¯p ) 6= 0 such that the Artin symbol F /E (here a ¯ = a ¯p ) correpa ¯

sponding to the prime pa¯ equals Con(σ). By Lemma 13.3.3, A0 is cofinite in A. For each p ∈ A0 there exists a prime Q of F¯ lying over pa¯ with these ¯ p| properties: for q = |K ¯ (3a) σx ≡ xq mod Q Qfor every x ∈ F integral with respect to Q; xi − x ¯j ), and, since d(f¯; t¯) = i6=j (¯ (3b) x ¯i 6≡ x ¯j mod Q for every i 6= j. Part C: Finding ap . For each i between 1 and n let j be an integer with ¯j . Then i 6= j. Hence, by (3a) and (3b), x ¯i ≡ 6 x ¯qi mod Q. That is, σ¯ xi = x

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241

¯ p. the polynomial f¯(¯ ap , X) has no roots in K ¯p by ϕ. If a ∈ OK Let ap be an element of Ok which is mapped to a satisfies a ≡ ap mod p, then a ¯ = a ¯p , and therefore f (a, b) 6= 0 for each b ∈ OK . Since f (a, X) is monic, this proves f (a, X) has no zero in K. Let K be a global field. An arithmetic progression of OK is a set of the form a + a, where a ∈ OK and a is an ideal of OK . Theorem 13.3.5: Let K be a global field and H a separable Hilbert subset of K. Then: (a) H contains an arithmetic progression of OK . (b) Let q0 , q1 , . . . , qn be distinct primes of K, b1 , . . . , bn elements of K, and ε > 0. Then there exists x ∈ H with |x − bi |qi < ε for i = 1, . . . , n and |x|q ≤ 1 for each prime of K not in {q0 , q1 , . . . , qn }. (c) The intersection of H with each arithmetical progression of OK is nonempty. Proof: By Lemma 13.1.2 there exist absolutely irreducible polynomials h1 , . . . , hm ∈ OK [T, X], monic and separable in X, with degX (hi ) > 1, 0 (h1 , . . . , hm ) ⊆ H. Apply Lemma 13.3.4 to find disi = 1, . . . , m, and HK tinct prime ideals p1 , . . . , pm of OK and elements a1 , . . . , am ∈ OK such 0 that x ∈ HK (hi ) for each x ∈ K with |x − ai |pi < 1, i = 1, . . . , m. The Chinese remainder theorem [Lang7, p. 94] produces an element a ∈ OK with a+pi = ai +pi for i = 1, . . . , m. Thus, with a = p1 . . . pm , we have a+a ⊆ H. This proves (a). For the proof of (b) choose p1 , . . . , pm above not in {q0 , q1 , . . . , qn }. Then use the strong approximation theorem for global fields [Cassels-Fr¨ohlich, p. 67] to find x ∈ K with |x − ai |pi < 1, i = 1, . . . , m, |x − bj |qj < ε, j = 1, . . . , n, and |x|q ≤ 1 for each prime q not in {p1 , . . . , pm , q0 , . . . , qn }. Then x ∈ H and |x|pi ≤ 1 for i = 1, . . . , m, as desired. Each arithmetical progression in OK has the form A = {x ∈ OK | vqi (x− a) ≥ ki , i = 1, . . . , r} for some a ∈ OK , ultrametric primes q1 , . . . , qr which are finite on OK , and positive integers k1 , . . . , kr . If K is a number field, choose q0 to be a metric prime. If K is a function field of one variable over a finite field, choose q0 to be a prime which is not finite on OK . Statement (b) gives x ∈ A with vq (x) ≥ 0 for all ultrametric primes q of K which are finite on OK . Thus, x ∈ OK .

13.4 Hilbertian Rings We call an integral domain R with quotient field K Hilbertian if every separable Hilbert subset of K r contains elements all of whose coordinates are in R. In this case, each overring of R is Hilbertian. In particular, K is Hilbertian.

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Proposition 13.4.1: Let R be an integral domain with quotient field K. Suppose either R is finitely generated over Z or R is finitely generated over a field K0 and K/K0 is transcendental. Then R is Hilbertian. Proof: Let R0 be either Z or K0 . In the former case let K0 = Q. By assumption, R = R0 [u1 , . . . , un ]. Assume without loss u1 , . . . , um with m ≤ n form a transcendental base for K/K0 . By Proposition 12.3.3, every separable Hilbertian subset of K r contains a separable Hilbertian subset of K0 (u1 , . . . , um ). We may therefore assume that u1 , . . . , un are algebraically independent over K0 . Consider an irreducible polynomial f ∈ K[T1 , . . . , Tr , X] which is separable in X. We have to prove that HK (f ) ∩ Rr 6= ∅ (Lemma 12.1.6). There are several cases to consider. Case 1: r = 1 and n = 0. Then R = R0 = Z. By Theorem 13.3.5 there exists a ∈ R such that f (a, X) is irreducible. Case 2: r = 1, n = 1, and R0 = K0 . If K0 is finite, then K is global and R = OK . Thus, Theorem 13.3.5 gives a ∈ R with f (a, X) irreducible. If K0 is infinite, then Proposition 13.2.1 gives a, b ∈ K0 such that f (a + bu1 , X) is irreducible. Case 3: r = 1, n = 1, and R0 = Z. Proposition 13.2.1 gives a nonempty Zariski Q-open subset U of A2 such that f (a + bu1 , X) is irreducible for all (a, b) ∈ U (Q). Since Z is infinite, we may choose (a, b) in U (Z). Case 4: r = 1 and n ≥ 2. Then R0 [u1 , . . . , un−1 ] is an infinite ring. Hence, by Proposition 13.2.1, it has elements a, b such that f (a + bu1 , X) is irreducible. Case 5: r ≥ 2. Consider f as a polynomial in Tr , X with coefficients in the infinite ring R0 [u, T1 , . . . , Tr−1 ]. As such, f is irreducible. Replace R0 and K0 in Cases 2, 3, and 4 by R0 [u, T1 , . . . , Tr−1 ] and K0 (u, T1 , . . . , Tr−1 ) to find g ∈ R0 [u, T1 , . . . , Tr−1 ] such that f (T1 , . . . , Tr−1 , g(T1 , . . . , Tr−1 ), X) is irreducible. Now use an induction hypothesis on r to find a1 , . . . , ar−1 ∈ R with f (a1 , . . . , ar−1 , g(a1 , . . . , ar−1 ), X) irreducible. Then ar = g(a1 , . . . , ar−1 ) is in R and f (a, X) is irreducible. The quotient fields of the rings mentioned in Proposition 13.4.1 are the classical Hilbertian fields. Theorem 13.4.2: Suppose K is a global field or a finitely generated transcendental extension of an arbitrary field K0 . Then K is Hilbertian. Moreover, each Hilbert set of K is nonempty. Proof: As a consequence of Proposition 13.4.1, K is Hilbertian. To prove the second statement, assume char(K) = p > 0. Then K is imperfect. Hence, by Uchida (Proposition 12.4.3), every Hilbertian set of K is nonempty. The proof of Proposition 13.4.1 gives another useful version of Theorem 13.4.2:

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243

Proposition 13.4.3: Let K be a finitely generated separable extension of a field K0 with m = trans.deg(K/K0 ) > 0. Let r ≤ m and H a separable Hilbert subset of K r . Then H contains a point (u1 , . . . , ur ) with u1 , . . . , ur algebraically independent over K0 . Proof: Choose a separating transcendence base t1 , . . . , tm for K/K0 . Then K/K0 (t) is a finite separable extension. By Corollary 12.2.3, H contains a separable Hilbert subset of K0 (t)r . We may therefore assume K = K0 (t). By Lemma 12.1.6, we may assume H = HK (f ) with an irreducible f ∈ K[T1 , . . . , Tr , X], monic and separable in X. For m = 1 and K0 finite, Theorem 13.3.5 gives u ∈ H transcendental over K0 . For m = 1 and K0 infinite, Proposition 13.2.1 does the job. Assume m ≥ 2. Then K 0 = K0 (t2 , . . . , tm , T2 , . . . , Tr ) is an infinite field, t1 is transcendental over K 0 , and K 0 (t1 ) = K(T2 , . . . , Tr ). Consider f as an irreducible polynomial in T1 , X over K 0 (t1 ). Proposition 13.2.1 gives nonzero g, h ∈ K0 [t2 , . . . , tm , T2 , . . . , Tr ] such that f1 (T2 , . . . , Tr , X) = f g(T2 , . . . , Tr ) + h(T2 , . . . , Tr )t1 , T2 , . . . , Tr , X

is irreducible over K. Let K1 = K0 (t1 ). Apply the induction hypothesis to K1 and f1 instead of to K and f and find u2 , . . . , ur in K algebraically independent over K1 such that f (g(u2 , . . . , ur ) + h(u2 , . . . , ur )t1 , u2 , . . . , ur , X) is irreducible over K. Put u1 = g(u2 , . . . , ur ) + h(u2 , . . . , ur )t1 . Then f (u, X) is irreducible over K and K0 (u1 , u2 , . . . , ur ) = K0 (t1 , u2 , . . . , ur ). Hence, u1 , . . . , ur are algebraically independent over K0 , as desired. Remark 13.4.4: More Hilbertian rings. Exercise 4 proves that each valuation ring of a Hilbertian field is Hilbertian. Geyer extends this result to several valuations of rank 1. He considers a Hilbertian field K, nonequivalent absolute values | |1 , . . . , | |n of K, and a separable Hilbert subset H of K r . Let a1 , . . . , ar be tuples in K r and let ε > 0. Then there exists x ∈ H such that |x − ai |i < ε, i = 1, . . . , n [Geyer1, Lemma 3.4]. In particular, if v1 , . . . , vn are valuations of rank 1, then their holomorphy ring, R = {a ∈ K | vi (a) ≥ 0, i = 1, . . . , n} is Hilbertian. Proposition 19.7 of [Jarden14] generalizes Geyer’s result. Here one considers valuations v1 , . . . , vm and orderings αi , i = 1, . . . , m, and −cj <j x − bj <j cj , j = 1, . . . , n. Moreover, the holomorphy ring of arbitrary finitely many valuations of K is Hilbertian [Jarden14, Proposition 19.6]. In contrast, holomorphy ring of infinitely many valuations of a field K may not be Hilbertian even if K is Hilbertian. For example, if K0 is a finite

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Chapter 13. The Classical Hilbertian Fields

field, the holomorphy ring of all valuations of K0 (t) is K0 . The latter field is not Hilbertian, although K0 (t) is Hilbertian. More interesting examples can be found in Example 15.5.6. The following Lemma improves Exercise 2 of Section 12: Lemma 13.4.5: Let m be a cardinal number and {Kα | α < m} a transfinite sequence of fields. Suppose that for each α < m the field S Kα+1 is a proper finitely generated regular extension of Kα . Then K = α<m Kα is a Hilbertian field. Proof: Let f ∈ K(T1 , . . . , Tr )[X] be a monic separable irreducible polynomial. By Lemma 12.1.6, it suffices to prove that HK (f ) 6= ∅. Indeed, the coefficients of f belong to Kα for some α < m. By Theorem 13.4.2, Kα+1 is Hilbertian. Find a1 , . . . , ar ∈ Kα+1 such that f (a, X) is separable and irreducible in Kα+1 [X]. Choose a zero b of f (a, X) in the separable closure of Kα+1 . By Corollary 2.6.5(d), K is a regular extension of Kα+1 . In particular, K is linearly disjoint from Kα+1 (b) over Kα+1 . Therefore, f (a, X) is irreducible in K[X]. The PAC and the Hilbert properties are in a sense opposing. Nevertheless there are fields that are both PAC and Hilbertian: Proposition 13.4.6 ([Fried3]): Every field K has a regular extension F which is PAC and Hilbertian. Proof: Wellorder the varieties of dimension at least 1 that are defined over K in a transfinite sequence {Vα | α < m}. Use transfinite induction to define, for each Q β < m, a function field Fβ for Vβ which is algebraically independent from α max char(K), [K(x) : K(u, t)] . Put ˜ ˜ y) : K(x)] = l. Hence, K(x, y)/K y = u1/l . Then, [K(x, y) : K(x)] = [K(x, is a regular extension. By Corollary 10.2.2(a), (x, y) generate a variety W0 over K. Let ϕ0 : W0 → V be the projection ϕ0 (x, y) = x. Then ϕ0 is a dominant separable rational map of degree l. By construction, (x0 , 0) ∈ W . If a ∈ V0 (K), then the specialization x0 → a extends to a specialization (x0 , 0) → (a, 0). Hence, (a, 0) ∈ W0 (K) and ϕ0 (a, 0) = a. Thus, ϕ0 (W0 (K)) = V0 (K). (g) Suppose V is a variety over K. For each i between 1 and m let hi ∈ K[X1 , . . . , Xn , Y ] be a polynomial which is monic and of degree at least 2 in Y . Let yi be a root of hi (x, Y ) in K(x)s . Suppose K(x, yi ) is a regular extension of K. Let g ∈ K[X] be a polynomial which does not vanish on V . Generalize the notation of Section 13.1 and define 0 HK,V = {a ∈ V (K) | g(a) 6= 0 and

m Y

hi (a, b) 6= 0 for all b ∈ K}.

i=1 0 6= 0. Indeed, Wi = V ∩ V (hi ) is a variety If V (K) is nonthin, then HK,V n+1 in A with generic point (x, yi ) over K . Let ϕi : Wi → V be the projection on the first n coordinates and let V0 = V (g). Then V (K) 6⊆ V0 (K) ∪ Sm ϕ (W (K)) (because V (K) is nonthin), so there is a a ∈ V (K) r V0 (K)∪ i i i=1 Sm 0 i=1 ϕi (Wi (K)) = HK,V , as claimed. 0 0 6= ∅ for all HK,V as above such that K(x, yi ) Conversely, suppose HK,V is a regular extension of K, i = 1, . . . , m. Giving ϕi and Wi as in (1), we may assume that Wi is absolutely irreducible (by (e)). Let Fi be the function field of Wi over K. Then Fi can be chosen to be a finite separable extension of K(x). Choose a primitive element yi for Fi /K(x) which is integral over K[x]. Let hi ∈ K[X, Y ] a polynomial such that hi (x, Y ) = irr(yi , K(x)). By (a), we may assume that (x, yi ) is a generic point of Wi over K and ϕi : Wi → V

13.5 Hilbertianity via Coverings

247

0 is the projection on the first n coordinates. Let V0 = V (g). Then HK,V 6= ∅ Sm is equivalent to V (K) 6⊆ V0 (K) ∪ i=1 ϕi (Wi (K)). (h) An (K) is nonthin if and only if for all absolutely irreducible polynomials h1 , . . . , hm ∈ K[X1 , . . . , Xn , Y ] which are separable and monic in Y of 0 degree at least 2 in Y and for all nonzero g ∈ K[X1 , . . . , Xn ], the set HK,A n is nonempty. Indeed, a generic point x = (x1 , . . . , xn ) for An over K consists of algebraically independent elements x1 , . . . , xn over K. If y ∈ K(x)s is integral over K and h ∈ K[X, Y ] is a monic polynomial in Y such that h(x, Y ) = irr(y, K(x)), then h is absolutely irreducible if and only if K(x, y) is a regular extension of K (Corollary 10.2.2). Thus, our statement is a special case of (g). (i) Suppose A1 (K) is a thin set. Then so is V (K) for every variety V over K. Indeed, suppose dim(V ) = r > 0. Let x be a generic point of V over K. Then K(x) is a regular extension of K of transcendence degree r. By (e) and (f), there exist varieties W1 , . . . , Wn over KSand dominant separable m maps ϕi : Wi → A1 , i = 1, . . . , m such that K = i=1 ϕi (Wi (K)). Use (a) to assume that Wi is a curve defined by gi (T, Y ) = 0, where gi ∈ K[T, Y ] is absolutely irreducible and is monic and separable in Y . Further, assume ϕi is the projection on the first coordinate. Proposition 13.4.3 gives a transcendental element t ∈ K(x) such that gi (t, Y ) is irreducible over K(x), i = 1, . . . , m. Choose yi ∈ K(t)s with gi (t, yi ) = 0. Then K(x) and K(t, yi ) are linearly disjoint over K(t). Let ψ: V → A1 be the K-rational map defined by ψ(x) = t. Then ψ is defined outside a Zariski K-closed subset V0 of V of dimension less than r. Let Wi0 be the K-variety generated by (x, yi ). Let ϕ0i :SWi0 → V be the projection m ϕ0i (x, yi ) = x. We prove that V (K) = V0 (K) ∪ i=1 ϕ0i (Wi0 (K)).

K(x)

K(x, yi )

V o

ϕ0i

ψi0

ψ

K(t)

K(t, yi )

A1 o

Wi0

ϕi

Wi

Indeed, let b ∈ V (K) r V0 (K). Then a = ψ(b) ∈ K. By assumption, there exist i between 1 and m and ci ∈ Wi (K) with ϕi (ci ) = a. By linear disjointness, K[x, yi ] = K[x] ⊗K[t] K[yi ]. Hence, (x, yi ) → (b, ci ) is a Kspecialization. In other words, (b, ci ) ∈ Wi0 (K) and ϕ0i (b, ci ) = b, as desired. In the language of schemes, Wi0 is the fiber product of ϕ: V → A1 and ϕi : Wi → A1 . All of this gives a new characterization of Hilbertian field. Proposition 13.5.3: The following conditions on a field K are equivalent: (a) K is Hilbertian.

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Chapter 13. The Classical Hilbertian Fields

(b) A1 (K) is nonthin. (c) There is a variety V over K such that V (K) is nonthin. Proof: By Lemma 13.1.2, K is Hilbertian if and only if the following condi0 tion holds: HK,A 1 (h1 , . . . , hm ; g) 6= ∅ for all absolutely irreducible polynomials hi ∈ K[X, Y ] which are separable and of degree at least 2 in Y , and for all nonzero g ∈ K[X]. By Remark 13.5.2(h), the latter condition is equivalent to A1 (K) being nonthin. Thus, (a) is equivalent to (b). Remark 13.5.2(i) asserts that (b) is equivalent to (c). Remark 13.5.4: Examples of thin sets. Even if A1 (K) is nonthin, there may be varieties V over K such that V (K) is thin. For example, each finitely generated extension K of Q is Hilbertian (Theorem 13.4.2). Hence, by Proposition 13.5.3, A1 (K) is nonthin. However, by Faltings, C(K) is finite for each curve C over K of genus at least 2 [Faltings-W¨ ustholz, p. 205]. In particular, C(K) is thin. Moreover, let A be an Abelian variety over K of dimension d. For each integer n ≥ 2, multiplication by n is a dominant rational map of degree n2d of A onto itself [Mumford1, pp. 42 and 64]. By Mordell-Weil, A(K)/nA(K) is a finite group [Lang3, p. 71]. Let a1 , . . . , am be representatives for the cosets of A(K) modulo nA(K). For each i, the map S ϕi (x) = ai + nx gives a morphism m of degree n2d of A onto A. Also, A(K) = i=1 ϕi (A(K)). Therefore, A(K) is thin. For example, let E be an elliptic curve over K with E(K) infinite. Then E(K) is Zariski K-dense in E but E(K) is thin. Thus, the converse of Remark 13.5.2(b) is not true. Proposition 27.3.5 characterizes fields K for which V (K) is nonthin for every variety V over K as ‘ω-free PAC fields’.

13.6 Non-Hilbertian g-Hilbertian Fields In case where A = A1 , one may try to bound the genera of the curves Wi appearing in (1) of Section 13.5. This leads to a weaker version of Hilbertianity. Definition 13.6.1: g-Hilbertian fields. Let K be a field and g a nonnegative integer. We say that K is g-Hilbertian if K is not a finite union of sets ϕ(C(K)) with C a curve over K of genus at most g and ϕ: C → A1 a dominant separable rational map of degree at least 2. By Proposition 13.5.3, K is Hilbertian if and only if K is g-Hilbertian for each g ≥ 0. One may ask whether being g-Hilbertian for large g suffices for K to be Hilbertian. We show here that this is not the case. Theorem 13.6.2 (Zannier): Let K be a global field and g a nonnegative integer. Then K has an infinite normal extension N which is g-Hilbertian but not Hilbertian.

13.6 Non-Hilbertian g -Hilbertian Fields

249

Proof: The construction of N starts with a prime number l 6= char(K) and another prime number q with (2)

q > max(char(K), 2g − 2 + 2l).

Extend K, if necessary, to assume ζl , ζq ∈ K. Then use the Hilbertianity of K (Theorem 13.4.2) to choose u ∈ K which is not a qth power. Thus, √ q u∈ / Kins . By Eisenstein’s criterion (Lemma 2.3.10), f (X, Y ) = Y l − X lq + u is an absolutely irreducible polynomial. Now define an ascending sequence √ of Galois extensions Kn of K inductively: K0 = K and Kn+1 = Kn ( l alq − u | a ∈ Kn ). Then Kn /K is a pro-l-extension. That is, Kn /K is Galois and [L : K] is an l-power for each finite subextension L/K of Kn /K. S∞ Let M (K) S= n=0 Kn . This is an infinite pro-l extension of K. Let N = N (K) = M (K 0 ) with K 0 ranging over all finite purely inseparable extensions of K. Then N is a perfect field which is infinite and normal over K. For each a ∈ N there exists b ∈ N such that bl − alq + u = 0. Hence, N is non-Hilbertian. We prove however that N is g-Hilbertian. Note: We may replace K in the proof by any finite extension L in N . Indeed, then M (L) is contained in M (K 0 ) for some finite purely inseparable extension K 0 of K. Hence, N = N (L). Claim A: Suppose M √ is a finite Galois extension of K in N . Then [M : K] is a power of l and M ( q u) is a Galois extension of M of degree q. Indeed, there is a finite purely inseparable extension K 0 of K with M ⊆ 0 ), so [M : K] = [M K 0 : K 0 ] is√ a power of q. By the choice of u, M (K √ q K( u)/K is Galois of degree 1, so M ( q u)/M is a Galois extension of degree q. Let P be the set of all ultrametric prime divisors p of K with vp (u) = 0, √ q vp (l) = 0, vp (q) = 0, and K( pu/K 6= 1. Claim B: Each p in P is unramified in M (K). Indeed, it suffices to consider a finite extension L of K in M (K) in which p is unramified, to take a prime divisor √ q of L over p and an element a ∈ L, and to prove that q is unramified in L( l alq − u). By Example 2.3.8, it suffices to prove that l divides vq (alq − u). Suppose first vq (a) < 0. Then vq (alq − u) = vq (alq ) = lqvq (a). Now suppose vq (a) ≥ 0. Denote reduction modulo q by a bar. The assumptions √ √ √ ¯ qu ¯) = K( q u). on q imply Op [ q u]/Op is a ring-cover. By Remark 6.1.7, K( √ √ q ¯ = q. Next, let M be the ¯ qu Since K( pu/K 6= 1, this implies [K( ¯) : K] ¯ : K] ¯ is a Galois closure of L/K. Then M ⊆ M (K). Hence, by Claim√A, [M q ¯ ¯ ¯ ¯ ¯) 6⊆ L, so u ¯ is power of l. Therefore, [L : K] is also a power of L. Thus, K( u ¯ In particular, a ¯. Consequently, vq (alq − u) = 0. not a qth power in L. ¯lq 6= u

250

Chapter 13. The Classical Hilbertian Fields

Claim C: Let L be a finite extension of K in M (K). Consider nonconstant polynomials h1 , . . . , hr ∈ L[X]. Suppose hi has an irreducible L-factor of fi which is separable, of multiplicity not divisible by l; suppose further the degree of each irreducible L-factor of hi is less p than q, i = 1, . . . , r. Then / N for all a ∈ A and OK has an arithmetical progression A with l hi (a) ∈ i = 1, . . . , r. Indeed, write hi = fik gi with fi , gi ∈ L[X], fi irreducible, l - k, and gi relatively prime to fi . Choose k 0 , l0 ∈ Z with k 0 ≥ 1 and kk 0 + ll0 = 1. Then p 0 0 0 0 0 / N if and only if hki = fikk gik = fi gik fi−ll . For every a ∈ K, l hi (a) ∈ p 0 0 l k fi (a)gi (a) ∈ / N . Each root of fi in Ks is a simple root of fi gik . Therefore, 0 we may replace hi by fi gik , if necessary, to assume hi has a simple root ai in Ks . Now extend L, if necessary, to assume L is Galois over K with Galois √ group G. By Claim A, Q L( q u)/L Q is a Galois extension of degree q. Let M r be the splitting field of i=1 σ∈G hσi (X) over L. Since the degree of each irreducible L-factor of h1 , . . . , hr is less than q, M is a finite Galois extension of K whose degree is divisible √ only by prime numbers smaller than q. Hence, q u). Denote the set of all ultrametric primes M is a proper subfield of M ( √ p of K which unramify in M ( q u) with vp (u) = 0, vp (l) = 0, vp (q) = 0, and √ √ M ( q u)/K ⊆ Gal(M ( q u)/M ) r{1} by P0 . Then P0 is a subset of P . By p Chebotarev (Theorem 6.3.1), P0 is infinite. Omit finitely many elements from P0 to assume that for each p ∈ P0 all nonzero coefficients of the hi are p-units and ai is a simple root of hi modulo p. Now choose distinct primes p1 , . . . , pr in P0 . Consider i between 1 and r. Let p = pi . Choose a prime q = qi of L over p, a prime q0 of M over q, and denote reduction modulo q0 by a bar. Then ¯ i is a polynomial with coefficients in K ¯ which decomposes into ¯ = K, ¯ so h M ¯ ¯ 0 (¯ ¯ linear factors. In particular, a ¯i ∈ K, hi (¯ ai ) = 0, and h i ai ) 6= 0. Choose bi ∈ K with vq0 (bi − ai ) > 0. Then, vq (hi (bi )) ≥ 1 and vq (h0i (bi )) = 0. If vq (hi (bi )) ≥ 2, choose πi ∈ K with vp (πi ) = 1. Since p is unramified in M , vq (πi ) = 1. Then vq hi (bi + πi ) − hi (bi ) − h0i (bi )πi ≥ 2. Hence, vq (hi (bi + πi )) = 1. Thus, replacing bi by bi + πi , if necessary, we may assume that in any case vq (hi (bi )) = 1. The weak approximation theorem (Proposition 2.1.1) gives b, π ∈ K with vpi (b − bi ) ≥ 2 and vpi (π − πi ) ≥ 2, i = 1, . . . , r. Let A = b + π 2 OK . Then vpi (hi (a)) = 1 for each i and all a ∈ A. Consider an arbitrary finite purely inseparable extension K 0 of K. Denote the unique extension of pi to K 0 by p0i . Then vp0i (hi (a)) = 1 if char(K) = 0 and vp0i (hi (a)) is a power of char(K) if char(K) > 0. In each case, l - vp0i (hi (a)). By Claim B, applied to K 0 instead of to K, p0i is unramified in M (K 0 ). Hence, l - vi (hi (a)) for each extension vi of vp0i to M (K 0 ), so p p l hi (a) ∈ / M (K 0 ). Consequently, l hi (a) ∈ / N , as claimed. Claim D: Let L be a finite extension of K and C a curve of genus at most

13.6 Non-Hilbertian g -Hilbertian Fields

251

Qs g over L. Suppose C is defined by Y l = c j=1 fj (X)mj , where c ∈ K × , f1 , . . . , fs are distinct monic irreducible polynomials in L[X], and l - mj , j = 1, . . . , s. Then deg(fj ) < q, j = 1, . . . , s. Indeed, replace mj by its residue modulo l, if necessary, to assume 1 ≤ − 1. Choose a transcendental element x over L and y ∈ L(x)s with mj ≤ l Q s y l = c j=1 fj (x)mj . Then F = L(x, y) is the function field of C over L. Denote the prime divisor of L(x)/L corresponding to fj by Pj (Section 2.2). Then deg(Pj )Q= deg(fj ), vPj (fj (x)) = 1, and vPj (fi (x)) = 0 for j 6= i. s Hence, vPj (c i=1 fi (x)mi ) = mj is not divisible by l. By Example 2.3.8, Pj totally ramifies in F . Thus, F has a unique prime divisor Qj lying over Pj , eQj /Pj = l, and deg(Qj ) = deg(Pj ) = deg(fj ). Since l 6= char(L), the ramification is tame. Hence, by Riemann-Hurwitz (Remark 3.6.2(c)), 2g − 2 ≥ 2 · genus(F ) − 2 ≥ −2l + (l − 1) deg(fj ). Therefore, by (2), deg(fj ) ≤ 2g − 2 + 2l < q, j = 1, . . . , s, as claimed. Claim E: N is g-Hilbertian. Sm Indeed, assume N = i=1 ϕi (Ci (N )) with Ci a curve over N of genus at most g and ϕi : Ci → A1 a dominant separable rational map over N of degree at least 2. Replace K by a finite extension in N , if necessary, to assume that Ci and ϕi are defined over K and the genus of Ci over K is the same as the genus of Ci over N . Choose a transcendental element x over K and a generic point zi for Ci over K with ϕi (zi ) = x, i = 1, . . . , m. Then Ei = K(zi ) is a finite separable extension of K(x) of degree at least 2. Denote the Galois closure of Ei by Fi . Next, list the cyclic extensions of K(x) of degree l which are contained in one of the fields Ei as D1 , . . . , Dr . For each j between 1 and r choose a primitive element yj for Dj /K(x) such that yjl = hj (x) with hj ∈ K[X]. Then hj (x) is not an lth power in K(x). Since each Ei is a regular extension of K, so is Dj /K. Hence, deg(hj ) ≥ 1. Since N is perfect, we may again replace a finite purely inseparable extension in N to assume that hj = Qsj Kmby hjkjk such that cj ∈ K × , hjk are distinct monic separable irreducible cj k=1 polynomials of positive degrees, and mjk are positive integers. If l|mjk for some j, k, replace yj by yj hjk (x)−mjk /l . Thus, assume without loss, l - mjk for all j and k. By Claim D, deg(hjk ) < q. Hence, by Claim C, there exists an arithmetip l / N for each a ∈ A, j = 1, . . . , r. cal progression Qm A of OK such that hj (a) ∈ Let F = i=1 Fi . Let H be the set of all a ∈ K such that each K-place ˜ ∪ {∞} with ψ(x) = a maps Fi onto F¯i ∪ {∞} and Ei onto ψ: F → K ¯ Ei ∪ {∞} with these properties: (3a) F¯i /K is a Galois extension and Gal(F¯i /K) ∼ = Gal(Fi /K(x)). ¯i = K(ci ) and [K(ci ) : K] = (3b) ψ is finite at zi and ci = ψ(zi ) satisfies E [Ei : K(x)]. By Lemma 13.1.1, H contains a separable Hilbert subset. Theorem 13.3.5(c) gives a ∈ A ∩ H. By assumption, a = ϕi (c) with c ∈ Ci (N ) for some i between 1 and m. Extend the K-specialization (x, zi ) → (a, c) to a

252

Chapter 13. The Classical Hilbertian Fields

˜ ∪ {∞}. Then (3) is true with ci = c. Since Fi is the Galois place ψ: F → K closure of Ei /K(x), F¯i is the Galois closure of K(c)/K). Since K(c) ⊆ N and N/K is normal, F¯i ⊆ N . By Claim A, Gal(F¯i /K) is an l-group. By ¯i ) is a proper subgroup of ¯i : K] = [Ei : K(x)] > 1. Thus, Gal(F¯i /E (3b), [E Gal(F¯i /K). Hence, Gal(F¯i /Ei ) is contained in a normal subgroup of index l of Gal(F¯i /K) [Hall, p. 45, Cor. 4.2.2]. By (3a), Gal(Fi /Ei ) is contained in a normal subgroup of Gal(Fi /K(x)) of index l, so K(x) has a cyclic extension of degree l in Ei . It is Dj for some j between 1 and r. Then b = ψ(yj ) satisfies ¯i ⊆ N . This contradicts the choice of a in A. Thus, our bl = hj (a) and b ∈ E Sr initial assumption N ⊆ i=1 ϕi (Ci (N )) is false, so N is g-Hilbertian.

13.7 Twisted Wreath Products Given finite groups A and G, there are several ways of constructing a new group out of them. We describe here two of them: the semidirect product and the wreath product. Definition 13.7.1: Semidirect products. Let A and G be profinite groups. Suppose G acts on A continuously from the right. That is, there is a continuous map G × A → A mapping (σ, a) onto aσ and satisfying these rules: (ab)σ = aσ bσ , a1 = a, and (aσ )τ = aστ for a, b ∈ A and σ, τ ∈ G. The semidirect product G n A consists of all pairs (σ, a) ∈ G × A with the product rule (σ, a)(τ, b) = (στ, aτ b). This makes G n A a profinite group −1 with unit element (1, 1) and (σ, a)−1 = (σ −1 , a−σ ). Identify each σ ∈ G (resp. a ∈ A) with the pair (σ, 1) (resp. (1, a)). This embeds G and A into G n A such that (1) A is normal, G ∩ A = 1, and GA = G n A. Each element of G n A has a unique presentation as σa with σ ∈ G and a ∈ A. The product rule becomes (σa)(τ b) = (στ )(aτ b) and the action of G on A coincides with conjugation: aσ = σ −1 aσ. The projection σa 7→ σ of G n A on G is an epimorphism with kernel A. Conversely, suppose H is a profinite group and A, G are closed subgroups satisfying (1). Then H is the semidirect product G n A. Likewise, consider a short exact sequence α

1 −→ A −→ H −→ G −→ 1 Suppose the sequence splits. That is, there exists a homomorphism α0 : G → H satisfying α(α0 (g)) = g for each g ∈ G. Then α0 is an embedding which we call a group theoretic section of α. Identifying G with α0 (G), we have H = G n A. Suppose ϕ1 : G → E and ϕ2 : A → E are homomorphisms of profinite groups and ϕ2 (aσ ) = ϕ2 (a)ϕ1 (σ) for all a ∈ A and σ ∈ G. Then ϕ(σa) = ϕ1 (σ)ϕ2 (a) is a homomorphism ϕ: G n A → E. Here is a Galois theoretic interpretation of semidirect products: Let K, L, E, F be fields, with L/K Galois, F/K Galois, EL = F , and E ∩L = K. Then Gal(F/K) = Gal(F/E) n Gal(F/L).

13.7 Twisted Wreath Products

253

Definition 13.7.2: Twisted wreath products. Let A and G be finite groups, G0 a subgroup of G, and Σ is a system of representatives for the right cosets G0 σ, σ ∈ G. Thus, [ [ G = · G0 σ = · σ −1 G0 . σ∈Σ

σ∈Σ

Suppose G0 acts on A from the right. Let σ0 IndG for all σ ∈ G and σ0 ∈ G0 }. G0 (A) = {f : G → A | f (σσ0 ) = f (σ) g Make IndG G0 (A) a group by the rule (f g)(σ) = f (σ)g(σ). Then f (σ) = f (σ)g(σ) , where f g denotes conjugation in IndG G0 (A) and the right hand side is conjugation in A. σ The group G acts on IndG G0 (A) by f (τ ) = f (στ ). This gives rise to the semidirect product G n IndG G0 (A), which we also denote by A wrG0 G. Each element of this group has a unique presentation as a product σf with σ ∈ G and f ∈ IndG G0 (A). The product and the inverse operation in A wrG0 G −1

are given by (σf )(τ g) = στ f τ g and (σf )−1 = σ −1 f −σ . The identification σ = σ · 1 and f = 1 · σ identifies G and IndG G0 (A) as subgroups of (A) is normal, G ∩ IndG A wrG0 G. Under this identification, IndG G0 G0 (A) = 1, and −1 (1 · f )(σ · 1) = 1 · f σ , conjugaG · IndG G0 (A) = A wrG0 G. Since (σ · 1) tion of f by σ in A wrG0 G coincides with the action of σ on f . The map σf 7→ σ is an epimorphism π: A wrG0 G → G with Ker(π) = IndG G0 (A). We call A wrG0 G the twisted wreath product of A and G with respect to G0 . Twisted wreath products are usually non-Abelian: Lemma 13.7.3: Let A be a nontrivial finite group, G a finite group, and G0 a proper subgroup of G. Then A wrG0 G is not commutative. σ Proof: Choose a ∈ A, a 6= 1. Define a function f ∈ IndG G0 (A) by f (σ) = a for σ ∈ G0 and f (σ) = 1 for σ ∈ G r G0 . Choose σ ∈ G r G0 . Then f σ (1) = f (σ) = 1 6= a = f (1). Hence, f σ 6= f . Consequently, A wrG0 G is not commutative.

The next result shows that not only twisted wreath products are in general non-Abelian but their centers are small: Lemma 13.7.4: Let π: A wrG0 G → G be a twisted wreath product of finite groups, H1 / A wrG0 G, and h2 ∈ A wrG0 G. Put I = IndG G0 (A) = Ker(π) and G1 = π(H1 ). Suppose A 6= 1. (a) Suppose π(h2 ) ∈ / G0 and (G1 G0 : G0 ) > 2. Then there is an h1 ∈ H1 ∩ I with h1 h2 6= h2 h1 .

254

Chapter 13. The Classical Hilbertian Fields

(b) Suppose G1 6≤ G0 and π(h2 ) ∈ / G1 G0 . Then there is an h1 ∈ H1 ∩ I with 0 / hh1 ih for all h0 ∈ π −1 (G1 G0 ). In particular, h1 h2 6= h2 h1 . hh1 2 ∈ Proof: Put σ2 = π(h2 ). Consider σ1 ∈ G1 and g ∈ I. By definition, there are f1 , f2 ∈ I with σ1 f1 ∈ H1 and h2 = σ2 f2 . Put h1 = g σ1 f1 g −1 . Then h1 = [σ1 f1 , g −1 ] ∈ [H1 , I] ≤ H1 ∩ I. For each τ ∈ G h1 (τ ) = (g σ1 )f1 (τ )g(τ )−1 = g(σ1 τ )f1 (τ ) g(τ )−1 . Hence, for all τ ∈ G and f 0 ∈ I we have: (2a)

hh1 2 (1) = hσ1 2 f2 (1) = h1 (σ2 )f2 (1) = g(σ1 σ2 )f1 (σ2 )f2 (1) g(σ2 )−f2 (1) , 0

(2b) hτ1 f (1) = h1 (τ )f (2c)

0

(1)

f1 (1)

h1 (1) = g(σ1 )

= g(σ1 τ )f1 (τ )f −1

g(1)

0

(1)

g(τ )−f

0

(1)

, and

.

We apply (2) in the proofs of (a) and (b) to special values σ1 and g. Choose a ∈ A, a 6= 1. Proof of (a): Since (G1 G0 : G0 ) > 2, there is a σ1 ∈ G1 with distinct cosets σ1−1 G0 , σ2 G0 , G0 . Thus, none of the cosets σ1 G0 , σ2 G0 , σ1 σ2 G0 is G0 . Therefore, by definition of I, there is a g ∈ I with g(σ1 ) = g(σ2 ) = g(σ1 σ2 ) = 1 and g(1) = a. By (2a), hh1 2 (1) = 1. By (2c), h1 (1) 6= 1. Consequently, hh1 2 6= h1 , as desired. σ −1

Proof of (b): Since Gσ1 2 = G1 6≤ G0 , we have G1 6≤ G0 2 . Hence, G1 ∩ G0 σ −1

and G1 ∩ G0 2

are proper subgroups of G1 . Their union is a proper subset σ −1

of G1 . Thus, there is an element σ1 ∈ G1 r G0 ∪ G0 2 . It follows that / σ1 σ2 G0 . By assumption, σ2 ∈ / G1 G0 . Therefore, there is a g ∈ I with σ2 ∈ g(G1 G0 ) = 1, g(σ1 σ2 ) = 1, and g(σ2 ) = a−1 . Consider τ ∈ G1 G0 and f 0 ∈ I. By (2a), hh1 2 (1) = af2 (1) 6= 1. By 0

0

(2b), hτ1 f (1) = 1. Hence, (hk1 )τ f (1) = 1 for all integers k. It follows that 0 hh1 2 ∈ / hh1 ih for all h0 ∈ π −1 (G1 G0 ). Remark 13.7.5: Decomposition of IndG G0 (A) into a direct product. a ∈ A associate a function fa : G → A: σ a if σ ∈ G0 fa (σ) = 1 if σ ∈ / G0 .

To each

These functions satisfy the following rules: fa fb = fab and g −1 fa g = fag(1) for all a, b ∈ A and g ∈ IndG G0 (A). Thus, the map a 7→ fa identifies A with r G0 ) = 1} of IndG the normal subgroup {f ∈ IndG G0 (A) | f (G G0 (A). Applying σ ∈ G on A gives the following normal subgroup of IndG G0 (A): r σ −1 G0 ) = 1}. Aσ = {f ∈ IndG G0 (A) | f (G

13.7 Twisted Wreath Products

255

Q An arbitrary element f ∈ IndG G0 (A) has a unique presentation f = σ∈Σ fσ , σ −1 −1 (σ σ0 ) = f (σ −1 σ0 ) with fσ ∈ A . Specifically, fσ (G r σ G0 ) = 1 and fσQ σ for all σ ∈ Σ and σ0 ∈ G0 . It follows that IndG G0 (A) = σ∈Σ A . The latter relation allows us to present A as a quotient of IndG G0 (A) in G various ways: Let N = {f ∈ IndG0 (A) | f (1) = 1}. For each σ ∈ G let N σ = −1 = 1}. Then the map f 7→ f (σ −1 ) {f σ | f ∈ N } = {f ∈ IndG G0 (A) | f (σ gives rise to a short exact sequence 1 → N σ → IndG G0 (A) → A → 1. Like in the preceding paragraph, we find that Y −1 ) = 1} = Aτ . (3) N σ = {f ∈ IndG G0 (A) | f (σ τ ∈Σ G0 τ 6=G0 σ

Note that G0 leaves N = N 1 invariant, so N / IndG G0 (A)G0 . We summarize some of the groups mentioned above in the following diagram: (4)

IndG G0 (A)

IndG G0 (A)G0

N

N G0

1

G0

A wrG0 G

G

Remark 13.7.6: Interpretation of twisted wreath products in Galois theory. (a) Let Fˆ /K be a finite Galois extension with Gal(Fˆ /K) ∼ = A wrG0 G and A 6= 1. View IndG G0 (A) and G under this isomorphism as subgroups of ˆ be the fixed fields in Fˆ , respectively, of the Gal(Fˆ /K). Let F , L, L0 , and K G G subgroups N , IndG0 (A), IndG0 (A)G0 , and G. Galois theory interprets the various relations among the subgroups of A wrG0 G as relations among fields: (5a) K ⊆ L0 ⊆ L ⊂ F ⊆ Fˆ . ˆ = K and LK ˆ = Fˆ . (5b) L ∩ K (5c) L/K, F/L0 , and Fˆ /K are finite Galois extensions. (5d) The Q fields F σ with σ ∈ Σ are linearly disjoint over L. Moreover, Fˆ = σ∈Σ F σ . (5e) There is a field F0 with L ∩ F0 = L0 and F = LF0 . L G0

L0

K

A

F G0

F0

N

Fˆ G0

ˆ L0 K

ˆ K

256

Chapter 13. The Classical Hilbertian Fields

The assertion “F/L0 is Galois” follows from “N / IndG G0 (A)G0 ”. Thus, Conˆ dition (5e) follows from (5a)-(5d) by taking F0 = F ∩ L0 K. ˆ satisfying Conditions (b) Conversely, consider fields K, L0 , L, F, Fˆ , K (5a)-(5d). Let ˆ F0 = F ∩ (L0 K),

ˆ ∼ G = Gal(Fˆ /K) = Gal(L/K),

ˆ ∼ G0 = Gal(Fˆ /L0 K)) = Gal(L/L0 ), = Gal(F/F0 ) ∼ A = Gal(F/L) ∼ = Gal(F0 /L0 ). In particular, Σ is a subset of Gal(Fˆ /K). Suppose also that (5d) holds. Then A / Gal(F/L0 ). Thus, G0 viewed as a subgroup of Gal(F/L0 ) acts on A by conjugation. We construct an isomorphism ϕ: A wrG0 G → Gal(Fˆ /K) which ˆ is the identity on G and maps IndG G0 (A) onto Gal(F /L). Construction of ϕ: By (5d), Fˆ /L is a Galois extension of degree |A||Σ| . ˆ n Gal(Fˆ /L). Hence, by (5b), Gal(Fˆ /K) = Gal(Fˆ /K) G For each σ ∈ G, the group IndG0 (A) acts on F σ by the rule zf =

(6)

zσ

−1

f (σ−1 ) σ

f ∈ IndG G0 (A),

,

z ∈ F σ. 0

This action does not depend on σ. Indeed, assume F σ = F σ with σ, σ 0 ∈ G. Write σ = σ0 τ and σ 0 = σ00 τ 0 with σ0 , σ00 ∈ G0 and τ, τ 0 ∈ Σ. Since F/F0 0 is Galois, F τ = F τ , so by (5a) and (5d), τ = τ 0 . Thus, σ 0 = ρσ with ρ = λ0 λ−1 ∈ G0 . Moreover, ρ−1 f (σ −1 ρ−1 )ρ = f (σ −1 ρ−1 )ρ = f (σ −1 ρ−1 ρ) = f (σ −1 ). Hence,

z (ρσ)

−1

f ((ρσ)−1 ) ρσ

=

zσ

−1

f (σ−1 ) σ

.

−1

f

If z ∈ L, then z = z, because f (σ ) as an element of A fixes L. σ Thus, the action (6) defines a homomorphism ϕσ : IndG G0 (A) → Gal(F /L). −1

−1

−1

−1

If ϕσ (f ) = 1, then (z σ )f (σ ) = z σ for each z ∈ F σ . Hence, z f (σ ) = z for each z ∈ F . Therefore, f (σ −1 ) = 1. It follows that Ker(ϕσ ) = N σ , with N σ as in (3). Using (5d), the ϕσ ’s, with σ ranging on Σ, define a homomorphism ϕ0 : IndG G0 (A) →

Y

Gal(F σ /L) = Gal(Fˆ /L)

σ∈Σ

Q T by ϕ0 (f ) = σ∈Σ ϕσ (f ). The kernel of ϕ0 is σ∈Σ N σ = 1 and |IndG G0 (A)| = |Σ| ˆ |A| = |Gal(F /L)|. Hence, ϕ0 is an isomorphism.

13.7 Twisted Wreath Products

257

Next we show that ϕ0 is compatible with the action of G. To this end −1 −1 let z ∈ F σ and τ ∈ G. Then z τ ∈ F στ and −1 ) τ −1 τ −1 −1 f (τ σ −1 ) (στ f = (z τ )(τ σ ) zτ = (z σ

−1

)f (τ σ

−1

) σ

=

zσ

−1

f τ (σ−1 ) σ

τ

= zf .

Hence, τ −1 ϕ0 (f )τ = ϕ0 (f τ ), as desired. This allows us to combine ϕ0 with the identity map of G and define the isomorphism ϕ: A wrG0 G → Gal(Fˆ /K). ˆ L0 , L, F, Fˆ realize the Having established ϕ, we say that the fields K, K, twisted wreath product A wrG0 G. We say that the fields K, L0 , L, F, Fˆ ˆ such that realize the twisted wreath product if there exists a field K ˆ ˆ , L, F, F realize the twisted wreath product. K, K, L0 ˆ L0 , L, F, Fˆ that realize A wrG G. Suppose E (c) Consider fields K, K, 0 ¯ = Gal(E/L) and with L ⊆ E ⊆ F . Let A is a Galois extension of L 0 σ ˆ =Q ˆ ˆ ˆ ˆ E σ∈Σ E . Then E is a Galois extension of K in F . Let J = E ∩ K. ∼ Gal(Fˆ /K) ∼ ˆK ˆ = Fˆ . Hence, Gal(E/J) ˆ ˆ = G and Gal(E/L ˆ 0 J) = Then E = ˆ ˆ = G0 . Moreover, L ∩ J = K and LJ = E. Gal(Fˆ /L0 K)

L

L0

E

E0

}} }} } }

} }} }}

F ˆ E

Fˆ ww w ww ww

F0

K ˆ realize A¯ wrG G. By (b), K, J, L0 , L, E, E 0

L0 J

J

ˆ L0 K w ww ww

ˆ vK vv v v vv

Remark 13.7.7: Wreath products. Suppose G0 is the trivial subgroup of G. Then the twisted wreath product A wrG0 G simplifies to the (usual) wreath G product A wr G. In this case IndG G0 (A) is just the group A of all functions f : G → A. Multiplication is carried out componentwise. Again, G acts on AG by the formula f σ (τ ) = f (στ ). Thus, A wr G is the semidirect product G n AG . Each element of G n AG has a unique representation as a product σf with σ ∈ G and f ∈ AG . The multiplication rule is σf · τ g = (στ )(f τ g). Identify each a ∈ A with the function Q fa : G → A given by fa (1) = a and σ fa (σ) = 1 for σ 6= 1. Then IndG G0 (A) = σ∈G A .

258

Chapter 13. The Classical Hilbertian Fields

ˆ is a field Now Suppose K ⊆ L ⊆ F ⊆ Fˆ is a tower of fields and K satisfying these conditions: (7a) L/K, F/L, and Fˆ /K are finite Galois extensions. ˆ = K and LK ˆ = Fˆ . Write G = Gal(Fˆ /K). ˆ (7b) L ∩ K Q (7c) The fields F σ with σ ∈ G are linearly disjoint over L and Fˆ = σ∈G F σ . By Remark 13.7.6(b), there is an isomorphism ϕ: A wr GQ→ Gal(Fˆ /K) which is the identity on G, maps AG onto Gal(Fˆ /L), and σ6=1 Aσ onto Gal(Fˆ /F ). By (7b), restriction to L maps G isomorphically onto Gal(L/K). ˆ L, F, Fˆ realize the wreath product A wr G. We say that the fields K, K,

13.8 The Diamond Theorem The diamond theorem proved in this section says all fields ‘captured’ between two Galois extensions of a Hilbertian field are Hilbertian. In particular, this theorem implies that a non-Hilbertian Galois extension N of a Hilbertian field K is not the compositum of two Galois extensions of K which are properly contained in N . For example, Ks is not the compositum of two proper subfields which are Galois over K. Lemma 13.8.1: Let K ⊆ L0 ⊆ L be fields with L/K finite and Galois. Let c1 , . . . , cn be a basis of L0 /K and t1 , . . . , tn algebraically independent elements over K. Let f ∈ L0 [U, X] be an absolutely irreducible polynomial which is monic in X and Galois over L(U ) with degX f ≥ 2. Put G = Gal(L/K) = Gal(L(t)/K(t)), G0 = Gal(L/L0 ) = Gal(L(t)/L0 (t)), and A = Gal(f (U, X), L(U )). Then G0 acts on A and there exist fields F and Fˆ with the following properties: (a) F P is a regular extension of L. Moreover, F = L(t, z) with irr(z, L(t)) = n f ( i=1 ci ti , Z) ∈ L0 [t, Z]. (b) K(t), L0 (t), L(t), F, Fˆ realize A wrG0 G. Proof: Fix x ∈ L(U )s with f (U, x) = 0. Since f is absolutely irreducible, [L(U, x) : L(U )] = [L0 (U, x) : L0 (U )], so L(U ) is linearly disjoint from L0 (U, x) over L0 (U ) and we may identify both Gal(L(U )/L0 (U ) and Gal(L(U, x), L0 (U, x)) with G0 via restriction. Since f (U, X) is Galois over L(U ) with respect to X, the extension L(U, x)/L(U ) is Galois with Galois group A. In addition, A 6= 1 because degX f ≥ 2. The fixed field in L(U, x) of the subgroup of Aut(L(U, x)) generated by A and G0 is L0 (U ). By Artin [Lang7, p. 264, Thm. 1.8], L(U, x)/L0 (U ) is Galois and Gal(L(U, x)/L0 (U )) ∼ = Gal(L(U, x)/L0 (U, x))nGal(L(U, x)/L(U )) ∼ = G0 nA. This defines an action of G0 on A. As in Section 13.7, choose a system of representatives Σ for the right cosets G0 σ of G0 in G. Let S = {G0 σ | σ ∈ Σ}. Choose algebraically independent elements uS , S ∈ S, over K. Write Q = L(uS | S ∈ S). For

13.8 The Diamond Theorem

259

each S ∈ S write cSi = cσi and f S = f σ for some σ ∈ S. Since ci ∈ L0 and f ∈ L0 (U, X), both cSi and f S are independent of σ. Then choose a root z S of f S (uS , Z) in Qs and write F S = Q(z S ). Since f S is absolutely irreducible, LS0 (uS , z S ) is a regular extension of LS0 (Example 2.6.11). Hence, L(uS , z S ) is a regular extension of L. Since L(uS , z S ), S ∈ S, are algebraically independent over L, they are linearly disjoint over L (Lemma 2.6.7). Moreover, their compositum Fˆ = L(uS , z S | S ∈ S) = Q(z S | S ∈ S) is a regular extension of L. By Lemma 2.5.11, in each rectangle of the following diagram, the fields lying in the left upper corner and the right lower corner are linearly disjoint over the field in the left lower corner and their compositum is the field in the right upper corner: 0

0

L(uS , z S | S 0 6= S)

0

0

L(u)(z S | S 0 6= S)

A

0

0

ˆ L(uS , z S | S 0 ∈ S) = F

A

L(u, z S ) = F S

L(uS )

A

L(uS , z S )

S LS 0 (u )

A

S S LS 0 (u , z )

L(uS | S 0 6= S)

L(t) = L(u) = Q

L

LS 0

It follows that, F S /Q is a Galois extension with Galois group isomorphic to A, the F S , S ∈ S, are linearly disjoint over Q, and Fˆ is their compositum. Thus, Fˆ /Q is Galois. The matrix (cSi ) ∈ Mn (L) is invertible [Lang7, p. 286, Cor. 5.4]. Hence, the system of equations (1)

n X

cSi Ti = uS ,

S∈S

i=1

has a unique solution t01 , . . . , t0n . It satisfies L(t01 , . . . , t0n ) = L(uS | S ∈ S) = Q. Since trans.deg(Q/K) = n, the elements t01 , . . . , t0n are algebraically independent over K. We may therefore assume that t0i = ti , i = 1, . . . , n, so Q = L(t). Use the linear disjointness of the F S ’s over Q to extend the action of G on L to an action on Fˆ : (uS )τ = uSτ and (z S )τ = z Sτ . In particular, τ permutes the equations of (1). Hence, (tτ1 , . . . , tτn ) is a solution of (1), so it coincides with (t1 , . . . , tn ). In other words, the action of G on Q is the unique extension of the given action on L that fixes t1 , . . . , tn . In particular, K(t) is the fixed field of G in Q. Now write u = uG0 ·1 , z = z G0 ·1 , andQF = F G0 ·1 . Then F S = F σ for σ ˆ all PnS ∈ S and each σ ∈ S. Hence, F = σ∈Σ F . Moreover, by (1), u = i=1 ci ti ∈ L0 (t), u is transcendental over K, f (u, z) = 0, and F = Q(z).

260

Chapter 13. The Classical Hilbertian Fields

Hence, f (u, Z) ∈ L0 (u)[Z], and f (u, Z) is Galois over L(u). Since Q = L(u) is a purely transcendental extension of L(u), f (u, Z) is irreducible and Galois over Q. Therefore, f (u, Z) = irr(z, Q) (this settles (a)), and F is a Galois extension of L0 (t). Finally, since Fˆ /L(t) and L(t)/K(t) are Galois and each τ ∈ G = Gal(L(t)/K(t)) extends to an automorphism of Fˆ , the extension Fˆ /K(t) ˆ be the fixed field of G in Fˆ . Then K ˆ ∩ L(t) = K(t) is Galois. Let K ˆ = Fˆ . By Remark 13.7.6(b), is the fixed field of G in L(t) and L(t)K ˆ L0 (t), L(t), F, Fˆ realize A wrG G. K(t), K, 0 Proposition 13.8.2 (Realization of Twisted Wreath Products): Let K ⊆ L0 ⊆ L be a tower of fields with K Hilbertian and L/K finite Galois. Consider an absolutely irreducible polynomial f (T, X) ∈ L0 [T, X] which is Galois over L(T ). Let G = Gal(L/K), G0 = Gal(L/L0 ), and A = Gal(f (T, X), L(T )). Then G0 acts on A and there are fields M , N , such that K, L0 , L, M, N realize A wrG0 G. Proof: Let t1 , . . . , tn be algebraically independent elements over K with n = ˆ F , and Fˆ such that K(t), K, ˆ L0 (t), [L0 : K]. Lemma 13.8.1 gives fields K, L(t), F, Fˆ realize A wrG0 G. Thus, Conditions (5a)-(5d) of Section 13.7 hold for K(t), L0 (t), L(t) instead of for K, L0 , L. Since K is Hilbertian, Lemma 13.1.1 gives a ∈ K r such that those conditions hold for the residue fields under each L-place ϕ of Fˆ with ϕ(K(t)) = K ∪ {∞}. Lemma 2.6.6 gives such a place. Moreover, the residue fields of L0 (t) and L(t) under ϕ are L0 ˆ 0 , F 0 , and Fˆ 0 be the residue fields of K, ˆ F , and Fˆ , and L, respectively. Let K respectively, under ϕ. By Remark 13.7.6(b), they realize A wrG0 G. Theorem 13.8.3 (Diamond Theorem [Haran4, Thm. 4.1]): Let K be a Hilbertian field, M1 and M2 Galois extensions of K, and M an intermediate field of M1 M2 /K. Suppose that M 6⊆ M1 and M 6⊆ M2 . Then M is Hilbertian. Proof: Corollary 12.2.3 allows us to assume that [M : K] = ∞. Part A of the proof strengthens this assumption: Part A: We may assume [M : (M1 ∩ M )] = ∞. Otherwise, [M : (M1 ∩ M )] < ∞. Then K has a finite Galois extension M20 such that M ⊆ (M1 ∩ M )M20 . Then M ⊆ M1 M20 and [M : M ∩ M20 ] = ∞. Replace M1 by M20 and M2 by M1 , if necessary, to restore our assumption. Part B: Reduction to an absolutely irreducible Galois polynomial. By Lemma 13.1.4, each separable Hilbert subset of M contains a subset of the form M ∩HM 0 (f ), where M 0 is a finite Galois extension of M and f ∈ M [T, X] is an absolutely irreducible polynomial which is monic and separable in X and f (T, X) is Galois over M 0 (T ). It suffices to find a ∈ M such that f (a, X) is irreducible over M 0 . We may assume M 0 ⊆ M1 M2 . Indeed, K has a finite Galois extension K 0 such that M 0 ⊆ M K 0 . If M 0 6⊆ M2 K 0 , replace M2 by M2 K 0 . If M 0 ⊆ M2 K 0 ,

13.8 The Diamond Theorem

261

replace M1 by K 0 . In the latter case we still have [M : (K 0 ∩ M )] = ∞, because [M : K] = ∞ and [K 0 : K] < ∞. Thus, we may assume that f (T, X) is Galois over M1 M2 . It suffices to produce a ∈ M such that f (a, X) is irreducible over M1 M2 . Part C: Finite Galois extensions. Write M0 = M and N = M1 M2 . Then N/K is Galois. For each finite Galois extension L of K in N let Li = Mi ∩ L, i = 0, 1, 2. Then Li /K is Galois, i = 1, 2. Use the assumptions M0 6⊆ Mi , i = 1, 2 and [M0 : M1 ∩ M0 ] = ∞ (Part A) to choose L large with L0 6⊆ Li for i = 1, 2, [L0 : L1 ∩ L0 ] > 2, and (2) f ∈ L0 [T, X] and f (T, X) is Galois over L(T ). The conditions on the fields Li translate into conditions on the groups Gi = Gal(L/Li ), i = 0, 1, 2: (3a) G1 , G2 6≤ G0 . (3b) (G0 G1 : G0 ) > 2. Part D: Twisted wreath products. Let A0 = Gal(f, L(T )) = Gal(f, Ks (T )). Choose a basis b1 , . . . , bn for L0 /K and algebraically independent elements t1 , . . . , tn over K. By (2) and Lemma 13.8.1, the group G0 acts on A0 . Moreover, there are fields F, Fˆ such that (4a) K(t), L0 (t), L(t), F, Fˆ realize A0 wr PGn0 G and (4b) F = L(t, z) with irr(z, L(t)) = f ( i=1 bi ti , X). Since K is Hilbertian, Lemma 13.1.1 gives c1 , . . . , cn ∈ K such that the specialization t → c extends to an L-place of Fˆ onto a Galois extension Fˆ 0 of K with Galois group isomorphic to Gal(Fˆ /K(t)). Thus, Fˆ 0 has a subfield F 0 with these properties: G and (5a) K, L0 , L, F 0 , Fˆ 0 realize A0 wrG0P n (5b) F 0 = L(z 0 ) with irr(z 0 , L) = f ( i=1 bi ci , X). Pn Let a = i=1 bi ci . Then a is in L0 , hence in M . If we prove N ∩ F 0 = L, it will follow from (5b) that [N (z 0 ) : N ] = [F 0 : L] = deg(f (a, X)). Thus, f (a, X) will be irreducible over N , as desired. Part E: Conclusion of the proof. Let E = N ∩ F 0 and A = Gal(E/L). By ˆ of K such Remark 13.7.6(c), G0 acts on A and there is a Galois extension E that ˆ realize H = A wrG G. (6) K, L0 , L, E, E 0 ˆ ⊂ N. In particular, all conjugates of E over K are in N . Hence, E ˆ ˆ Identify H with Gal(E/K) such that resE/L : Gal( E/K) → Gal(L/K) ˆ coincides with the projection π: H → G. Then π ◦ resN/Eˆ = resN/L . For i = 1, 2 let Hi = resN/Eˆ (Gal(N/Mi )). Then Hi / H and π(Hi ) = resN/L (Gal(N/Mi )) = Gi . Since Gal(N/M1 ) and Gal(N/M2 ) are normal subgroups of Gal(N/K) with a trivial intersection, they commute. Hence, H1 and H2 commute.

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By (3a) there exists h2 ∈ H2 with π(h2 ) ∈ / G0 . If A were not trivial, then by (3b) and Lemma 13.7.4 there would be h1 ∈ H1 which does not commute with h2 . We conclude from this contradiction that A = 1 and therefore N ∩ F 0 = L, as desired. We conclude this section with an application of the diamond theorem that solves Problems 12.18 and 12.19 of [Fried-Jarden3]: Corollary 13.8.4 ([Haran-Jarden5]): Let K be a Hilbertian field and let N be a Galois extension of K which is not Hilbertian. Then N is not the compositum of two Galois extensions of K neither of which is contained in the other. In particular, this conclusion holds for Ks .

13.9 Weissauer’s Theorem The most useful application of the diamond theorem is part (b) of the following result: Theorem 13.9.1 (Weissauer): Let K be a Hilbertian field. (a) Let M be a separable algebraic extension of K and M 0 a finite proper separable extension of M . Suppose M 0 is not contained in the Galois closure of M/K. Then M 0 is Hilbertian. (b) Let N be a Galois extension of K and N 0 a finite proper separable extension of N . Then N 0 is Hilbertian. (c) Let N be a Galois extension of a Hilbertian field K and L a finite proper separable extension of K. Suppose that N ∩ L = K. Then N L is Hilbertian. Proof: Statement (c) is a special case of Statement (b). Statement (b) is a special case of Statement (a). Statement (a) follows from Corollary 12.2.3 if [M : K] < ∞. Suppose therefore that [M : K] = ∞. Then K has a finite Galois extension L with M 0 ⊆ M L. Let N be the Galois closure of M/K. Then M 0 is contained neither in L nor in N . By Theorem 13.8.3, M 0 is Hilbertian. Proposition 13.9.3 below is a stronger version of Theorem 13.9.1(c) which gives information about the Hilbert sets of N L. That version implies Theorem 13.9.1(a), hence also Theorem 13.9.1(b). The proof of Proposition 13.9.3 depends on the following lemma rather that on the diamond theorem: Lemma 13.9.2: Let L = K(α) be a finite proper separable extension of a field ˆ be the Galois hull of L/K. Consider an absolutely irreducible K and let L polynomial h ∈ L[T, X] with degX (h) > 1 which is separable with respect to X. For u, v algebraically independent elements over K, let E be a Galois extension of K(u, v) such that E ∩ L(u, v) = K(u, v). Then the polynomial ˆ h(u + αv, X) has no root in the field E L. ˆ Then choose σ ∈ Gal(F/E) with Proof: Put t = u + αv and F = E L. 0 0 α = σα 6= α and put t = σt = u + α0 v and h0 = σh. Assume that there

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exists x ∈ F with h(t, x) = 0. Then x0 = σx ∈ F and h0 (t0 , x0 ) = 0. Since ˆ t0 ) = L(u, ˆ v). Hence, t, t0 are algebraically independent α 6= α0 , we have L(t, over K. ˆ t0 , x) ˆ x) F L(t, L(t, ˆ L(t)

ˆ t0 ) L(t,

ˆ t0 , x0 ) L(t,

ˆ L

ˆ 0) L(t

ˆ 0 , x0 ) L(t

ˆ Since ˆ x) and L(t ˆ 0 , x0 ) are algebraically independent over L. Therefore, L(t, ˆ they are regular extensions of L they are linearly disjoint (Lemma 2.6.7). By ˆ t0 , x0 ) are linearly disjoint over L(t, ˆ t0 ). In ˆ t0 , x) and L(t, Lemma 2.5.11, L(t, particular, (1)

ˆ t0 , x0 ) = L(t, t0 ). ˆ t0 , x) ∩ L(t, L(t,

ˆ t0 ), we have Gal(F/E0 ) = Gal(F/E)× On the other hand, with E0 = E ∩ L(t, ˆ t0 )). Hence, στ = τ σ for all τ ∈ Gal(F/L(t, ˆ t0 , x)). In particular, Gal(F/L(t, ˆ t0 , x). By (1), L(t, ˆ t0 , x0 ) = τ x0 = τ σx = στ x = σx = x0 . Therefore, x0 ∈ L(t, ˆ t0 ), a contradiction to degX (h0 ) > 1. L(t, Proposition 13.9.3: Let L be a finite proper separable extension of a Hilbertian field K and let M be a Galois extension of K such that M ∩L = K. Then the field N = M L is Hilbertian. Moreover, every separable Hilbert subset of N contains elements of L. Proof: By Lemma 13.1.2, it suffices to consider absolutely irreducible polynomials h1 , . . . , hm ∈ N [T, X], separable monic and of degree at least 2 in X and to find c ∈ L such that hi (c, X) has no root in N , i = 1, . . . , m. Let u and v be algebraically independent over K. Choose a primitive element α for L/K and let L0 be a finite extension of L which is contained in N and contains all coefficients of h1 , . . . , hm . Put K 0 = M ∩ L0 and let F be a finite Galois extension of K 0 (u, v) that contains L0 (u, v) and over which all polynomials h1 (u + αv, X), . . . , hm (u + αv, X) decompose into linear factors. Let g ∈ K[u, v] be the product of the discriminants of h1 (u + αv, X), . . . , hm (u + αv, X) with respect to X. Let B 0 be the set of all (a, b) ∈ (K 0 )2 with g(a, b) 6= 0 satisfying the following condition: (2) The L0 -specialization (u, v) → (a, b) extends to a place ϕ of F which induces an isomorphism ϕ0 : Gal(F/K 0 (u, v)) → Gal(F 0 /K 0 ) (with F 0 being the residue field of F ) such that (ϕ0 σ)(ϕx) = ϕ(σx) for all σ ∈ Gal(F/K 0 (u, v)) and x ∈ F with ϕx 6= ∞. In particular ϕ maps the set of all zeros of hi (u + αv, X) bijectively onto the set of zeros of hi (a + αb, X), i = 1 . . . , m. By Lemma 13.1.1 and Example 2.6.10, B 0 contains a separable Hilbert subset A0 of (K 0 )2 . By Corollary 12.2.3, A0 contains a separable Hilbert subset of K 2 .

