Algebraic Numbers and Algebraic Functions

Algebraic Numbers and Algebraic Functions EMIL ARTIN Late of Princeton University GI 18 GORDON AND BREACH SCIENCE P...

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Algebraic Numbers and Algebraic Functions

EMIL ARTIN Late of Princeton University

GI 18 GORDON

AND

BREACH

SCIENCE PUBLISHERS NEW YORK· LONDON' PARIS

Copyright © 1967 by Gordon and Breach, Science Publishers, Inc. 150 Flfth Avenue, New York, New York 10011

Library of Congress Catalog Card Number: 67-26811 Editorial Office for Great Britain: Gordon and Breach Science Publishers Ltd. 8 Bloomsbury Way London WCl, England Editorial Office for France: 7-9 rue Emile Dubois Paris 14" Distributed in France by: Dunod Editeur 92, rue Bonaparte Paris 6e Distributed in Canada by: The Ryerson Press 299 Queen Street West Toronto 2B, Ontario

Printed in Belgium by the Saint Catherine Press, Ltd., Tempelhof, Bruges

General Preface A large number of mathematical books begin as lecture notes; but, since mathematicians are busy, and since the labor required to bring lecture notes up to the level of perfection which authors and the public demand of formally published books is very considerable, it follows that an even larger number of lecture notes make the transition to book form only after great delay or not at all. The present lecture note series aims to fill the resulting gap. It will consist of reprinted lecture notes, edited at least to a satisfactory level of completeness and intelligibility, though not necessarily to the perfection which is expected of a book. In addition to lecture notes, the series will include volumes of collected reprints of journal articles as current developments indicate, and mixed volumes including both notes and reprints. JACOB

T.

SCHWARTZ

MAURICE

LEVY

Preface These lecture notes represent the content of a course given at Princeton University during the academic year 1950/51. This course was a revised and extended version of a series of lectures given at New York University during the preceding summer. They cover the theory of valuation, local class field theory, the elements of algebraic number theory and the theory of algebraic function fields of one variable. It is intended to prepare notes for a second part in which global class field theory and other topics will be discussed. Apart from a knowledge of Galois theory, they presuppose a sufficient familiarity with the elementary notions of point set topology. The reader may get these notions for instance in N. Bourbaki, Elements de Mathematique, Livre III, Topologie generale, Chapitres I-III. In several places use is made of the theorems on Sylow groups. For the convenience of the reader an appendix has been prepared, containing the proofs of these theorems. The completion of these notes would not have been possible without the great care, patience and perseverance of Mr. I. T. A. O. Adamson who prepared them. Of equally great importance have been frequent discussions with Mr. }. T. Tate to whom many simplifications of proofs are due. Very helpful was the assistance of Mr. Peter Ceike who gave a lot of his time preparing the stencils for these notes. Finally I have to thank the Institute for Mathematics and Mechanics, New York University, for mimeographing these notes. Princeton Universt"ty 1951

June

EMIL ARTIN

Contents GENERAL PREFACE PREFACE.

V Vll

Part I General Valuation Theory

Chapter 1 VALUATIONS OF A FIELD

1. 2. 3. 4. 5. 6.

Equivalent Valuations . The Topology Induced by a Valuation Classification of Valuations . The Approximation Theorem Examples Completion of a Field

3 5 6 8

10 17

Chapter 2 COMPLETE FIELDS

1. 2. 3. 4. 5. 6. 7.

Normed Linear Spaces . Extension of the Valuation . Archimedean Case . The Non-Archimedean Case Newton's Polygon . The Algebraic Closure of a Complete Field Convergent Power Series

19 21

24 28 37 43 47

Chapter 3 e, f AND n I. The Ramification and Residue Class Degree 2. The Discrete Case . 3. The General Case ix

53

56 60

x

CONTENTS

Chapter 4 RAMIFICATION THEORY

1. 2. 3. 4. 5. 6.

Unramified Extensions . Tamely Ramified Extensions Characters of Abelian Groups The Inertia Group and Ramification Group Higher Ramification Groups Ramification Theory in the Discrete Case

64 67 71 72 77

82

Chapter 5 THE DIFFERENT

1. 2. 3. 4.

The Inverse Different . Complementary Bases . Fields with Separable Residue Class Field The Ramification Groups of a Sub field

Part II

86 89 93 95

Local Class Field Theory Chapter 6

PREPARATIONS FOR LOCAL CLASS FIELD THEORY

1. 2. 3. 4.

Galois Theory for Infinite Extensions Group Extensions . Galois Cohomology Theory Continuous Cocycles

103 108 114 117

Chapter 7 THE FIRST AND SECOND INEQUALITIES

1. 2. 3. 4. 5.

Introduction U nramified Extensions The First Inequality The Second Inequality: A Reduction Step The Second Inequality Concluded

127 127

130 133 135

Chapter 8 THE NORM RESIDUE SYMBOL

1. The Temporary Symbol (c, K I KIT) 2. Choice of a Standard Generator c 3. The Norm Residue Symbol for Finite Extensions

144

153 158

CONTENTS

Xl

Chapter 9 THE EXISTENCE THEOREM

1. 2. 3.

Introduction . The Infinite Product Space I The New Topology in K* . 4. The Norm Group and Norm Residue Symbol for Infinite Extensions 5. Extension Fields with Degree Equal to the Characteristic 6. The Existence Theorem 7. Uniqueness of the Norm Residue Symbol.

165 165

170 174 180 181 187

Chapter 10 ApPLICATIONS AND ILLUSTRATIONS

1. Fields with Perfect Residue Class Field

190

2. The Norm Residue Symbol for Certain Power Series Fields

3. Differentials in an Arbitrary Power Series Field

4. 5. 6.

The Conductor and Different for Cyclic p-Extensions The Rational p-adic Field Computation of the Index (ex : exn )

193 200 203 206 209

Part III Product Formula and Function Fields in One Variable Chapter 11 PREPARATIONS FOR THE GLOBAL THEORY

1. The Radical of a Ring .

2. Kronecker Products of Spaces and Rings . 3. Composite Extensions . 4. Extension of the Valuation of a Non-Complete Field

215 216 218 223

Chapter 12 CHARACTERIZATION OF FIELDS BY THE PRODUCT FORMULA

1. PF-Fields.

2. Upper Bound for the Order of a Parallelotope 3. Description of all PF-Fields 4. Finite Extensions of PF-Fields

225 227 230 235

Xli

CONTENTS

Chapter 13 DIFFERENTIALS IN PF-FIELDS 1. 2. 3. 4. 5. 6.

Valuation Vectors, Ideles, and Divisors Valuation Vectors in an Extension Field Some Results on Vector Spaces . Differentials in the Rational Subfield of a PF-Field . Differentials in a PF-Field The Different.

238 241 244 245 251 255

Chapter 14 THE RIEMANN-RoCH THEOREM

1. Parallelotopes in a Function Field 2. First Proof 3. Second Proof .

260 262 265

Chapter 15 CONSTANT FIELD EXTENSIONS

1. 2. 3. 4.

The Effective Degree Divisors in an Extension Field Finite Algebraic Constant Field Extensions The Genus in a Purely Transcendental Constant Field Extension 5. The Genus in an Arbitrary Constant Field Extension

271 278 279 284 287

Chapter 16 ApPLICATIONS OF THE RIEMANN-RoCH THEOREM

1. 2. 3. 4. 5. 6. 7. 8.

Places and Valuation Rings Algebraic Curves . Linear Series . Fields of Genus Zero Elliptic Fields. The Curve of Degree n H yperelliptic Fields The Theorem of Clifford

293 297 300

302 306 311

312 317

CONTENTS

Xlll

Chapter 17 DIFFERENTIALS IN FUNCTION FIELDS

1. 2. 3. 4.

Preparations Local Components of Differentials Differentials and Derivatives in Function Fields Differentials of the First Kind

321 322 324 329

Appendix THEOREMS ON p-GROUPS AND SYLOW GROUPS

1. 2. 3. 4.

S-Equivalence Classes . Theorems About p-Groups . The Existence of Sylow Subgroups Theorems About Sylow Subgroups

INDEX OF SYMBOLS SUBJECT INDEX

334

335 336

337 339 343

PART ONE

General Valuation Theory

CHAPTER ONE

Valuations of a Field

A valuation of a field k is a real-valued function for all x E k, satisfying the following requirements:

I x I ~ 0; I x I = 0 if and only if x = 0, (2) I xy I = I x I I y I , (3) If I x I :::;;; 1, then I 1 + x I :::;;; c, where

I x I , defined

(1)

c

IS

a constant;

c~1.

(1) and (2) together imply that a valuation is a homomorphism of the multiplicative group k* of non-zero elements of k into the positive real numbers. If this homomorphism is trivial, i.e. if I x I = 1 for all x E k*, the valuation is also called trivial.

1. Equivalent Valuations

Let I 11 and I 12 be two functions satisfying conditions (1) and (2) above; suppose that I 11 is non-trivial. These functions are said to be equivalent if I a 11 < 1 implies I a 12 < 1. Obviously for such functions I a 11 > 1 implies I a 12 > 1; but we can prove more. Theorem 1: Let I 11 and I 12 be equivalent functions, and suppose I 11 is non-trivial. Then I a 11 = 1 implies I a 12 = 1. Proof: Let b =F 0 be such that I b 11 < 1. Then I anb 11 < 1; whence I anb 12 < 1, and so I a 12 < I b l2"l/n. Letting n -+00, we have I a 12 :::;;; 1. Similarly, replacing a in this argument by l/a, we have I a 12 ~ 1, which proves the theorem. 3

1.

4

VALUATIONS OF A FIELD

Corollary: For non-trivial functions of this type, the relation of equivalence is reflexive, symmetric and transitive. There is a simple relation between equivalent functions, given by Theorem 2: If I 11 and I 12 are equivalent functions, and I 11 is non-trivial, then I a 12 = I a 11" for all a E k, where ex is a fixed positive real number. Proof: Since I 11 is non-trivial, we can select an element e E k* such that I e 11 > I; then I e 12 > 1 also. Set I a 11 = I e 111', where y is a non-negative real number. If min > y, then I a 11 < Ie 11m/n, whence I an/em 11 < 1. Then I an/em 12 < 1, from which we deduce that I a 12 < Ie 12m/n. Similarly, if min < y, then I a 12 > I e 1m / n It follows that I a 12 = I c I~· Now, clearly, log I a 11 log 1a 12 'Y = log 1 c It = log I C 12 • o

This proves the theorem, with 0:=

log I c 12 log 1 cit

•

In view of this result, let us agree that the equivalence class defined by the trivial function shall consist of this function alone. Our third condition for valuations has replaced the classical "Triangular Inequality" condition, viz., I a b I ~ Ia I Ib I. The connection between this condition and ours is given by

+

+

Theorem 3: Every valuation is equivalent to a valuation for which the triangular inequality holds. Proof. (I) When the constant c = 2, we shall show that the triangular inequality holds for the valuation itself. Letlal~lbl·

Then

I: l~l~ll+: 1~2~la+bl~2Ibl = 2 max (I a

Similarly

I, I b\).

2.

5

THE TOPOLOGY INDUCED BY A VALUATION

and

n

Now given aI' a z , "', an' we can find an integer r such that 2 r < 2n. Hence

~

I a1 + ... + an I = I a1 + ... + an + 0 + ... + 0 I ~ 2r max I av I < 2n max I av I . v • In particular, if we set all the a. = 1, we have I n also weaken the above inequality, and write

I ~ 2n.

We may

Consider

I a + b In =

Ian + (7) a n- b + ... + bn I 1

~ 2(n + 1) 111 II a In + I (7) II a In-II b I + ... + I b Itll ~ 4(n + 1) II a In + (7) I a In-l I b I + ... + I bin! = 4(n

+ I){I a I + I b I}n.

Hence

I a + b I ~ V"4(n + 1) (I a I + I b\). Letting n ~oo we obtain the desired result. We may note that, conversely, the triangular inequality implies that our third requirement is satisfied, and that we may choose c= 2. (2) When c > 2, we may write c = 2"'. Then it is easily verified that I 11 /", is an equivalent valuation for which the triangular inequality is satisfied.

2. The Topology Induced by a Valuation Let I I be a function satisfying the axioms (1) and (2) for valuations. In terms of this function we may define a topology in k by

1.

6


prescribing the fundamental system of neighborhoods of each element Xo E k to be the sets of elements x such that I x - Xo I < E. It is clear that equivalent functions induce the same topology in k, and that the trivial function induces the discrete topology. There is an intimate connection between our third axiom for valuations and the topology induced in k. Theorem 4: The topology induced by and only if axiom (3) is satisfied.

I I is Hausdorff if

Proof: (I) If the topology is Hausdorff, there exist neighborhoods separating 0 and -I. Thus we can find real numbers a and b such that if I x I ~ a, then I I + x I ?: b. Now let x be any element with I x I ~ I; then either I 1 + x I ~ Ija or I I + x I > Ija. In the latter case, set Y = - x/(1 + x); then 1 a-

~-l=a;

hence

Il+YI=ll!xl~b, i.e.

11

I 1 + x I ~ lib. We conclude, therefore, that if I x I ~ 1, then + x I ~ max (1Ia, lib), which is axiom (3).

(2) The converse is obvious if we replace I I by the equivalent function for which the triangular inequality holds. It should be remarked that the field operations are continuous in the topology induced on k by a valuation.

3. Classification of Valuations If the constant c of axiom (3) can be chosen to be I, i.e. if + x I ~ I, then the valuation is said to be non-archimedean. Otherwise the valuation is called archimedean. Obviously the valuations of an equivalence class are either all archimedean or all non-archimedean. For nonarchimedean valuations we obtain a sharpening of the triangular inequality:

I x I ~ I implies I I

3. Theorem 5:

For non-archimedean valuations, 1

11

7

CLASSIFICATION OF VALUATIONS

a + b 1 ~ max 1, I a.1 < I a 1 used in the following form:

I. This last result is frequently

Corollary 5.2: Suppose it is known that I a. I ~ I a 1 I for all I a1 an I < I a 1 I. Then for some v > 1, I a.1 = I a 1 I· We now give a necessary and sufficient condition for a field to be non-archimedean:

v, and that

+ ... +

Theorem 6: A valuation is non-archimedean if and only if the values of the rational integers are bounded.

1.

8


Proof: (1) The necessity of the condition is obvious, for if the valuation is non-archimedean, then Iml=ll+l+···+II~lll·

(2) To prove the sufficiency of the condition we consider the equivalent valuation for which the triangular inequality is satisfied. Obviously the values of the integers are bounded in this valuation also; say I m I ~ D. Consider

I a + b In = I (a + b)n I =

Ian + (~) an-1b + ... + bn I

~ I 1 I I an I + I (~) II a In-l I b I + ... + 11 I \ b In

D{I a In + I a In-l I b I + ... + I bin} ~ D(n + l){max(1 a I, I b Inn. ~

Hence

I a + b I ~ vD(n + 1) max (\ a I, I b\). Letting n -+00, we have the desired result. Corollary: A valuation of a field of characteristic p > 0 IS non-archimedean. We may remark that if ko is a subfield of k, then a valuation of k is (non-)archimedean on ko is (non-)archimedean on the whole of k. In particular, if the valuation is trivial on ko, it is nonarchimedean on k. 4. The Approximation Theorem

Let {an} be a sequence of elements of k; we say that a is the limit of this sequence with respect to the valuation if lim

n->-OO

Ia -

an I = 0 .

The following examples will be immediately useful: (a) If I a

I < 1, then lim an

n->-OO

=

O.

4.

9

THE APPROXIMATION THEOREM

I an - 0 I = I a In ---+ 0 as n ---+ (b) If I a I < 1, then

For

00.

. an hm an =0. l

n->OO

If I a

(c)

I>

+

1, then an lim - - = 1 . n->OO I an

+

For

/_ I ~_ll_/_l 1 + an - I + an -

(

I

(! r

a

1+

)n

~O

as

We now examine the possibility of finding a relation between nonequivalent valuations; we shall show that if the number of valuations considered is finite, no relation of a certain simple type is possible.

Theorem 7: Let Ill, ... , lin be a finite number of inequivalent non-trivial valuations of k. Then there is an element a E k such that I a 11 > 1, and I a I. < 1 (v = 2, ... , n). Proof. First let n = 2. Then since III and 112 are nonequivalent, there certainly exist elements b, C E k such that I b 11 < 1 and I b 12 1, while Ie 11 1 and Ie 12 < 1. Then a = c/b has the required properties. The proof now proceeds by induction. Suppose the theorem is true for n - 1 valuations; then there is an element b E k such that I b 11 > 1, and I b I. < 1 (v = 2, ... , n - 1). Let c be an element such that Ie 11 > 1 and I c In < 1. We have two cases to consider:

>

>

Case 1: I b In ~ 1. Consider the sequence af' = cbf'. Then I af' 11 = I C 11 I b 11f' > 1, while I af' In = I c In I b Inf' < 1; for sufficiently large T, I af' I. = I c Iv I b Ivf' < 1 (v = 2, ... , n - 1). Thus af'

is a suitable element, and the theorem is proved in this case. Case

2: I b In

>

1.

Here we consider the sequence cbf' Qr=l+bf'o

10

1.


This sequence converges to the limit c in the topologies induced by I 11 and I In· Thus af' = c + "If' where I "If' 11 and I "If' In ----+ 0 as T ----+00. Hence for T large enough, I af' 11 > 1 and I af' In < 1. For v = 2, "', n - 1, the sequence af' converges to the limit 0 in the topology induced by II •. Hence for large enough values of T, I af' I. < 1 (v = 2, "', n - 1). Thus af' is a suitable element, for T large enough, and the theorem is proved in this case also. Corollary: With the conditions of the theorem, there exists an element a which is close to 1 in III and close to 0 in I I. (v = 2, ... , n - 1).

Proof. If b is an element such that I b 11 > 1 and I b I. < 1 (v = 2, ... , n - 1), then af' = bf'/(1 + bf') satisfies our requirements for large enough values of T. Theorem 8: (The Approximation Theorem): Let Ill, ... , lin be a finite number of non-trivial inequivalent valuations. Given any E > 0, and any elements a. (v = 1, ... , n), there exists an element a such that I a - a.l. < E.

Proof.

We can find elements bi (i

= 1, "', n) close to 1 in

Iii and close to zero in II. (v =F i). Then a = albI + '" + anbn is

the required element. Let us denote by (k)i the field k with the topology of I Ii imposed upon it. Consider the Cartesian product (k)I X (k)z X ... X (k)n. The elements (a, a, "', a) of the diagonal form a field kD isomorphic to k. The Approximation Theorem states that kD is everywhere dense in the product space. The theorem shows clearly the impossibility of finding a non-trivial relation of the type n

II I x I~> =

1,

>=1

with real constants c•.

s.

Examples

Let k be the quotient field of an integral domain 0; then it is easily verified that a valuation I I of k induces a function 0 (which we still denote by I I), satisfying the conditions

5.

11

EXAMPLES

I a I ~ 0; I a I = 0 if and only if (2) I ab I = I a I I b I , (3) I a + b I ~ 0 max (I a I , I b I).

(1)

a

= 0,

Suppose, conversely, that we are given such a function on o. Then if x = alb (a, b E 0, X E k), we may define I x I = I a III b I; I x I is well-defined on k, and obviously satisfies our axioms (1) and (2) for valuations. To show that axiom (3) is also satisfied, let I x I ~ 1, i.e. I a I ~ I b I. Then = I a + b I ____ e max (I a I , I b \) =

\1 + x \

Ibl

~

Ibl

e.

Hence if k is the quotient field of an integral domain 0, the valuations of k are sufficiently described by their actions on o. First Example: Let k = R, the field of rational numbers; k is then the quotient field of the ring of integers o. Let m, n be integers > 1, and write m in the n-adic scale: m ( o ~ ai

= ao + a1n + ... + arnr

, we have I m I ~ {max (1, I n j)}logm/lOgn. There are now two cases to consider. Case 1:

I n I > 1 for all n > 1. Then I m I ~ I n IIOgm/logn : I m 11/IOgm ~ I n 11I1ogn.

Since I m I > 1 also, we may interchange the roles of m and n, obtaining the reversed inequality. Hence

I m 11/IOgm = I n 11llogn = ea. ,

1.

12 where

0:


is a positive real number. It follows that

I n 1= e"'!ogn= n", so that

I I is in this case equivalent to the ordinary

Ix I =

max (x, - x).

"absolute value",

Case 2: There exists an integer n > 1 such that / n / ::::;; 1. Then I m I ::::;; 1 for all m E o. If we exclude the trivial valuation, we must have / n I < 1 for some n E 0; clearly the set of all such integers n forms an ideal (p) of o. The generator of this ideal is a prime number; for if P = PtP2' we have I pi = I PI II P2/ < 1, and hence (say) I PI I < 1. This PI E (p), i.e. P divides PI; but PI divides P; hence P is a prime. If I PI = c, and n = pvb, (p, b) = 1, then / n I = cV. Every non-archimedean valuation is therefore defined by a prime number p. Conversely, let P be a prime number in 0, c a constant, 0 < c < 1. Let n = pvb, (p, b) = 1, and define the function II by setting I n I = CV. It is easily seen that this function satisfies the three conditions for such functions on 0, and hence leads to a valuation on R. This valuation can be described as follows: let x be a nonzero rational number, and write it as x = P'y, where the numerator and denominator are prime to p. Then I x I = C". Second Example. Let k be the field of rational functions over a field F: k = F(x). Then k is the quotient field of the ring of polynomials 0 = F[x]. Let II be a valuation of k which is trivial on F; I I will thus be non-archimedean. We have again two cases to consider. Case 1: I x I > 1.

Then if

f(x) = Co

+ cIX + ... + cnxn ,

cn =F 0, we have If(x)

I = I x In = I X Idegt(x) .

Conversely, if we select a number c

>

1 and set

If(x) I = cdegf ("') , our conditions for functions on

0

are easily verified. Hence this

5.

EXAMPLES

13

function yields a valuation of k described as follows: let a = f(x)fg(x), and define deg a = degf(x) - degg(x).

Then I a I = cdega • Obviously the different choices of c lead only to equivalent valuations. Case 2: I x I < 1. Then for any f(x) EO, I f(x) I ~ 1. If we exclude the trivial valuation, we must have I f(x) I < 1 for some f(x) Eo. As in the first example, the set of all such polynomials is an ideal, generated by an irreducible polynomial p(x). If I p(x) I = c, and f(x) = (p(x»· g(x), (p(x), g(x» = 1, then If(x) I = c' . Conversely, if p(x) is an irreducible polynomial, it defines a valuation of this type. This is shown in exactly the same way as in the first example. In both cases, the field k = F(x) and the field k = R of the rational numbers, we have found equivalence classes of valuations, one to each prime p (in the case of F(x), one to each irreducible polynomial) with one exception, an equivalence class which does not come from a prime. To remove this exception we introduce in both cases a "symbolic" prime, the so-called infinite prime, poo which we associate with the exceptional equivalence class. So I a lp", stands for the ordinary absolute value in the case k = R, and for cdega in the case k = F(x). We shall now make a definite choice of the constant c entering in the definition of the valuation associated with a prime p.

(I) k = R. (a) P =1= poo. We choose c = 1fp. If, therefore, 0, and a = pvb, where the numerator and denominator of b are prime to p, then we write I a /p = (lfp)v. The exponent v is called the ordinal of a at p and is denoted by v = ordp a. (b) P = poo. Then let I a Ipex> denote the ordinary absolute value. a =1=

(II) k = F(x). Select a fixed number d, 0 < d < 1. (a) P =1= poo, so that p is an irreducible polynomial; write c = ddegp. If a =1= 0 we write as in case I(a), a = pOb, v = ord, a, and so we define I a Ip = c· = ddegpoordva.

1.

14


(b) P = POCJ' so that I a I = cdega . where c > 1. We choose c = lid, and so define I alp., = d-deg a. In all cases we have made a definite choice of I a Ip in the equivalence class corresponding to p; we call this I a Ip the normal valuation at p. The case where k = F(x), where F is the field of all complex numbers, can be generalized as follows. Let D be a domain on the Gauss sphere and k the field of all functions meromorphic in D. If Xo ED, Xo *-w, andf(x) E k, we may write f(x)

where g(x)

E

= (x -

*- 0

k (g(xo)

or (0), and define a valuation by

If(x) 1"'0

If

Xo

=

00,

=

co

rd

.,!(",).

we write f(x) =

whereg(x)

xo)ord~/(:r) g(x),

E

k (g(w)

(~ )ord""(,,,) g(x) ,

*- 0 oroo), and define If(x) 100 = cord ""(,,,) .

This gives for each Xo E D a valuation of k - axioms (1) and (2) are obviously satisfied, while axiom (3) follows from

If Ixo

~

1 ~ f is regular at Xo ~ 1 f is regular at

+ ~ I 1 + f Ixo ~ 1.

Xo

The valuation Ilxo obviously describes the behavior of f(x) at the point Xo: If Iflxo < 1, or ordxo(f(x)) = n > 0, then f(x) has a zero of order n at Xo • If If Ixo > 1, or ordxJ(x) = n < 0, then f(x) has a pole or order -n at Xo. If If Ixo = 1, then f( x) is regular and non-zero at Xo • Should D be the whole Gauss sphere, we have k = F(x); in this case the irreducible polynomials are linear of type (x - x o). The valuation I f(x) Ix-xo as defined previously is now the valuation

5.

15

EXAMPLES

denoted by I f(x) Ixu; the valuation giv.en by I f(x) Ipeo is now denoted by If(x) leo. We see that to each pomt of the Gauss sphere there corresponds one of our valuations. It was in analogy to this situation, that we introduced in the case k = R, the field of rational numbers, the "infinite prime" and associated it with the ordinary absolute value. We now prove a theorem which establishes a relation between the normal valuations at all primes p: Theorem 9:

In both cases, k = Rand k = F(x), the product

TIp I a Ip = 1. We have already remarked that a relation of this form cannot be obtained for any finite number of valuations. Proof: Only a finite number of primes (irreducible polynomials) divide a given rational number (rational function). Hence I a Ip = 1 for almost all (i.e. all but a finite number of) primes p, and so the product TIp I a Ip is well-defined. If we write rP(a) = TIp I a Ip we see that rP(ab) = rP(a) rP(b)j thus it suffices to prove the result for a = q, where q is a prime (irreducible polynomial). For q E R,

For q E k(x),

This completes the proof. We notice that this is essentially the only relation of the form TI I a I~D = 1. For if ljJ(a) = TI I a I~~ = 1, we have for each prime q ljJ(q) = 1q 1:-1 q I:''''

But

.

I q Iq I q leo = 1 by the theorem; hence

1.

16


Thus eq = eco , and our relation is simply a power of the one established before. The product formula has a simple interpretation in the classical case of the field of rational functions with complex coefficients. In this case

cp(a) = d number of zeros-number of poles = 1; so a rational function has as many zeros as poles. Now that the valuations of the field of rational numbers have been determined, we can find the best constant c for our axiom (3).

Theorem 10:

For any valuation, we may take c=

Proof.

max (\1 I , 121) •

(1) When the valuation is non-archimedean, c=I=lll~121·

(2) When the valuation is archimedean, k must have characteristic zero; hence k contains R, the field of rational numbers. The valuation is archimedean on R, and hence is equivalent to the ordinary absolute value; suppose that for the rational integers n we have I n I = n i3 • Write c = 21%; then

1a + b 1~ 21% max (I ai, 1b I) . By the method of Theorem 3, we can deduce I

al

+ ... + am 1 ~ (2m)1% max I a

As a special case of this we have

Now

smce

v 1•

6.

17

COMPLETION OF A FIELD

Hence we have \a

+ b \'" ~ (2(m + 1))" 2

m

/l(max (\ a I , \ b \))'" •

Taking the m-th root, and letting m ---+co, we obtain

1a + b \ ~ 2/l max (I ai, \ b\). Since our theorem is proved in this case also. Since the constant c for an extension field is the same as for the prime field contained in it, it follows that if the valuation satisfies the triangular inequality on the prime field, then it does so also on the extension field.

6. Completion of a Field Let I I be a valuation of a field k; replace II, if necessary, by an equivalent valuation for which the triangular inequality holds. A sequence of elements {a.} is said to form a Cauchy sequence with respect to I I if, corresponding to every E > 0 there exists an integer N such that for fL, " ?:: N, I a", - a. I < E. A sequence {a.} is said to form a null-sequence with respect to I I if, corresponding to every E > 0, there exists an integer N such that for" ?:: N, I a.1 < E. k is said to be complete with respect to I I if every Cauchy sequence with respect to II converges to a limit in k. We shall now sketch the process of forming the completion of a field k. The Cauchy sequences form a ring P under term wise addition and multiplication: {a.}

+ {b.} = {a. + by} ,

{ay}{b y} = {a.b.} .

It is easily shown that the null-sequences form a maximal ideal N in P; hence the residue class ring PIN is a field k. The valuation I I of k naturally induces a valuation on k; we still denote this valuation by II. For if a E k is defined by the residue class of PIN containing the sequence {a v }, we define I ex I to be limy~O') I a. I. To justify this definition we must prove (a) that if {a.} is a Cauchy sequence, then so is

{I a, I},

18

1.


=

(b) that if {a.} {b.} mod N, then lim I a.1 = lim I b.1 , (c) that the valuation axioms are satisfied. The proofs of these statements are left to the reader. If a E k, let a' denote the equivalence class of Cauchy sequences containing (a, a, a, ... ); a' E k. If a' = b', then the sequence «a - b), (a - b), ... ) EN, so that a = b. Hence the mapping cp of k into k defined by cp( a) = a' is (I, 1); it is easily seen to be an isomorphism under which valuations are preserved: I a' I = I a I. Let k' = cp(k); we shall now show that k' is everywhere dense in k. To this end, let ct be an element of k defined by the sequence {a.}. We shall show that for large enough values of v, I ct - a: I is as small as we please. The elemnt ct is defined by the Cauchy sequence {(at - a.), (a 2 - a.), ... }. and

a:

I

(X -

a; I = lim I al' - a. I; I'---CO

but since {a.} is a Cauchy sequence, this limit may be made as small as we please by taking v large enough. Finally we prove that k is complete. Let {a.} be a Cauchy sequence in k. Since k' is everywhere dense in k, we can find a sequence {a.'} in k' such that I a.' - (X. I < l/v. This means that {(a.' - ct.)} is a null-sequence in k; hence {a.'} is a Cauchy sequence in k. Since absolute values are preserved under the mapping cp, {a.} is a Cauchy sequence in k. This defines an element f3 E k such that lim ......., I a.' - f3 I = O. Hence lim ..... o:> I ct. - f3 I = 0, i.e. f3 = lim ct. , and so k is complete. We now agree to identify the elements of k' with the corresponding elements of k; then k may be regarded as an extension of k. When k is the field of rational numbers, the completion under the ordinary absolute value ("the completion at the infinite prime") is the real number field; the completion under the valuation corresponding to a finite prime p ("the completion at p") is called the field of p-adic numbers.

CHAPTER TWO

Complete Fields 1. Normed Linear Spaces Let k be a field complete under the valuation II, and let S be a finite-dimensional vector space over k, with basis WI' W 2 , " ' , W" • Suppose S is normed; i.e. to each element ex E S corresponds a real number II ex ", which has the properties

(1)

"ex" ~ 0; " ex "

(2) (3)

II Xex II

"ex

=

0 if and only if ex = 0,

+ ,8 " ~ " ex II + " ,8 " , =

I x I II ex II for all x E k.

(We shall later specialize S to be a finite extension field of k; the norm" " will then be an extension of the valuation I I.) There are many possible norms for S; for example, if

f3

=

XIWI

+ ... +

XnWn ,

then

II f3llo

= max v

I Xv I

is a norm. This particular norm will be used in proving Theorem 1: All norms induce the same topology in S. Proof: The theorem is obviously true when the dimension n = 1. For then,8 = Xw and" ,8 " = I x III W " = c I x I (c =1= 0); hence any two norms can differ only in the constant factor c; this does not alter the topology. We may now proceed by induction; so we assume the theorem true for spaces of dimension up to n - 1. 19

2.

20

COMPLETE FIELDS

We first contend that for any I: > 0 there exists an 7) > 0 such that II (X II < 7) implies I x" I < 1:. For if not, there is an I: > 0 such that for every 7) > 0 we can find an element (X with II (X II < 7), but I x" I ~ 1:. Set f3 = (X/x,,; then and

Thus if we replace 7) by 7)1:, we see that for every 7) > 0 we can find an element f3 of this form with II f311 < 7). We may therefore form a sequence {f3v}:

with

II f3v II < 1/". Then

and

By the induction hypothesis, the norm II lion the (n - 1)dimensional subspace (wI' ... , w"-I) induces the same topology as the special norm II 110' That is,

II fJ. - f3,.11 small ~ II f3v - f3,. 110 small => Iy~.)

- y~,.)

I

small if ", fL are large => {y(v)} is a Cauchy sequence in k. Since k is complete, there exist elements E k such that

z,

Zi

= lim {y~.)} .

Set Then

+ ... + (Z"-l - y~~l) wn-lll yi III Wl II + ... + I Z,,-l - y~~l III W,,_l II
I (1 + a)2 + b 1 ~ c,

or, equivalently, that I a / :::;;; D for some D. Suppose that this is not the case; then for some a = a + ib, 11 + b2 /a 2 / :::;;; / I/a 2 / is arbitrarily small. Thus / x2 + I / takes on arbitrarily small values. We construct a sequence {xv} in k such that Then

or

hence one factor :::;;; 1/2'. We now adjust the signs of the {xv}; suppose this has been dome as far as xv; then we adjust the sign of X V+1 in such a way that / Xv - xV+l / :::;;; 1/2-. This adjusted sequence is a Cauchy sequence, for

I Xv -

x-+/'

I ~ / Xv -

Xv+!

I + I Xv+! -

I + ...

Xv+2

Since k is complete, this sequence has a limit j

E

k. Then

j2 + 1 = lim x~ + 1 = 0, '"

....

contradicting our hypothesis that proof.

V-l 1= k.

This completes the

24

2.

COMPLETE FIELDS

3. Archimedean Case The following theorem now completes our investigation in the case of complete archimedean fields. Theorem 4: The only complete archimedean fields are the real numbers and the complex numbers. Proof: Let k be a field complete under an archimedean valuation. Then k has characteristic zero, and so contains a subfield R isomorphic to the rational numbers. The only archimedean valuations of the rationals are those equivalent to the ordinary absolute value; so we may assume that the valuation of k induces the ordinary absolute value on R. Hence k contains the completion of R under this valuation, namely the real number field P; thus E = k(t) contains the field of complex numbers P(t)' We shall prove that E is in fact itself the field of complex numbers. We can, and shall, in fact, prove rather more than this-namely, that any complete normed field over the complex numbers is itself the field of complex numbers. In a normed field, a function II II is defined for all elements of the field, with real values, satisfying the following conditions:

II ex II ~ 0; II ex II = 0 if and only if ex = 0, (2) II ex + .811 ~ II ex II + 11.811, (3) II Zex II = I z III ex II for z EP(t), (4) II a;B II ~ II ex II . 11.811· (l)

We shall show that any such field E = P(t). The proof can be carried through by developing a theory of "complex integration" for E, similar to that for the complex numbers; the result follows by applying the analogue of Cauchy's Theorem. Here we avoid the use of the integral, by using approximating sums. Given a square (ZI , Z2 , Zs , Z4) in the complex plane, having c as center and as midpoints of the sides, we define an operator La by

'1' '2' 'S ,'4

3.

c

z,

25

ARCHIMEDEAN CASE

•

t,

It is easily verified that L is a linear homogeneous functional, such that L(z) = L(l) = 0, and hence vanishing on every linear function. 0 (f3 E E). First we show that IJx is continuous for x = f3 Let gEE and II gil < II f3- 1 11-1; then II gJf311 ~ II g IIII f3- 1 II < 1. Since II (gJf3)" II ~ II gJf3 11", the geometric series IJf3 ~~ g.J{3" converges absolutely; hence it converges to an element of E which is easily seen to be IJ(f3 - g). Now take II g II ~ til f3- 1 11-1. We have

"*

so that

I ~ ~ ~ - ~ II ~ II ~ II ~ II ~ II" II ~-111" ~ II ~ 1111 ~1 11

2 ,

which can be made as small as we please by choosing II g II small enough. This is precisely the condition that 1Jx be continuous at

x = f3.

26

2.

COMPLETE FIELDS

Suppose there exists an element ex of E which is not a complex number; then lj(z - ex) is continuous for every complex number z, and since

I z ~ 0: II =

rh 111 1

the function Ij(z - ex) approaches zero as

II z

1 0:

0: z

II,

I z 1-00 . The function

I

is continuous for every z on the Gauss sphere, and hence is bounded: say IIlj(z - ex) II < M. We have now L (

1 )

a ~

L - z

a

=

-L

-

+ (2c -

(c _ 0:)2

0:)

+ (z -

(z - C)2 0:) (c _ 0:)2

(z - C)2

a (z - 0:) (c - 0:)2 ,

since La vanishes on linear functions. Thus

where 8 is the length of the side of the square. Next consider a large square Q in the complex plane, with the origin as center. If the length of the side of Q is I, we can subdivide Q into n2 squares Q. of side lin. Let us denote by Z. the vertices, and by 3. the mid-points of the sides of the smaller squares, which lie on the sides of the large square Q. Then La

=

!.

coniour

(ZV+l - Z.)f(3 v)

3.

27

ARCHIMEDEAN CASE

is an approximating sum to the "integral" of J(z) taken round the contour formed by the sides of Q. We see easily that

Using the estimate for La [l/(z - ex)] we find

•

so that for a fixed

Q and n --00 we have

La (_1_)_0. z-rx Since 1 z -

1

rx -

Z =

1 Z2 •

rx

1 - rxJz '

we obtain

Therefore

~ IZ.+1-Z.I-1112A II L'(~)-LI(~)II~ z rx z contour 3. ~

2A "'" ---z2 "-t I Z.+1 contour

Z.

I=

8A

-1- .

Now L' (

!)- I

a

~=

27Ti

as n __ 00. Hence 27T ~ 8A/I, which is certainly false for large l. Thus there are no elements of E which are not complex numbers; so our theorem is proved. We see that the only archimedean fields are the algebraic number fields under the ordinary absolute value, since only these fields have the real or complex numbers as their completion.

28

2.

COMPLETE FIELDS

4. The Non-Archimedean Case We now go on to examine the non-archimedean case. Let k be a complete non-archimedean field, and consider the polynomial ring k[x]. There are many ways of extending the valuation of k to this ring, some of them very unpleasant. We shall be interested in the following type of extension: let I x I = c > 0, and if ~(x) =

ao + a1x

+ ... + a"x" E k[x],

define

\ ~(x) \ = max \ a"x'\

-

= max c-\ a,\.

