ALGEBRAIC NUMBERS AND ALGEBRAIC FUNCTIONS
Notes on Mathematics and I t s Applications General Editors: Jacob T. Schwar...
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ALGEBRAIC NUMBERS AND ALGEBRAIC FUNCTIONS
Notes on Mathematics and I t s Applications General Editors: Jacob T. Schwartx, Courant Institute of Mathematical Sciences and Maurice Lbvy, Universitk de Paris
Algebraic Numbers and
E. Artin, ALGEBRAIC NUMBERS AND ALGEBRAIC FUNCTIONS R. P. Boas, COLLECTED WORKS OF HIDEHIKO YAMABE M . Davis, FUNCTIONAL ANALYSIS M . Davis, LECTURES ON MODERN MATHEMATICS J. Eells, Jr., SINGULARITIES OF SMOOTH MAPS K. 0. Friedrichs, ADVANCED ORDINARY DIFFERENTIAL EQUATIONS K. 0. Friedrichs, SPECIAL TOPICS I N FLUID DYNAMICS K. 0. Friedrichs and H . N . Shapiro, INTEGRATION IN HILBERT SPACE M . Hausner and J. T. Schwartz, LIE GROUPS; LIE ALGEBRAS P . Hilton, HOMOTOPY THEORY AND DUALITY I;: John, LECTURES ON ADVANCED NUMERICAL ANALYSIS Allen M. Krall, STABILITY TECHNIQUES H . Mullish, AN INTRODUCTION TO COMPUTER PROGRAMMING J. T. Schwartx, w* ALGEBRAS A . Silverman, EXERCISES IN FORTRAN J . J. Stoker, NONLINEAR ELASTICITY
Algebraic Functions
EMIL ARTIN Late of Princeton University
Additional volumes in preparation
GI IB
G O R D O N A N D SCIENCE PUBLISHERS NEW YORK
BREACH
- LONDON . PARIS
Copyright 0 I967 by Gordon and Breach, Science Publishers, Inc. 150 Fifth Avenue, New York, New York 10011 Library of Congress Catalog Card Number: 67-26811 Editorial Ofice for Great Britain: Gordon and Breach Science Publishers Ltd. 8 Bloomsbury Way London WCI, England Editorial Ofice for France: 7-9 rue Emile Dubois Paris 14e 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.
Preface These lecture notes represent the content of a course given at Princeton University during the academic year 1950151. 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. I t 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, Eltments de Mathtmatique, Livre III, Topologie gtntrale, Chapitres 1-111. I n 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. 0. Adamson who prepared them. Of equally great importance have been frequent discussions with Mr. J. 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 University June 1951
Contents v vii 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 . . . .
. . . . . .
.
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 . . . .
. . . . . . .
.
. . . . .
. . . . .
. . .
. .
.
.
. . .
. . . . . .
. . . . . .
Chapter 2 COMPLETE FIELDS
1. 2. 3. 4. 5. 6. 7.
Chapter 3
1. The Ramification and Residue Class Degree 2. The Discrete Case . . . . . . 3. The General Case . . . . .
.
.
CONTENTS
CONTENTS
I. 2. 3.
4. 5. 6.
Chapter 4 RAMIFICATION THEORY Unramified Extensions . . . . Tamely Ramified Extensions . Characters of Abelian Groups . . . The Inertia Group and Ramification Group . Higher Ramification Groups Ramification Theory in the Discrete Case . Chapter 5 THEDIFFERENT
1. 2. 3. 4.
. . . The Inverse Different . Complementary Bases . . . . Fields with Separable Residue Class Field The Ramification Groups of a Subfield . Part ll
1. 2. 3. 4.
1. 2. 3.
4. 5.
Local Class Field Theory
Chapter 6 FOR LOCALCLASSFIELDTHEORY PREPARATIONS Galois Theory for Infinite Extensions . . Group Extensions . . . . Galois Cohomology Theory . . . . Continuous Cocycles . . . . Chapter 7 THEFIRSTAND SECOND INEQUALITIES Introduction . . . . . Unramified Extensions . . . The First Inequality . . . . The Second Inequality: A Reduction Step . The Second Inequality Concluded . . .
Chapter 8 THENORMRESIDUE SYMBOL 1. The Temporary Symbol (c, K I KIT) . . . 2. Choice of a Standard Generator c . . . 3. The Norm Residue Symbol for Finite Extensions
Chapter 9 THEEXISTENCE THEOREM 1. Introduction
.
.
.
2. The Infinite Product Space I
. . .
. . . . .
. .
. .
. . .
.
3. 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 . .
1. 2. 3.
4. 5. 6.
Chapter 10 APPLICATIONS AND ILLUSTRATIONS Fields with Perfect Residue Class Field . . . The Norm Residue Symbol for Certain Power Series . . . . . . . . . Fields Differentials in an Arbitrary Power Series Field . The Conductor and Different for Cyclic p-Extensions The Rational p-adic Field . . . . . . Computation of the Index ( a : an) . . . .
. . .
.
. .
Part III Product Formula and Function Fields in One Variable
Chapter 11
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
1.
2. 3. 4.
2 1 5 216 2 1 8 . 223
.
Chapter 12 CHARACTERIZATION OF FIELDSBY THE PRODUCT FORMULA . . . . . . . . 2 2 5 PF-Fields. Upper Bound for the Order of a Parallelotope . . . 227 . . . . . . 230 Description of all PF-Fields Finite Extensions of PF-Fields . . . . . . 235
xii
CONTENTS
CONTENTS
Chapter 17 IN FUNCTION FIELDS DIFFERENTIALS
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 . . . . .
.
.
. .
2 3 8 241 2 4 4 . 245 2 5 1 255
Chapter 14 THERIEMANN-ROCH THEOREM 1. Parallelotopes in a Function Field . . 2. First Proof 3. Second Proof . . .
.
. .
. .
. . .
. .
260 2 6 2 2 6 5
. . .
2 7 1 2 7 8 . 279
Chapter 15 CONSTANT FIELDEXTENSIONS
. . . 1. The Effective Degree . 2. Divisors in an Extension Field . . . 3 . Finite Algebraic Constant Field Extensions
. . .
4 . The Genus in a Purely Transcendental Constant Field Extension . . . . . . . . . 2 8 4 5. The Genus in an Arbitrary Constant Field Extension . 287 Chapter 16 OF THE RIEMANN-ROCH THEOREM APPLICATIONS 1. Places and Valuation Rings 2 . Algebraic Curves . 3. Linear Series . . 4 . Fields of Genus Zero . 5. Elliptic Fields . . . 6. The Curve of Degree n 7. Hyperelliptic Fields . 8. The Theorem of Clifford
.
.
.
. .
. . .
. .
. .
.
.
.
.
.
.
,
. .
. .
. .
. .
.
.
293 2 9 7 3 0 0 . 302 . 306 3 1 1 3 1 2 317
. . . . . . I Preparations . . . . . 2. Local Components of Differentials . 3 Differentials and .Derivatives in Function . Fields . . . 4. Differentials of the First Kind .
. . .
. .
321 322 . 324 3 2 9
Appendix THEOREMS ON p-GROUPSAND SYLOW GROUPS 1. 2. 3. 4.
. . S-Equivalence Classes . . Theorems About $-Groups . The Existence of Sylow Subgroups Theorems About Sylow Subgroups
. . .
.
. .
.
. .
.
. .
.
3 3 4 . 335 . 336 . 337
PART ONE
General Valuation Theory
CHAPTER ONE
Valuations of a Field A vahation of a field k is a real-valued function I x I for all x E k, satisfying the following requirements:
, defined
(I) I x I 2 0; ( x ( = 0 if and only if x = 0, (2) IXYI = 1x1 l y I, (3) If I x I < 1, then I 1 x I < c, where c is a constant; c 2 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 [ = 1 for all x E k*, the valuation is also called trivial. 1. Equivalent Valuations
Let I I, and I ,1 be two functions satisfying conditions (I) and (2) above; suppose that ( ,1 is non-trivial. These functions are said to be equivalent if I a ,1 < 1 implies I a 1, < 1. Obviously for such functions ( a more.
,1 > 1 implies 1 a ,1 > 1; but we can prove
Let I ,1 and I ,1 be equivalent functions, and I ,1 is non-trivial. Then 1 a 1, = 1 implies ( a ,1 = 1. Proof: Let b # 0 be such that I b 1, < 1. Then I anb 1, < 1; whence 1 anb ,1 < I, and so 1 a 1, < 1 b ];lln. Letting n +a, we have I a 1, < 1. Similarly, replacing a in this argument by l l a , we have I a 1, 2 1, which proves the theorem. Theorem 1:
suppose
3
1.
4
2.
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
n
Theorem 2: If j 1 , and / 1 , are equivalent functions, and I I I is non-trivial, then I a 1 , = I a lla for all a E k, where or is a fixed positive real number.
Proof: Since 1 1 , is non-trivial, we can select an element c E k* such that I c 1 , > 1; then I c 1 , > 1 also.
1 a 1,
THE TOPOLOGY INDUCED BY A VALUATION
Now given a , , a , , < 2n. Hence
< 2'
a*.,
a , , we can find an integer r such that
In particular, if we set all the a , = 1, we have I n also weaken the above inequality, and write
1
1
< 2n.
We may
c l,Y, where y is a non-negative real number. If 1 , < 1 c llmln, whence 1 an/cm1 , < 1. Then I an/cm1 , < 1, from which we deduce that I a 1 , < I c I,mln. Similarly, if m,n < y, then I a 1 , > 1 c I m Jn. It follows that I a ,1 = I c Jk. Now, clearly, log I a I , 1% l a I, Set
=
m/n > y, then 1 a
Y=logJcJ,=log~
This proves the theorem, with
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 b1 1 a 1 Ib 1 "Triangular Inequality" condition, viz., 1 a T h e connection between this condition and ours is given by
+
2 , we may write c = 2". Then it is easily verified that ( Illa is an equivalent valuation for which the triangular inequality is satisfied. 2. The Topology Induced by a Valuation
Similarly
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
3.
VALUATIONS OF A FIELD
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. The topology induced by and only if axiom (3) is satisfied. Theorem 4:
1 I
is HausdorfT if
Proof: (1) If the topology is HausdorfT, there exist neighborhoods separating 0 and - 1. Thus we can find real numbers a and b such that if I x 1 a, then I 1 x 1 >, b. Now let x be any element with I x 1 1; then either I 1 x 1 l l a or I 1 x 1 > l/a. In the latter case, set y = - xi(1 x); then
+
1,
Here we consider the sequence cb, a, = 1 b"
+
10
1.
5.
VALUATIONS OF A FIELD
This sequence converges to the limit c in the topologies induced by I 1, and I I , . Thus a, = c 7, where I q, ll and I q,1, -t 0 as r +a. Hence for r large enough, I a, 1, > 1 and I a, 1, < 1. For v = 2, ...,n - 1, the sequence a, converges to the limit 0 in the topology induced by I 1. . Hence for large enough values of r, I a, 1. < 1 (v = 2, ..-,n - 1). Thus a, is a suitable element, for r large enough, and the theorem is proved in this case also.
EXAMPLES
+
Corollary: With the conditions of the theorem, there exists
an element a which is close to 1 in I ,1 and close to 0 in n - 1).
