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
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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
+ <
+
.
Theorem 3: Every valuation is equivalent to a valuation for which the triangular inequality holds.
Proof. (1) When the constant c = 2, we shall show that the triangular inequality holds for the valuation itself. Let ] a ]G j b j . Then
Letting n +co we obtain the desired result. We may note that, conversely, the triangular inequality implies that our third requirement is satisfied, and that we may choose c = 2.
( 2 ) When c > 2 , we may write c = 2". Then it is easily verified that ( 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
+
<
+
+
< +
<
For non-archimedean valuations, Ia+bI <max(laI,(bI).
I
<
Proof. Let 1 a 1 1 b 1; then I 1, whence 1 alb
+
<
1 a/b (
< 1.
It follows that
I a + b I < I b I =max(Ia
<
by the Theorem. By hypothesis, I b / is not / a I, so that we have 1 b I 1 a b I . But using the theorem again we have
< +
Thusif la1 < J b ( , t h e n l a + b I = I b / . We notice that this equality does not necessarily hold when l a / = I bI;forexample,ifb = - a , w e h a v e I a + b j = O < l a [ . In general we have
hence
i.e. I1
Theorem 5:
CLASSIFICATION OF VALUATIONS
I 1 + x / < llb. We conclude, therefore, that if I x I < 1, then + x I < max (lla, llb), which is axiom (3).
(2) The converse is obvious if we replace I / by the equivalent function for which the triangular inequality holds. It should be remarked that the field operations are continuous in the topology induced on k by a valuation. 3. Classification of Valuations
If the constant c of axiom (3) can be chosen to be 1, i.e. if 1 implies I 1 x 1 1, then the valuation is said to be non-archimedean. Otherwise the valuation is called archimedean. Obviously the valuations of an equivalence class are either all archimedean or all non-archimedean. For nonarchimedean valuations we obtain a sharpening of the triangular inequality:
Ix 1
<
+
<
if, for every v > 1, I a, I < I a, used in the following form:
I.
This last result is frequently
<
Corollary 5.2: Suppose it is known that 1 a, I 1 a, I for all v, and that 1 a, --. a, 1 < 1 a, 1. Then for some v > 1, I a v I = Ia1I. We now give a necessary and sufficient condition for a field to be non-archimedean:
+ +
Theorem 6: A valuation is non-archimedean if and only if the values of the rational integers are bounded.
1.
8
VALUATIONS OF A FIELD
Proof: (1) The necessity of the condition is obvious, for if the valuation is non-archimedean, then
For I a n - 0 1 (b) If ( a 1
4.
THE APPROXIMATION THEOREM
=
( a I n e O as n-+co.
< 1, then an
lim --
(2) T o prove the sufficiency of the condition we consider the equivalent valuation for which the triangular inequality is satisfied. Obviously the values of the integers are bounded in this valuation also; say I m I D. Consider
<
+Ia
< D(I a In
In-l
Ib (
4-
t
+ an-
n --+a 1
(c) If 1 a
O.
1 > 1, then
For
We now examine the possibility of finding a relation between nonequivalent valuations; we shall show that if the number of valuations considered is finite, no relation of a certain simple type is possible.
l ) @ a x ( l a I , IbI))n.
Hence \a+bI
Theorem 7: Let I I,, 1 1, be a finite number of inequivalent non-trivial valuations of k. Then there is an element a E k such that 1 a 1 , > 1, and 1 a l Y < 1 (V= 2, ...,n). a*.,
Letting n e c o , we have the desired result. Corollary: A valuation of a field of characteristic
p > 0 is
non-archimedean. We may remark that if k, is a subfield of k, then a valuation of k is (non-)archimedean on k, is (non-)archimedean on the whole of k. In particular, if the valuation is trivial on k, , it is nonarchimedean on k.
Proof. First let n = 2. Then since I ,1 and I 1 , are nonequivalent, there certainIy exist elements b, c E k such that I b 1 , < 1 and I b 1 , >, 1, while IcI, 2 1 and I c 1 , < 1 . Then a = c/b has the required properties. The proof now proceeds by induction. Suppose the theorem is true for n - 1 valuations; then there is an element b E k such that I b 1 , > 1, and I b I , < 1 (v = 2, n - I). Let c be an element such that I c 1 , > 1 and I c 1, < 1. We have two cases to consider: a*.,
4. The Approximation Theorem
Let {a,} be a sequence of elements of k; we say that a is the limit of this sequence with respect to the valuation if lim l a - a , \
n--+OO
=0.
The following examples will be immediately useful: (a) If 1 a (
< 1, then lim an = 0
n-+CO
.
I b 1,
< 1.
Case 2: ( b 1,
> 1.
Case I :
Consider the sequence a, = cbr. Then while I a,I, = I cI,I b I n r < 1; for sufficiently large r, ( a, 1, = I c 1 , ( b I v r < 1 (v = 2, ..., n - 1). Thus a, is a suitable element, and the theorem is proved in this case. ( a , / , = IcI, I b
Ilr > 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
<
<
(v = 2,
.-a,
Proof. If b is an element such that I b 1, > 1 and I b 1, < 1 (v = 2, ..., n - I), then a, = br/(l + br) satisfies our requirements for large enough values of r. (The Approximation Theorem): Let I I,, .-.,I 1, be a finite number of non-trivial inequivalent valuations. Given n), there exists an any E > 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.
<
Second Example. Let k be the field of rational functions over a field F: k = F(x). Then k is the quotient field of the ring of polynomials o = F[x]. Let I I be a valuation of k which is trivial on F ; I I will thus be non-archimedean. We have again two cases to consider. Case I : 1x1 > 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.
<
Case 2: 1 x I < 1. Then for any f(x) E o, I f(x) I 1. If we exclude the trivial valuation, we must have If(x) I < 1 for some f(x) E D. As in the first example, the set of all such polynomials is an ideal, generated by an irreducible polynomial p(x). If I P(X) I = c, and f(x) = (P(x)Ivg ( 4 , (P(x), g@)) = 1, then If(4 I = c v . Conversely, if p(x) is an irreducible polynomial, it defines a valuation of this type. This is shown in exactly the same way as in the first example. I n both cases, the field k = F(x) and the field k = R of the rational numbers, we have found equivalence classes of valuations, one to each prime p (in the case of F(x), one to each irreducible polynomial) with one exception, an equivalence class which does not come from a prime. T o remove this exception we introduce in both cases a "symbolic" prime, the so-called infinite prime, p, which we associate with the exceptional equivalence class. So 1 a lDm stands for the ordinary absolute value in the case k = R, and for cdega in the case k = F(x). We shall now make a definite choice of the constant c entering in the definition of the valuation associated with a prime p. (I) k = A. (a) p #pm . We choose c = l/p. If, therefore, a # 0, and a = pvb, where the numerator and denominator of b . exponent v are prime to p, then we write I a ID = ( 1 1 ~ ) ~The is called the ordinal of a at p and is denoted by v = ordp a. (b) p = p, . Then let 1 a , I denote the ordinary absolute value. (11) k = F(x). Select a fixed number d, 0 < d < 1. (a) p # p a , so that p is an irreducible polynomial; write c = ddegp. If a # 0 we write as in case I(a), a = pvb, v = ord,,a, and so we define I a ,1 = cv = ddegp.ordya.
1.
14
(b) p = p , , so that I a I = @pa. where c > 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<2"max(Ial,Ibl). By the method of Theorem 3, we can deduce
I a, +
+ a,
I < ( 2 ~ ) ~ m1aa,x I .
I t is easily shown that the null-sequences form a maximal ideal
since
k.
The valuation I I of k naturally induces a valuation on A; we still denote this valuation by I I. For if a E k is defined by the residue class of PIN containing the sequence (a,}, we define I ar I to be lim,,, I a, I. To justify this definition we must prove (a) that if {a,} is a Cauchy sequence, then so is
{I
a,
I},
1.
VALUATIONS OF A FIELD
(b) that if {a,) E {b,) mod N, then lim I a, I = lim I b, (c) that the valuation axioms are satisfied.
I,
The proofs of these statements are left to the reader. If a E k, let a' denote the equivalence class of Cauchy sequences containing (a, a, a, ..-); a' E k. If a' = b', then the sequence ((a - b), (a - b), EN, SO that a = b. Hence the mapping c$ of k into k defined by +(a) = a' is (1, 1); it is easily seen to be an isomorphism under which valuations are preserved: I a' I = I a I. Let k' = +(k); we shall now show that kt is everywhere dense in k. T o this end, let a be an element of k defined by the sequence {a,). We shall show that for large enough values of v, I a - a: I is as small as we please. The elemnt a - a: is defined by the Cauchy sequence {(a, - a,), (a, - a,), ...), and -.a)
Ia--a;I=
lim l a , - a , ) ;
P--XI
but since {a,} is a Cauchy sequence, this limit may be made as small as we please by taking v large enough. Finally we prove that k is complete. Let {a,) be a Cauchy sequence in k. Since kt is everywhere dense in k, we can find a sequence {a,') in k' such that I a,' - a, I < llv. This means that {(a,' - a,)) is a null-sequence in k; hence {a,') is a Cauchy sequence in k. Since absolute values are preserved under the mapping 4, {a,) is a Cauchy sequence in k. This defines an element B E k such = O . Hence lim,,,la,-PI = 0 , i.e. that limV,,Ia,'--/3I /3 = lim a,, and so k is complete. We now agree to identify the elements of k' with the corresponding elements of k; then k may be regarded as an extension of k. When k is the field of rational numbers, the completion under the ordinary absolute value ("the completion at the infinite prime") is the real number field; the completion under the valuation corresponding to a finite prime p ("the completion at p") is called the jield of p-adic numbers.
CHAPTER TWO
Complete Fields 1. Normed Linear Spaces
Let k be a field complete under the valuation I I, and let S be a finite-dimensional vector space over k, with basis w, , w, , ..., w, . Suppose S is normed; i.e. to each element a E S corresponds a real number I I a 11, which has the properties
(We shall later specialize S to be a finite extension field of k; the norm 11 11 will then be an extension of the valuation I I.) There are many possible norms for S ; for example, if
then
is a norm. This particular norm will be used in proving Theorem 1:
All norms induce the same topology in S.
Proof: The theorem is obviously true when the dimension n = 1. For then B = xw and 1 1 /3 1 1 = I x I 11 o 11 = c I x I (c # 0); hence any two norms can differ only in the constant factor c; this does not alter the topology. We may now proceed by induction; so we assume the theorem true for spaces of dimension up to n - 1. 19
20
2.
2.
COMPLETE FIELDS
We first contend that for any E > 0 there exists an 9 > 0 such that 11 a 1 1 < r ] implies I xn I < E . For if not, there is an E > 0 such that for every 9 > 0 we can find an element a with 1 1 a / I < 9, but / xn 1 2 E . Set /3 = a/xn; then
if v is large enough. Hence
i.e. I I y I I is smaller than any chosen 7: I I y In other words, z;Wl
Thus if we replace 9 by TE, we see that for every 9 > 0 we can find an element /3 of this form with / I /3 1 1 < 9. We may therefore form a sequence {flu}:
with 11 /3, 11
< llv. Then
21
EXTENSION OF THE VALUATION
+ + z,,-lwn-l +
wn
I( = 0, and so y
= 0.
= 0,
which contradicts the linear independence of the basis elements , --., wn . This completes the proof of our assertion. From this we deduce at once that: For any E > 0, there exists an r ] > 0 such that if I I a I I < r ] , then II a 1 , < E. Thus the topology induced by norm I I II is stronger than that induced by the special norm 1 I 1 , . Finally, since ol
II a l l G l I II I I + -..+ l xn I I l I1 G I1 llo(ll I1 + --.+ I1 II), Xl
Wl
Wl
wn
Wn
the topology induced by I I 1, is stronger than that induced by This completes the proof of our theorem.
and
By the induction hypothesis, the norm 1 1 1 1 on the (n - 1)induces the same topology dimensional subspace (w, , as the special norm I I 1 lo . That is,
II I I.
Corollary: Let @,I, /3, = Z X ~ be ( a~ Cauchy ) ~ ~sequence in S. Then the sequences (i = 1, ..., n) are Cauchy sequences ink, and conversely.
..a,
small if v, p are large (y("))is a Cauchy sequence in k. Since k is complete, there exist elements zi E k such that xi = lim {y:")
.
2. Extension of the Valuation
We now apply these results to the case of an extension field. Let k be a complete field, E a finite extension of k. Our task is to extend the valuation of k to E. Suppose for the moment we have carried out this extension; then the extended valuation I I on E is a norm of E considered as a vector space over k, and we have
v--to3
Set y = zlwl
Then
+ ... +
Z,-~W~-,
+ . wn
Theorem 2:
E is complete under the extended valuation.
Proof: Let @,I be a Cauchy sequence in E: /3, = Zxi(v)wc. By the corollary to Theorem 1, each is a Cauchy sequence in k, and so has a limit yi E k. Hence
22
2.
2.
COMPLETE FIELDS
Thus E is complete. Now let a be an element of E for which a I < 1; then I or Iv-+ 0; hence {av}is a null-sequence. Thus if
I
each sequence (x*(~)) is a null-sequence in k. The norm of a, N(a), is a homogeneous polynomial in x, Hence
Hence we have proved that1 a ( < 1 3 1 N(a) wecanobtain 1 a 1 > 1 + 1 N(a) 1 > 1. Thus
1 < 1;
T o establish this we must treat the archimedean and non-archimedean cases separately. Theorem 3: If k is complete under the valuation I I, then the valuation can be extended to E = k(i) where i2 1 = 0.
+
, ..-,x,
.
Proof: (1) If i E k, then E = k, and the proof is trivial. (2) If i $ k, then E consists of elements a = a bi, (a, b E k), N(a) = a2 b2. Hence we must show that
similarly
Now consider any a E E, and set @ = an/N(a) where = deg(E
+
+
<
or, equivalently, that ( a I D for some D. Suppose that this is not the case; then for some a = a ib, I 1 b2/a2I I l/a2 I is arbitrarily small. Thus I x2 1 I takes on arbitrarily small values. We construct a sequence {x,} in k such that 1 Ix:-k1Id- 2.4' '
+
n
23
EXTENSION OF THE VALUATION
I k).
Then
+
<
Then
I x:
-%:+I1
I
= (x;
+ 1)
-@;+I
+
+ 1) 1 < F1 ,
hence I /3 I = 1. Therefore
<
We have proved that if it is possible to find an extension of the valuation to E, then E is complete under the extension; and the extended valuation is given by I a I = 1;/1 N(a) I . Hence to establish the possibility of extending the valuation, it will be sufficient l N(a) I coincides with I a I for a E k, and to show that f(a) = d satisfies the valuation axioms for a E E. Certainly if a E k,
hence one factor 1/2v. We now adjust the signs of the (x,}; suppose this has been dome as far as x,; then we adjust the sign of 1/2v. This adjusted sequence x, in such a way that I xv - x, I is a Cauchy sequence, for
<
Since k is complete, this sequence has a limit j and using the properties of the norm N(a), we can easily verify that axioms (1) and (2) are satisfied. Thus it remains to prove that for some C,
I N(a) I < 1 =. I N(l
+ a) 1 < C .
+
j2 1 = lim v+m x,Z
contradicting our hypothesis that proof.
E k.
Then
+ 1 = 0,
a$ k. This completes the
24
2.
3.
COMPLETE FIELDS
ARCHIMEDEAN CASE
3. Archimedean Case
The following theorem now completes our investigation in the case of complete archimedean fields. Theorem 4: The only complete archimedean fields are the real numbers and the complex numbers.
Proof: Let k be a field complete under an archimedean valuation. Then k has characteristic zero, and so contains a subfield R isomorphic to the rational numbers. The only archimedean valuations of the rationals are those equivalent to the ordinary absolute value; so we may assume that the valuation of k induces the ordinary absolute value on R. Hence k contains the completion of R under this valuation, namely the real number field P ; thus E = k(z] contains the field of complex numbers P(zJ We shall prove that E is in fact itself the field of complex numbers. We can, and shall, in fact, prove rather more than this-namely, UA that any complete normed field over the complex numbers is itself the field of complex numbers. In a normed field, a function I I I I is defined for all elements of the field, with real values, satisfying the following conditions:
It is easily verified that L is a linear homogeneous functional, such that L(x) = L(l) = 0, and hence vanishing on every linear function. First we show that l l x is continuous for x = /3 # 0 (B E E). ll 4 1 1 II B-I II < 1. Let4 E Eand II 4 II < I1 B-' 11-l; then 11 518 11 Since 1 ( (f/,9)v1 1 11 4/15 1 1: the geometric series l/P Zr p/flv converges absolutely; hence it converges to an element of E which is easily seen to be 1/(p - 4). Now take I I 4 I I 1 1 8-I 11-l. We have
<
<
<
We shall show that any such field E = P(z]. The proof can be carried through by developing a theory of "complex integration" for E, similar to that for the complex numbers; the result follows by applying the analogue of Cauchy's Theorem. Here we avoid the use of the integral, by using approximating sums. Given a square (x, ,x, , z, ,z,) in the complex plane, having c as center and t , , t,, t3,t4 as midpoints of the sides, we define an operator Lo by
so that
which can be made as small as we please by choosing I ( 5' I I small enough. This is precisely the condition that l l x be continuous at X =
p.
26
2.
3.
COMPLETE FIELDS
Suppose there exists an element a of E which is not a complex number; then l / ( z - a) is continuous for every complex number x, and since
27
ARCHIMEDEAN CASE
is an approximating sum to the "integral" of f(x) taken round the contour formed by the sides of Q. We see easily that
Using the estimate for L Q v [ l / (z a)] we find the function l / ( z - a) approaches zero as I x I +co
. The function so that for a fixed Q and n -tco we have
is continuous for every z on the Gauss sphere, and hence is bounded: say 11 l / ( x - a) 1 1 < M. We have now 1 LQ
=L~
Since
- z + (2c - a) ( 2 - c)" ( C - a)z -I- ( z - a) ( c - a)z
1 1 ---=-.z-a z
1
a
1-a/~'
z2
we obtain
since LQ vanishes on linear functions. Thus
Therefore
2
1
~ ~ L ~ - ) - L ~ + ) ~ ~l G ~v+1-~vl7A contour
2A
G , where 6 is the length of the side of the square. Next consider a large square Q in the complex plane, with the origin as center. If the length of the side of Q is 1, we can subdivide Q into n2 squares Q , of side l/n. Let us denote by Zv the vertices, and by 3 , the mid-points of the sides of the smaller squares, which lie on the sides of the large square Q. Then
Lb
=
2
contour
- ZJf(3V)
2
contour
I3vI
lzv+l-sl=-
8A I '
Now
<
as n -too. Hence 2 ~ r 8A/l, which is certainly false for large I. Thus there are no elements of E which are not complex numbers; so our theorem is proved. We see that the only archimedean fields are the algebraic number fields under the ordinary absolute value, since only these fields have the real or complex numbers as their completion.
28
2.
4.
COMPLETE FIELDS
4. The Non-Archimedean Case
We now go on to examine the non-archimedean case. Let k be a complete non-archimedean field, and consider the polynomial ring k[x]. There are many ways of extending the valuation of k to this ring, some of them very unpleasant. We shall be interested in the following type of extension: let I x I = c > 0, and if +(x) = a,
+ alx + + a,xn E k[x],
THE NON-ARCHIMEDEAN CASE
29
We now prove the classical result, known as Hensel's Lemma, which allows us, under certain conditions, to refine an approximate factorization of a polynomial to a precise factorization. Theorem 5: (Hensel's Lemma). Let f(x) be a polynomial in k[x]. If (1) there exist polynomials $(x), $(x), h(x) such that
(2) $(x) # 0 and has absolute value equal to that of its highest term,
define
I+(x) I = max I upv I = rnax cV I a,, 1. v v
(3) there exist polynomials A(x), B(x), C(x) and an element d E k such that
The axioms (1) and (3) of Chapter I can be verified immediately. T o verify axiom (2) we notice first that
since
ICS.
a,b, xk
I
< rnax ( a,x"
max I bjxj 1
.
then we can construct polynomials @(x), Y(x)
E k[x]
such that
+
Next we write $(x) = $,(x) $,(x) where $,(x) is the sum of all the terms of $(x) having maximal valuation; I $2(x) I < I 4 1 ( ~ )I . Similarly we write $(x) = $,(x) f $,(x). Then
Proof: As a preliminary step, consider the process of dividing a polynomial We see at once that the last three products are smaller in valuation than I$,(x) I I $,(x) 1; and of course
The term of highest degree in $,(x) $,(x) is the product of the highest terms in $,(x) and in &(x); so its value is I $,(x) I I &(x) I. Therefore I $,(x) $,(x) I = I$l(x) I I ICI1(x) I. Using the non-archimedean property we obtain
+ b,x + + bmxm
+(x) = a,
+ alx + ... + a,xn,
by
where by hypothesis, I $(x) I = I anxn I. The first stage in the division process consists in writing
Now Thus we have, in fact, defined an extended valuation.
g(x) = b,
30
2.
4.
COMPLETE FIELDS
<
hence we have I gl(x) / I g ( x ) I. We repeat this argument at each stage of the division process; finally, if g ( x ) = q(x) +(x) f r(x), we obtain I r ( x ) I 1 g ( x ) I . As another preliminary we make a deduction from the relation
<
THE NON-ARCHIMEDEAN CASE
31
Referring now to our preliminary remark about division processes, we have
We see that this yields Hence using the given bounds for I B ( x ) I, I $(x) I, I C ( x ) 1. Since I d 1. we have I A ( x ) +(x) I Now multiply the relation (*) by h(x)/d; we obtain
<
I
< 1,
and
Now we write
B(x) d
= q(x)Q(x)
+ PI(-.), Now we define
Then
Then since I &(x)
I < I +(x) I,
I +1(x) I
< deg +(x),
and deg Bl(x)
I
= IQ(4 = I a s n
I
we have
9
and
We now give estimates of the degrees and absolute values of the polynomials we have introduced. We have, immediately, deg &(x), deg k ( x ) < deg +(x). Further,
We shall show first that +,(x) $,(x) is a better approximation to f(x) than is +(x) $(x); and then that the process by which we obtained
32
2.
4. THE NON-ARCHIMEDEAN CASE
COMPLETE FIELDS
41(~),i,bl(x) may be repeated indefinitely to obtain approximations which are increasingly accurate. Let f ( 4 - A(x) *1(4 = h(x). Then
33
We have now recovered the original conditions of the theorem, stated now for #,(x), +,(x), hl(x) and Cl(x). Thus the whole process may be repeated, yielding new polynomials a2(x), B2(x), h,(x), + 2 ( ~ ) i,b2(x). , We shall also obtain the estimate
I h&) I < K l I hl(4 I
9
where Hence
KI
= max
It is clear that we may now proceed indefinitely, obtaining sequences of polynomials {a&)), {&(x)), {hv(x)),{4v(x)),{i,bv(x))where and where we write Now x < 1 by the conditions of the theorem. Thus $,(x) #,(x) is a better approximation than $(x) i,b(x). Let A(x) +1(4 B(x) #1(4 = d C1(x).
+
Then Cl(4 = C(x)
as v -+aand
+
+ 4 4 Bl(4 + B ( 4 g(x).