-

The axioms (1) and (3) of Chapter I can be verified immediately. To verify axiom (2) we notice first that \ ~(x) r/J(x) \ ~ \ ~(x)

II r/J(x) \,

slllce

I( ~

a_bl") Xk

I"+v=k

I~ max I aixi I max I b;x; \ •

Next we write cp(x) = CPl(X) + CP2(X) where CPl(X) is the sum of all the terms of cp(x) having maximal valuation; ICP2(X) I < I CPl(X) I . Similarly we write rP(x) = rPl(X) + rP2(X). Then ~(x)

r/J(x) = ~l(X) r/Jl(x)

+ ~l(X) r/J2(X) + ~2(X) r/Jl(x) + ~2(X) r/J2(X).

We see at once that the last three products are smaller in valuation than I CPl(X) I I rPl(X) I; and of course

I~l(X) r/Jl(X) \ ~

\ ~l(X) \\ r/Jl(X) \.

The term of highest degree in CPl(X) rPl(X) is the product of the highest terms in CPl(X) and in rPl(X); so its value is I CPl(X) II rPl(X) I. Therefore I CPl(X) rPl(X) I = Icpl(x) II rPl(X) I. Using the non-archimedean property we obtain \ ~(x) r/J(x) \ =

I ~l(X) \\ r/Jl(x) \ =

\ ~(x)

II r/J(x) \.

Thus we have, in fact, defined an extended valuation.

4.

29

THE NON-ARCHIMEDEAN CASE

We now prove the classical result, known as Hensel's Lemma, which allows us, under certain conditions, to refine an approximate factorization of a polynomial to a precise factorization.

Theorem 5: (Hensel's Lemma). Let f(x) be a polynomial in k[x]. If (1) there exist polynomials cp(x), r/J(x) , h(x) such that f(x) = .p(x) r/J(x)

+ h(x),

(2) cp(x) =1= 0 and has absolute value equal to that of its highest term, (3) there exist polynomials A(x), B(x), C(x) and an element d E k such that

+ B(x) r/J(x) = d + C(x), I r/J(x) I, I B(x) I ~ 1, I C(x) I < I d I ~ 1, I h(x) I < I d 121.p(x) I; A(x) .p(x)

(4)

then we can construct polynomials tl>(x) , P(x)

I cti(x) -

.p(x)

E

k[x] such that

f(x) = cti(x) 1J'(x)

(1)

deg cti(x) = deg .p(x)

(2)

I < I d 11.p(x) I;

11J'(x) - r/J(x)

I < Id I

(3)

Proof: As a preliminary step, consider the process of dividing a polynomial

by

where by hypothesis, 1cp(x) 1 = 1 anxn I. The first stage in the division process consists in writing gl(X)

Now

= g(x) -

b .p(x) ~ xm-n. an

2.

30

COMPLETE FIELDS

hence we have \ gl(X) \ ~ \ g(X) \. We repeat this argument at each stage of the division process; finally, if g(x) = q(x) cp(x) r(x), we obtain \ r(x) \ ~ \ g(x) \. As another preliminary we make a deduction from the relation

+

A(x) cp(x)

+ B(x) r/J(x) = d + C(x)

(*)

We see that this yields \ A(x)CP(x) \ ~ max {\ B(x) r/J(x) \, \ d \' \ C(x) \} ~ max {I, \ d \, I d \}

using the given bounds for \ B(x) \, \ ,p(x) \, \ C(x) \. Since \ d \ ~ 1, we have \ A(x) cp(x) \ ~ 1. Now multiply the relation (*) by h(x)/d; we obtain

A(x~h(x) cp(x)

+ B(x~h(x) r/J(x) =

h(x)

+ C(x~h(x) •

Now we write

B(x~h(x) = q(x)CP(x) + f11(X), C(X~h(X)

= ql(X)CP(X)

+ k(x).

Then

(A(x~h(x) + q(x) r/J(x) -

ql(X») cp(x)

+ f11(X) r/J(x) =

h(x)

+ k(x)

or, briefly,

cx1(x) cp(x)

+ f11(X) r/J(x) = h(x) + k(x).

We now give estimates of the degrees and absolute values of the polynomials we have introduced. We have, immediately, deg fJl(X), deg k(x) < degcp(x). Further, deg (cp(x) r/J(x»

~

max {degf(x), deg h(x)},

and deg (cx1(x)CP(x»

~

max {deg (fJl(X) r/J(x», deg h(x), deg k(x)}

~

max {deg (cp(x) r/J(x», deg h(x), degcp(x)} max {degf(x), deg h(x)}.

~

4.

31


Referring now to our preliminary remark about division processes. we have

1Pt(X) 1~ I B(i~~(X) 1~ 1~~? 1 < I d 114>(X) I ~ 14>(X) I. 1k(x) I ~ 1C(i~~(X) 1~ 1h(x) I. Hence

I ~(X) 4>(X) I ~ max {I Pt(X) r/J(X) I. / h(x) /. 1k(x) j} ~ max {j Pt(X)

I, I h(x) I , I k(x) j}

I

\ 1h(x) 1 1 h(x) 1 1d 1 ,I h(x) 1• 1h(x) 1 = -I-d-I

~ max I and

I h(x) I I ~(x) I ~ 1d 114>(x) I ~ I d I ~ 1. Now we define

4>t(x) =

4>(x)

Then since 1fJl(X)

+ Pt(x)

and

r/Jt(x) = r/J(x)

I < I cp(x) I. and deg fJl(X) < l4>t(x) 1= 14>(x) I = 1a"x" 1,

+ cxt(x).

deg cp(x), we have

and

also

1r/Jt(x) 1~ max {I r/J(x) I, I CXt(x) I} ~

1

and deg (4)t(x) r/Jt(x)) ~

max {deg (4)(x) r/J(x)), deg (~(x) 4>(x)), deg (r/J(x) Pt(x)), deg (cxt(x) Pt(x))}

~

max {degf(x), deg h(x)}.

We shall show first that CPl(X) rPl(X) is a better approximation to f(x) than is cp(x) rP(x)j and then that the process by which we obtained

2.

32

COMPLETE FIELDS

CPl(X), rPl(X) may be repeated indefinitely to obtain approximations which are increasingly accurate. Let f(x) - CPl(X) 1/Jt(X) =

~(x).

Then h1(x) = h(x) - (h(x)

+ k(x)) -

O!&x) f31(X) = - k(x) - CXl(X) f31(X).

Hence deg h1(x)

~

max {deg k(x), deg O!t(x) f31(X)}

~


Also,

I h1(x) I ~ max {I k(x) I , I cxl(X) f31(X) I} ::;::

I C(x) h(x) I I h(x) I I h(x) I I I d I 'I d II cp(x) I I d I Ie I h(x) I,

--..:: max =

where we write Ie

=

I

I

I C(x) I I h(x) \ max -I-d- I '\ d 121 cp(x) \ ;

I

x < 1 by the conditions of the theorem. Thus CPl(X) rPl(X) is a better approximation than cp(x) rP(x). Let A(x) CPl(X) + B(x) IPt(X) = d + C1(x). Then C1(x) = C(x)

+ A(x) f31(X) + B(x) O!t(x).

Hence \ C1(x) I ~ max {I C(x) \ , \ A(x) f31(X) \, \ B(x) O!t(x) =

max II C(x)

I

!

I}

I \ A(x) cp(x) II f31(X) \ I B(x) \\ O!t(x) cp(x) II ' Icp(x) I ' Icp(x) I I h(x) I

I h(x) I

I

~ max \ C(x) I , I d II cp(x) I ' I d II cp(x) I = Ie I d I < I d I using the results obtained above.

4.


33

We have now recovered the original conditions of the theorem, stated now for cPt(x) , ,pt(x), ht(x) and Ct(x). Thus the whole process may be repeated, yielding new polynomials a2(x), [32(X), h2(x), CP2(X), ,p2(X). We shall also obtain the estimate

I h2(X) I ~ Kl 1 h1(x) I , where

It is clear that we may now proceed indefinitely, obtaining sequences of polynomials {a.(x)}, (f3.(x)}, {h.(x)}, {cp.(x)}, {,p.(x)} where CP. = '"

+ Pt + + ... + P. ) (32

,p. = r/J + ~ + (X2 + ... + (X.

and

f - CP.r/J. =

h••

Now

I h.(x) I ~ K'I h(x) 1- 0 as v-oo and deg h.(x) ~ max {degf(x), deg h(:c)};

ICP.(x) I = Icp(x) 1 •

Also

as v-oo; and

as v-oo. From these considerations it follows that {cp.(x)} and {,p.(x)} are Cauchy sequences of polynomials; these polynomials are of bounded degree, since for every v we have deg CP.(x)

=

deg cp(x),

2.

34

COMPLETE FIELDS

and deg (cp,(x) r/J,(x))

~


Let cti(x)

=

lim {cp.(x)}

and

P(x)

= lim {r/J.(x)};

these limit functions are polynomials, and deg cti(x) = deg cp(x). Finally, f(x) - cti(x) P(x) = lim {h.(x)}

thus f(x)

= 0;

= JL2 > ... ). Suppose Iv joins the points of the Newton diagram corresponding to the (iv_l)-th and (iv)-th terms of J(x). Then we have just seen how to construct polynomials cpv(x) of degree iv, whose roots are all the roots y/v) of J(x) for which ord yp(v) ~ JL, . Obviously ord y/v) > JLV+1' so that yp(v) is also a root of CPv+1(x); hence cpv(x) divides cpv+1(x). We see also that the roots y ofJ(x) for which ord y = JLV+1 are those which are roots of CPv+1(x) but not of cpv(x).

6. The Algebraic Closure of a Complete Field Let k be a complete non-archimedean field, and let C be its algebraic closure. Then we extend the valuation of k to C by defining I 0: I for 0: E C to be I 0: I as defined previously in the finite

2.

44

COMPLETE FIELDS

extension k( a} The verification that this is in fact a valuation for C is left to the reader; it should be remarked that the verification is actually carried out in subfields of K which are finite extensions of k. C is not necessarily complete under this extended valuation. For instance the algebraic closure of the field of dyadic numbers does not contain the element

The valuation induces a metric in C, and since the valuation is non-archimedean, we have the stronger form of the triangular inequality: 1ex - fJ 1 ~ max {I ex I, 1fJ I}. Spaces in which this inequality holds are called by Krasner ultrametric spaces. Consider a triangle in such a space, with sides a, b, c. Let a = max (a, b, c); then, since a ~ max (b, c) = b, say, we have a = b, and c ~ a. Thus every triangle is isosceles, and has its base at most equal to the equal sides. The geometry of circles in such spaces is also rather unusual. For example, if we define a circle of center a and radius r to consist of those points x such that 1 x - a 1 < r, it is easily seen that every point inside the circle is a center. We now use this ultrametric geometry to prove: Theorem 8:

Let ex

E

C be separable over k, and let

r = min I a(ex) a¢l

ex

I,

where the a are the isomorphic maps of k(ex). Let fJ be a point, i.e. an element of K, inside the circle with center ex and radius r. Then k( ex) C k(fJ).

Proof:

Take k(fJ) as the new ground field; then f(x)

= Irr (ex, k, x)

is separable over k(fJ). If cf;(x) = Irr (ex, k(f3), x), then cf;(x) 1 f(x). Let a be any isomorphic map of k(ex, f3) k(fJ). Since 1

a(f3 - ex)

=

f3 - a(ex),

and conjugate elements have the same absolute value (they have the same norm) we deduce that 1fJ - a(ex) 1= 1fJ - ex 1 < r. Consider the triangle formed by fJ, ex, a(ex); using the ultra metric

6.

THE ALGEBRAIC CLOSURE OF A COMPLETE FIELD

45

property we have as above I (X - a«(X) I ~ I fJ - (X I < r, i.e. a«(X) lies inside the circle. Hence a«(X) = (x, and so, since (X is separable, the degree [k«(X, fJ) : k(fJ)] is equal to 1, and k«(X) C k(fJ). Let J(x) be a polynomial in k[x] with highest coefficient 1. In C[x], we have J(x) = (x - (Xl) ... (x - (Xn)' Suppose that in the valuation of k[x] induced by taking I x I = 1, we have IJ(x) I ~ A, where A ~ 1. Then if (X E C, and I (X I > A, we see that (Xn is the dominant term in

cannot be a root of J(x) = O. Hence if IJ(x) I ~ A and (Xn are the roots of J(x) = 0, then I (Xv I ~ A. Consider now two monic polynomials J(x), g(x) of the same degree, n, such that IJ(x) - g(x) I < €. Let fJ be a root of g(x), (Xl' " ' , (Xn the roots of J(x). Then Hence

(X

(Xl' " ' ,

f(f3) = f(f3) - g(f3),

If(f3) I = If(f3)

and

- g(f3) I ~ €An,

where A is the upper bound of the absolute values of the coefficients, and hence of the roots, of J(x) and g(x). Hence

I f3 -

(Xl

II f3 -

and so one of the roots

(Xi ,

(X21 ...

say

(Xl

I f3

-

(Xn

I ~ €An,

must satisfy the relation

1f3-(Xll~A~.

Thus by suitable choice of €, each root fJ of g(x) may be brought as close as we wish to some root (Xi of J(x). Similarly by interchanging the roles ofJ(x) andg(x), we may bring each root (Xi ofJ(x) as close as we wish to some root of g(x). Let us now assume that € has been chosen such that every fJ is closer to some (Xi than min I (Xi - (Xj I; (Xi =F- (Xj in this way the fJ's are split into sets "belonging" to the various (Xi • Suppose for the moment that J(x) is irreducible and separable. Then, since I fJ - (Xi I < min I (Xi - (Xi I, the preceding theorem gives k(fJ) :) k«(X); but J(x) and g(x) are of the same degree, whence Theorem 9: If J(x) is a separable, irreducible monic polynomial of degree n, and if g(x) is any monic polynomial of degree n such that IJ(x) - g(x) I is sufficiently small, then J(x) and g(x)

2.

46

COMPLETE FIELDS

generate the same field and g(x) is also irreducible and separable. If f(x) is not separable, let its factorization in C[x] be f(x)

with distinct

()(i •

=

(x -

()(lYl

(x -

()(2Y' ... (x

-

()(,Y'

In this case we can establish the following result:

Theorem 10: If g(x) is sufficiently close to f(x), then the number of roots f3i of g(x) (counted in their multiplicity) which belong to ()(I IS )II •

Proof: If the theorem is false, we can construct a Cauchy sequence of polynomials withf(x) as limit for which we do not have )Ii roots near ()(i • From this sequence we can extract a subsequence of polynomials for which we have exactly I-ti roots near ()(i (l-ti =F- )Ii for some i). Since k is crimplete, the limit of this sequence of polynomials is f(x), and the limits of the sets of roots near ()(t are the ()(i • Hence we have (x -

()(lYl

(x -

()(2YI •••

(x - ()(rY' = f«x) = (x -

~tl

(x - ()(2Y'" ..• (x - ()(rY" •

This contradicts the unique factorization in C[x], so our theorem is proved. In a similar manner we can prove the result Theorem 11: If f(x) is irreducible, then any polynomial sufficiently close to f(x) is also irreducible. Proof: If the theorem is false, we can construct a Cauchy sequence of reducible polynomials, with f(x) as limit. From this sequence we can extract a subsequence {gn(x)}: gn(x) = hn(x) mn(x) for which the polynomials hn(x) have the same degree, and have their roots in the same proximity to the roots of f(x). Then the sequence {hn(x)} tends to a limit in k[x], whose roots are the roots of f(x). This contradicts the irreducibility of f(x). N ow although k is complete, its algebraic closure C need not be complete; the completion of C, C, is of course complete, but we can prove more:

7. Theorem 12:

47

CONVERGENT POWER SERIES

C is

algebraically closed.

Proof: We must consider the separable and inseparable polynomials of C[x] separately. (1) Letf(x) be a separable irreducible polynomial in C[x]. The valuation of k can be extended to a valuation for the roots of f(x). We can then approximate f(x) by a polynomial g(x) in C[x] sufficiently closely for the roots of f(x) and g(x) to generate the same field over C. But g(x) does not generate any extension of C; hence the roots of f(x) lie in C. (2) If the characteristic of k is zero, there are no inseparable polynomials in C[x]. Let the characteristic be p #- 0; if ex is an element of C, ex is defined by a Cauchy sequence {a.} in C. But if {a.} is a Cauchy sequence, so is {a// P }, and this sequence defines ex l / P , which is therefore in C. Hence there are no proper inseparable extensions of C. 7. Convergent Power Series Let k be a complete field with a non-trivial valuation I I. Convergence of series is defined in k in the natural way: ~:m a. is said to converge to the sum a if for every given € > 0 we have I ~:,. a. - a I < € for all sufficiently large n. The properties of the ordinary absolute value which are used in the discussion of real or complex series are shared by all valuations I I. Hence the arguments of the classical theory may be applied unchanged to the case of series in k. In particular we can prove the Cauchy criterion for convergence, and its corollaries: 1. The terms of a convergent series are bounded in absolute value. 2. If a series is absolutely convergent (i.e. if ~:m I a. I converges in the reals) then the series is convergent. Let E be the field of all formal power series with coefficients in k: E consists of all formal expressions f(x) = ~.: a.x·, with a. E k, where only a finite number of the terms with negative index v are non-zero. We define a valuation I Ix in E such that I x Ix < 1, and I Ix is trivial on k. If an element is written in the form ~'" a • XV 00 with avo #- 0, then I ~.o avx' Ix = I X' o Ix = (I x Ix)vo. This ~

2.

48

COMPLETE FIELDS

valuation I Ix on E and the valuation I I on k are therefore totally unrelated. An element f(x) = ~ a_x' in E is said to be convergent for the value x = C E k (c #- 0), when ~ ave' converges in k. We shall say simply that f(x) is convergent if it is convergent for some c #- O. Theorem 13: If f(x) is convergent for x = c #- 0, then f(x) is convergent also for x = dE k, whenever I d I < I c I.

Proof: Let I c I = lja. Since ~ ave' is convergent, we have I a_ I : : :; Ma·. If I d I < I c I, then

I a_d-I < Ma-I d I' =

I ave I : : :; M, V

whence

M(a I d I)"·

Since a I d I < 1, the geometric series ~M(a I d I)' is convergent. Thus ~ I a_d' I is convergent; since absolute convergence implies convergence, this proves the theorem. Now let F be the set of all formal power series with coefficients in k which converge for some value of x E k (x #- 0). It is easy to show that F is a subfield of E. Notice, however, that F is not complete in I Ix. Theorem 14: F is algebraically closed in E.

Proof: Let e = ~:m a.x· be an element of E which is algebraic over F. We have to prove that e lies in F. It will be sufficient to consider separable elements e. For if e is inseparable and p#-O is the characteristic, then ep ' is separable for high enough values of r. But ep' = ~:. a'P'x vp', and the convergence of this series implies the convergence of e. Let f(y) = Irr (e, F, y); we shall show that we may confine our attention to elements e for which f(y) splits into distinct linear factors in the algebraic closure A of E. Let the roots of f(y) in A be e1 = e, e2 , ... , en . Since the valuation I Ix of E can be extended to A, we can find I ei Ix. Let T~r

I e1 - e. I.: = s.

Now set N

y=z+ ~>.x" -m

7.


49

Then N

fey) = / (z

+ 'Z a.-x

v

)

= g(z)

-m

is an irreducible polynomial in z with roots \ = (}i - ~: a.,x"; in particular '\ = ~:+l a.,x·, and if Al is convergent, so is (}1 = (). Obviously and

for i > 1. Hence if N is large enough, I Al Ix will be as small as we please, so that we have I \ Ix ~ 8 for i > 1. Now consider the equation whose roots are ILi = \/x N ' The root ILl = Al/XN is a power series of the form ~~ a.,xv , hence I ILl Ix ~ I x Ix; the other roots may have absolute values as large as we please by suitable choice of N. Since the convergence of ILl obviously implies the convergence of AI' and hence of (}l , it follows that we need only consider elements () which are separable over F, and whose defining equationf(y) has roots (}1 = (), (}2 , •.. , (}n in A with the property that I (}l I ~ I x Ix and I (}i Ix > 1 for i > 1. We may further normalize f(y) so that none of the power series appearing as coefficients have terms with negative indices, while at least one of these series has non-zero constant term, i.e.

where at least one bl'o #- O. If we draw the Newton diagram of f(y), it is obvious from the form of the coefficients that none of the points in the diagram lie below the x-axis, while at least one of the points lies on the x-axis; further, since f(y) is irreducible, its Newton polygon starts on the y-axis. The first side of the Newton polygon of f(y) corresponds to the roots y off(y) for which ord y is greatest, and it has slope - ord y. Only (}1 = () has the maximum ordinal; hence the first side of the polygon joins the first and second points in the Newton diagram, and its slope is - ord (}1 < O. The other sides of the polygon correspond to the remaining roots (}i' and have slopes - ord (}i > O. Hence the Newton diagram has the form shown:

2.

50

COMPLETE FIELDS

/"

"

•

". /

~

.

~------------------------------------

Thus

where bio =1= O. Since y = () is a root of f(y) , we have

whence

But we know thatf(y) has a root given by 00

f)

=

I-a"x·.

_1

7.


51

From these two formulas we obtain a recursion relation: (1)

These polynomials !fm have two important properties: (a) They are universal; i.e. the coefficients do not depend on the particular ground field k. (b) All their coefficients are positive. since the series :E c""xv are convergent, we have all (x" E k having I (x" I < some fixed l/a". Since the number of series occurring as coefficients in f(y) is finite, we can write I c"v I < Ma', where Now

I c"v(X/ I < M" for

and

We now go over to the field of real numbers, ko , and the field of convergent power series Fo over k o . Let cP be a root of the equation 4>2 =

1 MaVxV+ (1 Mavxv) 4> + (1 Mavxv) 4>2 + ... 1

1

(2)

1

where the sum on the right is infinite: 4>

= M ax + M ax 4> + M ax 4>2 + ... , I-ax

I-ax

I-ax

whence

.J._1.±~1. 2 4 -M - - ax -·

'f' -

1- ax

The roots are analytic functions of x near zero; one of them, which we call CPl' vanishes at x = O. cPI may be expanded as a power series convergent in a certain circle round the origin, say CPI = :E~ (X"x'. We shall now show that all the coefficients (Xv are positive, and I a,1 ~ (Xv' We proceed by induction; (Xl is easily

52

2.

COMPLETE FIELDS

shown to be positive and I a l I < CXI ; hence we assume cx, ~ 0 and I ai I < cx, for i ::::;; m. Since CPI satisfies the equation (2). its coefficients satisfy

Since only positive signs occur in !fm. we have am +l we already have expression (1) and hence

CXm +1

Again using the fact that only positive signs occur in

~

O. For

!fm • we have

Hence I a. I ::::;; cx., where the cx. are coefficients of a convergent power series. Thus () = :E a"x" is convergent. This completes the proof of !he theorem.

CHAPTER TH REE

e, f and n

1. The Ramification and Residue Class Degree The value group of a field under a valuation is the group of non-zero real numbers which occur as values of the field elements. From now on, unless specific mention is made to the contrary, we shall be dealing with non-archimedean valuations. For these we have:

Theorem 1: If k is a non-archimedean field, then k has the same value group as k.

k its completion,

Proof: Let ex a non-zero element of k; ex is defined by a Cauchy sequence {a v} of elements of k, and / ex / = lim / av /. The sequence {/ a v /} converges in a quite trivial way: All its terms eventually become equal to / ex /. We have

I al' I = I 0: + (al' - ll:) I ~ max (Ill: / , I al' - l : J) = Ill: I since / ex / =t- 0 and / a" - ex / can be made as small as we please, in particular less than / ex / by choosing fL large enough. Thus / ex / = / al' / for large enough fL. This proves the theorem. Now let k be a non-archimedean field, not necessarily complete; and let E be a finite extension of k. If we can extend the valuation of k to E, we may consider the value group $ E of E. Then the value group $k of k is a subgroup, and we call the index e = ($ E : $k) the ramification of this extended valuation. Consider the set of elements ex E k such that / ex / < 1. It is easily shown that this set is a ring; we shall call it the ring of integers and denote it by o. The set of elements ex E 0 such that / ex / < 1 53

54

3. e,

f

AND

n

forms a maximal ideal p of 0; the proof that p is an ideal is obvious. To prove it is maximal, suppose that there is an ideal a #- p such that pea C o. Then there exists an element (X E a which is not in p; that is I (X I = 1. Hence if (3 E 0, (31(X E 0 also, and (3 = (X ((3I(X) Ea. That is, a = o. Hence p is a maximal ideal and the residue class ring o/p is a field, which we denote by 'k. If k is the completion of k, we can construct a ring of integers 0, a maximal ideal p, and a residue class ring o/p = k. We have the result expressed by: Theorem 2:

There is a natural isomorphism between 'k and k.

We have seen that if (X E 0, then (X = lim a v E 0 and v large enough. Thus 0 is the limit of elements of 0; i.e. 0 is the closure of o. Similarly p is the closure of p. Consider the mapping of o/p into o/p given by a p_ a p. This mapping is certainly well-denned, and is an "onto" mapping since, given any (X E 0, we can find an a E 0 such that I (X - a I < 1; then (X p = a p, which is the image of a p. The mapping is (1, 1) since if a, b E 0 and a = b mod p we have I a - b I < 1; hence a = b mod p. It is easily verified that the mapping is homomorphic; hence our theorem is completed. From now on we shall identify the residue class fields 'k, k under this isomorphism. N ow let E be an extension field of k, with ring of integers D and prime ideal ~. Then D :) 0 = D () k and ~ :) p = ~ () k. Proof:

I I = I a v I for (X

+

+

+

+

+

Theorem 3: There is a natural isomorphism of the residue class field 'k onto a subfield of the residue class field E.

+ +

+

Proof: Consider the mapping a p- a ~ (a E k). This is well-defined since a p ~ = a ~; it is easily seen to be a homomorphism of k into E. Finally, the mapping is (1, 1) onto the image set, for

+ +

a

+ ~ = b + ~ => I a - b I < 1 => a + p = b + p.

We shall now identify 'k with its image under this isomorphism and so consider E as an extension field of k. We denote the degree of this extension by [E: k] = f.

1.

THE RAMIFICATION AND RESIDUE CLASS DEGREE

...

55

Let WI' W Z ' "', Wr be representatives of residue classes of E which are linearly independent with respect to k; that is, if there exist elements Cl' C2 , " ' , Cr in 0 such that C1Wl + ... + CrW r lies in ~, then all the Ci lie in p. Consider the linear combination C1Wl + ... + CrW r , with Ci E 0; if one of the Ci , say C1 , lies in 0 but not in p, then C1Wl + ... + CrW r =1= 0 mod ~, and hence

Now consider the linear combination d1w 1 + ... + drw r , with di E k, and suppose d1 has the largest absolute value among the di . Then

I diw i + ... + drwr I = I dl II WI + ~: W z + .. ' + ~: Hence if then

WI' • ", Wr

Wr

I= Idll .

are linearly independent with respect to k,

Let now 7Tl ,7T2' "', 7Ts be elements of E such that l7Tl I, "', l7Ts l are representatives of different cosets of 5D E/5Dk • We shall use these Wi , 7Tj to prove the important result of

Theorem 4: If E is a finite extension of k of degree n, then

ef::::;: n. Proof:

Our first contention is that

,.,. I c.,.7T.1 • Ik,.,. c.,.W,.7T.1 = max Certainly

and since the 1 7T i 1 represent different co sets of 5D E/5Dk' the 1 :E,. cww,, 7T i 1 are all different. Hence

as required.

3. e, j

56

It follows that the elements

AND

WI'TT.

n

are linearly independent, since

~ CI'.WI'TT. = 0 ~ \ ~ C"vW"TT.\ = 0 => max \ TT.C"v \ = 0 => c"v = O. Hence TS ~ deg (E I k). Thus if deg (E I k) = n is finite, then e and j are also finite, and ~ n. The equality ej = n will be proved in the next section for a very important subclass of the complete fields, which includes the cases of algebraic number theory and function theory. Nevertheless, the equality does not always hold even for complete fields, as is shown by the following example. Let R2 be the field of 2-adic numbers, and let k be the completion of the field R 2(y'2, V"2, ~2, ... ). Then deg (k(V-l) : k) = 2, but the corresponding e and j are both 1. This is caused by the fact that we may write .r---1

v-I

2

- 2

= 1 + v2 + V"2 + ... mod 2.

This series is not convergent, but if we denote by the first n terms, and by Ui the i-th term, we have

\ V-1 -

Sn \

Sn

the sum of

< \ Un +1 \ •

2. The Discrete Case We have already introduced the function ord a = - loge 1 a I, 1. The set of values taken up by ord a for a #- 0, a E k forms an additive group of real numbers; such a group can consist only of numbers which are everywhere dense on the real axis or which are situated at equal distances from each other on the axis. Hence we see that the value group, which is a multiplicative group of positive real numbers, must be either everywhere dense, or else an infinite cyclic group. In this latter case the valuation is said to be discrete. In the discrete case, let TT be an element of k such that I TT I takes the maximal value < 1. Then the value group consists of the numbers 1TT I'; given any non-zero element (X E k, there is a positive or negative integer (or zero) v such that I (X I = I TT Iv. Then

c

>

2.

I a/1T l = 1 =

57

THE DISCRETE CASE

I; hence a/1T' and its inverse are both elements of 0, i.e. a/1T' is a unit € of o. For every a =1= 0 we have a factorization a = 1T v€, where I € I = I; hence there is only one prime, namely 1T. Since k and its completion 'h have the same value group, the same element 1T can be taken as prime for 'h; thus if a complete field 'h is the completion of a subfield k, the prime Jor 'h may be chosen in k. Now let k be a complete field under a discrete valuation. If a E k can be written as a = 1T v€, we shall take ord a = v; in other words, we select C = 1/11T I. We now suppose that for every positive and negative ordinal v an element 1Tv has been selected such that 11T, I = 11T lv-obviously 1T, = 1T would suffice, but we shall find it useful to use other elements. Suppose further that for each element in k we have selected a representative c in 0, the representative of the zero residue class being zero. Then we prove p

l1T v/a

V

Theorem 5: Let k be a complete field with discrete valuation. Every a E k can be written in the form a = ~: Cv1T., where n = ord a and Cn $. 0 mod p. Proof: When a = 0, there is nothing to prove, so we suppose =F O. Since ord a = n, we have I a I = l1Tn I, hence a/1Tn is a € mod p. unit €. Its residue class modulo p is represented by Cn Thus I € - Cn I < 1, whence I €1Tn - Cn1Tn I < l 1Tn I; I.e. a = €1Tn = Cn1Tn + a' where I a' I < l1Tn I. We may repeat the procedure with a', and so on, obtaining at the m-th stage a

=

where I a(m) I < l1Tm I; thus a(m) - 0 as m _00. This proves the theorem. If the representatives Ci and the 1T v are chosen once for all, then the series representation of a is unique. It is important to notice that the c. and 1T v are chosen from the same field k; thus the characteristic of the field containing the c. is the characteristic of k, not of k. We shall illustrate this remark by considering the valuation induced on the rational field R by a finite prime p. Any element a E R can be written as a = po (b/c) where band C are prime to p; the ring of integers 0 consists of

3. e,

58

f

AND

n

these elements a for which v ~ 0, and the prime ideal p consists of the a such that v > O. Let ~ be the completion of R under this valuation, 5, P the corresponding ring and prime ideal. Since 5/p = o/p, we can choose representatives for the residue classes in R, and in fact in o. The residue classes are represented by the elements blc E R where band c are prime to p; we can solve the congruence C(X 1 mod p in integers. Then bd is a representative of the residue class containing blc, since

=

.!!-. _ bd = c

b(l - cd) c

== 0 mod p.

Hence every residue class contains an integer, and a complete set of representatives is given by 0, 1, 2, "', p - 1. This set of elemen ts does not form a field since it is not closed under the field operations of R. Now every p-adic number (X can be written (X = ~: c.r; we see that the field of p-adic numbers, ~, has characteristic zero. It can be shown that the field ~ contains the (p - l)-th roots of unity; it is often convenient to choose these as representatives of the residue classes. This analysis of complete fields may also be used to prove that the field of formal power series k = F{x} over any field F is complete. Any element (X E k can be written as (X = ~; c.x·, Cv E F. The ring of integers 0, respectively the prime ideal p are made up of the elements (X for which n ~ 0, respectively n > O. We can choose x = 7T; and as representatives of o/p the elements of F; hence the completion of k consists of power series in x with coefficients in F; i.e. k is complete. Now let k be a complete field with discrete valuation; let E be a finite extension with degree n and ramification e. The finiteness of e shows that the extended valuation is discrete on E. Let II, 7T be primes in E, k respectively; then the value groups are ~E = {I II Iv}, ~k = {17T Iv}. Since (~E: ~k) = e, I II Ie = 17T I; hence 7T = € lIe. We shall find it more convenient to represent ~E as {17T vIIfJ I} where -00 ~ v ~oo and 0 ~ f.L ~ e - 1. Theorem 6: If the valuation is discrete, then ef = n. Proof: Let W l , W 2 , "', WI be elements of E which represent a basis of the residue class field EpI'kp. Thus the generic residue

2.

59

THE DISCRETE CASE

+ ... + cfwf,

class is represented by C1Wl + C2W 2 Then if 0: is an element of E, 0: can be written

with

Ci E

o.

Thus every element 0: E E can be expressed as a linear combination of the ef elements wJII' with coefficients in the ground field k. Thus the degree of the extension Elk is at most ef: n ~ ej. We have already seen that for all extensions, whether the valuation is discrete or not, n ~ ej. Hence n = ef and our theorem is proved. Theorem 7: The elements {wJII'} (p = 1, I) form a basis Dover o.

···,f;

f.L = 0, ... ,

e-

Proof: We have already seen, in the course of the last proof, that {wplIl'} form a basis. Let 0: be an integer of E. We can write 0: = ~!"p d!,pwpJII', and since I Wp I = 1, and the I JI!' I are all distinct, we have

Hence Now Thus

Now 7T was defined as having the largest absolute value less than 1. Hence 1/7T has the smallest absolute value greater than 1. Thus \ d...p

\

(x) if;(x) mod p, where ¢>(x) and relatively prime. For if f(x) t/J(x) are relatively prime, there exist polynomials A(x), B(x) such such that A(x) cfo(x) + B(x) .fi(x) = 1 mod p.

=

Hence A(x) cfo(x)

+ B(x) .fi(x) = 1 + C(x) ,

3.

THE GENERAL CASE

61

where I C(x) I < 1. And since f(x) == cp(x) !{lex) mod~, we have f(x) = cp(x) !{lex) + hex), with I hex) I < 1. By Hensel's Lemma, this situation implies a factorization of the irreducible polynomial f(x), which is impossible. Hence if f(x) splits modulo ~, it must do so as a power of an irreducible polynomial: f(x) = [P(x)]!'. But deg a = deg P(x), and deg 0: = degf(x), hence deg a divides deg 0: as required. Thus if a is any residue class of Ep , and 0: E E is any representative of a, then deg a divides deg 0:. But since deg 0: = [k(o:) : k] divides n, this implies that deg a can be divisible only by primes which divide n. Let Ep' be the separable part of Ep. The degree of Ep I Ep' is some power pv of the characteristic of k p . Iff = f'pv, then Ep' = k p(a), where deg a = f'; f' is divisible by all the primes dividing f except possible the prime p. Hence, by our preceding remarks, these primes must divide n. Finally, if a is an inseparable element, then p divides deg a; hence p must divide n. This completes the proof. Theorem 9: If deg (E I k) = q, where q is a prime not equal to the characteristic p of the residue class field, then ef = q.

Proof: Since q is not the characteristic of the residue class field, q does not lie in the prime ideal; hence I q I = 1. Furthermore, q is not the characteristic of k, and hence Elk is separable. Let E = k(o:), and let f(x) = Irr (0:, h, x) = xq + a1xq - 1 + ... + aq • We may apply the transformation x = y + al/q, since q is not the characteristic; f(x) assumes the form y'1 + b2yq-l + .. , + bq • Thus we may assume at the outset that a l = O. We remark that

If I 0: I does not lie in the value group mk , we have e ~ q, since 10: Iq E mk . Since ef:::;:; q, we obtain e = q andf = 1; hence ef = q. If, on the other hand, I 0: I lies in mk , we have I 0: I = I a I where a E k. We may write {:3 = a{:3, and {:3 will satisfy an equation with second coefficient zero; hence we may assume without loss of generality that I 0: I = 1. In the course of Theorem 8 we saw that f(x) either remains

3. e,

62

f

AND

n

irreducible modulo p or splits as a power of an irreducible polynomial; since q is a prime the only splitting of this kind must be j(x) == (x - c)q mod p. We show now that this second case is not possible: since aq -- cq , mod p, c ~ 0; hence the second term of (x -- c)q, namely qcq-\ is -¥= 0 mod p. This contradicts our assumption about f(x). Thus c.: satisfies an irreducible congruence of degree q; f ;::? q. Hence f = q, e = I, and ef = q. This proves the theorem. In order to prove that ef divides n, we shall require the Lemma: If E :J F :J k, then e(E I k) = e(E I F) e(F I k), and similarly for f, and hence for ef. The verification of this is left to the reader. It follows at once that for a tower of fields with prime degree unequal to the characteristic of the residue class field, ef = n. We shall now prove the main result. Theorem 10: ef divides n; the quotient characteristic of the residue class field.

IS

a power of the

Proof: We shall prove the result first for the case where E is a separable extension of k. Let deg (E I k) = n = qim, where q is a prime unequal to the characteristic of k and (q, m) = I. We assert that ef = qis where (q, s) = 1.

t

H

!

K

I

/E2

+ t

Q

E

G

I

/EI k

1

r

3.

THE GENERAL CASE

63

Let K be the smallest normal extension of k containing E, and let G be its Galois group; let H be the subgroup of G corresponding to E. Let Q be a q-Sylow subgroup of H, E2 the corresponding field. Now Q, considered as subgroup of G, may be embedded in a q-Sylow subgroup Q' of G; the corresponding field El is a sub field of E 2 • The degree of E2 I E is prime to q since Q is the biggest q-group of H. Similarly the degree of El I k is prime to q, since Q' is the biggest q-group in G. Now the degree of E2 I El is a power of q. which must be qi, for [E2: E1] [El: k]

=

[E2: k]

=

[E: k] [E2: E].

Now between the groups Q and Q' there are intermediate groups, each of which is normal in the one preceding and each of which has index q. Thus the extension E2 I El may be split into steps, each of degree q. The product ef for the extension E2 I El is equal to the corresponding product for each step of degree q. From this we see that ef = qi. But this is also the q-contribution of ef by the extension Elk, since the degree of E2 I E is prime to q. We obtain this result for every prime divisor of n unequal to the characteristic of k. Hence ef can differ from n only by a power of the characteristic. Since ef ~ n, we must have efo = n where o is a power of the characteristic. This proves the theorem for separable extensions. If Elk contains inseparable elements, let Eo be the largest separable part. Then n[E : Eo] must be a power of the characteristic of k; if this is non-zero it is also the characteristic of k. We have seen that ef[E: Eo] can be divisible only by this prime. Hence for the inseparable part we have again njef = a power of the characteristic of k. This completes the proof of the theorem. Corollary: If the residue class field has characteristic zero, then ef= n. We may write n = eft : 0 is called the defect of the extension.

CHAPTER. FOUR.