I 1,
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 111 b I; I x I is well-defined on k, and obviously satisfies our axioms (1) and (2) 1, for valuations. T o show that axiom (3) is also satisfied, let I x I i.e. 1 a 1 1 b 1. Then
0, and any elements a, (v = 1, .--, element a such that I a - a, 1, < E . Theorem 8:
n) close to 1 in Proof. We can find elements b, (i = 1, I li and close to zero in I 1, (v # i). Then a = a,b, + + anbn is the required element. Let us denote by (k), the field k with the topology of I 1, imposed x (k), . upon it. Consider the Cartesian product (k), x (k), x The elements (a, a, a) of the diagonal form a field k, isomorphic to k. The Approximation Theorem states that k, is everywhere dense in the product space. The theorem shows clearly the impossibility of finding a non-trivial relation of the type .-a,
Hence if k is the quotient field of an integral domain o, 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:
(0 < a, < n; nT< m, i.r r
Let I I be a valuation of R; suppose I I replaced, if necessary, by the equivalent valuation for which the triangular inequality holds. Then I a, I < n, and we have
a*.,
with real constants c,
m . < -)log log n
Im /
< (- log m + 1) n .{max (1, I n
l)}logmeogn.
Using this estimate for I m Is, extracting the sth root, and letting s -KO, we have I m 1 < {max (1, I n I)}logmlogn. There are now two cases to consider.
.
Case 1: In1
> 1 for alln > 1.
Then
5. Examples
Let k be the quotient field of an integral domain o; then it is easily verified that a valuation I I of k induces a function o (which we still denote by I I), satisfying the conditions
Since I m 1 > 1 also, we may interchange the roles of m and n, obtaining the reversed inequality. Hence
1.
12
5.
VALUATIONS OF A FIELD
where a is a positive real number. It follows that
so that 1 1 is in this case equivalent to the ordinary "absolute value", I x I = max (x, - x). Case 2: There exists an integer n > 1 such that 1 n 1 ,< 1. Then 1 m I 1 for all m E o. If we exclude the trivial valuation, we must have 1 n 1 < I for some n E o; 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 = p d , , we have 1 p I = I PI I I p2 I < 1, and hence (say) I p1 I < 1. This p, E (p), i.e. p divides p,; but p, divides p; hence p is a prime. If I p I = c, and n = pvb, (p, b) = 1, then I n I = c'. Every non-archimedean valuation is therefore defined by a prime number p. Conversely, letp be a prime number in D, c a constant, 0 < c < 1. Let n = pvb, (p, b) = 1, and define the function I I by setting I n I = cV.It is easily seen that this function satisfies the three conditions for such functions on o, 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 = cv.
1. Then if c, # 0,we have
Conversely, if we select a number c
> 1 and set
our conditions for functions on o are easily verified. Hence this
13
EXAMPLES
function yields a valuation of k described as follows: let a = f(x)/g(x), and define deg a
= degf (x)
- deg g(x)
.
Then [ a I = cdega. Obviously the different choices of c lead only to equivalent valuations.
1. We choose = d-dega. c = lld, and so define I a I n all cases we have made a definite choice of 1 a 1, in the equivalence class corresponding to p; we call this 1 a ,1 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 x, E D, X, f m , and f(x) E k, we may write
where g(x) E k (g(x,) # 0 or m), and define a valuation by
If
X,
5.
VALUATIONS OF A FIELD
=a,we write f(x) = ( $ ) o r d m f ( x ) g(x) ,
where g(x) E k (~(co)# 0 oroo), and define
EXAMPLES
15
denoted by / f(x) 1,; the valuation given by 1 f ( x ) , I is now denoted by I f(x) I, . We see that to each point 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 = R and k = F(x), the product I I p l a I p = 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 ,1 = 1 for almost all (i.e. all but a finite number of) primes p, and so the product II, I a , 1 is well-defined. If we write +(a) = IT, 1 a ,1 we see that $(ab) = +(a) $(b); thus it suffices to prove the result for a = q, where q is a prime (irreducible polynomial). For q E R,
This gives for each x, E D a valuation of k - axioms (1) and (2) are obviously satisfied, while axiom (3) follows from
1 f lxo
< 1 + f is regular at x, 1 + f is regular at x, 11 + f L 0 < 1-
For q E k(x),
3
3
The valuation / 1., obviously describes the behavior of f(x) at the point x,: If 1 f 1%. < 1, or ordxo(f(x)) = n > 0, then f(x) has a zero of order n at x, . If I f 1%. > 1, or ordx0f(x) = n < 0, then f(x) has a pole or order -n at x, . If I f lxo = 1, then f(x) is regular and non-zero at x, . Should D be the whole Gauss sphere, we have k = F(x); in this case the irreducible polynomials are linear of type (x - x,). The valuation I f(x) lx-xo as defined previously is now the valuation
This completes the proof. We notice that this is essentially the only relation of the form n 1 a 12 = 1. For if #(a) = II 1 a 12 = 1, we have for each prime q
But I q
1,
I q 1,
= 1 by the theorem; hence
1.
16
VALUATIONS OF A FIELD
Thus e, = e,, 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. I n this case
Hence we have Taking the m-th root, and letting m +a,we obtain
+(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
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.
c=m=(IlI,(20* Proof.
(1) When the valuation is non-archimedean, c=1=111>,12).
6. Completion of a Field
As a special case of this we have
Let I I be a valuation of a field k; replace I I, 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 p, v 2 N, I ap - 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 v 2 N, I a, I < E. k is said to be complete with respect to / I if every Cauchy sequence with respect to I I 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 termwise addition and multiplication:
Now
N in P; hence the residue class ring PIN is a field
(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 / = no. Write c = 2 ~ then ; (a+bI av~'xvp',and the convergence of this series implies the convergence of 8. Let f(y) = Irr (8, F, y); we shall show that we may confine our attention to elements 6' 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 8, = 8, 8, , ..., 8, . Since the valuation I 1, of E can be extended to A, we can find I Oi I , . Let min ( 8, - fId ,1 = 6. a >1
Now set
CONVERGENT POWER SERIES
Then
N
is an irreducible polynomial in x with roots A, = Oi - E- a$'; in a particular A, = I:,,, a,&" and if A, is convergent, so is 8, = 8. Obviously
for i > 1. Hence if N is large enough, 1 A, 1, will be as small as we please, so that we have I A, 1, >, 6 for i > 1. Now consider the equation whose roots are p, =,hi/xN. The root p, = A,/x, is a power series of the form I:, a $ ' , I x I,; the other roots may have absolute values as hence I p, 1, large as we please by suitable choice of N. Since the convergence of pl obviously implies the convergence of A,, and hence of 8,, it follows that we need only consider elements 8 which are separable over F, and whose defining equation f(y) has roots 8, = 8,8, , .-.,8, in A with the property that 1 8, 1 ,< I x 1, and I 8, 1, > 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.
0. Hence the Newton diagram has the form shown:
2.
7.
COMPLETE FIELDS
CONVERGENT POWER SERIES
51
From these two formulas we obtain a recursion relation:
These polynomials
tL, 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. Now since the series EcP,xV are convergent, we have
I ~ , , ~ aI , < , ~ MPfor all q, E k having I or, I < some fixed lla, . Since the number of series occurring as coefficients in f(y) is finite, we can write I c,, I < Mav, where M
= max (M,)
and
P
a
= max (a$. P
We now go over to the field of real numbers, ko , and the field of convergent power series Fo over k,. Let 4 be a root of the equation
Thus
where the sum on the right is infinite:
where blo # 0. Since y = 8 is a root of f(y), we have
M a x =-.1-ax
1 I-$'
whence whence
But we know that f(y) has a root given by
The roots are analytic functions of x near zero; one of them, may be expanded as a which we call 4,, vanishes at x = 0. power series convergent in a certain circle round the origin, say 4, = 2; a,xY. We shall now show that all the coefficients orv are positive, and I a, I a,. We proceed by induction; a, is easily
ordor
a" E B j
Now if
p # 0, ord p > 0, so that j > i. Thus every element a of Bi has (modulo this B j ) period p. Since Bi/Bi 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 { B i ) the factor group is abelian and of type (p, p, p, provided that the discontinuity does not occur at B,, . I n the case of non-discrete valuations, no information can be obtained about discontinuities at 9+,; this difficulty does not arise in the discrete case where B+, = B, . Let us examine the special case when e = f = 1. Then if a E E, there is an element b E F such that 1 a 1 = I b 1 ; thus I a/b I = 1 and alb c mod p where c E F (since f = 1). This implies that I a - bc / :< 1 b I = / a 1, or, writing bc = a, that / a - a / < / a / . Hence a = a + / 3 where 1 / 3 1 < 1 a i . Then .a*),
Similarly
Hence also ord ((a7 - ~ aa) 2 ord a
+ i +j.
We can replace a by u - ~ T - without ~ ~ altering the ordinal; this yields ord (a~a-lr-l - 1) a 2 ord a i j.
+ +
-
since a
E F.
We now obtain
Thus if a E Bi , T E B j , then the commutator of a and T lies in B c b i . I n particular, when i = j, we see that the factor group B,/B,, is abelian. If a discontinuity occurs at B, , the factor group is certainly contained in B,/B,,; thus the factor group at a jump is
whence
abelian. We shall now show that the factor group is of type (p, p, p, ..-) where p is the characteristic of the residue class field. Since p = 0 => B,, = 1, (for there are no inseparable extensions of fields of characteristic zero), we consider the case p #O. We examine
since / P/a/ < 1. Thus to every element 01 E E we can find another element /3 such that
a"
- 1 = ((a - 1)
+ 1)P - 1
Thus if a E Br where r is a real number we cannot have
Thus for any a
E
E. Thus we have proved:
4.
82
RAMIFICATION THEORY
If e = f
Theorem 10:
=
1, then 8,= B,+, where r is any
real number. 6. Ramification Theory in the Discrete Case
We shall now assume that the valuation is discrete; then we have shown the existence of elements .rr E k, 17E E such that every element a E k, respectively a E E, can be written a = emv, respectively a = EIPwhere the E are units; further, we know that I rr I = I 17 le, where e is the ramification number. If a = EIP we shall make the natural definition, ord a = v; in particular, ord 17= 1, ord m = e. These ordinals depend on the field E. In this discrete case, nothing new can be added about the Galois group of T I k. But since the value group BE is now cyclic, and the Galois group of V I T is isomorphic to the non-critical part of BE, it follows that V / T is a cyclic extension field, of degree eo , where e = pye, , (eo ,p) = 1. Hence it is easily verified that V = T ( V 6 ) where .rrl E T. The study of the higher ramification groups %, is simplified in the discrete case, since now we need consider only integral values of i-for the ordinal, as defined above, takes only integral values. I n particular we notice that B,, = %+,, , and especially B+, = %, . The sequence of groups is now B1 3 23, 3 3 (I), where Bi/B2* is abelian, and B4/Bjis of type (p, p, p, -..) where j
= min
(pi, ordp + i) 2 i + 1.
Thus the group of E I T is solvable (this is true also for the nondiscrete case); and any insolvable step in E I k comes from the residue class field (since the group of T / k is isomorphic to that of E 16). We have already remarked that we need examine the effect of the elements a E 6 only for the generators of the group of non-zero elements of E; thus it will be sufficient to consider their effect on all elements 17 for which ord 17 = 1. Quotients of these give the units of the field, which, along with one of the elements 17, give all elements of E. We shall now make the further restriction that the residue class field E I be separable. This is certainly true in the important cases of algebraic number fields and of fields of func-
tions over finite ground fields. Only in this case do the finer parts of the theory appear; for instance, only under this assumption was Herbrand able to find the inertia group and ramification groups for an arbitrary normal subfield of E 1 k. We remark that if E 16 is separable, then E I T is purely ramified with degree e. Theorem 11: If the valuation is discrete, and if the residue class field E I k is separable, then the integers D in E have a minimal basis relative to the integers o in k consisting of the powers of an integer. I n other words, there is an element a E E such that every 8 E D can be expressed as
where all the xi
E O.
