Introduction to Galois cohomology and its applications

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LONDON MATHEMATICAL SOCIETY LECTURE NOTE SERIES Managing Editor: Professor M. Reid, Mathematics Institute, University of Warwick, Coventry CV4 7AL, United Kingdom The titles below are available from booksellers, or from Cambridge University Press at www. cambridge.org/mathematics 225 A mathematical introduction to string theory, S. ALBEVERIO et al 226 Novikov conjectures, index theorems and rigidity I, S. C. FERRY, A. RANICKI & J. ROSENBERG (eds) 227 Novikov conjectures, index theorems and rigidity II, S. C. FERRY, A. RANICKI & J. ROSENBERG (eds) 228 Ergodic theory of Z d -actions, M. POLLICOTT & K. SCHMIDT (eds) 229 230 231 232 233 234 235 236 237 238 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 267 268 269 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306

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London Mathematical Society Lecture notes series: 377

An Introduction to Galois Cohomology and its Applications ´ g o ry b e r h u y gre Universit´ e Joseph Fourier, Grenoble

cambridge university press Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, S˜ ao Paulo, Delhi, Dubai, Tokyo, Mexico City Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521738668 c G. Berhuy 2010 c J.-P. Tignol 2010 Foreword This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2010 Printed in the United Kingdom at the University Press, Cambridge A catalog record for this publication is available from the British Library ISBN 978-0-521-73866-8 Paperback Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

To my dear friend Frédérique

Contents

Foreword Introduction Part I I

page xi 1

An introduction to Galois cohomology

11

Infinite Galois theory I.1 Reminiscences on field theory I.2 Galois theory I.2.1 Definitions and first examples I.2.2 The Galois correspondence I.2.3 Morphisms of Galois extensions I.2.4 The Galois group as a profinite group Exercises

13 13 17 17 18 19

II

Cohomology of profinite groups II.3 Cohomology sets: basic properties II.3.1 Definitions II.3.2 Functoriality II.3.3 Cohomology sets as a direct limit II.4 Cohomology sequences II.4.1 The case of a subgroup II.4.2 The case of a normal subgroup II.4.3 The case of a central subgroup II.5 Twisting II.6 Cup-products Exercises

26 26 26 36 41 45 46 51 52 56 61 64

III

Galois cohomology III.7 Warm-up III.7.1 Digression: categories and functors

69 69 69

vii

21 24

Contents

viii III.7.2 Algebraic group-schemes III.7.3 The Galois cohomology functor Abstract Galois descent III.8.1 Matrices reloaded III.8.2 Actions of group-valued functors III.8.3 Twisted forms III.8.4 The Galois descent condition III.8.5 Stabilizers III.8.6 Galois descent lemma III.8.7 Hilbert’s Theorem 90 First applications of Galois descent III.9.1 Galois descent of algebras III.9.2 The conjugacy problem III.9.3 Cup-products with values in μ2 Exercises

77 85 96 97 99 101 103 104 106 110 117 117 121 126 130

IV

Galois cohomology of quadratic forms IV.10 Algebraic group-schemes associated to quadratic forms IV.10.1 Quadratic forms over rings IV.10.2 Orthogonal groups IV.10.3 Clifford groups and spinors IV.11 Galois cohomology of quadratic forms IV.11.1 Galois cohomology of orthogonal groups IV.11.2 Galois cohomology of spinors IV.12 Cohomological invariants of quadratic forms IV.12.1 Classification of quadratic forms over Q IV.12.2 Higher cohomological invariants Exercises

134 134 134 136 139 145 145 147 152 152 154 158

V

´ Etale and Galois algebras ´ V.13 Etale algebras V.14 Galois algebras V.14.1 Definition and first properties V.14.2 Galois algebras and Galois cohomology Exercises

160 160 164 164 170 177

VI

Group extensions, Galois embedding problems and Galois cohomology VI.15 Group extensions VI.16 Galois embedding problems Exercises

179 179 185 187

III.8

III.9

Part II VII

Applications

Galois embedding problems and the trace form VII.17 The trace form of an étale algebra

189 191 192

Contents VII.18 VII.19

ix Computation of e∗ (sn ) Applications to inverse Galois theory Exercises

VIII Galois cohomology of central simple algebras VIII.20 Central simple algebras VIII.21 Algebras with involutions VIII.21.1 Basic concepts VIII.21.2 Hyperbolic involutions VIII.21.3 Similitudes VIII.21.4 Cohomology of algebras with involution VIII.21.5 Trace forms Exercises

196 200 205 207 207 217 217 222 226 228 231 245

IX

Digression: a geometric interpretation of H 1 (−, G) IX.22 Reminiscences on schemes IX.23 Torsors Exercises

249 249 253 259

X

Galois cohomology and Noether’s problem X.24 Formulation of Noether’s problem X.25 The strategy X.26 Residue maps X.27 An unramified cohomological invariant X.28 Proof of Theorem X.24.1

261 261 262 265 271 272

XI

The rationality problem for adjoint algebraic groups XI.29 R-equivalence groups XI.30 The rationality problem for adjoint groups XI.31 Examples of non-rational adjoint groups Exercises

274 275 278 281 287

XII

Essential dimension of functors XII.32 Essential dimension: definition and first examples XII.33 First results XII.34 Cohomological invariants and essential dimension XII.35 Generic objects and essential dimension XII.36 Generically free representations XII.37 Some examples XII.38 Complements and open problems Exercises

290 290 292 296 298 300 301 306 308

References Index

310 314

Foreword Like an idea whose time has come, nonabelian Galois cohomology burst into the world in the mid 50’s. There had been harbingers, of course. Chˆ atelet’s méthode galoisienne for genus 1 curves and Weil’s observations on homogeneous spaces had opened the way, and it was a small step to write down the basic operations so that they make sense in a noncommutative situation. Within a few years, several pioneers realized almost simultaneously that the formalism of Galois cohomology could be used to classify various algebraic structures and to illuminate the definition of some of their invariants. This simple and remarkably penetrating idea, soon popularized by Serre’s famous monograph Cohomologie galoisienne, immediately took hold. Galois cohomology is indeed algebra at its best: a few formal basic operations with a broad spectrum of far-reaching applications. Grégory Berhuy’s monograph provides a very welcome introduction to Galois descent techniques and nonabelian Galois cohomology, aimed at people who are new to the subject. Beginners will find here a thorough discussion of the technical details that are usually left to the reader. Together with advanced readers, they will appreciate a tasteful tour of applications, including some to which the author, himself an avid cocyclist, has contributed. (Incidentally, the title of Section III.8.1 also offers a glimpse into his taste in movies.) As may be expected, the list of applications discussed here is far from exhaustive, and in the last chapters the exposition is more demanding. It strikes a nice balance between a thorough account and a survey, and it provides a unique introduction to several of the exciting developments of the last decade, such as essential dimension and new advances on rationality problems. The many who did not have the good fortune to take his course at the University of Southampton will be thankful to Grégory Berhuy for making available the text of his lectures. I trust it will give to a large audience an idea of the scope and beauty of the subject, and inspire many of them to contribute to it in their turn. Jean-Pierre Tignol

xi

Introduction

A recurrent problem arising in mathematics is to decide if two given mathematical structures defined over a field k are isomorphic. Quite often, it is easier to deal with this problem after scalar extension to a bigger field Ω containing k, for example an algebraic closure of k, or a finite Galois extension. In the case where the two structures happen to be isomorphic over Ω, this leads to the natural descent problem: if two k-structures are isomorphic over Ω, are they isomorphic over k? Of course, the answer is no in general. For example, consider the following matrices M, M0 ∈ M2 (R) : M0 =

0 1

−2 0

,M =

0 2 −1 0

.

It is easy to see that they are conjugate by an element of GL2 (C), since √ 2, and therefore are both similar to they have same eigenvalues ±i √ i 2 0√ . In fact we have 0 −i 2

i 0 0 −i

M

i 0

0 −i

−1 = M0 ,

so M and M0 are even conjugate by an element of SL2 (C). A classical result in linear algebra says that M and M0 are already conjugate by an element of GL2 (R), but this is quite obvious here since 1

2

Introduction

the equality above rewrites 1 0 1 M 0 −1 0

0 −1

−1 = M0 .

However, they are not conjugate by an element of SL2 (R). Indeed, it is easy to check that a matrix P ∈ GL2 (R) such that P M = M0 P has the form a 2c P = . c −a Since det(P ) = −(a2 + 2c2 ) < 0, P cannot belong to SL2 (R). Therefore, M and M0 are conjugate by an element of SL2 (C) but not by an element of SL2 (R). Hence, the descent problem for conjugacy classes of matrices has a positive answer when we conjugate by elements of the general linear group, but has a negative one when we conjugate by elements of the special linear group. So, how could we explain the difference between these two cases? This is where Galois cohomology comes into play, and we would like now to give an insight of how this could be used to measure the obstruction to descent problems on the previous example. If k is a field, let us denote by G(k) the group GL2 (k) or SL2 (k) indifferently. Assume that QM Q−1 = M0 for some Q ∈ G(C). The idea is to measure how far is Q to have real coefficients, so it is natural to consider the −1 difference QQ , where Q is the matrix obtained from Q by letting the complex conjugation act coefficientwise. Indeed, we will have Q ∈ G(R) −1 if and only if Q = Q, that is if and only if QQ = I2 . Of course, −1 if QQ = I2 , then M and M0 are conjugate by an element of G(R), but this is not the only case when this happens to be true. Indeed, if we assume that P M P −1 = M0 for some P ∈ G(R), then we easily get that QP −1 ∈ G(C) commutes with M0 . Therefore, there exists C ∈ ZG (M0 )(C) = {C ∈ G(C) | CM0 = M0 C} such that Q = CP . We then easily have Q = C P = CP , and therefore −1

QQ

= CC

−1

for some C ∈ ZG (M0 )(C).

Conversely, if the equality above holds then P = C −1 Q is an element of G(R) satisfying P M P −1 = M0 . Indeed, we have P =C

−1

Q = C −1 Q = P,

Introduction

3

so P ∈ G(R), and P M P −1 = C −1 QM Q−1 C = C −1 M0 C = M0 C −1 C = M0 . Thus, M and M0 will be congugate by an element of G(R) if and only if −1

QQ

= CC

−1

for some C ∈ ZG (M0 )(C). −1

Notice also for later use that QQ ∈ G(C) commutes with M0 , as we may check by applying complex conjugation on both sides of the equality QM Q−1 = M0 . If we go back to our previous example, we have Q = −1

i 0

0 −i

, and

= −I2 . Easy computations show that we have z −2z for some z, z ∈ C . ZG (M0 )(C) = C ∈ G(C) | C = z z

therefore QQ

−1

−1

Therefore, we will have C ∈ ZG (M0 )(C) and CC = QQ = −I2 if and only if iu −2iv C= for some u, v ∈ R, (u, v) = (0, 0). iv iu Notice that the determinant of the matrix above is −(u2 + 2v 2 ) < 0. Thus, if G(C) = GL2 (C), one may take u = 1 and v = 0, but if −1 −1 G(C) = SL2 (C), the equation CC = −I2 = QQ has no solution in ZG (M0 )(C). This explains a bit more conceptually the difference between the two descent problems. In some sense, if QM Q−1 = M0 −1 for some Q ∈ G(C), the matrix QQ measures how far is M to be conjugate to M0 over R. Of course, all the results above remain valid if M and M0 are square matrices of size n, and if G(k) = GLn (k), SLn (k), On (k) or even Sp2n (k). If we have a closer look to the previous computations, we see that the reason why all this works is that C/R is a Galois extension, whose Galois group is generated by complex conjugation. Let us consider now a more general problem: let Ω/k be a finite Galois extension, and let M, M0 ∈ Mn (k) be two matrices such that QM Q−1 = M0 for some Q ∈ G(Ω).

4

Introduction

Does there exist P ∈ G(k) such that P M P −1 = M0 ? Since Ω/k is a finite Galois extension, then for all x ∈ Ω, we have x ∈ k if and only if σ(x) = x for all σ ∈ Gal(Ω/k). If now Q ∈ G(Ω), then let us denote by σ ·Q ∈ G(Ω) the matrix obtained from Q by letting σ act coefficientwise. Then we have Q ∈ G(k)

⇐⇒ ⇐⇒

σ·Q = Q for all σ ∈ Gal(Ω/k) Q(σ·Q)−1 = I2 for all σ ∈ Gal(Ω/k).

As before, applying σ ∈ Gal(Ω/k) to the equality QM Q−1 = M0 , we see that Q(σ·Q)−1 ∈ ZG (M0 )(Ω). We therefore get a map

αQ :

Gal(Ω/k) −→ ZG (M0 )(Ω) σ −→ Q(σ·Q)−1 .

Arguing as at the beginning of this introduction, one can show that M and M0 will be conjugate by an element of G(k) if and only if there exists C ∈ ZG (M0 )(Ω) such that αQ = αC , that is if and only if there exists C ∈ ZG (M0 )(Ω) such that Q(σ·Q)−1 = C(σ·C)−1 for all σ ∈ Gal(Ω/k). To summarize, to any matrix M ∈ Mn (k) which is conjugate to M0 by an element of G(Ω), we may associate a map αQ : Gal(Ω/k) −→ ZG (M0 )(Ω), which measures how far is M to be conjugate to M0 by an element of G(k). This has a kind of a converse: for any map α:

Gal(Ω/k) −→ ZG (M0 )(Ω) σ −→ ασ

such that α = αQ for some Q ∈ G(Ω), one may associate a matrix of Mn (k) which is conjugate to M0 by an element of G(k) by setting Mα = Q−1 M0 Q. To see that Mα is indeed an element of Mn (k), notice first that we have σ·(CM C −1 ) = (σ·C)(σ·M )(σ·C)−1 for all C ∈ G(Ω), M ∈ Mn (Ω), σ ∈ Gal(Ω/k). Thus, for all σ ∈

Introduction

5

Gal(Ω/k), we have σ·Mα

= = = = =

(σ·Q)−1 M0 (σ·Q) Q−1 Q(σ·Q)−1 M0 (σ·Q) Q−1 M0 Q(σ·Q)−1 (σ·Q) Q−1 M0 Q Mα ,

the third equality coming from the fact that ασ = Q(σ · Q)−1 lies in ZG (M0 )(Ω). Not all the maps α : Gal(Ω/k) −→ ZG (M0 )(Ω) may be written αQ for some Q ∈ G(Ω). In fact, easy computations show that a necessary condition for this to hold is that α is a cocycle, that is αστ = ασ σ·ατ for all σ, τ ∈ Gal(Ω/k). This condition is not sufficient in general. However, it happens to be the case if G(Ω) = GLn (Ω) or SLn (Ω) (this will follow from Hilbert 90). Notice that until now we picked a matrix Q ∈ G(Ω) which conjugates M into M0 , but this matrix Q is certainly not unique. We could therefore wonder what happens if we take another matrix Q ∈ G(Ω) which conjugates M into M0 . Computations show that we have Q Q−1 ∈ ZG (M0 )(Ω). Therefore, there exists C ∈ ZG (M0 )(Ω) such that Q = CQ, and we easily get that

ασQ = CασQ (σ·C)−1 for all σ ∈ Gal(Ω/k). Two cocycles α, α : Gal(Ω/k) −→ ZG (M0 )(Ω) such that ασ = Cασ (σ·C)−1 for all σ ∈ Gal(Ω/k) for some C ∈ ZG (M0 )(Ω) will be called cohomologous. Being cohomologous is an equivalence relation on the set of cocycles, and the set of equivalence classes is denoted by H 1 (Gal(Ω/k), ZG (M0 )(Ω)). If α is a cocycle, we will denote by [α] the corresponding equivalence class. Therefore, to any matrix M ∈ Mn (k) which is conjugate to M0 by an element of G(Ω), one may associate a well-defined cohomology class [αQ ], where Q ∈ G(Ω) is any matrix satisfying QM Q−1 = M0 . It is important to notice that the class [αQ ] does not characterize M −1 completely. Indeed, for every P ∈ G(k), it is easy to check that αQP =

6

Introduction

αQ . In particular, the cohomology classes associated to the matrices M and P M P −1 are equal, for all P ∈ G(k).

Conversely, if α = αQ and α = αQ are cohomologous, it is not too difficult to see that P = Q−1 C −1 Q ∈ G(k), and that the corresponding matrices Mα and Mα satisfy P Mα P −1 = Mα . Thus the previous considerations show that, in the case where every cocycle α : Gal(Ω/k) −→ ZG (M0 )(Ω) may be written α = αQ for some Q ∈ G(Ω), the set H 1 (Gal(Ω/k), ZG (M0 )(Ω)) is in one-to-one correspondence with the set of G(k)-conjugacy classes of matrices M ∈ Mn (k) which are conjugate to M0 by an element of G(Ω). Many situations can be dealt with in a similar way. For example, reasoning as above and using Hilbert 90, one can show that the set of isomorphism classes of quadratic forms q which are isomorphic to the quadratic form x21 + . . . + x2n over Ω is in one-to-one correspondence with H 1 (Gal(Ω/k), On (Ω)).The case of k-algebras is a little bit more subtle, but one can show that the set of isomorphism classes of k-algebras which are isomorphic to a given k-algebra A over Ω is in one-to-one correspondence with H 1 (Gal(Ω/k), AutΩ−alg (A ⊗k Ω)). Quite often, algebraic structures can be well understood over a separable closure ks of k. In the best cases, they even become isomorphic over ks . Therefore, it is useful to extend this setting to the case of infinite Galois field extensions. To do this, we will introduce the notion of a profinite group in Chapter 1, and recollect some facts on infinite Galois theory. Then in Chapter 2 we define the cohomology sets H i (Γ, A) for any profinite group Γ and any Γ-group A, and study their functorial properties and their behavior with respect to short exact sequences. We also introduce the cup-product, which is useful to construct higher cohomology classes. Chapter 3 deals with Galois cohomology and the central part of this chapter is devoted to formalize Galois descent and to give applications. We then come back to the conjugacy problem for matrices and compute the total obstruction in an example. In Chapter 4, we study Galois cohomology of quadratic forms and give a cohomological interpretation of some classical invariants attached to quadratic forms, such as the determinant or the Hasse invariant. In Chapter 5, we obtain an algebraic interpretation of Galois field extensions with Galois group G in terms of H 1 (Gal(ks /k), G). In Chapter 6, we give a cohomological

Introduction

7

obstruction of the following Galois embedding problem: given a group ˜ −→ G −→ 1, where A is a central subgroup extension 1 −→ A −→ G ˜ of G, and given a Galois field extension E/k with Galois group G, does ˜ with Galois group G ˜ such that there exists a Galois field extension E/k ˜ A = E? E

The next chapters describe various applications of Galois cohomology. Chapter 7 is devoted to the study of a certain Galois embedding problem with kernel A = Z/2Z. In this particular case we prove a formula of Serre which computes the obstruction in terms of the classical invariant of the trace form of E, and we give simple applications. We then study Galois cohomology of central simple algebras with or without involutions in Chapter 8. As an application of Galois cohomology techniques, we compute the Hasse invariant of certain quadratic forms attached to these algebras. In Chapter 9, we briefly introduce the notion of a G-torsor, which gives a geometric interpretation of Galois cohomology. We apply this point of view to derive some results on cohomological invariants of algebraic groups. In Chapter 10, we describe applications of Galois cohomology to the so-called Noether’s problem: given a field k and a finite group G, is there a linear faithful representation V of G such that the field extension k(V )G /k is purely transcendental ? This is known to be true when G is abelian and k ⊃ μn , but false for G = Z/8Z and k = Q. We will introduce the residue maps in Galois cohomology and use their properties to prove that Noether’s problem has a negative solution when G = Z/2m Z, m ≥ 3 and k = Q. To do so, we attach to each Galois extension of group G over a field K ⊃ k a non-vanishing cohomological obstruction. In Chapter 11, we study another kind of rationality problem: given a linear algebraic group G over k, is the underlying variety rational? This is known to be true for classical groups when k is algebraically closed. We will show that the answer is negative in general when k is an arbitrary field. We will focus on the case where G is an automorphism group of some algebra with a symplectic involution. Once again, the answer will come from the existence of a non-zero cohomological obstruction. Finally in Chapter 12, we introduce the notion of essential dimension of a functor, which is an active research topic, for which substantial progress has been made recently. If G is a finite group, the essential dimension of the Galois cohomology functor H 1 (− , G) will be the number of independent parameters needed to describe a Galois extension of group G.

8

Introduction

This introduction to Galois cohomology does not pretend to be complete. For example, we are aware that an historical introduction to the subject is missing. The curious reader is referred to [30], p. 446-449, as well as [58] and [59] for more information and numerous references. Moreover, we tried to reduce the prerequisites necessary to read these notes to the minimum. Only some basic knowledge on Galois theory and algebra (definition of group, ring, field, k-algebra, notion of tensor product) is required. Also it was impossible to cover all the ‘hot topics’ (such as Serre’s conjecture II, Hasse principle, Rost invariants) or applications of Galois cohomology. Once again, we refer to [30], [58] and [59]. More advanced material on Galois cohomology may be found in [25],[26], [30] or [58], each of these references focusing on a different aspect of the theory: cohomological invariants (including the construction of Rost invariants) and applications to Noether’s problem in [25], MerkurjevSuslin’s theorem in [26], algebras with involution in [30] or cohomology of algebraic groups over fields of small cohomological dimension in [58]. This book is an extended version of notes of some postgraduate lectures on Galois cohomology that we gave at the University of Southampton, which included originally Chapters 1-7. The main goal of these lectures was to introduce enough material on Galois cohomology to fully understand the proof of Serre’s formula [61] aiming at an audience having a minimal background in algebra, and to give applications to Galois embedding problems. The method we chose to establish this formula differs a bit from the original one. It was suggested as an alternative proof by Serre himself in [61]. Moreover, it was a good occasion to introduce classical tools such as exact sequences in cohomology, Galois descent, Hilbert 90 and some standard results such as Springer’s cohomological interpretation of the Hasse invariant. Consequently, the material introduced in Part I is really basic, but is sufficient to obtain beautiful applications to inverse Galois theory or to the conjugacy problem. We also took a particular care to make the first half of this book self-contained, with an exception made for the section on infinite Galois theory and for Proposition III.7.23. Let us also mention the existence of lectures notes [2] presenting a shortened and simplified exposition of the material introduced in Chapters II and III (in these notes, all Galois extensions considered are finite, only the first cohomology set is presented and the functorial aspect of the theory is not treated). The second part of the book gives an insight of how Galois cohomology may be useful to solve some algebraic problems, and presents active research topics, such as ra-

Introduction

9

tionality questions or essential dimension of algebraic groups and often requires more advanced material. Therefore, proofs of the most difficult results are skipped. We hope that these notes will help the reader willing to study more advanced books on this subject, such as those cited above. This book could not have been written without the encouragements and the support of Gerhard Roerhle, and we would like to thank him warmly. We are also grateful to our colleagues and friends Vincent Beck, Jérˆ ome Ducoat, Jean Fasel, Nicolas Grenier-Boley, Emmanuel Lequeu, Frédérique Oggier, Gerhard Roerhle and Jean-Pierre Tignol, who took time to read partly or integrally some earlier versions of the manuscript, despite the fact they certainly had better things to do. Their careful reading, judicious comments and remarks permitted to improve significantly the exposition and to detect many misprints or inaccuracies. The whole LATEX support team of Cambridge University Press deserves a special mention for its efficiency and its patience. Finally, we would like to thank Roger Astley, Caroline Brown and Clare Dennison for their helpfulness in the whole editing process.

Part I An introduction to Galois cohomology

I Infinite Galois theory

In the introduction we explained how Galois cohomology could be used to classify mathematical structures defined over k which become isomorphic over a finite Galois extension, and why it would be useful to extend this setting to arbitrary Galois extensions (not necessarily finite). In this chapter, we would like to briefly recall some standard facts on infinite Galois theory. The reader may refer to [42] for details. §I.1 Reminiscences on field theory Definition I.1.1. Let k be a field. A field extension of k is a pair (K, ε), where K is a field and ε : k −→ K is a ring morphism (necessarily injective since k is a field). In other words, K is an extension of k if it contains a subfield isomorphic to k. We will also say that K is an extension of k. We will denote it by K/k if ε is clear from the context. A morphism ι : (K1 , ε1 ) −→ (K2 , ε2 ) of field extensions of k is a ring morphism u : K1 −→ K2 such that u ◦ ε1 = ε2 . We will also say that ι is a k-embedding of K1 into K2 . An isomorphism of field extensions is a morphism which is bijective. If ι : (K1 , ε1 ) −→ (K2 , ε2 ) is an isomorphism of field extensions of k, we will also say that ι is a k-isomorphism of K1 onto K2 , or that K1 and K2 are k-isomorphic. For the rest of this section, we will identify k with its image ε(k), and therefore consider that we have an inclusion k ⊂ K. Definition I.1.2. If K/k is a field extension, then K has a natural structure of a k-vector space. The degree of K/k is the dimension of K as a k-vector space, and is denoted by [K : k]. 13

14

Infinite Galois theory

Definition I.1.3. Let K/k be a field extension, and let A be a subset of K. The subring of K generated by A over k is the smallest subring of K containing A and k. It is denoted by k[A]. The subextension of K generated by A over k is the smallest subfield of K containing A and k. It is denoted by k(A). If A = {α1 , . . . , αn }, we denote them by k[α1 , . . . , αn ] and k(α1 , . . . , αn ) respectively. Remark I.1.4. It is not difficult to see that k[α1 , . . . , αn ] = {P (α1 , . . . , αn ) | P ∈ k[X1 , . . . , Xn ]}, and that k(α1 , . . . , αn ) is the field of fractions of k[α1 , . . . , αn ]. Definition I.1.5. Let K/k be a field extension and let K1 , K2 be two subfields of K containing k. The compositum of K1 and K2 is the subfield generated by K1 ∪ K2 . It is denoted by K1 K2 . Definition I.1.6. Let K/k be a field extension, and let α ∈ K. We say that α is algebraic over k if there exists a non-zero polynomial P ∈ k[X] such that P (α) = 0. We say that α is transcendental over k otherwise. A field extension K/k is called algebraic if every element of K is algebraic over k. Proposition I.1.7. Let K/k be a field extension, and let α ∈ K. The set Iα = {P ∈ k[X] | P (α) = 0} is an ideal of k[X]. It is a non-zero ideal if and only if α is algebraic over k. In this case, there exists a unique monic irreducible polynomial μα,k such that Iα = (μα,k ). Definition I.1.8. The polynomial μα,k is called the minimal polynomial of α over k. Remark I.1.9. From the definition of the minimal polynomial, it follows that if P ∈ k[X] satisfies P (α) = 0, then μα,k |P ; if moreover P is monic and irreducible then P = μα,k . Theorem I.1.10. Let K/k be a field extension. Then α ∈ K is algebraic over k if and only if k(α)/k has finite degree. In this case, a k-basis of k(α) is given by 1, α, . . . , αd−1 , where d = deg(μα,k ). In particular, k(α) = k[α] and we have the equality [k(α) : k] = deg(μα,k ).

I.1 Reminiscences on field theory

15

Remark I.1.11. It follows easily that if α1 , . . . , αn ∈ K are algebraic over k, then k(α1 , . . . , αn )/k has finite degree. Definition I.1.12. We say that a field k is algebraically closed if every non-constant polynomial with coefficients in k has a root in k. An algebraic closure of a field k is an algebraic field extension kalg /k such that kalg is algebraically closed. One can show that every field k has an algebraic closure, and that two algebraic closures are k-isomorphic. Definition I.1.13. Let k be a field, and let kalg be a fixed algebraic closure of k. A polynomial f ∈ k[X] of degree n is separable over k if it has n distinct roots in kalg . If K/k is a field extension, we say that x ∈ K is separable over k if x is algebraic over k and its minimal polynomial over k is separable. Finally we say that K/k is separable if every element of K is separable over k. Remark I.1.14. One can show that the compositum of two separable extensions is again separable and that k(α1 , . . . , αn )/k is separable if and only if α1 , . . . , αn are separable over k. Definition I.1.15. The separable closure of k in kalg is the maximal subfield ks of kalg such that ks /k is separable. It is exactly the subfield of elements of kalg xhich are separable over k. We continue by stating some results on extensions of morphisms which will be useful in the sequel. First, we need a definition. Definition I.1.16. Let K and K be two fields, let L/K and L /K be two field extensions, and let ι : K −→ K be a ring morphism. We say that a ring morphism ϕ : L −→ L is an extension of ι if the diagram LO K

ϕ

/ L O

ι

/ K

commutes. Notation: Let K and K be two fields, and let ι : K −→ K be a ring morphism. If P ∈ K[X], P = an X n + an−1 X n−1 + . . . + a0 , we denote by ι(P ) the element of K [X] defined by ι(P ) = ι(an )X n + ι(an−1 )X n−1 + . . . + ι(a0 ).

16


Lemma I.1.17. Let K and K be two fields, let L/K be a field extension, and let ι : K −→ K be a ring morphism. Finally, let α ∈ L. For of ι, ϕ(α) is a root of ι(μα,K ). every extension ϕ : L −→ Kalg Proof. Write μα,K = X n + an−1 X n−1 + . . . + a0 . Since ϕ is an extension of ι, we have ι(μα,K ) = X n + ϕ(an−1 )X n−1 + . . . + ϕ(a0 ). Since ϕ is a ring morphism, we then get ι(μα,K )(ϕ(α)) = ϕ(μα,K (α)) = ϕ(0) = 0. Hence ϕ(α) is a root of ι(μα,K ) as claimed. Proposition I.1.18. Let K, K be two fields, let ι : K −→ K be a ring , then there morphism, and let α ∈ Kalg . If β is a root of ι(μα,K ) in Kalg exists a unique extension ϕ : K(α) −→ Kalg of ι such that ϕ(α) = β. In particular, the set of extensions of ι is in bijection with the set of roots of ι(μα,K ). Theorem I.1.19. Let K be a field, let L/K be an algebraic field extension. Let E be an algebraically closed field, and let τ : K −→ E be a ring morphism. Then there exists a ring morphism σ : L −→ E such that σ|K = τ . In other words, there exists an extension σ : L −→ E of τ. Corollary I.1.20. Let K and K be two fields, and let ι : K −→ K be a ring morphism. Then there exists an extension ϕ : Ks −→ Ks of ι. Proof. Let τ be the composition of ι with the inclusion K ⊂ Kalg . By Theorem I.1.19, there exists ϕ : Ks −→ Kalg such that ϕ|K = τ . Let α ∈ Ks , so that μα,K has no multiple roots. Let α = α1 , . . . , αn be its distinct roots in Kalg . Let P = τ (μα,K ). Since μα,K = μαi ,K for all i, Lemma I.1.17 implies that ϕ(α1 ), . . . , ϕ(αn ) are roots of P . Since P and μα,K have the same degree and ϕ is injective, this implies that P is separable. Since ϕ(α) is a root of P , we have μϕ(α),K |P . Hence μϕ(α),K has no multiple roots and ϕ(α) is separable over K . In other words, ϕ(Ks ) ⊂ Ks ; this concludes the proof.

I.2 Galois theory

17

§I.2 Galois theory I.2.1 Definitions and first examples Definition I.2.1. We say that a field extension Ω/k (contained in kalg ) is a Galois extension if it is separable and for every k-linear embedding σ : Ω −→ kalg we have σ(Ω) = Ω (so σ is a k-automorphism of Ω). In this case, the group Gal(Ω/k) of all k-automorphisms of Ω/k is called the Galois group of Ω/k. Notice that we did not assume Ω/k to be finite in the previous definition. Example I.2.2. Let α ∈ kalg be a separable element over k, and let Ω = k(α1 , . . . , αn ), where α1 = α, . . . , αn are the n distinct roots of μα,k in kalg . Then Ω/k is a finite Galois extension. Indeed, since the αi ’s have the same minimal polynomial, they are all separable over k, and k(α1 , . . . , αn )/k is separable by Remark I.1.14. Now let σ : Ω −→ kalg be a k-embedding. By Lemma I.1.17, σ(αi ) is one of the αj ’s, and thus is an element of Ω. This proves that σ(Ω) ⊂ Ω. Since Ω is a finite dimensional k-vector space and σ : Ω −→ Ω is k-linear and injective, then σ is bijective and we are done. Lemma I.2.3. Let Ω/k be a Galois extension. Let α ∈ Ω, and let α1 = α, . . . , αn be the roots of μα,k in kalg . Then αi ∈ Ω for i = 1, . . . , n. Proof. For i = 1, . . . , n, there exists a unique k-embedding τi : k(α) −→ kalg satisfying τi (α) = αi by Proposition I.1.18. Each τi extends to a k-embedding σi : Ω −→ kalg by Theorem I.1.19. Since Ω/k is a Galois extension, σi (Ω) = Ω. In particular, σi (α) = τi (α) = αi ∈ Ω. The next lemma is very useful and will be frequently used without further reference. Lemma I.2.4. Let Ω/k be a Galois extension, and let α ∈ Ω. Then there exists a finite Galois subextension of Ω/k containing α. Proof. Let α1 = α, . . . , αn be the roots of μα,k in kalg . The previous lemma implies that k(α1 , . . . , αn ) ⊂ Ω. Since k(α1 , . . . , αn )/k is a finite Galois extension containing α by Example I.2.2, we are done. We now give an example of an infinite Galois extension. Lemma I.2.5. Let ks be the separable closure of k in a fixed algebraic closure of k. Then the extension ks /k is Galois.

18


Proof. First, ks /k is separable. Now let σ : ks −→ kalg be a k-linear embedding. Let x ∈ ks and let L/k be a finite Galois extension containing x (which exists by Example I.2.2). Then σ|L : L −→ kalg is a k-embedding of L into kalg . Since L/k is a Galois extension, we have σ(x) ∈ L. In particular σ(x) is separable, since L/k is separable. Therefore, we have proved that σ(ks ) ⊂ ks . To prove the missing inclusion, let x ∈ ks and let L /k be a finite Galois extension containing x . Since L /k is a Galois extension, σ|L is a k-automorphism of L . Hence, there exists x ∈ L ⊂ ks such that σ|L (x) = x . Thus we have x = σ|L (x) = σ(x), and therefore, σ(ks ) = ks . I.2.2 The Galois correspondence We would like now to understand better the structure of the Galois group of a Galois extension Ω/k, not necessarily of finite degree over k. In particular, we would like to have a Galois correspondence between subfields of Ω and subgroups of Gal(Ω/k) as in the case of finite Galois extensions. Unfortunately, the following example shows that this correspondence does not hold in the infinite case. √ Example I.2.6. Let Ω = Q( p, p prime ). Then Ω/Q is a Galois extension, as the reader may check. For a prime number p, let σp the √ unique element of Gal(Ω/Q) which fixes p if p = p, and which maps √ √ p onto − p. Now consider the subgroup H of Gal(Ω/Q) generated by the σp ’s. Notice that H = Gal(Ω/Q) since H does not contain the element σ ∈ Gal(Ω/Q) √ √ which maps p onto − p for all prime numbers p. However, we have ΩH = ΩGal(Ω/Q) = Q. Indeed, any element x ∈ Ω is contained in some subfield E of the form √ √ E = Q( p1 , . . . , pr ). Notice that E/Q is a finite Galois extension. Now assume that x ∈ ΩH . Since σp1 , . . . , σpr ∈ H, and since they generate Gal(E/Q), we conclude that x ∈ Q by classical Galois theory. In order to get a Galois correspondence, we define a topology on the Galois group of Ω/k. Definition I.2.7. Let Ω/k be a Galois extension. The Krull topology on Ω/k is the unique topology such that for all σ ∈ Gal(Ω/k), the familly of subsets {σGal(Ω/L) | σ ∈ Gal(Ω/k), L/k a finite Galois extension, L ⊂ Ω}

I.2 Galois theory

19

is a basis of open neighbourhoods of σ. We may now state the fundamental theorem of Galois theory. Theorem I.2.8 (Fundamental theorem of Galois theory). Let Ω/k be a Galois extension. Then there exist one-to-one correspondences between the following sets: (1)

The set of subfields K of Ω containing k and the set of closed subgroups of Gal(Ω/k).

(2)

The set of subfields K of Ω containing k such that [K : k] < +∞ and the set of open subgroups of Gal(Ω/k).

(3)

The set of subfields K of Ω containing k such that K/k is a finite Galois extension and the set of open normal subgroups of Gal(Ω/k).

In all cases, the correspondence is given by K ΩH

−→ Gal(Ω/K) ←− H.

Moreover, if H is an open normal subgroup of Gal(Ω/k), then we have Gal(ΩH /k) Gal(Ω/k)/H. In particular, for any finite Galois subextension L/k of Ω/k, we have Gal(Ω/k)/Gal(Ω/L) Gal(L/k). All these results and their proofs may be found in [42], Chapter IV. See also [26], Chapter 4.

I.2.3 Morphisms of Galois extensions To continue this section on Galois theory, we would like to have a closer look at morphisms of Galois extensions. Proposition I.2.9. Let K and K be two fields, let Ω/K two Galois extensions (not necessarily finite), and let ι : a ring morphism. Assume that there exist two extensions of ι,i = 1, 2. Then for all τ ∈ Gal(Ω /K ), there exists Gal(Ω/K) such that

and Ω /K be K −→ K be ϕi : Ω −→ Ω a unique τ ∈

τ ◦ ϕ1 = ϕ2 ◦ τ. In particular, there exists ρ ∈ Gal(Ω/K) such that ϕ1 = ϕ2 ◦ ρ.

20


Proof. Let ϕ1 , ϕ2 as in the statement of the proposition, and let τ ∈ Gal(Ω /K ). Let x ∈ Ω. We have to show that there exists y ∈ Ω such that τ (ϕ1 (x)) = ϕ2 (y). Notice that if such a y exists, it is unique since ϕ2 is injective. Let x1 , . . . , xn be the n distinct roots of μx,K in Kalg . Since Ω/K is Galois, xi ∈ Ω for i = 1, . . . , n by Lemma I.2.3. Since ϕ1 is an extension of ι and τ is K -linear, we easily deduce that ι(μx,K )(τ (ϕ1 (x))) = τ (ϕ1 (μx,K (x))) = 0. . Now since ϕ2 is an extension Thus, τ (ϕ1 (x)) is a root of ι(μx,K ) in Kalg of ι, ϕ2 (xi ) is a root of ι(μx,K ) by Lemma I.1.17. Since ϕ2 is injective, it follows that ϕ2 (x1 ), . . . , ϕ2 (xn ) are the n distinct roots of ι(μx,K ). Therefore, τ (ϕ1 (x)) = ϕ2 (xi ) for some i. We then set y = xi .

Hence we have shown that there is a unique map τ : Ω −→ Ω such that τ ◦ ϕ1 = ϕ2 ◦ τ. We have to check that τ ∈ Gal(Ω/K). If x, x ∈ Ω and λ ∈ K, we have τ (ϕ1 (λx + x )) = λτ (ϕ1 (x)) + τ (ϕ1 (x )) = λϕ2 (τ (x)) + ϕ2 (τ (x )). Since ϕ2 is K-linear, we get τ (ϕ1 (λx + x )) = ϕ2 (λτ (x) + τ (x )). But we also have τ (ϕ1 (λx + x )) = ϕ2 (τ (λx + x )). By injectivity of ϕ2 , we get τ (λx + x ) = λτ (x) + τ (x ). Similarly, we can check that τ (xx ) = τ (x)τ (x ) and τ (1) = 1. It remains to show that τ is bijective, but this follows immediately from the fact that Ω/K is Galois. The last part of the proposition is an immediate application of the first one. Corollary I.2.10. Let K and K be two fields, let Ω/K and Ω /K be two Galois extensions, and let ι : K −→ K be a ring morphism. Let ϕ : Ω −→ Ω be an extension of ι. For all τ ∈ Gal(Ω /K ), let ϕ(τ ) be the unique element of Gal(Ω/K) such that τ ◦ ϕ = ϕ ◦ ϕ(τ ). Then the map ϕ : Gal(Ω /K ) −→ Gal(Ω/K) is a continuous group morphism. Moreover, if ϕ is another extension of ι, then there exists ρ ∈ Gal(Ω/K) such that ϕ = ϕ ◦ ρ, and we have ϕ = Int(ρ) ◦ ϕ.

I.2 Galois theory

21

Proof. Let τ1 , τ2 ∈ Gal(Ω /K). By definition of ϕ, we have (τ1 ◦ τ2 ) ◦ ϕ = τ1 ◦ ϕ ◦ (ϕ(τ2 )) = ϕ ◦ (ϕ(τ1 ) ◦ ϕ(τ2 )). Since ϕ(τ1 ◦ τ2 ) is the unique element of Gal(Ω/K) satisfying (τ1 ◦ τ2 ) ◦ ϕ = ϕ ◦ ϕ(τ1 ◦ τ2 ), we have ϕ(τ1 ◦ τ2 ) = ϕ(τ1 ) ◦ ϕ(τ2 ). This proves that ϕ is a group morphism. The continuity is left to the reader. Now, if ϕ is another extension of ι, then by the previous proposition, there exists ρ ∈ Gal(Ω/K) such that ϕ = ϕ ◦ ρ. Therefore, for every τ ∈ Gal(Ω /K ), we have τ ◦ ϕ = (τ ◦ ϕ) ◦ ρ−1 = ϕ ◦ (ϕ(τ ) ◦ ρ−1 ) = ϕ ◦ (ρ ◦ ϕ(τ ) ◦ ρ−1 ). We conclude as before.

I.2.4 The Galois group as a profinite group Let Ω/k be a Galois extension. Lemma I.2.4 shows in particular that an element σ ∈ Gal(Ω/k) is completely determined by its restrictions to finite Galois subextensions L/k of Ω/k. Intuitively, Gal(Ω/k) then should be completely determined by the finite groups Gal(L/k). This is indeed the case, and in order to make this statement more precise, we need to introduce the concept of an inverse limit. Definition I.2.11. A directed set is a partially ordered set (I, ≤), such that for all i, j ∈ I, there exists k ∈ I such that i ≤ k and j ≤ k. Examples I.2.12. The reader will easily convince himself that the following sets are examples of directed sets: (1)

The set N with the order relation ≤.

(2)

The set N∗ = {1, 2, . . .} with the divisibility relation.

Definition I.2.13. A projective system of sets (groups, rings, etc) is a family of sets (groups, rings, etc) (Xi )i∈I , indexed by a directed set I, together with maps (resp. group morphisms, ring morphisms, etc) πij : Xj −→ Xi for any i, j ∈ I, i ≤ j, satisfying the following properties: (1)

πii = IdXi for all i ∈ I.

(2)

For all i, j, k ∈ I, i ≤ j ≤ k, we have πij ◦ πjk = πik .

Examples I.2.14. We now give some examples of projective systems indexed by the directed sets introduced above:

22


(1)

Let p be a fixed prime number. For any n ∈ N, let Xn = Z/pn Z, and let πmn : Z/pn Z −→ Z/pm Z be the natural projection for m ≤ n. Then we obtain a projective system of rings.

(2)

For any n ∈ N∗ , let Xn = Z/nZ, and let πmn : Z/nZ −→ Z/mZ be again the natural projection for all m|n. Then we obtain once again a projective system of rings.

Definition I.2.15. If ((Xi )i∈I , (πij )) is a projective system of sets (groups, rings, etc), the inverse limit lim ←−Xi is the subset (subgroup, i∈I

subring, etc) lim ←−Xi =

(xi )i∈I ∈

i∈I

Xi | πij (xj ) = xi for all i ≤ j

.

i∈I

Recall now a definition from topology. Definition I.2.16. Let (Xi )i∈I be a family of topological spaces. The product topology on Xi is the unique topology such that a basis i∈I

of open neighbourboods of (xi )i∈I consists of the subsets

Ui , where

i∈I

Ui ⊂ Xi is an open neighbourhood of xi and Ui = Xi for all but finitely many i ∈ I. If ((Xi )i∈I , (πij )) is a projective system of topological spaces (groups, rings, etc), then the inverse limit is also a topological space (group, ring, etc) with respect to the topology induced by the product topology. In particular, if each Xi is finite, it may be endowed with the discrete topology, and in this case we get a natural structure of a topological space on the inverse limit ← lim −Xi . i∈I

Examples I.2.17. The projective systems introduced previously allow us to define two topological rings by taking the corresponding inverse limits: (1)

Let p be a fixed prime number. The topological ring n lim Zp = ← − Z/p Z n∈N

is called the ring of p-adic integers. One can show that the ring Zp defined above is homeomorphic to the completion of Z with respect to the p-adic valuation as a topological ring.

I.2 Galois theory (2)

23

The topological ring ˆ = lim Z/nZ Z ←−∗ n∈N

is called the profinite completion of Z. Both of them play an important role in number theory. We can now elucidate the structure of the Galois group as a topological group. If k is a field, the set of all finite Galois subextensions of Ω/k with the partial order relation ‘⊂’ is a directed set, since the compositum of two finite Galois extensions is a finite Galois extension. Moreover, for any finite Galois extension L/k, let XL = Gal(L/k), and for any L/k, L /k such that L ⊂ L , let πL,L be the group morphism defined by πL,L :

Gal(L /k) −→ Gal(L/k) σ −→ σ|L .

We obtain in this way a projective system of groups. Therefore, the following statement makes sense: Theorem I.2.18. Let Ω/k be a Galois extension. Then we have an isomorphism of topological groups Gal(Ω/k) lim ←−Gal(L/k), L

where L/k runs over all finite Galois subextensions of Ω/k. Proof. Let us consider the map Θ:

Gal(Ω/k) −→ lim ←−Gal(L/k) L

σ −→ (σ|L )L . This is clearly an abstract group morphism. Now assume that σ ∈ ker(Θ), and let x ∈ Ω. Pick any finite Galois subextension L/k of Ω/k containing x. By assumption σ|L is the identity, so we get σ(x) = σ|L (x) = x. Hence σ = IdΩ and Θ is injective. Now let (σ (L) )L ∈ (L) lim ←−Gal(L/k). For any x ∈ Ω, we set σ(x) = σ (x), where L/k is a L

finite Galois subextension of Ω/k containing x. We claim that the result does not depend on the choice of L. Indeed, assume that L1 , L2 are two finite Galois subextensions of Ω/k containing x. Then L = L1 L2 is a

24


finite Galois subextension of Ω/k containing x, and Li ⊂ L. Therefore, by assumption, we have σ (L) (x) = (σ (L) )|Li (x) = σ (Li ) (x). Now it is clear that σ ∈ Gal(Ω/k), and that it is a preimage of (σ (L) )L by Θ. Hence Θ is surjective as well. It is easy to check that Θ is bicontinuous (it follows from the definition of the Krull topology), so we are done. Definition I.2.19. A topological group Γ is profinite if it is isomorphic as a topological group to an inverse limit of finite groups (each of them being endowed with the discrete topology). In view of the previous result, the Galois group of an arbitrary Galois extension is profinite. We now list some properties of profinite groups without proof. We refer the reader to [12] for more details. A profinite group Γ is compact and totally disconnected (that is the only non-empty connected subsets are one-point subsets). In particular, one-point subsets are closed, every open subgroup is also closed and has finite index. Moreover, every neighbourhood of 1 contains an open normal subgroup (hence of finite index). If Γ is a closed subgroup of Γ, then Γ is profinite, and if moreover Γ is normal, so is Γ/Γ . Finally, if N denotes the set of open normal subgroups of Γ, the map θ:

Γ −→ lim ←− Γ/U U ∈N

g −→ (gU )U is an isomorphism of topological groups.

Exercises 1. Show that a finite group (endowed with the discrete topology) is profinite. 2. Show that the algebraic definition of Zp given in this chapter coincides with the classical analytic definition. 3. Prove that there exists an isomorphism of topological groups ˆ Z Zp , p

where p runs over the set of prime numbers.

Exercises

25

4. Let q = pr , r ≥ 1, where p is a prime number, and let k = Fq the ˆ finite field with q elements. Show that Gal(ks /k) Z. 5. Let Qab ⊂ C be the maximal abelian subextension of Q. Determine Gal(Qab /Q). Hint: Use the fact that every finite abelian extension of Q is contained in a cyclotomic extension (Kronecker-Weber’s theorem).

II Cohomology of profinite groups

In this chapter, we define the cohomology sets associated to a profinite group Γ, and establish fundamental properties which will be crucial when studying Galois cohomology. From now on, if Ω/k is a Galois extension, we will write GΩ for its Galois group whenever k is clear from the context. §II.3 Cohomology sets: basic properties II.3.1 Definitions In the introduction, we ‘solved’ the descent problem for conjugacy classes of matrices associated to a finite Galois extension Ω/k of Galois group GΩ . We would like now to investigate the case where Ω/k is an infinite Galois extension. The main idea is that the problem locally boils down to the previous case. Let us fix M0 ∈ Mn (k) and let us consider a specific matrix M ∈ Mn (k) such that QM Q−1 = M0 for some Q ∈ SLn (Ω). If L/k is any finite Galois subextension of Ω/k with Galois group GL containing all the entries of Q, then Q ∈ SLn (L) and the equality above may be read in Mn (L). Therefore, for this particular matrix M , the descent problem may be solved by examining the corresponding element in H 1 (GL , ZSLn (M0 )(L)). Now if we take another finite Galois subextension L /k such that M ∈ Mn (L ), we obtain an obstruction living in H 1 (GL , ZSLn (M0 )(L )). But the fact that M is conjugate or not to M0 by an element of SLn (k) is an intrinsic property of M and of the field k, and should certainly not depend on the chosen Galois extension L/k. Therefore, we need to find a way to patch these local obstructions together. If we try to imitate the method followed in the introduction, we need 26

II.3 Cohomology sets: basic properties

27

first an appropriate action of GΩ on Mn (Ω) and SLn (Ω). Since we want to patch together the local obstructions, we need this action to coincide with the local actions on the various sets Mn (L) and SLn (L). Setting σ·(mij ) = (σ(mij )) gives rise to an action on Mn (Ω) and SL(Ω) which satisfies the desired properties. Indeed, if M ∈ Mn (Ω) (or SLn (Ω)) and if L/k is a finite Galois extension containing all the entries of M , then the action of σ on M is nothing but the action of σ|L on M , when M is viewed as an element of L. The reason why this works here is that GΩ is a profinite group, isomorphic to the projective limit of the finite groups GL . In particular, an element σ ∈ GΩ is completely determined by its restrictions to finite Galois subextensions. Moreover, and maybe more importantly, the stabilizer of a given matrix M ∈ Mn (Ω) for the action of GΩ is equal to the open subgroup Gal(Ω/K), where K is the subfield of Ω generated by the entries of M . Indeed, σ ∈ GΩ will act trivially on M if and only if it acts trivially on the entries of M , that is if it restricts to the identity on K. Therefore, this stabilizer contains an open normal subgroup Gal(Ω/L) (where L is a finite Galois subextension containing the entries of M ), and consequently the action of GΩ induces an action on GΩ /Gal(Ω/L), and thus on GL since these two groups are isomorphic by Theorem I.2.8. These considerations generalize to arbitrary profinite groups, and lead to the following definitions: Definition II.3.1. Let Γ be a profinite group. A left action of Γ on a discrete topological space A is called continuous if for all a ∈ A, the set StabΓ (a) = {σ ∈ Γ | σ·a = a} is an open subgroup of Γ. One can show that this is equivalent to ask for the map Γ × A −→ A (σ, a) −→ σ·a to be continuous. Discrete topological spaces with a continuous left action of Γ are called

28

Cohomology of profinite groups

Γ-sets. A group A which is also a Γ-set is called a Γ-group if Γ acts by group morphisms, i.e. σ·(a1 a2 ) = (σ·a1 )(σ·a2 ) for σ ∈ Γ, a1 , a2 ∈ A. A Γ-group which is commutative is called a Γ-module. A morphism of Γ-sets (resp. Γ-groups, Γ-modules) is a map (resp. a group morphism) f : A −→ A satisfying the following property: f (σ·a) = σ·f (a) for all σ ∈ Γ and all a ∈ A. Examples II.3.2. (1)

Assume that Γ is a finite group. Then any discrete topological set A on which Γ acts on the left is a Γ-set. Indeed such an action is continuous since any finite set is open for the discrete topology.

(2)

Any discrete topological set A on which Γ acts trivially is a Γ-set.

(3)

Let Ω/k be a Galois extension of group GΩ . Then the map GΩ × Ω −→ Ω (σ, x) −→ σ(x) endows Ω with the structure of a GΩ -module.

(4)

Let V be a k-vector space, and let us denote by VΩ the tensor product V ⊗k Ω. Then the action of GΩ on VΩ defined on elementary tensors by σ·(v ⊗ λ) = v ⊗ (σ·λ) for all v ∈ V, λ ∈ Ω is continuous, and therefore endows VΩ with the structure of a GΩ -module.

(5)

Let V and W be two k-vector spaces of dimension n and m respectively. If Ω/k is a Galois extension of group GΩ , then GΩ acts on HomΩ (VΩ , WΩ ) as follows: for all σ ∈ GΩ and f ∈ HomΩ (VΩ , WΩ ), set (σ·f )(x) = σ·(f (σ −1 ·v)) for all x ∈ VΩ . The choice of bases induces an isomorphism HomΩ (VΩ , WΩ ) Mm×n (Ω), and the corresponding action of GΩ on Mm×n (Ω) is simply the action entrywise. Therefore, the action defined above is continuous. This remains true if we replace V and W by finite dimensional k-algebras, and if we consider morphisms of Ωalgebras.


29

(6)

The action in the previous example also induces an action of GΩ on GL(VΩ ). One may check easily that it is an action by group automorphisms, so that GL(VΩ ) is a GΩ -group. In particular, GLn (Ω) is a GΩ -group. The same is true for other matrix groups such as SLn (Ω) or On (Ω).

(7)

Let μn (Ω) be the group of nth roots of 1 in Ω. Then μn (Ω) is a GΩ -module.

As already observed, a matrix M ∈ Mn (Ω) may be viewed as an element of Mn (L) for a suitable finite Galois subextension L/k of Ω/k. In other words, Mn (Ω) = Mn (L) = Mn (Ω)Gal(Ω/L) = Mn (Ω)U , L/k

U ∈N

L/k

where N is the subset of open normal subgroups of GΩ . This equality is not specific to Mn (Ω). In fact, it characterizes more generally Γ-sets, where Γ is a profinite group. Lemma II.3.3. Let Γ be a profinite group, and let A be a discrete topological set on which Γ acts on the left. Then the action of Γ on A is continuous if and only if we have A= AU , U ∈N

where N denotes the set of open normal subgroups of Γ. Proof. Assume that the action of Γ is continuous, and let a ∈ A. Then StabΓ (a) is an open subgroup, which contains 1. Hence, it contains some U ∈ N . In particular, a ∈ AU . It follows that we have A= AU . U ∈N

Conversely, assume that the equality above holds, and let a ∈ A. By assumption, there exists U ∈ N such that a ∈ AU . Therefore, for all σ ∈ U , we have σ·a = a. If now τ ∈ StabΓ (a), then

Thus

τ σ·a = τ ·a = a for all σ ∈ U. τ U ⊂ StabΓ (a). Since 1 ∈ U , the other inclusion holds

τ ∈StabΓ (a)

as well, and we get StabΓ (a) =

τ ∈StabΓ (a)

τ U.

30


It follows that StabΓ (a) is open. This concludes the proof. At this point, we may define the 0th -cohomology set H 0 (Γ, A). Definition II.3.4. For any Γ-set A, we set H 0 (Γ, A) = AΓ . If A is a Γ-group, this is a subgroup of A. The set H 0 (Γ, A) is called the 0th cohomology set of Γ with coefficients in A. Remark II.3.5. We will use this notation only episodically in this book, and will prefer the notation AΓ . We would like now to define the main object of this chapter, namely the first cohomology set H 1 (Γ, A). We first need an appropriate definition of a 1-cocycle. Let us go back to our conjugacy problem. Now that we have a suitable action of GΩ on SLn (Ω), we can mimick the reasoning made in the introduction and obtain a map αQ :

GΩ −→ ZSLn (M0 )(Ω) σ −→ Q(σ·Q)−1

which measures the obstruction to the conjugacy problem for the pair of matrices M and M0 . This map satisfies the cocycle condition stated in the introduction, but has the extra property to contain all the local obstructions we wanted to patch together. Indeed, let L/k be a finite Galois subextension of Ω/k containing all the entries of Q. As already observed at the very beginning of this chapter, we have σ ·Q = σ|L ·Q, so ασQ ∈ ZSLn (M0 )(L). Moreover let σ, σ ∈ Gal(Ω/L), and assume that σ = στ for some τ ∈ Gal(Ω/L). Since L contains all the entries of Q, we have τ ·Q = Q and thus ασQ = ασQ . Taking into account that we have a group isomorphism GΩ /Gal(Ω/L) GL induced by restriction to L, we see that the map αQ factors through a map GL −→ ZSLn (M0 )(L) σ −→ Q(σ·Q)−1 which is the local obstruction obtained when considering Q as an element of SLn (L). The crucial point here is that for all σ ∈ GΩ , αQ is constant on an open


31

neighbourhood of σ. This is equivalent to say that αQ is a continuous map, as the next proposition shows: Proposition II.3.6. Let Γ be a profinite group, let A be a Γ-set and let n ≥ 1 be an integer. For any map α : Γn −→ A, the following conditions are equivalent: (1)

α is continuous

(2)

α is locally constant, that is for every s = (σ1 , . . . , σn ) ∈ Γn , there exists an open neighbourhood of s on which α is constant

(3)

There exist U ∈ N and a map α(U ) : (Γ/U )n −→ AU such that (U )

ασ1 ,...,σn = ασ1 ,...,σn for all σ1 , . . . , σn ∈ Γ. Proof. For every s = (σ1 , . . . , σn ) ∈ Γn , we will write αs instead of ασ1 ,...,σn . (1) ⇒ (2) Assume that α is continuous, and let s = (σ1 , . . . , σn ) ∈ Γn . Then the set Us = α−1 ({αs }) is an open neighbourhood of s, since {αs } is open in A and α is continuous. By definition, α is constant on Us . (2) ⇒ (3) Assume that α is locally constant. For all s = (σ1 , . . . , σn ) ∈ Γn , let Us be an open neighbourhood of s on which α is constant. By definition of the product topology, one may assume that Us = Vs(1) × · · · × Vs(n) , (i)

where Vs is an open neighbourhood of σi in Γ. The family (Us )s∈Γn is an open covering of Γn . Since Γ is compact, so is Γn , and thus there exists a finite subset T of Γn such that Γn = Ut . t∈T (i)

For every t = (τ1 , . . . , τn ) ∈ T , notice that Ut neighbourhood of 1, and that we have (1)

Ut = τ1 Ut

= τi−1 Vt

(i)

is an open

(n)

× · · · × τn Ut .

By Lemma II.3.3, for every t ∈ T , there exists an open subgroup Ut of Γ such that αt ∈ AUt . Since T is finite, the set U0 =

i,t

(i)

Ut ∩

Ut

t

is an open neighbourhood of 1 in Γ. Since Γ is a profinite group, there

32


exists a normal open subgroup U of Γ contained in U0 . Now let s = (σ1 , . . . , σn ) ∈ Γn . By choice of T , there exists t ∈ T such that s ∈ Ut , (i) so for i = 1, . . . , n we may write σi = τi ui for some ui ∈ Ut . We then (i) (i) get σ1 U ×· · ·×σn U ⊂ Ut , since U ⊂ Ut for all i and Ut is a subgroup. Hence for all s ∈ σ1 U × · · · × σn U , we have

αs = αt ∈ AUt , since α is constant on Ut . Moreover, since U ⊂ U0 ⊂ Ut , we have AUt ⊂ AU . Therefore, the map α(U ) :

(Γ/U )n −→ AU (σ 1 , . . . , σ n ) −→ ασ1 ,...,σn

is well-defined, and satisfies the required conditions. (3) ⇒ (1) Assume that we have an open normal subgroup U and a map α(U ) : (Γ/U )n −→ AU satisfying (3). Let V be an open subset of A. We have to prove that α−1 (V ) is open in Γ. Since α−1 (V ) = α−1 ({v}), v∈V

it is enough to show that α−1 ({v}) is open for every v ∈ V . If v does not lie in the image of α, this is obvious, so we can assume that α−1 ({v}) is not empty. Let s = (σ1 , . . . , σn ) ∈ α−1 ({v}). The assumption implies that for all t ∈ σ1 U × · · · × σn U , we have (U )

αt = ασ1 ,...,σn = αs = v. Thus α−1 ({v}), contains an open neighbourhood of s. Hence α−1 ({v}) is open, and we are done. Taking the previous observations into consideration, it is natural to set the following definition: Definition II.3.7. Let A be a Γ-group. A 1-cocycle of Γ with values in A is a continuous map α : Γ −→ A such that αστ = ασ σ·ατ for σ, τ ∈ Γ. We denote by Z 1 (Γ, A) the set of all 1-cocycles of Γ with values in A. The constant map Γ −→ A σ −→ 1 is an element of Z 1 (Γ, A), which is called the trivial 1-cocycle. Notice also that for any 1-cocycle α, we have α1 = 1.


33

Remark II.3.8. If Γ acts trivially on A, a 1-cocycle is just a continuous morphism α : Γ −→ A. In order to define the cohomology set H 1 (Γ, A), we need now an appropriate notion of cohomologous cocycles, which coincides with the one defined in the introduction in a particular case. This will be provided by the following lemma: Lemma II.3.9. Let Γ be a profinite group, let A be a Γ-group and let α : Γ −→ A be a 1-cocycle. Then for all a ∈ A, the map α :

Γ −→ A σ −→ aασ σ·a−1

is again a 1-cocycle. Proof. Let σ, τ ∈ Γ. We have ασ σ·ατ = (aασ σ·a−1 )σ·(aατ τ ·a−1 ). Since Γ acts on A by group automorphisms, we get ασ σ·ατ = aασ (σ·ατ )στ ·a−1 = aαστ στ ·a−1 = αστ .

It remains to prove that α is continuous. Let V be an open subset −1 −1 of We have to prove that α (V ) is open in Γ. Since α (V ) = A. −1 −1 α ({v}), it is enough to show that α ({v}) is open for every v∈V

v ∈ V . If v does not lie in the image of α , this is obvious, so we −1 −1 can assume that α ({v}) is not empty. Let σ ∈ α ({v}). We then have ασ = v. By assumption, α is continuous. Since {1} is open in A, α−1 ({1}) is an open subgroup of Γ. Moreover, StabΓ (a) is open since Γ acts continuously on A. Therefore, U = α−1 ({1}) ∩ StabΓ (a) is open in Γ, and so is σU . Now for every τ ∈ U , we have αστ

−1

= = = = = =

aαστ στ ·a−1 aασ σ·ατ στ ·a−1 aασ στ ·a−1 aασ σ·a−1 ασ v. −1

Therefore for all σ ∈ α ({v}), α ({v}) contains an open neighbour−1 hood of σ and thus α ({v}) is open. This concludes the proof.

34


This leads to the following definition: Definition II.3.10. Two 1-cocycles α, α are said to be cohomologous if there exists a ∈ A satisfying ασ = aασ σ·a−1 for all σ ∈ Γ. It is denoted by α ∼ α . Remark II.3.11. The symbol ‘σ · a−1 ’ may seem ambiguous at first sight, since it could denote (σ ·a)−1 as well as σ ·(a−1 ). However, these two elements are equal in our setting, since Γ acts on A by group automorphisms; we will keep this notation throughout. Definition II.3.12. Let Γ be a profinite group, and let A be a Γgroup. The relation ‘∼’ is easily checked to be an equivalence relation on Z 1 (Γ, A). We denote by H 1 (Γ, A) the quotient set H 1 (Γ, A) = Z 1 (Γ, A)/∼ . It is called the first cohomology set of Γ with coefficients in A . The set H 1 (Γ, A) is not a group in general. However, it has a special element which is the class of the trivial cocycle. Therefore, H 1 (Γ, A) is a pointed set in the following sense: Definition II.3.13. A pointed set is a pair (E, x), where E is a nonempty set and x ∈ E. The element x is called the base point. A map of pointed sets f : (E, x) −→ (F, y) is a set-theoretic map such that f (x) = y. We will often forget to specify the base point when it is clear from the context. Example II.3.14. The set H 1 (Γ, A) is a pointed set, and any abstract group G may be considered as a pointed set, whose base point is the neutral element. Remark II.3.15. If A is a Γ-module, the set Z 1 (Γ, A) is an abelian group for the pointwise multiplication of functions. This operation is compatible with the equivalence relation, hence it induces an abelian group structure on H 1 (Γ, A). We would like now to define higher cohomology groups. Let A be a Γ-module (written additively here) and n ≥ 0. We set C 0 (Γ, A) = A, and if n ≥ 1, we denote by C n (Γ, A) the set of all continuous maps from Γn to A.


35

We now define a map dn : C n (Γ, A) −→ C n+1 (Γ, A) by the formulas: d0 (a) = σ·a − a and for all n ≥ 1

dn (α)σ1 ,...,σn+1 =

σ1 · ασ2 ,...,σn+1 +

n

(−1)i ασ1 ,...,σi σi+1 ,...,σn+1

i=1

+(−1)n+1 ασ1 ,...,σn . Definition II.3.16. An n-cocycle of Γ with values in A is a map α ∈ C n (Γ, A) satisfying: (1)

dn (α) = 0

(2)

ασ1 ,...,σn = 0 whenever σi = 1 for some i.

A map α ∈ C n (Γ, A) is an n-coboundary of Γ with values in A if there exists β ∈ C n−1 (Γ, A) such that: (1)

α = dn−1 (β)

(2)

βσ1 ,...,σn−1 = 0 whenever σi = 1 for some i.

Notice that for n = 1, condition (2) is empty. The set of n-cocycles and n-coboundaries are abelian subgroups of C n (Γ, A), denoted by Z n (Γ, A) and B n (Γ, A) respectively. One can check that dn dn−1 = 0, so B n (Γ, A) is a subgroup of Z n (Γ, A), and we may define H n (Γ, A) = Z n (Γ, A)/B n (Γ, A). The group H n (Γ, A) is called the nth cohomology group of Γ with coefficients in A. Two n-cocycles are said to be cohomologous if they have the same image in H n (Γ, A), i.e. if they differ by an n-coboundary. The constant map Γn −→ A σ −→ 1 is an element of Z n (Γ, A), which is called the trivial n-cocycle. Remarks II.3.17.

36


(1)

In the existing literature, n-cocycles and n-coboundaries are often defined to be elements of ker(dn ) and im(dn−1 ) respectively, and cocycles and coboundaries satisfying the extra condition (2) are called normalized. However, one can show that the two quotient groups obtained with these two different definitions are canonically isomorphic.

(2)

If Γ is a finite abstract group, we recover the classical definition of the cohomology groups associated to a finite group (modulo the previous remark).

(3)

Assume that A is a Γ-module (denoted additively). Since a 1cocycle α always satisfies α1 = 0, Definition II.3.12 is consistent with Definition II.3.16.

Now write A multiplicatively (it is more convenient for our purpose), and let us give explicitly the formulas defining a 2-cocycle. A 2-cocycle is a continuous map α : Γ × Γ −→ A satisfying σ·ατ,ρ ασ,τ ρ = αστ,ρ ασ,τ for σ, τ, ρ ∈ Γ, and σ1,τ = σσ,1 = 1 for all σ, τ ∈ Γ, and two 2-cocycles α, α are cohomologous if there exists a continuous map ϕ : Γ −→ A satisfying ϕ1 = 1 and = (σ·ϕτ )ϕ−1 ασ,τ στ ϕσ ασ,τ for all σ, τ ∈ Γ.

In the following, we will always assume implicitly that, when we write H n (Γ, A), the set A has the appropriate structure.

II.3.2 Functoriality We now start to study the functorial properties of cohomology sets, which are their main interest. In the sequel, as well as in the other paragraphs, we will not check that the various maps we consider are continuous, and leave the verifications as an exercice for the reader. Definition II.3.18. Let Γ, Γ be two profinite groups. Let A be a Γ-set and A be a Γ -set. Moreover, let ϕ : Γ −→ Γ be a morphism of profinite groups (in particular, ϕ is continuous), and let f : A −→ A be a map. If A and A are groups, we require that f is a group morphism.


37

We say that f and ϕ are compatible if f (ϕ(σ )·a) = σ ·f (a) for σ ∈ Γ , a ∈ A. Notice that it follows from the very definition that if a is fixed by Γ, then f (a) is fixed by Γ . Hence f induces by restriction a map of pointed sets f∗ : H 0 (Γ, A) −→ H 0 (Γ , A ). The following proposition shows that this is also true for higher cohomology sets. Proposition II.3.19. Let Γ, Γ , A, A as above, and let ϕ : Γ −→ Γ and f : A −→ A be two compatible maps. For any n-cocycle α ∈ Z n (Γ, A), the map f∗ (α) :

Γn −→ A (σ1 , . . . , σn ) −→ f (αϕ(σ1 ),...,ϕ(σn ) )

is an n-cocycle, and the map f∗ :

H n (Γ, A) −→ H n (Γ , A ) [α] −→ [f∗ (α)]

is a well-defined map of pointed sets (resp. a group morphism if A and A are abelian). Proof. We only prove the result for n = 1 in the case of Γ-groups. The remaining cases may be proved similarly. Let α ∈ Z 1 (Γ, A) and set β = f∗ (α). By definition, we have βσ = f (αϕ(σ ) ) for all σ ∈ Γ . Hence βσ τ = f (αϕ(σ )ϕ(τ ) ), since ϕ is a group morphism. Since α is a 1-cocycle, we get βσ τ = f (αϕ(σ ) ϕ(σ )·αϕ(τ ) ) = f (αϕ(σ ) )f (ϕ(σ )·αϕ(τ ) ). By compatibility, we get that βσ τ = f (αϕ(σ ) )σ ·f (αϕ(τ ) ) = βσ σ ·βτ . Hence β is a 1-cocycle. Now we have to show that if α and α are cohomologous, then the corresponding β and β are also cohomologous, so assume that ασ = aασ σ · a−1 for all σ ∈ Γ,

38


for some a ∈ A. Applying this relation to σ = ϕ(σ ) and taking f on both sides gives βσ = f (aαϕ(σ ) ϕ(σ )·a−1 ). Since f is a group morphism which is compatible with ϕ, we get βσ = f (a)f (αϕ(σ ) ) σ ·f (a)−1 = f (a)βσ σ ·f (a)−1 . Hence β and β are cohomologous. Finally, it is clear from the definition that f∗ maps the trivial class onto the trivial class. Example II.3.20. Assume that Γ = Γ and ϕ = IdΓ . Then a compatible map f : A −→ A is just a morphism of Γ-sets (or Γ-groups, etc), and the map f∗ just sends the cohomology class of α onto the cohomology class of f ◦ α. Moreover, if g : A −→ A is a morphism of Γ-sets (or Γ-groups, etc), we have (g ◦ f )∗ = g∗ ◦ f∗ . Example II.3.21. Assume that Γ = Γ , ϕ = Int(ρ), for a fixed element ρ ∈ Γ and let f:

A −→ A a −→ ρ−1 ·a.