Hence
as v -+a;
and
as v+m. From these considerations it follows that {$,(x)) and {#,(x)) are Cauchy sequences of polynomials; these polynomials are of bounded degree, since for every v we have using the results obtained above.
34
2.
4.
COMPLETE FIELDS
and deg ($,(x) *,(x))
< m=
@(x)= lim {$,(x))
and
{degf(x), deg h(x)}.
THE NON-ARCHIMEDEAN CASE
35
(3) there exist polynomials A(x), B(x), C(x) and an element d E k such that
Let Y(x) = lim {+,(x));
these limit functions are polynomials, and deg @(x)= deg $(x). Finally, f (x) - O(X)Y(x) = lim {hv(x)}= 0; thus f(x) = @(x)Y(x). Now we have only to notice that
and deg @(x)= deg $(x). f (x) = @(x)Y(x) For the remainder of this section we restrict ourselves to this special valuation and this form of Hensel's Lemma. We digress for a moment from our main task to give two simple illustrations of the use of Hensel's Lemma. Example I : Let a = 1 mod 8 (a is a rational number). We shall show that a is a dyadic square, i.e. that x2 - a can be factored in the field of 2-adic numbers:
Further (x t This completes the proof of Hensel's Lemma. For our present purpose of proving that a non-archimedean valuation can be extended, we use the special valuation of k[x] induced by taking I x I = 1, i.e. the valuation given by
Using this special valuation, Hensel's Lemma takes the following form:
We have Thus the conditions of Hensel's lemma are satisfied, and our assertion is proved. We shall see later that this implies that in R(+), the ideal (2) splits into two distinct factors. Exemple 2: Let a be a quadratic residue modulo p, where p is an odd prime; i.e. a r b2 mod p, where (b,p) = 1. Then we shall show that a is a square in the p-adic numbers. We have: x2 - a = (x - b) (x
Let f(x) be a polynomial in k[x], and let the valuation in k[x] be the special valuation just described. If Theorem 5a:
(1) there exist polynomials 4(x), $(x), h(x) such that
f ( 4 = $(x) *(x) + h(x),
+ b) + (b2- a);
h(x) = b2 - a;
and We have I C(x) I = 0 < 1 d 1, and I h(x) 1 ,< lp 1 < 1 d l2 = 1 since (p, d) = 1. The conditions are again satisfied, so our assertion is proved.
2.
36
COMPLETE FIELDS
Hensel's Lemma can now be applied to give us a Reducibility Criterion. Theorem 6:
Consider
where a, #O. If I f(x) 1 = 1, I an 1 i > 0, then f(x) is reducible. Proof:
< 1,
but
I ai I
=
1 for some
it will be sufficient to prove the assertion for Nk(cc),k(a). Let f(x) = Irr (a, k, x), the irreducible monic polynomial in k[x] of which a is a root, be a, a,-,x -.. xn (ai E k). Since a, = fN(a), we see that
+
+ +
-.
&
using the corollary to Hensel's Lemma. Now f(x - 1) = Irr (a
Let i be the maximal index for which ( ai ( = 1. Set
+ 1, k, x),
then By assumption there is at least one term (anxn) in h(x), and I h(x) 1 < 1. Further,
whence
I N(1 + I < 1. ff)
This completes the proof of Hence all the conditions of Hensel's Lemma are satisfied, and we have a factorization f(x) = g,(x)g,(x), where deg (g,(x)) = i. Corollary: If ] f(x) 1 = 1, and f(x) is irreducible, with I a, then I ai 1 < 1 for all i > 0. Consider now
1 < 1,
Theorem 7: Let k be a field complete under:a non-archimedean valuation I 1; let E be an extension of k of degree n. Then there is a unique extension of I I to E defined by
E is complete in this extended valuation. this is irreducible if f(x) is irreducible. By the corollary just stated, if I a, 1 < 1, then / a$ / < 1 for i < n. Hence if 1 f(x) 1 = 1 and f(x) max (I a, I , I a, I), is irreducible, then for 1 i n - 1 , I ai I and the equality sign can hold only if j a, I = or, I. This enables us to complete the proof that we can extend the valuation. Let us recall that all that remains to be proved is that if or E E, then
< <
<
5. Newton's Polygon Let k be a complete non-archimedean field, and consider the polynomial f(x) = a, alx .-. anxnE k[x].
+
+ +
Let y be an element of some extension field of k, and suppose 1 y I is known. Let c be a fixed number > 1, and define ord or = - log, 1 or I when or # 0; when a = 0 we write ord a = +a.We shall now show how to estimate I f(y) I. We map the term a,xv of f(x) on the point (v, ord a,) in the Cartesian plane; we call the set of points so obtained the Newton
38
2.
COMPLETE FIELDS
diagram. The convex closure of the Newton diagram is the Newton polygon of the polynomial (see fig. 1) Now ord a,yv = ord av v ord y.
+
Thus the point in the Newton diagram corresponding to a# lies on the straight line 1,; y x ord y = ord avyu
+
with slope - ord y. Now
o the
intercept cut off on the y-axis by lV1is less than that cut off by lVz. Thus if I aNyNI = max, I a,yv 1, then 1, cuts off the minimum intercept on the y-axis; thus 1, is the lower line of support of the Newton polygon with slope - ord y. Thus to find the maximum absolute value of the terms a,yv we draw this line of support, and measure its intercept r] on the y-axis. Then max I avyv1 = c-q. If only one vertex of the polygon lies on the line of support, then only one term avyvattains the maximum absolute value; hence we have If(y) ( = max I avyvI = c-". If, on the other hand, the line of support contains more than one vertex (in which case it is a side of the polygon), then there are several maximal terms, and all we can say is that
(See figure 2.) We shall now find the absolute values of the roots of
Let y be a root of f(x): f(y) = 0.Now 0 = I f(y)
FIG.1. The absence of a point of the diagram for x in xa is missing; i.e. ord a, = m .
=
2 means that the term
I < rnax ( avyuI ,and
if there is only one term with maximum absolute value. Hence if y is a root there must be at least two terms avyvwith maximum absolute value. The points in the Newton diagram corresponding to these terms must therefore lie on the line of support of the Newton polygon with slope - ord y. Hence the points are vertices of the Newton polygon and the line of support is a side. Hence we have established the preliminary result that if y is a root of f(x) then ord y must be the slope of one of the sides of the Newton polygon of f(x).
2.
COMPLETE FIELDS
Let the last vertex on I be that which corresponds to aixi. We define ~(x)=ao+a,x+--.+a,xi; f(x) - $(x) $(x)
= h(x) = ~
+(x)=l;
+ + a,xn.
,+~x~+l
Then
A(x) = C(x) = 0,
and
B(x) = d
=
1.
The conditions of Hensel's Lemma are satisfied since
to prove the last statement we have only to notice that
FIG.2. Th: dotted lines have slope -ord f i = -ord y, = 0.
4;
the solid lines have slope
since we included in $ ( x ) all the terms with maximum absolute value; finally, $(x) has absolute value equal to that of its highest term aixi. Hensel's Lemma yields an exact factorization f ( x ) = $,(x) +,(x) where $,(x) is a polynomial of degree i; hence +,(x) is a polynomial of degree n - i. From the last part of Hensel's Lemma we have
and also Let us now consider one of the sides of the polygon, say I; let its slope be -p. We introduce the valuation of k[x] induced by setting I x I = C-p, i.e.
1 f(x) I = max I a , , ~ - ~I .' The vertices of the polygon which lie on I correspond to terms a,xv which have the maximal absolute value in this valuation.
I*(x)
-*o(x)I < Id1 = 1Thus +,(x) is dominated by its constant term, 1. We notice that if a polynomial is irreducible its Newton polygon must be a straight line; this condition, however, is not sufficient. Let us now examine the roots of $,(x), which are, of course, also roots of f(x). Since
42
2.
COMPLETE FIELDS
the Newton polygon of +,(x) cannot lie below the side of the Newton polygon of f(x) which we are considering; and since, by Hensel's Lemma, +,(x) and +(x) have the same highest term, the Newton polygon of 4,(x) must terminate at the point representing a,xi (see figure 3). By the same reasoning as was used above we find that if y' is a root of +,(x) then - ord y' is the slope of one of the sides of the Newton polygon of +,(x). All these sides have slopes not greater than the slope of 1. Hence if y' is a root of +,(x), ord y' 2 p. We examine also the roots of #,(x). Since +,(x) is dominated by its constant term in the valuation induced by 1, its Newton polygon has its first vertex at the origin. The origin is the only vertex of the polygon on the line of support with slope - p, and all the sides
FIG. 4.
The Newton Polygon of $,(x) must be situated like ODE.
of the polygon have a greater slope than that of 1. Hence, by the same arguments as before, if y" is a root of #,(x), ord y" < p. Now let the sides of the Newton polygon of f(x) be I,, I,, .--, with slopes - p, , - p, , (pl > p, > Suppose 1, joins the points of the Newton diagram corresponding to the (iv-,)-th and (iv)-th terms of f(x). Then we have just seen how to construct polynomials +,(x) of degree iv, whose roots are all the roots yp(v) of f(x) for which ord yp(v))Ip, . Obviously ord yp(,) > pv+l , SO that yp(")is also a root of +,+,(x); hence +,(x) divides +,,(x). We see also that the roots y of f(x) for which ord y = pv+, are those which are roots of +,,(x) but not of +,(x). --a).
6. The Algebraic Closure of a Complete Field
FIG. 3. The chosen side 1 is the third side in fig. 1. The Newton Polygon of 4,(x) must:be situated like ABCD.
Let k be a complete non-archimedean field, and let C be its algebraic closure. Then we extend the valuation of k to C by defining I a I for a E C to be I a I as defined previously in the finite
44
2.
COMPLETE FIELDS
extension k(a). The verification that this is in fact a valuation for C is left to the reader; it should be remarked that the verification is actually carried out in subfields of K which are finite extensions of k. C is not necessarily complete under this extended valuation. For instance the algebraic closure of the field of dyadic numbers does not contain the element
The valuation induces a metric in C, and since the valuation is non-archimedean, we have the stronger form of the triangular max {I a 1, 1 /3 I). Spaces in which this inequality: I a - /3 1 inequality holds are called by Krasner ultrametric spaces. Consider a triangle in such a space, with sides a, b, c. Let a = max (a, b, c); then, since a max (b, c) = b, say, we have a = b, and c a. Thus every triangle is hosceles, and has its base at most equal to the equal sides. The geometry of circles in such spaces is also rather unusual. For example, if we define a circle of center a and radius r to consist of those points x such that I x - a I < r, it is easily seen that every point inside the circle is a center. We now use this ultrametric geometry to prove:
<
<
<
Theorem 8:
Let a E C be separable over k, and let r = min I u(a) - CL ( , a# 1
where the a are the isomorphic maps of k(a). Let /3 be a point, i.e. an element of K, inside the circle with center a and radius r. Then k(a) C k(/3). Proof:
Take k(/3) as the new ground field; then f(x) = Irr (a,k, x)
is separable over k(/3). If $(x) = Irr (a, k(/3), x), then $(x) I f(x). Let o be any isomorphic map of k(a, /3) I k(/3). Since
and conjugate elements have the same absolute value (they have the same norm) we deduce that 1 /3 - o(a) 1 = I - a 1 < r. Consider the triangle formed by /3, a, u(a); using the ultrametric
<
I /3 - a I < r, i.e. o(a) property we have as above I a - @(a)I lies inside the circle. Hence o(a) = a, and so, since a is separable, the degree [k(a, /3) : k(@] is equal to 1, and k(a) C k(/3). Let f(x) be a polynomial in k[x] with highest coefficient 1. In C[x], we have f(x) = (x - a,) (x - a,). Suppose that in the valuation of k[x] induced by taking I x I = 1, we have I f(x) I A, where A 1. Then if a E C, and I a I > A, we see that an is the dominant term in
<
f(a)
= Clr.
+ upn-I + ... + a,.
<
Hence a cannot be a root of f(x) = 0. Hence if I f(x) I A and , ---, an are the roots of f(x) = 0, then I a, I A. Consider now two monic polynomials f(x), g(x) of the same degree, n, such that I f(x) - g(x) 1 < E. Let /3 be a root of g(x), a, , , an the roots of f(x). Then
<
a,
+ - -
where A is the upper bound of the absolute values of the coefficients, and hence of the roots, of f(x) and g(x). Hence
and so one of the roots ai , say a, must satisfy the relation
IB-.,I
Thus by suitable choice of E, each root /3 of g(x) may be brought as close as we wish to some root ol, of f(x). Similarly by interchanging the roles of f(x) and g(x), we may bring each root ai of f(x) as close as we wish to some root of g(x). Let us now assume that E has been chosen such that every /3 is closer to some a( than min / ai - O L ~1; ai # O I ~ in this way the /3's are split into sets "belonging" to the various q . Suppose for the moment that f(x) is irreducible and separable. Then, since I /3 - ai I < min I ai - olj I, the preceding theorem gives k(/3) 3 k(a); but f(x) and g(x) are of the same degree, whence Theorem 9: If f(x) is a separable, irreducible monic polynomial of degree n, and if g(x) is any monic polynomial of degree n such that I f(x) - g(x) I is sufficiently small, then f(x) and g(x)
46
2.
generate the same field and g(x) is also irreducible and separable. If f(x) is not separable, let its factorization in C[x] be
with distinct % . I n this case we can establish the following result: Theorem 10: If g(x) is sufficiently close to f(x), then the number of roots ,tId of g(x) (counted in their multiplicity) which belong to ffl 1s Vl .
Proof: If the theorem is false, we can construct a Cauchy sequence of polynomials with f(x) as limit for which we do not have vc roots near oli . From this sequence we can extract a subsequence of polynomials for which we have exactly pi roots near cq (pi f v$ for some i). Since k is complete, the limit of this sequence of polynomials is f(x), and the limits of the sets of roots near ai are the ari . Hence we have (X
- alp (X - a
p
...
(X
7.
COMPLETE FIELDS
- aJr
= f((x) = (x
- al)"l (x - ~ r ~(x -) a,)"' ~ .~
This contradicts the unique factorization in C[x], so our theorem is proved. In a similar manner we can prove the result Theorem 11: If f(x) is irreducible, then any polynomial sufficiently close to f(x) is also irreducible.
Proof: If the theorem is false, we can construct a Cauchy sequence of reducible polynomials, with f(x) as limit. From this sequence we can extract a subsequence {g,(x)): g,(x) = h,(x) m,(x) for which the polynomials h,(x) have the same degree, and have their roots in the same proximity to the roots of f(x). Then the sequence {h,(x)) tends to a limit in k[x], whose roots are the roots of f(x). This contradicts the irreducibility of f(x). Now although k is complete, its algebraic closure C need not be complete; the completion of C, c , is of course complete, but we can prove more:
Theorem 12:
CONVERGENT POWER SERIES
47
f?is algebraically closed.
Proof: We must consider the separable and inseparable polynomials of C[x] separately. (1) Let f(x) be a separable irreducible polynomial in c[x]. The valuation of k can be extended to a valuation for the roots of f(x). We can then approximate f(x) by a polynomial g(x) in C[x] sufficiently closely for the roots of f(x) and g(x) to generate the same field over f?. But g(x) does not generate any extension of I?; hence the roots of f(x) lie in C. (2) If the characteristic of k is zero, there are no inseparable polynomials in c[x]. Let the characteristic be p # 0; if ar is an element of C, a is defined by a Cauchy sequence {a,) in C. But if {a,) is a Cauchy sequence, so is {a,llp), and this sequence defines Hence there are no proper inseparable orllp, which is therefore in extensions of
c.
c.
7. Convergent Power Series
Let k be a complete field with a non-trivial valuation / I. Convergence of series is defined in k in the natural way: Zzm a, is said to converge to the sum a if for every given E > 0 we have I Zil, a, - a I < E for all sufficiently large n. The properties of the ordinary absolute value which are used in the discussion of real or complex series are shared by all valuations I I . Hence the arguments of the classical theory may be applied unchanged to the case of series in k. In particular we can prove the Cauchy criterion for convergence, and its corollaries:
1. The terms of a convergent series are bounded in absolute value. 2. If a series is absolutely convergent (i.e. if Emm I av 1 converges in the reals) then the series is convergent. Let E be the field of all formal power series with coefficients in k: E consists of all formal expressions f(x) = Z z a,xv, with a, E k, where only a finite number of the terms with negative index v are non-zero. We define a valuation I , 1 in E such that I x , 1 < 1, and I , 1 is trivial on k. If an element is written in the form E: a,xv This with avo# 0, then 1 E t a,xv, 1 = 1 xvo 1, = (1 x I,).o.
48
2.
7.
COMPLETE FIELDS
valuation ( I, on E and the valuation I I on k are therefore totally unrelated. An element f(x) = I:avxvin E is said to be convergent for the value x = c E k (c # 0), when E a c Vconverges in k. We shall say simply that f(x) is convergent if it is convergent for some c # 0. Theorem 13: If f(x) is convergent for x = c # 0, then f ( x ) is convergent also for x = d E k, whenever I d 1 < I c 1.
Proof: Let I c 1 = lla. Since I:a,,cv is convergent, we have l a v l <Mav. If [ d l < I c l , then
1 a,,cvI
< M,
whence
Since a I d I < I, the geometric series E M(a I d I)vis convergent. Thus E I a,& ( is convtirgent; since absolute convergence implies convergence, this proves the theorem. Now let F be the set of all formal power series with coefficients in k which converge for some value of x E k (x # 0). It is easy to show that F is a subfield of E. Notice, however, that F is not complete in I ,1 . Theorem 14: F is algebraically closed in E.
Proof: Let 8 = I:zmafiv be an element of E which is algebraic over F. We have to prove that 8 lies in F. I t will be sufficient to consider separable elements 8. For if 8 is inseparable and p # 0 is the characteristic, then 8p' is separable for high enough values of r. But 8pT= I:> 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.
<
where at least one b,, # 0. If we draw the Newton diagram of f(y), it is obvious from the form of the coefficients that none of the points in the diagram lie below the x-axis, while at least one of the points lies on the x-axis; further, since f(y) is irreducible, its Newton polygon starts on the y-axis. The first side of the Newton polygon of f(y) corresponds to the roots y of f(y) for which ord y is greatest, and it has slope - ord y. Only 8, = 8 has the maximum ordinal; hence the first side of the polygon joins the first and second points in the Newton diagram, and its slope is - ord 8, < 0. The other sides of the polygon correspond to the remaining roots 8,, and have slopes - ord Bi > 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
<
2.
52
COMPLETE FIELDS
shown to be positive and / a, / < al;hence we assume a,. 2 0 and I a, I < a, for i m. Since 4, satisfies the equation (2), ~ t coeffis cients satisfy
<
am+l
= $rn(al,
.", a m ; Mav).
CHAPTER THREE
Since only positive signs occur in $ m , we have a,,,,, a,,, we already have expression (1) and hence
2 0.
Again using the fact that only positive signs occur in #m
, we have
For
e, f and n 1. The Ramification and Residue Class Degree
<
Hence I a , I a,, where the a, are coefficients of a convergent power series. Thus 8 = E a 3 ' is convergent. This completes the proof of the theorem.
The value group of a field under a valuation is the group of non-zero real numbers which occur as values of the field elements. From now on, unless specific mention is made to the contrary, we shall be dealing with non-archimedean valuations. For these we have: Theorem 1:
then
If k is a non-archimedean field,
& its completion,
A has the same value group as k.
Proof: Let a a non-zero element of &; a is defined by a Cauchy sequence {a,} of elements of k , and I a I = lim I a , I. The sequence { I a, I } converges in a quite trivial way: All its terms eventually become equal to I or I. We have
since I a I # 0 and I a,, - a I can be made as small as we please, in particular less than I a I by choosing p large enough. Thus I a I = I a, 1 for large enough p. This proves the theorem. Now let k be a non-archimedean field, not necessarily complete; and let E be a finite extension of k . If we can extend the valuation of k to E, we may consider the value group BE of E. Then the value group B, of k is a subgroup, and we call the index e = (BE : B,) the ramiJication of this extended valuation. Consider the set of elements a E k such that I a I 5 1. I t is easily shown that this set is a ring; we shall call it the ring of integers and denote it by o. The set of elements a E o such that 1 a I < 1 53
1. forms a maximal ideal p of o; the proof that p is an ideal is obvious. T o prove it is maximal, suppose that there is an ideal a # p such that p C a C o. Then there exists an element a E a which is not in p; that is I a 1 = 1. Hence if /3 E 0, /3/a E o also, and = a (/3/a) E a. That is, a = o. Hence p is a maximal ideal and the residue class ring o/p is a field, which we denote by li. If & is the completion of k, we can construct a ring of integers 5, a maximal ideal p, and a residue class ring 5/ii = fi. We have the result expressed by: Theorem 2:
There is a natural isomorphism between ?t and
Proof: We have seen that if a
IaI
E 6,
then a
=
fi.
55
Let w1 , w , , -.., or be representatives of residue classes of E which are linearly independent with respect to K ; that is, if there exist elements c, , c, , c, in a such that c1wl c,w, lies in Q, then all the ci lie in p. Consider the linear combination c,w, ... c,w, , with ci E o; if one of the ci , say c, , lies in o but not in p, then c,w, -.. c,w, 0 mod $'3, and hence
+ +
+ +
+ +
+
+ +
... d p , , with Now consider the linear combination dlwl di E k, and suppose dl has the largest absolute value among the di . Then
lim a, E o and
I a, I for
v large enough. Thus 5 is the limit of elements of o; i.e. 6 is the closure of 0. Similarly p is the closure of p. Consider the mapping of o/p into ii/p given by a p -+ a 6. =
THE RAMIFICATION AND RESIDUE CLASS DEGREE
+
+
This mapping is certainly well-delined, and is an "onto" mapping since, given any a E 5, we can find an a E o such that I a - a I < 1; then a f~= a @,which is the image of a p. The mapping b mod j3 we have I a - b I < 1; is (1, 1) since if a, b E o and a hence a r b mod p. It is easily verified that the mapping is homomorphic; hence our theorem is completed. From now on we shall identify the residue class fields K, fZ under this isomorphism. Now let E be an extension field of k, with ring of integers L) and prime ideal p. Then L) 3 o = L) n k and Q 3 p = Q n k.
+
-
+
+
Hence if wl then
,
..a,
w,
I d,wl
are linearly independent with respect to K,
+ + d,w, / = max I d, ( . v
Let now n1 , n 2 , ns be elements of E such that I n, I, -.., I .rr, I are representatives of different cosets of We shall use these wi , 9 to prove the important result of Theorem 4:
ef
< n.
Proof:
If E is a finite extension of k of degree n, then
Our first contention is that
Theorem 3: There is a natural isomorphism of the residue class field ?t onto a subfield of the residue class field E.