Ramification Theory

1. Unramified Extensions Let k be a complete field, C its algebraic closure. Let the corresponding residue class fields be k, C; under the natural isomorphism, h may be considered as a subfield of C. The canonical image in C of an integer 0: in C shall be denoted by Ii; that of a polynomial cp{x) in C[x] by g;(x). A given polynomiall{l(x) in C[x] is always the image of some polynomial cp{x) of C[x]; cp(x) may be selected so that it has the same degree as I{l{x). If the leading coefficient of I{l(x) is 1 we may assume that the leading coefficient of cp{x) is also 1. In the sequel these conventions about the degree and leading coefficients will be tacitly assumed. Let I{l(x) = ;P(x) be an irreducible polynomial in k[x], with leading coefficient 1. We may factor cp(x) in C[x]: o/(x) = (x - f31) (x - f32) ••• (x - f3n).

Since all the roots f3i are integers, we may go over to C[x], where ;P(x) = o/(x) = (x - ~1) ... (x - ~,,).

This shows that I{l{x) splits into linear factors in C[x]; hence C is algebraically closed. Since ;P(x) is irreducible in hex], cp(x) is irreducible in k[x], and hence F = k(f31) has degree n over k. The residue class field F contains the sub field h(PI)' which is of degree n over h; hence f = deg (F I h) ~ n. But if e is the ramification, we have ef ~ n. Hence e = 1, f = n, and F = k{~l). This shows that every simple extension h(PI) of k is the residue class field of a subfield F of C, with the same degree as h(PI) I k, and ramification 1. 64

1.

UNRAMIFIED EXTENSIONS

65

Since all finite extensions of k may be obtained by repeated simple extensions, we have proved Theorem 1: Every finite extension of k is the residue class field for a finite extension of k with ef = nand e = 1. We return to the case of a simple extension, and assume now that if;(x) is separable. Then Pi i= Pj for i i= j, and hence I f3i - f3j I = I for i i= j. Let 0: be an integer of C such that if;( Ci) = 0; say Ci = PI· Then I 0: - f31 I < 1; that is, I 0: - f31 I is less than the mutual distance of the f3i. Theorem 7 of Chapter 2 shows that k(f31) C k(a). Let E be any subfield of C such that E => k(Pl). Then E contains an element 0: such that Ci = PI; hence k(f31) C k( 0:) C E. If we assume in particular that deg (E I k) = deg (E I k) and that E = k(Pl)' then k(f31) C E, but deg (E I k)

hence E

= deg (k(/Jl) I k) = deg (k(f31) I k);

= k(f31). Thus we have proved

Theorem 2: To a given separable extension k(Pl) of k, there corresponds one and only one extension Eo of k such that (a) deg (Eo I k) = deg (Eo I k) and (b) Eo = k(Pl). Corollary: If E is an extension of k such that E contains Eo, then E contains Eo. This discussion suggests the following definition: A field extension Elk of degree n with ramification e and residue class degree f is said to be unramified if it satisfies the following conditions:

(1) e=l, (2) ef= n, (3) Elk is separable. The third condition is inserted to exclude the critical behavior of the different when inseparability occurs. This difficulty does not arise in the classical case of the power series over the complex numbers, where the residue class has characteristic zero; nor in the case of number fields, where the residue class fields are finite. In neither of these cases is any inseparability possible.

66

4.

RAMIFICATION THEORY

Theorem 2 may now be restated in the following terms: Theorem 2A: There is a (1, 1) correspondence between the unramified subfields of C and the separable subfields of C. We see also that the unramified subfields of a given extension Elk are precisely those whose residue class fields are separable subfields of E. Furthermore, the lattice of unramified subfields of E is exactly the same as the lattice of separable subfields of E. Hence, in particular, there is a unique maximal unramified subfield T of E, which corresponds to the largest separable subfield of E; T is called the Inertia Field (Triigheitskorper) of E. Theorem 3: Let E = k(o:) where 0: satisfies an equation f(x) = 0 with integral coefficients such that all its roots modulo p are distinct. Then E is unramified over k.

Proof: It will be sufficient to prove the result when f(x) is irreducible. If f(x) is irreducible in k it is irreducible also in k, for we have seen that an irreducible polynomial can split in k only as a power of an irreducible polynomial-which would contradict the assumption that the roots modulo p are distinct. Hence deg (E I k) = deg (k(ii) I k); but k(ii) is contained in the residue class field of E. Since f ~ n, we have E = k(ii), and hence Elk is unramified. Corollary: If Elk is unramified, and if Q is a complete field containing k, then EQ I Q is unramified.

Proof: We can write E = k(o:) where 0: satisfies an irreducible equation cp(x) = 0 which is separable mod p. But EQ = Q(o:), where 0: still satisfies an equation which is separable mod p. The result follows now from the theorem. Thus the translation of an unramified field by a complete field is again unramified. We can also prove, by obvious arguments, that an unramified extension of an unramified extension is unramified over the ground field, and that any subfield of an unramified extension is again unramified. We conclude this section with two examples.

2.

TAMELY RAMIFIED EXTENSIONS

67

Example 1: Let k = F{t}, the field of formal power series over the field of constants F. We shall show that the unramified extensions of k are produced by separable extensions of F. We have shown previously that the residue class field of k is isomorphic to F under a natural mapping. Let E be a finite extension of k, E its residue class field. Then if ci E E, ci satisfies an irreducible equation ifJ(x) = 0 in k. Let ifJ(x) = ¢(x); since every coefficient of t/>(x) may be replaced by a congruent one, we can assume that all the coefficients of t/>(x) lie in F. Suppose now that Elk is unramified, and that Elk, which must therefore be separable, is generated by ci. Then we have seen that E is generated by a root (3 of t/>(x). The condition e = 1 implies that t is prime in E; hence E = F1{t}, where FI = F({3). It is easy to prove the converse: That if FI is separable over F then FI{t} is unramified over F{t}. Example 2: Let k be a finite field of q elements; this is the case which arises in algebraic number theory. We shall show that the unramified extensions of k are uniquely determined by their degree, and are obtained from k by adjoining certain roots of unity. The first statement follows from the fact that a finite field has only one extension of given degree. N ow let Elk be an unramified extension of degree n. The qn - 1 non-zero elements of E satisfy the equation xqn- l = 0, which is therefore separable, having qn - 1 distinct roots in E. Hence if t is a primitive (qn - l)-th root of unity, kal is unramified. Further, since ~n_l - 1 splits in k(t), it splits in k(t); hence E C k(t). By the Corollary to Theorem 2, E C k(t). On the other hand, let ex E E lie in the same residue class as t; then I ex - t I < 1, whereas the mutual distance of the roots of ~n_l _ 1 is exactly 1. Hence, by Theorem 7 of Chapter 2, k(t) C k(ex) C E, and we have E = k(t).

2. Tamely Ramified Extensions A finite extension Elk of degree n with ramification e and residue class degree f is said to be tamely ramified if it satisfies the following conditions:

68 (1) (2) (3)

4.

RAMIFICATION THEORY

ef = n, £ I k is separable, e is not divisible by the characteristic p of k.

Clearly all unramified fields are tamely ramified. We can deduce at once from the definition that a subfield of a tamely ramified field is tamely ramified; and that a tower of tamely ramified fields is tamely ramified. We shall now give an important example of a tamely ramified extension. Let ex be a root of the polynomial t/>(x) = xm - a, where a E k, (p, m) = 1, and t/>(x) is not assumed to be irreducible. Certainly I ex 1m = 1 ex I lies in the value group of k; let d be the exact period of I ex 1 with respect to 5D k (i.e. the smallest integer d such that 1ex Id = 1b I, b E k); then d divides m. Let {1 = exd/b; thus 1{11 = 1, and {1m/d = a/b m/d. Hence (1 satisfies an equation ifi(x) = x m/d - c = 0 where 1 c 1 = 1; ifi'({1) = (mid) (1m/d-\ so that 1 ifi'({1) I = 1; hence ifi(x) is separable modulo p. Thus k({1) 1 k is unramified. We now consider k({1) as the ground field; over k({1) ex satisfies the equation xd - b{1 = O. k({1) has the same value group as k; hence d is the period of I ex I with respect to the value group of k({1). Thus k(ex) I k({1) has ramification ~ d but the degree of k( ex) I k({1) is ~ d. It follows that k( ex) 1 k({1) is fully ramified, and that x d - b{1 is irreducible in k({1). In the course of this discussion we have verified all the conditions for k(ex) to be a tamely ramified extension, and we have shown that it is made up of an unramified extension followed by a fully ramified extension. Now let Elk be a finite extension; we shall construct the largest tamely ramified subfield of E. If T is the inertia field, T is clearly contained in this subfield, so we may confine our attention to the tamely ramified extensions of T. Since T is the largest separable subfield of £, £ I T is purely inseparable. Let p be the characteristic of the residue class field, and write e = eop", where (eo, p) = 1. 5D E /5D T is a finite Abelian group of order e. Let us choose a basis for this group, and let ex be an element of E which represents one of the basis elements with period d prime to p; that is, 1 ex Id = I a I, with a E k (since 5D T = 5D k ). Hence ex d = €a where I € 1 = 1. Since £1' 1 T l' is purely inseparable, some power €P'" represents a residue

2.

69

TAMELY RAMIFIED EXTENSIONS

class in T. Let the corresponding power aPiJ. be written CXl; since PI' is prime to d, cx and CXl generate the same cyclic group, so that CXl may be taken instead of cx. We have now aId = €la l where El C E T mod p; thus cxld = cal fl = b fl where fl E E and 1fll < 1b I· We now study the equation c/>(x) = xd - b = 0; d is prime to the characteristic of F, hence to that of k; thus c/>(x) is separable over k:

=

+

+

Obviously 1 Yi 1

= Wi =

1 (Xl 1

and

1 Yi

- Y; 1 ~ 1 YI I·

Further, however, c/>'(Yl) = dYld- l , so that 1 (g, h); then Xg(h) is a character of H, which is trivial (i.e. takes the value 1) on Ho. Thus Xg(h) may be regarded as a character of the factor group HjHo. Hence g -+ Xg(h) is a homomorphian of G into (HjHo)*; the kernel is clearly Go . Thus we have

GjGo :::: some subgroup of (HjHo)*;

similarly HjHo '" some subgroup of (GjG o)* .

Since HjHo is finite, by hypothesis, we have HjHo::::::' (HjHo)*; so GjGo '" some subgroup of HjHo.

Thus GjGo is also finite, and hence isomorphic to (GjG o)*; so HjHo '" some subgroup of GjGo '

Hence we have the result of the theorem: GjGo "-' HjHo. 4. The Inertia Group and Ramification Group Let Elk be a finite extension field. In sectIOns 2 and 3 of this chapter we have defined two important subfields of E: T, the Inertia Field (Tragheitskorper), which is the largest unramified subfield of E, and V, the Ramification Field (Verzweigungskorper), which is the largest tamely ramified subfield of E. When Elk is a normal extension, with Galois group G, the subfields T and V correspond to subgroups ~ and mof G, which are called respectively the Inertia Group and the Ramification Group of Elk. In this section we shall describe these two subgroups. At first, however, we do not assume that Elk is normal. Instead, we let C be the algebraic closure of k, and consider the set of isomorphic maps of E into C which act like the identity on k. This set of maps, of course, does not form a group; but it is known from Galois Theory that the number of such maps is equal to the degree of the largest separable subfield of E. A separable subfield of E may be described by giving the set of maps which

4.

THE INERTIA GROUP AND RAMIFICATION GROUP

73

act like the identity on the subfield. Our immediate task IS to describe the inertia field and ramification field in this way. First let Elk be an unramified extension. We have seen 1ll Section 1 that E = k(fll) and E = k(Pl)' where Irr (/31 , k, x) Irr( ex, T) be the residue class of Texlex mod p. We shall show first that 4>(ex, T) depends only on I ex I. Let € be a unit of E; then since T E:J, T€ = € mod p, whence T€/€ = I mod p. Thus we have

4>(ex€, T) = T(ex€) = T(ex) . Td ex€ ex €

= T€ex = 4>(ex, T),

which proves our assertion. Further, we have

and

since 1T2ex I = I ex I. Thus 4>( ex, T) is a pairing operation as described in Section 4. We have now to find the kernels under the operation. The :J-kernel K3 consists of the automorphisms T such that Texlex = I mod p for all ex E E; since T: = 1 mod p

~ I T:

- 1 I< 1

we see that the kernel K'J

~ I TCX

= m. If the

- ex

:BE

I < I ex I ~ T E m,

kernel is K E

,

we have

:JIm '"'-' mEIKE. Also KE must certainly contain the value group mk

of k, since Tex - ex for ex E k. But since no element whose period is a power of the characteristic p can occur in :JIm, no such element can occur in mEl KE. Hence KE consists of all the elements of mE whose periods modulo mk are powers of p. Thus we may say that :JIm is isomorphic to the "non-critical part" of the value group, i.e. to the co sets of mE modulo mk which have period prime to p.

5.

77

HIGHER RAMIFICATION GROUPS

So far 4>(ex, T) has denoted a residue class, i.e. an element of E. It is natural to ask whether there is an element in E, lying in this residue class, which can be naturally selected to represent it. If e = pr1.e o , where (P, eo) = 1, the order of 'J/f8 is eo; hence [4>(ex, T)]eO 1 mod p. The equation xeo - 1 = 0 has eo roots in the algebraic closure A of F, of absolute value 1 and mutual distance 1. Since 1 (TiX/ex)e O - 1 1 < 1, we have TiX/ex nearer to one root' than to any other; hence E -:JF(TiX/ex) -:JF(,). Thus' is a root of unity in E, congruent to TiX/ex modulo p. We now choose 4>( ex, T) to be represented by this ,; this is well-defined, since if 4>(ex, T) = 4>({3, T') mod p, the corresponding roots of unity are congruent: = mod p, whence 1 1 < 1; thus = since the mutual distance of the roots is 1. Thus the operation 4>(ex, T), which formerly had values in E, may now be regarded as a pairing operation on f8 E and 'J into E. Now G/f8 is an extension group of 'J/f8. Let u be an element of G; the mapping T E 'J ~ UTU- I gives an automorphism of 'J modulo f8. We shall give a description of 4>(ex, um-I); this gives the character values of UTU- l in E, and this will be sufficient to describe UTU-I. We have

=

'1 '2

'1 '2 ,

since 1 u-lex 1 = in E; hence

1 ex I.

'1 - '2

But 4>( ex, um-I ) and 4>( ex, T) are roots of unity

In particular, if G is abelian, UTU- I U(4)(rx, 'T))

= T, so that

= 4>(rx, 'T);

thus 4>( ex, T) is a root of unity in k.

s.

Higher Ramification Groups

+

Let i denote either a real number, or a "real number zero", in a sense to be made more precise in a moment. We define the sets f8 i to consist of those automorphisms U E G such that ord (arx - rx) ;;;, ord rx

+ i,

4.

78

RAMIFICATION THEORY

or uex - ex) ___ . or d ( - - : ; : / l ex

ex E E.

for all

When i is a real number T, this shall mean simply that uex - ex) ;> r; ord (-ex-

when i =

T

+ 0, uex - ex) ord ( -ex- ;> i

shall mean uex - ex) ord ( -ex> r.

We now consider the operation which, acting on ex, produces (uex - ex)/ex; this bears a certain resemblance to logarithmic differentiation:

+ f3(u 1) f3 + (u -

(u - 1) exf3 = u(ex) (u - 1) f3 (u - 1) exf3 _ u(ex) (u exf3 ex f3

1) ex,

1) ex ex'

whence we have

!

I (u ~;) exf31 ~ max I (u ~ 1) ex I ' I (u ~ 1) f311. Further, (u-l)ex

-1

1 uex

1 ex

uex - ex 1 ex uex

=---=---'-,

(u - 1) ex-I (u - 1) ex ex ex-I = ex • uex

whence

I (u ~!! ex-I I = I (u ~ 1) ex I. Thus, when we examine the effect of 5Bi on the elements of a group we can restrict our examination to the generators of the group,

5.

79


Let us first notice, however, that the sets mi are invariant subgroups of GJ; let a, T E mi , a E E: (aT - 1) ex = (a - 1) Tex

+ (T

(aT - 1) ex = (a - 1) 'Tct Tex ex

Tex

- 1) ex,

+ (T

- 1) ex ,

ex

ex

whence 1 (aT

-: 1) ex 1 .( max 1 (a

~ 1) ct 1 ' 1 (T

-ex 1) ex I.

Thus mi is a subgroup of GJ; that it is an invariant subgroup follows from the invariant form of the definition. Let E = k( ao); then for each a E G we define

If a *- 1, then aao *- a o , and hence i" i,,; hence for i > max" ia , the group mi consists of the identity automorphism alone. It follows from the definition that for i > j, mi £ mj ; we must now examine the discontinuities in this descending sequence of groups. We consider first the subsequence {m r } where the indices T are restricted to be real numbers. Suppose there are discontinuities at T1 , T 2 , "', Tk (k .( n - 1 where n is the order of G). Then: T

E

lBr for all

T

58+0 = 1, (for there are no inseparable extensions of fields of characteristic zero), we consider the case p =1= O. We examine aP - 1 = «a - 1)

=

(a -

+ 1)P - 1 1)P + p(a-I )P-l + ... + p(a-I).

Thus (a P - 1) ex

= (a -

1)P ex

+ p(a -

I)P-l a

+ ... + p(a -

1) ex.

5. Hence, if u

Thus u

E

E

81


ID, , we have

ord (uP

-

IDi ~ uP

E

1) a

~

ord a

+ min (pi, ordp + i).

ID j where j = min (pi, ord p

+ z).

P *- 0, ord p > 0, so that j > i. Thus every element

Now if

u of IDi has

(modulo this ID j ) period p. Since IDi/IDj contains the factor group at the jump, we have completed the proof of our Theorem 9: At a discontinuity in the sequence of ramification groups {IDi} the factor group is abelian and of type (p, p, p, ... ), provided that the discontinuity does not occur at ID+o . In the case of non-discrete valuations, no information can be obtained about discontinuities at ID+o; this difficulty does not arise in the discrete case where ID+o = IDl . Let us examine the special case when e = f = 1. Then if ex E E, there is an element b E F such that 1 a 1 = 1 b I; thus 1 a/b 1 = 1 and alb -- c mod p where e EF (since f = 1). This implies that 1 a - be 1 1, and ordyll2 > 1; but ord llf'(a) = 1 (since f'(a) =1= 0 mod p); hence ordf(f3) = 1, and since f3 lies in the same

4.

84

RAMIFICATION THEORY

residue class as ex, it may be taken as representative of the generating residue class. Thus we can suppose without loss of generality that ordf(ex) = 1; thus f(ex) can be taken as our element n. Since I, &, ... , &,-1 form a basis for Elk, we obtain a minimal basis for Elk by taking {exv(f(ex))"}; thus every element in D can be expressed as a polynomial in ex with integer coefficients. Thus 1, ex, ... , ex n- 1 form a minimal basis for Elk. This completes the proof of the theorem. N ow let () E D be given by

Let a be an isomorphic map of ElF into the algebraic closure C. Then (a - 1) 8 =

c1(aex -

+ c (aex

ex)

2

2

-

ex2)

+ '" + cn_1(ao:

n

-

1-

exn - 1)

= (aex - 0:) {3, where f3 is integral, i.e.

I f3 I ~ 1.

Hence

1(a - 1) 81 .:( 1(a - 1) ex 1 ,

and so

I (a or, ord

1) ex

I=

max I (a - 1) 8 I , OED

«a - 1) ex) = min ord «a - 1) 8). OED

The ideal generated in C by (a - 1) ex, i.e. ((a - 1) ex . Dc) called (by Hilbert) an "element" of ElF. Let now ElF be normal. Then we see that ord

«a - 1) ex) > 0

IS

== 0: mod p

aex

a leaves all residue classes fixed

aE~,

the inertia group. Suppose, then, that a E~; since ~ leaves the inertia field T fixed, we can take T as our ground field. Thus the powers of ex, or of any element n with ord n = 1, form a minimal basis for E 1 T. If a E~, we have ord (a - 1) ex

= ord (a

- 1) II

6.

RAMIFICATION THEORY IN THE DISCRETE CASE

85

for all such elements n; thus in examining the effect of an automorphism a, we need examine its action on only one such n. Suppose that a E 58i but a €f- 5B i +1; this means that ord

(a - 1) ex ____ . 9

ex

t,

but

ord (a - 1) ex ;;f> i ex

and

ord (a - 1) ex

+ 1.

Hence ord

(a-l)ex ex

.

= t,

=

i

+ 1.

We see at once that if ord (a - 1) ex = 0, then a is not an element of the inertia group; if ord (a - 1) ex = 1, then a E.;) but a €f- 58 1 , and we obtain immediately the classical definition of the higher ramification groups: a E 58 j

-¢>

all == II mod pHI.

CHAPTER FIVE

The Different Throughout this chapter we shall be dealing with finite separable extensions E of a field k which is complete under a discrete valuation. As usual, we shall denote by D, 5p, II and 0, P, 7T the rings of integers, prime ideals, and primes in E and k respectively. The trace from E to k will be denoted by S Elk or simply by S. 1. The Inverse Different

Let T be any set in E; its complementary set T' is defined by A E T'

¢>

S(AT) C o.

It is easily seen that if TI C T z then TI' :) T z'. In particular, when T = D, we obtain the complimentary set D': AE D'

¢>

S(AD) Co.

D' is called the inverse different. We now introduce the notion of a fractional ideal in E. Let 111 be any additive group in E such that I1lD = 111. If a E 111, and I f3 I ~ I a I, then f3/a ED, and so f3 = a . f3/a E 111. This means that 111 contains, along with a, any element with ordinal ~ ord a. There are thus two possibilities: (I) ord a is not bounded for a E 111; then clearly 111 = E; (2) ord a is bounded; let CX o be an element with maximal ordinal in Ill-this exists since the valuation is discrete; then 111 = aoD = llvD where v = ord a o . In the second case we call 111 a fractional ideal (or an ideal for short). Theorem 1: D' is a fractional ideal.

Proof:

Clearly D' is not empty, since D' :) D. 86

1.

THE INVERSE DIFFERENT

87

On the other hand, .0' =1= E. For since Elk is separable, there is at least one element 0: E E such that S(o:) = 0: =1= O. Then S( 0:/7T v • 1) = a/7T and this does not lie in 0 if v is large enough; in other words, for large enough v, O:/7T P does not lie in .0'; hence .0' =1= E. If 8('\.0) Co and S(p.D) Co, then S(('\ p.)o) C 0; thus .0' is closed under addition. If S(AD) Co, then S(AD' D) Co, since .0.0 = .0; hence '\.0 CD'. Thus .0' is a fractional ideal as described above. From this result it follows that we may express .0' as II-iD, where j ~ 0, since .0' :::l D. The different !l is defined to be the inverse of .0' : !l = .0'-1 = IIiD. The fundamental property of the different is given by P

,

+

Theorem 2: is unramified. Proof:

!l

=

.0, and hence .0'

=

.0, if and only if Elk

There are three cases to consider:

Case 1: Elk is unramified. Then 7T = II and we have seen that every map of Elk into C comes from a map of Elk into C; thus the trace in Elk comes from the trace in Elk. More precisely, if 0: is a representative in E of a residue class ti in E, then 8 E1k(ti) = Silk(ti). Since Elk is separable (E I k is unramified), the trace 8 ifl k is not identically zero. Hence there is an element 0: in such that SElk(O:) ~ 0 mod p. Now let f3 = 7T- i 8 be any element of E with negative ordinal (i> 0; 8 a unit). Then S

(,8 ~) =

S(7T-i o:) = 7T-iS(o:) ¢'. 0;

since 0:/8 ED, this shows that then .0 = .0'.

f3 ¢'. .0'.

Hence if Elk is unramified,

Case 2: e > 1. Then 7T = IIe. Let a be any map of E into the algebraic closure C; if 0: E Ill, then I 0: I < 1, whence I ao: I < I and so I 8(0:) I < 1. This shows that S(Ill) C p. Hence

S

(! Ill) = S (! III

.0 ) C o.

5.

88

THE DIFFERENT

Thus I/7T \l3 CD', and in particular II/7T, which has ordinal- (e - 1) is contained in .0'. Hence .0' contains .0 as a proper subset. We remark that in many cases - (e - 1) is the largest negative ordinal occurring in the inverse different.

Case 3: e = 1, but Elk is inseparable. Here we use the transitivity of the trace: SElk(ex) = STlk(SEIT(ex». We propose to show that S(D) C p, so it will suffice to show that I SEIT(ex) I < 1. This reduces our investigation to the case where the residue class field is totally inseparable. Let ex E E: If a is any map of E I T into C, then aex ex mod p. Hence S(ex) nex 0 mod p since the degree of an inseparable extension is divisible by the characteristic. Thus SeD) C Pi hence S(I/7T D) Co, so that I/7T ED' and .0' contains .0 as a proper subset. This completes the proof of the theorem. A second important property of the different is contained in

=

= =

Theorem 3: If E -:JF -:J k, then !lElk Proof: ,\ E

Let !lFlk

=

S-l!lF.

=

!lEIF!lFlk.

We have

!lE~k - SEli,\DE) Co - SFliSEIF(,\DE)) C 0 -

SFlk[(SEIF(AD E)) .oF]

-

1\

\

~-1",-1

Eo

""'EIF

=

Co - SEIF(,\DE) C !l'Ftk

",-1 ",-1

....,Flk....,EIF·

Thus ",-1

""'Elk

=

",-1 ",-1

""'Flk""'EIF;

hence

Corollary: If T is the inertia field of Elk, then Proof:

!l Elk =

!l EI T •

We have only to recall that since T is un ramified,

!lTlk = D T •

2.

89

COMPLEMENTARY BASES

2. Complementary Bases Let k be any field, E a separable extension of degree n. Let WI' W2 , ••• , Wn form a basis for Elk. We examine whether there exists an element gEE such that S(w"g) = 0 (i = 1, 2, ... , n). If we write

we see that the equations S(Wit) = X1S(WIWi)

+ ... + xnS(wnw;) = 0

form a system of n homogeneous equations in n unknowns. Multiplying the equations in turn by arbitrary elements ai E k and adding, we obtain S((a1Wl + ... + anwn) Y) = 0; thus S(Eg) = 0, which is impossible unless g = 0 smce Elk is separable. Since the system of homogeneous equations S(wig) = 0 has only the trivial solution, it follows from the theory of systems of linear equations that any non-homogeneous system S(Wig) = bi (i = 1,2, ... , n), with bi E k, has exactly one solution. In particular there is exactly one element W'j E E such that S(WiW'j) = 8ij • The set of elements W'IW'2' ... , w'n is called the complementary basis to WI> W 2 , ••• , W n • To justify this name we must showthatthew'iarelinearlyindependent;soletxlw'l +xnw'n=O. Multiply by Wi' and take the trace; this yields XiS(WiW/) = Xi = O. Thus the W'i are linearly independent. It is easy to see that if g = XIW I XnWn> then Xi = S(gw'i)' and that if 7J = ylw'l + ... Yn w' n' then Yi = S( 7J Wi)· We can prove

+ ...

+

+ ... +

Theorem 4: Let WI' ... , Wn be a basis (not necessarily minimal) for Elk; let w\ , ... , w'n be the complementary basis. If T= Wlo + ... + Wno, then T' = W'IO w'no.

+ ... +

Proof:

Any element ,\

E

E may be expressed as

90

5.

THE DIFFERENT

Then ,\ E T'

-=> S('\T) Co-=> S('\Wi) Co (i = 1, ... , n) -=> x, E o.

Hence T' =

w~o

+ ... + w~o •

We now go on to examine the special case of a basis formed by the powers of a single element of E. We require the following preliminary result, due to Euler: Lemma:

Let E = k(ex); f(x) = Irr (ex, k, x). Then

S (f(x) ~) = x - ex f'(ex)

Xi

for

i = 0,1, ... , n - 1.

Proof: Since Elk is separable, the roots exo = distinct. Now

cx, "', CXn-l

are

S (f(x) ~) _ "" f(x) ~ x - ex f'(ex) - ~ X - exv f'(ex.) ,

•

which is a polynomial of degree We have

~

n - 1.

but

Hence

-f:-)] "'=1% = cx.i [s (M.. X ex f (ex)

.

for

v = 1, ... , n.

Thus a polynomial of degree ~ n - 1 has n common zeros with Xi. It follows that it must be identical with xi. It is now easy to compute the complementary basis to 1, ex, ex 2, "', exn - 1 • The result is given by

2.

91

COMPLEMENTARY BASES

Theorem 5: If f(x)

= (x -

ex) (b o + b1x

+ ... + bn_ xn1

1 ),

then the complementary basis is formed by bo b1 bn- 1 f'(ex) , f'(ex) , .•• , f'(ex) •

Proof:

Hence rxib· )

s ( f'(:) = 8i ; • This is the precise condition that bjJf'(rx) should form a complementary basis to the rxi. We shall now evaluate the coefficients bi • Let f(x)

" = kavxv

(a" = 1).

v=o

Then, since f(rx)

= 0, we have

f(x) =f(x) - f(ex) = ~ a XV - exV x-ex X-ex k.( v x-ex v=1 =

k" av(xv + rxxv+2 + ... + exv+1

v=l

Hence

We may write this symbolically as bi =

[f~:~ ] x

"'~'"

,

1 ).

92

5.

THE DIFFERENT

where the symbol [AJ denotes the integral part of A in the obvious sense. If ex is an integer, we may simplify those results even more. For then the coefficients ai of f(x) are all integers of k, and we may replace the basis elements bn _ l f'(CY.)

1

f'(CY.) ,

bn _ 3 1 f'(CY.) = an - 2 f'(CY.)

+ an -

cy'2

CY.

l

+ f'(CY.) ,

f'(CY.)

and so on by the equivalent basis 2 n l 1 CY. cy' cy' f'(CY.) , f'(CY.) , f'(CY.) , ••• , f'(CY.).

Theorem 6: If ex is an integer such that E T

= 0 + OCY. + ... +

then the complementary set T' f'(ex) . D.

=

ocy'

n

-

l

= k(ex), and

,

TJf'(ex). Furthermore !l divides

Proof: The first statement follows at once from the preceding discussion and Theorem 4. To prove the second statement we have only to notice that since ex is an integer TeD; hence .0' = !l-I C T' = TJf'(ex). It follows that f'(ex) C !IT C !lD, which proves the result. Finally, we may apply these results to the case where the residue class field Elk is separable; then E has a minimal basis consisting of the powers of an integer ex. (Theorem 10, Chapter 4). Theorem 7: If Elk is separable, then !l = f'(ex) . .0, where ex is the element whose powers form a minimal basis. Proof: Since the powers of ex form a minimal basis, the set T of Theorem 6 is exactly the ring of integers D. Hence .0'

=

!l-l

= f~CY.)

,

3.

FIELDS WITH SEPARABLE RESIDUE CLASS FIELD

93

and so !l = !,(rJ.) .

n.

3. Fields with Separable Residue Class Field We shall now give a description of all extensions Elk where the valuation is discrete and the residue class field is separable. Let Elk be such an extension, T the inertia field. Then, as we have remarked earlier, !lElk = !lEIT' We have seen also that 1, II, II2, ... , Ile-l forms a minimal basis for E I T. Hence, if f(x) = Irr (II, T, x), we have !l = f'(II) . n. Now f(x) is a polynoao; a o = Nil, so that mial of degree e: f(x) = x e + a1x e- 1 I II I = V'Ta;;l, whence I ao I = I II Ie = ITT I; thus TT I a o , but TT2 *- a o • Further, since the coefficients a. (v = 1, ... , e) are the elementary symmetric functions of the roots of f(x), we have TT I a•. Thus f(x) is a polynomial satisfying the Eisenstein criterion. Conversely, we shall show that such a polynomial gives rise to a completely ramified extension. Le f(x) be an Eisenstein polynomial in T[x], II a root of the polynomial. Then

+ ... +

n e + aine- + ... + ao = 0; I

since all the

ai

I II I < 1. Further, since (v = 1, ... , e - 1), have I Ille = ITT I. Thus the ramifica-

are integers,

and I a o I = ITT I, we must tion of the extension defined by f(x) must be at least e; but the degree is at most e. Hence both ramification and degree are equal to e, and there is no residue class field extension. We have also proved that the polynomial f(x) is irreducible. The proceding analysis has shown

Theorem 8: All possible extensions of k with separable residue class field consist of an unramified extension (constructed by making a separable extension of the residue class field 'k) followed by an Eisenstein extension. We shall now compute the different of such an extension. We have !l = f'(II) . n = IIan where a = ordf'(II). Now !,(n) = en·-I

+ (e -

1) aI n

e- 2

+ ... + a._I;

5.

94

THE DIFFERENT

since j( x) is an Eisenstein polynomial, the ordinals of the coefficients a. are divisible bye. Hence the ordinals of the non-zero terms are all incongruent mod e. Thus a = min ord (e[]e-I, (e - I) a l []e-2, , a e-

l )·

If the field is tamely ramified, e is not divisible by the characteristic of the residue class field, so ord e = O. Thus a = e - I, and !l = []e-ID. This is analogous to the case of function theory: for if a Riemann surface has a winding point with e leaves, the winding number is e - 1. If, on the other hand, the ramification is wild, we have a ~ e. We now ask whether, for a given value of e, there is a bound on the indices a arising from all Eisenstein equations. If e =1= 0, we have a ~ ord (e[]e-l) = e - 1 -I- ord e. Thus if the characteristic of E does not divide e, a is bounded by e - 1 + ord e; but if the characteristic of E divides e, e = 0, ord e = 00, then a is unbounded, since the ordinals of the a v (v ~ e - I) can be made as large as we please. We conclude this section by computing a explicitly when Elk is normal. We recall that if Elk is a normal extension, and E = k(ex) where ex is an integer such that the powers of ex form a minimal basis, then the position of the automorphisms a of ElF in the higher ramification groups In l , 1n 2 , ••• is determined by

ord (aex - ex) = i

+ 1 -=> a E In

i

but

This has been established for i = 1, 2, .... It is easy to verify that if we define Ino to be the inertia group 3 and In_I to be the whole Galois group , then the result holds also for i = - I and O. Now let us define, for each a E , the index i(a), given by i(a) + 1 = ord (aex - ex). Then - I ~ i(a) ~oo, and i(O) =00 only for a = 1. Then our criterion for the position of a in the groups In j may be written i(a) = j

-=>

a E In; ,

Now ifj(x) = Irr (ex, k, x), we have I(x)

= II (x UEb = e. Consequently (A U B) n ~a n ~b = e, so that A U B is not a member of @. In general, if ~a = At U A2 U ... U An , where the Ai are disjoint closed sets, then exactly one of the Ai is in @. In order to prove that na. ~a. is not empty we must exhibit an element a E G which lies in each ~a • Let ex be any element of Q; we shall define the effect of a on F(ex). If U is the group of Q I F(ex), then G='TtU(= U)U'T2UU ···U'TnU; this union is finite since F(ex) I F is finite. Now G is closed (since it is a Galois group); hence G E @. The co sets TiU are closed and disjoint; hence exactly one of them, say 'TiU, is a member of @. Let a have the effect on F(ex) of 'TiU. We have now to show that the mapping a we have constructed is in fact a well-defined element of G. To this effect let E t :::) E2 be finite subfields of Q with groups Ut , U 2 : U 2 :::) U t . Let a have the effect on E t of 'TtUt E @, and on E2 of 'T 2U 2 E @. Since 'TtUt and 'T2U2 E @, hence 'TtUt n 'T 2U 2 Thus 'TtUt C'T2U2 and the effect of 'T 2U2 on E t is the same as that of 'TtUt . Hence a is welldefined. Finally we must prove that a E na. ~a.. For any subgroup U, a has the effect of some coset 'TU E @ on the fixed field of U. Hence aU = TU E@. Thus aU n ~a. (for every ~a. E @). Hence every neighborhood of a contains an element of ~a.; thus a E;P(1. = ~a. (for every ~a. E @). This completes the proof. We now illustrate the use of this theory by an interesting special case. Let p be a prime number. Consider the sequence of fields

e

e

* e.

*e

108

6.


F C FI C",, where Fn I F is cyclic of degree pn. Let Q = UnFn. Let 0: E Q; then 0: E some Fn, i.e. F(o:) CFn; hence F(o:) = Fr for some T ~ n. It follows that the only finite subfields of Q are the Fr; the only infinite subfield is Q itself. Let a be an element of the group G of Q IF which acts on FI like a generator of the group of FI I F. If a(o:) = 0: for an element 0: E Q, the field F( 0:) = F r remains fixed. If 0: 1= F, then F( 0:) ::l F I , so FI remains fixed, contrary to the definition of a. Hence 0: E F. Thus the fixed field of H = {a v} is F itself. By Lemma 3, G = H. Now if the group of Q I Fn is Un we have UI ::l U2 ::l ... , so that G satisfies the first countability axiom. Hence, if'T E G, we can express 'T as lim.-->CIJ aa. where the a. are integers. Now we examine the conditions under which a sequence {aa.} converges. For any given neighborhood Un' there must be an index N such that for 11-, v ;;:, N, aa.-a/" E Un , i.e. aa.-a/" leaves Fn fixed; la is a generator of the group of Fn I F; hence a v - al' = 0 mod pn. Hence {aa.} convergent ~ {a.} convergent in the p-adic topology to a p-adic integer 0:. We may write symbolically 'T = arI-. This is well-defined, since if 0: = lim a. = lim b., we have a. - b. - 0, hence aa.-b. leaves high Fn fixed and so lim aa. = lim ab". It is obvious that arI-a P = arI-+{J. Hence the group G of Q I F is isomorphic to the additive group of p-adic integers. 2. Group Extensions

The problem of group extensions is the following: Given a group G and an abelian group A, we wish to find a group G which contains A as a normal subgroup, and such that GIA '" G. Let us assume first of all that such an extension exists. Then there is an isomorphism between the elements a of G and the cosets of G module A; we denote this isomorphism by

where u" is an element selected from the coset to which a corresponds. Let x E G; then since A is a normal subgroup of G, xAx- I = A, and a _ xax- I is an automorphism of A. This automorphism of A

2.

109

GROUP EXTENSIONS

depends only on the coset of C modulo A in which x lies. For let b be an element of A; then (bx) a(bx)-l = b(xax-1 ) b-1 = xax-l,

since xax- 1 , b, b-1 lie in A, and A is abelian. The automorphism is therefore defined by G E G, and we display the fact by writing or We now examine the rules of combination of these automorphisms: (aT = u,,(!l7au;1)

U-;l

= (u"uT)a(u"u,r 1.