It is easy to see that f and ϕ are compatible. Claim: For n ≥ 0, the induced map f∗ : H n (Γ, A) −→ H n (Γ, A) is the identity. For n = 0, this is clear. Again, we only prove the claim for n = 1 in the case of Γ-groups. We have to show that α and β = f∗ (α) are cohomologous for every 1-cocycle α ∈ Z 1 (Γ, A). First notice that applying the cocyclicity relation to τ = σ −1 gives σ −1 ·ασ = ασ−1 −1 for σ ∈ Γ. Now we have βσ = ρ−1 ·αρσρ−1 = ρ−1 ·(αρ (ρ·ασρ−1 )) = (ρ−1 ·αρ )ασρ−1 , since Γ acts on A by group automorphisms. Hence we get βσ = (ρ−1 ·αρ )(ασ σ·αρ−1 ) = αρ−1 −1 ασ σ·αρ−1 . Setting a = αρ−1 −1 then shows that α and β are cohomologous.


39

Example II.3.22. Let G, A be finite groups, where A is a G-group, and let H be a subgroup of G, acting on A by restricting the action of G. Now let ϕ : H −→ G be the inclusion and let f = IdA . These two maps are compatible, and then we get a map Res : H n (G, A) −→ H n (H, A), called the restriction map from G to H. If [α] ∈ H n (G, A), a cocycle representing Res([α]) is α|H n . Example II.3.23. Let G, A be finite groups, and assume that G acts trivially on A. Let Γ be a profinite group acting continuously on G and acting trivially on A. Let ϕ : Γ −→ G be a morphism of profinite groups, and let f = IdA . Then ϕ and f are compatible, and we get a map ϕ∗ : H n (G, A) −→ H n (Γ, A), called the inverse image with respect to ϕ. If [α] ∈ H n (G, A), a cocycle β representing ϕ∗ ([α]) is given by β:

Γn −→ A (σ1 , . . . , σn ) −→ αϕ(σ1 ),...,ϕ(σn ) .

We would like to observe now that the map ϕ∗ depends on ϕ only up to conjugation. For, let ρ ∈ G and set ψ = Int(ρ) ◦ ϕ. Then ψ ∗ ([α]) is represented by the cocycle γ defined by γ:

Γn −→ A (σ1 , . . . , σn ) −→ αρϕ(σ1 )ρ−1 ,...,ρϕ(σn )ρ−1 .

Keeping in mind that G acts trivially on A, we see that the equality ψ ∗ ([α]) = ϕ∗ ([α]) is a consequence of Example II.3.21. Assume now that Γ also acts trivially on G, so that H 1 (Γ, G) is nothing but the set of conjugacy classes of continuous morphisms ϕ : Γ −→ G. The previous observations then imply that for each class [α] ∈ H n (G, A), we have a well-defined map H 1 (Γ, G) −→ H n (Γ, A) [ϕ] −→ ϕ∗ ([α]). Example II.3.24. Let Γ be a profinite group acting continuously on A, and let U, U ∈ N be two normal open subgroups of Γ, U ⊃ U . Then Γ/U and Γ/U acts continuously on AU and AU respectively. Let us denote by πU : Γ −→ Γ/U and πU : Γ −→ Γ/U the canonical

40


projections. One can easily check that the maps ϕ : Γ/U −→ Γ/U and f : AU −→ AU are compatible, so we get an inflation map

inf U,U : H n (Γ/U, AU ) −→ H n (Γ/U , AU ). If [α] ∈ H n (Γ/U, AU ), the cohomology class inf U,U ([α]) is represented by the cocycle β:

(Γ/U )n −→ AU (πU (σ1 ), . . . , πU (σn )) −→ απU (σ1 ),...,πU (σn ) .

Example II.3.25. Let Γ be a profinite group acting continuously on A, and let U ∈ N be a normal open subgroup of Γ. Then the maps πU : Γ −→ Γ/U and the inclusion AU −→ A are compatible, and give rise to a map fU : H n (Γ/U, AU ) −→ H n (Γ, A). Proposition II.3.26. For i = 1, . . . , 4, let Ai be a Γi -set (Γi -group,etc). Assume that we have two commutative diagrams A1

f1

f3

A4

/ A2 f2

f4

/ A3

and ΓO 1 o

ϕ1

ϕ3

Γ4 o

ΓO 2 ϕ2

ϕ4

Γ3

where ϕi is a morphism of profinite groups compatible with fi , for i = 1, . . . , 4. Then the diagram H n (Γ1 , A1 )

f1∗

f3∗

H n (Γ4 , A4 )

f4∗

/ H n (Γ2 , A2 )

f2∗

/ H n (Γ3 , A3 )

is commutative. Proof. If n = 0, the result is clear, so we assume that n ≥ 1. Let


41

[α] ∈ H n (Γ1 , A1 ). By definition, f1∗ ([α]) is represented by the cocycle Γn2 −→ A2 (σ1 , . . . , σn ) −→ f1 (αϕ1 (σ1 ),...,ϕ1 (σn ) ). Therefore, (f2∗ ◦ f1∗ )([α]) is represented by the cocycle Γn3 −→ A3 (σ1 , . . . , σn ) −→ f2 (f1 (αϕ1 (ϕ2 (σ1 )),...,ϕ1 (ϕ2 (σn )) )). Similarly, (f4∗ ◦ f3∗ )([α]) is represented by the cocycle Γn3 −→ A3 (σ1 , . . . , σn ) −→ f4 (f3 (αϕ3 (ϕ4 (σ1 )),...,ϕ3 (ϕ4 (σn )) )). Since f2 ◦ f1 = f4 ◦ f3 and ϕ1 ◦ ϕ2 = ϕ3 ◦ ϕ4 by assumption, we get the desired result. Convention: From now on, if f : A −→ B is a morphism of Γ-sets (Γgroups,etc), the symbol f∗ will denote the map in cohomology obtained when taking ϕ = IdΓ as in Example II.3.20, unless specified otherwise.

II.3.3 Cohomology sets as a direct limit In this paragraph, we would like to relate the cohomology of profinite groups to the cohomology of its finite quotients. The key ingredient to do this is Proposition II.3.6, which says more or less that an n-cocycle α : Γn −→ A is locally defined by a family of n-cocycles α(U ) : (Γ/U )n −→ AU , where U runs through the set of open normal subgroups of Γ. In view of the relation (U )

ασ1 ,...,σn = ασ1 ,...,σn for all σ1 , . . . , σn ∈ Γ, it easily implies that for all U, U ∈ N , U ⊃ U , we have

inf U,U ([α(U ) ]) = [α(U ) ]. Conversely, we will see a family of cohomology classes satisfying this coherence condition may be patched together to define a cohomology class in H n (Γ, A). We now go into details, and start with the definition of a direct limit.

42


Definition II.3.27. A directed system of sets (groups, rings, etc) is a family of sets (groups, rings, etc) (Xi )i∈I , indexed by a directed set I, together with maps (resp. group morphisms, ring morphisms,etc) ιij : Xi −→ Xj for any i, j ∈ I, i ≤ j, satisfying the following properties: (1)

ιii = IdXi for all i ∈ I.

(2)

For all i, j, k ∈ I, i ≤ j ≤ k, we have ιjk ◦ ιij = ιik .

Example II.3.28. Let Γ be a profinite group, and let A be a Γ-set. If U ∈ N , set XU = AU . For all U, U ∈ N , U ⊃ U , we denote by ιU,U the inclusion AU ⊂ AU . It is easy to check that we get a directed system of sets. Definition II.3.29. Let ((Xi )i∈I , (ιij )) be a directed system of sets (groups, rings, etc). We define an equivalence relation on the disjoint union Xi as follows: if i, j ∈ I, i ≤ j, xi ∈ Xi , xj ∈ Xj , we say that i∈I

xi ∼ xj if there exists k ∈ I such that k ≥ i, j and ιjk (xj ) = ιik (xi ). The direct limit of the sets (groups, rings, etc) ((Xi )i∈I , (ιij )), denoted by lim X , is the set (groups, ring, etc.) of equivalence classes Xi /∼ . i −→ i∈I

i∈I

The following lemma give a nicer description of direct limits in a particular case. Lemma II.3.30. Let ((Xi )i∈I , (ιij )) be a directed system of sets (groups, rings, etc). Assume that we have injective maps (group morphisms, ring morphisms, etc) fi : Xi −→ X such that fi = fj ◦ ιij for all i ≤ j. Then lim X fi (Xi ) ⊂ X. i −→ i∈I

i∈I

In particular, if the Xi ’s are subsets (subgroups, subrings, etc) of a same set (group, ring, etc) X satisfying Xi ⊂ Xj for all i ≤ j, we have lim X Xi . i −→ i∈I

i∈I

Proof. To see this, we define a map f : − lim →Xi −→ for xi ∈ Xi , set

i∈I

fi (Xi ) as follows:

i∈I

f (xi /∼ ) = fi (xi ). If xi ∼ xj , i ≤ j, then there exists k ∈ I, k ≥ i, j such that ιjk (xj ) = ιik (xi ) and therefore fk (ιjk (xj )) = fk (ιik (xi )), that is fj (xj ) = fi (xi )


43

by assumption. Hence f is well-defined, and one can check easily that f is a group (ring, etc) morphism if the Xi ’s are. We claim that f is bijective. Since surjectivity is clear by definition of f , we just need to check that f is injective. Assume that f (xi /∼ ) = f (xj /∼ ) for some i, j ∈ I, xi ∈ Xi , xj ∈ Xj . Let k ∈ I, k ≥ i, j. By assumption on the fi ’s, we have f (xi /∼ ) = fi (xi ) = fk (ιik (xi )), and similarly f (xj /∼ ) = fk (ιjk (xj )). Now using the equality f (xi /∼ ) = f (xj /∼ ) and the injectivity of fk , we get ιik (xi ) = ιjk (xj ), and thus xi /∼ = xj /∼ . This concludes the proof. Example II.3.31. Let Γ be a profinite group, and let A be a Γ-set. It U follows from Lemma II.3.3 and the previous result that A lim −→ A . U ∈N

Let Γ be a profinite group, and let A be a Γ-group (resp. a Γ-module if n ≥ 2). Recall from Examples II.3.24 and II.3.25 that we have maps

inf U,U : H n (Γ/U, AU ) −→ H n (Γ/U , AU ), for all U, U ∈ N , U ⊃ U and a map fU : H n (Γ/U, AU ) −→ H n (Γ, A). The following lemma follows from direct computations. Lemma II.3.32. The sets H n (Γ/U, AU ) together with the maps inf U,U form a directed system of pointed sets (resp. of groups if A is abelian). Moreover, we have fU = fU ◦ inf U,U . We now come to the main result of this section. Theorem II.3.33. Let Γ be a profinite group, and let A be a Γ-group. Then we have an isomorphism of pointed sets (resp. an isomorphism of groups if A is abelian) n U n lim −→ H (Γ/U, A ) H (Γ, A).

U ∈N

If [ξU ] ∈ H n (Γ/U, AU ), this isomorphism maps [ξU ]/∼ onto fU ([ξU ]). Proof. We first prove that there exists a well-defined map n U n f: − lim → H (Γ/U, A ) −→ H (Γ, A), U ∈N

which sends the equivalence class of [ξU ] ∈ H n (Γ/U, AU ) onto fU ([ξU ]).

44


Let U, U ∈ N , [ξU ] ∈ H n (Γ/U, AU ) and [ξU ] ∈ H n (Γ/U , AU ) such that [ξU ]/∼ = [ξU ]/∼ . By definition of the direct limit, there exists V ∈ N such that U ⊃ V, U ⊃ V and inf U,V ([ξU ]) = inf U ,V ([ξU ]). Applying fV on both sides and using the previous lemma, we obtain that fU ([ξU ]) = fU ([ξU ]), proving that f is well-defined. Clearly, f maps the class of the trivial cocycle to the class of the trivial cocycle. Moreover if A is abelian, one can check easily that f is a group morphism. Let us prove that f is bijective. For U, U ∈ N , U ⊃ U , let us denote by πU : Γ −→ Γ/U and πU,U : Γ/U −→ Γ/U the canonical projections. Let [α] ∈ H n (Γ, A). By Proposition II.3.6, there exists U ∈ N and a map α(U ) : (Γ/U )n −→ AU such that (U )

ασ1 ,...,σn = απU (σ1 ),...,πU (σn ) for all σ1 , . . . , σn ∈ Γ. Notice that by definition of the action of Γ/U on AU , we have πU (σ)·a = σ·a for all σ ∈ Γ, a ∈ AU . It follows easily that α(U ) is an n-cocycle. Moreover, by definition of fU and α(U ) , we have f ([α(U ) ]/∼ ) = fU ([α(U ) ]) = [α]. Therefore, f is surjective. It remains to prove the injectivity of f . Let us also denote by fU and inf U,U the inflation maps at the level of cocycles. It is easy to check that we still have fU = fU ◦ inf U,U for all U ⊃ U . Notice also that for all U ∈ N , the map fU : Z 1 (Γ/U, AU ) → Z 1 (Γ, A) is injective. Indeed, if ξ, ξ ∈ Z 1 (Γ/U, AU ) are two cocycles such that fU (ξ) = fU (ξ ), then we have ξπU (σ) = ξπ U (σ) for all σ ∈ Γ. Since πU : Γ → Γ/U is surjective, we get ξ = ξ . We are now ready to prove that f is injective. Let [ξU ] ∈ H n (Γ/U, AU ) and [ξU ] ∈ H n (Γ/U , AU ) such that f ([ξU ]/∼ ) = f ([ξU ]/∼ ). Let U ∈ N be a normal open subset of Γ contained in U and U . Since

II.4 Cohomology sequences

45

[inf U,U ([ξU ])]/∼ = [ξU ]/∼ and [inf U ,U ([ξU ])]/∼ = [ξU ]/∼ , we may assume without loss of generality that U = U . The equality f ([ξU ]/∼ ) = f ([ξU ]/∼ ) then rewrites fU ([ξU ]) = fU ([ξU ]). Therefore, there exists a ∈ A such that fU (ξU )σ = a fU (ξU )σ σ·a−1 for all σ ∈ Γ. By Lemma II.3.3, there exists U0 ∈ N such that a ∈ AU0 . Now let V ∈ N contained in U and U0 . In particular, we have AU ⊂ AV and AU0 ⊂ AV . Let η ∈ Z 1 (Γ/V, AV ) be the cocycle defined by η:

Γ/V −→ AV πV (σ) −→ a inf U,V (ξU )πV (σ) πV (σ)·a.

Taking into account that we have σ · a = πV (σ) · a for all σ ∈ Γ, the previous equality yields fV (inf U,V (ξU )) = fV (η). By injectivity of fV , we get inf U,V (ξU ) = η ∈ Z 1 (Γ/V, AV ). Since η and inf U,V (ξU ) are cohomologous by construction, we have inf U,V ([ξU )] = [η] = inf U,V ([ξU ]), and thus [ξU ]/∼ = [ξU ]/∼ . This concludes the proof. §II.4 Cohomology sequences If G is a finite group and A is a finite G-module, the groups H n (G, A) are known to have interesting properties with respect to exact sequences of G-modules (see [11] or [26] for an account on cohomology of finite groups). We now proceed to show that similar properties hold in our more general setting. Definition II.4.1. Let f : A −→ B be a map of pointed sets. The kernel of f is the preimage by f of the base point of B. A sequence of pointed sets A

f

/B

g

/C

is called exact at B if imf = ker g. A sequence of pointed sets A0 −→ A1 −→ · · · −→ Ai−1 −→ Ai −→ Ai+1 −→ · · ·

46


is called exact if it is exact at Ai for all i ≥ 1. An exact sequence of groups (resp. of Γ-groups, resp. of Γ-modules) is an exact sequence of pointed sets such that all the maps involved are group morphisms (resp. morphisms of Γ-groups, resp. of Γ-modules). For example, the sequence B

g

/1

/C

is exact if and only if g is surjective, and the sequence /A

1

f

/B

is exact if and only if f has trivial kernel. This does not imply that f is injective, unless A and B are groups. Assume that we have an exact sequence 1

/A

f

/B

g

/C

/1

of pointed Γ-sets. The goal of the next paragraphs is to derive some exact sequences in cohomology, under some reasonable conditions on A, B and C. We will keep this notation throughout.

II.4.1 The case of a subgroup Assume that A and B are Γ-groups, that f is a group morphism (hence f is injective), and that g induces a bijection of Γ-sets B/f (A) C, where B/f (A) is the set of left cosets modulo f (A). In other words, g is surjective and for all b, b ∈ B we have g(b) = g(b ) ⇐⇒ b = bf (a) for some a ∈ A. For instance, these conditions are satisfied in the following cases: (1)

A is a Γ-subgroup of B, C = B/A, f is the inclusion and g is the natural projection.

(2)

C is a Γ-group and g is a group morphism (this will be the case in the next subsection).

As pointed out previously in Example II.3.20, f and g induce maps on fixed points by restriction, namely f∗ : AΓ −→ B Γ and g∗ : B Γ −→ C Γ . Our next goal is to define a map of pointed sets δ 0 : C Γ −→ H 1 (Γ, A).


47

Let c ∈ C Γ , and let b ∈ B any preimage of c under g, i.e. g(b) = c. By assumption, we have c = σ·c for all σ ∈ Γ. Therefore, we have g(σ·b) = σ·g(b) = σ·c = c = g(b). By assumption on g, there exists a unique element ασ ∈ A such that f (ασ ) = b−1 σ·b. Lemma II.4.2. The map α : Γ −→ A is a 1-cocycle, and its class in H 1 (Γ, A) does not depend on the choice of b ∈ B. Proof. Let us prove that α is a cocycle. By definition of α, for all σ, τ ∈ Γ, we have f (αστ ) = b−1 στ · b = b−1 σ·(bb−1 τ ·b) = (b−1 σ·b) σ·(b−1 τ · b). Hence we have f (αστ ) = f (ασ )σ·f (ατ ) = f (ασ )f (σ·ατ ) = f (ασ σ·ατ ). By injectivity of f , we get αστ = ασ σ·ατ . Let us prove now that the cohomology class of α does not depend on the choice of b. Let b ∈ B be another preimage of c under g. We then have g(b ) = c = g(b), so b = bf (a−1 ) = bf (a)−1 for some a ∈ A by assumption on g, and let α be the corresponding 1-cocycle. We then have f (ασ ) = f (a)b−1 σ·(bf (a−1 )) = f (a)f (ασ ) σ · f (a−1 ) = f (aασ σ·a−1 ), so by injectivity of f , this implies that α and α are cohomologous. This concludes the proof. Notice that a preimage under g of the base point of C Γ is the neutral element 1 ∈ B (since g is a morphism of pointed sets). Since Γ acts by group automorphisms on B, we have σ·1 = 1 for all σ ∈ Γ, and therefore the base point of C Γ is mapped onto the trivial cohomology class. We then have constructed a map of pointed sets δ0 :

C Γ −→ H 1 (Γ, A) c −→ [α],

where the cocycle α is defined by the relations f (ασ ) = b−1 σ·b for all σ ∈ Γ, for an arbitrary preimage b ∈ B of c.

48


Definition II.4.3. The map δ 0 : C Γ −→ H 1 (Γ, A) is called the 0th connecting map. Proposition II.4.4. The sequence of pointed sets

1

/ AΓ

/ BΓ

f∗

g∗

/ CΓ

δ0

/ H 1 (Γ, A)

f∗

/ H 1 (Γ, B)

is exact. Proof. The exactness of the sequence 1

/ AΓ

f∗

/ BΓ

g∗

/ CΓ

is left to the reader. Exactness at C Γ : We need to check that im(g∗ ) = ker(δ 0 ). Let c ∈ C G , and let us denote by α the cocycle representing δ 0 (c), as defined above. Assume first that c ∈ im(g∗ ), that is c = g(b) for some b ∈ B Γ . Then by definition α is the trivial cocycle, and c is mapped onto the trivial class. Therefore, im(g∗ ) ⊂ ker(δ 0 ). Conversely, assume that δ 0 (c) is trivial, that is ασ = a σ·a−1 for all σ ∈ Γ, for some a ∈ A. Let b ∈ B be a preimage of c under g. We then have f (a σ·a−1 ) = b−1 σ·b for all σ ∈ Γ, so f (a)σ·f (a)−1 = b−1 σ·b for all σ ∈ Γ. Hence bf (a) ∈ B Γ , and we have c = g(b) = g(bf (a)) ∈ im(g∗ ). Hence ker(δ 0 ) = im(g∗ ), which is what we wanted to prove. Exactness at H 1 (Γ, A): We need to prove that im(δ 0 ) = ker(f∗ ). Let c ∈ C G and let b ∈ B satisfying c = g(b). Then by definition of f∗ and δ 0 (c), f∗ (δ 0 (c)) is the class of the 1-cocycle Γ −→ B σ −→ b−1 σ·b, which is cohomologous to the trivial cocycle. Hence im(δ 0 ) ⊂ ker(f∗ ). Now if [α] ∈ H 1 (Γ, A) satisfies f∗ ([α]) = 1, then f (ασ ) = b−1 σ · b for some b in B. Therefore, we have σ·g(b) = g(σ·b) = g(bf (ασ )) = g(b) for all σ ∈ Γ. Hence c = g(b) lies in C Γ . Thus b ∈ B is a preimage of c ∈ C Γ under g and [α] = δ 0 (c) by definition of δ 0 . This concludes the proof.


49

Before continuing, we need to define an action of B Γ on C Γ . Let β ∈ B Γ and c ∈ C Γ . Let b ∈ B be a preimage of c under g, and set β ·c = g(βb) ∈ C. Let us check that it does not depend on the choice of b. If b ∈ B is another preimage of c under g, then b = bf (a) for some a ∈ A, hence g(βb ) = g(βbf (a)) = g(βb) by assumption on g. Hence β · c does not depend the choice of b. We now show that β ·c ∈ C Γ . For σ ∈ Γ, we have σ·(β ·c) = σ·g(βb) = g(σ·(βb)) = g((σ·β)(σ·b)). Since β ∈ B Γ , we get σ ·(β ·c) = g(β(σ ·b)) for all σ ∈ Γ. Now g(σ ·b) = σ · g(b) = σ · c = c since c ∈ C Γ , so σ · b is also a preimage of c. Since β ·c does not depend on the choice of a preimage of c, we conclude that σ·(β ·c) = β ·c for all σ ∈ Γ, so β ·c ∈ C Γ . It is then clear that the map BΓ × C Γ (β, c)

−→ C Γ −→ β ·c

gives rise to an action of B Γ on C Γ . We will denote by C Γ /B Γ of the group B Γ in C Γ . Notice that this is a pointed set, whose base point is the orbit of 1. The next result identifies this orbit set. Corollary II.4.5. There is a natural bijection of pointed sets between the orbit set C Γ /B Γ and ker(H 1 (Γ, A) −→ H 1 (Γ, B)). More precisely, the bijection sends the orbit of c ∈ C Γ onto δ 0 (c). Proof. By Proposition II.4.4, we have ker(H 1 (Γ, A) −→ H 1 (Γ, B)) = im(δ 0 ). Hence we have to construct a bijection between C Γ /B Γ and im(δ 0 ). Let c, c ∈ C Γ lying in the same orbit, that is c = β·c for some β ∈ B Γ . Then c = g(βb), for some preimage b ∈ B of c, and βb is a preimage of c . Since we have (βb)−1 σ·(βb) = b−1 β −1 (σ·β)(σ·b) = b−1 σ·b, it turns out that δ 0 (c ) = δ 0 (c). Therefore, the map ϕ:

C Γ /B Γ −→ im(δ 0 ) B Γ ·c −→ δ 0 (c)

is a well-defined surjective map. It remains to prove its injectivity. Let

50


c, c ∈ C Γ such that δ 0 (c ) = δ 0 (c), and let α and α be the cocycles representing δ 0 (c) and δ 0 (c ) respectively. By assumption, there exists a ∈ A such that ασ = a ασ σ·a−1 for all σ ∈ Γ. If b (resp. b ) is a preimage of c (resp. c ) in B, applying f to this last equality implies that b−1 σ·b = f (a)b−1 (σ·b)(σ·f (a))−1 . It easily turns out that β = b f (a)b−1 ∈ B Γ . Hence c = g(b ) = g(b f (a)) = g(βb) = β · c. Therefore, c and c lie in the same orbit, showing that ϕ is injective. Finally, the orbit of 1 is mapped onto [1]. This concludes the proof. Proposition II.4.6. Let A, B, C be Γ-sets, and let A , B , C be Γ -sets. Assume that we have a commutative diagram with exact rows 1

/A

1

/ A

/B

f

h1

/C

g

h2

/ B

f

/1

h3

/ C

g

/1

satisfying the conditions explained at the beginning of the section. Let us denote by δ 0 and δ 0 the respective connecting maps. If ϕ : Γ −→ Γ is compatible with h1 , h2 and h3 , the diagram CΓ

C

δ0

h3∗ Γ

/ H 1 (Γ, A) h1∗

δ

0

/ H 1 (Γ , A )

is commutative. Proof. Let c ∈ C Γ , and let b ∈ B be any preimage of c under g. The cohomology class δ 0 (c) is represented by the cocycle α defined by the relations f (ασ ) = b−1 σ·b for all σ ∈ Γ. Therefore, h1∗ (δ 0 (c)) is represented by the cocycle β:

Γ −→ B σ −→ h1 (αϕ(σ ) ).

By commutativity of the diagram, we have f (βσ ) = f (h1 (αϕ(σ ) )) = h2 (f (αϕ(σ ) )) = h2 (b−1 ϕ(σ )·b),


51

for all σ ∈ Γ . Since ϕ and h2 are compatible (so h2 is in particular a group morphism), we get f (βσ ) = h2 (b)−1 σ ·h2 (b) for all σ ∈ Γ . On the other hand, h3∗ (c) = h3 (c) by definition. By commutativity of the diagram, we have h3 (c) = h3 (g(b)) = g (h2 (b)).

Therefore, h2 (b) is a preimage of h3 (c) under g and thus δ 0 (h3∗ (c)) = δ 0 (h3 (c)) is represented by the cocycle β defined by f (βσ ) = h2 (b)−1 σ ·h2 (b) for all σ ∈ Γ . Hence we have f (βσ ) = f (βσ ) for all σ ∈ Γ . Since f is injective, this yields β = β. In other words, h1∗ (δ 0 (c)) and δ 0 (h3∗ (c)) are represented by the same cocycle. This concludes the proof.

II.4.2 The case of a normal subgroup We now assume that we have an exact sequence of Γ-groups 1

/A

f

/B

g

/C

/1

so A can be identified with a normal subgroup of B. Proposition II.4.7. The sequence of pointed sets 1 −→ AΓ −→ B Γ −→ C Γ −→ H 1 (Γ, A) −→ H 1 (Γ, B) −→ H 1 (Γ, C) is exact. Proof. By Proposition II.4.4, only the exactness at H 1 (Γ, B) needs a proof. If [β] = f∗ ([α]) for some [α] ∈ H 1 (Γ, A), then we have g∗ ([β]) = g∗ (f∗ ([α])) = (g ◦ f )∗ ([α]) = 1, since g ◦ f is trivial by assumption. Hence im(f∗ ) ⊂ ker(g∗ ). Conversely, let [β] ∈ H 1 (Γ, B) such that g∗ ([β]) = 1. Then there exists c ∈ C such that g(βσ ) = c−1 σ·c for all σ ∈ Γ. Write c = g(b). We then have g(βσ ) = g(b−1 σ·b), so βσ = b−1 (σ·b)f (aσ ),

52


for some aσ ∈ A. Since f (A) is normal in B, (σ·b)f (aσ )(σ·b)−1 ∈ f (A), so βσ = b−1 f (ασ ) σ·b for some ασ ∈ A, and thus bβσ σ·b−1 = f (ασ ) for all σ ∈ Γ. The fact that the map Γ −→ B σ −→ bβσ σ·b−1 is a 1-cocycle and the injectivity of f imply easily that α is a cocycle. Moreover, by construction of α, we have [β] = f∗ ([α]) ∈ im(g∗ ). This concludes the proof.

II.4.3 The case of a central subgroup Assume now that A is abelian and that f (A) is a central subgroup of B. Our next goal is to define a map of pointed sets δ 1 : H 1 (Γ, C) −→ H 2 (Γ, A). Given a 1-cocycle γ ∈ Z 1 (Γ, C), we will denote by βσ ∈ B a preimage of γσ under g. We have −1 −1 g(βσ (σ·βτ )βστ ) = γσ σ·γτ γστ = 1, −1 so βσ (σ·βτ )βστ = f (ασ,τ ) for some (unique) ασ,τ ∈ A.

Lemma II.4.8. The map α : Γ × Γ −→ A is a 2-cocycle, whose class in H 2 (Γ, A) only depends on [γ]. Proof. Let us check that α is a 2-cocycle. First of all, we have f (ασ,1 ) = βσ (σ·β1 )βσ−1 = βσ βσ−1 = 1, since β1 = 1. By injectivity of f , we get ασ,1 = 1 for all σ ∈ Γ. Similar arguments show that we have α1,τ = 1 for all τ ∈ Γ as well. Moreover, we have −1 f ((σ·ατ,ρ )ασ,τ ρ ) = f (σ·ατ,ρ )f (ασ,τ ρ ) = f (σ·ατ,ρ )βσ (σ·βτ ρ )βστ ρ.

Since f (A) is central in B, we get −1 −1 f ((σ·ατ,ρ )ασ,τ ρ ) = βσ f (σ·ατ,ρ )(σ·βτ ρ )βστ ρ = βσ (σ·f (ατ,ρ ))(σ·βτ ρ )βστ ρ .

Since σ acts by group automorphisms on B and f (ατ,ρ ) = βτ (τ ·βρ )βτ−1 ρ, this yields −1 f ((σ·ατ,ρ )ασ,τ ρ ) = βσ (σ·βτ )(στ ·βρ )(σ·βτ−1 ρ )(σ·βτ ρ )βστ ρ .


53

Thus we get −1 f ((σ·ατ,ρ )ασ,τ ρ ) = βσ (σ·βτ )(στ ·βρ )βστ ρ.

On the other hand, we have −1 f (αστ,ρ ασ,τ ) = f (ασ,τ αστ,ρ ) = f (ασ,τ )f (αστ,ρ ) = βσ (σ·βτ )(στ ·βρ )βστ ρ.

Hence f ((σ ·ατ,ρ )ασ,τ ρ ) = f (αστ,ρ ασ,τ ), and the injectivity of f shows that α is a 2-cocycle. Now take another preimage βσ of γσ under g for each σ, and denote by α the corresponding 2-cocycle. Since g(βσ ) = g(βσ ), we have βσ = βσ f (ϕσ ) for some ϕσ ∈ A. Since β1 = β1 = 1 and f is injective, we get that ϕ1 = 1. One can easily see, using the fact that f (A) is central in B, that f (ασ,τ ) = f ((σ·ϕτ )ϕ−1 στ ϕσ ασ,τ ) = f (d1 (ϕ)σ,τ ασ,τ ).

Using the injectivity of f , we get that α and α are cohomologous. Finally, let γ be a 1-cocycle cohomologous to γ, so γσ = cγσ σ ·c−1 for some c ∈ C. Writing c = g(b), one sees that βσ = bβσ σ·b−1 is a preimage of γσ . Then we have ) = f (ασ,τ = = =

−1 −1 (bβσ σ·b−1 )(σ·b σ·βτ στ ·b−1 )(στ ·b βστ b ) −1 −1 bβσ σ·βτ βστ b bf (ασ,τ )b−1 f (ασ,τ ),

since f (A) is central. Hence we get α = α in this case. This concludes the proof. If γ = 1, we may take βσ = 1 for all σ ∈ Γ. In this case, α is the trivial 2-cocycle, and thus δ 1 ([1]) = [1]. Therefore, we get a map of pointed sets δ 1 : H 1 (Γ, C) −→ H 2 (Γ, A) which sends [γ] ∈ H 1 (Γ, C) to the class [α] ∈ H 2 (Γ, A), where α is defined by the relations −1 f (ασ,τ ) = βσ (σ·βτ )βστ for all σ, τ ∈ Γ,

where βσ ∈ B is any preimage of γσ under g. Definition II.4.9. The map δ 1 : H 1 (Γ, C) −→ H 2 (Γ, A) is called the first connecting map.

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Proposition II.4.10. The sequence of pointed sets 1 → AΓ → B Γ → C Γ → H 1 (Γ, A) → H 1 (Γ, B) → H 1 (Γ, C) → H 2 (Γ, A) is exact. Proof. We keep the notation above. By Proposition II.4.7, we only have to prove the exactness at H 1 (Γ, C). Assume that [γ] = g∗ ([β]) for some [β] ∈ H 1 (Γ, B). Then there exists c ∈ C such that γσ = cg(βσ ) σ·c−1 for all σ ∈ Γ. Let b ∈ B be a preimage of c under g. We then have γσ = g(b)g(βσ ) σ·g(b)−1 = g(bβσ σ·b−1 ) for all σ ∈ Γ. Therefore, replacing β by a cohomologous cocycle if necessary, we may assume without loss of generality that γσ = g(βσ ) for all σ ∈ Γ. In this case, βσ is a preimage of γσ under g, and δ 1 ([γ]) is represented by the 2-cocycle α : Γ × Γ −→ A defined by the relations −1 f (ασ,τ ) = βσ σ·βτ βστ for all σ, τ ∈ Γ.

Since β is a 1-cocycle, we get f (ασ,τ ) = 1 for all σ, τ ∈ Γ, and the injectivity of f implies that α = 1. Therefore, δ 1 ([γ]) = 1 and we get im(g∗ ) ⊂ ker(δ 1 ). Conversely, assume that δ 1 ([γ]) = 1 is the trivial class, so that ασ,τ = ϕσ (σ ·ϕτ )ϕ−1 στ , for some continuous map ϕ : Γ −→ A satisfying ϕ1 = 1. We then have −1 f (ασ,τ ) = βσ (σ·βτ )βστ = f (ϕσ )(σ·f (ϕτ ))f (ϕστ )−1 .

It easily follows from the fact that f (A) is central that the map β :

Γ −→ B σ −→ βσ f (ϕσ )−1

is a cocycle. Moreover, the cohomology class g∗ ([β ]) is represented by the cocycle Γ −→ C σ −→ g(βσ ).


55

Now we have g(βσ ) = g(βσ )g(f (ϕσ ))−1 = g(βσ ) = γσ , since g ◦ f = 1. Therefore [γ] = g∗ ([β ]) ∈ im(g∗ ) and we are done. Proposition II.4.11. Let A, B, C be Γ-groups, and let A , B , C be Γ -groups. Assume that we have a commutative diagram with exact rows 1

/A

1

/ A

f

h1

/B

g

h2

f

/ B

/C

/1

h3

g

/ C

/1

Assume that f (A) and f (A ) are central subgroups of B and B respec tively, and let us denote by δ 1 and δ 1 the respective first connecting maps. If ϕ : Γ −→ Γ is compatible with h1 , h2 and h3 , the diagram H 1 (Γ, C)

δ1

h3∗

H 1 (Γ , C )

/ H 2 (Γ, A) h1∗

δ

1

/ H 2 (Γ , A )

is commutative. Proof. Let [γ] ∈ H 1 (Γ, C). For all σ ∈ Γ, let βσ ∈ B be a preimage of γσ under g. The cohomology class δ 1 ([γ]) is represented by the cocycle α : Γ × Γ −→ A defined by the relations −1 for all σ, τ ∈ Γ. f (ασ,τ ) = βσ (σ·βτ )βστ

Hence h1∗ (δ 1 ([γ])) is represented by the cocycle η:

Γ × Γ −→ B (σ , τ ) −→ h1 (αϕ(σ ),ϕ(τ ) ).

By commutativity of the diagram, we have −1 f (ησ ,τ ) = h2 (f (αϕ(σ ),ϕ(τ ) )) = h2 (βϕ(σ ) (ϕ(σ )·βϕ(τ ) )βϕ(σ )ϕ(τ ) ),

for all σ , τ ∈ Γ . Since ϕ is a group morphism, we get −1 f (ησ ,τ ) = h2 (βϕ(σ ) (ϕ(σ )·βϕ(τ ) )βϕ(σ τ ) ) for all σ , τ ∈ Γ .

Now since ϕ and h2 are compatible (so h2 is in particular a group morphism), we get −1 f (ησ ,τ ) = h2 (βϕ(σ ) )(σ ·h2 (βϕ(τ ) ))h2 (βϕ(σ τ ) ) for all σ , τ ∈ Γ .

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On the other hand, h3∗ ([γ]) is represented by the cocycle Γ −→ A σ −→ h3 (γϕ(σ ) ). By commutativity of the diagram, we have h3 (γϕ(σ ) ) = h3 (g(βϕ(σ ) )) = g (h2 (βϕ(σ ) )). Therefore, h2 (βϕ(σ ) ) is a preimage of h3 (γϕ(σ ) ) under g for all σ ∈ Γ , and thus the cohomology class δ 1 (h3∗ ([γ])) is represented by the cocycle η :

Γ × Γ −→ B (σ , τ ) −→ ησ ,τ ,

where ησ ,τ satisfies −1 f (ησ ,τ ) = h2 (βϕ(σ ) )(σ ·h2 (βϕ(τ ) ))h2 (βϕ(σ τ ) ) for all σ , τ ∈ Γ .

Therefore, f (ησ ,τ ) = f (ησ ,τ ) for all σ , τ ∈ Γ .

By injectivity of f , we get η = η. Hence h1∗ (δ 1 ([γ])) and δ 1 (h3∗ ([γ])) are represented by the same cocycle. This concludes the proof. §II.5 Twisting Let f : A −→ B be a morphism of Γ-groups. Even if the map of pointed sets f∗ : H 1 (Γ, A) −→ H 1 (Γ, B) may have trivial kernel in some cases, it does not mean that it will be injective. To study injectivity of f∗ , the main idea is to identify the fiber f∗−1 ([β]) of an element [β] ∈ H 1 (Γ, B) under f∗ to the kernel of some appropriate induced map in cohomology. For this, we need the method of twisting. Lemma II.5.1. Let A be a Γ-group, and let α ∈ Z 1 (Γ, A) be a cocycle. Then the map Γ × A −→ A (σ, a) −→ σ∗a = ασ σ·a ασ−1 endows A with the structure of a Γ-group. Proof. We first prove that ∗ is indeed an action of Γ on A. Let a ∈ A.

II.5 Twisting

57

Since α1 = 1, it is clear that 1 ∈ Γ acts trivially on a. Now if σ, τ ∈ Γ, we have −1 στ ∗a = αστ στ ·a αστ = ασ σ·ατ στ ·a σ·ατ−1 ασ−1 = ασ (σ·(ατ τ ·aατ−1 )ασ−1 = ασ σ·(τ ∗a) ασ−1 = σ∗(τ ∗a). Moreover, the action ∗ is an action by group automorphisms. Indeed, if a, a ∈ A and σ ∈ Γ, we have σ∗(aa ) = = = =

ασ σ·aa ασ−1 ασ (σ·a)(σ·a ) ασ−1 ασ σ·a ασ−1 ασ (σ·a ) ασ−1 (σ∗a)(σ∗a ).

It remains to prove that this action is continuous. For, let σ ∈ Γ such that σ ∗ a = a. We need to prove that the stabilizer of a with respect to the action ∗ contains an open neighbourhood of σ. Let U = StabΓ (a) ∩ α−1 ({1}), where StabΓ (a) is the stabilizer of a under the untwisted action of Γ. Then U is an open subgroup of Γ since α is continuous and A is a Γ-group. For all τ ∈ U , we then have στ · a = σ·(τ ·a) = σ·a, and αστ = ασ σ·ατ = ασ σ·1 = ασ . Putting things together, for all τ ∈ U , we have στ ∗a = σ∗a = a. Hence the stabilizer of a under the action ∗ contains σU . This concludes the proof. Definition II.5.2. The Γ-group A described in the previous lemma will be denoted by Aα in the sequel. We will say that Aα is obtained by twisting A by the cocycle α. We then have the following proposition: Proposition II.5.3. Let A be a Γ-group and α ∈ Z 1 (Γ, A). Then the map θα :

H 1 (Γ, Aα ) −→ H 1 (Γ, A) [γ] −→ [γα]

58


is a well-defined bijection which maps the class of the trivial cocycle of H 1 (Γ, Aα ) onto [α]. Proof. Let γ ∈ Z 1 (Γ, Aα ). We first have to check that γα ∈ Z 1 (Γ, A). For all σ, τ ∈ Γ, we have γστ = γσ σ∗γτ , so by definition of the action of Γ on Aα , we have γστ = γσ ασ (σ·γτ )ασ−1 . Thus γστ αστ = γσ ασ (σ·γτ )(σ·ατ ) = γσ ασ σ·(γτ ατ ), and therefore γα is a cocycle with values in A. Now assume that γ ∈ Z 1 (Γ, Aα ) is cohomologous to γ, so that γσ = aγσ σ∗a−1 for some a ∈ Aα . We then have γσ = aγσ ασ (σ · a−1 )ασ−1 , and so γσ ασ = aγσ ασ σ · a−1 . Therefore, γ α and γα are cohomologous, proving that θα is well-defined. To check that θα is a bijection, observe that α−1 is a 1-cocycle with values in Aα . Indeed, for all σ, τ ∈ Γ, we have −1 αστ

= = = =

(ασ σ·ατ )−1 σ·ατ−1 ασ−1 ασ−1 ασ σ·ατ−1 ασ−1 ασ−1 σ∗ατ−1 .

We may then twist Aα by α−1 . By definition, we have (Aα )α−1 = A, so we get a map θα−1 :

H 1 (Γ, A) −→ H 1 (Γ, Aα ) [ξ] −→ [ξα−1 ]

.

It is then easy to check that θα and θα−1 are mutually inverse. Remark II.5.4. If A is commutative, then Aα = A as Γ-modules and θα is just the translation by [α]. We now describe the functorial properties of twisting. First, assume that f : A −→ B is a morphism of Γ-groups, and let β = f∗ (α) ∈ Z 1 (Γ, B). Then the map f , considered as a map fα : Aα −→ Bβ , is a morphism of Γ-groups and the diagram H 1 (Γ, Aα )

θα

/ H 1 (Γ, A)

θβ

/ H 1 (Γ, B)

(fα )∗

H 1 (Γ, Bβ )

f∗

is commutative. In particular, we get immediatly that θα induces a bijection between ker((fα )∗ ) and the fiber f∗−1 ([β]). Hence we get the following lemma:

II.5 Twisting

59

Lemma II.5.5. Let f : A −→ B be a map of Γ-groups. The induced map f∗ : H 1 (Γ, A) −→ H 1 (Γ, B) is injective if and only if (fα )∗ has trivial kernel for every α ∈ Z 1 (Γ, A). Before continuing, we would like to generalize the definition of a twisted Γ-group. Let A be a Γ-group. We let Γ act on Aut(A) as follows: if f ∈ Aut(A) and σ ∈ Γ, we set (σ·f )(a) = σ·(f (σ −1 ·a)) for all a ∈ A. This is an action by group automorphisms. We will assume in the sequel that this action is continuous. Let us point out that this is not necessarily the case (see the exercises for a counterexample). Let ξ ∈ Z 1 (Γ, Aut(A)). As above, we may define a new continuous action ∗ of Γ on A by σ ∗ a = ξσ (σ·a) for all σ ∈ Γ, a ∈ A. This new Γ-group is denoted by Aξ . Now if α ∈ Z 1 (Γ, A), we denote by α ∈ Z 1 (Γ, Aut(A)) the image of α under the map Int : A −→ Aut(A). Hence the Γ-group Aα is nothing but the Γ-group Aα defined above. However, we will keep the notation Aα in the sequel for this particular twisted Γ-group. Now assume that we have an exact sequence of Γ-groups 1

/A

f

/B

g

/C

/1

where A is abelian and f (A) is a central subgroup of B. Let γ ∈ Z 1 (Γ, C). If c ∈ C, let b ∈ B such that c = g(b). Then the inner automorphism Int(b) ∈ Aut(B) does not depend on the choice of b, since f (A) is a central subgroup of B. We then get a well-defined map ψ : C −→ Aut(B). Let β = ψ∗ (γ) ∈ Z 1 (Γ, Aut(B)). Finally, set [ε] = δ 1 ([γ]), where δ 1 is the connecting map δ 1 : H 1 (Γ, B) −→ H 2 (Γ, A). Proposition II.5.6. With the previous notation, the sequence of Γgroups 1

/A

/ Bβ

/ Cγ

/1

60


is exact and the diagram H 1 (Γ, Cγ )

θγ

δγ1

H 2 (Γ, A)

/ H 1 (Γ, C) δ1

μ

/ H 2 (Γ, A)

commutes, where δγ1 is the connecting map with respect to the exact sequence above, and μ is multiplication by [ε]. Proof. We let the reader check that the morphisms involved in the exact sequence respect the new actions of Γ. Let α ∈ Z 1 (Γ, Cγ ). Let xσ be a preimage of ασ in Bβ under gβ = g and let yσ ∈ B be a preimage of γσ under g. −1 By definition of ε, we have f (εσ,τ ) = yσ (σ · yτ )yστ . Since xσ yσ is a 1 1 preimage of ασ γσ for all σ ∈ Γ, δ (θγ (α)) = δ ([αγ]) is represented by the cocycle η : Γ × Γ −→ A satisfying −1 −1 xστ , f (ησ,τ ) = xσ yσ (σ·(xτ yτ ))yστ

for all σ, τ ∈ Γ. Hence we get f (ησ,τ ) = xσ yσ (σ·xτ )yσ−1 f (εσ,τ )x−1 στ . By definition of β and Bβ , we have σ∗a = yσ σ·a yσ−1 , and therefore, −1 f (ησ,τ ) = xσ (σ∗xτ )x−1 στ f (εσ,τ ) = xσ (σ∗xτ )f (εσ,τ )xστ .

Now δγ1 ([α]) is represented by the cocycle η : Γ × Γ −→ A satisfying ) = xσ (σ∗xτ )x−1 f (ησ,τ στ for all σ, τ ∈ Γ.

Therefore, we get )f (εσ,τ ) = f (ησ,τ εσ,τ ), f (ησ,τ ) = f (ησ,τ

for all σ, τ ∈ Γ. By injectivity of f , this means that η = η ε. In other words, δ 1 (θγ ([α])) = μ(δγ1 ([α])). This completes the proof of the proposition.

II.6 Cup-products

61

§II.6 Cup-products In this section, we introduce briefly the cup-product, which is a useful tool to construct higher cohomology classes. Let A and B be two Γ-modules. Since A and B are abelian groups, that is Z-modules, we may consider A ⊗Z B. Then Γ acts continuously on A ⊗Z B as follows: σ·(a ⊗ b) = (σ·a) ⊗ (σ·b) for all σ ∈ Γ, a ∈ A, b ∈ B. This endows A ⊗Z B with a structure of a Γ-module. We will keep this notation until the end of the section. Proposition II.6.1. Let p, q ≥ 1 be two integers. Let α ∈ Z p (Γ, A) and β ∈ Z q (Γ, B) be two cocycles. The map α ∪ β:

Γp+q −→ A ⊗Z B (σ1 , . . . , σp+q ) −→ ασ1 ,...,σp ⊗ σ1 · · · σp ·βσp+1 ,...,σp+q

is a (p + q)-cocycle, whose cohomology class only depends on the cohomology classes [α] and [β], and the map ∪:

H p (Γ, A) × H q (Γ, B) −→ H p+q (Γ, A ⊗Z B) ([α], [β]) −→ [α ∪ β]

is Z-bilinear. Moreover, after identifying canonically the Z-modules B ⊗Z A and A ⊗Z B, we have [α] ∪ [β] = (−1)pq [β] ∪ [α] for all [α] ∈ H p (Γ, A), [β] ∈ H q (Γ, B). Proof. The Γ-modules A and B will be denoted additively. We will only sketch a proof of the first part, and refer to [11] or [26] for a detailed proof. Tedious but straightforward computations show that we have the formula dp+q (α ∪ β) = dp (α) ∪ β + (−1)p α ∪ dq (β), for all continuous maps α : Γp −→ A and β : Γq −→ B. In particular, if α and β are cocycles, so is α ∪ β. To prove that the cohomology class of α ∪ β only depends on [α] and [β], we have to prove that if α or β is a coboundary, so is α ∪ β. Assume for example that α is a coboundary, so α = dp−1 (γ) for some continuous map γ : Γp−1 −→ A. Then we have dp−1+q (γ ∪ β) = α ∪ β + (−1)p−1 γ ∪ dq (β) = α ∪ β,

62


since β is a cocycle. Hence α ∪ β lies in the image of dp+q−1 , and is therefore a coboundary. The other case is proved in a similar way.

Remark II.6.2. The last part of the proposition is a property which is difficult to establish at the level of cocycles, and its proof needs shifting techniques in cohomology (see [26] for example for a detailed proof). However, we will use it only in the case where Γ acts trivially on A and B and 2A = 2B = 0. In this particular case, it is immediate. Definition II.6.3. The map ∪ is called the cup-product. Let C be a third Γ-module and let θ : A × B −→ C be a Z-bilinear map satisfying θ(σ·a, σ·b) = σ·θ(a, b) for all σ ∈ Γ, a ∈ A, b ∈ B. Such a map will be called a bilinear map of Γ-modules. It induces a map of Γ-modules A ⊗Z B −→ C, still denoted by θ. We therefore get an induced map θ∗ : H n (Γ, A ⊗Z B) −→ H n (Γ, C). Definition II.6.4. The map ∪θ : H p (Γ, A) × H q (Γ, B) −→ H p+q (Γ, C) defined by [α] ∪θ [β] = θ∗ ([α] ∪ [β]) = [θ ◦ (α ∪ β)] is called the cup-product relative to θ. Remark II.6.5. It follows from the properties of the cup-product that ∪θ is also Z-bilinear and satisfies [α] ∪θ [β] = (−1)pq [β] ∪θ [α] for all [α] ∈ H p (Γ, A), [β] ∈ H q (Γ, B). We end this chapter by studying the functorial properties of the cupproduct. Proposition II.6.6. Let Γ and Γ be two profinite groups, and let θ : A1 × A2 −→ A3 and θ : A1 × A2 −→ A3 be two bilinear maps of Γ-modules and Γ -modules respectively. Assume that ϕ : Γ −→ Γ is compatible with fi : Ai −→ Ai for i = 1, 2, 3, and that the diagram

II.6 Cup-products / A3

θ

A1 × A2

f3

(f1 ,f2 )

A1 × A2

63

θ

/ A3

is commutative. Then for all p, q ≥ 0, the diagram ∪θ

H p (Γ, A1 ) × H q (Γ, A2 ) ((f1 )∗,(f2 )∗ )

/ H p+q (Γ, A3 ) (f3 )∗

H p (Γ , A1 ) × H q (Γ , A2 )

∪θ

/ H p+q (Γ , A ) 3

is commutative. In other words, for all [α] ∈ H p (Γ, A1 ) and all [β] ∈ H q (Γ, A2 ), we have (f3 )∗ ([α] ∪θ [β]) = (f1 )∗ ([α]) ∪θ (f2 )∗ ([β]). Proof. Notice first that the diagram A1 ⊗Z A2

θ

f1 ⊗f2

A1 ⊗Z A2

/ A3 f3

θ

/ A3

is commutative. Let α ∈ Z p (Γ, A1 ) and let β ∈ Z q (Γ, A2 ). Then [α]∪θ [β] is represented by the cocycle Γp+q −→ A1 ⊗Z A2 (σ1 , . . . , σp+q ) −→ θ(ασ1 ,...,σp ⊗ σ1 · · · σp ·βσp+1 ,...,σp+q ), and (f3 )∗ ([α] ∪θ [β]) is thus represented by the cocycle ξ : Γ defined for all σ1 , . . . , σp+q ∈ Γ by

p+q

−→ A3

ξσ1 ,...,σp+q = f3 (θ(αϕ(σ1 ),...,ϕ(σp ) ⊗ ϕ(σ1 ) · · · ϕ(σp )·βϕ(σp+1 ),...,ϕ(σp+q ) )).

On the other hand, (f1 )∗ ([α]) ∪θ (f2 )∗ ([β]) is represented by the cocycle p+q −→ A3 defined for all σ1 , . . . , σp+q ∈ Γ by ξ : Γ = θ (f1 (αϕ(σ1 ),...,ϕ(σp ) ) ⊗ σ1 · · · σp ·f2 (βϕ(σp+1 ξσ 1 ,...,σp+q ),...,ϕ(σp+q ) )).

64


To prove the proposition, it is enough to check that ξ = ξ. ∈ Γ . Since ϕ is compatible with f2 , we have σ1 , . . . , σp+q

Let

ξσ 1 ,...,σp+q = θ (f1 (αϕ(σ1 ),...,ϕ(σp ) ) ⊗ f2 (ϕ(σ1 · · · σp )·βϕ(σp+1 ),...,ϕ(σp+q ) )).

Using the fact that ϕ is a group morphism and the commutativity of the diagram above, we get the desired equality. Example II.6.7. Let Γ and Γ be two profinite groups, acting trivially on three abelian groups A, B, C. Let θ : A × B −→ C be a bilinear map, and let ϕ : Γ −→ Γ be a morphism of profinite groups. If [α] ∈ H n (Γ, A), we will denote by [α]Γ ∈ H n (Γ , A) its image under the map H n (Γ, A) −→ H n (Γ , A) induced by ϕ and IdA (and similarly for the other modules). For all [α] ∈ H p (Γ, A), [β] ∈ H q (Γ, B), we then have ([α] ∪θ [β])Γ = [α]Γ ∪θ [β]Γ ∈ H p+q (Γ , C). Notes Our treatment of cohomology of profinite groups follows [58], which is the standard reference on this topic, and [30], from which we borrowed notation. The reader will find more results on twisting in [58], as well as an account on cohomology of groups in [57] or [26].

Exercises 1. Let Γ be a profinite group, and let A be a discrete set on which Γ acts on the left. Show that Γ acts continuously on A if and only if the map Γ × A −→ A (σ, a) −→ σ·a is continuous. 2. Let Γ be a profinite group, let n ≥ 0 be an integer, and let A, A , B be three Γ-groups (resp. three Γ-modules if n ≥ 2).

Exercises (a)

65

Show that the action of Γ on A × B defined by σ · (a, b) = (σ · a, σ · b) for all σ ∈ Γ, a ∈ A, b ∈ B endows A×B with the structure of a Γ-group, and that we have a bijection of pointed sets (resp. a group isomorphism if n ≥ 2) H n (Γ, A × B) H n (Γ, A) × H n (Γ, B).

(b)

Show that if the two Γ-groups A and A are isomorphic, then we have a bijection of pointed sets (resp. a group isomorphism if n ≥ 2) H n (Γ, A) H n (Γ, A ).

3. Let α ∈ Z 1 (Γ, A) and let β ∈ Z 1 (Γ, B) be the image of α under the map induced by the inclusion A ⊂ B. Show that there is a natural bijection between the fiber of H 1 (Γ, A) −→ H 1 (Γ, B) over [β] and the orbit set of the group BβΓ in (Bβ /Aα )Γ . 4. Let E be the additive group of complex sequences u = (un )n∈Z with finite support, considered as a topological discrete group, and let p be a prime integer. For x ∈ Zp and u ∈ E, denote by x·u the element of E defined by if n ≤ 0 un (x·u)n = 2iπx e pn un if n ≥ 1. 2iπxn

2iπx

Here e pn has to be understood as e pn , where xn ∈ Z is any representative of the class of x modulo pn+1 Zp . (a)

Show that the map Zp × E −→ E u −→ x·u endows E with a structure of a Zp -module.

(b)

Let f ∈ Aut(E) be the automorphism of E defined by f:

E −→ E (un )n∈Z −→ (un−1 )n∈Z .

Show that the stabilizer of f for the action of Zp induced by its action on E is {0}. (c)

Deduce that the action of Zp on Aut(E) is not continuous.

66

Cohomology of profinite groups 5. Let B a Γ-group, let A be a Γ-subgroup and let C = B/A. Assume that A is normal in B. In particular, C is a Γ-group. (a)

If c = bA ∈ C Γ , and α ∈ Z 1 (Γ, A), show that βσ = bασ σ · b−1 ∈ A for all σ ∈ Γ, and that the map β : Γ −→ A is a 1-cocycle, whose cohomology class only depends on c and [α].

(b)

Show that the map C Γ × H 1 (Γ, A) −→ H 1 (Γ, A) (c, [α]) −→ [β] endows H 1 (Γ, A) with a left action of C Γ , and that there is a natural bijection between ker(H 1 (Γ, B) −→ H 1 (Γ, C)) and the orbit set of the group C Γ in H 1 (Γ, A).

(c)

Let β ∈ Z 1 (Γ, B). Let γ ∈ Z 1 (Γ, C) be the image of β under the map induced by the canonical projection B −→ C, and let α ∈ Z 1 (Γ, Aut(A)) be the image of β under the map induced by the conjugation map B −→ Aut(A). Show that there is a natural bijection between the fiber of H 1 (Γ, B) −→ H 1 (Γ, C) over [γ] and the orbit set of the group CγΓ in H 1 (Γ, Aα ).

6. Let B a Γ-group, let A be a Γ-subgroup and let C = B/A. Assume that A is central in B. (a)

Show that H 1 (Γ, A) acts naturally on H 1 (Γ, B) by [α] · [β] = [αβ].

(b)

Show that there is a natural bijection between the kernel of the connecting map δ 1 : H 1 (Γ, C) −→ H 2 (Γ, A) and the orbit set of the group H 1 (Γ, A) in H 1 (Γ, B).

(c)

Let γ ∈ Z 1 (Γ, C). The conjugation map C −→ Aut(B) induces a 1-cocycle β ∈ Z 1 (Γ, B). Let [ε] = δ 1 ([γ]). Show that there is a natural bijection between the fiber of δ 1 : H 1 (Γ, C) −→ H 2 (Γ, A) over [ε] and the orbit set of the group H 1 (Γ, A) in H 1 (Γ, Bβ ).

Exercises

67

7. Let A be a Γ-group. A principal homogeneous space over A is a non-empty Γ-set P endowed with a simply transitive right action ∗ of A which is compatible with the left action of Γ, that is σ·(x∗a) = (σ·x)∗(σ·a), for all σ ∈ Γ, a ∈ A, x ∈ P. A morphism of principal homogeneous spaces is a map which is A-equivariant and Γ-equivariant. We denote by Tors(Γ, A) the set of isomorphism classes of principal homogeneous spaces over A. (a)

Check that a morphism of principal homogeneous spaces is an isomorphism.

(b)

Let α ∈ Z 1 (Γ, A) be a 1-cocycle, and let Pα be the set A, endowed with the following actions of Γ and A: Γ × Pα −→ Pα (σ, x) −→ ασ (σ·x) Pα × A −→ Pα (x, a) −→ xa. Show that Pα is a principal homogeneous space, and that for all α, α ∈ Z 1 (Γ, A), we have α ∼ α ⇒ Pα Pα .

(c)

Let P be a principal homogeneous space, and let x ∈ P . For every σ ∈ Γ, justify the existence of a unique ασ ∈ A such that σ·x = x ∗ ασ . Show that the map α : Γ −→ A is a 1-cocycle, whose cohomology class does not depend on the choice of x.

(d)

Conclude that H 1 (Γ, A) Tors(Γ, A), that is H 1 (Γ, A) classifies principal homogeneous spaces up to isomorphism.

8. Let f : A −→ B be a morphism of Γ-groups. (a)

If P is a principal homogeneous space over A, consider P × B endowed with the diagonal action of Γ. Show that the groups A and B acts on P by (P × B) × A −→ P × B ((x, b), a) −→ (x∗a, f (a−1 )b)

68

Cohomology of profinite groups (P × B) × B −→ P × B ((x, b), b ) −→ (x, bb ), and that these two actions commute. (b)

Show that the induced action of B on the set of A-orbits f∗ (P ) = (P × B)/A endows f∗ (P ) with a structure of a principal homogeneous space over B and that the isomorphism class of f∗ (P ) only depends on the isomorphism class of P . We then get a map f∗ : Tors(Γ, A) −→ Tors(Γ, B).

(c)

Show that diagram H 1 (Γ, A)

/ H 1 (Γ, B)

Tors(Γ, A)

/ Tors(Γ, B)

is commutative.

III Galois cohomology

In this chapter, we study the specific properties of the cohomology sets associated to the Galois group of a Galois field extension. Our main goal is to express the obstruction of a Galois descent problem associated to algebraic objects in terms of Galois cohomology. Of course, this kind of question only makes sense if the algebraic objects may be ‘defined over an arbitrary field’, and if we have a notion of ‘scalar extension’ of our objects. For example, to each field extension K/k of a field k, we can consider the set Mn (K) of matrices with coefficients in K, and if ι : K −→ L is a morphism of extensions of k, one can associate a map Mn (K) −→ Mn (L) (mij ) −→ (ι(mij )). Another example may be obtained by considering the set Algn (K) of K-algebras of dimension n over K. In this case, the map Algn (K) −→ Algn (L) associated to a morphism K −→ L is given by the tensor product ⊗K L. In both cases, scalar extension maps satisfy some natural properties: the scalar extension map induced by the trivial extension is the identity map, and extending scalars from K to L, then from L to M is the same as extending scalars from K to M . We therefore start this chapter by formalizing this situation and introducing the concept of a functor between two categories. §III.7 Warm-up III.7.1 Digression: categories and functors In this section, we recall some basic facts on categories and covariant functors. The reader is referred to [35] for details. Roughly speaking, the notion of a category is here to palliate to the (mostly inconvenient) fact 69

70

Galois cohomology

that there is no ‘set of all sets’. More seriously, when studying various objects such as sets, groups, rings, algebras, varieties, a common setting appears constantly. We define special kinds of maps between them, namely morphisms, sharing some similar properties: the composition of two morphisms (when defined) is a morphism, the identity map is a morphism, the composition of morphisms is associative. The notion of a category allows us to axiomatize this common setting. As we will see, the axioms of a category are essentially related to the morphisms between the objects. Definition III.7.1. A category C consists of: (1)

A collection of objects Ob(C).

(2)

For two objects A, B ∈ Ob(C) a set MorC (A, B) (possibly empty), called the set of morphisms f : A −→ B from A to B.

(3)

For three objects A, B, C ∈ Ob(C) a law of composition MorC (B, C) × MorC (A, B) −→ MorC (A, C) (g, f ) −→ g ◦ f satisfying the following axioms: (a)

Two sets MorC (A, B) and MorC (A , B ) are disjoint unless A = A and B = B , in which case they are equal.

(b)

For all A ∈ Ob(C), there is a morphism IdA ∈ MorC (A, A), called the identity morphism, acting as left and right identity on the elements of MorC (A, B) and MorC (B, A) respectively.

(c)

The law of composition is associative (when defined).

We may then define the notion of isomorphism, endomorphism and automorphism in an obvious way. Remark III.7.2. If C is a category, then for all A ∈ Ob(C), the identity morphism IdA is unique. Indeed, assume that IdA is another identity morphism. Then we have IdA = IdA ◦ IdA = IdA , by definition of the identity morphisms. Examples III.7.3. The following categories will be used frequently in the sequel:

III.7 Warm-up

71

(1)

The category Sets is the category whose objects are sets, and whose morphisms are set-theoretic maps.

(2)

The category of pointed sets Sets∗ is the category whose objects are pointed sets and whose morphisms are maps of pointed sets.

(3)

The category of groups Grps is the category whose objects are groups and whose morphisms are group morphisms. The category of abelian groups AbGrps is the category whose objects are abelian groups and whose morphisms are group morphisms.

(4)

Let k be a field. The category Algk is the category whose objects are (associative unital) commutative k-algebras and whose morphisms are k-algebra morphisms.

(5)

Let k be a field. The category Ck is the category whose objects are the field extensions (K, ι) of k and whose morphisms are morphisms of extensions of k.

(6)

If Γ is a profinite group, then Γ-sets, Γ-groups and Γ-modules, together with the appropriate morphisms, form three categories that we denote by SetsΓ , GrpsΓ and ModΓ respectively.

We now introduce the notion of a subcategory. Definition III.7.4. Let C be a category. A subcategory of C is a category C such that (1)

Ob(C ) is a subcollection of Ob(C).

(2)

For every pair of objects A, B ∈ Ob(C), we have MorC (A, B) ⊂ MorC (A, B).

(3)

For all A ∈ Ob(C ), IdA ∈ MorC (A, A).

(4)

The composition law in C is the restriction of the composition law of C.

Example III.7.5. The categories Grps and Algk are subcategories of Sets. Definition III.7.6. Let C, C be two categories. A covariant functor is a rule F : C −→ C which associates to each object A ∈ Ob(C) an object F(A) in Ob(C ) and to each morphism f : A −→ B of C associates a morphism F(f ) : F(A) −→ F(B) of C such that: (1)

For all A in Ob(C), we have F(IdA ) = IdF(A) .

72 (2)

Galois cohomology If f : A −→ B and g : B −→ C are two morphisms of C, then F(g ◦ f ) = F(g) ◦ F(f ).

The definition is perhaps a bit more enlightening when C = Ck and C = Sets. In this case, a functor may be viewed (roughly speaking) as a family of sets associated to field extensions, together with ‘scalar extension maps’. Examples III.7.7. Here are some classical examples of functors: (1)

The forgetful functor F : Grps −→ Sets, which maps a group G to itself (viewed as a set), and a morphism of groups to itself (viewed as a map between sets).

(2)

Let k be a field. For any field extension L/k, set F(L/k) = Mn (L). If ϕ : L −→ L is a morphism of field extensions of k, we define F(ϕ) to be the map F(ϕ) :

Mn (L) −→ Mn (L ) (mij ) −→ (ϕ(mij )).

Clearly, axioms (1) and (2) are satisfied, so we obtain a functor from the category Ck to the category Sets, that we denote by Mn . (3)

If C is a subcategory of a category C, the obvious rule F : C −→ C is a functor.

(4)

Let n ≥ 1, and let k be a field. Let C = Algk and C = Grps. For any commutative k-algebra R, set F(R) = GLn (R). If ϕ : R −→ R is a morphism of k-algebras, we define a map F(ϕ) by F(ϕ) :

F(R) −→ F(R ) (mij ) −→ (ϕ(mij )).

It is easy to check that this map is a group morphism, and that axioms (1) and (2) are once again satisfied. Hence we obtain a functor GLn : Algk −→ Grps. If n = 1, it is denoted by Gm . (5)

Let C be a category and let A ∈ Ob(C) be an object of C. For every B, C ∈ Ob(C), and every f : B −→ C ∈ MorC (B, C), let hA (B) = MorC (A, B) and hA (f ) :

MorC (A, B) −→ MorC (A, C) ϕ −→ f ◦ ϕ.

We then obtain a covariant functor hA : C −→ Sets.

III.7 Warm-up (6)

73

From Example II.3.20, we see that we have covariant functors H 1 (Γ,− ) : GrpsΓ −→ Sets∗ and H n (Γ,− ) : ModΓ −→ AbGrps.

Definition III.7.8. Let C, C be two categories, and let F1 : C −→ C and F2 : C −→ C be two covariant functors. A morphism (or natural transformation) of functors Θ : F1 −→ F2 is a rule assigning to each object A ∈ Ob(C) an element ΘA ∈ MorC (F1 (A), F2 (A)), ΘA : F1 (A) −→ F2 (A), such that for every morphism f : A −→ B of C the diagram F1 (A)

ΘA

F1 (f )

F1 (B)

ΘB

/ F2 (A) F2 (f )

/ F2 (B)

is commutative. Example III.7.9. Let k be a field. For every commutative k-algebra R, let detR be the group morphism detR :

GLn (R) −→ R× M −→ det(M ).

The properties of the determinant of matrices imply immediately that the rule det : GLn −→ Gm obtained in this way is a natural transformation of functors. Clearly, the composition of two morphisms is a morphism, and this composition is associative. Moreover, for any functor F : C −→ C , there is an identity element 1F for the composition, which assigns to each A ∈ Ob(C) the morphism IdF(A) . Therefore, with this notion of morphism, the covariant functors from C −→ C form a category. In particular, we get the notion of isomorphism of functors. We now define the notion of a subfunctor of a functor. Definition III.7.10. Let C be a category, let C be a subcategory of Sets, and let F : C −→ C be a functor. A functor F : C −→ C is a subfunctor of F if the following properties hold: (1)

For all A ∈ Ob(C), we have F (A) ⊂ F(A).

74 (2)

Galois cohomology For all A, B ∈ Ob(C), and every map f ∈ MorC (A, B), the induced morphism F (f ) : F (A) −→ F (B) is the restriction of F(f ) : F(A) −→ F(B). In other words, the diagram F (A) F(A)

F (f )

F(f )

/ F (B) / F(B)

commutes. Example III.7.11. The functor of n × n of matrices of trace zero is a subfunctor of Mn . Definition III.7.12. Let F : C −→ Sets be a covariant functor. We say that F is representable if F hA for some object A ∈ Ob(C), where hA is the functor defined in Example III.7.7(5). In this case, we will say that F is represented by A. The reader may wonder if the object A above is uniquely defined up to isomorphism. The answer is affirmative, and is given by Yoneda’s Lemma. Lemma III.7.13 (Yoneda’s Lemma). Let C be a category. For every pair of objects A, B ∈ Ob(C), there is a one-to-one correspondence between the set of morphisms ϕ : B −→ A and the set of natural transformations Θ : hA −→ hB . More precisely, let ΨA,B and ΨA,B be the maps defined as follows: if ϕ : B −→ A is a morphism of C, we define ΨA,B (ϕ) : hA −→ hB by ΨA,B (ϕ)C :

MorC (A, C) −→ MorC (B, C) f −→ f ◦ ϕ

for all C ∈ Ob(C). If Θ : hA −→ hB is a natural transformation of functors, we define ΨA,B (Θ) : B −→ A to be the morphism ΘA (IdA ) ∈ hB (A) = MorC (B, A). Then ΨA,B and ΨA,B are mutually inverse. Moreover, for all A, B, C ∈ Ob(C), all ϕ ∈ MorC (B, A) and all ϕ ∈ MorC (C, B) we have: (1)

ΨA,A (IdA ) = 1hA .

(2)

ΨA,C (ϕ ◦ ϕ ) = ΨB,C (ϕ ) ◦ ΨA,B (ϕ).