Proof: Consider the mapping a is well-defined since a p 3 =a
+ +
homomorphism of image set, for
+ p -+a f Q ( a k). This + Q; it is easily seen to be a E
Certainly
6 into E. Finally, the mapping is (I, 1) onto the
We shall now identify ?t with its image under this isomorphism and so consider E as an extension field of A. We denote the degree of this extension by [E : K] = f.
and since the I ni 1 represent different cosets of I Z, c,w,ri I are all different. Hence
as required.
the
2. I t follows that the elements w , ~ , are linearly independent, since
<
deg ( E / k). Thus if deg (E I k) = n is finite, then e Hence rs and f are also finite, and n. The equality ef = n will be proved in the next section for a very important subclass of the complete fields, which includes the cases of algebraic number theory and function theory. Nevertheless, the equality does not always hold even for complete fields, as is shown by the following example. Let R, be the field of 2-adic numbers, and let k be the completion of the field R,(& Y2, $9, Then deg (A(-) : k) = 2, but the corresponding e and f are both 1. This is caused by the fact that we may write
<
.a+).
THE DISCRETE CASE
57
I a/+ I = 1 = I r v / aI; hence a/.rrV and its inverse are both elements of D,i.e. a / r Vis a unit E of o. For every a # 0 we have a factorization a = rVe,where I E I = 1; hence there is only one prime, namely T. Since k and its completion & have the same value group, the same element T can be taken as prime for &; thus if a complete field & is the completion of a subfield k, the prime 'for & may be chosen in k. Now let k be a complete field under a discrete valuation. If a E k can be written as a = rive, we shall take ord a = v; in other words, we select c = 111 T I. We now suppose that for every positive and negative ordinal v an element n, has been selected would suffice, but we such that I n, I = I T Iv-obviously T, = rV shall find it useful to use other elements. Suppose further that for each element in h we have selected a representative c in o, the representative of the zero residue class being zero. Then we prove
This series is not convergent, but if we denote by s, the sum of the first n terms, and by ui the i-th term, we have
Let k be a complete field with discrete valuation. Every a E k can be written in the form a = C : c , ~ , , where n = ord a and c, f 0 mod p.
2. The Discrete Case
Proof: When a = 0, there is nothing to prove, so we suppose 0. Since ord a = n, we have I a I = I T, I, hence a/r, is a unit E. Its residue class modulo p is represented by c, E E mod p . Thus I e - c n I < 1, whence I E T , - C , . ~ ~ ,
Theorem 5:
a #
We have already introduced the function ord a = - log, I a I, c > 1. The set of values taken up by ord a for a # 0, a E k forms an additive group of real numbers; such a group can consist only of numbers which are everywhere dense on the real axis or which are situated at equal distances from each other on the axis. Hence we see that the value group, which is a multiplicative group of positive real numbers, must be either everywhere dense, or else an infinite cyclic group. In this latter case the valuation is said to be discrete. I n the discrete case, let T be an element of k such that I T I takes the maximal value < 1. Then the value group consists of the numbers I T Iv;given any non-zero element a E k, there is a positive or negative integer (or zero) v such that I OI 1 = 1 T 1'. Then
+ 0 as m +a. This proves the where I a ( m ) I < I rmI; thus a(m) theorem. If the representatives ci and the T, are chosen once for all, then the series representation of a is unique. It is important to notice that the c, and T, are chosen from the same field k; thus the characteristic of the field containing the c, is the characteristic of k, not of h. We shall illustrate this remark by considering the valuation induced on the rational field R by a finite prime p. Any element a E R can be written as a = pu(blc) where b and c are prime to p; the ring of integers o consists of
3. e, f
58
AND
2.
n
these elements a for which v 2 0, and the prime ideal p consists of the a such that v > 0. Let R, be the completion of R under this valuation, 5, 6 the corresponding ring and prime ideal. Since a/+? = o/p, we can choose representatives for the residue classes in R, and in fact in o. The residue classes are represented by the elements b/c E R where b and c are prime t o p ; we can solve the congruence ca = 1 modp in integers. Then bd is a representative of the residue class containing blc, since
59
THE DISCRETE CASE
+ +
+
class is represented by c p , c202 .-- cfwf, with c4 E o. Then if a is an element of E, a can be written
Thus every element a E E can be expressed as a linear combination of the ef elements w p with coefficients in the ground field k. Thus the degree of the extension E I k is at most ef: n ef. We have already seen that for all extensions, whether the valuation is discrete or not, n 2 ef. Hence n = ef and our theorem is proved.
<
Hence every residue class contains an integer, and a complete set of representatives is given by 0, 1,2, --.,p - 1. This set of elements does not form a field since it is not closed under the field operations of R. Now every p-adic number a can be written a = E:cqv; we see that the field of p-adic numbers, R, , has characteristic zero. It can be shown that the field R, contains the (p - 1)-th roots of unity; it is often convenient to choose these as representatives of the residue classes. This analysis of complete fields may also be used to prove that the field of formal power series k = F{x) over any field F is complete. Any element a E k can be written as a = Er c,xV, c,, EF. The ring of integers o, respectively the prime ideal p are made up of the elements a for which n 2 0, respectively n > 0. We can choose x = .rr; and as representatives of o/p the elements of F; hence the completion of k consists of power series in x with coefficients in F; i.e. k is complete. Now let k be a complete field with discrete valuation; let E be a finite extension with degree n and ramification e. The finiteness of e shows that the extended valuation is discrete on E. Let 17, .rr be primes in E, k respectively; then the value groups are B E = {I 17lV}, Bk = { I n - .),1 Since (BE:23,) = e, 1171e= I T 1; hence .rr = E De.We shall find it more convenient to represent B E as {I .rrVncl1) where --a v
<
Theorem 6:
<
If the valuation is discrete, then ef = n.
Proof: Let w, , w, , wf be elements of E which represent a basis of the residue class field Ep/kp.Thus the generic residue
Theorem 7: The elements (w$7~}(P = 1, e - 1) form a basis D over o.
f ; p = 0,
--*,
Proof: We have already seen, in the course of the last proof, that {wp17/1}form a basis. Let a be an integer of E. We can write a = E,,, d,,&,lIfl, and since I w, I = 1, and the ( 17, ( are all distinct, we have
IaI
I
= max dPPIP[
.
Thus
Now .rr was defined as having the largest absolute value less than 1. Hence l/.rr has the smallest absolute value greater than 1. Thus 1 I d , I < / ~ ll d ~ l C l .
.-a,
But this is exactly the condition for {wpZI~) to be a basis of O over o.
3. 3. The General Case
In this section we shall prove that if k is complete under an arbitrary valuation, and if E is a finite extension of degree n, with ramification e and residue class degree f, then ef divides n, and the quotient 6 is a power of the characteristic of the residue class field k, . This section is added for the sake of completeness, and the result will not be used in the sequel, Our first step is the proof of a weaker result: Theorem 8: The only primes which can divide e and f are these which divide n.
Proof: (1) The proof for e is simple. Since I or I = N(or)lJn, I a In lies in the value group of k. The factor group B,/B, is abelian of order e. Hence e can contain only primes dividing n.
(2) T o prove the analogous result for f, we introduce the notion of the degree, deg a, of an element a E E; by this we shall mean the degree of the irreducible equation in k of which a is a root; hence deg or = [k(u) : k]. Similarly the degree of a residue class 6 E Ep is the degree of the irreducible congruence in k of which 6i is a root; hence deg 2 = [K(&) : El. We shall prove that if a is an integer of E, and 6i the residue class in which it lies, then deg & divides deg a. Let f(x) = Irr (or, k, x) = xn alxn-l .-. a,.
+
<
+ + / < 1. By the corollary
Since or is an integer, ( a I 1, and so I a, to Theorem 6, Chapter 11, this implies that all the coefficients of 1. Now Hensel's Lemma may be used f(x) are integers: I a4 I to show that f(x) cannot split modulo p into two factors which are relatively prime. For if f(x) = +(x) $(x) mod p , where +(x) and +(x) are relatively prime, there exist polynomials A(x), B(x) such such that A(x) +(x) B(x) $(x) = 1 mod p.
<
+
61
THE GENERAL CASE
where ( C(x) I < 1. And since f(x) r +(x) #(x) mod p , we have h(x), with I h(x) I < 1. By Hensel's Lemma, f(x) = +(x) $(x) this situation implies a factorization of the irreducible polynomial f(x), which is impossible. Hence if f(x) splits modulo p, it must do so as a power of an irreducible polynomial: f(x) r [P(x)]P. But deg Z = deg P(x), and deg or = deg f(x), hence deg & divides deg or as required. Thus if & is any residue class of E, , and a E E is any representative of E, then deg 2 divides deg a. But since deg a = [k(a) : k] divides n, this implies that deg 2 can be divisible only by primes which divide n. Let E,' be the separable part of E,. The degree of E, I E,' is some power pvof the characteristic of kp . Iff = f'pv, then E,' = A,(&), where deg & = f'; f ' is divisible by all the primes dividing f except possible the prime p. Hence, by our preceding remarks, these primes must divide n. Finally, if 2 is an inseparable element, then p divides deg 15; hence p must divide n. This completes the proof.
+
Theorem 9: If deg (E I k) = q, where q is a prime not equal to the characteristic p of the residue class field, then ef = q.
Proof: Since q is not the characteristic of the residue class field, q does not lie in the prime ideal; hence ( q 1 = 1. Furthermore, q is not the characteristic of k, and hence E I k is separable. Let E = k(a), and let f(x) = Irr (a, k, x) = XI + a1xq-l + .-. a,.
+
+
a,/q, since q is not the We may apply the transformation x = y b,yq-l bq . characteristic; f(x) assumes the form y'l Thus we may assume at the outset that a, = 0. We remark that
+
+ +
If I cx I does not lie in the value group 13, , we have e 2 q, since (Q E Bk . Since ef q, we obtain e = q and f = 1; hence ef = q. If, on the other hand, I a I lies in Bk , we have I or ( = 1 a 1 where a E k. We may write = a& and will satisfy an equation with second coefficient zero; hence we may assume without loss of generality that I or 1 = 1. In the course of Theorem 8 we saw that f(x) either remains
I or
<
3. irreducible modulo p or splits as a power of an irreducible polynomial; since q is a prime the only splitting of this kind must be j(x) = ( x - c)q mod p. We show now that this second case is not cq, mod p, c 0; hence the second term possible: since a, of (x -- c),, namely qcq-l, is 0 mod p . This contradicts our assumption about f(x). Thus n satisfies an irreducible congruence of degree q; f 2 q. Hence f = q, e = 1, and ef = q. This proves the theorem. In order to prove that ef divides n, we shall require the
-
+
+
Lemma: If E 3 F 3 k, then e(E 1 k) = e(E I F) e(F 1 k), and
similarly for f, and hence for ef. T h e verification of this is left to the reader. I t follows at once that for a tower of fields with prime degree unequal to the characteristic of the residue class field, ef = n. We shall now prove the main result. Theorem 10: ef divides n; the quotient is a power of the characteristic of the residue class field.
Proof: We shall prove the result first for the case where E is a separable extension of k. Let deg (E / k) = n = qinz, where q is a prime unequal to the characteristic of k and jq, 7%) = I . We assert that ef = qis where (q, s) = 1.
THE GENERAL CASE
63
Let K be the smallest normal extension of k containing E, and let G be its Galois group; let H be the subgroup of G corresponding to E. Let Q be a q-Sylow subgroup of H, E, the corresponding field. Now Q, considered as subgroup of G, may be embedded in a q-Sylow subgroup Q' of G; the corresponding field El is a subfield of E, . T h e degree of E, 1 E is prime to q since Q is the biggest q-group of H. Similarly the degree of El I k is prime to q, since Q' is the biggest q-group in G. Now the degree of E, / El is a power of q, which must be qi, for [E, : EJ [El: k ] = [E, : k ] = [E : k ] [E, : El.
Now between the groups Q and Q' there are intermediate groups, each of which is normal in the one preceding and each of which has index q. T h u s the extension E, I El may be split into steps, each of degree q. T h e product ef for the extension E, I El is equal to the corresponding product for each step of degree q. From this we see that ef = qi. But this is also the q-contribution of ef by the extension E 1 k, since the degree of E, I E is prime to q. We obtain this result for every prime divisor of n unequal to the characteristic of A. Hence ef can differ from n only by a power n, we must have efS = n where of the characteristic. Since ef 6 is a power of the characteristic. This proves the theorem for separable extensions. If E I k contains inseparable elements, let Eo be the largest separable part. Then n[E : Eo]must be a power of the characteristic of k; if this is non-zero it is also the characteristic of A. We have seen that ef[E : Eo] can be divisible only by this prime. Hence for the inseparable part we have again nlef = a power of the characteristic of A. This completes the proof of the theorem.
<
Corollary: If the residue class field has characteristic zero, then ef = n. We may write n = ef6 : 6 is called the defect of the extension.
1.
UNRAMIFIED EXTENSIONS
65
Since all finite extensions of & may be obtained by repeated simple extensions, we have proved
CHAPTER FOUR
Ramification Theory 1. Unramified Extensions
Let k be a complete field, C its algebraic closure. Let the corunder the natural isomorresponding residue class fields be k, phism, k may be considered as a subfield of C. The canonical image in C of an integer a in C shall be denoted by 5; that of a polynomial $(x) in C[x] by $(x). A given polynomial $(x) in €[XI is always the image of some polynomial $(x) of C[x]; +(x) may be selected so that it has the same degree as $(x). If the leading coefficient of #(x) is 1 we may assume that the leading coefficient of +(x) is also 1. In the sequel these conventions about the degree and leading coefficients will be tacitly assumed. Let $(x) = $(x) be an irreducible polynomial in &[XI, with leading coefficient 1. We may factor +(x) in C[x]:
e;
Since all the roots
are integers, we may go over to €[XI, where
Theorem 1: Every finite extension of is the residue class field for a finite extension of k with ef = n and e = 1. We return to the case of a simple extension, and assume now that #(x) is separable. Then pi # Pj for i # j, and hence j pi - pi I = 1 for i # j. Let ar be an integer of C such that $ ( C ) = O ; say & = P I . Then Iar-/3,j < 1; that is, (a--/3,I is less than the mutual distance of the & . Theorem 7 of Chapter 2 shows that k(/3,) C k(a). Let E be any subfield of C such that 3 &(PI).Then E contains an element a such that E = PI; hence k(pl) C k(a) C E. If we assume in particular that deg (E 1 k) = deg ( E I A) and that E = k(pl), then k(/3,) C E, but
hence E = A(/?,). Thus we have proved Theorem 2: T o a given separable extension k ( a ) of 6, there corresponds one and only one extension Eo of k such that (a) deg (Eo I k) = deg (Eo 1 k) and (b) Eo = k(p1). Corollary: If E is an extension of k such that E contains Eo, then E contains Eo . This discussion suggests the following definition: A field extension E I k of degree n with ramification e and residue class degree f is said to be unramijied if it satisfies the following conditions:
(I) This shows that $(x) splits into linear factors in €[XI; hence is algebraically closed. Since $(x) is irreducible in Qx], $(x) is irreducible in k[x], and hence F = k(/3,) has degree n over k. The residue class field P contains the subfield k(&), which is of degree n over k; hence f = deg (F 1 5) 2 n. But if e is the ramification, we have ef n. Hence e = 1, f = n, and F = K(pl). This shows that every simple extension k(pl) of k is the residue class field of a subfield F of C, with the same degree as K(pl) I h, and ramification 1.
<
e =
1,
(2) ef = n, (3) E 1 k is separable. The third condition is inserted to exclude the critical behavior of the different when inseparability occurs. This difficulty does not arise in the classical case of the power series over the complex numbers, where the residue class has characteristic zero; nor in the case of number fields, where the residue class fields are finite. I n neither of these cases is any inseparability possible.
66
4.
RAMIFICATION THEORY
Theorem 2 may now be restated in the following terms: Theorem 2A: There is a (1, 1) correspondence between the unramified subfields of C and the separable subfields
of
e.
We see also that the unramified subfields of a given extension E 1 k are precisely those whose residue class fields are separable subfields of E. Furthermore, the lattice of unramified subfields of E is exactly the same as the lattice of separable subfields of E. Hence, in particular, there is a unique maximal unramified subfield T of E, which corresponds to the largest separable subfield of E; T is called the Inertia Field ( Tragheitskorper) of E. Theorem 3: Let E = k(a) where a satisfies an equation f(x) = 0 with integral coefficients such that all its roots modulo p are distinct. Then E is unramified over k.
Proof: It will be sufficient to prove the result when f(x) is irreducible. If f(x) is irreducible in k it is irreducible also in A, for we have seen that an irreducible polynomial can split in K only as a power of an irreducible polynomial-which would contradict the assumption that the roots modulo p are distinct. Hence deg (E I k) = deg (A(&)I A); but A(G) is contained in the residue n, we have E = K(G), and hence E I k class field of E. Since f is unramified.
<
Corollary: If E I k is unramified, and if SZ is a complete field containing k, then El2 I SZ is unramified.
Proof: We can write E = k(a) where a satisfies an irreducible equation +(x) = 0 which is separable mod p . But ESZ = SZ(a), where a still satisfies an equation which is separable mod p. The result follows now from the theorem. Thus the translation of an unramified field by a complete field is again unramified. We can also prove, by obvious arguments, that an unramified extension of an unramified extension is unramified over the ground field, and that any subfield of an unramified extension is again unramified. We conclude this section with two examples.
2.
TAMELY RAMIFIED EXTENSIONS
67
Example I : Let k = F{t}, the field of formal power series over the field of constants F. We shall show that the unramified extensions of k are produced by separable extensions of F. We have shown previously that the residue class field of k is isomorphic to F under a natural mapping. Let E be a finite extension of k, E its residue class field. Then if G E E, a! satisfies an irreducible equation #(x) = 0 in R. Let #(x) = $(x); since every coefficient of $(x) may be replaced by a congruent one, we can assume that all the coefficients of $(x) lie in F. Suppose now that E 1 k is unramified, and that E I K, which must therefore be separable, is generated by G. Then we have seen that E is generated by a root fi of $(x). The condition e = 1 implies that t is prime in E; hence E = Fl{t}, where Fl = F(/3). It is easy to prove the converse: That if Fl is separable over F then Fl{t} is unramified over F{t}. Example 2: Let k be a finite field of q elements; this is the case which arises in algebraic number theory. We shall show that the unramified extensions of k are uniquely determined by their degree, and are obtained from k by adjoining certain roots of unity. The first statement follows from the fact that a finite field has only one extension of given degree. Now let E I k be an unramified extension of degree n. The qn - 1 non-zero elements of E satisfy the equation xqn-l = 0, which is therefore separable, having qn - 1 distinct roots in E. Hence if 5 is a primitive (qn - 1)-th root of unity, k(5) is unramified. Further, since xqn-l - 1 splits in k({), it splits in k(5); hence E C k(5). By the Corollary to Theorem 2, E C k(5). On the other hand, let a E E lie in the same residue class as 5 ; then ( a - 5 1 < 1, whereas the mutual distance of the roots of xun-l - 1 is exactly 1. Hence, by Theorem 7 of Chapter 2, k(5) C k(a) C E, and we have E = k(5).
2. Tamely Ramified Extensions
A finite extension E I k of degree n with ramification e and residue class degree f is said to be tamely ramijied if it satisfies the following conditions:
4.
68
2.
RAMIFICATION THEORY
(1) ef= n, (2) E I R is separable, (3) e is not divisible by the characteristic p of 5. Clearly all unramified fields are tamely ramified. We can deduce at once from the definition that a subfield of a tamely ramified field is tamely ramified; and that a tower of tamely ramified fields is tamely ramified. We shall now give an important example of a tamely ramified extension. Let a be a root of the polynomial $(x) = xm - a, where a E k, (p, m) = 1, and +(x) is not assumed to be irreducible. Certainly I a Im = I a I lies in the value group of k; let d be the exact period of 1 a I with respect to B, (i.e. the smallest integer d such that I a Id = I b 1, b E k); then d divides m. Let /? = ad/b; thus 1 p 1 = 1, and pmld= a/bmld. Hence /3 satisfies an equation #(x) = xmld - c = 0 where I c I = 1; #(p) = (mld) pmld-l, SO that I $'(p) / = 1 ; hence #(x) is separable modulo p. Thus k(p) 1 k is unramified. We now consider k(/3) as the ground field; over k(p) a satisfies the equation xd - b/3 = 0. k(p) has the same value group as k; hence d is the period of I a I with respect to the value group of k@). Thus k(a) I k(p) has ramification 3 d but the degree of d. It follows that k(a) I k(p) is fully ramified, and k(a) I k(P) is that xd - b/3 is irreducible in k@). In the course of this discussion we have verified all the conditions for k(a) to be a tamely ramified extension, and we have shown that it is made up of an unramified extension followed by a fully ramified extension. Now let E / k be a finite extension; we shall construct the largest tamely ramified subfield of E. If T is the inertia field, T is clearly contained in this subfield, so we may confine our attention to the tamely ramified extensions of T. Since '2 is the largest separable subfield of E, E I T is purely inseparable. Let p be the characteristic of the residue class field, and write e = eopv,where (e, ,p) = 1. BE/%, is a finite Abelian group of order e. Let us choose a basis for this group, and let a be an element of E which represents one of the basis elements with period d prime top; that is, I a Id = I a I, with a E k (since B, = Bk). Hence ad = €a where / E I = 1. Since E, I T, is purely inseparable, some power EP' represents a residue
69
TAMELY RAMIFIED EXTENSIONS
class in T. Let the corresponding power up' be written a,; since p r is prime to d, a and a, generate the same cyclic group, so that a, may be taken instead of a. We have now ald = ~ , a , where E , = c E T mod p; thus ald = ca, f l = b /3 where E E and I B l < lbl. We now study the equation 4(x) = xd - b = 0;d is prime to the characteristic of F, hence to that of k; thus 4(x) is separable over k:
+
+
+(XI= (X- 71) "' (X - Yd). Obviously
Further, however, 4'(y1) = dy,d-l, so that
whence I yi Now
- y,. I
I
= yi
/ for all i # j.
<
It follows that for at least one of the yi, say y,, we have I a, - y, I < I y, I; that is, a, is closer to y, than the mutual distance of the roots yi . Hence by Theorem 7 of Chapter 2 we have T(y,) C T(al) C E; by the nature of its construction, T(y,) is tamely ramified over T and hence over k. I a, I was a representative of a coset of B,/B,, and since I a, I = I y, I, we see that a field element y, can be chosen such that yld lies in T, and I y, I represents this coset. yT of this type to represent Choose representatives y, , y, , all the basis elements with period prlme to p. Then the field V = T (y, , y, , ..., y,) is a tamely ramified subfield of E, since a tamely ramified tower is tamely ramified. Now if di denotes the period of the basis element represented by yi , the ramification of V is 3 d1d2 ...d, = 0,; but V is a tamely ramified subfield of E, and so its ramification eo . Hence me.,
<
70
4.
3.
RAMIFICATION THEORY
e(V I k) = e,, the largest possible ramification for a tamely ramified subfield of E. Next, since = T is the largest separable subfield of E, f(V I k) = deg (2'1 k) is the largest possible residue class degree for a subfield of E with separable residue class field. Hence, since V is tamely ramified, deg (V 1 k) = e(V I k) f(V I k) is the largest possible degree of any tamely ramified subfield of E. We shall now prove that all tamely ramified subfields of E are contained in V. T o do this we require the Lemma: The translation of a tamely ramified extension by a complete field is again tamely ramified.