-
Proof: Let w, , o , , .-.,wf be representatives (in E) of a basis for the residue class field E I 6; thus if 8 E E is a representative of one of the residue classes, we have 8 xlwl xtwf mod p. We have seen earlier (Chapter 3, Theorem 7) that a field basis for f ; v = 0, -.-, e - 1); and this E I k is given by {winY) (i = 1, is a minimal basis (i.e. when the integers of E are represented in terms of the basis, the coefficients in k are integers). When E is unramified (E = T), we have e = 1, f = n. Let a be a representative in E of the residue class which generates E : E = h(6). Then 1, 5, ..-,(Yf-l is a basis for E; hence 1, a, at-l is a minimal basis for E 1 k. When E is purely ramified (k = T), we have e = p, f = 1. Then a basis for E is given by a unit w,; hence a minimal basis for Re-l. E I k is given by 1, 17, In the general case, we let E = R(&), and let f(x) be a polynomial in k[x] such that j(x) = Irr (E, K, x). Since E is separable, we have 0 mod p, where a is a representative f(a) E 0 mod p, but f'(a) of (Y in E. Since f(a) r 0 mod p we have ord f(a) 2 1. Suppose ord (f(a)) > 1; let 17 be any element with ord 17 = 1, and set a 17= 8. Then
+ +
ma-,
me.,
a*.,
+
+
> 1; but ord ITf'(a) = 1 (since Now ord f(a) > 1, and ord 0 mod p); hence ord f(P) = 1, and since 8 lies in the same
f'(a)
+
84
4.
RAMIFICATION THEORY
residue class as a,it may be taken as representative of the generating residue class. Thus we can suppose without loss of generality that ord f(a) = 1; thus f(a) can be taken as our element 17. Since form a basis for E 1 A, we obtain a minimal basis for 1, c, ..-,&f-l E I k by taking {av(f(a))p); thus every element in O can be expressed as a polynomial in a with integer coefficients. Thus 1, a, an-' form a minimal basis for E I k. This completes the proof of the theorem. Now let 0 E D be given by
Let o be an isomorphic map of E IF into the algebraic closure C. Then (a - 1) 0 = cl(aa - a) + c2(aar2- a2) = (aa - a) p, where /?is integral, i.e. I /3 /
+ ... + C , - ~ ( U ~-~ -an-') '
< 1. Hence
ord ((a - 1) a) = min ord ((a - 1) 8). 0cD
The ideal generated in C by (a - 1) a, i.e. ((a - 1) a . D,) is called (by Hilbert) an "element" of E I F. Let now E I F be normal. Then we see that ord ((a - 1) a) > 0 c- oa r a mod p c- o leaves all residue classes fixed the inertia group. Suppose, then, that a E 3; since 3 leaves the inertia field T fixed, we can take T as our ground field. Thus the powers of a, or of any element 17 with ord 17 = 1, form a minimal basis for E I T. If a E 3, we have ord (a - 1) a = ord(a - 1 ) l l
for all such elements 17; thus in examining the effect of an automorphism a, we need examine its action on only one such I7. Suppose that a E Bi but a # Bi+,; this means that ord (0 - 1) a , a
but
ord (a - 1) cr > i + l .
and
ord (a - 1) or = i
or
Hence ord
(a - 1) a a
.
=t,
+ 1.
We see at once that if ord (a - 1) a = 0, then a is not an element of the inertia group; if ord (a - 1) a = 1, then u € 3 but a # B1 , and we obtain immediately the classical definition of the higher ramification groups:
1.
CHAPTER FIVE
THE INVERSE DIFFERENT
87
On the other hand, 0' # E. For since E / k is separable, there is at least one eIement a E E such that S(a) = a # 0. Then S(a/.rr, . 1) = a/.rrv,and this does not lie in o if v is large enough; in other words, for large enough v, a/rVdoes not lie in D'; hence D' # E. If $(AD) C o and S(@) C 0, then S((h p)o) C o; thus D' is closed under addition. If S(hD) C o, then S(XD D) C o, since DD = D; hence AD C Dr. Thus Or is a fractional idea1 as described above. From this result it follows that we may express Dr as If-%, where j 2 0, since D' 3 D. The diflerent 3) is defined to be the inverse of a' : 3) = Dr-I = If@. The fundamental property of the different is given by
+
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, V, n a n d o, p, n- the rings of integers, prime ideals, and primes in E and k respectively. The trace from E to k will be denoted by S,,, or simply by S. 1. The Inverse Different
Let T be any set in E; its complementary set T is defined by
I t is easily seen that if TI C T, then TI' 3 T,'. I n particular, when T = D, we obtain the complimentary set D':
Theorem 2: is unramified.
3)
= D, and hence D' = D, if and only if E
Ik
Proof: There are three cases to consider: Case I: E 1 k is unramified. Then n- = 17 and we have seen that every map of E I k into C' comes from a map of E ( k into C; thus the trace in E I k comes from the trace in E I k. More precisely, if a is a representative in E of a residue class 6 in E, then s ~ ( E )= SE,,-(ti). Since E I is separable (E I k is unramified), the trace SElcis not identically zero. Hence there is an element a in such that SElk(a) 0 mod p. Now let ,B = n-is be any element of E with negative ordinal ( i > 0; 6 a unit). Then
+
D' is called the inverse different. We now introduce the notion of a fractional ideal in E. Let % ' be any additive group in E such that 2ID = (u. If a E 2l, and I a I, then P/a E D, and so = a .P/a E (u. This means [ I that (u contains, along with a, any element with ordinal 2 ord a. There are thus two possibilities: (1) ord a is not bounded for a E %; then clearly '% = E; (2) ord a is bounded; let a, be an element with maximal ordinal in '%-this exists since the valuation is discrete; then '% = a& = P D where v = ord a,. I n the second case we call (u a fractional ideal (or an ideal for short).
1. Then n- = 17e. Let a be any map of E into the algebraic closure C; if a E 9, then I a I < 1, whence I aa I < 1 and so I S(a) I < 1. This shows that S(p) C p. Hence 1 ~ ( ~ ' =s(&o) 4 ) ?T
Co.
88
5.
Thus l/n )P CD', and in particular 17/n, which has ordinal - (e - 1) is contained in D'. Hence D' contains D 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 E 1 & is inseparable. Here we use the transitivity of the trace: SElk(a)= STlk(SEIT(a)). We propose to show that S(D) C p, so it will suffice to show that I SEIT(a)I < 1. This reduces our investigation to the case where the residue class field is totally inseparable. Let a E E: If o is any map of E ( T into C, then oa E a mod p. Hence S(a) r na 0 mod p since the degree of an inseparable extension is divisible by the characteristic. Thus S(D)C p; hence S ( l / n D) C o, so that l / n ED' and D' contains D as a proper subset. This completes the proof of the theorem. A second important property of the different is contained in
-
Theorem 3:
2.
THE DIFFERENT
If E 3 F 3 k, then DEIk= DEIFDFIk .
Proof: Let DFlk= P D F . We have
89
COMPLEMENTARY BASES
2. Complementary Bases
Let k be any field, E a separable extension of degree n. Let , w, , .-.,onform a basis for E / k. We examine whether there exists an element 6 E E such that S(wi[) = 0 (i = 1, 2, n). If we write
w,
-.a,
we see that the equations
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 anwn) .Y) = 0; S((alw1
+ +
thus S(E5) = 0, which is impossible unless 5 = 0 since E I k is separable. Since the system of homogeneous equations S(w&) = 0 has only the trivial solution, it follows from the theory of systems of linear equations that any non-homogeneous system S(w& = b, (i = 1, 2, n), with b, E k, has exactly one solution. In particular there is exactly one element E E such that S(wiwrj) = aii. The set of elements w',wt2, w', is called the on.To justify this name we must complementary basis to a,, w,, showthat the~'~are1inearlyindependent ;so letx1w1, xnwl,=O. Multiply by wi , and take the trace; this yields xiS(wpil) = xi = 0. Thus the w', are linearly independent. ..- x,w,, then x, = S(fwl$), It is easy to see that if [ = x,w, and that if 7 = y1w1, ... ynwl,, then yi = S(rlwi). We can prove -.a,
..a,
a*.,
+ +
Corollary: If T is the inertia field of E I k, then DEIk= BE,,.
Proof: We have only to recall that since T is unramified, IDTlk
=D ~ .
+ +
+ +
Theorem 4: Let w, , , w, be a basis (not necessarily minimal) for E 1 k; let w', , ..-, w', be the complementary basis. If T = wlo + wno , then T' = wl,o ... $ wl,o.
+
+
Proof: Any element h E E may be expressed as
5.
90
2.
THE DIFFERENT
Theorem 5:
Then
COMPLEMENTARY BASES
If
f(x) = (x - a) (bo
A E T ' o S(AT) C o e-S(Awi) C o ( i = 1, .--,n) -=-xi€o.
Hence
+ b,x + + bn-l~n-l),
then the complementary basis is formed by
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:
Proof:
Let E = k(a); f(x) = Irr ( a , k, x). Then
s ( f(x) - .i" x -a
f (4
for
=xi
i = 0 , I,
.-.,n - 1 . Hence
Proof: Since E 1 k is separable, the roots a, = a, distinct. Now
-..,an-,
are This is the precise condition that bjlf'(a) should form a complementary basis to the a i . We shall now evaluate the coefficients bi . Let
which is a polynomial of degree We have
< n - 1. Then, since f ( a ) = 0,we have
but
Hence Hence
+ a,a + -..+ anan-,, ... bn-, = an = 1 . b, = a, + a,a + --.+ ancP2, bo = a,
Y
EXTENSIONS
109
depends only on the coset of G modulo A in which x lies. For let b be an element of A; then (bx) a(bx)-l
= b(xax-l)
b-I
= xax-l,
since xax-l, b, b-l lie in A, and A is abelian. The automorphism is therefore defined by u E G, and we display the fact by writing
We now examine the rules of combination of these automorphisms: (a')"
= u,(zi,au;')
u,'
=
(u,~,)a(u,u,)-~.
Now since u -t Au, is an isomorphism, u,u, lies in the coset Au,, and hence produces the automorphism known as am. Thus
Since u p , lies in the coset Au,, such that uuu,
, there
= a,,,
is an element a,,,
E
A
UUT.
The elements u, lie in the group G,and so must obey the associative law: Expressing this fact, we have
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 € which contains A as a normal subgroup, and such that € / A r G. Let us assume first of all that such an extension exists. Then there is an isomorphism between the elements u of G and the cosets of G module A; we denote this isomorphism by at) Au,,
where u, is an element selected from the coset to which u corresponds. Let x E G; then since A is a normal subgroup of (7, xAx-l = A, and a -+ xax-I is an automorphism of A. This automorphism of A
Thus if € 3 A and €/A gz G, every element u E G defines an automorphism a -+ a Oof A, such that (a30 = aoT;further there is defined for every pair of elements o, T E G an 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 o E G
2. GROUP defines an automorphism a -+ a, of A such that (aT), = a,'. Let there be given also, for each pair of elements a, T E G, an element a,., E A satisfying the relations
Such a set of elements {a,,,} is called a factor set. We shall show that under these conditions there exists a group € 3 A, such that €/A g G. Consider the pairs (a, u) with a E A, a E G;let these pairs be multiplied according to the following rule:
(Heuristically: (a, u) corresponds to the element "au,". Thus (a, o) (b, 7) = "au," "bu," = ab~u,u, = ab~a,,,~,,.) 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
EXTENSIONS
Then e
- ( 6 , ~=) (a;,\b1a1,,
,7) = (b, 7);
for if we set u = T = 1, p = T in the factor set relations, we obtain al,lal,T = ~1~ , t z , whence ,~, a,,, = a,,, . Thus e is an identity element. Similarly, if we set T = p = 1 in the associativity relation, we obtain a,,, = @,, . Using the result we prove the existence of an inverse to each element:
[Heuristically, au, bu, = a;;l,u,; hence abua,,,uu, = a;,iul and , a = a;l,a;,tb-o]. Hence the set of pairs (a, a) form a group G. Consider now the mapping (a, a) -t 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 . a;,: , 1) = 6 Then u = T-I
thus the kernel is an invariant subgroup A isomorphic to A. Hence
On the other hand,
G / A r G. Now define the elements u, = (1, a). For these we have the multiplication rules
whence (since a,,,, = a,.,). Hence in virtue of the relations between the a,,:. Thus the multiplication is associative. Next we show the existence of an identity element. Let
Every element of
G can be expressed as Zu,:
For
(a, a) = ( a . u z , 1) (1, a) = i u , . [Heuristically: u1 E A and so u p l = al,ul a& = 11.