III.7 Warm-up

75

In particular, for any pair of objects A, B ∈ Ob(C), we have hA hB ⇐⇒ A B. Proof. The fact that ΨA,B (ϕ) is actually a natural transformation is left as an exercise for the reader. Let us prove that ΨA,B and ΨA,B are mutually inverse. For any morphism ϕ : B −→ A, we have ΨA,B (ΨA,B (ϕ)) = ΨA,B (ϕ)A (IdA ) = IdA ◦ ϕ = ϕ, so ΨA,B ◦ ΨA,B is the identity map. Now if Θ : hA −→ hB is a natural transformation, for any object C ∈ Ob(C) and any morphism f ∈ MorC (A, C), we have ΨA,B (ΨA,B (Θ))C (f ) = f ◦ ΨA,B (Θ) = f ◦ ΘA (IdA ). Since Θ is a natural transformation, the diagram hA (A)

ΘA

hA (f )

hA (C)

ΘC

/ hB (A) hB (f )

/ hB (C)

is commutative. In particular, we have hB (f )(ΘA (IdA )) = ΘC (hA (f )(IdA )). By definition of hA (f ) and hB (f ), this reads f ◦ ΘA (IdA ) = ΘC (f ). Hence we get ΨA,B (ΨA,B (Θ))C (f ) = ΘC (f ) for all C ∈ Ob(C) and all f ∈ MorC (A, C), that is ΨA,B (ΨA,B (Θ)) = Θ. Thus ΨA,B ◦ ΨA,B is the identity map as well; this concludes the proof of the first part. Relations (1) and (2) are clear, in view of the definition of ΨA,B . It remains to prove the last statement. If ϕ : B −→ A is an isomorphism, relations (1) and (2) readily imply that ΨA,B (ϕ) : hA −→ hB is an isomorphism of functors, whose inverse is the natural transformation ΨB,A (ϕ−1 ) : hB −→ hA . Conversely, if Θ : hA −→ hB is an isomorphism

76

Galois cohomology

of functors, write Θ = ΨA,B (ϕ) for some ϕ ∈ MorC (B, A), and Θ−1 = ΨB,A (ϕ ) for some ϕ ∈ MorC (A, B). Now we have ΨB,B (IdB ) = 1hB = Θ ◦ Θ−1 = ΨA,B (ϕ) ◦ ΨB,A (ϕ ) = ΨB,B (ϕ ◦ ϕ). Since ΨB,B is a bijection, we get ϕ ◦ ϕ = IdB . Similarly, ϕ ◦ ϕ = IdA , so ϕ : B −→ A is an isomorphism. We now give a fundamental example of a representable functor: Lemma III.7.14. Let k be a field, let A = k[X1 , . . . , Xn ]/I be a finitely generated k-algebra. Then the functor V (I) : Algk −→ Sets defined by V (I)(R) = {(r1 , . . . , rn ) ∈ Rn | f (r1 , . . . , rn ) = 0 for all f ∈ I} is isomorphic to hA . Proof. Let R be a commutative unital k-algebra. For r = (r1 , . . . , rn ) ∈ V (I)(R), let evr :

k[X1 , . . . , Xn ] −→ R Xi −→ ri

be the evaluation at r. By definition of V (I), the kernel of evr contains I, so it induces a morphism ϕR (r) :

k[X1 , . . . , Xn ]/I −→ R X i −→ ri .

We then get a map ϕR : V (I)(R) −→ hk[X1 ,...,Xn ]/I (R). It is easy to check that the corresponding rule ϕ : V (I) −→ hk[X1 ,...,Xn ]/I is a natural transformation of functors. Conversely, let u : k[X1 , . . . , Xn ]/I −→ R be a k-algebra morphism. Then ψR (u) = (u(X 1 ), . . . , u(X n )) ∈ V (I)(R) since we have f (u(X 1 ), . . . , u(X n )) = u(f (X 1 , . . . , X n )) = u(f ) = u(0) = 0, for all f ∈ I. We then get a map ψR : hk[X1 ,...,Xn ]/I (R) −→ V (I)(R). Once again, it is easy to check that the rule ψ : hk[X1 ,...,Xn ]/I −→ V (I) is a natural transformation of functors. Clearly, ϕ and ψ are mutually inverse. This concludes the proof.

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77

III.7.2 Algebraic group-schemes Let K/k be a field extension, and let Ω/K be a Galois extension of K with Galois group GΩ . In order to define cohomology sets of the profinite group GΩ , we need GΩ -groups. The first step is to obtain groups on which GΩ acts by group automorphisms. We proceed to show that a natural way to achieve this is to consider Ω-points of group-valued functors. Let F : Ck −→ Sets be a covariant functor. If K/k is a field extension, we will denote F(K/k) simply by F(K). If K −→ K is a morphism of extensions of k, for every x ∈ F(K), we will denote by xK ∈ F(K ) the image of x under the map F(K) −→ F(K ) if there is no ambiguity on the choice of the map K −→ K . Notice that since F is a functor, every σ ∈ GΩ induces a map F(σ) : F(Ω) −→ F(Ω). Lemma III.7.15. The map GΩ × F(Ω) −→ F(Ω) (σ, x) −→ σ·x = F(σ)(x) gives rise to an action of GΩ on F(Ω). If Ω/K and Ω /K are two Galois extensions such that Ω ⊂ Ω , we have σ ·xΩ = (σ|Ω ·x)Ω for all x ∈ F(Ω), σ ∈ GΩ . Moreover, if F : Ck −→ Grps is a group-valued functor, the action above is an action by group automorphisms, that is σ·(xy) = (σ·x)(σ·y) for all σ ∈ GΩ , x, y ∈ F(Ω). Proof. Since F is a functor, we have F(IdΩ ) = IdF(Ω) . Therefore, IdΩ ·x = x for all x ∈ F(Ω). Now let σ, τ ∈ GΩ . Since F is a covariant functor, we have F(σ ◦ τ ) = F(σ) ◦ F(τ ), and thus (σ ◦ τ )·x = σ · (τ ·x) for all x ∈ F(Ω). This proves the first part of the lemma. Let Ω/K and Ω /K be two Galois extensions such that Ω ⊂ Ω , and let

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Galois cohomology

F(ι) : F(Ω) −→ F(Ω ) be the map induced by the inclusion ι : Ω ⊂ Ω . For all σ ∈ GΩ , the diagram σ|

Ω

Ω

/Ω

/ Ω

ι

ι

Ω

σ

is commutative. Therefore, the induced diagram F(Ω)

F(σ| ) Ω

/ F(Ω)

F(ι)

F(Ω )

F(ι)

F(σ ) / F(Ω )

commutes. In other words, we have F(ι) ◦ F(σ|Ω )(x) = F(σ ) ◦ F(ι)(x) for all x ∈ F(Ω), that is (σ|Ω ·x)Ω = σ ·xΩ for all x ∈ F(Ω). Finally, if F is a group-valued functor, F(σ) is a group morphism and the last part follows. This concludes the proof. Unfortunately, there is no reason for this action to be continuous. However this is the case for a representable functor F : Algk −→ Sets under some mild assumption, as we proceed to show now. Lemma III.7.16. Let F : Algk −→ Sets be a representable functor, and let A be a commutative k-algebra such that F hA . Then the following properties hold: (1)

For every Galois extension Ω/K, the map F(K) −→ F(Ω) is injective and induces a bijection (resp. a group isomorphism if F is a group-valued functor) F(K) F(Ω)GΩ .

(2)

Assume that A is finitely generated over k, and let Ω/K be a Galois extension. For every finite Galois extension L/K contained in Ω, denote by ιL : F(L) −→ F(Ω) the map induced by the inclusion L ⊂ Ω. Then GΩ acts continuously on F(Ω), and we have F(Ω) = ιL (F(L)). L⊂Ω

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∼

Proof. Let Ψ : F −→ hA be an isomorphism of functors. Notice first that since Ψ is a natural transformation of functors, then for every morphism of k-algebras ϕ : R −→ S, we have a commutative diagram F(R)

ΨR

/ Homk−alg (A, R)

ΨS

/ Homk−alg (A, S)

F(ϕ)

F(S)

where the second vertical map is left composition by ϕ. In other words, if a ∈ F(R) and f = ΨR (a) ∈ Homk−alg (A, R) is the corresponding morphism, then F(ϕ)(a) corresponds to the morphism ϕ ◦ f ∈ Homk−alg (A, S). We may now start the proof of the lemma. Let Ω/K be a Galois extension and let ε : K −→ Ω be the corresponding injective morphism. The previous considerations show that the map F(ε) : F(K) −→ F(Ω) identifies to the map Homk−alg (A, K) −→ Homk−alg (A, Ω) f −→ ε ◦ f. The injectivity of F(ε) then comes from the injectivity of ε. We now prove that the image of F(ε) is F(Ω)GΩ . For, let σ ∈ GΩ , let a ∈ F(Ω) and let f ∈ Homk−alg (A, Ω) be the corresponding morphism, that is f = ΨΩ (a). If a lies in the image of F(ε), then there exists f ∈ Homk−alg (A, K) such that f = ε ◦ f . Now the action of σ on F(Ω) corresponds to left composition by σ, so σ ·a corresponds to the morphism σ ◦ f = (σ ◦ ε) ◦ f = ε ◦ f = f, the second equality coming from the K-linearity of σ. Hence σ · a = a for all σ ∈ GΩ . Conversely, assume that a ∈ F(Ω) satisfies σ·a = a for all σ ∈ GΩ . We then have σ ◦ f = f for all σ ∈ GΩ . It implies that for all x ∈ A, f (x) is fixed by GΩ , that is f (x) ∈ K since

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Galois cohomology

Ω/K is a Galois extension. Consider now the k-algebra morphism f :

A −→ K x −→ f (x),

and let a be the corresponding element of F(K). Since f = ε ◦ f by definition, a is the image of a under the map F(ε). This proves (1). Let us prove (2). Let a ∈ F(Ω), let f ∈ Homk−alg (A, Ω) be the corresponding morphism and let σ ∈ GΩ . The equality σ · a = a is equivalent to σ ◦ f = f . Let α1 , . . . , αn be a set of generators of A, and let K = K(f (α1 ), . . . , f (αn )). This is a finite extension of K contained in Ω. Now σ ◦ f = f if and only if σ restricts to the identity on K , that is σ ∈ Gal(Ω/K ). Hence the stabilizer of a is Gal(Ω/K ), which is an open subgroup of GΩ since K /K is finite (this follows from the definition of the Krull topology). Therefore, the action of GΩ on F(Ω) is continuous. Using Lemma II.3.3 and the first point, we get immediately the equality F(Ω) = ιL (F(L)). L⊂Ω

This concludes the proof. Remark III.7.17. The two previous lemmas show in particular that if G : Algk −→ Grps is represented by a finite-dimensional k-algebra, then for every Galois extension Ω/K, the group G(Ω) is a GΩ -group. Such a functor deserves a special name. Definition III.7.18. Let k be a field. A group-scheme defined over k is a covariant functor G : Algk −→ Grps. An affine group-scheme defined over k is a group-scheme G : Algk −→ Grps which is representable as a functor Algk −→ Sets. An algebraic group-scheme defined over k is an affine group-scheme G which is represented by a finitely generated k-algebra A. If furthermore Akalg is reduced, we say that G is an algebraic group defined over k. Examples III.7.19. (1)

Let V be a finite dimensional k-vector space. For any commutative unital k-algebra R, set GL(V )(R) = GL(VR ), where VR = V ⊗k R. If V = k n , we just denote it by GLn (this

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81

is consistent with the definition given in the first paragraph). We obtain an algebraic group-scheme GL(V ). Indeed, if we choose a basis of V , then we can see that GL(V ) is isomorphic to GLn as a group-scheme. Hence, this is enough to check that GLn is an algebraic group-scheme. Let G be the group-scheme defined by 2

G(R) = {(x11 , x12 , . . . , xnn , y) ∈ Rn

+1

| det((xij ))y = 1}

for any commutative k-algebra R. Then the natural transformation ϕ : GLn −→ G defined by ϕR :

GLn (R) −→ G(R) (mij ) −→ (m11 , m12 , . . . , mnn , det((mij ))−1 )

is an isomorphism. Now by Lemma III.7.14, the group-scheme G (and therefore GLn ) is isomorphic as a functor Algk −→ Sets to the representable functor hA , where A = k[Xij , Y ]/(det((Xij ))Y − 1). It follows that GLn is an algebraic group-scheme. The reader may show as an exercice that GLn is in fact an algebraic group. (2)

If A is a finite dimensional k-algebra, the functor GL1 (A) defined by GL1 (A)(R) = A× R is an algebraic group-scheme. To see this, for all a ∈ AR , let us denote by a the endomorphism of right R-modules given by left multiplication by a. Then a ∈ A× R if and only if a is an isomorphism. Since AR is a free R-module, this is equivalent to det(a ) ∈ R× . Now if e1 , . . . , en is a k-basis of A, let us denote by PA ∈ R[X1 , . . . , Xn ] the polynomial PA = det(e1 ⊗1 X1 + . . . + en ⊗1 Xn ). It is easy to check that PA ∈ k[X1 , . . . , Xn ]. Moreover, a = n × ei ⊗ λi ∈ A× R if and only if PA (λ1 , . . . , λn ) ∈ R . Arguing as i=1

in the previous example, we see that GL1 (A) is represented by the k-algebra k[X1 , . . . , Xn , Y ]/(PA Y − 1). In particular, if A = End(V ), we recover the result of (1).

82 (3)

Galois cohomology The additive group-scheme Ga is defined by Ga (R) = R. By Lemma III.7.14, it is represented by k[X]. Clearly, this is an algebraic group.

(4)

The multiplicative group-scheme Gm is defined by Gm (R) = R× . By (1), this is an affine group-scheme represented by k[X, Y ]/(XY − 1), and therefore an algebraic group.

(5)

The group-scheme of nth -roots of unity μn is defined by μn (R) = {r ∈ R | rn = 1}. By Lemma III.7.14, it is an affine group-scheme, represented by k[X]/(X n −1). This is an algebraic group if and only if n is prime to the characteristic of k.

(6)

Let G be an abstract finite group of order n. If R is a commutative k-algebra, we will index the coordinates of an element r of Rn by the elements of G. Set

n G(R) = r ∈ R rg = 1, rg rh = δg,h rg for all g, h ∈ G . g

We define the product of r = (rg )g∈G and r = (rg )g∈G by (r.r )g = rh rh −1 g . h∈G

One can check that G(R) is a group for this multiplication law. Notice that if R has no non-trivial idempotents, then G(R) is isomorphic to the abstract group G via the isomorphism ∼

G −→ G(R) g −→ (δg,h )h∈G . One can show that the corresponding functor is an affine group scheme, still denoted by G, called the constant group scheme (associated to) G. The reader will show that the k-algebra representing G is Map(G, k). In particular, G is an algebraic group.

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83

If A is a finite dimensional k-algebra (not necessarily commutative), we define the group-scheme Autalg (A) by Autalg (A)(R) = AutR−alg (AR ), where AR = A ⊗k R. The reader will check as an exercise that this group-scheme is affine.

If G : Ck −→ Grps is a group-scheme, the action GΩ on G(Ω) may seem a bit mysterious. We would like to make it more explicit in the case of algebraic group-schemes. We start with an example. Example III.7.20. Let G = GLn . In this case, G is represented by the finite dimensional k-algebra k[Xij , Y ]/(det(Xij )Y − 1) by Example III.7.19 (1). An invertible matrix M ∈ GLn (Ω) corresponds to the morphism f : A −→ Ω, which sends X ij to mij and Y to 1/ det(mij ). Then σ · M corresponds to the morphism f = σ ◦ f , as pointed out at the beginning of the proof of Lemma III.7.16. By definition of f , this morphism sends X ij to σ(mij ) and Y to σ(1/ det(mij )). Since we have σ(1/ det(mij )) = 1/ det(σ(mij )), it follows that f corresponds to the matrix (σ(mij )). Therefore, the action of GΩ on GLn (Ω) is just the usual action coefficient by coefficient. To describe this action in the general case, we need first to introduce the concept of a closed subgroup. Definition III.7.21. Let G and H be two affine group-schemes defined over k, represented by A and B respectively. We say that H is a closed subgroup of G if there exists a natural transformation Θ : H −→ G such that the morphism of k-algebras ϕ : A −→ B corresponding to Θ via Yoneda’s lemma is surjective. In this case Θ is called a closed embedding. Remark III.7.22. In this setting, Θ is injective, that is ΘR : H(R) −→ G(R) is injective for all R. In particular, H(R) identifies to a subgroup of G(R) for all R. Indeed, if f1 , f2 ∈ Homk−alg (B, R) satisfy f1 ◦ ϕ = f2 ◦ ϕ, then we have f1 = f2 by surjectivity of ϕ. But by Yoneda’s Lemma the map Homk−alg (B, R) −→ Homk−alg (A, R) f −→ f ◦ ϕ is ΘR , so we are done.

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Galois cohomology

This explains the term ‘subgroup’ in the previous definition. We now would like to explain in which sense H is ‘closed’ in G when G is algebraic. In this case, since A is finitely generated, we may write A = k[X1 , . . . , Xn ]/I. Since ϕ is surjective, B is isomorphic to a quotient of A, so B k[X1 , . . . , Xn ]/J for some ideal J of k[X1 , . . . , Xn ] containing I. Recall now that G V (I) and H V (J) by Lemma III.7.14, and notice that V (J) is a subfunctor of V (I). Now if we identify H(kalg ) and G(kalg ) to V (J)(kalg ) and V (I)(kalg ) respectively, then H(kalg ) is a closed subset of G(kalg ) for the Zariski topology. As a final remark, let us mention that if G and H are both algebraic group-schemes, then a natural transformation Θ : H → G is a closed embedding if and only if ΘR is injective for all R (see [30, Proposition 22.2] for example). Before going further, we need the following result, which is proved in [69]: Proposition III.7.23. Any algebraic group-scheme G defined over k is a closed subgroup of GLn , for some n ≥ 1. We are now ready to describe the action of GΩ on G(Ω) in more explicit terms. Lemma III.7.24. Let G be an algebraic group-scheme defined over k represented by B, and let Θ : G −→ GLn be a closed embedding. Let K/k be a field extension, let Ω/K be a Galois extension, and assume that GΩ acts naturally on GLn (Ω). Then for every σ ∈ GΩ and every g ∈ G(Ω), σ·g is the unique element g ∈ G(Ω) satisfying ΘΩ (g ) = σ·ΘΩ (g). Proof. Notice that there is at most one element g ∈ G(Ω) satisfying the equality above, since ΘΩ is injective by Remark III.7.22. Now if g ∈ G(Ω) and σ ∈ GΩ , the naturality of Θ shows that we have ΘΩ (G(σ)(g)) = ΘΩ (σ)(g), that is ΘΩ (σ·g) = σ·ΘΩ (g). The results follows. Remark III.7.25. The previous lemma and Example III.7.20 make the natural Galois action on an algebraic group-scheme G totally explicit.

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In particular, it shows that if G ⊂ GLn , the action of GΩ on G(Ω) is simply the restriction of the natural action of GΩ on matrices. Example III.7.26. Let us describe the action of GΩ on Mn (Ω). The map Mn −→ GL2n Θ: In M M −→ 0 In is easily seen to be a closed embedding. It follows from the previous lemma that the action of GΩ on Mn (Ω) is nothing but the action of GΩ entrywise.

III.7.3 The Galois cohomology functor We now introduce the general setting in which we are going to work, in order to get some functorial properties of cohomology of Galois groups of field extensions. Notice first that if G is an algebraic group-scheme defined over a field k, then by Remark III.7.17 the group G(Ω) is a GΩ -group for every Galois extension Ω/k. Therefore, one may consider the set H n (GΩ , G(Ω)). Moreover, for every finite Galois subextension L/k of Ω/k, G(Ω)Gal(Ω/L) identifies to G(L) by Lemma III.7.16 (1). Using this property and the fact that GΩ /Gal(Ω/L) is isomorphic to GL , Theorem II.3.33 shows that H n (GΩ , G(Ω)) is the direct limit of the sets H n (GL , G(L)), a property we were looking for in order to generalize our approach of the conjugacy problem of matrices to infinite Galois extensions. However, limiting ourselves to consider cohomology sets of algebraicgroup schemes is a bit too restrictive, since the group-schemes involved in other descent problems may not be representable (even if it will be true in most of the cases) and may not even be defined on the category Algk . The idea is then to consider group-schemes G having the properties listed in Lemma III.7.16, which are really the only ones used to obtain cohomology sets behaving well. Definition III.7.27. A group-scheme G : Ck −→ Grps is a Galois functor if for every field extension K/k and every Galois extension Ω/K, the following conditions are satisfied: (1)

The map G(K) −→ G(Ω) is injective, and induces a group iso-

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Galois cohomology morphism

(2)

G(Ω) =

G(K) G(Ω)GΩ .

ιL,Ω (G(L)), where L/K runs over the set of finite

L⊂Ω

Galois subextensions of Ω and ιL,Ω : G(L) −→ G(Ω) is the map induced by the inclusion L ⊂ Ω. Example III.7.28. An algebraic group-scheme, viewed as a functor from Ck to Grps, is a Galois functor by Lemma III.7.16. We will see other examples of Galois functors in the next paragraph. Remark III.7.29. Let G : Ck −→ Grps be a Galois functor. By Lemma II.3.3, conditions (1) and (2) imply that for every field extension K/k and every Galois extension Ω/K, the profinite group GΩ acts continuously on G(Ω) via GΩ × G(Ω) −→ G(Ω) (σ, g) −→ σ·g = G(σ)(g). Lemma III.7.15 ensures that this action is an action by group automorphisms, so G(Ω) is a GΩ -group. We may then consider the pointed set H n (GΩ , G(Ω)). Let G : Ck −→ Grps be a Galois functor. We would like now to use Theorem II.3.33 to relate the Galois cohomology of GΩ to the Galois cohomology of its finite Galois subextensions. For every (not necessarily) finite Galois subextensions L/K and L /K of Ω/K such that L ⊂ L , by Lemma III.7.15, the maps

rL,L :

GL −→ GL σ −→ σ|L

and ιL,L : G(L) −→ G(L ) are compatible. We then get a map ρL,L : H n (GL , G(L)) −→ H n (GL , G(L )). It is easy to check that we obtain this way a directed system of pointed sets. When L = Ω, we will simply denote these three maps by rL , ιL and ρL respectively. We then have the following theorem.

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Theorem III.7.30. Let G : Ck −→ Grps be a Galois functor. For every field extension K/k and every Galois extension Ω/K, we have an isomorphism of pointed sets (resp. an isomorphism of groups if G takes values in AbGrps ) n n lim −→ H (GL , G(L)) H (GΩ , G(Ω)),

L⊂Ω

where L/K runs through the finite Galois subextensions of Ω/K. If [α] ∈ H n (GL , G(L)), this isomorphism maps [α]/∼ onto ρL ([α]). Proof. We start with some preliminary remarks. Since G is a Galois functor, the map ιL induces a group isomorphism ∼

G(L) −→ G(Ω)Gal(Ω/L) , that we still denote by ιL . Moreover, we have an isomorphism ∼

θL :

GΩ /Gal(Ω/L) −→ GL σ −→ σ|L

induced by rL . The maps θL and ιL are compatible. Indeed, for every ∼ σ ∈ GΩ /Gal(Ω/L) −→ GL and g ∈ G(L) we have ιL (θL (σ)·g) = ιL (σ|L ·g) = (σ|L ·g)Ω . By Lemma III.7.15, we get ιL (θL (σ)·g) = σ·gΩ = σ·gΩ , the last equality coming from the definition of the action of σ on G(Ω)Gal(Ω/L) . This reads ιL (θL (σ)·g) = σ·ιL (g), which is what we wanted to prove. Therefore, we get a bijection ∼

ηL : H n (GL , G(L)) −→ H n (GΩ /Gal(Ω/L), G(Ω)Gal(Ω/L) ) for all L/K induced by the two previous compatible maps. Let us denote by fL : H n (GΩ /Gal(Ω/L), G(Ω)Gal(Ω/L) ) −→ H n (GΩ , G(Ω)) the map induced in cohomology by the two compatible maps πL : GΩ −→ GΩ /Gal(Ω/L) and the inclusion G(Ω)Gal(Ω/L) −→ G(Ω). Notice that the two diagrams G(L)

ιL

/ G(Ω)

ιL

G(Ω)Gal(Ω/L)

/ G(Ω)

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Galois cohomology

and o GL O

GΩ

rL

θL

GΩ /Gal(Ω/L) o

πL

GΩ

are commutative. Applying Proposition II.3.26, we get the equality fL ◦ ηL = ρL . We are now ready to prove the theorem. First of all, Theorem II.3.33 shows that we have n Gal(Ω/L) ) H n (GΩ , G(Ω)), lim −→ H (GΩ /Gal(Ω/L), G(Ω)

L⊂Ω

since it follows from the Galois correspondence that open normal subgroups of GΩ have the form Gal(Ω/L), where L/K is a finite Galois subextension of Ω/K. If [ξ] ∈ H n (GΩ /Gal(Ω/L), G(Ω)Gal(Ω/L) ), this isomorphism maps [ξ]/∼ onto fL ([ξ]). Now for every finite Galois extensions L/K and L /K such that L ⊂ L , the corresponding inflation map will be denoted by inf L,L . Using again Proposition II.3.26, one can check that we have a commutative diagram / H n (GΩ /Gal(Ω/L), G(Ω)Gal(Ω/L) )

H n (GL , G(L)) ρL,L

H n (GL , G(L ))

inf L,L

/ H n (GΩ /Gal(Ω/L ), G(Ω)Gal(Ω/L ) )

where the horizontal maps are the bijections defined above. It follows easily that we have a well-defined bijection ∼

n n Gal(Ω/L) ). u: − lim −→ H (GΩ /Gal(Ω/L), G(Ω) → H (GL , G(L)) −→ lim L⊂Ω

L⊂Ω

If [α] ∈ H (GL , G(L)), then u maps [α]/∼ onto ηL ([α])/∼ . Details are left for the reader as an exercise. Composing u with the previous map, we get a bijection n

n n lim −→ H (GL , G(L)) H (GΩ , G(Ω)),

L⊂Ω

mapping the equivalence class of [α] onto fL (ηL ([α])). Since fL ◦ηL = ρL , we are done.

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We now give an application of this result to the computation of the Galois cohomology of the algebraic group-scheme Ga . Proposition III.7.31. Let k be a field. For every n ≥ 1 and every Galois extension Ω/k, we have H n (GΩ , Ω) = 0. Proof. By Theorem III.7.30, it is enough to prove H n (GL , L) = 0 for every finite Galois extension L/k. Let L/k be such an extension. By Dedekind’s lemma, the elements of GL are linearly independent over L σ = 0. (see [42] for a proof of this fact, for example). In particular, Let y ∈ L such that

σ(y) = 0. Notice that

σ∈GL

σ∈GL

σ(y) ∈ k since it is

σ∈GL

fixed by every element of GL . Then the element x = (

σ(y))−1 y ∈ L

σ∈GL

satisfies

σ·x =

σ∈GL

σ(x) = 1.

σ∈GL

Let α ∈ Z n (GL , L), and let GLn−1 −→ L a : (σ , . . . , σ ασ1 ,...,σn−1 ,ρ σ1 · · · σn−1 ρ·x. 1 n−1 ) −→ ρ∈GL

We are going to prove that α = dn−1 ((−1)n a), which will imply that α is cohomologous to the trivial cocycle. By definition, the element dn−1 (a)σ1 ,...,σn is equal to

σ1 ·ασ2 ,...,σn ,ρ +

ρ∈GL

+(−1)n

n−1

(−1)i ασ1 ,...,σi σi+1 ,...,σn ,ρ

σ1 · · · σn ρ·x

i=1

ασ1 ,...,σn−1 ,ρ σ1 · · · σn−1 ρ·x.

ρ∈GL

Performing the change of indices σn ρ ↔ ρ in the second sum, we see that we have dn−1 (a)σ1 ,...,σn =

dn (α)σ1 ,...,σn ,ρ − (−1)n+1 ασ1 ,...,σn σ1 · · · σn ρ·x.

ρ∈GL

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Since α is a n-cocycle, we have dn (α) = 0 and we get σ1 · · · σn ρ·x dn−1 (a)σ1 ,...,σn = (−1)n ασ1 ,...,σn =

(−1)n ασ1 ,...,σn

ρ∈G L

ρ·x

ρ∈GL

=

(−1)n ασ1 ,...,σn .

This concludes the proof. We now establish some functoriality properties of Galois cohomology. In the sequel, G will denote a Galois functor. Let ι : K −→ K be a morphism of field extensions of k. Let Ω/K and Ω /K be two Galois extensions, and assume that we have a morphism ϕ : Ω −→ Ω of field extensions of k which extends ι. Let ϕ : GΩ −→ GΩ the continuous group morphism associated to ϕ by Corollary I.2.10. Lemma III.7.32. The maps ϕ : GΩ −→ GΩ and G(ϕ) : G(Ω) −→ G(Ω ) are compatible. Proof. We need to prove that for all σ ∈ GΩ and all g ∈ G(Ω), we have G(ϕ)(ϕ(σ )·g) = σ ·G(ϕ)(g). By definition of the action of GΩ on G(Ω), we have G(ϕ)(ϕ(σ )·g) = (G(ϕ) ◦ G(ϕ(σ )))(g). Since G is a functor, this yields G(ϕ)(ϕ(σ )·g) = G(ϕ ◦ ϕ(σ ))(g). By definition of ϕ, we have ϕ ◦ ϕ(σ ) = σ ◦ ϕ. We then get G(ϕ)(ϕ(σ )·g) = G(σ ◦ ϕ)(g) = G(σ )(G(ϕ)(g)) = σ ·G(ϕ)(g). This proves the lemma. In view of the previous lemma and Proposition II.3.19, we have an induced map Rϕ : H n (GΩ , G(Ω)) −→ H n (GΩ , G(Ω )). Proposition III.7.33. Let ϕ : Ω −→ Ω” be an extension of ι. Then the map Rϕ : H n (GΩ , G(Ω)) −→ H n (GΩ , G(Ω )) only depends on ι.

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91

Proof. Let ϕ : Ω −→ Ω be another extension of ι. By Corollary I.2.10, there exists ρ ∈ GΩ such that ϕ = ϕ ◦ ρ, and we have ϕ = Int(ρ) ◦ ϕ. In particular, we have G(ϕ )(g) = G(ϕ ◦ ρ−1 )(g) = (G(ϕ) ◦ G(ρ−1 ))(g) = G(ϕ)(ρ−1 ·g), for all g ∈ G(Ω). We then have two commutative diagrams G(Ω)

ρ−1 ·

G(ϕ )

G(Ω )

o GΩ O ϕ

GΩ

/ G(Ω) G(ϕ)

G(Ω )

Int(ρ)

GΩ O ϕ

GΩ

for wich all the corresponding maps are compatible, by the previous lemma and Example II.3.21. By Proposition II.3.26 and the same example, we get H n (GΩ , G(Ω)) Rϕ

H n (GΩ , G(Ω ))

H n (GΩ , G(Ω)) Rϕ

H n (GΩ , G(Ω ))

and this completes the proof of the proposition. We are ready to apply these results to define a Galois cohomology functor H n (− , G) : Ck −→ Sets∗ . Let K/k and L/k be two field extensions, and let ι : K −→ L be a morphism of extensions. Now let Kalg and Lalg be algebraic closures of K and L respectively. Let Ks and Ls be the corresponding separable closures. Finally, let ϕ : Ks −→ Ls be any extension of ι (such an extension exists by Corollary I.1.20), and let ϕ : GLs −→ GKs be the

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Galois cohomology

continuous group morphism associated to ϕ by Corollary I.2.10. By Proposition III.7.33, we get a map Rϕ : H n (GKs , G(Ks )) −→ H n (GLs , G(Ls )), which only depends on ι. Remark III.7.34. If K = L and ι = IdK , we obtain the identity map, since we may take ϕ = IdKs . Lemma III.7.35. Let K/k, L/k and M/k be three field extensions of k, and let two morphisms of extensions ι : K −→ L and η : L −→ M . Let ϕ : Ks −→ Ls and ψ : Ls −→ Ms be extensions of ι and η respectively. Then we have Rψ◦ϕ = Rψ ◦ Rϕ . Proof. The map ψ ◦ ϕ : Ks −→ Ms is an extension of K −→ M . Moreover, it is easy to check that we have ψ ◦ ϕ = ψ ◦ ϕ. Now, consider the commutative diagrams G(Ks )

G(ϕ)

G(ψ◦ϕ)

/ G(Ls ) G(ψ)

G(Ms )

G(Ms )

and GKs o O ψ◦ϕ

GMs

ϕ

GLs O ψ

GMs

and apply Proposition II.3.26. Corollary III.7.36. Let K be a field, let Kalg be an algebraic closure of K and let Ks be the corresponding separable closure. Then the set H n (GKs , G(Ks )) does not depend on the choice of Kalg , up to canonical bijection. (1)

(2)

Proof. Let Kalg and Kalg be two algebraic closures of K. We denote by (1)

(2)

(1)

(2)

Ks and Ks the respective separable closures. Let ϕ : Kalg −→ Kalg be a K-isomorphism. Since ϕ maps separable elements to separable elements (we have already seen this during the proof of Corollary I.1.20),

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93

we obtain a K-embedding ϕs : Ks(1) −→ Ks(2) . Now the inverse map (2)

(1)

ψs : Kalg −→ Kalg (2)

induces a K-embedding ψs : Ks

ϕs ◦ ψs ψs ◦ ϕs

(1)

−→ Ks . By definition, we have = =

IdK (2) , s IdK (1) . s

Applying Lemma III.7.35 and Remark III.7.34, we get that that ϕs∗ is bijective with inverse ψ s∗ . This bijection is canonical, since it only depends on the restriction of ϕ to K, which is the identity. Definition III.7.37. Let G : Algk −→ Grps be a Galois functor, and let K/k be a field extension. We define the nth Galois cohomology set of G by H n (K, G) = H n (GKs , G(Ks )). If G is abelian (i.e. G(R) is an abelian group for all R), it is a commutative group. If ι : K −→ L is a morphism of field extensions of k, the corresponding map H n (K, G) −→ H n (L, G) is called the restriction map, and is denoted by ResL/K . It follows from Remark III.7.34 and Lemma III.7.35 that if ι = IdK , ResK/K is the identity map, and that for any tower of field extensions K −→ L −→ M , we have ResM/K = ResM/L ◦ ResL/K . Therefore, we get a functor H 1 (− , G) : Ck −→ Sets∗ . If G is abelian, the restriction map is a group morphism, and for n ≥ 1, we obtain a functor H n (− , G) : Ck −→ AbGrps. Remark III.7.38. The restriction map is easier to describe if L/K is separable: indeed, in this case Ks = Ls and GLs is a closed subgroup of GKs ; therefore applying the restriction map is just restricting the cocycles. We would like to continue this section by a translation of some results of the previous chapter in our situation. Until the end, G, G , H, H , N and N will denote Galois functors.

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Galois cohomology

Theorem III.7.39. Let K/k and L/k be two field extensions of k, let K −→ L be a morphism of field extensions, and let Ks −→ Ls be any extension to the separable closures. (1)

Assume that we have an exact sequence of GKs -groups 1 −→ N (Ks ) −→ G(Ks ) −→ H(Ks ) −→ 1. Then the exact sequence of pointed sets 1 −→ N (K) −→ G(K) −→ H(K) may be extended to / H(K)

0 δK

/ H 1 (K, G)

/ H 1 (K, N )

/ H 1 (K, H).

If moreover N (Ks ) identifies to a central subgroup of G(Ks ), then the exact sequence above may be extended further to / H 1 (K, H) (2)

1 δK

/ H 2 (K, N ).

Assume that the diagram G(Ks )

/ H(Ks )

G (Ks )

/ H (Ks )

commutes. Then the diagram H n (K, G)

/ H n (K, H)

H n (K, G )

/ H n (L, H )

commutes. (3)

Assume that the diagram G(Ks )

/ H(Ks )

G(Ls )

/ H(Ls )

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95

commutes. Then the diagram H n (K, G)

/ H n (K, H)

H n (L, G)

/ H n (L, H)

commutes, where the vertical maps are the restriction maps. (4)

Assume that we have a commutative diagram with exact rows 1

/ N (Ks )

/ G(Ks )

/ H(Ks )

/1

1

/ N (Ks )

/ G (Ks )

/ H (Ks )

/1

0 0 and let us denote by δK and δK the respective 0th connecting maps. Then the diagram 0 δK

H(K) H (K)

0 δK

/ H 1 (K, N ) / H 1 (K, N )

is commutative. If moreover N (Ks ) and N (Ks ) identify to central subgroups of G(Ks ) and G (Ks ) respectively, then the diagram 1 δK

H 1 (K, H) H 1 (K, H )

1 δK

/ H 2 (K, N ) / H 2 (K, N )

1 1 is commutative, where δK and δK denote the respective first connecting maps.

(5)

Assume that we have a commutative diagram with exact rows 1

/ N (Ks )

/ G(Ks )

/ H(Ks )

/1

1

/ N (Ls )

/ G(Ls )

/ H(Ls )

/1

96

Galois cohomology 0 0 and let us denote by δK and δL the respective 0th connecting maps. Then the diagram

H(K) H(L)

0 δK

/ H 1 (K, N ) / H 1 (L, N )

0 δL

is commutative. If moreover N (Ks ) and N (Ls ) identify to central subgroups of G(Ks ) and G(Ls ) respectively, then the diagram H 1 (K, H) H 1 (L, H)

1 δK

1 δL

/ H 2 (K, N ) / H 2 (L, N )

1 1 and δL denote the respective first conis commutative, where δK necting maps.

(6)

For all n ≥ 1, we have a bijection of pointed sets (resp. a group isomorphism if G is abelian) n H n (K, G) lim −→ H (GL , G(L)), L

where L/K runs through the finite Galois extensions of K. Proof. (1) is a direct application of Propositions II.4.7 and II.4.10, together with Lemma III.7.16 (2). To prove points (2) − (5), apply Propositions II.3.26, II.4.6 and II.4.11, as well as Lemma III.7.32 for points (3) and (5). Finally, (6) is just an application of Theorem III.7.30 to Ω = Ks . §III.8 Abstract Galois descent Now that the decor is set and the actors are in place, we are ready to expose the theory of Galois descent. As explained in the introduction, Galois cohomology measures in which extent two objects defined over k are isomorphic, provided they are isomorphic over a field extension of k. We would like to generalize the approach used to solve the Galois descent problem for conjugacy of matrices to arbitrary algebraic objects. We start this section by having a closer look at this case.

III.8 Abstract Galois descent

97

III.8.1 Matrices reloaded In this paragraph, we would like to extract the essential arguments of our solution to the conjugacy problem, and rewrite them in a more concise and formal way, in order to find a method to attack the general Galois descent problem. Let us first reformulate the result we obtained. In fact, we have proved in the introduction that the set of G(k)-conjugacy classes of matrices M ∈ Mn (Ω) which are G(Ω)-conjugate to a given matrix M0 ∈ Mn (k) is in one-to-one correspondence with the set of cohomology classes [α] ∈ H 1 (GΩ , ZG (M0 )(Ω)), which may be written [α] = [αC ] for some C ∈ G(Ω), where αC is the cocycle αC :

GΩ −→ ZG (M0 )(Ω) σ −→ C(σ·C)−1 .

This set of cohomology classes is nothing but the kernel of the map H 1 (GΩ , ZG (M0 )(Ω)) −→ H 1 (GΩ , G(Ω)) induced by the inclusion ZG (M0 )(Ω) ⊂ G(Ω). This observation will allow us to give a more conceptual (and less miraculous) explanation of our result. Notice first that the congugacy class of a matrix may be reinterpreted as an orbit under the action of G ⊂ GLn by conjugation. This action will be denoted by ∗ in the sequel. The next crucial observation is then the following one: if M0 ∈ Mn (k), we may rewrite ZG (M0 )(Ω) as ZG (M0 )(Ω) = {C ∈ G(Ω) | C ∗ M0 = M0 }. In other words, ZG (M0 )(Ω) is nothing but the stabilizer of M0 (viewed as an element of Mn (Ω)) with respect to the action of G(Ω) on Mn (Ω). The second important point is that the action of GΩ on G(Ω) restricts to an action on ZG (M0 )(Ω). To see this, recall that the action of GΩ on a matrix S = (sij ) ∈ Mn (Ω) is given by σ·S = (σ(sij )). In particular, the following properties hold: (i) Mn (Ω)GΩ = Mn (k) (ii) For all S ∈ Mn (Ω), C ∈ G(Ω) and σ ∈ GΩ , we have σ·(C ∗ S) = (σ·C) ∗ (σ·S).

98

Galois cohomology

We have in fact an even more general property. If ι : K −→ L is a morphism of field extensions of k and S ∈ Mn (K), set ι·S = (ι(sij )). We then have (ii ) For all morphisms of extensions ι : K −→ L, S ∈ Mn (K) and all C ∈ G(K), we have ι·(C ∗ S) = (ι·C) ∗ (ι·S). If now C ∈ ZG (M0 )(Ω) and σ ∈ G(Ω), we have (σ·C) ∗ M0 = (σ·C) ∗ (σ·M0 ) by (i), since M0 ∈ Mn (k). Using (ii), we then get (σ·C) ∗ M0 = σ·(C ∗ M0 ) = σ·M0 = M0 , the second equality coming from the fact that C ∈ ZG (M0 )(Ω). Hence the action of GΩ on G(Ω) restricts to an action on ZG (M0 )(Ω) as claimed. Now it follows from elementary group theory that we have a bijection G(Ω)/ZG (M0 )(Ω) G(Ω) ∗ M0 . Equivalently, we have an exact sequence 1 −→ ZG (M0 )(Ω) −→ G(Ω) −→ G(Ω) ∗ M0 −→ 1, which may be easily seen to be an exact sequence of GΩ -sets satisfying the conditions explained in § II.4.1. The apparition of ker[H 1 (GΩ , ZG (M0 )(Ω)) −→ H 1 (GΩ , G(Ω))] is not a real surprise then, in view of Corollary II.4.5. The same corollary says that this kernel is in one-to-one correspondence with the orbit set of G(Ω)GΩ in (G(Ω) ∗ M0 )GΩ . Let us now check that this orbit set is precisely the set of G(k)-conjugacy classes of matrices which become G(Ω)-conjugate to M0 , at least in the cases considered in the introduction. Therefore, assume until the end of this paragraph that G = GLn or SLn . In this case, we have (iii) G(Ω)GΩ = G(k). Notice now that the action G(Ω)GΩ = G(k) on (G(Ω) ∗ M0 )GΩ defined before Corollary II.4.5 is simply the restriction of the action of G(k) on Mn (k) by conjugation. Indeed, if M ∈ Mn (k) has the form M = Q ∗ M0


99

for some Q ∈ G(Ω), and if P ∈ G(k), Q is a preimage of M under the map G(Ω) → G(Ω) ∗ M0 , and therefore we have P · M = (P Q) ∗ M0 = P ∗ (Q ∗ M0 ) = P ∗ M. Using (i) and (iii), it follows that the orbit set (G(Ω) ∗ M0 )GΩ /G(Ω)GΩ is nothing but the set of G(k)-conjugacy classes of matrices M ∈ Mn (k) which become G(Ω)-conjugate to M0 . Therefore, we have proved that our solution the conjugacy problem for matrices was nothing but an application of Corollary II.4.5, and we have identified three important properties which make this actually work. Notice that (i) and (iii) may seem redundant a priori, but this is only due to our specific example. In more abstract situations, both conditions may be of different nature. For example, one may replace matrices by quadratic forms of dimension n and study the Galois descent problem for isomorphism classes of quadratic forms. In this case, we see that properties (i) and (iii) do not concern the same objects. Our next goal is now to reformulate this new approach in our functorial context, and derive a general solution for abstract Galois descent problems. In particular, we will need to find appropriate substitutes for ZG (M0 )(Ω) and properties (i) − (iii).

III.8.2 Actions of group-valued functors If we analyze the Galois descent problem for conjugacy classes of matrices, some key ingredients are needed in order for this question to make sense. First of all, matrices with coefficients in a field form a functor Mn : Ck −→ Sets, so that for a given Galois extension Ω/k, we may consider the sets Mn (k) and Mn (Ω), and we have an induced map Mn (k) −→ Mn (Ω) which allows us to extend scalars. Therefore, the algebraic objects we are going to consider will be points of a covariant functor. Let k be any field, and let F : Ck −→ Sets be a functor. We will write F(K) instead of F(K/k). If K −→ L is a morphism of field extensions and a ∈ F(K), recall that we denote by aL the image of a under the induced map F(K) −→ F(L) if there is no ambiguity in the choice of the map K −→ L. To set up the Galois descent problem for conjugacy classes of matrices,

100

Galois cohomology

we also need an action of some subfunctor of GLn on matrices. As we have seen in the previous paragraph, this action has some nice functorial properties. This leads to the following definition: Definition III.8.1. Let G : Ck −→ Grps be a group-valued functor. We say that G acts on F if the following conditions hold: (1)

For every field extension K/k, the group G(K) acts on the set F(K). This action will be denoted by ∗.

(2)

For every morphism ι : K −→ L of field extensions, the following diagram is commutative: G(K) × F(K) (G(ι),F(ι))

G(L) × F(L)

/ F(K) F(ι)

/ F(L)

that is F(ι)(g ∗ a) = G(ι)(g) ∗ F(ι)(a) for all a ∈ F(K), g ∈ G(K). In other words, for every field extension K/k, we have a group action of G(K) on F(K) which is functorial in K. Notice that the last condition rewrites (g ∗ a)L = gL ∗ aL for all a ∈ F(K), g ∈ G(K) for a given field extension L/K if we use the short notation recalled at the beginning of the paragraph. Examples III.8.2. (1)

Let G ⊂ GLn be an algebraic group-scheme and let F be the functor defined by F(K) = K n for every field extension K/k. Then G acts by left multiplication on F.

(2)

If G ⊂ GLn is an algebraic group-scheme and F = Mn , then G acts on F by conjugation.

Remark III.8.3. Let K/k be a field extension, and let Ω/K be Galois extension. Recall from Lemma III.7.15 that, given a covariant functor F : Ck −→ Sets, we have a natural action of GΩ on F(Ω) defined by GΩ × F(Ω) −→ F(Ω) (σ, a) −→ σ·a = F(σ)(a).


101

If G is a group-valued functor acting on F, we have by definition F(σ)(g ∗ a) = G(σ)(g) ∗ F(σ)(a) for all a ∈ F(Ω), σ ∈ GΩ , which rewrites as σ·(g ∗ a) = (σ·g) ∗ (σ·a) for all a ∈ F(Ω), σ ∈ GΩ . We would like to continue by giving a reformulation of the general Galois descent problem. For this, we need to introduce the concept of a twisted form.

III.8.3 Twisted forms Let G be a group-valued functor acting on a functor F : Ck −→ Sets. This action of G allows us to define an equivalence relation on the set F(K) for every field extension K/k by identifying two elements which are in the same G(K)-orbit. For example, in the case of matrices, two matrices of Mn (K) will be equivalent if and only if they are G(K)conjugate. More precisely, we have the following definition: Definition III.8.4. Let G be a group-valued functor defined over k acting on F. For every field extension K/k we define an equivalence relation ∼K on F(K) as follows: two elements b, b ∈ F(K) are equivalent over K if there exists g ∈ G(K) such that b = g ∗ b . We will denote by [b] the corresponding equivalence class. We may now formulate a general descent problem. Galois descent problem: let F : Ck −→ Sets and let G : Ck −→ Grps be a group-scheme acting on F. Finally, let Ω/k be a Galois extension and let a, a ∈ F(k). Assume that aΩ ∼Ω aΩ . Do we have a ∼k a ? Notice that the answer to this question only depends on the G(k)equivalence class of a and a . We now give a special name to elements of F which become equivalent to a fixed element a ∈ F(k) over Ω. Definition III.8.5. Let a ∈ F(k), let K/k be a field extension and let Ω/K be a Galois extension. An element a ∈ F(K) is called a twisted K-form of a if aΩ ∼Ω aΩ . Clearly, if a ∈ F(K) is a twisted K-form of a and a ∼K a , then a is also a twisted K-form of a, so the action of G(K) restricts to the set of twisted K-forms of a.

102

Galois cohomology

We denote by Fa (Ω/K) the set of K-equivalence classes of twisted Kforms of a, that is Fa (Ω/K) = {[a ] | a ∈ F(K), aΩ ∼Ω aΩ }. Notice that Fa (Ω/K) always contains the class of aK , so it is natural to consider Fa (Ω/K) as a pointed set, where the base point is [aK ]. We now would like to define a functor Fa : Ck −→ Sets∗ . Let ι : K −→ K be a morphism of field extensions of k, let Ω/K and Ω /K be two Galois extension, and assume that we have an extension ϕ : Ω −→ Ω of ι. If a ∈ F(K) is a twisted K-form of a, then aK is a twisted K -form of a as well. Indeed, functorial properties of F imply that (aK )Ω = (aΩ )Ω (where the last scalar extension to Ω is obtained via F(ϕ) : F(Ω) −→ F(Ω )). Now if g ∈ G(Ω) satisfies g ∗ aΩ = aΩ , then we have gΩ ∗ (aK )Ω = gΩ ∗ (aΩ )Ω = (g ∗ aΩ )Ω , by definition of the action of G on F. We then get gΩ ∗ (aK )Ω = (aΩ )Ω = aΩ , showing that aK is a twisted K -form of a. Notice that it does not depend on the choice of the extension ϕ of ι. Therefore, the map F(K) −→ F(K ) induces a map Fa (Ω/K) −→ Fa (Ω /K ) [a ] −→ [aK ]. In particular, we obtain a functor Fa : Ck −→ Sets∗ by setting Fa (K) = Fa (Ks /K) for every field extension K/k, the map induced by a morphism of field extension ι : K −→ K being the map Fa (ι) :

Fa (K) −→ Fa (K ) [a ] −→ [aK ].

Example III.8.6. As pointed out before, if F = Mn , then G ⊂ GLn acts on F by conjugation, and two matrices M, M ∈ Mn (K) are then equivalent if and only if they are G(K)-conjugate. Moreover, if M0 ∈ Mn (k), then FM0 (Ω/K) is the set of G(K)-conjugacy classes of matrices M ∈ Mn (K) which are G(Ω)-conjugate to M0 . Using the notation introduced previously, the Galois descent problem may be reinterpreted in terms of twisted forms as follows: given a ∈ F(k) and a Galois extension Ω/k, do we have Fa (Ω/k) = {[a]} ?


103

We would like to describe the functor Fa in terms of Galois cohomology of a suitable group-scheme associated to a, under some reasonable conditions on F and G. To do this, we will continue to try to generalize the approach described in the first paragraph of this section.

III.8.4 The Galois descent condition One of the crucial property we used to solve the conjugacy problem is the equality Mn (Ω)GΩ = Mn (k), where we let GΩ act on S ∈ Mn (Ω) coefficientwise. By Example III.7.26, this action is nothing but the action of GΩ induced by the functorial properties of Mn , as described in Lemma III.7.15. Now let us go back to our more general setting. For every field extension K/k and every Galois extension Ω/K, we have an action of GΩ on the set F(Ω) given by σ·a = F(σ)(a) for σ ∈ GΩ and a ∈ F(Ω). The second part of Lemma III.7.15, applied to the Galois extensions K/K and Ω/K, then yields σ·aΩ = aΩ for all σ ∈ GΩ , a ∈ F(K). However, contrary to the case of matrices, an element of F(Ω) on which GΩ acts trivially does not necessarily comes from an element of F(K). Example III.8.7. Let us consider the functor F : Ck −→ Sets defined as follows: for a field extension K/k, set F(K) =

{0} {0, 1}

if [K : k] ≤ 1 if [K : k] ≥ 2

the map induced by a morphism K −→ K being the inclusion of sets. In particular, for every Galois extension Ω/k, the Galois group GΩ acts trivially on F(Ω). However, if [Ω : k] > 1, the element 1 ∈ F(Ω) does not come from an element of F(k). These considerations lead to the following definition: Definition III.8.8. We say that a functor F : Ck −→ Sets satisfies the Galois descent condition if for every field extension K/k and every

104

Galois cohomology

Galois extension Ω/K the map F(K) −→ F(Ω) is injective and induces a bijection F(K) F(Ω)GΩ . Example III.8.9. The functor Mn satisfies the Galois descent condition, as well as every representable functor by Lemma III.7.16, or as any Galois functor by definition.

III.8.5 Stabilizers It follows from the considerations of the previous paragraph that it is reasonable to consider Galois descent problems for elements of a functor satisfying the Galois descent condition. Now that we have set a suitable framework for the general Galois descent problem, we need an appropriate substitute for the set ZG (M0 )(Ω). As noticed before, denoting by ∗ the action of G ⊂ GLn on matrices by conjugation, the subgroup ZG (M0 )(Ω) may be reinterpreted as the stabilizer of M0 with respect to the action of G(Ω) on Mn (Ω), that is ZG (M0 )(Ω) = StabG (M0 )(Ω) = {C ∈ G(Ω) | C ∗ M0 = M0 }. Since in our setting we have a group-scheme acting on F, it seems sensible to introduce the following definition: Definition III.8.10. Let G : Ck −→ Grps be a group-valued functor acting on F. For a ∈ F(k), and every field extension K/k, we set StabG (a)(K) = {g ∈ G(K) | g ∗ aK = aK } for all K/k. If K −→ K is a morphism of field extensions, the map G(K) −→ G(K ) restricts to a map StabG (a)(K) −→ StabG (a)(K ). Indeed, for every g ∈ StabG (a)(K), we have gK ∗ aK = (g ∗ aK )K = (aK )K = aK . We then get a subfunctor StabG (a) : Ck −→ Grps of G, called the stabilizer of a. Example III.8.11. If F = Mn , G ⊂ GLn and M0 ∈ Mn (k), then StabG (M0 )(K) = ZG (M0 )(K) for all K/k. Remark III.8.12. Let K be a field, and let Ω/K be a Galois extension. By definition, the map StabG (a)(σ) : StabG (a)(Ω) −→ StabG (a)(Ω) is obtained by restriction of the map G(Ω) −→ G(Ω). Hence, the natural action of GΩ on G(Ω) restricts to an action on StabG (a)(Ω).


105

We have now to ensure that the action of GΩ on StabG (a)(Ω) is continuous. Contrary to the case of matrices, the functor StabG (a) may not be representable even if G is, so we may not conclude to the continuity of the action of GΩ this way. However, if G is a Galois functor, so is StabG (a), as we proceed to show now. Lemma III.8.13. Let G : Ck −→ Grps be a Galois functor acting on a functor F satisfying the Galois descent condition. Then for all a ∈ F(k), StabG (a) is a Galois functor. In particular, for every field extension K/k and every Galois extension Ω/K, StabG (a)(Ω) is a GΩ -group. Proof. Let K/k be a field extension and let Ω/K be a Galois extension. Since StabG (a) is a subfunctor of G, we have a commutative diagram StabG (a)(K)

/ StabG (a)(Ω)

G(K)

/ G(Ω)

where the vertical maps are inclusions. Therefore, the injectivity of G(K) −→ G(Ω) implies the injectivity of the map StabG (a)(K) −→ StabG (a)(Ω). Let g ∈ StabG (a)(Ω)GΩ . Since G is a Galois functor, we have g = gΩ for some g ∈ G(K). We have to check that g ∈ StabG (a)(K). But we have ∗ (aK )Ω = g ∗ aΩ = aΩ = (aK )Ω . (g ∗ aK )Ω = gΩ

Since the map F(K) −→ F(Ω) is injective, we get g ∗ aK = aK , so g ∈ StabG (a)(K). Hence we have a group isomorphism StabG (a)(K) StabG (a)(Ω)GΩ . Finally, if g ∈ StabG (a)(Ω), there exists a finite Galois subextension . Once again, we have to L/K of Ω/K and g ∈ G(L) such that g = gΩ check that g ∈ StabG (a)(L), which can be done as before. Thus we have StabG (a)(L). StabG (a)(Ω) = L⊂Ω

Thus StabG (a) is a Galois functor. The last part of the lemma follows from Remark III.7.29.

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Remark III.8.14. To establish the result above, only the injectivity of the map F(K) −→ F(Ω) for all Galois extensions Ω/K was needed. However, the condition on fixed points will be essential to prove the Galois descent lemma, as for the case of matrices. Using the results of the previous section, we then obtain a Galois cohomology set H 1 (GΩ , StabG (a)(Ω)) for any Galois extension Ω/K, as well as a functor H 1 (− , StabG (a)) : Ck −→ Sets∗ . III.8.6 Galois descent lemma We are now ready to state and prove the Galois descent lemma: Theorem III.8.15 (Galois descent lemma). Let F : Ck −→ Sets be a functor satisfying the Galois descent condition, let G : Ck −→ Grps be a Galois functor acting on F, and let a ∈ F(k). Then for every field extension K/k and every Galois extension Ω/K, we have a bijection of pointed sets ∼

Fa (Ω/K) −→ ker[H 1 (GΩ , StabG (a)(Ω)) −→ H 1 (GΩ , G(Ω))] which is functorial in Ω. More precisely, let ι : K −→ K be a morphism of field extensions of k, let Ω/K and Ω /K be two Galois extensions, and assume that we have an extension ϕ : Ω −→ Ω of field extensions of ι. Then the diagram ∼

Fa (Ω/K)

/ ker[H 1 (GΩ , StabG (a)(Ω)) −→ H 1 (GΩ , G(Ω))] Rϕ

Fa (Ω /K )

/ ker[H 1 (GΩ , StabG (a)(Ω )) −→ H 1 (GΩ , G(Ω ))]

∼

is commutative. In particular, we have an isomorphism of functors from Ck to Sets∗ ∼

Fa −→ ker[H 1 (− , StabG (a)) −→ H 1 (− , G)]. Therefore if H 1 (− , G) = 1, we have an isomorphism of functors ∼

Fa −→ H 1 (− , StabG (a)). Remark III.8.16. Saying that we have a bijection of pointed sets means that it preserves the base points. Hence for every field extension K/k, the class of [aK ] corresponds to the class of the trivial cocycle.


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Proof of Theorem III.8.15. The key ingredient of the proof is Corollary II.4.5. First, by Remark III.8.12, the action of GΩ on G(Ω) restricts to an action on StabG (a)(Ω), which is continuous by Lemma III.8.13. Moreover, we have an exact sequence 1 −→ StabG (a)(Ω) −→ G(Ω) −→ G(Ω) ∗ aΩ −→ 1 which satisfies the conditions of § II.4.1. By Corollary II.4.5, the kernel of H 1 (GΩ , StabG (a)(Ω)) −→ H 1 (GΩ , G(Ω)) is in one-to-one correspondence with the orbit set of G(Ω)GΩ in (G(Ω) ∗ aΩ )GΩ . Notice that G(Ω) ∗ aΩ is simply the set of elements of F(Ω) which are equivalent to aΩ . Since F satisfies the Galois descent condition, it implies that (G(Ω) ∗ aΩ )GΩ is equal to the set {aΩ | a ∈ F(K), aΩ ∼Ω aΩ }. In other words, (G(Ω) ∗ aΩ )GΩ is the image of the set of twisted K-forms of a under the map F(K) −→ F(Ω). Now since G is Galois functor, G(Ω)GΩ is the image of G(K) under the map G(K) −→ G(Ω). Now we claim that if gΩ ∈ G(Ω)GΩ and aΩ ∈ (G(Ω) ∗ aΩ )GΩ , then we have gΩ · aΩ = (g ∗ a )Ω , where ‘·’ denotes here the action defined before Corollary II.4.5. Indeed, since a is a twisted form of a, we may write aΩ = g ∗aΩ for some g ∈ G(Ω). Then g is a preimage of aΩ under the map G(Ω) → G(Ω)∗aΩ and thus gΩ · aΩ

= (gΩ g ) ∗ aΩ = gΩ ∗ (g ∗ aΩ ) = gΩ ∗ aΩ = (g ∗ a )Ω .

We then get the G(Ω)GΩ -orbit of aΩ in (G(Ω) ∗ aΩ )GΩ is the image of G(k) ∗ a under the map F(k) → F(Ω). Hence the map F(k) → F(Ω) induces a bijection between Fa (Ω/k) and the orbit set of G(Ω)GΩ in (G(Ω) ∗ aΩ )GΩ . This proves the first part of the theorem. Before proving the functorial properties of the bijection, we would like to make it a bit more explicit. If [a ] ∈ Fa (Ω/K), the corresponding orbit of G(Ω)GΩ in (G(Ω) ∗ aΩ )GΩ is the orbit of aΩ . By definition of a twisted K-form, we may write g ∗ aΩ = aΩ for some g ∈ G(Ω). Thus g −1 is a preimage of aΩ under the map G(Ω) −→ G(Ω) ∗ aΩ . By Corollary II.4.5, the corresponding cohomology class is δ 0 (g −1 · aΩ ), that is the

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cohomology class represented by the cocycle α:

GΩ −→ StabG (a)(Ω) σ −→ g σ·g −1 .

Conversely, if [α] ∈ ker[H 1 (GΩ , StabG (a)(Ω)) −→ H 1 (GΩ , G(Ω))], then there exists g ∈ G(Ω) such that ασ = g σ·g −1 for all σ ∈ GΩ . In other words, we have [α] = δ 0 (g −1·aΩ ), and the corresponding element in Fa (Ω/K) is represented by the unique element a ∈ F(K) satisfying aΩ = g −1 ∗ aΩ . We now prove the functoriality of the bijection. Let ι : K −→ K be a morphism of field extensions of k, let Ω/K and Ω /K be two Galois extensions, and assume that we have an extension ϕ : Ω −→ Ω of ι. Let ϕ : GΩ −→ GΩ the continuous group morphism associated to ϕ by Corollary I.2.10. Let us denote by η and η the inclusions StabG (a)(Ω) ⊂ G(Ω) and StabG (a)(Ω ) ⊂ G(Ω ) respectively. We first show that the map Rϕ : H 1 (GΩ , StabG (a)(Ω)) −→ H 1 (GΩ , StabG (a)(Ω )) restricts to a map Rϕ : ker(η∗ ) −→ ker(η∗ ). Let [ξ] ∈ ker(η∗ ), so that there exists g ∈ G(Ω) such that ξσ = gσ·g −1 for all σ ∈ GΩ . We are going to prove that Rϕ ([ξ]) is represented by the cocycle GΩ −→ StabG (a)(Ω ) −1 σ −→ gΩ σ ·gΩ .

In particular, this will show that Rϕ ([ξ]) ∈ ker(η∗ ). By definition, Rϕ ([ξ]) is represented by the cocycle ξ :

GΩ −→ StabG (a)(Ω ) σ −→ StabG (a)(ϕ)(ξϕ(σ ) ).

Now StabG (a)(ϕ) is the restriction of G(ϕ) by definition. Hence for all σ ∈ GΩ , we have ξσ = G(ϕ)(gϕ(σ )·g −1 ) = [G(ϕ)(g)][G(ϕ)(ϕ(σ )·g)]−1 ,


109

since G(ϕ) is a group morphism. By compatibility of ϕ and G(ϕ), we get −1 ξσ = [G(ϕ)(g)][σ ·G(ϕ)(g)]−1 = gΩ σ ·gΩ for all σ ∈ GΩ .

Now let [a ] ∈ Fa (Ω/K). Then [a ]K ∈ Fa (Ω /K ) is just the class [aK ] by definition. The cohomology class corresponding to [a ] is represented by the cocycle GΩ −→ StabG (a)(Ω)

α:

σ −→ gσ·g −1 ,

where g ∗ aΩ = aΩ . Since we have gΩ ∗ aΩ = (g ∗ aΩ )Ω = (aΩ )Ω = aΩ , the cohomology class corresponding to [aK ] is represented by β:

GΩ −→ StabG (Ω ) −1 σ −→ gΩ σ ·gΩ .

We have to check that Rϕ ([α]) is represented by the cocycle β, which follows from the computations above. This proves the functoriality of the bijection. The second part of the theorem follows from an application of the previous point to Ω = Ks , and from the definition of the restriction maps. Remark III.8.17. Quite often, it is useful in the computations to explicitly know how the correspondence works, so we describe it one more time: (1)

If [a ] ∈ Fa (Ω/K) is the equivalence class of a twisted K-form a ∈ F(K) of a, pick g ∈ G(Ω) such that g ∗ aΩ = aΩ . The corresponding cohomology class in the kernel of the map H 1 (GΩ , StabG (a)(Ω)) −→ H 1 (GΩ , G(Ω)) is the class of the cocycle α:

(2)

GΩ −→ StabG (a)(Ω) ασ −→ ασ = g σ·g −1 .

If [α] ∈ ker[H 1 (GΩ , StabG (a)(Ω)) −→ H 1 (GΩ , G(Ω))], pick g ∈ G(Ω) such that ασ = g σ ·g −1 for all σ ∈ GΩ ; the corresponding class in Fa (Ω/K) is the equivalence class of the unique a ∈ F(K) satisfying aΩ = g −1 ∗ aΩ .

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Galois cohomology

Remark III.8.18. Let L/K be a finite Galois extension, and let ρL : H 1 (GL , StabG (a)(L)) −→ H 1 (K, StabG (a)) be the corresponding map of pointed sets. Applying the last part of the Galois descent lemma to K = K , Ω = L and Ω = Ks shows that, if [α] ∈ ker[H 1 (GL , StabG (a)(L)) −→ H 1 (GL , G(L))] corresponds to [a ] ∈ Fa (L/K), then ρL ([α]) corresponds to [a ] ∈ Fa (K). To apply the Galois Descent Lemma in a more convenient way, we need examples of Galois functors G satisfying H 1 (− , G) = 1. This will be provided by Hilbert 90 and this is the topic of the next section.

III.8.7 Hilbert’s Theorem 90 To prove the so-called Hilbert’s Theorem 90, we will need some preliminary results on semi-linear actions. Definition III.8.19. Let Ω/k be a Galois extension, and let U be a (right) vector space over Ω with an action ∗ of GΩ on U . We will denote by ‘·’ the standard linear action of GΩ on Ω. We say that GΩ acts by semi-linear automorphisms on U if we have for all u, u ∈ U, λ ∈ Ω σ ∗ (u + u ) = σ ∗ u + σ ∗ u ; σ ∗ (uλ) = (σ ∗ u)(σ·λ). Examples III.8.20. (1)

Let V be a k-vector space, and let U = VΩ . The action of GΩ on U defined on elementary tensors by σ ∗ (v ⊗ λ) = v ⊗ (σ·λ) for all v ∈ V, λ ∈ Ω is a continuous action by semi-linear automorphisms.

(2)

Let U = Ωn , and let GΩ act in an obvious way on each coordinate. We obtain that way a continuous action by semi-linear automorphisms. Morever, U GΩ = k n , and we have a canonical isomorphism of Ω-vector spaces ∼

U GΩ ⊗k Ω −→ U, which sends u ⊗ λ onto uλ. This isomorphism is also an isomorphism of GΩ -modules, as the reader may check. The following lemma generalizes the previous example.


111

Lemma III.8.21 (Galois descent of vector spaces). Let U be a vector space over Ω. If GΩ acts continuously on U by semi-linear automorphisms, then U GΩ = {u ∈ U | σ ∗ u = u for all σ ∈ GΩ } is a k-vector space and the map f:

U GΩ ⊗k Ω −→ U u ⊗ λ −→ uλ

is an isomorphism of Ω-vector spaces. Proof (borrowed from [30]): It is clear that U GΩ is a k-vector space. We first prove the surjectivity of f . Let u ∈ U . Since the action of GΩ on U is continuous, the stabilizer of u under the action ∗ is open, hence has the form Gal(Ω/K) for some finite extension K/k. Let L/k be the Galois closure of K/k in Ω. Then L/k is a finite Galois extension such that Gal(Ω/L) acts trivially on u. Let (λi )1≤i≤n be a k-basis of L and let σ1 = IdΩ , σ2 , . . . , σn be a set of representatives of the left cosets of Gal(Ω/L) in GΩ , so that the orbit of u in U consists of σ1 ∗ u = u, σ2 ∗ u, . . . , σn ∗ u (we have exactly n = [L : k] cosets, since GΩ /Gal(Ω/L) GL ). Let ui =

σj ∗ (uλi ).

j

We are going to show that ui ∈ U GΩ . For any σ ∈ GΩ , we have σσj = σ σ for some ∈ {1, . . . , n} and some σ ∈ Gal(Ω/L) by choice of the σj ’s. Hence we have (σσj ) ∗ (uλi ) = (σ σ ) ∗ (uλi ) = σ ∗ (σ ∗ (uλi )). Since σ ∈ Gal(Ω/L) acts trivially on u ∈ U by choice of L and λi ∈ L, we have σ ∗ (uλi ) = (σ ∗ u)(σ ·λi ) = uλi . Thus (σσj ) ∗ (uλi ) = σ ∗ (uλi ). The action of σ on ui =

σj ∗ (uλi )

j

then just permutes the terms of the sum, so ui ∈ U GΩ . Since (σ1 )|L , . . . , (σn )|L are precisely the n k-automorphisms of L/k, they are linearly independent over L in Endk (L) (this is Dedekind’s Lemma; see [42] for instance). Hence the matrix M = (σj ·λi )i,j lies in GLn (L). By definition of uj , we have uj = (σk ∗ u)(σk ·λj ). Now if k

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Galois cohomology

M −1 = (mij ), from the equation M −1 M = In , we get m1j (σk ·λj ) = δ1k for all k = 1, . . . n, j

by comparing first rows. Hence we have uj m1j = (σk ∗ u)(σk ·λj )m1j = (σk ∗ u)δ1k = σ1 ∗ u = u, j

j

k

k

the last equality coming from the fact that σ1 = IdΩ . Therefore, we have ⎞ ⎛ u= uj m1j = f ⎝ uj ⊗ m1j ⎠ , j

j

which proves the surjectivity of f . To prove its injectivity, it is enough to prove the following: Claim: Any vectors u1 , . . . , ur ∈ U GΩ which k-linearly independent remain Ω-linearly independent in U . Indeed, assume that the claim is proved, and let x ∈ ker(f ). One may write x = u1 ⊗ μ1 + . . . + ur ⊗ μr , for some μ1 , . . . , μr ∈ Ω and some u1 , . . . , ur ∈ U GΩ which are linearly independent. By assumption, f (x) = 0 = u1 μ1 + . . . + ur μr . Now the claim implies that μ1 = . . . = μr = 0, and thus x = 0, proving the injectivity of f . It remains to prove the claim. We are going to do it by a way of contradiction. Assume that we have k-linearly independent vectors u1 , . . . , ur ∈ U GΩ for which there exist μ1 , . . . , μr ∈ Ω which are not all zero, such that u1 μ1 + . . . + ur μr = 0. We may assume that r is minimal, r > 1 and μ1 = 1. By assumption, / k. Choose the μi ’s are not all in k, so we may also assume that μ2 ∈ σ ∈ GΩ such that σ·μ2 = μ2 . We have σ∗ ui μi = (σ ∗ ui )(σ·μi ) = ui (σ·μi ) = 0 i

and therefore we get

i

i

ui (σ ·μi − μi ) = 0, a non-trivial relation with

i≥2

fewer terms. This is a contradiction, and this concludes the proof.


113

Remark III.8.22. If we endow U GΩ ⊗k Ω with the natural semi-linear action as in Example III.8.20 (1), we claim that the isomorphism f above is an isomorphism of GΩ -modules, that is f is equivariant with respect to the two semi-linear actions. To check this, it is enough to do it on elementary tensors. Now for all u ∈ U GΩ , λ ∈ Ω and σ ∈ GΩ , we have f (σ ∗ (u ⊗ λ)) = u(σ·λ) = (σ ∗ u)(σ·λ) = σ ∗ (uλ) = σ ∗ f (u ⊗ λ), hence the claim. We may then rephrase the lemma above by saying that any Ω-vector space U endowed with a semi-linear action of GΩ is ‘defined over k’, hence the name of ‘Galois descent of vector spaces’. Before continuing, we need to recall a few definitions. Definition III.8.23. A simple k-algebra is an associative finite dimensional unital k-algebra which has no proper two-sided ideals. A semi-simple k-algebra is a k-algebra which is isomorphic to the direct product of finitely many simple algebras. A separable k-algebra is a k-algebra A such that AK is semi-simple for every field extension K/k. For example, a finite separable extension L/k is a separable k-algebra in that sense. We are now ready to state Hilbert’s theorem 90. Proposition III.8.24 (Hilbert 90). Let A be a semi-simple k-algebra. For every Galois extension Ω/k, we have H 1 (GΩ , GL1 (A)(Ω)) = 1. In particular, the following properties hold: (1)

For any separable k-algebra A,we have H 1 (− , GL1 (A)) = 1.