Proof: Let E I k be a tamely ramified extension. Let F be a complete field containing k. E is obtained from k by constructing first an unramified extension, then adjoining certain radicals. E F is constructed from F in precisely the same way, and hence is tamely ramified. Now suppose there is a tamely ramified subfield of E which is not contained in V; by the lemma, its translation by V is again tamely ramified. But this translation is of higher degree than V, contrary to our result that V has the largest possible degree of any tamely ramified subfield of E. Hence V is the unique maximal tamely ramified subfield of E: V is called the Ramijication Field ( Verxweigungsk6rper). We conclude this section with an Example: Let k be the field of formal power series F{t}. The unramified part of a tamely ramified extension E of k has the form Fl{t) where Fl is a separable extension of F (Section 1, Example 1). Tamely ramified extensions of Fl{t} are obtained by adjoining roots of elements which represent the basis elements of BE 1 Bk , which have period prime to the characteristic of F. Since the valuation is discrete, BE/Bkis cyclic, and we may choose as representative of a basis element a series a = ct(1 -..); E =Fl{t}(.i;/ol), where m is prime to the characteristic of F. Since, however, the m-th root of a series 1 can be extracted in Fit}, it suffices to adjoin 72.If F is algebraically closed, the only possible extensions are by q--t.
+
+
CHARACTERS OF ABELIAN GROUPS
71
3. Characters of Abelian Groups
Let G be a finite Abelian group with basis elements a, , a, , a, having periods el , e, , -.-,e, . A character of G is a homomorphic mapping x of G into the non-zero complex numbers. If x is a character, [X(a,)]ev= 1; hence X(a,) = 4, is an e,-th root of unity; and if a = aTa: a:, then X(a) =:E:E Thus we see that a character x is described by a set of roots of unity, (6, , E, , -.. E,), where each E, is an e,-th root of unity. Conversely, any such set of roots of unity defines a character. Hence there are in all ele2 e, distinct characters, i.e. as many as the order of the group. If X(a) and xf(a) are characters, then = X(a) X1(a) is also a character, which we denote by XXf(a). If X, Xf are described by (el, E,), (elf, E,'), then xX' is described by (qelf, ..., ere,'). Thus the characters form a group G* of the same order as G; G* is called the dual group of G. G* is clearly a direct product of groups of order ei; hence . - a ,
€7.
- a * ,
G*
(el) x (e,) x
... x (e,) r G.
Thus we have established Theorem 4: A finite Abelian group is isomorphic to its dual group. Consider now the following situation: G and H are Abelian groups. Z is a finite cyclic group. By a pairing operation on G and H into Z we shall mean a function q5 which maps the product G x H into Z such that
4k1g2 h) ?
= 4(g,
, h) 4(g2, h),
Let Go be the G-kernel, i.e. the set of elements g E G such that +(g, h) = 1 for all h E H ; similarly let Ho be the H-kernel, i.e. the set of elements h E H such that 4(g, h) = 1 for all g E G. For this situation we prove Theorem 5:
GIGo r H/Ho.
If H/Ho is finite, then GIG, is also finite, and
72
4.
4.
RAMIFICATION THEORY
Proof: We may regard Z as a group of roots of unity. For a fixed element g E G, we write xg(h) = #(g, h); then xg(h) is a character of H, which is trivial (i.e. takes the value 1) on Ho. Thus x,(h) may be regarded as a character of the factor group H/Ho. Hence g -+ xg(h) is a homomorphian of G into (H/Ho)*; the kernel is clearly Go . Thus we have GIGor some subgroup of (H/Ho)*;
.
H/Hog some subgroup of (GIGo)*
.
Thus GIGo is also finite, and hence isomorphic to (GIGo)*; so some subgroup of GIG,.
Hence we have the result of the theorem: GIGo r H/H,
act like the identity on the subfield. Our immediate task is to describe the inertia field and ramification field in this way. First let E I k be an unramified extension. We have seen in Section 1 that E = k(/3,) and E = &(PI),where
The isomorphic maps of E 1 k into C I k carry /3, into /3, ,p, , -..,&; these clearly induce maps of E I & into / k, namely maps which j3, . Since E I k is unramified, &x) is carry Pl into P I , P 2 , separable, and so these f maps of E I & are distinct. Since they are equal in number to the degree of E I k, these are all the isomorphic maps of E 1 E into 1 6. From this discussion we see that a map of E 1 k into C ( k is uniquely determined by its effect on E 1 5. We are now in a position to prove ..a,
Since H/Hois finite, by hypothesis, we have H/H, g (H/Ho)*; so
H/Ho
73
e
similarly
GIG, g some subgroup of H/Ho
THE INERTIA GROUP AND RAMIFICATION GROUP
.
4. The Inertia Group and Ramification Group
Let E ( k be a finite extension field. I n sectlons 2 and 3 of this chapter we have defined two important subfields of E: T, the Inertia Field (Tragheitskorper), which is the largest unrarnified subfield of E, and V, the Ramification Field (Verzweigungskorper), which is the largest tamely ramified subfield of E. When E ( k is a normal extension, with Galois group G, the subfields T and V correspond to subgroups 3 and % of G, which are called respectively the Inertia Group and the Ramification Group of E ( k. I n this section we shall describe these two subgroups. At first, however, we do not assume that E I k is normal. Instead, we let C be the algebraic closure of k, and consider the set of isomorphic maps of E into C which act like the identity on k. This set of maps, of course, does not form a group; but it is known from Galois Theory that the number of such maps is equal to the degree of the largest separable subfield of E. A separable subfield of E may be described by giving the set of maps which
Theorem 6: Let T be the inertia field of E 1 k. Then the isomorphic maps of E into C which act like the identity on T are precisely these which act like the identity on all the residue classes of E.
Proof: It is clear that the maps which leave T fixed also leave fixed all the separable residue classes of E. Consider, therefore, the inseparable residue classes: let a E E be a representative of one of these. Then for some power pvof the characteristic of A, DLP" represents a separable residue class, which is left fixed by the maps which leave T fixed. Since p-th roots are unique in fields of characteristic p, this implies that the residue class represented by a is left fixed. Hence every isomorphic map of E 1 k which leaves T fixed also leaves every residue class fixed. Conversely, if a is a map which leaves fixed every residue class of E, then it leaves fixed every residue class of T,and hence, since T is unramified, it acts like the identity on T. This completes the proof. We must now carry out a similar analysis for the ramification field V. First we consider a tamely ramified extension E 1 k. Let u ( # 1) be an isomorphic map of E / k into C. We consider two cases :
74
75
4 . RAMIFICATION THEORY
4 . THE INERTIA GROUP AND RAMIFICATION GROUP
(a) o does not leave fixed all residue classes of E. Select a E E as representative of one of the residue classes which is changed by a. Then, since ua and a are both integers, we have
Hence I oap - ap 1 = 1 a lp. This follows immediately when p is odd. When p = 2 we have
(b) o leaves fixed all the residue classes of E. Then a leaves T fixed, and so may be considered as an isomorphic map of E I T. We have seen that E = T("i/& , VG, ...) where the a, are suitably chosen elements of T and the mi are prime to the characteristic p of A. Consider the action of a on one of the generators, say a = &i; a satisfies the equation xm - a = 0, hence so does oa, and we have already shown that for the roots of such an equation,
Thus in either case we have found a non-zero element a E E for which I ua - a I = I a I. Consequently, if for every a # 0 in E we have / oa - a I < ( a I, then a = 1. We can now prove Theorem 7: Let V be the ramification field of E I k. Then the isomorphic maps of E into C which act like the identity on V are precisely those maps a for which I o(a) - a I < I a I for all non-zero a E E.
Proof: If o is a map such that 1 o(a) - a 1 < 1 a 1 for all non-zero a E E, it clearly has this property for every element of V. Since V is tamely ramified it follows from the preceding discussion that o acts like the identity on V. Conversely, let o be a map such that I o(a) - a I = I a I for some non-zero element a E E. We shall show that o cannot leave V fixed. Let p be the characteristic of k, . Then
I a(. - a)' I = Now since
I
001'
+ (-
I (;) I < 1,
a)'
+ 2 (,P) v=1
we have
(- a)' (.a)'-'
I
= I a'
I.
but since 1 2# 1 < 1 oIP I, this gives 1 uap - @ 1 = 1 a lp. Thus in all cases, 1 oaP - ap I = I a IP, and repeating the process, we obtain I a d ' - OLP' I = I apv 1. The period of I upv, I with respect to kp is prime t o p if we take v large enough. Thus for large enough v, there exists an element /I E V such that I ap' I = I /I 1, i.e. apU//Iis a unit. The residue class represented by this unit may be inseparable; but a suitable pp-th power will be separable. There is therefore no restriction of generality if we assume that the residue class is already separable. I t may thus be represented by an integer y E T, i.e. apV//I= y mod p . Hence ( apV//3- y I < 1, which implies that
Thus ap' = /Iy
+ 6 , where a.''
where I a6 I = 16
1 6 1 < 1 /Iy I. = a@)
Similarly
+ us,
1. Now consider
Since /Iy E V, we have shown that o cannot act like the identity on V. This completes the proof. Let now E be a normal extension field of F, with Galois group 8. The inertia group 3, which corresponds to the inertia field T, consists of those elements a of 6 which leave the residue classes of E fixed. The ramification group B, corresponding to the ramification field V, consists of those elements o of 8 for which I ua - or I < I a I for all a E E. It is easily verified that 3 and D are invariant subgroups of 6. We shall now try to describe the factor groups 813 and Q / B .
4.
76
5.
RAMIFICATION THEORY
The group 613 is the Galois group of T/F. If we map the group 6 onto its effect in the residue class field, we obtain a homomorphism between 6 and the group of E I A with kernel 3. Hence we have Theorem 8: The Galois group of the Inertia Field TI k is isomorphic to the Galois group of the residue class field E / h. In order to describe the group 6/23, we consider first the subgroup 3/23. This can be examined by constructing a pairing operation between 3 and the value group 23, of E, as follows: Let 7 be an element of 3, a E E; let +(a, T) be the residue class of 7a/a mod p. We shall show first that +(a, 7) depends only on I a I . Let E be a unit of E; then since 7 E 3, TE E mod p, whence TC/E 1 mod p. Thus we have
-
-
which proves our assertion. Further, we have
77
HIGHER RAMIFICATION GROUPS
So far +(a, 7) has denoted a residue class, i.e. an element of E. It is natural to ask whether there is an element in E, lying in this residue class, which can be naturally selected to represent it. If e = po'eo, where (p, e,) = 1, the order of 3/23 is e,; hence [+(a, 7)le0E 1 mod p . The equation xeo- 1 = 0 has e, roots in the algebraic closure A of F, of absolute value 1 and mutual distance 1. Since I ( ~ a / a ) ~-o 1 I < 1, we have 7q'a nearer to one ) F(5). Thus 5 is a root 5 than to any other; hence E 3 F ( ~ a / a 3 root of unity in E, congruent to ~ a / amodulo p. We now choose +(or, 7) to be represented by this 5; this is well-defined, since 7') mod p , the corresponding roots of unity if +(a, 7) = +(/I, r C2 mod p, whence 1 5, < 1; thus are congruent: = C2 , since the mutual distance of the roots is 1. Thus the operation +(a, T), which formerly had values in E, may now be regarded as a pairing operation on 23, and 3 into E. Now GI23 is an extension group of 3/23. Let a be an element of G; the mapping 7 E 3 -+ma-I gives an automorphism of 3 modulo 23. We shall give a description of +(a, o~a-I);this gives the character values of a ~ o - Iin E, and this will be sufficient to describe a ~ o - l .We have ma-la T(U-la) +(a, mu-') 0 = .($(a, 7)) mod p, a
cl
- (-)
and
since 1 r2a 1 = 1 a 1. Thus +(a, 7) is a pairing operation as described in Section 4. We have now to find the kernels under the operation. The %kernel Kg consists of the automorphisms T such that 7a/a 1 mod p for all a E E; since
-
since / a-la I = / a in E; hence
I. But +(a, m-u-l) and +(a, 7) are roots of unity +(a, u7u-l) = u(+(a, 7)).
In particular, if G is abelian, u ~ u - l= 7, so that .(+(a, 7)) = +(a, 7); thus #(or, 7) is a root of unity in k.
we see that the kernel K3 = D. If the BE kernel is I(, , we have 3/23 BE/KE.Also K E must certainly contain the value group 23, of k, since ra a for a E k. But since no element whose period is a no such element can power of the characteristicp can occur in 3/%, occur in 23,/K,. Hence K, consists of all the elements of 23, whose periods modulo 23, are powers of p. Thus we may say that 3/23 is isomorphic to the "non-critical part" of the value group, i.e. to the cosets of 23, modulo 23, which have period prime to p.
--
5. Higher Ramification Groups
+
Let i denote either a real number, or a "real number zero", in a sense to be made more precise in a moment. We define the sets %, to consist of those automorphisms a E G such that ord (ua - a) 3 ord a + i,
78
4.
or ord
5.
RAMIFICATION THEORY
aa - a
(_)
2i
for all
a EE
HIGHER RAMIFICATION GROUPS
79
Let us first notice, however, that the sets Bi are invariant subgroups of 6; let a , T E %, , ol E E:
When i is a real number r , this shall mean simply that ord when i = r
aa - a
Y;
+ 0,
whence ord
aa - a (_)
2i
shall mean ord
aa - a (_)
> r.
We now consider the operation which, acting on a, produces ( o a - a)/a;this bears a certain resemblance to logarithmic differentiation:
whence we have
T h u s %$ is a subgroup of 6; that it is an invariant subgroup follows from the invariant form of the definition. Let E = k(ao); then for each o E G we define ia = ord
T
E
jTa
8,. for all r < ri o ord --a) + ord
whence
Thus, when we examine the effect of 2.3, on the elements of a group we can restrict our examination to the generators of the group,
_
If a # 1, then aa, f a, , and hence i, cco.Clearly o cannot occur in Bi for i > i,; hence for i > max, i, , the group %, consists of the identity automorphism alone. I t follows from the definition that for i >j, Bi E %,.; we must now examine the discontinuities in this descending sequence of groups. We consider first the subsequence {B,) where the indices r are restricted to be real numbers. Suppose there are discontinuities n - 1 where n is the order of G). Then: at r , , r , , ..., rk (k
<
Further,
( aaOaT ) .
7a - a (_--)
r for all r
< r, and all a t E
r, for all a t E
T h u s the real line may be split into intervals rip, < r ,< ri such that %, = Bri for ri-, < Y ,< ri . Since the whole sequence of groups %, is monotone, it is clear that %+,, = %, when ri-, < r < r, . We cannot, however, give a definite description of the groups I t may happen that illTi+, = Sri or = B r i f l ; but in general BTi+, lies between Dri and Bri+, . Hence in general a
4.
80
5.
RAMIFICATION THEORY
jump occurs in two stages-between Bpi and Bri+, , then between B,&, and Bri+, . Our next task is to analyze the factor groups at these jumps. Let a E Bi (i # O), 7 E Bi . Then
+
- 1 = ( a - 1) (T - 1) + (T - 1) + ( a - l), UT - TO = (a - 1) - 1) - (T - 1) (U - 1). ar
(T
Now
HIGHER RAMIFICATION GROUPS
Hence, if a E !I? we have (, ord(o'Thus a E Bi
+ min(pi,ordp + i). where j = min (pi, ord p + 21.
l ) a > 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).
<
Theorem 1: D' is a fractional ideal.
Proof:
Clearly D' is not empty, since D' 3 a. 86
since 0116 E D, this shows that t9 $ Dr. Hence if E ( k is unramified, then a = D'.
Case 2: e > 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
<
Thus a polynomial of degree n - 1 has n common zeros with xi. It follows that it must be identical with xi. I t is now easy to compute the complementary basis to 1, a , a2, ..., an-l. The result is given by
We may write this symbolically as
92
5.
3.
THE DIFFERENT
where the symbol [A] denotes the integral part of A in the obvious sense. If a is an integer, we may simplify those results even more. For then the coefficients ai of f(x) are all integers of k, and we may replace the basis elements
FIELDS WITH SEPARABLE RESIDUE CLASS FIELD
and so
a =f'(01)
93
. D.
3. Fields with Separable Residue Class Field
We shall now give a description of all extensions E 1 k where the valuation is discrete and the residue class field is separable. Let E I k be such an extension, T the inertia field. Then, as we have remarked earlier, a,,, = a,,, . We have seen also that 1, IT, 112, ne-I forms a minimal basis for E 1 T. Hence, if f(x) = Irr (R; T, x), we have a = f'(17) D. Now f(x) is a polynomial of degree e: f(x) = xe alxe-l a,; a, = NIT, so that 1 1 7 1 = Q M , whence l a o l = 1L7le= I T ] ; thus r r ( a , , but rr2 # a, . Further, since the coefficients a, (v = 1, e) are the elementary symmetric functions of the roots of f(x), we have rr I a, . Thus f(x) is a polynomial satisfying the Eisenstein criterion. Conversely, we shall show that such a polynomial gives rise to a completely ramified extension. Le f(x) be an Eisenstein polynomial in T[x], 17 a root of the polynomial. Then ma.,
+
and so on by the equivalent basis
+ --- +
as.,
Theorem 6:
If
LY
is an integer such that E = ~ ( L Yand ),
then the complementary set T' = T/f('cw). f'(LY) .D.
Furthermore 9 divides
Proof: The first statement follows at once from the preceding discussion and Theorem 4. T o prove the second statement we have only to notice that since a is an integer T C D; hence D' = 3 - I C T' = T/f'(a). It follows C )a T C BD, which proves the result. that ~ ' ( L Y Finally, we may apply these results to the case where the residue class field E I A is separable; then E has a minimal basis consisting of the powers of an integer a. (Theorem 10, Chapter 4). If I?' I is separable, then 3 = fl(a) D, where a is the element whose powers form a minimal basis. Theorem 7:
Proof: Since the powers of LY form a minimal basis, the set T of Theorem 6 is exactly the ring of integers D. Hence
since all the a, are integers,
I 17 1 < 1. Further,
since
and I a, I = I rr I, we must have I I7 le = I rr I. Thus the ramification of the extension defined by f(x) must be at least e; but the degree is at most e. Hence both ramification and degree are equal to e, and there is no residue class field extension. We have also proved that the polynomial f(x) is irreducible. The proceding analysis has shown Theorem 8: All possible extensions of k with separable residue class field consist of an unramified extension (constructed by making a separable extension of the residue class field A) followed by an Eisenstein extension. We shall now compute the different of such an extension. We =f'(17).D = ITaD where a = ord f'(17).Now have
5.
94
THE RAMIFICATION GROUPS OF A SUBFIELD
THE DIFFERENT
sincef(x) is an Eisenstein polynomial, the ordinals of the coefficients a, are divisible by e. Hence the ordinals of the non-zero terms are all incongruent mod e. Thus a = min ord (ene-,, (e - 1) a117e-2, ..-, ae-,). If the field is tamely ramified, e is not divisible by the characteristic of the residue class field, so ord e = 0. Thus a = e - 1, and = ne-ID. This is analogous to the case of function theory: for if a Riemann surface has a winding point with e leaves, the winding number is e - 1. If, on the other hand, the ramification e. is wild, we have a We now ask whether, for a given value of e, there is a bound on the indices a arising from all Eisenstein equations. If e # 0, we ord (ene-l) = e - 1 -t ord e. Thus if the characteristic have a of E does not divide e, a is bounded by e - 1 ord e; but if the characteristic of E divides e, e = 0, ord e = a , then a is unbounded, e - 1) can be made as large as since the ordinals of the a, (v we please. We conclude this section by computing a explicitly when E I k is normal. We recall that if E I k is a normal extension, and E = k(a) where a is an integer such that the pourers of a form a minimal basis, then the position of the automorphisms a of E IF in the is determined by higher ramification groups 23, , B 2 ,
<
+
95
Hence ord f'(a) = C, ord ( a - ua) = C, (1
+ i(u)).
,#I
a# 1
+
If a E 6 = 23-, , but a $ 23, , then i(a) 1 = 0. But if i(a) # - 1, 1 groups 23, , 23, , -.-,Bi(,, . Thus each a then a lies in the i(o) makes a contribution to ord f'(a) equal to the number of groups in which it occurs. Hence if # (23,) denotes the number of elements in 23, , we have
+
and so, finally,
<
ord (ua - a ) = i + 1
-
aE
Bi
but
u$ 23i+1
This has been established for i = 1, 2, .-.. It is easy to verify that if we define 23, to be the inertia group 3 and 23-, to be the whole Galois group 8, then the result holds also for i = - 1 and 0. Now let us define, for each a E 8 , the index i(o), given by i(a) 1 = ord (aa - a). Then - 1 i(u) < a , and i(0) =a only for a = 1. Then our criterion for the position of a in the groups 23,. may be written
+
<
4. The Ramification Groups of a Subfield
Let E be a separable extension of a ground field k, F an intermediate field. Let a, p be generators over k of the integers of E, F respectively. Let a, T be isomorphic maps of E, F respectively into the algebraic closure A, acting like the identity on k. We define the element &(a, E) = (aa - a) DA and similarly a(., F ) = (TP - p) DA, where DA is the ring of integers in A. We have already noticed (Ch. IV, $6) that if 0 is any integer of E, then &(a, E) I (a0 - 0) DA; a similar result holds for &(T,F). It is well known that for a given map T of F, there exist several maps a of E which have the effect of T on F: We call these the prolongations of T to F and write a ( T. We shall now deduce a relation between &(T,F ) and the elements %(a,E) where u 1 T. The following lemma holds in the case of fields with discrete valuation. Lemma:
&(T, F )
divides ll,,,&(a, E).
Now if f(x) = Irr (a,k, x), we have
f (4 =
(x oeQ
- aa)
and
f '(a) =
(a
- w).
Proof: The statement is obviously true for T = 1; for then (F(7,F ) = 0, and since the identity is one of the prolongations, no,, &(a, E) = 0 also.
5.
96
We suppose now that T # 1. Let S be the set of maps of E which act like the identity on F. Let u be a prolongation of T; then since T and a have the same effect on F, T-la E S; hence u = TA, where X E S. Thus
=7
THE RAMIFICATION GROUPS OF A SUBFIELD
THE DIFFERENT
97
The lemma shows that if T # 1, then 17,1TE(u, E)/E(T,F ) is an integral ideal of DA . We must prove that it is in fact D, itself. Consider the product
n
(ha - 7-10,) DA .
AES
where the product 17' extends over all maps u of E which are not identity on F. Hence
Now let f(x) = Irr (a, F, x):
Thus we can write
where fr(x) has the following meaning: If
then F(x) = Ta,
+ Talx + --- + xn. and
Then f'(a)
- f(a)
where the product I?' extends over all maps u of E, other than the identity, which are identity on F. Since the residue class fields are separable, we have
- a,)
= (~a,
+ (7% - al) a + + (ran-l - an-l) an-'.
Since a is an integer in E, the ai are integers in F. We have remarked a a) DA . Hence that for any integer a E F, E(T,F ) I ( ~ E(T,F ) I (f(a) - f(a)) DA . This proves the lemma. We now make the additional assumption that the residue class field E I K is separable, and obtain the stronger result: Theorem 9:
When I% 1
is separable, then
E(u, F) =
ITE(u, E). "
17
Proof: We have already remarked that the theorem is true when r is the identity.