,
whence we have
Finally, u g = (1, a) (a . a;,:,
-
1) = (ao,a) = a%,.
2. Thus if we define cfo = 2,we have
.
Let C: and G' be two extensions of A with factor group G. Let the cosets of G modulo A be written Au,, those of €' modulo A be written Av,. We say that G and €' are equivalent extensions if G is isomorphic to €' under a mapping which acts like the identity on A and carries the coset Au, onto the coset Av,. We now obtain necessary and sufficient conditions for two factor sets to yield equivalent extensions. Suppose a,,, , b,, , yield equivalent extensions G, G'. Then, since Au, is mapped onto Avo, we have u, mapped onto cov, (c, E A). Thus
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, -t cov, . Since the factor set a,,, 1 satisfies the associativity relation, we see that for given groups A and G we can always find at least one extension G containing A as normal subgroup and such that €/A g 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-l . Suppose G acts on A in a prescribed manner and let G be an extension of the type we are considering; the factor set ~ , P , , v may be written a,,, . Let the coset correa*.,
113
sponding to a E G be denoted by Au; then the remaining cosets can be written AuV.We have uVu@= ay,@uP,
Hence if we identify the subgroup A with the group A, we have constructed an extension with the required properties. The elements u, which represent the cosets of G module A satisfy the relation u s , = a,,,~o,
-
GROUP EXTENSIONS
where p is the remainder when p Hence we have a,,, = 1 if if =un
+ v is divided by n. p
+ v < n,
p+v),n.
We shall write un = a. Then au = ua (= un+l); but ua = a%, hence a = a,. 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 = a 4 For all p, v we define
It is a simple matter of computation to verify that the a,,, 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 (7 and G'; let the cosets of G, respectively €', modulo A be written AuV,respectively Avv. Suppose first that the groups G, (7' are equivalent; then there is a mapping from (7 to G' which acts like the identity on A and maps the coset Auv onto the coset Avv. Hence u is mapped onto cv (c E A). Thus where we write N(c) = el+,+ +an-', by analogy with the case in which ?i is an extension of a field A with Galois group G. Conversely if we are given two invariant elements connected by such a relation it is easy to see that they yield equivalent extensions under the isomorphism defined by u + cv. We deduce that there is a (1, 1) correspondence between the non-equivalent extensions (7 of our type and the cosets of the group of invariant elements of A modulo the sub-group of elements which arc norms.
3. 3. Galois Cohomology Theory
We shall now express these results in terms of the Galois cohomology theory which we shall develop. Let A, as before, be a multiplicative abelian group; G a group of operators on A. Thus for every a E G the mapping a + a n is a homomorphism of A into itself and (a7), = am. We now define cochains and coboundary operators for this system: The 0-cochains are the elements a E A, the 1-cochains are the functions a, mapping G into A, and the 2-cochains are the functions a,,, mapping G x G into A. The cochains may or may not be restricted by any continuity conditions. We now introduce the coboundary operator 8 which acts as follows:
We define cocycles as in topology: a cochain a... is a cocycle if its coboundary aa ... = 1. We can prove that every cochain which is a coboundary is a cocycle: i.e. for any cochain a...
The verification of this is left to the reader. I t is also easy to verify that the cochains form an abelian group, if we define multiplication of the functions by multiplying their values, and that the coboundary of a product is the product of the coboundaries. The cocycles form a subgroup of the group of cochains, and the coboundaries form a subgroup of the cocyles. We then define the cohomology group to be the factor group of the cocycles modulo the coboundaries. Using this new terminology we can sum up our results on group extensions in the following statement: Theorem 6: Let A be an abelian group, G a group of operators acting on A in a prescribed fashion. Then to every Zcocycle a,,, there corresponds an extension group (7 of the type described
GALOIS COHOMOLOGY THEORY
115
above; conversely, to every such extension corresponds a 2-cocycle. Further, two 2-cocycles yield equivalent extensions if and only if they are cohomologous. Hence there is a (I, 1) correspondence between the non-equivalent extension groups and the elements of the second cohomology group. When G is cyclic we have already established a (1,l) correspondence between the non-equivalent extensions and the factor group of invariant elements modulo norms. From this fact it is easy to deduce Theorem 7: When G is cyclic, the second cohomology group is isomorphic to the factor group of invariant elements modulo norms. We now continue with the cohomology theory. Let A and G be, as before, an abelian group, and a group of operators. Let H be a subgroup of G, with elements y. We can define the second cohomology group $,(A, H), and we shall study its relation to the group $,(A, G). (Nowadays this group is denoted by H2(G, A).) Let a,,, be a cocycle for the system (A, G). Then if we restrict a and 7 to lie in H, the resulting function a,l,ya is a cocycle for the system (A, H). Further, if a,,, is a coboundary for (A, G) we have
and restricting o and
T
to H we obtain
hence ayl,y2is a coboundary for (A, H). Thus we have defined a natural mapping from the second cohomology group $,(A, G) into $,(A, H);this mapping is easily shown to be a homomorphism: we call it the canonical homomorphism. In general it is neither an "onto" mapping nor an isomorphism; for it is not necessarily possible to extend a cocycle of (A, H ) to a cocycle of (A, G). I n the chapter on Local Class Field Theory we shall show that the mapping is "onto" when A is the multiplicative group of a local field and G is its Galois group relative to some ground field.
4. Now let A, be the subgroup of A which is left fixed by the elements of H: a E A, o ay = a. Then all elements of the same coset of G modulo H have the same action on A,. 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,. We can thus define the second cohomology group $,(A,, GIH), and we shall study its relation to the group $,(A, G). Let a,,,, be a cocycle for the system (A,, G/H); then if we define a,,, = a,,,, for all a, T E G, we obtain a cocycle a,,, for is a coboundary for (AH, GIH), the system (A, G). Further, if a, we have
,
But a,
= a, =
117
CONTINUOUS COCYCLES
1, and a,
= a,
since the coset Hr
= TH. Hence
Thus a, E A,, and a, is a cocycle for (A,, GIH). Similarly let a,,, be a Zcocycle for (A, G) such that a,,, depends only on the cosets of a, T mod H. We have the relation
We set a
= y,
and obtain
whence and defining b, = b , for all a E G we obtain a,,, = b,,bu,/b,, , so that a,,, is a coboundary for (A, G). In this way we have defined a mapping from the second cohomology group $,(AH,GIH) into the group $,(A, G). This is obviously a homomorphism; but it is not an "onto" mapping. We shall prove later that when the first cohomology group $,(A, H ) is trivial, then the mapping is an isomorphism. We say that a cocycle a,,, splits in H if a,,, is cohomologous to the identity cocycle when a and T are restricted to H. H is called a splitting group of the cocycle. Every cocycle has at least the trivial splitting group H = 1, for a,,, = 1 if we choose u, = 1 as representative of the coset A in the group extension defined by(A, a,,,, G). We notice that a cocycle a,,, such that a,,, = a,l,,ys for all y, , y, E H splits on H, for a,l,y2 = a,,, = 1. Obviously if a cocycle splits on H, so do all cocycles 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 Lemma 1 : There is a natural homomorphism from the second cohomology group $,(A,, G/H) into the subgroup of &,(A, G) consisting of elements which split on H. Let a, be a 1-cocycle for the system (A, G), and assume that a, depends only on the coset of a modulo H, i.e. a,, = a, for all y E H. Since a, is a cocycle we have a s , * = a,, . Set y = a; then = a,,
.
Recalling that for any cocycle we have a,,,
= al,,,
we deduce
Thus a,,, E A,, and a,,, is a cocycle for (AH, GIH). We sum up these remarks in Lemma 2: Every cocycle a, (a,,,) of (A, G) which depends only on the coset of a (a, 7) modulo H, is a cocycle for (A,, GIH).
4. Continuous Cocycles
From now on, let A be the multiplicative group of non-zero elements in a field 52 where D is a normal extension of a ground field F; let G be the Galois group of D I F. Let the topology in l(a be discrete, and let G have the Galois group topology described earlier. We shall restrict ourselves now to cochains which are continous functions; we can give an algebraic interpretation of this continuity condition. Consider the 1-cochain a, which maps G + A. If this map is continuous, then, given any neighborhood N of a, in A there exists a neighborhood o U of a in G such that for every element T E uU, the corresponding a, lies in N. Since the topology in l(a
4. is discrete we can choose N = a,. Hence if the cochain is continuous there exists a neighborhood oU of a such that for every element 7 E aU, a, = a,. The group G is covered by these neighborhoods; hence, since G is compact, it is covered by a finite number of them. We have then
where a, is constant on o,U, . The sets U, are groups corresponding to finite extensions of F. Hence U = U1 n U, n n U , is the group corresponding to the compositum E of these fields. Obviously U C U,, and (U, : U) is finite; thus each a$, splits into a finite number of cosets modulo U. Hence if a, is continuous, there exists a subgroup U such that G = UVn=, T,,U and a, 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,t p,V and a, is constant on each coset pvV. Similarly, if a,,, is a continuous map from G x G - t A, we can find a subgroup U, and even a normal subgroup V of finite index in G, such that a,,, = a,,,,,. In order to prove our next theorem we require the following lemma from the Galois theory of finite extensions.
CONTINUOUS COCYCLES
119
between the characters 8 -+ & which is known to be impossible. Thus a, = bl-O. 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 cocycle a, is already a coboundary. Since a, is a cocycle we have aa, = 1, i.e. a,a\ = a,. Since a, is continuous we can find a normal subgroup U of G such that a, is constant on every coset 7U. Let A E U; then aAa2= an, . Since A lies in the coset defined by 1, a, = a,; and since AT lies in the coset UT = TU, we have aAT= a,. Now alall = a, , so a, = 1; hence we obtain a: = a,. Thus a, lies in the field E which is fixed under U. E IF is a finite normal extension, with Galois group G/U. We define c,, = a, E E; then
These are the Noether equations for the field E I F ; by our lemma, their only solutions are C- = bl- b u ,
(b E E).
Lemma: If E IF is a finite normal extension with Galois group G, then a,aa, = a,, o a, = bl-a (b E E).
(The equations a,a% = a,, are known as Noether's equations.) Proof: (1) Obviously a, = bl-a satisfies the equations. (2) To prove that these are the only solutions we set
where 8 is any element of E. Then
since G is a finite group. Now b cannot be zero for all elements 0 E E, otherwise we should have a relation of linear dependence
thus a, is a boundary. A is still the multiplicative group of a normal extension 9 1 F , with Galois group G. Let H be a subgroup of G and let A, be the multiplicative group of the fixed field of H. If a cocycle a,,, splits on H , i.e. if a,,, is cohomologous to the identity cocycle when a, T are restricted to H, we call H a splitting group and A, 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,,, = a,,, for a, T E U; and there is a cocycle a;,, cohomologous to a,,, such that a;,, = 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 $,(A,, GIH) is isomorphic to the
4. subgroup of $,(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 $,(A, G) splits on a normal subgroup H, then it can be represented by a cocycle a,,, which takes values in A,, and which depends only on the cosets of a and T modulo H. We recall that an element {a) of $,(A, C ) is an equivalence class of cocycles; this equivalence class yields a group extension G{,) containing A as normal subgroup such that €/A E 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 $,(A, G) which splits on H, where H for the moment is any subgroup, not necessarily normal. Let a,,, be any cocycle in {a). Then there is a cochain a, on (A, H) such that
121
CONTINUOUS COCYCLES
We must show, however, that the new cocycle a,,, is continuous; for this it is sufficient to show that we can extend a, to a continuous cochain a,. Since a, is continuous on (A, H) there is a subgroup U of G such that a, is constant on the cosets of H modulo H n U. Let H = U, T,(H n U); then the cosets T,U are distinct, for
But 71-172 E H; hence TY'T~ E
H
nu
= ,
T~(H n u) = 7,(H
n u).