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Thus, there are a, b ∈ K satisfying (2). Let c = a + αb. Assume there is an i ¯ in N . Then between 1 and m such that the polynomial hi (c, X) has a root x x ¯ ∈ F 0 ∩ N . Since res: Gal(N/L0 ) → Gal(M/K 0 ) is an isomorphism, there exists a Galois extension E 0 of K 0 contained in M such that E 0 L0 = F 0 ∩ N . Since E 0 ∩ L0 = K 0 = M ∩ L0 , Gal(F 0 /E 0 ) · Gal(F 0 /L0 ) = Gal(F 0 /K 0 ) Gal(F 0 /E 0 ) ∩ Gal(F 0 /L0 ) = Gal(F 0 /F 0 ∩ N ).

(3)

Therefore, K 0 (u, v) has a Galois extension E in F such that ϕ0 (Gal(F/E)) = Gal(F 0 /E 0 ). From (3), E ∩ L0 (u, v) = K 0 (u, v) and ϕ0 (Gal(F/EL0 )) = Gal(F 0 /F 0 ∩ N ). Moreover, the polynomial hi (u + αv, X) has a root x such x) = x ¯= that ϕ(x) = x ¯. For each σ ∈ Gal(F/EL0 ) we have ϕ(σx) = ϕ0 (σ)(¯ ϕ(x). Hence, σx = x. In particular, x ∈ EL0 , a contradiction to Lemma 13.9.2. We deduce Theorem 13.9.1(a) from Proposition 13.9.3: Proposition 13.9.4: Let M be a separable algebraic extension of a Hilbertian field K and M 0 a proper finite separable extension of M . Suppose M 0 is not contained in the Galois closure of M/K. Then M 0 is Hilbertian. ˆ . With no loss replace M Proof: Denote the Galois closure of M/K by M ˆ ∩ M 0 . Choose a primitive element α for M 0 /M . Put L0 = K(α), and by M ˆ ∩ L0 . Then the conditions of Proposition 13.9.3 are satisfied for K0 = M 0 0 ˆ K , L , M replacing K, L, M . Let f ∈ M 0 [T, X] be an irreducible polynomial, monic and separable in X with degX (f ) > 1. By Lemma 13.1.2, there exist h1 , . . . , hm ∈ M 0 [T, X] 0 absolutely irreducible and separable in X such that HM 0 (h1 , . . . , hm ) ⊆ 0 ˆ HM 0 (f ). Since h1 , . . . , hm are irreducible in M [T, X], Proposition 13.9.3, ˆ (α)[X] and theregives a ∈ L0 with h1 (a, X), . . . , hm (a, X) irreducible in M 0 0 fore in M [X]. By Proposition 13.2.2, M is Hilbertian. ˜ whose finite proper extensions Example 13.9.5: Non-Hilbertian subfield of Q are Hilbertian. Denote the compositum of all finite solvable extensions of Q by Qsolv . It is not Hilbertian; there exists no a ∈ Qsolv such that X 2 − a is irreducible over Qsolv . However, Qsolv is a Galois extension of Q. Hence, by Corollary 13.9.1(b), every finite proper extension of Qsolv is Hilbertian. By Corollary 13.8.4, Qsolv is not the compositum of two Galois extensions of Q neither of which is contained in the other.

Exercises 1. [Fried10, §2] This exercise shows that it is not always possible to take m = 1 in Lemma 13.1.2. Consider an irreducible polynomial f ∈ Q[T, X], monic in X, and g1 , g2 ∈ Q[Y ], polynomials of positive degree for which f (gi (Y ), X)

Exercises

265

is reducible, i = 1, 2. In addition, suppose there exists an irreducible h ∈ Q[T, X] with degX (h) > 1 such that HQ0 (h) = {a ∈ Q | h(a, X) has no zero in Q} ⊆ HQ (f ). (a) Use Exercise 1 of Chapter 12 to conclude that h(gi (Y ), X) has a factor of degree 1 in X, say X − mi (Y ), where mi ∈ Q(Y ), i = 1, 2. (b) Use field theory to interpret (a): Let t be an indeterminate, yi a zero of gi (Y ) − t, and xi = mi (yi ), i = 1, 2. Note that x1 and x2 are both zeros of h(t, X), so that they are conjugate over Q(t). Conclude that [Q(y1 ) ∩ Q(y20 ) : Q(t)] > 1 for some conjugate y20 of y2 over Q(t). (c) Consider the case f (T, X) = X 4 + 2X 2 − T and g1 (Y ) = Y 4 + 2Y 2 , g2 (Y ) = −4Y 4 − 4Y 2 − 1. Prove that the splitting field of gi (Y ) − t over Q(t) has the dihedral group of order 8 as its Galois group over Q(t). Observe that since Q(y1 )/Q(t) is a nonnormal extension of degree 4, Q(y12 )/Q(t) is its only quadratic subextension. Prove that the prime of Q(t) corresponding to the specialization t → 0 is unramified in Q(y12 ) but ramified in Q(y22 ). Conclude that there is no irreducible h ∈ Q[T, X] with degX (h) > 1 and HQ0 (h) ⊆ HQ (f ). 2. Let f1 (T, X), . . . , fm (T, X), with degX (fi ) > 1, i = 1, . . . , m, be absolutely irreducible polynomials, separable in X, with coefficients in a global field K. Let t be transcendental over K and denote the splitting fields of f1 (t, X), . . . , fm (t, X), respectively, over E = K(t) by F1 , . . . , Fm . Assume F1 , . . . , Fm are linearly disjoint over E and let F = F1 . . . Fm . Observe that there is a σ ∈ Gal(F/E) that fixes none of the roots of f1 (t, X) . . . fm (t, X) and improve Lemma 13.3.4. Show there exists a prime ideal p of OK and an element ap ∈ OK such that for a ∈ OK , if a ≡ ap mod p, then fi (a, X) has no zeros in K, i = 1, . . . , m. 3.

Consider the three absolutely irreducible polynomials f1 (T, X) = X 2 − T, f2 (T, X) = X 2 − (T + 1), f3 (T, X) = X 2 − T (T + 1)

and let H = HQ (f1 , f2 , f3 ). Choose a prime number p and an integer a such that both a and a + 1 are quadratic nonresidues modulo p. Prove that a + pZ ⊆ H, even though F1 , F2 , and F3 (in the notation of Exercise 2) are not linearly disjoint over E. 4. Let K be a Hilbertian field with valuation v. Prove that every Hilbert subset H of K r is v-dense in K r . Hint: Let H = HK (f1 , . . . , fm ), where f1 , . . . , fm are irreducible polynomials in K(T1 , . . . , Tr )[X1 , . . . , Xn ]. For (a1 , . . . , ar ) ∈ K r , and γ = v(c) an element of the value group, each polynomial in the set {fi (a1 + cT1ε1 , . . . , ar + cTrεr , X) | 1 ≤ i ≤ m and ε1 , . . . , εr ∈ {±1}}, is irreducible in K(T)[X]. Substitute elements t1 , . . . , tr for T1 , . . . , Tr so that these polynomials remain irreducible in K(X). Thus, find (b1 , . . . , br ) ∈ H such that v(bi − ai ) ≥ γ, i = 1, . . . , r.

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5. Suppose K is the quotient field of a Hilbertian ring R. Let M1 , M2 be Galois extensions of K and M an extension of K in M1 M2 which is contained in neither M1 nor in M2 . Prove the integral closure of R in M is Hilbertian. Hint: In the proof of Theorem 13.8.3 choose b1 , . . . , bn integral over R. Then choose c1 , . . . , cn in R.

Notes Hilbert [Hilbert, p. 280] proves Lemma 13.1.1(b) for number fields. Our proof of Proposition 13.2.1 is a version of Inaba’s proof [Inaba]. Theorem 13.3.5(a) for number fields appears in [Eichler]. We follow [Fried5]. Lang [Lang3, p. 152] reproduces Franz’s power series expansion proof [Franz]. Theorem 13.6.2 for g = 0 and K a number field appears in [CorvajaZannier]. Generalization to arbitrary g and arbitrary countable Hilbertian field K of characteristic 0 appears in [Fried-Jarden4]. The proof uses a deep group theoretic result due to Guralnick, Thompson, et al. and the Riemann existence theorem. Thus, one cannot generalize the proof to positive characteristic. An elementary proof of the theorem for arbitrary g but still for K a number field can be found in [Zannier]. Our proof generalizes that of Zannier. We replace the polynomial Y 2 − X 2q + 2 which Zannier uses by the polynomial Y l − X lq + u in order to make the proof works for each global field K. Weissauer proves that every finite proper separable extension of a Galois extension of a Hilbertian field is Hilbertian [Weissauer, Satz 9.7]. Section 13.9 replaces Weissauer’s “nonstandard” proof by a simpler “standard” one still using Weissauer’s auxiliary variable trick as in Lemma 13.9.2, [Fried10, Thm. 1.3]. Haran’s diamond theorem 13.8.3 generalizes Weissauer’s theorem (Theorem 13.9.1) and Corollary 13.8.5 [Haran-Jarden5]. Its proof is an offspring of the proofs of the two earlier results. Proposition 13.4.1 says the ring of integers OK of a number field K is Hilbertian. Thus, OK ∩ HK (f ) is an infinite set for each irreducible polynomial f ∈ K[T, X]. One may further ask when OK r HK (f ) is finite. This is not always the case. For example, OK r HK (X 2 − T ) is the infinite set of uller proves the set Z r HQ (f ) is finite all squares in OK . But, for K = Q, M¨ in each of the following cases: degX (f ) is a prime number and the curve f (T, X) = 0 has positive genus or Gal(f (T, X), Q(T )) ∼ = Sn for some positive integer n 6= 5 [M¨ uller, Thm. 1.2]. The proof uses Siegel’s theorem about integral points on algebraic curves and classical results about finite groups.

Chapter 14. Nonstandard Structures A. Robinson invented “nonstandard” methods in order to supplement the Weierstrass ε, δ formalism of the calculus by a rigorous version of the classical calculus of infinitesimals in the spirit of Leibniz and other formalists. We will apply the nonstandard approach to algebra in Chapter 15 in order to find new Hilbertian fields. Its main virtue, from an algebraic point of view, is that it creates additional algebraic structures to which well known theorems can be applied. Here we present the basics of the nonstandard method: the higher order structure on a set M (Section 14.1); the concept of an enlargement M ∗ of M (Sections 14.2 and 14.3); and the existence of M ∗ (Section 14.4) via ultraproducts.

14.1 Higher Order Predicate Calculus Sentences that speak of arbitrary subsets, functions, relations, etc., are common in mathematics, even though they are usually outside the scope of first order languages. Here we introduce a language which includes such sentences. First the notion of a type (of a higher order object) is inductively defined by the following rules: (1a) The number 0 is a type. (1b) If n is a positive integer and τ (1), . . . , τ (n) are types, then the sequence (τ (1), . . . , τ (n)) is a type. Denote the set of all types by T . Given a set M , attach a set Mτ to each type τ as follows: (2a) M0 = M . (2b) If τ (1), . . . , τ (n) are types and τ = (τ (1), . . . , τ (n)), then Mτ = P(Mτ (1) × · · · × Mτ (n) ) = {all subsets of Mτ (1) × · · · × Mτ (n) }. Elements of Mτ are objects of type τ over M . We call them sets (or relations) of type τ if τ 6= 0. A higher order set is a set of type τ for some τ 6= 0. Call the system M = hMτ | τ ∈ T i the higher order structure over M . For each higher order set A of M introduce a sequence of variables XA1 , XA2 , . . . . Inductively define formulas of the higher order language L∞ (M ) as follows: (3a) XAi = XAj , XAi = a, and a = b are formulas for each higher order set A, all a, b ∈ A, and i, j ∈ N. (3b) If A(1), . . . , A(n) are higher order sets, A(0) is a subset of P(A(1) × · · · × A(n)), and for ν between 0 and n either tν is a variable XA(ν),i(ν) or an element of A(ν), then (t1 , . . . , tn ) ∈ t0 is a formula.

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(3c) Negations and disjunctions (hence also conjunctions and implications) of formulas are formulas. (3d) If ϕ is a formula, then (∃XAi ∈ A)[ϕ] (hence also (∀XAi ∈ A)[ϕ]) is a formula for all higher order set A and i ∈ N. Define free variables of a formula as usual (Section 7.1). In particular, a sentence is a formula without free variables. Likewise, a substitution is a function f that, for each higher order set A of M , replaces each XAi by an element xAi of A and fixes the elements of A. Interpret the truth of a formula ϕ under f by reading “(t1 , . . . , tn ) ∈ t0 ” as “(f (t1 ), · · · , f (tn )) belongs to f (t0 ).” Note that each object t of type τ = (τ (1), . . . , τ (n)) 6= 0 has two roles in the language L∞ (M ). First, it is a constant: t is equal or not equal to another object of the same type. Second, it is a relation between objects of types τ (1), . . . , τ (n). If the set M has a first order structure, embed this in the higher order structure of M in a natural way. For example, if M is a field, then addition corresponds to a subset A of M × M × M , and X1 + b = c in the first order language becomes (XM 1 , b, c) ∈ A. Remark: Let N∗ be a proper elementary extension of N. Then N(0) (resp. (N∗ )(0) ) is the collection of subsets of N (resp. N∗ ). The second order structure N∗ ∪ (N∗ )(0) is not an elementary extension of the second order structure N ∪ N(0) . For example the induction axiom (∀X ∈ N(0) ) [1 ∈ X ∧ (∀x ∈ N)[x ∈ X → x + 1 ∈ X]] → (∀x ∈ N)[x ∈ X] holds in N ∪ N(0) but it fails in N∗ ∪ (N∗ )(0) if we replace N and N(0) by N∗ and (N∗ )(0) . Indeed, the induction axiom does not hold for X = N. In order to restore the elementary extension property for higher order structures of the elementary extension N∗ of N, we must restrict quantification of subsets of N∗ to a proper subcollection of (N∗ )(0) , to be denoted N∗(0) , the “inner” subset of N∗ . We define these in the next section.

14.2 Enlargements We consider a set M together with its higher order structure and define an enlargement of M as a special model of the higher order theory of M satisfying Conditions I and II below and III of Section 14.3. The enlargement will be saturated with respect to all higher order relations. The underlying set M ∗ of this model has the property that to each object a of type τ of M there corresponds an object a∗ of M ∗ of the same type (i.e. a∗ is an element of (M ∗ )τ ). Call such an a∗ standard. In particular, Mτ itself, an element of M(τ ) , corresponds to an element (Mτ )∗ of (M ∗ )(τ ) . This means (Mτ )∗ is a subset of (M ∗ )τ . Call the elements of (Mτ )∗ the internal objects of type τ of M ∗ . All other elements of (M ∗ )τ are external.

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269

Note: For τ = 0 we have (M0 )∗ = (M ∗ )0 = M ∗ . Thus, every element of M ∗ is internal, although (M ∗ )(0) itself may have external subsets. To simplify notation write Mτ∗ for (Mτ )∗ . If τ (1), . . . , τ (n) are types and if R ⊆ Mτ (1) × · · · × Mτ (n) , then R ∈ Mτ for τ = (τ (1), . . . , τ (n)). Hence, R∗ ∈ Mτ∗ , that is, R∗ ⊆ (M ∗ )τ (1) × · · · × (M ∗ )τ (n) . We demand, however, that an enlargement satisfy a stronger condition: I. Internal condition on n-ary relations. If τ (1), . . . , τ (n) are types and if R ⊆ Mτ (1) × · · · × Mτ (n) , then R∗ ⊆ Mτ∗(1) × · · · × Mτ∗(n) . That is, the elements of a standard n-ary relation are n-tuples with internal coordinates. Call them internal n-tuples. Suppose τ is a type and A is a subset of Mτ . By Condition I, A∗ ⊆ Mτ∗ . So, each element of A∗ is an element of Mτ∗ , hence internal. Consequently, each element of a standard set is internal. On the other hand, each subset of A∗ is an element of (M ∗ )(τ ) . It is ∗ internal exactly when it belongs to M(τ ). To interpret a formula of L∞ (M ) in M ∗ consider only internal substitutions f . These satisfy the condition f (XAi ) ∈ A∗ for each higher order set A. Define the truth value of the formula ϕ under f first by placing an asterisk to the right of each constant or relation symbol that appears in ϕ to obtain a formula ϕ∗ . Then interpret the formula ϕ∗ (under f ) as usual. Note: If (∃X ∈ A) is part of ϕ, then (∃X ∈ A∗ ) is a part of ϕ∗ . Since each element of A∗ is internal, this means one quantifies only on internal objects. For example, consider a formula ϕ(X1 , . . . , Xn ) where Xν = XMτ (ν) ,i(ν) , ν = 1, . . . , n. Let R = {(a1 , . . . , an ) ∈ Mτ (1) × · · · × Mτ (n) | M |= ϕ(a1 , . . . , an )}. Then the sentence θ (∀X1 ∈ Mτ (1) ) · · · (∀Xn ∈ Mτ (n) )[(X1 , . . . , Xn ) ∈ R ↔ ϕ(X1 , . . . , Xn )] is true in M . The sentence θ∗ takes the form (∀X1 ∈ Mτ∗(1) ) . . . (∀Xn ∈ Mτ∗(n) )[(X1 , . . . , Xn ) ∈ R∗ ↔ ϕ∗ (X1 , . . . , Xn )]. It is reasonable to ask that θ∗ will be true in M ∗ . Since Condition I implies that R∗ contains only elements with internal coordinates, we may rephrase θ∗ as R∗ = {(a1 , . . . , an ) ∈ Mτ∗(1) × · · · × Mτ∗(n) | M ∗ |= ϕ∗ (a1 , . . . , an )}. Here is the condition guaranteeing this indeed holds:

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Chapter 14. Nonstandard Structures

II. Elementary extension condition. Let A1 , . . . , An be higher order sets, ϕ(XA1 ,i(1) , . . . , XAn ,i(n) ) a formula of L∞ (M ), and aν ∈ Aν , ν = 1, . . . , n. Then ϕ(a1 , . . . , an ) is true in M if and only if ϕ∗ (a∗1 , . . . , a∗n ) is true in M ∗ . For example, if a ∈ Mτ , then a∗ ∈ Mτ∗ . Thus, standard objects are internal. Also, if a, b ∈ Mτ and a 6= b, then a∗ 6= b∗ : the canonical map a 7→ a∗ of Mτ into Mτ∗ is injective. Therefore, we occasionally regard Mτ as a subset of Mτ∗ . In addition, for R ⊆ A1 × · · · × An , if a ∈ A1 × · · · × An belongs to R∗ , then a ∈ R. Thus, R∗ ∩ (A1 × · · · × An ) = R and the relation R∗ is an extension of R. In particular, if M has a first order structure, then it is an elementary substructure of a natural extension to M ∗ . Now consider the notion of an internal function. Let A and B be two higher order sets. We view a function f : A → B as a subset of A × B that satisfies these two conditions: (1a) (1b)

(∀a ∈ A)(∃b ∈ B)[(a, b) ∈ f ]; (∀a ∈ A)(∀b1 , b2 ∈ B)[(a, b1 ) ∈ f ∧ (a, b2 ) ∈ f → b1 = b2 ].

The subset f ∗ of A∗ × B ∗ satisfies the corresponding conditions: it is a function from A∗ to B ∗ . Call it a standard function. For F , the set of all functions from A to B, the elements of F ∗ are the internal functions from A∗ to B ∗ . Lemma 14.2.1: The image of each internal subset of A∗ under an internal function f : A∗ → B ∗ is an internal subset of B ∗ . Proof: For each f ∈ F the following statement holds in M : (3)

(∀f ∈ F )(∀X ⊆ A)(∃Y ⊆ B) [(∀a ∈ X) f (a) ∈ Y ]

∧ [(∀b ∈ Y )(∃a ∈ X) f (a) = b]

The close “∀X ⊆ A” is not part of the language L∞ (M ). However, it can be reinterpreted as “(∀X ∈ M(τ ) )(∀x ∈ Mτ )[x ∈ X → x ∈ A]”, assuming A ⊆ Mτ and with X = XM(τ ) ,1 and x = XMτ ,1 . The interpretation of this ∗ ∗ ∗ in M ∗ is “(∀X ∈ M(τ ) )(∀x ∈ Mτ )[x ∈ X → X ∈ A ]”. This means “for all ∗ internal subsets X of A ”. Similarly, “∃Y ⊆ B” interprets in M ∗ as “there is an internal subset of B ∗ ”. Thus, our claim is the interpretation of (3) in M ∗.

14.3 Concurrent Relations Suppose a higher order relation R of M is finite with elements a1 , . . . , an . Then M |= (∀X ∈ R)[X = a1 ∨ · · · ∨ X = an ], so that M ∗ |= (∀X ∈ R∗ )[X = a∗1 ∨ · · · ∨ X = a∗n ]. That is, each element of R∗ is standard. We impose a final condition on M ∗ guaranteeing R∗ will be a true “enlargement” of R if R is an infinite relation.

14.3 Concurrent Relations

271

Definition 14.3.1: Let A and B be two higher order sets of M . Call a binary relation R ⊆ A × B concurrent if for all a1 , . . . , an ∈ A there exists b ∈ B with (ai , b) ∈ R, i = 1, . . . , n. III. Compactness (or saturation) condition. If A, B are two higher order sets of M and if R ⊆ A × B is a concurrent relation, then there exists b ∈ B ∗ such that (a, b) ∈ R∗ for each a ∈ A. Corollary 14.3.2: Let A be a higher order set of M . (a) If A = {a1 , . . . , an } is a finite set, then A∗ = {a∗1 , . . . , a∗n } and A is identified with A∗ as higher order sets. (b) If A is infinite, then A is properly contained in A∗ . Wn XA,1 = a1 ] holds in M . Hence, Proof of (a): The sentence (∀XA,1 ∈ A)[ i=1 W n by Condition II of Section 14.2, (∀XA,1 ∈ A∗ )[ i=1 XA,1 = a∗1 ] is true in M ∗ . ∗ ∗ ∗ Thus, A = {a1 , . . . , an }. Proof of (b): For every a1 , . . . , an in A there exists b ∈ A with ai 6= b, i = 1, . . . , n. That is, the inequality relation on A is concurrent. Hence, there exists b ∈ A∗ such that a 6= b for each a ∈ A (although not for each a ∈ A∗ ). Remark 14.3.3: Warning. Let A be a higher order set of M . Suppose A is an element of another higher order set B. In Section 14.2, we have identified B with a subset of B ∗ . Under this identification we have identified the element A of B with the element A∗ of B ∗ . Corollary 14.3.2 shows that the latter identification identifies A with A∗ as higher order sets if and only if A is finite. The case N = M , the natural numbers, provides the first example of an external object. Replacing M with N ∪ M , if necessary, we tacitly assume from now on that N ⊆ M . Lemma 14.3.4: The set N is an external subset of N∗ . Proof: By Corollary 14.3.2, N∗ contains a nonstandard element c. Since there are no elements of N between n and n + 1, the same is true for N∗ . In the extension I0 > I1 > I2 > · · · is a strictly descending sequence. 15. (a) Prove that N∗ r N is not an internal subset of N∗ . (b) Prove that for each a ∈ N the set {x ∈ N∗ | x > a} is internal.

Notes This chapter is in the spirit of Chapter 2 of the [Robinson-Roquette], which contains a nonstandard proof of the Siegel-Mahler theorem.

Chapter 15. Nonstandard Approach to Hilbert’s Irreducibility Theorem We use the nonstandard methods of Chapter 14 to give a new criterion for a field K to be Hilbertian: There exists a nonstandard element t of an enlargement K ∗ of K such that t has only finitely many poles in K(t)s ∩ K ∗ . From this there results a second and uniform proof (Theorem 15.3.4) that the classical Hilbertian fields are indeed Hilbertian. In addition, a formal power series field, K0 ((X1 , . . . , Xn )) of n ≥ 2 variables over an arbitrary field K0 , is also Hilbertian (Example 15.5.2).

15.1 Criteria for Hilbertianity We give two criteria for a field K to be Hilbertian in terms of an enlargement K ∗ of K. The first is a straightforward application of the compactness property of K. Proposition 15.1.1 (Gilmore-Robinson): The field K is Hilbertian if and only if there exists t ∈ K ∗ r K such that K(t)s ∩ K ∗ = K(t). Proof: Suppose K is Hilbertian. If f1 , . . . , fm ∈ K[T, X] irreducible polynomials which are separable in X and g1 , . . . , gm ∈ K[T ] are nonzero polynomials, then there exists a ∈ K with fi (a, X) irreducible K[X] and gi (a) 6= 0, i = 1, . . . , m. The compactness property (Condition III of Section 14.3) gives t ∈ K ∗ such that for each irreducible f ∈ K[T, X] which is separable in X and each 0 6= g ∈ K[T ], the polynomial f (t, X) is irreducible in K ∗ [X] and g(t) 6= 0. The second condition implies t 6∈ K. Consider x ∈ K(t)s ∩ K ∗ . Let f ∈ K[T, X] be an irreducible polynomial which is separable in X with f (t, x) = 0. Since f (t, X) is irreducible over K ∗ , it is linear. Hence, x ∈ K(t). Conversely, suppose t is an element of K ∗ r K with K(t)s ∩ K ∗ = K. Let f1 , . . . , fm ∈ K[T, X] be irreducible polynomials which are separable, monic, and of degree at least 2 in X and let 0 6= g ∈ K[T ]. If for some i between 1 and m, fi (t, X) is reducible over K ∗ , the coefficients of its factors lie in K(t)s ∩ K ∗ = K(t). Thus, fi (t, X) is reducible over K(t). But since t is transcendental over K, this means fi (T, X) is reducible over K, contrary to our assumption. Finally, since K ∗ is an elementary extension of K, there exists a ∈ K such that g(a) 6= 0 and fi (a, X) is irreducible over K, i = 1, . . . , m. By Lemma 12.1.4, K is Hilbertian. One may reapproach the classical Hilbertian fields through Proposition 15.1.1 [Roquette2], but it is easier to apply the following weaker condition of Weissauer. Consider t ∈ K ∗ r K and let Ωt = K(t)s ∩ K ∗ . Then Ωt is a separable algebraic (possibly infinite) extension of K(t). Since Ωt is regular over K

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(Example 14.5.3) consider it as a generalized function field of one variable over K. It is a union of function fields of one variable over K. Call an equivalence class of valuations of Ωt which are trivial on K a prime divisor of Ωt /K. Refer to a prime divisor as a pole of t if t has a negative value. Finally, call t polefinite if t has only finitely many poles in Ωt . In other words, there is an integer m such that the number of poles of t in any function field F with K(t) ⊆ F ⊆ Ωt is at most m. Proposition 15.1.2 ([Weissauer, Folgerung 3.2]): If K ∗ contains a polefinite element, then K is Hilbertian. Proof: The existence of a polefinite element implies K 6= K ∗ . Hence, K is infinite (discussion preceding Corollary 14.3.2). Assume K is not Hilbertian. Then Proposition 13.2.2 gives an irreducible polynomial f ∈ K[T, X], separable, monic, and of degree at least 2 in X, such that f (a, X) is reducible over K for each a ∈ K. The same statement holds in K ∗ . In particular, for each t ∈ K ∗ r K, (1) f (t, X) is irreducible over K(t), but reducible over K ∗ . Consider now an element t of K ∗ r K. The remaining parts of the proof consider properties of the prime divisors of K(t). Part A: Removing ramification over ∞. Let F be the splitting field of ˜ ∪ {∞} let pa f (t, X) over K(t). Suppose degt (f ) = d. For each a ∈ K be the prime divisor of K(t)/K defined by t 7→ a. In particular, p∞ is the unique pole of t in K(t)/K. Since K is infinite, we may choose a ∈ K with pa unramified in F and with residue field F¯ Galois over K. Apply 1 to replace f (T, X) by the K-automorphism of K(t) defined by t 7→ t−a 1 d T g(T, X) = f ( T −a , X), the field F by the splitting field of g(t, X), and pa by p∞ (Section 2.2). Thus assume, along with (1), that (2) p∞ is unramified in F . Each a ∈ K defines a K-automorphism σa of K(t) by t 7→ t + a. It fixes p∞ . Extend σa to an automorphism of K(t)s . Denote σa (F ) by Fa . Thus, p∞ is unramified in Fa and F¯a is Galois over K. By Corollary 2.3.7(c), (3) p∞ is unramified in the compositum, F 0 , of Fa for all a ∈ K. Part B: The finiteness of F 0 over ∞. Let P be a prime divisor of F 0 over p∞ . Denote reduction modulo P by a bar. The residue fields F¯ and F¯a are conjugate over K (Section 2.3). Since both are Galois extensions of K (Lemma 6.1.1), they coincide. The compositum of the residue class fields of unramified extensions of K(t) is the residue class field of their compositum (Lemma 2.4.8). Thus, F¯ 0 = F¯ is a finite extension of K which is independent of P. Part C: The infinitude of [F 0 ∩ K ∗ : K(t)]. By (1), Fa is not linearly disjoint from K ∗ over K(t). Hence, Ea = Fa ∩ K ∗ , a regular extension of K contained in Ωt , is a proper extension of K(t). Let R(Ea ) (resp. R(Fa ))

15.2 Arithmetical Primes Versus Functional Primes

279

be the set of prime divisors of K(t)/K that ramify in Ea (resp. in Fa ). By Remark 3.6.2(b), R(Ea ) is finite and nonempty. Note: pc = pc0 if and only if c0 is conjugate to c over K. If R(F ) = {pc1 , . . . , pcn }, then R(Fa ) = {pc1 −a , . . . , pcn −a }. For L a finite separable extension of K(t) let R(L) be {pd1 , . . . , pdr }. Since K is an infinite field, we may choose a ∈ K with ci − a 6= τ (dj ) for all i and j and for every K(t)-isomorphism τ of L. Hence R(Ea ) ∩ R(L) ⊆ R(Fa ) ∩ R(L) = ∅. Since R(Ea ) 6= ∅, the field Ea is not contained in L. Therefore, the compositum E 0 of Ea for all a ∈ K is an infinite extension of K(t) contained in Ωt . Part D: Conclusion of the proof. Assume t has only m poles in Ωt . Use Part C to choose a finite extension N of K(t) in E 0 with [N : K(t)] > m[F¯ : K]. Let q1 , . . . , qk be all extensions of p∞ to N . Each of them extends to a ¯q ⊆ F¯ . Hence, by Proposition 2.3.2, pole of t in Ωt , so k ≤ m. By Part B, N i [N : K(t)] =

k X

¯q : K] ≤ m[F¯ : K]. [N i

i=1

This contradiction to the choice of N prove that t is not a polefinite element.

15.2 Arithmetical Primes Versus Functional Primes To apply Proposition 15.1.2 we start with a field K that carries arithmetic structure and extend this structure to an enlargement K ∗ of K. For each t ∈ K ∗ r K we consider finite extensions F of K(t) in K ∗ . Then we compare the function field structure of F/K with the arithmetic structure induced on F from K ∗ . The goal is to find conditions on t to be a polefinite element. Let S is a set of primes of K. Thus, S is a set of equivalent classes of absolute values of K (Section 13.3). For each p ∈ S choose a representative | |p . Define a map vp : K → R ∪{∞} by vp (a) = − log(|a|p ). Conditions (3a)– (3d) of Section 13.3 on the valuation translate into the following properties of vp : (1a) vp (a) = ∞ if and only if a = 0. (1b) vp (ab) = vp (a) + vp (b). (1c) vp (a + b) ≥ min(vp (a), vp (b)) − log 2. (1d) There is an a ∈ K × with vp (a) 6= 0. We call vp an additive absolute value of K. If p is ultrametric, then |a + b|p ≤ max(|a|p , |b|p ), hence vp (a + b) ≥ min(vp (a), vp (b)) for all a, b ∈ K. Thus, vp is a valuation of K. The following lemma gives a simple criterion for p to be metric: Lemma 15.2.1: A prime p of a field K is metric if and only if vp (2) < 0. Proof: Suppose first that p is metric. By Section 13.3, there is a positive integer n with |n|p > 1. Hence, char(K) = 0 and the restriction of | |p to Q is

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Chapter 15. Nonstandard Approach to Hilbert’s Irreducibility Theorem

a metric absolute value. By a theorem of Ostrowski, | |p is equivalent to the ordinary absolute value of Q. Thus, there is a positive real number c such that |x|p = |x|c for each x ∈ Q. In particular, |2|p = |2|c = |nlog 2/ log n |c > 1. Consequently, vp (2) < 0. Conversely, if vp (2) < 0, then |2|p > 1, so p is metric. We assume now that S satisfies the following finiteness condition: (2) If a ∈ K × , then {p ∈ S | vp (a) 6= 0} is a finite set. By Lemma 15.2.1, there are only finitely many archimedean primes in S. Consider again an enlargement of K ∗ of K. Then S extends to a set ∗ S of arithmetical primes of K ∗ . To each p ∈ S ∗ there corresponds a star-additive absolute value vp : (K ∗ )× → R∗ . In particular, each prime in S extends to an element p of S ∗ , a standard prime. The elements of S ∗ r S are the nonstandard primes. Condition (2) becomes: (2)∗ If a 6= a ∈ K ∗ , then {p ∈ S ∗ | vp (a) 6= 0} is starfinite (Example 14.5.1). In particular, if 0 6= a ∈ K, then the finite set {p ∈ S | vp (a) 6= 0} does not grow in the enlargement Thus, vp (a) = 0 for all p ∈ S ∗ r S and a ∈ K × . Therefore, by (2), there are only finitely many archimedean primes in S ∗ ; they are all standard. For an arbitrary prime p ∈ S ∗ consider the convex hull of vp (K × ): Γp = {r ∈ R∗ | ∃x ∈ K × with |r| ≤ vp (x)}. In the notation of Example 14.5.2, (3) Γp = Rfin if p ∈ S and Γp = 0 if p ∈ S ∗ r S. Unlike nonstandard primes, standard primes form valuations that are nontrivial on K. We modify them so that they will “behave” like the nonstandard primes. However, these “modified” primes will not be internal objects. Combining internal and external objects is the key to the nonstandard machinery. ˙ p = R∗ /Γp . By (3), R˙ p = For p ∈ S ∗ , consider the ordered group R ˙ (Example 14.5.2) if p is standard and R ˙ p = R∗ otherwise. Define the R ˙ p by v˙ p (x) = vp (x) + Γp . modified valuation v˙ p : (K ∗ )× → R Suppose p is standard. Then p may be metric or ultrametric. In the second case vp is a valuation, whereas in the first case it is not. In both cases, however, vp satisfies (3). Since log 2 ∈ R ⊆ Γp , v˙ p (a+b) ≥ min{v˙ p (a), v˙ p (b)}. Hence, v˙ p is a valuation. In addition, if 0 6= x ∈ K, then vp (x) ∈ R. Hence, v˙ p (x) = 0. In summary: ˙ p defined Lemma 15.2.2: For each p ∈ S ∗ the function v˙ p : (K ∗ )× → R by v˙ p (x) = vp (x) + Γp is a valuation which is trivial on K. Moreover, if p ∈ S ∗ r S, then v˙ p = vp . Consider an element t ∈ K ∗ r K. Let F be a finite extension of K(t) in K . By Example 14.5.3, F is a regular extension of K, hence a function field ∗

15.3 Fields with the Product Formula

281

over K. Consider p ∈ S ∗ with v˙ p not vanishing on F × . Then, by Lemma 15.2.2, the restriction of v˙ p to F is a valuation which is trivial valuation on K. Thus, it defines a prime P of F/K. We say P is induced from p. Conversely, we may ask if a prime divisor P of F/K (i.e. a functional prime) if it is induced by an arithmetical prime. This question is inspired by the following observation. Proposition 15.2.3: Let S be a set of primes of a field K with {p ∈ S | vp (a) 6= 0} being a finite set for each a ∈ K × . Let t be a nonstandard element of an enlargement K ∗ of K. Suppose S(t) = {p ∈ S ∗ | v˙ p (t) < 0} is a finite set and for each finite separable extension F of K(t) in K ∗ all poles of t in F are induced by arithmetical primes (i.e. elements of S ∗ ). Then t is a polefinite element. Proof: Let P be a pole of t in Ωt . For each finite extension F of K in Ωt the restriction of P to F is also a pole of t. Hence, there is a pF ∈ S ∗ which induces P |F . In particular, v˙ pF (t) < 0, so pF ∈ S(t). Since S(t) is finite, there is a p ∈ S(t) which induces P |F for each F as above (Use compactness of S(t).) Therefore, p determines P . Thus, the number of poles of t in Ωt is at most |S(t)|. Consequently, t is a polefinite element.

15.3 Fields with the Product Formula As in Section 15.2, suppose S is a nonempty set of primes of K. For each p ∈ S choose an absolute value | |p representing p and let vp be the corresponding additive absolute value. We say S satisfies a product formula if the following statement holds: for each p ∈ S there exists a positive real number λp with the following property: Q λ (1) For each a ∈ K × the set {p ∈ S | |a|p 6= 1} is finite and p∈S |a|p p = 1. In this case call K a field with a product formula. Example 15.3.1: Basic examples of product formulas. For K = Q and p a prime number, define | |p by |a|p = p−r for a = xy · pr with r ∈ Z and x, y ∈ Z relatively prime to p. The infinite absolute value | |∞ is the usual absolute value. This set S of primes satisfies the product formula with λp = 1 for each p ∈ S. For K0 (t), a rational function field over an arbitrary field K0 , choose a real number 0 < c < 1. For each irreducible polynomial p ∈ K0 [t] define the absolute value | |p by |u|p = c−r for u = fg ·pr with f, g polynomials relatively prime to p. Let λp be deg(p). For a quotient f /g of polynomials, define the infinite absolute value as |f /g|∞ = cdeg(f )−deg(g) , and let λ∞ = 1. Check that the corresponding set of primes of K0 (t) satisfies the product formula. If a set S of primes of a field K is a field with a the product formula and if S 0 is the set of primes of a finite extension K 0 /K lying over S, then

282

Chapter 15. Nonstandard Approach to Hilbert’s Irreducibility Theorem

S 0 satisfies the product formula [Lang3, p. 20]. We conclude from Example 15.3.1 that every global field is a field with a product formula. Again let K be any field with the product formula with respect to a nonempty set of primes S. Rewrite condition (1) additively: P (2) For each a ∈ K × , {p ∈ S | vp (a) 6= 0} is finite and p∈S λp vp (a) = 0. For an enlargement K ∗ of K, Condition (2) has a similar form: ∗ ∗ × ∗ (2) For P each a ∈ (K ) , the set {p ∈∗ S | vp (a) 6= 0} is starfinite and p∈S P∗ λp vp (a) = 0, where λp ∈ R and λp > 0 are independent of p and here is the starfinite summation (Example 14.5.1). Proposition 15.3.2: Let t be a nonstandard element of K ∗ and F a finite separable extension of K(t) in K ∗ . If there exists q ∈ S ∗ with v˙ q (t) < 0, then each functional prime of F is induced by an arithmetical prime of K ∗ . Proof: Denote the set of all functional primes of F which are induced by elements of S ∗ by P. Let D = {x ∈ F | vP (x) ≥ 0 for all P ∈ P} = {x ∈ F | v˙ p (x) ≥ 0 for all p ∈ S ∗ } be the holomorphy ring of P. The assumption v˙ q (t) < 0 implies that t 6∈ D and P 6= ∅, so D 6= F . Assume P does not contain all the functional primes of F . Then F = Quot(D) (Proposition 3.3.2(a)). We now show that D is a field. This will give a contradiction to D 6= F and establish the theorem. We need to prove that if x ∈ D× and x 6= 0, then x−1 ∈ D. Indeed, by (2)∗ the set S(x) = {p ∈ S ∗ | vp (x) < 0} is starfinite. For p ∈ S ∗ r S, Lemma 15.2.2 shows that vp (x) = v˙ p (x) ≥ 0. This means S(x) contains only standard primes. Hence, by Corollary 14.3.4(a), S(x) is a finite set. For p ∈ S(x) we have vp (x) < 0 and v˙ p (x) ≥ 0, so v˙ p (x) = 0. Now use the summation formula (2)∗ and the additivity of starfinite summation (Example 14.5.1): X X (3) λp vp (x) = ˙ λp vp (x) = ˙ 0. p∈S ∗ r S(x)

p∈S ∗

If p0 ∈ S ∗ r S(x) and v˙ p0 (x) 6= 0, then v˙ p0 (x) > 0. This means X ˙ 0, λp vp (x) ≥ λp0 vp0 (x) > p∈S ∗ −S(x)

a contradiction to (3). Thus, v˙ p (x) = 0 for each p ∈ S ∗ r S(x) and therefore for each p ∈ S ∗ . It follows that v˙ p (x−1 ) = 0 for each p ∈ S ∗ . Consequently, x−1 ∈ D, as desired. Theorem 15.3.3 ([Weissauer, Satz 6.2]): Every field K with the product formula is Hilbertian. Proof: Let S be a nonempty set of primes of a field K satisfying the product formula. Choose q ∈ S and a ∈ K with vq (a) < 0. In an enlargement

15.4 Generalized Krull Domains

283

K ∗ of K, pick a nonstandard positive integer ω ∈ N∗ r N. We prove the nonstandard element t = aω of K ∗ is polefinite. Indeed, S(t) = {p ∈ S ∗ | vp (t) < 0} = {p ∈ S ∗ | vp (a) < 0} = {p ∈ S ∗ | v˙ p (t) < 0}. Therefore, S(t) is a starfinite set that contains q. In particular, v˙ q (t) < 0. By Lemma 15.1.2, if p ∈ S ∗ r S, then v˙ p (a) = 0. Thus, S(t) consists only of standard primes, so, by Corollary 14.3.4(b), S(t) is a finite set. Suppose F is a finite separable extension of K(t) in K ∗ . By Propositions 15.3.2, each functional prime P of F/K(t) is induced by some p ∈ S ∗ . If P is a pole of t, then p ∈ S(t). By Proposition 15.2.3, t is polefinite. Consequently, by Proposition 15.1.2, K is Hilbertian. Corollary 15.3.4: Every number field and every function field of several variables over an arbitrary field is Hilbertian.

15.4 Generalized Krull Domains Non-standard methods produce new Hilbertian fields: The quotient field of each ‘generalized Krull domain of dimension at least 2’ is Hilbertian (Theorem 15.4.6). In particular, fields of formal power series of at least two variables over arbitrary fields are Hilbertian (Theorem 15.4.6). Let R be an integral domain with quotient field K. We call R a generalized Krull domain if K has a nonempty set S of primes satisfying the following conditions: (1a) For each p ∈ S, vp is a real valuation. (1b) The valuation ring Op of vp is the local ring of R relative to mp = {a ∈ R | vT p (a) > 0}. (1c) R = p∈S Op . (1d) For each a ∈ K × the set {p ∈ S | vp (a) 6= 0} is finite. The dimension of a ring R is the maximal integer n for which there is a sequence p0 ⊂ p1 ⊂ · · · ⊂ pn of n + 1 distinct prime ideals. Thus, dim(R) ≥ 2 if and only if (1e) R has a maximal ideal M which properly contains a nonzero prime ideal. Thus, R is a generalized Krull ring of dimension exceeding 1 if and only if it satisfies Condition (1). Lemma 15.4.1: Let R be an integral domain satisfying Condition (1). Then: (a) For each nonunit b of R, b 6= 0, there exists p ∈ S with vp (b) > 0. (b) For each p ∈ S, mp is minimal among nonzero prime ideals of R. (c) If p, q are distinct primes in S, then mp 6⊆ mq and mp 6= M . T Proof of (a): Otherwise b−1 ∈ p∈S Op = R. Proof of (2b): Let n ⊆ mp be a nonzero prime ideal. Choose 0 6= a ∈ n. Since vp is real, for each b ∈ mp there is a positive integer n with vp (a) ≤

284

Chapter 15. Nonstandard Approach to Hilbert’s Irreducibility Theorem n

n

nvp (b) = vp (bn ), so ba ∈ Op . By (1b), there are c, d ∈ R with ba = dc and d 6∈ mp , so dbn = ac ∈ n. Since d 6∈ n, we have b ∈ n. Consequently, n = mp . Proof of (c): By (1b), mp 6= mq . By (b), mp 6⊂ mq Hence, mp 6⊆ mq . In addition, by (b) and (1e), m 6= M . Lemma 15.4.2: The local ring RM of R at M is a generalized Krull domain of dimension exceeding 1 with respect to S 0 = {p ∈ S | mp ⊂ M }. Proof: Conditions (1a), (1b), (1d), and (1e) follow from T the basic definitions of the local ring RM . It remains to prove that RM = p∈S 0 Op . This follows if we show, for x ∈ K with vp (x) ≥ 0 for each p ∈ S 0 , that x ∈ RM . By (1d), the set T = {q ∈ S | vq (x) < 0} is finite. If q ∈ T , then q ∈ / S0, r so mq 6⊆ M . Choose aq ∈ mq M . By (1a) there is a positive integer n(q) Q n(q) n(q) and y = ax. Then a ∈ R r M with vq (aq ) > −vq (x). Let a = q∈T aq and vp (y) ≥ 0 for each p ∈ S. Therefore, by (1c), y ∈ R. Consequently, x ∈ RM . For the goal of this section - a proof that K is Hilbertian - we may replace R by RM to assume that (2) R is a local ring and M is its maximal ideal. Consider an enlargement K ∗ of K. Then R∗ is a local ring with maximal ideal M ∗ . Also, S ∗ has these properties: (1a)∗ For p ∈ S ∗ , vp is a valuation of K ∗ with values in R∗ . (1b)∗ The valuation ring Op∗ of vp is the local ring of R∗ relative to m∗p = {a ∈ R∗ | vp (a) > 0}. T ∗ (1c) R∗ = p∈S ∗ Op∗ . (1d)∗ For each a ∈ K ∗ r{0} the set {p ∈ S ∗ | vp (a) 6= 0} is starfinite. (1e)∗ M ∗ properly contains the prime ideals m∗p . As with fields with a product formula, consider the modified valuations v˙ p and their valuation rings O˙ p = {x ∈ K ∗ | v˙ p (x) ≥ 0}. The holomorphy T ring of these valuations, R˙ = p∈S ∗ O˙ p , contains both K (Lemma 15.2.2) and R∗ . Moreover: Lemma 15.4.3: The ring R˙ equals K · R∗ . Proof: We have to show only that each x ∈ R˙ lies in the composite K · R∗ . Indeed, v˙ p (x) ≥ 0 for each p ∈ S ∗ . From (1d)∗ and lemma 15.2.2 the starfinite set S(x) = {p ∈ S ∗ | vp (x) < 0} contains only standard primes. Hence, (Corollary 14.3.5(b)), it is actually finite. For each p ∈ S(x) there exists v˙ p (x) ≥ 0). By (1b) we may find 0 6= ap ∈ R rp ∈ R with vp (x) ≥ rp (because Q with vp (ap ) ≥ −vp (x). Let a = p∈S(x) ap and y = ax. Then vp (y) ≥ 0 for each p ∈ S ∗ . By (1c)∗ , y ∈ R∗ . Consequently, x = a−1 y ∈ K · R∗ . ˙ Suppose for each p ∈ S we Lemma 15.4.4: Let x and y be nonunits of R. ˙ have v˙ p (x) > 0 or v˙ p (y) > 0. Then αx + βy 6= 1 for all α, β ∈ R.

15.4 Generalized Krull Domains

285

Proof: Assume there exist α, β ∈ R˙ with αx + βy = 1. Apply Lemma 15.4.3 to write (3)

ax by + = 1 with a, b ∈ R∗ and c ∈ K × . c c

We show that both summands on the left hand side of (3) belong to M ∗ . This contradiction to (3) will conclude the proof of the lemma. Indeed, let p ∈ S ∗ with vp (c) > 0. Then v˙ p (c) = 0 and p is standard (Lemma 15.2.2). Hence, v˙ p (x) > 0 or v˙ p (y) > 0. If v˙ p (x) > 0, then v˙ p ( ax c )> by ax ˙ 0. So, by (3), vp ( c ) = 0. Similarly, if v˙ p (y) > 0, then 0. Hence, vp ( c ) > by ˙ 0 and vp ( ax vp ( c ) > c ) = 0. by Next consider p ∈ S ∗ with vp (c) = 0. Then vp ( ax c ) ≥ 0 and vp ( c ) ≥ 0. by ∗ It follows from (1c)∗ that both ax c and c belong to R . ˙ there exists p ∈ S ∗ with v˙ p (x) > 0. Hence Since x is a nonunit in R, ax ax ∗ v˙ p ( c ) > 0 and therefore c ∈ M ∗ . Similarly by c ∈M . Our next lemma is “standard”: Lemma 15.4.5: Let {p1 , . . . , pm } and {q1 , . . . , qn } be two disjoint finite subsets of S. Then there exists an element a ∈ R such that (5) vpi (a) = 0, i = 1, . . . , m, and vqj (a) > 0, j = 1, . . . , n. Proof: Proceed by induction on m. Suppose m = 1. By Lemma 15.4.1(c), mqj 6⊆ mp1 , 1 ≤ j ≤ n. Let aj ∈ mqj r mp1 . Then a = a1 · · · an satisfies (5). Suppose m > 1. The induction hypothesis gives a1 ∈ R with vp1 (a1 ) = · · · = vpm−1 (a1 ) = 0 and vpm (a1 ), vq1 (a1 ), . . . , vqn (a1 ) > 0. By the case m = 1, there exists a2 ∈ R with vpm (a2 ) = 0 and vp1 (a2 ), . . . , vpm−1 (a2 ), vq1 (a2 ), . . . , vqn (a2 ) > 0. The element a = a1 + a2 satisfies (5).

We now prove the main theorem of this section. Theorem 15.4.6 (Weissauer): The quotient field of a generalized Krull domain of dimension exceeding 1 is Hilbertian. Proof: As previously, let K, R, S, M satisfy (1)-(2). Let q ∈ S and choose b ∈ M r mq (by (Lemma 15.4.1(c)). By (1d) and Lemma 15.4.1(a), the set T (b) = {p ∈ S | vp (b) > 0} is finite and nonempty. For each finite set T with T (b) ⊆ T ⊆ S there exists aT ∈ R such that vp (aT ) = 0 for p ∈ T (b) and vp (aT ) > 0 for each p ∈ T r T (b) (Lemma 15.4.5). Proposition 14.3.6 gives a ∈ R∗ such that vp (a) = 0 for each p ∈ T (b) and vp (a) > 0 for each p ∈ S r T (b). By Lemma 15.2.2, vp (b) = 0 for each p ∈ S ∗ r S. Thus, (5) T (a) = {p ∈ S ∗ | vp (a) > 0} is disjoint from T (b) = {p ∈ S ∗ | vp (b) > 0}. Choose ω ∈ N∗ r N. Put x = aω and y = bω . We conclude the proof in parts, from Lemma 15.1.2, by showing that t = xy is polefinite.

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Part A: S(t) = {p ∈ S ∗ | vp (t) < 0} is a finite set. Indeed, as q ∈ S r T (b), we have vq (a) > 0. Hence, v˙ q (x) > 0. Next choose q0 ∈ T (b). Then vq0 (b) > 0 ˙ Moreover, if and therefore v˙ q0 (y) > 0. Thus, x and y are nonunits of R. p ∈ T (b), then v˙ p (y) > 0 and if p ∈ S r T (b), then v˙ p (x) > 0. By Lemma 15.4.4, ˙ αx + βy 6= 1 for all α, β ∈ R.

(6)

In particular, αt + β 6= 1 for all α, β ∈ K, so t ∈ / K. For each p ∈ S ∗ , (5) gives (7a) (7b)

vp (x) > 0 =⇒ vp (y) = 0 and vp (t) = vp (x); and vp (y) > 0 =⇒ vp (x) = 0 and vp (t) = −vp (y).

Therefore, S(t) = T (b) is a finite set. Part B: Application of Proposition 15.2.3. To complete the proof it suffices to show that if F/K(t) is a finite separable extension in K ∗ , then the functional primes of F are induced by arithmetical primes. Let P be the set of functional primes of F which are induced by arithT metical primes. Let D = P ∈P OP be the corresponding holomorphy ring in F . Since vq (t) = v˙ q (x) > 0, the set P is nonempty. Assume P excludes a functional prime of F . Then the holomorphy ring theorem (Proposition 3.3.2) says that D is a Dedekind domain. By (7), A = {z = {z B = {z = {z

∈ D| ∈ D| ∈ D| ∈ D|

v˙ p (z) ≥ v˙ p (x) for all p ∈ S ∗ } vP (z) ≥ max(vP (t), 0) for all P ∈ P}; and v˙ p (z) ≥ v˙ p (y) for all p ∈ S ∗ } vP (z) ≥ max(−vP (t), 0) for all P ∈ P}.

Every maximal ideal of D is the center of a prime P ∈ P (Proposition 3.3.2(d)). Thus, A is the ideal of zeros of t and B is the ideal of poles of t. That is, A = P1k1 · · · Prkr and B = Ql11 · · · Qlss , where P1 , . . . , Pr , Q1 , . . . , Qs are distinct maximal ideals of D and k1 , . . . , kr , l1 , . . . , ls are positive integers with vPi (t) = ki , i = 1, . . . , r and vQj (t) = −lj , j = 1, . . . , s. In particular, A and B are relatively prime ideals of D. Hence, A + B = D. Thus, there exist ˙ λ ∈ A and µ ∈ B such that λ + µ = 1. By definition, A ⊆ xR˙ and B ⊆ y R. ˙ a contradiction to (6). Thus, λ = αx and µ = βy with α, β ∈ R,

15.5 Examples Let R be an integral domain with quotient field K. Suppose S is a nonempty set of primes of K which satisfies Conditions (1b)–(1d) of Section 15.4. Suppose in addition vp is a discrete valuation, p ∈ S. Then R is a Krull domain, hence a generalized Krull domain.

15.5 Examples

287

Every Dedekind ring R is a Krull domain with S being the set of primes of K associated with the maximal ideals of R. Since each nonzero prime ideal of R is maximal, dim(R) = 1. Thus, R does not satisfy Condition (1e) of Section 15.4. Example 15.5.1: Polynomial rings over fields. Every unique factorization domain R is a Krull domain. Here S corresponds to the set of nonzero prime ideals of R. For example, the polynomial ring R = K0 [X1 , . . . , Xn ] over an arbitrary field K0 is a unique factorization domain [Zariski-Samuel1, p. 38, Thm. 13]. When n ≥ 2, R/(RX1 + RX2 ) ∼ = K0 [X3 , . . . , Xn ] is an integral domain. Hence, RX1 +RX2 is a prime ideal of R which properly contains each of the prime ideals RX1 and RX2 . Thus, dim(R) ≥ 2. (Indeed, dim(R) = n [Matsumura, p. 117, Thm. 15.4].) By Theorem 15.4.6, K0 (X1 , . . . , Xn ) is Hilbertian. This is a weaker result than Theorem 13.4.2 which says that K0 (X1 , . . . , Xn ) is Hilbertian for each n ≥ 1. Example 15.5.2 [Weissauer p. 203]: Formal power series over a field. Let K0 be a field and n ≥ 2. The ring of formal power series R = K0 [[X1 , . . . , Xn ]] is a local integral domain with the maximal ideal

M=

∞ X

fi | fi ∈ K0 [X1 , . . . , Xn ] is a form of degree i .

i=1

Denote its quotient field by K0 ((X1 , . . . , Xn )): the field of formal power series over K0 in X1 , . . . , Xn . The Weierstrass preparation theorem implies that R is a unique factorization domain [Zariski-Samuel2, p. 148]. For each prime element p of R one of the elements X1 or X2 does not belong to Rp. Thus, Rp is properly contained in M , so dim(R) ≥ 2. (Again, dim(R) = n [Zariski-Samuel2, p. 218 or Matsumura, p. 117, Thm. 15.4].) By Theorem 15.4.6, K is Hilbertian. This settles a problem of [Lang3, p. 142, end of third paragraph]. Example 15.5.3: Formal power series over a ring. Let A be a Krull domain and n ≥ 1. Then the ring of polynomials A[X1 , . . . , Xn ] and the ring of formal power series A[[X1 , . . . , Xn ]] in the variables X1 , . . . , Xn over A are Krull domains (see [Matsumura, p. 89] for the case n = 1; the general case follows by induction). Suppose in addition that A is not a field. Let p0 be a nonzero prime ideal of A. Then P =

∞ X

fi ∈ A[X1 , . . . , Xn ] is a form of degree i

i=1

P0 =

∞ X i=0

fi ∈ A[X1 , . . . , Xn ] is a form of degree i and f0 ∈ p0

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are nonzero prime ideals of R = A[[X1 , . . . , Xn ]] and P0 ⊂ P . Indeed, A/P ∼ = A and A/P ∼ = A/p0 are integral domains. Thus, dim(A[[X1 , . . . , Xn ]]) ≥ 2 (Actually, dim(A) = dim(A) + n [Matsumura, p. 117]). Consequently, by Theorem 15.4.6, the quotient field of A[[X1 , . . . , Xn ]] is a Hilbertian field. For example, the quotient fields of Z[[X1 , . . . , Xn ]] and O[[X1 , . . . , Xn ]], where O is a discrete valuation rings, are Hilbertian. Lemma 15.5.4: No Henselian field is Hilbertian. Proof: Let K be a Henselian field with valuation ring R and maximal ideal m. Choose m ∈ m, m 6= 0 and a prime number p 6= char(K). Consider the irreducible polynomials f (T, X) = X p + mT − 1 and g(T, X) = X p + T −1 − 1 of K(T )[X]. Assume K is Hilbertian. Then there exists a ∈ K × with both f (a, X) and g(a, X) irreducible in K[X]. In particular, none of them has a zero in K. But either a ∈ R or a−1 ∈ m. Suppose first a ∈ R. Then f (a, 1) ≡ ∂f (a, 1) 6≡ 0 mod m. Since K is Henselian, f (a, X) has a zero 0 mod m and ∂X in K. Similarly, if a−1 ∈ m, then g(a, X) has a zero in K. This contradiction to the preceding paragraph proves that K is not Hilbertian. Example 15.5.5: Qp and the formal power series K0 ((X)) are complete discrete valuation fields. Hence, they are Henselian (Proposition 3.5.2). By Lemma 15.5.4, they are not Hilbertian. Thus, the assumption “n ≥ 2” in Example 15.5.2 is necessary. Example 15.5.6 (Geyer): In the notation of Example 15.5.2, the ring R = K0 [[X1 , . . . , Xn ]] of formal power series is not Hilbertian. Indeed, let K = Quot(R) and check forPchar(K0 ) 6= 2 (resp. char(K0 ) 6= 3) that every power series f (X1 , X2 ) = 1+ i+j>0 aij X1i X2j is a square (resp. a cube) in R. Thus, the polynomial Z 2 −(1+X1 T ) (resp. Z 3 −(1+X1 T )) is irreducible in K[T, Z] but Z 2 − (1 + X1 t) (resp. Z 3 − (1 + X1 t)) is reducible for each t ∈ R. Lemma 15.5.7: Let R0 be a unique factorization domain with quotient field K0 . Consider a set P of unequivalent prime elements of R0 . For each p ∈ P let vp be the corresponding discrete valuation of K0 . Let K be an algebraic extension of K0 . Suppose each vp with p ∈ P is unramified in K. For each p ∈ P choose an extension wp of vp to K. Then the holomorphy ring R = {x ∈ K | wp (x) ≥ 0 for each p ∈ P } is a unique factorization domain with quotient field K. Proof: The assumptions imply wp (p) = 1 and wp0 (p) = 0 for all distinct p, p0 ∈ P . Consider x ∈ R. Then wp (x) ≥ 0 for each p ∈ P . Moreover, 0. Indeed, if f = irr(x, K0 ) there are only finitely many p ∈ P with wp (x) >Q and vp (f (0)) = 0, then wp (x) = 0. Thus, u = x p∈P p−wp (x) is an element ofQK and wp (u) = 0 for each p ∈ P . Hence, u is a unit of R and x = u p∈P pwp (x) is the desired decomposition of x. Finally, observe that R contains the integral closure of R0 in K. Therefore, K = Quot(R).

Exercises

289

The following example generalizes [Corvaja-Zannier, Theorem 1 (i) and (ii)]: Example 15.5.8: Unique factorization domain with a non-Hilbertian quotient field. Let R0 be either Z or F [t] for some finite field F . It is a unique factorization domain. Put K0 = Quot(R0 ). The proof of Theorem 13.6.2 (with K0 replacing K) gives prime numbers l and q, an element u ∈ K × , an infinite set P of nonequivalent prime elements of R0 , and a field extension K (in the notation of Theorem 13.6.2 equals to M (K0 )) with these properties: (1a) For every x ∈ K there is a y ∈ K with y l − xlq + u = 0. (1b) For each p ∈ P , the discrete valuation vp is unramified in K. Since Y l − X lq + u is absolutely irreducible, Condition (1a) implies K is not Hilbertian. Let wp , p ∈ P , and R be as in Lemma 15.5.7. Then K = Quot(R), {wp | p ∈ P } is an infinite set of discrete valuations, and R is a unique factorization domain. This answers negatively the questions posed in Problems 14.20 and 14.21 of [Fried-Jarden3]. The following problem asks for a generalization of Example 15.5.3: Problem 15.5.9: Let A be a generalized Krull domain which is not a field. (a) Is A[[X]] a generalized Krull domain? (b) Is Quot(A[[X]]) Hilbertian? By Theorem 15.4.6, an affirmative answer to Problem 15.5.9(a) will give an affirmative answer to Problem 15.5.9(b).

Exercises 1. Let K be a countable Hilbertian field. List the irreducible polynomials in K[T, X] as f1 (T, X), f2 (T, X), . . .. For each i choose ti ∈ K for which f1 (ti , Y ), . . . , fi (ti , Y ) are irreducible in K[Y ]. (a) Observe that the infinite set H = {t1 , t2 , . . .} has the universal Hilbert subset property: H r HK (g1 , . . . , gm ) is finite for every collection {g1 , . . . , gm } of irreducible polynomials in K[T, X]. (b) Let K ∗ be an enlargement of K. Prove that if t ∈ H ∗ r H, then K(t) is algebraically closed in K ∗ . 2. Let A = {1, 22! , 33! , . . .} and considerSan enlargement Q∗ of Q. Prove ∞ for each nonstandard element t ∈ A∗ that n=1 Q(t1/n ) ⊆ Q∗ . 3.

Let f1 , f2 , f3 , . . . be a sequence in Q(Y ). Put gn (Y ) = f1 (. . . (fn−1 (fn (Y ))).

Suppose that if y is transcendental over Q and x = gn (y), then x has at least n distinct poles in Q(y). Let A = {g1 (1), g2 (2), g3 (3), . . .}. Consider a nonstandard element t of A∗ . Prove that the equation t = gn (Y ) is solvable in Q∗ for each n. Conclude that t is not polefinite (Section 15.1).