N ow since G - Au" is an isomorphism, U"U lies in the coset Au", and hence produces the automorphism known as a aT. Thus 7

Since u"u .. lies such that

In

the coset Au".. , there

IS

an element a" ... E A

The elements u" lie in the group C, and so must obey the associative law: Expressing this fact, we have

Hence

Thus if C :> A and CIA:::: G, every element G E G defines an automorphism a _ a" of A, such that (aT)" = a"T; further there is defined for every pair of elements G, 'T EGan element a" ... E A, satisfying the relation

Let us suppose, conversely, that we are given an abelian group A, and an operator group G acting on A; thus every element G E G

110

6.


defines an automorphism a _ a" of A such that (aT)" = aaT. Let there be given also, for each pair of elements a, T E G, an element a".T E A satisfying the relations

Such a set of elements {a"'T} is called a factor set. We shall show that under these conditions there exists a group G:J A, such that G/A,....., G. Consider the pairs (a, a) with a E A, a E G; let these pairs be multiplied according to the following rule:

(Heuristically: (a, a) corresponds to the element "au,,". Thus (a, a) (b, T) = "au,," "bu/' = ab"uuUT = ab"a"'Tu"T') We shall now prove that the set of all such pairs is the required group G. First we verify that the multiplication is associative.

Hence

On the other hand, [(a, a) (b, '1')] (c, p)

= (abaoa,T' aT) (C, p);

whence [(a, a) (b, '1')] (c, p) = (abac"Ta",TaaT,p, aTp) = (a, a) [(b, T) (c, p)]

in virtue of the relations between the a,,:T' Thus the multiplication is associative. Next we show the existence of an identity element. Let e = (a;.~, 1).

[Heuristically: a1.~Ul = 1].

UI E

A and so

UIU I

=

al.lu l ,

whence we have

2.

111

GROUP EXTENSIONS

Then for if we set a = T = I, p = T in the factor set relations, we obtain a1.Ia1.T = aL.a1.T' whence a1.1 = al,T' Thus e is an identity element. Similarly, if we set T = P = I in the associativity relation, we obtain a".1 = aI.I' Using the result we prove the existence of an inverse to each element:

[Heuristically, au,,' bUT = allu I ; hence ab"a".Tu"T = al.~ul and -1 -Ib-a] a = T -1 , a = aI.la".T . Hence the set of pairs (a, a) form a group G. Consider now the mapping (a, a) -- a, which is easily verified to be a homomorphism of G onto G. The kernel (i.e. the set of elements mapped into I) consists of pairs which we choose to write as (a . al.~ , 1) = d. Then

db = (a.al:~ 1) (b.al~l 1) = (abal.~ 1) = db ; J

J

J

thus the kernel is an invariant subgroup A isomorphic to A. Hence G/A~ G. Now define the elements u" = (I, a). For these we have the multiplication rules U.,tlT = (1, a) (1, T) = (a".T , aT) = (a".Ta~~ , 1) (1, aT)

Every element of G can be expressed as au,,: For (a, a)

= (a . a;:~ , 1) (1, a) = du" .

=

a) (a . a-II) -- (a" , a) -- d>u a· 1.1'

Finally, Ua Ii -(1'

112

6.


Thus if we define ii a = aa, we have

Hence if we identify the subgroup A with the group A, we have constructed an extension C with the required properties. The elements U o which represent the co sets of C module A satisfy the relation Let C and C' be two extensions of A with factor group G. Let the cosets of C modulo A be written Aua, those of C' modulo A be written AVa. We say that C and C' are equivalent extensions if C is isomorphic to C' under a mapping which acts like the identity on A and carries the coset AU a onto the coset AVa. We now obtain necessary and sufficient conditions for two factor sets to yield equivalent extensions. Suppose aa,T' ba,T yield equivalent extensions C, C'. Then, since AUa is mapped onto Av", we have u" mapped onto c"v" (c" E A). Thus

But Hence

Conversely, if this relation holds between the factor sets, it is easy to see that they yield equivalent extensions, under the isomorphism defined by U,,- C"V", Since the factor set a",T = 1 satisfies the associativity relation, we see that for given groups A and G we can always find at least one extension C containing A as normal subgroup and such that CIA G. We now consider the special case in which G is a finite cyclic group of order n. Let a be a generator of G chosen once for all: then G = 1, a, ... , an-I. Suppose G acts on A in a prescribed manner and let C be an extension of the type we are considering; the factor set a"/.I,,," may be written aI'," Let the coset corre'"'-J

2.

113

GROUP EXTENSIONS

sponding to a E G be denoted by Au; then the remaining cosets can be written Au'. We have

where p is the remainder when Hence we have

jJ.,

+ v is divided by n.

if if

+ v < n, fL + v ~ n. fL

We shall write un = a. Then au = ua (= Un+l); but ua = aau, hence a = aa. Thus to every extension G corresponds an invariant element a E A defined in this manner. Conversely, let a be an invariant element of A: a = aa. For all jJ." v we define

It is a simple matter of computation to verify that the a.,11 so defined satisfy the associativity relations and so can be used to form a group extension. Finally we obtain the necessary and sufficient condition for two invariant elements a, b to yield equivalent extensions. Let a and b yield extension groups G and G'; let the co sets of G, respectively G', modulo A be written Au', respectively Av·. Suppose first that the groups G, G' are equivalent; then there is a mapping from G to G' which acts like the identity on A and maps the coset Au· onto the coset Av". Hence u is mapped onto ev (e E A). Thus a

=

un ---+ CV

• CV . . . . .

cv

=

c1+ aY = a. Then all elements of the same coset of G modulo H have the same action on A H • Hence if H is a normal subgroup of G, the factor group G/H acts in a natural way as group of operators on A H • We can thus define the second cohomology group fJ 2(A H , G/H), and we shall study its relation to the group fJ 2(A, G). Let aUH,TH be a cocycle for the system (AH' G/H); then if we define aU,T = aUH,TH for all 0, 'T E G, we obtain a co cycle aU,T for the system (A, G). Further, if auH,THis a coboundary for (AH' G/H), we have

and defining bu = bUH for all 0 E G we obtain aU,T = buhaT/baT' so that aU,T is a coboundary for (A, G). In this way we have defined a mapping from the second cohomology group fJ 2(A H , G/H) into the group fJ 2(A, G). This is obviously a homomorphism; but it is not an "onto" mapping. We shall prove later that when the first cohomology group fJI(A, H) is trivial, then the mapping is an isomorphism. We say that a co cycle aU,T splits in H if aU,T is cohomologous to the identity cocycle when 0 and 'T are restricted to H. H is called a splitting group of the cocycle. Every co cycle has at least the trivial splitting group H = 1, for aLI = 1 if we choose U I = I as representative of the coset A in the group extension defined by(A, au, .... G). We notice that a cocycle aU,T such that aU,T = aUYl'TYI for all YI , ')'2 E H splits on H, for ayl'Y' = aLI = 1. Obviously if a co cycle splits on H, so do all co cycles cohomologous to it; we say that the corresponding element of the cohomology group splits on H. We may now sum up our preliminary remarks in There is a natural homomorphism from the second cohomology group fJ 2(A H , G/H) into the subgroup of fJ 2(A, G) consisting of elements which split on H. Let au be a l-cocycle for the system (A, G), and assume that aa depends only on the coset of 0 modulo H, i.e. a uy = au for all 'Y E H. Since au is a cocycle we have a~TU = aUT' Set')' = OJ then Lemma 1:

a.yal =

ayT'

4.

117

CONTINUOUS CO CYCLES

But a y = a 1 = I, and ayr = aT since the coset H'T = 'TH. Hence

Thus aT E A H' and aT is a cocycle for (A H' G/H). Similarly let a".T be a 2-cocycle for (A, G) such that a".T depends only on the co sets of a, 'T mod H. We have the relation

We set a

=

y, and obtain

whence

Recalling that for any co cycle we have a1."

=

aLl' we deduce

aT,p = a~.p.

Thus aT,p E A H , and a p is a cocycle for (AH' G/H). We sum up these remarks In T•

Lemma 2: Every co cycle a" (a",T) of (A, G) which depends only on the coset of a (a, 'T) modulo H, is a cocycle for (AH' G/H).

4. Continuous Cocycles

From now on, let A be the multiplicative group of non-zero elements in a field Q where Q is a normal extension of a ground field F; let G be the Galois group of Q I F. Let the topology in Q be discrete, and let G have the Galois group topology described earlier. We shall restrict ourselves now to cochains which are continous functions; we can give an algebraic interpretation of this continuity condition. Consider the I-cochain a" which maps G _ A. If this map is continuous, then, given any neighborhood N of a" in A there exists a neighborhood aU of a in G such that for every element 'T E aU, the corresponding aT lies in N. Since the topology in Q

118

6.


is discrete we can choose N = aCT . Hence if the cochain is continuous there exists a neighborhood aU of a such that for every element T E aU, aT = aCT • The group G is covered by these neighborhoods; hence, since G is compact, it is covered by a finite number of them. We have then r

G

= U a.U., ~=l

where a o is constant on a.U •. The sets U. are groups corresponding to finite extensions of F. Hence U = U t n U 2 n ... n V,. is the group corresponding to the compositum E of these fields. Obviously U C U., and (U. : U) is finite; thus each a.U. splits into a finite number of cosets modulo U. Hence if aa is continuous, there exists a subgroup U such that G = U:=I T.U and aCT is constant on each coset T.U. If we replace U by the invariant subgroup V C U which corresponds to the smallest normal extension containing E we have G = U:=I p. V and aCT is constant on each coset p. V. Similarly, if aCT,T is a continuous map from G X G -+ A, we can find a subgroup U, and even a normal subgroup V of finite index in G, such that aa,T = aaV,TV' In order to prove our next theorem we require the following lemma from the Galois theory of finite extensions. Lemma: If ElF is a finite normal extension with Galois group G, then (b E E).

(The equations aaaaT = aaT are known as Noether's equations.)

Proof: (1) Obviously aCT = bl-a satisfies the equations. (2) To prove that these are the only solutions we set

where (} is any element of E. Then

aaba

=

~8aTaaaaT T

=

~8aTaCTT

= b,

T

since G is a finite group. Now b cannot be zero for all elements () E E, otherwise we should have a relation of linear dependence

4.

CONTINUOUS COCYCLES

119

between the characters (} -+ OT which is known to be impossible. Thus a" = hl -". We are now in a position to prove our theorem:

Theorem 8: With A and G as described above, the first cohomology group is trivial. Proof: We have to show that every co cycle a" is already a coboundary. Since a" is a cocycle we have oa" = 1, i.e. a"a"T = a"T' Since a" is continuous we can find a normal subgroup U of G such that a" is constant on every coset T U. Let A E U; then aJ.a~ = aJ.T . Since Alies in the coset defined by 1, a y = a l ; and since AT lies in the coset UT = TU, we have aJ.T = aT' Now ala l l = aI' so a l = I; hence we obtain a~ = aT' Thus aT lies in the field E which is fixed under U. ElF is a finite normal extension, with Galois group G/U. We define c"u = a" E E; then

These are the Noether equations for the field ElF; by our lemma, their only solutions are Call

=

bI-("U)

(b E E).

Hence

thus a" is a boundary. A is still the multiplicative group of a normal extension Q IF, with Galois group G. Let H be a subgroup of G and let AH be the multiplicative group of the fixed field of H. If a cocycle a",T splits on H, i.e, if a",T is cohomologous to the identity Co cycle when a, T are restricted to H, we call H a splitting group and AH a splitting field of the cocycle, In the case of a field, since we restrict ourselves to continuous cochains, we can find a normal subgroup U of G such that a",T = a1.1 for a, T E U; and there is a cocycle aG,T co homologous to a",T such that a~,l = 1. Hence we can always find a finite normal splitting field for any continuous cocycle. We now wish to prove that if H is a normal subgroup of G, then the second cohomology group D2(A H , GIH) is isomorphic to the

120

6.


subgroup of ~}z(A, G) which splits on H. This theorem can be proved when A is any group whose first cohomology group is trivial; we give the proof for the case that A is the multiplicative group in a field, since a large part of the theorem can be proved without assuming that H is a normal subgroup. Our first step is to prove Lemma 3: If an element of the cohomology group D2(A, G) splits on a normal subgroup H, then it can be represented by a cocycle aa.T which takes values in A H , and which depends only on the cosets of a and 7' modulo H. We recall that an element {a} of D2(A, C) is an equivalence class of cocycles; this equivalence class yields a group extension G{a} containing A as normal subgroup such thatG/A "-' G. The equivalent cocycles in {a} correspond to the different choices of representatives for the cosets of G modulo A. We now proceed to prove the lemma. Let {a} be an element of D2(A, G) which splits on H, where H for the moment is any subgroup, not necessarily normal. Let aa.T be any co cycle in {a}. Then there is a cochain ay on (A, H) such that

Extend ay to a cochain aa on (A, G), and write

Thus we have defined a cocycle cohomologous to a".T, i.e. with the property that

a~.T E

{a},

Dropping the accent, we see that we have chosen a co cycle a".T E {a} such that if u" are the corresponding coset representatives for the group extension, G, then

4.

CONTINUOUS COCYCLES

121

We must show, however, that the new co cycle aU,T is continuous; for this it is sufficient to show that we can extend a y to a continuous cochain au . Since ay is continuous on (A, H) there is a subgroup U of G such that a y is constant on the cosets of H modulo H n u. Let H = U. -r.(H n U); then the cosets -r.U are distinct, for

We now define au = ay on the co sets -r.U, and give au arbitrary constant values on the remaining cosets. Thus a y may be extended to a continuous cochain au. Let G be written as a sum of cosets modulo H: G

= Hu UUa.H. a.

We see that every element a form

E

G can be written uniquely in the

Now uu lies in the same coset modulo A as choice of coset representatives, writing

Then and hence

Further,

uua.Uy.

We make a new

122

6.


Thus in the group extension defined by {a} we can choose coset representatives u,,' such that

for all a E G, all y E H. Let a~,T be the cocycle in {a} determined by this choice of representatives; then

Thus a~.'T

=

a~.ry •

The proof that this new cocycle is continuous is left to the reader. We drop the accents again, and see that we have constructed a co cycle a",T in {a} such that the corresponding coset representatives u" satisfy the relation U"U y = U"Y • Consider now the associativity relation

We set a

=

y,

T

= y', and p

=

a, obtaining

But ay,y = 1, and hence

Next we define, for a

i

H, -I.

1.fJa

(8)

=

1 y 1 ~ k,j 8y- aY,a

y

where (} is an element of A. Then

Hence

,

4.


123

Now 8 may be chosen so that rPu(8) =I=- OJ for this value of 8 we have also rPy,,,(8) =I=- O. If we notice also that rP,,«()) = rP"Y«())' we see that if 8 is chosen so that rP,,( 8) = rP"Y( ()), we see that if () is chosen so that rP,,«()) =I=- 0, then rPy,,,y{8) =I=- 0 for all y, y' E H. We therefore choose suitable values 81 , ()2' "', ()r for the cosets Ha 1H, Ha 2H, , HarH, and define .

rP')I;;;uuy = Vy

V OY '

Let the cocycle associated with this new choice of coset representatives be denoted by C",r' We shall show that this is the co cycle we are looking for. First we have V,,(VrVy) (v"v r ) Vy

=

V"Vry

=

C",ryV"ry,

= C",rV"rVy = C",rV"ry'

Hence Secondly we have V,,(VyV r )

=

V"Vyr

=

c",YTvuyr,

(v"Vy) Vr

=

VuyVr

=

cuy,rvuyr •

Hence Thirdly we have

= VyC",rV"r = C",rV·,,,r, (VyV,,) Vr = VuyVr = Cvy,rVyur' Vy(V"V r)

124

6.


Hence Cya.T

=

C~ • .,.

•

Thus if {a} is an element of D2(A, G) which splits on any subgroup H, not necessarily normal, then there is a cocycle C".T in {a} with the properties

(These are the Brauer factor set relations for a non-normal splitting field.) In particular, when H is a normal subgroup,

Hence CO,T lies in A H , and clearly it depends only on the cosets of a and T modulo H. This completes the proof of Lemma 3. Combining the results of Lemmas 1, 2 and 3, we see that we have proved the existence of a homomorphism of D2(A H , G/H) onto the subgroup of D2(A, G) which splits on H. We shall now prove that this is an isomorphism. Let {a H } be an element of D2(A H , G/H) which is mapped into the identity of D2(A, G). We may choose a representative a OH ,7H of {a H } such that aH,H = 1. Suppose a"H,7H is mapped into the cocycle a",T of (A, G); then aLl = 1, and by hypothesis bjJ"T aa,T =-b-' OT

We wish to replace b" by a cohomologous cochain depending only on the coset of the argument modulo H. Set a = Yl' 'T = Y2 :

Since the first cohomology group is trivial on H, we can write

4.


Now define d" = boco-t, so that a",T that d y = b"C"-l = 1. Next, setting T

125

= d"d"T/d(JT' We notice first =

y, we have

but Hence

Thus do is a cochain depending only on the coset of a modulo H. Finally, if we set a = y, we have

but Hence

We sum up our results in Theorem 9: The second cohomology group f>2(AH' G/H) is isomorphic to the subgroup of f>2(A, G) whose elements split on H. We conclude this section with the following theorem: Theorem 10:

Let H be a subgroup of finite index n in G. Let

{a} be an element of the cohomology group f>2(A, G) which splits on H. Then {a}n = 1 where 1 is the identity of f>2(A, G). Proof: We saw in the proof of Lemma 3 that an element {a} of f>z(A, G) which splits on H can be represented by a cocycle a".T such that a".T = a".TY' This part of the proof of Lemma 3 did not require the triviality of the first cohomology group. Let G

n

n

v=l

v=!

= U T)/ = U TT)/.

126

6.


Consider the associativity relation

Set p = T, ,and take the product from v = 1 to n, using the fact that a".T = a",Ty' We obtain where n

ftl =

II a ,T," ,=1

CHAPTER SEVEN

The Fi rst and Second I neq ual ities

1. Introduction Let k be a complete field with discrete valuation and a finite residue class field of characteristic p. Let C be the separable part of the algebraic closure of k, r the Galois group of C I k. Roughly stated, the aim of Local Class Field Theory is to give a description of the subfields of C by means of certain objects in the ground field k. So far it has not been found possible to give such a description except for subfields K of C whose Galois group is abelian. In this abelian case we shall show how to set up a welldetermined isomorphism between the Galois group and the quotient group of k* modulo a certain subgroup. When K I k is finite, this subgroup consists of the norms of non-zero elements of K; when K. k is infinite we extend our definitions in a natural way so that the subgroup may still be considered as a norm group. Let E be an extension of k of finite degree n. Let S(E I k) be the subgroup of D2(C, F) which splits on E; denote the order of this subgroup by [E : k]. Our immediate task is to study S(E I k); in this chapter we shall show that for any extension of degree n, S(E I k) is a cyclic group of order n. 2. Unramified Extensions We recall the following facts from Chapter Four: Unramified extensions are completely determined by their residue class fields; namely, to every separable extension of the residue class field k there corresponds one and only one unramified extension of k. The Galois group of an unramified extension is isomorphic to the 127

128

7.

THE FIRST AND SECOND INEQUALITIES

Galois group of the residue class field extension. In the case under consideration the residue class field is finite, and of characteristic p. Let k have q = pr elements; it is well-known that for every integer n there exists one, and (up to an isomorphism) only one extension Elk of degree n, namely the splitting field of ;t!l" - x. The Galois group of Elk is cyclic, and has a canonical generator 0': 0'(0:) = afJ for all 0: E E. We immediately deduce Theorem 1: If k is a finite field, then all unramified extensions of k are cyclic. N ext we extablish the Lemma:

Every element a in k is the norm of an element x in E.

Proof:

The mapping x -- N(x) is a homomorphism of E* into k*. The kernel consists of these elements for which xl+q+ql+ ... +qn-l

Thus the order of the kernel is Hence

q"-l

=

x q-l

~

(qn - 1)/(q - 1).

. . number of Images obtamed =

=

1.

order of E* d fk I ~ q - 1. or er 0 erne

Since k* contains only q - 1 elements, the equality sign must hold. Hence every element of k* is a norm. Theorem 2: If T I k is an unramified extension, then every unit in k is the norm of some unit in T.

Proof: Let € be a given unit in k. By the preceding lemma, the residue class of k contallllllg € is the norm of a residue class in T; this residue class will contain a unit Eo . Since the formation of the norm in T I k is the same as in T I k, we see that € = NEo mod p. Hence €/NE o is a unit € l ' and €l I mod p.

=

2.

129

UNRAMIFIED EXTENSIONS

Now let €m be a unit of k such that construct a unit Em of T such that

€m

=

I mod pm; we shall

and

Such a unit Em' if it exists, will have the form Em = I + 7T mY m , where Ym is an integer of T, and 7T is a prime in k (hence in T). Let €m = I + 7Tm Xm, where Xm is an integer of k. If

we have N(I

+ 7Tmym) = (I + 7Tmym)1+o+... +on- 1 = 1 + 7Tm Xm mod pm+!.

Hence

Now not all elements of T have trace zero. Suppose x is an element of T such that S(x) = a =p 0 in k; let xm denote the residue class of k which contains Xm . Then S(xmx/a) = xm . Hence if Y m is any integer of T in the residue class xmx/a, then S(Ym) = Xm mod p. From these remarks it follows that we can construct a unit Em in T such that and

The theorem now follows easily. We have €=NEomodp;

Then we construct a unit El - I mod p such that

Similarly we construct a unit E2

and so on.

=

1 mod p2 such that

130

7.


Since the units Ei converge to 1, the product EoElE2 ... converges to a unit E' of T, and clearly € = NE'. This completes the proof of the theorem. We now introduce the convention of representing groups by their generic elements: for example, 0: shall denote the group of all non-zero elements 0: E h, NA the subgroup of 0: whose elements are norms of the non-zero elements A E E.

Lemma: If T I h is unramified, of degree n, then the factor group o:/NA is cyclic of order n. Proof: Since T I h is unramified, a prime 7T in h remains a prime in T. Thus any element A in T can be written A = 7T"E, where E is a unit in T. Then NA = 7T"nNE. We have just seen that the group NE coincides with the group of units in h. Thus the elements of h whose ordinals are divisible by n are precisily the norms NA. Hence o:/NA is cyclic of order n; the cosets are (NA, TTN A, ... , 7Tn - 1 NA). Let E I h be any finite normal extension. Let R be the subgroup of r which leaves E fixed; then rlR is the Galois group of E I h. We recall that the second cohomology group Sj2(E, rlR) is isomorphic to S(E I h). Hence, when E is a cyclic extension (in particular when E is unramified) we have

S(E I k) r v Sj2(E, r/H) ~ invari:::::ments = rx/NA.

3. The First Inequality Let E I h be any extension of degree n. The proof that S(E I h) is a cyclic group of order n proceeds in two stages. The first step is taken in our next theorem:

Theorem 3: If E I h is any extension of degree n, then S(E I h) contains a cyclic subgroup of order n, and [E : h] ? n.

3.

131

THE FIRST INEQUALITY

Proof: Case 1: E is an unramified extension: E = Tn. In this case our preceding remarks give us the result immediately. Unramified extensions are cyclic, and so we have

S(E I k)~ fJ 2(E, T/H) "" rx/NA.

But, by the lemma, rxl N A is cyclic for an unramified extension. Hence when E is unramified we have the even stronger statement that SeE I k) is itself a cyclic group of order nand [E: k] = n. Case 2: E is a totally ramified extension. Let Tn be the unramified extension of degree n; let U be the subgroup of r which leaves Tn fixed. We notice that E ( l Tn = k, since this intersection must be both unramified and totally ramified. Since Tn I k is a normal extension, U is a normal subgroup of hence Hu = UH, and so Hu is the group generated by Hand U and is therefore the subgroup corresponding to E ( l Tn = k; hence Hu = The intersection H ( l U corresponds to the compositum ETn. The Galois group of ETn I E is

r;

r.

r

HU

H

which is the Galois group of Tn I k. Hence ETn I E is cyclic of degree n, and it is easy to see that ETn I E is unramified. We shall now show that SeE I k) contains S(Tn I k), so let a".u."",u be a cocycle which splits on Tn. Since Tn is cyclic we may assume the cocyle has the form

= a[ I';V ]-[;]-[;], where a is an element of k. First we remark that we may choose a in H. For since aU C Hu = r, we can write a = hu; we then replace a by au-I. Thus we can assume that a is in H. Then a"ll

The restriction of

(l

H

=

a".u."",u

aVU

(l

aVH

=

aV(U

to H is therefore

(l

H).

a"v(unH)."",(unH) •

132

7.


This can be regarded as a cocycle for (ETn, HIU n H); we shall show that it is cohomologous to the identity cocycle. Since ETn I H is cyclic it suffices to prove that the element a lies in the group of norms NET"IE' If 7T is a prime in k, II a prime in E, then, since Elk is totally ramified, 7T = € IIn. Thus the ordinal of a with respect to II is a multiple of n, and since ETn I E is unramified, this is precisely the condition that a lie in NET"IE . Hence a"VlI,,,/1U splits on E, and so SeE I k) :l S(Tn I k). This proves the theorem in this case. Case 3: E is an arbitrary extension. Let T be the inertia field of Elk, and, as in Case 2, let Tn be the unramified extension of degree n, corresponding to the subgroup U of r. Then T is a subfield of Tn, corresponding to the subgroup HU, and E I T is totally ramified. Let a",7 be a cocycle of (C, T) which splits on Tn; if we restrict the subscripts of a".7 to lie in UH, we obtain a cocycle of (C, UH) which splits on Tn. Since E I T is totally ramified, we can apply the result of Case 2, showing that a",7 splits on E. Hence SeE I k) :l S(Tn I k). This proves the theorem in this final case. The result we have just proved, namely that for any extension Elk of degree n, [E: k] ~ n is known as the First Inequality. In Section 4 we shall prove the Second Inequality, which states that [E : k] ~ n. Before we proceed with this proof, however, we remark that it will have the following important consequence:

Theorem 4: If Elk is any extension of degree n, then SeE I k) is a cyclic group of order n. Further, if Tn is the unramified extension of degree n, then the cocycles of (C, T) which split on E are precisely those which split on Tn . We have already remarked that every cocycle has a splitting field of finite degree (Chapter 6, Section 4). Hence every cocycle has an unramified splitting field of the same degree. From this we shall deduce Theorem 5: ,v2(C, T) is isomorphic to the additive group of rational numbers modulo 1.

4.

THE SECOND INEQUALITY: A REDUCTION STEP

133

4. The Second Inequality: A Reduction Step The aim of this section and the next is to prove

Theorem 6: If Elk is any extension of degree n, then [E : k] ~ n. The first stage in our proof is a reduction to the case where Elk is an extension of prime degree. Let PI be any prime number. 8(E I k) is an abelian group, and so the elements of 8(E I k) with period a power of PI form a subgroup which we denote by 8 1'1 (E I k); let the order of 8 P1 (E I k) be [E: k]Pl' We first prove some elementary properties of the symbol [E : k]Pl . Let E' be a finite extension of E. It is easy to see that 8 P1 (E I k) is contained in the corresponding group 81>I(E' I k) of order [E' : k]Pl . We have the following results:

(I)

[E: k]Pl ~ [E' : k]Pl .

This follows at once from the definition, (2)

[E': k]Pl ~ [E' : E]Pl [E : k]Pl .

8 P1 (E' I E) is a subgroup of ~2(C, H) where H is the subgroup of r which leaves E fixed. Thus we can map 8 P1 (E' I k) into 81>I(E' I E) by the canonical homomorphism, i.e. by restricting the subscripts to H. Obviously the number of images is ~ [E' : E]Pl . The kernel consists of these elements of 81>I(E' I k) which are cohomologous to the identity when their subscripts are restricted to H; thus the kernel consists of the elements of 81>I(E' I k) which split on E, namely 8 P1 (E I k), which contains [E : k]Pl elements. Hence [E' : k]Pl ~ [E' : E]1'l [E : k]Pl .

(3) If PI does not divide deg (E I k), then [E : k]Pl = 1. For every element {a} E 8(E I k) we have {a}n = {I} (Chapter 6, Theorem 9). But for every element {a} E 8 P1 (E I k), we have {a}pl = {I}. If PI does not divide n, these statements imply that {a} = {I}. Hence [E : k]Pl = 1. Let K be a finite normal field containing E and let G be its Galois group; let G I be the subgroup of G which leaves E fixed.

134

7.


I

K

t

Q

!

G,

1E

Q,

G

E,

k Let Q be a PI-Sylow subgroup of G I , E2 the corresponding field. Now Q, considered as a pcsubgroup of G may be imbedded in a PI-Sylow subgroup Q' of G; the corresponding field EI is a subfield of E 2 • The degree of E2 I EI is prime to PI since Q is a maximal PI-subgroup of G I ; similarly the degree of EI I k is prime to PI, since Q' is a maximal PI-subgroup of G. Now the degree of E2 I EI is a power

5.

THE SECOND INEQUALITY CONCLUDED

135

of PI' which must be (E I k)p, , i.e. the PI-contribution to the degree (E I k) since (E21 E1 ) (El 1 k)

= (E2 I k) =

(E21 E) (E 1 k).

The extension E2 I El can thus be expressed as a cyclic tower of degree (E I k)Pl . Assuming that the second inequality is true for cyclic extensions of prime degree, and applying (2) above, we obtain Now (by (1) above) (by (2) above) (El 1 k) is prime to PI •

It follows that [E: k]Pl is finite for every prime PI. Hence the groups SP1(E I k) are finite for every prime PI , and are different from the identity only if PI divides n. Hence S(E I k) is itself a finite group, and the groups SP1(E I k) are its Sylow subgroups. Thus [E : k] = II [E : k]Vl ~ II (E 1 k)Vl = n. Hence we have proved the second inequality under the assumption that it is true for extensions of prime degree.

S. The Second Inequality Concluded We have now to consider a normal extension K I k of prime degree I. The Galois group of such an extension is cyclic, generated by an element u. We introduce the following notation, continuing our convention that groups shall be denoted by their generic elements: Let A denote the generic element of K*, (X the generic element of k*, E the generic unit of K, € the generic unit of k.

136

7.


Let p, 7T be the prime ideal and a prime in k. Let \p, JI be the prime ideal and a prime in K. Let JIB be an element of K in \p 8 ; i.e. ord JIB ? s. Let f3n be the generic element of k which is 1 mod n. Let Vn be the generic element of k of the form Vn = {3nNE (hence Vn NE mod pn: Vn is a norm residue mod pn). Finally, we denote the index of a group G2 in GI by (GI : G 2 ). First we prove the group-theoretical

=

=

Lemma: Let T be a homomorphism of a group 0: into some other group. We denote the image group by To:, and the kernel by O:T. Then, if {3 is a subgroup of 0:, (0:: fJ) = (To:: TfJ) (O:T: fJT). Proof: Since T(O:T{3) = T({3) , we have (0:: O:T(3) = (To: : T(3). Then (tl! : fJ) = (tl! : tl!l'fJ) (O:TfJ : fJ) = (To: : TfJ) (tl!T : tl!T n fJ)

=

(To:: TfJ) (tl!T: fJT).

Since K 1 k is cyclic, we know that the subgroup S(K 1 k) is isomorphic to the factor group o:jNA. Thus [K: k] = (0: : NA); hence we must prove that (0: : NA) ~ t. We have already established this for unramified extensions, so we shall assume that K is not unramified, and so, since t is a prime, K is totally ramified. The lemma above gives us a first reduction step; for if we map A -- 1A I, and hence 0: --+ 10: I, we obtain (tl! : NA) = (I 0: 1: 1NA i) (€ : NE).

Since K is completely ramified, have to prove (€ : NE) ~ t. We shall consider the index

(10:1: 1NA I)

= 1.

Thus we

(€ : vll ) = (€ : vI) (vI: V2) ••• (Vn-I : vn )

and show that only one of the factors on the right can be different from 1, and that this exceptional factor cannot exceed t. We shall also show that for a high enough value of n (which we can determine precisely) Vn = NE. This will, of course, prove the second inequality.

5.

137


First we consider (€ : VI). Since VI

= f3I N E:::> f31€!'

we have

Mapping the units € into their residue classes modulo p, and applying the lemma, we have (€ : vI) ~ (€ : f31€!) = (k~ : k~').

When I=p, the mapping

k: k:

\= 1 (€ : vI) I ~ I

--+

I

when when

is an isomorphism. Hence

l=p, l*"p.

Next we consider (vn : vn+l) (n ~ 1). Since /3n :::) /3n+l

+

, we

have

We shall now compute the norm of 1 xlls' where x integer in k. We have !-1 N(1 + xlI.) = (1 + xlI:").

IS

an

II "=0

Hence !

N(1

+ xlI") = 1 + k x"(E'lI~(I1)), .=1

where

~'

ranges over all polynomials 4>.(a)

=

ao + ala

+ ... + al-la-1

which have precisely V coefficients equal to 1, and the remaining 1 - V coefficients equal to zero. When v = 1, E'lI~(I1) = 11 B B

When v = I,

+ 1111s + ... + 11111-1 = S(lI). 8

8

138

7.


In general, if r/J.(a) is a polynomial of the type described, so is air/J.(a) for all indices i (0 ~ i ~ I - 1). For v < I, all the air/J.(a) are distinct. Indeed, if air/J.(a) = air/J.(a), we have ai-ir/J.(a) = r/J.(a). Let h be the minimum exponent such that ukr/J.(a) = r/J.(a); since a1r/J.(a) = r/J.(a), it is easy to see that h is a factor of I. Hence h = 1 (l is a prime). Thus ao + ala + ... + al_1a- 1 =r/J.(a) = ar/J.(a) = al-1 + aoa + ... + al_2az- 1. Hence all the ai are equal, and

4>( a) = 1 + a + a2

+ ... + aZ-l.

Hence for v < I all the associated polynomials air/J.(a) are distinct. Thus for 1 < v < I, each such set of associated polynomials contributes a term to the coefficient of x·, namely

Since the coefficient of x· is a sum of terms of this form, it is itself of this form, and we obtain finally 1-1

N(l

+ xIIB) = 1 + xS(IIB) + xZN(IIs) + I- x·S(II.B). _2

We shall show that the dominating factor in this expansion is either XS(ll8) or XW(ll8). The ordinal of the latter is easily computed; we must now estimate ord S(ll8). S(\PS) is obviously an ideal in h, for S(II~l»)

±

S(II~2»)

= S(IIsm ± II; 2»)

and for any integer (X E h, (XS(ll8) = S«Xll8). Thus S(\PS) = pr for some r, which we shall now determine. S(\ll8)

=

pr p-rS(\ll8)

=

0

=

and

p-(r+1) S(\ll8) =1= 0

and

S(p-(r+1) D\llB) =1= 0

S(D\llHr)

and

and

S(D\llHr+1lZ) ¢ 0 \ll8-(r+1lZ ¢ 1)-1,

S(p-rD\llB)

\llB-lr

0

C0 C 1)-1

5.

where 1) is the different. Let 1) may be calculated. Then S(\lI8)

= pr

139


\lI8-tm-Ir

=

\lim;

CD

s+m-Zr~O

m+s

r~-Z-

we have shown earlier how m

and

\lIs+m-(I+1lr

¢D

and

s + m - (Z

+ 1) r < 0

and

m+s r>-Z--1.

Finally we obtain the result that S(\lI8)

=

p

[m+s]

-I •

We now return to consider the index (Vn : vn+1) = (f3n : f3n n f3n+1NE),

where n ~ 1. We have to examine three cases separately, depending on the relation of n to I and m; the three cases are (1) (2) (3)

(n (n

+ 1) (/- 1) > m, + 1)(/- 1) < m,

(n+l)(I-l)=m.

First we show that cases 2 and 3 can occur only if I characteristic of the residue class field. Since n ~ 1, (n

+ 1) (Z -

1)

~

= p, the

m

implies 2(1 - 1) ~ m. Since I ~ 2, we have I ~ 2(/- 1); hence I ~ m. This implies that 1) = \lim C \lI1 = p, and hence that p-l C 1)-1. If 7T is an element of h, with ordinal 1, 7T-1 E p-l C 1)-1, and so, by the definition of the inverse different,

This means that 7T II, and hence, since I is a prime, I We now consider the three cases in turn. Case 1:

(n

+ 1) (1- 1) > m. = (n + 1) I - m -

Let us define s

1.

= p.

140

7.

Then s

~

n


+ 1;

hence N~'

=

p' C pMl.

Also, m+s=(n+ 1)/-1;

hence

[m i S] = [(n+ 1) -

+]

=

n,

and [

m+s+l] 1

= [n

+ 1] = n + 1.

Thus and

Hence there is an element IIs with ordinal exactly s such that S(IIs) = E7Tn . If x is an integer of k, we have 1-1

N(l

+ xll.) = 1 + xS(ll.) + x1N(ll.) + ~ x·S(ll•• ) 1'=2

= 1 + xS(ll.) mod pMl

=1 +

EX7T"

mod pn+l.

Every element fln may be expressed as 1 + EX7T n; hence every fln is congruent to the norm of a unit modulo pMl. Now R = NE mod pn+1 I"'n -

=:>

~ NE -=

1 mod p"+1

=:>

~ is a I"'n+1 R NE •

Thus we have fln C fln+1NE, and hence (vn : vn+1) = (f3n : f3n () f3n+1NE ) = 1.

Since (n + 1) (1- I) > m, we have (n + r + 1) (1- I) > m for all positive integers r; hence fln+r = fln+r+1En+r' where En+r = 1 X IIB+rl. Since limr->oo En+r = 1, the infinite product II~o En+r converges to a unit E, and fln = NE. Hence (fln: fln () NE) = 1, and so, by the first isomorphism theorem, (f3nNE : NE) = (v n : NE) = 1.

+

5.


141

When 1 is not equal to the characteristic p of the residue class field, this result holds for all n ;:;::, 1. Thus

~

We have already proved that when 1 7'= p, (E : VI) (ex: NA) = (E : NE)

~

1, and hence

I.

Thus we have proved the second inequality in this case. For cases 2 and 3 we have 1 = p. Case 2:

(n

+ 1) (1- 1) < m.

Then (n

+ 1) (1- 1) + 1 = (n + 1) 1- n ~ m,

and

(n

+ 1) I ~ m + n.

Hence

m+ n] ~n + 1, [--1and so

S(jpn) =

lr;-"J C pMl.

On the other hand, N(jpn) = pn. Hence if IIn is any element of jpn and x is any integer of k, we have 1-1

N(l

+ xIIn) = 1 + xS(IIn) + xW(IIn) + ~ x"S(IIvn ) "=2

=

1

+ xIN(IIn) mod pn+1.

Let us choose an element IIn with ordinal exactly n; then N(IIn} = E7Tn • Hence N(l

+ xIIn) = 1 +

EXl71'n

mod pn+1.

Let x run through a system of representatives (Xl' ... , Xq) of the residue classes modulo p. Then every element f3n is congruent to some 1 + X i 7Tn modulo pn+l. Since 1 = P (the characteristic of the residue class field), the mapping x __ Xl is an isomorphism of the residue class field, and so (XII, X21, ... , x q l ) is also a system of

142

7.


representatives for the residue classes. Hence every fln is congruent EXtl7Tn modulo pn+l. to some 1 Thus every element fln - NE mod pn+l. By exactly the same argument as in Case 1, we obtain the result that (vn : Vn+l) = 1.

+

+

Case 3: (n 1) (/- 1) Here we have (n

and hence

=

m.

+ 1) 1 -

1

= m + n,

[m i n] = [n + 1 -

+] = n.