Using the transitivity of the different, we have
whence the result follows since each of the factors is an integral ideal. We now assume that E 1 k, F I k are normal extensions. Let 6 be the Galois group of E I k, 8 the subgroup corresponding to F; 5 is an invariant subgroup and 618 is the Galois group of F I k. We take the view, however, that the Galois group of F I k is again
5.
98
THE DIFFERENT
99
THE RAMIFICATION GROUPS OF A SUBFIELD
8 where these elements are identified whose effects on F are the same. Let the sequence of ramification groups of E I k be
Now, by the definition of i(u), we know that a . Hence
E
B,,,,
but
a $ %,(+,
Let the corresponding sequences for E 1 F and F I k be, respectively, where 80,Uy= 1 when o may now be written
E BV and
The following result is at once obvious: Theorem 10:
For i = - 1 , 0 , 1,
me-,
v=0
Bi* = S j A Bi
+ +
formula
-0
since the groups (75 n Bi) form a decreasing sequence. Hence DO o 7 E B o g ; thus we have our first result.
T E
Theorem 11:
The inertia group
5 of F I k is given by
This statement is to be read with the understanding that elements of BOB are to be identified when their effects on F are identical. Since the higher ramification groups are contained in 5, we can now assume that T E BOB, i.e. 75 B0 # 0 . We first prove the
we can write
ordE(78 - p) =
hi)
6 Bv . Our
.
We now attack the more difficult problem of describing the groups Bi . Let a be an automorphism of E; we recall the definition of i ( a ) : i ( a ) 1 = ordE(ua - a), the ordinal being based on the prime in E. Similarly we define i ( ~ )where , T is an automorphism of F, by ;(T) 1 = ord, (TP - P), the ordinal now being based on the prime in F. From the fact that
whence
= 0 when a
2 OrdE(oa - a). 017
Now ordE( 7 p - /3) = e(E IF) .ord, (7/3 - /3) = #(3*) (i(r)
+ 1).
Proof: Then
Let ha be the fixed element in 5 , such that rXOE BV.
On the other hand every a 1 T can be expressed as a = TX where E Sj. Thus we may write the result
h
as asserted. Now 7 % written in the form
Bv # 0 o T
E BY$, SO
our formula may now be
5.
100 where 7 E Bm(.,$j but still further, writing
T
THE DIFFERENT
4 Bm(,,+,4j; 0 < m(7)
simplify
where the sum on the right is defined to be zero when m(7) = 0. Finally
: l/avwhere $(O) = 0, Thus L(T) is a sum of the form $ ( j ) =,z the a, are bounded, and a, I a, , a, I a, , ---, a, I a, . It is easy to see that the positive integers must occur among the values of $(j).We define the function @(z) by the relation
We conclude our investigation with the following theorem. Theorem 12: If @(i - 1) Hence Bi = Big.
<j
< @(t], then
T
E
o T E Bi
Proof:
Now since m ( ~ is ) itself @(K)for some integer K, we have
Thus T E B ~ ~ ~ ~ T E B , .
This completes the proof.
.
PART TWO
Local Class Field Theory
CHAPTER SIX
Preparations for Local Class
Field T h e o r y 1. Galois Theory for Infinite Extensions
We define a normal extension Q IF to be one in which every element is separable, and such that every irreducible polynomial in F[x] which has one root in SZ splits in Q[x]. The Galois group of such an extension is the group of all automorphisms of Q which act like the identity map on F. When the extension Q I F is infinite, we shall be unable to establish a (1, 1) correspondence between the subgroups of the Galois group G and the subfields of 52. But by introducing a topology into G we shall establish a (1, 1) correspondence between the closed subgroups of G and the subfields of Q. Lemma 1 : Every isomorphic map a: Q -+Q which leaves F fixed is an onto mapping, and hence is an element of G.
Proof: Every element of 52 lies in a finite normal subfield E C Q. a acts on E as an isomorphism; hence it maps E into E, and so onto E. Thus a maps every finite normal subfield onto itself, and hence maps Q onto Q. Lemma 2:
Let E be any intermediate field: F C E C 9. Let E into Q, in which F is left fixed. Then
a be an isomorphic map of a can be extended to $2.
Proof: Let a E Q. Then we shall prove that a can be extended to E(a). Let f(x) = Irr (a, F, x), +(x) = Irr (a,E, x). Then f(x) = +(x) q(x). Since of(%) = f(x), we have f(x) = acj(x) . aq(x). 103
1. Now since f(x) has one root, a, in Q, it splits completely in Q. Hence o+(x) has a root, say B, in 9. Then if oE = E l , o can be extended to an isomorphism T : E(a) -+ El@). Now we consider the set of all extensions T of o to higher fields. This set can be partially ordered by defining T, 2 7, (where T, is an extension of a to field E,) to mean that El 3 E, and that T1 is an extension of T, to El . The set is inductively ordered under this relation, for consider any totally ordered subset (7,): The fields on which these act, (E,), are also totally ordered. Let E' = u Ea , and define T' on E' to have the effect of T~ on E, . Obviously T' any T, , so there is a maximal element to the set (7,): By Zorn's Lemma, to any inductively ordered set there exists a maximal element T. It is clear that T is the required extension of o to a; for if there is an element a in SZ on which the action of T is not defined, we can extend T to E(a) by our earlier remarks: This would contradict the maximality of T. We now introduce a topology into the Galois group G of 9 I F. The neighborhoods of an element a E G are defined by referring to the finite subfields of SZ. Let E be a finite subfield of Q; we define the neighborhood NE of o to consist of the elements T E G which have the same effect on E as a. Thus if UE is the Galois group of Q 1 E, then NE = oU, . I t is easily shown that these neighborhoods define a topology in G. This topology is HausdorfT, for if a # T there is an element a E 9 such that o(a) # ~ ( a ) .Let U be the group of Q I F(a). Then oU and TU are obviously disjoint neighborhoods of a and T. Let H be a subgroup of G, E the fixed field under H. Then the Galois group of Q I E is If, the closure of H. Lemma 3:
a E E; U the group of Q I F(a). Since a lies in the closure of H, oU, which is a neighborhood of a, contains an element T E H. Thus oU = TU, whence o E TU. NOW the group U and the element T leave F(a) fixed; hence o(a) = a: o leaves every element of E fixed. Thus If is contained in the group of a I E. (b) Let o be an element of G which leaves E fixed, and let oU be a neighborhood of a, so that U is the group of SZ I F(a) for some a E Q. Let N I E be a finite normal extension of E containing E(a). The elements of H induce isomorphic maps of N into Q;
Proof: (a) Let o E If;
GALOIS THEORY FOR INFINITE EXTENSIONS
105
since N is normal, these maps are automorphisms of N, and under their action E is left fixed. Thus H induces on N elements of the group of N I E; since E is exactly the fixed field of H, H induces the whole group of N I E. Now o maps E(a) -+ E(af), say, which is contained in N since N is normal. Hence this mapping is produced by some automorphism of N I E (using the result of finite Galois theory); and this automorphism is in turn produced by an element T E H: o(a) = ~ ( a ) . Thus a = o-%(a), which means that a-I7 E U: T E uU. Thus every neighborhood of a contains an element T E H, i.e. o E If. This proves the inverse inclusion relation; hence our theorem is established. Lemma 4:
Let E be a subfield of 52, H the Galois group of
9 1 E. Then E is the fixed field of H. a
Proof: We have to show that if such that o(a) # a.
a
$ E then there is an element
EH
Let +(x) = Irr (a, E, x); since E(a) I E is a proper extension and 9 is separable, there is a root a' # a of +(x) in Q. The map of E(a) -+ E(af) obtained by leaving E fixed and mapping a -+ a' must be produced by some element o E G. Since a leaves E fixed, o E H. Since u(a) = a' # a, our lemma is proved. From Lemmas 3 and 4 we deduce immediately the Fundamental Theorem of Galois Theory for Infinite Extensions: Theorem 1 : Let Q 1 F be a normal extension; let G be its
Galois group, with the topology described above. Then there is a
(I, 1) correspondence between the subfields of 52 and the closed subgroups of G: Viz. (I) The Galois group H of 9 I E, where E is a subfield, is a closed subgroup of G, and E is exactly the field left fixed by H. (2) Every closed subgroup H is the Galois group of the field which it leaves fixed. We now prove several supplementary statements: Theorem 2: Let E be a subfield of SZ, H the group of Q I E. Then the topology of H considered as a Galois group is the same as the topology induced in it as a subgroup of G.
1. Proof: Let a be an element of 9. Then U, the group of Q 1 F(a), is a neighborhood of 1 in the group topology of G; similarly U*, the group of Q I E(a), is a neighborhood of 1 in the group topology of H. But U* = U n H ; hence U* is also a neighborhood in the induced topology. Conversely, every neighborhood in the induced topology is also a neighborhood in the group topology. Theorem 3: Let E be a normal subfield of Q, H the group of Q I E. Then H is an invariant subgroup of G and G/H is the group of E IF. Further, the group topology in G/H is the same as the topology which it inherits as a factor space.
Proof: Since E IF is normal, the automorphisms of G act as automorphisms on E, and all the automorphisms of E arise in this way from automorphisms of G. Thus the Galois group of E I F is G provided we identify those elements of G which have the same effect on E. Let o E G. Then oH has the same effect as a; i.e. oHo-l has the effect of the identity; hence oHo-l C H and H is a normal subgroup. The group of E IF is obviously G/H. The neighborhoods of 1 in the Galois group topology of G/H are the groups U* of finite subfields E I F(a). U* = U / H where U is the group of Q I F(a). But U I H is a neighborhood of 1 in the inherited topology of G/H. Conversely every neigborhood of 1 in the inherited topology is a neighborhood in the group topology. Theorem 4: Let (52,) be a set of normal extensions of F, with Galois groups {G,). If LIPn Qa = F, then the Galois group of Q = IT Q, is G = IT G,; the topology in G is the Cartesian product topology. The proof of this theorem is left to the reader. Theorem 5: With the topology described above, the Galois group G of Q I F is compact.
Proof: Let @ = (4,) be a family of indexed closed sets having the finite intersection property, i.e. every finite subfamily of @ has a non-empty intersection. We must show that the total intersection n $, is non-empty. The existence of a maximal family containing @ and having
GALOIS THEORY FOR INFINITE EXTENSIONS
107
the finite intersection property follows from an application of Zorn's Lemma. Hence we shall assume that @ is a maximal family: @ has therefore the additional properties: (1) The intersection of any finite number of members of @ is again a member of @. (2) Any closed set 4 which contains a member of @ is itself a member of @. Now suppose $ B = A v B, where A and B are closed sets; we shall show that either A or B is a member of @. For suppose A 4 @. Then A n (some finite intersection of 4), = A n $, = O (where O denotes the empty set); similarly, if B $ @, then B n $b = O. Consequently (A u B) n $, n $* = O, so that A v B is not a member of @. In general, if $, = A, u A, V u A,, where the Ai are disjoint closed sets, then exactly one of the Ai is in @. In order to prove that nu$, is not empty we must exhibit an element o E G which lies in each 4,. Let a be any element of Q; we shall define the effect of o on F(a). If U is the group of Q I F(a), u rnU; this union is finite since then G = rlU(= U) v r 2 U u F(a) IF is finite. Now G is closed (since it is a Galois group); hence G E @. The cosets r i U are closed and disjoint; hence exactly one of them, say riU, is a member of @. Let o have the effect on F(a) of rill. We have now to show that the mapping o we have constructed is in fact a well-defined element of G. T o this effect let El 3 E, be finite subfields of Q with groups Ul , U,: U, 3 Ul . Let o have the effect on El of rlU1 E @, and on E, of r2U2 E @. Since rlU1 and r 2 U 2 E @, hence rlU1 n r2U2 # O. Thus rlU1 C r 2 U 2 and the effect of 7,U2 on El is the same as that of rlU1. Hence a is welldefined. Finally we must prove that o E nu$,. For any subgroup U, o has the effect of some coset TU E @ on the fixed field of U. Hence oU = r U E @. Thus UU n $, # O (for every $, E @). Hence every neighborhood of a contains an element of 4;, thus o E 6, = $, (for every $, E @). This completes the proof. We now illustrate the use of this theory by an interesting special case. Let p be a prime number. Consider the sequence of fields
2. GROUP F C Fl C .-.,where F, / F is cyclic of degree pn. Let Q = v, F, . Let a E Q; then a E some F, , i.e. F(a) C F,; hence F(a) = F,
<
for some r n. It follows that the only finite subfields of Q are the F,; the only infinite subfield is 9 itself. Let u be an element of the group G of Q 1 F which acts on Fl like a generator of the group of Fl / F. If a(a) = a for an element a E SZ, the field F(a) = F, remains fixed. If a 6F, then F(a) 3 Fl , so Fl remains fixed, contrary to the definition of o. Hence a E F. Thus the fixed field of H = {av}is F itself. By Lemma 3, G = R. Now if the group of Q I F, is U, we have Ul 3 U, 3 ..., so that G satisfies the first countability axiom. Hence, if T E G, we can express T as lirn,,, oav where the a, are integers. Now we examine the conditions under which a sequence {oav} converges. For any given neighborhood U, , there must be an index N such that for p, v N, E U, , i.e. U ~ V - ~ Mleaves Fn fixed; la is a generator of the group of F, 1 F; hence a, - a, E 0 modpn. Hence {uav} convergent => {a,) convergent in the p-adic topology to a p-adic integer a. We may write symbolically T = ua. This is well-defined, since if a = lim a, = lim b, , we have a, - b, -+0, hence oav-b~leaves high Fn fixed and so lim uap = lim ub~r.I t is obvious that uaop = oa+p. Hence the group G of Q / F is isomorphic to the additive group of p-adic integers.
>
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)
<
Theorem 4: If E / k is any extension of degree n, then S(E I k) is a cyclic group of order n. Further, if Tn is the unramified extension of degree n, then the cocycles of (C, which split on E are precisely those which split on Tn . We have already remarked that every cocycle has a splitting field of finite degree (Chapter 6, Section 4). Hence every cocycle has an unramified splitting field of the same degree. From this we shall deduce
r)
Theorem 5: B,(C, r)is isomorphic to the additive group of rational numbers modulo 1.
4.
THE SECOND INEQUALITY: A REDUCTION STEP
133
4. The Second Inequality: A Reduction Step
The aim of this section and the next is to prove Theorem 6: If E 1 k is any extension of degree n, then [E : k] n. The first stage in our proof is a reduction to the case where E I k is an extension of prime degree. Let p, be any prime number. S(E I k) is an abelian group, and so the elements of S ( E j k) with period a power of p, form a subgroup which we denote by Spl(E I k); let the order of SPl(E I k) be [E : KIP,. We first prove some elementary properties of the symbol [E : KIP,. Let E' be a finite extension of E. It is easy to see that SPl(E I k) is contained in the corresponding group SP1(E1I k) of order [E' : k],, . We have the following results:
<
This follows at once from the definition.
<
(2) [E' : kIpl [E' : Elp1[E : k],, . SP1(E1I E) is a subgroup of $,(C, H ) where H i s the subgroup of r which leaves E fixed. Thus we can map SP1(E1I k) into Spl(E1 I E) by the canonical homomorphism, i.e. by restricting the subscripts to H. Obviously the number of images is < [E' : Elp1. The kernel consists of these elements of SP1(E1I k) which are cohomologous to the identity when their subscripts are restricted to H ; thus the kernel consists of the elements of SP1(E1I k) which split on E, namely Spl(E I k), which contains [E : RIP, elements. Hence [E" : k],, [E' : El,, [E :k],, .
<
(3) If p, does not divide deg (E I k), then [E : kIp1 = 1. For every element {a) E S(E I k) we have {a)" = {I) (Chapter 6, Theorem 9). But for every element {a) E SPl(E 1 k), we have {a)pl= (1). If p, does not divide n, these statements imply that {a) = (1). Hence [E : k],, = I. Let K be a finite normal field containing E and let G be its Galois group; let G, be the subgroup of G which leaves E fixed.
7.
THE FIRST AND SECOND INEQUALITIES
5.
THE SECOND INEQUALITY CONCLUDED
of p,, which must be (E / k),, degree (E I k) since
135
, i.e. the pl-contribution to the
(E, I El) (El I k) = (E, I k) = (E2 I E) (E I
4.
The extension E, I El can thus be expressed as a cyclic tower of degree ( E I k)pl. Assuming that the second inequality is true for cyclic extensions of prime degree, and applying (2) above, we obtain [E2 : E1lPl G (E I k)Dl ' (by (1) above)
< [E, :EIIP1[El : kl, , 5 (E 1 k )
since
(by (2) above)
(El ( k) is prime to p1 .
It follows that [E : kIp1 is finite for every prime p, . Hence the groups SPl(E / k) are finite for every prime p, , and are different from the identity only if p, divides n. Hence S ( E I k) is itself a finite group, and the groups SPl(E I k) are its Sylow subgroups. Thus [E : k]
=
r]:[E : k],, <
(E I k)P1= n.
pl
Hence we have proved the second inequality under the assumption that it is true for extensions of prime degree. 5. The Second Inequality Concluded
Let Q be a p,-Sylow subgroup of G1 , E, the corresponding field. Now Q, considered as a pl-subgroup of G may be imbedded in a pl-Sylow subgroup Q' of G; the corresponding field El is a subfield of E, . The degree of E, 1 El is prime to p, since Q is a maximal pl-subgroup of GI; similarly the degree of El 1 k is prime to pl , since Q' is a maximalp,-subgroup of G. Now the degree of E, 1 El is a power
We have now to consider a normal extension K / k of prime degree 1. The Galois group of such an extension is cyclic, generated by an element o. We introduce the following notation, continuing our convention that groups shall be denoted by their generic elements: Let A denote the generic element of K*, the generic element of k*, a! the generic unit of K, E the generic unit of k. E
136
V,
7.
5.
THE FIRST AND SECOND INEQUALITIES
Let p, rr be the prime ideal and a prime in k. Let Q,l7be the prime ideal and a prime in K. Let II, be an element of K in Vs;i.e. ord 17, 2 s. Let j3, be the generic element of k which is r 1 mod n. Let v, be the generic element of k of the form v, = &NE (hence E N E mod pn: vn is a norm residue mod pn). Finally, we denote the index of a group G, in Gl by (GI : G2). First we prove the group-theoretical
First we consider (E: vl). Since v1 = plNE 3 p1rz,
we have (€
(E
Proof: Since T(a,P)
= T(/3), we have (a : aTP) = (Ta : TP).
Since K / k is cyclic, we know that the subgroup S(K I k) is isomorphic to the factor group aINA. Thus [K : k] = (a : NA); hence we must prove that (a : NA) 6 I. We have already established this for unramified extensions, so we shall assume that K is not unramified, and so, since I is a prime, K is totally ramified. The lemma above gives us a first reduction step; for if we map A -+ I A I, and hence a -t I a I, we obtain ( a :NA) = (1 a
1 : 1 NA I) (c
Since K is completely ramified, have to prove (E : NE) < I. We shall consider the index
(1
a
: NE).
=1
l
]
Next we consider (vn : v,,)
(€
: plE1).
l (n
--+ k g z is
an isomorphism. Hence
when when
2 I).
1= p , l+p.
Since j3, 3
We shall now compute the norm of 1 integer in k. We have
+x&,
we have
where x is an
Hence
where C' ranges over all polynomials
1 : I N A I) = 1. Thus we
and show that only one of the factors on the right can be different from 1, and that this exceptional factor cannot exceed I. We shall also show that for a high enough value of n (which we can determine precisely) v, = NE. This will, of course, prove the second inequality.
<
< (E : P1el) = (kg : kgg).
:vl)
When I = p, the mapping kg v
Then
: VJ
Mapping the units E into their residue classes modulo p , and applying the lemma, we have
Lemma: Let T be a homomorphism of a group a into some
other group. We denote the image group by Ta, and the kernel by a,. Then, if j3 is a subgroup of a,
THE SECOND INEQUALITY CONCLUDED
+,(u) = a,
+ alu + .--+ al-lu-l
which have precisely v coefficients equal to 1, and the remaining I - v coefficients equal to zero. When v = 1,
When v = I,
138
7.
5.
THE FIRST AND SECOND INEQUALITIES
In general, if $,(a) is a polynomial of the type described, so is oi$,(u) for all indices i (0 i 1 - 1). For v < I, all the ai$,(u) are distinct. Indeed, if ai+,(o) = aA$,(a), we have ai-j$,(u) = $,(a). Let k be the minimum exponent such that ok$,(a) = $,(a); since U ~ $ ~ (= U )+,(a), it is easy to see that k is a factor of I. Hence k = 1 (I is a prime). Thus
< <
THE SECOND INEQUALITY CONCLUDED
139
where 3 is the different. Let 3 = 'pm;we have shown earlier how m may be calculated. Then s ( F ) = pr 0 'p8"-1r c D os+m--Ir>O u r < -m+s 1
,d and
$D s+m-(l+l)r
and
r > m7 +-s 1 .
vfm-(l+l)c
Finally we obtain the result that Hence all the ai are equal, and We now return to consider the index Hence for v < I all the associated polynomials ai$,(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
<
<
implies 2(1- 1) m. Since 1 2 2, we have I 2(1- 1); hence I m. This implies that 3 = 'pm C 'p' = p, and hence that p-1 C 3-l. If T is an element of k, with ordinal 1, r-l E p-l C 3-l, and so, by the definition of the inverse different,
<
and for any integer a E k, as(&) = S(a 17,). Thus S('pS)= pr for some r, which we shall now determine. S ( p ) = pr e p-'S((PL) = o e S ( P - ~ D '= ~ ~o ) e S(D(PL-lr) C o + 'pa-1r C 3-1
and and and and
p-('+l S('P" #
0
S(p-(?+l)D'p8)# o S(D!$P-(T+l)l) 4o 'pa- (r+l)Z 4 B-l,
This means that T / I, and hence, since I is a prime, I = p. We now consider the three cases in turn.
+
1) (I - 1) > m. Case I: (n Letusdefines=(n+ 1 ) l - m -
1.
7.
140
n
Then s
5.
THE FIRST AND SECOND INEQUALITIES
+ 1; hence
THE SECOND INEQUALITY CONCLUDED
When I is not equal to the characteristic 1. Thus field, this result holds for all n
N(P8 = pS C pn+l.
Also,
(E
m+s=(n+
1)l- 1;
p of the residue class
:NE) = (E:v,) (v, :NE) = ( E : v,).
We have already proved that when I # p, (E : v,)
hence
141
( a : N A ) = ( E : NE)
< I, and hence
< I.
Thus we have proved the second inequality in this case. For cases 2 and 3 we have I = p.
and
Case 2: Then
Thus S((PS)= pn
and
S ( p + l ) = pn+l.
Hence there is an element 17, with ordinal exactly s such that S(l7,) = e r n . If x is an integer of k, we have
+ x n S )= 1 + xS(.K) + xZN(&) + 2 xvS(17,,)
(n
+ 1) (1 - 1) < m.