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4. [Weissauer, Satz 2.3] Generalize the Gilmore-Robinson criterion as follows. Let K ∗ be an enlargement of a field K. Then K is Hilbertian if and only if there is a Hilbertian field F with K ⊆ F ⊆ K ∗ and F is separably closed in K ∗ . 5. Consider the set A = {2, 3!, 4!, 5!, . . .} and let t be a nonstandard element of A∗ . Show that the metric absolute values of Q induces via Q∗ (Section 15.2) the infinite prime, t → ∞ of Q(t), while every ultrametric absolute value of Q induces the prime, t → 0, of Q(t). Conclude from Proposition 15.3.2, that all other primes of Q(t) are induced by nonstandard arithmetical primes of Q∗ . 6. Consider the field K = K0 ((X1 , . . . , Xn )) of formal power series in n variables over a basic field K0 . Construct a K0 -place ϕ: K → K0 ∪ {∞} with ϕ(Xi ) = 0, i = 1, . . . , n. Conclude that K/K0 is a regular extension. Hint: The case n = 1 is easy. For arbitrary n embed K in the field of iterated power series L = K0 ((X1 ))((X2 )) · · · ((Xn )).

Notes The nonstandard characterization of Hilbertian fields appears in [GilmoreRobinson]. [Roquette2] has exploited the Gilmore-Robinson criterion through a nonstandard interpretation of the Siegel-Mahler theorem (compare with the remarks at the beginning of Section 13.3). Most of this Chapter is from Weissauer’s thesis [Weissauer]. The proof of 15.3.2 for number field that appear in [Robinson] based on nonstandard interpretation of the “distributions” that appear in Weil’s thesis [Weil1]. The influence of [Robinson-Roquette] in Section 15.2 and parts of Section 15.3 should be obvious. The polefinite property of Section 5.1 and its relation to Hilbert’s irreducibility theorem appears in a standard form based on p-adic analysis in the case K = Q in [Sprind¸zuk1]. Standard simplified proofs of both Sprind¸zuk’s result and Weissauer’s approach to Hilbertianity of fields with a product formula (Section 15.3), featuring their common concepts, appear in [Fried10] which also gives the concept of a universal Hilbert subset (e.g. in Exercise 1) whose existence is a consequence of the Gilmore-Robinson observation (Proposition 15.1). As far as we know [Sprind¸zuk2] is the first to give an explicit universal Hilbert subset (over Q): p 2 H = {[exp( log(log(m))] + m!2m }∞ m=1 . It has the property that H r HQ (g1 , . . . , gt ) is finite for every collection {g1 , . . . , gt } of irreducible polynomials in Q[T, X] [Fried10, Thm. 4.9]. Finally, [Klein] gives “standard” proofs of Weissauer’s results, Theorems 15.3.3 and 15.4.6.

Chapter 16. Galois Groups over Hilbertian Fields Given a field K, one may ask which finite groups occur as Galois groups over K. If K is Hilbertian, then every finite group that occurs over K(t), with t transcendental over K, also occurs over K. Moreover, suppose F/K(t) is Galois with Galois group G and F/K is regular. Then K has a linearly disjoint sequence of Galois extensions, L1 , L2 , L3 , . . ., with Gal(Li /K) ∼ = G, i = 1, 2, 3, . . . (Lemma 16.2.6). We prove that this set up occurs for symmetric groups (Corollary 16.2.7), Abelian groups (Proposition 16.3.5), and when char(K) = 0 for alternating groups (Proposition 16.7.6). If K is PAC (but not necessarily Hilbertian), every finite group is regular over K (Proposition 16.12.2). If K is Hilbertian, then Zp occurs over K (Corollary 16.6.7) but is not necessarily regular over K. For example, Zp is not regular over Q (Corollary 16.6.11). Realization of a finite nonsimple group over K is usually done in steps. First one realizes a quotient of the group and then embeds the solution field in a larger Galois extension with the given Galois group. The latter step is always possible when K is Hilbertian and the kernel is a product of nonAbelian simple groups each of which has a GAR realization over K (Sections 16.8 and 16.9). For example, An with n = 5 or n ≥ 7 is GAR over K when char(K) - (n − 1)n (Corollary 16.9.2). A Zp -extension N of K is an example of a Galois extension with finitely generated Galois groups. If K is Hilbertian, then so is N (Proposition 16.11.1). This is one ingredient of the proof that each Abelian extension of K is Hilbertian (Theorem 16.11.3). Finally, the regularity of Z/pZ over a Hilbertian field K has some implications to the structure of Gal(K). For example, Gal(K) has no closed prosolvable normal subgroup. In particular, the center of Gal(K) is trivial. Chapter 18 will exploit the result about the regularity of Sn .

16.1 Galois Groups of Polynomials We prove two theorems about preservation of Galois groups of polynomials under specializations of parameters. One of them (Proposition 16.1.5) is a polynomial analog of Lemma 13.1.1(b). It assumes the ground field to be Hilbertian. The other one (Proposition 16.1.4) is an application of BertiniNoether. Here the ground field is arbitrary but the polynomial in question is absolutely irreducible and Galois. We start with an analog of Lemma 13.1.1(a) for Galois groups of polynomials: Let f ∈ K[X] be a separable polynomial of degree n. By definition, f

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Chapter 16. Galois Groups over Hilbertian Fields

has n distinct roots x1 , . . . , xn in Ks . Thus, L = K(x1 , . . . , xn ) is a finite Galois extension. Restriction of elements of Gal(L/K) to {x1 , . . . , xn } gives an embedding of Gal(L/K) into the group of all permutations of {x1 , . . . , xn }, called a permutation representation. The image of Gal(L/K) under this representation is Gal(f, K). ¯ and a separable polynomial f¯ ∈ K[X] ¯ Next consider another field K of ∼ ¯ ¯ degree n. We write Gal(f, K) = Gal(f , K) to indicate that the two groups in question are isomorphic as permutation groups. Thus, there exists an ¯ as abstract groups and there isomorphism σ 7→ σ ¯ of Gal(f, K) and Gal(f¯, K) ¯ ¯n of the roots of f with σxi = σ ¯x ¯i , σ ∈ Gal(f, K), is a listing x ¯1 , . . . , x ¯ into i = 1, . . . , n. Similarly we speak about an embedding of Gal(f¯, K) ∼ Gal(f, K) as permutation groups. Finally, we write Gal(f, K) = G for a finite group G when Gal(f, K) and G are isomorphic as abstract groups. Lemma 16.1.1: Let L/K be a finite Galois extension, f ∈ K[X] a separable polynomial, and ϕ a place of L. Denote the residue field of K (resp. L) under ¯ (resp. L). ¯ Suppose L is the splitting field of f over K and f¯ = ϕ(f ) ϕ by K ¯ is a separable polynomial in K[X] with deg(f¯) = deg(f ). ∗ ¯ → Gal(f, K). (a) Then there is an embedding ϕ : Gal(f¯, K) (b) Suppose in addition, f is irreducible and Galois, and f¯ is irreducible. ¯ is the splitting field of f¯ Then f¯ is Galois, ϕ∗ is an isomorphism, and L ¯ over K. ¯ K ¯ is separable. Then L ¯ is the splitting field of (c) Alternatively, suppose L/ ¯ f¯ over K. Qn Proof of (a): Let a be the leading coefficient of f . Then f (X) = a i=1 (X − xi ) with distinct x1 , . . . , xn ∈ Ks and L = K(x1 , . . . , xn ) is the splitting field of f over K. Extend ϕ to a place of L(X) with the same notation ¯ ×. such that ϕ(X) = X. Since deg(f¯) = deg(f ), we haveQ a ¯ = ϕ(a) ∈ K n × ¯ and f¯(X) = a ¯ i=1 (X − x ¯i ). By Hence, x ¯i = ϕ(xi ), i = 1, . . . , n, are in K s ¯n are distinct. Therefore, the map xi 7→ x ¯i , i = 1, . . . , n, assumption, x ¯1 , . . . , x is bijective. Lemma 6.1.1 gives an epimorphism σ 7→ σ ¯ of the decomposition group ¯ K). ¯ It satisfies σ ¯ ¯ y¯ = σy for each y ∈ K with y¯ = ϕ(y) ∈ L. Dϕ onto Aut(L/ ¯x ¯i = x ¯i . Since σ permutes x1 , . . . , xn , we have Suppose σ ¯ = 1. Then σxi = σ σxi = xi , i = 1, . . . , n. Hence, σ = 1. ¯ K(¯ ¯ x1 , . . . , x ¯n )). Then there is a σ ∈ Dϕ with Moreover, let τ ∈ Aut(L/ ¯ K) ¯ → σ ¯ = τ . By the preceding paragraph, σ ¯ = 1. It follows that res: Aut(L/ ¯ ¯ ¯n )/K) is an isomorphism. Gal(K(¯ x1 , . . . , x Finally, consider each σ ∈ Gal(L/K) as a permutation of {x1 , . . . , xn }. ¯n }. Then σ ¯ 7→ σ is an Likewise, consider σ ¯ as a permutation of {¯ x1 , . . . , x ¯ → Gal(f, K). embedding ϕ∗ : Gal(f¯, K) Proof of (b): Since f is irreducible and Galois, deg(f ) = |Gal(f, K)|. Since

16.1 Galois Groups of Polynomials

293

¯ f¯ is irreducible, deg(f¯) ≤ |Gal(f¯, K)|. By (a), ¯ ¯ = [K(¯ ¯ x1 , . . . , x ¯n ) : K] deg(f ) = deg(f¯) ≤ |Gal(f¯, K)| ¯ : K] ¯ ≤ [L : K] = |Gal(f, K)| = deg(f ). ≤ [L ¯ = |Gal(f, K)| and deg(f¯) = [K(¯ ¯ x1 , . . . , x ¯ Thus, f¯ Hence, |Gal(f¯, K)| ¯n ) : K]. ¯ = K(¯ ¯ x1 , . . . , x ¯n ) is Galois is Galois and ϕ∗ is an isomorphism. In addition, L ¯ over K. ¯ K ¯ is normal and separable, it is Galois. By the proof Proof of (c): Since L/ ¯ = K(¯ ¯ x1 , . . . , x ¯ ¯ ¯n )) = 1. Therefore, L ¯n ). of (a), Gal(L/K(¯ x1 , . . . , x An essential assumption in Lemma 16.1.1 is the irreducibility of f¯. Lemma 16.1.4 below uses Bertini-Noether to satisfy this assumption. Proposition 16.1.5 applies Hilbert irreducibility theorem to achieve the same goal. Lemma 16.1.2: Let V be a variety in An over a field K and x a generic point of V over K. Then V has a nonempty Zariski K-open subset V0 with ˜ there is a K-place of K(x) with the following property: For each a ∈ V0 (K) ϕ(x) = a and with residue field K(a). Proof: By Corollary 10.2.2(a), K(x)/K is a regular extension. Assume without loss that x1 , . . . , xr form a separating transcendence base for K(x)/K. Example 2.6.10 produces a K-place ϕ0 of K(x1 , . . . , xr ) with ϕ(xi ) = ai , i = 1, . . . , r, and with residue field K(a1 , . . . , ar ). Use Remark 6.1.5 to find a nonzero polynomial g ∈ K[X1 , . . . , Xr ] such that K[x1 , . . . , xi+1 , g(x1 , . . . , xr )−1 ]/K[x1 , . . . , xi , g(x1 , . . . , xr )−1 ] is a ring cover, i = r, . . . , n − 1. Let V0 = {a ∈ V | g(a1 , . . . , ar ) 6= 0}. ˜ Let ϕ a place of K(x) which extends ϕ0 with Suppose a ∈ V0 (K). ϕ(x) = a. By Remark 6.1.7, the residue field of K(x) at ϕ is K(a). Remark 16.1.3: Simple points. The conclusion of Lemma 16.1.2 is actually true for each simple point a of V [Jarden-Roquette, Cor. A2]. Proposition 16.1.4: Let V be a variety over a field K0 in Am , u a generic point of V over K0 , K = K0 (u), and h1 , . . . , hk ∈ K0 [U1 , . . . , Um , T1 , . . . , Tr , X] polynomials. Suppose hj (u, T, X) are absolutely irreducible as polynomials in (T, X) over K and Galois as polynomial in X over K(T). Then there exists a nonempty Zariski K0 -open subset V0 of V with the following property: For each a ∈ V0 (K0 ), hj (a, T, X) is absolutely irreducible, Galois over K0 (T), and Gal(hj (a, T, X), K0 (T)) ∼ = Gal(hj (u, T, X), K(T)), j = 1, . . . , k. Proof: Bertini-Noether (Proposition 9.4.3) gives a nonempty Zariski K0 open subset V1 of V such that hj (a, T, X) are absolutely irreducible for each

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˜ 0 ), j = 1, . . . , k. Lemma 16.1.2, applied to V over K0 (T), gives a a ∈ V1 (K nonempty Zariski K0 (T)-open subset V20 of V satisfying this: For each a ∈ ^ V20 (K 0 (T)) there is a K0 (T)-place ϕ of K0 (u, T) with residue field K0 (a, T ). ˜ 0) ⊆ Choose a nonempty Zariski K0 -open subset V2 of V such that V2 (K ˜ 0 ). Finally, there is a nonempty Zariski K0 -open subset V3 of V with V20 (K hj (a, T, X) separable and degX (hj (a, T, X)) = degX (hj (u, T, X)) for each ˜ 0 ), j = 1, . . . , k. a ∈ V3 (K V0 = V1 ∩ V2 ∩ V3 is a nonempty Zariski K0 -open subset of V . For each a ∈ V0 (K) let ϕ be a K0 -place of K(T) with residue field K0 (T). By Lemma 16.1.1, hj (a, T, X) is Galois over K0 (T) and (1) holds. Proposition 16.1.5: Let K be a Hilbertian field and hj ∈ K[T1 , . . . , Tr , X] a separable polynomial in X, j = 1, . . . , k. Then K r has a separable Hilbert subset H with Gal(hj (a, X), K) ∼ = Gal(hj (T, K), K(T)) for each a ∈ H, j = 1, . . . , k. Proof: Denote the splitting field of hj (T, X) over K(T) by Fj . Proposition 13.1.1(a) gives a nonempty Zariski K-open subset U1 of Ar satisfying the following assertion: For each a ∈ U1 (K) there is a K-place ϕj of Fj extending T 7→ a such that the residue field F¯j of Fj under ϕj is Galois over K, j = 1, . . . , k. Proposition 13.1.1(b) gives a separable Hilbert subset H 0 of K r such that for each a ∈ H 0 ∩ U1 (K) and every K-place ϕ of Fj with residue field F¯j we have Gal(F¯j /K) ∼ = Gal(Fj /K(T)). Finally, there is a nonempty Zariski open subset U2 of Ar satisfying this: hj (a, X) is separable and degX (hj (a, X)) = degX (hj (T, X)) for each a ∈ U2 (K), j = 1, . . . , k. Put H = H 0 ∩ U1 (K) ∩ U2 (K). Consider a ∈ H and let ϕj be as above. Then Lemma 16.1.1 gives an isomorphism ϕ∗j : Gal(hj (a, X), K) → Gal(hj (T, X), K(T)), j = 1, . . . , k, as claimed.

16.2 Stable Polynomials Let K be a field and G a profinite group. Suppose K has a Galois extension L with Gal(L/K) ∼ = G. Then, G occurs (or is realizable) over K and L is a G-extension of K. If G belongs to a family G of profinite groups, we also say that L is a G-extension of K. For example, when G is Abelian, we say L is an Abelian extension of K. If G is an inverse limit of finite solvable groups, we say L is a prosolvable extension of K. The main problem of Galois theory is whether every finite group occurs over Q. Even if this holds, it is not clear whether the same holds for every Hilbertian field.

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295

To approach the latter problem, consider algebraically independent elements t1 , . . . , tr over K. If K is Hilbertian, G is finite, and G occurs over K(t), then G occurs over K (Lemma 13.1.1(b)). If we want G to occur more than once over K, we have to assume more, as we now explain. We say G is regular over K if there exist algebraically independent elements t1 , . . . , tr over K such that K(t) has a Galois extension F which is regular over K with Gal(F/K(t)) ∼ = G. This stronger property is inherited by all extensions of K: Lemma 16.2.1: Let K be a field and G a profinite group. Suppose G is regular over K. Then G is regular over every extension L of K. Proof: Let t = (t1 , . . . , tr ) and F be as above. Consider a field extension L of K. Assume without loss that t1 , . . . , tr are algebraically independent over L. Then F is linearly disjoint from L over K (Lemma 2.6.7). Hence, F L is a regular extension of L (Corollary 2.6.8), F is linearly disjoint from L(t) over K(t) (Lemma 2.5.3), and Gal(F L/L(t)) ∼ = G. Consequently, G is regular over L. The regular inverse Galois problem asks whether every finite group is regular over every field. By Lemma 16.2.1, it suffices to check the problem over Q and over each of the fields Fp . In this generality the problem is still wide open. Nevertheless, there are many special cases where groups G are proved to be regular over specific fields. This chapter discusses cases when this happens. Remark 16.2.2: Reinterpretation of ‘regularity’ by polynomials. Consider an irreducible polynomial f ∈ K[T1 , . . . , Tr , X], separable with respect to X. Call f X-stable over K if (1)

Gal(f (T, X), K(T)) ∼ = Gal(f (T, X), L(T))

for every extension L of K. In this case denote the splitting field of f (T, X) over K(T) by F . Then Gal(F/K(T)) ∼ = Gal(F L/L(T)). Hence, F is linearly disjoint from L over K (Lemma 2.5.3). Thus, F is a regular extension of K and G is regular over K. In addition, f is absolutely irreducible (Corollary 10.2.2). Conversely, suppose K(T1 , . . . , Tr ) has a Galois extension F which is regular over K and with Galois group G. Choose a primitive element x for F/K(T) and let f ∈ K[T, X] be an irreducible polynomial with f (T, x) = 0. Then f (T, X) is X-stable over K, Galois with respect to X over K(T), and Gal(f (T, X), K(T)) ∼ = G. Lemma 16.2.3: Let K be a field and f a polynomial in K[T1 , . . . , Tr , X], separable in X. Suppose Gal(f (T, X), K(T)) ∼ = Gal(f (T, X), Ks (T)). Then f is X-stable over K. Proof: Condition (1) holds when L is a purely inseparable or a regular extension of K. Hence, we have to consider only the case where L is a

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separable algebraic extension of K. In this case both maps res: Gal(f (T, X), Ks (T)) → Gal(f (T, X), L(T)) res: Gal(f (T, X), L(T)) → Gal(f (T, X), K(T)) are injective. By assumption, their compositum is bijective. Hence, each of them is bijective. Lemma 16.2.4: Let K be a field and N a Galois extension of K. (a) Suppose f is a polynomial in N [T1 , . . . , Tr , X] separable in X. Then K has a finite Galois extension L in N with f ∈ L[T, X] and Gal(f (T, X), L(T)) ∼ = Gal(f (T, X), N (T)). (b) If f is Galois over N (T), then f is Galois over L(T). (c) If f is X-stable over N , then f is X-stable over L. Proof of (a): Choose a finite extension L0 of K in N which contains the coefficients of f . Denote the splitting field of f (T, X) over L0 (T) by F . Then L1 = F ∩ N is a finitely generated extension of K (Lemma 10.5.1). In addition, L1 /K is a separable algebraic extension. Hence, L1 /K is finite. Also, Gal(F/L1 (T)) ∼ = Gal(F N/N (T)). Finally, let L be the Galois closure of L1 /K. Then L satisfies the requirements of (a). Proof of (b): Suppose f and L satisfy the conclusion of (a) and f is Galois over N (T). Let x1 , . . . , xn be the roots of f (T, X) in K(T)s . Consider σ ∈ Gal(f (T, X), L(T)). Suppose xσ1 = x1 . By (a), σ extends to an element σ in Gal(f (T, X), N (T)). Hence, xσi = xi , i = 1, . . . , n. Therefore, f is Galois over L(T). Proof of (c): By assumption, Gal(f (T, X), N (T)) ∼ = Gal(f (T, X), Ks (T)). Hence, by (a), Gal(f (T, X)), L(T)) ∼ = Gal(f (T, X), Ks (T)). By Lemma 16.2.3, f is X-stable over L. Example 16.2.5: Stable polynomials. (a) The general polynomial of degree n is f (T1 , . . . , Tn , X) = X n + T1 X n−1 + · · · + Tn . It satisfies Gal(f (T, X), K(T)) ∼ = Sn for every positive integer n and for every field K [Lang7, p. 272, Example 4]. Thus, f is X-stable over every field. (b) The polynomial X n −T satisfies Gal(X n −T, K(T )) ∼ = Z/nZ for every field K with char(K) - n that contains ζn . Thus, X n − T is X-stable over K. If however ζn 6∈ K, then Gal(X n − T, K(T )) ∼ = Gal(K(ζn )/K) n Z/nZ. Therefore, X n − T is not X-stable over K. (c) The polynomial X p − X − T satisfies Gal(X p − X − T, K(T )) ∼ = Z/pZ for every field K of characteristic p [Lang7, p. 290, Thm. 6.4]. Therefore, f is X-stable over K.

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297

(d) Suppose Gal(f (T, X), K(T)) is simple. Let F be the splitting field of f (T, X) over K(T). Suppose F 6⊆ Ks (T). Then F ∩ Ks (T) = K(T). Therefore, f is X-stable over K. (e) Suppose f ∈ K[T, X] is absolutely irreducible and Galois over K(T). Then f is X-stable over K. Indeed, let L be an extension of K. Then Gal(f (T, X), L(T)) is a subgroup of Gal(f (T, X), K(T)). On the other hand, f (T, X) is also Galois over L(T). Hence, |Gal(f (T, X), L(T))| = degX (f (T, X)) = |Gal(f (T, X), K(T))|. Thus, Gal(f (T, X), L(T)) ∼ = Gal(f (T, X), K(T)), as claimed.

We use stable polynomial to construct linearly disjoint sequences of Galois extensions over Hilbertian fields with a given Galois groups: Lemma 16.2.6: Let fi (T1 , . . . , Tr , X) be an X-stable polynomial over a Hilbertian field K and Gi = Gal(fi (T, X), K(T)), i = 1, 2, 3, . . . . Given a finite separable extension L0 of K, there is a sequence, L1 , L2 , L3 , . . . of Galois extensions of K with the following properties: (a) Gal(Li /K) ∼ = Gi , and fi has an Li -rational zero ci , i = 1, 2, 3, . . . . (b) c1 , c2 , c3 , . . . are distinct. (c) The sequence L0 , L1 , L2 , . . . is linearly disjoint over K. Proof: Suppose by induction, there are Galois extensions L1 , . . . , Ln of K such that Gal(Li /K) ∼ = Gi , i = 1, . . . , n, and L0 , L1 , . . . , Ln are linearly disjoint over K. In addition suppose there are ai ∈ K r and bi ∈ Li such that fi (ai , bi ) = 0 and a1 , . . . , an are distinct. Then L = L0 L1 · · · Ln is a finite separable extension of K. By assumption, Gal(fn+1 (T, X), L(T)) ∼ = Gal(fn+1 (T, X), K(T)) = Gn+1 . Apply Lemma 16.1.5 and Corollary 12.2.3 to find an+1 ∈ K r with Gal(fn+1 (an+1 , X), L) ∼ = Gal(fn+1 (an+1 , X), K) ∼ = Gn+1 and an+1 6= a1 , . . . , an . Let Ln+1 be the splitting field of f (a, X) over K. Then Gal(Ln+1 /K) ∼ = Gn+1 and Ln+1 is linearly disjoint from L over K. In particular, L0 , . . . , Ln+1 are linearly disjoint over K. Finally, choose a zero bn+1 of fn+1 (an+1 , X). Then bn+1 ∈ Ln+1 . Our next result is an immediate application of this lemma to the polynomials of Example 16.2.5(a),(b),(c): Corollary 16.2.7: Let K be a Hilbertian field and G a finite group. Then K has a linearly disjoint sequence of Galois extensions with G as Galois group in each of the following cases: (a) G = Sn , n ∈ N. (b) ζn ∈ K and G = Z/nZ, n ∈ N, char(K) - n.

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(c) char(K) = p > 0 and G = Z/pZ. We prove in Section 16.4 that every finite Abelian group A satisfies the conclusion of Corollary 16.2.7. But this simple corollary already has interesting implications to closed normal subgroups of Gal(K) (Section 16.12). The following result shows that in many cases it suffices to take r = 1 in the definition of “G is regular over K”: Proposition 16.2.8: Let K be an infinite field and G a finite group. Suppose G is K-regular. Then: (a) There is an X-stable polynomial h ∈ K[T, X] which is Galois with respect to X and with Gal(h(T, X), K(T )) ∼ = G. (b) If in addition K is Hilbertian, then K has a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions such that Gal(Li /K) ∼ = G for all i. Proof: Statement (b) follows from (a) by Lemma 16.2.6. We prove (a): Remark 16.2.2 gives a polynomial f ∈ K[T1 , . . . , Tr , X] which is Xstable over K with r ≥ 2 and Galois with respect to X over K(T) such that Gal(f (T, X), K(T)) ∼ = G. Assume without loss f is monic with respect to X. Let u1 , u2 be algebraically independent elements over K. Put L = K(u1 , u2 ), T0 = (T1 , . . . , Tr−1 ), and g(T0 , X) = f (T1 , . . . , Tr−1 , u1 + u2 Tr−1 , X). By Proposition 10.5.4, g(T0 , X) is absolutely irreducible. Extend the map T → (T0 , u1 + u2 Tr−1 ) to an L(T0 )-place of L(T) with residue field L(T0 ) (Lemma 2.2.7). By Lemma 16.1.1, G∼ = Gal(f (T, X), L(T)) ∼ = Gal(g(T0 , X), L(T0 )). By assumption, K is infinite. Hence, Proposition 16.1.4 gives b1 , b2 ∈ K with g¯(T0 , X) = f (T1 , . . . , Tr−1 , b1 + b2 Tr−1 , X) absolutely irreducible, Galois over g (T0 , X), K(T0 )) ∼ K(T0 ), and Gal(¯ = Gal(g(T0 , X), L(T0 )) ∼ = G. Finally, use induction on r to find an absolutely irreducible polynomial h ∈ K[T1 , X] with Gal(h(T1 , X), K(T1 )) ∼ = G. Problem 16.2.9:

Does Proposition 16.2.8 hold if K is finite?

16.3 Regular Realization of Finite Abelian Groups The inverse problem of Galois theory has an affirmative solutions for every finite Abelian group A and every Hilbertian field K (Corollary 16.3.6). Moreover, every finite Abelian group is even regular over K (Proposition 16.3.5). The proof of the latter result moves from the case where A is cyclic to the general case. In the cyclic case we have to distinguish between the case where char(K) does not divide the order of the group and where the order of the group is a power of char(K). In the general case, we distinguish between the cases where K is finite and K is infinite.

16.3 Regular Realization of Finite Abelian Groups

299

Lemma 16.3.1: Let K be a field, n a positive integer with char(K) - n, and t an indeterminate. Then K(t) has a cyclic extension F of degree n which is contained in K((t)). Proof: Choose a root of unity ζn of order n in Ks . Let L = K(ζn ) and G = Gal(L/K). Then there is a map χ: G → {1, . . . , n − 1} such that χ(σ) σ(ζn ) = ζn . Then gcd(χ(σ), n) = 1 and (1)

χ(στ ) ≡ χ(σ)χ(τ ) mod n

for all σ, τ ∈ G. By Example 3.5.1, K((t)) is a regular extension of K and L((t)) = K((t))(ζn ). Thus, we may identify G with Gal(L((t))/K((t))). Choose a primitive element c of L/K. Consider the element g(t) =

Y

χ(σ−1 )

1 + σ(c)t

σ∈G

of L[t]. Since char(K) - n, Hensel’s lemma (Proposition 3.5.2) gives an x ∈ Q −1 L[[t]] with xn = 1 + ct. Then y = σ∈G σ(x)χ(σ ) ∈ L[[t]] and y n = Q Q −1 n χ(σ −1 ) = σ∈G (1 + σ(c)t)χ(σ ) = g(t). Since ζn ∈ L, F = σ∈G σ(x ) L(t, y) is a cyclic extension of degree d of L(t), where d|n and y d ∈ L(t) [Lang7, p. 289, Thm. 6.2(ii)]. Since χ(σ −1 ) is relatively prime to n, we must have d = n. The Galois group Gal(F/L(t)) is generated by an element ω satisfying ω(y) = ζn y. By (1) there exist for each τ, ρ ∈ G a positive integer k(τ, ρ) and a polynoQ Q −1 −1 mial fτ (t) ∈ L[t] such that τ (y) = σ∈G τ σ(x)χ(σ ) = ρ∈G ρ(x)χ(ρ τ ) = Q Q χ(ρ−1 )χ(τ )+k(τ,ρ)n = y χ(τ ) ρ∈G (1 + ρ(c)t)k(τ,ρ) = y χ(τ ) fτ (t). It ρ∈G ρ(x) follows that G leaves F invariant. Let E be the fixed field of G in F . K((t)) E K(t) K

L((t)) F = L(t, y) L(t) L= K(ζn )

Denote the subgroup of Aut(F/K(t)) generated by G and Gal(F/L(t)) by H. Then the fixed field of H is K(t), so F/K(t) is a Galois extension with Gal(F/K(t)) = G · Gal(F/L(t)). Moreover, given τ ∈ G, put m = χ(τ ). Then τ ω(y) = τ (ζn y) = ζnm y m fτ (t) = ω(y)m fτ (t) = ω(y m fτ (t)) = ωτ (y). Thus, τ ω = ωτ , so G commutes with Gal(F/L(t)). Therefore, E/K(t) is a Galois extension with Gal(E/K(t)) ∼ = Gal(F/L(t)) ∼ = Z/nZ.

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Lemma 16.3.2: Suppose p = char(K). Let L be a cyclic extension of degree pn , n ≥ 1, of K. Then K has a Z/pn+1 Z-extension L0 which contains L. Proof: Define L0 to be L(x) where x is a zero of X p − X − a with a ∈ L. The three parts of the proof produce a, and then show L0 has the desired properties. Part A: Construction of a. Since L/K is separable, there is a b1 ∈ L with c = traceL/K (b1 ) 6= 0 [Lang7, p. 286, Thm. 5.2]. Put b = bc1 . Then traceL/K (b) = 1 and traceL/K (bp − b) = (traceL/K (b))p − traceL/K (b) = 0. With σ a generator of Gal(L/K), the additive form of Hilbert’s Theorem 90 [Lang7, p. 290, Thm. 6.3] gives a ∈ L with σa − a = bp − b.

(2)

Part B: Irreducibility of X p − X − a. Assume X p − X − a is reducible over L. Then x ∈ L [Lang7, p. 290, Thm. 6.4(b)]. Thus (3) (σx − x)p − (σx − x) − (bp − b) = (σx − x)p − (σx − x) − (σa − a) = (σxp − σx − σa) − (xp − x − a) = 0 Since b is a root of X p − X − (bp − b), there is an i with σx − x = b + i [Lang7, p. 290, Thm. 6.4(b)]. Apply traceL/K to both sides to get 0 on the left and 1 on the right. This contradiction proves X p − X − a is irreducible. Part C: Extension of σ to σ 0 that maps x to x+b. Equality (2) implies x+b is a zero of X p − X − σa. Thus, by Part B, σ extends to an automorphism σ 0 of L0 with σ 0 (x) = x + b. We need only prove that σ 0 has order pn+1 . Induction shows (σ 0 )j (x) = x + b + σb + · · · + σ j−1 b. In particular, n

(σ 0 )p (x) = x + traceL/K (b) = x + 1.

(4) n

Hence, (σ 0 )ip (x) = x + i, i = 1, . . . , p. Therefore, the order of σ 0 is pn+1 , as contended. Remark 16.3.3: Lemma 16.3.2 is a special case of a theorem of Witt. Suppose char(K) = p. Then AS(K) = {xp − x | x ∈ K} is a subgroup of the additive group of K and K/AS(K) is a vector space over Fp of dimension, say, r. Consider an embedding problem G → Gal(L/K) over K with G a finite p-group which is generated by r elements. Witt’s theorem says this problem is solvable (If r = ∞, there is no restriction on the number of generators of G.) The technique of Galois cohomology [Ribes, p. 257, Cor. 3.4] simplifies Witt’s original proof [Witt]. Lemma 16.3.4: Let K be a field, t an indeterminate, and A a finite cyclic group. Then K(t) has a Galois extension F such that Gal(F/K(t)) ∼ = A and F/K is regular. Proof: We put p = char(K) and divide the proof into three parts:

16.3 Regular Realization of Finite Abelian Groups

301

Part A: A ∼ = Z/mZ and p - m. By Lemma 16.3.1, K(t) has a cyclic extension Em of degree m which is contained in K((t)). By Example 3.5.1, K((t)) is a regular extension of K. Hence, so is Em (Corollary 2.6.5(b)). Part B: A ∼ = Z/pk Z. Assume without loss that k ≥ 1. By Eisenstein’s criterion and Gauss’ lemma, the polynomial X p − X − t is irreducible over ˜ K(t). Let x be a root of X p −X −t in K(t)s . Then, by Artin-Schreier, [Lang7, p. 290, Thm. 6.4(b)], K(x) is a cyclic extension of degree p of K(t). Lemma 16.3.2 gives a cyclic extension Epk of K(t) of degree pk which contains K(x). ˜ By the preceding paragraph, K(x) ∩ K(t) = K(t). Since Gal(Epk /K(t)) k is a cyclic group of order p , each subextension of Epk which properly contains ˜ K(t) must contain K(x). Hence, Epk ∩ K(t) = K(t). Thus, Epk is linearly ˜ disjoint from K(t) over K(t). By the tower property (Lemma 2.5.3), Epk is ˜ over K; that is, Epk /K is regular. linearly disjoint from K Part C: A ∼ = Z/nZ, n = mpk , p - m. The compositum En = Em Epk is a ˜ cyclic extension of K(t) of degree n. Moreover, En ∩ K(t) decomposes into a cyclic extension of K(t) of degree which divides m and a cyclic extension of K(t) degree dividing pk . By Parts A and B, both subextensions must be K(t). It follows that En is a regular extension of K. We generalize Lemma 16.3.4 from cyclic groups to arbitrary Abelian groups: Proposition 16.3.5: Let K be a field, t an indeterminate, and A a finite Abelian group. Then K(t) has a Galois extension F such that Gal(F/K(t)) ∼ = A and F/K is regular. Proof: The first two parts of the proof prove the proposition in the case where A = (Z/qZ)n , q = pr , p is a prime number, and n, r are positive integers. Part A: A = (Z/qZ)n is as above and K is infinite. Choose algebraically independent elements t1 , . . . , tn over K. For each i between 1 and n, Lemma 16.3.4 gives a finite cyclic extension Ei of K(ti ) of degree q which is regular over K. Then E1 , . . . , En are algebraically independent over K. Hence, by Corollary 2.6.8, E = E1 · · · En is a regular extension of K. In addition, by Lemma 2.6.7, E1 , . . . , En are linearly disjoint over K. Hence, by Lemma 2.5.11, E1 (t), . . . , En (t) are linearly disjoint over K(t) and each Ei (t) is a cyclic extension of K(t) of degree q. It follows that Gal(E/K(t)) ∼ = (Z/qZ)n . Consequently, by Proposition 16.2.8, K(t) has a cyclic extension F of degree q which is regular over K. Part B: A = (Z/qZ)n is as above and K is finite. Choose a transcendental element u over K(t). Lemma 16.3.4 gives a cyclic extension E of K(t, u) of degree q which is regular over K(t). Choose a polynomial h ∈ K(t)[u, X] which is X-stable with Gal(h(u, X), K(t, u)) ∼ = Z/qZ (e.g. use Proposition

302

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16.2.8). Denote the unique extension of K of degree q by Kq . By Theorem 13.4.2, K(t) is Hilbertian. Hence, by Lemma 16.2.6, K(t) has sequence F1 , F2 , F3 , . . . of cyclic extensions of degree q such that Kq (t), F1 , F2 , . . . , Fn are linearly disjoint over K(t). Then F = F1 F2 · · · Fn is a Galois extension of K(t) with Gal(F/K(t)) ∼ = (Z/qZ)n . Moreover, F ∩ Kq (t) = K(t). ˜ The group A = Gal(F ∩ K(t)/K) is cyclic quotient of Gal(F/K(t)), so ˜ A is of exponent q. Hence, [F ∩ K(t) : K(t)] = pj with j ≤ r. Therefore, ˜ ˜ F ∩ K(t) ⊆ Kq (t). It follows from the preceding paragraph that F ∩ K(t) = K(t). Consequently, F/K is regular. ∼ Qm (Z/qi Z)ni , where qi = pri , m, ni , ri are positive integers, Part C: A = i i=1 and p1 , . . . , pr are distinct prime numbers. Parts A and B give for each i a Galois extension Fi of K(t) which is regular over K and with Gal(Fi /K(t)) ∼ = (Z/qi Z)ni . Since q1 , . . . , qm are pairwise relatively prime, F1 , . . . , Fm are linearly disjoint over K(t). Hence, F = F1 · · · Fm is a Galois extension with Gal(F/K(t)) ∼ = A. ˜ Let E = F ∩ K(t). By the preceding paragraph, Fi ∩ E = K(t), so i = 1, . . . , m. Since q1 , . . . , qm are pairwise relatively prime, qini |[F : E] for Qm [F : K(t)] = i=1 q ni divides [F : E]. Hence, E = K(t). Consequently, F is a regular extension of K. Part D: A is an arbitrary finite Abelian group. Then A=

mi m Y Y

r

(Z/pi ij Z)nij

i=1 j=1

where p1 , . . . , pm are distinct prime numbers, m ≥ 0, and mi , nij are positive integers. For each i let ri = max(ri1 , . . . , ri,mi ) and ni = max(ni1 , . . . , ni,mi ). Part C gives a GaloisQextension Fˆ of K(t) which is regular over K and m with Gal(Fˆ /K(t)) ∼ = i=1 (Z/pri i Z)ni . By construction, A is a quotient of Gal(Fˆ /K(t). Hence, K(t) has a Galois extension F in Fˆ with Gal(F/K(t)) ∼ = A. Since Fˆ /K is regular, so is F/K. Corollary 16.3.6: Let K be a Hilbertian field and A a finite Abelian group. Then A is realizable over K. Moreover, K has a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions with Gal(Li /K) ∼ = A for each i. Proof: By Proposition 16.3.5, A is regular over K. Hence, by Proposition 16.2.8, K has a linearly disjoint sequence L1 , L2 , L3 , . . . of Galois extensions with Gal(Li /K) ∼ = A for each i.

16.4 Split Embedding Problems with Abelian Kernels Attempts to realize a finite Abelian group A over a Hilbertian field K usually lead to an extension of K with roots of unity. This gives a “split embedding problem with Abelian kernel”. The main result of this section is that each

16.4 Split Embedding Problems with Abelian Kernels

303

such problem is solvable. Consequently, A occurs over K. Indeed, for A to be regular over K it is not necessary that K is Hilbertian. We prove that it is true for arbitrary field. Definition 16.4.1: Embedding problems. Let L/K be a Galois extension, G a profinite group, and α: G → Gal(L/K) an epimorphism. The embedding problem associated with α consists of embedding L in a Galois extension N of K with an isomorphism β: Gal(N/K) → G satisfying α ◦ β = resN/L . Refer to α as an embedding problem over K, β its solution, and N the solution field. Call the problem finite if G is finite. The problem splits if α has a section, that is, an embedding α0 : Gal(L/K) → G with α ◦ α0 = idGal(L/K) . The latter case occurs when G = Gal(L/K) n Ker(α) and α is the projection of G onto Gal(L/K). Let t1 , . . . , tn be algebraically independent elements over K. Then res: Gal(L(t)/K(t)) → Gal(L/K) is an isomorphism. Hence, α: G → Gal(L/K) gives rise to an embedding problem αt : G → Gal(L(t)/K(t)) over K(t) with resL(t)/L ◦ αt = α. Refer to a solution of αt as a solution of α over K(t). Refer to a solution field F of αt as a regular solution of α if F/L is regular. We say α is regularly solvable if there are t1 , . . . , tn as above and αt has a solution field F which is regular over L. Lemma 16.4.2: Let K be a Hilbertian field, α: H → Gal(L/K) a finite embedding problem, t1 , . . . , tn algebraically independent elements over K, and M a finite separable extension of L. If α is solvable over K(t), then α is also solvable over K. If α is regularly solvable, then α has a solution field N over K which is linearly disjoint from M over L. Proof: Let F be a solution field of α over K(t). Thus, F is a Galois extension of K(t) which contains L and there is an isomorphism θ: Gal(F/K(t)) → H with α ◦ θ = resF/L . Lemma 13.1.1 gives a separable Hilbert subset A of K r having the following property: For each a ∈ A there is a K-place ϕ: F → ˜ ∪ {∞} satisfying these conditions: K (3a) ϕ(t) = a, K(t)ϕ = K, and L(t)ϕ = L. (3b) There is an isomorphism ϕ∗ : Gal(F¯ /K) → Gal(F/K(t)) with resF/L ◦ ϕ∗ = resF¯ /L . Here, F¯ = F¯ϕ . The map β = θ ◦ ϕ∗ solves embedding problem α. Suppose now F/L is regular. Then Gal(F M/M (t)) ∼ = Gal(F/L(t)). Lemma 13.1.1 gives a separable Hilbert subset A0 of M r satisfying this: For each a ∈ A ∩ A0 and each M -place ϕ of F M satisfying (3), Gal(F M/M (t)) ∼ = Gal(F¯ M/M ). By Corollary 12.2.3, there is an a ∈ A ∩ A0 . The preceding paragraph gives ϕ satisfying (3). The corresponding field F¯ is linearly disjoint from M over L. Let A and G be finite groups. Recall that AG is the group of all functions f : G → A and G acts on AG by the rule f τ (σ) = f (τ σ). The semidirect product G n AG is the wreath product A wr G (Remark 13.7.7).

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Chapter 16. Galois Groups over Hilbertian Fields

Lemma 16.4.3: Let G and A be finite groups with A Abelian and G acting on A. Then there is an epimorphism A wr G → G n A with kernel in AG . Q −1 Proof: Define a map α: AG → A by α(f ) = σ∈G f (σ)σ . Since A is Abelian, the right hand side is a well defined homomorphism. For each a ∈ A define a function fa : G → A by fa (1) = a and fa (σ) = 1 for σ ∈ G, σ 6= 1. Then α(fa ) = a, so α is surjective. Next consider τ ∈ G. Then Y Y Y −1 −1 −1 f τ (σ)σ = f (τ σ)σ = f (ρ)ρ τ = α(f )τ . α(f τ ) = σ∈G

σ∈G

ρ∈G

Thus, α respects the action of G. Therefore, α extends to an epimorphism α: G n AG → G n A satisfying α(σf ) = σα(f ) for all σ ∈ G and f ∈ AG . This gives rise to the following commutative diagram: 1

/ AG

/ G n AG

α

1

/A

By definition, Ker(α) ≤ AG .

/G

/1

/G

/1

α

/ GnA

Proposition 16.4.4: Let L/K be a finite Galois extension of degree n with Galois group G. Suppose G acts on a finite Abelian group A. Let π: GnA → G be the projection map. Then π is regularly solvable. Proof: By Proposition 16.3.5, K(T ) has a Galois extension N such that Gal(N/K(T )) ∼ = A and N/K is regular. By Proposition 16.2.8, there exists an X-stable polynomial h ∈ K[T, X] with Gal(h(T, X), K(T )) ∼ = A. In particular, h(T, X) is absolutely irreducible and Gal(h(T, X), L(T )) ∼ = A. Let π ˆ : G n AG → G be the projection on G. Let t1 , . . . , tn be algebraically independent elements over K with n = [L : K]. Lemma 13.8.1, with G0 trivial, gives a Galois extension Fˆ of K(t) which contains L and an isomorphism ˆ ◦ γ = res. Moreover, Fˆ is a regular γ: Gal(Fˆ /K(t)) → G n AG with π extension of L. Lemma 16.4.3 gives an epimorphism α: G n AG → G n A which is the identity map on G. Thus, π ˆ = π ◦ α. Let F be the fixed field in Fˆ of Ker(α ◦ γ). Then there is an epimorphism β: Gal(F/K(t)) → G n A with α ◦ γ = β ◦ resFˆ /F and π ◦ β = resF/L . Thus, β is a solution of π over K(t) with F being the solution field. By construction, L ⊆ F ⊆ Fˆ . Since Fˆ /L is regular, so is F/L (Corollary 2.6.5(b)). Proposition 16.4.5 ([Ikeda, p. 126]): Let K be a Hilbertian field. Then every finite split embedding problem over K with Abelian kernel is solvable Proof: Combine Propositions 16.4.2 and 16.4.4.

16.4 Split Embedding Problems with Abelian Kernels

305

Remark 16.4.6: Proposition 16.4.5 does not hold for an arbitrary profinite Abelian group. For example, Zp is regular over Q for no p (Proposition 16.6.10), but it is not known if every finite p-group is regular over Q. Nevertheless, every finite p-group occurs over Q. More generally, every finite solvable group occurs over every global field. This is a theorem of Shafarevich. Its proof does not use Hilbert irreducibility theorem but rather class field theory and complicated combinatorial arguments. See [Neukirch-SchmidtWingberg, Section 9.5] for a proof that uses cohomological arguments and for a reference to the original articles. Class field theory is not available over an arbitrary Hilbertian field K. Thus, Shafarevich’s proof does not apply to K. However, when p = char(K) > 0, every finite p-group occurs over K. This follows from a theorem of Witt (Remark 16.3.3), but it is not known if each finite p-group occurs over K when char(K) = 0. Amazingly enough, both realization problems raised in Remark 16.4.6 are easy consequences of Shafarevich’s theorem when char(K) > 0. This is the content of the following result: Theorem 16.4.7: Let G be a finite p-group and K a field of positive characteristic. Then G is regular over K. If in addition K is Hilbertian, then G is realizable over K. Proof: The second statement of the theorem follows from the first one by Hilbert (Lemma 13.1.1). In order to prove that G is regular over every field of positive characteristic, it suffices to prove that G is regular over every finite field K (Lemma 16.2.1). Assume without loss G is nontrivial. Let t be an indeterminate. By Shafarevich, G × G occurs over K(t). Thus, K(t) has linearly disjoint Galois extensions F1 and F2 with Gal(Fi /K(t)) ∼ = G, i = 1, 2. Assume none of them ˜ ∩ Fi is a cyclic extension of is regular over K. Then, for i = 1, 2, the field K K of degree pki with ki ≥ 1. Hence, Fi contains the unique extension Kp of K of degree p. Therefore, both F1 and F2 contain Kp (t) which is a proper extension of K(t). This contradiction to the linear disjointness of F1 and F2 over K(t) proves that one of them is regular over K. Corollary 16.4.8: Let K be a field, G a finite group, and A a finite Abelian group. Suppose G is regular over K and G acts on A. Then G n A is regular over K. Proof: There exists t1 , . . . , tn and a finite Galois extension E of K(t) such that Gal(E/K(t)) ∼ = G and E/K is regular. Choose an indeterminate u. Proposition 16.4.4 gives a Galois extension F of K(t, u) such that Gal(F/K(t, u)) ∼ = G n A and F is a regular extension of E. By Corollary 6.2.5, F is a regular extension of K. Thus, G n A is regular over K. Remark 16.4.9: Realization of p-groups of low order. The results of this section imply that each p-group of order at most p4 is regular over every field K.

306

Chapter 16. Galois Groups over Hilbertian Fields

To this end let A be the smallest family of all finite groups satisfying this: (4a) Every Abelian group belongs to A. (4b) Suppose H = G · A with G ∈ A and A Abelian and normal. Then H ∈ A. Note that the group H in (4b) is a quotient of G n A. Hence, if G is regular over K, so is H (Proposition 16.4.8). It follows by induction on the order of the group that each G ∈ A is regular over K. Thus, if K is Hilbertian, every G ∈ A occurs over K (Lemma 13.1.1(b)). The family A contains each finite group G which satisfies one of the following conditions: (5a) G has nilpotence class at most 2, i.e. [G, G] ≤ Z(G) (Thompson [MalleMatzat, p. 277, Prop. 2.9(a)]). (5b) G is solvable and every Sylow subgroup of G is Abelian (Thompson [Malle-Matzat, p. 277, Prop. 2.9(b)]). (5c) G is a p-group of order at most p4 (Dentzer [Malle-Matzat, p. 278, Cor. 2.10]. (5d) G is a 2-group of order 25 (Dentzer [Malle-Matzat, p. 278, Cor. 2.10]). However, there are groups of order p5 (for p 6= 2) and 26 which do not belong to G [Malle-Matzat, p. 278].

16.5 Embedding Quadratic Extensions in Z/2n Z-Extensions Nonsplit embedding problems with Abelian kernel √ over Hilbertian fields need not be solvable. For example Z/4Z → Gal(Q( −1)/Q) √ is not solvable. Otherwise, Q has a Galois extension N containing Q( −1) with Gal(N/Q) ∼ = √ Z/4Z. The only subfields of N are Q, Q( −1), and N . Hence, N ∩ R = Q. Hence, [N R : R] = 4, which is a contradiction. Here is a general criterion for a quadratic extension to be embeddable in a Z/4Z-extension. Proposition 16.5.1: Let K be a field with char(K) 6=√2 and a a nonsquare in K. Then the embedding problem Z/4Z → Gal(K( a)/K) is solvable if and only if there are x, y ∈ K with a = x2 + y 2 . √ Proof: By assumption, L = K( a) is a quadratic extension of K. Suppose K has a Galois extension N √containing L with Gal(N/K) ∼ = Z/4Z. Then u). Let σ be a generator of Gal(N/K). Then there is a u ∈ L with N = L( √ √ √ √ u). By √ Kummer theory, σ u = v u with v ∈ L. Hence, L(√σu) = L( √ of K σ 2 u = v · σv u = b u with b = v · σv. Thus, b is an element √ 2 generates Gal(L( u)/L), so which is a norm from L. On the other had, σ √ √ √ √ σ 2 u = − u. Therefore, b =√−1, v · σv u = − u, and v · σv = −1. Finally, write √ suppose d = 0. Then √ v =√c + d a with c, d ∈ K. First v ∈ K and v 2 u = − u. Hence, v 2 = −1 and −1 ∈ K. Therefore, the

16.5 Embedding Quadratic Extensions in Z/2n Z-Extensions

307

2 2 √ identity a = −1(1 − a4 ) + 1 + a4 yields a representation of a as a sum of two squares in K. Now suppose d 6= 0. Then −1 = v · σv = c2 − ad2 . 2 2 Hence, a = dc + d1 , as desired. Conversely, suppose a = x2 + y 2 with x, y ∈ K. Then x,√ y 6= 0 and −1 = c2 − ad2 , where c = xy and d = y1 . Put v = c + d a. Then normL/K v = c2 − ad2 = −1. Hence, normL/K v 2 = 1. Let σ be an element of Gal(K) whose restriction to L generates Gal(L/K). Hilbert’s Theorem 90 gives u ∈ L with v 2 = σu u [Lang7, p. 288, Thm. 6.1]. Thus, √ √ σ√ u2 σ√ u 2 and u = ±v. Replacing v by −v, if necessary, we may assume v = u √ √ √ √ √ √ √ √ √ u = − u, σ 3 u = −v u,√and σ 4 u = u. σ u =√v u,√σ 2 u√= v · σv √ Thus, u, σ u, σ 2 u, √ σ 3 u are distinct conjugates of u over K. All of of degree 4 them belong to N = L( u). Therefore, N is a cyclic extension √ of K which contains L. Consequently, Z/4Z → Gal(K( a)/K) is solvable. Remark 16.5.2: Embedding in cyclic extensions of higher order. It is possible to slightly generalize Proposition 16.5.1: Suppose K is a field and a is √ a nonsquare in K. Then K( a) can be embedded in a Z/8Z-extension of K if and only if a is a sum of two squares in K and there are x1 , x2 , x3 , x4 in K, not all zero, with x21 + 2x22 + ax23 − 2ax24 = 0. This result is proved in [Kiming, Thm. 3] in a slightly different form. The formulation we give appears in [Geyer-Jensen2, 20◦ ]. Let√K be a number field and a ∈ K. Suppose −1 is a nonsquare √ in K and K( −1) is embeddable in a Z/16Z-extension of K. Then K( −1) is embeddable in a Z/2n Z-extension of K for each positive integer n [GeyerJensen2, Thm. 4]. The proof requires class field theory. √ √ As an example, let K = Q( −14). Then K( −1) is embeddable in a Z/8Z-extension of K [Arason-Fein-Schacher-Sonn, p. 846, Cor. 4 or Geyer√ Jensen2, Remark 19◦ ]. But K( −1) is not embeddable in a Z/16Z-extension of K [Geyer-Jensen2, Example to Proposition 3]. Let OK be the ring of integers of K. Consider a nonzero prime ideal p of OK such that ˆ p. (1) −1 a nonsquare but −1 is the sum of two squares in the completion K Since X 2 +Y 2 +1 is absolutely irreducible. Hence, −1 is a sum of two squares ˆp in Fp for all large p. By Hensel’s lemma, −1 is a sum of two squares in K for all but finitely many p. Hence, by Chebotarev, Condition (1) holds with √ ˆ p ( −1) is embeddable in a Z/2n Z-extension of K ˆ p for all density 12 . Then K n [Geyer-Jensen2, Prop. 3]. Thus, there could be no criterion for embedding a quadratic extension of a field in a Z/16Z-extension similar to the one we gave above for embedding in a Z/8Z-extension. √ √ Consider now the field K = Q( −17). Then K( −1) is embeddable in a Z/2n Z-extension for each positive integer n but is not embeddable in a Z2 -extension [Geyer-Jensen1, p. 371].

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16.6 Zp -Extensions of Hilbertian Fields Lemma 16.3.2 implies that if K is a field of characteristic p, then every Z/pZ-extension of K can be embedding in a Zp -extension. This result does not generalize to the general case. Remark 16.5.2 gives an example of a quadratic extension L/K (with char(K) 6= 2) which can not be embedded in a Z2 -extension. So, we have to settle for less. We fix a field K, a prime number p, and let q = p if p = 2 and q = 4 if p = 2. We ask when Zp occurs over K. We prove this is the case when Z/qZ occurs over K. This condition is satisfied when K is Hilbertian. Lemma 16.6.1: Suppose p 6= char(K). Let L be a cyclic extension of K of degree pn with n ≥ 1. (a) Suppose ζpn+1 ∈ K. Then K has a Z/pn+1 Z-extension which contains L. (b) Suppose ζpk ∈ K for all k. Then K has a Zp -extension that contains L. Proof: Statement (b) follows from (a) by induction and taking inverse limit. It remains to prove Statement (a). n The theory of cyclic extensions gives a ∈ K with L = K(a1/p ) [Lang7, n+1 / K. The field L0 = K(a1/p ) p. 289, Thm. 6.2(i)]. In particular, a1/p ∈ contains L. By [Lang7, p. 289, Thm. 6.2(ii)], L0 is a cyclic extension of n+1−k ∈ K. If k ≤ n, then a1/p = degree pk for some k ≤ n + 1 and a1/p n+1−k n−k (a1/p )p ∈ K. We conclude from this contradiction that k = n + 1, as desired. Remark 16.6.2: Abelian pro-p groups as Zp -modules. Let A be a finite Abelian additive p-group of exponent pm . Then Z/pm Z acts on A through the rule (k + pm Z)a = ka, a ∈ A. Since Z/pm Z is a quotient of Zp , this defines a continuous action of Zp on A with the discrete topology. This action commutes with homomorphisms of finite Abelian p-groups. Hence, it defines a continuous action of Zp on projective limits of Abelian p-groups (also called Abelian pro-p groups which commutes with homomorphisms. r Consider elements a1 , . . . , ar of A. Then the map Pr Pr Zp → A given by compact, (z1 , . . . , zr ) 7→ i=1 zi ai is continuous. Its image i=1 Zp ai is P r hence closed. On the P other hand, one can approximate each sum i=1 zi ai n with zi ∈ Zp by sums i=1 ki ai with ki ∈ Z. Therefore, the closed subgroup of A generated by a1 , . . . , ar coincides with the Zp -submodule of A generated by a1 , . . . , ar . Lemma 16.6.3: Let 0 → B → A → Z/pZ → 0 be an exact sequence of ∼ Zp . Then either A ∼ Abelian pro-p groups with B = = Zp or A = B ⊕ A0 with A0 ∼ = Z/pZ. Proof: Consider A as a Zp -module (Remark 16.6.2). The exact sequence yields pA ≤ Zp . By Lemma 1.4.2(e), pA is generated by one element. On the other hand, A is generated by two elements, one which generates B and the other with image in Z/pZ generating that module. Assume A ∼ 6 Zp = and A ∼ 6 Zp ⊕ Z/pZ. Then, by Proposition 2.2.3, A ∼ = = Zp ⊕ Z/pk Z with k ≥ 2

16.6 Zp -Extensions of Hilbertian Fields

309

or A ∼ = Zp ⊕ Zp . In each cases pA ∼ = Z/pZ ⊕ Z/pZ, so pA is not generated by one element, contradicting the preceding paragraph. Therefore, A ∼ = Zp or A ∼ Z ⊕ Z/pZ. In the second case A has a subgroup A isomorphic to = p 0 Z/pZ. Since Zp contain no nontrivial closed subgroups of finite order (Lemma 1.4.2(d)), A0 is not contained in B. Consequently, A = B ⊕ A0 . Lemma 16.6.4: Suppose p 6= char(K). Let L = K(ζpn | n = 1, 2, 3, . . .). Then L/K is an Abelian extension and there is a field K∞ satisfying the following conditions: (a) K∞ (ζq ) = L and K∞ ∩ K(ζq ) = L. (b) Gal(L/K) = Gal(L/K(ζq )) × Gal(L/K∞ ). (c) Gal(L/K∞ ) ∼ = Gal(K(ζq )/K) and Gal(K(ζq )/K) is isomorphic to a subgroup of Z/(p − 1)Z (resp. Z/2Z) if p 6= 2 (resp. p = 2). (d) If L 6= K(ζq ), then Gal(K∞ /K) ∼ = Gal(L/K(ζq )) ∼ = Zp . Proof: Let n be a positive integer. Then K(ζpn ) is the splitting field of the n separable polynomial X p − 1 over K. Hence, K(ζpn )/K is Galois. Embed Gal(K(ζpn )/K) into (Z/pn Z)× by mapping each σ ∈ Gal(K(ζpn )/K) onto s(σ)

the element s(σ) of (Z/pn Z)× satisfying ζpσn = ζpn . Thus, K(ζpn )/K is Abelian. Hence, L/K is also Abelian. In particular, Gal(K(ζq )/K) is isomorphic to a subgroup of Z/(p−1)Z if p 6= 2 and of Z/2Z if p = 2. If K(ζq ) = L, then K∞ = K satisfies Conditions (a)-(c). Assume from now on K(ζq ) 6= L. Let m be the maximal positive integer with ζpm ∈ K(ζq ). Then ζpm 6= xp for all x ∈ K(ζq ). If p = 2, then m ≥ 2 and −1 = ζ42 is a square in K(ζq ), so ζpm 6= −4y 4 for all y ∈ K(ζ4 ). Let 1/pn−m

n−m

) and X p n ≥ m. Then K(ζpn ) = K(ζq )(ζpm K(ζq ) [Lang7, p. 297, Thm. 9.1]. So, (1)

− ζpm is irreducible over

[K(ζpn ) : K(ζq )] = pn−m .

Divide the rest of the proof into two parts. Part A: Suppose p 6= 2 and q = p. Then (Z/pn Z)× ∼ = Z/(p − 1)Z × s(σ) n−1 Z/p Z. Let σ ∈ Gal(K(ζpn )/K(ζp )). Then ζp = ζpσ = ζp . Hence, n−1

s(σ) ≡ 1 mod p. Therefore, s(σ)p ≡ 1 mod pn [LeVeque, p. 50, Thm. 45]. Thus, Gal(K(ζpn )/K(ζp )) is isomorphic to a subgroup of Z/pn−1 Z. It follows from (1) that Gal(K(ζpn )/K(ζp )) ∼ = Z/pn−m Z. By the beginning of the preceding paragraph, Gal(K(ζpn )/K) is cyclic of order dpn−m with d = [K(ζp ) : K]. Let τ be a generator of this group. n−m n−m Then ord(τ p ) = d. Denote the fixed field of τ p in K(ζpn ) by Kn . Then Kn ∩ K(ζp ) = K and Kn (ζp ) = K(ζpn ). Moreover, Kn is the unique field with these properties. Hence, Kn ⊆ Kn0 for n0 ≥ m. Now let K∞ = S ∞ n=m Kn . Then K∞ ∩ K(ζp ) = K and K∞ (ζp ) = L. It follows from the preceding paragraph that Gal(L/K∞ ) ∼ = Gal(K(ζp )/K) and Gal(K∞ /K) ∼ = ∼ Gal(L/K(ζp )) = Zp .

310

Chapter 16. Galois Groups over Hilbertian Fields n−3

Part B: Suppose p = 2 and q = 4. Let n ≥ max(3, m). Then 52 ≡1+ 2n−1 mod 2n and ord2n 5 = 2n−2 [LeVeque, p. 54]. In addition, −1+2n Z does not belong to the subgroup of (Z/2n Z)× generated by 5 + 2n Z. Otherwise, 5k ≡ −1 mod 2n for some 1 ≤ k < 2n−2 . Raising both sides to an odd l+1 power, we may assume k = 2l with 1 ≤ l ≤ n − 3. Then 52 ≡ 1 mod 2n . n−3 2 n ≡ −1 mod 2 . It follows that Hence, n − 2 ≤ l + 1, so l = n − 3 and 5 −1 ≡ 1 + 2n−1 mod 2n , which is a contradiction. Consequently, (Z/2n Z)× ∼ = Z/2Z × Z/2n−2 Z with −1 generating the first factor and 5 generating the second factor. s(σ) Consider σ ∈ Gal(K(ζ2n )/K(ζ4 )). Then ζ4 = ζ4σ = ζ4 . Hence, s(σ) ≡ 1 mod 4. By the preceding paragraph, s(σ) ≡ (−1)i 5j mod 2n with 0 ≤ i ≤ 1 and 0 ≤ j ≤ n−2. Then (−1)i ≡ 1 mod 4, so i = 0 and s(σ) ≡ 5j mod 2n . Thus, Gal(K(ζ2n )/K(ζ4 )) is isomorphic to a subgroup of Z/2n−2 Z. By (1), Gal(K(ζ2n /K(ζ4 ))) ∼ = Z/2n−m Z. Taking inverse limit, we get Gal(L/K(ζ4 )) ∼ = Z2 . This gives a short exact sequence of Abelian pro-2 groups 0 → Z2 → Gal(L/K) → Gal(K(ζ4 )/K) → 1 with Gal(K(ζ4 )/K) trivial or isomorphic to Z/2Z. By Lemma 16.6.3, Gal(L/K) = Gal(L/K(ζ4 )) × A0 with A0 trivial or of order 2. Denote the fixed field of A0 in L by K∞ . It satisfies Conditions (a)–(d). The case where p 6= char(K), and 1 < [L : K] < ∞ is the most complicated. We need some concepts and facts from group theory. Let G be a profinite group and x, y, z ∈ G. Define the commutator of x, y by [x, y] = x−1 y −1 xy. It satisfies the following identities: (2)

[x, y]−1 = [y, x], [x, y]z = [xz , y z ], [x, yz] = [x, z][x, y]z , [xy, z] = [x, z]y [y, z]

The commutator subgroup of G is [G, G] = h[x, y] | x, y ∈ Gi. Suppose N is a closed subgroup of G which contains [G, G]. For each n ∈ N and g ∈ G we have ng = n[n, g]. Hence, N / G and G/N is Abelian. Conversely, if N is a closed normal subgroup of G and G/N is Abelian, then [G, G] ≤ N . Consider now a profinite group C and an epimorphism g: G → C. Suppose A = Ker(g) is Abelian. Define an action of C on A in the following way: γ for a ∈ A. For each γ ∈ C choose γ˜ ∈ G with g(˜ γ ) = γ. Then let aγ = γ˜ −1 a˜ Since A is Abelian, this action is independent of γ˜ . P Suppose C is finite. The group ring Z[C] consists of all formal sums γ∈C kγ γ with kγ ∈ Z. Addition is defined componentwise. Multiplication in Z[C] is a linear extension of multiplication in C. Thus, X X X X kγ γ · lδ δ = kγ lδ ε. γ∈C

δ∈C

ε∈C

γδ=ε

The action of C on A naturally extends to an action of Z[C] on A: P Y akγ )γ . a γ∈C kγ γ = γ∈C

16.6 Zp -Extensions of Hilbertian Fields

311

Let γ be an element of C with γ m = 1. Put c = (1 − γ)c = 1 − γ m = 0. Hence, (3)

(1 − γ k )c =

k−1 X

γ i (1 − γ)c = 0,

Pm−1 i=0

γ i . Then

k = 0, . . . , m − 1

i=0

Lemma 16.6.5: Let p be a prime number, C a finite cyclic group, G a profinite group, and g: G → C an epimorphism. Suppose A = Ker(g) is an Abelian pro-p group and f0 : A → Zp is an epimorphism. Let q 0 be a power of p and let π: Zp → Z/q 0 Z an epimorphism. Put α = π ◦ f0 . Suppose one of the following conditions holds: (a) q 0 6= 1 and p - |C|. (b) p = |C| = 2 and q 0 ≥ 4. (c) p = q 0 = |C| = 2 and α extends to an epimorphism β: G → Z/4Z. Then there exists an epimorphism f : A → Zp with Ker(f ) / G such that G/Ker(f ) is Abelian. Proof: The first two parts of the proof are common to all cases. The rest of the proof handles each case separately. Part A: The commutator of G. As A is Abelian, C acts on A by lifting and conjugating. Extend this to an action of Z[C] on A. Since C is Abelian, [G, G] ≤ A. Moreover, [G, G] = ha1−γ | a ∈ A, γ ∈ Ci.