Thus S(\P n ) = pn, and also N(\P n ) = pn. Hence, if IIn is an element of K with ordinal exactly n and x is an integer of k, we have N(1

+ xIIn) = 1 + 7Tncf>(X) mod pn+l,

where r/J(x) is a polynomial of degree 1 = p. Let (Xl' X 2 , ••• , Xq) be a system of residue classes modulo p. Then, as in Case 2, every element fln is congruent to some one of the q elements 1 X i 7T n mod pn+l. Since a polynomial r/J(x) of degree I may take the same value for as many as I different X t , the elements (r/J(x l ), ••• , r/J(xq )) may represent only T of the residue classes, where q ;): T ;): q/ I. Now every fln which is congruent to some one of the T distinct elements 1 + r/J( Xi) 7T n is congruent to the norm of a unit mod n+l, and hence lies in the group fln+lNE. It follows that

+

(f3n : f3n+INE) = !L ~ 1; T

hence We have now proved the second inequality for the case 1= p. For we have When l=p,

5.


143

When the index n satisfies the inequality (n we have

+ 1) (1- 1) < m,

+ 1) (I -

1) = m, we have

no, we have (n

+ 1) (1- 1) > m,

When the index n = no is such that (no

Finally for any index n and by Case 1,

>

and

Thus if N

>

Vn

= NE.

no , and

Vn

= NE.

Hence, finally (€ : NE) :::;;; t. This completes the proof of the second inequality for completely ramified cyclic extensions of prime degree; this, as we have seen, is sufficient to prove the inequality in all cases. The fact that [K: k] = n means that when K I k is cyclic with degree equal to the characteristic of the residue class field, Case 3 actually occurs. That is to say, there is an integer no such that (no

+ 1) (I -

1) = m,

and

Then the elements fln o are not all norms of units, while all the elements flno+i (i = 1, 2, 3, ... ) are norms of units. The ideal \y = pno+l is called the conductor of the extension, and we see that

where

1)

is the different.

CHAPTER EIGHT

The Norm Residue Symbol 1. The Temporary Symbol (c. K I kj'r:) Let k be a complete field with discrete valuation and a finite residue class field. Let K be a finite normal extension with Galois group G. Let c be any element of the second cohomology group fJ 2(K, G), and let aU,T be a representative cocycle; then aU,T satisfies the associativity relation

We shall make repeated use of this relation. Let

then f(T)

E

k, for

[f(T)]P

= II a~'T = II ap,~p,UT = II apu,T ueG

ueG

a p ,UT

ueG

sInce

Hence [f(T)]P

= II apu., = II a U•T = f(T) , ueG

ueG

Now let us consider the effect of taking another representative co cycle for c; let

144

1.

THE TEMPORARY SYMBOL

145

Then

Let us denote the norm group of K I k by NKlkA; when we can do so without confusion we shall omit the subscript. We see that

II a"•.NA = II b",.NA, "

"

where a",T , ba,T are any representative co cycles of c. We now define the symbol (c, K IkIT) by writing

where a",T is any representative cocycle. Obviously, if C1 and C2 are two elements of fJ 2(K, G) we have

We have also the result

for ( C,

K\ k) = II aO,Tp NA = II a".Ta"T,P NA' au T

P

a

a

T,P

But

II aU"p = NK1kaT,p' a

Hence

=

(II a",.NA ) (II a"T,pNA) "

=

"

C' ~ \k)(c, ~ \k) .

146

8.

THE NORM RESIDUE SYMBOL

Hence the mapping T - - (e, K I kiT) is a homomorphism of the Galois group of K I k into the factor group cx.1 N Klk • Since the image group is commutative, the kernel of this mapping contains the commutator subgroup G' of G.

(1) Let Ko be a subfield of K which is normal over k. Let H be the subgroup of G corresponding to Ko; then H is a normal subgroup. We choose a fixed system of representatives for the cosets of G modulo H; we denote by ii the representative of the coset containing the element u. Thus if y is the generic element of H, we have iiy = yii = ii. Now let e be an element of the cohomology group fJ 2(K, G); let aU,T be a cocycle representing e. We shall now prove that if the degree of Kover Ko is m, then a':,T is cohomologous to a co cycle du ,. for (Ko, GIH), and hence that em may be considered as an element of fJ 2(Ko , GIH). Let

su =I1 aa",u )'

a.),

and define

We shall show that du,T is a co cycle for (Ko, GIH)

By the associativity relation, and hence and hence

Sunstituting these results, we obtain

1.


147

smce

by the normality of H. Now

smce and

Similarly d"y l' T = d" , T • Hence d",T is a co cycle for (Ko, GJH). (Cf. Chapter 6, Section 3, Lemma 2.) We can thus define the symbol [(em, Ko I k)/,rH], which we shall write simply as [(em, Ko I k)JT]. We have (

em, Ko T

I k) -_ II d.N _ ii, Kolk a

II a",iiaijy,f N _ a _ Kolk' a.Y

y,UT

Now as ii runs through a complete system of coset representatives. so does aT; hence

II ay,ii = II a y .';;" ii

a

Thus we have

Since the group N Klk is contained in the group N Kolk (this follows from the transitivity of the norm), we have

We may choose::;

=

T,

and so obtain finally

8.

148


(2) Let Ko be any subfield of K, not necessarily normal. Let us write as before, G = U iiH. By the symbol Res Ko (c) we shall mean the equivalence class of co cycles obtained by restricting to H the subscripts of each a in c. Let T be an element of H; then we can form the symbol U •T

(

ReS K o (c),

KI Ko) - IT -

'Y

T

=

a'Y •• NKIKo

IT a~.•NKlk iJ,T

using the transitivity of the norm. The associativity relation gives

hence ali

1'.'

Since

T

E

=

aij.-rOiiY.!.

a-u.j"r

.

H

IT aij,'Y = II aij,YT' 'Y

l'

so we have

Finally we have, for all NKoik

(

T

E

H,

ReSKO (c), 7'

KI Ko) _ (C, K. I k) 7'

(3) Let,.\ be an isomorphic map of K onto K\ under which k is mapped onto kA. If c is an element of 5 2(K) represented by aa .• , we define the element cAin 5 2 (KA) to be the class of cocycles represented by a~.T = bMrl.ATA-l • It is easily seen that

1.

149


(4) In order to describe the fourth important property of our symbol [(c, K I k)jT] we must first introduce a group-theoretical notion. Let G be a group, H a subgroup of finite index; let G', H' be the commutator subgroups of G and H respectively. Let us choose fixed representatives Ui for the cosets of G modulo H: G = U HUi; define a = Ui for every element U E HUi . Consider the element uiTUiT-1 where T is any element of G; since UiT E HUiT, we have uiT-1 E T- 1U(IH, and hence uiTUiT-I E H. We define the transfer (Vorlagerung) of T into H to be

IT U{rUiT-IH'.

VG->H(T) =

Uj

V(T) does not depend on the choice of coset representatives Ui. If we write UiT = Uj , we have V(T) =

IT U;TU-;IH'.

Replacing Ui by YiUi , we obtain V'(T) =

IT Yi a ( raiV;IH' = IT

i 1H' =

Ui TU

V(T)

since cosets of H modulo H' commute with one another. We can also write G as a union of left cosets modulo H: G = U u-;tH. Let us define a = uil for every element U E uilH. Then we have ii =

0;1

U E o;lH u-1 E Hu, 0---1 -

=

-

at 0--- 1-

1

= 0;1.

Hence a = u-c1 . We can define another transfer in terms of the by writing

a

Let uiT-1 = Urn; as i runs through all possible indices, so does m. We have uiT-1 E HUm, hence Ui E HUmT, and so UmT = Ui. Hence, finally,

Thus we may define the transfer by using either left or right coset representatives.

150

8.


It is fairly clear that the transfer is a homomorphic map of G into BIB'. For

where we write

Uj

=

U"T,

and then

Hence we have V(-rp) = V(T) V(P).

We notice also that

since cosets modulo B' are commutative. Thus the kernel under the transfer map contains the commutator subgroup G'; hence the transfer defines a homomorphism from GIG' into BIB'. Now let G be the Galois group of the extension K I k, and let B be the subgroup corresponding to an intermediate field K o ' We shall study the symbol (

ReSKO (c), K I K o) VG ....H(r) •

Since (

ReSKO (c), K I K o) _ H' -1,

we have immediately (

ReSKO (c), K I VG-+H(r)

KO) _- II (ReSKO(c), K I KO) -- II II a ai

1 u.-ru.-rI1

U

i

Y

We can use the associativity relation to show that

_ "I,ajTajr-

1

NKIK 0 •

1.


151

But II." a~,T,ujT-l lies in N KIKo. Hence

Now

If we replace UiT by Uj , then as Ui ranges over all coset representatives, so does Uj. Thus we have

We use the associativity relation again, and obtain ay,aj'Tayaju,jl

=

-raj

a Uj 1 ,ulra ya1,-r

To conclude this computation, we must show that the norm group N Klk is contained in N KIKo • We have NKlk(A)

= II A-ra, = II (II AU,)." y,ai

y

at

= NKIKo (II AU,) E NKIKo.

u,

Hence

8.

152


and so finally (

ResKO (c),

KI Ko) = (C, KIk) N

V G ...H () T

T

K1Ko •

Let K I k be a cyclic extension of degree nj let a be a generator of the Galois group G. We recall that each group extension defined by K and G can be described by a coset representative U a and an element a E k such that u an = a. The element c of the cohomology group Y)2(K, G) which corresponds to this extension is represented by a cocycle if v+p,. It follows that a C mI where m = ("', mp , .. -). On the other hand, a contains all ideals mpJpt -t mp/ P2 + ... + mp/ p ,; the set of these is everywhere dense in mI; so a is dense in mI. But a is closed, so a = mI. Now let G be a topological group with a Hausdorff topology defined by a fundamental system of neighborhoods of the identity given by certain subgroups of finite index. (An example of such a group is the Galois group G of A). It is easily shown that if G is complete in this topology, then it is compact. Let a be an element of G, and let (a) be the closure of the cyclic subgroup generated by a; we wish to find a description for (a). Clearly we have a homomorphism of I into (a) defined by v---+- a Y

•

This map is continuous in the inherited topology of I; for let V be a neighborhood of the identity in (a), with index j in (a). Then v _ 0 modj ~ a E V; but since the numbers v - 0 modj form a neighborhood of zero in I, this is precisely the statement of continuity. The mapping v ---+- a may now be extended from the dense subset I to the whole space 1; that is to say, we may define am for every m in 1. The extended mapping is continuous, and the extension is unique. We may also show that the mapping (a, m) ---+- am is continuous from G X 1 to G. Y

Y

168

9.

THE EXISTENCE THEOREM

The usual rules for exponentiation may be easily verified: (am)ml

=

a mm',

aT = Ta -¢> (aT)m = amTm. Since the map a -- am is continuous from 1 to G, and 1 is compact, its image is closed, and hence contains (7',,,) = • 1>(7'11) for all units

E.

E

=

(A7',,,I k) = (~) (A7'~k)

7'

7.

UNIQUENESS OF THE NORM RESIDUE SYMBOL

Hence if a =

E7T"

189

is any element of k,

which proves our assertion that (A I kiT) is uniquely determined by the three properties listed above.

CHAPTER TEN

Applications and Illustrations

1. Fields with Perfect Residue Class Field

Let k be a complete field under a non-archimedean valuation; let k be its residue class field. We consider the case in which k is perfect and of characteristic p > 0; this certainly includes the case we have been discussing in the previous three chapters, since all finite fields are perfect. We make a slight change in our usual notation: Il: shall denote the generic element of k. [Il:] shall denote the generic element of k; namely the residue class to which Il: belongs. We shall construct in k a particular system of representatives, to be denoted by Ii, of the residue classes [ex]. First we notice that if Il: = fJ mod p', then exP fJP mod p'+1. For if Il: = fJ Y1T', then

+

N ow consider the residue class [ex]; since k is perfect, the residue class [Il:]P-' is well-defined. Let Il:, be any element of [Il: ]P-', and consider the sequence {Il:~'}. We have Il:,-- a~~1' mod p, so that Il:~' = Il:~:::' mod p'+I'; hence {Il:~'} is a Cauchy sequence. Since all the terms of this sequence lie in the residue class [ex] so does its limit; we denote this limit by Ii. The limit Ii is independent of the choice of the elements Il:,; for if {1l:~p1 is another such sequence with limit Ii', we have Il:~P' = Il:~' mod p', and hence Ii = Ii'. We choose Ii as representative in k of the residue class [Il:]. 190

1.

191

FIELDS WITH PERFECT RESIDUE CLASS FIELD

Let [,8] be another element of k, with representative

p = lim~~·. Since

the representative

Thus the representatives Ii are multiplicatively isomorphic to the residue classes of k. Further, the Ii are uniquely determined by this property; for let Ii, /1, ... be another set with the same property. Then

We now make the additional assumption that the representatives Ii form a field. We shall show that this field must be isomorphic to k. For let Ii = y, and let IX, ,8, Y be elements of k such that IX E [IX], ,8 E [,8], y E [y]. Then IX = Ii, ,8 y = y mod p; hence IX ,8 - y mod p, and therefore [IX] [,8] = [y]. It follows that the representatives Ii can form a field only if k has the same charac-

+ /1

+

+

/1,

teristic as k. On the other hand, suppose k has characteristic p. Then [IX]

+ ~] = ex + ~ = lim y~V ,

where

Thus we may write Yv = IXv +,8v where IXv hence ex

E

v [IX]P- , ,8.

E

v [,8]P_ ;

+ ~ = lim (IXv + ~.)P" = lim (ex;" + ~;V) = Ii + p.

We can now sum up our discussion by giving a precise description of fields of characteristic p with perfect residue class field. We recall

192

to.

APPLICATIONS AND ILLUSTRATIONS

that if 7T is a prime element in k, then k consists of all power series 1: &(v)7T v • Hence we have Theorem 1: Fields of characteristic p > 0 with perfect residue class field are isomorphic to fields of formal power series over the residue class field as field of constants. We now restrict ourselves to the case in which the residue class field k is finite, containing q = pr elements. Theorem 2: k contains the (q - l)-st roots of unity, and no other roots of unity with period prime to p. Proof: The non-zero elements [IX] of k form a cyclic group of order q - 1. Hence the representatives ii in k also form a cyclic group of order q - 1; these representatives are therefore the (q - l)-st roots of unity. This proves the first assertion of our theorem. Suppose, next, that { is a primitive m-th root of unity in k, where (m, p) = 1. Then there is a positive integer 1 such that q 1 mod m, i.e. {qi = {. Suppose { lies in the residue class [IX]; the representative of [IX] is given by

=

~

any element of [a:Jr· •

Clearly we obtain the same limit if we restrict ourselves to the subsequence{a~;:V} = {a~I"} where a" E [1X]q-I". We may obviously r -I". choose IX" =..,q Hence ii

= lim (~q_I")ql,, =

~.

Now ii is an element of a cyclic group of order (q {q-l = 1, i.e. m divides (q - 1).

1). Hence

This completes the proof. The structure of complete fields k of characteristic zero with perfect residue class fields k of characteristic p > 0 has been investigated by Witt (Crelle's Journal, 176 (1936), p. 126). We consider here only the case in which the residue class field is finite, containing q = pi elements. k contains a subfield isomorphic to the rational numbers; and the valuation of k induces the p-adic valuation on this subfield.

2.


193

Hence, since k is complete, it contains a subfield isomorhic to the p-adic numbers ~ . k also contains the (q - l)-th roots of unity; hence k contains ~m where { is a primitive (q - I )-th root. Let 71" be a prime element in k; then k consists of all elements of the form

Let p = me. Then we may write

i.e. every element

Il:

E

k may be expressed as

Hence k is a finite extension of Rp: k

=

Rp({, 71").

2. The Norm Residue Symbol for Certain Power Series Fields

Let k be the field of formal power series over a field F of characteristic p > 0; let Fo be the prime field. If C is the separable part of the algebraic closure of k, r the Galois group of C I k, then the mapping T -- (C I kiT) is a (1, 1) correspondence between rand k. We denote the inverse mapping by cf>, i.e., if a = (C I kjT), then cf>(a) = T. We study expressions of the form

(~ty)

- (x:),

where we choose the same value for if (xyl ~h is another value,

(xy/~)

for both terms. Clearly

10.

194 where

II


is an element in the prime field. The expression

(:t

YI -

(~)

lies in the prime field, since both terms are roots of the equation gp - g = xy, and these roots differ only by elements of the prime field. Now let x and y be fixed elements of k, u any element of k. Consider the mapping

We shall denote this map by xdy and call it a differential. The image of u under the differential xdy will be denoted by ~ uxdy, i.e.

- (UXy)4>(YI -(uxY) -. f uxdy-~

~

We shall see later how the properties of the map xdy justify this notation and terminology. Two differentials are defined to be equal if and only if their effects coincide on every element of k, i.e. for all

uek.

The sum of two differentials is defined by the relation

We now deduce some elementary properties of these differentials: 1.

(Xl

+x

2)

dy

= xldy + x2dy.

This follows from the linearity of the operator 1/ fJ and the fact that cf>(y) is an isomorphism. 2. d(yz)

= ydz + zdy.

2. We have, for every

195

THE NORM RESIDUE SYMBOL U E

k,

f ud(yz) = (U;zt(Y)(Z) _ (U;Z) = [(U;ztZ) _ (U;Z)rY) + (U;ztY) _ (U;Z) y

= [f UYdzr + f uzdy )

=

f uydz + f uzdy,

smce ~ uydz lies in the prime field, and hence is unaffected by

CP(y)· 3. dy" = ndy..-l.

This follows by induction using (2) if n is positive. If n is negative, we write dy = d(y-n+1yn) = y-n+1dyn

+ y"(1 -

n) y-n dy (using 2).

Hence ndy = y-n+1dy",

i.e.

dyn = ny..-ldy.

From this result we deduce at once the formulas dyP =0, dyPz =yPdz.

Since the field of constants F is perfect, every a hence the differential map is homogeneous: xd(az) = axdz.

The linearity of the map follows directly from

Since

E

F is a p-th power;

10.

196


we must show that (y/ fJ)~(II) = (y/ fJ). This is obvious if Y/ fJ lies in k. If Y/ fJ = 8 is not in k, 8 satisfies the irreducible equation gP - g - Y = 0; hence Y E Nk(o)lk' From the properties of the norm residue symbol we deduce that cp(y) acts like the identity on k(8); hence pdy = in this case also. In particular we have pd(yz) = 0, so Pydz = - Pzdy. Then

°

Thus the differential map ydz is linear in its arguments. We now evaluate some special "integrals".

f (x

P -

dy (XP - X)~(Y) x) - = - y r

(XP - -X) = x(,) - x = 0 -P ,

since x lies in the ground field k. (2)

If

C E

F and y -::/= 0, (n -=1= 0)

(a) If n i= 0, let n = prm where (p, m) we may write c in the form c = epr • Then

(n

= 0).

=

l. Since F is perfect

using (1) above

(b) If n = 0,

2.

197


Set ~ = (cj fJ); since c is an element of F, F(~) I F is a cyclic extensions of the residue class field. Hence k(~) I k is an unramified extension. Thus ordq,(y)

=

I.e.

ordy,

We have

Thus

We can now raise this result to the q-th power; S FlFo(c) remams unaltered. Hence

I> q ord y

o

S

I> q OrdY_l _

-

0

-

()

FIFo c.

Summing these equations we obtain the required result

1.j c dy y = (9"C )4>(11) (3)

If

I xy I ~

- ( P = (ordy) SFIFo(c). C )

1, then pxdy

y f xdy -- (Xp

Now for any a

E

k such that

-a = (a p

= 0, )4>(II)

- (xY) - . p

I a I < 1, it is easily verified that

+ a + a + ...). P

P

2

Hence (xyj fJ) E k, and so (xyj p)fI(lI) = (xyj p). This proves the required result. We notice that this result implies the continuity of the integral in both its arguments, for

f (xdy provided

Ix -

xldYl) Xl

=

f (x -

Xl) dy

+ f xld(y -

Yl)

= 0,

I an I y - Yl I are sufficiently small.

198

10.


(4) If I y I < 1, and x is developed as a power series in y with coefficients in F: x = L cvy, then

Since L cvY converges, there exists an index N such that

I (k

cvyv) y

v>N

I < 1,

and hence

Thus

f xdy = f (k cvyv) dy = ~ f cvYv dy, lI~N

II~::::N

and using 2 above, we obtain

f xdy = ordy .

SFIF

(Cl)·

We may call Cl the residue of x with respect to y, and write = Res y x. In particular, if y = t is a uniformizing parameter, and v is any element of k, we have ~ vdt = S FIFo(Res t v), since ord t = 1. Again let !y I < 1, and let x be written as L cvY. Then we define formally the derivative dxjdy to be the series L vcvyv-l. We shall show that dx = (dxjdy) dy. Consider C_1

! udx = lim ! ud ( N~oo j

j

= }j~

k Cyyv)

II~N

f u ( k vcvyv-l) dy fI~N

=

f u (k vcvyV-l) dy = f u: dy.

This proves our assertion.

2.

199


In particular, if v and yare any elements of k, and t is a uniformizing parameter, we have

Specializing v to be x, we obtain

and, similarly, setting v

=

x/y, we obtain

! x y = Yy ! x dy Yyd dt dt =

SFIFo

( ( x dy )) Rest ydt

.

Rewriting these integrals, we obtain (

pXy)(Y) - (xY) p =

SFIFo

(

pX )(Y) - ( px ) =

SFIFo

,

(Rest( xdt dy )) ' (

Rest

(Xydt dy )) .

This describes the action of ~(y) on the cyclic p-extensions. We shall now discuss the conditions under which a differential ydx is zero. Clearly if y = 0, then ydx = 0, so we shall assume y i= O. Let t be a uniformizing parameter, and write x = ~ cJ". Then

Now ydx = 0 means that u = atr/y (a EF). Then

~uydx =

0 for all u

E

k. We may take

for all a E F and all integers r. As a ranges over F, so does aC r , and so we have either r = 0 (i.e. r = 0 mod p) or C_ r = 0 (since the trace is not identically

10.

200


zero). Hence x contains only powers of t P , and since F is perfect, this implies that x is a p-th power. We can now give the condition for y to be a norm for all cyclic p-extensions k(xl p) I k. For y is a norm for k(xl p) I k o¢> 1>(y) acts like the identity on (xl p) o¢> pxdy = O. Thus if y is a norm for all such extensions, then pxdy = 0 for all Xj hence dy = 0, and so y is a p-th power. 3. Differentials in an Arbitrary Power Series Field Let F be any field, and let k = F{t} be the field of formal power series in t with coefficients in F. If y = ~ cvt we define its derivative V

,

This is easily seen to be linear, F-homogeneous, and continuous in the valuation topology on k. One may establish without difficulty the formal rule d(yz) _ dz dt - Y dt

If

t1

+ z dy

dt·

is another uniformizing parameter, we may prove that

This result is immediate if y is a finite power series in n

tl:y =

kcA. --In

Then

The result then follows for arbitrary power series since the map dldt is continuous.

3.

DIFFERENTIAL IN AN ARBITRARY POWER SERIES FIELD

201

If y and z are any elements of k, and dz/dt i= 0, we may define dy

dz

dyJdt

=

dz/dt .

This clearly does not depend on the choice of the parameter t. We define the residue of y = ~ c vtV with respect to t to be

The residue is thus linear and F-homogeneous. We may notice the special cases: dy

Rest dt

= 0,

when n = - 1; = 0, when n i= - 1. We must now examine the effec1; on the residues of a change in the uniformizing parameter. We obtain

Theorem 3: If t and tl are uniformizing parameters, then Rest (y)

= Restl (y

::1)'

Proof: Since the residue is linear, F-homogeneous and continuous, it suffices to prove the theorem for y = tn. The result is obvious for the trivial change t = altl , so we may assume dt

+ ... • dt --1 +2at 21 1

We have now to show that

°

when n = - 1, and = when n i= - 1. When n ~ 0, the result is clearly true, since t n dt/dt J contains no negative powers of t.

202

10.

When n


= - 1, we have

so the result is true here also. When n < - 1 we consider first the case in which the characteristic of F is zero. In this case we can write

and this vanishes for n i= - 1. Now for any characteristic, and any fixed n

< - 1,

we have

1 dt t- n

dt1

where P(a 2 , as, ... ) is a polynomial constructed quite formally: that is, P( a2 , a3 , ••• ) is a universal polynomial of the av with rational integer coefficients. P( a 2 , a 3 , ••• ) is thus the &ame for all fields, and contains only a finite number of the coefficients a v • But we have just seen that for fields of characteristic zero this polynomial is the zero polynomial. Hence it is the zero polynomial also in fields of characteristic p > O. This completes the proof. To rid ourselves of the dependence of the residue on the uniformizing parameter, we introduce the notion of a differential ydz; this is purely formal. We say that ydz = Ytdzt if and only if dz/dz t = Yt/Y. We now define the residue of a differential: Res (ydz) = ReSt (Y ;:).

This does not depend on the uniformizing parameter, for ReSt (ydz)

=

ReSt (Y

= ReSt! (Y

~;) =

ReSt!

:~ ) =

(Y~:

ReSt!

::1)

(ydz).

4.

CONDUCTOR AND DIFFERENT FOR CYCLIC P-EXTENSIONS

203

In this notation, our earlier results would be written Xy)oi>\Y) (&J

- (xY) &J =

SFIFo

(Res (xdy)),

4. The Conductor and Different for Cyclic p-Extensions Let k be any field of characteristic P; let be a cyclic extension of degree p. We saw in Chapter 9, Section 5, that K = k(a/~) where a E k. We must now investigate the degree of freedom we have in choosing generators for fields of this type. So let

i.e. suppose there are two possible generators cx, ex V

ex =

-

fJ

such that

a,

We may choose the generator a of the Galois group such that acx = ex

Then

afJ = fJ + T,

It follows that

where 0

fJIT aC

We have

cx =

=

O. We have seen that we can replace x by an element involving no positive powers of t; hence wo have only to consider the case

10.

206


of an extension k(w/ I(J) I k where is a constant. An extension of this type, however, is unramified, and so f = !I = D.

s.

The Rational p-adic Field

Let k be the rational p-adic number field ~ , i.e. the completion of the rational field in the valuation induced by the rational prime p. Our aim in this section is to describe the maximal abelian extension A of k; we shall prove

Theorem 4:

A is obtained by adjoining to k all roots of unity.

Proof: We have already seen in Chapter 4 that all unramified extensions of k are obtained by adjoining m-th roots of unity with (m, p) = 1. Thus if we adjoin all such roots of unity we obtain the maximal unramified extension Too, which has norm group U, the group of units in k. Next we consider the field K = k(,), where' is a primitive pr-th root of unity; this is certainly an abelian extension. Now all the primitive pr-th roots of unity are roots of the polynomial p'

x Xpr--l _

1 = 1 + x p· -1 1

+ .. , + X(P-l)p·-l = IT (1

- ~)

of degree (p - l)pr-l = cp(pr). Thus deg (K I k) ::::;;; cp(pr). We notice that if we put x = 1 we obtain

Now if (and '" are two primitive pr-th roots of unity, we have

~ =-

f

= 1

+ ~ + ~2 + ... + ~"-l,

which is an integer of K. Since we can write , = ('")1' for some ft, we have

~ = 1 - (~.)I' = 1 + ~v 1-

~.

1-

~v

+ ~2v + ... + ~(v-l) '

5.

THE RATIONAL P-ADlC FIELD

207

which is also an integer of K. Hence (1 - ')/(1 - 'v) is a unit; so I 1 - , I = 11 I. This shows that

'v

I pi = II 11 -

1= 11 -

~v

nl>(p');

hence the ramification e(K 1 k) ? cp(pr). But the degree n(K 1 k) ::::;;; cp(pr); hence we have e = n = cp(pr) and f = 1. It follows that (1 is a prime in K, and that

n

Thus all powers of p are norms for K 1 k. The whole norm group is therefore NK1k=nP"Vr> vEl

where Vr is a certain subgroup of units such that

We shall now show that Vr is precisely the group of units which are congruent to I modulo pro Consider therefore the group Us consisting of elements 1 + aps (I a 1: : ; ; 1); we restrict ourselves to s ? 1 when p is 0dd and to s ? 2 when p = 2. Then we have (I

+ ap")P = (I + apS+1) mod pS+2,

and so we may write 1

+ aps+1 == (1 + apS)P (1 + bpSH).

Repeating the process, we obtain 1

+ ap·+1 =

(1

+ ap8)P (1 + bps+1)p (1 + Cps+3),

and so on; finally we arrive at the result (1

and hence US +1

= Ul. Ur

=

+ aPS+1) =

(1

+ a'p8)P,

By iteration we obtain UPr-1

=

Ur r-2

='"

=

Up,-l 1

208

10.


when p is odd,

when p = 2. We now consider the two cases separately. When p is odd, we have (1

+ apS)V-l = 1 + (p -

1) aps

== 1 -

ap' modp8+1.

Hence we may write 1 - ap8 = (1

+ apS)JJ-l (1 + bp8+!) •

By repetitlOn of this process we finally reach the result that U~-l = Us. Hence we have

and so U r When p

= Vr • = 2 we notice that U2 = Ua

5Ua = U~

u

5U:

since a number which is congruent to 1 mod 4 is congruent to or 5 mod. 8. Hence we have

Since U;r-1 C Vr , by the same argument as we used when p is odd, it will follow that U r C Vr if we can show that 5 2H is a norm. It is easily verified that 52r-. = N Klk(2 z). Hence Ur C V r , and it follows as before that U r = V r • In all cases, then, we have

+

If we adjoin all pr-th roots of unity, we obtain an extension which we may call K~oo, for which the norm group is U vE1 pv, since U r = 1.

nr

6.

COMPUTATION OF THE INDEX

(0:: 0:11)

209

It follows now that the compositum TooK 00 , which is obtained by adjoining all roots of unity to k, has nor~ group 1. TooK 00 is ,. therefore the maximal abelian extension A. This completes the proof of Theorem 4. 6. Computation of the Index (a: all)

Let k be a complete field, either (1) archimedean, or (2) nonarc hi me dean with finite residue class field. We define normal forms for the valuation in k as follows: Case 1: (a) k is the field of real numbers. Let the normal valuation be the ordinary absolute value. (b) k is the field of complex numbers. Let the normal valuation be the square of the ordinary absolute value. Case 2: If the residue class field of k contains q elements, we define the normal valuation by prescribing 17T I = l/q. For use in a later chapter we prove the following result:

Theorem 5: Let k be a field of the type described above. If n is prime to the characteristic of k, and if all the n-th roots of unity lie in k, then (0: : o:n) = n2 /1 n I where I n I is the normal valuation.

Proof: Case 1:

We must consider the two cases separately. k archimedean.

If k is the real field, the only possible values for n are I and 2 (the real field contains only the first and second roots of unity). We see at once that 12 (0: : 0:1 ) = 1 =

m'

(0:: 0:2 ) =- 2 =

22

121.

If k is the complex field, n may have any value, and every element of k is an n-th power. Hence n2 (0:: 0:") = 1 = mi.

210

10.

Case 2:


k non-archimedean.

We apply the Lemma of Chapter 7, Section 5, with the homomorphism T: 0: -+ 10:1. We have (ex: exn) = (I ex 1: 1ex In) (e : en)

= n{e : en).

We may therefore consider the index (e : en). First let r be so large that 1n7Tr+l 1 ~ 17T2r I, and consider the group U r of elements 1 a7T r (I a 1~ 1). Then

+

{I

+ a7Tr)n = 1 + an7Tr mod 7T 2r _ 1 + an7Tr mod n7Tr +1.

Hence we may write

and repeating the process, we have

Finally

Thus if ord n

'n

=

s, we have

U': =

Ur+••

Let denote the group of n-th roots of unity in k. Suppose r is so large that none of the =1= 1 lie in U Again we apply the Lemma of Chapter 7, Section 5, this time with the homomorphism T: e -+ en. We obtain

'n

r.

Hence

Now (Ur : U r+B) = number of residue classes modulo TT B = where q is the number of elements in the residue class field.

qs,

6.

COMPUTATION OF THE INDEX

Hence

Finally we have the result of the theorem:

(0:: 0:11-)

211

PART THREE

Product Formula and Function Fields in one Variable

CHAPTER ELEVEN

Preparations for the Global Theory

1. The Radical of a Ring

Let R be a commutative ring. An element 0: of R is said to be nilpotent if some power of 0:, say o:n = O. The set of all nilpotent elements of R is called the radical, N. Theorem 1:

The radical is the intersection of all prime ideals

of R.

Proof: Let S be a multiplicative semigroup in R not containing zero. We shall show that there exists a prime ideal of R which does not intersect S. Consider the set of all ideals of R which do not intersect Sj this set is not empty, since the zero ideal belongs to it. The conditions of Zorn's Lemma are easily verified, so there exists an ideal o which is maximal in this set. We contend that 0 is a prime ideal. Let 0:1 ¢ 0, 0:2 ¢ OJ then (0,0:1) respectively (0,0:2 ) are larger than 0 and so contain elements Sl respectively S2 of S. Thus

where n 1 , n 2 are integers, and r 1 , r 2 Hence

E

R.

It follows that 0:10:2 ¢ 0, and so 0 is a prime ideal. Now if 0: EN, o:n = 0 lies in all prime ideals p. On the other hand, if 0: is not in N, the elements 0:, 0: 2, 0:3 , ••• form a semigroup not containing zerOj thus there is a prime ideal p not containing 0:. This completes the proof. 215

216

XI.

PREPARATIONS FOR THE GLOBAL THEORY

2. Kronecker Products of Spaces and Rings

Let k be a commutative field, and let X and Y be vector spaces over k. We form the vector space V whose basis vectors are the elements (x, y) of the Cartesian product of X and Y; the elements of V are therefore of the form ~i=1 o:.(x. ,y.) with 0:. E k, x. E X, y. E Y. Consider the subspace N which consists of those elements of V such that n

~ cx)(x.) A(y.) = 0 i-I

for every linear map of X into k and every linear map A of Y into k. The Kronecker product of X and Y with respect to k is then defined to be the factor space of V mod N:

We may therefore regard the elements of X X k Y as being those of V, where equality is now defined to be congruence modulo N. Hence in X X k Y we have (x

+ x',Y) = (x,y) + (x',y), cx(x,y) = (cxx,y).

To prove the first of these results, we notice that for all linear maps I and A, we have

+ x') A(y) -

lex

lex) A(y) - lex') A(y) = 0

which means that (x

+ x', y) == (x, y) + (x', y) mod N.

Similarly, SInce cxl(x) A(y) = l(ax) A(y) = lex) A(CXY)

for all maps I and A, our second assertion is proved. From these remarks it follows that we may operate formally with the pairs (x, y) as if they were products xy-provided we

2.

KRONECKER PRODUCTS OF SPACES AND RINGS

217

maintain a strict distinction between the elements of the products which come from X and those which come from Y. Theorem 2: If YI' Y2' "', Yn are linearly independent in Y, then XIYI + X2Y2 + .,. + Xn.:V" = 0 in X

Xk

Y if and only if Xl

= .. , = Xn

= O.

Proof: The "if" part is trivial. To prove the "only if" part, we recall that a linear functional defined on a subspace of a vector space may be extended to the whole space. Hence, since the Yi are linearly independent, we can form linear maps Ai such that AlYj) = Dij . Hence, since XlYl

in X

Xk

+ ... + x"yn = 0

Y, we have

for all linear maps I of X; hence Xi This completes the proof.

=

O.

Corollary: In order to test whether I: cx"xvYv = 0 in X X k Y, we express the Yv which occur in terms of linearly independent vectors Y/; say Yv = I: CviY/' Then I: a"xvYv = I: avCvixvY/, which is zero if and only if all the elements I: CXvCviYv are zero. Theorem 3: If Xl' X2 , " ' , Xn are linearly independent in X andYl 'Y2 , "', Ym are linearly independent in Y, then the elements xiYj are linearly independent in X X k Y. Proof: Suppose I: CvpXvYp = 0 in X maps I and A we have

Xk

Y. Then for all linear

We may construct maps Ii , Aj such that li(xv) = Div , \(Yp) Hence

= DiP'

218

XI.


Corollary: If X and Y have finite k-dimension, then so does X X k Y; in fact, dim X

Xk

Y = dim X . dim Y.

Suppose now that X and Yare rings with unit elements, containing k in their centers. We may introduce a multiplication operation in the vector space V by defining (x,y) (x',y') = (xx',yy')

and extending the definition to V by requiring that the distributive law be satisfied. It is easily verified that Y forms a ring under this multi plication. N is a two-sided ideal of this ring. For if ~ cxv(xv ,Yv) EN, then ~ aJ(xv) A(Yv) = 0 for all linear maps. Consider in particular the maps I(x) = l'(ax), A(Y) = ,\'(by), where I', ,\' are arbitrarily given linear maps. We obtain ~ CXv l'(ax v) A'(byv) = 0

for all I', A; hence

~ cx.(axv , byv) = (a, b) ~ cx.(x" ,Yv) lies in N. Similarly [~ cxv(xv ,Yv)] (a, b) lies in N. From this remark it follows that the Kronecker product of X and Y is a ring, namely the residue class ring V! N. Since X and Y have unit elements, X X k Y contains sub rings isomorphic to X and Y, consisting respectively, of the elements (x, 1), (1, y). If we identify X and Y with these isomorphic subnngs, we see that xy = (x, 1) (1, y)

= (x, y) = (1, y) (x, 1)

= yx.

If X o , Yo are subrings of X and Y, we may form the Kronecker product Xo X k Yo; it is easily seen that this is imbedded in X X k Y in a natural way. 3. Composite Extensions Let A and B be arbitrary extension fields of k such that A (\ B = k. We define a composite extension of A and B by giving

3.

COMPOSITE EXTENSIONS

219

a pair of isomorphic mappings a, r which have the same effect on k, and which map A and B respectively into some field F. The composite extension is then the smallest subfield of F which contains both aA and rB; we denote this subfield by aA . rB. In general aA . rB will not be merely the product of aA and rB (i.e. the set of finite sums of products of elements from aA and rB); aA . rB will be the smallest subfield containing this product, in fact its quotient field. Two pairs of isomorphisms (a, r) and (aI' rl) are said to be equivalent, or to yield equivalent composite extensions if there exists an isomorphism A of aA . rB onto alA· rIB such that Aa

=

al

,

Ar

=

r1 •

There is a very intimate connection between these composite extensions and the Kronecker product A X k B which we shall now investigate. Suppose first we are given a composite extension defined by isomorphisms (a, r). Then we may map R = A X k B into aA . rB by mapping onto

We must show that this mapping is well-defined; so let ~ (Xiaib" be zero in R. We may express the elements bi in terms of linearly, independent elements

b: :

Then we have

~ cxiaibi t

=

~ cxif3i.aib~ = 0 i,Y

and hence for all v.

Since a is an isomorphism, this implies

~ a(cxi) a(f3i.) a(at) = O. i

220

XI.