(2)

For any finite dimensional k-vector space V , we have H 1 (− , GL(V )) = 1. In particular, H 1 (− , Gm ) = 1.

Proof (much inspired from [30]): Since A is semi-simple, we may write A A1 × · · · × Ar , where A1 , . . . , Ar are simple k-algebras. Then we have an isomorphism of GΩ -groups × × A× Ω (A1 )Ω × · · · × (Ar )Ω ,

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Galois cohomology

and therefore × × 1 1 H 1 (GΩ , A× Ω ) H (GΩ , (A1 )Ω ) × · · · × H (GΩ , (Ar )Ω ).

Thus we may assume without any loss of generality that A is a simple algebra. Let α ∈ Z 1 (GΩ , GL1 (A)(Ω)). The Galois group GΩ of Ω/k acts continuously on AΩ as follows: σ · (v ⊗ λ) = v ⊗ (σ·λ) for all a ∈ A, λ ∈ Ω. We now twist the action in a continuous action by semi-linear automorphisms as follows: σ ∗ a = ασ (σ·a) for all a ∈ AΩ , σ ∈ GΩ . Set U = {u ∈ AΩ | σ ∗ u = u for all σ ∈ GΩ }. In our particular case, scalar multiplication by an element λ ∈ Ω in the vector space AΩ is just right multiplication by 1 ⊗ λ. We then get an isomorphism ∼

f:

UΩ −→ AΩ u ⊗ λ −→ u(1 ⊗ λ)

from the previous lemma. Notice that for all a ∈ AΩ , a0 ∈ A and σ ∈ GΩ , we have σ ∗ (a(a0 ⊗ 1)) = (σ ∗ a)(a0 ⊗ 1), as we may see by checking it on elementary tensors. In particular, if a ∈ U then a(a0 ⊗ 1) ∈ U . Thus the external law U × A −→ U (u, a0 ) −→ u • a0 = u(a0 ⊗ 1) endows U with a structure of a right A-module. This endows UΩ with a structure of a right AΩ -module, and f turns out to be an isomorphism of AΩ -modules. Indeed, for all u ∈ U, a ∈ A and λ, λ ∈ Ω, we have f ((u ⊗ λ) • (a ⊗ λ ))

= = = = =

f (u • a ⊗ λλ ) (u • a)(1 ⊗ λλ ) u(a ⊗ 1)(1 ⊗ λλ ) u(1 ⊗ λ)(a ⊗ λ ) f (u ⊗ λ) • (a ⊗ λ),

which is enough to prove AΩ -linearity using a distributivity argument. Let I be a minimal right ideal of A. Since U and A are non-trivial


115

finitely generated A-modules (since they are finite dimensional over k), we have U I r and A I s as A-modules (see for example [53] for more details). Since UΩ AΩ , we have dimk (U ) = dimk (A) and thus r = s, meaning that U is isomorphic to A as a right A-module; in particular, U = u0 • A for some u0 ∈ U . The map ϕ:

AΩ −→ AΩ a ⊗ λ −→ f ((u0 • a) ⊗ λ)

is then an automorphism of AΩ -modules. Since ϕ is AΩ -linear, ϕ is simply the right multiplication by ϕ(1) = f (u0 ⊗1) = u0 . The bijectivity of ϕ then implies that u0 ∈ A× Ω . Now since u0 ∈ U , we have by definition of the twisted action that u0 = ασ σ·u0 for all σ ∈ GΩ . Hence ασ = u0 σ·u−1 0 for all σ ∈ GΩ . This shows that α is cohomologous to the trivial cocycle. This proves the first part of the proposition. The fact that H 1 (− , GL1 (A)) is the trivial functor whenever A is separable follows from the previous point and from the fact that AK is a semisimple K-algebra for any field extension K/k. Applying this to A = Endk (V ) yields H 1 (− , GL(V )) = 1. Remark III.8.25. Assume that Ω/k is a finite cyclic extension of degree n, and let γ be a generator of GΩ . If α ∈ Z 1 (GΩ , Ω× ) is a 1-cocycle, we have αγ n = αγ n−1 γ n−1 ·αγ = . . . = NΩ/k (αγ ). Since γ n = 1, we get NΩ/k (αγ ) = 1. Conversely, any element x ∈ Ω× of norm 1 determines completely a cocycle with values in Ω× by the formula αγ m =

m−1

γ i ·x for m = 0, . . . , n − 1.

i=0

Now let x ∈ Ω× satisfying NΩ/k (x) = 1, and let α be the corresponding cocycle. By Hilbert 90, we know that α is cohomologous to the trivial σ(z) for all σ ∈ GΩ . cocycle, so there exists z ∈ Ω× such that ασ = z γ(z) Applying this equality to σ = γ, we get x = , which is the classical z version of Hilbert 90.

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Corollary III.8.26. Let V be a finite dimensional k-vector space. For every field extension K/k and every Galois extension Ω/K, we have H 1 (GΩ , SL(V )(Ω)) = 1. In particular, H 1 (− , SL(V )) = 1. Proof. We have an exact sequence of GΩ -groups / GL(VΩ )

/ SL(VΩ )

1

/ Ω×

/1,

where the last map is the determinant. By Theorem III.7.39, we have an exact sequence in cohomology GL(V )

/ k×

δ0

/ H 1 (GΩ , GL(V )(Ω)) ,

/ H 1 (GΩ , SL(V )(Ω))

where the first map is the determinant. By Hilbert 90, we get an exact sequence δ0

/ k×

GL(V )

/1.

/ H 1 (GΩ , SL(V )(Ω))

Since the determinant map is surjective, and since the sequence above is exact at k × , it follows that the 0th connecting map is trivial, hence we get H 1 (GΩ , SL(V )(Ω)) = 1. Let n ≥ 2 be an integer. Recall that the algebraic group-scheme μn is defined by μn (R) = {r ∈ R | rn = 1}. Proposition III.8.27. Assume that char(k) does not divide n. Then we have a group isomorphism H 1 (k, μn ) k × /k ×n . Proof. Since char(k) does not divide n, the polynomial X n − a ∈ ks [X] is separable for all a ∈ ks× . Therefore, the map ks× −→ ks× x −→ xn is surjective and we have an exact sequence of Gks -groups / μn (ks )

1

/ k× s

.n

/ k× s

/1,

which induces an exact sequence in cohomology by Theorem III.7.39: k×

.n

/ k×

δ0

/ H 1 (k, μn )

/ H 1 (k, Gm ) .

Using Hilbert 90 and the exactness of this sequence, we get the desired isomorphism.

III.9 First applications of Galois descent

117

Remark III.8.28. The proof shows that the isomorphism is given by ∼

k n /k ×n −→ H 1 (k, μn ) a −→ δ 0 (a).

,

where δ 0 is the 0th connecting map associated to the exact sequence above. Therefore, the bijection can be made explicit as follows: σ(x) . Then if a ¯ ∈ k× /k ×n , let x ∈ ks such that xn = a and set ασ = x the map α : Gks −→ μn (ks ) is a cocycle, whose cohomology class does not depend on the choice of a. Conversely, if α is a cocycle with values in μn (ks ) ⊂ ks× , then there σ(x) for all σ ∈ Gks by Hilbert 90. Now exists x ∈ ks× such that ασ = x since ασn = 1, we get that σ ·xn = xn for all σ ∈ Gks , so a = xn ∈ k × . The class of a modulo k×n does not depend on the choice of x. §III.9 First applications of Galois descent III.9.1 Galois descent of algebras Let A be a finite dimensional k-vector space. For any field extension K/k, let F(K) be the set of unital associative K-algebras with underlying K-vector space AK . If ι : K −→ L is a morphism of field extensions of k, we define a map F(ι) by F(ι) :

F(K) −→ F(L) A −→ AL .

We then obtain a functor F : Ck −→ Sets. Now let f ∈ GL(AK ), and let A be a unital associative K-algebra. We will write x ·A y for the product of two elements x, y ∈ A. The map AK × AK −→ AK (x, y) −→ f (f −1 (x)·A f −1 (y)) endows AK with a structure of a unital associative K-algebra, that we will denote by f·A. Straightforward computations show that this induces an action of GL(A) on F. Notice that by definition, we have f (x) ·f·A f (y) = f (x·A y) for all x, y ∈ A, so that f is an isomorphism of K-algebras from A onto f ·A. It easily

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follows that two unital associative K-algebras A and B are equivalent if and only if there are isomorphic as K-algebras. Now fix a k-algebra A ∈ F(k). Then StabGL(A) (A) is nothing but Autalg (A). It is not difficult to check that F satisfies the conditions of the Galois descent lemma. Hence we get Proposition III.9.1. Let K/k be a field extension, and let Ω/K be a Galois field extension. For any k-algebra A, the pointed set H 1 (GΩ , Autalg (A)(Ω)) classifies the isomorphism classes of K-algebras which become isomorphic to A over Ω. Moreover, the class of the trivial cocycle corresponds to the isomorphism class of AK . Remark III.9.2. Let A be a k-algebra and let Ω/k be a Galois extension. If B is a k-algebra such that there exists an isomorphism ∼ f : BΩ −→ AΩ of Ω-algebras, the corresponding cohomology class is represented by the cocycle α:

GΩ −→ AutΩ (AΩ ) σ −→ f ◦ σ·f −1 .

Indeed, since f is a Ω-algebra isomorphism, we have x·f·BΩ y = f (f −1 (x)·AΩ f −1 (y)) = x·AΩ y for all x, y ∈ AΩ , and therefore f ·BΩ = AΩ . Remark III.8.17 then yields the result. Conversely, the k-algebra corresponding to [α] ∈ H 1 (GΩ , AutΩ−alg (AΩ )) is the isomorphism class of B = {a ∈ AΩ | ασ (σ·a) = a for all σ ∈ GΩ }, where the k-algebra structure is given by restriction of the algebra structure on AΩ . To see this, notice first that the isomorphism of Ω-vectors spaces ∼

f : BΩ −→ AΩ given by Lemma III.8.21 is in fact an isomorphism of Ω-algebras, so that f ·BΩ = AΩ . In view of Remark III.8.17, it is therefore enough to show that we have ασ = f ◦ σ·f −1 for all σ ∈ GΩ ,


119

that is ασ ◦ σ·f = f for all σ ∈ GΩ . Since the elements of B ⊗k 1 span BΩ as an Ω-vector space, it is enough to check this equality on the elements of the form x ⊗ 1, x ∈ B. Now for all x ∈ B and σ ∈ GΩ , we have ασ ((σ·f )(x ⊗ 1))

= = = = =

ασ (σ·(f (σ −1 ·(x ⊗ 1)))) ασ (σ·(f (x ⊗ 1))) ασ (σ·x) x f (x ⊗ 1),

which is the result we were looking for. We will use this remark in the sequel without any further reference. We now give an application of Galois descent to the case of central simple algebras. Definition III.9.3. Let K be a field. A central simple k-algebra is a simple k-algebra with center k. One can show that a k-algebra A is central simple if it becomes isomorphic to a matrix algebra over ks (resp. over a finite Galois extension Ω/k). See [19] for more details for example. In view of the definition, the dimension of a central simple algebra A over its center k is always the square of an integer and central simple algebras define a functor. We will denote by CSAn : Ck −→ Sets∗ the functor of isomorphism classes of central simple algebras of dimension n2 (where the base point is the matrix algebra). The integer n is called the degree of A, and is denoted by degk (A). Denoting by PGLn the group-scheme Autalg (Mn (k)), we get by Proposition III.9.1: Proposition III.9.4. For any field extension K/k, and every Galois extension Ω/K, the pointed set H 1 (GΩ , PGLn (Ω)) classifies the isomorphism classes of central simple K-algebras of degree n which become isomorphic to Mn (k) over Ω. In particular, we have an isomorphism of functors from Ck to Sets∗ CSAn H 1 (− , PGLn ). Matrix algebras are mapped onto the class of the trivial cocycle, and the restriction map ResL/K corresponds to the tensor product ⊗K L.

120

Galois cohomology

Remark III.9.5. The notation PGLn is deliberate, and is consistent with the definition of the group PGLn (K) where K is a field. Indeed, by definition PGLn (K) = AutK−alg (Mn (K)). It is well-known that every K-algebra automorphism of Mn (K) has the form Int(M ) :

Mn (K) −→ Mn (K) M −→ M M M −1

for some M ∈ GLn (K), and therefore we have PGLn (K) = {Int(M ) | M ∈ GLn (K)}. Now Int(M ) is the identity map if and only if M commutes with any matrix, which means that M = λIn for some λ ∈ k × . Therefore, we get a canonical isomorphism PGLn (K) GLn (K)/K × = PGLn (K). Notice also that PGLn (Ks ) SL(Ks )/μn (Ks ) if n is prime to char(K). Let us now consider the case of G-algebras. Definition III.9.6. Let k be a field and let G be an abstract group. A G-algebra over k is a k-algebra on which G acts faithfully by k-algebra automorphisms. Two G-algebras over k are isomorphic if there exists an isomorphism of k-algebras which commutes with the actions of G. It will be denoted by G . Let A be a finite dimensional k-vector space and let G be an abstract finite group. For any field extension K/k, let F(K) be the set of Galgebras over K with underlying vector space AK . If ι : K −→ L is a morphism of field extensions of k, we define a map F(ι) by F(ι) :

F(K) −→ F(L) A −→ AL ,

where the structure of G-algebra on AL is given on elementary tensors by g·(a ⊗ λ) = (g·a) ⊗ λ for all g ∈ G, a ∈ A, λ ∈ L. Now let f ∈ GL(AK ), and let A be a G-algebra over K. Consider the K-algebra f ·A as defined before. The map G × f ·A −→ f ·A (g, x) −→ f (g·f −1 (x))


121

endows f ·A with a structure of a G-algebra over K (details are left to the reader as an exercise). We then get an action of GL(A) on F. Moreover, two G-algebras over K are equivalent if and only if they are isomorphic as G-algebras. Once again, all the conditions of the Galois descent lemma are fulfilled, and we get: Proposition III.9.7. Let G be a finite abstract group and let A be a G-algebra. For any field extension K/k and every Galois extension Ω/K, the pointed set H 1 (GΩ , AutG−alg (A)(Ω)) classifies the isomorphism classes of G-algebras over K which become G-isomorphic to A over Ω. Moreover, the class of the trivial cocycle corresponds to the isomorphism class of AK . Remark III.9.8. Galois descent of algebras still works perfectly if we ask for the multiplication law to satisfy additional properties, such as commutativity or associativity. Il also works if the algebras are not unital nor associative.

III.9.2 The conjugacy problem We now study the following conjugacy problem: let G ⊂ GLn be a Galois functor and let M, M0 ∈ Mn (k) satisfying QM Q−1 = M0 for some Q ∈ G(ks ). Does there exist P ∈ G(k) such that P M P −1 = M0 ? Let us denote by ZG (M0 ) the centralizer of M0 in G, that is the groupscheme defined by ZG (M0 )(K) = {M ∈ G(K) | M M0 = M0 M } for all field extension K/k. By Galois descent, the G(k)-conjugacy classes over k of matrices M which are G(ks )-conjugate to M0 are in one-to-one correspondence with ker[H 1 (k, ZG (M0 )) −→ H 1 (k, G)]. Moreover, the G(k)-conjugacy class of M0 corresponds to the trivial cocycle. Therefore, the conjugacy problem has a positive answer for all matrices M if and only if the map H 1 (k, ZG (M0 )) −→ H 1 (k, G) has trivial kernel. In particular, if H 1 (k, G) = 1, the total obstruction to this problem is measured by H 1 (k, ZG (M0 )). We have seen in the introduction that the conjugacy problem has a negative answer for G = SLn . We would like to recover this fact by using Galois cohomology.

122

Galois cohomology

Definition III.9.9. Let E be a finite dimensional k-algebra, and let R be a commutative k-algebra. For all x ∈ E ⊗k R, we set NER /R (x) = det(x ), where x : ER −→ ER is the R-linear map induced by left multiplication by x. Notice that the definition above makes sense since ER is a free R-module of finite rank. The map NER /R : ER −→ R is called the norm map of ER . Example III.9.10. Let E = kn , n ≥ 1. If x = (x1 , . . . , xn ), then we have NE/k (x) = x1 · · · xn , since the representative matrix of x in the canonical basis of E is simply the diagonal matrix whose diagonal entries are x1 , . . . , xn . Definition III.9.11. If L is a semi-simple commutative k-algebra, we (1) denote by Gm,L the algebraic group-scheme over k defined by Gm,L (R) = {x ∈ L× R | NLR /R (x) = 1}, (1)

for every k-algebra R. (1)

We now compute H 1 (k, Gm,L ) in a special case. Lemma III.9.12. Assume that we have an isomorphism of ks -algebras Lks ksn for some n ≥ 1. Then we have H 1 (k, Gm,L ) k × /NL/k (L× ). (1)

(1)

Proof. The idea of course is to fit Gm,L (ks ) into an exact sequence of Gks -modules. We first prove that the norm map × NLks /ks : L× ks −→ ks ∼

is surjective. Let ϕ : Lks −→ ksn be an isomorphism of ks -algebras. We claim that we have NLks /ks (x) = Nksn /ks (ϕ(x)) for all x ∈ Lks . Indeed, if e = (e1 , . . . , en ) is a ks -basis of Lks , then we have easily Mat(ϕ(x) , ϕ(e)) = Mat(x , e),


123

where ϕ(e) = (ϕ(e1 ), . . . , ϕ(en )). The desired equality then follows immediately. Now for λ ∈ ks× , set xλ = ϕ−1 ((λ, 1, . . . , 1)). The equality above and Example III.9.10 then yield NL⊗k ks /ks (xλ ) = Nksn /ks ((λ, 1, . . . , 1)) = λ. Thus NLks /ks is surjective and we have an exact sequence of Gks -modules 1

/ G(1) (ks ) m,L

/ L×

ks

/ k× s

/1,

where the last map is the norm map NLks /ks . Notice now the assumption on L implies easily that L has no nilpotent elements, so L is a semi-simple k-algebra by [8, § 7, Proposition 5]. Since L× ks = GL1 (L)(ks ), applying Galois cohomology to this sequence and using Hilbert 90 yield the exact sequence (L ⊗k 1)× −→ k × −→ H 1 (k, Gm,L ) −→ 1, (1)

the first map being NLks /ks . Now it is obvious from the properties of the determinant that we have NLks /ks (x ⊗ 1) = NL/k (x) for all x ∈ L. The exactness of the sequence above then gives the desired result. Remark III.9.13. The proof above shows that the 0th -connecting map (1) δ 0 : k × → H 1 (k, Gm,L ) associated to the exact sequence of Gks -modules 1

/ G(1) (ks ) m,L

/ L×

ks

/ k× s

/1

is surjective with kernel NL/k (k × ). Thus, the correspondence works as follows: if a ¯ ∈ k × /NL/k (L× ), let z ∈ L× ks such that a = NLks /ks (z). Then the cohomology class corresponding to a ¯ is represented by the cocycle (1)

Gks −→ Gm,L (ks ) α:

σ −→

σ(z) . z (1)

Conversely, if α is a cocycle with values in Gm,L (ks ), then there exists σ(z) for all σ ∈ Gks . Now set a = NLks /ks (z). z ∈ L× ks such that ασ = z × We have a ∈ k , and the class of a modulo NL/k (L× ) is the one corresponding to [α].

124

Galois cohomology

Remark III.9.14. Algebras satisfying the condition of the previous lemma are called ´ etale, and will be studied in detail in a forthcoming chapter. Now let us go back to the conjugacy problem of matrices. If G = GLn , it is well-known that the conjugacy problem has an affirmative answer in this case, and this can be proved without any use of cohomology. The cohomological reinterpretation of the problem implies that the functor H 1 (− , ZG (M0 )) should be trivial, which is actually the case. We will not prove this in full generality, but just in a particular example. Assume that M0 = Cχ ∈ Mn (k) is a companion matrix of some monic polynomial χ ∈ k[X] of degree n ≥ 1. In this case, it is known that every matrix commuting with M0 is a polynomial in M0 , so ZG (M0 )(ks ) = ks [M0 ]∩G(ks ). Moreover, the minimal polynomial and the characteristic polynomial are both equal to χ. Set L = k[X]/(χ), so that we have an isomorphism of k-algebras ∼

L −→ k[M0 ] P −→ P (M0 ), which induces in turn a Galois equivariant isomorphism of ks -algebras ∼

f:

Lks −→ ks [M0 ] X ⊗ λ −→ λM0 .

In particular, f induces an isomorphism of Gks -modules × L× ks ks [M0 ] .

Notice now that if C ∈ GLn (ks ) commutes with M0 , then C −1 also commutes with M0 . Therefore, we have the equalities ZGLn (M0 )(ks ) = ks [M0 ] ∩ GLn (ks ) = ks [M0 ]× , so f induces an isomorphism of Gks -modules GL1 (L)(ks ) = L× ks ZGLn (M0 )(ks ). By Hilbert 90, we then get H 1 (GΩ , ZGLn (M0 )(Ω)) = 1, as expected. Now let us identify ZSLn (M0 )(Ω). Claim: We have det(f (x)) = NLks /ks (x) for all x ∈ Lks . To see this, set α = X ∈ L. Then e = (1 ⊗ 1, α ⊗ 1, . . . , αn−1 ⊗ 1) is

III.9 First applications of Galois descent a ks -basis of Lks . Let x =

n−1

125

αi ⊗ λi ∈ Lks , and let P =

i=0

n−1

λi X i .

i=0

Clearly, we have x = P (α⊗1 ). Now the matrix of α⊗1 in the basis e is easily seen to be Cχ = M0 , and so the matrix of x in the basis e is P (M0 ) = f (x). Therefore det(x ) = det(f (x)), and we are done. We then get an isomorphism of Gks -modules (1)

ZSLn (M0 )(ks ) Gm,L (ks ). In particular, if χ is separable, we have Lks ks [X]/(χ) ksn , and therefore by Lemma III.9.12 we have H 1 (k, ZSLn (M0 )) k × /NL/k (L× ), which is not trivial in general.

For example, assume that char(k) = 2. Let M0 = 0 −d M= , d ∈ k× . −1 0

0 1

d 0

and let

√ Then M0 is the companion matrix of χ = X 2 − d and thus L = k( d). i 0 −1 (where i is a Moreover, we have QM Q = M0 , with Q = 0 −i square root of −1 in ks ), so M and M0 are conjugate by an element of SL2 (ks ). However, they are not conjugate by an element of SL2 (k) in general. To see this, let us compute the class in k × /NL/k (L× ) corresponding to the conjugacy class of M . Notice first that Qσ·Q−1 is the identity matrix I2 if σ(i) = i and is -I2 otherwise. In other words, we have ασQ = (iI2 )−1 σ·(iI2 ) for all σ ∈ Gks . (1)

Via the isomorphism H 1 (Gks , Gm,L (ks )) H 1 (Gks , ZSLn (M0 )(ks )) induced by f∗ , the cohomology class [α(Q) ] correspond to the cohomology class of the cocycle (1)

β (Q) :

Gks −→ Gm,L (ks ) σ −→ (1 ⊗ i)−1 σ·1 ⊗ i.

Now NLks /ks (1 ⊗ i) = (1 ⊗ i)2 = −1, and thus the conjugacy class of M

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Galois cohomology

corresponds to the class of −1 in k × /NL/k (L× ) by Remark III.9.13. In particular, M and M0 are conjugate over k if and only if −1 ∈ NL/k (L× ). Therefore, to produce counterexamples, one may take for k any subfield of R and d < 0, as we did in the introduction.

III.9.3 Cup-products with values in μ2 Throughout this section, k will be a field of characteristic different from 2. Since μ2 (ks ) is a Gks -module, we have cohomology groups H m (k, μ2 ) for all m ≥ 1. These groups are abelian groups and we will write the group law additively. However, the group of coefficients μ2 (ks ) will be written multiplicatively as usual. Therefore, if [α], [β] ∈ H m (k, μ2 ), we have by definition [α] + [β] = [αβ]. In particular, 2[α] = 0 for all [α] ∈ H m (k, μ2 ). If a ∈ k × , we will denote by (a) the cohomology class in H 1 (k, μ2 ) representing the square-class a ∈ k × /k ×2 . Let xa ∈ ks× satisfying x2a = a. Recall from Remark III.8.28 that (a) is represented by the cocycle Gks −→ μ2 (ks ) αa :

σ −→

σ(xa ) . xa

If now b ∈ k × and xb ∈ ks× satisfies x2b = b, then the element xa xb ∈ ks× satisfies (xa xb )2 = ab. Therefore, (ab) is represented by the cocycle αa αb , and we have the equality (ab) = (a) + (b) for all a, b ∈ k × . Since μ2 (ks ) is an abelian group, hence a Z-module, we can form the tensor product μ2 (ks ) ⊗Z μ2 (ks ). The map ϕ:

μ2 (ks ) × μ2 (ks ) −→ μ2 (ks ) ((−1)m , (−1)n ) −→ (−1)nm

is Z-bilinear, so we may consider the cup-product ∪ϕ , that we will simply denote by ∪. We refer to Chapter 1, Section II.6 for the definition of the cup-product. Let us describe the cup product (a) ∪ (b) ∈ H 2 (k, μ2 ), for a, b ∈ ks× . For


127

any σ ∈ Gks , let εa (σ), εb (σ) ∈ {0, 1} defined by σ(xb ) σ(xa ) = (−1)εa (σ) , = (−1)εb (σ) . xa xb It follows from the definition that the cup-product is represented by the cocycle αa ∪ αb :

Gks × Gks −→ μ2 (ks ) σ, τ −→ (−1)εa (σ)εb (τ ) .

We are now going to establish some useful properties of the cup-product: Proposition III.9.15. For all a, b ∈ k × , the following properties hold: (1)

(a) ∪ (b) = (b) ∪ (a).

(2)

√ √ (a) ∪ (b) = 0 ⇐⇒ b is a norm of k( a)/k (where k( a) = k if a is a square).

(3)

(a) ∪ (1 − a) = 0.

(4)

(a) ∪ (−a) = 0.

(5)

(a) ∪ (a) = (a) ∪ (−1).

Remark III.9.16. The reader with an expert eye has certainly noticed the similarity between these properties and those of the quaternion algebra (a, b). This is not a coincidence. One can show that we have an isomorphism Br2 (k) H 2 (k, μ2 ), where Br2 (k) denotes the subgroup of the Brauer group of k consisting of elements of exponent at most 2 and that, under this isomorphism, the class of the quaternion algebra (a, b) corresponds to (a) ∪ (b). This isomorphism will be fully proved in Chapter VIII. Of course, theses relations may be established directly. However, as an illustration of Galois descent, we would like to supply another proof based on the properties of quaternion algebras. Let us first recall the definition of (a, b). Definition III.9.17. If a, b ∈ k × , we define (a, b) to be the k-algebra generated by two elements i, j and subject to the relations i2 = a, j 2 = b and ij = −ji. One can check that it is a central simple k-algebra of degree 2 over k, called a quaternion algebra. Remark III.9.18. One can show that every central simple algebra of degree 2 is isomorphic to a quaternion algebra. We will not need this fact here.

128

Galois cohomology

Proposition III.9.19. The connecting map δ 1 : H 1 (k, PGL2 ) −→ H 2 (k, μ2 ) associated to the sequence 1

/ μ2 (ks )

/ SL2 (ks )

/ PGL2 (ks )

/1

has trivial kernel and maps the isomorphism class of the quaternion algebra (a, b) to (a) ∪ (b). Proof. The fact that δ 1 has trivial kernel comes from the fact that H 1 (k, SL2 ) = 1. Now, to compute the image of Q = (a, b), we need first to find a cocycle α ∈ Z 1 (k, PGL2 ) corresponding to Q, and therefore we need an explicit isomorphism between Qks and M2 (ks ). One can check that the map f : Qks −→ M2 (ks ) defined by √ √ x + y a b(z + t a) √ √ f (x + yi + zj + tij) = x−y a z−t a is an isomorphism of ks -algebras; this is a possible way to prove that (a, b) is actually a central simple algebra. If E11 , E21 , E12 and E22 denote the elementary matrices of M2 (k), the inverse map f −1 : M2 (ks ) −→ Qks is given by

f

−1

(E11 ) =

f −1 (E12 ) =

1 1+ √ i , a 1 1 j + √ ij , 2b a 1 2

f

−1

(E22 ) =

f −1 (E21 ) =

1− 1 j− 2 1 2

1 √ i a 1 √ ij . a

For σ ∈ Gks , let xa , x−b , ∈ ks satisfying x2a = a, x2−b = −b, and for any σ ∈ Gks , let εa (σ), ε−b (σ) ∈ {0, 1} defined by σ(xa ) σ(x−b ) = (−1)εa (σ) , = (−1)ε−b (σ) . xa x−b We now compute a matrix Mσ ∈ SL2 (ks ) such that ασ = f σ·f −1 = Int(Mσ ) for all σ ∈ Gks . Assume first that εa (σ) = 0, that is σ(xa ) = xa . Then σ ·f −1 (Eij ) = f −1 (Eij ) for 1 ≤ i, j ≤ 2, so Mσ = I2 in this case. Now if εa (σ) = 1, that is σ(xa ) = −xa , then we have


σ·f

−1

(E11 ) =

σ·f −1 (E12 ) =

1 1 1− √ i , σ·f −1 (E22 ) = 2 a 1 1 j − √ ij , σ·f −1 (E21 ) = 2b a

129 1 1+ 2 1 j+ 2

1 √ i a 1 √ ij . a

Therefore we have 1 E21 , ασ (E21 ) = bE12 . b 0 x−b ∈ SL2 (ks ) in One can check that we can take Mσ = −x−1 0 −b this case. 0 x−b , we have Mσ = M εa (σ) for all σ ∈ Thus, setting M = −x−1 0 −b Gks . Then the cocycle ασ (E11 ) = E22 , ασ (E22 ) = E11 , ασ (E12 ) =

α:

Gks −→ PGL2 (ks ) σ −→ ασ = Int(M εa (σ) )

represents the isomorphism class of Q. For all σ ∈ Gks , a preimage of ασ in SL2 (ks ) is then M εa (σ) . By definition of the first connecting map, δ 1 (Q) is then represented by the cocycle γ : Gks × Gks −→ μ2 (ks ) defined by −1 Mσ σ·Mτ Mστ = γσ,τ Id2 for all σ, τ ∈ Gks .

Notice that σ·M = (−1)ε−b (σ) M for all σ ∈ Gks . Therefore, we get γσ,τ I2 = (−1)ε−b (σ)εa (τ ) M εa (σ)+εa (τ )−εa (στ ) . Since the map Gks −→ μ2 (ks ) σ −→ (−1)εa (σ) is a cocycle and Gks acts trivially on μ2 (ks ), we get that εa (σ) + εa (τ ) − εa (στ ) is even for all σ, τ ∈ Gks . Taking into account that M 2 = −I2 , we get γσ,τ = (−1)

εa (σ)+εa (τ )−εa (στ ) +ε−b (σ)εa (τ ) 2

.

Now if σ and τ both fix xa or both map xa onto −xa , then στ fixes xa . If only one of them fixes xa , then στ maps xa onto −xa . Therefore in εa (σ) + εa (τ ) − εa (στ ) = εa (σ)εa (τ ). Finally, we get all cases, we have 2 γσ,τ = (−1)εa (σ)εa (τ )+ε−b (σ)εa (τ ) ,

130

Galois cohomology

that is δ 1 (Q) = (a) ∪ (a) + (−b) ∪ (a). It follows from the identity (u) + (v) = (uv) ∈ H 1 (k, μ2 ) for all u, v ∈ k × and the bilinearity of the cup product that we have δ 1 (Q) = (a) ∪ (a) + (−1) ∪ (a) + (b) ∪ (a) = (−a) ∪ (a) + (b) ∪ (a). Since (a, 1) M2 (k), which represents the trivial cohomology class, applying this formula to b = 1 leads to 0 = (−a) ∪ (a), so δ 1 (Q) = (b)∪(a). Now (a, b) (b, a), so we get δ 1 (Q) = δ 1 ((b, a)) = (a)∪(b). Proof of Proposition III.9.15. Since (a, b) (b, a), we get (1) by applying δ 1 . Now δ 1 has trivial kernel, so (a) ∪ (b) = 0 ⇐⇒ (a, b) M2 (k), √ which is equivalent to say that b is a norm of k( a) by the well-known properties of quaternion algebras (see [53], for example), hence (2). A direct application of (2) gives (3) and (4). Now by bilinearity of the cupproduct and the fact that (−a) = (−1) + (a), (5) follows from (4). Remark III.9.20. Applying Example II.6.7 and the definition of the restriction map, we see that for every field extension L/K, we have ResL/K ([α] ∪ [β]) = ResL/K ([α]) ∪ ResL/K ([β]) for all [α] ∈ H p (K, μ2 ), [β] ∈ H q (K, μ2 ). Notes The proof of the fact that H n (k, G) is independent from the choice of an algebraic closure and the construction of the restriction map follow the arguments outlined in [58, Chapter II,§ 1.1]. Our exposition of Galois descent is more general than the versions that may be found in [58] or [30]. However, we are convinced that it follows from the general descent theory of Grothendieck. The reader may find other interesting applications of Galois descent in [30], where the first cohomology sets associated to classical algebraic groups are described in terms of various algebraic structures.

Exercises 1. Show that H 1 (− , Sp2n ) = 1. Hint: What is the automorphism group of a non-degenerate alternating bilinear form?

Exercises

131

2. Let k be a field, and let n, d ≥ 1 be two integers. For every field extension K/k, let us denote by Hn,d (K) the set of homogeneous polynomials of degree d in n variables with coefficients in K. We let act K × × GLn (K) on Hn,d (K) by ⎛ ⎞ X1 −1 ⎜ .. ⎟ ((λ, M ) · P )(X1 , . . . , Xn ) = λP (M ⎝ . ⎠). Xn If P, Q ∈ Hn,d (K), we say that P and Q are similar if they are in the same orbit under the action of K × × GLn (K). We set Sim(P )(K) = {(λ, M ) ∈ K × × GLn (K) | (λ, M ) · P = P }. If P0 ∈ Hn,d (k), we denote by GP0 the functor of similarity classes of twisted forms of P0 . (a)

Show that Hn,d is a functor on which Gm × GLn acts.

(b)

Let P0 ∈ Hn,d (k). Show that we have an isomorphism of functors H 1 (− , Sim(P0 )) GP0 . We now assume until the end that n = d = 3 and P0 = X13 .

(c)

For every field extension K/k, ⎧ ⎛ a 0 ⎨ Sim(P0 )(K) = (a3 , ⎝ b c ⎩ e f

check that ⎫ ⎞ 0 ⎬ d ⎠) ∈ K × × GL3 (K) . ⎭ g

(d)

Compute H 1 (k, Sim(P0 )). Hint: Show that the obvious morphism Sim(P0 )(ks ) −→ ks× × GL2 (ks ) is surjective, with kernel isomorphic to ks × ks .

(e)

What are the twisted forms of P0 , up to similarity?

3. We keep the notation of the previous exercise. We now let GLn act on Hn,d by ⎛ ⎞ X1 −1 ⎜ .. ⎟ (M · P )(X1 , . . . , Xn ) = P (M ⎝ . ⎠). Xn

132

Galois cohomology If P, Q ∈ Hn,d (K), we say that P and Q are isomorphic if they are in the same orbit under the action of GLn (K). We set Aut(P )(K) = {M ∈ GLn (K) | M · P = P }. If P0 ∈ Hn,d (k), we denote by FP0 the functor of isomorphism classes of twisted forms of P0 . (a)

Let P0 ∈ Hn,d (k). Show that we have an isomorphism of functors H 1 (− , Aut(P0 )) FP0 . We now assume until the end that n = d = 3 and P0 = X13 .

(b)

For all field extension K/k, check that ⎧⎛ ⎫ ⎞ ⎨ a 0 0 ⎬ Aut(P0 )(K) = ⎝ b c d ⎠ ∈ GL3 (K) | a ∈ μ3 (K) . ⎩ ⎭ e f g

(c)

Check that we have a split exact sequence of Gks -groups 1 → ks × ks → Aut(P0 )(ks ) → μ3 (ks ) × GL2 (ks ) → 1. Deduce that the map H 1 (k, Aut(P0 )) −→ H 1 (k, μ3 × GL2 ) is surjective. The goal of the next questions is to prove that this map is also injective. Let β : Gks → Aut(P0 )(ks ), let γ = π∗ (β) and let α : Gks → ks × ks be the image of β by the map induced by the conjugation map Aut(P0 )(ks ) −→ Aut(ks × ks ). 0 aσ , where aσ ∈ For all σ ∈ Gks , write β = X σ Mσ μ3 (ks ), X ∈ ks × ks and Mσ ∈ GL2 (ks ).

(d)

Check that for all X ∈ (ks × ks )α and all σ ∈ Gks , we have σ ∗ X = a−1 σ Mσ σ·X.

(e)

Justify the existence of P ∈ GL2 (ks ) such that a−1 σ Mσ = −1 for all σ ∈ Gks , and check that the map P σ·P ϕ:

ks × ks −→ (ks × ks )α X −→ P X

is an isomorphism of Gks -modules.

Exercises (f)

Conclude using Chapter II Exercise 5 (c).

(g)

Show that we have a bijection of pointed sets

133

H 1 (k, Aut(P0 )) k × /k ×3 . If a ∈ k × /k ×3 , what is the corresponding twisted form of P0 (up to isomorphim) ? 4. Let Ω/k be a Galois extension, and let B(Ω) be the subgroup of upper triangular matrices of GL2 (Ω). (a)

Check that P1 (Ω) is a GΩ -set for the obvious action of GΩ , and that P1 (Ω)GΩ = P1 (k).

(b)

Show that the pointed GΩ -set of left cosets GL2 (Ω)/B(Ω) is isomorphic to P1 (Ω).

(c)

Show that H 1 (GΩ , B(Ω)) = 1.

(d)

Check that we have an exact sequence of GΩ -groups 1 −→ Ω −→ B(Ω) −→ Ω× × Ω× −→ 1.

(e)

Recover the fact that the group H 1 (GΩ , Ω) is trivial.

IV Galois cohomology of quadratic forms

In this chapter, we define some algebraic group-schemes associated to quadratic forms. We also define some classical invariants of quadratic forms and give a cohomological interpretation of these invariants using Galois cohomology. §IV.10 Algebraic group-schemes associated to quadratic forms IV.10.1 Quadratic forms over rings Throughout this section, R is a commutative ring with unit satisfying 2 ∈ R× . We start with some definitions and results on quadratic forms over rings. We will not need the full theory, so we only define quadratic forms over free R-modules of finite rank. In this setting, the basic results which are well-known when R is a field remain true. We will refer to [53, Chapter 1, § 6] for proofs. A quadratic form over R is a pair (M, ϕ), where M is a free R-module of finite rank and a map ϕ : M → R satisfying the following conditions: (1)

ϕ(λx) = λ2 ϕ(x) for all λ ∈ R, x ∈ M .

(2)

The map bϕ :

M × M −→ R 1 (x, y) −→ (ϕ(x + y) − ϕ(x) − ϕ(y)) 2

is R-bilinear. Notice that bϕ completely determines ϕ, since bϕ (x, x) = ϕ(x). 134

IV.10 Group-schemes associated to quadratic forms

135

Two quadratic spaces (M, ϕ) and (M , ϕ ) are isomorphic if there exists an isomorphism f : M → M of R-modules such that ϕ (f (x)) = ϕ(x) for all x ∈ M . An automorphism f : M → M of R-modules such that ϕ(f (x)) = ϕ(x) for all x ∈ M is called an isometry. We say that (M, ϕ) is regular (or non-singular) if the adjoint map b∗ϕ : x ∈ M → bϕ (x, ·) ∈ M ∗ is an isomorphism of R-modules. Finally, if ρ : R → S is a ring morphism, we may consider the S-module MS = M ⊗R S. We then define ϕS : MS → S to be the quadratic form associated to the bilinear form bϕS (x ⊗ λ, y ⊗ μ) = λμρ(bϕ (x, y)), for all λ, μ ∈ S, x, y ∈ M. It is straightforward to check that ϕ is regular if and only if ϕS is regular for some extension S ⊃ R. If e = (e1 , . . . , en ) is an R-basis of M , the representative matrix of ϕ in this basis is by definition the matrix Mat(ϕ, e) = (bϕ (ei , ej )); if x = x1 e1 +. . .+xn en , then ϕ(x) = X t Mat(ϕ, e)X, where X is the column ⎡ ⎤ x1 ⎢ ⎥ vector ⎣ ... ⎦. If f ∈ GL(V ), then the representative matrix of ϕ with xn respect to the basis f (e) is M t Mat(ϕ, e)M , where M = Mat(f, e). It is easy to check that ϕ is regular if and only if det(Mat(ϕ, e)) ∈ R× . If ϕ is regular, we define the determinant of ϕ to be the square-class det(ϕ) = det(Mat(ϕ, e)) ∈ R× /R×2 . It does not depend on the choice of e, and only depends on the isomorphism class of ϕ. Moreover, for every ring morphism ρ : R → S, we have det(ϕS ) = ρ(det(ϕ)) ∈ S × /S ×2 . Two elements x, y ∈ M are orthogonal with respect to ϕ if bϕ (x, y) = 0. An orthogonal basis is a basis of M consisting of pairwise orthogonal elements. If R is a field, every quadratic form has an orthogonal basis. If (M, ϕ) is a regular quadratic space, the rank of the R-module M is called the dimension of ϕ, and is denoted by dim(ϕ). Notation: We denote by a1 , . . . , an the quadratic form Rn −→ R (x1 , . . . , xn ) −→ a1 x21 + . . . + an x2n .

136

Galois cohomology of quadratic forms

If a ∈ R and n ≥ 0 is an integer, we will denote by n × a then ndimensional quadratic form a, . . . , a. If (M, ϕ) and (M , ϕ ) are two quadratic spaces, we define a quadratic form ϕ ⊥ ϕ on M × M by ϕ ⊥ ϕ ((x, x )) = ϕ(x) + ϕ(x ) for all x ∈ M, x ∈ M . The quadratic space (M × M , ϕ ⊥ ϕ ) is called the orthogonal sum of (M, ϕ) and (M , ϕ ). It is easy to check that the orthogonal sum is an associative operation on the set of quadratic spaces. If n ≥ 0 is an integer, we denote by n × ϕ the orthogonal sum of n copies of ϕ. There is also a unique quadratic form ϕ ⊗ ϕ on M ⊗R M satisfying bϕ⊗ϕ (x ⊗ x , y ⊗ y ) = bϕ (x, y)bϕ (x , y ) for all x, y ∈ M, x , y ∈ M . The quadratic space (M ⊗R M , ϕ ⊗ ϕ ) is called the tensor product of (M, ϕ) and (M , ϕ ). If λ ∈ R, the tensor product of (R, λ) and (M, ϕ) is canonically isomorphic to the quadratic space (M, λϕ), where λϕ is defined by (λϕ)(x) = λϕ(x) for all x ∈ M. If (M, ϕ) and (M , ϕ) are regular and λ ∈ R× , so are the quadratic spaces (M × M , ϕ ⊥ ϕ ), (M ⊗R M , ϕ ⊗ ϕ ) and (M, λϕ).

IV.10.2 Orthogonal groups From now on, k will denote a field of characteristic different from 2, and (V, q) will denote a regular quadratic form of dimension n over k. We define the algebraic group-scheme O(q) as follows: O(q)(R) = {f ∈ GL(V )(R) | qR ◦ f = qR }. Definition IV.10.1. The algebraic group-scheme O(q) is called the orthogonal group of q. From the definition of the orthogonal group O(q), it is easy to see that for every k-algebra R and every f ∈ O(q)(R), we have det(f ) ∈ μ2 (R) (be careful, it does not imply that det(f ) = ±1). We therefore get a morphism of algebraic group-schemes det : O(q) → μ2 .


137

Therefore it is quite natural to consider the kernel of this morphism, that is the group-scheme whose set of R-points is ker(detR ): Definition IV.10.2. The special orthogonal group of q is the algebraic group-scheme O+ (q) defined by O+ (q)(R) = {f ∈ GL(V )(R) | qR ◦ f = qR , det(f ) = 1}. We now define a particular type of isometry, which will play an important role in the sequel. Definition IV.10.3. Let x ∈ V such that q(x) = 0. Let H be the hyperplane H = {x}⊥ = {y ∈ V | bq (x, y) = 0}. Let τx : M → M be the unique k-linear map satisfying τx (x) = −x and τ (y) = y for all y ∈ H. The map τx is called a reflection relative to the hyperplane H. One can check that we have bq (x, v) x for all v ∈ V. τx (v) = v − 2 bq (x, x) Moreover, from the definition, it is clear that τx ∈ O(q)(k) and that det(τx ) = −1. Remark IV.10.4. A vector x ∈ V, x = 0 is called an isotropic vector if q(x) = 0, and is called anisotropic otherwise. A quadratic form is called isotropic if q has at least one isotropic vector, and anisotropic otherwise. If q is isotropic, one can show that we have q q ⊥ 1, −1, for some suitable quadratic form q . We refer the reader to [53] for more details. The following proposition is classical: Proposition IV.10.5. Let (V, q) be a regular quadratic form over k. Then every isometry f is a product of reflections. Proof. We will prove the theorem by induction on n = dim V . If n = 1, then there exists a ∈ k × such that q(x) = ax2 . Therefore, the only isometries of q are ±Id, and the result is clear. Now assume that the result is true for n ≥ 1 and assume that dim V = n + 1. Let f be an isometry of q, and choose x ∈ V such that q(x) = 0, which is possible since q is regular, hence not identically zero.

138


Let us assume first that f (x) = x, and let W = {x}⊥ . Since q(x) = 0, V is the orthogonal sum of kx and W (it comes from the general theory of quadratic forms over fields, or may be proved directly). Then it is easy to check that f (W ) ⊂ W and that f = f|W is an isometry of q|W . Since dim(W ) = n, we may apply the induction hypothesis. Hence we may write f = τw 1 ◦ · · · ◦ τw r , where τw 1 , . . . , τw r are reflections of the quadratic space (W, q|W ). Now we have for all λ ∈ k and all w ∈ W τwi (λx + w) = λτwi (x) + τwi (w) = λx + τw i (w), the last equality coming from the fact that x is orthogonal to every w ∈ W . Hence we get (τw1 ◦ · · · ◦ τwr )(λx + w)

= λx + (τw 1 ◦ · · · ◦ τw r )(w) = λx + f (w) = f (λx + w),

since f is the restriction of f to W and f (x) = x. Hence we get f = τw1 ◦ · · · ◦ τwr and we are done. Let us go back to the general case and set y = f (x). Since f is an isometry, we have q(y) = q(f (x)) = q(x) = 0. Using the bilinearity of bq , it is easy to check that q(x + y) + q(x − y) = 2(q(x) + q(y)). Hence q(x + y) + q(x − y) = 4q(x) = 0, so either q(x + y) or q(x − y) is different from 0. Notice also that x + y and x − y are orthogonal, since we have bq (x + y, x − y) = bq (x, x) − bq (y, y) = q(x) − q(y) = 0. Assume first that q(x + y) = 0. Then we have τx+y (y) =

1 1 (τx+y (x + y) − τx+y (x − y)) = (−(x + y) − (x − y)) = −x 2 2

and thus (τx ◦ τx+y ◦ f )(x) = τx (−x) = x. By the previous case, τx ◦τx+y ◦f is a product of reflections, and therefore so is f . Now if q(x − y) = 0, we have 1 1 (τx−y (x + y) − τx−y (x − y)) = ((x + y) + (x − y)) = x. 2 2 Hence we have τx−y (y) =

(τx−y ◦ f )(x) = τx−y (y) = x and we conclude as before. We end this section with an easy lemma, whose proof is left to the reader:


139

Lemma IV.10.6. We have an exact sequence of Gks -groups 1

/ O+ (q)(ks )

/ O(q)(ks )

det

/ μ2 (ks )

/1.

IV.10.3 Clifford groups and spinors Definition IV.10.7. Let R be a ring, and let M be a free R-module of finite rank. Set T0 (M ) = R, Tn (M ) = M ⊗n if n ≥ 1. The tensor algebra T (M ) is the R-algebra $ T (M ) = Tn (M ), n≥0

where the product is defined on elementary tensors by (x1 ⊗ · · · ⊗ xn )·(y1 ⊗ · · · ⊗ ym ) = x1 ⊗ · · · ⊗ xn ⊗ y1 ⊗ · · · ⊗ ym . Definition IV.10.8. Let (M, ϕ) be a quadratic form over a ring R (2 ∈ R× ) and let I(ϕ) be the two-sided ideal of T (M ) generated by the set {x ⊗ x − ϕ(x) | x ∈ M }. The Clifford algebra of (M, ϕ), denoted by C(M, ϕ) is defined by C(M, ϕ) = T (M )/I(ϕ). The image of a vector x ∈ M under the canonical projection T (M ) → C(M, ϕ) is denoted by x. Clearly, C(M, ϕ) is generated by M = {x | x ∈ M } as an R-algebra. There is a canonical involution t on T (M ) defined on elementary tensors by (x1 ⊗ x2 ⊗ · · · ⊗ xn )t = xn ⊗ · · · ⊗ x2 ⊗ x1 . This involution stabilizes I(ϕ), and therefore induces an involution on C(M, ϕ), still denoted by t . The automorphism M −→ M x −→ −x induces an automorphism on T (M ) which stabilizes I(ϕ). Therefore, it also induces an automorphism γ : C(M, ϕ) → C(M, ϕ).

140


Notice that γ and t commute with scalar extensions. We then get a Z/2Z-grading of C(M, ϕ) by setting C0 (M, ϕ) C1 (M, ϕ)

= {s ∈ C(M, ϕ) | γ(s) = s}; = {s ∈ C(M, ϕ) | γ(s) = −s}.

Let (V, q) be a quadratic form over a field k. The following proposition collects the standard properties of the Clifford algebra. We refer to [53] once again for more details. Proposition IV.10.9. Let (V, q) be a regular quadratic space. (1)

For any k-algebra R, we have a canonical isomorphism C(V, q)R C(VR , qR ) which respects the grading. 1 (x·y + y·x). 2 In particular, if x and y are orthogonal (with respect to q), then x·y = −y·x in C(V, q).

(2)

If x, y ∈ V , then bq (x, y) =

(3)

If q(x) = 0, then x is invertible in C(V, q).

(4)

mn 1 If e1 , . . . , en is an orthogonal basis for q, the elements em 1 · · · en , where mi = 0, 1 for all i, form a k-basis of C(V, q). In particular, the map V → C(V, q) is injective.

In view of (4), we will omit the bar notation from now on. We now define the Clifford group of q. First, we need a lemma: Lemma IV.10.10. Let (V, q) be a quadratic form over a field k, and let f : V → V be an isometry of q. Then there exists an invertible element sf ∈ C(V, q) such that f (x) = γ(sf )xs−1 f for all x ∈ V . Proof. We start with the case of reflections. If f = τx , for some x ∈ V satisfying q(x) = 0, then for all y ∈ V , we have τx (y) = y − 2

bq (x, y) x = y − (xy + yx)x−2 x = −xyx−1 = γ(x)yx−1 . bq (x, x)

Now if f is any isometry of q, then f = τx1 ◦ · · · ◦ τxr by Proposition IV.10.5. Setting sf = x1 · · · xr , the previous case leads immediately to f (y) = γ(sf )ys−1 f for all y ∈ V . We now define a group scheme associated to (V, q).


141

Definition IV.10.11. Let R be a k-algebra. The Clifford group of (V, q) is the affine algebraic group scheme defined by −1 Γ(V, q)(R) = {s ∈ C(V, q)× = VR }. R | γ(s)VR s

For any element s ∈ Γ(V, q)(R), we can define an automorphism αs :

VR −→ VR x −→ γ(s)xs−1 ,

and then we get a morphism αR : Γ(R) → GL(V )(R). We therefore get a group-scheme morphism α : Γ(V, q) → GL(V ). Notice that we have a canonical isomorphism Γ(V, q)(R) Γ(VR , qR ). We now study more carefully the morphism α. We start with the following: Lemma IV.10.12. For every k-algebra R, we have ker(αR ) = R× . Proof. The inclusion R× ⊂ ker(αR ) is clear. Now assume that s ∈ −1 = x for all x ∈ VR , and write s = s0 +s1 , si ∈ C(V, q)× R satisfies γ(s)xs Ci (V, q)R . We get easily s0 x = xs0 s1 x = −xs1 . Let us introduce some notation. Let e1 , . . . , en be an orthogonal basis for q. Hence e1 ⊗ 1R , . . . , en ⊗ 1R is an orthogonal basis for qR . We will still write e1 , . . . , en for short. Notice that ei is invertible, since e2i = q(ei ) = 0 (since q is non-singular). For any subset I of {1, . . . , n}, set eI = ei1 · · · eik , where i1 < · · · < ik are the elements of I if I = ∅, and e∅ = 1. Then it is clear that the family of elements eI where I has an even (resp. odd) number of elements is a k-basis of C0 (V, q)R (resp. C1 (V, q)R ). Hence we may write λI eI . s0 = I,|I| even Let i ∈ {1, . . . , n}. If i ∈ / I, then eI ei = ei eI , since I has even cardinality. If i ∈ I, then ei commutes with itself, and anticommutes with the other

142


ej , j ∈ I − {i}; since I − {i} has odd cardinality, we get eI ei = −ei eI . From the equality s0 ei = ei s0 , we get (after dividing by ei ) λI eI = 0, Ii,|I| even and then λI = 0 for all I containing i. Since any non-empty subset I contains at least one element, the only remaining term is the constant term, so s0 ∈ R. λI eI , and let i ∈ Let us show that s1 = 0. Write s1 = I,|I| odd {1, . . . , n}. If i ∈ / I, then eI ei = −ei eI since I has odd cardinality. If i ∈ I, then ei commutes with itself, and anticommutes with the other ej , j ∈ I − {i}; since I − {i} has even cardinality, we get eI ei = ei eI . From the equality s1 ei = −ei s1 , we get (after dividing by ei ) λI eI = 0, Ii,|I| odd and then λI = 0 for all I containing i. Since any non-empty subset I contains at least one element, we get s1 = 0. Finally, s ∈ R, and since s is invertible, we get s ∈ R× . Definition IV.10.13. Let (M, ϕ) be a quadratic form over a ring R. If s ∈ C(M, ϕ), we set NR (s) = st s ∈ C(M, ϕ). The map NR : C(M, ϕ) → C(M, ϕ) is called the norm of C(M, ϕ). If x ∈ M , we have NR (x) = x2 = ϕ(x). Notice that NR commutes with scalar extensions (since the involution does).

t

Lemma IV.10.14. Let (V, q) be a quadratic form over a field k, and let R be a k-algebra. If s ∈ Γ(V, q)(R), then NR (s) = st s ∈ R× . The norm defines a group-scheme morphism N : Γ(V, q) → Gm , and NR (γ(s)) = NR (s). Proof. If s ∈ Γ(V, q)(R), then γ(s)VR = VR s. Applying the involution t to this equality, we get VR γ(s)t = st VR . It is easy to see from the definitions that γ and t commute, so we get VR γ(st ) = st VR . Applying the automorphism γ to this equality, we get VR st = γ(st )VR . Since st is invertible whenever s is invertible, we obtain that st ∈ Γ(V, q)(R). Hence NR (s) = st s ∈ Γ(V, q)(R).


143

To prove that NR (s) ∈ R× , it is enough to show that αR (st s) = Id by the previous lemma. Notice first that if v ∈ VR , we have γ(v t ) = γ(v) = −v. Now let s ∈ Γ(V, q)(R). Then for all x ∈ VR , we have γ(s)xs−1 ∈ VR and therefore γ(s)xs−1 = −γ((γ(s)xs−1 )t ) = −γ(s−t xγ(st )) = γ(s−t )xst . We then get αst s (x) = γ(st )γ(s)xs−1 s−t = γ(st )γ(s−t )xst s−t = x. Thus α(st s) = Id as claimed, proving that NR (s) ∈ R× . It remains to show the last part. If s, s ∈ Γ(V, q)(R), we have NR (ss ) = (ss )t (ss ) = st st ss . Since NR (s) = st s ∈ R× by the previous point, we get NR (ss ) = st st ss = st sst s = NR (s)NR (s ). Therefore N is a group-scheme morphism. Now for all s ∈ Γ(V, q)(R), we have NR (γ(s)) = γ(s)t γ(s) = γ(st s) = γ(NR (s)) = NR (s), the last equality coming from the fact that γ is the identity on R. Lemma IV.10.15. For all s ∈ Γ(V, q)(R), αs is an isometry of qR . Proof. We have qR (αs (x)) = αs (x)t αs (x), since by definition. Hence

t

is the identity on VR

qR (αs (x)) = s−t xt γ(s)t (γ(s)xs−1 ) = s−t xNR (s)xs−1 . Thus we get qR (αs (x)) = NR (s)s−t x2 s−1 = qR (x)NR (s)NR (s−1 ) = qR (x), since NR (s−1 ) = NR (s)−1 . Hence we get a group-scheme morphism α : Γ(V, q) → O(q) with kernel Gm . We then have the following result.

144


Theorem IV.10.16. For any field extension K/k, we have an exact sequence 1

/ Γ(V, q)(K)

/ K×

αK

/ O(q)(K)

/1.

Moreover, for any x ∈ V satisfying q(x) = 0, the preimages of τx in Γ(V, q)(K) are the elements λx, λ ∈ K × . Proof. Let us prove that αK is surjective. By Lemma IV.10.10 and its proof, we have αK (x) = τx . Since O(q)(K) is generated by reflections, the surjectivity follows. The equality ker(αK ) = K × is a particular case of Lemma IV.10.12. The last part of the theorem is then clear. If K/k is a field extension, it follows from the definition of the norm that we have NK (λ) = λ2 for all λ ∈ K × . Therefore the previous theorem implies that the norm induces a morphism SNK : O(q)(K) → K × /K ×2 satisfying SNK (τx ) = q(x) ∈ K × /K ×2 for all x ∈ V such that q(x) = 0. Definition IV.10.17. The morphism SNK is called the spinor norm. We now define another group-scheme associated to q. Definition IV.10.18. We denote by Pin(q) the algebraic group-scheme defined by Pin(q)(R) = {x ∈ Γ(V, q)(R) | NR (x) = 1}. Corollary IV.10.19. We have an exact sequence of Gks -groups 1

/ μ2 (ks )

/ Pin(q)(ks )

αks

/ O(q)(ks )

/1,

where the map μ2 (ks ) −→ Pin(q)(ks ) is the inclusion. If x ∈ Vks satis1 fies qks (x) = 0, the preimages of τx in Pin(q)(ks ) are ± % x. q(x) Proof. If s ∈ Pin(q)(ks ) satisfies αks (s) = 1, then s ∈ ks× by Lemma IV.10.12. But we have in this case 1 = NKs (s) = s2 , so s = ±1. Now let x ∈ V and let f ∈ O(q)(ks ) be an isometry of qks . By the computations −1 for all made in the proof of Lemma IV.10.10, we have τx (y) = γ(x)yx 1 y ∈ Vks . We have Nks (x) = x2 = qks (x), so Nks % x = 1 and q(x)

IV.11 Galois cohomology of quadratic forms %

1 q(x)

145

x ∈ Pin(q)(ks ). Moreover, we have %

τx (y) = γ

%

1

1

x y q(x)

%

−1

1 q(x)

for all y ∈ Vks .

x

x . It follows that %

1

x is a preimage of q(x) q(x) τx in Pin(q)ks , and since the kernel of αks is {±1}, there is only one 1 other preimage, which is − % x. Surjectivity follows from the fact q(x) that O(q)(ks ) is generated by reflections (cf. Proposition IV.10.5). The fact that the maps preserve the Gks -actions is left to the reader. Hence τx = αks

Notation: If q = 1, . . . , 1, we denote O(q) and Pin(q) simply by On and Pinn respectively. Definition IV.10.20. The inverse image of O+ (q) in Γ(V, q) by the morphism α is called the special Clifford group of q, and is denoted by Γ+ (V, q). The spinor group of q is the algebraic group-scheme Spin(q) = Pin(q) ∩ Γ+ (V, q). If q = 1, . . . , 1, we simply denote it by Spinn . The next lemma directly follows from the previous corollary and the definition of the spinor group. Lemma IV.10.21. We have an exact sequence of Gks -groups 1

/ μ2 (ks )

/ Spin(q)(ks )

αks

/ O+ (q)(ks )

/1.

§IV.11 Galois cohomology of quadratic forms IV.11.1 Galois cohomology of orthogonal groups Let V be a k-vector space of dimension n. For every field extension K/k, denote by F(K) the set of quadratic forms on VK . If ι : K → K is a morphism of field extensions, let F(ι) :

F(K) −→ F(K ) q −→ qK

146


be the corresponding scalar extension map. We then obtain a functor F : Ck −→ Sets, on which the functor GL(V ) acts as follows: for every field extension K/k, every f ∈ GL(V )(K) and every q ∈ F(K), set f ·q = q ◦ f −1 . Now let (V, q) be a regular quadratic form on V , and let Quadn (L) be the pointed set of isomorphism classes of regular quadratic forms on VL , the base point being the isomorphism class of qL . We then obtain a functor Quadn : Ck → Sets∗ . By definition of the action of GL(V ) on F, it is clear that the stabilizer of q is O(q). Since all regular quadratic forms are isomorphic over a separably closed field, the functor of twisted forms of q is Quadn , and an immediate application of the Galois descent lemma gives: Proposition IV.11.1. We have an isomorphism of functors from Ck to Sets∗ Quadn H 1 (− , O(q)). We continue with a cohomological interpretation of the determinant. Proposition IV.11.2. The map det∗ : H 1 (k, O(q)) → H 1 (k, μ2 ) sends the isomorphism class of a quadratic form q onto det(q) det(q )−1 . Proof. Let f ∈ GL(Vks ) such that f ·qk s = qks . A cocycle corresponding to q is given by α:

Gks −→ O(q)(ks ) σ −→ f ◦ σ·f −1 .

Let B, B denote the representative matrices of q, q in a fixed basis of V . By choice of f , we have qk s = qks ◦ f, that is qk s (x) = qks (f (x)) for all x ∈ Vks . From this equality, we get det(B ) = det(B) det(f )2 . Hence (det(f )−1 )2 = det(B) det(B )−1 . Now a cocycle representing det∗ ([α]) is Gks −→ μ2 (ks ) σ −→ det(ασ ) =

σ(det(f )−1 ) . det(f )−1

IV.11 Galois cohomology of quadratic forms

147

This cocycle corresponds to the square-class of det(B) det(B )−1 by Remark III.8.28, which is also the square-class of det(q) det(q )−1 . Corollary IV.11.3. The pointed set H 1 (k, O+ (q)) is in 1-1 correspondence with the set of isomorphism classes of quadratic forms (V, q ) satisfying det(q ) = det(q). Proof. The image of H 1 (k, O+ (q)) −→ H 1 (k, O(q)) is equal to the kernel of det∗ , so it is equal to the set of isomorphism classes of quadratic forms (V, q ) satisfying det(q ) = det(q). Now we want to prove that the map H 1 (k, O+ (q)) −→ H 1 (k, O(q)) is injective. Let [α] ∈ H 1 (k, O+ (q)) and let [β] ∈ H 1 (k, O(q)) be its image via the map induced by the inclusion O+ (q)(ks ) ⊂ O(q)(ks ). By Lemma II.5.5, it is enough to show that the kernel of H 1 (k, O+ (q)α ) −→ H 1 (k, O(q)β ) is trivial for every [α]. Now the sequence 1 −→ O+ (q)(ks )α −→ O(q)(ks )β −→ μ2 (ks ) −→ 1 is easily seen to be exact, so we have an exact sequence μ2 (k) −→ H 1 (k, O+ (q)(ks )α ) −→ H 1 (k, O(q)(ks )β ). Let us compute the connecting map δ 0 : μ2 (k) −→ H 1 (k, O+ (q)(ks )α ). A preimage of ε ∈ μ2 (ks ) in O(q)(ks ) is εId. Now we have σ ∗ εId = βσ (σ·εId)βσ−1 = εId, since ε ∈ k. Hence δ 0 (ε) = 1, so δ 0 is trivial. Therefore the map H 1 (k, O+ (q)(ks )α ) −→ H 1 (k, O(q)(ks )β ) has trivial kernel, which concludes the proof.

IV.11.2 Galois cohomology of spinors IV.11.2.1 The Hasse invariant of a quadratic form

In this section, we define a cohomology class of degree 2 attached to a quadratic form q over k. Assume that q a1 , . . . , an . We set w2 (q) = (ai ) ∪ (aj ) ∈ H 2 (k, μ2 ). 1≤i<j≤n

One can show that w2 (q) does not depend on the choice of a diagonalization of q, but only depends on the isomorphism class of q. This follows

148


from [53, Chapter 2, Remark 12.5] and the fact that the map k × × k × −→ H 2 (k, μ2 ) (a, b) −→ (a) ∪ (b) is a Steinberg symbol (see [53, Chapter 2, § 12] for a definition). Definition IV.11.4. The cohomology class w2 (q) is called the Hasse invariant of q. The following lemma may be proved by direct computation. Lemma IV.11.5. Let q, q be two quadratic forms. Then we have w2 (q ⊥ q ) = w2 (q) + (det(q)) ∪ (det(q )) + w2 (q ) and w2 (λq) = w2 (q) + (n − 1)(λ) ∪ (det(q)) +

n(n − 1) (−1) ∪ (λ), 2

where n = dim(q). The notation w2 (q), which may appear strange at first, is justified by the fact that one may define a family of cohomology classes (wm (q))m≥0 , called Stiefel-Whitney classes. We refer to [18] for more information. IV.11.2.2 Galois cohomology of spinor groups

We now give a cohomological interpretation of the Hasse invariant. Theorem IV.11.6. The first connecting map associated to the exact sequence 1

/ μ2 (ks )

/ Pin(q)(ks )

αks

/ O(q)(ks )

/1

is given by δ 1 (q ) = w2 (q) + w2 (q ) + (det(q)) ∪ (− det(q )). Proof. Let (V, q ) be a quadratic form on V . Let e = (e1 , . . . , en ), e = (e1 , . . . , en ) be two orthogonal bases of V with respect to q and q respectively, and set ai = q(ei ), bi = q (ei ), ci = bi a−1 i . In order to simplify notation, we will still denote by e1 , . . . , en and


149

e1 , . . . , en the corresponding orthogonal bases of Vks with respect to qks and qk s respectively. We will denote by τi ∈ O(q)(ks ) the reflection τei . Finally, write √ √ σ( ai ) σ( ci ) = (−1)si (σ) and = (−1)εi (σ) for all σ ∈ Gks . ci ai We first show that a cocycle ξ : Gks −→ O(q)(ks ) representing q is given by s (σ)

ξσ = τ1 1

◦ · · · ◦ τnsn (σ) for all σ ∈ Gks .

The map f:

Vks −→ Vks √ ei −→ ci ei

is easily seen to satisfy f ·qk s = qks . Then a cocycle representing q is ξ:

Gks −→ O(q)(ks ) σ −→ ξσ = f ◦ σ·f −1 .

For i = 1, . . . , n, we have ξσ (ei )

= f (σ·(f −1 (σ −1 ·ei ))) −1 (ei )) = f (σ·f 1 = f σ· √ ei ci 1 = f (−1)−si (σ) √ ei ci = (−1)si (σ) ei .

The fact that τi (ei ) = −ei and that the ej ’s are mutually orthogonal lead easily to the desired conclusion. We now compute a cocycle representing δ 1 (q ). By Corollary IV.10.19, a preimage of ξσ in Pin(q)(ks ) is given by 1 s (σ) e11 · · · esnn (σ) , ξ˜σ = √ s (σ) √ 1 a1 · · · an sn (σ) and we have

&

(−1) i

εi (σ)si (τ )

s1 (τ ) σ· ξ˜τ = √ s (τ ) · · · esnn (τ ) . √ sn (τ ) e1 1 a1 · · · an

150


Taking into account that ei and ej anticommute if i = j, we get that &

εi (σ)si (τ )+

&

si (σ)sj (τ )

(−1) s1 (σ)+s1 (τ ) · · · ensn (σ)+sn (τ ) . ξ˜σ σ· ξ˜τ = √ s (σ)+s (τ ) √ sn (σ)+sn (τ ) e1 1 1 a1 · · · an i

i<j

Now recall that the map Gks −→ μ2 (ks ), σ → (−1)si (σ) is a cocycle with values in μ2 (ks ). Since μ2 (ks ) is abelian and Gks acts trivially on it, it follows that this map is a group morphism. Therefore, we have si (σ) + si (τ ) − si (στ ) ∈ 2Z 1 (i) for all σ, τ ∈ Gks . Set mστ = (si (σ) + si (τ ) − si (στ )). 2 Then we have s (σ)+si (τ )

ei i

2m(i) σ,τ +si (στ )

m(i) σ,τ si (στ ) ei .

= ei

= ai

(i)

Since si (σ) + si (τ ) = 2mσ,τ + si (στ ), we have √

ai

and thus

si (σ)+si (τ )

&

√ m(i) σ,τ

= ai &

εi (σ)si (τ )+

ai

i<j

that is

& −1 = (−1) i ξ˜σ σ· ξ˜τ ξ˜στ

,

si (σ)sj (τ )

(−1) ξ˜σ σ· ξ˜τ = √ s (στ ) √ a1 1 · · · an sn (στ ) i

si (στ )

s (στ )

e11

εi (σ)si (τ )+

&

· · · ensn (στ ) ,

si (σ)sj (τ )

.

i<j

Finally, we get that δ 1 (q ) is represented by the cocycle &

Gks × Gks −→ μ2 (ks ), (σ, τ ) → (−1) Hence δ 1 (q ) =

(ai ) ∪ (bi a−1 i )+

i

εi (σ)si (τ )+

i

&

si (σ)sj (τ )

.

i<j

−1 (bi a−1 i ) ∪ (bj aj ). i<j

Taking into account that we work up to squares, we obtain δ 1 (q ) = (ai ) ∪ (ai bi ) + (ai bi ) ∪ (aj bj ). i

i<j

Using the bilinearity and the commutativity of the cup-product, we get δ 1 (q ) = (ai ) ∪ (ai ) + (ai ) ∪ (aj ) + (bi ) ∪ (bj ) + (ai ) ∪ (bj ). i

i<j

i<j

i,j


151

Now we have (ai ) ∪ (ai ) = (ai ) ∪ (−1) = (a1 · · · an ) ∪ (−1) = (det(q)) ∪ (−1). i

i

Morever we have (ai ) ∪ (bj ) =

⎞ ⎛ (ai ) ∪ ⎝ (bj )⎠ = (det(q)) ∪ (det(q )).

i,j

i

j

Hence, using bilinearity again, we finally get δ 1 (q ) = w2 (q) + w2 (q ) + (det(q)) ∪ (− det(q )). This completes the proof of the theorem. This result was originally proved by Springer in [63]. Notice however that Springer’s definition of the Hasse invariant is slightly different, and that the minus sign is missing in his formula. Applying the previous theorem to q = 1, . . . , 1, we get: Corollary IV.11.7. The connecting map δ 1 : H 1 (k, On ) −→ H 2 (k, μ2 ) associated to the exact sequence 1

/ μ2 (ks )

/ Pinn (ks )

αks

/ On (ks )

/1

is given by δ 1 (q) = w2 (q). Remark IV.11.8. The previous corollary provides another way to show that the Hasse invariant is well-defined, since we know that δ 1 (q) only depends on the isomorphism class of q. Remark IV.11.9. For every morphism K −→ L of field extensions of k, the diagram 1

/ μ2 (Ks )

/ Pinn (Ks )

/ On (Ks )

/1

1

/ μ2 (Ls )

/ Pinn (Ls )

/ On (Ls )

/1

is commutative. Hence Theorem III.7.39 (5) and Corollary IV.11.7 imply that we have w2 (qL ) = ResL/K (w2 (q)), for every quadratic form q over K.