(4)

Indeed, choose a generator γ0 of C and an element γ˜0 in G with g(˜ γ0 ) = γ0 . ˜ = 1 for all γ, δ ∈ C and For each i let γei = γ˜ i . Then [˜ γ , δ] 0

(5)

0

a1−γ = [˜ γ , a]

for each a ∈ A.

Thus, the right hand side of (4) (which we denote by G1 ) is contained in [G, G]. To prove the other inclusion, it suffices to prove [u, v] ∈ G1 for all u, v ∈ G. To this end, write u = a˜ γ and v = bδ˜ with a, b ∈ A and γ, δ ∈ C. Now use (2), (5) and the hypothesis that A is Abelian: ˜ = [a˜ ˜ γ , b]δ [u, v] = [a˜ γ , bδ] γ , δ][a˜ ˜ γ [˜ ˜ b]γδ [˜ ˜ γ , b δ ] ∈ G1 = [a, δ] γ , δ][a, γ , b]δ = [aγ , δ][˜ Pm−1 Part B: Twist of A. Let m = |C| and put c = i=0 γ0i . Let µ: A → A be the homomorphism given by µ(a) = ac . Since γ0m = 1, (3) implies (1−γ)c = 0 for each γ ∈ C. Hence, (a1−γ )c = a(1−γ)c = 1 for each a ∈ A. Therefore, by (4), µ(w) = wc = 1 for each w ∈ [G, G]. γ , a]) = 0. Hence, α(aγ ) = α(a) for all a ∈ A and By (5), α(a1−γ ) = α([˜ Pm−1 i γ ∈ C. Consequently, α(µ(a)) = i=0 α(aγ0 ) = mα(a).

312

Chapter 16. Galois Groups over Hilbertian Fields

Part C: Suppose q 0 > 1 and p - |C|. Then m is an invertible element of Zp and Z/q 0 Z. Let m−1 : Zp → Zp and m−1 : Z/q 0 Z → Z/q 0 Z be multiplications by m−1 . Define a homomorphism f : A → Zp by f = m−1 ◦ f0 ◦ µ. By Part B, [G, G] ≤ Ker(µ) ≤ Ker(f ). Thus, Ker(f ) / G and G/Ker(f ) is Abelian. By Part B, this establishes the following commutative diagram: f µ

A α

Z/q 0 Z

f0 /A / Zp DD DD DDα DD π D! m / Z/q 0 Z

'/

m−1

Zp π

−1

m

/ Z/q 0 Z

In particular, α = π ◦ f . Thus, π(f (A)) = α(A) = Z/q 0 Z. It follows from Lemma 1.4.3 that f is surjective. Part D: Suppose p = |C| = 2 and q 0 ≥ 4. In this case Part B gives the following commutative diagram: A α

Z/q 0 Z

µ

f0 /A / Z2 DD DD α DD DD π " 2 / Z/q 0 Z

Let H = f0 (µ(A)). By Remark 1.2.1(e), H is a closed subgroup of Z2 . It satisfies, π(H) = 2(Z/q 0 Z) = π(2Z2 ). Hence, H + q 0 Z2 = 2Z2 + q 0 Z2 = 2Z2 . By Lemma 1.4.2, H is trivial or H = 2n Z2 for some positive integer n. It follows from q 0 ≥ 4 that H = 2Z2 . Part E: Suppose p = q 0 = 2 = |C| and α extends to an epimorphism β: G → Z/4Z. Then (G : A) = |C| = 2. Hence, in the notation of Part B, a = γ˜02 ∈ A. So, aγ0 = a. Hence, µ(a) = aaγ0 = a2 . The existence of β gives a commutative diagram 1

0

/A

/G

α

β

/ Z/2Z

/ Z/4Z

g

/C

/1

β¯

π ¯

/ Z/2Z

/1

¯ γ0 )) = 1 + 2Z. Hence, with both rows exact. In particular, π ¯ (β(˜ γ0 )) = β(g(˜ 2 γ0 ) = 1 + 2Z. Therefore, by β(˜ γ0 ) = ±1 + 4Z. So, π(f0 (a)) = α(a) = β(˜ Lemma 1.4.3, hf0 (a)i = Z2 . Hence, hf0 (µ(a))i = hf0 (a2 )i = 2Z2 . It follows that f0 (µ(A)) = Z2 or f0 (µ(A)) = 2Z2 .

16.6 Zp -Extensions of Hilbertian Fields

313

Suppose first f0 (µ(A)) = Z2 . Then f = f0 ◦ µ maps A onto Z2 . For each w ∈ [G, G], Part B implies f (w) = f0 (µ(w)) = 1. Therefore, [G, G] ≤ Ker(f ), Ker(f ) / G, and G/Ker(f ) is Abelian. The case f0 (µ(A)) = 2Z2 (which we henceforth assume), will be handled in Part F. Part F: Conclusion of the proof in cases (b) and (c). Multiplication by 2 gives an isomorphism of Z2 onto 2Z2 . Hence, by Parts D and E, there is an epimorphism f : A → Z2 with 2f = f0 ◦ µ. For each w ∈ [G, G], Part B implies 2f (w) = f0 (µ(w)) = 0. Hence, f (w) = 0. Therefore, [G, G] ≤ Ker(f ), Ker(f ) / G, and G/Ker(f ) is Abelian. Theorem 16.6.6 ([Whaples, Thm. 2]): Let K be a field and p a prime number. Put q = p if p 6= 2 and q = 4 if p = 2. Suppose Z/qZ occurs over K. Then Zp occurs over K. Proof: Let K 0 be a Z/qZ-extension of K. Suppose first p = char(K). Lemma 16.3.2 embeds K 0 in a Zp -extension of K. Assume from now on, p 6= char(K). Let L = K(ζpn | n = 1, 2, 3, . . . ). When L = K, Lemma 16.6.1 gives a Zp -extension of K. When L 6= K(ζq ), Lemma 16.6.4 gives a Zp -extension of K. Assume from now on, L = K(ζq ) and L 6= K. Let C = Gal(L/K). Then C is a cyclic group of order which divides p − 1 if p 6= 2 and of order 2 of p = 2. Put L0 = LK 0 . Then L0 is a finite Abelian extension of K. Moreover, 0 L /L is a cyclic extension of degree q 0 which is a power of p. We distinguish between three cases: Case A: p 6= 2. Then K 0 /K is a cyclic extension of degree p and [L : K]|p − 1. Hence, q 0 = [L0 : L] = p and p - |C|. Case B: p = 2 and L ∩ K 0 = K. 4, |C| = 2, and q 0 = [L0 : L] = 4.

Then K 0 /K is a cyclic extension of degree

Case C: p = 2 and L ∩ K 0 6⊆ K. Then, L ⊆ K 0 , K 0 = L0 , and q 0 = |C| = 2. In each case Lemma 16.6.1 gives a Zp -extension F of L which contains L0 . Denote the compositum of all finite Abelian extensions of L of a ppower order by N . In particular, F ⊆ N . Since L/K is Galois, so is N/K. Put G = Gal(N/K) and A = Gal(N/L). Let f0 be resN/F : Gal(N/L) → Gal(F/L) and let π be resF/L0 : Gal(F/L) → Gal(L0 /L). Then α = π ◦ f0 = resN/L0 : Gal(N/L) → Gal(L0 /L). In Case C, Gal(K 0 /K) ∼ = Z/4Z and β = resN/L0 : Gal(N/K) → Gal(L0 /K) extends α. In each case, Lemma 16.6.5 gives an extension F 0 of L in N with F 0 /K Abelian and Gal(F 0 /L) ∼ = Zp . Suppose first p 6= 2. Then [L : K]|p − 1. So, Gal(F 0 /K) has a unique subgroup of order p − 1. This is a special case of a profinite version of the Schur-Zassenhaus theorem (Lemma 22.10.1). Its fixed field E in F 0 satisfies Gal(E/K) ∼ = Zp . Now suppose p = 2. Lemma

314

Chapter 16. Galois Groups over Hilbertian Fields

16.6.3 gives a Z2 -extension E of K in F 0 . Consequently, in each case K has a Zp -extension. Corollary 16.6.7: Let K be a Hilbertian field and p a prime number. Then Zp -occurs over K. Proof: By Corollary 16.3.6, Z/pZ and Z/p2 Z occur over K. Hence, by Theorem 16.6.6, Zp occurs over K. Remark 16.6.8: The assumption in Theorem 16.6.6 that Z/4Z rather than Z/2Z occurs over K is necessary for the theorem to hold. Indeed, Gal(C/R) ∼ = Z/2Z but Z2 does not occur over R. Proposition 16.6.10 below shows it is impossible to conclude that Zp is regular over K in Corollary 16.6.7. Lemma 16.6.9: Let E be a field, p a prime number, v a discrete valuation of ¯v ) and E ¯v (ζp , ζp2 , ζp3 , . . .) E, and F a Zp -extension of E. Suppose p - char(E ¯ is an infinite extension of Ev . Then v is unramified in F . Proof: Let w be a valuation of F extending v. Denote reduction at w by a bar. Assume w is ramified over E. Then its inertia group Iw/v is nontrivial. By Lemma 1.4.2, Iw/v is an open subgroup of Gal(F/E). Replace E by the fixed field of Iw/v in F , if necessary, to assume Iw/v = Gal(F/E). Let n be a positive integer. Denote the unique extension of E in F of degree pn by En . Let vn be a normalized valuation of En over v. It is totally ˆ of E under v. Then E ˆn = ramified over E. Consider the completion E ˆn /E) ˆ ∼ ˆ is the completion of En under vn (Proposition 3.5.3) and Gal(E En E = n ∼ Gal(En /E) = Z/p Z [Cassels-Fr¨ ohlich, p. 41, Prop. 3]. ˆn and x ∈ E ˆ with vn (y) = 1 and v(x) = 1. Then vn (x) = Choose y ∈ E n n ¯n = E, ¯ pn = vn (y p ). Hence, y p = ux with u ∈ En and vn (u) = 0. Since E 0 −1 0 there is an a ∈ E with a ¯=u ¯. Let u = ua and x1 = ax. Then vn (u − 1) > ¯v ), Hensel’s lemma (Proposition 3.5.2(a)) 0, and x1 ∈ E. Since p - char(E n pn gives u1 ∈ En with up1 = u0 . Put y1 = u−1 1 y. Then y1 = x1 . Since En /E is ¯v . cyclic of degree n, this implies ζpn ∈ En . Taking residues, we find ζpn ∈ E ¯ ¯ Thus, Ev (ζp , ζp2 , ζp3 , . . .) = Ev , contrary to the assumption of the lemma. Consequently, v is unramified in F . Proposition 16.6.10: Let K be a field and p a prime number. Suppose p 6= char(K) and K(ζp , ζp2 , ζp3 , . . .) is an infinite extension of K. Then Zp is not regular over K. Proof: Assume Zp is regular over K. Then there are algebraically independent elements t1 , . . . , tr over K and there is a Zp -extension F of E = K(t1 , . . . , tr ). Induction on r proves it suffices to consider the case where r = 1. Put t = t1 . Let E1 be the unique extension of E in F of degree p. Then E1 /K is regular. Remark 3.6.2(b) gives a prime divisor p of E/K which is ramified in E1 . The valuation vp of E associated with p is discrete (Section 3.1). Its

16.7 Symmetric and Alternating Groups over Hilbertian Fields

315

¯p is a finite extension of K. Hence, E ¯p (ζp , ζp2 , ζp3 , . . .) is an residue field E ¯v ) = char(K) - p. Hence, by infinite extension of K. In addition, char(E Lemma 16.6.9, p is unramified in F . In particular, p is unramified in E1 . It follows from this contradiction that Zp is not regular over K. Corollary 16.6.11: Let K be a field and p a prime number. Suppose p 6= char(K) and K is a finitely generated over its prime field. Then Zp is not regular over K.

16.7 Symmetric and Alternating Groups over Hilbertian Fields The Galois group of the general polynomial of degree n is Sn (Example 16.2.5(a)). There is a standard strategy to construct special polynomials with Galois groups Sn . Lemma 16.7.1: Let K be a field, t1 , . . . , tr algebraically independent elements over K, and F a separable extension of K(t) of degree n. Denote ∼ ˜ K(t)) ˜ the Galois closure of F/K(t) by Fˆ . Suppose Gal(Fˆ K/ = Sn . Then, ∼ ˆ Gal(F /K(t)) = Sn . Proof: Since [F : K(t)] = n, the group G = Gal(Fˆ /K(t)) is isomorphic to a ˜ : K(t)] ˜ subgroup of Sn , On the other hand, |Sn | = [Fˆ K ≤ [Fˆ : K(t)] = |G|. ∼ Consequently, G = Sn . Consider for example the case when r = 1. Put t = t1 . Let x be a primitive element of F/K(t), and f ∈ K[T, X] an absolutely irreducible polynomial which separable and of degree n in X with f (t, x) = 0. Then ˜ = Gal(f (t, X), K(t)) ˜ G acts transitively on the n distinct roots of f (t, X) in ] ˜ K(t) ˜ ˜ one K(t). Inspecting inertia groups of prime divisors of F K/ in Fˆ K, ˜ may prove that G contains cycles which generate Sn . Then, by Lemma 16.7.1, Gal(Fˆ /K(t)) ∼ = Sn . When the Galois group is Sn , one may consider the fixed field E of An in Fˆ . It is a quadratic extension of K(t). If we prove that E is rational over K, then An becomes regular over K. We are able to do it, for example, if char(K) 6= 2 and K(t)/K has at most three prime divisors which ramify in F , each of degree 1 (Lemma 16.7.5). We start with a valuation theoretic condition that yields e-cycles for the Galois group of a polynomial. By an e-cycle of Sn we mean a cycle of length e. Lemma 16.7.2: Let (E, v) be a discrete valued field, f ∈ Ov [X] a monic separable irreducible polynomial over E, and x a root of f in Es . Denote ¯ is separably closed, f¯(X) = (X −¯ reduction at v by a bar. Suppose E a)e η(X) ¯ ¯ with a ∈ Ov , e ≥ 1, char(E) - e, and η ∈ E[X] monic separable polynomial with η(¯ a) 6= 0. Suppose v extends to a valuation w of E(x) with w(x − a) > 0 and ew/v = e. Then Gal(f, E) contains an e-cycle.

316

Chapter 16. Galois Groups over Hilbertian Fields

ˆ vˆ). Embed Es into E ˆs and Proof: Denote the completion of (E, v) by (E, ˆ be the valuation ring of E. ˆ By ˆs . Let O extend vˆ to a valuation vˆ of E assumption, η(X) is relatively prime to (X − a ¯)e . Hence, Hensel’s lemma (Proposition 3.5.2(b)) gives a factorization f (X) = g(X)h(X) with g, h ∈ ¯ ˆ = η(X). O[X] monic, g¯(X) = (X − a ¯)e and h(X) Let g(X) = g1 (X) · · · gk (X) with g1 , . . . , gk distinct monic irreducible ˆ polynomials in E[X]. For each i choose a root yi of gi in Es . Then there is a σi ∈ Gal(E) with σi x = yi . The formula wi (z) = vˆ(σi z) for z ∈ E(x) ˆ i ) as completion defines a valuation wi of E(x) which extends v and with E(y (Proposition 3.5.3(a),(b)). Similarly, for each root y of h(X) there is a σ ∈ Gal(E) with σx = y and v 0 (z) = vˆ(σz) for z ∈ E(x) defines an extension of v. By assumption, ¯ y ) = 0 and h(¯ ¯ a) 6= 0. Hence, v 0 (x − a) = vˆ(y − a) = 0. Since w(x − a) > 0, h(¯ this implies, w 6= v 0 . Thus, w is one of the wi ’s, say w = w1 (Proposition 3.5.3(c)). Therefore, e = ew/v ≤

k X i=1

ewi /v ≤

k X i=1

ˆ i ) : E] ˆ = [E(y

k X

deg(gi ) = deg(g) = e.

i=1

ˆ List the roots of g in Es as It follows that k = 1 and g is irreducible over E. x1 , . . . , xe . ˆ ˆ = deg(g) = e. Since char(E) ¯ - e, By the preceding paragraph, [E(x) : E] 1/e ˆ ˆ ˆ ˆ ˆ E(x)/E is tamely ramified. Hence, E(x) = E(π ) for some π ∈ E [Lang5, ˆ ¯ is separably closed, it contains ζe . By Hensel’s lemma, ζe ∈ E. p. 52]. Since E ˆ ˆ ˆ Hence, E(x)/E is a cyclic group of order e. In other words, Gal(g, E) is a cyclic group of order e. It acts transitively on x1 , . . . , xe . Therefore, it is generated by an e-cycle, say (x1 x2 . . . xe ). ¯ Finally, by assumption, h(X) is separable. Hence, it is a product of ¯ distinct monic linear factors in E[X]. By Hensel’s lemma, h(X) Qn decomposes ˆ into distinct monic linear factors in E[X], say, h(X) = i=e+1 (X − xi ). ˆ is ˆ ˆ acts trivially on xe+1 , . . . , xn . Thus, Gal(f, E) Therefore, Gal(E(x)/ E) ˆ ≤ Gal(f, E), the group generated by the cycle (x1 x2 . . . xe ). Since Gal(f, E) Gal(f, E) contains an e-cycle Lemma 16.7.3: Let K be a field, q ∈ K[X] with deg(q) = n, and x a transcendental element over K. Suppose char(K) - n. Put t = q(x) and f (T, X) = q(X) − T . Then: ˜ (a) f (t, X) is irreducible over K(t), separable in X, and [K(x) : K(t)] = n. (b) Denote the derivative of q by q 0 . Let ϕ be a K-place of K(t). Denote the corresponding prime divisor of K(t)/K by p. Suppose ϕ(t) 6= ∞ and ϕ(t) 6= q(b) for each root b of q 0 (X). Then p is unramified in K(x). ˜ (c) Gal(f (t, X), K(t)) contains an e-cycle. ˜ such that q(X) − q(b) = (X − b)e η(X) with (d) Suppose there is a b ∈ K ˜ ˜ e ≥ 2, char(K) - e, η(X) ∈ K[X], and η(b) 6= 0. Then Gal(f (t, X), K) contains an e-cycle.

16.7 Symmetric and Alternating Groups over Hilbertian Fields

317

˜ λ ∈ K[X], ˜ (e) Suppose q 0 (X) = (X − b)λ(X) with b ∈ K, and λ(b) 6= 0. ˜ Then Gal(f (t, X), K(t)) contains a 2-cycle. Proof: Let c be the coefficient of X n in q. Replace x by cx and t by cn−1 t to assume q is monic. Proof of (a): Since f (T, X) is linear in T , it is absolutely irreducible. Hence, ˜ f (t, X) is irreducible over K(t) and over K(t). Therefore, [K(x) : K(t)] = n. The separability of f (t, X) follows from the assumption char(K) - n. Proof of (b): Extend ϕ to a place ψ of K(x). Then ϕ(t) = ψ(q(x)) = q(ψ(x)). Hence, by assumption, ψ(q 0 (x)) = q 0 (ψ(x)) 6= 0. Therefore, the value of ϕ at the product of all K(t)-conjugates of q 0 (x) is not 0. In other words, ϕ(NK(x)/K(t) q 0 (x)) 6= 0. By (a), f (t, X) = irr(x, K(t)). In addition, ∂f (t, x). It follows from Lemma 6.1.8 that p is unramified in K(x). q 0 (x) = ∂X Proof of (c): Let q(X) = X n + cn−1 X n−1 + · · · + c0 . Then xn + cn−1 xn−1 + ˜ ˜ with v∞ (t) = −1 and let K · · · + c0 = t. Let v∞ be the valuation of K(t)/ ˜ w be an extension of v∞ to K(x). Put e = ew/v∞ . By Example 2.3.11, w(x) = −1 and e = n. Now make a change of variables: u = t−1 and y = t−1 x. Put g(U, Y ) = ˜ ˜ ˜ ˜ = K(t), K(y) = K(x), Y n + cn−1 U Y n−1 + · · · + c0 U n − U n−1 . Then K(u) ˜ w(u) = n, w(y) = n − 1, g(u, Y ) is irreducible over K(u), separable in ˜ Y , g(u, y) = 0, and g(0, Y ) = Y n . By Lemma 16.7.2, Gal(g(u, Y ), K(u)) ˜ contains an n-cycle. Therefore, Gal(f (t, X), K(t)) which is isomorphic to ˜ Gal(g(u, X), K(u)) has an n-cycle. ˜ ˜ with va (t − Proof of (d): Put q(b) = a. Let va be the valuation of K(t)/ K e ˜ a) = 1. By assumption, f (a, X) = (X − b) η(X) with η ∈ K[X], η(b) 6= 0. ˆ (It is K((t ˜ ˆ be the ˜ − a)).) Let O Denote the completion of (K(t), va ) by E ˆ valuation ring of E. By Hensel’s lemma, f (t, X) = g(X)h(X) with g, h monic ˆ polynomials in O[X] whose residue at va are (X − b)e , η(X), respectively. ˆ ¯ = b. Let w be the corresponding extension Embed K(x) in Es such that x ˆ ˆ ≤ deg(g) = e. : E] of va to K(x). Then ew/va ≤ [E(x) By the preceding paragraph, (x − b)e η(x) = q(x) − a = t − a. Since η(b) 6= 0, we have w(η(x)) = 0. Hence, ew(x − b) = ew/va ≤ e. Thus, w(x − b) = 1 and ew/va = e. ˜ We conclude from Lemma 16.7.2 that Gal(f (t, X), K(t)) has an e-cycle. e ˜ Proof of (e): Write q(X) − q(b) = (X − b) η(X) with e ≥ 1, η ∈ K[X], and η(b) 6= 0. Then (1) q 0 (X) = (X − b)e−1 eη(X) + (X − b)η 0 (X) If e = char(K), then q 0 (X) = (X − b)e η 0 (X). Hence, e = 1, which is impossible. Therefore, e 6= char(K), so the second factor on the right hand side of (1) does not vanish in b. On the other hand, q 0 (X) = (X − b)λ(X) ˜ with λ(b) 6= 0. Hence, e = 2. Consequently, by (d), Gal(f (t, X), K(t)) has a 2-cycle.

318

Chapter 16. Galois Groups over Hilbertian Fields

Lemma 16.7.4: Let K be a field, n ≥ 2 an integer, x an indeterminate, and t = xn −xn−1 . Suppose char(K) - (n−1)n. Then X n −X n−1 −t is separable, ˜ irreducible over both K(t) and K(t), and ˜ = Sn . Gal(X n − X n−1 − t, K(t)) = Gal(X n − X n−1 − t, K(t)) Moreover, K(t) has at most three prime divisors which ramify in K(x). Each of them has degree 1. Proof: The polynomial q(X) = X n − X n−1 factors as q(X) = X n−1 (X − 1). ˜ By Lemma 16.7.3(d) (with b = 0), Gal(X n − X n−1 − t, K(t)) contains an 0 n−2 (X − n−1 (n − 1)-cycle. The derivative of q(X) is q (X) = nX n ). By n−1 n n−1 ˜ Lemma 16.7.3(e) (with b = n ), Gal(X − X − t, K(t)) contains a 2˜ cycle. Hence, by [Waerden1, p. 199], Gal(X n − X n−1 − t, K(t)) = Sn . It ˜ = Sn . follows from Lemma 16.7.1 that Gal(X n − X n−1 − t, K(t)) n−1 The derivative q 0 (X) has exactly two roots, 0 and n−1 =a n . Put q n and observe that q(0) = 0. Let ϕ be a K-place of K(t) and p the prime divisor of K(t)/K corresponding to ϕ. Suppose first that ϕ(t) 6= a, 0, ∞. By Lemma 16.7.3(b), p is unramified in K(x). Thus, p ramifies in K(x) in at most three cases, when ϕ(t) is a, or 0, or ∞. In each of these cases deg(p) = 1. Lemma 16.7.5: Let K be a field, t an indeterminate, and E a quadratic extension of K(t). Suppose char(K) 6= 2, E/K is regular, and K(t)/K has at most three prime divisors which ramify in E, each of degree 1. Then E = K(u) with u transcendental over K. Proof: Denote the genus of E by g. Let p1 , . . . , pk be the number of prime divisors of K(t)/K which ramify in E. By assumption, k ≤ 3 and deg(pi ) = 1, i = 1, . . . , k. Since [E : K(t)] = 2, each pi extends uniquely to a prime divisor qi of E/K of degree 1. Since char(K) 6= 2, the ramification of pi in E is tame. By Riemann-Hurwitz (Remark 3.6.2(c)), 2g−2 = −4+k ≤ −1. Hence, g ≤ 12 . This implies, g = 0 and k = 2. It follows from Example 3.2.4 that E = K(u) with u transcendental over K. Proposition 16.7.6: Let K be a field and an integer n ≥ 2 with char(K) (n − 1)n. Then An is regular over K. Specifically, there is a tower of fields K(t) ⊆ K(u) ⊆ F satisfying: F/K is regular, F/K(t) is Galois, Gal(F/K(t)) ∼ = Sn , Gal(F/K(u)) ∼ = An , and K(t)/K has at most three prime divisors which ramify in F , each of degree 1. Proof: Let t and x be transcendental elements over K with xn − xn−1 = t. By Proposition 16.7.4, K(x)/K(t) is separable. Denote the Galois closure of ∼ ˜ K(t)) ˜ K(x)/K(t) by F . By Proposition 16.7.4, Gal(F/K(t)) ∼ = Gal(F K/ = Sn . In particular, F/K is regular. In addition, K(t)/K has at most three prime divisors p1 , p2 , p3 which may ramify in K(x), each of degree 1. Denote the fixed field of An in F by E. By Corollary 2.3.7(c), p1 , p2 , p3 are the

16.7 Symmetric and Alternating Groups over Hilbertian Fields

319

only prime divisors of K(t)/K which may ramify in F . Hence, they are the only prime divisors of K(t)/K which may ramify in E. By Lemma 16.7.5, E = K(u) with u transcendental over K and Gal(F/K(u)) ∼ = An , as desired. Next we use the polynomial X n − X n−1 − t to solve embedding problems √ of the form Sn → Gal(K( a)/K) when K is Hilbertian. Remark 16.7.7: On the discriminant of a polynomial. Let K be aQfield and n f ∈ K[X] a monic separable polynomial of degree n. Write f (X) = i=1 (X− xi ) with xi ∈ Ks . Put N = K(x1 , . . . , xn ) and embed Gal(N/K) into Sn by σ(i) = j if σ(xi ) = xj . Formula (1) of Section 6.1 for the discriminant disc(f ) of f is (2)

disc(f ) = (−1)

n(n−1) 2

n Y n(n−1) Y 2 (xi − xj ) = (−1) f 0 (xj ), j=1

i6=j

with f 0 the derivative of f . It can be rewritten as disc(f ) = Thus, Y p disc(f ) = (xi − xj ).

Q

i<j (xi

− xj )2 .

i<j

For each σ ∈ Gal(N/K) we have Y p p (xσ(i) − xσ(j) ) = (−1)sgn(σ) disc(f ). σ( disc(f )) = i<j

where sgn(σ) is the number of pairs (i, j) with i < j and σ(i) > σ(j). If char(K) 6= 2, then (−1)sgn(σ) 6= 1 for odd σ and (−1)sgn(σ) = 1 for even σ. In this case, p Gal N/K( disc(f ) ) = An ∩ Gal(N/K). p In particular, if Gal(N/K) = Sn , then K( disc(f )) is the unique quadratic p extension of K in N and Gal(N/K( disc(f ))) = An . Qn−1 Suppose now char(K) - n. Then f 0 (X) = n j=1 (X − yj ) with yj ∈ ˜ Substituting in (2) and changing the order of multiplication gives an K. alternative formula for the discriminant: (3)

disc(f ) = (−1)

n(n−1) 2

nn

n−1 Y

f (yj ).

j=1

The proof of part (a) of the following result gives an explicit polynomial over K(t) with Galois group An if char(K) - (n − 1)n.

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Chapter 16. Galois Groups over Hilbertian Fields

Proposition 16.7.8 (Brink): Let K be a Hilbertian field, a a nonsquare in K, and n ≥ 2 an integer with char(K) - (n − 1)n. Then (a) An is regular over K and √ (b) the embedding problem Sn → Gal(K( a)/K) is solvable. Proof: Let f (t, X) = X n − X n−1 − t and F be the splitting field of f (t, X) over K(t). By Proposition 16.7.4, Gal(F/K(t)) ∼ = Sn . Since char(K) - n, we n−1 0 n−2 have f (t, X) = n X − n X . Hence, by (3), n − 1 f (0)n−2 n i h n − 1 n n − 1 n−1 n(n−1) = (−1) 2 nn (−t)n−2 − −t n n

disc(f ) = (−1)

= (−1)

n(n−1) 2

nn f

(n−1)(n+2) 2

[(n − 1)n−1 tn−2 + nn tn−1 ].

p By Remark 16.7.7, K( disc(f )) is a quadratic extension of K(t) in F and p Gal F/K( disc(f ) ) ∼ = An . Let ( u=

disc(f )t1−n = (−1)

(n−1)(n+2) 2

2−n

(n−1)(n+2) 2

disc(f )t

= (−1)

[(n − 1)n−1 t−1 + nn ] n−1

[(n − 1)

n

+ n t]

if n is odd if n is even

√ and u0 = u. Then K(u) = K(t), so both u and u0 are transcendental over K. pMoreover, u is a product of disc(f ) with a square of K(t), so K(u0 ) = K( disc(f ) ). Thus, Gal(F/K(u0 )) ∼ = An . This proves (a). Next, express t in terms of u and substitute in f (t, X) to get the irreducible polynomial for the roots of f (t, X) in terms of u:

g(u, X) =

X n − X n−1 −

X n − X n−1 −

(n−1)n−1 (−1)(n−1)(n−2)/2 u−nn

if n is odd

(−1)(n−1)(n−2)/2 u−(n−1)n−1 nn

if n is even.

Then Gal(g(u, X), K(u)) = Sn and Gal(g((u0 )2 , X), K(u0 )) = An . √ Let v = ua and v 0 = u0 a. The map u 7→ v extends to an automorphism of K(u) over K. This automorphism further extends to an isomorphism of K(u0 ) onto K(v 0 ) and further to an isomorphism of F onto a field F 0 . Thus, Gal(F 0 /K(u)) ∼ = Sn and Gal(F 0 /K(v 0 )) ∼ = An . Since a is not a square in K, there is no c ∈ K with v = c2 u. Hence, K(u0 ) and K(v 0 ) are distinct quadratic extensions of K(u). Therefore, K(u0 ) 6⊆ F 0 . Put F 00 = K(u0 )F 0 . Then 00 0 0 /K(u0 )) ∼ Gal(F√ = Sn and Gal(F 00 /K(u0 , v 0 )) ∼ = An . In addition, K(u √, v ) = 0 K(u , a). By Lemma 16.4.2, the embedding problem Sn → Gal(K( a)/K) is solvable.

16.8 GAR-Realizations

321

16.8 GAR-Realizations Let K be a field, L a finite Galois extension of K, and (1)

α

1 −→ C −→ G −→ Gal(L/K) −→ 1

an embedding problem over K with kernel C having a trivial center. We give a sufficient condition for the embedding problem to have a regular solution. As usual, Aut(C) denotes the group of all automorphisms of C. For each c ∈ C let ι(c) be the inner automorphism of C induced by conjugation with c. The map c 7→ ι(c) identifies C with the group Inn(C) of all inner automorphisms of C. Let F/K be an extension of fields. We say F is rational over K (or F is K-rational) if F = K(T ) with T being a set of algebraically independent elements over K. Definition 16.8.1: GAR-Realizations. Let C be a finite group with a trivial center. We say C is GA over K if there are algebraically independent elements t1 , . . . , tr over K satisfying: (2) K(t) has a finite extension F which is regular over K such that Aut(C) acts on F with K(t) being the fixed field of C. Denote the fixed field of Aut(C) in F by E. We say C is GAR over K if in addition to (2) this holds: (3) Every extension E 0 of E satisfying E 0 Ks = Ks (t) is a purely transcen dental extension of E 0 ∩ Ks . Remark 16.8.2: (a) The “G” in GAR abbreviates “Galois”, “A” abbreviates “Automorphisms”, and “R” abbreviates “Rational”. (b) The GAR-condition implies that the fixed field E of Aut(C) is regular over K. But, it does not require E to be rational over K. This is however the case when r = 1 (L¨ uroth’s theorem, Remark 3.6.2(a)). Lemma Qr 16.8.3: Let C1 , . . . , Cr be finite non-Abelian simple groups. Put C = i=1 Ci . Then: (a) Each normal simple subgroup N of C coincides with Ci for some i between 1 and r. ∼ (b) Suppose Qr αi : C1 → Ci is an isomorphism, i = 1, . . . , r. Then Aut(C) = Sr n i=1 Aut(Ci ). Proof of (a): Choose n ∈ N , n 6= 1. Write n = c1 · · · cr with ci ∈ Ci . Assume, without loss, c1 6= 1. Since C1 is simple, its center is trivial. Hence, there is a c01 ∈ C1 with c1 c01 6= c01 c1 . For each i ≥ 2 we have c01 ci = ci c01 . Therefore, nc01 6= c01 n and (c01 )−1 n−1 c01 n1 6= 1. It follows that, N ∩ C1 6= 1. Consequently, N = C1 . Proof of (b): To simplify notation identify Ci with C1 via αi . Thus, each element of C is an r-tuple c = (c1 , . . . , cr ) with ci ∈ C1 . Each Ci is then the

322

Chapter 16. Galois Groups over Hilbertian Fields

group of all c with cj = 1 for j 6= i. We embed Aut(C1 )r and Sr in Aut(C) by the rules: cγ = (cγ11 , . . . , cγr r )

and

(c1 , . . . , cr )σ = (c1σ−1 , . . . , crσ−1 )

This gives: (4)

σ −1 γσ = (γ1σ−1 , . . . , γrσ−1 ).

Thus, Sr normalizes Aut(C1 )r . In addition observe that Aut(C1 )r ∩ Sr = 1. Finally, consider γ ∈ Aut(C). By (a), there is a unique σ ∈ Sr with Ciγ = Ciσ , i = 1, . . . , r. So, there are γ1 , . . . , γr in Aut(C1 ) with γ = (γ1 , . . . , γr )σ −1 . Thus, Aut(C) is the semidirect product of Aut(C1 )r with Sr , where the action of Sr on Aut(C1 )r given by (4). Remark 16.8.4: Minimal normal subgroups. Let G be a finite group and C a normal subgroup. Suppose C is a minimal normal subgroup of G. Thus, C 6= 1, C / G, and G has no normal subgroup in C other than 1 and C. Let C1 be a simple normal subgroup of C. The conjugates C1 , C2 , . . . , Cr of C1 in G generate a nontrivial normal subgroup of G which is contained in C. Hence, they generate C. Suppose in addition, C1 is non-Abelian. Let k be Qkan integer between 0 and r − 1. Suppose by induction hC1 , . . . , Ck i = i=1 Ci . By Lemma Qk Qk+1 16.8.3(a), Qr Ck+1 6≤ i=1 Ci . Hence, hC1 , . . . , Ck+1 i = i=1 Ci . In particular, C = i=1 Ci . By construction, G acts transitively on the set {C1 , . . . , Cr } by conjugation. Lemma 16.8.5 (Semilinear Rationality Criterion): Let K, L, E, F be fields with L/K Galois, K ⊆ E, E ∩ L = K, and EL = F . Let V be a L-subspace of F . Consider the K-subspace U = {v ∈ V | v σ = v for all σ ∈ Gal(F/E)} of E. (a) Suppose V is Gal(F/E)-invariant. Then each K-basis of U is an L-basis of V . Thus, V ∼ = U ⊗K L. (b) Suppose, in addition to the assumption of (a), F = L(V ). Then E = K(U ). (c) Suppose, in addition to the assumptions of (a) and (b), dim(V ) is finite and equal trans.deg(F/L). Then F is L-rational and E is K-rational. Proof of (a): By assumption, E is linearly disjoint from L over K. Thus, each K-basis of U is linearly independent over L. Hence, it suffices to prove that U spans V over L. To this end consider v ∈ V . Choose a finite Galois extension F0 of E in F containing v. Let L0 = F0 ∩ L. By assumption, F0 L = F and res: Gal(F/E) → Gal(L/K) is an isomorphism. Hence, res(Gal(F/F0 )) = Gal(L/L0 ) and res: Gal(F0 /E) → Gal(L0 /K) is an isomorphism. Put m = [F0 : E] = [L0 : K] and G = Gal(F0 /E).

16.8 GAR-Realizations

323

P Choose a basis c1 , . . . , cm for L0 /K. For each i let ui = σ∈G cσi v σ . σ Then ui ∈ U . Since det(ci ) 6= 0 [Lang7, p. 286, Cor. 5.4] and |G| = m, each v σ is a linear combination of the ui with coefficients which are rational functions in cτi , τ ∈ G, i = 1, . . . , m. In particular, v is in the L-vector space spanned by U , as desired. Proof of (b): By the tower property (Lemma 2.5.3), E is linearly disjoint from L(U ) over K(U ). By (a), F = L(V ) = L(U ). Therefore, E = K(U ). Proof of (c): Let u1 , . . . , un be a K-basis of U . By (a), u1 , . . . , un is an L-basis of V . By (b), F = L(u1 , . . . , un ). Since n = trans.deg(F/L), the elements u1 , . . . , un are algebraically independent over L. Therefore, F is L-rational. By (b), E = K(u1 , . . . , un ). Since F/E and L/K are algebraic, trans.deg(E/K) = trans.deg(F/L) = n. Consequently, E is K-rational.

Proposition 16.8.6 (Matzat): Consider an embedding problem (1), where Qr C a minimal normal subgroup of G and C = i=1 Ci with Ci simple nonAbelian and conjugate to C1 in G, i = 1, . . . , r. Suppose C1 is GAR over K. Then embedding problem (1) is regularly solvable over K. Proof: We break the proof into several parts. Part A: Group theory. Gal(L/K) with (5)

The map g 7→ (ι(g), α(g)) embeds G into Aut(C)×

G ∩ Aut(C) = C,

α = pr2 ,

For each i let Di = NG (Ci ). Put D = (5), (6)

pr2 (G) = Gal(L/K).

Tr

i=1

Di . Then C ≤ D ≤ Di ≤ G. By

Di ∩ Aut(Ci ) = Ci .

For each σ ∈ G, Remark 16.8.4 gives a unique π ∈ Sr with Ciσ = Ciπ , i = 1, . . . , r. Thus, Diσ = Diπ , i = 1, . . . , r. Therefore, Qr D is normal in G. Finally, as in Lemma 16.8.3(b), we embed i=1 Aut(Ci ) in Aut(C) by the rule: (c1 , . . . , cr )(γ1 ,...,γr ) = (cγ11 , . . . , cγr r ). Part B: GA-realization of C. The assumption on C1 gives algebraically independent elements t1 , . . . , tm over K and fields E1 , F1 satisfying this: (7a) K ⊆ E1 ⊆ K(t1 , . . . , tm ) ⊆ F1 , F1 /K regular, F1 /E1 Galois, (7b) Gal(F1 /E1 ) = Aut(C1 ), and Gal(F1 /K(t1 , . . . , tm )) = C1 . (7c) Suppose M is an extension of E1 with M Ks = Ks (t1 , . . . , tm ). Then M is a field of rational functions over M ∩ Ks .

324

Chapter 16. Galois Groups over Hilbertian Fields

Choose algebraically independent elements tij , i = 1, . . . , r, j = 1, . . . , m over K with t1j = tj , j = 1, . . . , m. For each i let ti = (ti1 , . . . , tim ). i = tij , j = Also, let αi : K(t1 ) → K(ti ) be the K-isomorphism with tα j 1, . . . , m. Put Ei = E1αi and extend αi to an isomorphism of F1 on a field Fi . The latter induces an isomorphism of Gal(F1 /K(t1 )) and Gal(F1 /E1 ) onto Gal(Fi /K(ti )) and Gal(Fi /Ei ), respectively. They are given by γ 7→ αi−1 γαi . Identify Gal(Fi /K(ti )) and Gal(Fi /Ei ) through this isomorphism with Ci and Aut(Ci ), respectively. Hence, Ei , K(ti ), Fi satisfy (7) with i replacing 1. Also, F1 , . . . , Fr are algebraically independent over K and regular over K (by (7a)). Therefore, they are linearly disjoint over K (Lemma 2.6.7). Next let E = E1 · · · Er , t = (t1 , · · · , tr ), and F = F1 · · · Fr . Then (8a) K ⊆ E ⊆ K(t) ⊆ F , F/K is regular, F/E is Galois, Qr Qr (8b) Gal(F/E) = i=1 Aut(Ci ), and Gal(F/K(t)) = i=1 Ci . The group Sr acts on F by permuting the triples (Ei , ti , Fi ), i = 1, . . . , r. −1 Namely, xπ = xαi αj for π ∈ Sr with iπ = j and for x ∈ Fi . This induces an action on Gal(F/E) which coincides with the action given by (4). (Recall: αi with the To obtain (4) we identify Ci with C1 via αi and then identify Qr identity map.) So, by Lemma 16.8.3(b), the action of i=1 Aut(Ci ) on F extends to an action of Aut(C) on F over K. Denote the fixed field of Aut(C) in F by E0 . Part C: The field crossing argument. By 7(a), F is linearly disjoint from L over K. Put Q = F L. Then Gal(Q/E0 ) ∼ = Aut(C) × Gal(L/K) with pr2 = resQ/L . Also, Q/L is regular. Identify Gal(Q/E0 L) with Aut(C) and Gal(Q/L(t)) with C. Part A identifies G with a subgroup of Gal(Q/E0 ). Denote the fixed field of G in Q by P . By (5), Gal(Q/P ) = G and restriction of G into Gal(L/K) is surjective. (9)

P L = L(t)

and P ∩ L = K.

Thus, all that remains to be proved is the rationality of P over K. Part D: New transcendence basis for L(t)/L. The group D1 acts on L because L/K is Galois. Denote the fixed field by M0 . Each σ ∈ D1 satisfies C1σ = C1 . Hence, the permutation of {1, . . . , r} corresponding to σ fixes 1, so F1σ = F1 . Thus, D1 acts on F1 L. Denote the fixed field by M . By (6), D1 ∩Aut(C1 ) = C1 . The fixed field of C1 in F1 L is L(t1 ) and that of Aut(C1 ) in F1 L is E1 L. Therefore, M E1 L = L(t1 ). Next observe that the restriction of D1 to F1 maps D1 into Aut(C1 ), hence into Gal(F1 /E1 ). Hence, E1 ⊆ M . Thus, M L = L(t1 ), so M Ks = Ks (t1 ). By construction, M ∩ L = M0 . It follows from (7c) that there are algebraically independent elements v1 , . . . , vm over K with M = M0 (v1 , . . . , vm ). They satisfy (10)

L(v1 , . . . , vm ) = L(t1 ).

16.9 Embedding Problems over Hilbertian Fields

325

Here is a partial diagram of the fields involved: F1

F1 L

K(t1 )

E1

M

K

M0

L(t1 ) y y yy yy y y

L

For each i between 1 and r choose σi ∈ G with C1σi = Ci (Remark 16.8.4). Then D1σi = Di . Put vij = vjσi , j = 1, . . . , m, and vi = (vi1 , . . . , vim ). Then let v = (v1 , . . . , vr ). By (10), L(vi ) = L(ti ), i = 1, . . . , r, so L(v) = L(t). Hence, the vij are algebraically independent over K. Since vj ∈ M , we have vjδ = vj for all δ ∈ D1 and j = 1, . . . , m. If (σ δσ −1 )σi

δ = vj i i 1 ≤ i ≤ r and δ ∈ Di , then σi δσi−1 ∈ D1 , hence vij Tr δ = vij for all i and j. It follows for δ = D = i=1 Di that vij

σ

= vj j = vij .

Part E: Rationality of P . Denote the fixed field of D in Q by N . By Part δ A, N/P is Galois. Put N0 = N ∩ L. By Part D, vij = vij for each δ ∈ D and all i, j. Thus, N0 (v) ⊆ N . As [L(v) : N0 (v)] = [L : N0 ] = [L(v) : N ], we have N = N0 (v). By (9), P N0 = N . P

N

L(t)

K

N0

L

Q

Let V be the vector space spanned by the vij ’s over N0 . For each σ ∈ G σ −1 σi0

and each i there exists i0 with Ciσ = Ci0 = Ci i −1 (σσi−1 0 σi )σi σi0

. Hence, σσi−1 0 σi ∈ Di , so

σi−1 σi0

σ vij = vij = vij = vi0 ,j . Thus, Gal(N/P ) leaves V invariant. By Part D, dim(V ) = rm = trans.deg(P/K). It follows from Lemma 16.8.5 that P is K-rational.

16.9 Embedding Problems over Hilbertian Fields An affirmative solution to the inverse problem of Galois theory, the realization of all finite groups over Q, seems at present to be out of reach. Matzat’s method of GAR-realization gives an effective tool for a partial solution of

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the problem, namely for the of realization of finite groups with non-Abelian composition factors. Every finite group G has a sequence N0 , . . . , Nn of subgroups with N0 = G, Nn = 1, Ni / Ni−1 , and Ni−1 /Ni simple, i = 1, . . . , n. For each finite simple group C, the number of i with C ∼ = Ni−1 /Ni depends only on G and not on the sequence [Huppert, p. 64, Satz 11.7]. If this number is positive, C is a composition factor of G. Proposition 16.9.1 (Matzat): Let α: G → Gal(L/K) be a finite embedding problem over a field K. Suppose every finite embedding problem for G with an Abelian kernel is solvable. (a) If K is Hilbertian and each composition factor of Ker(α) is GAR over K, then α is solvable. (b) If each composition factor of Ker(α) is GAR over every extension of K, then α is regularly solvable. Proof: Assume without loss that Ker(α) is nontrivial. Choose a minimal normal subgroup C of G in Ker(α). Then α induces an isomorphism α0 : G/C → Gal(L/K). The rest of the proof splits into two parts. Part A: Proof of (a). An induction hypothesis gives a Galois extension L0 of K containing L and an isomorphism β 0 : Gal(L0 /K) → G/C with α0 ◦ β 0 = resL0 /L . 0 −1 Let π: G → G/C be the quotient map. Qr The kernel of (β ) ◦ π: G → 0 ∼ Gal(L /K) is C. By Remark 16.8.4, C = i=1 Ci with Ci simple non-Abelian and Ci conjugate to C1 in G, i = 1, . . . , r. In particular, C1 is a composition factor of G. By assumption, α is solvable if C1 is Abelian. Suppose C1 is nonAbelian. Then C1 is GAR over K. Hence, by Proposition 16.8.6, (β 0 )−1 ◦ π is regularly solvable. Since K is Hilbertian, (β 0 )−1 ◦ π is solvable (Lemma 16.4.2). In other words, K has a Galois extension N containing L0 and there is an isomorphism β: Gal(N/K) → G with (β 0 )−1 ◦ π ◦ β = resN/L0 . Then, α ◦ β = resN/L and β solves embedding problem α. Gal(N/K) res β 0 Gal(L /K) β 0 res α / Gal(L/K) G π α0 / Gal(L/K) G/C

16.9 Embedding Problems over Hilbertian Fields

327

Part B: Proof of (b). An induction hypothesis gives algebraically independent elements t1 , . . . , tm over K, a Galois extension E of K(t), and an epimorphism β 0 : Gal(E/K(t)) → G/C satisfying: E is a regular extension of L and α0 ◦ β 0 = resE/L . As above, let π: G → G/C be the quotient map. By assumption, each composition factor of C is regular over K(t). Proposition 16.8.6 gives algebraically independent elements u1 , . . . , un over K(t), a Galois extension F of K(t, u), and an isomorphism β: Gal(F/K(t, u)) → G satisfying: F is a regular extension of E and (β 0 )−1 ◦ π ◦ β = resF/E . Then F is a regular extension of L and β is a regular solution of embedding problem α. Proposition 16.9.2: Let K be a field and n a positive integer with char(K) - (n − 1)n and n 6= 2, 3, 6. Then An is GAR over K. Proof: Use the notation of Proposition 16.7.6. The assumption n 6= 2 implies that the center of Sn is trivial. The assumption n 6= 2, 3, 6 implies Sn = Aut(An ) [Suzuki, p. 299, Statement 2.17]. Hence, F/K(u) is a GArealization of An . We must still prove Condition (3) of Section 16.8. Consider an extension E 0 of K(t) with E 0 Ks = Ks (u). Let L = E 0 ∩ Ks . We have to prove E 0 is L-rational. To begin, note that E 0 ∩Ks (t) = L(t) and [E 0 : L(t)] = [Ks (u) : Ks (t)] = [K(u) : K(t)] = 2. In particular, E 0 is a function field of one variable over L. Since the genus of Ks (u) is 0, so is the genus of E 0 (Proposition 3.4.2(b)). Suppose p0 is a prime divisor of L(t)/L which ramifies in E 0 . Then deg(p0 ) = 1. Hence, there are exactly one prime divisor ps of Ks (t)/Ks over p0 (Proposition 3.4.2(e)) and ps ramifies in Ks (u) (Proposition 3.4.2(c)). Let p be the common restriction of ps and of p0 to K(t). Then p ramifies in K(u) (Proposition 3.4.2(c)). But there are at most three such p and each of them is of degree 1. Hence, there are at most three possibilities for p0 (Proposition 3.4.2(e)). It follows from Lemma 16.7.5 that E 0 is L-rational. The combination of Propositions 16.9.1 and 16.9.2 gives concrete solvable embedding problems with non-Abelian kernels. Proposition 16.9.3: Let K be a field and α: G → Gal(L/K) a finite embedding problem over K. Suppose each composition factor of Ker(α) is An with char(K) - (n − 1)n and n 6= 6. Then α is regularly solvable over K. If, in addition, K is Hilbertian, then α is solvable over K. Take L = K in Proposition 16.9.3: Corollary 16.9.4: Let K be a field and G a finite group. Suppose each composition factor of G is An with char(K) - (n − 1)n and n 6= 6. Then G is regular over K. If, in addition, K is Hilbertian, then G is realizable over K. Remark 16.9.5: More GAR-realizations. There is a long list of finite nonAbelian simple groups which are known to be GAR over Q. Beside An (with n 6= 6), this list includes PSL2 (Fp ) with p odd and p 6≡ ±1 mod 24 and

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all sporadic groups with the possible exception of M23 . See [Malle-Matzat, Thm. IV.4.3]. The list becomes longer over the maximal Abelian extension Qab of Q. In addition to the groups that are GAR over Q it contains A6 , PSL2 (Fp ) with p odd, and M23 . See [Malle-Matzat, Thm. IV.4.6]. It is still unknown whether every finite non-Abelian simple group is GAR over Qab . If this is the case, each finite embedding problem over Qab would be solvable, as we will see in Example 24.8.5. It is even unknown if every finite non-Abelian simple group is GAR over ˜ An affirmative answer to this question would each field K containing Q. ˜ 1 , t2 )) which is enable us to solve embedding problems over the field Q((t Hilbertian by Example 15.5.2. The lists over Q and Qab have been established in large part by using the Riemann Existence Theorem. This partially explains the lack of knowledge of GAR realizations in characteristic p. An exception to our lack of knowledge of GAR realization is the family An of alternative groups. Theorem 15 of [Brink] improves Proposition 16.9.2 and proves that An is GAR over every field K if n 6= 2, 6 and char(K) 6= 2. The case char(K) = 2 is left open.

16.10 Finitely Generated Profinite Groups Let S be a subset of a profinite group G. Denote the closed subgroup generated by S by hSi. We say S generates G if hSi = G. In this case each map ϕ0 of S into a profinite group H has at most one extension to a (continuous) homomorphism ϕ: G → H. A profinite group G is finitely generated if it has a finite set of generators. In this case, the minimal number of generators of G is the rank of G. Note that the rank of a quotient of G does not exceed the rank of G. Example 16.10.1: Consider the group (Z/pZ)n as a vector space over Fp of dimension n. Each group theoretic set of generators of (Z/pZ)n generates (Z/pZ)n as a vector space over Fp . Hence, rank(Z/pZ)n = n. Since (Z/pZ)n is a quotient of Znp , the rank of the latter group is at least n. For each i between 1 and n consider the element ei = (0, . . . , 1, . . . , 0) with 1 in the ith coordinate and 0 elsewhere. Then e1 , . . . , en generates Znp . It follows that rank(Znp ) = n. Lemma 16.10.2: A finitely generated profinite group G has, for each positive integer n, only finitely many open subgroups of index at most n. Proof: Each open subgroup M of G of index ≤ n contains an open normal subgroup N with G/N isomorphic to a subgroup of Sn . Indeed, suppose m = (G : M ) ≤ n. Then, B = {gM | g ∈ G} is a set of order m. Multiplication from the left with an element x of G induces a permutation π(x) of B. Specifically, π(x)(gM ) = xgM . Thus, π is a

16.10 Finitely Generated Profinite Groups

329

T homomorphism of G into Sn with Ker(π) = x∈G M x . Therefore, Ker(π) is an open normal subgroup of G which is contained in M . Moreover, G/Ker(π) is isomorphic to a subgroup of Sm , hence to a subgroup of Sn . Thus, it suffices to prove G has only finitely many open normal subgroups N with G/N isomorphic to a subgroup of Sn . The map α 7→ Ker(α) maps the set of all homomorphisms α: G → Sn onto the set of all open normal subgroups N of G such that G/N is isomorphic to a subgroup of Sn . Hence, the number ν of those N ’s does not exceed the number of the α’s. Let S be a finite set of generators of G. Then every homomorphism α: G → Sn is determined by its values on S. Therefore, ν ≤ (n!)|S| . We call a profinite group G small if for each positive integer n the group G has only finitely many open subgroups of index n. By Lemma 16.10.2, every finitely generated profinite group is small. Thus, each of the results we prove in this section for small profinite groups holds for finitely generated profinite groups. Remark 16.10.3: Small profinite groups and open subgroups. Let G be a profinite group. (a) Denote the intersection of all open subgroups of G of index at most n by Gn . Then Gn is a closed normal subgroup of G. Moreover, G is small if and only if Gn is open in G for all n. (b) Let H be an open subgroup of index m of G. Consider open subgroups G0 and H 0 of G and H, respectively. Then (G : H 0 ) = (G : H)(H : H 0 ) and (H : H ∩ G0 ) ≤ (G : G0 ). Thus, Gmn ≤ Hn ≤ Gn . By (a), G is small if and only if H is small. (c) Let G and H be as in (b). Suppose H is finitely generated, say by h1 , . . . , hd . Let g1 , . . . , gn be representatives of G/H. Then g1 , . . . , gn , h1 , . . . , hd generate G. Conversely, if G is finitely generated, then H is finitely generated (Exercise 7). We prove a qualitative version of this result in Section 17.6. (d) Let α: G → H be an epimorphism of profinite groups. If g1 , . . . , gn are generators of G, then α(g1 ), . . . , α(gn ) are generators of H. If G is small, so is H. Indeed, let n be a positive integer and H0 an open subgroup of H of index n. Then G0 = α−1 (H0 ) is an open subgroup of G of index n. The map H0 7→ α−1 (H0 ) is injective. Since there are only finitely many G0 ’s, there are only finitely many H0 ’s. Example 16.10.4: A small profinite group which is not finitely generated. Q p Let A = Zp with p ranging over all prime numbers. Consider an open Q subgroup N of A of index n with n = p≤m pkp . Then N contains the Q Q open subgroup p≤m pkp Zpp × p>m Zpp . Hence, there are only finitely many possibilities for N . Thus, A is small. On the other hand, rank(A) ≥ rank(Zpp ) = p for each p (Example 16.10.1). Therefore, A is not finitely generated.

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Remark 16.10.5: Characteristic subgroups. A closed subgroup N of a profinite group G is characteristic if it is invariant under every automorphism of G. In particular, N is normal in G. Suppose G is small. Then, Gn is a characteristic open subgroup. The decreasing sequence G ≥ G2 ≥ G3 ≥ · · · intersects in 1. Thus, it consists of a basis of open neighborhoods of 1 in G. Let N be an open characteristic subgroup of G. Then Aut(G/N ) is a finite group. For each α ∈ Aut(G) define αN ∈ Aut(G/N ) by αN (gN ) = α(g)N . The map α 7→ αN gives a homomorphism ΦN : Aut(G) → Aut(G/N ). By the preceding paragraph, the intersection of all these N is the trivial group. Hence, the ΦN combine to an embedding Aut(G) → lim Aut(G/N ) ←− which is actually surjective. Thus, Aut(G) is a profinite group. The most useful properties of small profinite groups are embodied in the following result: Proposition 16.10.6: Let G be a small profinite group. Then: (a) Every epimorphism of G onto itself is an automorphism. (b) Let α: G → H and β: H → G be epimorphisms. Then both α and β are isomorphisms. Proof of (a): Let θ: G → G be an epimorphism. Let Gn be the finite set of all open subgroups of G of index at most n. The map H 7→ θ−1 (H) maps Gn injectively into itself. Hence, it maps Gn onto itself. Therefore, in the notation of Remark 16.10.3 \ \ \ Gn = H= θ−1 (H) = θ−1 ( H) = θ−1 (Gn ). H∈Gn

If θ(g) = 1, then g ∈

H∈Gn

T∞

n=1

θ−1 (Gn ) =

H∈Gn

T∞

n=1

Gn = 1. Therefore, θ is injective.

Proof of (b): By (a), β ◦ α is an isomorphism. Hence, both α and β are injective. For a profinite group G, we denote the set of all finite quotients (up to an isomorphism) of G by Im(G). Proposition 16.10.7: Let G and H be profinite groups with G small. (a) If Im(H) ⊆ Im(G), then H is a quotient of G. (b) If Im(H) = Im(G), then H is isomorphic to G. Proof of (a): First we prove H is small. Indeed, let n be a positive number and B1 , . . . , Br distinct open subgroups of H of indexTat most n. Choose an r open normal subgroup N of H which is contained in i=1 Bi . Then H/N ∈ Im(H). By assumption, G has an open normal subgroup M with G/M ∼ = H/N . Hence, G has r open subgroups of index at most n. Consequently, r is bounded. In the notation of Remark 16.10.3(a), the finite group H/Hn belongs to Im(H), and therefore to Im(G). Thus, G has an open normal subgroup K

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331

with G/K ∼ = H/Hn . In particular, K is an intersection of open subgroups of index ≤ n. Hence, Gn ≤ K. Therefore, there is an epimorphism from G/Gn to H/Hn . Denote the finite nonempty set of all epimorphisms of G/Gn → H/Hn by Φn . Let ϕ: G/Gn+1 → H/Hn+1 be an epimorphism. It maps the set of all subgroups of G/Gn of index at most n onto the set of all subgroups of H/Hn of index at most n. Hence, ϕ(Gn /Gn+1 ) ≤ Hn /Hn+1 . Therefore, ϕ induces an epimorphism ϕ: ¯ G/Gn → H/Hn . This defines a map Φn+1 → Φn . By Corollary 1.1.4, lim Φn is nonempty. Each element in lim Φn gives ←− ←− a compatible system of epimorphisms βn : G/Gn → H/Hn . It defines an epimorphism β: G → H, as desired. Proof of (b): Statement (a) gives epimorphisms ϕ: G → H and ψ: H → G. Thus, ψ ◦ ϕ: G → G and ϕ ◦ ψ: H → H are epimorphisms. By (a), H is small. Hence, by Proposition 16.10.6, both ψ◦ϕ and ϕ◦ϕ are automorphisms. Consequently, both ϕ and ψ are isomorphisms. Corollary 16.10.8: Let α: G → H be an epimorphism of profinite groups. Suppose G is small and Im(G) ⊆ Im(H). Then α is an isomorphism. Proof: Since G is small, so is H (remark 16.10.3(d)). By Proposition 16.10.7(a), G is a quotient of H. Therefore, by Proposition 16.10.6(b), α is an isomorphism. Example 16.10.9: Small Galois groups. ˜ (Section 1.5). Thus, Gal(K) is (a) For each finite field K, Gal(K) ∼ =Z generated by one element. (b) Let p be a prime number. The local compactness of Qp and Krasner’s lemma imply that Gal(Qp ) is small [Lang5, p. 54, Prop. 14]. Deeper arguments show Gal(Qp ) is generated by 4 elements [Jannsen, Satz 3.6]. (c) Let K be a number field, OK its ring of integers, and S a finite number of prime ideals of OK . Denote the maximal algebraic extension of K unramified outside S by KS . It is a Galois extension of K (Corollary 2.3.7(c)). We prove Gal(KS /K) is a small profinite group. Let T be the set of all prime numbers which ramify in K or lie under a prime ideal belonging to S. Then T is finite and KS ⊆ QT . Suppose we already know that Gal(QT /Q) is small. Then, by Remark 16.10.3(b),(d), Gal(QT /K) is small. Hence, Gal(KS /K) is also small. We may therefore assume K = Q and S is a finite set of prime numbers. Suppose L is a finite extension of Q in QS of degree at most n. By [Serre5, p. 130, Proposition 6], X log p + n|S| log n. log discriminant(L/Q) ≤ (n − 1) p∈S

Thus, discriminant(L/Q) is bounded. By Hermite-Minkowski [Lang5, p. 121, Thm. 5], there are only finitely many extensions of Q a given discriminant. Consequently, there are only finitely many possibilities for L.

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Alternatively, one may follow [Serre9, p. 107] and first observe that d = discriminant(L/Q) is divisible only by p ∈ S. For each such p, [LQp : Qp ] ≤ n. By (b), there are only finitely many possibilities for LQp , hence for the pth part of the different of L/Q. Therefore, there are only finitely many possibilities for d. It is, however, not clear whether Gal(KS /K) is finitely generated [Shafarevich1, §3]. (d) There are small Q absolute Galois groups which are not finitely generated. The group A = Zpp of Example 16.10.4 is one example. To construct a field with absolute Galois group A, we start from a field K of characteristic 0 that contains all roots of unity. Then Gal(K((t))) ∼ = ˆ [Geyer-Jarden2, Cor. 4.2]. Thus, for each p there is an algebraic Gal(K) × Z extension of K((t)) with absolute Galois group Gal(K) × Zp . In particular, taking K to be algebraically closed, we find a field K 0 with Gal(K 0 ) ∼ = Zp . all prime numbers and Induction gives fields Kp,i with p ranging overQ i = 1, . . . , p satisfying these conditions: Gal(Kp,i ) = l

F1 > F2 > · · · of characteristic T∞ subgroups of F with rank(Fi ) ≥ m. By Levi’s result (Lemma 17.5.10), i=1 Fi = 1. Therefore, F is residually-C.

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Remark 17.5.12: Let F be the free abstract group of rank m ≥ 2 and let C be an infinite family of finite non-Abelian simple groups. By [Weigel1,2,3] or [Dixon-Pyber-Seress-Shalev], F is residually-C. Note that the assumption of the groups in C to be non-Abelian is essential. If C consists of all groups Z/pZ, then the intersection of all normal subgroups N of F with F/N ∈ C contains the commutator subgroup [F, F ] of F which is not trivial.