Multiply by T(b:), and sum over v, replacing a({Ji v) by T({Jiv) (a = T on k); this yields

~ u(<X;) r{~iv) u(at) T(b~) = ~ u(<Xj) U(<Xi) T(b;) = i,v

o.

t

This shows that the mapping is well-defined. It is clearly a homomorphism of R onto the product of aA and TB in F; since F is a field the image has no divisors of zero, and so the kernel of the mapping is a prime ideal j). Thus R/j) is isomorphic to the product of aA and TB; and hence uA . TB "'-' quotient field of R/j).

It is clear that if (aI' TI) is a pair equivalent to (a, T), then the equivalent composite extension alA· TIB corresponds to the same prime ideal of R. For if,\ is the map which links (a, T) and (aI' TI)' and if

then

Thus each equivalence class of composite extensions corresponds to a prime ideal in R. Conversely, let j) be a prime ideal in R; j) =I=- R. Let J-t be the natural homomorphism of R onto R/j). Then J-t maps A onto a homomorphic image J-tA; but a homomorphism of a field is either the zero map or an isomorphism, and J-t cannot be the zero map since J-tA = 0 implies J-t(A X k B) = 0, whence j) = R. Hence J-t maps A onto an isomorphic image J-tA. Similarly J-t maps B onto an isomorphic image J-tB. Thus J-t maps A and B into an integral domain R/j) which consists of linear combinations of products of elements in J-tA and J-tB. If we form the quotient field of R/j), we may form the compositum of J-tA and J-tB; in this way defines a composite extension. Clearly if j) is a maximal ideal, R/j) is already a field, and hence is itself a composite extension of A and B. In our applications, one of the fields-say B-will be an algebraic extension of k. Since we may consider A X k B = U A X k B'

3.

221

COMPOSITE EXTENSIONS

where the fields B' are the finite subfields of B, it suffices to consider the case where B I k is a finite extension. Let P be a prime ideal of R = A X k B; then R' = Rip may be considered as an integral domain of finite dimension over A. Suppose R' = AWl

+ ... + Aw,,;

then exR' = Aexwl

+ ... + Aexw" =

R',

since the Wi are assumed linearly independent. Thus R' is already a field, and so all the prime ideals in R = A X k B are maximal. We now prove the Theorem 4: Let R be any ring; let PI' P2' "', Pr be distincts maximal ideals. Then Rln Pi is isomorphic to the direct sum of the fields Rlpi . Proof: Let cP" be the natural map of R onto Rip". Consider the map of R given by

This is clearly a homomorphism of R into the direct sum of the Rlpi' with kernel Pi . It remains to show that the mapping is onto. Since Pi =1= PI (i = 2, "', r), there are elements ai which lie in Pi' but not in PI; hence cPl(ai ) =1= 0, but cPi(ai ) = O. Then if aI' = a2a3 ... ar , we have cPl(al ') =1= 0, while cPi(a,,') = 0 a' (i -- 2 , ... , r)', similarly we define a' 2' ...'r· Now let ii = (iiI' ii2 , "', iir) be an element of the direct sum. Since each cPi maps R onto the field Rlpi' there are elements Ai E R such that cPi( Ai) = iiilcPi( a/). Then clearly cP( ~ a/Ai) = ii. This completes the prooof. We apply this to the case where R = A X k B, and B is a finite extension of k of degree n. In this case each field Rip" is a composite extension aA . 7B of A and B. We now agree to identify A with its image aA in each of these extensions. Hence a composite extension of A and B is now defined by an isomorphism 7 of B into an extension field of A such that 7 acts like the identity on k. Another isomorphism 71 gives rise to an equivalent composite extension

n

222

XI.


if there is an isomorphism Awhich acts like identity on A such that

AT

=

TI'

Let PI' P2' ... , Pr be maximal ideals of R; we may view the direct sum of the fields R/Pi as a space over A. If the degree (R/pi : A)

= mj ,

we have r

~m( = dimR/nPi ~ (R: A) =

n.

i=1

This shows that the number of distinct maximal ideals in R IS finite. If we take all the maximal ideals in our decomposition (and these are all the prime ideals of R), we have

R/nPi = R/N ~ direct sum of all composite extensions. Hence we see that

~

mi = n if and only if N = (0).

To show that there may exist fields A and B for which A X k B has a non-zero radical, we consider the case where k = Ip(x) (Ip is the field withp elements), and let A = k(ex), B = k(f3) where exP = f3P = x. Then in the Kronecker product ex - f3 is clearly nilpotent: (ex - ~)P = ex P - W = o. Now let A be the algebraic closure of A, and restrict the isomorphisms T to be mappings of B into A. Certainly A contains k, the algebraic closure of k. Let Bo be the separable part of B; no = deg (Bo I k)j n = nops where p is the characteristic of k. Then it is known from Galois Theory that there are precisely no distinct maps T of B into k. Hence there are no maps of B into A. If two of these maps Ti , Tj yield equivalent composite extensions A . TiB and A . TjB, then A . TiB is isomorphic to A . TjB under an isomorphism A which leaves A fixed. Now A . TiBO is in the separable part of A . TiB over A. Since B I Bo is purely inseparable of degree ps, A· TiB is purely inseparable over A . TiBO with degree pSi:::;;; ps. Thus A . TiBO is precisely the separable part of A • TiB over Aj if

4.

VALUATION OF A NON-COMPLETE FIELD

223

then deg (A . TiB I A) = miP·i. By the same argument as above there are exactly m/ distinct maps ft., and hence m/ maps Tj equivalent to Ti • Thus ~' m/ = no , where the summation is restricted to the isomorphisms Ti which yield inequivalent composite extensions. Finally, since si ,;:;;; S, we have

once more. This shows that A X k B has radical zero if and only if all the Si = s. In particular A X k B has radical zero if B is a separable extension.

4. Extension of the Valuation of a Non-Complete Field Let k be a field with a valuation, not necessarily complete, and let E be a finite extension of k. Suppose first that we have obtained an extension of the valuation of k to E. The completion E of E under this valuation will contain the completion k of k under the original valuation; E also contains E, and hence the compositum Ek. This field Ek is a finite extension of k; hence (Theorem 2 of Chapter 2) it is complete. It follows that Ek contains E; hence Ek = E. On the other hand, if an extension to E is not yet known, we may construct the algebraic closure e of k (any sufficiently high finite extension would suffice); we can extend the valuation of ke. Since e contains the algebraic closure of k, there exist isomorphic maps a of Elk into elk. The valuation of e induces a valuation on aE which is an extension of the valuation of k. We may then define a valuation in E by writing I ex I = I aex I for ex E E. We notice that the completion of aE will be aE . k. There are several different maps of E into e, and we may ask when two of these maps give rise to the same extended valuation on E. So suppose a and T are maps of E into e which yield the same valuation. Then Ta- 1 is a map of aE onto TE which preserves the valuation of aE; this map can be extended to the completions of

224

XI.


aE and TE, that is to the composita aE . k and TE . k. Since Ta- I is identity on k, this extension is identity on k. Hence if a and T yield the same valuation on E, then they are equivalent. Conversely, suppose a and T are equivalent; let ,\ be the map of aE . k onto TEk such that ,\ is identity on k and T = '\a. Then a induces the valuation I ex 11 = I aex 1 and T induces the valuation I ex 12 = I Tex 1 = 1'\aex I. But aex and '\aex are conjugates over a complete ground field k; hence they have the same valuation1ex 11 = 1ex 12 . Referring to our previous discussion, we see that there is exactly one extension of the valuation of k to E for each prime ideal in the Kronecker product k X k E. Furthermore if we call the degree of aE . k over k the local degree, we see that the sum of the local degrees for all extensions is not greater than the degree of Elk.

CHAPTER TWELVE

Characterization of Fields by the Product Formula

1. PF-Fields

We saw in Chapter 1, Section 5, that the field of rational numbers and fields of rational functions over arbitrary ground fields satisfy a product formula IIp I a Ip = 1 for all non-zero elements a. We shall now study arbitrary fields with a product formula of this type. More precisely, let k be a field satisfying

Axiom 1: There is a set 9Jl of inequivalent non-trivial valuations k such that for every non-zero element ex of k, I ex Iv = 1 for all but a finite number of Ilv' and Ilv I ex Iv = 1. We denote by p the equivalence class of valuations defined by I Iv; and we call p a prime of k. In this chapter I Iv shall always denote the special valuation in the equivalence class p which occurs in the product formula of Axiom 1. For each non-archimedean prime p in 9Jl we may form a ring 0v of p-integers, consisting of those elements of k for which I ex Iv ~ 1. The elements ex for which I ex Iv < 1 form a prime ideal of 0v; we shall denote this ideal by p-no confusion will arise from this notation. Finally we denote the residue class field at p by kv :

I Iv of

Consider the set ko of elements of k such that I ex Iv ~ 1 for all p E 9Jl. Then clearly either ex = 0 and I ex Iv = 0 for all p E 9Jl; or else, because of the product formula, I ex Iv = 1 for all p E 9Jl. In the latter case we must distinguish two possibilities:

225

XII.

226 (1) for if

m

CHARACTERIZATION OF FIELDS

contains no archimedean prime. Then ko is a field,

I ex I" = 1,8 I" = 1 for all P Em, then I ex ±,8 I" ~ 1, I ex,8 I" = 1 and I ex-I I" = 1 for all p Em. ko is the largest subfield

of k on which all the primes of m are trivial: we call ko the field of constants. It is easily seen that ko may be isomorphically mapped into every residue class field k,,; under this map we may consider ko as a subfield of all k" . The degree of k" I ko will be denoted by f(p). (2) m contains an archimedean prime q. Then ko is not a field, for 1 E ko , but 1 1 = 2 ¢:. ko ,since I 2 Iq > 1. Since 12 Iq > 1 for all archimedean primes q, we conclude that there can be only a finite number of archimedean primes in m. We now introduce a second axiom, which guarantees the existence of at least one "reasonable" prime in m.

+

Axiom 2: There is at least one prime q of m of one of the following types:

is archimedean, q is discrete and kq is finite, q is discrete and f( q) = deg (k q I ko) is finite. Henceforth primes of these types will be called reasonable. Fields which satisfy Axioms 1 and 2 will be called Product Formula Fields, or PF-fields. We shall have to consider sets S which will be arbitrary when ko is not a field and ko-vector spaces when ko is a field. For such sets S we define the order of S as follows: (1) (2) (3)

q

(1) If there are archimedean primes, or if ko is finite, the order of S shall be the number of elements of S. (2) If ko is an infinite field, the order shall be cr where c is a fixed number, c > 1, and r is the maximal number of elements of S which are linearly independent with respect to ko . For example, let S be the set of residue classes modulo q where q is a reasonable prime. If S is a finite set, then ko (if it is a field) is finite (ko C S), so we have to use the first definition; if not, we have the order of S = ct(q). For reasonable non-archimedean primes we define the norm N q to be this order of the residue classes. We introduce a normal valuation /I /lq for reasonable primes q:

2.

UPPER BOUND FOR ORDER OF A PARALLELOTOPE

227

(1) If q is archimedean, and the completion of k under p is the real field, we define II ex II" to be the ordinary absolute value of ex. (2) If q is archimedean, and the completion of k under q is the field of complex numbers, we define II ex Ilq to be the square of the ordinary absolute value of ex. (3) If q is discrete, we define II ex Ilq to be (Nq)-ordq O.

2. Upper Bound for the Order of a Parallelotope Consider the set S con siting of elements ex such that I ex Iq ~ x" for all p E 9Jl where the x" are fixed real numbers. We may think of S as a parallelotope in the space obtained by forming the Cartesian product of k with itself as many times as there are primes in 9Jl. We wish to find an upper bound for the order M of S. First we prove Theorem 1: Let q be a reasonable prime. Let S be a set of elements of k with order M > 1. If I ex Iq ~ x for all ex E S, there exists a non-zero element () E k, which is either an element of S or a difference of two elements of S, such that I () Iq ~ Aqx/MP(q); Aq is a constant depending only on q. Proof:

We must consider several cases:

Case 1: q is archimedean and the completion is real. In this case the order M is the number of elements of S. Since I ex Iq ~ x, we have II ex Ii q ~ X1fp(q) for all ex E S. Hence if we decompose the interval [- X1fp(q), X1fp(q)] on the real line into M - 1 equal parts, then, by the Pigeon-holing Principle, at least one of these parts must contain two elements of S. Hence there is a non-zero element () E k, the difference of two elements of S, such that 2x1 / p (q) - 1.

11811q ~ M

228

XII.


Since M? 2, we have M - 1 ? M12, and hence 4,xl/p(q) 118I1q~~.

Finally we have

i () iq ::::;: AqXIMp(q)

where Aq = 4p(q,.

Case 2: q is archimedean, and the completion is complex. Here also M is the number of elements of S. If i (X iQ ::::;: x, we have the ordinary absolute value of (X ~ xt / 2p (q); hence (X lies in the square (in the complex plane) with center the origin and side 2Xl / 2p (Q). If we decompose this square into N2 equal squares, where N < VM::::;: N 1, i.e. N2 < M::::;: (N 1)2, then at least one of these smaller squares must contain two elements of S. Hence there is a non-zero element () E k, the difference of two elements of S such that the ordinary absolute value of

+

+

Now

hence ordinary absolute value of 8~

27 / 2Xl / 2P(q)

VM

.

Therefore, 1

Aqx 8 1q ~. MP(q)'

where Aq = 2 7p (Q). Case 3: q is discrete. Let (Xl be an element with maximal i (Xl iq in S. We replace the set S by S' = S/(Xt; then every element of S' satisfies i (x' i ::::;: 1, i.e. the elements of S' lie in the ring of q-integers uq. We can find an integer T such that

2.

UPPER BOUND FOR ORDER OF A PARALLELOTOPE

229

There are now two cases to distinguish: (a) Residue class field is finite. Then the ring oq/qr has (Nqy elements. It follows from the Pigeon-holing Principle that there is a non-zero element () in k such that ()/Cl.l is the difference of two elements of S' and

Hence ord

()/Cl.1 ~

r, and so

whence

and

where Aq = (Nq)p(q) .

(b) Residue class field kq is a finite extension of ko , of degree J. Then if CI. is any element of 0,

with ai E k q ; thus there cannot be more than fr elements linearly independent over ko modulo qr. But M = cdimS ', and (Nqy = cfr ; hence cfr < cdimS ', i.e. dim S' > fro Thus there are more than fr elements of S' linearly independent with respect to k o; these cannot be linearly independent modulo qr. Hence there is a non-zero element () of S such that ()/Cl.1 is a linear combination of more thanfr linearly independent elements of S', such that this combination is non-trivial modulo qr and hence such that ()/Cl.1 _ 0 mod qr. As in case (a) we deduce that I () Iq ~ Aqx/Mp(q). This completes the proof of Theorem I in all cases. We can now give the desired bound for the order of the elements in a parallelotope:

230

XII.


Theorem 2: Let S consist of elements ex such that lex II' ~ xl' for all P E un, where the xl' are fixed real numbers. Let q be a fixed reasonable prime in un. Then if M is the order of S, M'::;; Dq

(

IIxp

l/P(q) )

,

where Dq is a constant depending only on q. Proof:

By Theorem 1 there is an element 0 in k such that

*

Next we estimate I 0 II' for p q. If p is archimedean, there is no constant field k o; hence 0 must be a difference a - b of elements of S. One easily finds a constant fL such that I 0 II' ~ fLx p • If pis non-archimedean, whether 0 is a difference of two elements of S or is itself an element of S, we may conclude that I 0 II' ~ xl' . Using the product formula, we obtain 1=

II I

fLkA q (j

Ip ,::;;

IT xl'

J\Ilp(q)

,

1lE9Jl

where k is the number of archimedean primes p obtain M'::;; Dq

.

(

IT xl'

)l/P(q)

* q.

Finally we

,

pE9Jl

where Dq depends only on q.

3. Description of all PF·Fields Let k be a PF-field. We define a rational subfield R of k: Case 1: When there are archimedean primes denote the field of rational numbers.

In

un,

R shall

3.

DESCRIPTION OF ALL PF-FIELDS

231

Case 2: When there is a field of constants ko , let z be a fixed element of k which does not lie in k o; define R = ko(z). This is a transcendental extension of ko , for ko is algebraically closed in k; namely, all primes are trivial on ko , and hence on any algebraic extension k I , i.e. ki C ko . In both cases we have already determined all the valuations of R (see Chapter I, Section 5); we denote the equivalence classes of valuations of R by Latin letters, p, q, .... Each prime P in un induces on R a valuation p; we say that P divides p, and write pip. We restrict our attention to these primes P in un which induce non-trivial valuations in R. A given prime p in R may have several divisors P in un; but there can be only a finite number, for if I a Ip > I, then I a Ip > I for all p dividing p, and by Axiom I this is possible only for a finite number of p. Clearly if PI' P2' ... , P r divide p, all the valuations lip; are equivalent in R, and IIp1p.pE'J)l I Ip is a valuation in the equivalence class p. Let poo denote the infinite prime in R. Then un contains primes Poo which divide Poo: Case 1: If R is the field of rational numbers, the archimedean primes in un divide Poo . Case 2: If R = ko(z), not all primes are trivial on R; hence there is at least one prime Poo such that I zip", > 1. This prime Poo divides Poo . We now give a description of all PF-fields.

Theorem 3:

(I) (2)

A PF-field is either

an algebraic number field, or a finite extension of a field of rational functions.

Proof: Let (Xl' (X2' ... , (Xr be elements of k which are linearly independent over R. We shall show that r is necessarily finite. To this end we consider the set S consisting of elements

where the Vi range over all integers of the rational subfield R (i.e. rational integers or polynomials) such that I Vi IP", ~ I A IPO) where A is a given integer of R. Let M be the order of S.

232

M

XII.


We first obtain a lower bound for M, namely we prove that > /I A /I~",. To do this we must consider two cases-

Case 1: R is the field of rational numbers. Here there are 1 possible values for each Vi; hence there are exactly 2 II A Ill'", in all M = (2/1 A /II'", lY elements of S. Clearly M > /I A /I~",.

+

+

Case 2: R = ko(z). In this case S is a vector space over ko • If the degree of A is d, each Vi has degree ~ d. It follows that the I}, and hence M = crCd+l); but dimension of S over ko is r(d /I A /II'", = cd. Hence M > /I A /I~", . We now estimate the size of the parallelotope containing the set S. For all primes Pro, we have I (X II'", ~ B p ", I A II'"" where B p ", is a constant depending on the elements (Xl' (X2' ••• , (Xr. For the remaining primes p (which are certainly non-archimedean), we have

+

since I A Iq ~ 1; this value, max I (Xv II" is 1 for all but a finite number of primes p. Let q be a fixed reasonable prime; then by Theorem 2 we have

where E is a constant, depending on q and the (Xl but not on I Alp", . We now compare the upper and lower bounds for M, and obtain

I a II'",

For any element a E R we may write "(Poo) is a certain constant; hence

Now keeping T and the elements tend to infinity, we obtain T ,::;;

(Xi

=

/I a

,,~;;"') where

fixed, and letting

ptq) ~ }.(p",)

/I A /II'", (*)

3.

233

DESCRIPTION OF ALL PF-FIELDS

Thus r is bounded; hence k is a finite extension of R. This completes the proof.

Corollary 1:

No prime in Wl can induce a trivial valuation on R.

Proof: Suppose p E 9Jl induces a trivial valuation on R. Then p also induces a trivial valuation on every finite extension of R, in particular on k itself. This contradicts Axiom 1.

Corollary 2:

Every prime in 9Jl is reasonable.

Proof: All the primes in R are reasonable. It follows from the local theory that an extension of a reasonable prime to a finite extension is reasonable. Let n be the degree of R. By raising the product formula to a suitable power, we may assume that ~ .\(Poo) = n: from now on we shall assume that the product formula of Axiom 1 already has this property. Now set r = n in (*); we see that p(q) ~ 1 where q was a reasonable prime. By Corollary 2, p(p) ~ 1 for all p E UJl. Let p be a non-archimedean prime of UJl, which induces a (nontrivial) prime p of R. Let k*(p), R*(p) denote the completions of k and R respectively. Let n p , ep, fp denote respectively the degree, ramification and residue class degree of the extension k*(p) I R*(p). Then we have the following statements:

Np = (Np)fll ,

and, if a is any element of R,

Now we have already defined the normal valuation equivalence class p of R (see Chapter 1, Section 5)-

We have also

II lip

in the

XII.

234


Hence

The same result is easily verified when p is an archimedean prime. Hence for all primes p E un and all a E R, we have

We now apply the product formula of Axiom 1 to the elements of R:

= II II a II;\,P(P) = 1, p

where the accents mean that the product or sum is taken for all the primes p E Wl which divide p. We thus obtain a product formula

II II a II;'npp(p) =

1

1>

in the rational sub field R. In Chapter 1, Section 5, however, we saw that in such a formula (a) (b) Let

all primes p of R must appear, the exponents ~'npl>(p) are all equal.

~'npp(p)

= d;

d is a constant. When p

= Pao

we already have

Hence d=

k

Poo 11>""

npoop(poo) =

k

}.(Poo) = n.

p""lpoo

Since p(p) ~ 1, we have ~PI1> np ~ n. But we have already seen that ~ np ~ n, where this summation is extended over all primes in k (not only in un) such that pip. We conclude (1) un includes all primes p of k which divide p, and (2) p(p) = 1. We sum up our results in

4.

FINITE EXTENSIONS OF PF-FIELDS

235

Theorem 4: Let k be a PF-field, R its rational subfield. Then (a) the set un consists of all primes p in k which divide all the primes p of R; (b) for each such prime p, ~pIP np = n; (c) the valuations lip which occur in the product formula (raised to a power if necessary to make ~ .\(Poo) = n) are precisely the normal valuation II lip.

4. Finite Extensions of PF-Fields We shall now show that every finite extension of a PF-field is again a PF-field. Since the rational number field and fields of rational functions are PF-fields, this will prove the converse of Theorem 3, namely that all algebraic number fields and algebraic function fields are PF-fields. Lemma 1: Let k be a field which satisfies Axiom 1; let F be a subfield not consisting entirely of constants. Then Axiom 1 holds in F for the set 91 of primes p induced by those p in un which are non-trivial on F. Proof: We define I a Ip = II' I a Ip where a EF, p E91 and the accent denotes that the product is taken over all p E un which divide p. We remark that this cannot be an infinite product, and hence I a Ip is a well-defined valuation in the equivalence class p. Clearly

Thus Axiom 1 holds in F. We may remark that if kiF is a finite extension, no prime of 9Jl can induce a trivial valuation on F. In this case, therefore, 91 consists of all valuations induced on F by primes in 9Jl. Theorem S: Let k be a PF-field, Elk a finite extension. Then E is also a PF-field.

XII.

236


Proof: It follows at once from the Local Theory that an extension of a reasonable prime to a finite extension is also reasonable. Hence Axiom 2 holds in E. It will be sufficient to prove that Axiom I holds in E when E is an automorphic extension; for if Axiom 1 holds for an automorphic field containing E, then, by the Lemma, it holds in E also. So let Elk be automorphic, with defect pr and automorphisms CT, T, •••• If 0.: E E, then N (0.:)

= II (a:u)pr =

a E k.

u

If p

E

9Jl,

Let

\p

be some extension of p to E; then

for

I a: I'll = I a:

U

I~u

.

Then

thus Axiom 1 holds in E. This completes the proof. The following remark about the situation of Lemma 1 will be useful in the sequel: Lemma 2:

If Fo is the set of constants under 91, then

Proof: Clearly F () ko C Fo . On the other hand, if I a Ip ~ 1 for a E F and all p Em, then clearly I a II' ~ I for all p E Wl which divide primes of 91. Since the other primes in 9Jl are trivial on F, we have I a II' ~ 1 for all p E 9Jl, and hence a EF () ko .

4.

FINITE EXTENSIONS OF PF-FIELDS

237

Theorem 6: Let k be a PF-field with no archimedean primes. Let E be a finite extension of k. Then, if Eo , ko are the constant fields of E, k respectively, Eo is the algebraic closure of ko in E. Proof: Clearly every element of E which is algebraic over ko lies in Eo. On the other hand, k I ko has transcendence degree 1, and since E is a finite extension of k, E I ko also has transcendence degree 1. Thus if c is an element of E transcendental over ko , E I ko(c) is an algebraic extension. Hence if c E Eo, E I Eo is algebraic; and therefore, since all valuations of 9Jl(E) are trivial on Eo , they are also trivial on E contrary to Axiom 1. Hence Eo cannot contain any element transcendental over ko •

CHAPTER THIRTEEN

Differentials in PF-fields

1. Valuation Vectors, Ideles, and Divisors Let k be a PF-field. We form the completions k*(p) for all the primes p in the set m. Let P be the Cartesian product of all these completed fields; P consists of vectors

with one component ~p from each k*(p). P forms a ring under component-wise addition and multiplication. It is easy to see that the mapping is an isomorphism of k into P. If ( is any element of P, we define I ( Ip to be I (p Ip . We now define the subset V of P consisting of vectors ( for which for almost all primes p. This subset is a subring of P, and it contains the isomorphic replica of k. V is caIIed the ring of valuation vectors: we shall sometimes write V(k) instead of V when it is necessary to emphasize that we are dealing with the valuation vectors associated with a particular field k. The vectors a of V, for which for almost all primes p, for all primes p, 238

1.

VALUATION VECTORS, IDELES, AND DIVISORS

239

form a multiplicative subgroup I of V; the element of this subgroup are called ideles. I contains a subgroup isomorphic to the group of non-zero elements of k. We define a topology in the ring of valuation vectors by means of the ideles. If a is an idele, we define the parallelotope JIa of V to consist of the vectors g in V for which I g II' ~ I a II" These parallelotopes are taken to form a fundamental system of neighborhoods of zero in the additive group of V. This defines the socalled restricted direct product topology on V; this is not the same as the topology induced by the ordinary Cartesian product topology in P. We recall the definitions of some topological notions: A filter in V is a family IY of sets such that:

(I) Every set containing a set of IY is itself a set of IY. (2) Every finite intersection of sets of IY belongs to IY. (3) IY does not contain the empty set. IY is a Cauchy filter in V if, in addition, given any JIa , there exists a vector ga such that ga + ITa is a set of IY. A Cauchy filter IY is said to be convergent, with limit g, if there is a vector g such that g + JIa is a set of IY for all parallelotopes JIa . Finally, V is complete if every Cauchy filter converges. Theorem 1: topology.

V is complete in the restricted direct product

Proof: Let IY be a Cauchy filter in V; let IYp be the filter induced (by projection) on the component space k*(p). Since each k*(p) is complete, each filter IYp is convergent, with limit ex p • We show first that I ex p II' ~ I for almost all p. Let JIa be a fixed parallelotope with projection Np on k*(p); let ga be a vector of V such that ga + JIa is a set of IY. Then (ga) + Np is a set of IYp , and hence contains a p • By the definitions of parallelotopes and valuation vectors, Np is the unit circle for almost all p, and I ga II' ~ I for almost all p. Hence, for almost all p, the unit circle is a set of IYp and therefore I ex p II' ~l for all but a finite number of p. Thus ex = ( ... , ex p , ... ) is a valuation vector. We claim now that ex is the limit of IY. Let JIa be a fixed parallelotope, and let JIb be so chosen that JIb + JIb C JIa • Let g be a

240

XIII.

DIFFERENTIALS IN PF-FIELDS

vector such that g + IIb is a set of ~; then (g)p + (lIb)p contains a: p • It follows that a: E g + lIb, and so g E a: + llb . Now we have ex

+ lla ~ cx + llb + llb ~ g + llb ,

a set of~. Hence a: + lIa is a set of~; i.e. ~ converges. If a is an ideIe, we may define its absolute value

Ia I =

II I a Ip . p

Then I a I may be regarded as the volume of the parallelotope lIa . If a is an idele , a = (... , ap' ... ) , then a:a = ( ... , a:a P' ... ) is also an ideIe, and we have

I cxa I = II I ex Ip I a Ip = I a I p

by the product formula. Hence the parallelotopes lIa and IIa.a have the same volume. We see that when we describe a parallelotope by means of an idele only the absolute values of the components play a role, not the actual components. This suggests that in order to describe parallelotopes we need merely prescribe an absolute value, or an ordinal number, for each prime p. To this end we introduce formal symbols a=

II p'P

with vp = 0 for almost all primes p. In the case of algebraic function fields the product is extended over all primes p, and a is called a divisor. In the case of algebraic number fields the product extends only over the finite primes, and a is called an ideal. It is easy to see that each idele a defines a unique divisor (or ideal) which we may also denote by a: a-+-

II p

ordpa

•

Similarly every element a: E k defines a divisor (or ideal). If a is a fixed divisor (or ideal) of k, the set of all divisors (ideals) of the form a:a is said to form a divisor class (ideal class).

2.

VALUATION VECTORS IN AN EXTENSION FIELD

A divisor (or ideal) n = npl'p is said to be integral if ILp for all p. Let b = np·p be any divisor (ideal), and define

241 ~

0

Then b = b1b;1 = b l !b 2 ; we naturally call b l the numerator and the denominator of b. Thus any divisor (ideal) can be expressed as a quotient of integral divisors (ideals).

b2

2. Valuation Vectors in an Extension Field Let k be a PF-field; let K I k be a finite extension of degree n. We adopt the following notation: V(k), V(K) shall denote the rings of valuation vectors in k, K. n, III shall denote the ideles of k, K. lla, ll<Jl' shall denote the parallelotopes of V(k), V(K). V(k) can be mapped naturally into V(K) as follows: map g E V(k) onto the vector whose ~-component is gp, whenever ~ divides p. Since gp E k*(p) C K*(~), and since there is only a finite number of primes dividing a fixed p in k, the image is indeed a valuation vector. It is easily seen that this mapping is continuous in the restricted direct product topologies of V(k) and V(K). Our aim is to give a description of the space V(K) in terms of V(k). For this purpose we introduce the space

vn =

V(k) X V(k) X •.. X V(k)

and its subspace kn

=

k X k X ... X k.

We choose a fixed field basis for K! k: the mapping ifJ : V'" -- V(K) by writing

Then ifJ(V"') C V(K) and ifJ(kn ) tinuous mapping.

WI' W 2 , " ' , W n ,

and define

= K; ifJ is easily seen to be a con-

Lemma 1: ifJ(V"') is everywhere dense in V(K).

XIII.

242


Proof: Let X be an element of V(K), il~r' any parallelotope; we have to prove that there is an element Y in r/J( vn) such that X - Y E II~I" To this end we construct a non-empty set 8' of primes p in k, containing at least (I) all archimedean primes, (2) all primes p which possess an extension !p in K for which either I X 1'l1 > 1 or 121 Ip =1= 1. Let 8 be the set of all primes of K which are extensions of the primes in S'. 8 contains only a finite number of primes, so by the Approximation Theorem we can find an element ex E K such that

for all tl in 8. Let

then define vectors gl' gz, ... , gn in V(K) by writing for I,p E 8, and (gi)~ = 0 for I,p ¢; 8. Then

(~i)'l1 = ai

is a vector of V(K) such that X - Y Eil'll" This completes the proof. Lemma 2:

There exists an idele

b

such that ilb

+k =

V(k).

Proof: We prove the theorem first for the rational subfield R. Let g be a valuation vector; then gp E k*(p). We have seen that gp may be written as a power series in p with integral coefficients (when p is non-archimedean); when R is the field of rational numbers, gpo> = m 7]00 where m is an integer and 0 :;:;: 7]", < 1. We define the principal part of g at a prime p to be

+

Pr (l:) l' !>

= po a. + a l + ... + a1 po-I P v_

Prl'GO(0

(p non-archimedean),

=m.

Write a = Ep Prp(g). Then a E R. If g = a lies in the parallelotope ill . Hence VCR) = R the lemma for the rational field.

+ 7]

we see that TJ This proves

+ ill'

2.

VALUATION VECTORS IN AN EXTENSION FIELD

243

We now show that if k is any field in which the lemma holds, then it holds also in any finite extension K of k. Suppose therefore that in k we have V(k) = k + llb • Then

But cf>(llbn) is contained in a parallelotope II(I' of K, for

where M'l1 depends only on the W v ' Since cf>( fT'l) is everywhere dense in V(K) we have V(K) = cf>(vn) + ll(J and hence V(K) = rp(V7l)

where ll'ij' :::) ll(J' Lemma 3:

+ II~ C I1~ + II~ + K C niJ + K,

+ ll(J' • This completes the proof of the lemma.

The mapping cf> is bicontinuous.

Proof: We have already remarked that cf> is continuous. Let ll(J' be a fixed parallelotope in V(K) and let X be an element of cf>-l(lla' n cf>(vn»j i.e. X E Vn and cf>(X) Ella'. We have to show that X is contained in llan where lla is a parallelotope of V(k). Since V(k) = k + llb' we can write X Y + Z where Z E k'" and Y Ellon, so that cf>(Y) is in a fixed parallelotope ll'13" Now cf>(Z) lies in K, and rp(Z)

= rp(X) -

(Z) is either finite, or else a finite dimensional vector space over Ko • hence over ko • It follows that Z lies in a fixed lltn and hence

x=

Y

+ Z C JIb + II~C some II;.

Thus if

then

~1' •••• ~n Ella'

Finally let c be any idetej we see that if

XIII.

244


then egl' "', egn E ena .

This shows that rfo is bicontinuous. From this it follows that rfo is a (1, 1) mapping, and hence a homomorphism. For suppose glwl

+ ... + gnwn =

O.

Then, since rfo is bicontinuous, all the gi lie in every parallelotope, however small; hence, since V(k) is Hausdorff, all the g, are zero. Thus rfo is (1, 1) and bicontinuous. Since V(k) is complete, V" is complete and so rfo( vn) is complete. Hence rfo( V") is closed. But rfo(V") is everywhere dense. It follows that rfo(V") = V(K) and we have

Theorem 2: V(K) is isomorphic to V(k) X k K both topologically and algebraically.

3. Some Results on Vector Spaces Let X be a vector space over an arbitrary field ko • Let Y be a subspace of X. The ko-dimension of the factor space X/Y will be denoted by (X: Y)k o ' When we may do so without causing confusion we shall omit the subscript ko .

Theorem 3: Let A and C be subspaces of the same vector space. Let B be a subspace of A. Then (A : B)

= (A (\ C : B (\ C)

+ (A + C : B + C).

Proof: We map A onto (A + C)/C by mapping 0: E A onto the coset 0: + C. The kernel of this mapping is A (\ C. Hence (A: B

+ (A (\ C)) = «A + C)/C: (B + C)/C) = (A + C: B + C).

Further, (A : B)

= (A : B + (A (\ C» + (B + (A (\ C) : B) = (A + C : B + C) + (A (\ C : B (\ (A (\ C» = (A + C : B + C) + (A (\ C : B (\ C).

4.

DIFFERENTIALS IN THE RATIONAL SUBFIELD

245

This is the required result. Let V be any vector space over an arbitrary field ko, of finite dimension n. Then it is well known that the linear ko-homogeneous mappings of V into ko form a vector space 17, which is also of dimension n over k o; 17 is called the dual space of V.

Theorem 4:

Let K be a finite extension field of k o • Then

1? is a one dimensional K-space. Proof: 1? consists of the linear ko-homogeneous mapping of K into k o • Let Ao be an element of 1? such that Ao(g) is not identically zero. We shall show that every element of 1? can be expressed as A( g) = Ao( ex g) where ex is an element of K. Clearly every such function A( g) is an element of 1?. On the other hand, we shall show that if gl , g2' "', gn form a basis for K, then Ao(glo, Ao(g2o, "', Ao(gng) are linearly independent in Kover ko, and hence form a ko basis for 1? since dim 1? = dim K = n. To this end we notice that if Ao(exg) = 0 for aU g E K then ex = 0; for if ex =F 0 then ex-1g lies in K, and hence

for all g contrary to hypothesis. Now we have

Hence the Ao(gig) are linearly independent over ko • Thus every element A in 1? can be written in the form

This completes the proof.

4. Differentials in the Rational Subfleld of a PF-Field Let k be a PF-field. Then by Theorem 3 of Chapter 12, k is either:

246

XIII.


(a) a finite extension of the field of rational numbers (briefly, a number field); or (b) a finite extension of the field of rational functions in one variable x over a constant field ko (briefly, a function field). We introduce the following notation: R shall denote (a) for number fields, the field of rational numbers and (b) for function fields the field ko(x). (In both cases we shall call R the rational subfield as in Chapter 12.) P shall denote (a) for number fields, the field of real numbers; and (b) for function fields, the completion of ko(x) at Pco • r shall denote (a) for number fields, the additive group of real numbers modulo I, with the natural topology; and (b) for function fields, the constant field ko with the discrete topology. N shall denote (a) for number fields, a fixed sufficiently small neighborhood of zero in r. For definiteness we take (- f, -1-). (b) for function fields, the zero element of ko . A differential of k is now defined to be a continuous linear map A of the ring V(k) into r such that A vanishes for elements in k. When k is a function field, A is further required to be ko-homogeneous. In both the number field and function field cases this definition implies the existence of a parallelotope nn in V(k) such that A(nn) eN. In the number field case the elements of a parallelotope no do not form an additive group. We therefore define the set

where go is the valuation vector with component I at all nonarchimedean primes and component zero at all archimedean primes (thus for function fields nno = nn). The set nno is an additive group. Its image A(nnO) is also an additive group, and is contained in N. But N contains no subgroup of r other than the zero element, so A(nnO) = O. We must now prove separately for the number fields and function fields two lemma concerning the linear maps of Pinto r. Lemma A: In the number field case, let fL(X) be a continuous exx (mod I), where ex is linear map of Pinto r. Then fL(x) a fixed real number.

=-

4.

247


Proof: Let fL(1/2 n) = an/2 n , where is real. Since fL is continuous, lim 2a,.n =

n-->OO

t

~ an/2n ~

t

and an

o.

Now,

= fL (0) == 0 (mod 1). Thus (an+! - an)/2n is an integer. But

Hence an+! = an for all n ? some fixed N. We write an = - ex for n? N. Then fL(1/2 n) = - ex/2n, and by the additivity, fL(r/2 n) = - exr/2 n for all n ? N. Hence we have fL(I/2 n) = - ex/2 n for all n. It follows that fL(X) = - (XX for all rational dyadic fractions x, and hence by continuity fL(X)

= -

(XX

(mod 1)

for all real numbers x. Lemma B: In the function field case, let fL(g) be a continuous linear, ko-homogeneous function of Pinto r such that fL(g) = 0 if g is an integer of R. Then fL(g) = Residue at Poo of (f(x) g) where f(x) is a fixed polynomial. Proof: The continuity of fL implies that fL(X-n) vanishes for all large enough n, say n > N. Let

Then fL(g)

Set fL(X- i ) fLW

= fL(a,.x n + ... + a o) + a_lfL(x-1) + a_2fL(x-2) + .... = -

Ci.