(n+l)(I-1)+1
and
=(n+ 1)l-n<m,
(n+ 1)1<m+n.
Hence
1-1
N(1
-2
+ xS(17,) mod pn+l 1+ mod pa+,.
and so
=1 3
EX#
+
Every element Pn may be expressed as 1 EXT"; hence every is congruent to the norm of a unit modulo pn+l. Now fin zz NE mod pn+l
&!- = 1 mod pn+l
NE -
On the other hand, N('pn) = pn. Hence if 17,is any element of p, and x is any integer of k, we have
-
Pn
e-- is a fin+,.
NE
Thus we have Pn C Pn+,NE, and hence
+ +
+ +
1) (I - 1) > 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.
<
Hence, finally (E : NE) I. This completes the proof of the second inequality for completely ramified cyclic extensions of prime degree; this, as we have seen, is sufficient to prove the inequality in all cases. The fact that [K : k] = n means that when K I k is cyclic with degree equal to the characteristic of the residue class field, Case 3 actually occurs. That is to say, there is an integer no such that (no
+ 1) (I - 1) = m,
and
(vno: vno+l)= 1.
Then the elements pno are not all norms of units, while all the (i = 1, 2, 3, are norms of units. The ideal elements 5 = pno+l is called the conductor of the extension, and we see that ..a)
< 1; where 3 is the different.
We have now proved the second inequality for the case I =pa For we have (E :vN) = (E: vl) (vl :v2) "' (vN-1 : vN). When I = p, (E : vl) = 1.
> no ,
143
1.
THE TEMPORARY SYMBOL
Then
Let us denote the norm group of K I k by NKIkA;when we can do so without confusion we shall omit the subscript. We see that
CHAPTER EIGHT
The Norm Residue Symbol 1. The Temporary Symbol (c, K I k / ~ )
Let k be a complete field with discrete valuation and a finite residue class field. Let K be a finite normal extension with Galois group G. Let c be any element of the second cohomology group &(K, G),and let a,,, be a representative cocycle; then a,,, satisfies the associativity relation
where a,,, , b,,, are any representative cocycles of c. We now define the symbol (c, K lk/~)by writing
where a,,, is any representative cocycle. G)we have Obviously, if c, and c, are two elements of &.(K,
We have also the result
We shall make repeated use of this relation. Let
then f ( ~ E ) k, for
But
since
Hence
Hence
[f(.)lp
=
n mc
aw
=
J Ja,., = EG
f(7).
Now let us consider the effect of taking another representative cocycle for c ; let
1. Hence the mapping T -+ (c, K I k/r) is a homomorphism of the Galois group of K I k into the factor group a/NKlk. Since the image group is commutative, the kernel of this mapping contains the commutator subgroup G' of G.
(I) Let KObe a subfield of K which is normal over K. Let H be the subgroup of G corresponding to KO;then H is a normal subgroup. We choose a fixed system of representatives for the cosets of G modulo H ; we denote by 6 the representative of the coset containing the element a. Thus if y is the generic element of H, we have = 76 = 6. Now let c be an element of the cohomology group $,(K, G); let a,,, be a cocycle representing c. We shall now prove that if the degree of K over KOis m, then a:,, is cohomologous to a cocycle do,, for (KO,GJH), and hence that cm may be considered as an element of $,(KO, GIH). Let
THE TEMPORARY SYMBOL
since
since
TI=
.t
and
-
-
myl = 07.
Similarly daYl,,= do,, . Hence do,, is a cocycle for (KO, GIH). (Cf. Chapter 6, Section 3, Lemma 2.) We can thus define the symbol [(cm,KO1 k)/rH], which we shall write simply as [(cm,KOI k ) / ~ ] .We have
and define Now as a runs through a complete system of coset representatives, so does &; hence We shall show that d,,, is a cocycle for ( K O ,G/H) Thus we have
By the associativity relation, a,,pw,, = a;,sa,,,,
and hence
~U.@W.? a",, = ----,
Since the group NKIkis contained in the group NK,lk (this follows from the transitivity of the norm), we have
a 0 3 ~ ~
Sunstituting these results, we obtain
We may choose
? = T,
and so obtain finally
148
8.
(2) Let KObe any subfield of K , not necessarily normal. Let us write as before, G = U (TH. By the symbol ResKo(c) we shall mean the equivalence class of cocycles obtained by restricting to H the subscripts of each a,,, in c. Let T be an element of H ; then we can form the symbol
for any T
EH.
1. THE TEMPORARY
THE NORM RESIDUE SYMBOL
Then
149
SYMBOL
(4) I n order to describe the fourth important property of ] must first introduce a group-theoretical our symbol [(c,K I k ) / ~we notion. Let G be a group, H a subgroup of finite index; let G', H' be the commutator subgroups of G and H respectively. Let us choose fixed representatives ai for the cosets of G modulo H : G = U Ha,; define 6 = ad for every element a E H o d . Consider the element ~ ~ r o i 7where -l T is any element of G ; since u p E -, we have 6 7 - I E ~ - l o r l H ,and hence ui~a,7-l E H . We define the transfer (Vorlagerung) of T into H to be
V ( T )does not depend on the choice of coset representatives ad . = a, , we have If we write using the transitivity of the norm. The associativity relation gives Replacing o, by ydad, we obtain
Since
T E
since cosets of H modulo H' commute with one another. We can also write G as a union of left cosets modulo H: G = U uTIH. Let us define t.5 = oil for every element a E o i l H . Then we have
H
so we have
-
Hence t.5 = o-1-l. We can define another transfer in terms of the 6 by writing Finally we have, for all
T
E H,
n
N
V*(T)= ] IT U ~ ~ - ' T U=~ ~ H?TU;~H'. ' 7 ',
0-1
I
-
(3) Let h be an isomorphic map of K onto K" under which k
is mapped onto kL. If c is an element of & ( K ) represented by a,,, , we define the element c q n &(K" to be the class of cocycles It is easily seen that represented by a:,, = bAoL-l,LTA-l.
Let a , ~ - l= a,; as i runs through all possible indices, so does m. We have ad+ E H a , , hence a, E Hu,T, and so u,7 = a, Hence, finally, V*(T)= r ]:U ~ T U ~ T - '=H V(T). '
.
v
Om
Thus we may define the transfer by using either left or right coset representatives.
1 50
8.
1.
THE NORM RESIDUE SYMBOL
It is fairly clear that the transfer is a homomorphic map of G into H / H f . For
THE TEMPORARY SYMBOL
But D yu;~,,G-I lies in NKkK0.Hence
where we write ui = G, and then by ai , then as ai ranges over all coset representaIf we replace tives, so does ai . Thus we have
We notice also that We use the associativity relation again, and obtain since cosets modulo H' are commutative. Thus the kernel under the transfer map contains the commutator subgroup G'; hence the transfer defines a homomorphism from GIG' into H / H f . Now let G be the Galois group of the extension K 1 k, and let H be the subgroup corresponding to an intermediate field K O .We shall study the symbol
lies in NKIKo . Hence But 17,a~~i,,,,
T o conclude this computation, we must show that the norm group
NKIkis contained in N K I K o . We have we have immediately
We can use the associativity relation to show that
Hence
2.
CHOICE OF A STANDARD GENERATOR
c
153
is a generator of a/NA. Now let c be kept fixed, and consider the mapping oi -+
Let K I k be a cyclic extension of degree n; let a be a generator of the Galois group G. We recall that each group extension defined by K and G can be described by a coset representative u, and an element a E k such that uon = a. The element c of the cohomology group $,(K, G) which corresponds to this extension is represented by a cocycle (1 if v+p
Hence
and consequently
Conversely, if [(c, K I k)/o] is given, the element c may be represented by a cocycle of the form
where a is any element in the coset [(c, K I k)/u]. We recall that for any extension K I k of degree n, not necessarily normal, the group S ( K I k) is cyclic of order n. When K I k is normal, with group G, S ( K I k) is naturally isomorphic to S2(K, G); further, when K 1 k is cyclic,
(9) . =
This obviously maps the Galois group G onto a/NA; since both these groups are cyclic of order n, the mapping is an isomorphism.
2. Choice of a Standard Generator c
Our next task is to choose a standard generator c for B2(K, G). T o this end we shall associate with each element c a certain numerical invariant. We consider first the unramified extension T, / k of degree n. As we have seen earlier, the group G of T, I k is isomorphic to the group of the residue class field extension, and hence is cyclic of order n. We use this fact to single out a definite generator for G, determined by intrinsic properties; namely, we choose as generator, a, the isomorphism which acts on the residue class field by raising every element into the qth power. (q is the number of elements in the residue class field of k). We shall call this generator o the canonical generator. Now let c be an element of B,(T,, G). We have seen that c corresponds uniquely to a coset aNrn,, of a/Nrfi,, . The elements .of NTn,,are precisely the elements of k whose ordinals (with respect to the prime in k) are either zero or a multiple of n. Hence if a' is another representative of the coset a N T n l kW , e have ord a
z ord
a' modulo n.
Thus ord a (mod n) is an invariant of the element c. In order to have a uniform module we prefer to use the invariant I=-
Thus when K I k is cyclic of order n, a/NA is cyclic of order n. Let c be a generator of B,(K, G); then, if o is a generator of G,
aiNA
ord a mod 1, n
and we now write c as c,(T, I k). Obviously the invariant r can be any one of the fractions 0, lln, 2/n, ..., (n - I)/n. Theorem 1 : c,(T, I k) = criT,, fied extension of degree ns.
I k), where
T,, is the unrami-
2. Proof: Let a be the canonical generator for the group G of corresponding Tns I k. Let H be the subgroup (1, on, -.., on-lH), and the to Tn . Then the group of Tn I k is (H, uH, canonical generator is obviously OH. Now the element c,(Tn I k) of sj,(Tn, GIH) can be represented by a cocycle
where a
E k.
CHOICE OF A STANDARD GENERATOR c
155
By the previous theorem, we need not refer to the extension field explicitly, provided its degree is divisible by the denominator of r; hence we can write the element simply as c,(k). I t is easy to show that
4 4 cdk) = cr+,(k). Hence we have proved Theorem 5 of Chapter 7, namely that 4i2(C,r)is isomorphic to the additive group of rational numbers modulo 1. Furthermore, we have now given a description of the isomorphic mapping between the two groups. Before introducing the norm residue symbol we have one more theorem to prove.
Then
Theorem 2:
and
Let E I k be an extension of degree n. Then
ord a -= r mod 1. n
Consider next the cocycle for (T,,
, G), defined
by writing
This defines an element c of sj2(Tn,, G), for which we have
Hence the invariant for c is congruent to as ord -=--
ns
Proof: (1) First let E = Tf be an unramified extension of degree f. Let Tn be a further unramified extension, of degree n, such that c,(k) splits on Tn; that is to say, the denominator of r divides n, and c, may be regarded as an element of sj2(Tn). Let a be the canonical generator of Tn I k. Then c,(k) may be represented by a cocycle
where ord, a a ~ k , --
ord a - r mod 1. n
Thus c = c,(Tn, 1 k), and so our theorem is proved. Let C be, as before, the separable part of the algebraic closure of k, r the Galois group of C I k, and consider the second cohomology group of C I k. Since we consider only continuous cocylces, every element of sj,(C, r ) splits on some finite extension of k. The second inequality implies that every element of &,(C, r ) which splits on an extension of degree n, splits also on the unramified extension of the same degree; hence it is of the form c,(Tn I k).
n
= r mod 1,
and
Clearly of is the canonical generator of Tn I Tf , and ResTf (c,) can be represented by the cocycle aav,,alrf, Then we have
.
2. Thus the invariant of ResTf(c,(k)) is given by o r d ~a, - ford, a -----=fr mod 1. n lf
n
(2) Next we consider the case where E / k is a totally ramified extension of degree e. Let T, be an unramified extension of k, of degree n, such that c,(k) splits on T, . Let C be the separable part of the algebraic closure of k, I' its Galois group, and H and U the subgroups corresponding to E and T, respectively. Let the canonical generator of the group of T, / k be aU. When we considered this situation earlier (Theorem 3), we saw that there is no loss of generality if we take a E H , in which case
c
CHOICE OF A STANDARD GENERATOR
157
It is clear that u(U n H ) is indeed the canonical generator: for since E / k is totally ramified, the residue class field of E is the same as that of k. Since Tn splits c,(k), we may represent c,(k) by a cocycle
where ord, a n
--
a ~ k and
= r mod 1.
The restriction of c,(k) to E is then represented by the cocycle
-
a o v u ~ ,~ ua o~ v
( ~ n, o~P)( u n H )
which may be regarded as a cocycle for (ET,
I E).
Then
, = a.
We have seen also that ETn I E is an unramified extension of degree n; the cyclic group of this extension is obviously u(U n H). Hence the invariant of Res, (c,(k)) is given by
(3) Finally, let E / k be an extension of degree n with ramification e and residue class degree f. Let T be the inertia field: T I k is unramified of degree f, and E I T is totally ramified of degree e. Then
= Res, (cf,(T))
by part 1
Since ef = n, this proves the required result.
158
8.
3.
THE NORM RESIDUE SYMBOL
3. The N o r m Residue Symbol for Finite Extensions
THE NORM RESIDUE SYMBOL FOR FINITE EXTENSIONS
159
I n the same manner we prove the remaining two properties
We are at last in a position to define the norm residue symbol. Let K I k be a normal extension of degree n; let a be an element of the Galois group r of C 1 k. We define the norm residue symbol for K I k and a to be
Since the unramified extension of degree n splits c,,,(k), so does K , and hence the symbol on the right is defined. We can now obtain the properties of the norm residue symbol very easily. Let KObe a subfield of K ; let H and I& be the subgroups of r corresponding to K and KOrespectively.
( 1 ) If KOI k is normal, of degree m, and
T
E
r , we have
Elk (T)
Klk
=
(?)
N ~ l k
This is consistent with our previous definition, for if E is normal our definition is identical with Property 1 above. We must now show that the new symbol is well-defined, i.e. that it is independent of the choice of K . Consider first another normal extension K' containing K . Then
since N K I kC N E l k. Next, if K and K' are any normal extensions, each containing E, we have
Hence
(2) If KOI k is any extension, and
So far our definition of the norm residue symbol is restricted to the case where K I k is a normal extension. We now extend the definition for arbitrary extension fields E 1 k, as follows: Let K be any normal field containing E. Then define
T
E
Ho , we have
(T)K I k
NE, k
=
KK' ( k
NE k
(PT-)
=
Kt ( k
NE k
-
Hence (E I k / ~is) well-defined. The mapping T -+ (E I KIT) is obviously a homomorphism from the Galois group r of C I k into the group cr/N,,,A. We now list the properties of this general norm residue symbol and the mapping it defines. Hence
( 1 ) Let E, be any subfield of E. Then
3. For let K be any normal field containing E. We have
Thus
7
THE NORM RESIDUE SYMBOL FOR FINITE EXTENSIONS
acts on the residue class field CY
(2) Let T be an element in the subgroup corresponding to E,
.
Then
161
E = TIby sending
("1
+ CYQ' = dm, where m = ord -
.
This is well-defined since orq' = a. Our next task is to prove The homomorphism T +-(K I k / ~is) always onto. We have already seen that the theorem is true when K I k is cyclic; we now show it is true for cyclic towers by proving the Theorem 3:
For
Lemma: If E, 3 El 3 k, and the mappings T + (E2 I E1/7) and 7 + (El I KIT) are onto, then the mapping 7 -+(E, I k/r) is also onto.
(4) The property of the norm residue symbol involving the transfer does not necessarily hold for non-normal extensions. ( 5 ) If K I k is a cyclic extension, the mapping
is one-to-one and onto between the Galois group G of K I k and the factor group ol/NKlkA. (6) Let E I k be an arbitrary finite extension, T j its inertia field. Let T be an element of I'; if T acts on Tt= E (and hence on Tf)like a%,where a is the canonical generator for the Galois group of TfI A, we define the ordinal of 7 to be ord r = i.
Elk
),
=
(
Hence we have ord
T (k
) =(
-
(T)
T, I k
Then let
Multiplying by the group NEllkwe obtain
Hence alb E NEllk;say alb = NEllk(al)where arl esis, we can find an element 7, such that
Now, by property 1,
(
Proof: Let a be a given element of k, aNEzlk its coset modulo NEllk.We have to construct an element 7 E r which is mapped onto this coset. We can, by hypothesis, find an element 7, such that
O'd'
)
= +rd'N~,lk
. Take norms from El to k,
ord r mod f.
E El
. By hypoth-
Now we have
That is.
This proves the lemma. We have already remarked that this proves the theorem for cyclic towers; let us now apply this result to the proof for an arbitrary normal extension. This is obviously sufficient to prove the theorem in general, for if E is any extension, K a normal extension containing E, we have
Proof for Normal Extensions: Let K 1 k be normal of degree n = pmr, where (p, rp) = 1. Let Ep be the subfield of K corresponding to ap-Sylow subgroup of the Galois group. Then K / E, can be broken up into a tower of cyclic extensions of degree p. Hence if a is any element of k we can find an element T, such that
Taking the norm from Ep to k, we obtain
This construction may be carried out for every prime dividing n. Since the residual factors rpl , rpB, ..., rpk are clearly relatively prime, we can find integers xc such that 2 xirpi = 1. Hence
where This completes the proof of the theorem.
Having shown that the mapping T -+ (E I ~ / T ) is an onto homomorphism, our next problem is to find the kernel of the mapping. In the case of normal extensions with abelian groups the answer is provided by Theorem 4: Let K I k be an abelian extension. Let H be the subgroup of r which corresponds to K. Then H is the kernel of the mapping T -+ ( K I k / r ) .
Proof: The theorem has already been proved when K I k is cyclic. Suppose then that K I k is not cyclic, and proceed by induction on the degree of K ( k. Let r0 be an element of the kernel. Then (K I k / ~=~NKlk ) . Let K Obe an intermediate normal field, corresponding to subgroup Ho . Multiply by NKoI,,we obtain
By the induction hypothesis, T , E Ho . If we denote by 5, the effect of 7 , on K, we see that 5, lies in all proper subgroups of the abelian group r / H . Since we assumed that r / H is not cyclic, it follows that T~ must be the identity element of r / H ; i.e. r0 E H . Hence the kernel is contained in H ; and clearly H is contained in the kernel, so the theorem is proved. We have now our main theorem: Theorem 5: If K 1 k is an abelian extension with Galois group C, the mapping T -t ( K 1 k / ~is) an isomorphism between G and the group a / N A . Clearly ( K : k ) = ( a : N A ) . I n order to determine the kernel in the general case, let us examine first the maximal abelian extension A of k. Let U be the subgroup of corresponding to A . Then U is the smallest closed subgroup such that I'/U is abelian; that is to say, U is the closure of the commutator subgroup of I'. Let us define this closure as the actual commutator subgroup Then we have
r
rf.
Theorem 6: Let E I k be an arbitrary extension. Let H be the subgroup of corresponding to E. Then the kernel of the mapping T -t (E I k / r ) is the subgroup r f H .
r
164
8.
THE NORM RESIDUE SYMBOL
Prooj: Let the kernel be W C r,,, for
ro. Then clearly I" C T o . Further,
hence r ' H C ro. Now let Ea be the maximal abelian subfield of E : Ea = A n E, and Ea corresponds to the group I"H. By Theorem 5 we have
Hence ro= r ' H and our theorem is proved. As a result of this theorem we have
CHAPTER NINE
The Existence Theorem 1. Introduction
In this chapter k still denotes a complete field with discrete valuation and finite residue class field; k* is the multiplicative group of non-zero elements of k. Let A be the maximal abelian extension of k, G the Galois group of A ( k . Our aim is to extend the definitions of norm group and norm residue symbol to infinite extensions K of k; and for this purpose we are led to make a change in the topology of k*. k* is not compact in the valuation topology, so we introduce a new topology similar to that of the Galois group G. k* is now relatively compact, but no longer complete; we therefore form its completion A. Finally we construct a (1, 1) correspondence between K and G, which is both an isomorphism and a homeomorphism. We can then show that to every closed subgroup M of k, there corresponds a field K , such that the norm group NKMIk = M. This is the Existence Theorem. 2. The Infinite Product Space
i
Let I denote the ring of rational integers, I9 the ring of p-adic integers with the p-adic topology imposed on ~ t I,; is compact in I,; the this topology. We form the infinite direct product I = elements of I are the vectors
np
with one component for each rational prime. We impose the usual 165
166
9.
Cartesian product topology on 1; hence, by Tychonoff's Theorem,
f is compact.
I forms a ring under componentwise addition and multiplication, and it is easily verified that the ring operations are continuous in the product topology. A fundamental system of neighborhoods of zero in f is given by the ideals f1,where f E I. For suppose
then ff consists of all vectors m with m, 5 fIp; fIp= I, if p does not divide f, while fIpi= piviIpi. Thus fI describes the Tychonoff neighborhood of zero given by the local neighborhoods piuiIpi (i = 1, --.,r). If m is any vector of 1, the index of mf in I is the product of the indices of mpIp in I, . Hence this index is finite only if m, # 0 for all p, and m, = I for all but a finite number of p. It follows that the mf of finite index in f are the Tychonoff neighborhoods of zero, of the form ff with f E I , f f 0. Clearly I contains a subring isomorphic to I ; for there is evidently a (1, 1) correspondence between the rational integers r and the vectors (... r, r, r, of I. We shall therefore consider I as isomorphically imbedded in I. This imbedding gives rise to an induced topology in I: a fundamental system of neighborhoods of zero in the inherited topology is formed by the sets In f1,i.e. by the ideals fI. An element of I is therefore "near zero" in this topology if it is divisible by an integer containing high powers of many primes. I also contains an isomorphic replica of each I,, imbedded under the mapping . a - )
Theorem 1:
2.
THE EXISTENCE THEOREM
f is the completion of I in its inherited topology.
Proof: Since f is complete (because compact), it is sufficient to prove that I is everywhere dense in I. Let m be an element o f f : m = (.-., m, , let .-a);
THE INFINITE PRODUCT SPACE
f
167
We have to find an element r of I lying in the "f-neighborhood" of m; i.e. we must find a rational integer r such that r = mPimod pli
.
But, by the Chinese Remainder Theorem, this set of simultaneous congruences always possesses a solution. Hence our theorem is proved. Theorem 2:
The closed ideals of
I are principal
ideals.
Proof: Let a be a closed ideal of I. Then a 3 alp; alp is an ideal in I,, which is a principal ideal ring. Hence alp = mpIp !here m, is either zero or a prime power pv. It follows that a C mI where m = (..., m, , On the other ... mp,Ip,; the set hand, a contains all ideals mplIpl mp,Ip, of these is everywhere dense in mf; so a is dense in mI. But a is closed, so a = mI. Now let G be a topological group with a HausdorfT topology defined by a fundamental system of neighborhoods of the identity given by certain subgroups of finite index. (An example of such a group is the Galois group G of A). It is easily shown that if G is complete in this topology, then it is compact. Let a be an element of G, and let (a) be the closure of the cyclic subgroup generated by a; we wish to find a description for (a). Clearly we have a homomorphism of I into (a) defined by
+
m . 0 ) .