17.6 Open Subgroups of Free Profinite Groups We prove in this section that an open subgroup of a free profinite group is free and satisfies the Nielsen-Schreier formula for the ranks: Remark 17.6.1: Identifying F as a subgroup of Fˆ . Let C be a Melnikov formation of finite group and F the free abstract group on a set X. Suppose there is a nontrivial group C ∈ C with rank(C) ≤ |X|. Let FˆX (C) be the corresponding profinite completion. We use Proposition 17.5.11 to identify F with a dense abstract subgroup of FˆX (C) and to identify the canonical map θ: F → FˆX (C) with the inclusion map. Conversely, let Fˆ be a free pro-C group with a basis X. Lemma 17.4.3 extends the identity map X → X to an isomorphism FˆX (C) → Fˆ . The latter maps F onto the abstract subgroup F0 of Fˆ generated by X. It follows that F0 is free on X. Proposition 17.6.2: Let C be a Melnikov formation of finite groups, Fˆ a ˆ an open subgroup of Fˆ . Suppose H ˆ is pro-C (e.g. C free pro-C-group, and H ˆ / Fˆ ). Then H ˆ is a free pro-C group. Moreover, if e = rank(Fˆ ) is is full or H ˆ = 1 + (Fˆ : H)(e ˆ finite, then rank(H) − 1). If m = rank(Fˆ ) is infinite, then ˆ = m. rank(H) Proof: Let X be a basis for Fˆ and F the free abstract group on X. Consider the family N (X) of all normal subgroups N of F with X r N finite and F/N ∈ C. By Lemma 17.4.6(c), Fˆ is the profinite completion of F with respect to N (X). By Proposition 17.5.11, we may assume that F is contained in Fˆ and θ: F → Fˆ is the inclusion map. Lemma 17.2.1 gives a subgroup H ˆ is the closure of H in Fˆ . of F and a M ∈ N (X) such that M ≤ H and H ˆ is finite and H ˆ = lim H/N , where N ranges Moreover, (F : H) = (Fˆ : H) ←− over N0 (X) = {N ∈ N (X) | N ≤ H}. Let Y , R, and ϕ be as in Proposition 17.5.6. Then R is finite, H is free, Y is a basis of H, and Y r N is finite for each N ∈ N0 (X). By Proposition 17.4.2, the free pro-C-group on Y is isomorphic to the inverse limit lim H/N , ←− where N ranges over the family N (Y ) of all open normal subgroups N of H such that Y r N is finite and H/N ∈ C. We show that N0 (X) and N (Y ) are ˆ is the free cofinite in each other. By Lemma 17.2.3, this will prove that H pro-C-group on Y . ˆ . By Lemma Indeed, let N ∈ N0 (X). Denote the closure of N in Fˆ by N ∼ ˆ ˆ ˆ ˆ ˆ 17.2.1, N / H and H/N = H/N . Since H is pro-C, we have H/N ∈ C, so

17.6 Open Subgroups of Free Profinite Groups

359

N ∈ N (Y ). r Conversely, let N ∈ N T T (Y ). Then H/N ∈ C and Y N is finite. Let N0 = r∈R N r and H0 = r∈R H r . Then N0 and H0 are normal subgroups of F , M ≤ H0 , and N0 ≤ H0 . Since N H0 /N is normal in H/N , it belong to C. Since H0 /(N ∩ H0 ) is isomorphic to N H0 /N , it belongs toTC. Hence, H0 /(N r ∩H0 ) ∈ C for each r ∈ R. Since R is finite, H0 /N0 = H0 / r∈R (N r ∩ H0 ) ∈ C (Condition (1b) of Section 17.3)). In addition, F/H0 ∈ C because F/M ∈ C and F/H0 is a quotient of F/M . Finally, since C is Melnikov, F/N0 ∈ C. F

N0

N

N H0

N ∩ H0

H0

H

M If we prove that X r N0 is finite, it will follow that N0 ∈ N0 (X). This will complete the proof of the cofiniteness of N (Y ) and N0 (X) in each other. Indeed, X r M , Y r N , and R are finite sets. Thus, almost all x ∈ X belong to M . For each x ∈ X ∩ M and r ∈ R we have ϕ(r, x) = rxr −1 (Proposition 17.5.6(d)). For each r ∈ R the map x 7→ rxr−1 of X ∩ M into Y is injective. Hence, only finitely many x ∈ X ∩ M satisfy rxr−1 ∈ Y r N . Therefore, X0 = {x ∈ X ∩M | ϕ(r, x) = rxr−1 and rxr−1 ∈ N for all r ∈ R} is cofinite in X. It follows that almost every y ∈ Y can be written as y = vxv −1 with x ∈ X0 and v ∈ R. For each u ∈ R there exist r ∈ R and h ∈ H with uv = hr. Therefore, uyu−1 = uvxv −1 u−1 = hrxr−1 h−1 ∈ hN h−1 = N , so, y ∈ N u . It follows that almost every y ∈ Y belongs to N0 . By Lemma 17.5.4(d), X r N0 = (X r H) ∪ (X ∩ H r N0 ) ⊆ (X r H) ∪ (Y r N0 ) is finite, as claimed. The rank formula for the finitely generated case now follows from Proposition 17.5.7. When m is infinite, use Corollary 17.1.5. Corollary 17.6.3: Let G be a profinite group of rank at most e and H an open subgroup of G of index n. Then rank(H) ≤ 1 + n(e − 1). Proof: There exists an epimorphism θ: Fˆe → G and (Fˆe : θ−1 (H)) = n. By Proposition 17.6.2, rank(θ−1 (H)) = 1 + n(e − 1). Hence, by Corollary 17.1.4, rank(H) ≤ 1 + n(e − 1). The following result is used in Section 25.7 to prove that a closed norˆ of a free pro-C group is free pro-C if N ˆ contains nontrivial mal subgroup N elements of the underlying abstract free group. Proposition 17.6.4: Let C be a Melnikov formation, Fˆ a free pro-C group of rank m ≥ 2, X a basis of Fˆ , F the abstract group of Fˆ generated by X, ˆ a closed normal subgroup of Fˆ . Suppose N = F ∩ N ˆ 6= 1. Then: and N

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Chapter 17. Free Profinite Groups

ˆ containing N ˆ with a Schreier basis Y (a) Fˆ has an open normal subgroup E with respect to X such that Y ∩ N 6= ∅. ˆ containing N ˆ with (b) If m < ∞ than Fˆ has an open normal subgroup E ˆ ˆ ˆ ˆ rank(E/N ) < 1 + (F : E)(m − 1). ˆ , t 6= 1. Denote the set of all open normal Proof of (a): Let t ∈ F ∩ N ˆ by H. For each E ˆ ∈ H put E = F ∩ E. ˆ subgroups of Fˆ which contain N ˆ ∈ H, a Schreier basis Y of E with respect to X (Notation 17.5.3), and Find E a presentation t = s1 s2 · · · sn as a word in Y with a minimal length among all presentations of this type. We show that s1 ∈ N . ˆ is the intersection of all H ˆ ∈ H. Assume s1 ∈ / N . By Lemma 1.2.3, N ˆ ˆ ˆ ˆ / H. By Proposition Hence, there is an H ∈ H such that H ≤ E and s1 ∈ 17.5.6(b), H has a Schreier basis Z with respect to Y with lengthZ (t) < lengthY (t), a contradiction. ˆ Proof of (b): Continuing the proof of (a), we conclude from s1 ∈ Y ∩ N ˆ ˆ ˆ ˆ ˆ and from Nielsen-Schreier that rank(E/N ) < rank(E) = 1 + (F : E)(m − 1). Corollary 17.6.5: Let C be a Melnikov formation and Fˆ a free pro-C-group of positive rank. Then Fˆ is infinite. Proof: Assume Fˆ is finite. Then the trivial group is a free pro-C-group of positive rank (Proposition 17.6.2). This is a contradiction. Example 26.1.10 shows that Corollary 17.6.5 does not hold for arbitrary formations.

17.7 An Embedding Property Here is a characterization of free pro-C groups of finite rank in terms of their finite quotients: Lemma 17.7.1: Let C be a formation of finite groups, e a positive integer, and F a pro-C group. Then F ∼ = Fˆe (C) if and only if F satisfies the following conditions: (a) Every finite homomorphic image of F is generated by e elements. (b) Every C-group of rank at most e is a homomorphic image of F . Proof: A compactness argument shows that (a) holds if and only if rank(F ) ≤ e. Now apply Proposition 16.10.7(b). We conclude this section by establishing an “embedding property” of Fˆe (C). It depends on a surprising lemma that allows us to “lift generators” of a homomorphic image: Lemma 17.7.2 (Gasch¨ utz): Let π: G → H be an epimorphism of profinite groups with rank(G) ≤ e. Let h1 , . . . , he be a system of generators of H.

Exercises

361

Then there exists a system of generators g1 , . . . , ge of G such that π(gi ) = hi , i = 1, . . . , e. Proof (Roquette): We start with the crucial case: G is a finite group. For each subgroup C of G satisfying π(C) = H and all systems of generators a1 , . . . , ae of H denote the number of e-tuples c ∈ C e that generate C and satisfy π(c) = a by ϕC (a). We prove by induction on |C| that ϕC (a) is independent of a. Assume ϕB (a) is independent of a for every proper subgroup B of C |C| . Then there are exactly me elements satisfying π(B) = H. Let m = |H| b ∈ C e with π(b) = a. Each such b generates a subgroup B of C satisfying π(B) = H. Hence, (1)

me = ϕC (a) +

X0 B 1 use Nielsen-Schreier to prove for each x1 , . . . , xf ∈ Fˆe −1 that the inequality (Fˆe : hxi) < ∞ implies (Fˆe : hxi) ≤ fe−1 . Conclude for x1 , . . . , xe ∈ Fˆe that either hxi = Fˆe or (Fˆe : hxi) = ∞.

Notes Ershov [Ershov2, Lemma 1, and Exercise 9 of Chapter 21] suggests an alternative proof of Douady’s theorem (Proposition 17.1.1). The open subgroup theorem (Proposition 17.6.2) for a free profinite group is a special case of the subgroup theorem for free products of profinite groups appearing in [Binz-Neukirch-Wenzel]. The latter, is based on the deeper Kurosh subgroup theorem for free products of abstract groups [Specht, p. 189, Satz 8] while our proof of Theorem 17.6.2 is based on the simpler Schreier construction of Section 17.5. Gasch¨ utz’s proof of Lemma 17.7.2 [Gasch¨ utz] provides a (complicated) formula for ϕG (a). The proof of Levi’s Lemma 17.5.10 depends on Schreier’s construction of a basis of a subgroup of a free abstract group rather than on Nielsen’s construction as in [Lyndon-Schupp, p. 14, Prop. 3.3]. The latter proof considers a nontrivial element w in Fi (notation as in Lemma 17.5.10) and a basis X of F and proves that lengthX (w) ≥ i. Our proof constructs a basis Y of Fi such that lengthY (w) ≤ max(0, lengthX (w) − i).

Chapter 18. The Haar Measure It is well known that every locally compact group admits a (one sided) translation invariant Haar measure. Applications of the Haar measure in algebraic number theory to local fields and adelic groups appear in [Cassels-Fr¨ohlich, Chap. II] and [Weil6]. Here we use it to investigate absolute Galois groups of fields. Since these groups are compact, the Haar measure is a two sided invariant. We provide a simple direct proof of the existence and uniqueness of the Haar measure of profinite groups (Sections 18.1 and 18.2). We normalize the Haar measure µ of a compact group G so that µ(G) = 1. Thus, µ is a probability measure. When G = Gal(K) is the absolute Galois group of a field K, µ-independence of sets is related to linear disjointness of fields (Section 18.5). For K a Hilbertian field, we prove that hσ1 , . . . , σe i ∼ = Fˆe for almost all σ ∈ Gal(K)e . In addition, if K is countable, then the fixed fields Ks (σ1 , . . . , σe ) are PAC for almost all σ ∈ Gal(K)e (Section 18.6). In the uncountable case we provide examples where the set of σ ∈ Gal(K)e such that Ks (σ1 , . . . , σe ) is PAC, is nonmeasurable. The complete proof that every field is stable (Definition 18.9.1) lies unfortunately outside the scope of this book. Section 18.9 outlines the main ingredients and steps of the proof. Once this is done, we are able to construct for each countable Hilbertian field an abundance of Galois extensions which are PAC (Theorem 18.10.2).

18.1 The Haar Measure of a Profinite Group Let G be a profinite group. We define the completed Haar measure of G and prove its uniqueness. In the next section we prove the existence of the Haar measure. Consider a collection A of subsets of G. The σ-algebra generated by A is the smallest collection A0 of subsets of G which contains A and is closed under taking complements and countable unions. When A is a Boolean algebra, A0 is also the smallest collection of subsets of G which contains A and is monotone. That is, A0 is closed under countable increasing unions and countable decreasing intersections [Halmos, p. 27, Thm. B] The Borel field of G is the σ-algebra generated by all closed (= compact) subsets of G. We denote it by B or also by B(G) if reference to G is needed. Consider a function µ: B → R satisfying: (1a) 0 ≤ µ(B) ≤ 1 for each B ∈ B. (1b) µ(∅) = 0 and µ(G) = 1. S∞ (1c) P Let B1 , B2 , B3 , . . . be pairwise disjoint Borel sets. Then µ( i=1 Bi ) = ∞ i=1 µ(Bi ) (σ- additivity).

364

Chapter 18. The Haar Measure

(1d) If B ∈ B and g ∈ G, then µ(gB) = µ(Bg) = µ(B) (translation invariance). (1e) For all B ∈ B and each ε > 0 there are an open set U and a closed set C satisfying C ⊆ B ⊆ U and µ(U r C) < ε (regularity). Condition (1) has the following immediate consequences: (2a) Let A ⊆ B be Borel sets. Then µ(B r A) = µ(B) − µ(A). In particular, µ(G r A) = 1 − µ(A). S∞ P∞ (2b) Let B1 , B2 , B3 , . . . be Borel sets. Then µ( i=1 Bi ) ≤ i=1 µ(Bi ). Sn−1 0 r (Write Bn = Bn i=1 Bi and apply (1c).) , B , . . . be Borel sets satisfying µ(Bi ) = 0, i = 1, 2, 3, . . . . (2c) Let B1 , B 2 3 S∞ Then µ( i=1 Bi ) = 0 (use 2b). (2d) Let B1 , B T2∞, B3 , . . . be Borel sets satisfying µ(Bi ) = 1, i = 1, 2, 3, . . . . Then µ( i=1 Bi ) = 1 (use (2a) and (2c)). (2e) SupposeSA1 ⊆ A2 ⊆ A3 ⊆ · · · is an increasing sequence of Borel sets. ∞ Then µ( i=1 Ai ) = limi→∞ µ(Ai ) (use (1c)). (2f) Suppose TA1 ⊇ A2 ⊇ A3 ⊇ · · · is a decreasing sequence of Borel sets. ∞ Then µ( i=1 Ai ) = limi→∞ µ(Ai ) (take complements and use (2a) and (2e)). 1 (2g) Suppose S n H is an open subgroup of G of index n. Then µ(H) = n (write G = · i=1 gi H and apply (1c) and (1d)). Lemma 18.1.1: 1 . (a) If H is a closed subgroup of G, then µ(H) = (G:H) ¯ (b) Let N be an open normal subgroup of G, A a subset of G/N , and A = ¯ ¯ Then µ(A) = |A| {g ∈ G | gN ∈ A}. . (G:N )

(c) Finally, if U is a nonempty open subset of G, then µ(U ) > 0. Sn Proof of (a): First P suppose (G : H) = n is finite and write G = · i=1 gi H. n By (1), 1 = µ(G) = i=1 µ(gi H) = nµ(H). Hence, µ(H) = n1 , as claimed. Now assume H has infinite index. Then H is contained in the intersection of a decreasing sequence of open subgroups, H1 > H2 > . . . . Thus, 1 µ(H) ≤ limi→∞ (G:H = 0. i) Proof of (b): Use (a) and (1d). Proof of (c): By definition, U contains a coset gH, where H is an open 1 subgroup of G. By (a) and (1d), µ(U ) ≥ µ(gH) = µ(H) = (G:H) > 0. Definition 18.1.2: Zero sets. Call a subset A of G a zero set (with respect to (B, µ)) if A ⊆ B ∈ B with µ(B) = 0. Let Bˆ be the σ-algebra of G generated by B and the zero sets. Write Bˆµ , ˆ ˆ of Bˆ is the B(G), or Bˆµ (G) if a reference to µ or G is needed. Each set B union of a set B ∈ B and a zero set. Extend µ to a function µ ˆ: Bˆ → R by ˆ defining µ ˆ(B) to be µ(B). The extended measure µ ˆ has properties (1) and (2) with B replaced by Bˆ and also the following:

18.1 The Haar Measure of a Profinite Group

365

ˆµ (3a) Bˆ contains all zero sets with respect to (B, ˆ) (ˆ µ is complete). (3b) For each B ∈ Bˆ there are B1S, B2 ∈ B with B1 ⊆TB ⊆ B2 and ∞ ∞ µ(B2 r B1 ) = 0 (Take B1 = n=1 B1,n and B2 = n=1 B2,n with B1,n , B2,n ∈ B satisfying B1,n ⊆ B1,n+1 ⊆ B ⊆ B2,n+1 ⊆ B2,n and µ(B2,n r B1,n ) < n1 .) The following result shows that µ ˆ is determined by its values on openclosed sets. A compactness argument shows that each open-closed set is a union of finitely many left cosets of open normal subgroups of G. The latter union can be made disjoint. Hence, µ ˆ is unique. Proposition 18.1.3: Let B0 be the Boolean algebra of open-closed sets in G and B1 the σ-algebra generated by B0 . Then: (a) For every U ∈ Bˆµ there exist A, B ∈ B1 with A ⊆ U ⊆ B and µ(B r A) = 0. (b) Suppose µ and ν are functions from B to R satisfying (1). Then (Bˆµ , µ ˆ) = (Bˆν , νˆ). S Proof of (a): It suffices to prove (a) for U open. We write U = i∈I xi Mi with openSnormal subgroups Mi of G and xi ∈ G and let α be the supremum of all µ i∈I 0 xi Mi where I 0 is a countable subset of I. For each n we S choose a countable subset Jn of I such that α − µ i∈Jn xi Mi < n1 . Let S∞ S J = n=1 Jn . Then A = j∈J xj Mj ⊆ U , A ∈ B1 , andTµ(A) = α. Now we consider the closed normal subgroup N = j∈J Mj of G. Then G/N has a countable basis for its topology. Let π: G → G/N be the quotient map. The sets π(xi M ) = xi Mi N/N are open in G/N and their union is π(U ). By a lemma of Lindel¨ of [Hocking-Young, p. 66], I has a countable S subset K such that π(U ) = k∈K π(xk Mk ). In addition, π −1 (π(A)) = A and π(U r A) = π(U ) r π(A). Hence, U r A ⊆ π −1 (π(U r A)) =

[

(xk Mk N r A).

k∈K

The right hand side of this equality, which we denote by B0 , belongs to B1 . If we prove that µ(B0 ) = 0, then B = A ∪ B0 ∈ B1 and µ(B r A) = 0, as needed. It suffices to prove that µ(xk Mk N r A) = 0 for each k ∈ K. r Sr k Mk N A) > 0. Write N = S r Assume there exists k ∈ K with µ(x · ρ=1 (N ∩ Mk )nρ . Then xk Mk N r A = ρ=1 (xk Mk nρ r A), so there exists ρ with µ(xk Mk nρ r A) > 0. Note that Anρ = A, because nρ ∈ Mj for each j ∈ J. Hence, µ(xk Mk r A) = µ((xk Mk r A)nρ ) = µ(xk Mk nρ r A) > 0. S It follows that µ x M ∪ x M = µ(A) + µ(xk Mk r A) > α. This j j k k j∈J contradiction to the definition of α concludes the proof of (a). Proof of (b): Denote the collection of all Borel sets on which µ and ν coincide by B2 . Lemma 18.1.1 implies S each open subgroup of G belongs to B2 . Hence, r µ and ν coincide on all sets · i=1 gi N with N open normal in G. In other

366

Chapter 18. The Haar Measure

words, µ and ν coincide on B0 . By (2e) and (2f), B2 is monotone. It follows that µ and ν coincide on B1 . Consider now a closed subset C of G. The proof of (a) gives a set D ∈ B1 with C ⊆ D and µ(C) = µ(D). By the preceding paragraph, ν(D) = µ(D). Thus, ν(C) ≤ ν(D) = µ(C). By symmetry, µ(C) ≤ ν(C). Hence, µ(C) = ν(C), so C ∈ B2 . It follows that B = B2 . In particular, µ(B) = 0 if and only if ν(B) = 0 for each B ∈ B. Hence, a subset A of G is a zero set with respect to (B, µ) if and only if A is a zero set with respect to ˆ = νˆ. (B, ν). Therefore, Bˆµ = Bˆν and µ Call each set in Bˆ measurable. Example 18.1.4: Haar measure on finite groups. Let G be a finite group equipped with the discrete topology. Then each subset B of G is open and µ(B) = |B| |G| is the unique Haar measure of G. Example 18.1.5: Infinite profinite groups have nonmeasurable subsets. If G is an infinite profinite group, it has a countable abstract subgroup H. Let R of G modulo H. Then G = S be a set of representatives for the left cosetsP · h∈H hR. Assume R is measurable. Then 1 = h∈H µ(hR). Since µ(hR) = µ(R) for each h ∈ H, this is impossible. Therefore, R is nonmeasurable.

18.2 Existence of the Haar Measure Proposition 18.1.3 gives the uniqueness of the Haar measure on a profinite group. The definition of µ on the Boolean algebra B0 of open-closed sets in G is a straightforward application of the invariance property of µ. A theorem of Caratheodory [Halmos, p. 42, Thm. A] then extends µ to the σ-algebra Bˆ0 generated by B0 . If rank(G) ≤ ℵ0 , Bˆ0 contains every open set and therefore every Borel set of G. In the general case one has to work harder. The proof that we give here for arbitrary G avoids Caratheodory’s theorem. Proposition 18.2.1: Every profinite group G has a unique Haar measure. Proof: Proposition 18.1.3 asserts the uniqueness of µ. The proof of the existence of µ divides into five parts. It uses the notation of Proposition 18.1.3. Sm Part A: A measure on B0 . Represent B ∈ B0 as B = · i=1 gi N with N m an open normal subgroup. Define µ0 (B) to be (G:N ) . We have to show that this is independent of the choice of N . Suppose N 0 is an open S n normal subgroup of G contained in N . Put n = (N : N 0 ). Then N = · j=1 hj N 0 Sm Sn and B = · i=1 · j=1 gi hj N 0 . The computation of µ0 (B) with respect to N 0 mn m gives (G:N 0 ) = (G:N ) . Thus, µ0 (B) is well defined. By definition, µ0 (G) = 1. Finite additivity and translation invariance of µ0 on B0 are easy exercises.

18.2 Existence of the Haar Measure

367

Part B: Extension of µ0 to open sets.

For each open subset U of G let

µ1 (U ) = sup(µ0 (A) | A ∈ B0 and A ⊆ U ). The function µ1 extends µ0 and has these properties: (1a) U1 and U2 open =⇒ µ1 (U1 ∪ U2 ) ≤ µ1 (U1 ) + µ1 (U2 ). (1b) U1 and U2 open disjoint =⇒ µ1 (U1 ∪ U2 ) = µ1 (U1 ) + µ1 (U2 ). ). (1c) U1 ⊆ U2 open =⇒ µ1 (U1 ) ≤ µ1 (U2S P∞ ∞ (1d) Ui is open, i = 1, 2, 3, . . . =⇒ µ1 ( i=1 Ui ) ≤ i=1 µ1 (Ui ). (1e) U open and x ∈ G =⇒ µ1 (xU ) = µ1 (U x) = µ1 (U ). To prove (1a), denote U1 ∪ U2 by U . For ε > 0 let A ∈ B0 with A ⊆ U and µ1 (U ) < µ0 (A) + ε. The open set Ui is a union of sets in B0 , i = 1, 2. Since A is compact, A is contained in a union B1 ∪ · · · ∪ Bn such that for each i, Bi ∈ B0 and Bi ⊆ U1 or Bi ∈ B0 and Bi ⊆ U2 . Let C1 be the union of all Bi which are contained in U1 . Let C2 be the union of all Bi which are contained in U2 . Then C1 ∪ C2 = B1 ∪ · · · ∪ Bn and µ1 (U ) − ε < µ0 (A) ≤ µ0 (C1 ∪ C2 ) ≤ µ0 (C1 ) + µ0 (C2 ) ≤ µ1 (U1 ) + µ1 (U2 ). Since this inequality holds for each ε > 0, µ1 (U ) ≤ µ1 (U1 ) + µ1 (U2 ). To prove (1b), let ε > 0. Choose Ai ∈ B0 , Ai ⊆ Ui such that µ1 (Ui ) < µ0 (Ai ) + 2ε , i = 1, 2. Since µ0 is additive, µ1 (U1 ) + µ1 (U2 ) < µ0 (A1 ) + µ0 (A2 ) + ε = µ0 (A1 ∪ A2 ) + ε ≤ µ1 (U1 ∪ U2 ) + ε. Hence, µ1 (U1 ) + µ1 (U2 ) ≤ µ1 (U1 ∪ U2 ). Combining this with (1a) gives (1b). For (1c), let ε > 0 and choose A ∈ B0 such that A ⊆ U1 and µ1 (U1 ) < µ0 (A) + ε. Since A ⊆ U2 , µ1 (U 1 ) < µ1 (U2 ) + ε. Hence, µ1 (U1 ) ≤ µ1 (U2 ). S∞ To prove (1d), let U = i=1 Ui and let ε > 0. Choose A ∈ B0 with A ⊆S U and µ1 (U ) < µ0 (A) + ε. Since A is compact, there exists n with n A ⊆ i=1 Ui . Hence, by (1c) and (1a), µ1 (U ) ≤ µ1 (

n [

Ui ) + ε ≤

i=1

n X

µ1 (Ui ) + ε ≤

i=1

∞ X

µ1 (Ui ) + ε,

i=1

and (1d) follows. Finally, multiplication from the left (resp. right) with an element x ∈ G is a homeomorphism of G onto itself. Therefore, (1e) follows from the translation invariance of µ0 . Part C: An outer measure.

For each E ⊆ G let

µ2 (E) = inf(µ1 (U ) | U open and E ⊆ U ). The function µ2 extends µ1 and has these properties: (2a) E1 ⊆ E2 =⇒ µ2 (E1 ) ≤ µ2 (E2 ).

368

Chapter 18. The Haar Measure

S∞ P∞ (2b) µ2 ( i=1 Ei ) ≤ i=1 µ2 (Ei ). (2c) µ2 (xE) = µ2 (Ex) = µ2 (E), x ∈ G. Both (2a) and (2c) are easy exercises. For (2b), let ε > 0. Choose Ui open such that Ei ⊆ Ui and µ1 (Ui ) ≤ µ2 (Ei ) + 2εi . By (1d) and (2a) µ2 (

∞ [

i=1

Ei ) ≤ µ1 (

∞ [

Ui ) ≤

i=1

∞ X

µ1 (Ui ) ≤

i=1

∞ X

µ2 (Ei ) + ε,

i=1

and (2b) follows. Part D: Measurable sets. Call a subset E of G measurable if for each A ⊆ G we have µ2 (A) = µ2 (A ∩ E) + µ2 (A r E). The following rules hold: (3a) E is measurable =⇒ G r E is measurable. (3b) E1 and E2 are measurable =⇒ E1 ∪ E2 is measurable. (3c) µ2 (E) = 0 =⇒ E is measurable. Sn (3d) P E1 , . . . , En are measurable and mutually disjoint =⇒ µ2 (A∩ i=1 Ei ) = n i=1 µ2 (A ∩ Ei ) for each A ⊆ G. S∞ and mutually disjoint =⇒ E = i=1 Ei (3e) E1 , E2 , E3 , . . . are measurable P∞ is measurable and µ2 (E) = i=1 µ2 (Ei ). (3f) Each B ∈ B0 is measurable. (3g) Each open set is measurable. (3h) E is measurable and x ∈ G =⇒ xE and Ex are measurable. Rule (3a) follows from the definition. To prove (3b), let A ⊆ G. By (2b) µ2 (A) ≤ µ2 (A ∩ (E1 ∪ E2 )) + µ2 (A r(E1 ∪ E2 )) = µ2 ((A ∩ E1 ) ∪ ((A r E1 ) ∩ E2 )) + µ2 ((A r E1 ) r E2 ) ≤ µ2 (A ∩ E1 ) + µ2 ((A r E1 ) ∩ E2 ) + µ2 ((A r E1 ) r E2 ) = µ2 (A ∩ E1 ) + µ2 (A r E1 ) = µ2 (A). Thus, these inequalities are in fact equalities. Consequently, E1 ∪ E2 is measurable. To prove (3c), let A ⊆ G. By (2a) and (2b), µ2 (A) ≤ µ2 (A ∩ E) + µ2 (A r E) ≤ µ2 (E)+µ2 (A) = µ2 (A). Hence, µ2 (A) = µ2 (A∩E)+µ2 (A r E). Thus, E is measurable. The proof of (3d) is done by induction. It is trivial for n = 1. Assume it holds for n − 1. Then n n n [ [ [ µ2 A ∩ Ei = µ2 A ∩ Ei ∩ En + µ2 A ∩ Ei r E n i=1

i=1

i=1

= µ2 (A ∩ En ) + µ2 A ∩

n−1 [

Ei

i=1

= µ2 (A ∩ En ) +

n−1 X i=1

µ2 (A ∩ Ei ).

18.2 Existence of the Haar Measure

For (3e), (3b) implies that Fn = Hence, for each A ⊆ G, (3d) implies

369

Sn

i=1

Ei is measurable, n = 1, 2, 3, . . . .

µ2 (A) = µ2 (A ∩ Fn ) + µ2 (A r Fn ) ≥ µ2 (A ∩ Fn ) + µ2 (A r E) =

n X

µ2 (A ∩ Ei ) + µ2 (A r E).

i=1

Therefore, by (2b), µ2 (A) ≥

∞ X

µ2 (A ∩ Ei ) + µ2 (A r E) ≥ µ2 (A ∩ E) + µ2 (A r E) ≥ µ2 (A).

i=1

It follows that E is measurable. Sn Pn P∞ By (3d), µ2 (E) ≥ µ2 ( i=1 Ei ) = i=1 µ2 (Ei ). Hence, by (2b), µ2 (E) = i=1 µ2 (Ei ). To prove (3f), let A ⊆ G and U be an open set that contains A. Then U ∩ B and U r B are open disjoint sets. By (2a) and (1b) µ2 (A ∩ B) + µ2 (A r B) ≤ µ2 (U ∩ B) + µ2 (U r B) = µ1 (U ). Take the infimum on all U to conclude that µ2 (A) ≤ µ2 (A∩B)+µ2 (A r B) ≤ µ2 (A). Hence, B is measurable. Now we prove (3g): Let U be an open set. Consider A ⊆ G and ε > 0. Since µ2 (U ) = µ1 (U ) (Part C), there is a B ∈ B0 with B ⊆ U and µ2 (U ) − µ2 (B) < ε. Since B and U r B are disjoint open sets, (1b) implies µ2 (U ) = µ2 (B) + µ2 (U r B). Hence, µ2 (U r B) < ε. By (3f), µ2 (A) = µ2 (A ∩ B) + µ2 (A r B). Therefore, µ2 (A) ≤ µ2 (A ∩ U ) + µ2 (A r U ) ≤ µ2 (A ∩ (U r B)) + µ2 (A ∩ B) + µ2 (A r B) ≤ µ2 (U r B) + µ2 (A) < ε + µ2 (A) It follows that µ2 (A) = µ2 (A ∩ U ) + µ2 (A r U ). Thus, U is measurable. ˆ Let Bˆ be the collection of all measurable sets. Part E: The Baire field B. By (3a), (3b) and (3e), Bˆ is a σ-algebra. Since Bˆ contains each open set, it contains each Borel set. Denote the restriction of µ2 to Bˆ by µ. By (3e), (3h), and (2c), µ is a σ-additive invariant measure. For each E ∈ Bˆ and every ε > 0 there is an open set U with E ⊆ U and µ2 (U r E) < ε. Apply this to G r E to conclude the existence of a closed set C ⊆ E such that µ(E r C) < ε. T Thus, µ is regular. In particular, each E ∈ Bˆ is contained in a ∞ Borel set F = i=1 Ui with Ui open, i = 1, 2, 3, . . . such that µ(F r E) = 0. ˆ µ) is Combining this with (3c), we conclude that µ is complete on Bˆ and (B, the completion of the Borel field B with respect to the restriction of µ to B. Thus, µ is the desired Haar measure of G.

370

Chapter 18. The Haar Measure

Proposition 18.2.2: Let π: G → H be an epimorphism of profinite groups and µG , µH the corresponding Haar measures. Then µH (B) = µG (π −1 (B)) for each measurable subset B of H. Proof: The map B 7→ π −1 (B) maps closed subgroups of H onto closed subgroups of G. In addition, it commutes with complements and unions. Hence, it maps B(H) into B(G). The function µG ◦ π −1 satisfies Condition (1) of Section 18.1 for B ∈ B(H). This is clear for (1a)-(1d) of Section 18.1. To prove (1e) of Section 18.1, consider B ∈ B(H) and ε > 0. Then G has a closed subset C with C ⊆ π −1 (B) and µG (π −1 (B) r C) < ε. Then π(C) is closed in H and π(C) ⊆ B. Thus, π −1 (π(C)) ⊆ π −1 (B) and µG (π −1 (B r π(C))) ≤ µG (π −1 (B) r C) < ε. Applying this result to H r B and taking complements, we get an open subset U of H with B ⊆ U and µG ◦ π −1 (U r B) < ε. By Proposition 18.1.3, the completion of (B(H), µG ◦π −1 ) coincides with ˆ (B(H), µH ). ˆ If It remains to prove that µH (B) = µG (π −1 (B)) for each B ∈ B(H). A is a µH -zero set, then A ⊆ A1 for some A1 ∈ B(H) with µH (A1 ) = 0. Then µG (π −1 (A)) ≤ µG (π −1 (A1 )) = µH (A1 ) = 0, so µG (π −1 (A)) = 0. An ˆ arbitrary B ∈ B(H) differs from a set in B(H) by a µH -zero set. Hence, −1 µH (B) = µG (π (B)) holds for B. Example 18.2.3: Finite quotients. Let π be an epimorphism of a profinite group G onto a finite group H. Then, by Example 18.1.4 and Proposition |B| for each subset B of H. 18.2.2, µG (π −1 (B)) = |H| Proposition 18.2.4: Let H be an open subgroup of a profinite group G. ˆ Then µH (B) = (G : H)µG (B) for each B ∈ B(H). ˆ is a Haar measure of H, so it Proof: The restriction of (G : H)µG to B(H) coincides with µH .

18.3 Independence Now that we have a unique normalized Haar measure µ on G, we may regard G as a probability space. Recall thatT a family {A Qi | i ∈ I} of measurable subsets of G is µ-independent if µ( i∈J Ai ) = i∈J µ(Ai ) for each finite subset J of I. LemmaS18.3.1: The hold Pnfollowing P for measurable subsets A1 , . . . , An of G: n k−1 (a) µ i=1 Ai ) = k=1 (−1) 1≤i1 0 for each i, Q∞ Q∞ ∞ and i=1 µ(Ai ) = 0. Then µ( i=n Ai ) = i=n µ(Ai ) = 0 for each n. So, µ(A0 ) = 0. Remark 18.3.6: (a) Chebyshev’s inequality [R´enyi, p. 391] sharpens (b) of Lemma 18.3.5 by allowing one to assume only that the sets A1 , A2 , . . . are pairwise µindependent. (b) The most frequently used cases of 18.3.5 are when PLemma ∞ (b1) µ(Ai ) is a positive constant, where i=1 µ(Ai ) obviouslyP diverges, or 1 (b2) µ(Ap ) = p1e where p ranges over all prime numbers; then pe diverges for e = 1 (Example 18.3.3) and converges for e ≥ 2. Sequences A1 , A2 , . . . with µ(Ai ) = 1 (resp. µ(Ai ) = 0), i = 1, 2, . . . are clearly µ-independent. The next lemma ties independence to the group structure on G: Lemma 18.3.7: Open subgroups H1 , . . . , Hn of a profinite group G are µindependent if and only if (2)

(G :

n \

Hi ) =

i=1

n Y

(G : Hi ).

i=1

Suppose H1 , . . . , Hn are µ-independent open subgroups of G. For each i let A¯i be a set of left cosets of Hi in G. Put Ai = {g ∈ G | gHi ∈ A¯i }. Then A1 , . . . , An are µ-independent and (3)

µ(

n \

i=1

Ai ) =

n Y

|A¯i | . (G : Hi ) i=1

374

Chapter 18. The Haar Measure

Proof: Suppose (2) holds and let I be a subsetT of {1, . . . , n}. WithTno loss m n assume I = {1, . . . , m} with m ≤ n. Let K = i=1 Hi and H = i=1 Hi . ) embeds the coset space G/K into The canonical map gK 7→ (gH1 , . . . , gHmQ m G/H1 × · · · × G/Hm . Hence, (G : K) ≤ i=1 (G : Hi ). Similarly, (K : H) ≤

n Y i=m+1

(K : K ∩ Hi ) ≤

n Y

(G : Hi ).

i=m+1

Combining these with the Qm(G : H) = (G : K)(K : H) and Tm(2), we conclude Qm equality (G : K) = i=1 (G : Hi ). Thus, (2) gives µ( i=1 Hi ) = i=1 µ(Hi ), as desired. Necessity of (2) for µ-independence is clear. Now suppose H1 , . . . , Hn are µ-independent. Put A = A1 ∩ · · · ∩ An ¯ then gH = aH for some and A¯ = {aH | a ∈ A}. If g ∈ G and gH ∈ A, ¯ a ∈ A. Hence, gHi ∈ Ai , so g ∈ Ai , i = 1, . . . , n and g ∈ A. It follows ¯ List the elements of A¯ as a1 H, . . . , am H. Then that S A = {gH | gH ∈ A}. m ¯ A = · i=1 ai H and µ(A) = |A|µ(H). Next define a map α: G/H → G/H1 × · · · × G/Hn by α(gH) = (gH1 , . . . , gHn ). By (2), α is bijective. In particular, α maps A¯ bijectively onto A¯1 × · · · × ¯ = |A¯1 | · · · |A¯n |. In addition, µ(H) = µ(H1 ) · · · µ(Hn ). Hence, A¯n , so |A| Qn Qn ¯ µ(A) = |A|µ(H) = i=1 |A¯i |µ(Hi ) = i=1 µ(Ai ). Thus, A1 , . . . , An are µ-independent and (3) holds. Example 18.3.8: Relatively prime indices. Let H1 , . . . , Hn be open subgroups of a profinite group G. Suppose (G : H1 ), . . . , (G : Hn ) are relatively primes in pairs. Then (2) holds. Hence, by Lemma 18.3.7, H1 , . . . , Hn are µ-independent. Part (b) of the following lemma generalizes Lemma 16.8.3(a). Q Lemma 18.3.9: Let G = i∈I Si be a direct product of finite non-Abelian simple groups Si and N a closed normal subgroup of G. For each i ∈ I let πi : G → Si be the projection on the ith factor. Then the following hold: = Sj if and only if Sj ≤ N . (a) πj (N )Q (b) N = i∈I0Q Si , where I0 = {i ∈ I | Si ≤ N }. Moreover, G = N × N 0 0 with N = i∈I r I0 Si . (c) Let α be an automorphism of G with α(N ) = N . Then α(N 0 ) = N 0 . (d) Suppose G is a normal subgroup of a profinite group F and N / F . Then N0 / F. Proof of (a): First suppose Sj ≤ N . Since πj is the identity map on Sj , we have πj (N ) = Sj . Now suppose Sj 6≤ N . Since Sj is simple, Sj ∩ N = 1. Hence, [n, s] = 1 for all n ∈ N and s ∈ Sj . Therefore, [πj (n), s] = 1. Since Sj is non-Abelian, its center is trivial. Consequently, πj (n) = 1.

18.3 Independence

375

Q Proof of (b): By definition, i∈I0 Si ≤ N . QConversely, let n ∈ N . If j ∈ I r I0 , then πj (n) = 1, by (a). Hence, n ∈ i∈I0 Si . Q Proof of (c): Si ≤ N if and only if α(Si ) ≤ N , so α(N 0 ) = Si 6≤N α(Si ) = Q 0 Si 6≤N Si = N . Proof of (d): Apply (c) to conjugation of G by elements of F . Qr Lemma 18.3.10: Let G = i=1 Si be a direct product of r finite simple nonAbelian groups Si . Then G has exactly r normal subgroups N with G/N simple. Proof: Let N be a normal subgroup of G such that G/N is simple. By Q Lemma Q 18.3.9, N = i∈I0 Si , where I0 is a subset of {1, 2, . . . , r}. Then Since G/N is simple, I r I0 = {j} for some j between G/N = i∈I r I0 Si . Q 1 and r. Thus, N = i6=j Si . Consequently, there are exactly r possibilities for N . Lemma 18.3.11: Let G be a profinite group and λ an ordinal number. For each α < λ let Nα be an open normal subgroup of G. SupposeTG/Nα is a simple non-Abelian group and Nα 6= Nα0 if α 6= α0 . Then G/ α 0. Assume x1 , . . . , xr is a transcendence base for K(x)/K. If r = n, take K 0 = K. Otherwise, compute a nonzero polynomial f ∈ K[X1 , . . . , Xr+1 ] of minimal degree with (1)

f (x1 , . . . , xr+1 ) =

X

i

r+1 ai xi11 · · · xr+1 = 0,

with ai ∈ K.

i∈I

Let q be the maximal power of p that divides all exponents of (1). Put 1/q K1 = K(ai | i ∈ I). Then x1 , . . . , xr+1 satisfy the following irreducible relation over K1 : X 1/q i /q ir+1 /q ai x11 · · · xr+1 = 0. i∈I

One of the variables, say x1 , appears in this relation in a monomial which is not a pth power. Hence, x1 is separably algebraic over K1 (x2 , . . . , xn ). Now use induction on n to compute a finite purely inseparable extension K 0 of K1 such that K 0 (x2 , . . . , xn )/K 0 is separable. Then K 0 (x)/K 0 is also separable. In addition, find a basis w100 , . . . , wl0000 for K 0 /K1 .

426

Chapter 19. Effective Field Theory and Algebraic Geometry 1/q

Now order the ai ’s in a sequence a1 , . . . , ar . Examine which of the aj 1/q

1/q

K-linearly depends on a1 , . . . , aj−1 to find a basis w10 , . . . , wl00 for K1 /K. Then {wi0 wj00 | 1 ≤ i ≤ l0 , 1 ≤ j ≤ l00 } is a basis for K 0 /K which can be w ordered as w1 , . . . , wl . Finally, replace wj , if necessary, by w1j to assure w1 = 1. Lemma 19.7.2: Let (x1 , . . . , xn , z) be a presented (n + 1)-tuple over K with z separably algebraic over K(x). Then we can effectively find a polynomial g ∈ K[X] with these properties: (2) g(x) 6= 0, and K[x, g(x)−1 ] is integrally closed and presented in its function field. (3) The ring K[x, g(x)−1 , z] is a presented cover of the ring K[x, g(x)−1 ] with z a primitive element. Proof: Suppose we have found g ∈ K[X] satisfying (2). Present irr(z, K(x)) as a quotient of a polynomial in K[x, Z] by a polynomial in K[x]. Then multiply g by the product of the denominator and the discriminant of irr(z, K(x)). By the remarks leading up to Lemma 19.6.1, the ring K[x, g(x)−1 , z] is presented over K. Also (Definition 6.1.3), z is a primitive element for the cover K[x, g(x)−1 , z]/K[x, g(x)−1 ]. The argument that finds g satisfying (2) breaks into two cases. Case A: K(x)/K is separable. Reorder x1 , . . . , xn according to Lemma 19.2.4 to assume xi is separable over K(x1 , . . . , xi−1 ), i = 1, . . . , n. Use induction on n to effectively find a polynomial g1 ∈ K[X1 , . . . , Xn−1 ] with g1 (x) 6= 0 and K[x1 , . . . , xn−1 , g1 (x)−1 ] integrally closed. If xn is transcendental over K(x1 , . . . , xn−1 ), then K[x1 , . . . , xn , g1 (x)−1 ] is also integrally closed [Zariski-Samuel2, p. 85 or p. 126]. Otherwise, xn is separable algebraic over K(x1 , . . . , xn−1 ) and from the above, with z = xn , we can effectively find a multiple g ∈ K[X1 , . . . , Xn−1 ] of g1 such that K[x1 , . . . , xn , g(x)−1 ] is a ring cover of K[x1 , . . . , xn−1 , g(x)−1 ]. In particular K[x, g(x)−1 ] is integrally closed. Case B: K(x)/K is general. Assume without loss char(K) 6= 0. Apply Lemma 19.7.1 to find a finite purely inseparable extension K 0 of K with K 0 (x)/K 0 separable. Lemma 19.7.1 gives a linear basis w1 , . . . , wl with w1 = 1 for K 0 /K and a power q of char(K) with wq ∈ K for each w ∈ K 0 . The wi ’s need not be linearly independent over K(x). Find among them a basis, say w1 , . . . , wk , for K 0 (x)/K(x) and compute a polynomial a ∈ K[X] with a(x) 6= 0 and wi ∈ K[x, a(x)−1 , w1 , . . . , wk ], i = 1, . . . , l. Note: K(x) ∩ K 0 [x] ⊆ K[x, a(x)−1 ]. Use Case A to find h ∈ K 0 [X] with h(x) 6= 0 and K 0 [x, h(x)−1 ] integrally closed. Now let g = ahq . Then K(x) ∩ K 0 [x, h(x)−1 ] is contained in the ring K[x, g(x)−1 ], which is therefore integrally closed. Combine Lemma 19.7.2 with the stratification lemma:

Exercises

427

Proposition 19.7.3: There is an effective procedure for producing a basic normal stratification of a given constructible set.

Exercises 1. (a) Use the inequality pn+1 ≤ (p1 p2 · · · pn ) − 1 (Euclid) to show for each n n ∈ N that p1 p2 · · · pn ≤ 22 , where p1 < p2 < · · · is the sequence of primes. (b) Use (a) to show for each r ∈ N that {pj11 · · · pjnn | n ∈ N, 1 ≤ j1 , . . . , jn ≤ r} is a primitive recursive subset of N. 2. The racetrack problem. Consider the real points C1 (R) in the ellipse X 2 + 2Y 2 = 1 as the inside rail of a racetrack. Let C2 (R) be the locus of points (U, V ) traced out by the points on the outside of the ellipse that are 1 unit distance from the ellipse along a perpendicular to the ellipse. Let W ⊆ A4 with coordinates (X, Y, U, V ) be given by the equations f1 = X 2 + 2Y 2 − 1 = 0, f2 = XV − 2Y U + XY = 0, f3 = (U − X)2 + (V − Y )2 − 1 = 0. Let π: A4 → A2 with π(X, Y, U, V ) = (U, V ) be the projection onto the last coordinates. (a) Show that C2 (R) = π(W )(R). (b) Show that π(W ) = V (h), where h is a polynomial of degree 8 (= deg(f1 ) deg(f2 ) deg(f3 )) that generates the ideal Q[U, V ] ∩ I(f1 , f2 , f3 ) of Q[U, V ]: (b1) First eliminate X to show that Y, U, V satisfy: g1 = (1 − 2Y 2 )(V + Y )2 − 4(Y U )2 = 0, g2 = (U V − Y U )2 + (V − Y )2 − 1 (V + Y )2 = 0. Notice that both g1 and g2 are polynomials of degree 4 in Y with coefficients in Z[U, V ] and with leading coefficients −2 and 1, respectively. (b2) Now apply the proof of Theorem 9.2.1 (expression (1)) to g1 and g2 to eliminate Y . Alternatively, apply the resultant to g1 and g2 . 3. Prove that the number of monomials in X1 , . . . , Xn of degree at most d is n+d d . Hint: Given integers i1 , . . . , in between 0 and d define j1 = i1 + 1, j2 = i1 + i2 + 2, . . ., jn = i1 + i2 + · · · + in + n. Prove that the map (i1 , . . . , in ) 7→ (j1 , . . . , jn ) defines a bijective map between the set of all monomials in X1 , . . . , Xn of degree at most d and the set of all subsets with n elements of the set {1, 2, . . . , n + d}.

428

Chapter 19. Effective Field Theory and Algebraic Geometry

Notes This Chapter is an elaboration of [Fried-Haran-Jarden, §2]. Most of it is due to Kronecker. The common feature is the use of the “method of indeterminates.” Our sources include [Waerden3] for the proofs of Lemmas 19.1.3, 19.2.2, and 19.3.2; [Waerden1] for the proof of Lemma 19.5.3; a model for the proof of Lemma 19.2.4 from [Lang4]; [Zariski-Samuel2] for the proofs of Lemmas 19.3.1 and 19.5.1; and an elaboration of [Waerden2] for the proof of Proposition 19.5.6. Kronecker’s algorithm is no longer the state of the art for a deterministic procedure for factoring polynomials with coefficients in Z (Lemma 19.1.3). Indeed, [Lenstra-Lenstra-Lov´ asz] gives a bound on the time of factorization which is a polynomial in the height of the polynomial and [Lenstra] extends this to polynomials in several variables. No one has yet tested the effect of these procedures on such practical applications as the algorithm of Proposition 19.5.6. But this would be a key ingredient in producing a computer program that actually accomplishes the Galois stratification procedure and its corollaries of Chapters 30 and 31. L. v. d. Dries and K. Schmidt [v.d.Dries-Schmidt] use nonstandard approach and the notion of faithful flatness to prove a stronger result than that achieved in Remark 19.5.8: Given positive integers d and n, there is a positive integer e such that for every field K and all polynomials f1 , . . . , fm ∈ K[X1 , . . . , Xn ] of degree at most d, the ideal I generated by f1 , . . . , fm is prime if and only if for all g, h ∈ K[X1 , . . . , Xn ] of degree at most e, the relation gh ∈ I implies g ∈ I or h ∈ I. The idea of checking absolute irreducibility over the algebraic closure of a field K first and then using elimination of quantifiers to descend to K was communicated to the authors by v. d. Dries.

Chapter 20. The Elementary Theory of e-Free PAC Fields This chapter presents one of the highlights of this book, the study of the elementary theory of e-free PAC fields. We apply the elementary equivalence theorem for arbitrary PAC fields (Theorem 20.3.3) to the theory of perfect e-free PAC fields containing a fixed countable base field K. If K is finite and e = 1 or K is Hilbertian and e ≥ 1, then this theory coincides with ˜ the theory of all sentences with coefficients in K that are true in K(σ), for e almost all σ ∈ G(K) (Section 20.5). In particular, if K is explicitly given with elimination theory, then this theory is recursively decidable. In the special case where K is a global field and e = 1 we prove a transfer theorem ˜ (Theorem 20.9.3): A sentence θ of L(ring, OK ) is true among the fields K(σ) with probability equal to the probability that θ is true among the residue fields of K. Finally, we prove that the elementary theory of finite fields is recursively decidable (Theorem 20.10.6).

20.1 ℵ1 -Saturated PAC Fields We start with a result that strengthens the PAC property of a field which is also ℵ1 -saturated (Section 7.7): Lemma 20.1.1: Let K be an ℵ1 -saturated PAC field and R an integral domain which is countably generated over K. (a) Suppose R is contained in a field F regular over K. Then there exists a K-homomorphism ϕ: R → K. (b) Suppose in addition, char(K) = p > 0. Let S be a subset of R, pindependent over F p with |S| ≤ min(ℵ0 , [K : K p ]). Then ϕ can be chosen such that ϕ(S) is p-independent over K p . Proof: By assumption, R = K[x1 , x2 , x3 , . . .]. Denote the ideal of all polynomials in K[X1 , . . . , Xn ] that vanish at (x1 , . . . , xn ) by In . Let fn,1 , . . . , fn,rn be a system of generators for In . Then I1 ⊆ I2 ⊆ I3 · · ·. Since K[X1 , . . . , Xn ]/In ∼ =K K[x1 , . . . , xn ] and K(x1 , . . . , xn ), as a subfield of F , is regular over K, V (In ) is a variety defined over K (Corollary 10.2.2(a)). Since K is PAC, there exist elements a1 , . . . , an ∈ K with fnj (a1 , . . . , an ) = 0, j = 1, . . . , rn . Each fij with 1 ≤ i ≤ n and 1 ≤ j ≤ ri is a linear combination of fn1 , . . . , fnrn . Hence, (1)

ri n ^ ^ i=1 j=1

fij (a1 , . . . , ai ) = 0.

430

Chapter 20. The Elementary Theory of e-Free PAC Fields

The saturation property of K gives b1 , b2 , b3 , . . . in K with fnj (b1 , . . . , bn ) = 0 for all n and j. The map xi 7→ bi , i = 1, 2, 3, . . . extends to a Khomomorphism ϕ: R → K. Now assume char(F ) = p > 0 and S is as in (b). The essential case occurs when [K : K p ] ≥ ℵ0 and S = {s1 , s2 , s3 , . . .} is infinite. Write sn = gn (x1 , . . . , xkn ) with gn ∈ K[X1 , . . . , Xkn ]. Without loss assume that n ≤ kn ≤ kn+1 for all n. For each n Proposition 11.4.1 gives a1 , . . . , akn in K such that, in addition to (1), gi (a1 , . . . , aki ), i = 1, . . . , n, are p-independent over K p . Then proceed as before.

20.2 The Elementary Equivalence Theorem of ℵ1 -Saturated PAC Fields Conditions for the elementary equivalence of two PAC fields E and F are central to the model theory of PAC fields. If E and F are elementarily equivalent, then E and F have the same characteristic and thus the same prime field K. Since a polynomial f ∈ K[X] has a zero in E if and only if ˜ ∼ ˜ (Lemma 20.6.3). Assume it has a zero in F , it follows that E ∩ K =F ∩K ˜ ˜ therefore that L = E ∩ K also equals F ∩ K. More generally (Section 23.4), if L0 is a finite Galois extension of L and E 0 is a finite Galois extension of E containing L0 , then there exists a finite Galois extension F 0 of F , containing L0 , and a commutative diagram (1)

Gal(F 0 /F ) Gal(E 0 /E) o II II vv II vv v I v res II v res $ zvv 0 Gal(L /L) ϕ

with ϕ an isomorphism. If E and F are countable, then an inverse limit gives a similar diagram with absolute Galois groups replacing corresponding relative Galois groups. The next lemma shows this last condition to be essentially sufficient for the elementary equivalence of E and F . The basic element is Lemma 20.2.2 whose main ingredients, the field crossing argument and the use of existence of rational points on varieties, have already appeared in the proof of the Chebotarev density theorem and will appear again in Proposition 24.1.1. Let L, M be fields and Φ: Ls → Ms be an embedding with Φ(L) ⊆ M . Then Φ induces a homomorphism ϕ: Gal(M ) → Gal(L) with ϕ(σ)x = Φ−1 (σΦ(x)) for all σ ∈ Gal(M ) and x ∈ Ls . If Φ is an isomorphism with Φ(L) = M , then ϕ is an isomorphism. Lemma 20.2.1: Let E/L be a regular extension of fields. Then Es /Ls is regular. ˜ over L, so ELs is linearly Proof: By assumption, E is linearly disjoint from L ˜ ˜ disjoint from L over Ls . Since Es /ELs is separable and E L/EL s is purely

20.2 The Elementary Equivalence Theorem of ℵ1 -Saturated PAC Fields

431

˜ over ELs . It follows from the inseparable, Es is linearly disjoint from E L ˜ over tower property of linear disjointness that Es is linearly disjoint from L Ls . In other words, Es /Ls is regular. Lemma 20.2.2 (Embedding Lemma [Jarden-Kiehne, p. 279]): Let E/L and F/M be separable field extensions satisfying: E is countable and F is PAC and ℵ1 -saturated. In addition, suppose there are an isomorphism Φ0 : Ls → Ms with Φ0 (L) = M and a commutative diagram (2)

Gal(E) o

ϕ

res

Gal(L) o

Gal(F ) res

ϕ0

Gal(M )

where ϕ0 is the isomorphism induced by Φ0 and ϕ is a homomorphism. If char(L) = p > 0, add the assumption that [E : E p ] ≤ [F : F p ]. Then there exists an extension of Φ0 to an embedding Φ: Es → Fs which induces ϕ with F/Φ(E) separable. ˜ is a separable extension, E ∩ L ˜ = E ∩Ls . Similarly, F ∩ Proof: Since E ∩ L/L ˜ and resM (Gal(F )) = ˜ M = F ∩ Ms . In addition, resLs (Gal(E)) = Gal(E ∩ L) s ˜ ˜ ˜ . Therefore, replace Gal(F ∩ M ). The hypotheses thus give Φ0 (E∩ L) = F ∩ M ˜ and F ∩ M ˜ to assume that E/L and F/M L and M , respectively, by E ∩ L are regular extensions. By Lemma 20.2.1, Es /Ls and Fs /Ms are regular. Without loss assume L = M , Φ0 and ϕ0 are the identity isomorphisms, and E is algebraically independent from F over L. It follows from Corollary 2.6.8 that (3) EF is a separable (and regular) extension of both E and F . Also, by Lemma 2.6.7, Es is linearly disjoint from Fs over Ls . We divide the rest of the proof into three parts to separate out the use of the PAC property. Part A: The field crossing argument. From (2), ϕ(σ)x = σx for each σ ∈ Gal(F ) and x ∈ Ls . Thus, by Lemma 2.5.5, each σ ∈ Gal(F ) extends uniquely to a σ ˜ ∈ Gal(Es Fs /EF ) with σ ˜x =

ϕ(σ)x if x ∈ Es σx if x ∈ Fs .

˜ = σ. The map σ 7→ σ ˜ embeds Gal(F ) into Gal(Es Fs /EF ) and resFs Es /Fs σ Let D be the fixed field of the image of Gal(F ). Then res: Gal(Es Fs /D) → Gal(F ) is an isomorphism, so D ∩ Fs = F and DFs = Es Fs . Since D/EF is separable, (3) implies that D is a separable extension of both E and F . Hence, by Lemma 2.6.4, (4) D is a regular extension of F .

Chapter 20. The Elementary Theory of e-Free PAC Fields

432

Part B: Use of the PAC property. Note that Es Fs is an algebraic extension choose yi ∈ Fs of D. Hence, Es ⊆ Es Fs = D[Fs ] = Fs [D]. For each x ∈ Es P and di ∈ D, where i ranges over a finite set Ix such that x = i∈Ix yi di . Let D0 = E ∪ {di | x ∈ Es , i ∈ Ix }. Then, Es ⊆ Fs [D0 ] and since E is countable, so is D0 . We illustrate the relations among all rings and fields mentioned so far in the following diagram. (5) E

L

Es

| |||

Ls

Fs [D0 ] GGG r r GG r r

E s Fs yy y y F [D0 ] D E KK EEE KK E KK KK Fs KK KK KK wwwww F

If p = char(L) > 0, let S be a p-basis of E over E p . Since F (D0 )/E is separable, S is p-independent over F (D0 )p . Also, |S| ≤ [E : E p ] ≤ [F : F p ]. Applying Lemma 20.1.1 using the hypothesis and (4), we conclude that there exists an F -homomorphism Ψ: F [D0 ] → F with Ψ(S) p-independent over F p (if p > 0). Since E is a field, Ψ is injective on E, so Ψ(S) is a p-basis of Ψ(E) over Ψ(E)p . Therefore, F/Ψ(E) is separable. Since D is linearly ˜ Fs [D0 ] → Fs disjoint from Fs over F , Ψ extends to an Fs -homomorphism Ψ: (Lemma 2.5.5). Part C: Conclusion of the proof. (6)

Check that

˜ σ x) = σ Ψ(x) ˜ Ψ(˜ for each σ ∈ Gal(F )

and each x ∈ Fs ∪ D0 . Thus, (6) holds for each x ∈ Fs [D0 ]. Since Es ⊆ ˜ to Es by Φ. Fs [D0 ], (6) holds for each x ∈ Es . Denote the restriction of Ψ Then Φ is an Ls -embedding of Es into Fs which satisfies the conclusion of the Lemma. Recall that if K is a field, then L(ring, K) denotes the first order language of the theory of rings augmented with constant symbols for the elements of K (Example 7.3.1). We apply Skolem-L¨ owenheim (Proposition 7.4.2) and the Cantor back and forth argument to improve Lemma 20.2.2: Lemma 20.2.3: Let E/L and F/M be separable field extensions with both L and M countable and containing a given field K. Assume E and F have the same imperfect degree, they are PAC and ℵ1 -saturated, and there exists a K-isomorphism Φ0 : Ls → Ms with Φ0 (L) = M . Assume in addition that (2) is a commutative diagram with ϕ an isomorphism. Then E is K-elementarily equivalent to F . Proof: Skolem-L¨ owenheim gives a countable K-elementary subfield M1 of F that contains M . Since M1 /M is separable, we may apply Lemma 20.2.2

20.3 Elementary Equivalence of PAC Fields

433

to the diagram Gal(E)

res◦ϕ−1

/ Gal(M1 )

res

Gal(L)

res

ϕ−1 0

/ Gal(M )

to conclude that there is an extension of Φ−1 0 to an embedding Ψ1 : M1,s → Es with these properties: E/Ψ1 (M1 ) is separable and the diagram Gal(E)

res◦ϕ−1

/ Gal(M1 )

res

Gal(M10 )

res

ψ1

/ Gal(M1 )

is commutative, where M10 = Ψ1 (M1 ) and ψ1 is the isomorphism induced by Ψ1 . Now reverse the roles of E and F to find a countable K-elementary subfield L1 of E that contains M10 and an embedding Φ1 : L1,s → Fs that extends Ψ−1 1 appropriately. Proceed by induction to construct two towers of countable fields L ⊆ L1 ⊆ L2 ⊆ · · · ⊆ E

and M ⊆ M1 ⊆ M2 ⊆ · · · ⊆ F

and embeddings Φi : Li → Mi+1 , and Ψi : Mi → Li satisfying: Li (resp. Mi ) is a K-elementary subfieldSof E (resp. F ), theSmap Φi extends Ψ−1 i , and Ψi ∞ ∞ extends Φ−1 i−1 . Let L∞ = i=1 Li and M∞ = i=1 Mi . By Lemma 7.4.1(b), L∞ and M∞ are K-elementary subfields of E and F , respectively. Moreover, the Φi combine to give a K-isomorphism Φ∞ : L∞ → M∞ . Consequently, E ≡K F . Remark 20.2.4: Suppose E and F have the same cardinality, say m and they are m+ -saturated. Then with m+ -saturated replacing ℵ1 -saturated in Lemma 20.2.3, we may conclude that E ∼ =K F . The introduction of the m+ saturated concept in this approach usually forces one to use the continuum hypothesis 2ℵ0 = ℵ1 . Of course, a general principle of set theory asserts that arithmetical theorems proved using the continuum hypothesis hold even without assuming it. Our exposition achieves the same result directly.