We now define a, = a, on the cosets T,U, and give a, arbitrary constant values on the remaining cosets. Thus a, may be extended to a continuous cochain a,. Let G be written as a sum of cosets modulo H:
We see that every element a E G can be written uniquely in the form
Now u, lies in the same coset modulo A as uoUuy.We make a new choice of coset representatives, writing
Extend a, to a cochain a, on (A, G), and write
Thus we have defined a cocycle cohomologous to a,,, with the property that
, i.e. a:,,
E {a),
Then U&
= uouu 1 = u,,~
and
u; = ulu, = u,;
hence Dropping the accent, we see that we have chosen a cocycle a,,, E {a) such that if u, are the corresponding coset representatives for the group extension, G, then
.
u; = u & ,
Further,
4. 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:,, be the cocycle in {a} determined by this choice of representatives; then
u'Ju,uY) = u f y & = a&,u',,, I
J
.
We also define +, = 1 for all y E H. The calculation above shows that
K =a
Y
7
= yJ, and p = a, obtaining
But a,,, = 1, and hence
Next we define, for o 6H,
where 8 is an element of A. Then
Hence
- - ayay0= 1.
and
$w
+YYp
We now make a final choice of coset representatives, by writing v, = $;lu',
.
Then v, = UY , vuvy= f$;lu,uY
We set a = y,
123
Now 8 may be chosen so that +,(8) f 0; for this value of 8 we have also +,,,(8) # 0. If we notice also that +,(8) = +,,(8), we see that if 8 is chosen so that +,(8) = +,,(8), we see that if 8 is chosen so that +,(8) # 0, then +,,,,(8) # 0 for all y, y' E H. We therefore choose 8, for the cosets HolH, Ha,H, ..., suitable values el, 8, , Ho,H, and define =+Yug,'(ei)-
Thus a:,, = a& 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 cocycle a,,, in {a} such that the corresponding coset representatives U, satisfy the relation u,u, = u,, . Consider now the associativity relation
CONTINUOUS COCYCLES
=uw $:
= VOY,
Let the cocycle associated with this new choice of coset representatives be denoted by c,,, . We shall show that this is the cocycle we are looking for. First we have
4.
CONTINUOUS COCYCLES
125
Now define d, = boco-l, so that a,,, = d,da,/d, . We notice first that d, = b,ca-I = 1. Next, setting T = y, we have Thus if {a) is an element of s2(A, G) which splits on any subgroup H, not necessarily normal, then there is a cocycle c,,, in {a) with the properties co..rv - C0,T
G.YT - Cv.r
but a,,, = a,,, = a,, = 1.
, 9
Hence d,=d,.
(These are the Brauer factor set relations for a non-normal splitting field.) In particular, when H is a normal subgroup,
Hence c,,, lies in AH, 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 $,(AH, GIH) onto the subgroup of $,(A, G) which splits on H. We shall now prove that this is an isomorphism. Let {a,) be an element of $,(A,, GIH) which is mapped into the identity of $,(A, G). We may choose a representative a,,,,, of is mapped into the {a,) such that a,,, = 1. Suppose a,,,,, cocycle a,,, of (A, G); then a,,, = 1, and by hypothesis
We wish to replace b, by a cohomologous cochain depending only on the coset of the argument modulo H. Set a = yl, 7 = 7,:
Since the first cohomology group is trivial on H, we can write
Thus do is a cochain depending only on the coset of a modulo H. Finally, if we set a = y, we have
but
We sum up our results in Theorem 9: The second cohomology group $,(AH, GIH) is isomorphic to the subgroup of $,(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 $,(A, G) which splits on H. Then {a)" = 1 where 1 is the identity of $,(A, G).
Proof: We saw in the proof of Lemma 3 that an element {a) of $,(A, G) which splits on H can be represented by a cocycle a,,, such that a,,, = a,,,, . This part of the proof of Lemma 3 did not require the triviality of the first cohomology group. Let
Consider the associativity relation
Set p = r V,and take the product from v = 1 to n, using the fact that a,,, = a,, . We obtain
CHAPTER SEVEN
T h e First and Second Inequalities where 1. Introduction
Thus a:,,
af,;
i.e. at,, E {I).
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 1 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 ( k) be the subgroup of B,(C, r ) 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 h there corresponds one and only one unramified extension of k. The Galois group of an unramified extension is isomorphic to the 127
128
7.
2.
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 h have q = pr elements; it is well-known that for every integer n there exists one, and (up to an isomorphism) only one extension E I k of degree n, namely the splitting field of d"- x. The Galois group of E I & is cyclic, and has a canonical generator a: a(a) = aq for all a E E. We immediately deduce Theorem 1 : If h is a finite field, then all unramified extensions of k are cyclic. Next we extablish the Lemma: Every element a in & is the norm of an element x in
E.
Now let E, be a unit of k such that construct a unit Em of T such that E,
129
UNRAMIFIED EXTENSIONS
= 1 mod pm
and
E,
E
1 mod pm; we shall
= NE, mod pm+l.
E,
+
Such a unit Em, if it exists, will have the form Em = 1 rmYm, where Y, is an integer of T, and rr is a prime in k (hence in T). Let en, = 1 $ rrmxm,where x, is an integer of k. If E,
= NEmmod pm+l,
we have N(l Hence
+ amY,)
= (1
Proof:
+ vmym)l+~+...+~n-l1 + +nx,
mod pm+l.
nmS(Y,,) = amxmmod pm+l, S(Ym)r x, mod p.
The mapping x -+ N(x) is a homomorphism of kernel consists of these elements for which
Thus the order of the kernel is Hence
E* into A*. The
< (qn - l)/(q - 1).
number of images obtained =
order of E* order of kernel 2 4 - 1 .
Now not all elements of T have trace zero. Suppose x is an element of T such that S(x) = a # 0 in &; let 5, denote the residue class of 6 which contains xm . Then S(Z,x/a) = 5, . Hence if Y, is any integer of T in the residue class gmx/a,then S(Ym)E X, mod p. From these remarks it follows that we can construct a unit E, in T such that Em = 1
Proof: Let E be a given unit in k. By the preceding lemma, the residue class of A containing E 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 1 & is the same as in T I k, we see that E 3 NEo mod p. Hence CINE, is a unit el , and 3 1 mod p.
and
E,
= NE,
mod pm+l.
The theorem now follows easily. We have
Since R* contains only q - 1 elements, the equality sign must hold. Hence every element of &* 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.
+ amYm= 1 mod pm E 3
NEo mod p;
Then we construct a unit El el
-
-!.-
-
NEo -
'
1 mod p such that
= NEl mod p2;
€1 -
NEl - E2 '
Similarly we construct a unit E2 r 1 mod p2 such that e2 3 NE2 mod pa,
and so on.
130
7.
3.
THE FIRST AND SECOND INEQUALITIES
Since the units E, converge to 1, the product E,E,E, converges to a unit E' of T, and clearly E = NE'. This completes the proof of the theorem. We now introduce the convention of representing groups by their generic elements: for example, a shall denote the group of all non-zero elements a E k, NA the subgroup of a whose elements are norms of the non-zero elements A E E. Lemma: If T I k is unramified, of degree n, then the factor group a/NA is cyclic of order n.
Proof: Since T I k is unramified, a prime .rr in k remains a prime in T. Thus any element A in T can be written A = .rrvE,where E is a unit in T. Then NA = .rrvnNE. We have just seen that the group N E coincides with the group of units in k. Thus the elements of k whose ordinals are divisible by n are precisily the norms NA. Hence a/NA is cyclic of order n; the cosets are (NA, nNA, ..-, .rrn-lNA). Let E I k be any finite normal extension. Let H be the subgroup of I' which leaves E fixed; then r / H is the Galois group of E I k. We recall that the second cohomology group B,(E, r / H ) is isomorphic to S(E I k). Hence, when E is a cyclic extension (in particular when E is unramified) we have S(E I k) r M E , r l H ) r
invariant elements norms
= (UINA.
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
But, by the lemma, a/NA is cyclic for an unramified extension. Hence when E is unramified we have the even stronger statement that S(E I k) is itself a cyclic group of order n and [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 I' which leaves Tn fixed. We notice that E n 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 P, hence HU = UH, and so HU is the group generated by H and U and is therefore the subgroup corresponding to E n T, = k; hence HU = I'. The intersection H n U corresponds to the compositum ETn . The Galois group of ET, I E is
which is the Galois group of Tn I k. Hence ET, I E is cyclic of degree n, and it is easy to see that ETn I E is unramified. We shall now show that S ( E I k) contains S(T, I k), so let auv,,,,, be a cocycle which splits on Tn . Since T, is cyclic we may assume the cocyle has the form
3. The First Inequality
Let E I k be any extension of degree n. The proof that S ( E I k) is a cyclic group of order n proceeds in two stages. The first step is taken in our next theorem:
where a is an element of k. First we remark that we may choose a in H. For since oU C HU = I', we can write a = hu; we then replace a by uu-l. Thus we can assume that a is in H. Then
Theorem 3: If E / k is any extension of degree n, then S ( E I k) contains a cyclic subgroup of order n, and [E : k] 2 n.
The restriction of a0,,,,,
to H is therefore a,v~un,,,,~un,,
.
7.
132
THE FIRST AND SECOND INEQUALITIES
This can be regarded as a cocycle for (ET, , H/U n H); we shall show that it is cohomologous to the identity cocycle. Since ETn 1 H is cyclic it suffices to prove that the element a lies in the group of norms N E T n I E . If .rr is a prime in k, IT a prime in E, then, since E 1 k is totally ramified, n = €I?'< Thus the ordinal of a with respect to IT is a multiple of n, and since ET, I E is unramified, this is precisely the condition that a lie in N E T n J E . Hence a , splits on E, and so S(E 1 k) 3 S(T, I k). This proves the theorem in this case. Case 3: E is an arbitrary extension. Let T be the inertia field of E I k, and, as in Case 2, let T, be the unramified extension of degree n, corresponding to the subgroup U of r. Then T is a subfield of T, , corresponding to the subgroup HU, and E I T is totally ramified. which splits on T,; if we restrict Let a,,, be a cocycle of (C, the subscripts of a,,, to lie in UH, we obtain a cocycle of (C, UH) which splits on Tn . Since E / T is totally ramified, we can apply the result of Case 2, showing that a,,, splits on E. Hence S(E I k) 3 S(T, I k). This proves the theorem in this final case. The result we have just proved, namely that for any extension E I k 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:
r)
O u r < -m+s 1
,d and
$D s+m-(l+l)r m7 +-s 1 .
vfm-(l+l)c
Finally we obtain the result that Hence all the ai are equal, and We now return to consider the index Hence for v < I all the associated polynomials ai$,(u) are distinct. Thus for 1 < v < I, each such set of associated polynomials contributes a term to the coefficient of xV,namely
where n 2 1. We have to examine three cases separately, depending on the relation of n to I and m; the three cases are
Since the coefficient of xuis a sum of terms of this form, it is itself of this form, and we obtain finally First we show that cases 2 and 3 can occur only if I = p, the characteristic of the residue class field. Since n 2 1, We shall show that the dominating factor in this expansion is either xS(17,) or ~ ~ N ( 1 7The ~ ) . ordinal of the latter is easily computed; we must now estimate ord S(17,). S('pS) is obviously an ideal in k, for
m, we have (n r 1) (I - 1) > m Since (n for all positive integers r ; hence Pn+, = /In+,+,En+, , where En+, = 1 xDs+,, . Since lim,,, En+, = 1, the infinite product En+, converges to a unit E, and /In = NE. Hence (/In :8 ,n NE) = 1, and so, by the first isomorphism theorem, (finNE : NE) = (v, : NE) = 1.