+ +
This map is continuous in the inherited topology of I ; for let V be a neighborhood of the identity in (a), with index j in (a). Then v = 0 mod j =- a VE V; but since the numbers v = 0 mod j form a neighborhood of zero in I , this is precisely the statement of continuity. The mapping v -+a" may now be extended from the dense subset I to the whole space I ; that is to say, we may define am for every m in f. The extended mapping is continuous, and the extension is unique. We may also show that the mapping (a, m) -+ omis continuous from G x f to G.
168
9.
The usual rules for exponentiation may be easily verified:
Since the map o + omis continuous from 1' to G, and I is compact, its image is closed, and hence contains (a); on the other hand, all the om lie in (a), so the image space is precisely (0). The kernel of the map is the inverse image of the identity; hence it is a closed ideal in 1'. Since closed ideals are principal, the kernel is of the form d1', where d = d, , ...), d, = pvp; od = 1, and ov= 1 + v I d. We call d the period of a, and we have (a) Ild1'. We may describe the mapping o -t om more explicitly. (ma-,
Case 1: G is a finite group. In this case G is discrete, and (o) is the ordinary cyclic group generated by o. Since (o) r1'/d1', it follows that d lies in I , and is the ordinary period of o; hence om = 1 o d I m. Since I is everywhere dense in 1, given any m there exists a rational integer r such that r m mod d1'; then om = or.
-
Case 2: G is any group. Let V be a neighborhood of the identity. Then omVis a continuous extension of ordinary exponentiation in the factor group G/V. and m r r, This group is finite; hence omV = urVv,where Y,EI mod d v I (d, denotes the ordinary period of o in G/V). Since omE o r V v for all V, we see that om E fl, orVv.Since the topology in G is assumed to be HausdorfT, we have finally om = fl, orVv. We shall now make two important applications of the technique we have just developed. First let k be a finite field containing q elements. Let 9 be the algebraic closure of k, G the Galois group of 9/k. If E I k is any finite extension, E I k is normal, and has a cyclic Galois group generated by the mapping o : a -+an. This mapping u may be considered as an automorphism of 9. Theorem 3:
2.
THE EXISTENCE THEOREM
G = (0).
Proof: Let H be the ordinary cyclic subgroup generated by a; we shall show that H is everywhere dense in G.
THE INFINITE PRODUCT SPACE
f
169
Let r be any element of G; let TV be a neighborhood of T: V is a subgroup of G which leaves fixed a certain finite subfield E. Hence every element of rV, especially 7, acts on E like an element of its Galois group, i.e. like some power uv of o. ~ E fixed, i.e. r-lov E V; hence ovE TV. It follows that T - ~ Uleaves Thus H is everywhere dense in G; hence R = (a) = G. For our second application let k be a complete field with discrete valuation and finite residue class field of characteristic p. The multiplicative group k* of non-zero elements is a topological group in which the fundamental system of neighborhoods is given by the subgroups U, of elements a 1 mod pr (r 2 1). Let Uo be the group of all units; the subgroups Ur are all of finite index in Uo , and form a fundamental system for Uo . U, is complete in this topology, and hence compact. We shall apply the technique developed above to describe the group (6) where e is a unit in Ul . First we notice that if E lies in Ul , then ep" lies in Uv+, , and This result follows from the fact that if E hence epV+ 1 as Y -+a. lies in Uv, we can write
-
Then
which lies in UV+,since rr divides p. Next we remark that if m €1 has coordinate 0 at p, and has arbitrary coordinates elsewhere, then em = 1. For let j = pvc, where (p, c) = 1; then we may find an integer I E I lying in a j-neighborhood of m, i.e. I = m mod j. In particular, I 0 modpv; hence eqies in UV+,. As j becomes "highly divisible" and hence v +a, el -+em; but by our previous remark el-+ 1. Hence our result is proved. From this it follows that the period ideal of ( E ) will be generated by an element d = (..., 1, l,pv, 1, 1, -.-) where v may be infinite, i.e. pv= 0. We now show that v is finite if and only if l is a pv-th root of unity. Clearly
-
9.
170
THE EXISTENCE THEOREM
lies in the period ideal d1; hence epv = 1. On the other hand, for any E we have
3.
THE NEW TOPOLOGY IN
k*
171
U, . Clearly U is not the whole of V, since Uo is not of finite index in k*:
if also cpV= 1 we have To describe V we let crrf be an element of V with least positive ordinal f ; then Hence if E is not apv-throot of unity, its period ideal is generated by d = 1, 1,0, 1, 1, hence (E) E1/dI, which is isomorphic to the ordinary p-adic integers. (-.a,
..a);
3. The New Topology in k*
Let k be a field of the type described in Section 1, E a finite extension. Then we have Theorem 4:
The norm group NElkis open in k*.
Proof: It is clearly sufficient to show that NElk(UE)is open in k*, where UE is the group of units in E. But UE is a compact group, and the norm is a continuous function; hence NEI,(UE)is closed in Uo , the group of units in k. The cosets of Uo modulo NEIk(UE)are therefore also closed; and hence, sincc NEIk(UE)is of finite index in Uo , the union X of the cosets # NEIk(UE)is closed. Hence NEIk(UE),which is the complement of X, is open in Uo , and hence in k*. By this fact, and by the structure of the Galois group of the algebraic closure of k, we are led to introduce a new topology in k*, defined as follows: A fundamental system of neighborhoods of the identity shall be given by the subgroups of finite index which are open in the valuation topology. k* is not complete in this new topology, so we shall have to form its completion. First, however, we describe the neighborhoods V of the identity in k* by means of subgroups of Uo . Certainly V n Uo is an open subgroup, say U, of Uo . Then Uo 3 U 3 U,; and since (Uo : U,) is finite, U is the union of a finite number of cosets of Uo modulo
We obtain "small enough" neighborhoods V by taking U = U, with r large enough. We notice that the topology inherited by Uo from the new topology in k* is the same as the original valuation topology, for in both cases the fundamental neighborhoods are the subgroups U, . Next we form the completion A of k* with respect to the new topology; A will be compact. Since the old and new topologies are identical on Uo , and since Uo was complete in the old topology, Uo is complete in the new topology, and hence compact in A. It follows that the elements of which are not in k* must be obtained from the prime n, not from the units Uo , We can certainly form the symbolic powers rrm for m €1; let us now find the period of rr. The result is given by Theorem 5: The period of rr modulo the units is zero, i.e. nm is a unit o m = 0. A fortiori, rrm = 1 o m = 0.
Proof: Let m be an element of 1. Then rrm is close to a unit o is close to a unit, where I is a rational integer close to m ;i.e. m E I mod j, where j is divisible by high prime powers. Such an element rr' is close to a unit e rrz E UoV, where V is a )~ sufficiently small neighborhood of the identity. Let V = U ( ~ r r fU. 0 mod f. Then r r b U rrfv Uo , where f is sufficiently large. Hence I Thus rr' can be a unit o I = 0; hence, finally, rrm can be a unit om=0. This result enables us to give a complete description of the
-
9.
172
group &.Since n has period zero modulo Uo , the cosets nVUo(V€1) are all distinct, and the set k, defined by
lies in
A.
3.
THE EXISTENCE THEOREM
ko contains k* and Uo as subgroups, and clearly
so that ko/Uo is compact. Since Uo is compact, it follows that ko is compact, and therefore complete. Hence the completion k of k* is in k,; hence k, = &, i.e.
The elements of & are therefore of the form a = env where E is one of the original units and v €1and v E 1; it is natural to call v the ordinal of a: v = ord E. Computation with these elements is carried out by using sufficiently high approximations enr where r is integral. Clearly if a = ~ ~ n/3~=1E,~ + , then
THE NEW TOPOLOGY IN
k*
173
We may therefore take as a fundamental system of neighborhoods of the identity the subgroups
An equivalent fundamental system is given by the subgroups
where nf is a fixed element such that ord nf = f, and I is a rational integer. T o show the equivalence of the two systems we remark:
v' v
is also of the form (1) Every neighborhood of the form to exhibit this we have only to write rrfl = m f l . (2) Every neighborhood of the form contains one of form For
v: v'.
If I is chosen so large that ( ~ n f )is' contained in nflU, then I t is clear that the expression of & in the form k = UvcInvUois quite independent of the choice of the prime n in k*. Now let us find a simple expression for a fundamental system of neighborhoods of the identity in A. In k* such a system was given by the subgroups
Thus the two fundamental systems are equivalent. Now let E be a finite extension of k. We introduce a new topology in E*, and form the completion i? in exactly the same manner as for k. Then a fundamental system of neighborhoods of the identity in E is given by the subgroups (E)
Let (k* : V) = m; m
av.
a
=Um"uT
(El
vei
since rr = De . Clearly is contained in i?; and its inherited topology is precisely the topology we constructed in &: The neighborhoods in the inherited topology are (E)
V
(E)
n K" = U mfv(UK,n K") vef
=
U vef
(k) 'rrrv~K
,
174
9.
THE EXISTENCE THEOREM
Next we define the norm residue symbol (K I KIT) to be
since r
1 mod pK"
o E
1 mod pk.
4. The Norm Group and Norm Residue Symbol for Infinite Extensions
We must show that this intersection is non-empty; this follows ) from the compactness of our space &. For the cosets ( E I k / ~ have the finite intersection property: if El , E2 , E, are finite extensions of k, let a * - ,
We notice that & has lost the field structure of k, and is now merely a multiplicative group. I t is clear, however, that a field isomorphism of k which preserves the valuation can be extended to give a group isomorphism of &. Let E 1 k be a finite extension, and let oi be the isomorphic maps of E into the algebraic closure of k; then each ai can be extended to give a map of .?? onto KE. We can therefore define the norm of an element & in .?? to be
I t is easily verified that NEIL(&) is a subgroup of
6, and that
From now on we shall use the symbol NEl,to denote the generalized norm group N E I LThe . norm residue symbol will now be used to denote cosets of k modulo this new norm group. It is easily verified that all the formulas involving the norm residue symbol are unchanged by this redefinition. Now let K 1 k be an arbitrary (i.e. possibly infinite) algebraic extension. Denote by E the generic finite subfield of K. We now define the norm group N K l kof the extension K / NKI,=
n N E .~ ~ ~ E
This is clearly consistent with the ordinary meaning of NK,, when K is a finite extension. The intersection is certainly nonempty, since the identity is always a norm from any extension. The groups NEI, are all open and closed (every open subgroup is closed); but if K I k is infinite, NEIkis no longer open, though it is still closed, and therefore compact.
E
= E,E2
... I&;
then
I t follows from the compactness that the total intersection is nonempty; it is also closed and so compact. Clearly if a G (K / KIT), then a E ( E I k / ~ for ) all finite subfields E of K. Hence ( E / KIT) = a N E l k , and so (K / k / ~ = ) aNKjk. Now let C be the separable part of the algebraic closure of k, r its Galois group. We have the following analogue of the results of Chapter Eight: Theorem 6: The map T -t (K / k / ~ is ) a homomorphism of r onto KIN^,^. The kernel is given by HKr', where H, is the subgroup of r corresponding to K and r' is the closure of the commutator subgroup of I'.
Proof: The mapping is certainly a homomorphism. Now to prove the mapping is onto, let a N K I kbe a coset of );/N,~, . Let E be a finite subfield of K. Then we can find an element rE in I' such that (E I KIT,) = a N E l k .Not only T E , but every element in the set rEHErlwill have this property, and these are the only elements with the property. The sets rEHErlare closed in r , and they have the finite intersection property. Hence flErEHErl is non-empty. Let T be an element of this total intersection; then, for every E C K, we have ( E I k / ~ = ) a N K l k, and ) aNKlk. Thus the mapping is onto. hence ( K I k / ~ =
176
9.
THE EXISTENCE THEOREM
T o find the kernel we must determine the set of T such that ( K I KIT) = NKIk. For such an element T, we observe that NKIk C (E I KIT) for every finite subfield E. Multiplying by the norm group N E l k ,we obtain
Hence (E I KIT) = N E l k .It follows that T lies in the kernel HEY for every finite subfield E; the converse is clearly true also. Hence HEP. This certainly contains HKY. the kernel of our map is On the other hand, if we interpret H K P , HErfas the groups corresponding to the maximal abelian subfields of K, E respectively, we see at once that
nE
Conversely, let a be an element of k. We form the inverse image of a under the mapping NEI,, where E is a finite extension containing k; this consists of elements AE such that NElkAE = a. Since a is a closed set and NE,, is a continuous mapping, the set of all AE is closed, and hence compact, in I?. The set NEIKJAE)is the continuous image of a compact set, and hence is compact in KO; a = NKOlk(NELKJAE)). For different fields E the sets NEIKJAE)have the finite intersection property; indeed
E, 3
*NE~IE~(A is E an~ )A E ,~
nE
Thus the kernel is HKrf. Next we require the following result on the transitivity of the norm group. Theorem 7:
Let K I k be any extension, KO a finite subfield
NEIKo(AE) C NKIKo , and Hence there is an element B in # 0.This completes the proof. NKoIk(B) We can now show that the properties enjoyed by the norm residue symbol for finite extensions are carried over to this new symbol. (I) Let k C KOC K where KO is an arbitrary intermediate field. Then
of K. Then ECK
Proof: In the definition of the norm group N K l k , we can restrict the finite subfields E to those containing K O . For all such fields E we have
where the E I k, E' I k are finite extensions. Thus
for every E C K and every E' C KO. In particular,
C NEIKo , and hence, since NKIKo N K~ N I KIK C JNEIR . Taking the intersection over all such E, we obtain
for every E' C KO Hence
178
9.
4.
THE EXISTENCE THEOREM
But ( K I k/r)NKolkand (KO/ k/r) are both cosets of NKolk; hence they are equal, and we have the result
modulo
THE NORM GROUP AND NORM RESIDUE SYMBOL
Now we have
where E runs through all finite extensions contained in T. Hence
(2) Let KOI k be a finite extension. Then
ECT
)7(
c such nE
(7)
Elk
=
ord
Klk
(--)
.
Now ( K I KO/r)is a coset of KOmodulo NKIKo; by the transitivity of I KO/r)is a coset of 21 modulo N K l k .SO also is the norm NKoIk(K ( K I KIT); hence we have the relation
(3) Again let KO be a finite extension. Then, replacing K by KKo in the result just obtained we have
Multiplying by the norm group N K l kand applying the first result we have
(4) Let K I k be an arbitrary extension, T the inertia field. The group of T I k is isomorphic to the group of the residue class field T I which consists of symbolic powers of the canonical generator o:a + aq. If 7 acts on T(=E) like am(m E I), we define ord T = m.
ECT
Thus we have
for all finite extensions E between KOand K. Hence K 1 KO
179
(T)K l k
= m = ord r .
Let A be the maximal abelian extension of k; let G be its Galois group. Let K I k be any abelian extension, corresponding to the subgroup H of G. We know that the mapping of G into KINKlk given by
is an onto homomorphism, with kernel H. Thus the mapping defines an onto isomorphism from the Galois group of K 1 k (which consists of the cosets of G modulo H) onto &/N,~,. We now assert that this is an isomorphism not only algebraically, but also topologically; i.e. that the mapping is bicontinuous. Let HE be a neighborhood of the identity in the Galois group; HE leaves fixed a subfield E. Then
since HE is the whole Galois group of K I E, and hence ( K I E/HE)= Now NElkis an open subgroup of finite index in A, and hence is a neighborhood of the identity in the new topology. Thus we have established that the inverse map, from K / N ~ to the Galois group is continuous. But this is a ( I , 1 ) continuous
e.
~ ~
180
9.
THE EXISTENCE THEOREM
map from the compact space &/N,,, to the Hausdorff space G. Hence, by a well-known theorem in topology, the mapping is a homeomorphism. 5. Extension Fields with Degree Equal to the Characteristic
Let k be an arbitrary field of characteristic p # 0, and let K be a normal extension of degree p. The Galois group is cyclic; let a be the generator. Since K is separable, it can be generated from k by the adjunction of a single element; we wish to find a generator which satisfies an especially simple equation. Since K is separable, the trace is not identically zero; hence there is an element 8 # 0 in K such that S(8) = b # 0 (b E k). e does not lie in k, for if a E k, /i? - UP= S(8) = b # 0.
Conversely, let us examine the polynomials in k[x] of this form; so let f(x) = x P - x - a. Let K be the a splitting field of f(x), and let or be a root. Then oiP - a = a, and hence
+
+
Hence the roots of f(x) are a, a 1, a ( p - l), and so K = k(a). If f(x) has one root in k, then it splits into p linear factors in k, and a has the form CP - C , with c E k. On the other hand, if f(x) is irreducible in k, then K I k is a normal extension of degree p, hence cyclic. The elements of the Galois group map each of the roots into one of the others. Hence u,a = a v (v = 0, 1, -..,p - 1); we may choose as generator u the element 1. which maps a onto a me-,
+
+
6. The Existence Theorem
Hence K = k(8). Now consider
Let k be a complete field of the type we considered in Sections 1-4. Let A be its maximal abelian extension. Our aim in this section is to prove
Hence
Theorem 8: There is a (1, 1) correspondence between the subfields K of A and the closed subgroups M of & such that if K Mis the field corresponding to M then M = NKM!, . The proof of this theorem proceeds in two parts.
j? - uj? = S(8) = b # 0.
Set a = - P/a; then uol # a, and so a does not lie in k. It follows that K = k(a); further, ua = a 1. Thus we have constructed a generator a of K on which the Galois group has a particularly simple action. Now consider the irreducible equation satisfied by a; we know that a(& - a) = (aa)P - ua = (a l ) p - (a 1) = a9 - a.
+
+
+
Thus ap - a = a, where a lies in the ground field k. I t follows that Irr (a, k, x )
= xP - x
- a.
The roots of this equation are denoted by alp; so we have
K = k(a/p). We may remark that
Part I: If K I k is any extension field, then there is a (I, 1) correspondence between the closed subgroups of k containing NKlkand the abelian subfields of K. Part 2: The norm group of the maximal separable extension C is 1. These two assertions together give the theorem. Proof of Part I: Clearly to each abelian subfield KOof K there exists a closed subgroup containing N K l k, namely NKoIk . Conversely, let M be a closed subgroup containing N K l k .Let HMbe the inverse image of M under the norm residue mapping, i.e. (K I k/HM) = M. Since the mapping is a homeomorphism, HM is a closed subgroup of G. Let K M be the fixed field under HM:HMis then the Galois group of K I K M .
182
9.
THE EXISTENCE THEOREM
Let E denote the generic finite subfield of KM; let HE be the Galois group of K I E. Clearly H, C 0 HE.On the other hand, if TE H E , then T leaves every finite subfield E fixed; but every element of K, lies in some finite subfield, hence T leaves K M fixed. HE. Since the norm Finally we have the result that H, = residue mapping is an isomorphism we have
nE
nE
Hence
M
=
n NEJ = E
N K ~ ,
.
Thus we have shown how to construct a field K M C K such that M . Clearly the construction leads to a is the norm group N,,,, unique abelian subfield K M , since its HMis given by
Case I : p is not the characteristic of k. Let k, = k(5), where 5 is ap-th root of unity. Since
it follows that a = Nkl,,P where /3 E Ncikl; hence ord /3 = 0. It is clearly sufficient to prove is a p-eh power. Hence we may assume that k already contains 5. Let y be any element of k*; then we assert that y is a norm for $ k, the extension k ( v q ) I k. This is clear if q q E k. If then it satisfies the irreducible equation xp y = 0. If p = 2, this fact implies that y is a norm. If p is odd, it implies that - y is a norm; but so is (- 1 ) P = - 1, and hence y is itself a norm.
.Q/q
+
Similarly ay is a norm for the extension k ( 4 - ay) I k; but by our assumption on a, a is also a norm for this extension. Hence y 9 is a norm for the extension k ( 4 - cry) 1 k. 9
But y is a norm for the extension k ( Y 7 ) I k. By the corollary to the preceding theorem, we see that y is a norm for the composite extension k ( V z y , 4- Y). Hence y is a norm for the simple extension k(Vi). Since y may be any element in k*, we have obtained the result that k* = N,(T;,,~.But 9-
This result shows us that the lattice of subfields of K I k is the dual of the lattice of closed subgroups of k containing NKlk;i.e. each is obtained from the other by turning the lattice diagram upside-down. In particular we have the Corollary: If M and N are two closed subgroups containing NKlk , corresponding to subfields K, and K, , then (1) M n N corresponds to the compositum KMKN, (2) MN corresponds to the intersection K M nK, .
We shall use the first part of this Corollary in the Proof of Part 2: We must show that if a is an element of k which lies in the norm group NEIkof every finite extension E I k, then a = 1. , E First we can prove that a is a unit in k. For if a E N E I k where is the unramified extension of degree f, we have f I ord a. Since this holds for every positive integer f, we must have ord a = 0, i.e. a is a unit. Next we prove that for any prime p, a is a p-th power. We have to distinguish two cases:
(k(.ir/;;) : k ) = (k* :N k & j l k ) = 1 . Thus A(&)
= k;
i.e. a is a p-th power of an element of k.
Case 2: p is the characteristic of k. Since a is a unit which is a norm for every finite extension, in particular it is a norm for every cyclic extension of degree p. Let q = pS be the number of elements in the residue class field of k; then or may be written in the form
Hence if
is a p-th power, so is a. But we have ar=orppmodp
and, of course, or
E
aq
mod p.
184
9.
THE EXISTENCE THEOREM
Hence IS = 1 mod p ; and clearly /I is a norm from every extension. There is therefore no loss of generality if we assume to start with that a =-- 1 mod p. Suppose now that it is possible, for every positive integer n, to express ol in the form
These equations generate respectively the fields
a. = 13:yn,
where y,
= 1 mod pn.
Then y, -t 1, as n +GO; and hence /I,p -+ a. But since p is the characteristic of k the convergence of the sequence {p,p) entails that of {IS,), say /I, -t /3. Then a = P P and our assertion is proved. We have therefore to show that for every n we can express ol as a = /3,py, where y, = 1 mod pn . Suppose that this is not possible; then there is a maximal n for which it is possible. The corresponding y, is, like a, a norm in all cyclic p-extensions, and is of the form y,, = 1 A+,
+
where X must be a unit. First we show that p does not divide this maximal n. For if does divide n we can construct the element
K , is contained in the compositum
; By,/(l - y,) We see that 8/(1 - y,) E N K I I ksimilarly But by assumption y, E N K s l khence ;
E NKSlk.
p Thus
and so = 13SP~n+1,
in contradiction of the maximal property of n. Now let 19 be a unit of k, and consider the equations
Next we show how to choose 8 such that K , I k is unramified. Since the residue class field is finite, it has an extension field (which is normal and cyclic) of every degree, in particular of degree p. The extension of degree p is defined by an irreducible congruence of the form (8, an integer of k). zP - z - 0, E 0 mod p The irreducibility implies that 8, is a unit of k. It follows that the same polynomial, xp - z - do, is irreducible in k, and that it gives rise to an unramified extension K, I k of degree p. Hence, k ordinal divisible by p. when 8 = 8, , every element of N K S Ihas But ord 8, = 1, and = n; ord ( 1 - yn) = ord (kn)
186
9.