152


Corollary IV.11.10. The first connecting map associated to the exact sequence 1

/ μ2 (ks )

/ Spin(q)(ks )

αks

/ O+ (q)(ks )

/1

is given by δ 1 (q ) = w2 (q) + w2 (q ). Proof. We have a commutative diagram 1

/ μ2 (ks )

/ Spin(q)(ks )

1

/ μ2 (ks )

/ Pin(q)(ks )

αks

αks

/ O+ (q)(ks )

/1

/ O(q)(ks )

/1

which induces a commutative square in cohomology H 1 (k, O+ (q))

/ H 2 (k, μ2 )

H 1 (k, O(q))

/ H 2 (k, μ2 )

by Theorem III.7.39 (4). Now it suffices to apply the previous theorem and the fact that the pointed set H 1 (k, O+ (q)) classifies quadratic forms with same dimension and determinant as q (cf. Corollary IV.11.3), taking into account that (a) ∪ (−a) = 0 for all a ∈ k× . §IV.12 Cohomological invariants of quadratic forms IV.12.1 Classification of quadratic forms over Q We would like now to give a complete classification result of quadratic forms over Q in terms of invariants. Before doing this, we need to define the signature of a quadratic form. Lemma IV.12.1. Let (V, q) be a (regular) quadratic space over R. Then there exist unique integers r, s ≥ 0 such that q r × 1 ⊥ s × −1. Proof. We know that q a1 , · · · , an , for some ai ∈ R× . Now each ai is congruent to ±1 modulo R×2 . This proves the existence part. Now assume that we have q r × 1 ⊥ s × −1 r × 1 ⊥ s × −1.

IV.12 Cohomological invariants of quadratic forms

153

Let e1 , . . . , er , er+1 , . . . , en be the orthogonal basis corresponding to the first diagonalization, and let e1 , . . . , er , er +1 , . . . , en be the orthogonal basis corresponding to the second one. Let V+ be the subspace of V generated by e1 , . . . , er and let V− be the subspace of V generated by er +1 , . . . , en . By definition, we have dim V+ = r and dim V− = n − r . It is easy to check that we have q(x) > 0 for all x ∈ V+ , x = 0 and that q(x) < 0 for all x ∈ V− , x = 0. In particular, V+ ∩ V− = {0}. Now assume that r = r , that is for example r > r . In this case, we would have dim V+ + dim V− = r + n − r > n, and therefore V+ ∩V− = {0}, a contradiction. Hence r = r and therefore s = s since r + s = r + s = n. Definition IV.12.2. Let q be a quadratic form over R, and write q r × 1 ⊥ s × −1. The signature of q is the integer r − s ∈ Z, and is denoted by sign(q). If q is a quadratic form over Q, the signature of q is by definition the signature of qR , and is still denoted by sign(q). We say that q is indefinite if we have |sign(q)| < dim(q). Remark IV.12.3. Notice that if q q , then qR qR , and by uniqueness of the integers r and s defined in the previous theorem, we get sign(q) = sign(q ). In particular, the signature only depends on the isomorphism class of a given quadratic form. Remark IV.12.4. Given two quadratic forms q, q over Q or R, it is straightforward to check that we have sign(−q) = −sign(q), sign(q ⊥ q ) = sign(q) + sign(q ). We are now ready to state our classification theorem. Theorem IV.12.5. Two quadratic forms over Q are isomorphic if and only if they have same dimension, determinant, Hasse invariant and signature.

154


We refer the reader to [53, Chapter 6, Corollary 6.6] for a proof. Notice that in [53] the Hasse invariant is defined in terms of Brauer classes of quaternion algebras. However, in view of the group isomorphism Br2 (k) H 2 (k, μ2 ) for every field k, which maps the Brauer class of (a, b) onto (a) ∪ (b), we see that the theorem above is an equivalent formulation of the result proved in [53].

IV.12.2 Higher cohomological invariants We now would like to briefly mention the existence of other invariants of quadratic forms with values in H d (− , μ2 ). Definition IV.12.6. A cohomological invariant of degree d of F : Ck −→ Sets with values in μ2 is a natural transformation ι : F −→ H d (− , μ2 ) of functors from Ck to Sets. A cohomological invariant ι of degree d is constant if there exists [α] ∈ H d (k, μ2 ) such that for all K/k and all [ξ] ∈ H 1 (K, G) we have ιK ([ξ]) = ResK/k ([α]). ∗

If F : Ck −→ Sets , a cohomological invariant ι : F −→ H d (− , μ2 ) is normalized if, for every field extension K/k, ιK maps the base point of F(K) onto 0. Example IV.12.7. The Hasse invariant is a cohomological invariant of degree 2 with values in μ2 , in view of Remark IV.11.9. If K/k is a field extension, we define an equivalence relation on the set of regular quadratic spaces as follows: if q, q are two regular quadratic forms on K, we say that q and q are Witt equivalent if there exist two integers r, r ≥ 0 such that q ⊥ r × 1, −1 q ⊥ r × 1, −1. We denote it by q ∼ q . We denote by [q] the corresponding equivalence class. The set of Witt equivalence classes is denoted by W (K). One can show that the operations [q] + [q ] = [q ⊥ q ], [q]·[q ] = [q ⊗ q ] and − [q] = [−q] are well-defined, and induce a structure of a commutative ring on W (K), with neutral elements [0] and [1] (see [53], for example).


155

It is then easy to check that the map e0 :

W (K) −→ Z/2Z [q] −→ dim(q)

is a well-defined ring morphism. Its kernel I(K) is called the fundamental ideal of W (K). In other words, we have I(K) = {[q] ∈ W (K) | dim(q) is even }. If n ≥ 1 is an integer, and a1 , . . . , an ∈ K × , we set a1 , . . . , an = 1, −a1 ⊗ · · · ⊗ 1, −an . Such a quadratic form is called a n-fold Pfister form. Since [a, b] = [−a] − [b], I(K) is additively generated by the classes of 1-fold Pfister forms, and its nth -power I n (K) is additively generated by the classes of n-fold Pfister forms for all n ≥ 1. By convention, we will set I 0 (K) = W (K). If K → L is a morphism of field extensions of k, the maps W (K) −→ W (L), [q] → [qL ] and I n (K) −→ I n (L), [q] → [qL ] are respectively welldefined ring and group morphisms, and we get functors W (− ) : Ck −→ Rings and I n (− ) : Ck −→ Abgrps. Let K/k be a field extension. For n ≥ 1, and for a1 , . . . , an ∈ K × , we set en ([a1 , . . . , an ]) = (a1 ) ∪ · · · ∪ (an ) ∈ H n (K, μ2 ). Theorem IV.12.8. For all n ≥ 1, the map en extends in a unique way to a group morphism en : I n (K) −→ H n (K, μ2 ) and induces a group isomorphism I n (K)/I n+1 (K) H n (K, μ2 ). This is an easy exercise for n = 1 and is well-known for n = 2 (see [53, Chapter 2, § 12] for more details). The case n = 3 has been proved by Arason in [1], and by Jacob and Rost in [27] for n = 4. For arbitrary n, this follows from the work of Voevodsky et al. (see [45], [66] and [43]). The proof of this theorem is extremely difficult and earned Voevodsky a Fields medal in 2002. In the next chapters, we will only need the existence of en for n = 0, 1, 2 and 3.

156


Remark IV.12.9. This theorem says in particular that we have en (λq) = en (q) for all λ ∈ K × , q ∈ I n (K). Remark IV.12.10. The group morphisms en are easily seen to be compatible with scalar extension maps. In other words, we get comohological invariants en : I n (− ) −→ H n (− , μ2 ). The isomorphism described in Theorem IV.12.8 shows that a class [q] ∈ W (k) lies in I n (k) if and only if ej ([q]) = 0 for j = 0, . . . , n − 1. This was proved for n = 3 by Merkurjev in [39].

I n (k) = (0) (see [53] p.156), we get the following general clasSince n≥1

sification theorem for quadratic forms: Theorem IV.12.11. Let k be a field of characteristic different from 2, and let q, q be two regular quadratic forms over k. Then q q if and only if en ([q ⊥ −q ]) = 0 for all n ≥ 0. We end this section by some elementary results on the invariants en . Lemma IV.12.12. Let k be a field of characteristic different from 2, and let n, m ≥ 0 be two integers. If [q] ∈ I n (k) and [q ] ∈ I m (k), then [q] · [q ] ∈ I n+m (k) and we have en+m ([q] · [q ]) = en ([q]) ∪ em ([q ]). Proof. Since en , em and en+m are group morphisms, a distributivity argument shows that it is enough to prove it when q and q are Pfister forms. The result being clear in this case, we get the desired result. Lemma IV.12.13. Let q be a regular quadratic form of dimension 2m. Then we have e1 ([q]) = ((−1)m det(q)) ∈ H 1 (k, μ2 ). Proof. Write q a1 , b1 , . . . , am , bm . For all a, b ∈ k × , we have [a, b] = [−a] − [b]. The result follows easily. Remark IV.12.14. The invariant en may be difficult to compute for n ≥ 2, since an explicit decomposition of [q] into sum of classes of the


157

form ±[a1 , . . . , an ] is needed. However for n = 2, this is not necessary. Indeed, one may define another invariant attached to a quadratic form q as follows: if n = dim(q), set ⎧ 0 ⎪ ⎪ ⎨ (−1) ∪ (− det(q)) c(q) = w2 (q) + ⎪ (−1) ∪ (−1) ⎪ ⎩ (−1) ∪ (det(q))

if if if if

n ≡ 1, 2 mod 8 n ≡ 3, 4 mod 8 n ≡ 5, 6 mod 8 n ≡ 7, 8 mod 8.

The invariant c(q) is called the Clifford invariant (or Witt invariant) of q. One can show that c(q) only depends on the Witt class of q, and that its restriction to I 2 (k) induces a group morphism c : I 2 (k) −→ H 2 (k, μ2 ). We refer to [53, Chapter 2, p.81] for more details. Easy computations show that we have c([a, b]) = (a) ∪ (b). Hence the restriction of c to I 2 is nothing but the invariant e2 . Let L/k be a finite field extension and let s : L −→ k be a non-zero k-linear map. If z ∈ L, recall that NL/k (z) is the determinant of the endomorphism of k-vector space z :

L −→ L x −→ zx.

If (V, q) is a regular quadratic space, the map s∗ (q) :

V −→ k x −→ s(q(x))

is a regular quadratic form over k (where V is viewed as a k-vector space). Moreover, this induces a well-defined group morphism s∗ :

W (L) −→ W (k) [q] −→ [s∗ (q)].

See [53] for more details for example. We can now state the next result. Lemma IV.12.15. Let L/k be a finite extension, and let z ∈ L× . Then we have e1 (s∗ (1, −z)) = (NL/k (z)) ∈ H 1 (k, μ2 ).

158


Proof. Let v1 , . . . , vn be a k-basis of L, and let B = (s(vi vj )) the corresponding representative matrix of s∗ (1). Let M = (mij ) be the matrix of z in the basis v1 , . . . , vn . In other words, we have zvj =

n

mij vi .

i=1

Notice that by definition, we have det(M ) = NL/k (z). We have s(zvi vj ) =

n

mri s(vr vj ),

r=1

so the representative matrix of s∗ (z) is M t B. Hence, we get det(s∗ (z)) = det(M ) det(B) = NL/k (z) det(s∗ (1)). If [L : k] = m, we get det(s∗ (1, −z))

= det(s∗ (1) ⊥ −s∗ (z)) = (−1)m det(s∗ (1)) det(s∗ (z)) = (−1)m NL/k (z).

Using Lemma IV.12.13, we get the desired result. Corollary IV.12.16. Let [q] ∈ I n (k), and let L/k be a finite field extension such that [qL ] = 0 ∈ W (L). Then for all z ∈ L× , we have (NL/k (z)) ∪ en (q) = 0 ∈ H n+1 (k, μ2 ). Proof. Let s : L −→ k be any non-zero linear map. Since [qL ] = 0, we have [1, −z⊗qL ] = [1, −z]·[qL ] = 0, and therefore [s∗ (1, −z⊗qL )] = 0. It is easy to check using definitions that we have s∗ (1, −z ⊗ qL ) s∗ (1, −z) ⊗ q. Therefore, we get en+1 ([s∗ (1, −z)] · [q]) = 0. Using Lemma IV.12.12 and Lemma IV.12.15, we get the desired result.

Exercises 1. Let q be a quadratic form over k. Show that the connecting map δ 0 : O+ (q)(k) −→ H 1 (k, μ2 ) associated to the exact sequence 1 −→ μ2 (ks ) −→ Spin(q)(ks ) −→ O+ (q)(ks ) −→ 1 is given by the spinor norm SN .

Exercises

159

2. Let q be a quadratic form over k. (a)

Show that H 1 (k, Spin(q)) = 1 if and only if the spinor norm is surjective and every quadratic form q satisfying dim(q ) = dim(q), det(q ) = det(q) and w2 (q ) = w2 (q) is isomorphic to q.

(b)

(c)

Show that the spinor norm is surjective in the following cases: (i)

q is universal, i.e. q represents every element of k ×

(ii)

q represents 0.

Assume that the spinor norm is surjective. Show that the map H 1 (k, Spin(q)) −→ H 1 (k, O+ (q)) has trivial kernel, but is not injective in general.

3. Prove Lemma IV.11.5. 4. Let r ∈ Q+∗ . By comparing their invariants, show that we have an isomorphism r, r, r, r 1, 1, 1, 1, and deduce that every positive rational number is a sum of 4 squares in Q. 5. Let (V, q) be a quadratic space. We keep the definitions of Chapter II, Exercise 7. (a)

For a ∈ k× , show that Xa = {x ∈ Γ+ (V, q)(ks ) | Nks (x) = a} is a principal homogeneous space over Spin(q)(ks ), whose isomorphism class only depends on the square-class of a.

(b)

Show that Xa Xb if and only if b ≡ a mod k ×2 .

(c)

Show that the map H 1 (k, μ2 ) −→ H 1 (k, Spin(q)) sends (a) onto the isomorphism class of Xa .

6. Describe H 1 (k, Spin(q)) in terms of twisted forms of tensors. 7. Describe the map H 1 (k, μn2 ) −→ H 1 (k, On ) induced by the natural injection μ2 (ks )n → On (ks ).

V ´ Etale and Galois algebras

´ §V.13 Etale algebras The goal of this section is to give an interpretation of separable field extensions in terms of Galois cohomology. First of all, notice that if E/k is a separable field extension, and if L/k is a field extension, EL is not necessarily a field. For example, if k = R, E = L = C, we have EL C × C. Hence, the collection of separable extensions is not a functor, so we have no chance to classify them using Galois cohomology. To do so, we need a broader class of algebras. The following result is really classical, and is proven in [8]. Theorem V.13.1. For a finite dimensional commutative k-algebra E, the following conditions are equivalent: (1)

E E1 ×· · ·×Er , where E1 , . . . , Er are finite separable extensions of k.

(2)

Eks ks × · · · × ks .

(3)

|X(E)| = dimk E, where X(E) = Homk−alg (E, ks ). If k is infinite, the conditions above are also equivalent to:

(4)

E k[X]/(f ) for some separable polynomial f ∈ k[X].

Definition V.13.2. A finite dimensional commutative k-algebra satisfying the equivalent conditions above is called ´ etale. Examples V.13.3. (1)

The k-algebra kn is an étale algebra, called the split étale algebra.

(2)

A finite separable extension E/k is an étale algebra. 160

´ V.13 Etale algebras

161

It follows directly from the definition that if E is étale over K and K −→ L is a morphism of field extensions, then EL is an étale algebra over L. ´ n : Ck −→ Sets∗ the functor of isomorphism classes We denote by Et of étale algebras of dimension n, the base point being the isomorphism class of the split étale algebra. Notice the twisted K-forms of K n are by definition étale K-algebras of dimension n, so by Galois descent, étale algebras of dimension n are classified by H 1 (− , Autalg (k n )). Lemma V.13.4. For any field K, we have an isomorphism of abstract groups AutK−alg (K n ) Sn . Proof. If f : K n −→ K n is an automorphism, then it maps an idempotent to an idempotent. Denote by ei the element of K n defined by ei = (δij ) 1≤j≤n , where δij is the Kronecker symbol. For I ⊂ {1, . . . , n}, let eI = ei (so e∅ = 0). It is clear that the idempotents of K n are i∈I

the eI ’s. From the relation ei ej = δij ei , it follows that for two subsets I, J of {1, . . . , n}, we have eI eJ = eI∩J . Now let f (ei ) = eIi . Each Ii is non-empty since f is injective and f (0) = 0. Moreover, the previous relation implies that the subsets Ii ⊂ {1, . . . , n} are pairwise disjoint. Hence I1 , . . . , In is a partition of {1, . . . , n} and each Ii contains exactly one element, so f permutes the ei ’s. Conversely, any permutation of the ei ’s defines a K-algebra automorphism of K n . It is then easy to check that we get a group isomorphism Sn AutK−alg (K n ). Now we have Autalg (k n )(Ks ) = AutKs −alg (Ksn ) Sn ; the reader will easily check that GKs acts trivially on AutKs −alg (Ksn ), so using Proposition III.9.1 we deduce: Proposition V.13.5. We have an isomorphism of functors from Ck to Sets∗ ´ n H 1 (− , Sn ), Et where Sn is considered as a trivial GKs -group for every field extension K/k. We now take a closer look at this correspondence. By Remark III.9.2, a cocycle α whose class in H 1 (k, Sn ) corresponds to a given étale k-algebra

´ Etale and Galois algebras

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E may be constructed as follows: take any isomorphism of ks -algebras ∼ f : Eks −→ ksn , and for every σ ∈ Gks , set ασ = f ◦ σ·f −1 . However, we are going to see that in the case of étale k-algebras, we have a natural choice for f . For any χ ∈ X(E), we denote by χ : Eks −→ ks the morphism of ks -algebras defined by χ (u ⊗ λ) = χ(u)λ for all λ ∈ ks and all u ∈ E. Notice that the absolute Galois group Gks acts naturally on X(E) by composition on the left. Lemma V.13.6. For all χ ∈ X(E) and all σ ∈ Gks , we have σ·χ (σ·x) = σ(χ (x)) for all x ∈ Eks . Proof. It is enough to prove the equality for x = u ⊗ λ, u ∈ E, λ ∈ ks . The equality follows in this case by direct computation. X(E)

Lemma V.13.7. Let E be an étale k-algebra and f : Eks −→ ks the canonical map defined by f (x) = (χ (x))χ∈X(E) for all x ∈ Eks . Then f is an isomorphism of ks -algebras.

Proof. It is clear that f is an morphism of ks -algebras. In order to show that f is a bijection, it is enough to prove that f is injective. ( ker(χ ). It is well-known that this intersection is Now ker(f ) = χ∈X(E)

reduced to 0 if and only if the linear forms χ , χ ∈ X(E) span the dual space Ek∗s of ks -linear forms on Eks . Since we have dimks (Ek∗s ) = dimks (Eks ) = dimk (E) = |X(E)|, this is equivalent to saying that the linear forms χ , χ ∈ X(E) are linearly independent over ks . Assume the contrary and consider a dependence relation of minimal length r ≥ 1: a1 χ1 + . . . + ar χr = 0, where χ1 , . . . , χr are distinct elements of X(E) and a1 , . . . , ar are non-zero elements of ks . Then we have necessarily r ≥ 2. Let u0 ∈ E such that χ1 (u0 ) = χr (u0 ) (such a u0 exists since χ1 = χr ). Since the

´ V.13 Etale algebras

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χi ’s are algebra morphisms, applying (u0 ⊗ 1)x for all x ∈ Eks to this relation gives a1 χ1 (u0 )χ1 + . . . + ar χr (u0 )χr = 0. Multiplying by χr (u0 ) each side of the initial dependence relation then yields a1 χr (u0 )χ1 + . . . + ar χr (u0 )χr = 0. Subtracting the two relations gives a1 (χ1 (u0 ) − χr (u0 ))χ2 + . . . + ar−1 (χr−1 (u0 ) − χr (u0 ))χr−1 = 0. We then produce another dependence relation of length < r, which contradicts the minimality of r. Proposition V.13.8. Let E be an étale k-algebra, dimk (E) = n, and let ϕ : Gks −→ Sn be a continuous group morphism representing E. Let N ⊂ ks be the compositum of the subfields χ(E) of ks , χ ∈ X(E). Then N/k is a finite Galois extension and we have ker(ϕ) = Gal(ks /N ). In particular, Im(ϕ) GN . Proof. Let us prove the first part. Assume that E = E1 × · · · × Er , where Ei is a finite separable field extension of k. Clearly the map θ : X(E1 ) × . . . × X(Er ) −→ X(E) defined by θ(χ1 , . . . , χr )(x1 , . . . , xr ) = (χ1 (x1 ), . . . , χr (xr )) for all xi ∈ Ei is bijective, with inverse map X(E) −→ X(E1 ) × . . . × X(Er ) χ −→ (ι1 ◦ χ, . . . , ιr ◦ χ), where ιi : Ei −→ E is the canonical injection. Let Ni be the compositum of all the subfields χ (Ei ), where χ describes X(Ei ). Since Ei /k is a separable field extension, Ni is just the Galois closure of Ei in ks . It then follows from the description of X(E) that the compositum N of the subfields χ(E), χ ∈ X(E) is the compositum of N1 , . . . , Nr . Since the compositum of Galois extensions of k is a Galois extension of k, we are done. Now set X = X(E). Let f : Eks −→ ksX be the canonical isomorphism


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of ks -algebras defined in Lemma V.13.7, and denote by α : Gks −→ Autks −alg (ksX ) the continuous morphism defined by ασ = f ◦ (σ·f −1 ), for all σ ∈ Gks . Composing α on the left by an isomorphism Autks −alg (ksX ) Sn , we obtain a continuous morphism Gks −→ Sn which represents E. Since two morphisms representing E are conjugate, they have same kernel, so we have ker(ϕ) = ker(α). Hence it is enough to prove that ker(α) = Gal(ks /N ). Let σ ∈ Gks . Then σ ∈ ker(α) if and only if σ ·f = f . Since these two maps are ks -linear, this is equivalent to (σ·f )(u ⊗ 1) = f (u ⊗ 1) for all u ∈ E. Now we have (σ·f )(u ⊗ 1) = σ·(f (σ −1 ·(u ⊗ 1))) = σ·f (u ⊗ 1). By definition f (u ⊗ 1) = (χ(u))χ∈X , so we obtain that σ ∈ ker(α) if and only if σ ·χ(u) = χ(u) for all u ∈ E. Since N is the compositum of the subfields χ(E), χ ∈ X, this is equivalent to say that σ ∈ Gal(ks /N ). The last part of the proposition comes from the first isomorphism theorem and the isomorphism Gks /Gal(ks /N ) GN . This completes the proof. §V.14 Galois algebras In this section, we would like to describe a cohomological interpretation of finite Galois extensions with a given Galois group G. As previously, the difficulty is that the category of Galois extensions is not closed under scalar extension, so we need first to generalize the notion of Galois extension in an appropriate way. The right notion appears to be that of a Galois G-algebra.

V.14.1 Definition and first properties Recall that a G-algebra over k is a k-algebra on which G acts faithfully by k-algebra automorphisms. Notice that if L is a G-algebra over k, then

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G acts naturally on X(L) = Homk−alg (L, ks ) on the right as follows: for all χ ∈ X(L), all g ∈ G and all u ∈ L, we set (χ·g)(u) = χ(g·u). We start with a little lemma. Lemma V.14.1. Let G be an abstract finite group and let L/k be a Galgebra. Assume that L/k is a finite separable field extension and that |G| = dimk (L). Then L/k is a Galois extension if and only if the natural right action of G on X(L) is transitive. In this case, we have GL G. Proof. Notice first that, since L/k is separable of degree n, X(L) has n elements, where n = dimk L. Let ι ∈ X(L) be the inclusion L ⊂ ks , and consider the map ρ:

GL → X(L) σ −→ ι ◦ σ.

Notice that by construction, the image of ρ is the set of morphisms χ ∈ X(L) satisfying χ(L) ⊂ L. Let us consider the group morphism ϕ:

G −→ GL g −→ [L −→ L, x −→ g·x]

induced by the action of G on L, and consider the map ψ = ρ ◦ ϕ : G −→ X(L). By definition, we have ψ(g)(x) = ι(g·x) for all g ∈ G, x ∈ L. In other words, we have ψ(g) = ι·g. Assume first that G acts transitively on X(L), that is ψ is surjective. Since |G| = |X(L)| by assumption, ψ is then bijective and thus ρ is bijective as well. Therefore, χ(L) ⊂ L for all χ ∈ X(L) and L/k is a Galois extension. Now the bijectivity of ψ also implies the bijectivity of ϕ. Hence ϕ induces a group isomorphism G GL . Thus, we have proved that L/k is a Galois extension with Galois group isomorphic to G. Conversely, assume that L/k is Galois. In this case, |GL | = n = |G|


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and ϕ is an isomorphism. Now let χ, χ ∈ X(L). By assumption, there exist σ, σ ∈ GL such that χ = ι ◦ σ and χ = ι ◦ σ . We then have χ = χ ◦ (σ −1 ◦ σ ). Hence, GL G acts transitively on X(L). Definition V.14.2. Let G be an abstract group of order n, and let L be a G-algebra of dimension n over k. We say that L is a Galois G-algebra if L is étale and the right action of G on X(L) is transitive. Remark V.14.3. If L is a Galois G-algebra, then the action of G on X(L) is simply transitive since |G| = dimk L = |X(L)|. Remark V.14.4. In view of Lemma V.14.1, every Galois extension L/k is a Galois GL -algebra, and there is only one group G (up to isomorphism) for which L has a Galois G-algebra structure. We continue by giving another standard example of a Galois G-algebra. Example V.14.5. Let k be a field, and let G be a finite group of order n. Let us index the coordinates of the elements of k n with the elements of G, and let (eg )g∈G the corresponding n idempotents. If g ∈ G, let fg ∈ Autk−alg (k n ) be the unique k-algebra automorphism such that fg (eh ) = egh for all h ∈ G. In other words, fg is the automorphism fg :

L0 −→ L0 (xh )h∈G −→ (xg−1 h )h∈G .

It is easy to check the action of G on k n given by G × k n −→ k n (g, x) −→ fg (x) n

endows k with a structure of a Galois G-algebra, that we will denote by L0 . Lemma V.14.6. Let k be a field. Then for any finite group G of order n, any structure of a G-Galois algebra on k n is isomorphic to L0 . Proof. Assume that k n is endowed with a structure of a Galois Galgebra, that we will denote by L. As before, let us index the coordinates of the elements of k n with the elements of G, and let (eg )g∈G be the corresponding n idempotents. Since G acts by k-algebra automorphisms, g permutes the idempotents eh , h ∈ G by Lemma V.13.4 and its proof. Let us prove that this action is simply transitive. For all g ∈ G, denote


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by πg the projection on the g th -coordinate. The elements πg , g ∈ G are exactly the elements of X(L). By assumption, for all g ∈ G, there exists an unique g ∈ G such that πg = π1·g . Applying eg to this equality yields 1 = π1 (g ·eg ). Since π1 (eh ) = δh for all h ∈ G, we get that g ·eg = e1 , that is eg = g −1 ·e1 . This proves that the action of G permutes simply transitively the idempotents. In particular, we have a unique k-algebra ∼ automorphism f : kn −→ k n such that f (eg ) = g·e1 for all g ∈ G. It is easy to check that f is G-equivariant. Thus f induces an isomorphism of G-algebras L0 G L, and this concludes the proof. Remark V.14.7. If L is a Galois G-algebra over k and K/k is any field extension, then it is straightforward to check that LK is a Galois G-algebra over K, the action of G on LK being defined by: g·(u ⊗ λ) = (g·u) ⊗ λ for all g ∈ G, u ∈ L, λ ∈ K. We denote by G−Gal : Ck −→ Sets∗ the functor of isomorphism classes of Galois G-algebras, where the base point is the isomorphism class of the split Galois G-algebra. Lemma V.14.8. Let V be a k-vector space and let K/k be a field extension. Let G be a finite group acting on V linearly. Then we have (VK )G = (V G )K , where the action of G is extended to VK by K-linearity. Proof. One inclusion is clear. To prove the other one, let x = v1 ⊗ λ1 + . . . + vr ⊗ λr ∈ VK . One can always assume that λ1 , . . . , λr are linearly independent over k. In this case, the distributivity property of tensor product with respect to direct sums shows that for every w1 , . . . , wr ∈ V , we have w1 ⊗ λ1 + . . . + wr ⊗ λr = 0 ⇒ w1 = . . . = wr = 0. Now assume that x ∈ (Vk )G . For all g ∈ G, we get 0 = g·x − x = (g·v1 − v1 ) ⊗ λ1 + . . . + (g·vr − vr ) ⊗ λr . We then get that g ·vi − vi = 0 for all g and all i, so vi ∈ V G for all i and we are done.


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Theorem V.14.9. Let k be a field. Let L be a commutative G-algebra of dimension n over k, and assume that |G| = dimk (L). Then the following conditions are equivalent: (1)

L is a Galois G-algebra.

(2)

Lks is G-isomorphic to the Galois G-algebra (L0 )ks .

(3)

L is étale and LG = k.

Proof. (1) ⇒ (2) Assume that L is a Galois G-algebra. Then Lks is a Galois G-algebra over ks . Since L is étale, Lks ksn as a ks -algebra. Now transporting the action of G to ksn endows ksn with a structure of Galois G-algebra, which is isomorphic to the Galois G-algebra (L0 )ks by Lemma V.14.6. We then get an isomorphism Lks G (L0 )ks . (2) ⇒ (1) If Lks is isomorphic to the Galois G-algebra (L0 )ks , then in particular L is an étale algebra of rank n. Now we have to check that G acts transitively on X(L). Let ϕ, ϕ ∈ X(L). By assumption, Lks is a Galois G-algebra, so there exists g ∈ G such that ϕks ·g = ϕks . Applying this equality to u ⊗ 1, u ∈ L, we get ϕ(g · u) = ϕ (u) for all u ∈ L, that is ϕ·g = ϕ , so the action of G on X(L) is transitive. ∼

(2) ⇒ (3) If ψ : Lks −→ (L0 )ks is an isomorphism of G-algebras, then L is étale. Since G acts by automorphisms of k-algebras, we have k ⊂ LG . Now let u ∈ LG . Then x = ψ(u ⊗ 1) ∈ ((L0 )ks )G , since ψ is compatible with the G-actions. Consider e1 = (1, 0, . . . , 0) ∈ (L0 )ks . Since the action of G on (L0 )ks permutes transitively e1 , . . . , en , any element x may be written in a unique way as x = xg (g ·e1 ), for xg ∈ ks . For every g ∈ G, we have g ·x =

g∈G

g∈G

xg (g g·e1 ) =

xg −1 g (g·e1 ).

g∈G

Since g · x = x, comparing the first components yields xg −1 = x1 for all g ∈ G. Hence x = λ1(L0 )ks for some λ ∈ ks . This implies that u ⊗ 1 = 1 ⊗ λ. Now applying σ ∈ Gks gives u ⊗ 1 = 1 ⊗ (σ ·λ). Hence λ = σ ·λ for all σ ∈ Gks , so λ ∈ k. We then get u ⊗ 1 = 1 ⊗ λ = λ ⊗ 1, so u = λ ∈ k. ∼

(3) ⇒ (2) Since L is étale, we have an isomorphism ψ : Lks −→ ksn of ks -algebras. Now transporting the action of G on ksn endows ksn with


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a structure of G-algebra, and ψ is then G-equivariant with respect to the G-actions. We are going to prove that the action of G permutes transitively the idempotents e1 , . . . , en . Assume the contrary, and let g ·e1 ∈ ksn . By assumption, x does not have all its coordinates x= g∈G

equal, and hence x ∈ / ks 1ksn . Notice that by construction, we have x ∈ (ksn )G . Therefore, to obtain the desired contradiction, it is enough to show that (ksn )G = ks 1ksn . Since the isomorphism ψ is G-equivariant, it is enough to prove that (Lks )G = ks (1 ⊗ 1). In view of the assumption on L, it remains to prove that (Lks )G = (LG )ks , which follows from the previous lemma. This concludes the proof. We now explain how to construct Galois G-algebras from Galois Halgebras, where H is a subgroup of G. Let G be a finite group, and let H be a subgroup of G. For any Galois H-algebra M over k, we set IndG H M = {f ∈ M ap(G, M ) | f (hg) = h·f (g) for all h ∈ H, g ∈ G}. Notice that G acts on IndG H M by (g·f )(g ) = f (g g) for all g, g ∈ G, f ∈ IndG H M. Lemma V.14.10. The k-algebra IndG H M is a Galois G-algebra. Proof. Let us prove first that IndG etale k-algebra of dimension H M is an ´ |G|. Let Hg1 , . . . , Hgr the r right cosets of G modulo H. Clearly, any f ∈ IndG H M is uniquely determined by its values on g1 , . . . , gr , and therefore we get an isomorphism ∼

ϕ:

r IndG H M −→ M

f −→ (f (g1 ), . . . , f (gr )).

Since M is étale, M r is étale too and so is IndG H M . Now we have dimk IndG H M = r dimk M = r|H| = |G|. We now prove that IndG H M is a G-algebra. It is clear that G acts by k-algebra automorphisms. Let us prove that the action of G is faithful. Let g ∈ G, g = 1. We have to show that there exists f ∈ IndG HM G such that g · f = f . It is enough to find some f ∈ IndH M satisfying f (g) = f (1). Assume first that g ∈ / H. Then g and 1 represents two different right cosets of G modulo H. In view of the isomorphism above, one can easily

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construct an element f ∈ IndG H M such that f (g) = f (1). If now g ∈ H, since H acts faithfully on M , there exists x ∈ M such that h · x = x. Now let f ∈ IndG H M such that f (1) = x. Then we have f (g) = g·f (1) = g·x = x = f (1). It remains to prove that the G-algebra IndG H M is Galois. By Theorem G G V.14.9, it remains to check that (IndH M )G = k. If f ∈ (IndG HM) , then for all g, g ∈ G, we get f (g ) = (g.f )(g ) = f (g g), and therefore f (g) = f (1) for all g ∈ G. But we have f (1) = f (h) = h·f (1) for all h ∈ H, so f (1) ∈ M H = k. This concludes the proof. Definition V.14.11. The Galois G-algebra of IndG H M is called the Galois G-algebra induced by M .

V.14.2 Galois algebras and Galois cohomology We now relate Galois G-algebras and cohomology. By Theorem V.14.9, the G-algebras which become isomorphic to the G-algebra L0 over ks are exactly the Galois G-algebras over k. To use Galois descent, we need to determine AutG−alg (L0 ). Lemma V.14.12. For every field extension K/k, we have an isomorphism of abstract groups AutG−alg ((L0 )K ) G. Proof. With any loss of generality, we may assume that K = k. For any g ∈ G, let ϕg be the unique automorphism of k-algebras such that ϕg (eh ) = ehg−1 for all h ∈ G. It is easy to check that ϕg is an automorphism of G-algebras, and that the map ϕ:

G −→ AutG−alg (L0 ) g −→ ϕg

is a group morphism. Let us prove the bijectivity of ϕ. If g ∈ G lies in ker(ϕ), then we have ehg−1 = eh for all h ∈ G, which is only possible if g = 1. Thus ϕ is injective. We now prove its surjectivity. If f ∈ AutG−alg (L0 ), then f (e1 ) is an idempotent of L0 , say eg . Since f is G-equivariant we then get f (eh ) = f (h·e1 ) = h·f (e1 ) = h·eg = ehg for all h ∈ G.


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Therefore f = ϕg−1 = ϕ(g −1 ) and ϕ is surjective. This concludes the proof. In particular, we get a group isomorphism AutG−alg ((L0 )Ks ) G. One may easily check that this is an isomorphism of GKs -groups, where the action of GKs on G is trivial. By Proposition III.9.7, we then get: Proposition V.14.13. Let G be a finite abstract group. We have an isomorphism of functors from Ck to Sets G − Gal H 1 (− , G), where G is considered as a trivial GKs -group for every field extension K/k. Example V.14.14. If char(k) n and k contains a primitive nth -root of 1, then μn (ks ) Z/nZ as Gks -modules, and therefore, we get H 1 (k, Z/nZ) k × /k ×n , using Proposition III.8.27. This isomorphism is just a cohomological translation of Kummer theory of cyclic extensions in the presence of roots of unity. Let L be a Galois G-algebra over k and ϕ : Gks −→ G a continuous group morphism representing L. Since ϕ is continuous, ker(ϕ) is a closed normal subgroup of Gks . The Galois correspondence then shows the existence of a finite Galois extension M/k such that Gal(ks /M ) = ker(ϕ). Lemma V.14.15. With the previous notation, for all χ ∈ X(L), we have χ(L) = M . Proof. Since G acts transitively on X(L), we have ξ(L) = χ(L) for all ξ, χ ∈ X(L). Now applying Proposition V.13.8, we get that ker(ϕ) = Gal(ks /χ(L)), hence χ(L) = M by Galois correspondence. Let L be a Galois G-algebra over k. Set X = X(L) and let us fix χ0 ∈ X. Consider the ks -algebra ksX , and let us index the components of x ∈ ksX using elements of X. In particular, eχ , χ ∈ X will denote the canonical basis of ksX as a k-vector space (that is the primitive idempotents). We now define an action of G by ks -automorphisms on ksX as follows: g·eχ = eχ·g−1 for all g ∈ G, χ ∈ X.


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Since G permutes transitively the elements of X, G permutes transitively the eχ ’s, χ ∈ X, so we obtain a structure of Galois G-algebra (which is the only possible one up to isomorphism). We will assume from now on that ksX is endowed with this action of G. ∼

Let f : Lks −→ ksX be the canonical isomorphism of ks -algebras defined in Lemma V.13.7. We claim that f is an isomorphism of Galois Galgebras. Indeed we have f (u ⊗ λ) = χ(u)λeχ , χ

and therefore g·(f (u ⊗ λ)) =

χ(u)λeχ·g−1 =

χ

Hence we have f (g·(u ⊗ λ)) = f (g(u) ⊗ λ) =

(χ·g)(u)λeχ .

χ

(χ·g)(u)λeχ = g·(f (u ⊗ λ)), χ

proving the claim. Let α : Gks −→ AutG−alg (ksX ) be the continuous group morphism defined by ασ = f ◦ (σ·f −1 ) for all σ ∈ Gks . We identify AutG−alg (ksX ) to G via the group isomorphism η : G −→ AutG−alg (ksX ) defined by: η(g)(eχ0·h ) = eχ0 ·(gh) for all h ∈ G. Then ϕ = η −1 ◦ α represents the Galois G-algebra L by Galois descent. We are now going to give a more natural description of this cocycle. Since G acts simply transitively on X on the right, we define a continuous group morphism ψ : Gks −→ G by mapping σ ∈ Gks onto the unique element ψσ ∈ G satisfying σ·χ0 = χ0 ·ψσ . Lemma V.14.16. Keeping the notation above, we have ψ = ϕ. Proof. We first show the relation ασ (eχ ) = eσ·χ for all σ ∈ Gks and χ ∈ X. For χ ∈ X, let eχ = f −1 (eχ ). Then the eχ , χ ∈ X form a basis of the


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ks -vector space Lks , whose dual basis is χ , χ ∈ X. Indeed, by definition of f , we have f (eξ ) = (χ (eξ ))χ∈X for all ξ ∈ X. Since we have f (eξ ) = eξ by definition of eξ , we get that χ (eξ ) = δχ,ξ for all χ, ξ ∈ X. Let σ ∈ Gks and χ ∈ X. We have σ ·eχ = eσ·χ . This results from the relation ξ (σ·eχ ) = σ(σ−1·ξ (eχ )) = δξ,σ·χ , for all ξ ∈ X, the first equality following from Lemma V.13.6. Therefore, we get ασ (eχ ) = f (σ·f −1 (σ −1 ·eχ )) = f (σ·f −1 (eχ )) = f (σ·eχ ) = f (eσ·χ ) = eσ·χ . From this, we get ασ (eχ0·g ) = eσ·(χ0·g) = e(σ·χ0 )·g for all g ∈ G. By definition of ψ, we have σ·χ0 = χ0 ·ψσ . We then get ασ (eχ0 ·g ) = e(χ0 ·ψσ )·g = eχ0 ·(ψσ g) = η(ψσ )(eχ0 ·g ) for all g ∈ G. Therefore, α = η ◦ ψ, that is ψ = ϕ, and this concludes the proof. We thus have proved the following theorem: Theorem V.14.17. Let L be a Galois G-algebra over k, and let χ0 ∈ X(L). For all σ ∈ Gks , let ψσ ∈ G be the unique element of G satisfying σ·χ0 = χ0 ·ψσ . Then the continuous morphism ψ : Gks −→ G represents the Galois G-algebra L. Let M/k be a finite Galois extension. For every σ ∈ Gks , σ|M is a kembedding of M into ks , so its image is equal to M , since M is a Galois extension. Therefore, one can write σ|M = ι ◦ sσ for some sσ ∈ GM , where ι is the inclusion M ⊂ ks . Lemma V.14.18. The continuous group morphism s : Gks −→ GM is surjective and represents the Galois GM -algebra M . Proof. The surjectivity of s comes from the fact that M/k is a Galois field extension. To prove the rest of the lemma, notice that the inclusion ι is an element of X(M ). Now for all σ ∈ Gks , we have σ·ι = σ|M = ι·sσ . The result then follows from the previous theorem.


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We continue by giving a cohomological interpretation of the induced Galois algebra. Proposition V.14.19. Let G be a finite abstract group on which Gks acts trivially, and let H ⊂ G be a subgroup. The map H 1 (k, H) −→ H 1 (k, G) induced by the inclusion H ⊂ G corresponds to M −→ IndG HM. Proof. Let M be a Galois H-algebra over k, and let L = IndG H (M ). Let χ0 ∈ X(M ), and let ψ be the continuous morphism corresponding to M by Theorem V.14.17. Then the map χ0 :

IndG H M −→ ks f −→ χ0 (f (1))

is an element of X(L). We are going to show that ψ satisfies σ·χ0 = χ0 ·ψσ for all σ ∈ Gks , which will prove the desired result in view of Theorem V.14.17. For all σ ∈ Gks and all f ∈ L, we have σ·χ0 (f ) = σ(χ0 (f (1))) = (σ·χ0 )(f (1)) = (χ0 ·ψσ )(f (1)). Since f ∈ L and ψσ ∈ H, we get σ·χ0 (f ) = χ0 (ψσ ·f (1)) = χ0 (f (ψσ )). Now we have χ0 ·ψσ (f ) = χ0 (ψσ ·f ) = χ0 ((ψσ ·f )(1)). By definition of the action of G on L, we get χ0 ·ψσ (f ) = χ0 (f (ψσ )) = χ0 (ψσ ), and this concludes the proof. We are now able to prove that every Galois G-algebra is isomorphic to an induced Galois algebra. Theorem V.14.20. Let L be a G-Galois algebra over k, and let α : Gks −→ G be a continuous morphism representing L. Then L G IndG H M, where M = χ(L) for any χ ∈ X(L) and H = Im(α) GM .


175

Proof. Let M/k be the finite Galois field extension satisfying ker(α) = Gal(Gks /M ). The first isomorphism theorem shows that H = Im(α) Gal(M/k). Lemma V.14.18 and Proposition V.14.19 then show that L G IndG HM. Corollary V.14.21. Let L be a Galois G-algebra L, and let α : Gks −→ G be any continuous morphism representing L. Then L is a field if and only if the corresponding continuous morphism α is surjective, and L is isomorphic to k n if and only if α is trivial. Proof. By the previous theorem, L is a field if and only if [G : H] = 1, that is if and only if any continuous morphism representing L is surjective. Similarly, L is isomorphic to L0 if and only if [G : H] = n, that is if and only if H = {1}, or equivalently if and only if any continuous morphism representing L is trivial (this can also be deduced from Galois descent). We now come back to étale algebras. Notice that H 1 (k, Sn ) has two different interpretations. A cohomology class corresponds on the one hand to an étale k-algebra E, and to a Galois Sn -algebra L over k on the other hand. The following proposition describes the relation between E and L when E is a field. Proposition V.14.22. Let E/k be a separable field extension of degree n, let E gal /k be its Galois closure, and let e : Gks −→ Sn be a continuous morphism corresponding to E. Then Gks −→ Im(e) represents the Galois extension E gal /k. In particular, the Sn -Galois algebra represented by e is IndSGnE E gal , where GE = Im(e) is the Galois group of E gal /k. Proof. Apply Proposition V.13.8 and Proposition V.14.19. Let L be a G-algebra. If H is a normal subgroup of G, then LH is a G/H-algebra for the action of G/H defined by gH ·u = g·u for all gH ∈ G/H and all u ∈ LH . We then have the following proposition. /G π /N / 1 be an /H Proposition V.14.23. Let 1 exact sequence of abstract groups. Then for any Galois G-algebra L over k, the k-algebra LH is a Galois N -algebra. Moreover, the induced map π∗ : H 1 (k, G) −→ H 1 (k, N ) corresponds to L −→ LH .

176


Proof. Let L be a Galois G-algebra of dimension n over k, and let ∼ f : Lks −→ ksn be an isomorphism of G-algebras. Then f induces ∼ an isomorphism of N -algebras f : (Lks )H −→ (ksn )H , and therefore an ∼ H n H isomorphism f : (L )ks −→ (ks ) by Lemma V.14.8. Let r = [G : H]. We will index the components of an element of (ks )r using the elements of N , and the components of an element of (ks )n using the elements of G, and we let act G and N on (ks )n and (ks )r respectively by left translation on the indices. It is then easy to show that (ksn )H = {(xg )g∈G | xgh = xg for all g ∈ G, h ∈ H}. Therefore, we get an isomorphism of N -algebras ∼

θ : ksr −→ (ksn )H . Composing with f gives rise to an isomorphism of N -algebras ∼

f : (LH )ks −→ ksr , so LH is a Galois N -algebra. To prove the last part of the proposition, let χ0 ∈ X(L), and let ψ be the corresponding continuous morphism described in Theorem V.14.17. The map χ0 = (χ0 )|LH is an element of X(LH ). Moreover, for all σ ∈ Gks and all u ∈ LH , we have σ·χ0 (u) = σ(χ0 (u)) = χ0 (ψσ ·u) = χ0 (π(ψσ )·u) = χ0 (π∗ (ψ)σ ·u). Now use Theorem V.14.17 to conclude. We finish this section by giving an interpretation of the map H 1 (k, G) −→ H 1 (k, Sn ) induced by the inclusion G ⊂ Sn . Lemma V.14.24. Let G ⊂ Sn be a group of order n. Then the map H 1 (k, G) −→ H 1 (k, Sn ) induced by the inclusion maps the isomorphism class of a Galois G-algebra L onto the isomorphism class of the étale algebra L/k. Proof. An isomorphism f : Lks −→ (L0 )ks is also an isomorphism of étale algebras f : Lks −→ ksn . Now use Galois descent to conclude.

Exercises

177

Exercises 1. Let k be a field of characteristic p, where p is a prime number. Let k −→ k ℘: x −→ xp − x. (a)

Check that ℘ is a group morphism.

(b)

Show that H 1 (k, Z/pZ) k/℘(k).

(c)

Given a class a ∈ k/℘(k), show that the Galois Z/pZalgebra corresponding to a under the previous isomorphism is isomorphic to k[X]/(X p − X − a).

2. Let k be a field of characteristic different from 2. (a)

Show that the first connecting map associated to the exact sequence 1 −→ Z/2Z −→ Z/4Z −→ Z/2Z −→ 1 maps a square class (d) onto (−1) ∪ (d).

(b)

√ Deduce that a quadratic étale k-algebra k[ d]/k can be embedded in a Galois Z/4Z-algebra L/k such that LZ/2Z = √ k[ d] if and only if d is a sum of two squares of k.

3. Let G be a finite group, let H, H be two subgroups of G, and let M/k and M /k be Galois field extensions with Galois groups isomorphic to H and H respectively. Show that if IndG H M and G IndH M are G-isomorphic, then M and M are k-isomorphic and H = gHg −1 for some g ∈ G. 4. Let K/k be a field extension. For all σ ∈ Sn , we will denote by fσ : K n → K n the K-algebra automorphism defined by fσ (x1 , . . . , xn ) = (xσ−1 (1) , . . . , xσ−1 (n) ), for all x1 , . . . , xn ∈ K. Let E, E be two étale K-algebras, let ϕ ∈ AutK−alg (E) and let ϕ ∈ AutK−alg (E ). A morphism f : (E, ϕ) → (E , ϕ ) is a morphism of K-algebras such that f ◦ ϕ = ϕ ◦ f . (a)

For every pair (E, ϕ), where dimK (E) = n, show that there exists a permutation σ ∈ Sn , unique up to conjugation, such that (EKs , ϕ ⊗ IdKs ) (Ksn , fσ ).


178

We will say that (E, ϕ) is an étale K-algebra of type σ. (b)

Let C(σ) be the centralizer of σ in Sn . Show that H 1 (− , C(σ)) classifies isomorphism classes of étale algebras of dimension n of type σ.

5. Let L/k be a Galois field extension with Galois group G. For any subgroup H of G, prove that we have an isomorphism of Galois G-algebras H IndG H L G L ⊗k L ,

where L is considered as a Galois H-algebra over LH on the lefthand side. Hint: Write the corresponding cocycles.

VI Group extensions, Galois embedding problems and Galois cohomology

In this chapter, we give an interpretation of the cohomology group H 2 (G, A) in terms of group extensions. We then use this interpretation to give a complete obstruction of a Galois embedding problem in terms of Galois cohomology. §VI.15 Group extensions Let E be a finite group, and let A be an abelian normal subgroup of E. We set G = E/A, so we have an exact sequence 1 −→ A −→ E −→ G −→ 1. Since A is a normal subgroup of E, E acts on A by conjugation E × A −→ A (x, a) −→ x·a = Int(x)(a) = xax−1 , so we have a group morphism Int :

E −→ Aut(A) x −→ Int(x)|A .

Morever, if x ∈ A, then x · a = a for all a ∈ A, since A is abelian. Therefore, this action of E on A induces an action of G on A by group automorphisms, defined by G × A −→ A (π(x), a) −→ π(x)·a = xax−1 where π : E −→ G denotes the canonical projection. We denote by ϕ : G −→ Aut(A) the corresponding group morphism. Notice that we have ϕ ◦ π = Int by definition. This motivates the following definition. 179

180

Group extensions and Galois embedding problems

Definition VI.15.1. Let G, A be abstract finite groups, with A abelian, and assume that G acts by group automorphisms on A, via the group morphism ϕ : G −→ Aut(A). In other words, A is a G-module. A group extension of G by A, or more precisely a group extension of (G, A, ϕ), is an exact sequence 1

/A

ι

/E

π

/G

/1

such that the action of E by inner automorphisms on A extends the action of G on A. More precisely, we require: xι(a)x−1 = ι(ϕ(π(x))(a)) for all x ∈ E, a ∈ A. Remark VI.15.2. If no confusion is possible, the action of G on A will be denoted by G × A −→ (g, a) → g·a. The previous condition then may be rewritten xι(a)x−1 = ι(π(x)·a) for all x ∈ E, a ∈ A. Things are a bit complicated by the fact that A is not a subgroup of E, but only isomorphic to a subgroup of E via ι. However, we cannot avoid this problem since standard group extensions are constructed on the set E = A × G, as we will see later on. Example VI.15.3. Examples of group extensions are given by direct products and semi-direct products, but they are not the only ones. For instance, Z/8Z is an extension of Z/4Z by Z/2Z, which is not a semidirect product. Consider an extension of (G, A, ϕ) given by the exact sequence 1

/A

ι

/E

π

/G

/ 1.

This exact sequence allows us to define a bijection between the sets A×G and E as follows: let us choose a set-theoretic section s : G −→ E, that is a map such that π◦s = IdG . We also ask for the extra condition s(1) = 1. Now if x ∈ E, set g = π(x). We have π(xs(g)−1 ) = gg −1 = 1, so there exists a unique element a ∈ A such that ι(a) = xs(g)−1 . Therefore x = ι(a)s(g), and a and g are uniquely determined, once s is chosen. The map A × G −→ E (a, g) −→ ι(a)s(g) is then bijective. Of course, this is not a group isomorphism if A × G is endowed with the standard group law.

VI.15 Group extensions

181

Let us figure out what happens when we multiply two elements of E. Let x, x ∈ E and let π(x) = g, π(x ) = g . We have π(xx ) = gg , so the previous considerations show that we should be able to write xx = ι(a )s(gg ). But we have xx = ι(a)s(g)ι(a )s(g ) = ι(a)s(g)ι(a )s(g)−1 s(g)s(g ), hence xx = ι(a)(s(g)ι(a )s(g)−1 )(s(g)s(g )s(gg )−1 )s(gg ). By definition of the action of G on A, we have s(g)ι(a )s(g)−1 = ι(g·a ), since s is a section of π. Now it is easy to check that s(g)s(g )s(gg )−1 ∈ (s) ker(π) for all g, g ∈ G. Therefore, there exists a unique element αg,g ∈ A such that ι(αg,g ) = s(g)s(g )s(gg )−1 for all g, g ∈ G. (s)

To sum up, the group law on E is given by the rule (ι(a)s(g))(ι(a )s(g )) = ι(αg,g a g·a )s(gg ). (s)

Straightforward computations show that the map α(s) :

G × G −→ A (g, g ) −→ s(g)s(g )s(gg )−1

is a 2-cocycle. These considerations lead naturally to the following example. Example VI.15.4. Let G and A be finite groups, where A is abelian, and assume that G acts by group automorphisms on A. Let α ∈ Z 2 (G, A) be a 2-cocycle, and define an operation on the set A × G as follows: (a, g)(a , g ) = (αg,g ag·a , gg ), for all a, a ∈ A, g, g ∈ G. One can show that this defines a group structure on the set A × G (the nasty thing to check being the associativity, which comes from the cocyclicity condition). The neutral element is (1, 1), and we have an exact sequence (for this particular group law on A × G) 1

/A

ι1

/ A×G

π2

/G

/1,

182


where ι1 (a) = (a, 1) and π2 (a, g) = g. Moreover, one can check that we have (a, g)(a0 , 1)(a, g)−1 = (g·a0 , 1), so we get a group extension, denoted by A ×α G. If α is the trivial cocycle, we just recover the external semi-direct product A ×ϕ G. With this notation, we see that we have a group isomorphism ∼

f:

A ×α(s) G −→ E (a, g) −→ ι(a)s(g),

,

and that the diagram 1

/A

1

/A

/ A ×α(s) G

ι1

π2

/G

/1

/G

/1

f

/E

ι

π

is commutative Definition VI.15.5. Two extensions E, E (for short) of (G, A, ϕ) are ∼ equivalent if there exists a group isomorphism f : E −→ E such that the diagram 1

/A

ι

/E

/A

ι

/ E

π

/G

/1

π

/G

/1

f

1 commutes.

Thus we have proved that any group extension of (G, A, ϕ) is equivalent to A ×α G for some suitable cocycle α ∈ Z 2 (G, A). Now let us see what happens if we choose another set-theoretic section of π. Let t : G −→ E be another section satisfying t(1) = 1, and let α(t) be the corresponding 2-cocycle. For all g ∈ G, we have s(g)t(g)−1 ∈ ker(π) and there exists a unique element γg ∈ A such that s(g) = ι(γg )t(g). Thus we have ι(αg,g ) = ι(γg )t(g)ι(γg )t(g )t(gg )−1 ι(γgg )−1 . (s)

We then get ι(αg,g ) = ι(γg )[Int(t(g))(ι(γg ))]t(g)t(g )t(gg )−1 ι(γgg )−1 (t) = ι(γg )ι(g·γg )ι(αg,g )ι(γgg )−1 , (s)

VI.15 Group extensions

183

that is, taking into account that A is abelian and ι is an injective group morphism −1 αg,g = αg,g γg g·γg γgg . (s)

(t)

Moreover, we have γ1 = 1 since s(1) = t(1) = 1. This means that α(s) and α(t) differ by a 2-coboundary, so they represent the same class in H 2 (G, A). Therefore, we have associated to an extension E a cohomology class in H 2 (G, A). Moreover, one may check that two equivalent extensions yield the same cohomology class. Conversely, it is easy to check that if α and α are two cohomologous cocycles, then the group extensions A×α G and A ×α G are equivalent. Thus, a cohomology class in H 2 (G, A) defines an equivalence class of extensions. Morever, the two constructions are easily seen to be mutually inverse. Hence we have (almost) proved: Theorem VI.15.6. Let G, A be two finite groups, and assume that A is a G-module via ϕ : G −→ Aut(A). Then H 2 (G, A) is in one-toone correspondence with the equivalence classes of group extensions of (G, A, ϕ). Remark VI.15.7. Let us explain once again how this correspondence works. If [α] ∈ H 2 (G, A), the equivalence class of A ×α G only depends /A ι /E π /G / 1 is a group on [α]. Conversely, if 1 extension and s : G −→ E is a set-theoretic section of π satisfying s(1) = 1, the cohomology class of the cocycle α(s) : G × G −→ A defined by ι(αg,g ) = s(g)s(g )s(gg )−1 for all g, g ∈ G only depends on the equivalence class of the extension. Notice from the correspondence that the cocycle corresponding to a given extension 1 −→ A −→ E −→ G −→ 1 is trivial if and only if there exists a group-theoretic section s : G −→ E, that is, if the exact sequence 1 −→ A −→ E −→ G −→ 1 splits. In this case, the extension is equivalent to 1 −→ A −→ A ×ϕ G −→ G −→ 1. Notice also that if two extensions are equivalent, then the corresponding groups E, E are isomorphic but the converse is not necessarily true. If G acts trivially on A, that is, if ϕ(g) = IdA for all g ∈ G, then for any group extension E of G by A, A identifies with a central subgroup of E

184


(it is enough to check it on the extensions A ×α G). Conversely, assume that /1 /A ι /E π /G 1 is a group extension such that A (or more precisely ι(A)) is a central subgroup of E. By definition of a group extension, we get xι(a)x−1 = ι(a)xx−1 = ι(a) = ι(π(x)·a) for all x ∈ E, a ∈ A. Using the surjectivity of π and the injectivity of ι, we get that g·a = a for all a ∈ A, g ∈ G, that is G acts trivially on A. Definition VI.15.8. A group extension 1

/A

ι

/E

π

/G

/1

is central if ι(A) is a central subgroup of E. The previous considerations show that an extension of G by A is central if and only if G acts trivially on A. Therefore, we get the following result. Corollary VI.15.9. Let G, A be finite groups, where A is abelian, considered as a trivial G-module. Then H 2 (G, A) is in one-to-one correspondence with equivalence classes of central extensions of G by A. Example VI.15.10. The group H 2 (Sn , {±1}) is perfectly well-known (the action of Sn on {±1} being necessarily the trivial one). Remember that the signature is a morphism Sn −→ {±1}, and therefore gives rise to an element εn ∈ H 1 (Sn , {±1}). We may then consider the cup product εn ∪ εn ∈ H 2 (Sn , {±1}) induced by the Z-bilinear map {±1} −→ {±1} ((−1) , (−1)n ) −→ (−1)nm . m

If n = 2, 3, then H 2 (Sn , {±1}) Z/2Z and it is generated by εn ∪ εn . If n ≥ 4, H 2 (Sn , {±1}) Z/2Z × Z/2Z and it is generated by εn ∪ εn and another class sn . The extension corresponding to sn is denoted by Sñ . This extension can be characterized as follows: any element of Sñ whose image in Sn is a transposition (resp. a product of two transpositions with disjoint supports) is an element of order 2 (resp. of order 4). To the presentation of Sn with n − 1 generators ti , 1 ≤ i ≤ n − 1 subject to the relations t2i = 1, (ti ti+1 )3 = 1, ti tj = tj ti if |j − i| ≥ 2

VI.16 Galois embedding problems

185

corresponds a presentation of Sñ with n generators t˜i and ω subject to the relations t˜2i = 1, ω 2 = 1, ω t˜i = t˜i ω, (t˜i t˜i+1 )3 = 1, t˜i t˜j = ω t˜j t˜i if |j − i| ≥ 2. The maps ι : {±1} −→ Sñ and π : Sñ −→ Sn in the exact sequence / {±1}

1

ι

/ S˜ n

/ Sn

π

/1

are characterized by the following relations ι(−1) = ω π(t˜i ) = ti for all i. The reader will refer to [54] for a proof of these facts. For convenience, we will set sn = 0 if n = 1, 2, 3. Notice that if G is a subgroup of Sn , we can construct central extensions of G by {±1} using the restriction map Res : H 2 (Sn , {±1}) −→ H 2 (G, {±1}). ˜ corresponding to Res(sn ). In particular, we can define the extension G ˜ is a subgroup of Sñ and that we have a It is then easy to check that G commutative diagram 1

/ {±1}

/G ˜

/G

/1

1

/ {±1}

/ S˜ n

/ Sn

/1

where the two vertical maps are just inclusions. In particular, if G is ˜ of a finite group of order n, we may construct a central extension G G by {±1} after identifying G to a subgroup of Sn . For example, let ˜ = Z/8Z, and that if n = 4. One can check that if G = Z/4Z, then G ˜ G = Z/2Z × Z/2Z, we get G = Q8 . If G = An , we obtain a group Añ , which is a central extension of An by {±1}. §VI.16 Galois embedding problems Let G be a finite group, and let A be a finite abelian group on which G acts trivially. Let us fix a central extension 1

/A

ι

/ G

π

/G

/1.

Now let k be a field. We consider the following lifting problem: given

186


a continuous morphism f : Gks −→ G, does there exists a continuous morphism f : Gks −→ G such that the diagram Gks

f

/ G π

Gks

f

/G

is commutative? It is straightforward to see that the solvability of the problem only depends on the conjugacy class of f and on the isomorphism class of the extension. Let Gks act trivially on G, G and A. Since A identifies to a central subgroup of G via ι, we have an exact sequence in cohomology H 1 (k, G )

π∗

/ H 1 (k, G)

δ1

/ H 2 (k, A) ,

and by definition of π∗ , the lifting problem is equivalent to ask [f ] to be in the image of π∗ , that is in the kernel of the connecting map δ 1 . Let s : G −→ G be a set-theoretic section of π satisfying s(1) = 1. Since s(fσ ) is a preimage of fσ under π and Gks acts trivially on G, a 2-cocycle γ : Gks × Gks −→ A representing δ 1 ([f ]) is uniquely determined by the equalities ι(γσ,τ ) = s(fσ )s(fτ )s(fστ )−1 = s(fσ )s(fτ )s(fσ fτ )−1 for all σ, τ ∈ Gks . Now a 2-cocycle α : G × G −→ A representing the extension G is uniquely determined by the relations ι(αg,g ) = s(g)s(g )s(gg )−1 for all g, g ∈ G, by the correspondence between equivalence classes of extensions and H 2 (G, A). Hence we get that δ 1 ([f ]) is equal to f ∗ ([α]), where f ∗ ([α]) is the inverse image of [α] by f , as defined in Example II.3.23. Thus we have proved: Proposition VI.16.1. Let 1 −→ A −→ G −→ G −→ 1 be a central extension whose equivalence class corresponds to a class [α] ∈ H 2 (G, A), and let f : Gks −→ G be a continuous group morphism. Then the associated lifting problem has a solution if and only if f ∗ ([α]) = 0 ∈ H 2 (k, A). In practice, it is enough to consider the case where f is a surjective morphism. Indeed, let H = Im(f ), let H = π −1 (H) ⊂ G and let

Exercises

187

˜ with h : Gks −→ H be a lifting for h : Gks −→ H. Then composing h the inclusion H ⊂ G gives rise to a lifting of f . Now assume that f : Gks −→ G is surjective. Definition VI.16.2. A solution f˜ to the lifting problem for f is said to be proper if f˜ is surjective as well, and improper otherwise. The following lemma is left as an easy exercise. Lemma VI.16.3. Assume that A has prime order p, and assume that [α] = 0. Let f : Gks −→ G be a surjective continuous morphism. Then a solution to the associated lifting problem (if any) is proper. If we are only looking for proper solutions to the lifting problem, the question may be reformulated as a Galois embedding problem, in view of Corollary V.14.21: given a Galois field extension L/k with Galois group G, does there exists a Galois extension L /k with Galois group G such that LA = L? This is a classical question in inverse Galois theory. The previous results show that such a field extension L /k exists if and only if f ∗ ([α]) = 0, where f : Gks −→ G is any continuous morphism corresponding to L/k. We will use this result in the next chapter to solve completely a particular Galois embedding problem associated to separable field extensions. Example VI.16.4. Let p be a prime number, and consider a central extension 1 −→ Z/pZ −→ G −→ G −→ 1. For any field k of characteristic p and for any Galois extension L/k of group G, the associated Galois embedding problem has a solution. Indeed, we have H 2 (k, Z/pZ) = 0 in this case. This can be seen by applying Galois cohomology to the Artin-Schreier exact sequence 0 −→ Z/pZ −→ Ga −→ Ga −→ 0, and using the fact that H m (k, Ga ) = 0 for all m ≥ 1 (see Proposition III.7.31). Proposition VI.16.1 then yields the result.

Exercises 1. Fill the gaps in the proof of Theorem VI.15.6.

188

Group extensions and Galois embedding problems 2. Prove Lemma VI.16.3. 3. Let G be a finite group acting trivially on a finite abelian group A. Show that H 2 (G, A) is killed by |G| and |A|. In particular, if G and A have relatively prime orders, any central extension of G by A is trivial. 4. Let G = A = Z/3Z. For i = 1, 2, let Ei = Z/9Z, and let πi : Ei −→ G be the canonical projection. Consider the group morphisms ι1 :

A −→ E1 m + 3Z −→ 3m + 9Z

and ι2 :

A −→ E2 m + 3Z −→ −3m + 9Z.

(a)

Show that the central extensions E1 , E2 of G by A are not isomorphic (although the abstract groups E1 and E2 are isomorphic, and even equal).

(b)

Let [αi ] ∈ H 2 (G, A) be the cohomology class corresponding to Ei . Show that [α2 ] = −[α1 ], and recover the result of the previous question. Hint: Use Exercice 3.

5. Let τ1 , . . . , τk be k transpositions with disjoint supports in Sn . Show that we have 2

τ1 · · · τk = ω

k(k−1) 2

∈ Sñ .

˜ Z/8Z. 6. Let G be a cyclic subgroup of S4 of order 4. Show that G 7. Let G be a cyclic subgroup of Sn of order n = 2r ≥ 8. Show that Res(sn ) = 0. 8. Let n = 2r ≥ 4, and let G = D2n . After a suitable numbering of the elements of G, show that the action of G on itself by left translation identifies G to the subgroup of S2n generated by σ = (1 . . . 2n − 1) and τ = (2 2n)(3 2n − 1) · · · (n n + 2), and deduce that Res(s2n ) = 0. 9. Let G = Q8 be the quaternion group of order 8. Show that Res(s8 ) = 0.

Part II Applications

VII Galois embedding problems and the trace form

In this chapter, we are going to use the material introduced in the first part of the book to give a full answer to the following Galois embedding problem: Galois embedding problem: let E/k be a separable field extension of degree n and let E gal /k be its Galois closure (in a fixed separable ˜ E be closure of k). Let GE be the Galois group of E gal /k, and let G 2 the group extension corresponding to Res(sn ) ∈ H (GE , {±1}), where sn ∈ H 2 (Sn , {±1}) is the cohomology class corresponding to the central extension Sñ of Sn . Assume that Res(sn ) = 0 (otherwise, the problem ˜ with trivially has a solution). Does there exist a Galois extension E/k {±1} gal ˜ ˜ =E ? Galois group GE such that E Let us reformulate the problem. Let e : Gks −→ Sn be a continuous morphism representing E. Since E is a field, the compositum of the fields χ(E), χ ∈ X(E), is E gal . By Proposition V.14.22, we then have Im(e) = GE . Moreover, the continuous morphism f : Gks −→ GE represents E gal /k by Lemma V.14.18. Using the definitions, it is easy to check that f ∗ (Res(sn )) = e∗ (sn ) ∈ H 2 (k, {±1}). Hence the Galois embedding problem associated to E has a solution if and only if e∗ (sn ) = 0. Therefore, everything boils down to compute e∗ (sn ) in more explicit terms. We will suppose from now on that char(k) = 2 (this is not a huge restriction by Example VI.16.4 ). In this case, we may consider e∗ (sn ) as an element of H 2 (k, μ2 ). The crucial idea of this computation is that Sn may be identified to a subgroup of On (ks ), and that this identification may be used to view Sñ as a subgroup of Pinn (ks ). The computation of e∗ (sn ) is then achieved using the computation of the first connecting map associated to the 191

192

Galois embedding problems and the trace form

exact sequence 1 −→ μ2 (ks ) −→ Pinn (ks ) −→ On (ks ) −→ 1. With this in mind, let us denote by ι : Sn −→ On (ks ) the identification of Sn to a subgroup of On (ks ) by permutation matrices. Notice that ι is a morphism of Gks -groups. This map induces a map ι∗ : H 1 (k, Sn ) −→ H 1 (k, On ). Using the results of the previous chapters, we see ι∗ may be used to attach to each étale k-algebra E of rank n a quadratic form of dimension n, whose isomorphism class only depends on the isomorphism class of E. We will then start with the identification of this quadratic form. §VII.17 The trace form of an ´ etale algebra Definition VII.17.1. Let E be an étale algebra of rank n. For all a ∈ E, we denote by TrE/k (a) the trace of left multiplication by a in E. The trace form of E is the quadratic form TE defined by TE :

E −→ k x −→ TrE/k (x2 ).

Example VII.17.2. If E = k n , then it is easy to check that the canonical basis e1 , . . . , en of k n is an orthogonal basis for TE , and that TE (ei ) = 1 for 1 ≤ i ≤ n. Therefore, TE 1, . . . , 1. Lemma VII.17.3. Let E and E be two étale k-algebras and let K/k be a field extension. Finally, let a ∈ E and a ∈ E . Then the following properties hold: (1)

TrE×E /k ((a, a )) = TrE/k (a) + TrE /k (a ).

(2)

TrEK /K (a ⊗ 1) = TrE/k (a).

(3)

If ϕ : E −→ E is an isomorphism of k-algebras, then TrE/k (a) = TrE /k (ϕ(a)).

(4)

If σ1 , · · · , σn are the distinct elements of X(E), then TrE/k (a) = σ1 (a) + . . . + σn (a).

∼

Proof. For a ∈ E, we will denote by a the endomorphism of left multiplication by a in E. (1)

Let e = (e1 , . . . , en ) and e = (e1 , . . . , em ) be k-bases of E and E respectively. Then e = ((e1 , 0), . . . , (en , 0), (0, e1 ), . . . , (0, em )) is

VII.17 The trace form of an étale algebra

193

a k-basis of E×E . Now if M = Mat(a , e) and M = Mat(a , e ), then we have M 0 Mat((a,a ) , e ) = . 0 M The result follows immediately. (2)

If e = (e1 , . . . , en ) is a k-basis of E, then e = (e1 ⊗ 1, . . . , en ⊗ 1) is a K-basis of EK . Clearly, we have Mat(a⊗1 , e ) = Mat(a , e), hence the result.