20.3 Elementary Equivalence of PAC Fields We wish to remove the restriction of saturation put on the fields E and F from Lemma 20.2.2. This requires, however, a strengthening of (2) of Section 20.2 to an ultraproduct statement:

Chapter 20. The Elementary Theory of e-Free PAC Fields

434

Lemma 20.3.1: Let {Ei | i ∈ I} and {Fi | i ∈ I} be families of fields. For ) → Gal(Ei ) be each i ∈ I let ϕi : Gal(FiQ Q an isomorphism. Q Let D be an Fi /D, and ϕ∗ = ϕi /D. Then: ultrafilter of I. Put E ∗ = QEi /D, F ∗ = Q Gal(Fi )/D → Gal(Ei )/D induces an isomor(a) The isomorphism ϕ∗ : phism ϕ: Gal(F ∗ ) → Gal(E ∗ ). (b) Suppose in addition, there exists a field K which is contained in all Ei and Fi . For each i, let Li be a finite Galois extension of K such that the diagram (1)

Gal(Fi ) Gal(Ei ) o FF FF xx x F x res FFF xxres # {xx Gal(Li /K) ϕi

is commutative. Assume {i ∈ I | L ⊆ Li } ∈ D for each finite extension L of K. Then the diagram (2)

Gal(E ∗ ) o Gal(F ∗ ) CC CC {{ CC {{ { res CC { res ! }{{ Gal(K) ϕ

is commutative. Proof of (a): of F ∗ of degree n. Then there Q Let N be a Galois extension ∗ exists y ∈ Fi,s /D such that N = F (y). Choose a representative {yi | i ∈ I} of y. Then there exists D ∈ D such that N Qi = Fi (yi ) is a Galois extension of Fi of degree n for all i ∈ D, and N = Ni /D. Denote the fixed field the isomorphism of Gal(Ni /Fi ) in Ei,s of ϕi (Gal(Ni )) by Mi . Let ϕN,i be Q Mi /D is a Galois extension of onto Gal(Mi /EQ i ) induced by ϕi . Then M = E ∗ and ϕN = ϕN,i /D is an isomorphism of Gal(N/F ∗ ) onto Gal(M/E ∗ ) which is induced by ϕ∗ . If N 0 is a finite Galois extension of F ∗ that contains N , then the extension M 0 of E ∗ corresponding to N 0 contains M and ϕN 0 induces ϕN on the group Gal(N/F ∗ ). Thus, the inverse limit of the ϕN ’s defines an isomorphism ϕ: Gal(F ∗ ) → Gal(E ∗ ) which is induced by ϕ∗ . Proof of (b): Consider y ∈ Ks . Let L be a finite Galois extension of K which contains y. By assumption, Q D = {i ∈ I | L ⊆ Li } ∈ D. For each Ni /D. For each i ∈ D and each σi ∈ i ∈ I let Ni = LFi and N = Gal(Ni /Fi ) the commutativity of (1) implies ϕN,i (σi )(y) = σi (y). Hence, for each σ ∈ Gal(N/F ∗ ), ϕN (σ)(y) = σ(y). Therefore, ϕ(σ)(y) = σ(y) for each σ ∈ Gal(F ∗ ). Thus, (2) is commutative. In the special case where Ei = E, Fi = F and ϕi = ϕ0 for each i ∈ I, E and F are canonically embedded in E ∗ and F ∗ , respectively, and ϕ∗ induces ϕ0 on Gal(F ):

20.3 Elementary Equivalence of PAC Fields

435

Corollary 20.3.2: Let E and F be fields, ϕ: Gal(F ) → Gal(E) an isomorphism, and D an ultrafilter of I. Put E ∗ = E I /D and F ∗ = F I /D. Then there exists a commutative diagram (3)

Gal(E ∗ ) o

ϕ∗

res

Gal(E) o

Gal(F ∗ ) res

ϕ

Gal(F )

where ϕ∗ is an isomorphism. Theorem 20.3.3 (Elementarily Equivalence Theorem): Let E/L and F/M be separable field extensions with both L and M containing a field K. Suppose E and F are PAC fields having the same imperfect degree. In addition, suppose there exist a K-isomorphism Φ0 : Ls → Ms with Φ0 (L) = M and a commutative diagram (4)

Gal(E) o

ϕ

res

Gal(L) o

Gal(F ) res

ϕ0

Gal(M ) ,

where ϕ is an isomorphism and ϕ0 is the isomorphism induced by Φ0 . Then E is K-elementarily equivalent to F . Proof: Assume first L is countable. Choose a countable set I and a nonprincipal ultrafilter D of I. Then, in the notation of Corollary 20.3.2, combine the diagrams (3) and (4) to get a commutative diagram (5)

Gal(E ∗ ) o

ϕ∗

res

Gal(L) o

Gal(F ∗ ) res

ϕ0

Gal(M )

where ϕ∗ is an isomorphism. The fields E ∗ and F ∗ are PAC (Corollary 11.3.3) and ℵ1 -saturated (Lemma 7.7.4). By Lemma 20.2.3, E ∗ ≡K F ∗ . Therefore, E ≡K F . For the general case, let θ be a sentence of L(ring, K) which is true in E. There are only finitely many elements of K, say x1 , . . . , xn , that occur in θ. Let K0 be a countable subfield of K that contains x1 , . . . , xn . By Skolem-L¨owenheim (Proposition 7.4.2), L contains an elementary countable subfield L0 that contains K0 . Let M0 = Φ0 (L0 ). Then L/L0 and M/M0

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Chapter 20. The Elementary Theory of e-Free PAC Fields

are separable extensions and θ is a sentence of L(ring, K0 ). Also, there is a commutative diagram Gal(L) o

ϕ0

Gal(L0 ) o

ϕ0

Gal(M ) Gal(M0 )

where both horizontal arrows are the isomorphisms induced by Φ0 . The first part of the proof gives E ≡K0 F . Hence, θ is true in F . A special case of Theorem 20.3.3 is useful in investigating model completeness of PAC fields: Corollary 20.3.4: Let F/K be a separable extension of PAC fields such that F and K have the same imperfect degree. Suppose res: Gal(F ) → Gal(K) is an isomorphism. Then F is an elementary extension of K. The elementary equivalence theorem for separably closed fields is another special case of Theorem 20.3.3. Corollary 20.3.5 ([Ershov1, Prop.]): Let E and F be separably closed fields of characteristic p > 0 with the same imperfect degree. Then E and F are elementarily equivalent. ˜ p in Theorem 20.3.3. Proof: Set L = M = F

20.4 On e-Free PAC Fields In Theorem 20.3.3 we may assume without loss that E/L and F/M are regular extensions (e.g. as in the proof of Lemma 20.2.2). It is the relation between the groups Gal(E) (resp. Gal(F )) and Gal(L) (resp. Gal(M )) via restriction that complicates applications. When Gal(E) and Gal(F ) are isoutz’ lemma morphic to the free profinite group, Fˆe , on e generators, Gasch¨ (Lemma 17.7.2) comes to our aid: Proposition 20.4.1: Let E and F be e-free PAC fields with the same imper˜ ∩ E and F/K ˜ ∩F fect degree and with a common subfield K. Suppose E/K ∼ ˜ ˜ are separable extensions and K ∩ E =K K ∩ F . Then E ≡K F . ˜ ∩ E. By Proposition 17.7.3, there exists an isomorphism Proof: Put L = K ϕ: Gal(F ) → Gal(E) satisfying resEs /Ls ◦ ϕ = resFs /Ls . Hence, by Theorem 20.3.3, E ≡K F . ˜ ∩ E is also a perfect field and therefore E/K ˜ ∩E If E is perfect, then K is a separable extension. This simplifies Proposition 20.4.1: Corollary 20.4.2: Let K be a subfield of perfect e-free PAC fields E and ˜ ∩E ∼ ˜ ∩ F , then E ≡K F . F . If K =K K

20.4 On e-Free PAC Fields

437

Corollary 20.4.3: Let E and F be e-free PAC fields with the same imperfect degree. If F is a regular extension of E, then F is an elementary extension of E. We axiomatize the concept of e-free PAC fields for model-theoretic applications: Proposition 20.4.4: Let K be a field and e a positive integer. Then there exists a set Ax(K, e) of axioms in the language L(ring, K) such that a field extension F of K satisfies the axioms if and only if it is perfect, PAC and e-free. The axioms are sentences that interpret the field axioms, perfectness axioms, [p 6= 0] ∨ (∀X)(∃Y )[Y p = X], as p ranges over primes, the positive diagram of K (Example 7.3.1) and the following axioms: (a) PAC axioms: Every absolutely irreducible polynomial f (X, Y ) of degree d has a zero, d = 1, 2, . . . . (b) e-free axioms: The finite groups which appear as Galois groups over F are exactly the groups of rank bounded by e. Proof: Section 11.3 translates the PAC axioms into elementary statements. Thus, it suffices to translate the e-free axioms into elementary statements. For this, consider a polynomial f (X) = X n + u1 X n−1 + · · · + un with indeterminate coefficients u1 , . . . , un and a subgroup G of Sn which is given by its action on {1, 2, . . . , n}. Suppose the following assertion is an elementary statement on u1 , . . . , un : “The polynomial f is irreducible, separable, normal and has G as a Galois group.” Then consider all subgroups G1 , . . . , Gr of Sn which are generated by e elements. Restate axiom (b): “For each n and for every irreducible, separable, and normal polynomial f of degree ≤ n, the Galois group of f is isomorphic to one of the groups G1 , . . . , Gr ; and for each group G of order ≤ n and of rank ≤ e, there exists an irreducible, separable, normal polynomial of degree at most n with Galois group isomorphic to G .” The normality condition on f means that a root z of f (X) (in the algebraic closure), gives all other roots as polynomials in z of degree at most n − 1 with coefficients in F . To eliminate the reference to F˜ , use congruences modulo f (X) as follows: There exist polynomials p1 (Z) = Z, p2 (Z), . . . , pn (Z) of degree at most n − 1 with (1)

f (X) ≡

n Y

(X − pi (Z)) mod f (Z).

i=1

Of course, (1) is actually n congruence conditions on the coefficients of the powers of X on both sides. For example, equating the free coefficients on both sides gives the condition un ≡ (−1)n p1 (Z) · · · pn (Z) mod f (Z),

Chapter 20. The Elementary Theory of e-Free PAC Fields

438

which is equivalent to the existence of a polynomial g of degree at most n(n − 1) with un = (−1)n p1 (Z) · · · pn (Z) + g(Z)f (Z). Thus, the normality condition on f is elementary. The condition “Gal(f, F ) is isomorphic to G as a permutation group” may be shown to be elementary by considering the action on the roots of f . Then eliminate the reference to F˜ as before. Indeed, suppose f is monic, irreducible, separable, normal, and p1 (z) = z = z1 , p2 (z) = z2 , . . ., pn (z) = zn are the roots of f with pi ∈ F [X] a polynomial of degree at most |Gal(f, F )| − 1. Suppose σ ∈ Gal(f, F ). Then, for each i, σzi = pi (σz). Hence, zσ(i) = pi (zσ(1) ). Hence, pσ(i) (z) = pi (zσ(1) ). Conversely, suppose σ ∈ Sn satisfies pσ(i) (z) = pi (zσ(1) ) for i = 1, . . . , n. Let τ be the unique element of Gal(f, F ) with τ (1) = σ(1). Then pσ(i) (z) = pi (zσ(1) ) = pi (zτ (1) ) = pτ (i) (z). Hence, σ(i) = τ (i) for each i. Therefore, σ = τ ∈ Gal(f, F ). Consequently, “Gal(f, F ) ∼ = G00 is equivalent to n ^ ^ σ∈G i=1

[pσ(i) (z) = pi (zσ(1) )] ∧

^

n _

[pσ(i) (z) 6= pi (zσ(1) )].

σ∈Sn rG i=1

Remark 20.4.5: (a) If, in an application, K is the quotient field of a distinguished subring R, we may replace the positive diagram of K by the positive diagram of R. (b) When desired, axioms indicating that the imperfect exponent of F is m (0 ≤ m ≤ ∞) may replace the perfect axioms. (c) If K is presented with elimination theory, then Ax(K, e) can be effectively presented. (d) Let K be a field and G a group of order n. Then “G occurs as a Galois group over K” is an elementary statement on K. Indeed, the proof of Proposition 20.4.4 presents the equivalent statement “there is a monic Galois polynomial f ∈ K[X] with Gal(f, K) ∼ = G” as an elementary statement. Proposition 20.4.6 ([Klingen1]): Let K and L be elementarily equivalent fields. Suppose Gal(K) is a small profinite group. Then Gal(K) ∼ = Gal(L). Proof: By Remark 20.4.5, a finite group G occurs as a Galois group over K if and only if G occurs as a Galois group over L. Thus, Gal(K) and Gal(L) have the same finite quotients. It follows from Proposition 16.10.7 that Gal(K) ∼ = Gal(L).

20.5 The Elementary Theory of Perfect e-Free PAC Fields We interpret the elementary theory of perfect e-free PAC fields that contain a field K in the following cases: (1a) K is finite and e = 1.

20.5 The Elementary Theory of Perfect e-Free PAC Fields

439

(1b) K is countable and Hilbertian, and e ≥ 1. In each of these cases (K, e) is called a Hilbertian pair. ˜ For σ = (σ1 , . . . , σe ) in Gal(K)e , let K(σ) = Ks (σ)ins be the maximal ˜ of the unique purely inseparable extension of Ks (σ). It is the fixed field in K ˜ ˜ have the extension of σ to automorphisms of K. The fields Ks (σ) and K(σ) ˜ ˜ is a same absolute Galois group. If Ks (σ) is PAC, so is K(σ). But K(σ) perfect field. Apply Corollary 18.5.9 and Proposition 18.6.4 to case (1a) and Theorems 18.5.6 and 18.6.1 to case (1b): ˜ Theorem 20.5.1: Suppose (K, e) is a Hilbertian pair. Then K(σ) is a e perfect e-free PAC field for almost all σ ∈ Gal(K) . Recall the regular ultrafilters of Section 7.6 when the index set S is Gal(K)e and “small sets” are the subsets of Gal(K)e of measure zero. In particular, a regular ultrafilter of Gal(K)e contains all subsets of Gal(K)e of measure 1. We will compare an arbitrary e-free PAC field with a regular ˜ ultraproduct of the fields K(σ). ˜ Denote the theory of all sentences of L(ring, K) which are true in K(σ) e for almost all σ ∈ Gal(K) by Almost(K, e). Lemma 20.5.2: Let (K, e) be a Hilbertian pair. (a) A field F is a model of Almost(K, e) if and only if it is K-elementarily ˜ equivalent to a regular ultraproduct of the fields K(σ). ˜ (b) Every regular ultraproduct of the fields K(σ) is a perfect e-free PAC field. Proof: Statement (a) is a special case of Proposition 7.8.1(b). Statement (b) follows from Proposition 20.4.4 and Theorem 20.5.1. We define the corank of a field K as the rank of Gal(K). Lemma 20.5.3: Let K be a field and e a positive integer. Then, for every perfect field F of corank at most e that contains K, there exists a regular ˜ ˜ ∩E ∼ ˜ ∩ F. ultraproduct E of the K(σ)’s with K =K K Proof: Let τ1 , . . . , τe be generators of Gal(F ). For each finite Galois extension L of K the set S(L) = {σ ∈ Gal(K)e | resL (σ) = resL (τ )} has a positive Haar measure, so S(L) is not small. If L is contained in a larger finite Galois extension L0 of K, then S(L0 ) ⊆ S(L). By Lemma 7.6.1 there exists a regular ultrafilter D of Gal(K)e which contains the sets S(L) as L runs over all finite Galois extensions of K. Q ˜ Let E = K(σ)/D. Then L∩E = L∩F for every finite Galois extension ˜ ∩ E and K ˜ ∩ F are perfect, L of K. Therefore, Ks ∩ E = Ks ∩ F . Since K they are equal. Theorem 20.5.4: If (K, e) is a Hilbertian pair, then Ax(K, e) (Proposition 20.4.4) is a set of axioms for Almost(K, e). Specifically, a field F is perfect,

440

Chapter 20. The Elementary Theory of e-Free PAC Fields

e-free, PAC, and contains K if and only if it satisfies each sentence θ which ˜ is true in K(σ) for almost all σ ∈ Gal(K)e . Proof: Suppose that F |= Ax(K, e). By Lemma 20.5.3 there exists a regular ˜ ˜ ∩E ∼ ˜ ∩ F . By Lemma ultraproduct E of the K(σ)’s such that K =K K 20.5.2(b), E |= Ax(K, e). Corollary 20.4.2 now gives E ≡K F . From Lemma 20.5.2(a), F |= Almost(K, e). Theorem 20.5.1 gives the converse.

20.6 The Probable Truth of a Sentence Let K be a field and let e be a positive integer. For a sentence θ of L(ring, K), consider the truth set of θ: (1)

˜ |= θ}. S(K, e, θ) = {σ ∈ G(K)e | K(σ)

Refer to the case where K has elimination theory (Definition 19.2.8) and e and θ are explicitly given, as the explicit case. Regard the measure of ˜ S(K, e, θ) (if it exists) as the probability that θ is true among the K(σ)’s. Call a sentence λ of the form (2)

P ((∃X)[f1 (X) = 0], . . . , (∃X)[fm (X) = 0])

with f1 , . . . , fm ∈ K[X] separable polynomials and P a Boolean polynomial (Section 7.6), a test sentence. In this case it is fairly easy to describe the set S(K, e, λ). Indeed, the splitting field L of the polynomial f1 · · · fm is a finite Galois extension of K. Denote the set of all τ ∈ Gal(L/K)e with L(τ ) |= λ by S0 . Then (3)

S(K, e, λ) = {σ ∈ G(K)e | resL (σ) ∈ S0 }.

Indeed, if λ has the form (∃X)[f1 (X) = 0], then S0 consists exactly of the τ ∈ Gal(L/K)e for which L(τ ) contains at least one root of f1 (X). Since L contains all roots of f1 (X), this gives (3). An induction on the structure of P gives (3) in general. From (3) (4)

µ(S(K, e, λ)) =

|S0 | . [L : K]e

In the explicit case the right hand side of (4) can be computed effectively from Lemma 19.3.2. Lemma 20.6.1: Let K be a field, e be a positive integer, and λ a test sentence. Then µ(S(K, e, λ)) is a rational number which, in the explicit case, can be effectively computed. The reduction of arbitrary sentences to test sentences depends on a general result of field theory:

20.6 The Probable Truth of a Sentence

441

Lemma 20.6.2: Let K ⊆ L ⊆ L0 be a tower of fields. Suppose L0 /K is algebraic and L ∼ =K L0 . Then L = L0 . Proof: Consider x ∈ L0 . Let f = irr(x, K). Denote the set of all zeros of f in L (resp. L0 ) by Z (resp. Z 0 ). Then Z ⊆ Z 0 and |Z| = |Z 0 |, so Z = Z 0 . Therefore, x ∈ L. Lemma 20.6.3: Let E and F be fields having a common subfield K. (a) Suppose each irreducible polynomial f ∈ K[X] which has a root in E ˜ ∩ E into K ˜ ∩ F. has a root in F . Then there exists a K-embedding of K (b) Suppose an irreducible polynomial f ∈ K[X] has a root in E if and only ˜ ∩ F. ˜ ∩E ∼ if it has a root in F . Then K =K K Proof: Statement (b) follows from (a) by Lemma 20.6.2. Thus, it suffices to prove (a). Assume without loss E and F are algebraic over K. If K is finite and x ∈ E, then K(x) is a Galois extension of K. By assumption, irr(x, K) has a root in F . Hence, K(x) ⊆ F . Therefore, E ⊆ F . Now assume K is infinite. Let L be a finite extension of K in E. Choose ˆ ∩ F . List the ˆ of K which contains L. Put L0 = L a finite normal extension L ˜ as σ1 , . . . , σn . Put Li = σi (L0 ), i = 1, . . . , n. By K-isomorphisms of L0 into K assumption, for each x ∈ L the polynomial f = irr(x, K) has a root x0 ∈ L0 . ˆ Then τ −1 |L0 = σi for Extend the map x 7→ x0 to a K-automorphism τ of L. some i, so x = τ −1 (x0 ) = σi (x0 ) ∈ Li . It follows that L ⊆ L1 ∪ · · · ∪ Ln . Now consider each of the fields L, L1 , . . . , Ln as a subspace of the finite dimensional ˆ over K. Since K is infinite, there exists j with L ⊆ Lj . vector space L Therefore, σj−1 (L) ⊆ L0 ⊆ F . Denote the finite nonempty set of all K-embeddings of L into F by I(L). If L is contained in another finite extension L1 of K, contained in E, then restriction defines a canonical map of I(L1 ) into I(L). Take the inverse limit of the I(L)’s to establish the existence of a K-embedding of E into F (Corollary 1.1.4), as desired. Corollary 20.6.4: Let E and F be perfect fields with a common subfield K. (a) Suppose each separable irreducible polynomial f ∈ K[X] which has a ˜ ∩ E can be K-embedded into K ˜ ∩ F. root in E has a root in F . Then K (b) Suppose a separable irreducible polynomial f ∈ K[X] has a root in E if ˜ ∩ F. ˜ ∩E ∼ and only if it has a root in F . Then K =K K Proof: Lemma 20.6.3 covers the case where char(K) = 0, so we assume char(K) = p > 0. Let f ∈ K[X] be an irreducible polynomial. Then there exists a separable irreducible polynomial g ∈ K[X] and a power q of p such that f (X) = g(X q ). Since E (resp. F ) is perfect, f has a root in E (resp. F ) if and only if g has one, as well. Therefore, we may apply Lemma 20.6.3 to conclude both (a) and (b). We combine Corollary 20.4.2 with Corollary 20.6.4:

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Chapter 20. The Elementary Theory of e-Free PAC Fields

Lemma 20.6.5: Let (K, e) be a Hilbertian pair and E and F models of Almost(K, e). Then E ≡K F is and only if E and F satisfy exactly the same test sentences. Proof: Suppose E and F satisfy the same test sentences. Then a separable polynomial f ∈ K[X] has a root in E if and only if f has a root in F . Since E and F are models of Almost(K, e), they are perfect. Hence, by Corollary ˜ ∩E ∼ ˜ ∩ F . Thus, by Corollary 20.4.2, E ≡K F . 20.6.4, K =K K Proposition 20.6.6: Let (K, e) be a Hilbertian pair. For each sentence θ of L(ring, K) there exists a test sentence λ satisfying: (a) The sets S(K, e, θ) and S(K, e, λ) differ only by a zero set; the sentence θ ↔ λ belongs to Almost(K, e). (b) There is a formal proof (δ1 , . . . , δn ) of θ ↔ λ from the set of axioms Ax(K, e); both λ and (δ1 , . . . , δn ) can be found in the explicit case in a recursive way by checking all proofs from Ax(K, e). Proof: Proposition 7.8.2 gives a test sentence λ satisfying (a). By Theorem 20.5.4, Ax(K, e) is a set of axioms for Almost(K, e). Therefore, by Corollary 8.2.6, Ax(K, e) ` θ ↔ λ, which is the first part of (b). The second part of (b) follows from Proposition 8.7.2. We use Proposition 20.6.6 to generalize Lemma 20.6.1 to arbitrary sentences: Theorem 20.6.7 ([Jarden-Kiehne]): Let (K, e) be a Hilbertian pair and θ a sentence of L(ring, K). Then µ(S(K, e, θ)) is a rational number which, in the explicit case, can be recursively computed. In particular, the theory Almost(K, e) is recursively decidable. Proof: Indeed, µ(S(K, e, θ)) = 1 if and only if the sentence θ belongs to Almost(K, e). Remark 20.6.8: Chapter 31 uses algebraic geometry to prove that Almost(K, e) is primitive recursive.

20.7 Change of Base Field Suppose θ is a sentence of L(ring, K) and K 0 is a field containing K. Then θ is also a sentence of L(ring, K 0 ). It is therefore natural, to relate S(K, e, θ) and S(K 0 , e, θ). More generally, we relate S(K, e, θ) and S(K 0 , e, θ) when θ is an “infinite sentence”. To define the latter concept, we adjoin the symbol W∞ i=1 to the language and define the set of infinite sentences to be the smallest set of strings that satisfy the following rules: (1a) Every sentence of L(ring, K) is an infinite sentence. (1b) If θ is an infinite sentence, then so is ¬θ. W∞ (1c) For θ1 , θ2 , θ3 , . . . a sequence of infinite sentences, i=1 θi is an infinite sentence.

20.7 Change of Base Field

443

The interpretation of infinite sentences is given by the following rule: W∞ For F an extension field of K, F |= i=1 θi if F |= θi for some integer i. If θ is an infinite sentence, use (1) of Section 20.6 to define its truth set, S(K, e, θ). Observe that the operator S(K, e, ∗) commutes with infinite disjunctions. Also, if fields E and F containing K are K-elementarily equivalent, then the same infinite sentences are true in both of them. Consider a regular extension K 0 of K. Let ρ: Gal(K 0 )e → Gal(K)e be the ˜ ˜ ˜ ∩K ˜ 0 (σ) = K(ρ(σ)) = K(σ) for each σ ∈ Gal(K 0 )e . restriction map. Then K 0 e 0 Denote the measure of Gal(K ) by µ and use the rule µ0 (ρ−1 (A)) = µ(A) for each measurable subset A of Gal(K)e (Proposition 18.2.2). Theorem 20.7.1 ([Jacobson-Jarden1, Thm. 1.1]): Let (K, e) and (K 0 , e) be Hilbertian pairs such that K 0 is a regular extension of K. Then: ˜ ˜ 0 (σ). (a) For almost all σ ∈ Gal(K 0 )e , K(σ) ≺K (b) For each infinite sentence θ of L(ring, K), S(K 0 , e, θ) and ρ−1 (S(K, e, θ)) differ only by a zero set. (c) µ0 (S(K 0 , e, θ)) = µ(S(K, e, θ)). ˜ eProof: Denote the set of all σ in Gal(K)e (resp. in Gal(K 0 )e ) with K(σ) free and PAC by S (resp. S 0 ). By Theorem 20.5.1, µ(S) = 1 and µ0 (S 0 ) = 1. ˜ is an elementary Hence, µ0 (ρ−1 (S) ∩ S 0 ) = 1. By Corollary 20.4.3, K(σ) ˜ 0 (σ) for every σ ∈ ρ−1 (S) ∩ S 0 . This completes the proof of (a). subfield of K Statement (b) follows from (a); and (c) follows from (b). Consider the special case where K 0 = K(t1 , . . . , tr ) is the field of rational functions over K in the variables t1 , . . . , tr . Let R = K[t1 , . . . , tr ] be the corresponding ring of polynomials. Regard a sentence of L(ring, R) as a formula θ(t1 , . . . , tr ) of L(ring, K) involving the variables t1 , . . . , tr . If a1 , . . . , ar ∈ K, then θ(a) is a sentence of L(ring, K). The next theorem generalizes Hilbert’s irreducibility theorem (Exercise 4): Theorem 20.7.2: Let K be a countable Hilbertian field, e a positive integer, θ(t1 , . . . , tr ) a sentence of L(ring, R). Then there exists a separable Hilbert subset H of K r such that (2)

µ0 (S(K 0 , e, θ(t))) = µ(S(K, e, θ(a)))

for each a ∈ H. Proof: Using test sentences divides the proof into two parts. Part A: Reduction to test sentences. Proposition 20.6.6 gives a test sentence λ(t) for θ(t) of the form P (∃X)[f1 (t, X) = 0], . . . , (∃X)[fm (t, X) = 0] with f1 , . . . , fm ∈ K(t)[X] separable polynomials and P a boolean polynomial. Moreover, there exists a formal proof (δ1 (t), . . . , δn (t)) of θ(t) ↔ λ(t) from Ax(K 0 , e). The axioms of the positive diagram of K 0 (Example 7.3.1) involved in this proof have the form r1 (t)+r2 (t) = r3 (t) or r1 (t)·r2 (t) = r3 (t)

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Chapter 20. The Elementary Theory of e-Free PAC Fields

where r1 , r2 , r3 ∈ K(t). Define U to be the K-Zariski open set of all a ∈ Ar at which none of the denominators of the fi ’s and the rj ’s vanishes. Thus, ˜ K(σ) |= δi (a) for each a ∈ U (K), i = 1, . . . , n. Therefore, for each a ∈ U (K) ˜ ˜ e-free and PAC, K(σ) |= θ(a) ↔ λ(a). and for each σ ∈ Gal(K)e with K(σ) 0 Now apply Theorem 20.5.1 to both K and K to obtain (3a) (3b)

µ0 (S(K 0 , e, θ(t))) = µ0 (S(K 0 , e, λ(t)) µ(S(K, e, θ(a))) = µ(S(K, e, λ(a)))

and for each a ∈ U (K).

Part B: Test sentences. Let L0 be the splitting field of f1 (t, X) · · · fm (t, X) over K 0 . Take a primitive element z for L0 /K 0 which is integral over R and let g(t, X) = irr(z, K 0 ). Put h = gf1 · · · gm . Denote the set of all a ∈ U (K) such that the discriminants (and therefore also the leading coefficients) of f1 , . . . , fm , g remain nonzero under the specialization t → a by H 0 . Make H 0 smaller, if necessary, to assume that the specialization t → a induces an isomorphism of Gal(L0 /K 0 ) onto Gal(L/K), where L is the splitting field of h(a, X) over K, that preserves the operation on the roots of f1 , . . . , fn (Lemma 13.1.1(a)). For a ∈ H 0 , the number of σ 0 ∈ Gal(L0 /K 0 )e with L0 (σ 0 ) |= λ(t) is equal to the number of σ ∈ Gal(L/K)e such that L(σ) |= λ(a). Hence µ0 (S(K 0 , e, λ(t))) = µ(S(K, e, λ(a))). Therefore, (2) follows from (3). By Lemma 13.1.1, H 0 contains a Hilbert subset of K r . The theorem follows. Remark: With K fixed, Section 30.6 analyzes the effect of a change in e on µ(S(K, e, θ)).

20.8 The Fields Ks (σ1 , . . . , σe ) The free generators theorem (Theorem 18.5.6) and the PAC Nullstellensatz (Theorem 18.6.1) establish properties satisfied by almost all fields Ks (σ). As applications, however, the last sections developed a theory of properties ˜ shared by almost all fields K(σ). The next section explains this shift. Comparison of the theory of these fields to the theory of finite (and therefore ˜ perfect) fields forces us to consider the perfect fields K(σ) rather than the imperfect fields Ks (σ). ˜ Nevertheless, if we replace K(σ) by Ks (σ) and make some obvious changes, the results in Sections 20.4, 20.5, and 20.6 as well as their proofs remain valid. First, the analog of Corollary 20.4.2: Proposition 20.8.1: Let E and F be separable extensions of a field K with the same imperfect degree. Suppose E and F are e-free, PAC, Ks ∩ E ∼ =K Ks ∩ F . Then E ≡K F . Now replace Ax(K, e) by a set of axioms Ax0 (K, e). A field extension F of K satisfies Ax0 (K, e) if and only if F is PAC, e-free, [F : F p ] = [K : K p ],

20.8 The Fields Ks (σ1 , . . . , σe )

445

and F/K is separable. To express the separability of F/K by sentences of L(ring, K), choose a p-basis B of K over K p . Then “F/K is separable” if and only if “B0 is p-independent over F p ” for all finite subsets B0 of B. Having done this, we write the analog of Theorem 20.5.1: Theorem 20.8.2: Suppose (K, e) is an Hilbertian pair. Then, for almost all σ ∈ Gal(K)e , the field Ks (σ) is e-free, PAC, separable over K, and [Ks (σ) : Ks (σ)p ] = [K : K p ]. We denote the theory of all sentences of L(ring, K) which are true in Ks (σ) for almost all σ ∈ Gal(K)e by Almost0 (K, e). Then the analog of Lemma 20.5.2 holds: Lemma 20.8.3: Let (K, e) be an Hilbertian pair. (a) A field F is a model of Almost0 (K, E) if and only if it is K-elementarily equivalent to a regular ultraproduct of the fields Ks (σ). (b) Every regular ultraproduct of the fields Ks (σ) is e-free and PAC, has the same imperfect degree as K, and is separable over K. Now we present the analog of Lemma 20.5.3: Lemma 20.8.4: Let K be a field, e a positive integer, and F a field of corank at most e which is separable over K. Then there exists a regular ultraproduct E of the Ks (σ)’s with Ks ∩ E ∼ =K Ks ∩ F . This gives the analog of Theorem 20.5.4: Theorem 20.8.5: Let (K, e) be an Hilbertian pair. Then Ax0 (K, e) is a set of axioms for Almost0 (K, e). Finally, for a sentence θ of L(ring, K) let S 0 (K, e, θ) = {σ ∈ Gal(K)e | Ks (σ) |= θ}. Then the following analog of Theorem 20.6.7 holds: Theorem 20.8.6: Let (K, e) be an Hilbertian pair and θ a sentence of L(ring, K). Then µ(S 0 (K, e, θ)) is a rational number which, in the explicit case, can be recursively computed. In particular, the theory Almost0 (K, e) is recursively decidable. The results of Section 20.7 are in general false for the fields Ks (σ). Suppose for example, K = Fp (t) and K 0 = Fp (t, u) with t, u algebraically independent elements over Fp . Then the separable extensions of K have imperfect exponent 1 while the separable extensions of K 0 have imperfect exponent 2. Therefore, no Ks0 (σ) is elementarily equivalent to Ks (σ).

446

Chapter 20. The Elementary Theory of e-Free PAC Fields

20.9 The Transfer Theorem This section connects the elementary theory of finite fields with the elemen˜ tary theory of the fields K(σ). Let K be a global field and OK the ring of integers of K. Denote the set of nonzero prime ideals of OK by P (K). It is equipped with the Dirichlet density δ (Section 6.3). We consider models of the language L(ring, OK ) ¯ p , for that are field extensions either of K or of one of the residue fields K p ∈ P (K). ¯ p for almost By Proposition 7.8.1, a sentence θ of L(ring, OK ) is true in K all p ∈ P (K) if and only if θ is true in every nonprincipal ultraproduct of ¯ p /D is one of these ultraproducts, then, by Proposition ¯ p ’s. If F = Q K the K 7.9.1 and Corollary 11.3.4, F is a perfect, 1-free, PAC field that contains K. Since K is a global field, it is Hilbertian (Theorem 13.4.2). Hence, by Theorem 20.5.4, F is a model of Almost(K, 1). We have therefore proved: ˜ Lemma 20.9.1: If a sentence θ of L(ring, OK ) is true in K(σ) for almost all ¯ σ ∈ Gal(K), then θ is true in Kp for almost all p ∈ P (K). For a given sentence θ of L(ring, OK ) we compare the sets ˜ S(θ) = S(K, 1, θ) = {σ ∈ Gal(K) | K(σ) |= θ} and ¯ A(θ) = A(K, θ) = {p ∈ P (K) | Kp |= θ}, using the Dirichlet density δ of P (K) and the Haar measure µ of Gal(K). Lemma 20.9.2: Let λ be the test sentence (1)

p (∃X)[f1 (X) = 0], . . . , (∃X)[fm (X) = 0] ,

where f1 , . . . , fm ∈ K[X] are separable polynomials and p is a Boolean polynomial. Let B be the set of all p ∈ P (K) such that all coefficients of fi are p-integral and the leading coefficient and the discriminant of the fi ’s are p-units, i = 1, . . . , m. Denote the splitting field of f1 · · · fm over K by L. Then: (a) For each p ∈ B, every P ∈ P (L) over p, every σ ∈ DP , and every field ¯ p satisfying L ¯P ∩ F = L ¯ P (¯ extension F of K σ ) (where σ ¯ is the image of ¯ ¯ σ under the map DP → Gal(LP /Kp ) induced by P) we have L(σ) |= λ ⇐⇒ F |= λ. ¯ ¯ (b) Let S(λ) = {σ ∈ Gal(L/K) | L(σ) |= λ}. Then δ(A(λ)) = |S(λ)| . [L:K]

Proof of (a): First note that B is a cofinite set, because each fi is separable. Suppose p ∈ B. Then p is unramified in L (Section 6.2). If λ is (∃X)[fi (X) = 0], statement (a) is a reinterpretation of Lemma 6.1.8(a). The general case follows by induction on the structure of λ. Proof of (b): Use (a) and the Chebotarev density theorem (Theorem 6.3.1).

20.9 The Transfer Theorem

447

Theorem 20.9.3 (The Transfer Theorem): Let OK be the ring of integers of a global field K and θ a sentence of L(ring, OK ). Then S(θ) is measurable, A(θ) has a Dirichlet density, and µ(S(θ)) = δ(A(θ)). Proof: Proposition 20.6.6 provides a test sentence λ of the form (1) with ˜ θ ↔ λ true in K(σ) for almost all σ ∈ Gal(K). Without loss assume the coefficients of the polynomials f1 , . . . , fm belong to OK . Thus (Lemma 20.9.1), ¯ p , for almost all p ∈ P (K). Hence, S(θ) ≈ S(λ) (i.e. S(θ) θ ↔ λ is true in K and S(λ) differ by a set of measure zero) and A(θ) ≈ A(λ) (i.e. A(θ) and A(λ) differ by a finite set). Therefore, it suffices to prove the theorem for λ, rather than for θ. Let L be the splitting field of the polynomial f1 . . . fm . Then L is a ¯ finite Galois extension of K, S(λ) = {τ ∈ Gal(L/K) | L(τ ) |= λ} is a union ¯ of conjugacy classes of Gal(L/K), and S(λ) = {σ ∈ Gal(K) | resL σ ∈ S(λ)}. Hence, µ(S(λ)) =

¯ |S(λ)| [L:K] .

By Lemma 20.9.2,

(2)

and δ(A(λ)) =

A(λ) ≈ {p ∈ P (K) | ¯ |S(λ)| [L:K] .

L/K p

¯ ⊆ S(λ)}

Consequently, µ(S(λ)) = δ(A(λ)).

¯ If δ(A(λ)) = 0, then µ(S(λ)) = 0 and S(λ) is empty. Therefore, by (2), A(λ), hence A(θ), are finite sets. Therefore, Theorem 20.6.7 gives the following supplement to Theorem 20.9.3: Theorem 20.9.4 ([Ax1, p. 161, Cor.]): Let θ be a sentence of L(ring, OK ). Then δ(A(θ)) is a rational number which is positive if A(θ) is infinite. In the explicit case δ(A(θ)) can be recursively computed. The special case K = Q gives the decidability of the theory of sentences which are true in Fp for almost all p. The next result represents this theory in two more ways. Consider the set Q of all powers of prime numbers. Call a subset B of Q small if only finitely many prime numbers divide the elements of B. Proposition 20.9.5: The following three statements about a sentence θ of L(ring) are equivalent: ˜ (a) Q(σ) |= θ for almost all σ ∈ Gal(Q). (b) Fp |= θ for almost all p ∈ P (Q). (c) Fq |= θ for almost all q ∈ Q (i.e. for all but a small set). Proof: The equivalence “(a) ⇐⇒ (b)” is a special case of the transfer theorem. The implication “(c) =⇒ (b)” follows from the definitions. Finally, the implication “(a) Q =⇒ (c)” follows from Lemma 20.5.2(a), because every regular ultraproduct Fq /D, where D contains all complements of small sets of Q, is a 1-free PAC field of characteristic zero (Corollary 11.3.4).

448

Chapter 20. The Elementary Theory of e-Free PAC Fields

Corollary 20.9.6 ([Ax2, p. 265]): Let K be a given global field. Then the ¯ p for almost all theory of all sentences in L(ring, OK ) which are true in K p ∈ P (K) is recursively decidable. Proof: By Theorem 20.9.4, the set of all sentences θ of L(ring, K) with δ(A(θ)) = 1 is recursive. If δ(A(θ)) = 1, then there are only finitely many p ∈ P (K) for which θ ¯ p . The proof of the next theorem gives a recursive procedure may be false in K for displaying these primes: Theorem 20.9.7 ([Ax2, p. 264]): Let K be a given global field. Then the ¯ p for all p ∈ P (K) is recursively decidable. theory of all sentences true in K Proof: We follow the pattern of proof of Theorem 20.9.3. Part A: Finding a test sentence. Let θ ∈ L(ring, OK ). Proposition 20.6.4 recursively gives a test sentence λ of the form (2) of Section 20.6 and a formal proof (δ1 , . . . , δn ) from Ax(K, 1) (in the language L(ring, K)) of θ ↔ λ. Let A0 be the set of all p ∈ P (K) that divide one of the denominators of the elements of K involved in δ1 , . . . , δn . If a field F of characteristic not in A0 contains a homomorphic image of OK and if the axioms among the δi are true in F , then (δ1 , . . . , δn ) is a valid proof in F . In particular, θ ↔ λ is true in F . Part B: Reduction modulo p of the PAC axioms. Let 1 ≤ i ≤ n. If δi is the axiom “each absolutely irreducible polynomial f (X, Y ) of degree d has ¯ p | ≤ (d − 1)4 . By a zero”, define Ai to be the set of all p ∈ P (K) with |K ¯ Corollary 5.4.2, if p ∈ P (K) r Ai , then Kp |= δi . Otherwise, let Ai = A0 . ¯ p is perfect and 1-free, K ¯ p |= θ ↔ λ for Let B1 = A0 ∪ A1 ∪ · · · ∪ An . Since K each p ∈ P (K) r B1 . Part C: Exceptional primes for λ. Construct the splitting field L of the product of the polynomials f1 , . . . , fm appearing in λ and check if there exists ¯p τ ∈ Gal(L/K) with L(τ ) 6|= λ. In this case, λ (and therefore θ) is false in K L/K for almost all p ∈ P (K) with τ ∈ p (Lemma 20.9.2). By Chebotarev, infinitely many primes p satisfy the latter condition. Assume therefore that L(τ ) |= λ for each τ ∈ Gal(L/K). Compute a finite subset B2 of P (K) including all p which are either ramified in L or divide one of the discriminants ¯ p |= λ (or the leading coefficients) of f1 , . . . , fm . Let B = B1 ∪ B2 . Then K ¯ r (Lemma 20.9.2), so Kp |= θ for each p ∈ P (K) B. ¯ p |= θ for each of the Complete the procedure by checking whether K finitely many exceptional primes p ∈ B.

20.10 The Elementary Theory of Finite Fields We conclude with a discussion of the theory of finite fields and its decidability. ˆ (i.e. 1-free), Call a field F pseudo finite if F is perfect, Gal(F ) ∼ =Z and PAC.

20.10 The Elementary Theory of Finite Fields

449

Lemma 20.10.1: Every nonprincipal ultraproduct of distinct finite fields is pseudo finite. Proof: Let F1 , F2 , F3 , . . . be distinct finite fields. Consider a nonprincipal Q ultrafilter D of N and let F = Fi /D. Then F is perfect, 1-free, and PAC (Proposition 20.4.4). Thus, F is pseudo finite. The following result is a special case of Corollary 20.4.2 in the case e = 1: Proposition 20.10.2: Let E and F be pseudo finite fields. ˜ ∼ ˜ Then (a) Suppose E and F contain a common field K and E ∩ K =K F ∩ K. E ≡K F . (b) Suppose F is a regular extension of E. Then E ≺ F . Lemma 20.10.3: Let K be a finite field and L an algebraic extension of K. For each positive integer n denote the unique extension ofQK of degree n by Kn . Then there exists a nonprincipal ultraproduct D = Kn /D such that ˜ = L. D∩K Proof: For each positive integer d the set Ad = {n ∈ N | Kn ∩ Kd = L ∩ Kd } is infinite. If d|d0 , then Ad0 ⊆ Ad . Thus, given d1 , . . . , dr , we put d = lcm(d1 , . . . , dr ) and conclude from the relation Ad ⊆ Ad1 ∩ · · · ∩ Adr that the latter intersection is an infinite set. By Lemma 7.5.4, there Q exists a nonprincipal ultrafilter D on N which contains each Ad . Let D = Kn /D be the corresponding ultraproduct. ˜ = L: Consider a positive integer d ∈ D, let x be We prove that D ∩ K a primitive element for Kd over K, and put f = irr(x, K). Then f is Galois over K. If Kd ⊆ L, then Kd ⊆ Kn for each n ∈ Ad , so f has a root in Kn for each n ∈ Ad . By Loˇs (Proposition 7.7.1), f has a root in D, so D ⊆ L. If Kd 6⊆ L, then L ∩ Kd ⊂ Kd , so f has no root in Kn for each n ∈ Ad . By ˜ = L. Loˇs, f has no root in D, so Kd 6⊆ D. It follows that D ∩ K Proposition 20.10.4: Let θ be a sentence of L(ring). Then θ is true in almost all (i.e. all but finitely many) finite fields if and only if θ is true in every pseudo finite field. Proof: First suppose θ is true in almost all finite fields. Consider a pseudo finiteQ field F . By Lemma 20.10.3, there exists a nonprincipal ultraproduct ˜p = D ∩ F ˜ p . By Lemma 20.10.1, D is pseudo D = Fpn /D such that F ∩ F finite. Hence, by Proposition 20.10.2, F ≡K D. By Loˇs (Proposition 7.7.1), θ is true in D. Therefore, θ is true in F . . of distinct finite Conversely, suppose θ is false in a sequence F1 , F2 , F3 , . .Q fields. Choose a nonprincipal ultrapower D of N. Put F = Fi /D. By Loˇs, θ is false in F . By Lemma 20.10.1, F is pseudo finite. Corollary 20.10.5 ([Ax2, p. 240]): A field F is pseudo finite if and only if F is an infinite model of the theory of finite fields. Proof: First suppose F is pseudo finite. Then, by Proposition 20.10.4, each sentence θ which holds in every finite field holds in F .

450

Chapter 20. The Elementary Theory of e-Free PAC Fields

Conversely, suppose F is an infinite model of the theory of finite fields. ˆ (Proposition 20.4.4). For each positive Then F is perfect and Gal(F ) ∼ =Z integer d, let θd be the sentence “there are at most (d − 1)4 distinct elements or every absolutely irreducible polynomial in the variables X, Y of degree at most d has a zero”. By Corollary 5.4.2(b), θd holds in each finite field. Hence, θd holds in F . But F is infinite. Hence, every absolutely irreducible polynomial of degree d has a zero in F . Consequently, F is PAC and therefore pseudo finite. Like the theory of all residue fields of a given global field, the theory of finite fields is decidable. This was a problem raised by Tarski and solved by Ax: Theorem 20.10.6 ([Ax2, p. 264]): The theory of all sentences of L(ring) which are true in every finite field is recursively decidable. Proof: Consider θ ∈ L(ring). Follow the proof of Theorem 20.9.7 in the case K = Q to check if θ is true in Fp for all p ∈ P (Q). In the affirmative case, choose an integer n greater than (d − 1)4 for all d’s that appear in Part B and greater than of the primes belonging to Part C of that proof. Then conclude from Lemma 20.9.2 at the end of Part C of Theorem 20.9.7 that Fpi |= θ for each p ≥ n and each i. We therefore only need to check if, for a given prime p, the sentence θ is true in Fpi for all i ∈ N. This is equivalent to checking if θ is true in Fp (t)p for each p ∈ P (Fp (t)). This, again is a special case of Theorem 20.9.7. Now, the proof is complete. Example 20.10.7: Pseudo finite fields. (a) If a sentence θ of L(ring) holds in infinitely many finite fields, then it holds in every nonprincipal ultraproduct of those fields. Hence, by Lemma 20.10.1, θ holds in every pseudo finite field. Similarly, let K be a global field and θ a sentence of L(ring, OK ). Sup¯ p with p ranging over an infinite set A of prime divisors of pose θ holds in K ¯ p ’s. Reduction K. Then θ holds in each nonprincipal ultraproduct F of the K modulo p embeds OK in F . Thus, F extends K. ˜ (b) Let K be a global field. Then K(σ) is pseudo finite for almost all σ ∈ Gal(K) (Theorem 20.5.1 for e = 1). Consider a sentence θ of L(ring, OK ) ¯ p for infinitely many prime p of K. By Theorem 20.9.4, which holds in K δ(A(θ)) > 0. Hence, by the Transfer Theorem, µ(S(θ)) > 0. In particular, ˜ ˜ there is a σ ∈ Gal(K) such that K(σ) is pseudo finite and θ holds in K(σ). (c) Let F be an infinite algebraic extension of Q Fp . Then F is perfect and PAC (Corollary 11.2.4). Moreover, Gal(F ) ∼ = l∈S Zl for some set S of prime numbers (Exercise 8 of Chapter 1). Suppose F has an extension of ˆ Hence, F is a pseudo degree l for every prime number l. Then Gal(F ) ∼ = Z. finite field.

Exercises

451

Theorem 20.10.8: K be a global field. Denote the set of all nonprinciQ Let ¯ p /D, where p ranges over P (K), by F. Let T be the pal ultraproducts K set of all sentences θ ∈ L(ring, OK ) which hold in almost all residue fields ¯ p . Then: K (a) Each F ∈ F is pseudo finite. (b) A sentence θ of L(ring, OK ) belongs to T if and only if θ holds in every F ∈ F. (c) Every model of T is elementarily equivalent in L(ring, OK ) to some F ∈ F. ˜ ∼ ˜ (d) For every σ ∈ Gal(K) there is an F ∈ F with F ∩ K =K K(σ). Proof: Statement (a) is a special case of Lemma 20.10.1. Statement (b) is a special case of Proposition 7.8.1(a). Statement (c) is a special case of Proposition 7.8.1(b). Statement (d) can be proved directly from the Chebotarev density theorem, but we deduce it here from Lemma 20.5.3. By that lemma, ˜ ˜ )’s with E ∩ K ˜ ∼ By there is a regular ultraproduct E of the K(τ =K K(σ). Lemma 20.5.2, E is a pseudo finite field. By (c), there is an F ∈ F which is elementarily equivalent in L(ring, OK ) to E. It follows from 20.6.3, that ˜ ∼ ˜ Consequently, F ∩ K ˜ ∼ ˜ E∩K =K F ∩ K. =K K(σ).

Exercises 1. Give an example showing that the hypothesis of Corollary 20.3.4 can hold nontrivially, by taking K to be a PAC field for which Gal(K) is finitely generated. Then, let F be a nonprincipal ultraproduct of countably many copies of K. 2. Let λ be the test sentence (Section 20.6) (∃X)[f1 (X) = 0 ∨ f2 (X) = 0 ∨ f3 (X) = 0] with fi (X) = X 2 − ai and ai ∈ Z for i = 1, 2, 3, nonzero integers. What is the exact condition on a1 , a2 , a3 such that µ(S(K, 1, λ)) = 1? Now answer the same question for µ(S(K, 2, λ)) = 1. 3. [Jarden8, p. 149] It is a consequence of Proposition 20.6.6(a) that if (K, e) is a Hilbertian pair and θ is a sentence of L(ring, K), then S(K, e, θ) is a measurable set. We outline an alternative proof, valid for every countable field K and every positive integer e: ˜ define Let ϕ(X1 , . . . , Xn ) be a formula of L(ring, K). For x1 , . . . , xn ∈ K ˜ ˜ and K(σ) |= ϕ(x)} S(K, e, ϕ(x)) = {σ ∈ Gal(K)e | x1 , . . . , xn ∈ K(σ) (a) Use induction on structure to show that S S(K, e, ϕ(x)) is a Borel set. Hint: Use the identity S(K, e, (∃Y )ϕ(x, Y )) = y∈K˜ S(K, e, ϕ(x, y)). (b) Deduce that S(K, e, θ) is a Borel set for each infinite sentence of L(ring, K).

452

Chapter 20. The Elementary Theory of e-Free PAC Fields

4. Prove the converse of Theorem 20.7.2: Let K be a field and t an indeterminate. Consider the ring R = K[t] and the field K 0 = K(t). Suppose for each positive integer e and each sentence θ(t) of L(ring, R) there exists a nonempty subset H of K with µ0 (S(K 0 , e, θ(t))) = µ(S(K, e, θ(a))) for each a ∈ H. Prove that K is Hilbertian. Hint: Let f ∈ K[X] be separable. Then f is irreducible over K if and only if for all large e there exists S ⊆ Gal(K)e of positive measure such that ˜ f is irreducible over K(σ) for all σ ∈ S. 5. ([Ax2, p. 260]) Let K be a global field τ ∈ Gal(K). Prove that there Q and ˜ ∩F ∼ ˜ ). ¯ p /D with K exists a nonprincipal ultraproduct F = K = K(τ Hint: Either reproduce Ax’s direct application of the Chebotarev density theorem, or combine the Transfer theorem, Lemma 20.5.3 and Proposition 7.8.1. 6. Give an example of an infinite sentence of L(ring) which is true in each ˜ field Q(σ) but is false in each field Fp . Thus, the Transfer theorem does not generalize to infinite sentences. Indeed, the Haar measure is σ-additive while the Dirichlet density is only finitely-additive. 7. Verify the Transfer theorem over Q for the test sentence (∃X)[X 2 = a], where a is an integer, by using quadratic reciprocity and the Dirichlet density theorem (Corollary 6.3.2) for primes in arithmetic progressions. 8. Let K be a global field and f ∈ K[X] be separable and irreducible with deg(f ) = n > 1. Denote the set of prime numbers p such that f has no zero mod p by A. Put G = Gal(f, K) and let S be the set of all σ ∈ G which fix no zero of f . |S| (a) Apply the Transfer theorem to prove: δ(A) = |G| . Alternatively, use the Chebotarev density theorem to prove the equality directly. 1 1 1 (b) Suppose Gal(f, K) ∼ − 3! + · · · + (−1)n n! . = Sn . Prove: δ(A) = 2! 9. Let f ∈ Z[X, Y ] be an absolutely irreducible polynomial. Prove that for almost all primes p there exists a ∈ Fp such that f (a, Y ) decomposes into linear factors in Fp [Y ]. Hint: Let F = Q(x, y) be the function field of V (f ). Choose a stabilizing element t for the extension F/Q (Theorem 18.9.3). Let Fˆ be the Galois hull of F/Q(t). One way to complete the proof is to specialize t to elements of Q infinitely often so that x, y are integral over the corresponding local ring and such that the residue fields of Fˆ form a linearly disjoint sequence of Galois extensions of Q (as in the proof of Theorem 18.10.3). Use BorelCantelli (Lemma 18.3.5) to prove that for almost all σ ∈ Gal(K) there exists ˜ a ∈ Q(σ) such that f (a, Y ) decomposes into linear factors. Now use the transfer principle. Alternatively, let z be a primitive element for the extension Fˆ /Q(t). Let g ∈ Z[T, Z] be an absolutely irreducible polynomial with g(t, z) = 0. Now find an h ∈ Z[X], h 6= 0 such that for almost all p and for each b ∈ Fp

Notes

453

with h(b) 6= 0 and g(b, T ) has zeros in Fp the specialization t → b extends to a specilization (t, x) → (b, a) and the polynomial f (a, Y ) decomposes into linear factors. Thus, the result follows directly from Corollary 5.4.2. 10. Show that the small sets of Q in Section 20.9 cannot be replaced by finite sets, by giving a sentence θ ∈ L(ring) for which θ is true for all p ∈ P (Q) but the q ∈ Q for which Fq 6|= θ is infinite. 11. Let K be a number field. Denote the set of all prime ideals of OK by P (K). For each finite extension L of K let Splt(L/K) be the set of all p ∈ P (K) which split completely in L. (a) Prove that there exists an ultrafilter L of P (K) which contains Splt(L/K) for each finite extension L of K and each subset of P (K) of Dirichlet density 1. Q ˜ ⊆ F. ¯ p /L. Prove that Q (b) Let L be as in (a). Put F = p∈P (K) K 0 (c) Let LQbe another ultrafilter of P (K) satisfying the conditions of (a). Put ¯ p /L0 . Prove that F 0 ≡ F . F0 = K

Notes Ax initiated the investigation of free PAC fields in [Ax1] and [Ax2] by connecting the theory of finite fields with the theory of pseudo finite fields. This connection is based on the special case of the embedding lemma (Lemma 20.2.2) where E and F are 1-free and Gal(L) is procyclic (Ax’s proof [Ax2, p. 248] is very complicated) and on Exercise 5. The transfer theorem [Jarden2] strengthens this connection. The treatment of perfect e-free PAC fields can be found in [JardenKiehne]. [Cherlin-v.d.Dries-Macintyre] introduces the treatment of imperfect fields that are e-free and PAC. Finally, we follow [Jacobson-Jarden1] and include finite fields, with e = 1 as base fields, in addition to the countable Hilbertian fields. In [Ax1, Lemma 5], Ax proves Lemma 20.6.3(b) essentially for the case ˜ ∩E/K and K ˜ ∩F/K are separable extensions. His reduction of where both K the general case to the separable case is vague. The argument which appears in our proof and which uses vector spaces is due independently to B. Poizat [Poizat] and to H. W. Lenstra.

Chapter 21. Problems of Arithmetical Geometry We apply the model theory - measure theory technique of Chapter 20 to concrete field arithmetic problems. The transfer principles between properties ˜ of finite fields and properties of the fields K(σ) can often be accomplished through direct application of the Chebotarev density theorem. Usually, however, application of the transfer theorem avoids a repetition of arguments. This chapter includes the theory of Ci -fields, Kronecker conjugacy of global field extensions, Davenport’s problem on value sets of polynomials over finite fields, a solution of Schur’s conjecture on permutation polynomials, and a solution of the generalized Carlitz’s conjecture on the degree of a permutation polynomial in characteristic p. Each of these concrete problems focuses our attention on rich historically motivated concepts that could be overlooked in an abstract model theoretic viewpoint.

21.1 The Decomposition-Intersection Procedure The classical diophantine concern is the description of the Q-rational (resp. Zrational) points of a Zariski Q-closed set A. The decomposition-intersection procedure reduces this concern to the study of a union A∗ of subvarieties (Section 10.2) of A defined over Q. We start with an arbitrary base field K. To each nonempty Zariski K-closed set A in affine space, An , or in the projective space, Pn , there corresponds a canonical K-closed subset A0 defined as follows: First, decompose S ˜ A into its K-components, A = i Vi . Then decompose each Vi into its KS components, Vi = j Wij . For a fixed i, {Wij }j is a complete set of conjugate ˜ (Section 10.2). Thus, the intersection Ui = T Wij varieties, defined over K j S is invariant under the action of Gal(K): Ui is a K-closed set. Denote i Ui by A0 . Continue the procedure to obtain a descending sequence A ⊇ A(1) ⊇ (2) A ⊇ · · · ⊇ A(m) ⊇ A(m+1) ⊇ · · · of K-closed sets, with A(m+1) = (A(m) )0 . Hilbert’s basis theorem (Lemma 10.1.1) gives an integer m with A(r) = A(m) for all r ≥ m. Denote A(m) by A∗ . ˜ of A, A(1) , . . . , A(m) . The Let W1 , . . . , Ws be a list of all K-components 0 compositum, L , of the fields of definition of W1 , . . . , Ws is a finite normal extension of K. The maximal separable extension L of K in L0 is the smallest Galois extension of K such that each of W1 , . . . , Ws is L-closed. Call L the Galois splitting field of A over K. Lemma 21.1.1: Let M be a field extension of K with L ∩ M = K. Then: (a) A∗ contains all M -closed subvarieties of A. In particular, A∗ (M ) = A(M ).

21.2 Ci -Fields and Weakly Ci -Fields

455

(b) A∗ is nonempty if and only if A contains an M -closed nonempty subvariety. Proof of (a): Let W be a nonempty M -closed subvariety of A. Choose be a generic point x of W over M . Then x is a generic point of W over LM . Also, there exist i and j with x ∈ Wij . Hence, W ⊆ Wij . Each automorphism of L/K extends to an automorphism of LM/M . Since, for aTfixed i, the Wij ’s are conjugate under the action of Gal(L/K), we have W ⊆ j Wij = Ui ⊆ A0 . Proceed by induction to show that W ⊆ A∗ . In particular, each M -rational point of A belongs to A∗ . Therefore, A(M ) = A∗ (M ). Proof of (b): Suppose first A∗ is nonempty. Assume without loss that A = A∗ . Then A = A0 and A is nonempty. For each i we have Ui ⊆ Vi . Conversely, let x be a generic point of Vi over K. Then, there exists an i0 such that x ∈ Ui0 . Hence, x ∈ Vi0 , so Vi = Vi0 . Thus, i = i0 and Vi ⊆ Ui . It follows that Ui = Vi . This implies that Vi is absolutely irreducible. Otherwise there would be atTleast two Wij ’s and hence we would have the contradiction dim(Ui ) = dim( j Wij ) < dim(Vi ) (Lemma 10.1.2). Thus, Vi is a variety which is K-closed, hence also M -closed. The converse follows from (a).

21.2 Ci -Fields and Weakly Ci -Fields Recall that a form f (X0 , . . . , Xn ) with coefficients in a field K defines a projective Zariski K-closed set in Pn (Section 10.7). If x = (x0 , . . . , xn ) 6= (0, . . . , 0) is a nontrivial zero of f , then the (n + 1)-tuple (ax0 , . . . , axn ) defines the same point of Pn for each a ∈ K × . Since projective hypersurfaces are the simplest projective sets, they occupy a special place in diophantine investigations. We follow the lead of a problem due to Artin: Does each form f ∈ Fq [X0 , . . . , Xd ] of degree d have a nontrivial zero? Chevalley’s affirmative solution [Chevalley1] motivated Lang [Lang1] to explore the concept of a Ci -field: Definition 21.2.1: The field K is called Ci,d if each form f ∈ K[X0 , . . . , Xn ] of degree d with di ≤ n has a nontrivial zero in K n+1 . Call it Ci if it is Ci,d for each d ∈ N. In this case, K is Cj for each j ≥ i. Example 21.2.2: Every algebraically closed field K is C0 . Indeed, every form f ∈ K[X0 , . . . , Xn ] of positive degree with n ≥ 1 has a nontrivial zero in K n+1 . Lemma 21.2.3 (Chevalley-Warning): Let f1 , . . . , fm ∈ Fq [X1 , . . . , Xn ] be polynomials with deg(f1 · · · fm ) < n. Put, A = V (f1 , . . . , fm ). Then |A(Fq )| ≡ 0 mod p. Qm Proof: Consider g(X) = j=1 1−fj (X)q−1 . If x ∈ A(Fq ), then fj (x) = 0, j = 1, . . . , m. Hence, g(x) = 1. If x ∈ Fnq r A(Fq ), then fj (x) 6= 0 for some

456

Chapter 21. Problems of Arithmetical Geometry

j. Therefore, fj (x)q−1 = 1 and g(x) = 0. This gives X (1) |A(Fq )| + pZ = g(x). x∈Fn q

P Rewrite g(X) as ci X1i1 · · · Xnin where I is a finite set and ci ∈ F× q for each i ∈ I. Then X X X X (2) g(x) = ci xi11 · · · xinn . x∈Fn q

i∈I

x1 ∈Fq

xn ∈Fq

P Let i ∈ I. If i = (0, . . . , 0), then x1 ∈Fq xi11 = 0+pZ. Otherwise, i1 +· · ·+in ≤ Pm deg(g) ≤ j=1 deg(fj )(q − 1) < n(q − 1), so there is a j with 1 ≤ ij < q − 1. Put k = ij . Since the polynomial X k − 1 has at most k roots in F× q , there is × k an a ∈ Fq with a 6= 1. Hence, X x∈Fq

xk =

X

(ax)k = ak

x∈Fq

X

xk ,

x∈Fq

P so x∈Fq xk = 0. Thus, the right hand side of (2) is 0. By (1), |A(Fq )| ≡ 0 mod p. If f1 , . . . , fm are forms, (0, . . . , 0) ∈ A(Fq ). Hence, by Lemma 21.2.3, V (Fq ) has at least p points. An application of this argument to the case when m = 1 gives Chevalley’s result: Proposition 21.2.4: Every finite field is C1 . Ax applied ultraproducts to deduce from Chevalley’s result that each perfect PAC field with Abelian absolute Galois group is C1 (Theorem 22.9.6). He left unsolved this question: Problem 21.2.5 (Ax): Is every perfect PAC field C1 ? If Ax’s problem has an affirmative solution, then each form of degree d over Q in d+1 variables has a nonempty Q-closed subvariety (Lemma 21.3.1). This motivates the introduction of the weak Ci condition. We use it to prove some results about PAC fields which are Ci . Definition 21.2.6: A field K is called weakly Ci,d , if for each form f ∈ K[X0 , . . . , Xn ] of degree d, with di ≤ n, the Zariski K-closed set V (f ) of Pn contains a subvariety W which is Zariski K-closed. If K is weakly Ci,d for each d ∈ N, we say that K is weakly Ci . Remark 21.2.7: By definition, every Ci,d field is also weakly Ci,d . In addition, every perfect PAC field K which is weakly Ci,d is also Ci,d . Perfectness here guarantees that a Zariski K-closed variety is defined over K (Lemma 10.2.3).

21.2 Ci -Fields and Weakly Ci -Fields

457

Example 21.2.8: Separably closed fields. Let K be a field of positive characteristic p. Suppose K has an infinite sequence a1 , a2 , a3 , . . . of p-independent elements over K p . Then, a1 , a2 , a3 , . . . are p-independent over Ksp (Lemma Pqi 2.7.3). Hence, for each i and for all x0 , . . . , xqi ∈ Ks the relation j=0 aj xpj = 0 implies x0 , . . . , xqi = 0. Therefore, Ks is not Ci although Ks is PAC (Section 11.1). On the other hand, each form f ∈ K[X0 , X1 ] has a nontrivial zero ˜ 2 . Hence, Ks is weakly C0 . (x0 , x1 ) ∈ K For example, take K = Fp (t1 , t2 , t3 , . . .) with t1 , t2 , t3 , . . . algebraically independent over Fp . Then t1 , t2 , t3 , . . . are p-independent over K p (proof of Lemma 2.7.2). Thus, Ks is weakly Ci for no positive integer i. We explore the behavior of the weakly Ci property under field extensions. Recall: An extension F of K is primary if Ks ∩ F = K (Lemma 2.6.13). Lemma 21.2.9: A field K is weakly Ci,d if and only if every form f ∈ K[X0 , . . . , Xn ] of degree d with di ≤ n has a nontrivial zero x such that K(x) is a primary extension of K. Proof: Suppose first K is weakly Ci,d . Let f be as stated in the lemma. Then V (f ) contains a K-closed variety W . It is defined over a purely inseparable extension K 0 of K. Let p be a generic point of W over K 0 . Then p is represented by a zero (x0 , . . . , xn ) of f where x0 , say, is transcendental over K 0 (p) = K 0 ( xx10 , . . . , xxn0 ) and K 0 (p)/K 0 is regular. Therefore, Ks ∩ K(x0 , . . . , xn ) = K. Conversely, suppose that f (x) = 0 and that K(x)/K is a primary extension. Denote the K-closed subset of Pn generated by x by W . Since K(x) is linearly disjoint from Ks over K, the set W is Ks -irreducible (lemma 10.2.1). It follows that W is a variety (Lemma 10.2.4). The form that appears in the next lemma might aptly be called a weakly normic form. Lemma 21.2.10: Assume the field K is not separably closed. Let e0 be an integer. Then there exist an integer e > e0 and a form h ∈ K[Y1 , . . . , Ye ] of degree e satisfying this: Suppose K(y1 , . . . , ye ) is a primary extension of K. Then (3)

h(y1 , . . . , ye ) = 0 =⇒ y1 = · · · = ye = 0.

Proof: By assumption, there exists a nontrivial Galois extension L/K. Denote its degree by l and its Galois group by G. Let w1 , . . . , wl be a basis for L/K and consider the form Y (w1σ X1 + · · · + wlσ Xl ), g(X) = σ∈G

of degree l and with coefficients in K. Suppose K(y)/K is a primary extension and g(y) = 0. Then there exists σ ∈ G with w1σ y1 + · · · + wlσ yl = 0. Moreover, K(y) is linearly disjoint from L over K. Hence, y1 = · · · = yl = 0.