Then

= C1 Resp W + C2 Resp (xg) 00

= Resp (f(x) g), 00

00

+ ... + CN Res

p

00

(xn-1g)

248

XIII.


where I(x)

=

c1

+ c x + ... + 2

1 CN X N - •

We investigate first of all the differentials in the rational field R. Let g be a valuation vector of V(R), gp its component at a finite prime p. We may write gp in the form gp

=

HpW

+ 1]~,

where I TJp' Ip ::;;; 1. Hp(g) = 0 for all but a finite number of primes p. Thus a = 1: Hp( g) is an element of R. We define for each finite prime p YJp = gp - a

= YJ~ -

!- Hq{g)

q*"

and form the valuation vector "I which has components TJp at the finite primes and components 0 at the infinite prime. We may now write

g=

a

+ YJ +~,

with , component 0 at the finite primes and suitable component at the infinite prime. Suppose g may also be written in the form

'00

g = a'

+ YJ' + ~'.

Then (a - a') has at every finite prime the same component as "I' - 1], i.e.

\ a - a' \1'

= \1]' -

YJ

\1' ~ 1

for every finite p. Hence a - a' is an integer of R. Conversely, if m is any integer of R, we can write

g = (a -

m)

+ (YJ + m') + (~ + m"),

where m' has component m at every finite prime and component zero at the infinite prime, while mil has component m at the infinite prime and zero component elsewhere. It follows that the mapping is defined uniquely modulo 1.

4.


249

N ow let I-' be a differential, and let a be an idele such that I-'(lla) eN. We can find an element a E R such that ordp a = ordp a at all finite primes p. Hence lla = llau where u is a valuation vector with component I at all finite primes and suitable component at infinite prime. We define p(g) = I-'(ag). Then p is also a differential and

Clearly P(lllO) = 0 and hence p(TJ) = O. Since a E Rand p is a differential, p(a) = O. Hence p(g) = p(a)

+ p('T]) + p(~) = p(~),

since {<Xl is only unique modulo I, and p( g) is unique. It follows that pm is a function of {<Xl such that p(m) = 0 for all integers m. p({) is linear and continuous since the map of g -+ , is linear and continuous. Hence we may apply Lemmas A and B. Lemma B gives us immediately for the function field case p(~) =

Res (f(x)

~<Xl).

In the number field case, Lemma A yields p(~)

== -

C\!~<Xl

(mod 1).

Here we must apply the additional condition that p vanishes on 0 (mod 1), i.e. ex must be integers. Putting '<Xl = 1 we see that ex an integer. We now define the mapping .\(g)

=

>.(g)

== -

>.(g)

= Res (~"')

~<Xl

(mod 1)

(for number fields), (for function fields).

Then .\ is a differential. It is easily seen that .\ vanishes on field elements, and the continuity of'\ is proved as follows: (a) For the number field case, consider the parallelotope ll., , where e' is the idele with component 1 at all finite primes and component E at the infinite prime. For g Ell., we have with

XIII.

250


Then A(g) = A(tco), which can be made as smaIl as we please by suitable choice of e. Thus A is continuous. (b) In the function field case, A clearly vanishes on the parallelotope II.. , where e' has component 1 at all finite primes and ordp e' ~ 2. W'; have therefore proved the existence of differentials for the rational field R. We now deduce two important properties:

Property I: The differentials of R form a I-dimensional R-space. Let I-' be any differential; then we have

where m is an integer and a E R. Thus I-'(g) = A(mja g). This proves Property I.

Property II: There is an upper bound to the parallelotopes IIQ for which I-'(IIQ) eN, where I-' is any differential of R. By Property I, it is clearly sufficient to consider the special differential A. Suppose A(IIQ) eN. Then, since A(II10) = 0, we have A(IIQ II10) eN. Now if g E IIQ II1o, then ordp g ~ min (0, ordp a) for all finite primes. Thus we may already assume that or~ ~ 0 for all finite primes. Suppose or~ a < 0 for a certain finite prime. Then clearly, if m is any integer of R, the vector

+

+

g=

(; , ;

, ... , ;

,0)

(with component zero at the infinite prime) lies in IIQo. But t

~

=

(!!!.!!!.- ... !!!-) _ (0"0 " ... 0 !!!.-) p ' p ' 'p p'

and hence A(g) = mjp (mod 1) in the number field case, and

i\W =

Res ( ;(~j

)

in the function field case. In the first case we can certainly choose m such that mjp > l In the second case Res xt-1jp(x) =F 0 where f is the degree of p(x). In both cases we have obtained a contradic-

5.

251

DIFFERENTIALS IN A PF-FIELD

tion to the statement that g EIIao. Thus we must have ordp a = 0 for all finite primes. Hence IIa C IIIO (0,0, . ", E) where E is chosen so that >..(IIa) C N. In the number field case, I E I ~ -l and in the function field case or~ E ~ 2. Thus in the function field case IIa C IIp~ . This proves P~operty II.

+

Corollary: >"(I1aO) = O.

There is a maximal parallelotope IIa such that

S. Differentials in a PF-Field

Now let k be a PF-field. We shall prove the existence of differentials in k, and shall show that the differentials in k also satisfy Properties I and II. The proofs do not use the special properties of PF-fields, and hold for any finite extension k of a field R in which differentials exist and satisfy the two properties. So let k be a finite extension of R. In the function field case we assume that k and R have the same constant field. Let WI' W2' " ' , Wn be a basis for kjR. Then if X E V(k) we can write

where the gi are valuation vectors of V(R). Let I-' be a differential of k. Then

The functions I-'(giwi) are continuous linear (and in the case of function fields ko-homogeneous) maps of V(R) into Since aWi E k, JL(awi) = 0 and hence JL(gwi) is a differential of R. Thus I-'(gwi) = >..(aig) where ai E R. Finally we have

r.

/L(X) = '\(a1g1

+ a g + ... + angn )· 2 2

Conversely for arbitrary ai E R, the map

is a differential of k. Indeed JL is certainly a linear map, and clearly vanishes for elements of k. I-' is continuous since the topology on

252

XIII.


V(k) is the Cartesian product topology on V(R)n. (See section 2) We have therefore proved the existence of differentials in k. Theorem 5: If k is a PF-field, then the differentials of k form a I-dimensional k-space. Proof: Let 10 be a non trivial linear, R-homogeneous map of k into R. Such maps certainly exist. For example, the maps

are of this type. When klR is separable we have even an invariant map of this type, the trace SklR . Let 1 be any linear R-homogeneous map of k into R. Then by Theorem 4, we have I(cx) = lo(f3cx) were f3 is an element of k depending on I. Suppose that I( wi) = ai . Then

We can now extend 1 to a linear, V(R)-homogeneous map of V(k) into V(R), which we still denote by I. We define l(glWl

+ ... + gnwn) = algi + ... + ~gn .

This extended mapping is unique, and clearly I(X) = 10(f3X). On the other hand every linear, V(R)-homogeneous map of V(k) into V(R) can obviously be obtained in this way. Now we have seen that if I-' is a differential of k, then I-'(X) = A(l(X» where 1 is a map of the type we have been considering. Hence /L(X) = A(lo(f3X)).

This proves Theorem 5. Theorem 6: If k is a PF-field and I-' is any differential of k, then there is an upper bound for the parallelotopes IIa' such that I-'(IIa') eN, and hence I-'(II~O) = o. Proof: Let WI , W2' ••• , Wn be a fixed basis for kl R. We have seen that I-'(X) = A(l(X» where 1 is a certain linear map of V(k) into V(R). Let

5.

DIFFERENTIALS IN A PF-FIELD

253

Then the maps IkX) = gi may be written m the form liX) = l(exiX) with exi E k. Now suppose /L(II~)

=

>'(l(II~)) eN.

Since /(X) = li(exiIX), this implies that for all £

Suppose X E exil lIa' for £ = 1, ... , n. Then X Ill' C exil lIa' for all £, where Ill' denotes the unit parallelotope in V(k). It follows that

Now Ill' gi is a parallelotope in V(R). From our previous investigation it follows that gi is contained in a fixed parallelotope of V(R). Hence

x= ~giwiCII~, where 58 is fixed. We have now

na:ilII&, C n

II~ •

i=l

Hence

Thus lIa' C lIe', where lIe' is a fixed parallelotope of V(k). This completes the proof of Theorem 5. If g is a valuation vector in V(k), we may write

where gp is the valuation vector with the same p-component as and the component zero at all other primes.

g

XIII.

254


Let tt be a differential of k. Then we define tt 1'( g) to mean tt(g1')' Theorem 7:

Proof: Since tt is a differential there is a parallelotope IIQ such that tt(IIQ) eN and hence tt(IIQO) = O. We may write ~ = ~Pl

+ ~1'2 + ... + ~Pr + "I

with TJ

E

II~;

For P =1= Pi , g1' lies in IIQo and hence tt(g1') = O. Thus /L(TJ) =

k

1'*1',

/L(g1') =

k

1'*1'1

/L1'W =

o.

Now n

/LW =

k /L(g1'J + /L(TJ) = k /L(g1') = k /L1'W· i~l

l'

V

This completes the proof. Let P be a fixed prime, and suppose tt1'(g) = tt(g1') = 0 for all g having ord 1' g1' ~ v1" Changing the p-component of a to p"1' we obtain a set II~o on which tt (IIbO) = O. Since the parallelotopes IIb such that tt (IIbO) = 0 are bounded, it follows that v1' is bounded from below. Thus tt1'(g) is not identically zero. From this we deduce Theorem 8: If A and tt are differentials, then A(g) = tt(g) for all g E V(k) if and only if A1'(g) = tt1'(g) for all g E V(k) and one prime p. Theorem 9:

go

E

If tt(go) = 0 for all differentials tt of k, then

k.

Proof: Let A be a fixed non trivial differential. Then A(gog) is also a differential-it is clearly a continuous linear map, and it

6.

THE DIFFERENT

255

vanishes for field elements, for if g = ex E k, we have A(exgo) = 0 since g -- A( exg) is a differential. Hence there is an element f3 E k such that A(gog) = A(f3g). By the previous theorem,

for all primes p, i.e.

.\«go -

(3)p gp)

=0

for all p and hence (go - f3)p = O. So go = f3.

6. The Different Let k be a PF-field. K(k a finite extension. In the function field case we shall assume that K and k have the same constant field. Let WI , W2 , " ' , Wn be a basis for K/k. Then every valuation vector X E V(K) can be written uniquely in the form

where the gi are valuation vectors of V(k). Let A be a fixed non trivial differential of k. Then if I-' is any differential of K, there exists a continuous linear V(k)-homogeneous map lof V(K) into V(k) such that I(K) C k and fLeX)

= '\(l(X)).

Since K is also a PF-field we may write fLeX) = ~ fL~(X)

= ~ ~ fL~(X).

ill

p 0, then m(a) =

o.

If there is a field element ex =F 0 in the parallelotope a;): ordp a for all p. Hence n(ex) ;): n(a). But by the product formula, n(ex) = O. If n(a) > 0 it follows that there can be no non-zero field element in the parallelotope ITa' i.e. m(a) = O. Now let a divide b, so ITa :J ITb • We apply Theorem 3 of Chapter 13 to the case where A = ITa , B = lIb and C = k. This yields Proof:

ITa, then ord p

(JIa : Ih)

= (na n

k : lIb n k)

+ (lIa + k : lIb + k).

Hence we have n(b) - n(a)

= m(a) -

m(b)

+ (lIa + k : lIb + k)

(*)

From this formula we proceed to prove the Riemann-Roch Theorem.

2. First Proof For each divisor

b

we define the function l(b): l(b)

= (V: lIb + k).

2.

263

FIRST PROOF

We must first show that I(b) is finite for all divisors b. To this end we use the result (see Chapter 13, section 2) that there exists an ideIe ao such that iloo + k = V. We select a divisor a such that ilo = iloo + ilb • Then certainly a divides b and we have, by formula (*) n(b) - n(a)

=

m(a) - m(b)

= m(a) - m(b)

+ (V: lIb + k) + I(b).

Thus l(b) is finite, as we set out to prove. We may therefore rewrite the formula (*) as follows: n(b) - n(a) = m(a) - m(b)

+ I(b) -I(a).

Hence when a divides b we have n(a)

+ m(a) -I(a) = n(b) + m(b) -I(b).

Now let a and b be any two divisors, b their greatest common divisor. Since b I a and bib we have n(a)

+ m(a) -I(a) =

+ m(b) -I(b) = n(b) + m(b) -I(b).

We see that the value of n(a) of the field k. We write n(a)

n(b)

+ m(a) -

+ m(a) -

l(a) must be an invariant

I(a) = 1 - g

and we define g to be the genus of the field. If we consider the parallelotope defined by the unit element of k we have n( 1) = 0 and m( 1) = 1. The latter result is a consequence of the product formula which shows that the only field elements in ill are the elements of ko . We have therefore 1 - 1(1)

= 1 - g;

whence g = 1(1) = (V: ill + k). This shows III particular that the genus is a non-negative integer. Since m(a) ~ 0, we have I(a) ~ n(a) g - 1. Hence by choosing a such that n(a) is large enough we can make I(a) as large as we please.

+

XIV.

264

THE RIEMANN-ROCH THEOREM

We have now to interpret the function l(a). Let U denote the k). Then l(a) = dim U = dim where factor space V/(flo. is the dual space. If,\ is any linear map of U into ko we may regard A as a linear map of V into ko such that A vanishes on FIo. k, i.e. we may regard A as a differential of k vanishing on FIo.. Conversely each differential which vanishes on FIn gives rise to a linear map of V into ko • Thus lea) = dim = dimension (over ko) of those differentials of k which vanish on IIo. • Now let A(t) be any non trivial differential. By Theorem 6 of Chapter 13, there is a maximal parallelotope fl'X!-l on which vanishes: A(fl'X!-l) = O. By Theorem 5 of Chapter 13 any other differential p.(t) may be written in the form p.(g) = A(ag) with a E k. Clearly the maximal parallelotope on which A(ag-) vanishes is FII1.-1'X!-1 • Thus the divisors describing the maximal parallelotopes on which differentials vanish belong to a fixed divisor class, which is called the canonical class, or the class of differentials. Now let p.a) = A(ag) be a differential which vanishes on FII). . Then we have FII). C FIoc-l'X!-l , i.e. a-1 ::o-1 divides a. Hence aa::O is an integral divisor, so a E flo.-l'X!-l n k. Conversely for each a E flo.-l'X!-l n k, A( at) is a differential which vanishes on FIo.. It follows that

a

+

a

+

a

lea)

= m (a~).

We summarize the above results formally in If a is any divisor in a field of

Theorem 1: (Riemann-Roch). genus g, then n(a)

+ mea) =

m

C~) + 1 -g,

where ::0 is a divisor of the canonical class. The theorem has some immediate consequences: Let a = I, so that m(1) = I, n(I) = 0, whence m

Let

Q

(~) =g.

= 1/::0, then

m

(~) + n (~) =

2 -g.

3.

Hence n

265

SECOND PROOF

(~) = 2 -

2g.

We now recall Lemma 2 which shows that if n(l/a::o) if n(a) < 2 - 2g, then m (1/a::O) = O. This yields

> 0, i.e.

+

Corollary: If n(a) < 2 - 2g, then n(a) m(a) = 1 - g. This result is known as the Riemann part of the Riemann-Roch Theorem. 3. Second Proof

The second proof starts from the formula (*) of section 1. Again our first step is to define l(a):

l(a) = (V : ITa

+ k).

We have to prove l(a) is finite, and to this end we study the function r(a) = - m(a) - n(a). It is clear that r(a) is a class function, and if a I b then r(a) > r(b). (The first remark is obvious, and the second follows from formula (*).) Lemma 3:

Proof:

r(a) is bounded from above.

Let R

= ko(x) be a rational subfield of k. Then = n we have the missing implication in the proof of Lemma 4, namely that A(rxIIa.)

=0

¢>

IIXl - , :) II"a. •

It follows that l(a) = m(1ja:D) where !> is the special divisor. As in the first proof we can show that :D belongs to a fixed divisor class W. We have therefore proved again the Riemann-Roch Theorem: m(a)

+ n(a) = m (a~) + 1 -

g.

CHAPTER FIFTEEN

Constant Field Extensions

1. The Effective Degree Let K be a PF-field with a field of constants Ko. Then K is a finite extension of Ko(x) where x is an element of K not in Ko . We now adopt the following notation: ~ shall denote the generic prime in K. k shall denote a subfield of K in which Axiom 2 holds. (By section 4 of Chapter 12, Axiom 1 also holds in k.) Kjk is not necessarily finite. p shall denote the generic prime of k induced by a ~ which is non trivial on k. ko shall denote the constant field under all p. By section 4 of Chapter 12, ko = Ko " k. f(~) shall denote the degree of the residue class field K'll over Ko . f(p) shall denote the degree of the residue class field hI' over ko . f'll shall denote the degree of K'll over hI' whenever finite. e'll shall denote (ord'll a)j(ord p a) for a E k. n'll = e'lli'll = degree of K*(~) over k*(p) whenever finite. We have already seen that Axiom 1 holds in k for the primes p if we use the valuation I a I? =

IT II a II'll·

'lll? Since k is a PF-field, it follows that the valuation I I? is either the normal valuation II III' or a constant power of it (the same for all p). Let us agree to use the same constant c for defining the normal valuations in both K and k. Then we may write

I a II' =

II \I a II'll = 'llill 271

\I a 11;'(Klkl.

XV.

272

CONSTANT FIELD EXTENSIONS

We call the exponent m(K/k) the effective degree of K/k. It is necessarily finite, but need not be an integer. We shall see that in a certain sense it measures the size of the extension K/k, the constant field extension being disregarded.

Theorem 1:

1 m(KJk) = f(p) . ~ e\l3f(jp)

for all primes p. Proof:

Since the same c is used for

II III'

and

II

11\13 we have

Hence we have

II a 11;'(K'k) = IT II a IllP =

- ~ '\I3!(\I3)ordp a

c \1311'

\l3lp

and therefore m(KJk)f(p) =

k e\l3!(\I3) • \1311'

If F :) K :) k, where F, K, and k are PF-fields,

Theorem 2: then

m(FJk) = m(FJK) m(KJk). Proof: We denote the generic prime in F by element of k. Then

II a 11;'(F'k) = IT II a II~ = mil'

=

Let a be an

IT (IT II a IIIll) = IT II a 111Ii(F'K) \1311' ml\13

II a Ilm(F'K)m(K'k) p

and this proves the theorem.

!lJ.

\1311'

1.

THE EFFECTIVE DEGREE

273

We shall need two lemmas about the rational case: Lemma 1:

Proof: Let Po be the valuation in ko(x) defined by the irreducible polynomial x. The residue class field 'kPo = ko and so f(po) = 1. If \Po I Po in Ko(x), \Po is also the valuation corresponding to x. Hence f(\p o) = e'llo = 1. From Theorem 1 it follows that m(Ko(x)/ko(x)) Lemma 2:

=

1.

If Ko/k o is finite, then

Proof: Since

we have

To prove the equality it will be sufficient to show: If W l , W2' "', Wn are elements of K o , linearly independent over ko, then they are linearly independent over ko(x). Suppose therefore

where each fi(x) E ko(x). Then, multiplying by the denominators of the fi(X) we obtain a relation with polynomials as coefficients. So we may suppose the fi are already polynomials. If ai. is the coefficient of x· in fi(X), then

as coefficient of x' on the left side. Hence ai • = 0 for all i, v or fi(x) = O. We are now in a position to determine m(K/k) if K is a finite extension of k. To do this we first prove

XV.

274


Theorem 3: If K/k is finite, then

~ nIP

=

for all

deg (Kjk)

p.

(t) is a factor of f(t) in k[t) with highest coefficient 1. The other coefficients of cf>( t) are symmetric functions of the roots of cf>(t). These roots are also roots of f(t) and therefore algebraic over ko . It follows that the coefficients of cf>( t) are algebraic over ko . Since they lie in k, they must already lie in ko and cf>(t) E ko[t). Since f(t) is irreducible in ko[t) it follows that cf>(t) = f(t) and the Theorem is proved. Theorem 12: Let ko be algebraically closed in k, tX as before and F a field between ko and k such that deg (kiF) is finite. Then deg (k(tX)/F(tX)) = deg (k/F).

Proof: ko is then algebraically closed in F as well as in k and Theorem 11 gives deg (ko(tX)/ko)

= deg (F(tX)/F) = deg (k(tX)/k).

Now deg (k(cx)/k) . deg (k/F) = deg (k(cx)/F) = deg (k(cx)/F(cx)) • deg (F(cx)/F), or deg (F(cx)/F) • deg (k/F) = deg (k(cx)/F(cx)) • deg (F(cx)F) and our theorem follows. We specialize now the field F. It is well known that ko is algebraically closed in a pure transcendental extension ko(x). By induction one shows that ko is algebraically closed in F = kO(Xl , X 2 , ... , x r ) where the Xi are algebraically independent over k o • For an tX algebraic over ko it follows from the same fact that ki tX) is algebraically closed in F( tX). Let now k be an algebraic extension of such an F for which it is known that ko is still algebraically closed in k. The preceding theorems show that deg (k(cx)/F(cx)) = deg (k/F).

3.

FINITE ALGEBRAIC CONSTANT FIELD EXTENSIONS

281

But in spite of the fact that ko(rx) is algebraically closed in F(rx) it is not true in general that ko(rx) will be algebraically closed in k(rx). We now seek conditions under which this additional result will hold. Suppose f3 E k(rx), f3 algebraic over ko(rx). Then deg (k(cr.)/F(cr.,~))

~

deg (k(cr.)/F(cr.))

=

deg (kIF).

Let us assume that ko(rx, f3) can be generated by a single element: ko(rx, f3) = ko(Y). Then F(cr.,~)

= F(y)

k(cr.) = k(y).

and

Hence deg (k(cr.)/F(cr.,~))

=

deg (k(y)/F(y))

=

deg (k/F)

=

deg (k(cr.)/F(cr.)),

and we see that F(a, f3) = F(rx). This means that f3 lies already in F( a). Since f3 is algebraic over ko(rx) and k o(rx) is algebraically closed in F(rx), it follows that f3 lies in ko(rx), i.e. ko(rx) is algebraically closed in k(a). Now ko(rx, f3) can be generated by a single element when either rx or f3 is separable over ko (see van der Waerden, Modern Algebra, section 40). It follows that ko(rx) is algebraically closed in k(rx) in the following two cases: (1) (2)

rx is separable over ko ; k is separable over K (i.e. k is separably generated over k o).

In the second case we conclude as follows: kjF separable =;.. k( rx)jF( rx) separable =;.. f3 separable over F( rx). Now f3 was algebraic over ko(a). ko(rx) is algebraically closed inF(rx). cfo(t) = Irr (f3, ko(rx), t) then cfo(t) remains irreducible in F(rx) according to Theorem 11. So cfo(t) is separable, therefore f3 is separable over ko(rx). A certain power f3p. will be separable over ko and still generate the field ko(rx, f3) over ko(rx). Summing up we have Theorem 13: With the notation as described, ko(rx) is algebraically closed in k(rx) if one of the following conditions is fulfilled:

(1) (2)

ko(a)/ko is separable; k is separably generated over ko .

282

XV.


We specialize now to the case of transcendence degree r = I, when ko is the constant field of a PF-field. Whenever ko(Ci.) is algebraically closed in k(Ci.) it follows from earlier discussions that ko(Ci.) is the constant field Ko of k(Ci.) = K. Theorem 11 shows now that m(K/k) = 1. Theorem 14: Let Ci. be algebraic over ko, K = k(Ci.). We have m(K/k) = I and Ko = ko(Ci.) in either of the two cases:

(I) (2)

k o(Ci.)/ ko is separable; k is separably generated over ko •

If Ko/k o is finite and purely inseparable (characteristic p > 0), then K/k is purely inseparable. Theorem 4 shows that m(K/k) is a power of p. If Ko/k o is an arbitrary finite extension, let Kl = ko(Ci.) be the separable part of Ko . We obtain from Theorem 14 m(K/k) = m(K/K1k) m(K1k/k) = m(K/K1k) = power of p.

Theorem 15: If Ko/k o is finite, K = Kok, then m(K/k) is I if the characteristic is 0 and of the form I/pv (v ~ 0) if the characteristic is p > O. In order to treat infinite constant field extensions we use the following Definition: A family K(a) of constant field extensions of k shall be called a C-family if to any two fields K(a), K({J) a third field K(Y) of the family can be found that contains both of them. The union of the elements of all K(a) shall be denoted by K. It is again a constant field extension of k and and one shows easily K o =Ua K(a) o' Theorem 16: If m(Kla)/k) = I for all Ci., then m(K/k) = 1. Proof: Kri a) and k are linearly disjoint over ko • Therefore Ko = Ua Kria) and k are linearly disjoint over k o . Let ko be the algebraic closure of ko and k = kok. Since ko is algebraically closed, it is the constant field of k. Let kri a) stand for all finite algebraic extension of ko and put K(a) = kria)k. Then K(a) is a C-family with k as union.

3. If

k~a.)

FINITE ALGEBRAIC CONSTANT FIELD EXTENSIONS

283

C k~fJ), then m(K(fJ)jk)

=

m(K(I3)jK(a.») m(K(a.)jk)

~

m(K(a.)jk).

On the other hand

Theorem 15 shows that the numbers m(K(a.)/k) take on their minimum value for a certain K(Y). For all K(a.) J K(Y) we have m(K(a.)/k) ::;:;;; m(K(Y)/k) and therefore m(K(a.)/k) = m(K(Y)/k) or m(K(a.) jK(Y») = 1. If we apply Theorem 16 to all K(a.) J K(Y), we obtain m(k/k(Y») = 1 or m(k/k) = m(K(Y)jk) = a power of p. Let now Ko be an arbitrary (not necessarily algebraic) extension of ko , K = Kok and 1?0 the algebraic closure of Ko. We have

Since kok has an algebraically closed constant field, Theorem 7 shows that m(1?ok/kok) = 1. So m(kokjk) m(Kjk) = m(f?okjK) = a power of p.

Since m(1?ok/K) ::;:;;; I, m(Kjk) ~ m(kokjk) = m(K(Y)jk).

Theorem 17: If the characteristic is 0, thenm(K/k) = 1 for all constant field extensions. If it is P > 0, then m(K/k) = lip" (v ~ 0) and the minimum value is taken on by a finite algebraic extension K(Y). Theorem 18: Let K(a.) by any C-family above k. There exists a K(Y) such that m(K/k) = m(K(Y)/k). For any K(a.) containing K(Y) we have

Proof: m(Kjk) = m(KjK(a.») m(K(a.)jk)

~

m(K(fJ)jk).

xv.

284


Since m(K(a.'lk) = 1lpv, there exists a K(Yl for which the minimum value is achieved. For K(r:t.l J K(Yl one has

whence

and therefore

m(K(a.l/K(Y») = 1. This implies m(KIK(Y') = 1.

4. The Genus in a Purely Transcendental Constant Field Extension We consider the case K = k(t) where t is transcendental over k. The primes \l3 of K are those that are trivial on ko(t). We have to distinguish two kinds of primes: (1) \l3 is also trivial on k. They are then among the well-known valuations of the rational field k(t) which come from irreducible polynomials p(t) E k[t] (with highest coefficient 1. The infinite prime would not be trivial on ko(t)). But we have to single out those p(t) which give rise to the trivial valuation on ko(t). To find them, suppose \l3 induces a non trivial valuation on k()[t] that comes from the irreducible polynomial Po( t) E ko[t]. Since ko is algebraically closed in k we know from Theorem 11 that Po(t) remains irreducible in k. Since p(t) I Po(t), we have p(t) = Po(t)· The primes \l3 in question are therefore those irreducible polynomials p(t) E k[t] with highest coefficient 1 that have at least one non constant coefficient. (2) The primes that are non trivial on k, \l3 I p. Consider first a polynomial g(t) = Xl + xlt + ... + xrt r, Xi E k.

I t l!ll = 1 and Max I Xv II'· v

Since \l3 is trivial on ko(t) we have Ig(t)

l!ll ~

4.

285

PURELY TRANSCENDENTAL CONSTANT FIELD EXTENSION

Suppose the strict inequality holds. Dividing g(t) by the Xi with the maximal absolute value we obtain a new polynomial where Max v I XV I" = 1, and where I g(t) I" < 1 or g(t) = 0 (mod IP). This congruence would mean that t (as element of the residue class field Kw,) is algebraic over the residue class field h". Since h" is algebraic over ko , t would be algebraic over ko . But the residue class field Kw, contains an isomorphic replica of ko(t) where t is transcendental over ko • We have therefore Ig(t)

Iw, =

Max v

I Xv I,,·

This shows that there is only one IP I p and that eW, = 1, An arbitrary element X E K can be written in the form

p

=

IP.

g(t) X = x h(t)'

where X E k and g(t) and h(t) are relatively prime polynomials with highest coefficient 1. We first ask for those X that satisfy I X Iw, ~ 1 for all IP that are trivial on k. If h(t) were divisible by an irreducible polynomial P(t) with non constant coefficients, our condition would be violated at that prime since g(t) is relatively prime to h(t). This shows that h(t) E ko[t]. If conversely h(t) E ko[t] it is a constant of K and our condition is obviously satisfied. This allows us first to determine Ko. If namely I X Iw, = 1 for all IP then both g(t) and h(t) have to be in ko[t]. Therefore . X I" = I X I" = I which means that x E ko, so that X E ko(t). This means Ko = ko(t). Let now Xl' X2' "', X,. be elements of k linearly independent over k o ' Suppose

Clearing denominators, we may assume that the fi( t) are polynomials. Collecting terms we obtain an equation for t. Since t is transcendental over k, all coefficients have to vanish. Since the Xi are independent over ko this means all fi(t) = O. k and Ko are therefore linearly disjoint over ko' and this means m(K/k) = 1. Let now a be a divisor of k, lla the parallelotope of V(k). a may

286

XV.


be viewed as divisor of K and let ITa' be the parallelotope of V(K). We ask for the elements X E K such that X E ITa'. Then I X l!ll :(; 1 for all \l3 that are trivial on k. Therefore X = g(t)jh(t) where h(t) is in Ko. Let

We have

So g(t)

E

ITa' if and only if Xi

E

ITa. Hence

is a linear combination of elements of ITa n k with coefficients in Ko. This gives (because of the linear disjointness) mK(a) = mk(a).

Theorem 9 shows nK(a) = nk(a). Let g(k) be the genus of k, g(K) that of K. Let a be a divisor of k such that nk(a)

be a non trivial place of h. In the case of a function field we restrict ourselves to those places 1> which set like the identity map on the constant field ho • In this case we select an element x E h, x rf- ho such that 1>(x) oFro. (This can always be done, for if 1>(x) =ro, then 1>(I/x) = 0, oFro.) Let R be the rational subfield. R is the field of rational numbers if h is an algebraic number field and R = ho(x) where 1>(x) oFro in the case of a function field. Then 1>(a) oFro where a is any integer of R. In the number field case this follows from the fact that 1>(1) = 1. In the function field case it follows from our choice of x. We now show that 1> is not the identity on the integers of R. Since 1> is non trivial there is an element 7T oF 0 such that 1>(7T) = O. This 7T satisfies an irreducible equation ar",r

+ ar_l",r-l + ... + ao =

0

with integral coefficients in R. Applying the homomorphism 1> we obtain 1>(ao) = 0, since 1>(7T) = O. But a o oF 0 since the equation is assumed to be irreducible. Hence 1> is not the identity on the integers of R. It follows that 1> maps the integers of R into a field. The kernel is a non zero prime ideal (p) of R. If a is any element of R, and a = pvh, where p does not divide the denominator, or the numerator of h, then I a I = I p Iv. Hence the value group V R of II over the rational field R is cyclic. Now let ex be any integer of h. ex satisfies an equation f(o:) = ano: n + a.._lo:n - 1 + ... + ao = O. Since If(ex) I = 0, two of the terms must have maximal absolute msince value, say I ai exi I = I ajex j I. Then I exi - j I = I aj/ai I = I p 1 the ai lie in R. Since i - j may vary as ex varies, we introduce a uniform exponent by writing Hence if Vk is the value group of I lover h, we see that v~, is a subgroup of the cyclic group V R • But since Vk is an ordered group,

2.

299

ALGEBRAIC CURvES

the mapping I (X 1- I (X In' is 1 - 1 and order preserving, and so V~' = Vk' Thus Vk is a cyclic group of positive real numbers. Hence we may regard the mapping (X _ I (X I as a valuation of k. In particular, we have

Theorem 4: To every point on an algebraic curve corresponds a valuation of its function field. Let k = ko(x, y) where x and yare connected by the polynomial relation F(x, y) = 0 (not necessarily irreducible). k is then the function field of the curve defined by F(x, y). Let (x o , Yo) be an algebraic poInt on an irreducible constituent of this curve, i.e. an algebraic zero of an irreducible factor of F(x, y). Then (xo , Yo) gives rise to a valuation of k. Theorem 5: If of/ox and of/oy are not both zero at (xo' Yo), then the valuation induced by (xo ,Yo) is unique and the residue class field under this valuation is ko(xo ,Yo), Proof: Suppose of/oy (xo, Yo) oF O. Let p(x) = Irr (x o , k o , x). If 1> denotes the mapping defined by (x o , Yo), i.e. the mapping described by cP(o:)

=

cP(x)

=

for

0:

xo ,

0:

cP(y)

E ko,

= Yo

I

then 1>(P(x)) = p(xo) = O. Hence if p denotes any valuation induced by 1>, we see that I p(x) Iv < 1. Thus p induces on R = ko(x) the valuation defined by p = p(x). The map 1> can be extended to each completion k*(p) and hence also to R*(p). Let FI(x, t) be the irreducible factor of F(x, t) in R of which y is a root. Then the different inequivalent completions k*(p) are obtained by adjoining to R*(p) roots of the distinct irreducible factors of FI(x, t) in R*(p). Suppose fl(t) is Irr (y, R*(P), t). Then fl(t) is one of these factors. Since y - Yo oFro, we have I y Iv ~ 1, so Y is an integer and hence all the coefficients of fl(t) are integers (cf. Chapter 2, section 5). Since y is a root of F(x, t) = 0 we have F(x, t)

= fl(t) g(t)

300

XVI.

APPLICATIONS OF THE RIEMANN-ROCH THEOREM

and g(t) has integer coefficients. Then

BF at = fl(t) g'(t) + f~(t) g(t), of at (x, y) = g(y)f{(y), since fl(Y) = O. Now apply the map 1> which has been extended to the completions R*(p) and k*(p); we obtain (g(yo)) (f{(yo))

of

= at (xo ,Yo) i=- O.

Hence

But if there were any irreducible factors of F1(x, t) distinct from fl(t) these would be contained in g(t) and we should obtain 1>(g(yo» = O. Thus the valuation p induced by 1> is unique. Let fo(t) = 1>(fI(t». fo(t) is a polynomial in the residue class field Rp = ko(ko(xo). Since fl(t) is irreducible over R it follows that fo(t) is either irreducible or a power of an irreducible polynomial (cf. Chapter 3, Theorem 8). Since !o'(Yo) = 1>(N(yo» =1= 0, Yo is not a multiple root of fo( t) and hence fo( t) is irreducible in ~ . Hence deg (ko(xo , yo)/ko(x o))

= deg (k*(p)/R*(p)).

Now the residue class field kp of k certainly contains ko(xo) and Yo' Hence we have deg (k*(p)/R*(p)) ~ deg (kp/Rv) ~ deg (ko(xo, yo)/ko(xo)) ~

deg (k*(p)/R*(p)).

Thus we have the required result that kp = ko(x o , Yo). We notice also that the residue class field is separable over ko(xo). The extension ko(xo)jko may however be inseparable. 3. Linear Series

Let k be an algebraic function field in one variable. Let f be a fixed divisor class of k, ao a fixed divisor in f. Then if ai' a 2 , ... a r

3.

301

LINEAR SERIES

are divisors of t we have ai/aO We define the expressions

=

(Xi

where

(Xi

is an element of k. (*)

(with

Ci

in the constant field k o) to mean

The totality of all such expressions is called a linear series. The maximum number of linearly independent functions (Xi is called the dimension of the linear series. We see that the linear series does not depend on the choice of ao . For if we choose ao' as a new fixed divisor in t, we have

Hence

Thus the totality of all expressions (*) is unaltered by the change. Now let the ai be integral divisors. We have 1 a. · mtegra •

o¢>

ex. = -ain E -1 • ao ao

o¢>

1• extao· mtegra

Hence ca +···+ca 11 rr=cex +···+cex 1 1 r r aO

Enao-1·

The set of expressions (*) constructed using all the integral divisors of t is called a complete linear series. The dimension of f is defined to be the dimension of the integral divisors of f, i.e. dim t

The Riemann-Roch Theorem,

=

m(a(jl).

302

XVI.


can now be written in the form dim f

= n(f) + g -

1

+ dim (~) ,

where W is the canonical class. 4. Fields of Genus Zero In our first theorem we give a complete description of the fields of genus zero. Theorem 6: A field has genus zero if and only if it is of one of the following types: (a) a field of rational functions, (b) the function field of a conic, i.e. a field ko(x, y), where x and yare related by an equation ay2

Proof:

+ (b + ex) y + (d + ex + fx 2) = o.

We divide the proof into three parts.

Part 1. k = ko(x) is a rational field. We show it has genus zero. Consider the parallelotope IIa where a = p;;;" n(a) = - s. To find m(a) we notice that ex = c/>(x) E IIa if and only if c/>(x) is a polynomial of degree ~ s. Hence m(a) ~ s 1. It follows that 1 ~ l(a) + 1 - g, or g ~ l(a). But l(a) is zero if - s < 2 - 2g (by the Riemann part of the Riemann-Roch Theorem). Hence

+

g=O. Part 2.

k = ko(x, y), where ay2

+ (b + ex) y + (d + ex + fx 2) = o.

We show that k has genus zero. If this equation is of the first degree, or is reducible or yields a constant field extension, then k is a rational field and g = 0 by Part 1. We may suppose none of these is the case and that a 0:/= 0, say a = 1. Consider again IIa where a = p;s. In this case n(a) = - 2s since the degree of k over ko(x) is 2. We contend that

4.

303

FIELDS OF GENUS ZERO

+

lla contains all polynomials of the form c/>(x) yrp(x) where deg c/>(x) ~ sand deg rp(x) ~ s - 1. Hence m(a) ~ 2s 1. To show this we have only to remark that for finite primes p, I y Iv - 1, and that I y Ip ~ I x Ip . These results follow from an examination of the do~inant te~s in the equation of the conic. We now have 1 ~ l( a) 1 - g as in Part 1, and the result follows as before.

+

+

Part 3. k is a field of genus zero. We show it is one of the types described above. If b is a divisor of the canonical class, n(b)

= 2g -

2

=-

2.

Therefore m(b)

= 1-

n(b) = 3.

Hence the parallelotope llb contains three linearly independent elements of k, say ex, {3, y. If we set x = (3/ex, y = y/ex, and a-I = ex-I!) we see that lla-1 contains, 1, x, y. Since 1 E lla-1, a is an integral divisor. Since x E lla-1, ax is an integral divisor, b say, so that x = b/a. Then since deg (k/ko(x» = n (denominator of x), we have that deg (k/ko(x» ~ n(a)

= n(b-1 ) = 2.