(3)

If e = (e1 , . . . , en ) is a k-basis of E, then the family ϕ(e) = (ϕ(e1 ), . . . , ϕ(en )) is a k-basis of E . It is easy to check that we have Mat(ϕ(a) , ϕ(e)) = Mat(a , e), and the result follows.

(4)

Assume first that E = k n . In this case, the matrix of left multiplication by a = (a1 , · · · , an ) in the canonical basis is the diagonal matrix ⎛ ⎞ a1 ⎜ ⎟ .. ⎝ ⎠. . an Therefore, TrE/k (a) = a1 +. . .+an in this case. Since the elements of X(E) are the n projections, we are done. Now assume that E is an arbitrary étale k-algebra of rank n. By definition, we have a ks -algebra isomorphism ∼

f : Eks −→ ksn . Using (2) and (3), we get TrE/k (a) = TrEks /ks (a ⊗ 1) = Trksn /ks (f (a ⊗ 1)). Moreover, if σ1 , . . . , σn are the distinct elements of X(E), then the morphisms (σ1 )ks ◦ f −1 , . . . , (σn )ks ◦ f −1 are the n elements of X(ksn ). Using the previous case, we get TrE/k (a) =

n i=1

((σi )ks ◦ f −1 )(f (a ⊗ 1)) = σ1 (a) + . . . + σn (a).

194

Galois embedding problems and the trace form This concludes the proof.

The next result is an easy consequence of the previous lemma, and its proof is left to the reader. Lemma VII.17.4. Let E and E be two étale algebras of rank n over k. Then the following properties hold: (1)

TE×E TE ⊥ TE .

(2)

For any field extension K/k, we have TEK (TE )K .

(3)

If E E , then TE TE .

Corollary VII.17.5. For any étale k-algebra E, the trace form is a regular quadratic form whose isomorphism class only depends on the isomorphism class of E. Proof. It only remains to prove that TE is regular. But this property is invariant by scalar extension. Since we have (TE )ks TEks Tksn 1, . . . , 1, we are done. Recall now that for any field K, we have an injective morphism Sn −→ GLn (K), which sends a permutation s ∈ Sn to the matrix of the automorphism of ϕs of K n which sends ei to es(i) . In fact, this automorphism preserves the quadratic form 1, . . . , 1, so we get a morphism Sn −→ On (K). In particular, we get an injective group morphism ι : Sn −→ On (ks ), and it is easy to check that it is a morphism of Gks -groups. Lemma VII.17.6. The map H 1 (k, Sn ) −→ H 1 (k, On ) induced by the injection Sn −→ On (ks ) is the map E → TE . ∼

Proof. Let f : Eks −→ ksn be an isomorphism of ks -algebras, so that the map

α:

Gks −→ Sn σ −→ f ◦ σ·f −1

is a cocycle representing E. We saw that f induces an isomorphism TEks Tksn , that is an isomorphism (TE )ks 1, . . . , 1. Hence, by Galois descent, the map ι ◦ α is also a cocycle representing TE .

VII.17 The trace form of an étale algebra

195

Definition VII.17.7. The discriminant of E is the square-class dE = det(TE ) ∈ k × /k ×2 . Notice that the discriminant commutes with scalar extensions. Lemma VII.17.8. Let ε : Sn −→ {±1} = μ2 (ks ) be the signature morphism. Then ε∗ : H 1 (k, Sn ) −→ H 1 (k, μ2 ) is the map E −→ dE . Proof. Since ε is the composite map Sn

/ On (ks )

det

/ μ2 (ks ) ,

the previous lemma and Proposition IV.11.2 give the result. Lemma VII.17.9. Let E/k be a separable field extension of degree n, let α be a primitive element of E and let α1 = α, . . . , αn be its conjugates. Then we have (αi − αj )2 ∈ k × /k ×2 . dE = i<j

Proof. A k-basis of E is 1, α, . . . , αn−1 , so the representative matrix of TE in this basis is given by B = (TrE/k (αi+j−2 ))i,j . Since by Lemma VII.17.3 (4) we have TrE/k (αm ) = α1m + . . . + αnm , it is easy to see that we have B = M t M , where M = (αij−1 )i,j . Now apply the Vandermonde determinant formula to get the desired equality. Corollary VII.17.10. Let E/k be a Galois extension of degree n, and let G be its Galois group. Let us identify G to a subgroup of Sn by letting G act on itself by left translations. Then dE is a square if and only if G ⊂ An . Proof. Let σ1 = IdE , σ2 , . . . , σn be the elements of G. Let us choose a primitive element α of E/k, and let αi = σi (α), i = 1, . . . , n. The previous lemma shows that dE is a square in k if and only if the element (αi − αj ) = (σi (α) − σj (α)) ∈ E x= i<j

i<j

lies in k, which means that it is fixed by all the elements of G. But we have σ·x = (σ ◦ σi (α) − σ ◦ σj (α)) = ε(sσ )x, i<j

where sσ ∈ Sn is the permutation induced on the elements of G by left multiplication by σ. Thus x ∈ k if and only if G is a subgroup of An .

196


Remark VII.17.11. One can easily show that a group G of order n, identified to a subgroup of Sn by acting on itself by left translations, is a subgroup of An if and only if it has odd order or its Sylow 2-subgroup is not cyclic. We would like to end this section by explaining how the trace form may be useful to count the number of real roots of a separable polynomial of Q[X]. Proposition VII.17.12. Let f ∈ Q[X] be a non-zero separable polynomial, and let E = Q[X]/(f ). Let r1 be the number of real roots over f , and let r2 be the number of pairwise non-conjugate complex roots of f . Then we have (TE )R r1 × 1 ⊥ r2 × 1, −1. In particular, the signature of the trace form TE is the number of real roots of f . Proof. Without loss of generality, one can assume that f is monic. By Lemma VII.17.4, we have (TE )R TER . Now we have ER R[X]/(f ). Let f1 , . . . , fm ∈ R[X] be the distinct monic irreducible factors of f . By the Chinese Remainder Theorem, we have E ⊗Q R R[X]/(f1 ) × · · · × R[X]/(fm ). If fi has degree 1, then R[X]/(fi ) R. If fi has degree 2, then R[X]/(fi ) C. Therefore, we have an isomorphism of R-algebras E ⊗Q R Rr1 × Cr2 , where r1 is the number of real roots of f and r2 is the number of pairwise non-conjugate complex roots of f . Now it is easy to check that TR 1 and TC 1, −1. By Lemma VII.17.4 (4), we get (TE )R r1 × 1 ⊥ r2 × 1, −1. This concludes the proof. §VII.18 Computation of e∗ (sn ) This subsection is devoted to the proof of the following theorem, originally proved in [61] in a different way. Theorem VII.18.1 (Serre’s formula). Let E/k be an étale algebra represented by e : Gks −→ Sn . Then we have e∗ (sn ) = w2 (TE ) + (2) ∪ (dE ).

VII.18 Computation of e∗ (sn )

197

We will assume that n ≥ 4 and leave the remaining cases to the reader. Recall that ι : Sn −→ On (ks ) denotes the orthogonal representation of Sn by permutation matrices. Let X ⊂ Pinn (ks ) be the preimage of ι(Sn ) under the map αks : Pinn (ks ) −→ On (ks ), so that we have a commutative diagram 1

1

/ μ2 (ks )

/ μ2 (ks )

/X

/ Pinn (ks )

π

/ Sn

αks

/ On (ks )

/1

ι

/1

where π is just the composite of αks with ι−1 : ι(Sn ) −→ Sn . Recall for later use that the maps μ2 (ks ) −→ Pin(q)(ks ) and μ2 (ks ) −→ X are just inclusions. √ σ( 2) = (−1)χ(σ) . In particular, the map For any σ ∈ Gks , write √ 2 Gks −→ μ2 (ks ) σ −→ (−1)χ(σ) is a cocycle representing (2) ∈ H 1 (k, μ2 ). We also write ε(s) = (−1)ν(s) for any s ∈ Sn . The next lemma describes X as a Gks -group. Lemma VII.18.2. The extension of abstract groups 1 −→ {±1} −→ X −→ Sn −→ 1 is equivalent to 1 −→ {±1} −→ Sñ −→ Sn −→ 1. Moreover, for all σ ∈ Gks and all x ∈ X, we have σ·x = (−1)χ(σ)ν(π(x)) x. In particular, X is not isomorphic to Sñ as a Gks -group. Proof. Let e1 , . . . , en be the canonical basis of ksn , and let q0 = 1, . . . , 1, so that this basis is an orthogonal basis for q0 . Let (ij) ∈ Sn be a transposition, so its image in On (ks ) is the map fij which switches ei and ej and fixes em if m = i, j.

198


Let us check that we have fij = τei −ej . It is enough to prove that these maps coincide on a basis. First observe that q0 (ei − ej ) = q0 (ei ) − 2bq0 (ei , ej ) + q0 (ej ) = 2, so we have τei −ej (x) = x − bq0 (x, ei − ej )(ei − ej ). If m = i, j, we have fij (em ) = em . Since em is orthogonal to ei and to ej , hence to ei − ej , we have τei −ej (em ) = em = fij (em ). Moreover, since bq0 (ei , ej ) = 0 and bq0 (ei , ei ) = 1, we have τei −ej (ei ) = ei − (ei − ej ) τei −ej (ej ) = ej + (ei − ej )

= ej = ei

= =

fij (ei ); fij (ej ).

1 Hence the preimages of fij in Pinn (ks ) are ± √ (ei − ej ) by Corollary 2 IV.10.19. Now we have

2

1 √ (ei − ej ) 2

=

1 q0 (ei − ej ) = 1, 2

so these elements have order 2. Now let (ij)(kl) ∈ Sn be a product of two transpositions with disjoint supports. The image of this element under ι is τei −ej ◦ τek −el , and the corresponding preimages in Pinn (ks ) 1 are ± (ei − ej )(ek − el ). Since ei − ej and ek − el are orthogonal, they 2 anticommute in the Clifford algebra, so

2

1 (ei − ej )(ek − el ) 2

1 = − (ei − ej )2 (ek − el )2 = −1, 4

and thus these elements have order 4. Hence we proved that if x ∈ X is mapped under π onto a transposition (respectively a product of two transpositions with disjoint supports), then x has order 2 (respectively has order 4). These properties characterize Sñ . Now let us describe the action of Gks on X. Let x ∈ X and write π(x) = (i1 j1 ) · · · (ir jr ) as a product of transpositions with disjoint supports. 1 Then x = ± √ (ei1 − ej1 ) · · · (eir − ejr ). It is then clear that σ ·x = ( 2)r (−1)χ(σ) x if r is odd and that σ ·x = x if r is even, which is equivalent to the last statement of the lemma.

VII.18 Computation of e∗ (sn )

199

Proof of Theorem VII.18.1. The commutative diagram / μ2 (ks )

1

/X

/ μ2 (ks )

1

π

/ Sn

αks

/ On (ks )

/1

ι

/ Pinn (ks )

/1

induces a commutative diagram in cohomology H 1 (k, Sn )

δ1

/ H 2 (k, μ2 )

ι∗

H 1 (k, On )

/ H 2 (k, μ2 )

by Theorem III.7.39 (4). By Lemma VII.17.6, the first vertical map sends an étale algebra E to its trace form TE , and the second horizontal map sends a quadratic form q to w2 (q) by Corollary IV.11.7. Hence we have δ 1 (E) = w2 (TE ). Now we have to compute δ 1 (E) in a different way. Let e : Gks −→ Sn be a continuous morphism representing E. Let t : Sn −→ X be a section of π satisfying t(Id) = 1. By Theorem VI.15.6, the cocycle α : X × X −→ {±1} defined by αs,s = t(s)t(s )t(ss )−1 for all s, s ∈ Sn represents the extension 1 −→ {±1} −→ X −→ Sn −→ 1. Since this extension is equivalent to the extension 1 −→ {±1} −→ Sñ −→ Sn −→ 1 by Lemma VII.18.2, α is cohomologous to sn by the same theorem. Let xσ = t(eσ ), so that π(xσ ) = eσ for all σ ∈ Gks . By construction, δ 1 (E) is represented by the cocycle β : Gks −→ μ2 (ks ) defined by βσ,τ = xσ σ·xτ x−1 στ for all σ, τ ∈ Gks . By Lemma VII.18.2, we get βσ,τ = (−1)χ(σ)ν(eτ ) xσ xτ x−1 στ for all σ, τ ∈ Gks . Now we have −1 = αeσ ,eτ = e∗ (α)σ,τ . xσ xτ x−1 στ = t(eσ )t(eτ )t(eστ )

Therefore, βσ,τ = (−1)χ(σ)ν(eτ ) e∗ (α)σ,τ = (−1)χ(σ)ν(eτ ) e∗ (α)σ,τ . But the cocycle Gks −→ μ2 (ks ), σ −→ (−1)ν(eσ ) is just ε ◦ e, that is

200


ε∗ (e), which represents (dE ) ∈ H 1 (k, μ2 ) by Lemma VII.17.8. Hence the cocycle Gks × Gks −→ μ2 (ks ) (σ, τ ) −→ (−1)χ(σ)ν(eσ ) represents (2) ∪ (dE ). Now since α and sn are cohomologous, e∗ (α) is cohomologous to e∗ (sn ). Therefore, we get δ 1 (E) = e∗ (sn ) + (2) ∪ (dE ), hence the result. Remark VII.18.3. A generalization of Serre’s formula has been found by Fr¨ ohlich [24] in the context of orthogonal representations. This has been proved very useful to solve Galois embedding problems associated to central extensions of the type 1 −→ {±1} −→ X −→ G −→ 1. For more details and numerous references on the applications of this formula, the reader will refer to [16] and [24]. §VII.19 Applications to inverse Galois theory In this section, we would like to apply the previous result to the Galois inverse problem. Galois inverse problem: let k be a field. Describe all the finite groups which may occur as the Galois group of a field extension of k. This problem is not very thrilling if k is algebraically closed, real closed or even finite. However, this problem is particularly interesting (and far from being solved) if k = Q. It is conjectured that every finite group G may occur as a Galois group over a given number field. The most striking result in this direction is a deep theorem of Shafarevich which asserts that every finite solvable group occurs as a Galois group of a given finite field extension of Q (see [62]). The case of non-solvable finite groups is still open. We would like to apply the result of the previous sections in the case of central extensions with kernel {±1}. Let us start with the case of cyclic extensions of degree 4. Example VII.19.1. Let E/k be a Galois extension of group GE = ˜ E = Z/8Z. Then one can show that Z/4Z and discriminant d, so G 2 2 × d = a + b for some a, b ∈ k , and that there exists q ∈ k × such that

VII.19 Applications to inverse Galois theory 201 ) √ E = k( q(d + a d)) (see [20]). It is a good exercise on quadratic forms to check that TE 1, d, q, q (and in particular dE = d). Hence w2 (TE ) = (q) ∪ (q) = (q) ∪ (−1), and E/k can be embedded in a Galois extension of group Z/8Z if and only if (q) ∪ (−1) = (2) ∪ (d). Unfortunately, this does not generalize for cyclic groups of higher order. Indeed, one can show that Res(s2m ) ∈ H 2 (Z/2m Z, {±1}) is trivial for m ≥ 3, and so the group extension obtained is not Z/2m+1 Z (see [20], for example). ˜ E = Q8 . Now let E/k be a Galois extension of group GE = (Z/2Z)2 , so G √ √ × −1 ×2 / k , and the trace We have E = k( a, b), where a, b, ∈ k , ab ∈ form is TE 1, a, b, ab, so here dE = 1. It is easy to check that we have w2 (TE ) = (a) ∪ (b) + (−1) ∪ (ab), so E/k can be embedded in a Galois extension of group Q8 if and only if (a) ∪ (b) + (−1) ∪ (ab) = 0. Notice that (a) ∪ (b) + (−1) ∪ (ab) is also the Hasse-Witt invariant of the quadratic form a, b, (ab)−1 . Since quadratic forms of dimension 3 are classified by dimension, determinant and Hasse-Witt invariant (see [53, Chapter 2,Theorem 13.5]), one can reformulate the previous result as follows: √ √ Proposition VII.19.2. Let E = k( a, b)/k be a biquadratic field ˜ extension. Then E/k can be embedded in a Galois extension E/k of group Q8 if and only if we have an isomorphism of quadratic forms a, b, (ab)−1 1, 1, 1. This result is due to Witt (see [70]). Moreover, from an isomorphism ˜ between these quadratic forms, Witt gives an explicit construction of E. We now describe it briefly. Set q = a, b, (ab)−1 and q0 = 1, 1, 1. Let ∼ f : k3 −→ k 3 be an isomorphism of k-vector spaces satisfying q ◦ f = q0 . If P = (pij ) denotes the representative matrix of f in the canonical basis of k 3 , we have ⎛ ⎞ a ⎠ P = I3 . Pt ⎝ b (ab)−1 Therefore, we have det(P )2 = 1, that is det(f ) = ±1. Composing f on the right by a permutation endomorphism, we then may assume that

202


˜ defined det(f ) = 1. In this case, for any r ∈ k × the field extension E/k by * √ √ p33 ˜ E=k r 1 + p11 a + p22 b + √ √ a b is a Galois extension of Galois group Q8 . Moreover, every such extension may be obtained in this way (cf. [70]). Let us continue with more general considerations. If k is an arbitrary field, then e∗ (sn ) = 0 if dE = 1 and w2 (TE ) = 0. We would like to give examples of situations when this occurs. First we need a little lemma. Lemma VII.19.3. Let k be a field. Let r1 , r2 ≥ 0 be two integers, and let qr1 ,r2 be the quadratic form r1 × 1 ⊥ r2 × 1, −1. (1)

If r2 ≡ 0 mod 4, then det(qr1 ,r2 ) = 1 and w2 (qr1 ,r2 ) = 0.

(2)

If −1 is not a sum of two squares of k, the converse also holds.

Proof. Clearly, det(qr1 ,r2 ) = (−1)r2 . Moreover, we have w2 (qr1 ,r2 ) = w2 (r2 × −1) =

r2 (r2 − 1) (−1) ∪ (−1). 2

r2 (r2 − 1) are both even, hence the If r2 is a multiple of 4, then r2 and 2 first part of the lemma. Now assume that −1 is not a sum of two squares of k; in particular −1 ∈ / k ×2 . If r2 is odd, then dE = −1 ∈ k × /k ×2 , r2 (r2 − 1) is odd, so w2 (qr1 ,r2 ) = so dE = 1. Finally, if r2 ≡ 2 mod 4, 2 (−1) ∪ (−1). Since −1 is not a sum of two squares of k, this is non-zero by Proposition III.9.15 (2). This completes the proof of the lemma. The next result follows from a direct application of this lemma. Corollary VII.19.4. Let E/k be a separable field extension of finite degree. Assume that we have an isomorphism of quadratic forms TE qr1 ,r2 for some r1 , r2 ≥ 0. If r2 ≡ 0 mod 4, then dE = 1 and e∗ (sn ) = 0. In particular, this holds if TE n × 1. The following proposition describes exactly in which cases the Galois embedding problem has a solution for k = Q, provided that GE ⊂ An . It was originally proven in [61].

VII.19 Applications to inverse Galois theory

203

Proposition VII.19.5. Let E/Q be a finite extension of degree n. Denote by r1 (resp. r2 ) the number of real embeddings (resp. the number of pairwise non-conjugate complex embeddings) of E into C; in particular n = r1 + 2r2 . Then the following conditions are equivalent: (1)

dE = 1 and e∗ (sn ) = 0.

(2)

r2 ≡ 0 mod 4 and TE qr1 ,r2 .

Proof. The implication (2) ⇒ (1) follows from the previous corollary. Now assume that dE = 1 and e∗ (sn ) = 0. Therefore, dE = 1 and w2 (TE ) = 0. In particular, we have dE = 1 ∈ R× /R×2 . Since we have det((TE )R ) = det(TE ) ∈ R× /R×2 , we get det((TE )R ) = 1 ∈ R× /R×2 . We also have ResR/Q (w2 (TE )) = 0. Since the Hasse invariant commutes with scalar extension by Remark IV.11.9, we get w2 ((TE )R ) = 0. By Proposition VII.17.12, we have (TE )R qr1 ,r2 , and the two equalities above then yield det(qr1 ,r2 ) = 1 ∈ R× /R×2 and w2 (qr1 ,r2 ) = 0 ∈ H 2 (R, μ2 ). Since −1 is not a sum of squares in R, Lemma VII.19.3 implies that r2 ≡ 0 mod 4. This implies now that the quadratic form qr1 ,r2 over Q has trivial determinant and Hasse invariant. Notice also that dim(qr1 ,r2 ) = r1 + 2r2 = n. Hence we have the following equalities: dim(TE ) det(TE ) w2 (TE ) sign(TE )

= n = 1 = 0 = r1

= dim(qr1 ,r2 ) = det(qr1 ,r2 ) = w2 (qr1 ,r2 ) = sign(qr1 ,r2 ),

the last one coming from Proposition VII.17.12, and from the fact that the number of real embeddings of E equals the number or real roots of a polynomial defining E. By Theorem IV.12.5, this implies that TE qr1 ,r2 . This result is particularly interesting for the following reason. If the Galois embedding problem associated to E/k has a solution and TE √ qr1 ,r2 , then an element γ ∈ E gal such that E gal ( γ) is a Galois extension ˜ E may be computed by explicit methods. We refer with Galois group G to reader to [16] and the associated references for more details.

204


Even if the trace form TE may be computed and diagonalized by algorithmic methods, it is not always straightforward to decide whether or not (2) ∪ (dE ) + w2 (TE ) = 0, or if TE is isomorphic to qr1 ,r2 , especially over general fields. Therefore, it is preferable to find a nice diagonalization of this quadratic form, for example a diagonalization containing a lot of hyperbolic planes 1, −1. This can be done if a polynomial f defining E has sufficiently many zero coefficients. For example in [65], Vila computed the trace form of the étale algebra defined by a polynomial of the form f (X) = X n + a(bX + c)k , k odd, and used her results to construct Galois extensions of Q with group Añ for infinitely many values of n. If f (X) = X n + aX + b, the trace form has been computed by Serre in [61]. He obtains in particular the following result. Proposition VII.19.6. Assume that n is even and that char(k) n. Assume that the k-algebra E = k[X]/(X n + aX + b) is étale. Then we have n−2 × 1, −1. TE n, −(−1)n/2 ndE ⊥ 2 Proof. Let E = {x ∈ E | TrE/k (x) = 0}. Since TrE/k (1) = n = 0, it follows easily that we have E = k ⊥ E . Therefore, we have TE n ⊥ (TE )|E . Let α be the image of X in E, so 1, α, . . . , αn−1 is a k-basis of E. Let us denote by α1 = α, · · · , αn the images of α under the various k-algebra morphisms σi : E −→ ks . Notice that αi is a root of f (X) = X n +aX +b in kalg . By Lemma VII.17.3 (4), we have TrE/k (αj ) = α1j + . . . + αnj for all j ≥ 0. Since α1 , · · · , αn are also the distinct roots of f , using Newton identities, it is easy to check that we have TrE/k (αj ) = 0 for 1 ≤ j ≤ n − 2. Hence the linear subspace of E generated by α, · · · , α(n−2)/2 is totally isotropic. Therefore, we get (TE )|E c ⊥

n−2 × 1, −1. 2

Thus, we have TE n ⊥ c ⊥

n−2 × 1, −1. 2

Exercises

205

Comparing determinants, we see that c = −(−1)n/2 ndE ∈ k × /k ×2 . This concludes the proof. Corollary VII.19.7. Assume that n = 2(2m+1)2 for some m ≥ 1, and that char(k) n. Assume that the k-algebra E = k[X]/(X n + aX + b) is étale. Then w2 (TE ) = (2) ∪ (dE ). In particular, if E/k is a field, the associated Galois embedding problem has a solution. Proof. Assume that n = 2(2m + 1)2 for some m ≥ 1. In this case, n = 8r + 2 (with r = m2 + m) and the previous proposition implies that we have TE 2, 2dE ⊥ 4r × 1, −1. By Lemma IV.11.5, we have w2 (TE ) = w2 (2, 2dE ) + (dE ) ∪ ((−1)4r ) + w2 (4r × 1, −1). This yields w2 (TE ) = (2) ∪ (2dE ) +

4r(4r − 1) (−1) ∪ (−1) = (2) ∪ (2dE ). 2

Since 2 is a sum of two squares, we have (2) ∪ (2) = (2) ∪ (−1) = 0, by Proposition III.9.15. We now conclude the proof using the bilinearity of the cup-product. Example VII.19.8. Let n ≥ 1, and let f (X) = X n − X − 1 ∈ Q[X]. Set E = Q[X]/(f ). By a result of Selmer [56], f is irreducible over Q and Gal(E gal /Q) Sn . Therefore, if n = 2(2m + 1)2 , the previous ˜ corollary implies that there exists a Galois extension E/Q with Galois group Sñ containing E. Further results in inverse Galois theory and Galois embedding problems may be found in [60].

Exercises 1. Let E/k be a Galois extension of group G. If n = |G| is odd, show that TE n × 1.

206

Galois embedding problems and the trace form 2. Let E and E be two étale k-algebras. Show that TE⊗k E TE ⊗ TE . 3. Let E/k be a separable field extension. (a)

Assume that E is the compositum of pairwise linearly disjoint field extensions E1 , . . . , Er of even degree. Show that [TE ] ∈ I r (k).

(b)

Assume that E/k is a Galois extension of group G. Let r be the 2-rank of G, that is the largest integer r ≥ 0 such that G contains a subgroup isomorphic to (Z/2Z)r . Show that [TE ] ∈ I r (k). Hint: If K/k is a separable field extension, and if [q] ∈ I r (K), then [TrK/k (q)] ∈ I r (k) (see [53, Corollary 14.9, p. 93]). Moreover, if k ⊂ K ⊂ E, then TrE/k = TrK/k ◦ TrE/K .

(c)

Assume that E/k is a Galois extension of group G, and let r be the 2-rank of G. If r ≥ 3, deduce carefully from the previous questions that e∗ (sn ) = 0, that is the associated Galois embedding problem has a solution.

4. Prove Serre’s formula for n = 1, 2, 3. 5. Let Sˆn be the central extension of Sn by {±1} corresponding to εn ∪ εn ∈ H 2 (Sn , {±1}). Let E/k be a separable field extension. ˆ E be the Let GE ⊂ Sn be the Galois group of E gal /k, and let G central extension of GE corresponding to Res(εn ∪ εn ). Compute the obstruction to the associated embedding problem.

VIII Galois cohomology of central simple algebras

In this chapter, we study the cohomology of central simple algebras with or without involution. The first half is devoted to the proof of the isomorphism Br(k) H 2 (k, Gm ). In the second half, we will define several quadratic forms attached to an algebra with involution. As an illustration of Galois cohomology techniques, we will compute their Hasse invariants. These quadratic forms will be useful later when dealing with rationality problems of algebraic groups. §VIII.20 Central simple algebras Let us recollect some results on central simple k-algebras. Let k be a field. If A is a central simple k-algebra, we denote by Aop the opposite k-algebra . As a set, Aop = {aop | a ∈ A}, and we define the various operations on Aop as follows: for all a, b ∈ A, λ ∈ k, we have aop + bop = (a + b)op , λaop = (λa)op , aop bop = (ba)op . Then Aop is also a central simple k-algebra. The tensor product of two central simple k-algebras is again a central simple k-algebra. In particular, if A is a central simple k-algebra, so is Mr (A) Mr (k) ⊗k A for all r ≥ 1. We may then define an equivalence relation on the set of central simple k-algebras as follows. We say that two central simple 207

208

Galois cohomology of central simple algebras

k-algebras A and B are Brauer-equivalent if there exist two integers r, s ≥ 1 such that Mr (A) Ms (B). The Brauer-equivalence class of A is denoted by [A], and the set of Brauer-equivalence classes is denoted by Br(k). One can show that the operations [A] + [B] = [A ⊗k B], −[A] = [Aop ] endow Br(k) with the structure of an abelian group, whose neutral element is the class of k. The group Br(k) is called the Brauer group of k. If K is a field and L/K is a field extension, then for every central simple K-algebra A, the class [AL ] only depends on [A]. Therefore, we have a group morphism ResL/K : Br(K) −→ Br(L). We then get a functor Br(− ) : Ck −→ AbGrps. Finally, a field extension L/k is called a splitting field of A if AL Mn (L) for some n ≥ 1. We also say that L/k splits A or that A is split by L/k. Every central simple k-algebra can be split by a finite Galois field extension L/k. A maximal commutative subfield of A is a subalgebra L of A which is also a field, and such that degk (A) = [L : k] (that is dimk (A) = [L : k]2 ). A maximal commutative subfield L of A is a splitting field of A. More precisely, we have an isomorphism of L-algebras ∼

f:

AL −→ EndL (A) a ⊗ λ −→ (z −→ azλ),

where A is viewed as a right L-vector space. One can show that for any central simple k-algebra A, there exists a central simple k-algebra A which is Brauer-equivalent to A, and which contains a Galois extension L/k as a maximal commutative subfield. If u ∈ A× , we denote by Int(u) the automorphism Int(u) :

A −→ A x −→ uxu−1 .

Such an automorphism is called inner. By Skolem-Noether’s theorem, every k-automorphism of A is inner. We refer to [19] for proofs of these facts.

VIII.20 Central simple algebras

209

If A is a central simple k-algebra, we denote by PGL1 (A) the algebraic group-scheme of automorphisms of A. In particular, we have PGL1 (Mn (k)) = PGLn . If K/k is a field extension, as for the case of matrix algebras, we have × PGL1 (A)(K) A× K /K ,

since every automorphism of AK is inner. For any commutative k-algebra R, recall that the algebraic group-scheme GL1 (A) is defined by GL1 (A)(R) = A× R for any commutative k-algebra R. In particular, GL1 (Mn (k)) = GLn . If Ω/k be a Galois extension, and A is a central simple k-algebra which is split by Ω, Galois descent shows that A is obtained by twisting Mn (Ω) by an appropriate cocycle. We are going to show that the same is true at the level of invertible elements and automorphisms. In fact, this happens to be true for any pairs of k-algebras which become isomorphic over Ω. Let A and B be two arbitrary finite dimensional k-algebras such that AΩ BΩ , and let γ be a cocycle with values in Autalg (A)(Ω) representing the isomorphism class of B. Since any automorphism of the Ω-algebra AΩ induces a group automorphism of A× Ω , γ induces by restriction a cocycle of GΩ with values in the automorphism group of A× Ω, that we will denote by β. Lemma VIII.20.1. Keeping the notation above, we have isomorphisms of GΩ -groups GL1 (A)(Ω)β GL1 (B)(Ω) and Autalg (A)(Ω)γ Autalg (B)(Ω). Proof. Twisting by two cohomologous cocycles giving rise to two isomorphic GΩ -groups, one may assume that we have γσ = ϕ σ·ϕ−1 for all σ ∈ GΩ , ∼

where ϕ : BΩ −→ AΩ is an isomorphism of Ω-algebras. The map ϕ × ∼ −→ (A× induces by restriction a group isomorphism ϕ : BΩ Ω )β , since × × (AΩ )β and AΩ are equal as abstract groups. Therefore, it remains to × . By check that ϕ is GΩ -equivariant. Let σ ∈ GΩ , and let b ∈ BΩ × definition of the GΩ -action on (AΩ )β , we have σ ∗ ϕ(b) = βσ (σ·ϕ(b)) = γσ (σ·ϕ(b)) = γσ ((σ·ϕ)(σ·b)).

210


Thus we have σ ∗ ϕ(b) = ϕ(σ·b), meaning that ϕ is GΩ -equivariant. This concludes the proof of the first part. Now we have a group isomorphism ∼

ψ:

Aut(BΩ ) −→ Aut(AΩ ) f −→ ϕ ◦ f ◦ ϕ−1 .

We are going to show that ψ is GΩ -equivariant. For, let σ ∈ GΩ , and let f ∈ Aut(AΩ ). By definition, we have σ ∗ ψ(f ) = γσ ◦ (σ·ψ(f )) ◦ γσ−1 . Since we have σ·ψ(f ) = σ·(ϕ ◦ f ◦ ϕ−1 ) = (σ·ϕ) ◦ (σ·f ) ◦ (σ·ϕ)−1 , we get σ ∗ ψ(f )

= ϕ ◦ (σ·ϕ)−1 ◦ (σ·ϕ) ◦ (σ·f ) ◦ (σ·ϕ)−1 ◦ (σ·ϕ) ◦ ϕ−1 = ϕ ◦ σ·f ◦ ϕ−1 = ψ(σ·f ).

Hence ψ is an isomorphism of GΩ -groups. This concludes the proof. The rest of this section is devoted to establishing an isomorphism of functors from Ck to AbGrps Br(− ) H 2 (− , Gm ). Our first goal is to define a natural transformation θ : Br(− ) −→ H 2 (− , Gm ). Let n ≥ 1 be an integer. For every field extension K/k, let 1 δn,K : H 1 (K, PGLn ) −→ H 2 (K, Gm )

be the first connecting map associated to the exact sequence 1 −→ Ks× −→ GLn (Ks ) −→ PGLn (Ks ) −→ 1. By the last part of Theorem III.7.39, these maps give rise to a natural transformation of functors δn1 : H 1 (− , PGLn ) −→ H 2 (− , Gm ). We denote by θn : CSAn −→ H 2 (− , Gm ) the natural transformation obtained by composing δn1 with the bijection CSAn H 1 (− , PGLn ) obtained by Galois descent.


211

Lemma VIII.20.2. Let K/k be a field extension. Let A and A be central simple K-algebras of degree n and m respectively. Then the following properties hold: (1)

We have θnm,K (A ⊗k A ) = θn,K (A) + θm,K (A ).

(2)

If A and A are Brauer-equivalent, then θn,K (A) = θm,K (A ). ∼

Proof. Let ϕ : AKs −→ Mn (Ks ) be an isomorphism of Ks -algebras, and let GKs −→ PGLn (Ks ) α: σ −→ ϕ σ·ϕ−1 . Write ασ = Int(Mσ ), for some Mσ ∈ GLn (Ks ). Then by definition, θn,K (A) is the cohomology class represented by the cocycle β ∈ Z 2 (GKs , KS× ) defined by −1 Mσ σ·Mτ Mστ = βσ,τ In for all σ, τ ∈ GKs . ∼

Similarly, let ϕ : AKs −→ Mm (Ks ) be an isomorphism of Ks -algebras, let GKs −→ PGLm (Ks ) α : σ −→ ϕ σ·ϕ−1 , and write ασ = Int(Mσ ), for some Mσ ∈ GLm (Ks ), so that θn,K (A ) is the cohomology class represented by the cocycle β ∈ Z 2 (GKs , KS× ) defined by

Mσ σ·Mτ Mστ−1 = βσ,τ In for all σ, τ ∈ GKs . ∼

Let u : Mn (Ks ) ⊗Ks Mm (Ks ) −→ Mnm (Ks ) be the isomorphism of Ks -algebras given by the Kronecker product of matrices. The map u ◦ (ϕ ⊗ ϕ ) : (A ⊗K A ) ⊗K Ks −→ Mnm (Ks ) is then an isomorphism of Ks -algebras. Straightforward computations show that we have σ·(u ◦ (ϕ ⊗ ϕ )) = u ◦ (σ·ϕ ⊗ σ·ϕ ) for all σ ∈ GKs . Hence a cocycle representing A ⊗K B is given by α :

GKs −→ PGLnm (Ks ) σ −→ u ◦ (ασ ⊗ ασ ) ◦ u−1 .

Now we have u ◦ (Int(Mσ ) ⊗ Int(Mσ )) = u ◦ (Int(Mσ ⊗ Mσ )) = Int(u(Mσ ⊗ Mσ )) ◦ u,

212


and thus ασ = Int(Mσ ), where Mσ = u(Mσ ⊗ Mσ ) for all σ ∈ GKs . We easily have

Mσ σ·Mτ Mστ−1

= u(βσ,τ In ⊗ βσ,τ In ) = βσ,τ βσ,τ u(In ⊗ Im ) Inm . = βσ,τ βσ,τ

It follows that θnm (A ⊗K A ) is represented by the cocycle ββ , that is θnm,K (A ⊗K A ) = [ββ ] = [β] + [β ] = θn,K (A) + θm,K (A ). Assume now that A and A are Brauer-equivalent. Let r, s ≥ 1 be two integers such that Mr (A) Ms (A ). Notice that we have rn = sm, as we can see by comparing dimensions. To prove the desired result, it is enough to check that θn,K (A) = θnr,K (Mr (A)) for all r ≥ 1. Indeed, we will have θn,K (A) = θnr,K (Mr (A)) = θsm,K (Mr (A)) = θsm,K (Ms (A )), that is θn,K (A) = θm,K (A ). Notice that we implicitely used here the fact that θn,K maps K-isomorphic algebras onto the same cohomology class, which is obvious from its definition. Now θr,K (Mr (K)) = 0, since Mr (K) corresponds to the class of the trivial cocycle in H 1 (K, PGLr ). By the first point, we have θnr,K (Mr (A)) = θnr,K (Mr (K) ⊗K A) = θr,K (Mr (K)) + θn,K (A), and therefore θnr,K (Mr (A)) = θn,K (A). This concludes the proof. Let [A] ∈ Br(K), and let n = degk (A). We set θK ([A]) = θn,K (A). Lemma VIII.20.2 implies immediately that the map θK : Br(K) −→ H 2 (K, Gm ) is a well-defined group morphism. The various maps θK then induce a natural transformation of functors θ : Br(− ) −→ H 2 (− , Gm ). Indeed, if L/K is a field extension and A is a central simple K-algebra of degree n, then AL has also degree n over L, and we have θL ([AL ]) = θn,L (AL ) = ResL/K (θn,K (A)) = ResL/K (θK ([A])).


213

Lemma VIII.20.3. The natural transformation of functors θ : Br(− ) −→ H 2 (− , Gm ) is injective. Proof. Let K/k be a field extension and let [A] ∈ Br(K) such that θK ([A]) = 0. Let n = degK (A). We then have 1 ([α]) = 0, θK ([A]) = θn,K (A) = δn,K

where [α] ∈ H 1 (K, PGLn ) is the cohomology class corresponding to A. Applying cohomology to the exact sequence 1 −→ Ks× −→ GLn (Ks ) −→ PGLn (Ks ) −→ 1 1 has trivial kernel. Hence, we get [α] = 1 and Hilbert 90 show that δn,K and thus A Mn (K). Therefore, [A] = 0 and θK is injective.

Lemma VIII.20.4. Let L/K be a finite Galois extension of degree n. The connecting map H 1 (GL , PGLn (L)) −→ H 2 (GL , L× ) associated to the exact sequence of GL -groups 1 −→ L× −→ GLn (L) −→ PGLn (L) −→ 1 is bijective. Proof. Let [c] ∈ H 2 (GL , L× ). Consider the product V =

L of n

g∈GL

copies of L, indexed by the elements of GL , and let (eg )g∈GL be the corresponding canonical basis of this product, considered as a right Lvector space. For g ∈ GL , let ag ∈ GL(V ) defined by ag (eg ) = egg cg,g for all g ∈ GL . For all g, g , g ∈ G, we have ag (g·ag (eg )) = ag (eg g g·cg ,g ) = egg g cg,g g g·cg ,g . Since c is a 2-cocycle, we have g·cg ,g cg,g g = cgg ,g cg,g , and therefore ag (g·ag (eg )) = egg g cgg ,g cg,g = agg (eg )cg,g . We then have ag ◦ g·ag ◦ a−1 gg = cg,g IdV . Let Mg ∈ GLn (L) be the matrix of ag in the basis (eg )g∈GL . It follows from the previous computation that the map α:

GL × GL −→ L× g −→ Int(Mg )

214


is a cocycle with values in PGLn (L) whose cohomology class maps onto 1 : H 1 (GL , PGLn (L)) −→ H 2 (GL , L× ). [c] under the connecting map δL Let us prove the injectivity of this connecting map. Notice first that, 1 as before, δL has trivial kernel. Now let γ ∈ Z 1 (GL , PGLn (L)) be a cocycle. The cocycle β ∈ Z 1 (GL , Aut(GLn (L))) deduced from γ as explained just before Proposition II.5.6 is easily seen to be GL −→ Aut(GLn (L)) σ −→ γσ|GL

n (L)

.

By the first part of Proposition II.5.6, we have an exact sequence 1 −→ L× −→ GLn (L)β −→ PGLn (L)γ −→ 1, and by Lemma VIII.20.1, this is nothing but that the exact sequence 1 −→ L× −→ GL1 (A)(L) −→ PGL1 (A)(L) −→ 1 where A is the central simple K-algebra corresponding to [γ]. We denote 1 the corresponding connecting map. Since H 1 (GL , GL1 (A)(L)) = by δA,L 1 by Hilbert 90, applying cohomology to the exact sequence above shows 1 1 has trivial kernel. By Lemma II.5.5, δL is injective, and this that δA,L concludes the proof. Remark VIII.20.5. By Galois descent, for any c ∈ Z 2 (GL , L× ), the cohomology class [c] ∈ H 2 (GL , L× ) corresponds to the isomorphism class of the central simple K-algebra Ac = {M ∈ Mn (L) | Mg σ·M = M Mg for all g ∈ GL }. We now have the following result: Theorem VIII.20.6. The natural transformation of functors θ : Br(− ) −→ H 2 (− , Gm ) is an isomorphism. Moreover, for every field extension K/k and every central simple K-algebra A of degree n, the connecting map 1 : H 1 (K, PGLn ) −→ H 2 (K, Gm ) δn,K

corresponds to the map CSAn (K) −→ Br(K) A −→ [A].


215

Proof. Let K/k be a field extension. By Lemma VIII.20.3, θK is injective. Let us prove the surjectivity of θK . Let [ξ] ∈ H 2 (K, Gm ). By Theorem III.7.30, there exists a finite Galois extension L/K and [c] ∈ H 2 (GL , L× ) such that [ξ] = ρL ([c]), where ρL : H 2 (GL , L× ) −→ H 2 (K, Gm ) is the map induced in cohomology by the compatible maps GΩ −→ GL and L× −→ Ks× . Abusing notation, we will also denote by ρL the map H 1 (GL , PGLn (L)) −→ H 1 (K, PGLn ). By Lemma VIII.20.4, there exists [α] ∈ H 1 (GL , PGLn (L)) which is mapped onto [c] by the connecting map H 1 (GL , PGLn (L)) −→ H 2 (GL , L× ). It is easy to check that the diagram H 1 (GL , PGLn (L))

/ H 2 (GL , L× )

H 1 (K, PGLn )

/ H 2 (K, Gm )

is commutative. Therefore, ρL ([α]) is mapped onto ρL ([c]) = [ξ] by 1 . Notice that the isomorphism class of Ac then corresponds to the δn,K cohomology class ρL ([α]) ∈ H 1 (K, PGLn ) by Remark VIII.20.5 and Remark III.8.18. Thus we get θK ([Ac ]) = [ξ] and θK is surjective. We now prove the last statement of the theorem. Let A be a central simple K-algebra of degree n, and let [α] ∈ H 1 (K, PGLn ) be the corre−1 1 (δn,K ([α])) = [A], sponding cohomology class. We have to prove that θK 1 that is δn,K ([α]) = θK ([A]). But this comes from the definition of θK . This concludes the proof. Remark VIII.20.7. This result admits a slight generalization. For any Galois field extension Ω/K, we have a group isomorphism Br(Ω/K) H 2 (GΩ , Ω× ), where Br(Ω/K) = ker(Br(K) −→ Br(Ω)), and the connecting map H 1 (GΩ , PGLn (Ω)) −→ H 2 (GΩ , Ω× ) corresponds to A −→ [A].

216


The arguments needed to prove this result are almost identical to the ones used to establish the previous theorem. Since we will not really need this generalization, we leave the details to the reader. Corollary VIII.20.8. Let n ≥ 2, and let k be a field whose characteristic does not divide n. We have a group isomorphism H 2 (k, μn ) Brn (k), where Brn (k) is the n-torsion part of Br(k). Proof. The exact sequence / μn

1

/ k× s

×·n

/ ks

/1

induces an exact sequence in cohomology (using Hilbert 90) 1 −→ H 2 (k, μn ) −→ H 2 (k, Gm ) −→ H 2 (k, Gm ), where the last map is induced by raising to the nth -power. Hence the abelian group H 2 (k, μn ) identifies to the n-torsion part of H 2 (k, Gm ), and therefore to the n-torsion part of Br(k) by the previous theorem. Remark VIII.20.9. The isomorphism above is described as follows: if [A] ∈ Brn (k) ⊂ Br(k), then the corresponding cohomology class [α] ∈ H 2 (k, μn ) is the unique element of H 2 (k, μn ) which satisfies ι∗ ([α]) = θk ([A]), where ι∗ : H 2 (k, μn ) −→ H 2 (k, Gm ) is the map induced by the inclusion μn (ks ) ⊂ ks× . Corollary VIII.20.10. Let k be a field of characteristic different from 2. The isomorphism Br2 (k) H 2 (k, μ2 ) maps the Brauer class of a quaternion algebra (a, b) onto (a) ∪ (b). Proof. The commutative diagram 1

/ μ2

/ SL2 (ks )

/ PGL2 (ks )

/1

1

/ k×

/ GL2 (ks )

/ PGL2 (ks )

/1

s

VIII.21 Algebras with involutions

217

induces a commutative diagram / H 2 (k, μ2 )

H 1 (k, PGL2 )

H 1 (k, PGL2 )

δ1

/ H 2 (k, Gm )

by Theorem III.7.39 (4). Let [ξ] ∈ H 1 (k, PGL2 ) be the cohomology class corresponding to the isomorphism class of Q. By Proposition III.9.19 and the commutativity of the diagram above, we have θk ([Q]) = δ 1 ([ξ]) = ι∗ ((a) ∪ (b)), the first equality coming from the definition of θ. The result follows. §VIII.21 Algebras with involutions In this section, k is a field, and A is a central simple k-algebra. To simplify the exposition, we will assume that char(k) = 2.

VIII.21.1 Basic concepts Definition VIII.21.1. An involution σ : A −→ A is a ring antiautomorphism of order dividing 2. In other words, an involution is a map σ : A −→ A satisfying for all x, y ∈ A: (1)

σ(x + y) = σ(x) + σ(y).

(2)

σ(1) = 1.

(3)

σ(xy) = σ(y)σ(x).

(4)

σ(σ(x)) = x.

For example, the transposition is an involution on Mn (k). An element x ∈ A will be called symmetric if σ(x) = x, and skewsymmetric if σ(x) = −x. We denote by Sym(A, σ) the set of symmetric elements, and by Skew(A, σ) the set of skew-symmetric elements. We also set Sym(A, σ)× = Sym(A, σ) ∩ A× and Skew(A, σ)× = Skew(A, σ) ∩ A× . It easy to check that for every λ ∈ k, σ(λ) lies in the center of A, that is k. Hence σ|k is an automorphism of order dividing 2 of k.

218


We set k0 = {λ ∈ k|σ(λ) = λ}. We say that σ is an involution of the first kind if σ|k = Idk , that is if k = k0 , and an involution of the second kind (or unitary) otherwise. In this latter case, k/k0 is a quadratic field extension, and σ|k is the unique non-trivial k0 -automorphism of k/k0 . Remark VIII.21.2. Let A be a central simple k-algebra. If A carries an involution of the first kind, then [A] ∈ Br2 (k). Indeed, if σ is such an involution, the map f:

A −→ Aop a −→ (σ(a))op

is easily checked to be an isomorphism of k-algebras. In particular, we have [A] = [Aop ] = −[A], that is 2[A] = 0. From now on, we will only consider involutions of the first kind. Proposition VIII.21.3. Let σ, σ be two involutions of the first kind on A. Then there exists u ∈ A× such that σ = Int(u) ◦ σ and σ(u) = ±u. Moreover, u is uniquely determined up to multiplication by an element of k × . Proof. Notice that σ ◦ σ −1 is a k-algebra automorphism of A. By Skolem-Noether’s theorem, there exists u ∈ A× such that σ ◦ σ −1 = Int(u), that is σ = Int(u) ◦ σ. Easy computations show that we have σ◦Int(u) = Int(σ −1 (u))◦σ. Thus, we have σ 2 = Int(u) ◦ Int(σ −1 (u)) ◦ σ 2 = Int(uσ −1 (u)) ◦ σ 2 . Since σ and σ have order dividing 2, we get IdA = Int(uσ −1 (u)), so uσ −1 (u) lies in the center of A. Hence there exists λ ∈ k such that u = λσ(u). We then have u = λσ(λσ(u)) = λuσ(λ) = λσ(λ)u, since σ(λ) lies in the center of A. Since u ∈ A× , we get λσ(λ) = λ2 = 1. Therefore λ = ±1, and it follows that σ(u) = ±u.


219

Finally, if σ = Int(u1 )◦σ = Int(u2 )◦σ, where u1 , u2 are both symmetric or skew-symmetric, we have Int(u2 u−1 1 ) ◦ σ = σ. × It implies that Int(u2 u−1 1 ) = IdA . Thus, as before, there exists λ ∈ k such that

u2 = λu1 . This concludes the proof. Remark VIII.21.4. This proposition shows in particular that involutions of first kind on Mn (k) have the form σB = Int(B)◦t , where B is an invertible symmetric or skew-symmetric matrix. Lemma VIII.21.5. Let σ be an involution on A, and let u ∈ A× such that σ(u) = ±u. Set σ = Int(u) ◦ σ. (1)

If σ(u) = u, we have Sym(A, σ ) = uSym(A, σ) = Sym(A, σ)u−1 and Skew(A, σ ) = uSkew(A, σ) = Skew(A, σ)u−1 .

(2)

If σ(u) = −u, we have Sym(A, σ ) = uSkew(A, σ) = Skew(A, σ)u−1 and Skew(A, σ ) = uSym(A, σ) = Sym(A, σ)u−1 .

Proof. Assume that σ(u) = εu. For all x ∈ A, we have σ (x)

= = = =

uσ(x)u−1 ε(εu)σ(x)u−1 εσ(u)σ(x)u−1 εσ(xu)u−1 .

Similarly, we have σ (x) = εuσ(u−1 x). Therefore, we have σ (x) = ε x ⇐⇒ σ(xu) = εε xu ⇐⇒ σ(u−1 x) = εε u−1 x. The lemma follows easily. We now introduce the concept of isomorphic algebras with involutions.

220


Definition VIII.21.6. We say that two central simple k-algebras with involutions (A, σ) and (A , σ ) are isomorphic if there exists an isomor∼ phism of k-algebras f : A −→ A such that σ ◦ f = f ◦ σ. In this case, it is easy to check that σ and σ are involutions of the same kind. Moreover, f then induces isomorphisms of k-vector spaces Sym(A, σ) Sym(A , σ ) and Skew(A, σ) Skew(A , σ ). Two involutions σ, σ on A are conjugate if (A, σ) and (A, σ ) are isomorphic. Since every automorphism of A is inner, it follows that σ and σ are conjugate if and only if there exists a ∈ A× such that σ = Int(aσ(a)) ◦ σ. Let σ be an involution of the first kind on A, let L/k be a splitting field of A, and let ∼

f : AL −→ Mn (L) be an isomorphism of L-algebras. Let us denote by σL the involution σ ⊗ IdL . Then f ◦ σL ◦ f −1 is an involution of the first kind on Mn (L). In view of Remark VIII.21.4, f ◦ σL ◦ f −1 = σB for some invertible symmetric or skew-symmetric matrix. In other words, for every splitting field L/k, there exists B ∈ GLn (L) and B t = ±B such that (AL , σL ) (Mn (L), σB ). The next result allows us to decide in which case B is symmetric or skew-symmetric. Proposition VIII.21.7. Assume that σ is an involution of the first kind on a central simple k-algebra A of degree n. Then we have dimk (Sym(A, σ)) =

n(n − 1) n(n + 1) or . 2 2

The second case only holds if n is even. n(n + ε) if and only if for every 2 splitting field L/k of A, (AL , σL ) is isomorphic to (Mn (L), σB ) for some B ∈ GLn (L) satisfying B t = εB. Moreover, we have dimk (Sym(A, σ)) =


221

Proof. For every field extension L/k, it is easy to check that we have Sym(AL , σL ) = Sym(A, σ)L . Assume that L/k is a splitting field of A, so (AL , σL ) is isomorphic to (Mn (L), σB ) for some invertible symmetric or skew-symmetric matrix B. We then have Sym(A, σ)L = Sym(AL , σL ) Sym(Mn (L), σB ). Therefore, we get dimk (Sym(A, σ)) = dimL (Sym(A, σ)L ) = dimL (Sym(Mn (L), σB )). Assume first that B is symmetric. By Lemma VIII.21.5, we have Sym(Mn (L), σB ) = B Sym(Mn (L), t), and therefore we get n(n + 1) . 2 If B is skew-symmetric, by Lemma VIII.21.5, we have dimk (Sym(A, σ)) = dimL (Sym(Mn (L), t)) =

Sym(Mn (L), σB ) = B Skew(Mn (L), t), and therefore we get dimk (Sym(A, σ)) = dimL (Skew(Mn (L), t)) =

n(n − 1) . 2

Moreover in this case, we have det(B) = det(B t ) = det(−B) = (−1)n det(B). Since B is invertible, we get 1 = (−1)n , which implies that n is even. This concludes the proof. Definition VIII.21.8. We say that an involution σ of the first kind on a central simple k-algebra A of degree n is orthogonal (or of type 1 ) if for any splitting field L of A, the involution σL is isomorphic to σB for some symmetric invertible matrix B ∈ Mn (L). It is equivalent n(n + 1) . We say that σ is symplectic to say that dimk (Sym(A, σ)) = 2 (or of type −1) if, for any splitting field L of A, the involution σL is isomorphic to σB for some skew-symmetric invertible matrix B ∈ Mn (L). n(n − 1) . If A carries a It is equivalent to say that dimk Sym((A, σ)) = 2 symplectic involution, then n is necessarily even.

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Examples VIII.21.9. (1)

The involution σB on Mn (k) is orthogonal if B is symmetric and symplectic if B is skew-symmetric.

(2)

Let Q = (a, b)k . The map γ:

Q −→ Q x + yi + zj + tij −→ x − yi − zj − tij

is a symplectic involution on Q. Let K be a field. If ε = ±1 and n ≥ 1, we denote by CSAεn (K) the set of isomorphism classes of algebras with involution (A, σ), where degK (A) = n and σ has type ε. If L/K is a field extension and (A, σ) ∈ CSAεn (K), then (AL , σL ) ∈ CSAεn (L). We then get a functor CSAεn : Ck −→ Sets.

VIII.21.2 Hyperbolic involutions In this section, we define the concept of a hyperbolic involution, which generalizes in some sense the notion of a hyperbolic quadratic form or hyperbolic alternating form. Let us start with some easy considerations. Lemma VIII.21.10. Let σbe an involution on Mn (k). (1)

If σ is symplectic, then there exists E ∈ Mn (k) satisfying E 2 = E and σ(E) = In − E.

(2)

If σ = σB is orthogonal, then there exists E ∈ Mn (k) satisfying E 2 = E and σ(E) = In − E if and only if the quadratic form associated to B is hyperbolic.

n In both cases, such an element E satisfies rank(E) = . In particular 2 n is even. Proof. Write σ = σB for some B ∈ GLn (k) satisfying B t = ±B, and let b:

k n × k n −→ k (X, Y ) −→ X t BY

be the bilinear form associated to B. Assume first that an idempotent E of Mn (k) such that σ(E) = In − E


223

exists. Since E 2 = E, the endomorphisms corresponding to E and In −E are projectors. Thus we have k n = Im(E) ⊕ Im(In − E). But σ(E) = In −E and E have the same rank (since the rank is preserved by transposition and multiplication by an invertible matrix). Therefore n = 2rank(E). Assume now that B t = −B, so that σ is symplectic. In particular, n = 2m. It is known in this case that there exists a basis E1 , F1 , . . . , Em , Fm of k n such that b(Ei , Fi ) = 1, b(Fi , Ej ) = b(Fi , Fj ) = 0 for all i = j. It follows that there exists P ∈ GLn (k) such that 0 Im t P. B=P −Im 0 Im 0 Let E = P t (P t )−1 . It is easy to check that we have 0 0 E 2 = E and σB (E) = In − E. Assume that B t = B, so that σ is orthogonal and b is symmetric. Suppose that the quadratic form q:

k n −→ k X −→ X t BX

is hyperbolic. Then it is known that in this case that n = 2m, and that there exists a basis E1 , F1 , . . . , Em , Fm of k n such that b(Ei , Fi ) = 1, b(Fi , Ej ) = b(Fi , Fj ) = 0 for all i = j. Similar computations to those done in the previous case show that there exists E ∈ Mn (k) satisfying E 2 = E and σB (E) = In − E. Conversely, assume that such an element E ∈ Mn (k) exists. We have to prove that q has a totally isotropic subspace of dimension m. Let W be the image of E t . Then dimk (W ) = rank(E) = m. We claim that W is totally isotropic. To check it, it is enough to show that EBE t = 0.

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By assumption, we have E 2 = E and BE t B −1 = In − E. Therefore we have EBE t B −1 = E − E 2 = 0, and therefore EBE t = 0. Hence q is hyperbolic. This result motivates the following definition: Definition VIII.21.11. Let (A, σ) be a central simple k-algebra with an involution of the first kind. We say that σ is hyperbolic if there exists an idempotent e ∈ A such that σ(e) = 1 − e. Remark VIII.21.12. Notice that if A is a division algebra, then A does not carry any hyperbolic involution. Indeed, the only idempotents e of A are 0 and 1 in this case, which clearly do not satisfy the relation σ(e) = 1 − e for any involution σ of the first kind. In fact, if A carries a hyperbolic involution, one can show that A = M2 (B) for some central simple k-algebra B. See [30] for more details. Example VIII.21.13. Let σ = σB be an involution of the first kind on Mn (k). By Lemma VIII.21.10, if σ is symplectic, then σ is hyperbolic, and if σ is orthogonal, σ is hyperbolic if and only the quadratic form q:

k n −→ k X −→ X t BX

is hyperbolic. It is known that hyperbolic quadratic forms are isomorphic. This easily implies that all hyperbolic involutions on Mn (k) of same type are conjugate. We would like to generalize this fact to hyperbolic involutions on an arbitrary central simple k-algebra. We start with a lemma: Lemma VIII.21.14. Let A be a central simple k-algebra. If A carries a hyperbolic involution, then A has even degree. Moreover, if σ, σ are two hyperbolic involutions of the same type on A and e, e ∈ A are idempotents such that σ(e) = 1 − e and σ (e) = 1 − e , then there exists a ∈ A× such that e = aea−1 . Proof. Assume first that A = Mn (k), and let σ be a hyperbolic involution on A. In this case, we know from Lemma VIII.21.10 that n = 2m, where m is the rank of any idempotent E satisfying σ(E) = In − E. Now let σ be another hyperbolic involution of the same type, and let E be an idempotent satisfying σ (E ) = In − E . We then have


225

rank(E ) = rank(E) = m. Hence there exist two invertible matrices P and P such that Im 0 . P EP −1 = P E P −1 = 0 0 We then have (P this case.

−1

P )E(P

−1

P )−1 = E , and the result is proved in

Let us go back to the general case. If k is finite, A is isomorphic to a matrix algebra, and the result is already known. Hence one may assume ∼ that k is infinite. Let ϕ : Aks −→ Mn (ks ) be an isomorphism of ks algebras, let σs = ϕ ◦ σks ◦ ϕ−1 and let E = ϕ(e ⊗ 1). Then E 2 = E and σs (E) = In − E. Hence σs is hyperbolic, and therefore n is even by the previous point. Since n = degk (A), this proves the first part. Now let σs = ϕ ◦ σk s ◦ ϕ−1 and let E = ϕ(e ⊗ 1). Since σs is also hyperbolic, the first point shows that there exists an invertible matrix M ∈ Mn (ks ) such that M EM −1 = E . Let a = ϕ−1 (M ). Then a ∈ A× ks and e ⊗ 1 = a(e ⊗ 1)a−1 . Let W be the affine k-variety defined by W (R) = {a ∈ AR | a(e ⊗ 1) = (e ⊗ 1)a} for any commutative k-algebra R. This variety is not empty and isomorphic to an affine space, since it is the variety associated to a finite dimensional k-vector space. Therefore W is an affine rational variety. The subset U ⊂ W defined by U (R) = W (R) ∩ A× R is an open subset of W , and the previous considerations show that U (ks ) is not empty. Since k is infinite, it follows that U (k) = ∅, which is exactly what we wanted to prove. For those who are not familiar with this kind of geometric argument, one may also argue more explicitely as follows. Let us keep the notation of Example III.7.19 (2). Let w1 , . . . , wr be a k-basis of W , let e1 , . . . , em be a k-basis of A (where m = n2 ), and write wj =

m

aij ei for all j = 1, . . . , r.

i=1

Then a =

r j=1

wj ⊗ μj =

m i=1

r (ei ⊗ 1)( aij μj ) ∈ U (R) if and only if j=1

Q(μ1 , . . . , μr ) = 0, where Q ∈ k[Y1 , . . . , Yr ] is the polynomial

226

Galois cohomology of central simple algebras r r Q = PA ( a1j Yj , . . . , amj Yj ). j=1

j=1

Since U (ks ) is not empty, the polynomial Q is non-zero. Since k is infinite, this implies that Q does not vanish on k r , and therefore U (k) is not empty. We are now ready to prove the following result: Theorem VIII.21.15. Let A be a central simple k-algebra. Then two hyperbolic involutions of same type on A are conjugate. Proof. We are grateful to J.-P.Tignol for providing us the following arguments. Let σ, σ be two hyperbolic involutions of same type on A, and let e, e be the corresponding idempotents of A. By the previous lemma, there exists a ∈ A× such that e = aea−1 . Let σ = Int(a)−1 ◦σ ◦Int(a). It is enough to show that σ and σ are conjugate. We have σ (e) = σ (a−1 e a) = σ ◦ Int(a)−1 (e ) = Int(a)−1 (σ (e )). Thus, we get σ (e) = Int(a)−1 (1 − e ) = 1 − e = σ(e). By Proposition VIII.21.3, there exists u ∈ Sym(A, σ)× such that σ = Int(u) ◦ σ. Since σ(e) = σ (e) = uσ(e)u−1 = σ(u−1 eu), we have u−1 eu = e, that is ue = eu. Set v = eue + (1 − e). It easily follows from the relations e2 = e and ue = eu that we have vσ(v) = u. Therefore, we get σ = Int(vσ(v)) ◦ σ, and σ and σ are conjugate. This concludes the proof.

VIII.21.3 Similitudes In this paragraph, we define several group-schemes attached to a central simple k-algebra with involution (A, σ), where σ is an involution of the first kind. If R is a commutative k-algebra, we set σR = σ ⊗k IdR .


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Definition VIII.21.16. The group-scheme of similitudes of (A, σ), denoted by Sim(A, σ), is defined by × Sim(A, σ)(R) = {g ∈ A× R | gσR (g) ∈ R }.

If g ∈ Sim(A, σ)(R), the element μ(g) = gσ(g) ∈ R× is called the multiplier of the similitude g. If σ is orthogonal, we denote it by GO(A, σ), and if σ is symplectic, we denote it by GSp(A, σ). The group-scheme of isometries of (A, σ), denoted by Iso(A, σ), is defined by Iso(A, σ)(R) = {g ∈ A× R | gσR (g) = 1}. If σ is orthogonal, we denote it by O(A, σ), and if σ is symplectic, we denote it by Sp(A, σ). We also define the group-scheme Aut(A, σ) of automorphisms of (A, σ) by Aut(A, σ)(R) = {f ∈ Aut(AR ) | f ◦ σ = σ ◦ f }. If σ is orthogonal, we denote it by PGO(A, σ), and if σ is symplectic, we denote it by PGSp(A, σ). Notice that we have a morphism of group-schemes Int : Sim(A, σ) −→ Aut(A, σ), defined by IntR :

Sim(A, σ)(R) −→ Aut(A, σ)(R) g −→ Int(g).

Indeed, if gσR (g) ∈ R× , we have Int(gσR (g)) = IdAR , and thus we have Int(g) ◦ σR ◦ Int(g)−1 = Int(g) ◦ Int(σR (g)) ◦ σ = Int(gσR (g)) ◦ σR = σR . Moreover, Int(g) = IdAR if and only if g ∈ R× , so we get an exact sequence 1 −→ R× −→ Sim(A, σ)(R) −→ Aut(A, σ)(R). If K is a field, every automorphism of AK is inner. If f = Int(g) is such an automorphism, the previous computations show that f is an automorphism of (AK , σK ) if and only if Int(gσ(g)) = IdAK , that is if and only if gσ(g) ∈ K × . Therefore, IntK is surjective for every field K, so we get an exact sequence

228

Galois cohomology of central simple algebras 1 −→ K × −→ Sim(A, σ)(K) −→ Aut(A, σ)(K) −→ 1.

If we restrict the morphism Int to Iso(A, σ), we easily see that we have an exact sequence 1 −→ μ2 (R) −→ Iso(A, σ)(R) −→ Aut(A, σ)(R). If K is a field and Ks denotes a separable closure of K, the map Int is surjective on the Ks -points. Indeed, if f = Int(g) ∈ Aut(A, σ)(Ks ), 1 then g is a similitude and g = % g is an isometry of (AKs , σKs ) μ(g) satisfying f = Int(g ). We then get an exact sequence of GKs -groups 1 −→ μ2 (Ks ) −→ Iso(A, σ)(Ks ) −→ Aut(A, σ)(Ks ) −→ 1. VIII.21.4 Cohomology of algebras with involution We now give a description of the first Galois cohomology set of the group-schemes in the previous section. We start with a lemma. Lemma VIII.21.17. Let (A, σ) and (A , σ ) be two central simple kalgebras with involution of same type. Then we have (Aks , σks ) (Aks , σk s ). ∼

Proof. Let f : Aks −→ Mn (ks ) be an isomorphism of central simple ks algebras, and let σ0 = f ◦ σks ◦ f −1 . We know that we have σ0 = σB for some invertible matrix B ∈ Mn (ks ) satisfying B t = ±B. By definition, we have (Aks , σks ) (Mn (ks ), σB ). ⎛ 0 1 ⎜ −1 0 ⎜ ⎜ .. Set B0 = In if B t = B and B0 = ⎜ . ⎜ ⎝ 0 1 −1 0

⎞ ⎟ ⎟ ⎟ ⎟ if B t = −B. ⎟ ⎠

The map ksn × ksn −→ ks , (X, Y ) −→ X t BY is a regular symmetric bilinear form if B t = B and a regular skew-symmetric form if B t = −B. The theory of symmetric forms and skew-symmetric bilinear forms shows that in both cases, there exists P ∈ GL(ks ) such that B = P t B0 P . We then have σB = Int(P t ) ◦ σB0 ◦ Int(P t )−1 .


229

In particular, we get (Aks , σks ) (Mn (ks ), σB ) (Mn (ks ), σB0 ). The lemma follows easily. Reasoning as in the proof of Proposition III.9.7, we get the following result: Proposition VIII.21.18. Let k be a field, and let (A, σ) be a central simple k-algebra with an involution of the first kind of type ε. Then we have an isomorphism of functors Ck −→ Sets∗ H 1 (− , PSim(A, σ)) CSAεn , where the base point of CSAεn is chosen to be the isomorphism class of (A, σ). Now we would like to give a description of the functor H 1 (− , Iso(A, σ)). We define a functor Sym(A, σ)× : Ck −→ Sets∗ by setting Sym(A, σ)× (K) = Sym(AK , σK )× , for every field extension K/k, the base point being 1AK . Notice now that GL1 (A) acts on Sym(A, σ)× as follows: GL1 (A)(K) × Sym(A, σ)× (K) −→ Sym(A, σ)× (K) (a, u) −→ auσK (a). We denote by Sym(A, σ)× /∼ the corresponding functor of equivalence classes, and by u/∼ the equivalence class of an invertible symmetric element u. Now by definition of the action of GL1 (A), we have StabGL1 (A) (1A )(K) = Iso(A, σ)(K) for every field extension K/k. If u ∈ Sym(A, σ)× (K) and L/K is a field extension, we will denote by uL the element u ⊗ 1 ∈ Sym(A, σ)× (L). We claim that for all u ∈ Sym(AK , σK )× , we have uKs ∼Ks 1. Indeed, by Lemma VIII.21.17, the involutions Int(uKs ) ◦ σKs and σKs on AKs are conjugate. Therefore, there exists a ∈ A× Ks such that Int(a(uKs )σKs (a)) ◦ σKs = σKs .

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We deduce that there exists λ ∈ Ks× such that auKs σKs (a) = λ. Let √ a λ ∈ Ks× be a square-root of λ. Setting a = √ , we see that we have λ a uKs σKs (a ) = 1, that is uKs ∼ 1. It is easy to check that all the hypotheses of the Galois descent lemma are fulfilled. Since H 1 (− , GL1 (A)) = 1 by Hilbert 90, the Galois descent lemma yields Proposition VIII.21.19. We have an isomorphism of functors from Ck to Sets∗ H 1 (− , Iso(A, σ)) Sym(A, σ)× /∼ . We end this paragraph by identifying some maps in cohomology. Since an automorphism of an algebra with involution (A, σ) is an automorphism of A, we have an inclusion Aut(A, σ) ⊂ PGL1 (A). The proof of the following lemma is left to the reader as an exercise. Lemma VIII.21.20. For every field extension K/k, the map H 1 (K, Aut(A, σ)) −→ H 1 (K, PGL1 (A)) induced by the inclusion maps the isomorphism class of (A , σ ) onto the isomorphism class of A . Lemma VIII.21.21. For every field extension K/k, the map Int∗,K : H 1 (K, Iso(A, σ)) −→ H 1 (K, Aut(A, σ)) induced by the group morphism Int : Iso(A, σ) −→ Aut(A, σ) maps u/∼ onto the isomorphism class of (AK , Int(u) ◦ σK ). Proof. Let a ∈ AKs such that auKs σKs (a) = 1. Galois descent shows that a cocycle α representing u/∼ is given by α:

GKs −→ Iso(A, σ)(Ks ) ρ −→ a ρ·a−1 .

Notice that Int(a) induces an isomorphism of algebras with involution between (AKs , Int(uKs ) ◦ σKs ) and (AKs , σKs ). Indeed, we have Int(a) ◦ Int(uKs ) ◦ σKs ◦ Int(a)−1

= = = =

Int(auKs ) ◦ σKs ◦ Int(a)−1 Int(auKs ) ◦ Int(σKs (a)) ◦ σKs Int(auKs σKs (a)) ◦ σKs σ Ks .


231

By Galois descent, a cocycle β representing (AK , Int(u) ◦ σK ) is then given by β:

GKs −→ Aut(A, σ)(Ks ) ρ −→ Int(a) ρ·Int(a)−1 .