458

Chapter 21. Problems of Arithmetical Geometry

Next consider the form g2 = g(g(X1 , . . . , Xl ), g(Xl+1 , . . . , X2l ), . . . , g(X(l−1)l+1 , . . . , Xl2 )). It is of degree l2 and it has property (3) for e = l2 . Iterate this procedure to obtain, finally, a form h of degree exceeding e0 . The next results treat Ci -fields and weakly Ci -fields simultaneously: Lemma 21.2.11: Let K be a Ci -field (resp. weakly Ci -field) and f1 , f2 , . . . , fr forms over K of degree d in n variables with n > rdi . Then f1 , f2 , . . . , fr have a common nontrivial zero in K (resp. some primary extension of K). Proof: First suppose K is separably closed. Then K is Ci for no i but is weakly C0 (Example 21.2.8). By the projective dimension theorem (Section 10.7), V (f1 , . . . , fr ) is a nonempty Zariski closed subset of Pn−1 . Thus, ˜ which is a primary extension of K. f1 , . . . , fr have a nontrivial zero in K, Assume now K is not separably closed. Then Lemma 21.2.10 gives a weakly normic form h ∈ K[Y1 , . . . , Ye ] of degree e > r satisfying n h e i (4) n −1 ≥ n + 1. r r For each positive integer j, let Xj = (Xj1 , . . . , Xjn ) be a vector of variables. Let h1 (X1 , . . . , Xq ) = h(f1 (X1 ), . . . , fr (X1 ), . . . , f1 (Xq ), . . . , fr (Xq ), 0, . . . , 0) with q = re . Then h1 is a form of degree d1 = de in n1 = n re variables. If (x1 , . . . , xq ) is a nontrivial zero with coordinates in K (resp. a primary extension of K), then there exists j between 1 and q with xj 6= 0 and f1 (xj ) = · · · = fr (xj ) = 0. Replace h by h 1 in the above definition and define a form h2 of degree d2 = d2 e in n2 = n nr1 variables. Continue by induction to define, for every variables. positive integer k, a form hk of degree dk = dk e in nk = n nk−1 r Every nontrivial zero of hk in K (resp. in a primary extension E of K) defines a common nontrivial zero of f1 , . . . , fr in K (resp. in E). By assumption, a zero of that type exists if k satisfies nk > dik . Thus, it suffices to prove that nk > dik for all large k. The lemma is done, then, if we choose k such that nk > dik . Indeed, hn i n k > nk − n. (5) nk+1 = n r r Use (5) for nk rather than for nk+1 and substitute in the right hand side of (5) to obtain nk+1 > ( nr )2 nk−1 − n( nr + 1). Continue in this manner, inductively, to obtain the inequality n k n k−1 n k−2 n1 − n + + ··· + 1 nk+1 > r r r (6) h i e n k 1 nb −n +n , = b r r

21.2 Ci -Fields and Weakly Ci -Fields

459

with b = nr − 1. Use (4) to see that the right hand side of (6) bounds nk+1 1 n k > di1ei b (( rdni )k + dnik ). Since n > rdi , the b (( r ) + n). Therefore, di k+1

right side tends to infinity with k. Consequently, nk > dik for all large k, as claimed. Proposition 21.2.12: Let K be a Ci -field (resp. weakly Ci -field) and E is an extension of K of transcendence degree j. Then E is Ci+j (resp. weakly Ci+j ). Proof: It suffices to prove the proposition for simple transcendental extensions and then for finite separable extensions. Part A: E = K(t), t is transcendental over K. Let f ∈ K(t)[X0 , . . . , Xn ] be a form of degree d with n ≥ di+1 . Assume without loss the coefficients of f lie in K[t]. Denote the maximum of the degrees of the coefficients of f by r. Let s be a positive integer with (n − di+1 + 1)s > di (r + 1) − (n + 1). Consider the (n + 1)(s + 1) variables Yjk , j = 0, . . . , n and k = 0, . . . , s. There exist forms f0 (Y), . . . , fds+r (Y) over K of degree d with f(

s X

Y0k tk , . . . ,

k=0

s X

Ynk tk ) =

ds+r X

k=0

fl (Y)tl .

l=0

Since (n + 1)(s + 1) > di (ds + r + 1), Lemma 21.2.11 gives a nonzero y in K (n+1)(s+1) (resp. K(y)/K Ps is a primary extension) with f0 (y) = · · · = fds+r (y) = 0. Put xj = k=0 yjk tk , j = 0, . . . , n. These elements are not all zero and f (x) = 0. When K is Ci , x is K(t)-rational. Therefore, K(t) is Ci+1 . When K is weakly Ci , we may choose y with K(y) algebraically independent from K(t) over K. By Lemma 2.6.15(a), K(y, t) is a primary extension of K(t). Hence, K(x, t)/K(t) is primary. It follows from Lemma 21.2.9 that K(t) is weakly Ci+1 . Part B: E/K is finite. Let f ∈ E[X0 , . . . , Xn ] be a form of degree d, with n ≥ di . With w1 , . . . , we a basis for E/K, introduce new variables Zjk , j = 0, . . . , n and k = 1, . . . , e. Then there exist forms f1 (Z), . . . , fe (Z) over K of degree d with f(

e X k=1

Z0k wk , . . . ,

e X k=1

Znk wk ) =

e X

fk (Z)wk .

k=1

Since (n + 1)e > edi , Lemma 21.2.11 again gives a nonzero z such that f1 (z) = · · · = fP e (z) = 0 and K(z) = K (resp. K(z)/K) is primary). The e elements xj = k=1 zjk wk , j = 0, . . . , n, satisfy f (x) = 0 and not all of them are zero. If K is Ci , then E(x) = E. Therefore, E is Ci . If K is weakly Ci , then E(z)/E is primary (Lemma 2.6.15(a)). Hence, E(x)/E is primary. Consequently, E is weakly Ci .

460

Chapter 21. Problems of Arithmetical Geometry

Weakly Ci -fields have a large class of subfields that are also weakly Ci . Note, however, that this result does not hold for Ci fields (Exercise 5). Proposition 21.2.13: If L is a primary extension of a field K and L is weakly Ci , then K is also weakly Ci . Proof: Since a primary extension L(x) of L is, under the hypotheses, a primary extension of K, this is immediate from Lemma 21.2.9.

21.3 Perfect PAC Fields which are Ci First we note a relation between weakly Ci -fields, Ci -fields and the fields ˜ K(σ): Lemma 21.3.1: Let K be a countable Hilbertian field. Then the following are equivalent: (a) K is weakly Ci . ˜ (b) For each e ∈ N and for almost all σ ∈ Gal(K)e , K(σ) is Ci . Proof: If (a) holds, then (Proposition 21.2.12) every algebraic extension of K is weakly Ci . ˜ By Theorem 18.6.1, K(σ) is a perfect PAC field for almost all σ ∈ e Gal(K) . Since a weakly Ci PAC field is Ci (Remark 21.2.7), this gives (b). Assume now (b) holds. Let f ∈ K[X0 , . . . , Xn ] be a form of degree d with di ≤ n. Apply the decomposition-intersection procedure to the subset A = V (f ) of Pn (Section 21.1). Let L be the Galois splitting field of A over K. Choose generators σ10 , . . . , σe0 for Gal(L/K). Then there exist σ1 , . . . , σe ∈ ˜ is Ci . In Gal(K) that respectively extend σ10 , . . . , σe0 , and for which K(σ) ˜ ˜ particular, L ∩ K(σ) = K and A has a K(σ)-rational point. By Lemma 21.1.1, A∗ is nonempty. With K replacing M in Lemma 21.1.1 conclude that A contains a nonempty subvariety which is K-closed. Hence, K is weakly Ci . We do not expect a general field extension of a Ci -field to be Ci (Exercise 6). But, under simple conditions this is true for weakly Ci fields: Proposition 21.3.2: The following conditions on a countable Hilbertian field K are equivalent: (a) K is weakly Ci . (b) Every field extension F of K is weakly Ci . (c) Every perfect PAC field extension F of K is Ci . Proof of “(a) =⇒ (b)”: Assume without loss F is also countable. If F/K is algebraic, then F is weakly Ci (Proposition 21.2.12). Suppose F/K is transcendental. Let K 0 be a purely transcendental extension of K with F algebraic over K 0 . By Theorem 13.2.1, K 0 is Hilbertian. By Lemma 21.3.1, ˜ the field K(σ) is Ci for each e ∈ N and almost all σ ∈ Gal(K)e . Hence, by ˜ 0 (σ) is Ci for each e ∈ N and almost all σ ∈ Gal(K 0 )e . It Theorem 20.7.1, K

21.3 Perfect PAC Fields which are Ci

461

follows from Lemma 21.3.1 that K 0 is weakly Ci . Consequently (Proposition 21.2.12), F is weakly Ci . Proof of “(b) =⇒ (c)”: Use Remark 21.2.7. ˜ Proof of “(c) =⇒ (a)”: By Theorem 18.6.1 and assumption (c), K(σ) is Ci e for each e ∈ N and for almost all σ ∈ Gal(K) . Hence, by Lemma 21.3.1, K is weakly Ci . Proposition 21.2.4 says that every finite field is C1 . Hence, by Proposition 21.2.12, every algebraic extension of a finite field is C1 . Every other field contains either Q or Fp (t) for some p and transcendental element t. Each of the latter fields is countable and Hilbertian (Theorem 13.4.2). An application of Proposition 21.3.2 to i = 1 therefore gives this reformulation of Ax’s problem 21.2.5: Corollary 21.3.3: The following conditions are equivalent: (a) Every field is weakly C1 . (b) Each of the fields Q and Fp (t) is weakly C1 . (c) Every perfect PAC field is C1 . Lemma 21.3.4: Let K be a weakly Ci -field and F an extension of K. Then F is weakly Ci+1 . If, in addition, F is perfect and PAC, then F is Ci+1 . Proof: The second statement follows from the first by remark 21.2.7. To prove the first statement, suppose first F/K is algebraic. By Proposition 21.2.12, F is weakly Ci . Hence, F is weakly Ci+1 . Now suppose F/K is transcendental. Use Proposition 21.2.13 and replace K by a smaller field to assume K is countable. Choose t ∈ F transcendental over K. Then K(t) is Hilbertian (Theorem 13.4.2) and weakly Ci+1 (Proposition 21.2.12). It follows from Proposition 21.3.2 that F is weakly Ci+1 . Lemma 21.3.5: Let F be a field. (a) Suppose F contains an algebraically closed field. Then F is weakly C1 . (b) Suppose F has a positive characteristic. Then F is weakly C2 . (c) Suppose Gal(F ) is procyclic. Then F is weakly C1 . ˜ (d) Suppose F contains Q(σ) with σ ∈ Gal(Q). Then F is weakly C2 . Proof of (a): By Example 21.2.2, each algebraically closed field is C0 . Hence, by Lemma 21.3.4, F is weakly C1 . Proof of (b): By assumption, F contains a finite field. The latter is C1 (Proposition 21.2.4). Hence, by Lemma 21.3.4, F is weakly C2 . Proof of (c): Let K0 be the prime field of F . Consider a form f ∈ F [X0 , . . . , Xd ] of degree d. The coefficients of f are algebraic over a finitely ˜ ∩ F ) is procyclic. Suppose V (f ) generated extension K of K0 . Also, Gal(K ˜ has a subvariety W which is K ∩ F -closed. Then W is also F -closed. So,

462

Chapter 21. Problems of Arithmetical Geometry

assume without loss, F is algebraic over K. Finally, replace F by Fins , if necessary, to assume F is perfect (Proposition 21.2.13). Now K is either a finite field, or algebraic over Q, or transcendental over K0 . In all cases (K, 1) is a Hilbertian pair in the sense of Section 20.5 (Theorem 13.4.2). By Lemma 20.5.3, there is a regular ultraproduct E of the ˜ ˜ ∩E ∼ fields K(σ) with K =K F . By Lemma 20.5.2(b), E is pseudo finite. For each positive integer d the C1,d property of a field is elementary. Since every finite field has this property (Proposition 21.2.4), so does every pseudo finite field (Proposition 20.10.2). Therefore, E is C1 . Consequently, by Proposition 21.2.13, F is weakly C1 . Proof of (d): Use (c) and Lemma 21.3.4.

Suppose F in Lemma 2.3.5 is PAC and perfect. Then we may replace “weakly Ci ” by “Ci ” in each of the statement of that lemma. This gives the main result of this section: Theorem 21.3.6: Let F be a perfect PAC field. ˜ Then F is C1 . (a) Suppose F contains an algebraically closed field K. (b) Suppose F has a positive characteristic. Then F is C2 . (c) Suppose Gal(F ) is procyclic. Then F is C1 . ˜ (d) Suppose F contains a field Q(σ), where σ ∈ Gal(Q). Then F is C2 . Remark 21.3.7: Theorem 22.9.6 strengthens part (c) of Theorem 21.3.6 relaxing the condition “Gal(F ) is procyclic” to “Gal(F ) is Abelian.” J´anos Koll´ar proves in [Koll´ ar3, Thm. 1] that every PAC field of char acterisitc 0 is C1 . This settles Problem 21.2.5 in characterisitic 0.

21.4 The Existential Theory of PAC Fields In Section 28.10 it is shown that the theory of PAC fields is undecidable. It is therefore of interest to observe that the decomposition-intersection procedure gives a decision procedure for the existential theory of PAC fields: Definition 21.4.1: (1)

Call a sentence of L(ring) existential if it has the form h_^

(∃X1 ) · · · (∃Xn )

i

i [fij (X) = 0 ∧ gi (X) 6= 0]

j

with fij , gi ∈ Z[X], and i, j range over finite sets.

Replace each inequality gi (X) 6= 0 in (1) by the equivalent formula (∃Yi )[Yi gi (X) − 1 = 0] to assume the sentence has the form (2)

h_^

(∃X1 ) · · · (∃Xn )

i

j

i fij (X) = 0 .

21.5 Kronecker Classes of Number Fields

463

The bracketed expression in (2) defines a Zariski Q-closed subset A of An . Rewrite (2) as (3)

(∃X1 ) · · · (∃Xn )[X ∈ A].

To test if (3) is true in every PAC field of characteristic 0, apply the decomposition-intersection procedure (Section 21.1, effective by Proposition 19.5.6). Let L be the Galois splitting field of A over Q and τ1 , . . . , τe generators for Gal(L/Q). The set of σ ∈ Gal(Q)e whose restriction to L is τ has positive measure ˜ (= 1/[L : Q]e ). Hence, by Theorem 18.6.1, there is a σ such that M = Q(σ) is PAC and L ∩ M = Q. If A∗ is empty, then A has no M -rational point (Lemma 21.1.1). Thus, there is a PAC field of characteristic 0 for which (3) is false. If A∗ is nonempty, then (Lemma 21.1.1) each Q-component W of A∗ is a subvariety of A defined over Q (effectively computable). That variety has an M -rational point in each PAC field M of characteristic 0. In this case, (3) is true in every PAC field of characteristic 0. Now we check the PAC fields of positive characteristic. Proposition 10.4.2 gives us a finite set of primes S, such that for each p 6∈ S the variety W (as above) is well defined and remains a variety when considered as a Zariski Fp -closed set. For each p ∈ S, repeat the decomposition-intersection procedure for A over Fp . If A∗ is nonempty in each of these cases, then (3) is true in every PAC field. If, however, there exists p ∈ S for which A∗ is empty over Fp , then, as above, there exists a PAC field M such that A(M ) is empty. In this case (3) fails in some PAC field. We summarize: Theorem 21.4.2: Both the existential theory of PAC fields of a given characteristic and the existential theory of all PAC fields are primitive recursively decidable.

21.5 Kronecker Classes of Number Fields Our next example discusses the “Kronecker conjugacy” of polynomials. We place it in the framework of classical algebraic number theory and mention some results and open problems, with partial reformulation in terms of group theory. It is customary to credit Kronecker (in 1880) for the impetus to investigate the decomposition of primes in number field extensions [Jehne]. Throughout this section fix a global field K. Definition 21.5.1: Let L be a finite separable extension of K. Denote the set of all p ∈ P (K) that have a prime divisor P ∈ P (L) of relative degree ¯ P = Kp) ¯ by V (L/K). Call a finite separable extension M of K 1 (i.e. L Kronecker conjugate to L (over K) if V (L/K) and V (M/K) differ by

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only finitely many elements. The set of all finite separable extensions M of K which are Kronecker conjugate to L is called the Kronecker class of L/K. We denote it by K(L/K). Lemma 21.5.2 (Dedekind, Kummer): Let R be a Dedekind ring with quotient field K. Consider a finite separable extension L = K(x) of K. Suppose x is integral over R and let f = irr(x, K). If Rp [x] is the integral closure of Rp in L (by lemma 6.1.2, this holds for almost all p) and f (X) ≡ f1 (X)e1 · · · fr (X)er mod p is the factorization of f (X) as a product of powers of distinct monic irreducible polynomials modulo p, then pS = Pe11 · · · Perr where P1 , . . . , Pr are distinct prime ideals of S with f (Pi /p) = deg(fi ), i = 1, . . . , r. Proof: See [Lang5, p. 27] or [Janusz, p. 32].

The arithmetic condition for Kronecker conjugacy is equivalent to a Galois theoretic condition: Lemma 21.5.3: Let L1 and L2 be finite separable extensions of K and L a finite Galois extension of K which contains both L1 and L2 . Then L1 and L2 are Kronecker conjugate over K if and only if, with G = Gal(L/K), (1)

[ σ∈G

Gal(L/L1 )σ =

[

Gal(L/L2 )σ .

σ∈G

Proof: For i = 1, 2 let xi ∈ OLi be a primitive element of the extension Li /K, and let fi = irr(xi , K). By Lemma 21.5.2, V (Li /K) differs from the ¯ p by only finitely many set of primes p for which (∃X)[fi (X) = 0] is true in K primes. Let θ be the sentence (∃X)[f1 (X) = 0] ↔ (∃X)[f2 (X) = 0]. Then, Kronecker conjugacy of L1 and L2 over K is equivalent to the truth of ¯ p . By the transfer theorem (Theorem 20.9.3) applied to test θ in almost all K sentences, this is equivalent to the truth of θ in L(σ), for all σ ∈ Gal(L/K). But this is just a restatement of (1). If Li /K is Galois, i = 1, 2, and L1 and L2 are Kronecker conjugate over K, Lemma 21.5.3 implies L1 = L2 . This is Bauer’s theorem (see also Exercise 5 of Chapter 6). But, in general, Kronecker conjugacy does not imply conjugacy of the fields L1 and L2 over K. This is demonstrated in the following example:

21.5 Kronecker Classes of Number Fields

465

Example 21.5.4: Kronecker conjugate extensions of Q of different degrees q [Schinzel, Lemma 4]. Let L = Q(cos 2π 2 cos 2π . Then 7 ) and M = Q 7 L/Q is a cyclic extension of degree 3, M/L is a quadratic extension, M is not Galois over Q, but M is Kronecker conjugate to L over Q. We give proofs of these statements. 3 2 2πi/7 and η = Part A: irr(2 cos 2π 7 , Q) = X + X − 2X − 1. Let ζ = e 2π −1 2 cos 7 = ζ + ζ . Then Q(ζ)/Q is a cyclic extension of degree 6 and η is fixed by complex conjugation. Hence, Q(η)/Q is a cyclic extension of degree 3. The conjugates of η over Q are ηi = ζ i + ζ −i , i = 1, 2, 3 and they satisfy

η1 η2 η3 = 1,

(2)

η1 η2 + η1 η3 + η2 η3 = −2,

η1 + η2 + η3 = −1

Hence, irr(η, Q) = X 3 + X 2 − 2X − 1. √ Part B: η 6∈ L. Assume η = θ2 with θ in L. Then irr(θ, Q) is a polynomial of degree 3 with integral coefficients. The element θ is a zero of f (X) = X 6 + X 4 − 2X 2 − 1. Therefore, irr(θ, Q) divides f (X): X 6 + X 4 − 2X 2 − 1 = (X 3 + aX 2 + bX + 1)(X 3 + cX 2 + dX − 1) with a, b, c, d ∈ Z. Thus, a + c = 0, d − b = 0, ad + bc = 0, ac + b + d = 1 and c − a + bd = −2. Eliminate b and c to come down to the equations 2d = a2 + 1 and 2a = d2 + 2. Then d ≥ a from the first and a > d from the second, a contradiction. ˆ /Q). By Part B, Part C: The structure of the Galois group G = Gal(M √ √ M = Q( η1 ) is a quadratic extension of L. Similarly, Q( η2 ) is a quadratic √ extension of L which we claim to be different from Q( η1 ). Otherwise, there exists x ∈ L such that η2 = x2 η1 . By (2), η3−1 = (xη1 )2 , a contradiction to q √ ˆ = L(√η1 , √η2 ) is an extension Part B. Also η3 = η1−1 η2−1 implies that M ˆ /L) ∼ of L of degree 4 with Gal(M = Z/2Z × Z/2Z. The generator of the ˆ /L) by permuting the three subgroups cyclic group Gal(L/Q) acts on Gal(M of order 2 in a cyclic way. So, G is the semidirect product of Z/3Z with Z/2Z × Z/2Z. Part D: The Kronecker conjugacy of L and M . [ σ∈G

ˆ /M )σ = Gal(M

3 [ i=1

By part C we have

[ ˆ /L) = ˆ /L)σ . ˆ /Q(√ηi )) = Gal(M Gal(M Gal(M σ∈G

Hence, by Lemma 21.5.3, M and L are Kronecker conjugate.

Lemma 21.5.3 immediately gives L1 ⊆ L2 if L1 /Q is Galois and L1 and L2 are Kronecker conjugate — just as we saw in Schinzel’s example. Call a field M of a Kronecker class K(L/K) minimal if it contains no proper subfields of the same class.

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Lemma 21.5.5: Let L1 and L2 be minimal fields of a Kronecker class over ˆ 2 over K are equal. ˆ 1 and L K. Then their Galois closures L ˆ 1 and L ˆ 2 . By Proof: Let N be a finite Galois extension of K that contains L Lemma 21.5.3 [ [ Gal(N/L1 )σ = Gal(N/L2 )σ , σ∈G

σ∈G

ˆ 2 to conclude that with G = Gal(N/K). Restrict both sides to L [ σ∈G2

ˆ 2 /L1 ∩ L ˆ 2 )σ = Gal(L

[

ˆ 2 /L2 )σ , Gal(L

σ∈G2

ˆ 2 /K). By Lemma 21.5.3, L1 ∩ L ˆ 2 is Kronecker conjugate to with G2 = Gal(L ˆ2 ⊆ L ˆ2. L2 and therefore to L1 . Minimality of L1 implies that L1 = L1 ∩ L ˆ ˆ ˆ ˆ Therefore, L1 ⊆ L2 ; and, by symmetry, L1 = L2 . Corollary 21.5.6: Each Kronecker class contains only finitely many minimal fields. Remark 21.5.7: On the size of Kronecker classes. Fix a number field K. (a) Call the number of nonconjugate minimal fields of K(L/K) the width and denote it by ω(L/K). [Jehne, §4] gives L with ω(L/K) arbitrarily large. The proof incorporates Exercise 7 and the theorem of ScholzShafarevich that each p-group is a Galois group over K. (b) On the other hand, [Jehne, Theorem 3] gives examples in the number field case where K(L/K) is infinite. Indeed, this is the case if Aut(L/K) contains either a nontrivial automorphism of odd order, or a cyclic group of order 8, or a quaternion group of order 8. (c) For a quadratic extension L/K [Jehne, §6] conjectures that K(L/K) consists only of L. This conjecture has been proved by Saxl. To prove the conjecture, assume M is another field in K(L/K). By Lemma 21.5.3, L ⊂ M . Replace M by a smaller field to assume M is a minˆ be the Galois closure of M/K, G = Gal(M ˆ /K), imal extension of L. Let M ˆ ˆ H = Gal(M /L), and U = S Gal(M /M ). Then (G : H) = 2, U is a maximal subgroup of H, and H = g∈G U g . By [Jehne, Thm. 5], H has a unique minimal normal subgroup N which is non-Abelian and simple. Moreover, U N = H. So, V = U ∩ N is a proper subgroup of N . The minimality of N r implies N / G. Choose σ ∈ G S H. ViewSσ as an automorphism of N acting as conjugation. Then, N = h∈H V h ∩ h∈H V σh . But this contradicts the main result of [Saxl]: Let N be a finite simple group and V a proper subgroup. Then there exists a ∈ N lying in no Aut(N )-orbit of V . Saxl’s proof uses the classification of finite simple groups, proving the result case by case. The case where N = PSL(2, pr ) with p 6= 2 appears already in [Jehne, Thm. 5]. The case N = An with n ≥ 5 is due to [Klingen3].

21.6 Davenport’s Problem

467

(d) Saxl’s solution of Jehne’s conjecture has an interesting arithmetic consequence: Let d be a square√free integer and f ∈ Z[X] an irreducible polynomial without a root in Q( d). Then there are infinitely many prime numbers p such that dp = 1 but f (X) ≡ 0 mod p has no solution. √ ˜ M = Q( d, x), M ˆ the Galois Otherwise, let x be a root of f (X) in Q, ˆ /Q). Then, for almost p with d = 1 the closure of M/Q, and G = Gal(M p equation f (X) ≡ 0 mod p has a solution. By the transfer principle, [ √ ˆ /M )σ . ˆ /Q( d)) = Gal(M Gal(M σ∈G

√ Thus, M ∈ K(Q( d)/Q), in contrast to Jehne-Saxl.

Another problem of Jehne is still open (see the notes at the end of the chapter for a profinite version): Problem 21.5.8 ([Jehne, §7]): Are there fields K ⊂ L ⊂ M , with K global, L/K finite separable, M/K infinite separable, such that M0 ∈ K(L/K) for every intermediate field L ⊂ M0 ⊂ M of finite degree over K?

21.6 Davenport’s Problem Let K be a global field of characteristic p and let f, g ∈ K[Y ] be nonconstant polynomials which are not p-powers. For each prime p ∈ P (K) for which f is defined modulo p, consider the value set of f : ¯ p }. Vp (f ) = {f (y) | y ∈ K If g(Y ) = f (aY + b) with a, b ∈ K and a 6= 0, then g and f are said to be strictly linearly related. In this case, (1) Vp (f ) = Vp (g) for almost all p ∈ P (K). Call the pair (f, g) exceptional if f and g satisfy (1) but they are not strictly linearly related. Remark 21.6.1: Vp (X 8 ) = Vp (16X 8 ) for every prime number p. To prove this statement, verify the identity (2) X 8 − 16 = (X 2 − 2)(X 2 + 2) (X − 1)2 + 1 (X + 1)2 + 1 . −1 = p4 = 1 for By the multiplicativity of the Legendre symbol, p2 −2 p p each odd prime number p. Hence, at least one of the first three factors on the right hand side of (2) is divisible by p for some value of X. It follows that 16 is an 8th power modulo every p. Thus, for each x ∈ Fp there is a y ∈ Fp with x8 = 16y 8 , as claimed. Consider now a polynomial h(X) = cn X n + cn−1 X n−1 + · · · + c0 in Q[X] with n ≥ 1 and cn 6= 0. Then Vp (h(X 8 )) = Vp (h(16X 8 )) for each p which

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Chapter 21. Problems of Arithmetical Geometry

divides no denominator of c0 , . . . , cn−1 , cn . Assume there were a, b ∈ Q with a 6= 0 and h(16X 8 ) = h((aX + b)8 ). Comparing the coefficients of X 8n on both sides this would give 16 = a8 , which is a contradiction. It follows, (h(X 8 ), h(16X 8 )) is an exceptional pair. Problem 21.6.2: (a) Are all exceptional pairs in Q[X] of the form (h(X 8 ), h(16X 8 )) with h ∈ Q[X] (M¨ uller)? (b) For a given global field K find all exceptional pairs (f, g) with f, g ∈ K[X] (Davenport). A complete solution of this problem is still unavailable. Transfer principles, however, allow us to reduce (1) to a condition on a finite Galois extension of K(t), in complete analogy with the Galois theoretic version of Kronecker conjugacy of global field extensions. From this point we derive certain relations between f and g (e.g. deg(f ) = deg(g) if the degrees are relatively prime to char(K)). In the special case where K = Q and deg(f ) is a prime, we prove there is no g ∈ Q[X] such that (f, g) is exceptional. But we mention the existence of exceptional pairs (f, g) with deg(f ) prime over other number fields. Clearly (1) is equivalent to the truth of the sentence (3) (∀T ) (∃X)[f (X) = T ] ↔ (∃Y )[g(Y ) = T ] ¯ p . By Theorem 20.9.3, this is equivalent to the truth of in almost all fields K ˜ (3) in K(σ), for almost all σ ∈ Gal(K). First we eliminate (∀T ) to reduce the ] problem to a test sentence over K 0 = K(t). To this end we choose x, y ∈ K(t) with f (x) = t and g(y) = t and put E = K(t, x) and F = K(t, y). Assume that E/K(t) and F/K(t) are separable extensions. When p = char(K) > 0 ˜ this means, f (X) and g(X) are not a pth powers in K[X]. Lemma 21.6.3: Let N a finite Galois extension of K(t) containing E and F . Put G = Gal(N/K(t)). Then (1) is equivalent to [

(4)

ν∈G

Gal(N/E)ν =

[

Gal(N/F )ν .

ν∈G

˜ Proof: Put K = K(t). If (3) holds in K(σ) for almost all σ ∈ Gal(K), then ˜ 0 (σ) for almost all σ ∈ Gal(K 0 ) (Theorem 20.7.2). In particular, (3) holds in K formula in the outer brackets of (3) hold for t. That is, the sentence 0

(5)

(∃X)[f (X) = t] ↔ (∃Y )[g(Y ) = t]

˜ 0 (σ) for almost all σ ∈ Gal(K 0 ). Therefore, (5) holds in N (σ) for is true in K all σ ∈ G. Consequently, (4) is true. Conversely, suppose that the elements τ ∈ G that fix a root of f (X) − t are exactly the elements that fix a root of g(Y ) − t.

21.6 Davenport’s Problem

469

˜ We prove (3) holds in K(σ) for all σ ∈ Gal(K). Indeed, let σ ∈ Gal(K) ˜ and a ∈ K(σ). Extend the K-specialization t → a to a homomorphism ˜ The residue field N ¯ = ϕ of the integral closure R of K[t] in N into K. ϕ(R) is a normal extension of K(a). Denote the decomposition group {τ ∈ G | τ (Ker(ϕ)) = Ker(ϕ)} of ϕ by H. Then ϕ induces an epimorphism ϕ∗ of ¯ /K(a)) through the formula (Section 6.1) H onto Aut(N (6)

ϕ∗ (τ )(ϕ(z)) = ϕ(τ (z)) for each τ ∈ H and z ∈ R.

Observe that the roots of f (X) = t and g(X) − t are in R. Hence, ϕ maps the roots of f (X) − t (resp. g(Y ) − t) onto the roots of f (X) − a ˜ (resp. g(Y ) − a). Therefore, if b ∈ K(σ) satisfies f (b) = a, then there exists a zero x of f (X) = t with ϕ(x) = b. Note that H ∩ Gal(N/K(x)) is the decomposition group of ϕ over K(x). Hence, by Lemma 6.1.1(a), the ¯ /K(b)). restriction of ϕ∗ to H ∩Gal(N/K(x)) maps that subgroup onto Aut(N In particular, there exists τ ∈ H ∩ Gal(N/K(x)) with (7)

ϕ∗ (τ ) = resN¯ (σ).

Since τ (x) = x, the assumptions give a zero y of g(Y ) = t with τ (y) = y. Thus, c = ϕ(y) is a zero of g(Y ) = a for which (6) and (7) give σ(c) = c. Remark 21.6.4: The switch from the surjectivity of ϕ∗ on H to its surjectivity on H ∩ Gal(N/K(x)) is a subtlety in the proof of Lemma 21.6.3. From the surjectivity of ϕ∗ on H alone we could choose τ ∈ H such that (7) holds. It follows that there exist zeros x and x0 of f (X) = t with τ (x) = x0 and ϕ(x) = ϕ(x0 ) = b. If f (X) − a is separable, then this implies that x = x0 , and conclude the proof easily. Nothing, however, in the assumptions guaranties separability of f (X)−a. This same subtlety occurs in the stratification procedure of Chapter 30. Following Section 21.5 we say that fields E and F (and the polynomials f and g) satisfying (4) are Kronecker conjugate over K(t). This is independent of N . If E and F are minimal extensions of K(t) with this property, then as in Lemma 21.5.5, they have the same Galois closure over K(t). The next results, however, allow us the same conclusion under a less restrictive assumption than minimality. Throughout let deg(f ) = m, deg(g) = n. Corollary 21.6.5: Suppose E and F are Kronecker conjugate over K(t) ˆ (resp. Fˆ ) and [E : K(t)] and [F : K(t)] are not divisible by char(K). Let E be the Galois closure of E/K(t) (resp. F/K(t)). Then [E : K(t)] = [F : K(t)] ˆ = Fˆ . and E Proof: By assumption, p∞ is tamely ramified in all fields conjugate to E or to F . Hence, p∞ is tamely ramified in the composite N of all these fields [Cassels-Fr¨ohlich, p. 31]. Let P∞ be a prime divisor of N that lies over p∞ .

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Chapter 21. Problems of Arithmetical Geometry

Then the decomposition group of P∞ is cyclic [Cassels-Fr¨ohlich, p. 30]. Let τ be a generator of it. Then p∞ is unramified in N (τ ). Since p∞ is totally ramified in E (Example 2.3.11), the two fields are linearly disjoint over K(t) (Lemma 2.5.8). In particular, the polynomial f (X) − t (a root of which generates E over K(t)) remains irreducible over N (τ ). Thus, the group hτ i operates transitively on the roots x1 , . . . , xm of f (X) − t. Therefore, τ is an m-cycle on x1 , . . . , xm . By symmetry, τ is an n-cycle on y1 , . . . , yn , the roots of g(X) − t. If m < n, then τ m would fix each of x1 , . . . , xm but would move each of y1 , . . . , yn . This contradicts (4). Thus, m ≥ n; and by symmetry, m = n. Next restrict relation (4) to Fˆ to conclude that E ∩ Fˆ and F are Kronecker conjugate over K(t). By the first part of the corollary, [E ∩ Fˆ : K(t)] = ˆ ⊆ Fˆ . By [F : K(t)] = [E : K(t)]. Therefore, E = E ∩ Fˆ ⊆ Fˆ . Hence, E ˆ = Fˆ . symmetry E Lemma 21.6.6: Assume the hypotheses of Corollary 21.6.5. Then the polynomial f (X) − g(Y ) is reducible in K[X, Y ]; and there are constants a, b ∈ K and distinct roots x1 , . . . , xk of f (X) − t with 1 ≤ k < deg(f ) and ay + b = x1 + · · · + xk .

(8)

Proof: Assume f (X) − g(Y ) is irreducible in K[X, Y ]. Then, since y is transcendental over K, f (X)−t = f (X)−g(y) is irreducible in F [X]. Lemma 13.3.2 gives σ ∈ Gal(N/F ) that moves each root of f (X) − t. This is a contradiction, since σ(y) = y. Thus, f (X) − g(Y ) is reducible. Now consider a nontrivial factorization f (X) − g(Y ) = p(X, Y )q(X, Y ) in K[X, Y ]. By Corollary 21.6.5, degX (p) + degX (q) = degX (f (X) − g(Y )) = deg(f (X) − g(Y )) = deg(p) + deg(q). Since degX (p) ≤ deg(p) and degX (q) ≤ deg(q), it follows that degX (p) = deg(p) = k, with 1 ≤ k < deg(f ). Write p(X, Y ) as (9)

c00 X k + (c10 Y + c11 )X k − 1 + p2 (Y )X k−2 + · · · + pk (Y ),

with c00 , c10 , c11 ∈ K, c00 6= 0, and p2 , . . . , pk ∈ K[Y ]. Since, f (X) − t = p(X, y)q(X, y), the k roots of p(X, y), say x1 , . . . , xk , are roots of f (X) − t. Since f (X) − t is irreducible and separable over K(t), x1 , . . . , xk are distinct. −1 By (9), ay+b = x1 +· · ·+xk , where a = −c−1 00 c10 and b = −c00 c11 , as claimed. We summarize: Proposition 21.6.7: Let K be a global field and f, g ∈ K[X] nonconstant polynomials of degree not divisible by char(K). Suppose Vp (f ) = Vp (g) for almost all p ∈ P (K). Then:

21.6 Davenport’s Problem

471

(a) deg(f ) = deg(g). (b) f (X) − t and g(X) − t are Kronecker conjugate over the rational function field K(t). (c) The splitting fields of f (X) − t and g(X) − t over K(t) coincide. (d) f (X) − g(Y ) is reducible in K[X, Y ]. (e) There exist constants a, b ∈ K and distinct roots x1 , . . . , xk of f (X) − t with 1 ≤ k < deg(f ) and ay + b = x1 + · · · + xk . Note that conditions (a)-(e) are all necessary for f and g to be strictly linearly related. Here is a case when they are also sufficient. Let K be a field. Call a polynomial f ∈ K[X] decomposable over K if f (X) = f1 (f2 (X)) with f1 , f2 ∈ K[X], deg(f1 ) > 1, and deg(f2 ) > 1. (Then, deg(f ) = deg(f1 ) deg(f2 ).) Otherwise, f is indecomposable over K (e.g. if f is of prime degree). Theorem 21.6.8 ([Fried4]): Let f, g ∈ Q[X]. Suppose Vp (f ) = Vp (g) for almost all prime number p and f is indecomposable. Then g and f are strictly linearly related. Proof: We give an elementary proof in the case that deg(f ) = l is a prime. The proof in the general case depends on deeper group theoretic assertions, specifically the theory of doubly transitive permutation groups [Fried4]. Suppose without loss f is monic and let c be the leading coefficient of g. By Proposition 21.6.7(d) f (X) − g(Y ) = r(X, Y )s(X, Y )

(10)

Pm with r, s P ∈ Q[X, Y ] and m = deg(r) ≥ 1, n = deg(s) ≥ 1. Write r = i=0 ri n and s = i=0 si with ri , si ∈ Q[X, Y ] homogeneous of degree i. By (10), X l − cY l = rm (X, Y )sn (X, Y ).

(11) l

−1 is irreducible in Q[X] [Lang7, p. 184], one of the factors rm or Since ZZ−1 sm must be linear. Thus, either m = 1 or n = 1. Suppose for example m = 1. Then r(X, Y ) = a1 X + a2 Y + a3 with a1 , a2 , a3 ∈ Q and, say, a1 6= 0. Let a = − aa21 and b = − aa31 . Substituting X in (10) by aY + b gives f (aY + b) = g(Y ), as claimed.

Let n be one of the integers 7, 11, 13, 15, 21, or 31. [Fried8] and [Fried11] give indecomposable polynomials f and g of degree n and a number field K such that (f, g) is an exceptional pair over K. For indecomposable polynomials, these numbers are the only exceptional degrees: a consequence of the classification of finite simple groups.

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Chapter 21. Problems of Arithmetical Geometry

21.7 On Permutation Groups We gather in this section various definitions and results about permutation groups which enter the proof of the Schur Conjecture in the next section. Remark 21.7.1: The affine linear group. (a) Let G be a group. Suppose G acts on a set X from the left. For each x ∈ X put Gx = {σ ∈ G | σx = x}. Call G transitive if for all x, y ∈ X there is a σ ∈ G with σx = y. Call G regular if Gx = 1 for each x ∈ X. Note that ifTG is both transitive and regular, then |G| = |X|. Finally, call G faithful if x∈X Gx = 1. (b) Let G be a permutation group of a set X and H a subgroup of G. Suppose H is regular and transitive. Then H ∩Gx = 1 and G = HGx = Gx H for each x ∈ X. Indeed, if η ∈ H ∩ Gx , then ηx = η, so η = 1. For each σ ∈ G there is an η ∈ H with ηx = σx. Hence, σ = η · η −1 σ ∈ HGx . Thus, G = HGx . Applying the same argument to σ −1 we find that G = Gx H. If in addition H is normal in G, then G = Gx n H. (c) Denote the cyclic group of order n by Cn and the dihedral group of order 2n by Dn . The latter group is generated by elements σ, τ with defining relations τ 2 = 1, σ n = 1, and σ τ = σ −1 . (d) Let l be a prime number. Denote the group of all affine maps of Fl by AGL(1, Fl ). It consists of all maps σa,b : Fl → Fl with a ∈ F× l and b ∈ Fl defined by σa,b (x) = ax + b. Thus, |AGL(1, Fl )| = (l − 1)l. The subgroup L = {σ1,b | b ∈ Fl } is the unique l-Sylow subgroup of AGL(1, Fl ). Thus, L∼ = Cl and L is normal. Its action on Fl is transitive and regular. Hence, by (b), AGL(1, Fl ) = AGL(1, Fl )x n L. Moreover, AGL(1, Fl )x ∼ = F× l . Thus, AGL(1, Fl ) is a metacyclic group. It particular, AGL(1, Fl ) is solvable. Claim: Each σ ∈ AGL(1, Fl ) r L fixes exactly one x ∈ Fl . Indeed, b . σ = σa,b with a 6= 1. Hence, σx = x if and only if x = 1−a The following lemma establishes a converse to Remark 21.7.1(d): Lemma 21.7.2: Let l be an odd prime number and G a nontrivial finite solvable group acting faithfully and transitively on a set X of l elements. Let L be a minimal normal subgroup of G. Then: (a) L is transitive and regular. (b) L ∼ = Cl and G = Gx n L for each x ∈ X. (c) L is its own centralizer in G. (d) G/L is a cyclic group of order dividing l − 1 and Gx ∼ = G/L for each x ∈ X. (e) L is the unique l-Sylow subgroup of G. (f) G is isomorphic as a permutation group to a subgroup of AGL(1, Fl ). (g) Each ρ ∈ G r L belongs to exactly one Gx . (h) Let I be a subgroup of G which contains L and P a p-Sylow subgroup of I with p 6= l. Suppose P / I. Then P = 1.

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(i) Suppose (G : L) = 2 and let x ∈ X. Then G is the dihedral group Dl generated by the involution τ of Gx and a generator λ of L with the relation λτ = λ−1 . Proof of (a) and (b): Since G is solvable, L is Abelian. S Choose a system P Y of representatives of the L-orbits of X. Then X = · y∈Y Ly and l = y∈Y |Ly|. Given y, y 0 ∈ Y , choose σ ∈ G with σy = y 0 . Then σ(Ly) = σLσ −1 σy = Ly 0 . Thus, all L-orbits have the same length n. Therefore, l = |Y |n. By definition, L 6= 1. Choose λ ∈ L, λ 6= 1. Then there is an x ∈ X with λx 6= x, because G is faithful. Hence, Lx < L. Thus, n = |Lx| = (L : Lx ) > 1. Since l is prime, n = l and |Y | = 1. Therefore, L is transitive. Next consider λ ∈ L and x ∈ X with λx = x. For each x0 ∈ X choose 0 λ ∈ L with λ0 x = x0 . Then λx0 = λλ0 x = λ0 λx = λ0 x = x0 , so λ = 1. Thus, Lx = 1 and L is regular. Hence, by Remark 21.7.1(b), G = Gx n L. In addition, |L| = (L : Lx ) = l. Therefore, L ∼ = Cl . Proof of (c): Assume σ ∈ CG (L). Fix x ∈ X and write σ = λρ with λ ∈ L and ρ ∈ Gx . Since L is Abelian, ρ ∈ CG (L). For each x0 ∈ X choose λ0 ∈ L with λ0 x = x0 . Then ρx0 = ρλ0 x = λ0 ρx = λ0 x = x0 . Therefore, ρ = 1 and σ ∈ L, as asserted. Proof of (d): Conjugation of L by elements of G embeds G/L into Aut(L) (by (c)). The latter is isomorphic to F× l . Hence, G/L is a cyclic group of order dividing l − 1. By (b), Gx ∼ = G/L. Proof of (e): By (d), l2 - |G|, so L is an l-Sylow subgroup of G. Since L is normal, it is the unique l-Sylow subgroup of G. Proof of (f) and (g): Let λ be a generator of L. Choose x0 ∈ X. By (a) and (b), X = {x0 , λx0 , . . . , λl−1 x0 }. The bijection X → Fl mapping λb x0 onto b for b = 0, . . . , l −1 identifies L as a permutation group of Fl such that λb 0 = b for all b ∈ Fl . For an arbitrary x ∈ Fl we have λb x = λb λx 0 = λb+x 0 = x + b. Let σ ∈ G and a ∈ Fl with σλσ −1 = λa . Suppose first σ0 = 0. Then σx = σλx 0 = σλx σ −1 0 = λax 0 = ax. In the general case there is a b ∈ Fl with σ0 = λb 0. Then (λ−b σ)λ(λ−b σ)−1 = λa , so by the preceding case, λ−b σx = ax. Therefore, σx = ax + b. This gives an embedding of G into AGL(1, Fl ) as permutation groups. In particular, each σ ∈ G r L belongs to exactly one Gx (Remark 21.7.1(d)). Proof of (h): Under the assumptions of (h), P ∩ L = 1 and P, L / I. Hence, P ≤ CG (L) = L. Therefore, P = 1. Proof of (i): Suppose (G : L) = 2. Choose a generator λ of L and a generator τ of Gx . Then τ 2 = 1, λl = 1, and λτ = λt for some t ∈ Fl satisfying t2 = 1. By (c), τ ∈ / L = CG (L). Hence, t = −1. Consequently, G ∼ = Dl .

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Definition 21.7.3: Primitive permutation groups. Let G be a transitive permutation group of a nonempty set X. A block of G is a subset Y of X with the following property: For each σ ∈ G either σY = Y or Y ∩ σY = ∅. Let GY = {σ ∈ G | σY = Y }. Choose S a system of representatives R for the left cosets of GY in G. Then X = · ρ∈R ρY and G acts transitively on the set {ρY | ρ ∈ R} = {σY | σ ∈ G}. The set X itself, the empty set, and each {x} with x ∈ X are trivial blocks. Call G imprimitive if it has a nontrivial block, otherwise G is primitive. Lemma 21.7.4: Let G be a transitive permutation group of a set X and let x ∈ X. Then G is primitive if and only if Gx is maximal in G. Proof: Suppose first G is imprimitive. Then G has a nontrivial block Y . Since G is transitive, we may assume x ∈ Y . Then Gx ≤ GY . Since Y ⊂ X and G is transitive, GY < G. Now choose y ∈ Y , y 6= x, and σ ∈ G with σx = y. Then σY = Y but σx 6= x, so σ ∈ GY r Gx . Therefore, Gx is not maximal. Conversely, suppose G has a subgroup H with Gx < H < G. Put Y = Hx. Then H = GY . Moreover, Y is a G-block. Indeed, let σ ∈ G and y, y 0 ∈ Y with σy = y 0 . Then σηx = η 0 x for some 0 η, η ∈ H. Hence, (η 0 )−1 ση ∈ Gx < H, so σ ∈ H. Therefore, σY = Y . Further, σY 6= Y for each σ ∈ G r H. Hence, Y ⊂ X. By definition, x ∈ Y . In addition, σx 6= x for each σ ∈ H r Gx . Hence, Y contains more than one element. Thus, Y is a nontrivial G-block. Consequently, G is imprimitive. Lemma-Definition 21.7.5: Let G be a permutation group of a set A. We say G is doubly transitive if one of the following equivalent conditions hold: (a) G is transitive on A and there exists a ∈ A such that Ga is transitive on A r{a}. (b) G is transitive on A and for each a ∈ A, Ga is transitive on A r{a}. (c) Let a1 , a2 , b1 , b2 ∈ A with a1 6= a2 and b1 6= b2 . Then there is a τ ∈ G with τ a1 = b1 and τ a2 = b2 . Proof: Statements (a) and (b) are equivalent because G is transitive. Thus, it suffices to prove that (b) and (c) are equivalent: Suppose (b) holds. Let a1 , a2 , b1 , b2 be as in (b). Choose a ∈ A and α, β ∈ G with αa1 = a and βb1 = a. Then αa2 6= a and βb2 6= a. Hence, there is a σ ∈ Ga with σαa2 = βb2 . Put τ = β −1 σα. Then τ a1 = b1 and τ a2 = b2 . Conversely, suppose (c) holds. Let a ∈ A and b, b0 ∈ A r{a}. Then there is a σ ∈ G with σa = a and σb = b0 . Remark 21.7.6: Doubly transitive subgroups of AGL(1, Fp ). If x1 , x2 , y1 , y2 are elements in Fp with x1 6= x2 and y1 6= y2 , then there are unique a ∈ F× p and b ∈ Fp such that ax1 + b = y1 and ax2 + b = y2 . Thus, AGL(1, Fp ) is doubly transitive.

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475

Conversely, let G be a doubly transitive subgroup of AGL(1, Fp ). Then for each (a, b) ∈ (F× p , Fp ) there is a σ ∈ G with σ(1) = a + b and σ(0) = b. Then σ(x) = ax + b for all x ∈ Fp . It follows that G = AGL(1, Fp ). One of the main tools in the proof of Schur’s Proposition below is the group ring Z[H] of a finite Abelian group P H (Section 16.6). Each ρ ∈ Z[H] has a unique presentation as a sum η∈H cη η with coefficients cη ∈ Z. Define the support of ρ to be the set Supp(ρ) = {η ∈ H | cη 6= 0}. Then hSupp(ρ)i = hηP ∈ H | cη 6= 0i. 0 0 Let ρ0 = η∈H cη η be another element of Z[H]. Suppose cη , cη ≥ 0 P P 0 0 for all η. Then ρρ = τ ∈H ηη 0 =τ cη cη 0 τ and there are no cancellations among the coefficients of ρρ0 . Hence, Supp(ρρ0 ) = Supp(ρ)Supp(ρ0 ). 0 Consider a prime number p. We write ρ ≡ ρ0 mod Pp if cη ≡p cη mod p p for all η ∈ H. Since Z[H] is a commutative ring, ρ ≡ η∈H cη η mod p. Proposition 21.7.7 (Schur): Let G be a primitive permutation group of A = {1, 2, . . . , n}. Suppose G contains an n-cycle ν. Then G is doubly transitive or n is a prime number. Proof: We assume G is not doubly transitive, n is a composite number, and draw a contradiction. Denote the identity map of A by ε. Let H = hνi. Claim A: G = HG1 = G1 H and H ∩ G1 = {ε}. Indeed, H is transitive. Hence, for each σ ∈ G there is an η ∈ H with η(1) = σ(1). Therefore, σ = η · η −1 σ ∈ HG1 . Consequently, G = HG1 . Likewise, G = G1 H. It follows that each σ ∈ G can be uniquely written as σ = ησ1 with η ∈ H and σ1 ∈ G1 . Likewise, there are unique η 0 ∈ H and σ10 ∈ G1 with σ = σ10 η 0 P Claim B: Consider Z[H] as a subring of Z[G]. Let γ = σ∈G1 σ and R = Sr {ρ ∈ Z[H] | γρ = ργ}. Then there is a partition H = · j=1 Hj with H1 = {ε} Pr P and r ≥ 3 such that R = j=1 Zαj where αj = η∈Hj η. Let A1 , . . . , Ar with A1 = {1} be the G1 -orbits of A. Since G is not doubly transitive, G1 is not transitive on {2, . . . , n} (Lemma-Definition 21.7.5). Hence, r ≥ 3. Next define a map f : H → A by f (η) = η(1). Check that f is bijective and G1 ηG1 ∩ H = f −1 (G1 · η(1)) for each η ∈ H. For each j choose ηj ∈ H with S ηj (1) ∈ Aj . Let Hj = G1 ηj G1 ∩ H = f −1 (G1 · ηj (1)) = f −1 (Aj ). Then r H = · j=1 Hj and H1 = {ε}. P Now consider an element ρ = η∈H cη η in Z[H] with cη ∈ Z for each η ∈ H. Then ρ ∈ R if and only if X X X X (1) cη ση = cη ησ. σ∈G1 η∈H

σ∈G1 η∈H

By Claim A, the ση (resp. ησ) on the left (resp. right) hand side of (1) are distinct. Hence, (1) holds if and only if for all σ, σ 0 ∈ G1 and η, η 0 ∈ H the

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equality ση = η 0 σ 0 implies cη = cη0 . In other words, cη is constant as η ranges cηPis constantPon each Hj . Thus, there are on G1 ηG1 ∩ H. In particular, Pr r c1 , . . . , cr ∈ Z with ρ = j=1 cj η∈Hj η = j=1 cj αj . This concludes the proof of Claim B. Claim C: Let P ρ ∈ R. Put K = hSupp(ρ)i. Then K = {ε} or K = H. WritePρ = η∈H cη η. By Claim B, cη are constant on Hj , j = 1, . . . , r. Put κ = η∈Supp(ρ) η. Then Supp(κ) = Supp(ρ) and the coefficients of κ (which are 0 or 1) are also constant on each Hj . Hence, by Claim B, κ ∈ R. This means κγ = γκ. Since the coefficients of both κ and γ are nonnegative, Supp(κ)Supp(γ) = Supp(γ)Supp(κ). Therefore, KG1 = G1 K. This implies that KG1 is a subgroup of G which contains G1 . Since G is primitive, G1 is a maximal subgroup of G (Lemma 21.7.4). Hence, KG1 = G1 or KG1 = G. Thus, |K| · |G1 | = |G1 | or |K| · |G1 | = |G| = |H| · |G1 | (by Claim A). Therefore, K = {ε} or K = H, as claimed. Now choose a prime divisor p of n. Denote the unique subgroup of H of order p by P . Since n is composite, P is a proper subgroup of H. Claim D: For each j, Hj r P is a union of cosets of P . Indeed, P = {η ∈ H | η p = 1}. Hence, for each η0 ∈ H the set {η ∈ H | η p = η0 } is either empty or a coset of P . In the latter case, it consists of p elements. If we prove for each η0 ∈ H with η0 6= ε that (2) p|#{η ∈ Hj | η p = η0 }, then {η ∈ Hj | η p = η0 } is either empty or coincides with {η ∈ H | η p = η0 }. In the latter case it is a coset of P . This will P prove Claim D. To prove (2), consider an element ρ = η∈H cη η in Z[H]. For each η ∈ H integer between 0 and p − 1 satisfying c0η ≡ cη mod p. let c0η be the P unique 0 0 0 Put ρ = η∈H cη η. If cη is constant on each Hj , then so is cη . So, by 0 0 Claim B, ρ ∈ R implies ρ ∈ R. By definition, Supp(ρ ) ⊆ Supp(ρ). Also, 0 0 ρ1 ≡ ρ2 mod p, with ρ1 , ρP 2 ∈ Z[H], is equivalent to ρ1 = ρ2 . p 0 Specifically, let ρ = η∈Hj η . Then Supp(ρ ) ⊆ Supp(ρ) ⊆ {τ p | τ ∈ H}. The right hand side is a proper subgroup of H (because p|n). Hence, hSupp(ρ0 )i < H.

(3) In addition, ρ ≡

P

η∈Hj

η

p

0 ≡ αjp mod p. Hence, ρ0 = αjp . By definition,

R is a subring of Z[H]. As αj is in R (Claim B), so is αjp . Therefore, ρ0 ∈ R. It follows from (3) and Claim C that Supp(ρ0 ) ⊆ hSupp(ρ0 )i ≤ {ε}. In particular, the coefficient of η0 in ρ0 is 0. In other words, the coefficient of η0 in ρ is divisible by p. But the latter coefficient is exactly the right hand side of (2). Consequently, (2) is true. Claim E: Suppose 1 < j ≤ r and |Hj ∩ P | ≤ p−1 2 . Then αj η = αj for all η ∈ P. P P Put α = αj , π = η∈P η, and µ = η∈Hj ∩P η. By Claim D, Hj r P = Sm Pm · i=1 λi P with λ1 , . . . , λm ∈ H. Put λ = i=1 λi . Then

21.7 On Permutation Groups

477

P P P (4) λ ∈ Z[H] and α = η∈Hj η = η∈Hj r P η + η∈Hj ∩P η = λπ + µ For each η ∈ P , multiplication from the right by η yields a bijection of P onto itself. Hence, πη = π. Thus, in order to prove that αη = η, it suffices (by (4)) to prove that µ = 0. P P To this end observe that π 2 = η∈P πη = η∈P π = pπ. Similarly, πµ = kπ with k = |Hj ∩ P |. Since Z[H] is a commutative ring, this implies α2 − 2kα = pλ2 π + µ2 − 2kµ. Hence, (µ2 − 2kµ)0 = (α2 − 2kα)0 . By Claim B, α is in R. Therefore, by the proof of Claim D, (α2 − 2kα)0 ∈ R. Thus, (µ2 − 2kµ)0 ∈ R. In addition, Supp(µ2 − 2kµ)0 ⊆ Supp(µ2 − 2kµ) ⊆ P . Since p 6= n, P is a proper subgroup of H. Therefore, by Claim C, Supp (µ2 − 2kµ)0 ⊆ {ε}. It follows that (5)

µ2 − 2kµ = pρ + lε

Pk with ρ ∈ Z[H], ε ∈ / Supp(ρ), and l ∈ Z. By definition, µ = i=1 ηi , where η1 , . . . , ηk are distinct elements of H. Each row of the matrix B = (ηi ηj )1≤i,j≤k consists of distinct elements of H. Hence, each element of H appears at most k times as an entry of B and therefore as a summand in P k 2 2 i,j=1 ηi ηj = µ . Thus, the coefficients of µ − 2kµ lie between −2k and k. Since 2k < p, this implies that ρ in (5) is 0. It follows that (5) may be rewritten as µ2 = 2kµ + lε. Thus, ηi ηj ∈ {η1 , . . . , ηk , ε} for all i, j. This means that Supp(µ) ∪ {ε} = {η1 , . . . , ηk , ε} is a subgroup of P . But k + 1 < p. Hence, Supp(µ) ∪ {ε} = {ε}. As j > 1 ε∈ / Supp(µ) (Claim B). Thus, Supp(µ) = ∅. Consequently, µ = 0, as claimed. Pr Sr End of proof: By Claim B, · j=2 Hj ∩ P = P r{ε}. Hence, j=2 |Hj ∩ P | = p − 1. Since r ≥ 3, there is a j ≥ 2 with |Hj ∩ P | ≤ p−1 2 . Hence, by Claim E, P is contained in the subgroup L = {σ ∈ G | γαj σ = γαj } of G. By Claim B, γαj = αj γ. By definition of γ, γσ = γ for each σ ∈ G1 . Therefore, G1 ≤ L and P G1 ≤ L. But P ∩ G1 = 1. Hence, G1 < L. By the primitivity of G, L = G. In particular, (γαj )−1 ∈ L. Therefore, γαj = ε. It follows that εσ = ε for each σ ∈ G, that is G is trivial. This contradiction concludes the proof of the proposition. The following result completes Schur’s Proposition: Proposition 21.7.8 (Burnside): Let G be a transitive permutation group of a set A of p elements for some prime number p. Then G is doubly transitive or isomorphic (as a permutation group) to a subgroup of AGL(1, Fp ). Proof: For each a ∈ A we have (G : Ga ) = |A| = p. Hence, p divides |G|. By Cauchy’s theorem, G has an element π of order p. Decompose π as a product of disjoint cycles. The order of π, namely p, is the least common multiple of the lengths of these cycles. Thus, π must be a cycle of length p. We may therefore identify A with Fp such that π(a) = a + 1 for each a ∈ Fp . Denote the vector space of all functions from Fp to Fp by V . The operations in V are defined as follows: (f + g)(x) = f (x) + g(x) and (af )(x) =

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Pp−1 af (x). For each f ∈ V there is a unique polynomial h(X) = i=0 ai X i in Pp−1 Fp [X] with i=0 ai xi = f (x) for all x ∈ Fp . Indeed, the latter condition gives a Vandermonde system of equations for a0 , . . . , ap−1 which determines them uniquely. Define the degree of f to be deg(h) . Then let Vk be the subspace of all f ∈ V with deg(f ) ≤ k. The action of G on Fp naturally gives an action of G on V : (σf )(x) = f (σ −1 x). The rest of the proof is divided into four parts: Part A: A basis of Vk . Let k be an integer between 0 and p − 1. Then 1, X, . . . , X k give a basis of Vk . Thus, dim(Vk ) = k + 1. Conversely, suppose f0 , . . . , fk are elements of V with deg(fj ) = j, then they are linearly independent, so they form a basis of Vk . Suppose now 1 ≤ k ≤ p − 1. Consider f ∈ V of degree k given by f (x) = ak xk + ak−1 xk−1 + · · · + a0 with ai ∈ Fp and ak 6= 0. Then (πf )(x) = ak (π −1 x)k + ak−1 (π −1 x)k−1 + · · · + a0 = ak (x − 1)k + ak−1 (x − 1)k−1 + · · · + a0 = ak xk + (−kak + ak−1 )xk−1 + lower terms. Hence, f (x) − (πf )(x) = kak xk−1 + lower terms. Thus, deg(πf ) = k and deg(f − πf ) = k − 1. Inductively define fk = f and fj−1 = fj − πfj for j = k, k − 1, . . . , 1. Then deg(fj ) = j, so f0 , . . . , fk form a basis of V . Since f0 , . . . , fk are linear combinations of f, πf, . . . , π k f , the latter k functions also form a basis of V all of its elements have degree k. Part B: The fixed space of G0 . Assume from now on G is not doubly transitive. Then Fp has at least three G0 -orbits, A0 , A1 , A2 . Let gi : Fp → Fp be the characteristic function of Ai , i = 0, 1, 2. Then σgi = gi for each σ ∈ G0 and g0 , g1 , g2 are linearly independent. Thus, at most one of the functions g0 , g1 , g2 is of degree 0 and each of them belongs to the subspace U = {f ∈ V | σf = f for each σ ∈ G0 } of V . We prove that U contains a function f with 1 ≤ deg(f ) ≤ p − 2. If 1 ≤ deg(gi ) ≤ p − 2 for some i between 0 and 2, take f = gi . Otherwise, either deg(g0 ) = deg(g1 ) = deg(g2 ) = p−1 or, say, deg(g0 ) = 0 and deg(g1 ) = deg(g2 ) = p−1. In the former case there are b1 , b2 , c1 , c2 ∈ F× p with deg(b1 g0 + b2 g1 ) ≤ p − 2 and deg(c1 g0 + c2 g2 ) ≤ p − 2. Since the functions b1 g0 + b2 g1 and c1 g0 + c2 g2 are linearly independent, at least one of them is not of degree 0 (because dim(V0 ) = 1), so has degree between 1 and p − 2. In the latter case there are b1 , b2 ∈ F× p with 1 ≤ deg(b1 g1 + b2 g2 ) ≤ p − 2. Let k = deg(f ). Part C: The subspaces Vk and Vk−1 are G-invariant. By part A, f, πf, . . . , π k f form a basis of Vk . Hence, hπi leaves Vk invariant. In addition, hπi is transitive on Fp . Hence, G = hπiG0 = G0 hπi. Moreover, σf = f for each σ ∈ G0 . Therefore, G0 Vk = Vk . Consequently, GVk = Vk .

21.8 Schur’s Conjecture

479

Next consider the G-invariant subspace W =

nX σ∈G

cσ · σf |

X

o

cσ = 0

σ∈G

of V . We claim that W ⊆ Vk−1 . Indeed, write each σ ∈ G as σ = σ0 π i with σ0 ∈ G0 . Then, in the notation of Part A, (σfP )(x) = (π i f )(x) = −i k f (π Px) = f (x − i) = ak x + lower terms. Hence, if σ∈G cσ = 0, then deg( σ∈G cσ ·σf ) ≤ k−1, as claimed. In particular, dim(W ) ≤ dim(Vk−1 ) = k. Finally, note that f − πf, πf − π 2 f, . . . , π k−1 f − π k f are in W and they are linearly independent. Hence, dim(W ) = k. Therefore, W = Vk−1 . It follows that Vk−1 is G-invariant. Part D: End of proof. Define the product gh of two functions g, h ∈ V by (gh)(x) = g(x)h(x). Then, σ(gh) = σ(g)σ(h) for each σ ∈ G and V1 = {g ∈ V | gh ∈ Vk for all h ∈ Vk−1 }. Since both Vk−1 and Vk are G-invariant, so is V1 . In particular, applying σ −1 to the identity map of Fp , we find a ∈ F× p and b ∈ Fp with σ(x) = ax + b for all x ∈ Fp . This gives the desired embedding of G into AGL(1, Fp ). The combination of Schur’s result and Burnside’s result gives the following characterization of transitive proper subgroups of AGL(1, Fp ): Corollary 21.7.9: Let G be a transitive permutation group of a set A of n elements. Suppose G contains a cycle of length n but G is not doubly transitive. Then n is a prime number p and G is isomorphic (as a permutation group) to a subgroup of AGL(1, Fp ). In particular, G is solvable. Proof: By Schur (Proposition 21.7.7), n = p is a prime number. By Burnside (Proposition 21.7.8), G is isomorphic to a subgroup of AGL(1, Fp ). It follows from Remark 21.7.1(d) that G is solvable.

21.8 Schur’s Conjecture Let K be a field and f ∈ K[X]. We say that f permutes K, f is a permutation polynomial on K, or f is bijective on K if the map x 7→ f (x) is a bijection of K onto itself. Let R be a ring, f ∈ R[X], and P a maximal ideal of R. We say f is a permutation polynomial modulo P , if the reduction of f modulo P is a permutation polynomial on R/P . In this section we consider a global field K and a polynomial f ∈ OK [X] with char(K) - deg(f ) which is a permutation polynomial modulo infinitely many prime ideals of OK . We prove a conjecture of Schur: f is composed of linear polynomials and Dickson polynomials Dn (a, X). The latter are defined in Z[a, X] by induction on n: D0 (a, X) = 2, D1 (a, X) = X, and (1) Dn (a, X) = XDn−1 (a, X) − aDn−2 (a, X) for n ≥ 2.

480

Chapter 21. Problems of Arithmetical Geometry

Thus, D2 (a, X) = X 2 − 2a, D3 (a, X) = X 3 − 3aX, and D4 (a, X) = X 4 − 4aX 2 + 2a2 . Call Dn (a, X) the Dickson polynomial of degree n with parameter a. Lemma 21.8.1: Let K be a field, a ∈ K, and n a nonnegative integer. Use X and Z for variables. Then: (a) Dn (a, Z + aZ −1 ) = Z n + an Z −n . (b) Suppose f ∈ K(Z) satisfies f (Z + aZ −1 ) = Z n + an Z −n . Then f (X) = Dn (a, X). (c) bn Dn (a, X) = Dn (b2 a, bX). (d) For n ≥ 3 we have Dn (a, X) = X n − naX n−2 + g(X) with g ∈ Z[a, X] of degree at most n − 3. Proof of (a): Check (a) for n = 0 and n = 1. Next assume (a) by induction for n − 2 and n − 1. Then Dn (a, Z + aZ −1 ) = (Z + aZ −1 )Dn−1 (a, Z + aZ −1 ) − aDn−2 (a, Z + aZ −1 ) = (Z + aZ −1 )(Z n−1 + an−1 Z 1−n ) − a(Z n−2 + an−2 Z 2−n ) = Z n + an Z −n Proof of (b): Put X 0 = Z + aZ −1 . Then f (X 0 ) = Dn (a, X 0 ) (by (a)) and [K(Z) : K(X 0 )] ≤ 2. Hence, X 0 is a variable. Therefore, f (X) = Dn (a, X). Proof of (c) and (d): Carry out induction on n.