1
+ x~hT(II,)mod pn+l.
Let us choose an element ITn with ordinal exactly n; then N(l7,) = mn. Hence N(l
+ xI7,) = 1 + cxZnnmod pn+l.
Let x run through a system of representatives ( x , , ..., x,) of the residue classes modulo p. Then every element 13, is congruent to some 1 xprn modulo pn+l. Since I = p (the characteristic of the residue class field), the mapping x + x z is an isomorphism of x,') is also a system of the residue class field, and so (x,', x,',
+
a*.,
7.
142
5.
THE FIRST AND SECOND INEQUALITIES
representatives for the residue classes. Hence every p, is congruent to some 1 exilrn modulo pn+l. Thus every element p, E N E mod pn+l. By exactly the same argument as in Case 1, we obtain the result that (v, : v,,) = 1.
+
THE SECOND INEQUALITY CONCLUDED
When the index n satisfies the inequality (n we have (vn : v,+~)= 1. When the index n = no is such that (no
+
Case 3: (n 1) (I - 1) = m. Here we have (n+l)l--1 =m+n, and hence
Finally for any index n and by Case I, ( v :v ) Thus if N
Thus S ( v )= pn, and also N(Qn) = pn. Hence, if l7, is an element of K with ordinal exactly n and x is an integer of k, we have ~ ( +1xun) = 1
+ rrn+(x)mod pn+l,
where +(x) is a polynomial of degree I = p. Let (x, , x, , ..-,x,) be a system of residue classes modulo p. Then, as in Case 2, every element p, is congruent to some one of the q elements 1 xi+ mod pn+l. Since a polynomial +(x) of degree I may take the same value for as many as 1 different xi , the elements (+(xl), ..., +(x,)) may represent only r of the residue classes, where q 3 r 2 q/l. Now every /In which is congruent to some one of the r distinct elements 1 +(xi) rnis congruent to the norm of a unit mod and hence lies in the group pn+,NE. It follows that
+
+
@n
4 r
: Pn+lNE) = -
=1
+ 1) (I - 1) = m, we have
we have (n
and
+ 1) (I - 1) < m,
+ 1) (1 - 1) > m,
v, = NE.
> no,
~ )I, (E: vN)= (vno: v ~ , +$
and
vn = NE.
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: a shall denote the generic element of k. [a] shall denote the generic element of h; namely the residue class to which a belongs. We shall construct in k a particular system of representatives, to be denoted by G, of the residue classes [a]. First we notice that if a = /3 mod pv, then ap = ,l3p mod pV+l. For if a = /3 y r V , then
+
191
pZv.
Since
the representative
Thus the representatives ci are multiplicatively isomorphic to the residue classes of k. Further, the ci are uniquely determined by this ... be another set with the same property. property; for let &, Then
p,
- -
&. = lirn ([alp-')PY= lirn ([alp-")PV= lim & = &.
We now make the additional assumption that the representatives
A.
6 form a field. We shall show that this field must be isomorphic to For let G = 7, and let a , /3, y be elements of k such that a E [a], /3 E [PI, y E [y]. Then a r G, /3 r f?, y = 7 mod p ; hence a j3 y mod p, and therefore [a] [/I] = [y]. I t follows that the representatives &. can form a field only if k has the same charac-
+
-
teristic as Then
+p
+
K.
On the other hand, suppose k has characteristic p. [a]
+ @] = a+ p = limYFv,
where Now consider the residue class [ a ] ;since is perfect, the residue class [alp-" is well-defined. Let a , be any element of [alp-", and consider the sequence {a:'). We have a , = a::, mod p, so that pV - pv+p f f v = a'+P mod p v + p ; hence {a:') is a Cauchy sequence. Since all the terms of this sequence lie in the residue class [a] so does its limit; we denote this limit by G. The limit & is independent of the choice of the elements a,; for if is another such sequence with limit &', we have a, = a:' mod pv,and hence 6 = &'. We choose G as representative in k of the residue class [a]. 190 99'
Thus we may write y , = a , hence
+ /3,
where a, E [alp-",
/?,E [/3]p-';
We can now sum up our discussion by giving a precise description of fields of characteristicp with perfect residue class field. We recall
192
10.
APPLICATIONS AND ILLUSTRATIONS
that if T is a prime element in k, then k consists of all power series C C?(~)T". 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.
2.
THE NORM RESIDUE SYMBOL
193
Hence, since k is complete, it contains a subfield isomorhic to the p-adic numbers R, . k also contains the (q - 1)-th roots of unity; hence k contains k ( 5 ) where ( is a primitive (q - 1)-th root. Let T be a prime element in k; then k consists of all elements of the form
Let p = me. Then we may write k contains the (q - 1)-st roots of unity, and no other roots of unity with period prime top. Theorem 2:
Proof: The non-zero elements [a] of k form a cyclic group of order q - 1. Hence the representatives (Y in k also form a cyclic group of order q - 1; these representatives are therefore the (q - 1)-st roots of unity. This proves the first assertion of our theorem. Suppose, next, that 5 is a primitive m-th root of unity in k, where (m,p) = 1. Then there is a positive integer 1 such that I q E 1 mod m, i.e. (g = 5. Suppose 5 lies in the residue class [a]; the representative of [a] is given by 6 = lim a:";
% any element of
.
[ol]pv
Clearly we obtain the same limit if we restrict ourselves to the = {a?} where a,, E [a]q-'f We may obviously subsequence(a~~,l'} choose or, = 5@. Hence = iim ( ~ Q - ~ W ) Q ' , , = 5. (Y is an element of a cyclic group of order (q - 1). Hence 1, i.e. m divides (q - I). 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 = pf elements. k contains a subfield isomorphic to the rational numbers; and the valuation of k induces the p-adic valuation on this subfield.
Now
5q-I
=
i.e. every element a
Ek
may be expressed as s-1
a=
d
m
(d,, E R,(5)).
u=O
Hence k is a finite extension of R,: k
= R,((,
T).
2. The N o r m 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 F, 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 -t (C I k / ~ is ) a (I, 1) correspondence between r and k. We denote the inverse mapping by 4, i.e., if a = (C I k / ~ ) ,then +(a) = T . We study expressions of the form
where we choose the same value for (xylg) for both terms. Clearly if (xyl p), is another value,
194
10.
2.
APPLICATIONS AND ILLUSTRATIONS
where v is an element in the prime field. The expression
THE NORM RESIDUE SYMBOL
We have, for every u
E k,
lies in the prime field, since both terms are roots of the equation tp- 5 = 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 =
f uydz + f uady,
since $uydz lies in the prime field, and hence is unaffected by
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.
f uxdy =
(~r'')
- (??).
for all
3. dy* = ndyfl-l.
This follows by induction using (2) if n is positive. If n is negative, we write
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. xdy = xldyl y,9 f uxdy = f uxldyl
+w
r E k.
y ~~ )- ~ + l d +yyR(l ~ - n) y-n dy (using 2). dy = d ( ~ - ~ + l = Hence ndy =~ - ~ + l d y * ,
dyn=nyn-ldy.
1.e. '
From this result we deduce at once the formulas dyP = 0, dy% =y%h.
The sum of two differentials is defined by the relation
Since the field of constants F is perfect, every a E F is a p-th power; hence the differential map is homogeneous: xd(ax)
=
axdz.
We now deduce some elementary properties of these differentials:
The linearity of the map follows directly from
This follows from the linearity of the operator l / g and the fact that +(y) is an isomorphism.
Since
2. dbz) =ydz
+ zdy.
196
10.
2. THE
APPLICATIONS AND ILLUSTRATIONS
NORM RESIDUE SYMBOL
197
we must show that (y/p)d(Y)= ( y l p ) . This is obvious if y / p lies in k. If y / p = 8 is not in k, 8 satisfies the irreducible equation f* - 5 - y = 0; hence y E N k ( e ) ,., From the properties of the norm residue symbol we deduce that 4 ( y ) acts like the identity on k(8); hence 4; dy = 0 in this case also. I n particular we have 4 d(yz) = 0, so $ ydz = - 4; xdy. Then
Set 6 = (clg);since c is an element of F, F(6) I F is a cyclic extensions of the residue class field. Hence k(6) I k is an unramified extension. Thus
Thus the differential map ydz is linear in its arguments. We now evaluate some special "integrals".
remains We can now raise this result to the q-th power; SFlF0(c) unaltered. Hence
Summing these equations we obtain the required result since x lies in the ground field k.
(2) If c E F and y # 0,
(3) If IxyI
< 1, then
$ x d y = 0,
Now for any a E k such that If n f 0, let n = prm where (p, m) we may write c in the form c = epr. Then (a)
=
,
d~ f cyndr = f (cym)pr, =f c y,dy =
f d ( -mye m )
I a I < 1, it is easily verified that
1. Since F is perfect
using (1) above =O.
Hence (xy/g) E k, and so (xyl p)+(Y)= (xyl p). This proves the required result. We notice that this result implies the continuity of the integral in both its arguments, for
(b) If n = 0, provided
I x - xl I an I y - yl 1 are sufficiently small.
198
10.
2.
APPLICATIONS AND ILLUSTRATIONS
(4) If 1 y 1 c 1, and x is developed as a power series in y with coefficients in F: x = Z: cVyY, then
THE NORM RESIDUE SYMBOL
199
I n particular, if v and y are any elements of k, and t is a uniformizing parameter, we have d ~dt. f vdy = f v 2
Since Z: cVyvconverges, there exists an index N such that
Specializing v to be x, we obtain
f xdy = f r g d t = SFIF,,(Res, (x-
2))
and hence and, similarly, setting v
=
x/y, we obtain
Thus Rewriting these integrals, we obtain and using 2 above, we obtain
We may call c-, the residue of x with respect to y, and write c-, = Res, x. I n particular, if y = t is a uniformizing parameter, and v is any element of k, we have $vdt = S,,,o(Res, a), since ord t = 1. Again let 1 y 1 < 1, and let x be written as X cVyY. Then we define formally the derivative dxldy to be the series Z: vc,y+l. We shall show that dx = (dxldy) dy. Consider
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 # 0. Let t be a uniformizing parameter, and write x = C cVtu. Then
Now ydx = 0 means that u = atT/y(a E F). Then
This proves our assertion.
-
$ 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-, , and so we have either r = 0 (i.e. r 0 modp) or c-, = 0 (since the trace is not identically
200
10.
APPLICATIONS AND ILLUSTRATIONS
zero). Hence x contains only powers of t*, 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(x/p) I k. For y is a norm for k(x/p) I k e +(y) acts like the identity on (xlp) o $ xdy = 0. Thus if y is a norm for all such extensions, then $xdy = 0 for all x; 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 = Fit) be the field of formal power series in t with coefficients in F. If y = E cvtv,we define its derivative
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
If t, is another uniformizing parameter, we may prove that
This result is immediate if y is a finite power series in
3.