For all x ∈ AKs , we have (ρ·Int(a))(x) = ρ·(Int(a)(ρ−1 ·x)) = ρ·(a(ρ−1 ·x)a−1 ). Since ρ acts on AKs by algebra automorphisms, we get (ρ·Int(a))(x) = (ρ·a)x(ρ·a)−1 , hence ρ·Int(a) = Int(ρ·a). We then get βρ = Int(a ρ·a−1 ) = Int(αρ ) for all ρ ∈ GKs . This means that β represents Int∗,K (u/∼ ). This concludes the proof.

VIII.21.5 Trace forms In this paragraph, we associate some quadratic forms to central simple algebras with an involution of the first kind. We first define a characteristic polynomial for certain types of algebras. Definition VIII.21.22. Let R be a commutative ring with unit. Let A be an associative R-algebra with unit. Assume that there exist a commutative faithfully flat R-algebra S, a projective S-module P of finite rank n and an isomorphism of S-algebras ∼

ϕ : A ⊗R S −→ EndS (P ). For all a ∈ A, the polynomial χA,a = det(tId−ϕ(a⊗1S )) has coefficients in R and does not depend on the choice of (S, P, ϕ) (see [31] for a proof). It is called the reduced characteristic polynomial of a. Let us write χA,a = tn − an−1 tn−1 + . . . + (−1)n a0 . The elements an−1 and a0 are respectively called the reduced trace and reduced norm of a, and are denoted by TrdA (a) and NrdA (a). In other words, we have TrdA (a) = tr(ϕ(a ⊗ 1S )) and NrdA (a) = det(ϕ(a ⊗ 1S )).

232


Example VIII.21.23. Let A be a central simple k-algebra, let R be a commutative k-algebra, and let A = AR . Let L/k be any splitting field ∼ of A, and let ρ : AL −→ Mn (L) be an isomorphism of L-algebras. Then the R-algebra S = R ⊗k L is faithfully flat. Moreover, the S-module P = S n is free (hence projective) of rank n, and we have the following isomorphisms of L-algebras AR ⊗R S AL ⊗k S Mn (L) ⊗k S EndS (P ), the second one being induced by ρS . We then may define a characteristic polynomial for AR , and we have in particular reduced trace and reduced norm maps. Lemma VIII.21.24. Let A and B be two central simple k-algebras. Then the following properties hold: (1)

The map TrdA : A −→ k is k-linear. Moreover for all a, a ∈ A, we have TrdA (aa ) = TrdA (a a) and NrdA (aa ) = NrdA (a)NrdA (a ).

(2)

For all n ≥ 1, the maps TrdMn (k) and NrdMn (k) coincide with the usual trace and determinant maps.

(3)

For every field extension K/k, and every a ∈ A, we have TrdAK (a ⊗ 1) = TrdA (a), NrdAK (a ⊗ 1) = NrdA (a).

(4)

∼

If f : A −→ B is an isomorphism of k-algebras, then for all a ∈ A, we have TrdB (f (a)) = TrdA (a) and NrdB (f (a)) = NrdA (a).

(5)

For all a ∈ A, we have TrdAop (aop ) = TrdA (a) and NrdAop (aop ) = NrdA (a).

(6)

We have NrdA (a) = 0 if and only if a ∈ A× .

(7)

For all a ∈ A, b ∈ B, we have TrdA⊗k B (a ⊗ b) = TrdA (a)TrdB (b) and NrdA⊗k B (a ⊗ b) = NrdA (a)degk (B) NrdB (b)degk (A) .


233

Proof. Properties (1), (2), (3) and (7) come from the definitions and the properties of the trace and determinant maps. For example, let K/k be a field extension and let a ∈ A. Let L/k be a splitting field of A containing K (for example L = Ks ), and let ∼

ϕ : AL −→ Mn (L) be an isomorphism of L-algebras. Composing ϕ with the canonical isomorphism AK ⊗K L AL ∼

gives rise to an isomorphism of L-algebras ψ : AK ⊗K L −→ Mn (L) which satisfies ψ((a ⊗ 1) ⊗ 1) = ϕ(a ⊗ 1). Therefore, we get TrdAK (a ⊗ 1) = tr(ψ((a ⊗ 1) ⊗ 1)) = tr(ϕ(a ⊗ 1)) = TrdA (a). We will prove (4), (5) and (6) in detail. For the rest of the proof, let ∼ us fix a splitting field L of A, an isomorphism ϕ : AL −→ Mn (L) of L-algebras, and an element a ∈ A. ∼

Let us prove (4). Let f : A −→ B be an isomorphism of k-algebras. Then the map ϕ ◦ (f −1 ⊗ IdL ) : B −→ Mn (L) is an isomorphism of L-algebras. We then have TrdB (f (a)) = tr(ϕ ◦ (f −1 ⊗ IdL )(f (a) ⊗ 1)) = tr(ϕ(a ⊗ 1)) = TrdA (a), and similarly for NrdB (f (a)). This proves (4). We now prove (5). One can show that the map ψ:

(Aop )L −→ Mn (L) aop ⊗ λ −→ ϕ(a ⊗ λ)t

is a well-defined isomorphism of L-algebras. Therefore, we have TrdAop (aop ) = tr(ϕ(a ⊗ 1)t ) = tr(ϕ(a ⊗ 1)) = TrdA (a), and similarly for the second equality. It remains to prove (6). If a ∈ A× , then a ⊗ 1 ∈ A× L , and thus ϕ(a ⊗ 1) is an invertible matrix. Therefore NrdA (a) = det(ϕ(a ⊗ 1)) = 0.

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Now assume that a is not invertible. This means that the k-linear map a :

A −→ A a −→ aa

is not surjective. Since A is finite dimensional over k, it is not injective either, so there exists a ∈ A, a = 0 such that aa = 0. Hence a is either zero or a zero divisor, and therefore so is ϕ(a ⊗ 1) ∈ Mn (L). In particular, ϕ(a ⊗ 1) is not invertible. It follows that NrdA (a) = 0 by the properties of the determinant. This concludes the proof. Remark VIII.21.25. All these properties are true if we replace A by a R-algebra A satisfying the assumptions of Definition VIII.21.22. For ∼ example, assume that we have an isomorphism of R-algebras f : A −→ B, and let (S, P, ϕ) be a triple as in Definition VIII.21.22. We then have an isomorphism ∼

ϕ ◦ (f −1 ⊗ IdS ) : B −→ EndS (P ), and therefore TrdB (f (a)) = tr(ϕ ◦ (f −1 ⊗ IdS )(f (a) ⊗ 1)) = tr(ϕ(a ⊗ 1)) = TrdA (a). We are ready to define our first quadratic form. Definition VIII.21.26. Let A be a central simple k-algebra. The map TA :

A −→ k a −→ TrdA (a2 )

is a quadratic form, called the trace form of A. Proposition VIII.21.27. Let A be a central simple k-algebra, and let K/k be a field extension. Then the following properties hold: (1)

The isomorphism class of TA only depends on the isomorphism class of A.

(2)

TAK (TA )K .

(3)

TMn (K) n × 1 ⊥ n × 1, −1.

(4)

The quadratic form TA is regular. ∼

Proof. Let us prove (1). Let ϕ : A −→ B be an isomorphism of kalgebras. For all a ∈ A, we have TB (ϕ(a)) = TrdB (ϕ(a)2 ) = TrdB (ϕ(a2 )) = TrdA (a2 ) = TA (a).


235

Hence ϕ induces an isomorphism between the quadratic spaces (A, TA ) and (B, TB ). Point (2) is immediate using Lemma VIII.21.24 (3). Let us prove (3). The polar form bA associated to the trace form is bA :

A × A −→ k (a, a ) −→ TrdA (aa ).

Indeed, bA is bilinear and symmetric by Lemma VIII.21.24 (1), and satisfies bA (a, a) = TrdA (a2 ) for all a ∈ A. In particular, we get bMn (K) (M, M ) = tr(M M ) for all M, M ∈ Mn (K). Using this formula, it is easy to check that the elements Eij + Eji Eij − Eji , ,1 ≤ i < j ≤ n 2 2 form an orthogonal basis of A with respect to the trace form. The result follows by computing the value of TMn (K) at each element of this basis. It remains to prove (4). To check that TA is regular, it is enough to do it after extending scalars to kalg . By (2), we have Eii ,

(TA )kalg TAkalg . Since Akalg Mn (kalg ), (1) yields (TA )kalg TMn (kalg ) . Now apply (3) to conclude. We now associate some quadratic forms to algebras with involutions. Definition VIII.21.28. Let (A, σ) be a central simple k-algebra with an involution of the first kind. The map Tσ :

A −→ k a −→ TrdA (σ(a)a)

is a quadratic form called the trace form of (A, σ) . We denote respectively by Tσ+ and Tσ− the restriction of Tσ to Sym(A, σ) and Skew(A, σ). They are called the restricted trace forms of (A, σ). We now establish some auxiliary results on these various trace forms.

236


Lemma VIII.21.29. Let (A, σ) be a central simple k-algebra with an involution of the first kind. The following properties hold: (1)

For all a ∈ A, we have TrdA (σ(a)) = TrdA (a) and NrdA (σ(a)) = NrdA (a).

(2)

For all a ∈ Skew(A, σ), s ∈ Sym(A, σ), we have TrdA (as) = 0.

Proof. Observe first that the map A −→ Aop

ϕ:

a −→ σ(a)op

is an isomorphism of k-algebras. Therefore, for all x ∈ A, we have TrdA (x) = TrdAop (σ(x)op ) = TrdA (σ(x)) by Lemma VIII.21.24 (4) and (5). We then have TrdA (as)

= = = = =

TrdA (σ(as)) TrdA (σ(s)σ(a)) TrdA (−sa) −TrdA (sa) −TrdA (as).

We then get 2TrdA (as) = 0, and therefore TrdA (as) = 0. Proposition VIII.21.30. Let (A, σ) be a central simple k-algebra with an involution of the first kind. (1)

The isomorphism classes of Tσ , Tσ+ and Tσ− only depend on the isomorphism class of (A, σ).

(2)

We have TA Tσ+ ⊥ −Tσ− and Tσ Tσ+ ⊥ Tσ− .

(3)

For every field extension K/k, we have TσK (Tσ )K , Tσ+K (Tσ+ )K and Tσ−K (Tσ− )K .

(4)

The quadratic forms Tσ , Tσ+ and Tσ− are regular. ∼

Proof. Let us prove (1). Let ϕ : (A, σ) −→ (A , σ ) be an isomorphism of algebras with involution. For all a ∈ A, we have Tσ (ϕ(a))

= = = = =

TrdA (σ (ϕ(a))ϕ(a)) TrdA (ϕ(σ(a))ϕ(a)) TrdA (ϕ(σ(a)a)) TrdA (σ(a)a) Tσ (a).


237

Hence ϕ induces an isomorphism between the quadratic spaces (A, Tσ ) and (A , Tσ ). We have already noticed that ϕ induces isomorphisms of k-vector spaces Sym(A, σ) Sym(A , σ ) and Skew(A, σ) Skew(A , σ ). The rest follows from the equalities TrdA (ϕ(a)2 ) = TrdA (ϕ(a2 )) = TrdA (a2 ) for all a ∈ A. To prove (2), notice first that we have A = Sym(A, σ) ⊕ Skew(A, σ). Recall that the polar form of TA is A × A −→ k

bA :

(a, a ) −→ TrdA (aa ).

The previous lemma shows that the direct sum above is orthogonal with respect to the trace form. Moreover, for all s ∈ Sym(A, σ), a ∈ Skew(A, σ), we have TA (s) = TrdA (s2 ) = Tσ+ (s) and TA (a) = TrdA (a2 ) = −TrdA (−a2 ) = −Tσ− (a). We then get TA Tσ+ ⊥ −Tσ− . Notice now that the map bσ :

A × A −→ k (a, a ) −→ TrdA (σ(a)a )

is bilinear and symmetric. Indeed, for all a, a ∈ A, we have TrdA (σ(a)a ) = TrdA (σ(σ(a)a )) = TrdA (σ(a )a), the first equality coming from Lemma VIII.21.29 (1). Since bσ (a, a) = TrdA (σ(a)a), it turns out that bσ is the polar form of Tσ . It follows easily that Sym(A, σ) and Skew(A, σ) are orthogonal with respect to Tσ , and we get Tσ Tσ+ ⊥ Tσ− . Point (3) follows from Lemma VIII.21.24. To prove (4), notice that by definition of orthogonal sum and regularity, Tσ+ and Tσ− will be regular if and only if Tσ+ ⊥ −Tσ− is. By (2), this last orthogonal sum is isomorphic to TA , which is regular by Proposition VIII.21.27 (4). Therefore, Tσ+ and Tσ− are regular, and thus so is Tσ Tσ+ ⊥ Tσ− .

238


The following lemma is useful to compute the various trace forms. Lemma VIII.21.31. Let A and A be two central simple algebras. Then we have TA⊗B TA ⊗ TB . Moreover, if σ and σ are involutions of type ε and ε on A and A respectively, then σ ⊗ σ is an involution of type εε on A ⊗k B, and we have the following isomorphisms: (1)

Tσ⊗σ Tσ ⊗ Tσ .

(2)

+ + − + − Tσ⊗σ Tσ ⊗ Tσ ⊥ Tσ ⊗ Tσ .

(3)

− − + + − Tσ⊗σ Tσ ⊗ Tσ ⊥ Tσ ⊗ Tσ .

Proof. Let bA , bB and bA⊗k B the polar forms of TA , TB and TA⊗k B respectively. For all a, a ∈ A, b, b ∈ B, we have bA⊗k B (a ⊗ b, a ⊗ b) = TrdA⊗k B ((a ⊗ b)(a ⊗ b )) = TrdA⊗k B (aa ⊗ bb ) = TrdA (aa )TrdB (bb ) = bA (a, a )bB (b, b ). The first isomorphism follows. The isomorphism Tσ⊗σ Tσ ⊗ Tσ may be proved in a similar way. The reader will check that Sym(A⊗k B, σ⊗σ ) is the direct sum of the linear subspaces Sym(A, σ)⊗k Sym(A , σ ) and Skew(A, σ)⊗k Skew(A , σ ), and that Skew(A ⊗k B, σ ⊗ σ ) is the direct sum of the two linear subspaces Sym(A, σ) ⊗k Skew(A , σ ) and Skew(A, σ) ⊗k Sym(A , σ ). It follows in particular by dimension count that if σ and σ are involutions of type ε and ε on A and A respectively, then σ ⊗ σ is an involution of type εε on A⊗k A . Moreover, Lemma VIII.21.29 (2) implies easily that these direct sums are orthogonal. The rest of the lemma follows. Let us give some examples of computation of these trace forms. Examples VIII.21.32. (1)

Let A = Mn (k), and let t be the transposition. It is easy to check that the matrices Eij + Eji ,i < j Eii , 2


239

form an orthogonal basis of Sym(A, t ). Straightforward computations show that we have Tt+ n × 1 ⊥

n(n − 1) × 2. 2

Similarly, the matrices Eij − Eji ,i < j 2 form an orthogonal basis of Skew(A, t ), and we have Tt− (2)

n(n − 1) × 2. 2

Let Q = (a, b) be a quaternion k-algebra and let us consider the symplectic involution γ:

Q −→ Q x + yi + zj + tij −→ x − yi − zj − tij.

Then 1 is a basis of Sym(Q, γ), and i, j, ij is an orthogonal basis of Skew(Q, γ), so we have Tγ+ 2 and Tγ− −2a, −2b, 2ab. Therefore, we also get Tγ 2a, b. (3)

Let A = Mm (k) ⊗k Q, and let σ = t ⊗ γ. Lemma VIII.21.31 and the previous examples imply that we have Tσ+ m × 2 ⊥

m(m − 1) × 1, −a, −b, ab 2

and Tσ− m × −2a, −2b, 2ab ⊥ (4)

m(m − 1) × 1, −a, −b, ab. 2

Let A = M2m (k) and let σ = σB0 , where B0 is the matrix ⎛ ⎞ 0 1 ⎜ −1 0 ⎟ ⎜ ⎟ ⎜ ⎟ .. B0 = ⎜ ⎟. . ⎜ ⎟ ⎝ 0 1 ⎠ −1 0

240

Galois cohomology of central simple algebras ∼

Notice that we have an isomorphism ϕ : (1, −1) −→ M2 (k) of k-algebras such that 1 0 0 1 ϕ(i) = and ϕ(j) = . 0 −1 −1 0 It is easy to check this isomorphism induces an isomorphism of algebras of involutions ((1, −1), γ) (M2 (k), σH ), 0 1 where H = . We then get isomorphisms of algebras −1 0 with involution

(M2m (k), σ)

(Mm (k) ⊗k M2 (k), t ⊗ σH ) (Mm (k) ⊗k (1, −1), t ⊗ γ).

Therefore, the computations done in the previous case give Tσ+ m × 2 ⊥ m(m − 1) × 1, −1 and Tσ−

m × −2, 2, −2 ⊥ m(m − 1) × 1, −1 m × −2 ⊥ m2 × 1, −1.

We now would like to give a cohomological interpretation of the restricted trace form. Lemma VIII.21.33. Let (A, σ) be a central simple k-algebra with an involution of the first kind , let R be a commutative k-algebra and let f be an automorphism of (AR , σR ). Then the restriction of f to Sym(AR , σR ) is an automorphism of (Tσ+ )R . Proof. By Remark VIII.21.25, for all a ∈ AR and any automorphism f of AR , we have TrdAR (f (a)) = TrdAR (a). The lemma follows immediately. The previous result shows that we have a morphism of algebraic groups ρ : Aut(A, σ) −→ O(Tσ+ ), which is obtained by restricting an automorphism of a central simple algebra with involution to the corresponding set of σ-symmetric elements.


241

Lemma VIII.21.34. Let (A, σ) be a central simple k-algebra with an involution of the first kind. For every field extension K/k, the induced map ρ∗,K : H 1 (K, Aut(A, σ)) −→ H 1 (K, O(Tσ+ )) takes the isomorphism class of (A , σ ) onto the isomorphism class of Tσ+ . Proof. Let K/k be a field extension, let (A , σ ) be a central simple Kalgebra with an involution σ of same type as σ, and let ∼

ϕ : AKs −→ AKs be an isomorphism of Ks -algebras of (A , σ )Ks onto (A, σ)Ks . The map α:

GKs −→ Aut(A, σ)(Ks ) σ −→ ϕ σ·ϕ−1

is a cocycle representing (A , σ ). Now ϕ induces by restriction an isomorphism between the restricted trace forms of (A , σ )Ks and (A, σ)Ks , that is an isomorphism between (Tσ+ )Ks and (Tσ+ )Ks by Proposition VIII.21.30 (3). It follows that the map ρ ◦ α is a cocycle representing Tσ+ . The result follows. As an application of Galois cohomology, we are going to compute the Hasse invariant of Tσ+ when σ is a symplectic involution. The computation of the Hasse invariants of TA and Tσ will be done in the exercices. Before starting the proof, we would like to recall two facts on algebraic group-schemes, which we will use without further reference. (1)

Any morphism of algebraic group-schemes ρ : G −→ H induces by restriction a morphism G0 −→ H 0 , where G0 and H 0 denote the connected component of the neutral element of G and H respectively. Indeed, such a morphism ρ is continuous (for the Zariski topology) and therefore maps connected sets to connected sets.

(2)

If G and H are connected algebraic group-schemes, then any morphism of algebraic group-schemes ρ : G −→ H induces a ˜ −→ H ˜ between the universal covers of G and H morphism ρ˜ : G

242

Galois cohomology of central simple algebras (see [7, Proposition 2.24 (i),p.262]) such that the diagram ˜ G

πG

/G

πH

/H

ρ

ρ˜

˜ H

is commutative. We then get in particular an induced morphism τ : ker(πG ) −→ ker(πH ). Since PGSp(A, σ) is connected (see[30], for example), the group-scheme morphism ρ : PGSp( A, σ) −→ O(Tσ+ ) restricts to a group-scheme morphism (still denoted by ρ) ρ : PGSp(A, σ) −→ O+ (Tσ+ ). We then get the following result: Corollary VIII.21.35. Let (A, σ) be a central simple k-algebra with a symplectic involution. For any other symplectic involution σ on A, we have det(Tσ+ ) = det(Tσ+ ). Proof. This follows from the previous lemma and Corollary IV.11.3. We are now ready to prove the following result, due to Quéguiner[46]. Theorem VIII.21.36. Let (A, σ) be a central simple k-algebra of degree 2m with a symplectic involution. Then after identifying H 2 (k, μ2 ) and Br2 (k), we have the equality w2 (Tσ+ ) =

m(m − 1) ([(−1, −1)] + [A]). 2

⎛

0 1 ⎜ −1 0 ⎜ ⎜ Proof. Let B0 = ⎜ ⎜ ⎝

⎞ ..

.

⎟ ⎟ ⎟ ⎟, and denote by σ0 the invo⎟ 0 1 ⎠ −1 0

lution σB0 . We will write PGSp2m and Sp2m instead of PGSp(M2m (k), σ0 ) and Sp(M2m (k), σ0 ) respectively. It is known that the group Sp2m is the universal cover of PGSp2m (see [30] for example).


243

Let ρ : PGSp2m −→ O+ (Tσ+0 ) be the morphism of algebraic groups defined earlier, and let ρ˜ : Sp2m −→ Spin(Tσ+0 ) the induced morphism on the universal covers. Therefore, we also get an induced group morphism τ : μ2 −→ μ2 . For every field extension K/k, we then have a commutative diagram 1

/ μ2 (Ks )

1

/ μ2 (Ks )

/ Sp2m (Ks )

τ

/ PGSp2m (Ks )

ρ˜

/ Spin(Tσ+ )(Ks ) 0

/1

ρ

/ O+ (Tσ+ )(Ks ) 0

/1

which induces a commutative diagram Δ1K

H 1 (K, PGSp2m ) ρ∗,K

/ H 2 (K, μ2 ) τ∗,K

H 1 (K, O+ (Tσ+0 ))

1 δK

/ H 2 (K, μ2 )

by Theorem III.7.39 (4). Using Lemma VIII.21.34 and Corollary IV.11.10, we get the equality τ∗,K (Δ1K (A, σ)) = w2 (Tσ+0,K ) + w2 (Tσ+ ) for every K-algebra with a symplectic involution (A, σ). Let us now identify the map Δ1K . We have a commutative diagram 1

/ μ2 (Ks )

/ Sp2m (Ks )

/ PGSp2m (Ks )

/1

1

/ K×

/ GL2m (Ks )

/ PGL2m (Ks )

/1

s

which induces a commutative diagram

H 1 (K, PGSp2m ) H 1 (K, PGL2m )

Δ1K

/ H 2 (K, μ2 ) / H 2 (K, Gm )

by Theorem III.7.39 (4). Using Lemma VIII.21.20, Theorem VIII.20.6

244


and the commutativity of the diagram, we see that Δ1K maps the isomorphism class of (A, σ) onto [A]. Therefore, we get w2 (Tσ+ ) = w2 (Tσ+0,K ) + τ∗,K ([A]). By Example VIII.21.32 (4), we have Tσ+0,K m × 2 ⊥ m(m − 1) × 1, −1. Using Lemma IV.11.5, we have w2 (Tσ+0,K ) = w2 (m×2)+w2 (m(m−1)×1, −1)+(2m )∪((−1)m(m−1) ). Since m(m − 1) is even, the last term is 0. Moreover, w2 (m × 2) is a multiple of (2) ∪ (2) = (2) ∪ (−1). Since 2 is a sum of two squares, (2) ∪ (−1) = 0. We then have w2 (Tσ+0,K ) = w2 (m(m − 1) × 1, −1) = w2 (m(m − 1) × −1), and therefore w2 (Tσ+0,K ) =

m(m − 1)(m(m − 1) − 1) (−1) ∪ (−1). 2

m(m − 1)(m(m − 1) − 1) m(m − 1) and have 2 2 2 the same parity. Therefore, after identification of H (k, μ2 ) and Br2 (k), we get It is easy to check that

w2 (Tσ+ ) =

m(m − 1) [(−1, −1)] + τ∗,K ([A]) 2

for every field extension K/k. Now we need to identify the map τ∗,K . The map τK is either trivial or the identity map. Moreover, since μ2 (Ks ) = μ2 (ks ) and since the diagram μ2 (ks ) μ2 (Ks )

τk

/ μ2 (ks )

τK

/ μ2 (Ks )

commutes, we have τK = τk . Therefore, either τK is trivial for all K/k or equal to the identity for all K/k. Therefore, the induced map τ∗,K will be either trivial for every field extension K/k or equal to the identity map of Br2 (K) for every field extension K/k.

Exercises

245

Set K = k(t1 , t2 ). Then the quaternion K-algebra Q = (t1 , t2 ) is not trivial. Let us apply the previous equality to the K-algebra (A, σ) = (Mm (K) ⊗K Q, t ⊗ γ). By Example VIII.21.32 (3), we have m(m − 1) × 1, −t1 , −t2 , t1 t2 . 2 Similar computations to those performed before show that we have m(m − 1) + × −t1 , −t2 , t1 t2 . w2 (Tσ ) = w2 2 Tσ+ m × 2 ⊥

Applying Lemma IV.11.5 several times, we get m(m − 1) w2 (−t1 , −t2 , t1 t2 ). 2 The reader will check that we have w2 (Tσ+ ) =

w2 (−t1 , −t2 , t1 t2 ) = (−1) ∪ (−1) + (t1 ) ∪ (t2 ). m(m − 1) ([(−1, −1)] + [Q]) for this particular We then get w2 (Tσ+ ) = 2 algebra with involution, and it follows that we have τ∗,K ([Q]) =

m(m − 1) [Q]. 2

m(m − 1) is even, we get τ∗,K ([Q]) = 0. Thus τ∗,K is not the identity, 2 m(m − 1) and τ∗,K is trivial. If is odd, we get τ∗,K ([Q]) = [Q] = 0. 2 Thus τ∗,K is not trivial, and therefore is equal to the identity map. The previous considerations show that we have τ∗,K = 0 for all K/k if m(m − 1) m(m − 1) is even, and equal to the identity for all K/k if is 2 2 odd. Apply this to K = k, we get the desired result. If

Exercises 1. Let k be a field. Let Z/2Z act by group automorphisms on PGLn (ks ) as follows: if f = Int(M ), set 1·Int(M ) = Int((M −1 )t ). Let G be the corresponding semi-direct product.

246

Galois cohomology of central simple algebras (a)

Show that G is a Gks -group.

(b)

Show that H 1 (k, G) classifies isomorphisms classes of central simple algebras (B, τ ) with a unitary involution, whose center K is a quadratic étale k-algebra.

(c)

Consider the exact sequence of Gks -groups 1 −→ PGLn (ks ) −→ G −→ Z/2Z −→ 1. Describe the induced maps H 1 (k, PGLn ) −→ H 1 (k, G) and H 1 (k, G) −→ H 1 (k, Z/2Z).

2. Let A be a central simple k-algebra. For any commutative kalgebra R, set SL1 (A)(R) = {a ∈ AR | NrdAR (a) = 1}. (a)

For any field extension L/k, show that we have an exact sequence of GLs -modules 1 −→ SL1 (A)(Ls ) −→ GL1 (A)(Ls ) −→ L× s −→ 1.

(b)

Deduce that H 1 (L, SL1 (A)) L× /NrdAL (A× L ) for every field extension L/k.

3. Let A be a central simple k-algebra. For c ∈ k × , set Xc = {a ∈ Aks | NrdAks (a) = c}. Keeping the definitions of Chapter II, Exercice 7, prove that (a)

Xc is a principal homogeneous space over SL1 (A)(ks )

(b)

Any principal homogeneous space over SL1 (A)(ks ) is isomorphic to Xc for some c ∈ k ×

(c)

Xc Xd if and only if cd−1 ∈ NrdA (A× ).

4. Let n ≥ 1 be an integer, and let k be a field of characteristic different from 2. Let us denote by T the trace form of Mn (k). Show that for every central simple k-algebra A of degree n, we have det(TA ) = det(T ). 5. Let n ≥ 1 be an even integer, and let k be a field of characteristic different from 2. Let us denote by T the trace form of Mn (k). The main goal of this exercise is to compute the Hasse invariant of the trace form of a central simple algebra. We will follow the arguments given by Lewis and Morales in [34].

Exercises (a)

For every field extension K/k, show that the diagram 1

/ μn (Ks ) τ

/ μ2 (Ks ) 1 commutes. (b)

247

/ SLn (Ks )

/ PGLn (Ks )

/1

/ Spin(T )(Ks )

/ O+ (T )(Ks )

/1

Let d ∈ K × , and let M = diag(d, 1, . . . , 1) ∈ Mn (K). Check that we have Int(M ) = τv2 ◦ τw2 ◦ · · · ◦ τvn ◦ τwn , where vi = Ei1 − dE1i and wi = Ei1 − E1i for i = 2, . . . , n.

(c)

Show that the spinor norm of Int(M ) is d ∈ K × /K ×2 (see Chapter IV for a definition of the spinor norm).

(d)

Deduce that there exists a field extension K/k for which τK is not trivial.

(e)

Show that for every central simple k-algebra of degree n, we have w2 (TA ) = w2 (T ) +

(f)

n(n − 2) n n [A] = [(−1, −1)] + [A]. 2 8 2

Deduce the values of w2 (Tσ− ) and w2 (Tσ ) for any k-algebra (A, σ) with a symplectic involution.

6. Let k be a field, and let K/k be a quadratic étale k-algebra. If K = k × k, we say B is a central simple K-algebra if isomorphic to K and A A1 × A2 , where A1 and A2 are central simple k-algebras. We then extend the notion of a unitary involution on B in an obvious way. (a)

Let A be a central simple k-algebra. Show that the map ε:

A × Aop −→ A × Aop op (a1 , aop 2 ) −→ (a2 , a1 )

is a unitary involution on A × Aop . (b)

Assume that K = k × k and that (B, τ ) is a central simple K-algebra with a unitary involution. Show that there exists a central simple k-algebra A such that (B, τ ) (A × Aop , ε).

248

Galois cohomology of central simple algebras 7. Let (B, τ ) be a central simple K-algebra with a unitary involution. We say that τ is hyperbolic if there exists an idempotent e ∈ B such that τ (e) = 1 − e. (a)

Assume that K = k × k. Show that ε is hyperbolic.

(b)

Assume that K is a field, let k be the subfield of K consisting of τ -symmetric elements, and let ι be the non-trivial k-automorphism of K. If B = Mn (K), show that we have τ = Int(H) ◦ t, where H = (mij ) ∈ GLn (K) satisfies H ∗ = (ι(mji )) = H. Show that τ is hyperbolic if and only if the (K, ι)-hermitian form on K n represented by H is hyperbolic.

(c)

Show that all hyperbolic unitary involutions on B are conjugate.

IX Digression: a geometric interpretation of H 1(−, G)

This chapter is a preamble to the following ones, where we will need the notion of a G-torsor to derive some interesting properties of the functor H 1 (− , G). Roughly speaking, a G-torsor is a scheme-theoretic generalization of a principal homogeneous space over G(ks ). We will introduce this notion, as well as the notion of a generic G-torsor, which is a G-torsor which specializes densely to any other torsor. For example, √ k(t)( t)/k(t) is a generic Z/2Z-torsor. We will see in this chapter and the following ones that quite often generic G-torsors concentrate all the essential information on the functor H 1 (− , G). All of this will require the notion of a scheme. Since giving precise definitions would be too lengthy, and out of the scope of this book, we will just give an insight of what a scheme is, and leave the interested reader to refer to his/her favorite book on the subject. §IX.22 Reminiscences on schemes Let k be a field. By [17, Théorème de comparaison, p.18], a scheme X over k is uniquely determined by its functor of points, still denoted by X by abuse of notation, that is the functor

X:

Algk −→ Sets R −→ X(R) = Mork−Sch (Spec(R), X).

Moreover, morphisms of schemes X −→ Y correspond to natural transformations between the corresponding functors of points. Elements of X(R) are called R-points. For example, if A is a commutative k-algebra, the topological space 249

250

Digression: a geometric interpretation of H 1 (− , G)

Spec(A) of prime ideals of A (endowed with Zariski topology) is a kscheme (in fact, we also have to specify the sheaf of functions, but we will not do it here). Recall that Zariski topology is the topology whose closed subsets are the sets V (A) = {p ∈ Spec(A) | p ⊃ A}, where A is an ideal of A. The functor of points of Spec(A) is given by X:

Algk −→ Sets R −→ X(R) = Homk−alg (A, R).

Moreover, a morphism of k-schemes Spec(R) −→ Spec(A) corresponds to a morphism of k-algebras A −→ R, in view of Yoneda’s lemma. Notice that the k-algebra structure on A then gives rise to a morphism of k-schemes Spec(A) −→ Spec(k). If moreover A k[X1 , . . . , Xn ]/I is a finitely generated k-algebra, the functor of points of Spec(A) is hA V (I), where V (I) is the functor introduced in Lemma III.7.14. Such a k-scheme is called affine. If A is finitely generated, we say that Spec(A) is a k-scheme of finite type. If L/k is a field extension, an L-point corresponds to a morphism of k-algebras A −→ L. Since L is an integral domain, the kernel p of this morphism is a prime ideal, and we have a morphism of k-algebras Ap −→ L, where Ap is the localization of A at p. We then have a commutative diagram /L A@ O @@ @@ @@ Ap and thus the diagram / Spec(A) r8 r rr r r r rrr

Spec(L)

Spec(Ap )

commutes. If κ(p) = Ap /pAp Frac(A/p), the morphism Ap −→ L

IX.22 Reminiscences on schemes

251

gives rise to an injection κ(p) → L, and the two diagrams Ap O A

/ κ(p) /L

Spec(L)

/ Spec(A) O

Spec(κ(p))

/ Spec(Ap )

are commutative. Assume now that A is an integral domain, with field of fractions k(A). In this case, the topological space Spec(A) is irreducible, and the diagrams / k(A) AB O BB BB BB B! Ap / Spec(A) 8 q qq q q qq qqq Spec(Ap )

Spec(k(A))

are commutative. In general, given a k-scheme X, for any x ∈ X, one can define a local k-algebra OX,x , with residue field κ(x). If now L/k is a field extension and a : Spec(L) −→ X is an L-point, there exists a point x ∈ X such that we have an injective morphism κ(x) → L, and the two diagrams / u: X u u uu uu u uu Spec(OX,x ) Spec(L)

Spec(L)

/X O

Spec(κ(x))

/ Spec(OX,x )

252


are commutative. If X is irreducible, one can also define a function field k(X), and for any x ∈ X, the diagram / u: X u u uu uu u uu Spec(OX,x )

Spec(k(X))

commutes. If X = Spec(A) and x = p ∈ X, we have OX,x Ap , κ(x) κ(p) and k(X) k(A). If X = Spec(A) and A is an integral domain, the ideal (0) is prime, and the corresponding residue field is isomorphic to k(A). Since every non-empty Zariski open subset contains (0), (0) is dense in X. In general, if X is irreducible, there exists a unique point η ∈ X which is dense in X. Its residue field κ(η) is isomorphic to k(X). Such a point is called the generic point of X. We also have the notion of a k-scheme of finite type, which generalizes the one given in the affine case. A rational map X _ _ _/ Y is a morphism of k-schemes U −→ Y where U is a non-empty open subset of X. A rational map is dominant if its image is dense in Y . If Y is a k-scheme, a Y -scheme is a pair (X, f ), where X is a k-scheme and f : X −→ Y is a morphism of k-schemes. If X and X are Y schemes, a morphism of Y -schemes is a morphism of schemes X −→ X such that the diagram / X XB BB BB BB B Y commutes. We end this section by introducing the fibered product. If f : X −→ Y and f : X −→ Y are two morphisms of k-schemes, the fibered product X ×Y X (or pullback) is the k-scheme corresponding to the

IX.23 Torsors

253

functor of points Algk −→ Sets R −→ X(R) ×Y (R) X (R), where X(R) ×Y (R) X (R) = {(a, a ) ∈ X(R) × X (R) | fR (a) = fR (a )}. The fibered product has the following universal property: if g : T −→ X and g : T −→ X are two morphisms of schemes such that f ◦g = f ◦g , there exists a unique morphism ϕ : T −→ X ×Y X satisfying πX ◦ ϕ = g and πX ◦ ϕ = g , where πX and πX are projections onto X and X respectively. If X = Spec(A), X = Spec(A ) and Y = Spec(B) are affine k-schemes, then X ×Y X = Spec(A ⊗B A ), where the tensor product is formed with respect to the maps B −→ A and B −→ A . If f : X −→ Y is a morphism of k-schemes and y ∈ Y , we have a morphism f : Spec(κ(y)) −→ Y . The corresponding fibered product is called the fiber of y along f , and is denoted by f . If Y is irreducible, the generic fiber is f −1 (η), where η ∈ Y is the generic point of Y . §IX.23 Torsors We first extend the notion of a group-scheme. Definition IX.23.1. Let Y be a scheme. A group-scheme over Y is a functor G : Algk −→ Grps which is also the functor of points of a Y -scheme. We now give the definition of a G-torsor. Definition IX.23.2. Let G be an affine group-scheme over Y which is flat and locally of finite type over Y . We say that a morphism of schemes X −→ Y is a (flat) G-torsor over Y if G acts on X (on the right), the morphism X −→ Y is faithfully flat and locally of finite type, and the map ϕ : G ×Y X −→ X ×Y X defined by G ×Y X −→ X ×Y X (g, x) −→ (x, x·g) is an isomorphism. A morphism of G-torsors is just a G-equivariant morphism of Y -schemes.

254


One can show that a morphism between two Y -torsors is an isomorphism. If Y −→ Y is a morphism of schemes and X −→ Y is a G-torsor, then X ×Y Y is a G-torsor (in fact a G ×Y Y -torsor) over Y . The torsor Y × GY −→ Y is called the split G-torsor. Remark IX.23.3. We will not define the notion of a (faithfully) flat morphism locally of finite type. Let us just say that if X and Y are affine k-schemes of finite type, any morphism X −→ Y is locally of finite type. If X = Spec(A) and Y = Spec(B), then the morphism X −→ Y is (faithfully) flat if and only if the corresponding morphism B −→ A endows A with a structure of (faithfully) flat B-module. The reader will refer to [41] for the missing definitions and proofs on G-torsors. Example IX.23.4. If G is a linear algebraic group over k, and K/k is a field extension, then H 1 (K, G) classifies isomorphism classes of G-torsors over Spec(K), as well as isomorphism classes of principal homogeneous spaces over G(Ks ) (see Chapter II, Exercise 5). One direction of the correspondence between these two sets is given as follows (the other direction is given by Exercice 2): if X −→ Spec(K) is a G-torsor, since G is affine and smooth, X is affine and smooth by faithfully flat descent, hence X(Ks ) is not empty, and is a principal homogeneous space over G(Ks ). Let k be a field, and let G be a linear algebraic group over k. Since every k-algebra morphism R −→ S gives rise to a morphism of k-schemes Spec(S) −→ Spec(R), we get a functor H 1 (− , G) : Algk −→ Sets∗ by setting H 1 (R, G) = { isomorphism classes of GR − torsors over Spec(R)}, where the base point is the isomorphism class of the split G-torsor and the map H 1 (R, G) −→ H 1 (S, G) is given by H 1 (R, G) −→ H 1 (S, G), T −→ TS = T ×Spec(R) Spec(S). We would like now to introduce the notion of a generic torsor. We first explain how to specialize elements of a functor. Definition IX.23.5. Let k be a field, and let K/k, L/k be two field extensions. A pseudo k-place f : K L is a local morphism of kalgebras ϕf : Rf −→ L, where Rf is a local ring of K containing k.

IX.23 Torsors

255

Let F : Algk −→ Sets be a functor. Let K/k, L/k be two field extensions. Let a ∈ F(K) and b ∈ F(L). We say that b is a specialization of a if there exists a pseudo k-place f : K L and c ∈ F(Rf ) such that cK = a and b = F(ϕf )(c). Example IX.23.6. Let us illustrate the previous definition when F is the functor Quadn . Let K = k(X1 , . . . , Xn ) and let q = X1 , . . . , Xn . Now let R0 = 1 ]) and let Y = Spec(R0 ). Since X1 · · · Xn ∈ Spec(k[X1 , . . . , Xn ][ X1 ...X n × R0 , the quadratic form Q = X1 , . . . , Xn over R0 is regular. Moreover, we have QK q. Now if L/k is a field extension and if q is a regular quadratic form over L, we have q a1 , . . . , an , for some ai ∈ L× . Let p ∈ R0 be the kernel of the map R0 −→ L Xi −→ ai . This is a prime ideal of R0 , that is a point y of Y . The corresponding local ring OY,y is (R0 )p , and the map ϕ : OY,y −→ L which sends Xi onto ai defines a pseudo k-place K L. Let Q = QOY,y . Then Q is a regular quadratic form on OY,y satisfying QK q. By definition of scalar extension for quadratic forms, we have QL = ϕ(X1 ), . . . , ϕ(Xn ) = a1 , . . . , an q . Hence q is a specialization of q in the sense of the previous definition. Notice that, all in all, q is the quadratic form obtained from q after performing the substitution X1 = a1 , . . . , Xn = an , which agrees with the intuitive notion of specialization. Notice also that if L is infinite, replacing ai by ai λ2i provides a dense subset of Y such that the corresponding specialization is isomorphic to q . This example motivates the following definition. Definition IX.23.7. Let F : Algk −→ Sets be a functor. An element a ∈ F(K) is generic if for every field extension L/k with L infinite, every element b ∈ F(L) is a specialization of a (hence (a, K) is a versal pair in the sense of [3].)

256


Example IX.23.8. The previous example shows that X1 , . . . , Xn is a generic element of Quadn . One can show that the functors H 1 (− , On ) and Quadn are isomorphic as functors from Algk to Sets∗ (and not only as functors from Ck to Sets∗ as we already know from Galois descent). We are going to see that, more generally, the functor H 1 (− , G) has a generic element. Definition IX.23.9. Let k be a field and let G be a linear algebraic group over k. Let f : X −→ Y be a G-torsor with Y irreducible. We say that it is classifying for G if, for any field extension k /k with k infinite and for any G-torsor P over Spec(k ), the set of points y ∈ Y (k ) such that P is isomorphic to the fiber f −1 (y) is dense in Y . A generic G-torsor is the generic fiber of a classifying torsor. Examples IX.23.10. (1)

If ρ : G → GL(V ), then GL(V ) −→ GL(V )/ρ(G) is a classifying G-torsor.

(2)

Let V be a finite dimensional k-vector space. If G acts linearly and freely on a G-stable open subset of V , then by [64, Proposition 4.7] there exists a G-stable open subset U of V and a kscheme Y such that U −→ Y is a G-torsor. For example, assume G is a finite group acting linearlly and faithfully on V , and set U = V \ ker (g − 1). g=1

Then Spec(k[U ]) −→ Spec(k[U ]G ) is a classifying torsor. Notice that if k(V ) denotes the field of rational functions on the affine space V , the action of G on V induces an action of G on k(V ). If f : k(V ) −→ k is such a rational function, then we set g.f (v) = f (g −1 ·v) for all g ∈ G, v ∈ V. One can check that we get a linear faithful action of G on k(V ). The generic torsor corresponding to the above classifying torsor is then k(V )/k(V )G . See [25], Section 5.3 and Example 5.4. for a proof of these two facts.

IX.23 Torsors

257

Proposition IX.23.11. Let G be a linear algebraic group defined over k, and let P −→ Spec(K) be a generic G-torsor. Then for every field extension L/k with L infinite and every G-torsor T −→ Spec(L), there exist a regular locar ring R of K containing k, a morphism of local kalgebras R −→ L and a G-torsor P −→ Spec(R) such that PK P and PL T.

In particular, a generic G-torsor is a generic element of H 1 (− , G). Proof. Let f : X −→ Y be a classifying torsor whose generic fiber is isomorphic to P . Replacing Y by an open subset U , and X by f −1 (U ) if necessary, one can assume that Y is smooth since f −1 (U ) −→ U is also a classifying torsor. Take T −→ Spec(L) any torsor defined over L/k, L infinite. Since X −→ Y is a classifying torsor, there exists a L-rational point y : Spec(L) −→ Y such the diagram /X T Spec(L)

/Y

commutes, that is T is the pullback X ×Y Spec(L). Let OY,y be the local ring at the point y and let ϕ : Spec(OY,y ) −→ Y the canonical morphism. Notice that OY,y is a regular local ring, since Y is smooth. Consider P −→ Spec(OY,y ) the torsor obtained by pullingback X −→ Y along ϕ. The local ring OY,y is naturally a k-subalgebra of k(Y ) and using the universal property of the fibered product, we see that we have a diagram P TT TT T T)

P

7/ X ooo o o oo

/Y Spec(k(Y )) 8 r RRRR rr RR) rrrϕ Spec(OY,y ) showing that P −→ Spec(k(Y )) is nothing but the generic fiber of P −→ Spec(OY,y ). Moreover the morphism y : Spec(L) −→ Y factors through Spec(κ(y)). Let us denote by P −→ Spec(κ(y)) the torsor obtained by pulling-back P −→ Spec(OY,y ) along the morphism

258


Spec(κ(y)) −→ Spec(OY,y ). Then P −→ Spec(κ(y)) is the pull-back of P −→ Spec(OY,y ) along the morphism Spec(κ(y)) −→ Y . Using the universal property of the fibered product, one gets the existence of a map T → P such that the diagram dd/1 l l T ddddddddddddddodooo7 X l d o o uldlddddddddd / P o P d y Spec(L) ddr2/8 Y d d d d d d rr nn dddddd rrrϕ nv nndddddddddddd / Spec(OY,y ) Spec(κ(y)) commutes. This shows that T −→ Spec(L) comes from P −→ Spec(κ(y)). We deduce that T −→ Spec(L) is the pullback of P −→ Spec(OY,y ) along Spec(L) −→ Spec(OY,y ). This concludes the proof. We end this section by giving an application of generic torsors (see [25, Theorem 12.3]): Theorem IX.23.12. Let G be a linear algebraic group defined over k (char(k) = 2). Let ι, ι : H 1 (− , G) −→ H d (− , μ2 ) be two cohomological invariants, and let P −→ Spec(K) be a generic G-torsor. If ιK (P ) = ιK (P ), then ι = ι . Proof. We only give a partial proof in the case where k is infinite. The reader will refer to [25] to fill the gaps. Clearly, we can assume that ι = 0 without any loss of generality. Therefore, the assumption reads ι(P ) = 0. Let L/k be any field extension. Since k is infinite, so is L. By Proposition IX.23.11, for any G-torsor T −→ Spec(L), there exist a regular locar ring R of K containing k, a morphism of local k-algebras R −→ L and a G-torsor P −→ Spec(R) such that PK P and PL T.

By assumption, we have ιK (PK ) = ιK (P ) = 0.

Let m be the maximal ideal of R, and set k = R/m. By the Specialization Theorem [25, Theorem 12.2], we have ιk (Pk ) = 0. Since R −→ L

Exercises

259

is a morphism of local rings, it factors through R −→ k , so we have ιL (T ) = ιL (PL ) = ((ιk (Pk ))L = 0. Hence ι = 0, and this concludes the proof. Using this result, Serre computed the group of all cohomological invariants of H 1 (− , G) for some algebraic groups G. See [25] for more details. We will give other applications of this result in the next chapter.

Exercises 1. Let F, F : Algk −→ Sets be two functors, and let ϕ: F

/ / F

be a surjective natural transformation. Show that if a ∈ F(K) is a generic element of F, then ϕK (a) is a generic element of F . 2. Let k be a field, let G be an algebraic group over k represented by A, and let X be an affine k-scheme represented by B endowed with a right G-action. (a)

Express the condition ‘X −→ Spec(k) is a G-torsor’ in terms of A and B.

(b)

Let G be an abstract finite group, and let L be a Galois G-algebra over k. Using the previous question, show that Spec(L) −→ Spec(k) is a G-torsor.

3. Let k be a field, let G be an algebraic group over k represented by A, and let P be a principal homogeneous space over G(ks ). (a)

Define a natural left G(ks )-action on Aks , and check that G(ks ) act on the right on the set S = P × Aks by (x, a)·g = (x·g, g −1 ·a).

(b)

Fix an element x0 ∈ P . Show that any orbit of G(ks ) in S is represented in a unique way by an element of the form (x0 , a) for some a ∈ Aks .


260 (c)

Let C the orbit set of G(ks ) in S. We will denote by ξa the orbit of (x0 , a) ∈ S. Show that the operations C × C −→ C (ξa1 , ξa2 ) −→ ξa1 +a2 C × C −→ C (ξa1 , ξa2 ) −→ ξa1 a2 ks × C −→ C (λ, ξa ) −→ ξλa endows C with a structure of a ks -algebra.

(d)

Check that the diagonal action of Gks on S induces an action of Gks on C.

(e)

Let B = C Gks , and let X = Spec(B). Show that G acts on X on the right, that X −→ Spec(k) is a G-torsor, and that X(ks ) P .

4. Let X −→ Spec(k) be a G-torsor. Show that X is isomorphic to the split G-torsor if and only if X(k) = ∅.

X Galois cohomology and Noether’s problem

In this chapter, we give applications of Galois cohomology to Noether’s problem: given an infinite field k and a finite group G acting linearly and faithfully on a finite dimensional k-vector space V , is the field extension k(V )G /k purely transcendental? We will show that the answer is negative in general by constructing a non-zero cohomological obstruction. §X.24 Formulation of Noether’s problem Let k be an infinite field, and let G be a finite abstract group. Recall that a linear faithful representation of G is a finite dimensional k-vector space V on which G acts linearly and faithfully. Now, one can ask the following question: does there exist a linear faithful representation such that k(V )G /k rational (i.e. a purely transcendental extension) ? We will refer to this question as Noether’s problem for G over k, and we will denote it by (N oethG,k ). If we choose a basis e = (e1 , . . . , en ) of V , this may be reinterpreted in a more explicit way. The faithful action of G on V gives rise to an injective group morphism ρ : G → GLn (k). Moreover, k(V ) identifies to k(X1 , . . . , Xn ) via the isomorphism ∼

k(X1 , . . . , Xn ) −→ k(V ) + n , F −→ vi ei −→ F (v1 , . . . , vn ) . i=1

If g ∈ G and ρ(g)−1 = (aij )1≤i,j≤n , the action of g on F ∈ k(X1 , . . . , Xn ) 261

262

Galois cohomology and Noether’s problem

is then given by

⎛ g·F = F ⎝

n j=1

a1j Xj , . . . ,

n

⎞ anj Xj ⎠ .

j=1

Noether’s problem then reformulates as follows: does there exist an injective group morphism ρ : G → GLn (k) such that the field extension k(X1 , . . . , Xn )G /k is rational ? The answer is known to be positive for k = C when G is abelian by a theorem of Fischer [22]. However, it becomes false for k = Q and for various cyclic groups, for example G = Z/8Z (which is the smallest possible counterexample). In this section, we are going to prove the following theorem, due to Serre [25, Theorem 33.16]: Theorem X.24.1. Let G be a finite group with a 2-Sylow subgroup which is cyclic of order ≥ 8. Then Noether’s problem for G over Q has a negative answer. The case where G is abelian was proved by Endo-Miyata [21], Lenstra[33] and Voskrensenski˘ı [67]. Later on, Voskrensenski˘ı [68] and Saltman [51] proved the stronger property that there is no generic Galois extension of group G over a rational extension of Q. Notice that Saltman also proved in [52] that Noether’s problem has a negative answer over C in general when G is not abelian. In fact, he exhibits examples of finite groups G for which there is no generic Galois extension of group G over a rational extension of C, by considering unramified cohomology classes in the Brauer group (see the next section for the notion of an unramified cohomology class). We refer the reader to the survey of Colliot-Thélène and Sansuc [15] on rationality problems for more results on Noether’s problem and similar questions. §X.25 The strategy The main problem is to find a way to detect the non-rationality of a finitely generated field extension K/k. The answer is (sometimes) given by looking at the unramified cohomology of K/k. We will not be very precise here, since we will give details in the following paragraphs, but we would like to give a rough idea of the strategy of the proof. We assume that the reader is familiar with the language of valuations, and refer to [12], [44] or [71] for the missing definitions and basic properties of valued fields.

X.25 The strategy

263

Assume that char(k) = 2. Let υ be a discrete valuation of K which is trivial on k, and let ∂υ be the map defined by ∂υ :

H 1 (K, μ2 ) −→ Z/2Z (a) −→ υ(a)

where a is any representative of the square class corresponding to (a) via the isomorphism H 1 (K, μ2 ) K × /K ×2 . This map is a well-defined group morphism (if (a) = (b) in H 1 (K, μ2 ), then a and b differ by a non-zero square and υ(b) ≡ υ(a) (mod 2)), called the residue at υ. Say that a class (a) ∈ H 1 (K, μ2 ) is unramified at υ if ∂υ ((a)) = 0, 1 (K/k, μ2 ) the subgroup of all elements of H 1 (K, μ2 ) and denote by Hnr which are unramified at all discrete valuations of K which are trivial on 1 (K/k, μ2 ) since υ is trivial on k. Notice that if a ∈ k × , then (a) ∈ Hnr k. Hence the restriction map induces a group morphism 1 (K, μ2 ) H 1 (k, μ2 ) −→ Hnr

Elements lying in the image of this map are called constant. Assume that K = k(t) where t is an indeterminate over k. What is 1 1 (K/k, μ2 ) in this case ? Let (a) ∈ Hnr (K/k, μ2 ). Multiplying by Hnr a square if necessary, we may assume that a ∈ k[t]. Now let π be an arbitrary monic irreducible polynomial dividing a, and let υπ the corresponding valuation on k(t). By assumption, we will have ∂υπ ((a)) = υπ (a) = 0 ∈ Z/2Z. It implies that a = cP 2 for some c ∈ k × and P ∈ k[t]. 1 (K/k, μ2 ) consists of constant elements. It follows that (a) = (c), so Hnr We can even say a bit more: assume that a ∈ k × and that ResK/k (a) = 0 ∈ H 1 (K, μ2 ). It means that a ∈ k(t)×2 . Now if a = F (t)2 , F (t) ∈ k(t), P then write F = , where P, Q ∈ k[t], Q = 0 are relatively prime. We Q then have aQ(t)2 = P (t)2 . It easily implies that P (0) = 0 and Q(0) = 0. P (0)2 Therefore a = ∈ k ×2 , so (a) = 0. Consequently, we have proved Q(0)2 1 (k(t)/k, μ2 ) H 1 (k, μ2 ). Hnr

In fact, if K/k is any finitely generated extension, one can define residue morphisms ∂υ : H d (K, μ2 ) −→ H d−1 (κ(υ), μ2 ) for all d ≥ 1 and every discrete valuation of K which is trivial on k.

264


d In particular, we can define Hnr (K, μ2 ) as above for all d ≥ 1. It follows from [25, Theorem 10.1] that the restriction map induces a group isomorphism d (k(X1 , . . . , Xn ), μ2 ) for all d ≥ 1. H d (k, μ2 ) Hnr

In particular, the restriction map H d (k, μ2 ) −→ H d (k(X1 , . . . , Xn ), μ2 ) is injective. How can it be useful for our problem ? To explain this, we first need the notion of an unramified cohomological invariant. Definition X.25.1. Let k be a field, let G be an algebraic group-scheme defined over k. We say that a cohomological invariant ι : H 1 (− , G) −→ H d (− , μ2 ) is unramified if for all finitely generated field extensions K/k and all d (K/k, μ2 ). [α] ∈ H 1 (K, G) we have ιK ([α]) ∈ Hnr We now come to the key idea of the proof of Theorem X.24.1, which is given by the following proposition. Proposition X.25.2. Let k be a field, and let G be a finite group. If (N oeth)G,k has a positive answer, then every unramified normalized cohomological invariant of G is trivial. Proof. Let V be a linear faithful representation of G such that k(V )G /k is rational, and let ι : H 1 (− , G) −→ H d (− , μ2 ) be an unramified normalized cohomological invariant. By Example IX.23.10 (2), k(V )/k(V )G is a generic G-torsor. By assumption ιk(V )G (k(V )/k(V )G ) is an element d (k(V )G , μ2 ), and since k(V )G /k is rational, it means that there of Hnr exists a ∈ H d (k, μ2 ) such that ιk(V )G (k(V )/k(V )G ) = Resk(V )G /k (a). Let ι : H 1 (− , G) −→ H d (− , μ2 ) be the constant cohomological invariant corresponding to a. By Theorem IX.23.12, we deduce that ι = ι , that is ιK (L/K) = ResK/k (a) for every field extension K/k and every Galois G-algebra L/K. Applying

X.26 Residue maps

265

this equality to the split Galois G-algebra L0 , we get a = ιk (L0 ) = 0, since ι is normalized. It follows that ι = 0. Therefore, to prove that (N oethG,k ) has a negative answer, it is enough to produce a non-zero unramified normalized cohomological invariant of G. This is exactly the strategy we are going to use. Remark X.25.3. In fact, the conclusion of the previous proposition remains true if we only assume that H 1 (− , G) has a rational generic G-torsor, that is a generic G-torsor T −→ Spec(K), where K/k is rational. Hence producing a non-zero unramified normalized cohomological invariant of G leads to the stronger conclusion that H 1 (− , G) has no rational generic G-torsor. §X.26 Residue maps In this section, we introduce the so-called residue maps H n (K, μ2 ) −→ H n−1 (κ(υ), μ2 ), where (K, υ) is a valued field. All the valuations are supposed to be discrete and normalized. Let (K, υ) be a complete valued field. Since K is complete, the valuation υ extends uniquely to a valuation υs on Ks . The following result is certainly well-known, but we give a proof by lack of an appropriate reference. Lemma X.26.1. Keeping the previous notation, the residue field of υs is an algebraic closure of κ(υ). Proof. Let us denote by k and the residue fields of (K, υ) and (Ks , υs ) respectively. If Knr /K is the maximal unramified extension of K in Ks and υnr the unique extension of υ to Knr , then the residue field of (Knr , υnr ) is ks by [44, Chapter II, Prop.7.5], since a complete valued field is henselian. Let us also denote by Oυ , Onr and Os the valuation rings of υ, υnr and υs respectively and by mυ , mnr and ms the corresponding maximal ideals. Since K ⊂ Knr ⊂ Ks , /k is algebraic and we have ks ⊂ ⊂ kalg . If char(k) = 0, then we have kalg = ks = , and we are done. Now assume that char(k) = p > 0. Since /k is algebraic, it remains to prove that is algebraically closed. Let α ∈ alg = kalg . Then α is algebraic over k, and therefore over ks .

266


Since ks is separably closed, the minimal polynomial μα,ks of α over ks has the form r

μα,ks = X p − a ∈ ks [X], for some r ≥ 1 and some a ∈ ks ⊂ . Let b ∈ Os any lifting of a. Let c ∈ Os be any nonzero element, and consider the polynomial r

P = X p − cX − b ∈ Os [X] ⊂ Ks [X]. Since P = −c = 0, P and P are relatively prime, so P is separable over Ks . Since Ks is separably closed, P splits in Ks . Let β ∈ Ks be any root of P . Since a valuation ring is integrally closed, we have β ∈ Os . r pr Reducing the equality β p = cβ + b modulo ms , we get β = a ∈ and we have β

pr

r

= αp ∈ alg .

r

Hence (α − β)p = 0 and thus α = β ∈ . This concludes the proof. For all σ ∈ GKs , υs ◦ σ is a valuation on Ks extending υ, and therefore υs ◦ σ = υs . In particular, σ restricts to a K-automorphism Oυs −→ Oυs , which maps mυs onto mυs . Thus it induces by Lemma X.26.1 a κ(υ)automorphism σ : κ(υ)alg −→ κ(υ)alg , and therefore by restriction a κ(υ)-automorphism σ : κ(υ)s −→ κ(υ)s . We get in this way a morphism of profinite groups ϕυ :

GKs −→ Gκ(υ) σ −→ σ,

whose kernel is by definition the inertia group Iυ . We recall now the following result (see [25], Lemma 7.6): Lemma X.26.2. We have a split exact sequence 1

/ Iυ

/ GKs

ϕυ

/ Gκ(υ)s

/1.

X.26 Residue maps

267

From now on, we will assume that K and κ(υ) both have characteristic different from 2. For d ≥ 0, we will denote by jυ : H d (κ(υ), μ2 ) −→ H d (K, μ2 ) the group morphism induced by ϕυ and the identity. Lemma X.26.3. For d ≥ 1 and p, q ≥ 0, the following properties hold: (1)

The map jυ : H d (κ(υ), μ2 ) −→ H d (K, μ2 ) is injective.

(2)

If (L, w)/(K, υ) is an extension of complete valued fields, then κ(υ) ⊂ κ(w) and we have jw ◦ Resκ(w)/κ(υ) = ResL/K ◦ jυ .

(3)

For all [α] ∈ H p (κ(υ), μ2 ), [β] ∈ H q (κ(υ), μ2 ), we have jυ ([α] ∪ [β]) = jυ ([α]) ∪ jυ ([β]).

(4)

For all u1 , . . . , ud ∈ Oυ× , we have jυ ((u1 ) ∪ · · · ∪ (ud )) = (u1 ) ∪ · · · ∪ (ud ).

Proof. By Lemma X.26.2, the sequence 1

/ Iυ

/ GKs

ϕυ

/ Gκ(υ)s

/1

splits. Let ψυ : Gκ(υ)s −→ GKs be a morphism of profinite groups such that ϕυ ◦ ψυ = IdGκ(υ)s . Denoting by ρυ : H d (K, μ2 ) −→ H d (κ(υ), μ2 ) the group morphism induced by ψυ and the identity of μ2 , it is not difficult to see that we have ρυ ◦ jυ = IdH d (κ(υ),μ2 ) . In particular, jυ is injective. This proves (1). Point (2) simply follows from the fact that we have a commutative diagram / GKs GLs Gκ(w)

/ Gκ(υ)

and from the definitions of the various maps involved.

268


Point (3) being an immediate consequence of Example II.6.7, it remains to prove (4). By (3), it is enough to prove it for d = 1. Let u ∈ Oυ× , and 2 let y ∈ κ(υ)× s such that y = u. By Hensel’s lemma, y may be lifted to × a square root x ∈ Oυs of u. We then have x2 = u and x2 = u. A cocycle representing (u) is given by Gκ(υ)s −→ μ2 (κ(υ)s ) α:

τ −→

τ (x) , x

and therefore a cocycle representing jυ ((u)) is GKs −→ μ2 (Ks ) β:

σ −→

σ(x) . x

We now claim that σ(x) = x ⇐⇒ σ(x) = x for all σ ∈ GKs . Let σ ∈ GKs . If σ(x) = x, then σ(x) ≡ x mod mυ , that is σ(x) = x. Conversely, assume that σ(x) = x. Suppose that σ(x) = −x. Reducing modulo mυ , we would get x = −x, and thus 1 = −1 ∈ κ(υ), which is not possible since char(κ(υ)) = 2. Hence σ(x) = x as claimed. It follows that β is in fact the cocycle GKs −→ μ2 (Ks ) β:

σ −→

σ(x) , x

which is known to represent (u). This concludes the proof. We continue with the following theorem: Theorem X.26.4. Let (K, υ) be a complete valued field, and let π be a local parameter. Then for all d ≥ 1, every [α] ∈ H d (K, μ2 ) can be written in a unique way as [α] = jυ ([α0 ]) + (π) ∪ jυ ([α1 ]) where [αi ] ∈ H d−i (κ(υ), μ2 ). The class [α1 ] does not depend on the choice of π. Moreover, if [α1 ] = 0, the class [α0 ] does not depend on the choice of π. The reader will refer to [25], Chapter II for a proof of this result.

X.26 Residue maps

269

Remark X.26.5. If d = 1, then H d−1 (κ(υ), μ2 ) is canonically identified with Z/2Z, so [α1 ] = m, m = 0 or 1, and (π) ∪ jυ ([α1 ]) has to be understood as the class m(π). Assume now that (K, υ) is a valued field which is not necessarily complete, and let Kυ be the completion of K with respect to υ. Let [α] ∈ H d (K, μ2 ). By Theorem X.26.4, we may write ResKυ /K ([α]) = jυ ([α0 ]) + (π) ∪ jυ ([α1 ]). Definition X.26.6. Keeping the notation above, the residue of [α] at υ is the class ∂υ ([α]) = [α1 ] ∈ H d−1 (κ(υ), μ2 ). We say that [α] is unramified at υ if ∂υ ([α]) = 0 In this case, we define the specialization sυ ([α]) of [α] at υ by sυ ([α]) = [α0 ] ∈ H d−1 (κ(υ), μ2 ). Clearly, the maps ∂υ : H d (K, μ2 ) −→ H d−1 (κ(υ), μ2 ) and sυ : ker(∂υ ) −→ H d (κ(υ), μ2 ) are group morphisms. The following lemma immediately follows from the description of the map jυ in Lemma X.26.3 (4) and from the definition of the residue map: Lemma X.26.7. Let (K, υ) be a valued field. Let π ∈ Oυ be a local parameter, and assume that u1 , . . . , ud ∈ Oυ× . Then we have ∂υ ((u1 ) ∪ · · · ∪ (ud )) = 0, ∂υ ((u1 ) ∪ · · · ∪ (ud−1 ) ∪ (π)) = (u1 ) ∪ · · · ∪ (ud−1 ) and sυ ((u1 ) ∪ · · · ∪ (ud )) = (u1 ) ∪ · · · ∪ (ud ). The next proposition shows how ∂υ and sυ behave with respect to scalar extensions. Proposition X.26.8. Let (L, w)/(K, υ) be an extension of valued fields, and let d ≥ 1. Then for every [α] ∈ H d (K, μ2 ), we have ∂w (ResL/K ([α])) = e(w|υ)Resκ(w)/κ(υ) (∂υ ([α])).

270


Moreover, if [α] is unramified at υ, ResL/K ([α]) is unramified at w, and we have sw (ResL/K (α)) = Resκ(w)/κ(υ) (sυ ([α])). Proof. To compute the residue of ResL/K ([α]) at w, we need first to extend scalars to the completion Lw of L with respect to w. Let π be a local parameter for υ, and let π be a local parameter for w. By × . We have assumption, we have π = uπ e , where e = e(w|υ) and u ∈ Ow ResLw /L (ResL/K ([α])) = ResLw /K ([α]) = ResLw /Kυ (ResKυ /K ([α])). To simplify notation, set k = κ(υ), = κ(w) and [α ] = ResL/K ([α]). We have ResKυ /K ([α]) = jυ ([α0 ]) + (π) ∪ jυ ([α1 ]) where [αi ] ∈ H d−i (κ(υ), μ2 ), and thus ResLw /L ([α ]) = ResLw /Kυ (jυ ([α0 ]) + (π) ∪ jυ ([α1 ])). Using Lemma X.26.3 (2) and Remark III.9.20, we see that ResLw /L ([α ]) = jw (Res/k ([α0 ])) + ResL/K ((π)) ∪ jw (Res/k ([α1 ])). Now we have ResL/K ((π))

= (uπ e ) = (u) + e(π ) = jw ((u)) + e(π ),

the last equality coming from Remark X.26.3 (4). Using Remark X.26.3 (3), we get that ResLw /L ([α ]) is equal to jw (Res/k ([α0 ]) + (u) ∪ Res/k ([α1 ])) + (π ) ∪ jw (eRes/k ([α1 ])). By definition of the residue map, we then obtain ∂w ([α ]) = eResκ(w)/κ(υ) ([α1 ]) = e(w|υ)Resκ(w)/κ(υ) (∂υ ([α])). This equality shows in particular that if [α] is unramified at υ, then ResL/K ([α]) is unramified at w. Moreover, we get in this case ResLw /L ([α ]) = jw (Resκ(w)/κ(υ) ([α0 ])), that is sw ([α ]) = Resκ(w)/κ(υ) ([α0 ]) = Resκ(w)/κ(υ) (sυ ([α])). This concludes the proof.

X.27 An unramified cohomological invariant

271

Remark X.26.9. We have cheated a bit here. In fact, the residue map is defined in the hard way using spectral sequences and/or non-trivial results on cohomology of complete valued fields, and Theorem X.26.4 is a consequence of the properties of the residue map. The definitions above may be extended to cohomology classes with values in an arbitrary discrete GKs -module. See [1],[25] or [26] for more details. §X.27 An unramified cohomological invariant Let G be a finite group, and let k be a field of characteristic different from 2. For every field extension K and every Galois G-algebra L/K, we set ιK (L/K) = (2) ∪ (dL ) ∈ H 2 (K, μ2 ), where dL is the discriminant of L/K viewed as an étale algebra. It is not difficult to see that the cup-product commutes with scalar extensions. Moreover, since the same property holds for the discriminant, we get a cohomological invariant ι : H 1 (− , G) −→ H 2 (− , μ2 ), and this invariant is normalized since the discriminant of the split Galois G-algebra is trivial. Lemma X.27.1. If k = Q and G = Z/2m Z, m ≥ 3, the invariant ι is non-zero. m

Proof. Let n = 22 −1 and L = Q2 (ζn ). This extension is cyclic of group G by [57], Chapter IV, § 4, Proposition 16 and Corollaire 1. Notice that √ √ 3 | n, and therefore ζ3 ∈ L. In particular, −3 ∈ L and Q2 ( −3)/Q2 is the unique quadratic subextension of L/Q2 , since −3 is not a square in Q2 . On the other hand, the discriminant of L/Q2 is not trivial by Corollary VII.17.10 and Remark VII.17.11. Since dL is a polynomial expression in some elements of L by Lemma VII.17.9, we conclude that √ √ √ K( dL ) is a quadratic subfield of L. Hence Q2 ( dL ) = Q2 ( −3). Thus ×2 and ιQ2 (L/Q2 ) = (2) ∪ (−3). We now check that dL = −3 ∈ Q× 2 /Q2 2 (2) ∪ (−3) = 0 ∈ H (Q2 , μ2 ). Assume that (2) ∪ (−3) = 0. Then there exist a, b ∈ Q2 such that −3 = a2 − 2b2 , or equivalently there exists x, y, z ∈ Z2 , z = 0 such that −3z 2 = x2 −2y 2 . Dividing by a suitable power of 2, we may assume that x, y, z are not all lying in 2Z2 . Reducing modulo 2Z2 shows that x ≡ z mod 2Z2 . If x, z ∈ 2Z2 , we easily get that 2y 2 ∈ 4Z2 and thus y ∈ 2Z2 , contradicting our assumption. Hence we have x = 1 + 2x , z = 1 + 2z for some x , z ∈ Z2 . We then get

−3(1 + 4z + 4z 2 ) = 1 + 4x + 4x 2 − 2y 2 .