(The inequality is written because there may be cancellation between a and b, leaving a single prime in the denominator of x.) If deg(k/ko(x» = 1, then k is rational. If deg (k/ko(x» = 2, then the denominator of x is a. Since y E lla-1 we may write y = cia. It follows that y is integral at all finite primes dividing x. Hence if y can be expressed as a rational function of x, this function must be a polynomial. The only polynomials in x which can lie in lla-1 are linear polynomials. But y is not a linear polynomial in x since it is linearly independent of x. It follows that ko(x, y) is a proper extension of ko(x). Hence k = ko(x,y). In order to find the defining equation of this extension (which must be an irreducible quadratic) we consider the parallelotope lla-' . n( a-2 ) = 4, and consequently the following elements of k

304

XVI.


lie in IIa-2: 1, x, y, X2, xy, and y2. But m(a- 2 ) = 5; hence there must be a relation of linear dependence between these six elements: ay2

+ (b + ex) y + (d + ex + fx 2) = o.

This completes the proof. In order to distinguish between the two types of field of genus zero, we prove the following result: Theorem 7: A function field of genus zero is rational if and only if there exists a divisor of odd degree.

Proof: If k = ko(x), the denominator of x is a prime of degree 1. In addition, the numerator of x and the numerator of x + 1 are also primes of degree 1. Conversely, suppose k contains a divisor of odd degree. Since k contains a divisor of degree - 2 (any divisor of the canonical class), we can construct a divisor a such that n(a) = 1 and so m(a-1 ) = 2. Then IIa-l contains two linearly independent functions ex, {3. If we set x = (3lex we have 1 and x in IIrx-la-1 • Hence exa is integral, and n(exa) = n(a) = 1. Thus exa is a prime p. But xp is integral, so x = b/p. It follows that if ko is the constant field of k, then deg (k/ko(x))

= n(p) =

1.

Hence k = ko(x). Consider the algebraic curve C defined by an ideal a in the polynomial ring ko[x, y]. The algebraic points of C are the sets (xo' Yo) where x o , Yo lie in ko or some algebraic expression of ko , and are such that a is annihilated by (xo , Yo). A point (xo , Yo) on C with coordinates in ko is called a rational point. Theorem 8: The function field of a conic is rational if and only if the conic contains a rational point.

Proof:

Let the conic be ax2

+ bxy + ey2 + dx + ey + f

=

O.

If a = b = c = 0, then the conic is a straight line. The function field is rational, and the conic contains rational points. If one of a, b, c is non zero, we can make a change in the generators x, y if necessary, so that a oF 0, say a = 1.

4.

305

FIELDS OF GENUS ZERO

Suppose the function field k is rational, with constant field ko . Then there are three primes of degree one (cf. Theorem 7). Let them be PI ,1'2' Pa . If a is the denominator of y, n(a) = 2. Hence at least one of the Pi , say PI is not in the denominator of y. Hence ordp1 Y ? O. From the equation of the conic ordp1 x ? 0 also. It follows that in the homomorphic map of k into ko 00 defined by PI (see section 1), both x and y have images in ko ' This the conic contains a rational point. Conversely, suppose (xo , Yo) is a rational point on the conic. If we write x = xo u, y = Yo v we obtain a new equation of the form au! + buv + cv2 + du + ev = O.

+

+

+

This may be written in the form 2

a(: )

+b(: ) + c+ d(

! )(: )+ ! ) e(

=

o.

Thus l/v is a rational function of u/v (unless d = e = 0, in which case the equation yields a constant field extension). Hence v, and consequently u are rational functions of ulv. It follows that the function field k = ko(u/v), i.e. k is rational. Theorem 9: In a field of genus zero, every divisor of degree zero is a principal divisor.

Proof: Let a be a divisor such that n(a) = O. Then m(a) = 1, so IIa contains a single function ex. Hence IIa.-1a contains 1. It follows that exa-1 is integral. But n(ex-1a) = 0, and so ex-1a = 1. Hence a = ex.

Finally we prove the well-known Theorem 10: The only admissible generators of the field ko(x) are elements of the form (a bx)/(c dx) where a, b, c, dlie in k o .

+

+

Proof: Let PI , 1'2' "', be divisors of degree 1. Then x = 1'2/1'1 say. If u = Pa/Pl , then u is integral at all finite primes dividing x. Thus u is a polynomial in x. Since ord"l u = - 1, thus polynomial dx. is linear: u = a + bx. Similarly if v = 1'4/1'1' then v = c Hence

+

1'3/1'4

=

W

=

a C

+ bx + dx .

306

XVI.


Since n(P4) = 1, w is a generator of ko(x), and conversely, each generator is of this form.

5. Elliptic Fields

Fields of genus 1 are called elliptic fields. For such fields the Riemann-Roch Theorem has the form

n(a) where m(b-1)

+ m(a) = l(a) = m ( a~ ) ,

b belongs to the canonical class, n(b) = 2g - 2 = 0, = g = 1. There are several possibilities for the values of

m(a): (1) (2) (3a) (3b)

If n(a) > 0, then m(a) = O. If n(a) < 0, then m(a) = - n(a). If n(a) = 0, and a is a principal divisor, then m(a) = 1. If n(a) = 0, and a is not principal, then m(a) = O.

(1) has already been shown for all fields (Lemma 2 to the Riemann-Roch Theorem). To prove (2) we notice that

Hence m(a) = - n(a). In case (3), we see that if n(a) = 0 and m(a) > 0, then the parallelotope IIn contains at least one function ex. Hence 1 lies in IIrx-l n . It follows that exa-1 is integral. But n(exa-1) = O. Hence exa-1 = 1, a = ex. Since m(b-1) = 1 and n(b-1 ) = 0, it follows that b-1 , and hence b, is a principal divisor. Thus the canonical class is the unit class. A field of genus g 1 has a divisor of degree 2g - 2. For fields of genus 1, the minimal positive degree of a divisor can be arbitrarily large, (S. Lang, J. Tate, Principal homogeneous spaces over abelian varieties, Amer. J. Math., July, 1958, remark at end of introduction).

*"

Case J: There exists a divisor a with degree 1. This is certainly true for function fields over algebraically

5.

ELLIPTIC FIELDS

307

closed fields, because then all primes are of degree 1. We shall show later that it is true for function fields over finite fields. Since n(a) = 1, we have m(a-1 ) = 1 so lla-1 contains a function ex. Hence llu.-1a-1 contains 1; thus exa is integral and n( exa) = 1. It follows that we may assume our original a to be integral. Now m(a- 2 ) = 2 so lla-' contains two linearly independent functions 1 and x. Now xa2 is integral; hence if ko is the constant field of k, deg (k/ko(x)) = n denominator of x) = 2. We see that a2 = Pro the infinite prime in ko(x). Since m(a-3 ) = 3, lla-s contains three linearly independent functions 1, x, y. Now Y is integral at all finite primes dividing x. Hence if y is a rational function of x, this function is a polynomial. But lla-s contains no power of x higher than x itself, and y is not a linear polynomial in x since it is linearly independent of 1 and x. It follows that y does not lie in ko(x). Hence k = ko(x, y). To find the equation which defines this extension, we consider the parallelotope lla-8. This parallelotope contains the functions 1, x, X2, x 3 , xy, y, y2. But m(a-6 ) = 6, so there must be a relation of linear dependence between these 7 functions. Thus y2

+ (ax + b) y + c + dx + ex2 + fx!> = O.

If the characteristic of k is unequal to 2, 3, we can reduce this equation to the familiar Weierstrass form

We consider the following situation now: Let k be the field ko(x, y) where y2 = f(x), degf(x) = 3, the characteristic of k is unequal to 2, and f(x) is separable. Then k is the function field of the curve C F(x, y) = y2 - f(x) = O.

of/oy == 2y, hence of/oy (x o , Yo) oF 0 provided Yo oF 0, if Yo = 0 then and

since f(x) is assumed to be separable. It follows that of/ax and of/oy never vanish together on the curve. Hence by Theorem 5,

308

XVI.


each algebraic zero (xo ,Yo) of F(x, y) gives rise to a unique place. The residue class field under the corresponding valuation l' is ko(xo, Yo), and /(1') = deg (ko(xo, yo)/ko)· In particular if (xo ,Yo) is a rational point on C,/(p) = 1 = n(p). Conversely let l' be any prime of k of degree 1. Let 1> be the homomorphism defined by p. Then if .p(x) = Xo

=Foo,

.p(y) = Yo

=F oo ,

(Xo ,Yo) is a rational point of C. We saw in the preceding discussion that there exists an integral divisor a such that n(a) = 1. a has the property that a2 is the denominator of x in k. Hence a2 = PaJ' and a = PIX)' Hence n(p",) = 1. If 1>", is the homomorphism defined by 1'"" we have tP",(x) = 1>",(y) =00. We may call the pair (00,00) the point at infinity on C. We have the result: Lemma: There is a 1 - 1 correspondence between the primes of k of degree 1 and the rational points of C (including the point at infinity). Now let a be a divisor such that n(a) = 1. Then m(a-1) = 1. Thus IIa-l contains a single function ex, unique up to constant factors. Hence exa is integral and n(exa) = n(a) = 1. That is to say, exa is a prime p. Thus every divisor class of degree 1 contains a prime. This prime is clearly unique, since m(a-1) = 1. Further, if a defines a divisor class of degree 1, then a/p", defines a divisor class of degree O. If a defines a divisor class of degree 0, then ap", defines a divisor class of degree 1. Hence we have: Theorem 11: There are 1 - 1 correspondences between the elements of the following sets: The rational points on C and the point at infinity. The primes of degree 1. The divisor classes of degree 1. The classes of degree O. Corollary: The class number (= number of divisor classes of degree 1) is one more than the number of rational points on C. Clearly the point at infinity corresponds to the unit class.

5.

ELLIPTIC FIELDS

309

We now examine the multiplication of divisor classes of degree zero. Let (Xl' YI)' (X2 ,Y2) be rational points corresponding to the classes defined by divisors PI/Poo , P2/Poo respectively. Let the inverse of the product of PI/Poo and P2/Poo be represented by P3/Poo • Thus

must be a divisor of the unit class, i.e. must be a function (X E k. Now (X lies in the parallelotope IIp-3. Hence (X has the form 00

ex = a

+ bx + cy.

Further (X has zeros at PI , P2 and hence

+ bX1 + CYl = 0, a + bX 2 + CY2 = o.

a

Thus we may interpret (X geometrically as the straight line joining (Xl' YI) and (X2 ,Y2)· Unless (xl' YI) and (X2 ,Y2) coincide, these two equations are sufficient to determine (X up to a constant factor. When they do coincide, the argument must be modified slightly: (X becomes the tangent to the curve C at (Xl' YI). Clearly Ps corresponds to the point (xs , Ys) in which the straight line (X meets the curve agalll.

Case 2: There exists a divisor a of degree 2. Since n(a) = 2, m(a-I ) = 2, so IIa-l contains two linearly independent functions ex, 13. Hence, writing x = f3/(X we see that n,,-la-1 contains 1 and x. It follows that (Xa is integral. So, since n«(Xa) = 2, we may assume our original a is integral, and that IIa-l contains 1, x. Since xa is integral, we see that deg (kjko(x» = n(a) = 2, where ko is the constant field. Now m(a- 2 ) = 4, so IIa-2 contains 1, x, x 2 and a function y. By the argument of Case 1 (with the obvious modifications), we see that Y does not lie in ko(x), and hence k = ko(x, y). To find the defining equation, we must consider a parallelotope containg y2 so we examine IIa-, • This parallelotope contains the functions 1,

310

XVI.


X, X2, X3, X4, y, xy, X2y, y2. But m(a-4) = 8, so these 9 functions are connected by a relation of linear dependence: y2

+ (a + bx + cx

2

)

Y

+ (d + ex + fx + gx3 + hx3) = O. 2

Case 3: There exists a divisor a of degree 3. We can show as in the previous cases that a may be assumed to be integral. Hence, since m(a-1) = 3, IIa-l contains linearly independent functions 1, x, y. ax is integral. We may assume that a is the full denominator of x, for otherwise we recover either Case I or Case 2. Hence if ko is the constant field, deg (kjko(x)) = 3. By the obvious adaptation of the method of Case 1, we see that y is not contained in ko(x), and so k = ko(x, y). To find the defining relation we examine the parallelotope IIa--3' This contains the functions 1, x, y, X2, xy, y2, x3, x2y, xy2, y3. But m(a-3) = 9 so there is a relation of the form y3

+ (a + bx) + (c + dx + ex y2

2

) y

+ (f + gx + hx + mx 2

2

)

= O.

6.

THE CURVE OF DEGREE

n

311

6. The Curve of Degree n Let k be the function field of a curve of degree n, that is to say, let k = ko(x, y) where x and yare related by a polynomial equation F(x, y) = 0 of total degree n. We assume thatF(x, y) is irreducible, and that it does not give an extension of the constant field. (If F does give a constant field extension we may make this extension first; then F is reducible over the new constant field.) We also require that the coefficient of yn in F(x, y) be non zero. If this is not already the case, we make a change of variables as follows: Let ¢>(x, y) be the homogeneous part of F(x, y) of degree n. Make the transformation x = U + (Xv, y = v. Then ¢> becomes ¢>(u + (Xv, v). The coefficient of v n is ¢>((X, 1). We choose (X from ko (or from a separable extension, which does not change the genus) such that ¢>((X, 1) -=1= O. The equation F = 0 is now in the form we require for the proof of

Theorem 12: g(k) ~

ten -

1) (n - 2).

Proof: Let Poo be the infinite prime in ko(x). By examining the dominant terms in the equation F(x, y) = 0 we deduce that I y Ipro ~ I x Ipro for all p", dividing Poo. Further I y Ip ~ 1 for all finite primes p. Now consider the parallelotope IIp where a = p-;;,'. Since the effective degree is equal to the degree (there is no constant field extension) we have nk(a) = - ns. The parallelotope contains all functions of the form

.p,(X), y.p'_I(X), ... , y,,-I.ps_1I+I(X),

where ¢>i(X) denotes a polynomial of degree i. It follows that mea) ~ (s

+ 1) + s + ... + (s -

n

+ 2)

= ns

+1-

ten -

1) (n - 2).

We may choose s so large that mea)

+ n(a) =

Hence 1- g ~ 1-

t (n -

1 - g.

1)(n - 2)

and so we have the required result g ~

t (n -

1) (n - 2).

312

XVI.


7. Hyperelliptic Fields

Let k be any function field, and let JIa be a parallelotope, with m(a) > O. Lemma 1: If {3 E k and {3Ci. lies in the constant field ko .

E

JIa for all ex

E

JIa fI k, then {3

First Proof: Suppose Ci.1 , Ci. 2 , •", Ci. n form a ko-basis for the elements of k in JIa • Then {3Ci.i = ~ Ci.Ci.. (Ci. E k o). It follows that det (C - (3I) = 0 (C denotes the matrix [Cij] and I the unit matrix). Hence {3 is algebraic over ko . But ko is algebraically closed in k, and therefore {3 E ko . Second Proof: Let JIb be the smallest parallelotope containing Ci.1 , " ' , Ci. n . Then {3JIb is also the smallest. It follows that {3b = b. Hence, considered as a divisor, {3 = 1, i.e. (3 is a constant. Suppose now that JIb C JIa • Then a I band b = at where t is an integral divisor. If a and b have the property that m(a) = m(b) =F 0, i.e. if every field element which lies in JIa already lies in JIb we may say that a can be shrunk to b. Lemma 2:

If a can be shrunk to b, and b = ac, then m( ,-I) = 1.

Proof: Since t is an integral divisor, JIc-l contains the elements of k o • We must show it contains no other elements of k. Suppose {3 E JIc-1 fI k. Then {3c is integral. Let Ci. be any element of JIb fI k. Then Ci.b-I is integral. It follows that Ci.{3b-1c is integral and so Ci.{3 E JIbc-1 = JIa • By hypothesis, Ci.{3 E JIb , and hence, by Lemma I, (3 E ko . This completes the proof. We apply the Riemann-Roch Theorem to this divisor c-1 obtaining

Thus n(t)

+ m ( ~ ) = g.

7.

313

HYPERELLIPTIC FIELDS

It follows that 0 ~ n(c) ~ g. These results enable us to give simple examples of parallelotopes which can be shrunk, and other which cannot. First we show that if h is a divisor of the canonical class, then a = b-l cannot be shrunk. Suppose b = cb- l and m(a) = m(b). Thus

Hence n(c) = O. Since t is integral, it follows that c = 1. Thus cannot be shrunk. On the other hand let p be a prime of degree 1. We shall show that b-Ip-l can be shrunk to b-l . We have b-l

m(b-1p-l)

since n(p)

>

+ n(b-1p-l) =

m(p)

+1-

g = 1 - g,

0 implies m(p) = O. Hence m(b-1p-l)

+2 -

2g - 1 = 1 - g.

Thus

This shows that b-Ip-l can be shrunk. It is easily seen that if a is an integral divisor of degree > 1, then b-Ip-l cannot be shrunk. We shall now use these results to describe hyperelliptic fields: A function field of genus ~ 2 is called hyperelliptic if it is a quadratic extension of a field of genus zero. Let k be a function field with genus g ~ 2. If b is a canonical divisor, IIb-l contains linearly independent elements Xl , X 2 , ••• , Xg of k. If Q! is any element of k, IIrxb-l contains Q!XI , Q!X 2 , ••• , Q!Xg' Since the divisor class of b is an invariant of k, it follows that the subfield

is an invariant subfield of k.

Theorem 13:

If k # k', then k is hyperelliptic.

314

XVI.


Proof: Let deg (k/k') = p ~ 2. The effective degree m(k/k') = p also since k and k' have the same constant field. By choosing (X = xII we may assume 1, X2, "', Xg EIIX>_I and so k' = ko(X2 , "', Xg).

Now b-1 is the greatest common divisor of 1, X 2 , "', Xg in k, i.e. ord p b-1 = mini ordp xi; otherwise we could shrink b-1, which is impossible. It follows that b-1 is also a divisor in k'. Now let m'(a), n'(a), g', b' have the obvious meanings for k'. We see that n(b-1)

=

2 _ 2g,

n (b-1)

m(b-1) = g,

=

2 - 2g, p

m (b-1) = g,

this last equality holding because 1, X 2 , "', Xg E k', and so II;'-l. We apply the Riemann-Roch Theorem to b-1 in k' and get

(V) + 1 - g', 2-2g 1 ,(b), - p - - +g=m b' -g.

n (b-1)

+ m'(b-1) =

m'

From this we obtain p

(m' (V) - g') =

(p - 2) (g - 1) > O.

Hence m

,(b) b' >g

I

= m

,(1) b' .

But l/b' divides b/b', hence m

(V) ~ m (-}) =g'.

= m'(l/b'). If g' = 0 this means that we can shrink the parallelotope II(b')-l. But we have seen that this is impossible. Hence g' = 0, and m'(b/b') = g' = O. It follows that (p - 2) (g - 1) = O. Hence p = 2 and k is therefore a quadratic extension of k', i.e. k is a hyperelliptic field. For the converse of the theorem we need the following

It follows that m'(b/b')

7.

Lemma: If k is any field of genusg genus zero, then

~

2, and k' is a subfield of

+ h'X2 + ... + h'xg =F h.

h'xi

Proof:

315

HYPERELLIPTIC FIELDS

Since dimko (V' : ill'

we see that V' = Ill'

+ h') =

1'(1) = g'

=

0,

+ k. Now il~ Xi C il~ il b- 1 = ilb- 1

•

Hence Then (ill'

+ h) Xl + (ill' + h) x + ... + (ill' + h) x 2

Now

k

C ilb- 1

+ h.

+ h) = l(b- = m(l) = 1, so ilb- + h =F V. l

dimko (V :il b- 1

)

1

Hence But if xlh'

+ x~' + ... + xgh' = h,

some of the xi' say Xl , X 2 , by Chapter 13, Theorem 2 XIV'

"', Xr

form a basis for kjk', and hence

+ x 2 V' + ... + x,V' =

V.

This contradiction completes the proof of the lemma. Theorem 14:

If k is hyperelliptic, then h'

=ho(~, Xl

...,.:2) Xl

is the only quadratic subfield of genus zero. *

* For the sake of brevity we say that k' is a quadratic subfield of k if k is a quadratic extension of k'.

316

XVI.

Proof:


Let k' be any quadratic sub field of genus zero. Then k' =

X2

k'

+ ... + XII h' =F h,

Xl

Xl

since deg (k/k') = 2. If we adjoin to k' any element of k, not already in k', then we obtain all of k. It follows that Xi/Xl E k' for all i. Hence

By the preceding theorem, ko(1, subfield of k. Hence

X 2/X 1 , ... , Xu/Xl}

is a quadratic

Now let k be a hyper elliptic field. Let k' be its unique quadratic subfield of genus zero. If b is a divisor of the canonical class of k, b is also a divisor of k', and we have n' (b) = g - 1. Since k' is of genus zero every divisor of degree zero is principal. Hence all divisors of k' of the same degree belong to the same divisor class. It follows that b can be taken to be any divisor of degree g - 1 in k'. We notice that if k' is the function field of a conic without rational point, an hence has no divisors of odd degree, then the genus g of k must be odd. Case 1: k' is a rational field, k' = ko(x). Here we may take b = pr:.,-t, where p"" is the infinite prime in ko(x}. Then IIb-l contains the linearly independent elements 1, X, ... , xU-I. We now examine the parallelotopes IIa where ~ = p- 0, •

n(ai} = 2 - 2g - 2i < 2 - 2g, and hence by the Riemann part of the Riemann-Roch Theorem, n(ai} m(ai} = 1 - g, whence we have m(ai} = 2i - 1 g. When i = 1, m(ai} = g 1, and IIa, contains 1, X, X2, ... , xu, so we obtain no essentially new functions. When i = 2 however, we

+

+

+

8.

317

THE THEOREM OF CLIFFORD

+

have m(ai) = g 3 and so IIaa contains 1, x, X2, ... , .xU+!, y. By familiar arguments we can show that y does not lie in ko(x) and hence k = ko(x, y). We have then ordpooY ~ 2g 2 for at least one 1>00 dividing Pro . To find the defining equation of the extension kl k', which is quadratic, we must find a parallelotope containing y2. Hence we choose i such that - g + 1 - i = - 2g - 2, i.e. i- g . 3 The parallelotope IIQcontains 1 x , X2 " ... x 2uH " Y g+3' xy, ... , .xU+!y, y2. But m(au+s) = 3g + 5. Hence between these 3g + 6 functions there is a relation of linear dependence, which we may write

+

+

y2

+ tPHI(X) y + tP2H2(X) = 0,

where deg (Pix) :::;; i. Now y determined only up to a polynomial of degree g + 1 in x, so if the characteristic is not 2, we may assume y2 = J(x) where degJ(x) :::;; 2g 2. Since y appears for the first time in the parallelotope IIp ;,U-l, we must have degJ(x) > 2g. Hence the degree ofJ(x) is either 2g 1 or 2g 2. We notice also thatJ(x) must be square-free, otherwise we would obtain another y in an earlier parallelotope.

+

+

+

Case 2: k' is the function field of a conic without rational points. Then k' = kO(x l , X2) where axl 2 bX 22 + ... = 0 and a -=1= 0, b -=1= O. In this case we may take b = p!!-1)12 where Poo is the infinite prime in kO(x l ). Then IIb-1 contains the linearly independent functions 1, Xl' Xl 2 , ••• , xlU- 1 )/2, x 2 , X2XI ' ... , x 2xlU- 3)/2. The rest of the development of Case 2 is left to the reader.

+

8. The Theorem of Clifford Lemma 1: Let JI(X I , ... , x r ), ••• , Jm(x I , .•. , x r ) be non zero polynomials in kO[XI' ... , x r ]. Then there exists an integer N depending only on T, m, and the degrees of the Ji such that, if ko contains more than N elements, then there exist values Xl = a l , X 2 = a 2 , ••• , Xr = a r in ko such that

(i = 1, ... , m).

318

XVI.


Proof: The proof proceeds by induction on r. The result is clearly true when r = 0, since the polynomials fi are then non zero constants. Consider the polynomials fi as polynomials in x, with coefficients in kO[Xl , x 2 , ••• , X,_l]. Let the non zero coefficients of all thefi be denoted by g.(x1 , ... , X,_I). By induction hypothesis there exists an N such that if ko contains more than N elements we can achieve simultaneously g.(a l , ... , a,-I) i= 0 for all y, with elements ai E ko . Suppose ko contains enough elements for this result. Then consider the polynomialsfi(a l , ••• , a,_I' x,). Clearly if ko contains enough elements, we can choose x, = a, so that fi al , ... , a,) = 0 (i = 1, ... , m). This completes the proof. Lemma 2: Let Xl , X 2 , ••• , X, be elements of a field k linearly independent over its constant field k o . Let IIQ be the smallest parallelotope containing all the Xi; Cl = IIp·p. Let PI' ... , Pm be a finite set of primes. Then if ko contains enough elements, there C,X, (c i E ko) such that exists a linear combination X = ClX l ord p, X = y Pi (i = 1" . .. m).

+ ... +

Proof:

Suppose

We write y", = x",/xi. Then ord p , y", ;;::: 0, i.e. the y. are local integers at Pi' and y j = 1. The lemma will be proved if we can CrY, has ordinal find elements c" E ko such that y = ClYl zero at Pi for i = 1, ... , m. Let Y.(Pi) denote the homomorphic map of y. at the prime l"l . Then we must find c" E ko such that

+ ... +

(i= 1,2, ···,m).

According to Lemma 1 this can be accomplished if ko contains enough elements. Theorem 15:

If Cl and b are integral divisors, then

8.

THE THEOREM OF CLIFFORD

319

Proof: If IIa O-1 , IIb O- ] are the smallest parallelotopes containing all elements of IIa-1 n k, IIb-1 n k respectively, then

ao I a,

I

bo b,

and

Further,

whence

Consequently if the theorem is proved for ao and bo we have

We may therefore assume at the outset that IIa-l and IIb- l are the smallest parallelotopes containing IIa-l n k and IIb-l n k. If ko is an infinite field we have already enough constants for Lemma 2. If ko is a finite field, a sufficiently high separable extension will provide us with enough constants. Since m(a-l ), m(b-l ) do not change under a separable constant field extension, (see Theorems 21, 22, Chapter 15) we may assume that ko already contains enough elements. According to Lemma 2 we can find an element a: E IIa-l n k such that ord" a: = ord" a-I for all primes p dividing a and for all primes p dividing b. If a: = aI/a we see that al is relatively prime to a and b. We also see that

and (-1 a-lb-1) -ma _ (-lb-l ) . mal ( -lb-l) _-ma:

Hence we may assume that a and b are relatively prime, i.e

320

XVI.


Finally we have

+

f3 = 0, then a and f3 Hence if a E IIa-l n k, f3 E IIb- 1 n k, and a are constants. The dimension of the set(IIa-l n k) (IIb-l n k) is therefore m(cc1) m(b-1 ) - 1. But this set is contained in IIa-l b- 1 n k. Hence its dimension is at most m(n-1b-1). This gives the required result. If f is a divisor class, a any divisor in it, m(a- 1) is the dimension of f. In the cases where m(a-1 ) = 0 or m(a/b) = 0, the RiemannRoch Theorem gives a complete description of the dimensions of f and W/f. If m(a-1 ) 1= 0 then f contains integral divisors and we may assume that a is integral. If W/f contains an integral divisor b then W contains b = ab. Since m(b-1) = g we obtain the following complementary result to the Theorem of Riemann-Roch:

+

+

Theorem 16: (Clifford's Theorem). If b is an integral divisor of the canonical class W, and b = ab where a and b are also integral, then

Since by the Riemann-Roch Theorem m(a-1 )

= n(a) + 1 -

g + m(b-1 ),

we obtain 2m(n-1 )

= n(a) + 1 - g + m(a-1)

+ m(b-

1

)

~

n(n)

An equivalent statement to Theorem 16 is therefore Corollary:

If m(a-1 )

> 0 and m(a/b) > 0, then n(a)

~

2m(a-1 )

-

2.

+ 2.

CHAPTER SEVENTEEN

Differentials in Function Fields

1. Preparations

In this chapter we shall be dealing with function fields in the sense of Chapter 13, i.e. with fields of algebraic functions in one variable. In the present section we prove some preliminary results which will be used in the sequel. As usual, k shall denote a function field of our type; ko its field of constants. R = ko(x) shall denote a rational subfield. Theorem 1: There are infinitely many separable irreducible polynomials in ko[t].

First Proof: We may prove the theorem directly by considering two cases. If ko is an infinite field, all the linear polynomials t - ex (ex E ko) are separable. If ko is finite, the existence of such polynomials is well known. Second Proof: It is possible to give a unified proof for both cases by adapting Euclid's proof that the number of primes is infinite. Suppose Pl , P2, ... , Pr are separable polynomials. Let

f = PIP2 ••• Pr + 1. Then

f' = p{P• •.. Pr + PIP; ... Pr + .... f'

is not identically zero since the first term is not divisible by Pt and Pt' -=I=- 0, and the remaining terms are all divisible by Pl' Therefore f has an irreducible factor Pr+l which is separable. Now let p(x) be a separable polynomial of R = ko(x). Then 321

322

XVII.

DIFFERENTIALS IN FUNCTION FIELDS

I p(x) I < 1 and ! p'(x) Ip = 1. By Hensel's Lemma, the completion R*(P) contains a root Xo of the equation p(t) 0 and x Xo • The residue class field under lip is ki = ko(xo), and x - Xo is a prime element in R*(P). Hence by Theorem 5 of Chapter 3, R*(P) is isomorphic to the field of formal power series in x - Xo with coefficients in hI : R*(P) = ~{x - xo}.

=

Now let k be a finite extension of Rand p a prime in k dividing p. Then if k*(p)/R*(P) is unramified, the residue class field ko under I III is a separable extension of ki , and hence of ko . x - Xo is still a prime element, and hence

2. Local Components of Differentials In Chapter 13 we defined for a rational subfield R the normed differential A(g), describing A(~) by its infinite component \,)(~)

- Resp«lg", •

The maximal parallelotope on which A vanishes is I1p~. We now wish to examine the components of A( g) at the finite primes p (i.e. irreducible polynomials P(x». Thus we wish to find \,(~) A(~p). Clearly A(~p) 0 if I ~p I ~ 1. Any ~p E R*(P) can be written in the form ~v =

where deg 4>.(x)

co

I.

0 implies m(a) = O. Hence

+ 2g - 2 + 1 n(a) + g - 1.

m(a-1b-1) = n(a) =

g

We consider the special cases:

Case 1: n(a) = 1, a = p. In this case there are g linearly independent differentials with the required property. But the differentials of first kind have this property (they have no poles at all). Hence only the differentials of first kind have the required property. Thus there are no differentials which have exactly one pole of first order. Case 2: n(a) = 2, a = 1'11'2' Here there are g + 1 linearly independent differentials with the required property. Among these are the g differentials of first kind. It follows that there is a differential which has poles of order 1 at PI and 1'2 . Suppose now that ko is algebraically closed, and hence perfect. Then k is separably generated. Every prime is of first degree, and if t is a separating local uniformizing parameter for 1', we have

4.

DIFFERENTIALS OF THE FIRST KIND

333

Hence Res p (ydx

Since y

E

=

f l ' ydx p

=

f ydx). p

k, we have

f ydx = ~ f ydx = O. p

p

p

Thus ~ Res p (ydx)

=

O.

p

From this result we see again that a differential cannot have a single pole of order 1. We also see that if a differential ydx has poles of order I at two distinct primes PI and P2 and no other poles, then Res Pl (ydx)

=-

Res pB (ydx).

Altering ydx if necessary by a constant factor, we can ensure that Res Pl (ydx) =

+1

and

Now let PI , P2' "', Pr be given distinct primes and C1 , C2 , "', cr given elements of ko such that L C. = O. We shall show how to construct a differential which has simple poles at the given primes with residue Ci at Pi • To this end we construct differentials Yidx (i = 2, "', r) such that and

Then clearly ydx = C2Y2dx

+ ... + crYrdx

is a differential with the required property. This differential is not unique, but obviously two differentials with this property differ only by a differential of the first kind.

APPENDIX

Theorems on p-Groups and Sylow Groups 1. S.Equivalence Classes Definition 1: If C is a subset of a group G, then the set of all x E G satisfying xC = Cx (or xCx-1 = C) is called the normalizer Nc of C in G. One sees immediately that N c is a subgroup of G. Should C be a subgroup of G then C is contained in N c and is a normal subgroup of N c . Nc is then the largest subgroup of G having C as normal subgroup. Definition 2: Let S be a subgroup of G, C1 and C 2 subsets of G. We say that C i is S-equivalent to C 2 if C 2 = XC1X-1 for some XES. It is easy to see that this is an equivalence relation. Let C be a given subset of G. In order to find the number of sets S-equivalent to C we ask when xCx-1 = yCy-l for two elements x, YES. This equivalent to y-1xC = Cy-1x, hence to y-1x ENe. Since y-1x E S this means y-1x E S nNe or x E y . (S n N c). Hence x and y have to be in the same left coset of S modulo S nNe. This shows: Lemma 1: The number of sets in the S-equivalence class of C is the index: (1) (818 nNe) Should S = G this simplifies to (G/ N c). In this case we call the G-equivalent sets also conjugate sets. 334

2.

THEOREMS ABOUT p-GROUPS

335

If C consists of one element only, then the conjugates of C have also only one element. This gives a distribution of the elements of G into classes of conjugate elements. The number of elements in the class of a is (GINa). This number is 1 if Na = G, or if ax = xa for all x E G. The set Z of all these a forms an abelian normal subgroup of G, the center of G. Let G be a finite group of order n, Z the center of G, and z the order of Z. G is covered by its classes of conjugate elements. An element a outside Z lies in a class with (GINa) elements and (GINa) will be greater than 1. Counting the number of elements in G we obtain a formula of the type (2)

where each term in the sum is greater than 1.

2. Theorems About p-Groups

A group whose order n is a power pr of a prime is called a p-group. In formula (2) for n = pr each index of the sum is a power pB > 1. Therefore z must be divisible by p. This shows:

> 1.

Theorem 1:

A p-group G

-=1=

1 has a center Z of an order

Corollary 1: of order p.

A p-group G

-=1=

1 contains an invariant subgroup

Corollary 2:

One can find a chain

of invariant subgroups Gi of G such that each index (Gl+1 /Gi ) = p.

Proof: Corollary 1 shows the existence of an invariant subgroup G1 of order p. By induction one may assume the existence of a chain in GIG1 :

G1/G1 C G2/G1 C··· C GrlGI = GIG1 •

336

Ap.

THEOREMS ON p-GROUPS AND SYLOW GROUPS

The chain 1 = Go C G1 C G2 C··· C G r

has the desired properties. We prove: Theorem 2:

Let G be a p-group, S i= G a subgroup. Then

Nsi=S. Proof: Let Z be the center of G. We prove the theorem by induction. Case 1: If Z ¢ S, any element a of Z that is not in S shows that Ns i= S. Case 2: Let Z C S. Consider the natural homomorphism of G onto G/Z. S is mapped onto S/Z. Since Z esc N s , Ns is mapped onto Ns/Z and Ns/Z is the largest subgroup of G/Z that has S/Z as normal subgroup. We have to show Ns/Z i= S/Z. But this is our theorem in the group G/Z of smaller order.

Corollary 1: If S is of index p in G, then S is normal in G. Indeed N s can only be G itself. Corollary 2: If S C G, S i= G, then there exists a group SI between Sand G such that S is invariant subgroup of SI of index p.

Proof: S eNs and S i= N s . Let SI/S be a subgroup of Ns/S of order p. SI has the desired properties. Corollary 3:

There always exists a chain S = Go C G1 C ". C G. = G

such that 0t-l is normal in Gi of index p. The proof follows from the preceding corollary.

3. The Existence of Sylow Subgroups

Let G be a finite group of order n, p a prime and pr the exact power of p dividing n. A subgroup P of G of order pr is called a

4.

THEOREMS ABOUT SYLOW SUBGROUPS

337

Sylow subgroup of G. We prove that it exists always. If r = 0 the statement is trivial so we are concerned only with the case r >0. Lemma 2: Let G be abelian and element of order p.

pin. Then

G contains an

Proof: By induction. Select an element a -=1= 1 of G and let d be the period of a. If p I d then ad / p is the required element. Let p -r d and call S the cyclic subgroup generated by a. The factor group G/S is of order n/d < nand Pin/d. So there is an element bS in G/S of exact period p. Let e be the period of b: b2 = 1. Then (bS)e = S. Therefore Pie and be/p is the required element.

Theorem 3: Proof:

Every group G has a Sylow subgroup.

We proceed by induction on the order n.

Case 1: G has a subgroup S -=1= G of an index prime to p. The order of S is then divisible by exactly pr and is less than n. Therefore a Sylow subgroup P of S exists and is Sylow subgroup of G. Case 2 (actually only possible for p-groups). Every subgroup S -=1= G has an index divisible by p. We return to formula (2) and see that n as well as each term in the sum is divisible by p. Hence p I z the order of the center Z of G. Let S be the cyclic group of order p generated by an element a of order p of Z (use Lemma.2). S is normal subgroup of G. Let P/S be a Sylow subgroup (of order pr-l) of G/S. Then P is of order pr whence a Sylow subgroup of G. 4. Theorems About Sylow Subgroups Let G have order n, divisible by exactly pr, P a Sylow subgroup of G. Let S be any p-subgroup of G (not necessarily itself a Sylow subgroup of G). Lemma 3: S nNp=S

nP.

338

Ap.

THEOREMS ON p-GROUPS AND SYLOW GROUPS

Proof: That S n pes n N p is trivial. We show the converse. S n N p C S, hence S n N p = SI is a p-group. Since SI C N p we have xP = Px for each x E SI . Therefore SIP is a group and P is normal in it. We have for the index

So SIP is itself a p-group containing P. Since P is a maximal p-subgroup of G we get SIP = P whence SI C P. SI C S shows S n N p = SI C (S n P).

Consider all transforms of P: xPx-I, x E G. Their number is (G/N p ). Since P C N p , this index is prime to p. Distribute these transforms Pi into S-equivalence classes. The number of transforms in the S-equivalence class of Pi is given by the index (S/S n N p ) so in view of Lemma 3 by (S/S n Pi,). We obtain a formula of the type

(3) Each term on the right side is a power of p. Since the left side is prime to p it must happen for some Pi that (S/S n Pi) = 1 or S n Pi = S or S C Pi . Hence S is contained in some transform of P. Let for a moment S be itself a Sylow subgroup. Then S = Pi . So the transforms of P are all Sylow subgroup of G. Let S = P. Then

happens exactly one, namely for Pi = P. All other terms in (3) are divisible by p. (3) shows (G/N p ) = 1 (mod p).

Theorem 4: All Sylow subgroups P of a group G are conjugate. Their number, the index (G/N p ) is = 1 (mod p). Every p-subgroup S of G is contained in a Sylow subgroup of G.

Index of Symbols

{a}, 120

deg

Algebraic Numbers and Algebraic Functions

Algebraic Numbers and Algebraic Functions

Algebraic numbers and algebraic functions