DIFFERENTIAL I N A N ARBITRARY POWER SERIES FIELD
201
If y and z are any elements of k, and dxldt # 0, we may define
This clearly does not depend on the choice of the parameter t. We define the residue of y = C cvtvwith respect to t to be
The residue is thus linear and F-homogeneous. We may notice the special cases:
when n = - 1; = 0, when n # - 1. We must now examine the effect on the residues of a change in the uniformizing parameter. We obtain Theorem 3:
If t and t, are uniformizing parameters, then
Proof: Since the residue is linear, F-homogeneous and continuous, it sufficesto prove the theorem for y = tn. The result is obvious for the trivial change t = a,t, , so we may assume
We have now to show that Then
"
d --m
dt,
dy dt,
dt
dtl dt '
The result then follows for arbitrary power series since the map d/dt is continuous.
Res,,
)(;tini; dt
=1
when n = - 1, and = 0 when n # - 1. When n 2 0, the result is clearly true, since tn dtldt, contains no negative powers of t.
202
10.
APPLICATIONS AND ILLUSTRATIONS
When n = - 1, we have
In this notation, our earlier results would be written
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 4. The Conductor and Different for Cyclic p-Extensions
and this vanishes for n # - 1. Now for any characteristic, and any &xed n
+
1 dt - 1 2a2t, --t-" dt, tyn(l a,tl
+
< - 1, we have
+ - ... + P(a2 + ..-) -
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 / g ) where a E k. We must now investigate the degree of freedom we have in choosing generators for fields of this type. So let
a,, t -.) + ...
3
where P(a, , a , , is a polynomial constructed quite formally: that is, P(a2 ,a , , .-.)is a universal polynomial of the a, with rational integer coefficients. P(a2 ,a,, is thus the same for all fields, and contains only a finite number of the coefficients a,. 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 > 0. This completes the proof. To rid ourselves of the dependence of the residue on the uniformizing parameter, we introduce the notion of a diferential ydz; this is purely formal. We say that ydz = yldx, if and only if dzldz, = yl/y. We now define the residue of a differential: me-)
i.e. suppose there are two possible generators a, ap-a=@,
- - a )
'3.
Res (ydz) = Rest y -
(
Then
UP = /3 + r,
It follows that j3lr
dz dt
We have
(
dl)
= Res tl y - = Restl (ydz).
8"--/3=b.
We may choose the generator o of the Galois group such that
This does not depend on the uniformizing parameter, for Res, (ydz) = Rest
p such that
where 0
0. 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
206
10.
5.
APPLICATIONS AND ILLUSTRATIONS
of an extension k(o/p) j k where is a constant. An extension of this type, however, is unramified, and so f = = D.
5. The Rational p-adic Field
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 T , , which has norm group U, the group of units in k. Next we consider the field K = k(5), where 5 is a primitive pr-th root of unity; this is certainly an abelian extension. Now all the primitive p7-th roots of unity are roots of the polynomial
the ramification e(K / k) 3 4(pr). But the degree +(pr); hence we have e = n = $(pr) and f = 1. I t follows that (1 - 5 ) is a prime in K , and that
n(K 1 k)
1. Since ( 2 1, > 1 for all archimedean primes q, we conclude that there can be only a finite number of archimedean primes in XR. 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 'D of one of the following types: (1) q is archimedean, (2) q is discrete and A, is finite,
(3) q is discrete and f(q)
= deg
( R , I k,) 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-Jields. We shall have to consider sets S which will be arbitrary when k, is not a field and k,-vector spaces when k, is a field. For such sets S we define the order of S as follows: (1) If there are archimedean primes, or if k, is finite, the order of S shall be the number of elements of S. (2) If k, 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 k, . 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 k, (if it is a field) is finite (k, C S), so we have to use the first definition; if not, we have the order of S = cf'q). For reasonable non-archimedean primes we define the norm Nq to be this order of the residue classes. We introduce a normal valuation 11 11, for reasonable primes q:
(1) If q is archimedean, and the completion of k under p is the real field, we define 11 a [Ip to be the ordinary absolute value of a. (2) If q is archimedean, and the completion of k under q is 1he field of complex numbers, we define 11 a 11, to be the square of the ordinary absolute value of a. (3) If q is discrete, we define 11 a 11, to be (Nq)-Ordq~. We may now write the special valuation occurring in the product where formula of Axiom 1 as a power of 11 1 ,: I a 1, = 11 a! p(q) > 0. 2. Upper Bound for the Order of a Parallelotope
1, then I a 1, > 1 for all p dividingp, and by Axiom 1 this is possible only for a finite number of p . are equip , divide p, all the valuations I Clearly if p , , p , , I ,1 is a valuation in the equivalence valent in R, and 17,,,,,ESn class p. contains primes Let p , denote the infinite prime in R. Then p , which divide p , : Case 1: If R is the field of rational numbers, the archimedean primes in XR divide p, .
where k is the number of archimedean primes p # q. Finally we obtain
Case 2: If R = k,(z), not all primes are trivial on R ; hence > 1. This prime there is at least one prime p , such that I x p , divides p, . We now give a description of all PF-fields. Theorem 3:
where D,depends only on q.
A PF-field is either
(I) an algebraic number field, or (2) a finite extension of a field of rational functions. a, be elements of k which are linearly Proof: Let a,, a,, independent over R. We shall show that r is necessarily finite. T o this end we consider the set S consisting of elements ..a,
3. Description of all PF-Fields
Let k be a PF-field. We define a rational subjield R of k:
Case 1 : When there are archimedean primes in m, R shall denote the field of rational numbers.
where the v, range over all integers of the rational subfield R (i.e. I A 1,rational integers or polynomials) such that I vi lpm where A is a given integer of R. Let M be the order of S.
1) A 117pm. To do this we must consider two cases-
Case I : R is the field of rational numbers. Here there are exactly 2 11 A llPm 1 possible values for each vl; hence there are in all M = (2 I/ A (I, 1)' elements of S. Clearly M > I/ A llGm.
+
+
Case 2: R = k,(x). In this case S is a vector space over k, . If the degree of A is d, each v, has degree d. I t follows that the l); dimension of S over k, is r(d I), and hence M = ~ ~ ( ~ +but 11 A [Ipm = cd. Hence M > (1 A . We now estimate the size of the parallelotope containing the set S. For all primes pw , we have I a 1,BPmI A IPm, where BPmis a constant depending on the elements a, , a,, ..., a,. For the remaining primes p (which are certainly non-archimedean), we have
In the second case Res xf-l/p(x) # 0 where f is the degree of p(x). In both cases we have obtained a contradic-
a.
E R,
the map
is a differential of k. Indeed p is certainly a linear map, and clearly vanishes for elements of k. p is continuous since the topology on
252
XIII.
5.
DIFFERENTIALS IN PF-FIELDS
V(k) is the Cartesian product topology on V(R)n. (See section 2) We have therefore proved the existence of differentials in k.
DIFFERENTIALS I N A PF-FIELD
253
,
Then the maps ( X ) = may be written in the form li(X) = I(or,X) with or, E k. Now suppose
Theorem 5: If k is a PF-field, then the differentials of k form a 1-dimensional k-space.
Proof: Let I, be a non trivial linear, R-homogeneous map of k into R. Such maps certainly exist. For example, the maps
Since 1(X) = I,(or~lX),this implies that for all i
Suppose X E q-117,' for i = 1, n. Then X 17,' C arl 17,' for all i, where R;' denotes the unit parallelotope in V(k). It follows that a*.,
are of this type. When k/R is separable we have even an invariant map of this type, the trace Ski, . Let I be any linear R-homogeneous map of k into R. Then by Theorem 4, we have I(a) = lo(@) were /?is an element of k depending on I. Suppose that l(w,) = a, . Then
X(l,(XG)) = Xi(L!:li(X)) = X(rr,'[J C N.
Now 17,' & is a parallelotope in V(R).From our previous investigation it follows that & is contained in a fixed parallelotope of V(R). Hence
We can now extend I to a linear, V(R)-homogeneous map of V(k) into V(R), which we still denote by I. We define
x=~&Wtcl7~, where 23 is fixed. We have now
This extended mapping is unique, and clearly l(X) = I,(/?X). 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 p is a differential of k, then p(X) = X(I(X)) where I is a map of the type we have been considering. Hence
Hence
This proves Theorem 5. If k is a PF-field and p is any differential of k, then there is an upper bound for the parallelotopes IT,' such that ~(17,') C N, and hence p(17L0) = 0. Theorem 6:
Proof: Let w, , w , , ..., on be a fixed basis for k/R. We have seen that p(X) = A(l(X)) where I is a certain linear map of V(k) into V(R). Let
Thus n,' Cn;, where fl,' is a fixed parallelotope of V(k). This completes the proof of Theorem 5. If $. is a valuation vector in V(k), we may write
where $., is the valuation vector with the same p-component as and the component zero at all other primes.
5
254
XIII.
DIFFERENTIALS N I PF-FIELDS
Let p be a differential of k. Then we define p,(() to mean ~(5',).
vanishes for field elements, for if = a E k, we have A(a(,) = 0 since 4 -+ A(a() is a differential. Hence there is an element fl E k such that A(to() = A(/3(). By the previous theorem,
Theorem 7:
for all primes p, i.e. Proof: Since p is a differential there is a parallelotope II, such that p(I7,) C N and hence p(17,0) = 0. We may write
--
tp)
for all p and hence ( t o - B), = 0. So
=0
to= fl.
6. The Different
with
For p # pi
81,
, (,
lies in 17,0 and hence p((,) = 0. Thus
Now
This completes the proof. Let p be a fixed prime, and suppose p,([) = p((J = 0 for all f having ord, (, 2 v, . Changing the p-component of a to pl'v we obtain a set lAOon which p ( G o ) = 0. Since the parallelotopes 176 such that p ( a 0 ) = 0 are bounded, it follows that v, is bounded from below. Thus p,(f) is not identically zero. From this we deduce Theorem 8: If A and p are differentials, then A(() = p(() for all 5 E V(k) if and only if A,(() = pp(() for all 5 E V(k) and one prime p. Theorem 9:
If p(So) = 0 for all differentials p of k, then
50 E k. Proof: Let X be a fixed non trivial differential. Then A((,() is also a differential-it is clearly a continuous linear map, and it
Let k be a PF-field. K/k a finite extension. I n the function field case we shall assume that K and k have the same constant field. Let o, , o, , w, be a basis for K/k. Then every valuation vector X E V(K) can be written uniquely in the form ..a,
where the & are valuation vectors of V(k). Let A be a fixed non trivial differential of k. Then if p is any differential of K, there exists a continuous linear V(k)-homogeneous map I of V(K) into V(k) such that 1(K) C k and
Since K is also a PF-field we may write
NOWpV(X) = p(X@)where X q is the valuation vector with the same P-component as X and component zero at all other primes. We write
Xv = &wl
+
52%
+ + "'
Snwn
-
The Q-components of Xp are derived from the q-components of the &where f2 I q. Thus if any of the [$ have non zero q-components for q # p these may be replaced by zero without altering X 9
256
XIII.
DIFFERENTIALS IN PF-FIELDS
(which has zero component for q # 9). But since the representation of XI, in this form is unique, it follows that all the & have component zero except at p. Now
Hence a@?,is a fractional ideal in K*(p), namely V'Copvp. We can now define the 1-9-dilgerent Dpt:
where I ( o J = ai . Hence Xrp has zero component except at p. Then
Since a and a are divisors (or ideals), prpand v, are zero for all but a finite number of primes, i.e. a,,, = DFpfor all but a finite number of primes 9. We may therefore define the 1-different in the large:
Let K*(9), k*(p) denote the completions of K, k at 9 , p respectively. Let DI, , D, denote the rings of integers in K*((P), k*(p). We define the inverse I-!&different a$, by the relation XI, E z),$
if and only if
I(XqDcp)C o,
.