272


Thus −3(4z +4z 2 ) = 4+4x +4x 2 −2y 2 , so 2y 2 ∈ 4Z2 and thus y ∈ 2Z2 . If we set y = 2y , y ∈ Z2 , we get −3(z + z 2 ) = 1 + x + x 2 − 2y 2 . Now we get a contradiction, since z + z 2 and x + x 2 both lie in 2Z2 . Lemma X.27.2. If G = Z/2m Z, m ≥ 3, the invariant ι is unramified. Proof. Let K/k be a finitely generated field extension, let υ be a valuation on K which is trivial on k, and let L/K be a Galois G-algebra. Then ResKυ /K (ιK (L/K)) = ιKυ (LKυ /Kυ ). Set L = LKυ . This is a Galois G-algebra on Kυ . If L is not a field, then by Theorem V.14.20 L M r (as an étale algebra) for some Galois extension M/Kυ of group H ⊂ G, where H is a proper subgroup of G of index r. Since H is proper, r is a non-trivial power of 2, so we have dL = (dM )r = 1 ∈ Kυ× /Kυ×2 . Therefore (2) ∪ (dL ) = 0 and ιK (L/K) is unramified at υ. Assume now that L is a field. Then L /Kυ is a Galois extension of group G. Notice that by assumption, char(κ(υ)) = char(k) = 2. Thus, if L /Kυ is totally ramified, it is tamely ramified. Hence Kυ contains μ2m by [12, Chapter I, § 8, Proposition 1], so it contains a primitive 8th root of 1 since m ≥ 3, as well as a square root of 1. In particular 2 is a square in Kυ× , and we conclude as before. If L is not totally ramified, then the maximal unramified subfield L0 √ of LKυ is a non-trivial subfield and therefore contains Kυ ( dL ). Thus, √ Kυ ( dL )/Kυ is unramified. It implies that the valuation of dL is even, √ since the extension Kυ ( uπυ )/Kυ is totally ramified for any u ∈ Oυ× . Hence (2) ∪ (dL ) = (2) ∪ (u) for some u ∈ Oυ× . Notice that since υ is trivial on k, we have 2 ∈ Oυ× . It follows from Lemma X.26.7 that ∂υ ((2) ∪ (dL )) = 0. We then get that ιKυ (L /Kυ ) is also unramified at υ in this case. This concludes the proof. §X.28 Proof of Theorem X.24.1 We are now ready to conclude. Let G be a group with a cyclic 2-Sylow subgroup S of order at least 8. By [55, 6.2.11], S has a normal complement in G. In other words, there exists a surjective group morphism π : G −→ S and a group morphism s : S −→ G such that π ◦ s = IdS . In particular, π∗ ◦ s∗ = IdH 1 (− ,S) and π∗ : H 1 (− , G) −→ H 1 (− , S) is a surjective natural transformation of functors.

X.28 Proof of Theorem X.24.1

273

Let α : H 1 (− , G) −→ H 2 (− , μ2 ) be the normalized cohomological invariant of G defined by α = ι ◦ π∗ . Since ι is unramified by Lemma X.27.2, so is α. By Lemma X.27.1, ι is non-zero, and since π∗ is surjective, α is non-zero as well. By Proposition X.25.2, we conclude that (N oethG,Q ) has a negative answer. Remark X.28.1. The arguments exposed here are those used by Serre in [25] to prove Theorem X.24.1. Using the same method, Serre also proved that (N oethG,Q ) has a negative answer for subgroups of odd -7 . index of A

XI The rationality problem for adjoint algebraic groups

Let G be a connected linear algebraic group defined over k. In this chapter, we are interested in the following question: is G rational as an affine k-variety ? In other words, does there exist an open subset of G which is isomorphic to an open subset of an affine space ? This question may be reformulated in more algebraic terms as follows. Let A be the k-algebra representing G. Since G is connected, A is an integral domain (see [69]), so we can consider its field of fractions k(A). The problem now translates as: is k(A)/k a rational extension ? This is known to be true when k is algebraically closed, by a theorem of Chevalley [14, Cor.2]. When k is not algebraically closed, this is not true any more. We refer to the introduction of [38] and the associated references to have a brief account on the history of the problem. In [38], Merkurjev studies the rationality problem for classical adjoint algebraic groups. These groups may be viewed as the connected component of the identity element of automorphism groups of some algebras with involutions. He shows that these groups are not rational in general. To do so, he computes the R-equivalence group G(k)/R of such a group G, which measures in some sense how far G is from being rational. More precisely, a rational algebraic group satisfies G(L)/R = 1 for all field extensions L/k (see Section XI.29 for the definition of R-equivalence and a proof of this fact). He then exhibits infinite families of adjoint groups which have a non-trivial equivalence group over some field extension. The same method was applied by Chernousov and Merkurjev in [13] to construct infinite families of simply connected group which are not rational. In [5], Monsurr` o, Tignol and the author constructed new families of 274

XI.29 R-equivalence groups

275

non-rational adjoints groups. To do so, they constructed a non-zero cohomogical invariant G −→ H 4 (− , μ2 ), which turns out to factor through R-equivalence, giving rise to an invariant ι : G(− )/R −→ H 4 (− , μ2 ) satisfying ιk = 0. In particular, G(k)/R = 1 and G is not rational. In this chapter, we will explain their construction in the case where G is the automorphism group of an algebra with a symplectic involution, in order to simplify the arguments. The non-zero cohomological invariant will be constructed using the restricted trace forms introduced in Chapter VIII. §XI.29 R-equivalence groups We start by defining the notion of R-equivalence for algebraic groups. Let G be a group-scheme defined over k, let K/k be a field extension and let K[t]0,1 be the localization of K[t] with respect to the multiplicative subset generated by t and (t − 1). In other words, K[t]0,1 is the subalgebra of K(t) consisting of rational functions which are defined at 0 and 1. For i = 0, 1, we denote by evi : K[t]0,1 −→ K the evaluation map at i. We then get two induced group morphisms G(evi ) : G(K[t]0,1 ) −→ G(K). If g(t) ∈ G(K[t]0,1 ), we will denote by g(i) the image of g(t) under G(evi ). Since G(evi ) is a group morphism, for every g1 (t), g2 (t) ∈ G(K[t]0,1 ), we get (g1 (t)g2 (t)−1 )(i) = g1 (i)g2 (i)−1 . Notice also that any element g ∈ G(K) defines an element g(t) ∈ G(K[t]0,1 ) satisfying g(0) = g(1) = g, since the composition ev

i K K ⊂ K[t]0,1 −→

is the identity map. Such an element is called constant. Definition XI.29.1. We say that an element g ∈ G(K) is R-trivial if there exists g(t) ∈ G(K[t]0,1 ) such that g(0) = 1 and g(1) = g. Remark XI.29.2. This definition may be reformulated in more geometric terms: g ∈ G(K) is R-trivial if there exists a rational map f : A1K _ _ _/ G defined at 0 and 1 such that f (0) = 1 and f (1) = g.

276

The rationality problem for adjoint algebraic groups

Lemma XI.29.3. The subset RG(K) of G(K) consisting of R-trivial elements is a normal subgroup of G(K). Proof. Clearly 1 ∈ RG(K) since we can take g(t) = 1. Now if gj ∈ RG(K), j = 1, 2, then there exists gj (t) ∈ G(K[t]0,1 ) such that gj (0) = 1 and gj (1) = gj . The element g(t) = g1 (t)g2 (t)−1 ∈ G(K[t]0,1 ) then satisfies g(0) = 1 and g(1) = g1 g2−1 . Hence g1 g2−1 is R-trivial, and RG(K) is a subgroup of G(K). Now let g ∈ RG(K) and let h ∈ G(K). Let h(t) be the image of h under the morphism induced by K ⊂ K[t]0,1 , and let g(t) ∈ G(K[t]0,1 ) such that g(0) = 1 and g(1) = g. Set g (t) = h(t)g(t)h(t)−1 . Then g (0) = hh−1 = 1, and g (1) = hgh−1 . Hence hgh−1 is R-trivial. This shows that RG(K) is a normal subgroup of G(K). Definition XI.29.4. The R-equivalence group of G(K) is the group G(K)/R = G(K)/RG(K). Two elements g, g ∈ G(K) which differ by an element of RG(K) will be called R-equivalent. If L/K is a field extension, the diagram K[t]0,1 evi

K

/ L[t]0,1 evi

/L

commutes. It easily follows that the map K −→ L induces a map RG(K) −→ RG(L), so we get a group morphism G(K)/R −→ G(L)/R. We then obtain a functor G(− )/R : Ck −→ Grps. Clearly, if G, G are two group-schemes defined over k and K/k is a field extension, an element (g, g ) ∈ G(K) × G (K) is R-trivial if and only if g and g are R-trivial. Therefore, we have a canonical isomorphism of functors (G × G )(− )/R G(− )/R × G (− )/R. We say that a group-scheme G defined over k is R-trivial if G(− )/R = 1. We now relate the notions of R-triviality and rationality. First, we need a definition.

XI.29 R-equivalence groups

277

Definition XI.29.5. Let G be a connected algebraic group-scheme defined over k. We say that G is rational if there exists a non-empty open subset U of G which is isomorphic (as an affine variety) to an open subset of some affine space. We say that G is stably rational if there exists n ≥ 0 such that G × Ank is rational. The following result is the key ingredient of this chapter. Proposition XI.29.6. Let k be an infinite field, and let G be a connected algebraic group-scheme defined over k. If G is stably rational, then G is R-trivial. Proof. First, assume that G is rational. Let A be the k-algebra representing G, and let K/k be a field extension. The algebraic group-scheme GK :

AlgK −→ Grps R −→ G(R)

is represented by AK , and therefore is also connected and rational. The definitions imply that GK (K)/R = G(K)/R. Therefore, replacing G by GK , we may assume that K = k. Let U be an open subset of an affine space Ank and let U be an open ∼ subset of G such that we have an isomorphism f : U −→ U . Since k is infinite, U (k) is dense in G(k) since G is connected (hence irreducible as an affine k-variety). In particular, U (k) is not empty. Let g0 ∈ U (k). Replacing U (k) by g0−1 U (k) and U (k) by fk−1 (g0−1 U (k)), one may assume that 1 ∈ U (k). We denote by u0 ∈ U (k) the preimage of 1 under fk . Now let g ∈ G(k). Since U (k) and gU (k) are non-empty open subsets of G(k), we get U (k)∩gU (k) = 0. Hence there exist g1 , g2 ∈ U (k) such that g = g1 g2−1 . Let vi = fk−1 (gi ) ∈ U (k). Let UP (k) be an elementary open subset contained in U (k), where P ∈ k[X1 , . . . , Xn ] is a non-zero polynomial, so we have UP (k) = {(a1 , . . . , an ) ∈ k n | P (a1 , . . . , an ) = 0}. It is easy to see that there exists wi ∈ k n such that the polynomial P ((1 − t)u0 + tvi + t(1 − t)wi ) ∈ k[t] is not zero. Consequently, we have ui (t) = (1 − t)u0 + tvi + t(1 − t)wi ∈ U (k[t]0,1 ).

278


Moreover, we have by construction ui (0) = u0 and ui (1) = vi . Let gi (t) = fk[t]0,1 (ui (t)) ∈ U (k[t]0,1 ). Since we have a commutative diagram U (k[t]0,1 ) evi

U (k)

/ U (k[t]0,1 ) evi

/ U (k)

we get gi (0) = fk (ui (0)) = fk (u0 ) = 1 and gi (1) = fk (ui (1)) = fk (vi ) = gi . Therefore the element g(t) = g1 (t)g2−1 (t) ∈ G(k[t]0,1 ) satisfies g(0) = 1 and g(1) = g1 g2−1 = g. This shows that RG(k) = G(k), that is G(k)/R = 1. This proves that G(− )/R = 1. Assume now that G is stably rational. Then there exists n ≥ 0 such that G × Ank is rational. Notice that Ank is a rational connected algebraic group-scheme. Applying the previous point, we get G(− )/R G(− )/R × Ank (− )/R (G × Ank )(− )/R = 1. This concludes the proof. §XI.30 The rationality problem for adjoint groups We start this section by giving a description of the R-equivalence group of PGSp(A, σ). Let (A, σ) be a central simple k-algebra with a symplectic involution. Let Hyp(A, σ) be the multiplicative subgroup of k× generated by the elements of the form NL/k (z), z ∈ L× , where L/k runs over all finite field extensions such that σL is hyperbolic. Let also G(A, σ) be the multiplicative subgroup of k × defined by G(A, σ) = {μ(g) | g ∈ GSp(A, σ)(k)}. In [38], Merkurjev proved the following result: Theorem XI.30.1. We have a group isomorphism PGSp(A, σ)(k)/R G(A, σ)/k×2 Hyp(A, σ).

XI.30 The rationality problem for adjoint groups

279

As explained at the beginning of this chapter, we are going to construct a cohomological invariant of PGSp(A, σ) which factors through R-equivalence. We start with a lemma. Lemma XI.30.2. Let A be a central simple k-algebra of degree 2m. Let σ, σ0 be two symplectic involutions on A. If m is even, Tσ+ ⊥ −Tσ+0 ∈ I 3 k. Proof. By a celebrated theorem of Merkurjev ([39]), we have to prove that ei (Tσ+ ⊥ −Tσ+0 ) = 0 for i = 0, 1, 2. Since Tσ+ ⊥ −Tσ+0 is evendimensional, we have e0 ([Tσ+ ⊥ −Tσ+0 ]) = 0. By Lemma IV.12.13, we have e1 ([Tσ+ ⊥ −Tσ+0 ]) = (det(Tσ+ ) det(−Tσ+0 )), since m is even. Moreover, we have det(−Tσ+0 ) = (−1)m(2m−1) det(Tσ+0 ) = det(Tσ+0 ) = det(Tσ+ ), the last equality following from Corollary VIII.21.35. Therefore, we get e1 ([Tσ+ ⊥ −Tσ+0 ]) = 0. By Remark IV.12.14, we have e2 ([Tσ+ ⊥ −Tσ+0 ]) = c([Tσ+ ⊥ −Tσ+0 ]). Set r = m(2m + 1). The dimension of Tσ+ ⊥ −Tσ+0 is 2r = 2m(2m + 1). Since m is even, we have 2r ≡ 0 or 4 mod 8. Moreover, we have det(Tσ+ ⊥ −Tσ+0 ) = det(Tσ+ ) det(−Tσ+0 ) = 1, so we get if r ≡ 0[4] w2 (Tσ+ ⊥ −Tσ+0 ) + + e2 ([Tσ ⊥ −Tσ0 ]) = w2 (Tσ+ ⊥ −Tσ+0 ) + (−1) ∪ (−1) if r ≡ 2[4]. In other words, we have r e2 ([Tσ+ ⊥ −Tσ+0 ]) = w2 (Tσ+ ⊥ −Tσ+0 ) + (−1) ∪ (−1). 2 Now using Lemma IV.11.5, we get w2 (Tσ+ ⊥ −Tσ+0 ) = w2 (Tσ+ ) + w2 (−Tσ+0 ) + (det(Tσ+ )) ∪ (det(−Tσ+0 )). Applying this equality to σ0 , and substracting the two equations, we get w2 (Tσ+ ⊥ −Tσ+0 ) = w2 (Tσ+0 ⊥ −Tσ+0 ),

280


taking into account that the determinant and the Hasse invariant of Tσ+ do not depend of σ, by Corollary VIII.21.35 and Theorem VIII.21.36. Now, Tσ+0 ⊥ −Tσ+0 is a hyperbolic quadratic form of dimension 2r, so we have Tσ+0 ⊥ −Tσ+0 r × 1, −1, and thus we get w2 (Tσ+0 ⊥ −Tσ+0 ) = w2 (r × −1) =

r(r − 1) (−1) ∪ (−1). 2

Since m is even, r − 1 is odd, and we get r w2 (Tσ+ ⊥ −Tσ+0 ) = (−1) ∪ (−1). 2 Hence we obtain e2 ([Tσ+ ⊥ −Tσ+0 ]) = 0, and this concludes the proof. We are now ready to construct the cohomological invariant we are looking for. Proposition XI.30.3. Let A be a central simple k-algebra carrying a hyperbolic symplectic involution σ0 . For any symplectic involution σ on A, and every field extension K/k, the map ιK :

GSp(A, σ)(K) −→ H 4 (K, μ2 ) + ]) g −→ (μ(g)) ∪ e3 ([Tσ+ ⊥ −T(σ 0 )K

induces a well-defined group morphism θK : PGSp(A, σ)(K)/R −→ H 4 (K, μ2 ). These maps give rise to cohomological invariants ι : GSp(A, σ) −→ H 4 (− , μ2 ) and θ : PGSp(A, σ)(−)/R −→ H 4 (− , μ2 ). Proof. If g ∈ GSp(A, σ)(K) and λ ∈ K × , we have μ(λg) = λ2 μ(g). Thus ι induces a well-defined group morphism ιK : PGSp(A, σ)(K) −→ H 4 (K, μ2 ). We now prove that ιK factors through R-equivalence. By Theorem XI.30.1, it is enough to prove that ιK (λ2 ) = 0 for all λ ∈ K × , and that for every finite field extension L/K such that σL is hyperbolic and all z ∈ L× , we have ιK (NL/K (z)) = 0.

XI.31 Examples of non-rational adjoint groups

281

The first point is clear. Now let L/K and z ∈ L× as above. Since σL and σ0,L are hyperbolic involutions on AK , they are conjugate by Theorem VIII.21.15. Therefore, the corresponding trace forms are isomorphic by Proposition VIII.21.30 (1). The second part of the same proposition then + )L is hyperbolic. By Corollary IV.12.16, we implies that (Tσ+ ⊥ −T(σ 0 )K get + ]) = 0, (NL/K (z)) ∪ e3 ([Tσ+ ⊥ −T(σ 0 )K

which is what we wanted to prove. The functoriality part is left to the reader. Corollary XI.30.4. Assume that there exists a field extension K/k such that ιK is not identically zero. Then PGSp(A, σ) is not R-trivial, and in particular not (stably) rational. §XI.31 Examples of non-rational adjoint groups In order to use the previous corollary, we need some examples of algebras with involutions for which it is possible to compute this invariant explicitly. We will need some intermediate results. Proposition XI.31.1. Let (A, ρ) be a central simple k-algebra of even degree with an orthogonal involution, let u ∈ Sym(A, ρ)× , and let ρu = Int(u) ◦ ρ. Then we have det(Tρ+u ) = NrdA (u) det(Tρ+ ). Proof. Notice first that the map O(A, ρ)(ks ) −→ O(Tρ+ )(ks ) a −→ Int(a)|Sym(Ak

s

,ρk ) s

factors through PGO(A, ρ)(ks ). By Lemma VIII.21.21 and Lemma VIII.21.34, the induced map H 1 (k, O(A, ρ)) −→ H 1 (k, O(Tρ+ )) maps a class u/∼ onto the isomorphism class of Tρ+u . Claim: the map H 1 (k, O(A, ρ)) −→ H 1 (k, μ2 ) induced by the reduced norm sends u/∼ onto the square class of NrdA (u).

282


Indeed, by Galois descent, a cocycle α corresponding to the class u/∼ is given by α:

Gks −→ O(A, ρ)(ks ) τ −→ a τ ·a−1 ,

where a ∈ A× ks satisfies auks ρks (a) = 1. The image of u/∼ under the map induced by the reduced norm is then represented by the cocycle Gks −→ μ2 (ks ) β:

τ −→

NrdAks (a) . NrdAks (τ ·a)

Notice now that we have NrdAks (τ ·a) = τ ·NrdAks (a). ∼

Indeed, let ϕ : Aks −→ Mn (ks ) be an isomorphism of ks -algebras. Using the definition of the action of Gks on Aks , we see that we have ϕ(τ ·a) = τ ·ϕ(a). The desired equality follows from properties of the determinant (details are left to the reader). Hence we have βτ =

τ ·NrdAks (a−1 ) NrdAks (a) = for all τ ∈ Gks . τ ·NrdAks (a) NrdAks (a−1 )

The equality auks ρks (a) = 1 and the properties of the reduced norm imply that we have NrdAks (a−1 )2 = NrdAks (uks ) = NrdA (u). Hence β represents the square-class of NrdA (u), and this proves the claim. Since the map det∗ : H 1 (k, O(Tρ+ )) −→ H 1 (k, μ2 ) maps the isomordet(q) by phism class of a quadratic form q onto the square class det(Tρ+ ) Proposition IV.11.2, it is therefore enough to prove that the diagram


NrdAk

283

/ μ2 (ks ) O(A, ρ)(ks ) OOO O OOO OOO det OO' O(Tρ+ )(ks ) s

commutes to get the desired equality. By Lemma VIII.21.17, there exists an isomorphism ∼

f : (Aks , σks ) −→ (Mn (ks ), t) of ks -algebras with involution. Let a ∈ O(A, σ)(ks ). We have NrdAks (a) = det(f (a)). Moreover, let us denote by Int+ (a) and Int+ (f (a)) the restriction of Int(a) and Int(f (a)) to Sym(Aks , ρks ) and Sym(Mn (ks ), t) respectively. Now conjugation by f induces an isomorphism between the endomorphism ring of these vector spaces, which maps Int+ (a) ∈ O(Tρ+ )(ks ) onto Int+ (f (a)) ∈ O(Mn (k), t)(ks ). Since this isomorphism preserves determinants, we have det(Int+ (a)) = det(Int+ (f (a))), and the commutativity of the diagram reads det(Int+ (f (a))) = det(f (a)) for all a ∈ Aks . Notice that we have an equality O(Mn (k), t)(ks ) = On (ks ). Since reflections span On (ks ) by Proposition IV.10.5, it is enough to prove the equality det(Int+ (τx )) = det(τx ) = −1 for all x ∈ ksn which are anisotropic for the unit quadratic form n × 1. Since τλx = τx for all λ ∈ ks , one may assume that x is a unit vector. Let T be the matrix of τx in the canonical basis of ksn . Let e1 = x, e2 , . . . , en be an orthonormal basis of ksn with respect to the unit form (which exists since ks is separably closed), and let P be the corresponding base −1 change matrix. We then have the equality T = P T P , where T = −1 . In−1 Moreover, the matrix P is orthogonal by construction, so P −1 = P t ,

284


and Int(P ) induces an automorphism Int+ (P ) of Sym(Mn (ks ), t). Now we have Int+ (T ) = Int+ (P ) ◦ Int+ (T ) ◦ Int+ (P )−1 . It easily implies that we are reduced to check that det(Int+ (T )) = −1. Eij + Eji for all 1 ≤ i ≤ j ≤ n. It is easy to check that we 2 + have Int (T )(εij ) = −εij if i = 1 and j ≥ 2, and Int+ (T )(εij ) = εij otherwise. Hence we get Set εij =

det(Int+ (T )) = (−1)n−1 = −1, since n is even. This concludes the proof. Proposition XI.31.2. Let B be a central simple k-algebra of even degree, let ρ, ρ0 be two orthogonal involutions on B, and write ρ = Int(u) ◦ ρ0 , u ∈ Sym(B, ρ0 )× . Let Q = (a, b) be a quaternion k-algebra, and let γ be the corresponding symplectic involution. Let A = B ⊗k Q. Then the involutions σ = ρ ⊗ γ and σ0 = ρ0 ⊗ γ are symplectic, and we have e3 (Tσ+ ⊥ −Tσ+0 ) = (NrdB (u)) ∪ (a) ∪ (b). Proof. By Lemma VIII.21.31, σ and σ0 are symplectic, since ρ and ρ0 are orthogonal and γ is symplectic. According to the same lemma, we have the following equality in the Witt group of k: [Tσ+ ] = [Tρ+ ][Tγ+ ] + [Tρ− ][Tγ− ]. Using Proposition VIII.21.30, we get [Tσ+ ] = [Tρ+ ][Tγ+ ] + ([Tρ+ ] − [TB ])[Tγ− ] = [Tρ+ ]([Tγ+ ] + [Tγ− ]) − [TB ][Tγ− ] = [Tρ+ ][Tγ ] − [TB ][Tγ− ]. Since the same equality is true if we replace ρ by ρ0 , we get [Tσ+ ⊥ −Tσ+0 ] = [Tρ+ ⊥ −Tρ+0 ][Tγ ]. By Example VIII.21.32 (2), we have Tγ 2a, b.


285

Hence Tγ ∈ I 2 (k). By Remark IV.12.9, we have e2 ([Tγ ]) = e2 ([a, b]) = (a) ∪ (b). This may also be seen using the fact that e2 coincide with the Clifford invariant on I 2 (k). Since Tρ+ ⊥ −Tρ+0 is even dimensional, we have Tρ+ ⊥ −Tρ+0 ∈ I(k). By Lemma IV.12.12, we get e3 ([Tσ+ ⊥ −Tσ+0 ]) = e1 ([Tρ+ ⊥ −Tρ+0 ]) ∪ (a) ∪ (b). Notice that we have dim(Tρ+ ⊥ −Tρ+0 ) = n(n − 1), where n is the degree n(n − 1) n(n − 1)(n(n − 1) − 1) and have same parity. of B, and that 2 2 By Lemma IV.12.13, we then get e1 ([Tρ+ ⊥ −Tρ+0 ]) = ((−1)

n(n−1) 2

det(Tρ+ ⊥ −Tρ+0 )).

Now by Proposition XI.31.1, we have det(Tρ+ ⊥ −Tρ+0 ) = det(Tρ+ ) det(−Tρ+0 ) n(n−1) = (−1) 2 det(Tρ+ ) det(Tρ+0 ) n(n−1) = (−1) 2 NrdB (u) ∈ k× /k ×2 . The desired result follows. We are now ready to construct our family of examples. Theorem XI.31.3. Assume that −1 ∈ k ×2 , and that there exist elements a, b, c, d ∈ k × such that (a) ∪ (b) ∪ (c) ∪ (d) = 0 in H 4 (k, μ2 ). Let H = (a, b) and Q = (c, d). Let m, s ≥ 1 be two integers, with s odd and let ρ be the orthogonal involution on M2ms (H) (M2m (k) ⊗k H) ⊗k Ms (k) defined by ρ = [Int(diag(j, i, i, . . . , i) ⊗ 1)] ◦ ((t ⊗ γH ) ⊗ σB ), where B is any symmetric invertible matrix of Ms (k) and 1, i, j, ij is the standard basis of H. Finally, let A = M2ms (H) ⊗ Q and let σ = ρ ⊗ γQ . Then PGSp(A, σ) is not R-trivial, and in particular not stably rational.

286


Proof. Let ρ0 = (t ⊗ γH ) ⊗ σB and let σ0 = ρ0 ⊗ γQ . Then ρ0 is an orthogonal hyperbolic involution on M2m (H), and σ0 is a hyperbolic involution on A. Indeed, since −1 ∈ k ×2 , the unit quadratic form 2m × 1 is isomorphic to m × 1, −1, which is hyperbolic. Hence t = σIn is a hyperbolic involution, and therefore so are ρ0 and σ0 , since it is easy to see that the tensor product of an involution by a hyperbolic one is still hyperbolic. Now we have ρ = Int(u0 ⊗ 1) ◦ ρ0 , where u0 = diag(j, i, i, . . . , i). Since we have NrdM2ms (H) (u0 ⊗ 1)

= NrdM2m (H)⊗Ms (k) (u0 ⊗ 1) = NrdM2m (H) (u0 )s = (−b)s (−a)(2m−1)s = bs a(2m−1)s ,

we get e3 (Tσ+ ⊥ −Tσ+0 ) = (a(2m−1)s bs ) ∪ (c) ∪ (d) = (ab) ∪ (c) ∪ (d) by the previous proposition, since s and (2m − 1)s are odd. Now let g = diag(j, ωj, . . . , ωj), where ω ∈ k × satisfies ω 2 = −1, and let g = g ⊗ 1. Then we have ρ(g ) = diag(j, ωiji−1 , . . . , ωiji−1 ) = diag(j, −ωj, . . . , −ωj), and therefore gσ(g) = g ρ(g ) = b. Hence g ∈ GSp(A, σ)(k) and μ(g) = b. Therefore, we get ιk (g) = (b) ∪ (ab) ∪ (c) ∪ (d). But we have (b) ∪ (ab) = (b) ∪ (a) + (b) ∪ (b) = (a) ∪ (b) + (b) ∪ (−1) = (a) ∪ (b), since −1 is a square. Thus, we get ιk (g) = (a) ∪ (b) ∪ (c) ∪ (d) = 0, and we conclude using Corollary XI.30.4. To have explicit examples, we need to find fields satisfying the conditions of the theorem above. This will be provided by the following lemma.

Exercises

287

Lemma XI.31.4. Let k0 be any field of characteristic different from 2, and let k = k0 (t1 , . . . , tn ), where t1 , . . . , tn are independent indeterminates over k0 . Then we have (t1 ) ∪ · · · ∪ (tn ) = 0 in H n (k, μ2 ). Proof. We prove it by induction. If n = 1, this is clear since t1 is not a square in k(t1 )× . Assume the result is true for some n ≥ 1, and let us prove that (t1 ) ∪ · · · ∪ (tn+1 ) = 0 in H n+1 (k, μ2 ). Let v be the tn+1 -adic valuation on k0 (t1 , . . . , tn )(tn+1 ). Then we have ∂v ((t1 ) ∪ · · · ∪ (tn+1 )) = (t1 ) ∪ · · · ∪ (tn ) = 0 in H n (k0 (t1 , . . . , tn ), μ2 ). In particular, (t1 ) ∪ · · · ∪ (tn+1 ) = 0 and we are done. We may then take k = k0 (t1 , t2 , t3 , t4 ) and a = t1 , b = t2 , c = t3 , d = t4 in the previous theorem, where k0 is any field of characteristic different from 2 such that −1 ∈ k0×2 . More examples of non-rational automorphism groups of algebra with involutions constructed using similar arguments may be found in [5].

Exercises 1. Let K/k be a quadratic étale algebra. In this exercise, as well as the following ones, we denote by ι the non-trivial k-automorphism of K, (B, τ ) will denote a central simple K-algebra with a unitary involution (see Chapter VIII, Exercise 7 for a definition in the case where K = k × k) such that τ|K = ι. (a)

Check that if x ∈ Sym(B, τ ), then TrdB (x) ∈ k. We define a quadratic form over k by Tτ :

B −→ k x −→ TrdB (τ (x)x).

We denote by Tτ+ and Tτ− its restriction to Sym(B, τ ) and Skew(B, τ ) respectively.

288

The rationality problem for adjoint algebraic groups (b)

Let α ∈ K × such that τ (α) = −α, so that d = α2 ∈ k × . Check that Skew(B, τ ) = αSym(B, τ ), and deduce that we have Tτ− −dTτ+ .

(c)

If (B, τ ) = (A × Aop , ε), show that Tτ+ TA .

(d)

Deduce that Tτ , Tτ+ and Tτ− are regular quadratic forms.

2. Let τ be another unitary involution on B such that τ|K = ι, show that det(Tτ+ ) = det(Tτ+ ) and deduce that Tτ+ ⊥ −Tτ+ ∈ I 2 k. Hint : Let us denote by PGU(B, τ ) the automorphism group of (B, τ ). This a connected affine algebraic group-scheme defined over k. 3. Let A be a central simple k-algebra. Show that PGL1 (A) is rational. As for the case of involutions of the first kind, we may define the notion of a similitude for τ . We then obtain an affine algebraic group-scheme GU(B, τ ) defined over k. We then define a subgroup G(B, τ ) of k× by G(B, τ ) = {μ(g) | g ∈ GU(B, τ )(k)}. In [38], Merkurjev proved that PGU(B, τ )(k)/R G(B, τ )/NK/k (K × )Hyp(B, τ ), where Hyp(B, τ ) is the subgroup of k × generated by the elements NL/k (z), z ∈ L× , where L/k describes the finite field extensions such that τL is hyperbolic. 4. Let (A, ρ) be a central simple k-algebra with an involution of the first kind. If α ∈ K × satisfies τ (α) = −α, set d = α2 ∈ k × . Let B = A ⊗k K and let τ = ρ ⊗k ι. (a)

If ρ0 is another involution of the first kind of same type as ρ and τ0 = ρ0 ⊗ ι, show that we have Tτ+ ⊥ −Tτ+0 1, −d ⊗ (Tρ+ ⊥ −Tρ+0 ).

Exercises (b)

289

Assume that ρ0 is hyperbolic. For every field extension L/k, define a map ιL : GU(B, τ )(L) −→ H 3 (L, μ2 ) by ιL (g) = (μ(g)) ∪ e2 ([Tτ+ ⊥ −T(τ+0 )L ]). Show that the maps ιL induce a well-defined cohomological invariant θ : PGU(B, τ )(− )/R −→ H 3 (k, μ2 ).

(c)

Using the previous questions, find a field K and a central simple K-algebra with a unitary involution (B, τ ) such that PGU(B, τ ) is not stably rational. Hint: Assume that −1 ∈ k ×2 . Set A = M2m (H), where H is a quaternion algebra, and take ρ to be the orthogonal involution ρ = Int(diag(j, i, i, . . . , i)) ◦ (t ⊗ γH ).

XII Essential dimension of functors

When studying a class of mathematical objects, it is natural to ask how many independent parameters are needed to define them up to isomorphism, in order to measure their ‘degree of complication’. The notion of essential dimension has been defined to give a precise meaning to this number of parameters. Essential dimension was first introduced by Buhler and Reichstein in [10] for finite groups. It has then been extended to arbitrary algebraic groups by Reichstein [47]. Recently, Merkurjev [37] generalized this notion to arbitrary functors in some private notes. In the meantime, Rost also proposed a valuative approach to essential dimension [49]. In [3], a systematic study of essential dimension is developed, based on Merkurjev’s private notes. In particular, it is shown that the notions of essential dimension introduced by Buhler, Merkurjev, Reichstein and Rost all agree. In this chapter, we will follow the treatment of [3]. We will often state some results without proof. §XII.32 Essential dimension: definition and first examples Definition XII.32.1. Let F : Ck −→ Sets be a covariant functor, let K/k be a field extension and let a ∈ F(K). Finally, let E/k be a subextension of K/k. We say that a is defined over E if a lies in the image of F(E) −→ F(K). The essential dimension of a is the integer ed(a) defined by ed(a) = min{trdeg(E/k) | E ⊂ K, a is defined over E}. The essential dimension of F is the supremum of ed(a) for all a ∈ F(K) and for all K/k. The essential dimension of F will be denoted by edk (F). Let us give some examples. 290

XII.32 Essential dimension: definition and first examples

291

Examples XII.32.2. (1)

If K/k is algebraic, then ed(a) = 0 for all a ∈ F(K).

(2)

If char(k) = 2, let q be a non-degenerate quadratic form of dimension n over K/k. Then q a1 , . . . , an , for some ai ∈ K × . Set K = k(a1 , . . . , an ) ⊂ K and q = a1 , . . . , an . Then q qK and therefore ed(q) ≤ trdeg(K /k) = n. In particular, edk (Quadn ) ≤ n. It takes a bit more effort to prove that equality holds.

(3)

In the same spirit, if K/k is a field extension and L/K is an étale multiquadratic K-algebra of dimension 2n, we have √ √ L K[ a1 ] × · · · × K[ an ] L ⊗K K, √ √ where K = k(a1 , . . . , an ) and L = K [ a1 ] × · · · × K [ an ]. Hence ed(L/K) ≤ trdeg(K /k) = n. In fact, the essential dimension of multiquadratic algebras of dimension 2n is equal to n if char(k) = 2, as we will see later.

(4)

If char(k) = 2, the situation is quite different. In this case, every multiquadratic étale K-algebra of dimension 2n is isomorphic to K[℘−1 (a1 )] × · · · × K[℘−1 (an )] for some a1 , . . . , an ∈ K, where ℘ is the Weierstrass function ℘:

K −→ K x −→ x2 − x.

Thus, we still have ed(L/K) ≤ n. However, if k is large enough, we can have a strict inequality. For example, assume that k contains F4 , so there exists j ∈ k satisfying j 2 = j + 1. If L/K is a biquadratic extension we have L K[℘−1 (a)] × K[℘−1 (b)] K[℘−1 (j 2 λ)] × K[℘−1 (λ)], where λ = ℘(a) + j℘(b) ∈ K. To prove this, notice first that we have j 2 ℘(c) = j 2 (c2 − c) = (jc)2 − (1 + j)c = c + ℘(jc).

292

Essential dimension of functors Then we get j 2 λ = j 2 ℘(a) + ℘(b) = a + ℘(b + ja), and therefore K[℘−1 (j 2 λ)] K[℘−1 (a)]. We also have λ = ℘(a) + j℘(b) = ℘(a) + (1 + j 2 )℘(b) = ℘(a + b) + b + ℘(jb). Hence λ = b+℘(a+b+jb), and therefore K[℘−1 (λ)] K[℘−1 (b)]. Thus L/K is defined over k(λ), and we have ed(L/K) ≤ 1. We will see an explanation of this phenomenon later.

(5)

Let X be a scheme over k. It gives rise to a functor X : Ck −→ Sets, K/k −→ X(K) = Mor(Spec(K), X). If L/K is a field extension, the induced map X(K) −→ X(L) is just composition on the right by Spec(L) −→ Spec(K). Let us compute edk (X). Let K/k be a field extension, let a ∈ X(K) and let x ∈ X be the corresponding point. Assume that K/k as minimal transcendence degree over k, so that ed(a) = trdeg(K/k). The morphism a : Spec(K) −→ X factors through a : Spec(κ(x)) −→ X, so a = aK . Since K/k has minimal transcendence degree, we get ed(a) = trdeg(K/k) = trdeg(κ(x)/k). Hence we get edk (X) = sup trdeg κ(x) : k = dim(X). x∈X

If G is an algebraic group-scheme defined over k, we will write edk (G) for the essential dimension of the functor H 1 (− , G). Computing edk (G) is a particularly interesting problem because H 1 (− , G) often classifies algebraic objects up to isomorphism, as we have already seen in the previous chapters. §XII.33 First results Let F : Ck −→ Sets be a covariant functor, and let k /k be a field extension. If K/k is a field extension of k , we can associate the field extension K/k. Then one can view F as a functor Ck −→ Sets.

XII.33 First results

293

Lemma XII.33.1. Let k /k a field extension.Then edk (F) ≤ edk (F). Proof. If edk (F) = ∞, the result is obvious. Let edk (F) = n. Take K/k a field extension and a ∈ F(K). There is a subextension k ⊂ E ⊂ K with trdeg(E/k) ≤ n such that a is in the image of the map F(E) −→ F(K). The composite extension E = Ek then satisfies trdeg(E /k ) ≤ n and clearly a is in the image of the map F(E ) −→ F(K). Thus ed(a) ≤ n and edk (F) ≤ n. This lemma can be useful to give lower bounds (taking k = kalg for example). Remark XII.33.2. In general one does not have edk (F) = edk (F) for any field extension k /k. Indeed, let F : Ck −→ Sets be the functor defined by {0} if trdeg(K/k) ≤ 1 F(K) = {0, 1} if trdeg(K/k) ≥ 2 the map induced by a morphism K −→ K being the inclusion of sets. Clearly, edk (F) = 2, but edk (F) = 0 as soon as trdeg(k /k) ≥ 2 (since F becomes constant over k ). Lemma XII.33.3. Let ϕ : F −→ F be a natural transformation of functors. For all K/k and all a ∈ F(K), we have ed(a) ≥ ed(ϕK (a)). In particular, if ϕ is surjective, we have edk (F ) ≤ edk (F). Proof. Let K ⊂ K such that a = bK for some b ∈ F(K ) and ed(a) = trdeg(K /k). Then we have ϕK (a) = ϕK (bK ) = (ϕK (b))K . Hence ϕ(a) is defined over K and ed(ϕK (a)) ≤ trdeg(K /k) = ed(a). The last part is clear. The lemma above and Example XII.32.2 (5) then yield: Corollary XII.33.4. Let F be a functor, and let X be a k-scheme. Assume we have a surjective natural transformation X Then edk (F) ≤ dim(X).

//F.

294

Essential dimension of functors

Examples XII.33.5. (1)

If char(k) = 2, we have a surjection K ×n −→ Quadn (K) (a1 , . . . , an ) −→ a1 , . . . , an for all K/k, so we get a surjective natural transformation Gnm −→ Quadn . We then recover the inequality edk (Quadn ) ≤ n.

(2)

If char(k) = 2, we have a surjection Gnm (K) −→ H 1 (K, (Z/2Z)n ), which sends for all K/k the element (a1 , . . . , an ) ∈ (K × )n onto √ √ the K-algebra K[ a1 ] × . . . × K[ an ] . We then get a surjective natural transformation Gnm −→ H 1 (− , (Z/2Z)n ). Thus edk ((Z/2Z)n ) ≤ n.

(3)

If char(k) = 2, k ⊃ Fq , q = 2n , we have an exact sequence 0 −→ (Z/2Z)n −→ Ks −→ Ks −→ 0, for every field extension K/k, where the last map is Ks −→ Ks x −→ xq − x. Passing to cohomology, we get an exact sequence Ga −→ H 1 (− , (Z/2Z)n ) −→ H 1 (− , Ga ) = 0. Thus the map above is surjective, and we get edk ((Z/2Z)n ) ≤ 1.

Definition XII.33.6. We say that a functor F : Ck −→ Sets acts over a functor G : Ck −→ Sets if we have a natural transformation ϕ : F × G −→ G. For every field extension K/k, x ∈ F(K) and a ∈ G(K), we will write x·a instead of ϕ(x, a). We will say that the action of F on G is transitive if for every field

XII.33 First results

295

extension K/k and all a, a ∈ G(K), there exists x ∈ F(K) such that a = x·a. If π : G −→ H is a natural transformation of functors and K/k is an extension, each element a ∈ H(K) gives rise to a functor π −1 (a), defined over the category CK , by setting −1 π −1 (a)(L) = πL (a) = {x ∈ G(L) | πL (x) = aL }

for every extension L/K. / / H be a surjective natural transormation. We say Let π : G that a functor F is in fibration position for π if F acts transitively on each fiber of π. More precisely, for every extension K/k and every a ∈ H(K), we require that the functor F (viewed over the category CK ) acts transitively on π −1 (a). When F is in fibration position for π we π / / simply write F G H and call this a fibration of functors. Proposition XII.33.7. Let F G Then

π

/ / H be a fibration of functors.

edk (H) ≤ edk (G) ≤ edk (F) + edk (H). Proof. Let K/k be a field extension and a ∈ G(K). By definition there is a field extension E with k ⊂ E ⊂ K, satisfying trdeg(E : k) ≤ edk (H), and an element b ∈ H(E) such that bK = πK (a). Since πE is surjective there exists a ∈ G(E) such that πE (a ) = b . Now πK (aK ) = πK (a) and thus aK and a are in the same fiber. By assumption there exists an element c ∈ F(K) such that c·aK = a. Now there exists an extension E with k ⊂ E ⊂ K and trdeg(E : k) ≤ edk (F) such that c is in the image of the map F(E ) −→ F(K). Pick c ∈ F(E ) such that cK = c. Considering now the composite extension E = EE and setting d = cE ·aE ∈ G(E ) we have, since the action is functorial, dK = (cE ·aE )K = cK ·aK = c·aK = a, and thus ed(a) ≤ trdeg(E : k) ≤ trdeg(E : k)+trdeg(E : k) ≤ edk (H)+edk (F). Since this is true for an arbitrary element a the desired inequality follows. The other inequality is already known from Lemma XII.33.3. Corollary XII.33.8. We have max(edk (F), edk (G)) ≤ edk (F × G) ≤ edk (F) + edk (G).

296


Proof. The reader will check easily that we have two fibrations of functors F F×G

π

//G

G F×G

π

//F

and

The result then follows from the previous proposition. §XII.34 Cohomological invariants and essential dimension Definition XII.34.1. Let F : Ck −→ Sets∗ be a functor, and M be a torsion Gks -module M . We say that a cohomological invariant ι : F −→ H d (− , M ) is nontrivial if for all K/k, there exists L/K such that ιL = 0. Proposition XII.34.2. If F admits a non-trivial cohomological invariant of degree d, then edk (F) ≥ d. Proof. Assume to the contrary that edk (F) < d. Since edkalg (F) ≤ edk (F), one can assume that k is algebraically closed. Let K/k be a field extension, and let L/K be a field extension of K. We are going to show that ιL = 0, contradicting the assumption. Let a ∈ F(L). By assumption, a = bL for some b ∈ F(K ), where K ⊂ L and trdeg(K /k) < d. We then have ιL (a) = (ιK (b))L . But H d (K , M ) = 0 since trdeg(K /k) < d and k is algebraically closed by [58, Proposition 11,p.93], and therefore ιL (a) = 0. Thus ιL = 0, and this concludes the proof. Examples XII.34.3. (1)

Let k be a field of characteristic different from 2. Then we have edk ((Z/2Z)n ) = n. Indeed, we already know that we have edk ((Z/2Z)n ) ≤ n. To prove the other inequality, we define a cohomological invariant ι : H 1 (− , (Z/2Z)n ) −→ H n (− , μ2 ) as follows. For every field extension K/k, we set

XII.34 Cohomological invariants and essential dimension

297

√ √ ιK (K[ a1 ] × · · · × K[ an ]) = (a1 ) ∪ · · · ∪ (an ). Using Lemma XI.31.4, it is easy to see that ι is non-trivial, and we are done. (2)

Let k be a field, char(k) = 2. Then edk (Quadn ) = n. Indeed the nth Stiefel-Whitney class provides a cohomological invariant of Quadn of degree n, which is non-trivial for the same reason as in the previous example. We conclude as before.

This method can be quite limited as the next proposition shows. Proposition XII.34.4. Let k be a field, char(k) = 2, let Q be a division quaternion k-algebra, and let SL1 (Q) be the group of elements of reduced norm 1. Then edk (SL1 (Q)) = 1, but H 1 (− , SL1 (Q)) has no non-trivial cohomological invariants of degree 1. Proof. From Chapter VIII, Exercise 2, we have a bijection H 1 (K, SL1 (Q)) K × /NrdQK (Q× K) for every field extension K/k. Let us show that edk (SL1 (Q)) = 1. From the previous description of H 1 (− , SL1 (Q)), we have an obvious surjection / / H 1 (− , SL1 (Q)) , Gm and therefore edk (SL1 (Q)) ≤ 1 by Corollary XII.33.4. To prove the other equality, let t be an indeterminate over k. We now prove that ed(t) ≥ 1. Assume that t is defined over an algebraic subextension of k(t)/k. Since k(t)/k is purely transcendental, it means that t is defined over k. Therefore there exist u ∈ k and z ∈ Q× k(t) such that t = uNrdQk(t) (z). Write Q = (a, b) for some a, b ∈ k× . Then NrdQk(t) (z) is represented by the Pfister form a, b We then get t, a, b u, a, b by the well-known multiplicativity property of Pfister forms (see [53] for more details). Computing the invariants e3 on both sides, we get (t) ∪ (a) ∪ (b) = (u) ∪ (a) ∪ (b) ∈ H 3 (k(t), μ2 ).

298


Taking the residues corresponding to the t-adic valuation, we obtain (a) ∪ (b) = 0 ∈ H 2 (k, μ2 ). This is equivalent to say that [Q] = 0 ∈ Br(k) by Corollary VIII.20.10, which is a contradiction. Finally, let us prove that H 1 (−, SL1 (Q)) has no non-trivial cohomological invariants. Indeed, let ι : H 1 (− , SL1 (Q)) −→ H d (− , M ) be a cohomological invariant, and let K = kalg , so we have QK M2 (K). Let L/K be any field extension. Then we also have QL M2 (L). × We then have H 1 (L, SL1 (Q)) = 1, since NrdQL (Q× L ) = L . Therefore, ιL = 0 and ι is not non-trivial.

§XII.35 Generic objects and essential dimension Definition XII.35.1. Let F : Algk −→ Sets be a functor with a generic object. We say that a generic objet a ∈ F(K) is nice if for all K ⊂ K, a ∈ F(K ), a = aK ⇒ a is a generic object. Proposition XII.35.2. Assume that F : Algk −→ Sets has a nice generic object a ∈ F(K). Then edk (F) = ed(a). Proof. Since edk (F) ≥ ed(a) by definition, it remains to prove that for all L/k and all b ∈ F(L), we have ed(b) ≤ ed(a). If L is finite, then ed(b) = 0, and the inequality is clear. Assume that L is infinite. Let a ∈ F(K ) such that aK = a and ed(a) = trdeg(K /k). By assumption, a is generic, so b is a specialization of a . Therefore, there exists a pseudo k-place f : K L and c ∈ F(Rf ) such that a = cK and b = F(ϕf )(c). Since ϕf : Rf −→ L factors through κ(f ) = Rf /mf → L, we see that b = cL , where c ∈ F(κ(f )). Hence ed(b) ≤ trdeg(κ(f )/k) ≤ trdeg(K /k) = ed(a). This concludes the proof. We now give examples of functors with nice generic objects. Proposition IX.23.11 shows that a generic G-torsor P −→ Spec(K) is a generic object of H 1 (− , G). In fact, we are going to see that a generic torsor is a nice generic object for H 1 (− , G). We need a definition first.

XII.35 Generic objects and essential dimension

299

Definition XII.35.3. Let f : X −→ Y and f : X −→ Y be two G-torsors. We say that f is a compression of f if there is a diagram g X _ _ _/ X f

f

Y _ _h _/ Y

where g is a G-equivariant rational dominant morphism and h is a rational morphism too (necessarily dominant). Proposition XII.35.4. Let K be a field. Let T −→ Spec(K) be a Gtorsor, and let f : X −→ Y be a model for T , that is a G-torsor whose generic fiber is isomorphic to T . Let K ⊂ K, and let T −→ Spec(K ) . Then there exists a compression f : be a G-torsor such that T TK X −→ Y of f whose generic fiber is isomorphic to T . Proof. See [3] for a proof. Lemma XII.35.5. Let f : X −→ Y be a compression of a classifying torsor f : X −→ Y . Then f is also classifying. Proof. Let g X _ _ _/ X f

f

Y _ _h _/ Y

be such a compression. Let k /k be a field extension with k infinite and let P ∈ H 1 (k , G). Since f is classifying one can find a k -rational point y ∈ Y (k ) which lies in U , the open set on which h is defined, such that f −1 (y) P . Then the fiber of f at h(y) clearly gives a torsor isomorphic to P . We now have the following result: Theorem XII.35.6. A generic G-torsor T −→ Spec(K) is a nice generic object of H 1 (− , G). In particular, edk (G) = edk (T ). Proof. This follows from Proposition XII.35.4, Lemma XII.35.5 and Proposition XII.35.2.

300


Remark XII.35.7. This theorem reduces the computation of edk (G) to the computation of a generic element. In particular, if G is an abstract finite group, Example IX.23.10 shows that edk (G) is the essential dimension of k(V )/k(V )G , where V is any faithful linear representation of G. Theorem XII.35.8. Let G be an algebraic group and H a closed algebraic subgroup of G. Then edk (H) + dim(H) ≤ edk (G) + dim(G). In particular, if G is finite, we have edk (H) ≤ edk (G). Proof. Once again, we refer to [3] for a proof. §XII.36 Generically free representations Definition XII.36.1. Let G be an algebraic group defined over k. A linear representation V of G is generically free if there exists a G-stable open subset U of V on which G acts freely. Example XII.36.2. Any faithful linear representation V of a finite group G is generically free, since one may take

ker(g − 1). U =V \ g=1

Proposition XII.36.3. Let V be a generically free linear representation of G. Then edk (G) ≤ dim(V ) − dim(G). In particular, if G is a finite group and V is a faithful linear representation of G, then edk (G) ≤ dim(V ). Proof. By Example IX.23.10 (2), one can find a G-stable open subset U such that U −→ U/G is a classifying G-torsor. Thus its generic fiber T is a generic torsor, which is defined over k(U/G). Since trdeg(k(U/G)/k) = dim(V ) − dim(G), applying Theorem XII.35.6 yields the desired result. The last part follows from the previous point and Example XII.36.2. Proposition XII.36.4. Let G be a finite constant group scheme over k acting linearly and faithfully on a k-vector space V . Then the essential

XII.37 Some examples

301

dimension of G is the minimum of the integers trdeg(E/k), where E runs through the set of subfields of k(V ) on which G acts faithfully. Proof. By Theorem XII.35.6, edk (G) is equal to the essential dimension of k(V )/k(V )G , since this latter extension is a generic G-torsor by Example IX.23.10 (2). Let E ⊂ k(V ). We are going to prove that G acts faithfully on E if and only if k(V ) E ⊗E G k(V )G , which will be enough to get the desired result. Assume first that the isomorphism above holds. Since G acts faithfully on k(V ), it follows that G acts also faithfully on E. Conversely, assume that G acts faithfully on E ⊂ k(V ). Since we have E ∩ k(V )G = E G , then E/E G and k(V )G /E G are linearly disjoint and then Ek(V )G E ⊗E G k(V )G . We then have (Ek(V )G )G (E ⊗E G k(V )G )G = E G ⊗E G k(V )G k(V )G . Hence Ek(V )G /k(V )G is a Galois subextension of k(V )/k(V )G with Galois group G. We then get Ek(V )G = k(V ). Thus E ⊗E G k(V )G Ek(V )G = k(V ), and k(V )/k(V )G comes from E/E G . This concludes the proof. Remark XII.36.5. This result shows that the notion of essential dimension of a finite group G coincides with the one defined originally by Buhler and Reichstein defined in [10]. Recently, Karpenko and Merkurjev [29] proved the following remarkable result: Theorem XII.36.6. Let p be a prime number, let k be a field such that char(k) = p and μp ⊂ k. Finally, let G a finite p-group. Then edk (G) is the minimal dimension of a faithful linear representation V of G. Remark XII.36.7. This result is not true is G has composite order. §XII.37 Some examples In all the following examples, we suppose that char(k) = 2. We first would like to estimate the essential dimension of Sn , that is the essential dimension of étale algebras of dimension n. With the assumption on the ground field, Sn acts faithfully on the hyperplane H = {x ∈ k n | x1 + · · · + xn = 0} and thus on k(H) = k(X1 , . . . , Xn−1 ). Now k × acts on k(X1 , . . . , Xn−1 ) by λ · Xi = λXi for

302


all λ ∈ k × and all i = 1, . . . , n − 1. This action commutes with the action of Sn . We easily see that X1 Xn−2 Gm k(X1 , . . . , Xn−1 ) . =k ,..., Xn−1 Xn−1 Now, if n ≥ 3, the group Sn acts faithfully on the latter field. The X1 Xn−2 transcendence degree of k being equal to n − 2, one ,..., Xn−1 Xn−1 concludes that edk (Sn ) ≤ n − 2 for all n ≥ 3 by Proposition XII.36.4. In particular we find edk (S3 ) = 1 and edk (S4 ) = 2. The inequality above may be easily viewed as follows in terms of equations if char(k) n. Let K be an infinite field containing k and let E = K[X]/(f ) an étale K-algebra of rank n. Write f = X n + an−1 X n−1 + · · · + a0 , and let α = X + (f ). The an−1 induces an isomorphism E K[X]/(g), substitution α ↔ α − n where g is a separable polynomial of the form g = X n + bn−2 X n−2 + . . . + b1 X + b0 . If b1 = 0, and if we set β = X + (g), the substitution β ↔ bb01 β induces an isomorphism E K[X]/(h), where h is a separable polynomial of the form g = X n + cn−2 X n−2 + . . . + c1 X + c1 . Hence E is defined over k(c1 , . . . , cn−2 ), which has transcendence degree at most n − 2 over k. Assume now that n ≥ 5, and let us prove that edk (Sn ) ≤ n − 3 in this case. The group PGL2 (k) acts on k(X1 , . . . , Xn ) in the following way: .

a b c d

/ ·Xi =

aXi + b cXi + d

∀ i = 1, . . . , n.

If now i, j, k, are distinct, the cross-sections [Xi , Xj , Xk , X ] =

(Xi − Xk )(Xj − X ) (Xj − Xk )(Xi − X )


303

are PGL2 -invariant. Hence we have k([Xi , Xj , Xk , X ]) ⊂ k(X1 , . . . , Xn )PGL2 (k) where k([Xi , Xj , Xk , X ]) is a short notation for the field generated by the biratios [Xi , Xj , Xk , X ] for i, j, k, l all distinct. But it is easy to see that k([Xi , Xj , Xk , X ]) is generated by the biratios [X1 , X2 , X3 , Xi ] with i = 4, . . . , n. Hence k([Xi , Xj , Xk , X ]) k(Y1 , . . . , Yn−3 ). If n ≥ 5, every σ ∈ Sn \ {1} moves at least one of the [Xi , Xj , Xk , X ]’s. Consequently, since the above action commutes with the Sn -action, Sn acts faithfully on k(Y1 , . . . , Yn−3 ). Proposition XII.36.4 then shows that we have ed(Sn ) ≤ n − 3 for all n ≥ 5. Also, we have (Z/2Z)[n/2] ⊂ Sn , and therefore edk (Sn ) ≥ edk ((Z/2Z)[n/2] ) = [n/2]. Compare with [10, Theorem 6.2]. In particular we have edk (S5 ) = 2 and edk (S6 ) = 3. Remark XII.37.1. The question is still open concerning S7 . Do we have edk (S7 ) = 3 or 4 ? We are now going to compute the essential dimension of twisted forms of μ4 , that is algebraic groups which are isomorphic to μ4 after a suitable field extension. Since Aut(μ4 ) Z/2Z, each twisted form corresponds to a quadratic étale algebra K/k. For any commutative k-algebra R, the set of R-points of the associated twisted form μ4[K] can be described as μ4[K] (R) = {x ∈ (K ⊗k R)× | x4 = 1 and NK⊗k R/R (x) = 1}. √ If K = k × k, we just obtain μ4 , and if K = k( −1), we obtain Z/4Z. The following result is due to Rost (see [50]). Theorem XII.37.2. We have edk (μ4[K] ) = 1 if K = k × k and 2 otherwise. Proof. Since the case K = k × k is clear, we will assume that K is a field from now on. We first give a description of H 1 (− , μ4[K] ). Denoting by RK/k (Gm,K ) the algebraic group-scheme defined by RK/k (Gm,K )(R) = (K ⊗k R)×

304


for any commutative k-algebra R, let T be the kernel of the map Gm × RK/k (Gm,K ) −→ Gm given on the R-points by (x, y) −→ x4 NK⊗k R/R (y)−1 . Now define a group-scheme morphism θ : RK/k (Gm,K ) −→ T by θR (z) = (NK⊗k R/R (z), z 4 ). We can see that, for every field extension L/k, we have an exact sequence 1 −→ μ4[K] (Ls ) −→ RK/k (Gm,K )(Ls ) −→ T(Ls ) −→ 1. Since K ⊗k L is semi-simple, taking cohomology and applying Hilbert 90 yield an exact sequence (K ⊗k L)× −→ T(L) −→ H 1 (L, μ4[K] ) −→ 1. In particular, we get H 1 (L, μ4[K] )

{(x, y) ∈ L× × (K ⊗k L)× | x4 = NK⊗k L/L (y)} . {(NK⊗k L/L (z), z 4 ) | z ∈ (K ⊗k L)× }

The same exact sequence proves the existence of a surjective natural transformation / / H 1 (− , μ4[K] ) . T By Corollary XII.33.4, we get edk (μ4[K] ) ≤ dim(T) = 2. It remains to prove the missing inequality. Let σ be the unique nontrivial automorphism of K/k, and let σL = σ⊗IdL . The reader will check easily that the classical version of Hilbert 90 remains true if K ⊗k L L × L. Therefore, we get x4 = NK⊗k L/L (y) ⇐⇒ NK⊗k L/L (yx−2 ) = 1 ⇐⇒ y = x2

σL (λ) , λ

for some λ ∈ (K ⊗k L)× . Thus, we have H 1 (L, μ4[K] )

{(x, x2 σLλ(λ) ) | x ∈ L× , λ ∈ (K ⊗ L)× } . {(NK⊗k L/L (z), z 4 ) | z ∈ (K ⊗k L)× }

If x ∈ L× and λ ∈ (K ⊗k L)× , set [x, λ] = (x, x2

σL (λ) ). λ


305

Since μ4[K] is isomorphic to μ4 over some field extension and μ4 has essential dimension equal to 1, μ4[K] does not have any non-trivial cohomological invariant of degree 2. However, we are going to use non-zero invariants of degree 1 or 2 to get the result. √ Write K = k( d), and set η1,L :

H 1 (L, μ4[K] ) −→ H 1 (L, μ2 ) [x, λ] −→ (NK⊗k L/L (λ))

and η2,L :

H 2 (L, μ4[K] ) −→ H 2 (L, μ2 ) [x, λ] −→ (x) ∪ (d).

Let us check that these maps are well-defined. It follows from the definitions that we have σL (λ ) σL (λ) 4 z = x2 [x, λ] = [x , λ ] ⇐⇒ x = xNK⊗k L/L (z), x2 λ λ for some z ∈ (K ⊗k L)× . Since (NK⊗k L/L (z)) ∪ (d) = 0 by Proposition III.9.15 (2), it follows that η2,L is well-defined. Moreover, since we have x = xσL (z)z, we get 1=

x2 σL (z 2 )z 2 σL (λ λ−1 z 2 ) = , x2 λ λ−1 z 2

and therefore σL (λ λ−1 z 2 ) = λ λ−1 z 2 . Hence there exists u ∈ L× such that λ z 2 = λu. Taking norms on both sides, we obtain NK⊗k L/L (λ ) = NK⊗k L/L (λ) ∈ L× /L×2 . It follows that the map η1,L is well-defined. Clearly, these maps give rise to two cohomological invariants η1 , η2 (notice that η2 is not non-trivial). √ Let L0 = k(s, t) and let α = [s, 1 + t d]. To get the desired result, it is enough to show that ed(α) ≥ 2. Let α ∈ H 1 (L, μ4[K] ) be a torsor such = α (L ⊂ L0 ) with trdeg(L/k) = ed(α). We need to show that that αL 0 L/k has at least transcendence degree 2 over k. Let υ the s-adic valuation on k(t)(s), let υ be its restriction to L and let eυ |υ be the corresponding ramification index. We have η2 (α) = η2 (α )L0 . Notice that ∂υ (η2 (α)) = ∂υ ((s) ∪ (d)) = (d) = 0 ∈ H 1 (k(t), μ2 ),

306


since d ∈ k × is not a square. Now we have ∂υ (η2 (α)) = eυ |υ (∂υ (η2 (α )))k(t) by Proposition X.26.8. It follows that eυ |υ is necessarily odd. In particular, υ is not trivial. Hence it is enough to prove that trdeg(κ(υ )/k) ≥ 1 to get the desired conclusion. Indeed, we will have trdeg(L/k) > trdeg(κ(υ )) ≥ 1, and thus trdeg(L/k) ≥ 2. Assume that κ(υ )/k is an algebraic extension. Since κ(υ ) ⊂ k(t), we get κ(υ ) = k. Now we have ∂υ (η1 (α)) = ∂υ ((1 − dt2 )) = 0 = eυ |υ (∂υ (η1 (α )))k(t) , so we get eυ |υ (∂υ (η1 (α )))k(t) = 0. Since eυ |υ is odd, this yields (∂υ (η1 (α )))k(t) = 0. Since κ(υ ) = k and the restriction from k to k(t) is injective (see Chapter X, Section X.25), we get ∂υ (η1 (α )) = 0. Therefore, the specializations of η1 (α) and η1 (α ) are well-defined and we have sυ (η1 (α)) = (1 − t2 d) = (sυ (η1 (α )))k(t) by Proposition X.26.8. Let u ∈ k × such that sυ (η1 (α )) = (u). The previous equality shows that there exists f (t) ∈ k(t)× such that 1 − t2 d = uf (t)2 . Let π ∈ k[t] be an irreducible divisor of 1 − t2 d. Comparing π-adic valuations yields a contradiction. Hence trdeg(κ(υ )/k) ≥ 1, and this concludes the proof. Corollary XII.37.3. Assume that char(k) = 2. Then edk (Z/4Z) = 1 if −1 ∈ k×2 and 2 otherwise. Proof. If −1 ∈ k ×2 , Z/4Z μ4 , and the result is clear. If −1 ∈ / k ×2 , √ Z/4Z μ4[K] , where K = k( −1). Now apply the previous result. §XII.38 Complements and open problems We would like to end this chapter by giving some complements and stating some open problems concerning essential dimension. First of all, we would like to mention the existence of a local version of essential dimension of a functor at a prime p, defined as follows:

XII.38 Complements and open problems

307

Definition XII.38.1. Let F : Ck −→ Sets be a covariant functor, let p be a prime number, let K/k be a field extension and let a ∈ F(K). Finally, let E/k a subextension of K/k. The essential dimension at p of a is the integer ed(a; p) defined by ed(a; p) = min{ed(aL ) | [L : K] is prime to p}. The essential dimension at p of F is the supremum of ed(a; p) for all a ∈ F(K) and for all K/k. The essential dimension of F at p will be denoted by edk (F; p). If F = H 1 (− , G), we simply denote it by edk (G; p). When G is a finite abstract group and char(k) = p, one can show that edk (G; p) = edk (Gp ; p) = edk(μp ) (Gp ; p), where Gp is a p-Sylow subgroup of G. Therefore, the essential dimension at p of any finite group may be computed over an arbitrary field k satisfying char(k) = p in view of Theorem XII.36.6. See [29] for more details. Clearly, we have edk (G) ≥ sup(edk (G; p)). p

This inequality is known to be strict for finite groups (see the exercices for a counterexample), but no examples of connected groups for which the inequality is strict are known (at least to the author). Question 1: If G is a connected algebraic group defined over k, do we have edk (G) = sup(edk (G; p)) (at least if char(k) = 0) ? p

In ‘bad’ characteristic, only few results are known. For example, we have the following result, proved in [28]: Proposition XII.38.2. Let p be a prime number, let n ≥ 1, and let k be a field of characteristic p. Then we have 1 if |k| ≥ pn edk ((Z/pZ)n ) = 2 if |k| < pn . However, the essential dimension Z/pn Z over a field of characteristic p is not known for n ≥ 3. Question 2: Let k be a field of characteristic p > 0, and let G be a

308


finite p-group. Is edk (G) equal to the least dimension of a linear faithful representation of G? The computation of the essential dimension of a connected algebraic group G is a very interesting problem, far from being solved, even in the case where G is split and k = C. Recently, Brosnan, 0Reichstein and Vistoli [9] proved that edk (Spinn ) is n−1 equivalent to 2[ 2 when n −→ +∞. The number edC (PGLn ) is known only for few values of n. It is equal to 1 if n = 2, 3, 6 (see for example [47]) and is equal to 5 for n = 4 (Rost[49]; see also [36]). The essential of dimension PGL5 , or more generally of PGLp (p prime) is not known. An old conjecture states that every central simple algebra of degree p whose center contains C is isomorphic to a symbol algebra {a, b}p . If this is true, this would lead to the equality edC (PGLp ) = 2. Conversely, proving that edC (PGLp ) ≥ 3 (p ≥ 5) would immediately disprove this conjecture. It is interesting to note that Reichstein related the essential dimension of a given central simple algebra A to the essential dimension of higher trace forms. Later on, Florence generalized Reichstein’s result to separable algebras. See [48] and [23] for more details. Notes The reader will find more computations of essential dimension of some functors in [4],[6] and [49]. Essential dimension of finite groups is also studied in [28]. In fact, in [28], the authors are focusing mainly on the existence of generic polynomials for a given finite group G, that is on the existence of generic G-torsors over a rational extension of k.

Exercises 1. Let k be a field and L be an étale algebra over k. Recall that the (1) algebraic group-scheme Gm,L is defined by Gm,L (R) = {z ∈ L× R | NLR /R (z) = 1} (1)

for every commutative k-algebra R, and that we have H 1 (K, Gm,L ) K × /NLK /K (L× K) (1)

Exercises

309

for every field extension K/k. (a)

Let t be a transcendental element over k. Show that t belongs to NLk(t) /k(t) (L× k(t) ) if and only if L is isomorphic to a product of some finite separable field extensions of relatively prime degrees.

(b)

Deduce the value of edk (Gm,L ).

(1)

2. Let k be a field of characteristic different from 2. Show that edk (PGL2 ) = 2. Hint: Use the fact that every central simple algebra of degree 2 is isomorphic to a quaternion algebra. 3. Keeping√the notation of the proof of Theorem XII.37.2, show that [s, 1 + t d] is a generic μ4[K] -torsor. 4. Let G be a finite abstract group. (a)

Show that if edk (G) = 1, then G identifies to a subgroup of PGL2 (k). Hint: If E/k is a subextension of k(V )/k of transcendence degree 1, then E/k is purely transcendental.

(b)

Deduce that for all prime p ≥ 5, we have edQ (Z/pZ) ≥ 2.

(c)

Show that edQ (Z/5Z) = 2.

(d)

Assume that char(k) = p. Show that edk (Z/p2 Z) = 2.

5. Let G be a finite abstract group of order n. Show that for every prime integer p n, we have edk (G; p) = 0. Hint: Show that for any Galois G-algebra L/K, there exists a finite field extension M/K of degree dividing n such that LM is isomorphic to the split Galois G-algebra over M . 6. Let k be a field, and let p be prime number. (a)

Assume that p char(k). Show that edk (Z/pZ; p) = 1.

(b)

Compare edQ (Z/pZ) and sup(edQ (Z/pZ; )).

References

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Index

G-algebra, 120 R-equivalence group, 276 R-trivial element, 275 group-scheme, 276 Γ-group, 28 Γ-module, 28 Γ-set, 28 Γ-sets morphism, 28 k-embedding, 13 k-isomorphism, 13 ´ etale algebra, 160 affine scheme, 250 algebraic element, 14 extension, 14 algebraic closure, 15 algebraic group, 80 base point, 34 category, 70 of Γ-groups, 71 of Γ-modules, 71 of Γ-sets, 71 of pointed sets, 71 central simple algebra, 119 degree, 119 Clifford algebra, 139 Clifford group, 141 Clifford invariant, 157 closed embedding, 83 closed subgroup, 83 coboundary, 35 cocycle, 32 cohomologous, 34, 35 definition, 35 trivial, 32, 35

cohomological invariant, 154 constant, 154 normalized, 154 unramified, 264 connecting maps, 48, 53 cup-product, 62 direct limit, 42 directed set, 21 directed system, 42 essential dimension, 290 exact sequence, 46 associated to a normal subgroup, 51 associated to a central subgroup, 53 associated to a subgroup, 48 field extension, 13 degree, 13 Galois, 17 separable, 15 functor, 71 Galois, 85 representable, 74 Galois G-algebra, 166 induced, 170 Galois cohomology set, 93 Galois correspondence, 19 Galois descent condition, 103 Galois descent lemma, 106 Galois group, 17 generic element, 255 generically free representation, 300 group extension, 180 group-scheme, 80 affine, 80 algebraic, 80 Hasse invariant, 148

314

Index Hilbert 90, 113 inflation map, 40 inverse image, 39 inverse limit, 22 involution, 217 conjugate, 220 hyperbolic, 224 of the first kind, 218 of the second kind, 218 orthogonal, 221 symplectic, 221 type, 221 isometry, 135, 227 kernel, 45 Krull topology, 18 natural transformation, 73 Noether’s problem, 261 orthogonal basis, 135 orthogonal group, 136 pointed set, 34 profinite group, 24 cohomology sets, 30, 34, 35 projective system, 21 pseudo k-place, 254 quadratic form, 134 determinant, 135 indefinite, 153 regular, 135 quaternion algebra, 127 rational map, 252 reduced characteristic polynomial, 231 reduced norm, 231 reduced trace, 231 reflection, 137 residue, 269 restriction map, 39, 93 separable algebra, 113 separable polynomial, 15 signature, 153 similitude definition, 227 multiplier, 227 simple algebra, 113 special Clifford group, 145 special orthogonal group, 137 specialization of an element of a functor, 255 with respect to a valuation, 269 spinor group, 145

spinor norm, 144 stabilizer of an element, 104 subcategory, 71 subfunctor, 73 tensor algebra, 139 torsor, 253 classifying, 256 generic, 256 split, 254 trace form of a central simple algebra, 234 of an ´ etale algebra, 192 of an algebra with involution, 235 restricted, 235 transcendental element, 14 twisted Γ-group, 57 twisted form, 101 unramified element, 269 Yoneda’s Lemma, 74

315

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