Definition 21.8.2: Linear relation of polynomials. Let K be a field and f, g ∈ K[X] polynomials. We say f and g are linearly related over K if there exist a, b ∈ K × and c, d ∈ K with f (X) = ag(bX + c) + d. Lemma 21.8.3: Let K be a field, f ∈ K[X], K 0 an extension of K, and n a positive integer. Suppose char(K) - n and f (X) is linearly related over K 0 to Dn (a0 , X) for some a0 ∈ K 0 . Then f (X) is linearly related over K to Dn (a, X) for some a ∈ K. Proof: We prove the lemma only for n ≥ 3. The case n ≤ 2 may be checked directly. By assumption, there are α0 , β 0 ∈ (K 0 )× and a0 , γ 0 , δ 0 ∈ K 0 with f (X) = α0 Dn (a0 , β 0 X + γ 0 ) + δ 0 . Put a = (β 0 )−2 a0 , α = α0 (β 0 )n , γ = (β 0 )−1 γ 0 , and δ = δ 0 . By Lemma 21.8.1(c), f (X) = αDn (a, X + γ) + δ. By Lemma 21.8.1(d), f (X) = α(X + γ)n − nαa(X + γ)n−2 + lower terms = αX n + nαγX n−1 + α n2 γ 2 − na X n−2 + lower terms Hence, α ∈ K, nαγ ∈ K, and α n2 γ 2 − na ∈ K. By assumption, n 6= 0 in K. Therefore, γ ∈ K and a ∈ K. Finally, δ = f (0) − αDn (a, γ) ∈ K. Consequently, f (X) is linearly related to Dn (a, X) over K.

21.8 Schur’s Conjecture

481

Special cases of Dickson polynomials give rise to well known families of polynomials. For a = 0 and n ≥ 1 induction or Lemma 21.8.1(a) give Dn (0, X) = X n . The special case a = 1 gives the Chebyshev polynomials: Tn (X) = Dn (1, X). We extract properties of Chebyshev polynomials from Lemma 21.8.1: Lemma 21.8.4: Let K be a field and n a nonnegative integer. Then: (a) T0 (X) = 2, T1 (X) = X, and Tn (X) = XTn−1 (X) − Tn−2 (X) for n ≥ 2. (b) Tn (Z + Z −1 ) = Z n + Z −n . (c) Suppose f ∈ K(Z) satisfies f (Z + Z −1 ) = Z n + Z −n . Then f = Tn . (d) For each n ≥ 3 there is a g ∈ Z[X] of degree at most n − 3 satisfying Tn (X) = X n − nX n−2 + g(X). ˜ there are distinct (e) Suppose char(K) - n and n > 2. Then, for each a ∈ K ˜ x1 , x2 ∈ K with Tn (x1 ) = Tn (x2 ) = a. ˜ satisfying y 2 − ay + 1 = 0. Then y + y −1 = a. Proof of (e): Choose y ∈ K ˜ with z n = y, i = 1, . . . , n. Since char(K) - n, there are distinct z1 , . . . , zn ∈ K i Since n ≥ 3, there are i, j such that xi = zi + zi−1 and xj = zj + zj−1 are distinct. They satisfy Tn (xi ) = Tn (xj ) = a. Remark 21.8.5: Automorphisms of K(z). Let K be a field and z an indeterminate. Recall that PGL(2, K) is the quotient of GL(2, K) by the group of scalar matrices. Denote the image of a matrix ac db ∈ GL(2, K) 0 0 in PGL(2, K) by ac db . Thus, ac db = ac0 db 0 means that there is an 0 0 eb e ∈ K × with ac0 db 0 = ea 11.7.4 identifies PGL(2, K) with ec ed . Remark a b Aut(K(z)/K), where the action of c d on K(z) is defined by the formula: a b az+b c d z = cz+d . Lemma 21.8.6: Let K be an algebraically closedfield and n ≥ 3 an integer with char(K) - n. Consider the elements σ = 10 ζ0n and τ = 01 10 of PGL(2, K). Put ∆ = hσ, τ i. Then: (a) σ n = τ 2 = 1, σ τ = σ −1 , and ∆ ∼ = Dn . (b) Let ∆1 = hσ1 , τ1 i be a subgroup of PGL(2, K) with σ1n = τ12 = 1 and σ1τ1 = σ1−1 (so, ∆1 ∼ = Dn ). Then there is a λ ∈ PGL(2, K) with hσ1 iλ = λ hσi, τ1 = τ , and ∆λ1 = ∆. (c) Let z be an indeterminate. Then the fixed field of τ (resp. ∆) in K(z) is K(z + z −1 ) (resp. K(z n + z −n )). Proof of (a): Put ζ = ζn . All we need to do is to verify the relation σ τ = σ −1 :

0 1

1 0

1 0

0 ζ

0 1

1 ζ = 0 0

0 1 0 = . 1 0 ζ −1

Proof of (b): Denote the image of σ ∈ GL(2, K) in PGL(2, K) by [σ]. σ1 ] = σ 1 , Since K is algebraically closed, there are σ ˜1 , τ˜1 ∈ GL(2, K) with [˜ ˜1n = τ˜12 = 1. The minimal polynomial of σ ˜1 divides X n − 1. [˜ τ1 ] = τ1 , and σ ˜1 is conjugate Since n 6= char(K), X n − 1 has n distinct roots in K. Hence, σ

482

Chapter 21. Problems of Arithmetical Geometry

to a diagonal matrix a0 d0 with an = dn = 1. Therefore, [˜ σ1 ] is conjugate 1 0 −1 k to 0 a−1 d with a d = ζ and gcd(k, n) = 1. Replace ζ by ζ k , if necessary, to assume [˜ σ1 ] = σ. Let τ1 = wy xz with wy xz ∈ GL(2, K). Since τ1 σ = σ −1 τ1 , there is an a ∈ K × with w x w x 1 0 a 0 . = y z y z 0 ζ 0 aζ −1 Therefore, w = aw, xζ = ax, y = aζ −1 y, and zζ = aζ −1 z. Assume w 6= 0. Then a = 1, x = 0, y = 0, and z = 0 (use n 6= 2). This contradiction proves that w = 0. Hence, x 6= 0 and y 6= 0. Therefore, a = ζ and z = 0. Consequently, τ1 = y0 x0 . √ Put ρ = 0x √0y . Then, ρ−1 τ1 ρ = τ and ρ−1 σρ = σ. So, there exists λ ∈ PGL(2, K) as claimed. Proof of (c): Let x = z + z −1 and t = z n + z −n . Denote the fixed field of τ (resp. ∆) in K(z) by E1 (resp. E2 ). Then [K(z) : E1 ] = 2 and [K(z) : E2 ] = 2n. Deduce from τ z = z −1 and τ z −1 = z that τ x = x. Hence, K(x) is contained in E1 . On the other hand, z is a root of X 2 − xX + 1. Therefore, [K(z) : K(x)] ≤ 2. Combined with the preceding paragraph, this implies K(x) = E1 . Similarly, K(t) ⊆ E2 and z is a root of the equation X 2n − tX n + 1 = 0. Hence, K(t) = E2 . Proposition 21.8.7 (M¨ uller): Let K be an algebraically closed field of characteristic p, f ∈ K[X] a polynomial of prime degree l 6= p, and t an indeterminate. Suppose G = Gal(f (X) − t, K(t)) is solvable. Then either G∼ = Dl and f is linearly related to = Cl and f is linearly related to X l or G ∼ the Chebyshev polynomial Tl . Proof: By assumption, f (X) − t is a separable polynomial of degree l which is irreducible. List its zeros in K(t)s as x1 , . . . , xl . Then F = K(x1 , . . . , xl ) is the Galois closure of K(xj )/K(t), G = Gal(F/K(t)) acts faithfully and transitively on {x1 , . . . , xl }. When l = 2, G ∼ = C2 and f (X) is linearly related to X 2 (use that 2 6= p). So, assume from now on l > 2. Then, we may apply Lemma 21.7.2 with {x1 , . . . , xl } replacing X and Hj = Gal(F/K(xj )) replacing Gxj , j = 1, . . . , l. In particular, let L be a minimal normal subgroup of G. Then, by Lemma 27.7.2(b,e), L ∼ = Cl and L is the unique l-Sylow subgroup of G. Let m = [F : K(xj )] and g = genus(F/K). In the remaining parts of the proof we apply Riemann-Hurwitz to prove that g = 0 and m ≤ 2. We then show: If m = 1, then G ∼ = Cl and f is linearly related to X l . If m = 2, ∼ then G = Dl and f is linearly related to Tl .

21.8 Schur’s Conjecture

483

Part A: The Riemann-Hurwitz formula. Denote the set of all prime divisors of F/K which ramify over K(t) by R. It includes all P prime divisors ). Hence, Diff(F/K(t)) = of F/K which ramify over K(x j q∈R dq q and P Diff(F/K(xj )) = q∈R dq,j q (Section 3.6). Here dq (resp. dq,j ) are the different exponents of q over K(t) (resp. K(xj )). Since K is algebraically closed, deg(q) = 1 for each q ∈ R. Therefore, Riemann-Hurwitz formula for F/K(t) and F/K(xj ) becomes: X (2a) dq 2g − 2 = −2ml + q∈R

(2b)

2g − 2 = −2m +

X

dq,j ,

j = 1, . . . , l.

q∈R

Subtract the sum of the l equations (2b) from (2a): (3)

2(g − 1)(1 − l) =

X q∈R

dq −

l X

dq,j ).

j=1

Now consider q ∈ R. Denote the inertia group of q over K(t) by Iq . Then the inertia group of q over K(xj ) is Hj ∩ Iq . To compute the qth term in (3) we distinguish between two cases. Case A1: Iq acts intransitively on {x1 , . . . , xl }. By Lemma 21.7.2(a), L 6≤ Iq . Since L ∼ = Cl , we have L∩Iq = 1. Thus, Iq ∼ = LIq /L ≤ G/L. Since G/L is / L. Hence, by Lemma cyclic, Iq is cyclic. Choose a generator τ of Iq . Then τ ∈ 21.7.2(g), there is a k between 1 and l with τ ∈ Hk and therefore Iq ≤ Hk . This implies that q|K(xk ) is unramified over K(t). Therefore, dq = dq,k (Lemma 3.5.8). For j 6= k we have Iq ∩ Hj = 1 (Lemma 21.7.2(g)). Hence, q is unramified over K(xj ) and dq,j = 0 (Remark 3.5.7(b)). Consequently, the qth term in the right hand side of (3) is 0. Case A2: Iq acts transitively on {x1 , . . . , xl }. Then for each j we have (Iq : Iq ∩ Hj ) = l. Hence, l divides |Iq |. Therefore, by Lemma 21.7.2(e), L ≤ Iq . Let P be the trivial group if p = 0 and a p-Sylow subgroup of Iq if p 6= 0. Then P is a normal subgroup of Iq [Cassels-Fr¨ohlich, p. 29, Thm. 1(ii)]. Hence, by Lemma 21.7.2(h), P is trivial. Thus, Iq is cyclic [Cassels-Fr¨ohlich, p. 31, Cor. 1]. In particular, each Sylow subgroup of Iq other than L is normal, hence trivial (Lemma 21.7.2(h)). Therefore, Iq = L. Denote the fixed field of L, hence of Iq , in F by E. By Lemma 21.7.2(b), E/K(t) is Galois of degree m, K(xj )E = F , and K(xj ) ∩ E = K(t). By the preceding paragraph, q is totally ramified over E and q|E is unramified over K(t). Hence, q is unramified over K(xj ) and q|K(xj ) is totally and tamely ramified over K(t). Therefore, dq = l − 1 and dq,j = 0 (Remark 3.5.7). There are exactly m prime divisors of F/K lying over q|K(t) . Hence, they contribute m(l − 1) to the right hand side of (3).

484

Chapter 21. Problems of Arithmetical Geometry

Part B: Computation of g. Consider a prime divisor p of K(t)/K. Let q be a prime divisor of F/K over p. Suppose Iq acts transitively on {x1 , . . . , xl }. By Case A2, p totally ramifies in K(x1 ). Hence, there are only finitely many such p. Denote their number by r. Conversely, suppose Iq acts intransitively on {x1 , . . . , xl }. Case A1 gives k with q|K(xk ) unramified over K(t). Choose ρ ∈ G with ρxk = x1 . Then ρq|K(x1 ) is unramified over K(t). Hence, p is not totally ramified in K(x1 ). Thus, by Case A1, only the r prime divisors p with Iq transitive contribute to the right hand side of (3). The contribution of each of them is, by Case A2, m(l − 1). Thus, (3) simplifies to 2(l − 1)(1 − g) = rm(l − 1) and furthermore to (4)

2(1 − g) = rm.

By Example 2.3.11, the infinite prime divisor p∞ of K(t)/K totally ramifies in K(x1 ). Hence, r ≥ 1. Therefore, by (4), g = 0 and rm = 2. It follows that, either m = 1 and r = 2 or m = 2 and r = 1. Part C: Suppose m = 1 and r = 2. Then K(x1 ) = F is a Galois extension of degree l of K(t). Hence, G ∼ = Cl . Since r = 2, K(t) has a finite place pc , mapping t to c ∈ K, which totally ramifies in F . In particular, pc has only one prime divisor in F . Hence, f (X) − c has only one root in K. It follows that there are a ∈ K × and b ∈ K with f (X) − c = (aX + b)l . Thus, f (X) is linearly related to X l . Part D: Suppose m = 2 and r = 1. By Lemma 21.7.2(i), G is the dihedral group Dl generated by the involution τ of H1 and a generator σ of L with the relation σ τ = σ −1 . By part B, g = 0. Since K is algebraically closed, each prime divisor of F/K has degree 1. Hence, F = K(z) for some z (Example 3.2.4). Lemma with σ 0 = 1 0 21.8.6(b) gives an0 automorphism 0 λ1 of Aut(F/K) −1 −1 0 λσλ = 0 ζ (and ζ = ζl ) and τ = λτ λ = 1 0 . Put z = λz, x0 = λx1 , and t0 = λt. Then, K(x0 ) is the fixed field in K(z 0 ) of 01 10 and K(t0 ) is the D E fixed field in K(z 0 ) of 10 ζ0 , 01 10 . By Lemma 21.8.6(c), the former field is K(z 0 + (z 0 )−1 ) and the latter field is K((z 0 )l + (z 0 )−l ). Therefore, there are κ1 ∈ Aut(K(x0 )/K) and κ2 ∈ Aut(K(t0 )/K) with x0 = κ1 (z 0 + (z 0 )−1 ) and t0 = κ2 ((z 0 )l + (z 0 )−l ). obius transformations having coefficients in K. Identify λ, κ1 , κ2 with M¨ Put µ = λ−1 κ1 and ν = λ−1 κ2 . Then x1 = µ(z 0 + (z 0 )−1 ) and t = ν((z 0 )l + (z 0 )−l ). Hence, ν −1 ◦ f ◦ µ(z 0 + (z 0 )−1 ) = (z 0 )l + (z 0 )−l . Since z 0 is transcendental over K, Lemma 21.8.4(c) implies ν −1 ◦ f ◦ µ = Tl . Part E: Conclusion of the proof. It remains to prove that µ and ν are linear polynomials. Assume first ν is not a polynomial. Then ν −1 is not a polynomial. Thus, −1 −1 (∞) = ac . ν (X) = aX+b cX+d with a, b, c, d ∈ K and c 6= 0. It follows that ν Use Lemma 21.8.4(e) to find distinct e1 , e2 ∈ K with Tl (ei ) = ac . Then

21.8 Schur’s Conjecture

485

f (µ(ei )) = ν(Tl (ei )) = ν( ac ) = ∞. Since f is polynomial, µ(ei ) = ∞, i = 1, 2. This contradiction to the injectivity of µ on K ∪ {∞} proves that ν is a linear polynomial. Finally, assume that µ is not a polynomial. Then e = µ(∞) ∈ K. Hence, by the preceding paragraph, Tl (∞) = ν −1 (f (µ(∞))) = ν −1 (f (e)) ∈ K, which is a contradiction. Consequently, µ is a linear polynomial. Lemma 21.8.8: Let K be an algebraically closed field, f ∈ K[X] a polynomial of degree n, x a transcendental element over K, and t = f (x). Suppose char(K) - n. Then Gal(f (X) − t, K(t)) contains an n-cycle. Proof: By assumption, f (X) − t has n distinct roots x1 , . . . , xn in K(t)s . Put x = x1 and F = K(x1 , . . . , xn ). Let v∞ be the valuation of K(t)/K with v∞ (t) = −1. By Example 2.3.11, v∞ is totally ramified in K(x). Thus, v∞ has a unique extension w to K(x). ˆ (By Example 3.5.1, Denote the completion of E = K(t) at v∞ by E −1 ∼ ˆ and E(x)/ ˆ ˆ ˆ E E = K((t )).) By Lemma 3.5.3, f (X) − t is irreducible over E ˆ ˆ is a totally ramified extension of degree n. Since char(K) - n, E(x)/E is ˆ n ) is a compositum ˆ Then Fˆ = E(x ˆ 1 ) · · · E(x tamely ramified. Put Fˆ = F E. ˆ ˆ ˆ of tamely ramified extensions of E. Hence, F /E is tamely ramified [CasselsFr¨ohlich, p. 31, Cor. 2]. Since K is algebraically closed, the residue degree ˆ is 1. Therefore, Fˆ /E ˆ is totally and tamely ramified. Therefore, of Fˆ /E ˆ ˆ ˆ ˆ Gal(F /E) is cyclic [Cassels-Fr¨ ohlich, p. 31, Cor. 1]. It follows that E(x)/ E ∼ ˆ ˆ ˆ ˆ ˆ is a Galois extension. Hence, F = E(x) and Gal(f (X) − t, E) = Gal(F /E) ˆ is transitive. is cyclic of order n. It follows that Gal(f (X) − t, E) ˆ Choose a generator σ of Gal(f (X) − t, E). It decomposes into disjoint cycles whose lengths sum up to n. Since hσi is transitive, there is only one cycle. Thus, σ is an n-cycle. Finally, note that σ ∈ Gal(f (X) − t, K(t)). Let K be a field and f ∈ K[X] a polynomial with deg(f ) > 1. Then f = f1 ◦ f2 ◦ · · · ◦ fr with fi ∈ K[X] indecomposable and deg(fi ) > 1. Call each fi a decomposition factor of f over K. Lemma 21.8.9: Let K be a field, t an indeterminate, and f ∈ K[X] an indecomposable polynomial over K of degree n with char(K) - n. Then Gal(f (X) − t, K(t)) is a primitive group. Proof: Let x be a root of f (X) − t in K(t)s and F the Galois closure of K(x)/K(t). By Lemma 21.7.4, it suffices to prove that Gal(F/K(x)) is a maximal subgroup of Gal(F/K(t)). In other words, we have to prove that K(x)/K(t) is a minimal extension. Assume there is a field E which lies strictly between K(t) and K(x). Let d = [K(x) : E] and m = [E : K(t)]. Then d > 1 and m > 1. Let v∞ be the valuation of K(t)/K with v∞ (t) = −1. Then v∞ is totally ramified in K(x). Thus, v∞ has a unique extension w0 to K(x), and w0 satisfies w0 (x) = −1 and has the same residue field as v∞ , namely K (Example 2.3.11). Let w be uroth the restriction of w0 to E. Then the residue field of E at w is K. By L¨

486

Chapter 21. Problems of Arithmetical Geometry

(Remark 3.6.2(a)), E is a rational function field over K. Therefore, there is a y ∈ E with w(y) = −1 and v(y) ≥ 0 for all other valuations v of E/K. Therefore, w0 (y) = −d and v(y) ≥ 0 for all other valuations v of K(x)/K. It follows that y = h(x) for some h ∈ K[X]. The degree of the pole divisor of y as an element of K(x) is d. Therefore, [K(x) : K(y)] = d. Since K(y) ⊆ E, this implies E = K(y). Therefore, deg(h) = d. Repeat the above arguments for t, y replacing y, x to find a polynomial g ∈ K[X] of degree m with t = g(y). This gives f (x) = t = g(y) = g(h(x)). Hence, f (X) = g(h(X)). This contradiction to the indecomposability of f proves that K(x)/K(t) is minimal. Let K be a field and f ∈ K[X] a polynomial of positive degree. Put f ∗ (X, Y ) =

f (X) − f (Y ) . X −Y

Proposition 21.8.10: Let K be an algebraically closed field and f ∈ K[X] a polynomial of degree l. Suppose char(K) - l, f is indecomposable over K, and f ∗ (X, Y ) is reducible. Then l is prime and f (X) is linearly related to X l or to Tl (X). Proof: Let t be an indeterminate. Since char(K) - l, f (X) − t has l distinct roots x1 , . . . , xl in K(t)s . Hence, F = K(x1 , . . . , xl ) is a Galois extension of K(t). Put G = Gal(F/K(t) and Hj = Gal(F/K(xj )), j = 1, . . . , l. View G as a permutation group of {x1 , . . . , xl }. Since f (X) − t is irreducible in K(t)[X], G is transitive. Since f is indecomposable, G is primitive (Lemma 21.8.9). Further, by Lemma 21.8.8, G contains an l-cycle. Since f ∗ (X, Y ) is reducible, l−1 ≥ 2. By Gauss, f ∗ (x1 , Y ) is reducible in K(x1 )[Y ] of degree l − 1. Therefore, H1 is intransitive on the roots x2 , . . . , xl of f ∗ (x1 , Y ). This means that G is not doubly transitive (Lemma-Definition 21.7.5). It follows from Corollary 21.7.9 that G is solvable. Consequently, by Proposition 21.8.7, f is linearly related to X l or to Tl (X). Lemma 21.8.11 ([Fried-MacCrae, Thm. 3.5]): Let K be a field, f ∈ K[X] a polynomial, and L a field extension of K. Suppose char(K) - deg(f ) and f is decomposable over L. Then f is decomposable over K. Pn Pm Proof: There are g(X) = i=0 ai X i and h(X) = j=0 bj X j in L[X] with f (X) = g(h(X)) and deg(g), deg(h) < deg(f ). Assume without loss L is algebraically closed. Divide g and f by am , if necessary, to assume am = 1. −1/n Then replace X by bn X, if necessary, to assume bn = 1. It follows that f , g, and h are monic. Finally, choose β ∈ L with h(β) = 0. Then replace X by X + β to assume b0 = 0. We prove that g, h ∈ K[X]. Indeed, f (X) = (X n + bn−1 X n−1 + · · · + b1 X)m (5)

+

m−1 X i=0

ai (X n + bn−1 X n−1 + · · · + b1 X)i .

21.8 Schur’s Conjecture

487

The highest power of X involved in the sum on the right hand side of (5) is X (m−1)n . Hence, for each k between 1 and n − 1, only the first term on the right hand side of (5) contributes to the coefficient of X mn−k . Therefore, this coefficient is mbn−k + pk (bn−k+1 , . . . , bn−1 , 1) with pk ∈ Z[Xn−k+1 , . . . , Xn ]. Observe that deg(f ) = mn, so char(K) - m. Induction on k proves that bn−k ∈ K. Thus, h ∈ K[X]. Finally, suppose am−1 , . . . , am−r+1 have been proved to be in K. Then am−r h(X)m−r + am−r−1 h(X)m−r−1 + · · · + a0 is a polynomial in K[X]. Its leading coefficient is am−r . Hence, am−r ∈ K. Consequently, g ∈ K[X]. Let K be a field and f ∈ K[X]. We say f is injective (resp. surjective) on K if the map x 7→ f (x) of K into itself is injective (resp. surjective). Proposition 21.8.12: Let K be a perfect field and f ∈ K[X] a polynomial of degree l with char(K) - l. (a) Suppose K is PAC and f is injective on K and indecomposable over K. Then l is a prime number and f is linearly related over K to a Dickson polynomial of degree l. (b) Suppose K is pseudo finite and f is injective or surjective on K. Then each composition factor of f is linearly related over K to a Dickson polynomial of a prime degree. Proof of (a): By assumption, there are no x, y ∈ K with f (x) = f (y) and x 6= y. Hence, there are no x, y ∈ K with f ∗ (x, y) = 0 and x 6= y. By ˜ Therefore, by Proposition Proposition 11.1.1, f ∗ (X, Y ) is reducible over K. ˜ to X l or to Tl (X). Both 21.8.10, l is prime and f is linearly related over K l X and Tl (X) are Dickson polynomials. It follows from Lemma 21.8.3 that f is linearly related over K to a Dickson polynomial of degree l. Proof of (b): Every injective (resp. surjective) polynomial on a finite field F of degree n is bijective. Since this is an elementary statement, it holds for every pseudo finite field (Proposition 20.10.4). In particular, f is bijective on K. Suppose f (X) = g(h(X)) with g, h ∈ K[X]. Then g is surjective on K and h is injective. By the preceding paragraph both g and h are bijective on K. It follows that each composition factor q of f over K is bijective on K. By (a), each composition factor of f is linearly related to a Dickson polynomial of a prime degree. Theorem 21.8.13 (Schur’s Conjecture): Let K be a global field and f ∈ K[X] a polynomial with char(K) - deg(f ). Suppose f is a permutation polynomial modulo p for infinitely many prime divisors p of K. Then each composition factor of f is linearly related over K to a Dickson polynomial of a prime degree. Proof: The statement “f is a permutation polynomial” is elementary. By Example 20.10.7(a) there is a pseudo finite field F containing K such that f is

488

Chapter 21. Problems of Arithmetical Geometry

a permutation polynomial on F . By Proposition 21.8.12(b), each composition factor of f is linearly related over F , hence over K (Lemma 21.8.3) to a Dickson polynomial of a prime degree. Theorem 21.8.14: Let K be a finite field and f ∈ K[X] a polynomial with char(K) - deg(f ). Suppose f permutes infinitely many finite extensions L of K. Then each composition factor of f is linearly related over K to a Dickson polynomial of a prime degree. Proof: Take a nonprincipal ultraproduct F of those finite extensions of K on which f is a permutation polynomial. Then f is a permutation polynomial on F . By Example 20.10.7(a), F is pseudo finite and K ⊆ F . By Proposition 21.8.12(b), each composition factor of f is linearly related over F , hence over K, to a Dickson polynomial of a prime degree. We end this section with examples which give converses to Proposition 21.8.12 and Theorem 21.8.14: Example 21.8.15: Cyclic polynomials. Let n be a positive integer and F a finite field of order q. Then F× q is a cyclic group of order q − 1. Hence, the map x 7→ xn is bijective on F if and only if gcd(n, q − 1) = 1. Suppose n ≥ 3 is odd. Let q1 , . . . , qr be the prime divisors of n. Then both 1 and 2 are relatively prime to qi , i = 1, . . . , r. Choose an integer a with a ≡ 1 mod qi , i = 1, . . . , r. Then both a and a + 1 are relatively prime to n. Dirichlet’s theorem (Corollary 6.3.2) gives infinitely many prime numbers p ≡ a+1 mod n. By the preceding paragraph, X n is a permutation polynomial modulo p for each of these p. The next lemma will be used to give more examples of Dickson polynomials which are permutation polynomials modulo infinitely many primes: Lemma 21.8.16: Let n be a positive integer, q a power of a prime number, and a ∈ Fq . Suppose gcd(n, q 2 − 1) = 1. Then Dn (a, X) permutes Fq . Proof: Put K = Fq and L = Fq2 . Then L× is a cyclic group of order q 2 − 1. Hence, the map z 7→ z n is bijective on L. Now consider x1 , x2 ∈ K with Dn (a, x1 ) = Dn (a, x2 ). Choose z1 , z2 ∈ L with zi + azi−1 = xi , i = 1, 2. Then Dn (a, z1 + az1−1 ) = Dn (a, z2 + az2−1 ). By Lemma 21.8.1(a), z1n + an z1−n = z2n + an z2−n . Denote the common value of the two sides of the latter equality by b. The product of the two solutions of the equation X + an X −1 = b is an . Both z1n and z2n are solutions. Hence, z1n = z2n or z1n = an z2−n . By the preceding paragraph, z1 = z2 or z1 = az2−1 . In both cases, x1 = x2 . Consequently, Dn (a, X) is a permutation polynomial on K. Example 21.8.17: Permutation polynomials modulo infinitely many primes. (a) Let n be a positive integer, K a number field, and a ∈ OK . Suppose gcd(6, n) = 1 and K ∩ Q(ζn ) = Q. Then Dn (a, X) is a permutation polynomial modulo infinitely many prime ideals of OK .

21.9 Generalized Carlitz’s Conjecture

489

Indeed, list the prime divisors of n as q1 , . . . , qr . Choose an integer b with b ≡ 2 mod qi , i = 1, . . . , r. Then gcd(b, n) = 1 and gcd(b2 − 1, n) = 1. Denote the Galois closure of K(ζn )/Q by L. Then choose σ ∈ Gal(L/K) with σζn = ζnb . Chebotarev density theorem (Theorem 6.3.1) gives infinitely many prime ideals P of OL with L/Q = σ. Let P be one of them. Then P the decomposition group of P over Q is in Gal(L/K). Put p = P|K . Then p = N p is a prime number. By the proof of Corollary 6.3.2, p ≡ b mod n. ¯ p |2 − 1, n) = 1. Consequently, by Lemma Hence, by the choice of b, gcd(|K 21.8.16, Dn (a, X) is a permutation polynomial modulo p. (b) Let n be a positive integer, p a prime number, q a power of p, K a finite separable extension of Fq (t) which is regular over Fq , and a ∈ OK . Suppose gcd(p(q 2 − 1), n) = 1. Then Dn (a, X) is a permutation polynomial modulo infinitely many prime divisors of K/Fq . Indeed, let L be the Galois closure of K/Fq (t), Fqm the algebraic closure of Fq in L, and ϕ(n) the Euler totient function. Then q ϕ(n) ≡ 1 mod n. Consider a large positive integer k. Choose τ in Gal(L/K) whose restriction to Fqm coincides with the restriction of Frobkϕ(n)+1 . Denote the conjugacy q class of τ in Gal(L/Fq (t)) by Con(τ ). Proposition 6.4.8 gives a prime divisor L/Fq (t) p of Fq (t) with deg(p) = kϕ(n) + 1 and = Con(τ ). Hence, there is a p L/Fq (t) prime divisor Q of L with = τ . Denote the restriction of Q to K by Q ¯ q |2 − 1 = q 2(kϕ(n)+1) − 1 ≡ q. Then deg(q) = deg(p) = kϕ(n) + 1. Hence, |K 2 q − 1 mod n. Consequently, by Lemma 21.8.16, Dn (a, X) is a permutation polynomial modulo q.

21.9 Generalized Carlitz’s Conjecture Let q be a power of a prime number p and f ∈ Fq [X] a polynomial of degree n. Suppose f is a permutation polynomial of Fqk for infinitely many k but f is not a p-power in Fq [X]. Theorem 21.8.14 describes the composition factors of f when p - n. In the general case the generalized Carlitz’s Conjecture states that gcd(n, q − 1) = 1. In particular, for p 6= 2, the conjecture predicts that n is odd (Carlitz’s Conjecture). The paper [Fried-Guralnick-Saxl] gives more precise information about permutation polynomials than Carlitz’s Conjecture does. Not only that it proves that conjecture for p > 3, but it gives information about Gal(f (X) − ˜ q (t)). Suppose without loss that f is indecomposable over Fq . Then three t, F are three cases: (a) p - n and G is cyclic or isomorphic to the dihedral group Dn . This is essentially contained in Section 21.8. (b) n = pm and G ∼ = H n Fm p , where H is a subgroup of GL(m, Fp ) acting m linearly on Fp . a

a

(c) p ∈ {2, 3}, n = p (p2 −1) with a ≥ 3 odd, and G normalizes the simple group PSL(2, Fpa ).

490

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The proof that no other cases arise uses the classification of finite simple groups and is beyond the scope of this book. One can find an elementary proof of Carlitz’s Conjecture in [Lenstra]. That proof argues with decomposition and inertia groups over Fq (t). Here we follow [Cohen-Fried] which, following a suggestion of [Lenstra, Remark 1], takes place over Fq ((t)). The first step in the proof translates the assumption “f permutes infinitely many finite extensions of Fq ” into a statement of a general nature: Lemma 21.9.1: Let f ∈ Fq [X] be a polynomial which permutes infinitely (Y ) many finite extensions of Fq . Then f ∗ (X, Y ) = f (X)−f has no absolutely X−Y irreducible factor which belongs to Fq [X, Y ]. Proof: Assume h ∈ Fq [X, Y ] is an absolutely irreducible factor of f ∗ (X, Y ). By Corollary 5.4.2 there is an n0 such that for every integer n > n0 there are distinct x, y ∈ Fqn with h(x, y) = 0. Then f (x) = f (y). Hence, f does not permutes Fqn , in contradiction to our assumption. The converse of Lemma 21.9.1 is also true (see Exercise 14). However, it is not used in proof of the generalized Carlitz Conjecture. Lemma 21.9.2: Let N/M be a finite cyclic extension, σ an element of Gal(M ) whose restriction to N generates Gal(N/M ), and h ∈ M [X] a separable irreducible polynomial which becomes reducible over N . Then σx 6= x for every root x of h. Proof: Assume h has a root x with σx = x. By assumption, N ∩ Ms (σ) = M . Hence, N ∩ M (x) = M , so N and M (x) are linearly disjoint over M . Therefore, [N (x) : N ] = [M (x) : M ]. It follows that h = irr(x, M ) is irreducible over N , a contradiction. Lemma 21.9.3: Let L/K be a finite cyclic extension, f ∈ K[X] a polynomial of degree at least 2 with a nonzero derivative f 0 , and z an indeterminate. (Y ) Suppose each irreducible factor of f ∗ (X, Y ) = f (X)−f in K[X, Y ] is reX−Y ducible over L. Then f (X) − z is separable and each ρ ∈ Gal(K(z)) whose restriction to L generates Gal(L/K) fixes exactly one root of f (X) − z. Proof: By assumption, (f (X) − z)0 = f 0 (X) 6= 0. All roots of f (X) − z are transcendental over K while all roots of f 0 (X) are algebraic over K. Hence, f (X) − z is relatively prime in K(z)[X] to its derivative. Therefore, f (X) − z is separable. Let Fˆ be the finite Galois extension of E = K(z) generated by L(z) and all roots of f (X) − z. Choose a generator ρ0 of L/K and let G∗ = {ρ ∈ Gal(Fˆ /E) | ρ|L = ρ0 }. ∗ ∗ For each root S x ∗of f (X) − z let Gx = {ρ ∈ G | ρx = x}. We have to prove ∗ that G = · x Gx , where x ranges over all roots of f (X) − z. We divide the rest of the proof into two parts.

21.9 Generalized Carlitz’s Conjecture

491

Part A: Disjointness. Let x and y be distinct roots of f (X)−z in Fˆ . Then f (x) = z and f (y) = z, so x, y are transcendental over K. By the above, (y) = f (X)−z f ∗ (X, y) = f (X)−f X−y X−y ∈ K(y)[X] is a separable polynomial and x is ∗ a root of f (X, y). Let h ∈ K[y, X] be an irreducible factor of f ∗ (X, y) with h(x) = 0. In particular, h is primitive. By assumption, h(X) is reducible in L[y, X]. Hence, h(X) is reducible in L(y)[X]. Now consider ρ ∈ G∗y . Then ρ ∈ Gal(Fˆ /K(y)) and ρ|L(y) generates the cyclic group Gal(L(y)/K(y)). By Lemma 21.9.2, ρ fixes no root of h. In particular, ρx 6= x. Thus, G∗x ∩ G∗y = ∅. Part B: Every ρ ∈ G∗ fixes a root of f (X) − z. Indeed, extend ρ0 in the unique possible way to a generator ρ∗0 of Gal(L(z)/K(z)). Then G∗ = {ρ ∈ Gal(Fˆ /E) | ρ|L(z) = ρ∗0 }. Hence, |G∗ | = [Fˆ : L(z)]. Next let x be a root of f (X)−z in Fˆ and extend ρ∗0 in the unique possible way to a generator ρ∗1 of Gal(L(x)/K(x)). Then G∗x = {ρ ∈ Gal(Fˆ /K(x)) | ρ|L(x) = ρ∗1 }. Hence, |G∗x | = [Fˆ : L(x)]. K(x) n

L(x)

Fˆ

n

K(z)

L(z)

K

L

Finally, let x1 , . . . , xn be the distinct roots of f (X) − z.S Then n = n deg(f ) = [K(xi ) : K(z)] = [L(xi ) : L(z)]. By Part A, | i=1 G∗xi | = Pn ∗ n[Fˆ : L(xi )] = [L(xi ) : L(z)][Fˆ : L(xi )] = [Fˆ : L(z)] = |G∗ |. i=1 |Gxi | =S n ∗ Hence, G = i=1 G∗xi . Thus, each ρ ∈ G∗ fixes some xi . Lemma 21.9.4: Let K be a field, f ∈ K[X] a monic polynomial of degree n > 1, r a divisor of n with char(K) - r, and z an indeterminate. Put ˜ Then E(x) contains an E = K((z√−1 )) and let√x be a root of f (X) − z in E. rth root r z and [E( r z) : E] = r. Proof: Let v be the unique discrete complete valuation of E/K such that v(z −1 ) = 1 (Example 3.5.1). Extend v to E(x) in the unique possible way. By Lemma 3.5.2, both E and E(x) satisfy Hensel’s lemma. Now write f (X) = X n + an−1 X n−1 + · · · + a0 . Put g(Y ) = 1 + an−1 Y + · · · + a1 Y n−1 + a0 Y n and y = x−1 . Then xn g(y) = f (x). Consider the polynomial h(T ) = T r −g(y) ∈ E(x)[T ]. Then h(1) = −an−1 y−· · ·−a1 y n−1 − a0 y n and h0 (1) = r 6= 0. By assumption, xn +an−1 xn−1 +· · ·+a0 = z. Hence,

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v(x) < 0, so v(y) > 0. It follows that, v(h(1)) > 0 and v(h0 (1)) = 0. Hensel’s lemma gives t ∈ E(x) with tr = g(y). Finally, let w = xn/r t. Then w ∈ E(x) and wr = xn tr = xn g(y) = f (x). Moreover, E(w)/E is a totally ramified extension of degree r (Example 2.3.8). Theorem 21.9.5 (Lenstra): Let p be a prime number, q a power of p, and f ∈ Fq [X] a polynomial of degree n > 1. Suppose f 6= f1p for all f1 ∈ Fq [X] and f permutes infinitely many finite extensions of Fq . Then gcd(n, q − 1) = 1. Proof: Put K = Fq . Since f is not a pth power of a polynomial in K[X] (Y ) and K is perfect, the derivative of f is nonzero. Put f ∗ (X, Y ) = f (X)−f . X−Y ∗ By Lemma 21.9.1, no absolutely irreducible factor of f (X, Y ) belongs to K[X, Y ]. Choose a finite, necessarily cyclic, extension L such that all absolutely irreducible factors of f ∗ (X, Y ) belong to L[X, Y ]. Then every irreducible factor of f ∗ (X, Y ) over K is reducible over L. Let z be an indeterminate. Put E = K((z −1 )) and F = L((z −1 )). Then E/K is a regular extension and EL = F (Example 3.5.1). Hence, res: Gal(F/E) → Gal(L/K) is an isomorphism. By Lemma 21.9.3, f (X) − z is separable. Choose a finite Galois extension Fˆ of E which contains F and all roots of f (X) − z in Es . Now assume gcd(n, q − 1) > 1. Choose a common prime divisor r of n and q √ − 1. Then p 6= √ r and K contains all roots of 1 of order r. Put E 0 = E( r z) and F 0 = F ( r z). By Lemma 21.9.4, E 0 /E and F 0 /F are cyclic extensions of degree r. Therefore, both maps res: Gal(F 0 /F ) → Gal(E 0 /E) and res: Gal(F 0 /E 0 ) → Gal(F/E) are isomorphisms. In addition, Lemma 21.9.4 implies that E 0 , F 0 ⊆ Fˆ . We may therefore choose σ, τ ∈ Gal(Fˆ /E) such that hresF 0 σi = Gal(F 0 /E 0 ) and hresF 0 τ i = Gal(F 0 /F ). Put ρ = στ . Then resL ρ generates Gal(L/K). E(x)

F (x)

E0

F0

E

F

K

L

Fˆ

By Lemma 21.9.3, ρ fixes a root x of f (X) − z in Fˆ . By Lemma 21.9.4, E(x) contains a root of z of order r, so E 0 ⊆ E(x). Hence, resE 0 ρ = id. On the other hand, resE 0 ρ = resE 0 τ 6= id. This contradiction proves that gcd(n, q − 1) = 1.

Exercises

493

Exercises 1. Let K be a perfect field and V a Zariski K-closed set with V (K) infinite. Use the decomposition-intersection procedure and Corollary 10.5.3 to prove that V contains a curve defined over K. 2. For each positive integer i, give an example of a PAC field K which is Ci+1 but not Ci . Hint: Take K to be imperfect, follow Example 21.2.8 and use Propositions 21.2.4 and 21.2.12. 3. Let f ∈ Q[X0 , . . . , Xl ] be an irreducible form of prime degree l. Prove that V (f ) contains a variety defined over Q. Hint: The intersection of l hyperplanes in Pl is nonempty. ˜ 4. Prove that almost all σ ∈ Gal(Q) satisfy this: Let f ∈ Q(σ)[X 0 , . . . , Xn ] ˜ be a polynomial of degree d ≤ n with a Q(σ)-zero. Then V (f ) contains a ˜ curve defined over Q(σ). Hint: Use Chevalley-Warning, the transfer theorem and Exercise 1. 5. Give an example of fields K0 and K such that K0 is algebraically closed ˜ ) where τ is the in K, K is C1 but K0 is not. Suggestion: Take K0 to be Q(τ conjugation on C. Then use Lemmas 20.5.3 and 21.3.5. 6. Show that Fp (t), with t a transcendental, is not C1 . Suggestion: Choose a nonsquare a ∈ Fp and prove that X 2 + aY 2 + tZ 2 has no nontrivial zero in Fp (t). 7. Let M and L be finite separable extensions of a global field K. Suppose that for almost all p ∈ P (K) the number of primes P ∈ P (L) lying over p and having a relative degree 1 is equal to the number of primes P ∈ P (M ) lying over p and having relative degree 1. Prove that [L : K] = [M : K] ˆ of L/K is equal to the Galois closure M ˆ of M/K and the Galois closure L [Kronecker]. Hint: Follow these steps. (a) Choose primitive elements x and y for L/K and M/K, respectively, integral over OK . Put f = irr(x, K) and g = irr(y, K). Use Lemma 21.5.2 ¯p to prove that for almost all p ∈ P (K) the numbers of roots of f and g in K are equal. (b) Let N be a finite Galois extension of K that contains both L and M . Use the transfer theorem (or alternatively the Chebotarev density theorem) to prove that every σ ∈ Gal(N/K) fixes the same numbers of roots of f and of g. ˆ and the elements τ ∈ (c) Apply (b) to the elements σ ∈ Gal(N/L) ˆ Gal(N/M ) and prove that deg(f ) = deg(g). ˆ all roots of g belong to N (σ), so M ˆ ⊆ L. ˆ (d) Observe: For σ ∈ Gal(N/L), 8. S LetxG be S a finitex group and H, I, N subgroups. Suppose G = H n N and I = = I n N . Deduce that H is not a proper x∈G x∈G H . Prove that GS subgroup of I. Hint: Observe that x∈G (IN )x = G.

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S 9. (a) Let A / B ≤ G be a tower of finite groups. Prove that | x∈G Ax | ≤ (G : B)|A|. (b) Let K ⊆ L be a finite separable extension of a global field. Observe: The Dirichlet S density δ of the set V (L/K) (Section 20.5) is equal to the Haar measure of σ∈Gal(K) Gal(L)σ . (c) Use (a), (b) and Lemma 21.5.3 to prove that if M is a finite Galois extension of L which is Kronecker conjugate to L over K, then [M : L] ≤ δ −1 [Klingen2]. 10. Exercise 1(c) of Chapter 13 gives a pair of polynomials f (X) = X 4 + 2X 2 , g(X) = −4X 4 − 4X 2 − 1 for which f (X) − g(Y ) is reducible. Show however, that f (X) − t and g(Y ) − t are not Kronecker conjugate over Q(t). Hint: Use Corollary 21.6.5. 11. Prove for nonzero integers a, b1 , . . . , bn that the following two statements are equivalent. (a) There exist ε1 , . . . , εn ∈ {0, 1} and c ∈ Z satisfying a = bε11 . . . bεnn c2 . (b) For almost all primes p, if b1 , . . . , bn are quadratic residues modulo p, then so is a. Hint: Combine the transfer theorem (or directly the Chebotarev density theorem) with Kummer theory for quadratic extensions over Q. 12. Let n be a positive integer, q a power of a prime number p, and a ∈ Fq . Suppose gcd(n, q − 1) > 1. Find distinct x, y ∈ Fq with Dn (a, x) = Dn (a, y). Combine this with Theorem 21.8.13 to supply a proof of the generalized Carlitz’s Conjecture (Theorem 21.9.5) in the case where p - deg(f ). Hint: Choose a common prime divisor l of n and q − 1. Then choose x = z + az −1 and y = ζz + a−1 ζz with ζ, z ∈ F1 , ζ l = 1, and ζ 6= 1. 13. Let p, l be distinct prime numbers, q a power of p, and a ∈ F× q . Suppose l|p2 − 1. Then Dl (a, X) permutes only finitely many fields Fqk . Hint: Use the hint of Exercise 12 to reduce to the case l|q + 1 and l 6= 2. Then (1)

(l−1)/2 Y Dl (a, X) − Dl (a, Y ) = (X 2 − αi XY + Y 2 + βi2 a), X −Y i=1

where αi = ζli + ζl−i and βi = ζli − ζl−i . [Schinzel2, p. 52]. Observe that ζlq = ζl−1 , so αi , βi2 ∈ Fq . Next note that each of the factors on the right hand side of (1) is absolutely irreducible. Therefore, it has Fqk -zeros off the diagonal if k is large. 14. Prove the following generalization of a theorem of MacCluer (See also [Fried6, Thm. 1]): Let K be a pseudo finite field, f ∈ K[X] a separable polynomial of degree n > 0, x an indeterminate. Put t = f (x). Suppose no (Y ) K-irreducible factor of f ∗ (X, Y ) = f (X)−f is absolutely irreducible. Then X−Y f permutes K.

Notes

495

Hint: Let x1 , . . . , xn be the n distinct roots of f (X) = t. Put F = ˆ be the algebraic closure of K in F , T = {τ ∈ K(x1 , . . . , xn ). Let K Gal(F/K(t) | P resKˆ τ = resKˆ σ}, and Ti = T ∩ Gal(F/K(xi )), i = 1, S n. . . , n. n Prove: |T | = i=1 |Ti | and Ti ∩ Tj = ∅ for i 6= j. Conclude that T = · i=1 Ti . ˆ ˜ ∪ {∞} with ψ(t) = a. Then Now consider a ∈ K. Find a K-place ψ: F → K choose σ in the decomposition group of ψ which belongs to T . There is an i with σxi = xi . Hence, b = ψ(xi ) is in K and satisfies f (b) = a. Consequently, f is surjective on K. Since K is pseudo finite, f is also injective.

Notes The decomposition-intersection procedure has been used by several authors. For example, [Greenleaf] uses it to prove that for each d > 0 almost all fields Qp are C2,d . Also, it appears in the first version [Fried-Sacerdote], of the stratification procedure. [Klingen4] gives a comprehensive survey on Kronecker classes of number fields. Problem 21.5.8 has a negative solution if the following statement holds for each infinite profinite group S G of rank ≤ ℵ0 : For each closed subgroup H of G of infinite index the set g∈G H g does not contain an open neighborhood of 1. There is an attempt in [Klingen2, Thm. 6] to give a negative answer to Problem 21.5.8. Unfortunately there is an error in the proof: A closed subgroup H of a profinite group G of countable rank may have uncountably many, rather that countably many (as claimed in [Klingen2]), conjugates in G. For example, there are uncountably many involutions in G(Q), all conjugate. Thus, Problem 21.5.8 is still open. Davenport’s original problem [Fried-Jarden3, Problem 19.26] asked if Vp (f ) = Vp (g) for nonconstants f, g ∈ Q[X] and almost all p implies f and g are linearly related. Remark 21.6.1 supplies counter examples to that problem. Davenport’s present problem 21.6.2 is a modification of the older one. Lemma 21.7.2 overlaps with a result attributed to Galois [Huppert, p. 163, Satz 3.6]. More on Dickson’s polynomials can be found in [Schinzel2, §1.4-1.5] and [Lidl-Mullen-Turnwald]. The original proof of Proposition 21.7.7 appears in [Schur2]. Our proof is an elaboration of [Lidl-Mullen-Turnwald, p. 126]. Likewise, the original proof of Proposition 21.7.8 appears in [Burnside1]. We have elaborated [LidlMullen-Turnwald p. 127]. [Schur1] proves that every polynomial f ∈ Z[X] of prime degree which is a permutation polynomial modulo infinitely many prime numbers p is a ˜ to Cyclic polycomposition of polynomials which are linearly related over Q nomials or Chebyshev polynomials. Schur conjectured that his result holds for an arbitrary number field K. [Fried1] uses the theory of Riemann surfaces to prove Schur’s conjecture. [Turnwald, Thm. 2] uses algebraic methods

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Chapter 21. Problems of Arithmetical Geometry

to generalize Fried’s result to arbitrary global field (See also [Lidl-MullenTurnwald]). Our proof of Schur’s conjecture (Theorem 21.8.13) is based on a result of M¨ uller (Proposition 21.8.7). Our version of the proof of Cohen-Fried proof of the generalized Carlitz’s Conjecture (Theorem 21.9.5) uses improvements of Bensimhoun and Haran. Partially building on ideas of [Lenstra], [Guralnick-M¨ uller, §8] further generalizes the generalized Conjecture to nonconstant separable morphisms between smooth projective curves over Fq .

Chapter 22. Projective Groups and Frattini Covers Every profinite group is a Galois group of some Galois extension (Leptin, Proposition 2.6.12). However, not every profinite group is an absolute Galois group. For example, the only finite groups that appear as absolute Galois groups are the trivial group and Z/2Z (Artin [Lang7, p. 299, Cor. 9.3]). This raises the question of characterizing the absolute Galois groups among all profinite groups. This question is still wide open. A more restrictive question finds a complete solution in projective groups: By Theorem 11.6.2, the absolute Galois group of a PAC field is projective. Conversely, for each projective group G there exists a PAC field K such that Gal(K) ∼ = G (Theorem 23.1.2). Thus, a profinite group G is projective if and only if it is isomorphic to the absolute Galois group of a PAC field. In this chapter we define a projective group as a profinite group G for which every embedding problem is weakly solvable. By Gruenberg’s theorem (Lemma 22.3.2), it suffices to weakly solve only finite embedding problems. This leads to the second characterization of projective groups as those profinite groups which are isomorphic to closed subgroups of free profinite groups (Corollary 22.4.6). Projective groups also appear as the universal Frattini covers of profinite groups (Proposition 22.6.1). Thus, as a preparation for decidability and undecidability results about families of PAC fields, we introduce in this chapter the basis properties of Frattini covers.

22.1 The Frattini Group of a Profinite Group Consider a profinite group G. Denote the intersection of all maximal open subgroups of G by Φ(G). Here we call a subgroup M maximal if there is no subgroup M 0 of G such that M < M 0 < G. The characteristic (and therefore normal) closed subgroup Φ(G) is called the Frattini group of G. We characterize the elements of Φ(G) as “dispensable generators” of G: Lemma 22.1.1: Let G be a profinite group. An element g ∈ G belongs to Φ(G) if and only if there is no proper closed subgroup H of G for which hH, gi = G. Also, if H is a closed subgroup for which H · Φ(G) = G, then H = G. Proof: Let g ∈ Φ(G) and let H be a closed subgroup of G for which hH, gi = G. If H 6= G, then H is contained in a maximal open subgroup M of G. Therefore, G = hH, gi ≤ M < G, a contradiction. Conversely, suppose an element g ∈ G satisfies the above condition. Let M be a maximal open subgroup of G. If g 6∈ M , then G = hM, gi.

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Chapter 22. Projective Groups and Frattini Covers

Hence, M = G, a contradiction. Therefore, g ∈ M , and since M is arbitrary, g ∈ Φ(G). Finally, let H be a closed subgroup of G such that H · Φ(G) = G. If H < G, then H is contained in a maximal open subgroup M of G and M = M · Φ(G) = G, a contradiction. Thus, H = G. Lemma 22.1.2: The Frattini group of a profinite group G is pronilpotent. Proof: Assume first G is finite. Let P be a p-Sylow group of Φ(G). For g ∈ G the group P g is also a p-Sylow subgroup of Φ(G). Hence, there exists a ∈ Φ(G) such that P g = P a . Therefore, ga−1 is in the normalizer, NG (P ), of P in G. Thus, G = NG (P )Φ(G). By Lemma 22.1.1, G = NG (P ). Hence, P is normal in G, and therefore in Φ(G). It follows that Φ(G) is the direct product of its p-Sylow groups. That is, G is nilpotent [Huppert, p. 260]. In the general case observe that for any open normal subgroup N of G, Φ(G)/N ∩ Φ(G) ∼ = N Φ(G)/N ≤ Φ(G/N ). Since Φ(G/N ) is a nilpotent finite group, so is Φ(G)/N ∩ Φ(G). Therefore, Φ(G) = lim Φ(G)/N ∩ Φ(G) is a ←− pronilpotent group. Lemma 22.1.3: Let A be a minimal normal subgroup of a finite group G. If A ≤ Φ(G), then A is p-elementary Abelian for some prime number p. Proof: By Lemma 22.1.2, A is a nilpotent group. Hence, the center Z(A) of A is nontrivial [Huppert, p. 260] and normal in G. Thus, Z(A) = A and A is Abelian. As such, A is a direct product of its Sylow subgroups and each of them is normal in G. Therefore, A is an Abelian p-group for some prime number p. The subgroup of A consisting of all elements x with xp = 1 is normal in G. Consequently, A = (Z/pZ)m for some m ≥ 0. As an operation, taking the Frattini subgroup of a group has functorial properties: Lemma 22.1.4: (a) Let θ: G → H be an epimorphism of profinite groups. Then θ(Φ(G)) ≤ Φ(H). If Ker(θ) ≤ Φ(G), then θ(Φ(G)) = Φ(H). (b) Suppose U is a closed subgroup of G and N is a closed subgroup of Φ(U ), normal in G. Then N ≤ Φ(G). (c) Let N beQ a closed normal subgroup of G. Then Φ(N ) ≤ Φ(G). (d) Q Let G = i∈I Gi be a direct product of profinite groups. Then Φ(G) = i∈I Φ(Gi ). Proof of (a): The inverse image of each maximal open subgroup of H is a maximal open subgroup of G. Hence, θ(Φ(G)) ≤ Φ(H). Proof of (b): Assume N 6≤ Φ(G). Then G has a maximal open subgroup M with N 6≤ M . Thus, M N = G. Hence, U = U ∩ M N = (U ∩ M )N . Since N ≤ Φ(U ), Lemma 22.1.1 implies U = U ∩ M . Therefore, N ≤ M , contrary to the assumption.

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Proof of (c): Apply (b) to N and Φ(N ) instead of to U and N . Proof of (d): Each Q Gi is a closed normal subgroup of G. Therefore, (c) gives Φ(Gi ) ≤ Φ(G) and i∈I Φ(Gi ) ≤ Φ(G). Q Conversely, let M be a maximal open subgroup ofQ Gj . Then M × i6=j Gi is a maximal open subgroup of G. Thus Q Φ(G) ≤ M × i6=j Gi . Running over ) × all M , this gives Φ(G) ≤ Φ(G j i6=j Gi . Since this is true for each j ∈ I, Q we conclude that Φ(G) ≤ i∈I Φ(Gi ). Remark 22.1.5: The map θ: Φ(G) → Φ(H) in Lemma 22.1.4(a) need not be surjective. Let G = hbi n hai be the semidirect product of a cyclic group hai of order 5 with a cyclic group H = hbi of order 4 with the relation ab = a2 . Let θ: G → H be the quotient map. Both hai and hbi are maximal subgroups of G, so Φ(G) = 1. On the other hand, H has order 4, so Φ(H) is of order 2. Thus, θ(Φ(G)) < Φ(H).

22.2 Cartesian Squares The usual direct product of group theory has a useful generalization: Let α: B → A and γ: C → A be homomorphisms of profinite groups. Let B ×A C = {(b, c) ∈ B × C | α(b) = γ(c)}. Since A is Hausdorff, B ×A C is a closed subgroup of B × C. It is therefore a profinite group called the fiber product of B and C over A. There are natural projection maps prB : B ×A C → B and prC : B ×A C → C defined by prB (b, c) = b and prC (b, c) = c, respectively. It is possible to change A, B, and C such that the projection maps will be surjective: Put A0 = {a ∈ A | ∃(b, c) ∈ B × C: α(b) = γ(c) = a}, B0 = α−1 (A0 ), and C0 = γ −1 (A0 ). Then A0 (resp. B0 , C0 ) is a closed subgroup of A (resp. B, C), B0 ×A0 C0 = B ×A C, and the projections maps of B0 ×A0 C0 are surjective. Proposition 22.2.1: Consider a commutative diagram of profinite groups (1)

D

δ

γ

β

B

/C

α

/A

where β, α, γ, and δ are homomorphisms. The following statements are equivalent: (a) There exists an isomorphism θ: D → B ×A C with β ◦ θ−1 = prB and δ ◦ θ−1 = prC . (b) Let G be a profinite group and ϕ: G → B and ψ: G → C homomorphisms. Suppose α◦ϕ = γ◦ψ. Then there exists a unique homomorphism

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π: G → D which makes the following diagram commutative: (2)

G0 @PPP 00@@ PPP ψ 00 @@@πPPPP PPP 00 @ δ P'/ 0 C ϕ 0 D 00 00 β γ 0 B α /A

Proof of “(a)=⇒(b)”: Without loss assume D = B ×A C, β = prB , and δ = prC . Define a map π: G → D by π(g) = (ϕ(g), ψ(g)). Then π is the unique homomorphism which makes (2) commutative. Proof of “(b)=⇒(a)”: Let G = B ×A C, ϕ = prB , and ψ = prC . Then (b) gives a unique π that makes (2) commutative. Define θ: D → B ×A C by θ(d) = (β(d), δ(d)) for d ∈ D. Apply the uniqueness property to maps from D to D and from G to G. It implies the map π ◦ θ (resp. θ ◦ π) is the identity, and θ satisfies (a). Definition 22.2.2: A commutative diagram (1) satisfying the conditions of Proposition 22.2.1 is said to be a cartesian square. Lemma 22.2.3: Let (1) be a cartesian square of homomorphisms of profinite groups. (a) If b ∈ B and c ∈ C satisfy α(b) = γ(c), then there exists a unique d ∈ D with β(d) = b and δ(d) = c. (b) If α (resp. γ) is surjective, then δ (resp. β) is surjective. Proof of (a): Assume without loss that D = B ×A C. If b ∈ B and c ∈ C satisfy α(b) = γ(c), then d = (b, c) is the unique element of D satisfying β(d) = b and δ(d) = c. Proof of (b): Consider c ∈ C. By assumption, there is a b ∈ B with α(b) = γ(c). By (a), there is a d ∈ D with δ(d) = c. Thus, δ is surjective. Lemma 22.2.4: Let (1) be a commutative square of epimorphisms of profinite groups. Then (1) is cartesian if and only if Ker(α ◦β) = Ker(β)×Ker(δ). Proof: First suppose (1) is a cartesian square. Clearly Ker(β) · Ker(δ) ≤ Ker(α ◦ β). Assume without loss, D = B ×A C, β = prB , and δ = prC . Let (b, c) ∈ Ker(α ◦ β). Then α(b) = 1 = γ(c). Hence, (1, c) and (b, 1) belong to D. Moreover, (b, c) = (1, c)·(b, 1) ∈ Ker(β)·Ker(δ). Finally, Ker(β)∩Ker(δ) = 1. Consequently, Ker(α ◦ β) = Ker(β) × Ker(δ). Now suppose Ker(α ◦ β) = Ker(β) × Ker(δ). Define a homomorphism θ: D → B ×A C by θ(d) = (β(d), δ(d)). If θ(d) = (1, 1), then d ∈ Ker(β) ∩ Ker(δ) = 1. Thus, θ is injective.

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To prove that θ is surjective, consider (b, c) ∈ B ×A C. Then α(b) = γ(c). There exist d1 , d2 ∈ D with β(d1 ) = b and δ(d2 ) = c. Then α(β(d1 d−1 2 )) = −1 = d α(β(d1 )) · γ(δ(d2 ))−1 = α(b)γ(c)−1 = 1. By assumption, d1 d−1 2 3 d4 for some d3 ∈ Ker(β) and d4 ∈ Ker(δ). Let d = d3 d1 = d4 d2 . Then β(d) = b and δ(d) = c. Therefore, θ(d) = (b, c). Lemma 22.2.5: Let (1) be a cartesian square of epimorphisms of profinite groups. Then β maps Ker(δ) isomorphically onto Ker(α). Proof: Apply Lemma 22.2.4: Ker(α) = β(Ker(α ◦ β)) = β(Ker(β) × Ker(δ)) = β(Ker(δ)). In addition, Ker(δ) ∩ Ker(β) = 1. Hence, β: Ker(δ) → Ker(α) is an isomorphism. Lemma 22.2.6: Let (1) and (2) be commutative diagrams of homomorphisms of profinite groups. Suppose (1) is cartesian and all maps except possibly π are surjective. Then: (a) If π is surjective, then Ker(α ◦ ϕ) = Ker(ϕ)Ker(ψ). (b) If Ker(α ◦ ϕ) ≤ Ker(ϕ)Ker(ψ), then π is surjective. Proof of (a): The condition α◦ϕ = γ ◦ψ implies Ker(ϕ)Ker(ψ) ≤ Ker(α◦ϕ). Now suppose π is surjective. Let g ∈ Ker(α ◦ ϕ). Put b = ϕ(g). Then α(b) = 1. Hence, there is a d ∈ D with β(d) = b and δ(d) = 1. Choose an h ∈ G with π(h) = d. Then ψ(h) = δ(π(h)) = δ(d) = 1 and ϕ(gh−1 ) = bβ(π(h))−1 = bβ(d)−1 = 1. Therefore, g = gh−1 · h ∈ Ker(ϕ)Ker(ψ). Proof of (b): Suppose Ker(α ◦ ϕ) ≤ Ker(ϕ)Ker(ψ). Consider d ∈ D. Put b = β(d) and c = δ(d). Choose g, h ∈ G with ϕ(g) = b and ψ(h) = c. Then α(ϕ(gh−1 )) = α(ϕ(g))γ(ψ(h))−1 = α(b)γ(c)−1 = α(β(d))γ(δ(d))−1 = 1. Thus, there are g1 ∈ Ker(ϕ) and h1 ∈ Ker(ψ) with gh−1 = g1 h1 . The element g1−1 g = h1 h of G satisfies ϕ(g1−1 g) = b and ψ(h1 h) = c. Therefore, by Lemma 22.2.3(a), π(g1−1 g) = d. Example 22.2.7: Fiber products. Here are four examples where fiber products naturally appear. The verification that the occurring commutative squares are cartesian follows from Lemma 22.2.4: (a) Let M and M 0 be Galois extensions of a field K. Put L = M ∩ M 0 and N = M M 0 . Then Gal(N/K) is the fiber product of Gal(M/K) and Gal(M 0 /K) over Gal(L/K) with respect to the restriction maps. (b) Similarly, let G be a profinite group and K, L, M, N closed normal subgroups. Suppose K ∩L = N and KL = M . Then G/N = G/K ×G/M G/L with respect to the quotient maps. (c) Let ϕ: B → A be an epimorphism of profinite groups, C = Ker(ϕ), ¯ = B/B0 , and B0 a closed normal subgroup of B. Suppose B0 ∩C = 1. Put B ¯ ¯ ¯ A0 = ϕ(B0 ), A = A/A0 , α: A → A and β: B → B the canonical maps, and ¯ → A¯ the map induced from ϕ. Then B ∼ ¯ ×A¯ A. ϕ: ¯ B =B

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Indeed, if b ∈ B satisfies α(ϕ(b)) = 1, then ϕ(b) ∈ A0 . Hence, there exists b0 ∈ B0 with ϕ(b0 ) = ϕ(b). Therefore, b = b0 · b−1 0 b ∈ Ker(β)Ker(ϕ). ¯ A. It follows from Lemma 22.2.4 that B ∼ B × = ¯ A Suppose now B0 ∩ C is not necessarily trivial. Then by the preceding ¯ ×A¯ A. Hence, for each a ∈ A and b ∈ B satisfying paragraph, B/B0 ∩ C ∼ =B ϕ(β(b)) ¯ = α(a) there is a b0 ∈ B (not necessarily unique) with β(b0 ) = β(b) and ϕ(b0 ) = a. (d) Condition (1b) of Section 17.3 on a family C of finite groups is equivalent to “C is closed under fiber products”. Thus, a formation is a family of finite groups which is closed under taking quotients and fiber products. The homomorphism π of diagram (2) need not be surjective even if all other maps are surjective. Here is a condition for this to happen: Lemma 22.2.8: Let ϕ: G → B and ψ: G → C be epimorphisms of profinite groups. Then there is a commutative diagram (2), unique up to an isomorphism such that (1) is a cartesian square with β, α, γ, δ, π epimorphisms. Proof: Let M = Ker(ϕ)Ker(ψ), N = Ker(ϕ) ∩ Ker(ψ), A = G/M , D = G/N , and π: G → G/N the quotient map. Now find β, α, γ, δ that make (1) cartesian and (2) commutative. The following lemma gives a useful criterion for δ in the cartesian diagram (1) to have a group theoretic section: Lemma 22.2.9: Let