Now let P(X) = h(I(X)). Let 17,-1 ,17,-1 be the maximal parallelotopes in V(K), V(k) such that p(17&1) = 0, and A(17:-1) = 0. (It will become clear in a moment why we denote the parallelotopes by lI,,-, &-, rather than IT, 17,). Let 'U = 17VW, a = ll pvv. Then
This statement defines precisely the power 9 % of 9 occurring in 3. We have now
Thus Dl is a divisor in the function field case, an ideal in the number field case. We have
hence % = aDl. We have therefore proved Theorem 10: If P(X) = h(l(X)) as above, and if 17& .l7:-1 are the maximal sets of this form on which p, h respectively vanish, then % = aD,. We now consider the case of a separable extension K/k. The trace SKlk is here a non trivial linear map of K into k and may be extended to V(K).We now show that the S-9-different, defined as above, is identical with the different of the local extension K*(p)/k*(p) as we defined it in Chapter 5. T o this end we denote SK*('4)lk*(p) by SS,and prove Theorem 11: Let K/k be a finite extension, p a prime in k. If or is any element of K,
Proof: If K/k is inseparable, the result is trivial. We therefore suppose that K/k is separable.
258
XIII.
DIFFERENTIALS IN PF-FIELDS
We notice first that if the result holds for extensions Elk and K/E then it holds for KIA. We denote the primes in E by q, SE*(g)lk*(p) by Sg , and SK*(q)/E*(rg) by S,'. Then we have
Hence [S(X)]=
2 (53, i
~,(%) @IP
I n particular we obtain the result It is therefore sufficient to prove the theorem for K(a). Let f(x) = Irr (a, k, x ) and let the decomposition of f(x) into irreducible factors in k*(p) be
If gi(x) is of degree ni and has a root ,!Ii , then k*(p)(pi) is one of the fields K*(P) with 9 dividing p . Each field K*('@)is obtained from one of the factors of f(x). Since I;,,, ni = n, it follows that all vi = 1. The theorem then follows by comparing the second coefficient on both sides of equation (*). By comparing the constant terms on both sides of (*) we obtain another result which we shall require later: Theorem l l a :
Corollary: If S is extended to a map of V ( K )into V(k),then
Proof:
If
then we define
Now consider the S-!@-different
We have
where Dv is the diierent of K*(q)/k*(p)as defined in Chapter 5. Thus P9.s = 9'0.
By the argument of Theorem 10 it follows that 3 , = DIDfor almost all primes p, and hence by Theorem 2 of Chapter 5, we have Theorem 12: If K/k is a separable extension, there is only a finite number of ramified primes. The S-different of K/k which is simply called the diflerent of the extension is 3) = Up9%.Hence 3) is a divisor (ideal) of K.
1.
26 1
PARALLELOTOPES IN A FUNCTION FIELD
is the least common multiple of a and b. Similarly
&+&=a, where CHAPTER FOURTEEN
The Riemann-Roch Theorem In this chapter we offer two proofs of the Riemann-Roch Theorem. We shall see that both proofs depend largely on three results:
+
(1) There exists an idde b such that k IIb = V ( k ) . (2) The differentials in a field k from a 1-dimensional k-space. (3) There is an upper bound to the sets 172 on which a given differential vanishes. In the first proof we make use of the fact that we have already proved those statements in Chapter 13, both for number fields and function fields. The second proof is entirely self contained and includes alternative proofs of these results for the function field case. 1. Parallelotopes in a Function Field
Let k be a function field, k, its constant field. Let a be a divisor: a= pVp. a defines a parallelotope I7, consisting of vectors 6 E V (= V(k)) such that ord, 5, 3 v, . If or is an element of k, or, a , -..) lies in I&if and only if the valuation vector a = ord, or 2 v, for all p. It is natural to say that a divides or. Similarly, let b = P ~ Pbe a second divisor, defining the parallelotope I&. Then 17, contains I& if and only if p, 2 up for all p. We say a divides b, written a 1 b. I t is easily seen that
n,
(a*.,
where
nannb= n c ,
is the greatest common divisor of a, b. We may define the absolute value of a divisor a to be
where
n(a) is called the
degree of a. Clearly 4a,a2) = 4%)
+ n(a21,
and by the product formula, n ( a ) = 0 for all or in k. Hence
that is n(a) is an invariant of the divisor class of a. We notice also that if a divides b , then n(a) n(b). The ring of valuation vectors V , the field k, and all parallelotopes II, may be regarded as vector spaces over the constant field A,. If A and B are any two spaces over KO we shall denote by (A : B) the KO-dimensionof the factor space AIB.
0, then m(a) = 0.
If there is a field element or # 0 in the parallelotope 2 ord, a for all p . Hence n(a) 2 n(a). But by the product formula, n(a) = 0. If n(a) > 0 it follows that there can be no non-zero field element in the parallelotope 4 ,i.e. m(a) = 0. Now let a divide b, so 3 lib . We apply Theorem 3 of Chapter 13 to the case where A = &, B = I& and C = k . This yields Proof:
Hence when a divides b we have
Now let a and b be any two divisors, b their greatest common divisor. Since b ( a and b I b we have
lIa , then ord, a
Hence we have
From this formula we proceed to prove the Riemann-Roch Theorem.
2. First Proof For each divisor b we define the function I@):
+
Z(b) = (V :nb k).
We see that the value of n(a) of the field k. We write
+ m(a) - l(a)
must be an invariant
and we define g to be the genus of the field. If we consider the parallelotope defined by the unit element 1 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 17,are the elements of k , . We have therefore
+
whence g = l ( 1 ) = ( V :17, k ) . This shows in particular that the genus is a non-negative integer. Since m(a) 2 0, we have l(a) 2 n(a) g - 1. Hence by choosing a such that n(a) is large enough we can make l(a) as large as we please.
+
264
XIV.
THE RIEMANN-ROCH
THEOREM
3.
We have now to interpret the function l(a). Let U denote the factor space V/(& k). Then [(a) = dim U = dim 0 where Z? is the dual space. If h is any linear map of U into ko we may regard A as a linear map of V into k, such that A vanishes on 17, k, i.e. we may regard h as a differential of k vanishing on 17,. Conversely each differential which vanishes on & gives rise to a linear map of V into ko . Thus l(a) = dim tf = dimension (over KO) of those differentials of k which vanish on IT, . Now let A(.$) be any non trivial differential. By Theorem 6 of Chapter 13, there is a maximal parallelotope 17,-1 on which vanishes: A(n,-l)= 0. By Theorem 5 of Chapter 13 any other differential p(5) may be written in the form p(() = A(&() with a E k. Clearly the maximal parallelotope on which A(&) vanishes is I7,-1,-1. Thus the divisors describing the maximal parallelotopes on which differentials vanish belong to a fked divisor class, which is called the canonical class, or the class of dz~erentials. Now let p(.$) = h(cr5) be a differential which vanishes on IT,. Then we have 17, Cn,-,,-, , i.e. c l 9 - l divides a. Hence aaD is an integral divisor, so cr E IT,-I,-I n k. Conversely for each a E ~ , - I , - , n k , h(a5) is a differential which vanishes on IT,. It follows that
+
+
We summarize the above results formally in
SECOND PROOF
Hence
We now recall Lemma 2 which shows that if n(l/aD) > 0, iae. if n(a) < 2 - 2g, then m (l/aD) = 0. This yields
+
If n(a) < 2 - 2g, then n(a) m(a) = I - g. This result is known as the Riemann part of the Riemann-Roch Theorem. Corollary:
3. Second Proof
The second proof starts from the formula (*) of section I. Again our first step is to define 2(a):
We have to prove i(a)is finite, and to this end we study the function ( a ) = - m a ) - ( a ) . I t is clear that r(a) is a class function, and if a 1 b then r(a) > ~(b). (The first remark is obvious, and the second follows from formula (*).) Lemma 3:
r(a) is bounded from above.
Proof: Let R = ko(x) be a rational subfield of k. Then Theorem 1: (Riemann-Roch). If a is any divisor in a field of
genus g, then
where 3 is a divisor of the canonical class. The theorem has some immediate consequences: Let a = 1, so that m(1) = 1, n(1) = 0, whence
deg (k/k,(x)) = n 0, -as,
+
+
Case 2: k' is the function field of a conic without rational points. Then k' = ko(xl , x,) where ax12 bxg2 .-.= 0 and a # 0, # b 0. In this case we may take b = pz,g-1)'2where p, is the infinite prime in ko(xl). Then ITb-1 contains the linearly independent functions 1, xl , x12 , ..., x : ~ - ~ ,) /x2 ~ , x2x1, -,x ~ x ~ ~. The -~)/~ rest of the development of Case 2 is left to the reader.
+
--
8. The Theorem of Clifford Lemma 1 : Let fl(xl , ..., x,), fm(xl , x,) be non zero polynomials in ko[xl , xr]. Then there exists an integer N depending only on r, m, and the degrees of the f, such that, if k, contains more than N elements, then there exist values xl = a,, x2 = a 2 , ..., X, = a, in k, such that - a * ,
..a,
and hence by the Riemann part of the Riemann-Rocb Theorem, n(a,) m(a,) = 1 - g, whence we have m(ai) = 2i - 1 g. When i = 1, m(ai) = g 1, and IT,' contains 1, x, x2, -.., xg, so we obtain no essentially new functions. When i = 2 however, we
+
+
+
+
as-,
318
XVI.
8.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
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 k,[x, ,x, , ...,x,-,I. Let the non zero coefficients of all x,-,). By induction hypothesis there thef, be denoted b y g , ( x , , exists an N such that if k , contains more than N elements we can achieve simultaneously gv(al , a,-,) 0 for all v, with elements ai E ko . Suppose k , contains enough elements for this result. Then consider the polynomials fi(al , ...,a,-, , x,). Clearly if k , contains a,) = 0 enough elements, we can choose x, = a , so that fi(al , m). This completes the proof. (i = 1, a*.,
THE THEOREM OF CLIFFORD
319
Proof: If q o - 1 , IIbo-1 are the smallest parallelotopes containing all elements of 17,-1 n k , IIb-1 n k respectively, then
Further,
+
whence
am-,
Let x, , x, , ..., x, be elements of a field k linearly independent over its constant field k , . Let 17, be the smallest parallelotope containing all the xi; a = Qvv. Let pl , ..., p, be a finite set of primes. Then if k , contains enough elements, there .-. c,x, (ci E k,) such that exists a linear combination x = clxl ordp, x = vPi (i = 1, m).
Consequently if the theorem is proved for a, and b, we have
Lemma 2:
+ +
me-,
Proof:
Suppose min ordprx,, = vpt = ordPi X3. . Ir
We write y , = x , / x j . Then ord, 0 and m(a/b) > 0, then
f =P1Pz ...P,
+ 1.
Then
f' is not identically zero since the first term is not divisible by p, and p,' # 0, and the remaining terms are all divisible by p, . Therefore f has an irreducible factor p,,, which is separable. Now let p(x) be a separable polynomial of R = k,(x). Then 321
322
XVII.
2.
DIFFERENTIALS IN FUNCTION FIELDS
I p(x) I < 1 and I p'(x) ,1
1. By Hensel's Lemma, the completion R*(p) contains a root x, of the equation p(t) = 0 and x = x, . The residue class field under I , 1 is kl = k,(x,), and x - x, 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 - x, with coefficients in k, : R*@) = kl{x - x,). =
Now let k be a finite extension of R and p a prime in k dividing p. Then if k*(p)/R*(p) is unramified, the residue class field ko under I ,1 is a separable extension of kl , and hence of KO. x - xo is still a prime element, and hence
2. Local Components of Differentials
LOCAL COMPONENTS OF DIFFERENTAILS
323
Now