THE EXISTENCE THEOREM
hence
7. UNIQUENESS
OF THE NORM RESIDUE SYMBOL
187
As a Corollary to Part 2, we have the following property of the norm residue symbol for the field C. Let KObe any finite extension of k; then
and we have proved that n is not divisible by p. This contradiction arose from our assumption that a cannot be written in the form a = &Py, for every n; this assumption is therefore false. Hence, as we proved above, a is the p-th power of an element of k. We can now complete the proof of the theorem. Since
r
where H is the subgroup of which leaves KOfixed. T o prove this we recall that
where E is any finite extension, we have a = N,,$ where /3 E N C I E . For any prime p we can express a as a = yp; by the same argument we have /3 = Sp. Then
where E runs through all finite subfields of C. We can obviously restrict these fields E to be finite normal extensions K of k, containing K O . Then
I t follows that NElkS= CPy, where 5, is one of the p-th roots of unity lying in k. Thus if Z is the set of all p-th roots of unity in k, the set
using the property already proved for finite normal extensions. = 1, we have the required result. Since NClKo
YZ nNEIk
= Z(E)
is non-empty and compact. Clearly the family {Z(E)), as E ranges over all finite extensions, has the finite intersection property. Hence the total intersection is non-empty. That is to say, there is ap-th root of unity 5, in k such that
. We may now repeat our Hence a = yP = yOp, with yo E NClk argument, obtaining as a result that a can be expressed as an m-th power of an element 7 of k for any integer m. Since a is a unit, so is 7; a is therefore an m-th power of a unit for any m. This means that a lies in every neighborhood of 1; hence, since the topology in is Hausdo&, this implies that a = 1. This completes the proof of Part 2, and hence Theorem 8 is completely proved.
7. Uniqueness of the N o r m Residue Symbol
We now sum up our results so far: Let A be the maximal abelian extension of k. Then we have shown the existence of a mapping T -t (A I KIT) with the following properties: (I) The mapping is an isomorphism, both algebraic and topological, of the Galois group of A I k onto k. (2) I t affords a canonical map by which, given any closed subgroup M of k, we can find a field K Msuch that N,,, = M; namely, if H , is the inverse image of M under the mapplng, then KM is the fixed field under H M . (3) If T is an element of the Galois group of A 1 k, then ord T = ord
("!"
9.
188
THE EXISTENCE THEOREM
We contend that these three properties specify the mapping completely. So let 4 be a mapping of the Galois group of A I k onto such that:
( I ) 4 is both an isomorphism and a homeomorphism. (2) If M is a closed subgroup of k, K M the fixed field under +-l(M), then M = NKMllc . (3) ord r = ord 4(r). We shall show that $(r)= ( A I KIT). Certainly if +(T)
= a,
ord a = ord T = ord (A!k)
Now let M = (a), i.e. the closure of the cyclic group generated by a. Let HM= +-l(M), and let KM be the fixed field under HM;then M = N K I M iHence k. ( A jk/HM)= M, and so, in particular, we have ( A I k / r ) E M . Now all the elements of M have the form a Y( v E I).Hence, if ( A I k / ~=) avo, and ord a = m, we have vom = m, i.e. (v, - 1) m = 0. Thus if ord a is not a divisor of zero in I, we have the desired result
In particular, if a = T, we have
and if a = ET, where
Hence, since both
for all units
E.
E
+(T)
is any unit,
and (A 1 k / r ) are isomorphisms, we have
7.
UNIQUENESS OF THE NORM RESIDUE SYMBOL
Hence if a = errv is any element of
189
k,
which proves our assertion that ( A I KIT) is uniquely determined by the three properties listed above.
1. Let
[/?I
FIELDS WITH PERFECT RESIDUE CLASS FIELD
be another element of k, with representative = lirn
CHAPTER TEN
Applications and Illustrations 1. Fields with Perfect Residue Class Field
Let k be a complete field under a non-archimedean valuation; let R be its residue class field. We consider the case in which A is perfect and of characteristic p > 0; this certainly includes the case we have been discussing in the previous three chapters, since all finite fields are perfect. We make a slight change in our usual notation: 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
-
1, and so
- a = c lies in the ground field k, for
204
10.
APPLICATIONS AND ILLUSTRATIONS
Hence
I t is clear that no further changes may be made except in the constant w . Consider the extension k(x/p) 1 k, where x = E" cvtv,C-, # 0, We shall find the conductor of this extension. T o do this we must determine the smallest index r such that all elements y of the form 1 tra (I a 1 I) are norms for k(x/p) I k. We know that y is a norm for this extension if
b = r ( a + c p - c);
thus
+
Now let k be a complete field with characteristic p and perfect residue class field. Let t be a uniformizing parameter. We consider 00 an extension of the form k(a/p) where a = C-, cvtV.Our first task is to simplify the form of the generator a by changes of the type just described. We may first replace a by another element which has no terms in t vwhere v < 0 and p divides v. For if we have cv # 0 for v < 0, v = pp, we may replace a by a - (c;lp tp)P (cf;lptp), for which the coefficient of tv = 0. We have, of course, introduced a nonnegative coefficient for tp, but if p divides p, this may be removed by a repetition of the process. Indeed, since the number of terms with v < 0 is finite, a finite number of repetitions of this process will eventually yield an element
+
<
Let r = m
+ 1: y = 1 + tm+]a. Then
which may be written dy = tm/3dt. Thus SFIFn (Re. x
$)
= SFIF,, (Res
-m
+
this is convergent since
p divides
I b I < 1. Then
d t ) ) = 0. cvtv tmB 1 + tm+b
Hence all elements y = 1 tm+la are norms for this extension. On the other hand, not all elements of the form y = 1 tma are norms. For consider y = 1 ctm(c E F ) ; dy = mctm-ldt. Then
+
where w is a constant, I b 1 < 1, and if In particular p does not divide m. Now let
(2
+
v, then dv = 0.
Since m is not divisible by p, m #=0 in F ; as c ranges over F, so is not identically does c-,mc, and hence, since the trace SFIFo zero, we have S,,,Jc-,mc) # 0 for some value of c. Hence not all elements y = 1 tma are norms. It follows from the definition of the conductor f that f = pm+l, and hence the different
+
so c, - c = - b. Hence a' may be replaced by This completes the discussion only when x = E" cvtvand m > 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)
<
Thus all powers of p are norms for K / k. T h e whole norm group is therefore
where V, is a certain subgroup of units such that
We shall now show that V7 is precisely the group of units which are congruent to 1 modulo pr. Consider therefore the group Us consisting of elements 1); we restrict ourselves to s 1 when p is add 1 upS (1 a I and to s 2 when p = 2. Then we have
+
of degree (p - l)pr-I = 4(pr). Thus deg ( K ( k) notice that if we put x = 1 we obtain
Now if (.and
5. are two primitivepr-th
< 4(pr).
roots of unity, we have
<
(I
We
+ up), = (1 + apS+l)mod pS*,
and so we may write
Repeating the process, we obtain 1
+ ups"
= (1
+ ups), (I + bpS+l)"(1 + CP"+~),
and so on; finally we arrive at the result which is an integer of K. Since we can write 5 = (5;)~for some p, we have
207
which is also an integer of K. Hence (1 - <)/(I - 5') is a unit; so 11 - 5 1 = I 1 - 5' 1. This shows that
hence
Let k be the rational p-adic number field R, , i.e. the completion of the rational field in the valuation induced by the rational primep. Our aim in this section is to describe the maximal abelian extension A of k; we shall prove
THE RATIONAL P-ADIC FIELD
(1
+ upsf1) = (1 + alps)*,
and hence Us+, = Usp. By iteration we obtain
208
10.
6.
APPLICATIONS AND ILLUSTRATIONS
when p = 2. We now consider the two cases separately. When p is odd, we have
COMPUTATION OF THE INDEX (a:an)
I t follows now that the compositum T,K,, , which is obtained by adjoining all roots of unity to k, has norm group 1. T,KU, is therefore the maximal abelian extension A. This completes the proof of Theorem 4. 6. Computation of the Index
Hence we may write 1 - ups = (1
+ apS)pl (1 + bp8+l).
By repetition of this process we finally reach the result that u%1=u,. Hence we have
.
Hence Ur = NKIkUl,and therefore U,V , But
(U : U,)
= &')
= (U : V,),
and so U, = V , . When p = 2 we notice that
209
(a:
an)
Let k be a complete field, either (1) archimedean, or (2) nonarchimedean with finite residue class field. We define normal forms for the valuation in k as follows:
Case I : (a) k is the field of real numbers. Let the normal valuation be the ordinary absolute value. (b) k is the field of complex numbers. Let the normal valuation be the square of the ordinary absolute value. Case 2: If the residue class field of k contains q elements, we define the normal valuation by prescribing I .rr 1 = l/q. For use in a later chapter we prove the following result: Let k be a field of the type described above. If n is prime to the characteristic of k, and if all the n-th roots of unity lie in k, then (a : an) = n2/1n I where I n 1 is the normal valuation. Theorem 5:
since a number which is congruent to 1 mod 4 is congruent to 1 or 5 mod. 8. Hence we have
Since u:-' C Vr , by the same argument as we used when p is odd, it will follow that U pC Vr if we can show that 5"'- is a norm. I t is easily verified that 52r-2= NKIk(2 2). Hence U pC V,., and it follows as before that U , = Vr . In all cases, then, we have
+
If we adjoin all pr-th roots of unity, we obtain an extension which we may call KD,, for which the norm group is UVElpv,since n, u, = 1.
Proof:
We must consider the two cases separately.
Case I : k archimedean. If k is the real field, the only possible values for n are 1 and 2 (the real field contains only the first and second roots of unity). We see at once that 12
( a : al) = 1 = -
111'
If k is the complex field, n may have any value, and every element of k is an n-th power. Hence
210
10.
6.
APPLICATIONS AND ILLUSTRATIONS
Case 2: k non-archimedean. We apply the Lemma of Chapter 7, Section 5, with the homomorphism T : a -+ 1 a 1 . We have
COMPUTATION OF THE INDEX (a: an)
Hence n
(E
: en) = qSn= -
I4
.
Finally we have the result of the theorem: We may therefore consider the index ( E : en). First let r be so large that I nnr+l 1 I rr2' 1, and consider the group Ur of elements 1 anr ( 1 a I 1). Then
+
(1
-
+ anT)"
<
1
+ annr mod n2'
= 1 + annTmod nnT+l.
Hence we may write
and repeating the process, we have 1
+ annr
= (1
+ anr)" (1 + W + l ) n (1 + cnnT+2),
Finally 1
+ annT= (1 + a'n')".
Thus if ord n = s, we have
Let Cn denote the group of n-th roots of unity in k. Suppose r is so large that none of the 5, # 1 lie in Ur . Again we apply the Lemma of Chapter 7, Section 5, this time with the homomorphism T : E + en. We obtain
Hence
Now ( U , : U,,,) = number of residue classes modulo n8 = p8, where q is the number of elements in the residue class field.
PART THREE
Product Formula and Function Fields in one Variable
CHAPTER ELEVEN
Preparations for the Global Theory 1. The Radical of a Ring
Let R be a commutative ring. An element a of R is said to be nilpotent if some power of a, say an = 0. The set of all nilpotent elements of R is called the radical, N. Theorem 1:
The radical is the intersection of all prime ideals
of R. Proof: Let S be a multiplicative semigroup in R not containing zero. We shall show that there exists a prime ideal of R which does not intersect S. Consider the set of all ideals of R which do not intersect S ; this set is not empty, since the zero ideal belongs to it. The conditions of Zorn's Lemma are easily verified, so there exists an ideal a which is maximal in this set. We contend that a is a prime ideal. Let a, $ a, a, $ a; then (a, a,) respectively (a, a,) are larger than a and so contain elements s, respectively s, of S. Thus s1 = nlal
+ rial mod a,
s, zz
n2a,
+ r2a2mod a,
where n, , n, are integers, and r, , r, E R. Hence sls2 = nln,crla, + r3a1a2mod a. It follows that a,a, $ a, and so a is a prime ideal. Now if a EN, an = 0 lies in all prime ideals p. On the other form a semigroup hand, if a is not in N, the elements a, a 2 , 013, not containing zero; thus there is a prime ideal p not containing a. This completes the proof. 215
216
XI.
2.
PREPARATIONS FOR THE GLOBAL THEORY
2. Kronecker Products of Spaces and Rings
Let k be a commutative field, and let X and Y be vector spaces over k. We form the vector space V whose basis vectors are the elements (x, y) of the Cartesian product of X and Y; the elements of V are therefore of the form av(xv,y,) with a, E k, x, E X, y, E Y. Consider the subspace N which consists of those elements of V such that
for every linear map of X into k and every linear map A of Y into k. The Kronecker product of X and Y with respect to k is then defined to be the factor space of V mod N:
217
KRONECKER PRODUCTS OF SPACES AND RINGS
maintain a strict distinction between the elements of the products which come from X and those which come from Y. Theorem 2:
If yl , y,
, ..., yn are
then x1y1-t
in X
X,
x2y2
linearly independent in Y,
+ ... + xny, = o
Y if and only if x1 =
... = xn = 0.
Proof: The "if" part is trivial. T o prove the "only if" part, we recall that a linear functional defined on a subspace of a vector space may be extended to the whole space. Hence, since the y, are linearly independent, we can form linear maps A, such that A,(yj) = aij. Hence, since
,
in X x Y, we have We may therefore regard the elements of X x , Y as being those of V, where equality is now defined to be congruence modulo N. Hence in X x k Y we have
T o prove the first of these results, we notice that for all linear maps I and A, we have l(x
+ x') h(y) - l(x) h(y) - l(xr)X(y) = 0
which means that
Similarly, since
for all maps I and A, our second assertion is proved. From these remarks it follows that we may operate formally with the pairs (x, y) as if they were products xy-provided we
for all linear maps 1of X; hence X$ = 0. This completes the proof. Corollary: In order to test whether C ol,,xvyv= 0 in X x Y, we express the y, which occur in terms of linearly independent vectors y,'; say y, = C cdy,'. Then C a j v y v= C a,,cvixvy,', which is zero if and only if all the elements C a,,cviyvare zero. Theorem 3: If xl ,.x2, -..,xn are linearly independent in X and yl ,y, , .-.,y, are lmearly independent in Y, then the elements x,yj are linearly independent in X x Y.
,
Proof: Suppose C cv,xvy, = 0 in X X , Y. Then for all linear maps I and A we have
We may construct maps li ,Aj such that l,(xv) = a,, ,Ai(y,) Hence c,,l,(x) Xj(y)= ci* = 0.
=
aiP .
21 8
XI.
3.
PREPARATIONS FOR THE GLOBAL THEORY
Corollary: If X and Y have finite k-dimension, then so does X x, Y; in fact,
Suppose now that X and Y are rings with unit elements, containing k in their centers. We may introduce a multiplication operation in the vector space V by defining and extending the definition to V by requiring that the distributive law be satisfied. I t is easily verified that Y forms a ring under this multiplication. N is a two-sided ideal of this ring. For if 2 or,(x,, y,) EN, then Z: a,l(x,) h(y,) = 0 for all linear maps. Consider in particular the maps l(x) = Z'(ax), X(y) = A'(by), where l', A' are arbitrarily given linear maps. We obtain
2 % l1(axV)hl(by,) = 0 lies in N. Similarly [Z: aV(x,,y,)] (a, b) lies in N. From this remark it follows that the Kronecker product of X and Y is a ring, namely the residue class ring V / N . Since X and Y have unit elements, X x k Y contains subrings isomorphic to X and Y, consisting respectively, of the elements (x, I), (1,y). If we identify X and Y with these isomorphic subrings, we see that If Xo , Yoare subrings of X and Y, we may form the Kronecker product Xo x Yo; it is easily seen that this is imbedded in X x k Y in a natural way.
,
3. Composite Extensions
Let A and B be arbitrary extension fields of k such that A n B = k. We define a composite extension of A and B by giving
COMPOSITE EXTENSIONS
219
a pair of isomorphic mappings a, T which have the same effect on k, and which map A and B respectively into some field F. The composite extension is then the smallest subfield of F which contains both aA and TB; we denote this subfield by aA TB. In general UA . TB will not be merely the product of oA and TB (i.e. the set of finite sums of products of elements from aA and TB); oA . TB will be the smallest subfield containing this product, in fact its quotient field. Two pairs of isomorphisms (a, T) and (a,, 7,) are said to be equivalent, or to yield equivalent composite extensions if there exists an isomorphism h of aA TB onto olA . T,B such that Xu = u1 , AT = T, . There is a very intimate connection between these composite extensions and the Kronecker product A X, B which we shall now investigate. Suppose first we are given a composite extension defined by isomorphisms (a,7). Then we may map R = A X , B into aA TB by mapping
.
We must show that this mapping is well-defined; so let Z: aiaaba be zero in R. We may express the elements 6%in terms of linearly, independent elements b: :
Then we have
and hence
C; a&ai d
=0
for all v.
Since a is an isomorphism, this implies
218
XI.
3.
PREPARATIONS FOR THE GLOBAL THEORY
Corollary: If X and Y have finite k-dimension, then so does
X x, Y; in fact, Suppose now that X and Y are rings with unit elements, containing k in their centers. We may introduce a multiplication operation in the vector space V by defining and extending the definition to V by requiring that the distributive law be satisfied. It is easily verified that Y forms a ring under this multiplication. N is a two-sided ideal of this ring. For if X a,(xv, y,) E N, then Z a,l(x,) A(y,) = 0 for all linear maps. Consider in particular the maps l(x) = Z'(ax), A(y) = h'(by), where ,'Z A' are arbitrarily given linear maps. We obtain
2
l1(axV) A'(by,) = 0
a pair of isomorphic mappings a, T which have the same effect on k, and which map A and B respectively into some field F. The composite extension is then the smallest subfield of F which contains both oA and TB; we denote this subfield by UA .7B. In general oA . TB will not be merely the product of oA and TB (i.e. the set of finite sums of products of elements from oA and TB); oA TB will be the smallest subfield containing this product, in fact its quotient field. Two pairs of isomorphisms (o, T) and (a,, 7,) are said to be equivalent, or to yield equivalent composite extensions if there exists an isomorphism A of oA TB onto olA . T,B such that Ao = a, , AT = T~ . There is a very intimate connection between these composite extensions and the Kronecker product A x B which we shall now investigate. Suppose first we are given a composite extension defined by isomorphisms (a, 7). Then we may map R = A x B into oA TB by mapping
-
,
-
for all I f , A; hence
onto
lies in N. Similarly [Z a,(x,, y,)] (a, b) lies in N. From this remark it follows that the Kronecker product of X and Y is a ring, namely the residue class ring V / N . Since X and Y have unit elements, X x Y contains subrings isomorphic to X and Y, consisting respectively, of the elements (x, I), (1, y). If we identify X and Y with these isomorphic subrings, we see that
We must show that this mapping is well-defined; so let X %a,b4 be zero in R. We may express the elements bd in terms of linearly, independent elements b: :
,
If Xo , Yo are subrings of X and Y, we may form the Kronecker product Xo x Yo; it is easily seen that this is imbedded in X x Y in a natural way.
Then we have
and hence
2 q&a, = 0
for all v.
6
3. Composite Extensions
Let A and B be arbitrary extension fields of k such that = k. We define a composite extension of A and B by giving
AnB
219
COMPOSITE EXTENSIONS
Since o is an isomorphism, this implies
220
XI.
3.
PREPARATIONS FOR THE GLOBAL THEORY
Multiply by ~(b,'), and sum over v, replacing a(#&,,) by T(&) (a = T on k); this yields
This shows that the mapping is well-defined. It is clearly a homomorphism of R onto the product of uA and TB in F ; since F is a field the image has no divisors of zero, and so the kernel of the mapping is a prime ideal p. Thus R/p is isomorphic to the product of aA and TB; and hence a A TB
quotient field of Rip.
It is clear that if (0, , T ~is) a pair equivalent to (a, T), then the equivalent composite extension o,A . T,B corresponds to the same prime ideal of R. For if h is the map which links (a, 7) and (u,, T,), and if .(ai) o(ai) 7(bi) = 0,
2
COMPOSITE EXTENSIONS
221
where the fields B' are the finite subfields of B, it suffices to consider the case where B / k is a finite extension. Let p be a prime ideal of R = A x k B; then R' = R/p may be considered as an integral domain of finite dimension over A. Suppose R' = Awl
+ ... + Aw,;
then aR' = Aawl
+ + Aawn = R',
since the o,are assumed linearly independent. Thus R' is already a field, and so all the prime ideals in R = A x k B are maximal. We now prove the Theorem 4: Let R be any ring; let p, , p, , ..-,p, be distincts maximal ideals. Then R/fl p, is isomorphic to the direct sum of the fields Rip,.
Proof: Let $, be the natural map of R onto R/pi. Consider the map of R given by
then
2 XU(%)hu(a,) hi(bi)
=
q ( a J al(ai) ~ ~ ( b=i 0. )
Thus each equivalence class of composite extensions corresponds to a prime ideal in R. Conversely, let p be a prime ideal in R; p # R. Let p be the natural homomorphism of R onto Rlp. Then p maps A onto a homomorphic image pA; but a homomorphism of a field is either the zero map or an isomorphism, and p cannot be the zero map since PA = 0 implies p(A x B) = 0, whence p = R. Hence p maps A onto an isomorphic image pA. Similarly p maps B onto an isomorphic image pB. Thus p maps A and B into an integral domain Rlp which consists of linear combinations of products of elements in pA and pB. If we form the quotient field of Rlp, we may form the compositum of pA and pB; in this way defines a composite extension. Clearly if p is a maximal ideal, R/p is already a field, and hence is itself a composite extension of A and B. In our applications, one of the fields-say B-will be an algebraic extension of k. Since we may consider A x B = U A x B'
,
This is clearly a homomorphism of R into the direct sum of the p, . It remains to show that the mapping is Rlp, , with kernel onto. Since pi # p, (i = 2, r), there are elements a, which lie in p, , but not in p,; hence $,(a,) # 0, but $,(a,) = 0. Then if a,' = a,a, .-.a, , we have $,(a,') # 0, while $,(a,') = 0 (i = 2, -.-,r); similarly we define a,', a,'. 6,) be an element of the direct sum. Now let 6 = (6, ,6, , Since each 4, maps R onto the field Rlp,, there are elements A, E R such that $,(A,) = &,/$,(a,'). Then clearly $(C a('A,) = 6. This completes the prooof. We apply this to the case where R = A x, B, and B is a finite extension of k of degree n. In this case each field R/p, is a composite extension oA . TB of A and B. We now agree to identify A with its image UA in each of these extensions. Hence a composite extension of A and B is now defined by an isomorphism T of B into an extension field of A such that 7 acts like the identity on k. Another isomorphism 71 gives rise to an equivalent composite extension
n
.-a,
..a,
em.,
XI.
222
if there is an isomorphism X which acts like identity on A such that AT
.
= T1
4.
PREPARATIONS FOR THE GLOBAL THEORY
Let p, ,p, , p, be maximal ideals of R ; we may view the direct sum of the fields R/pi as a space over A. If the degree . a * ,
we have $mr=dim~/r)pi<(~:~)=n.
VALUATION OF A NON-COMPLETE FIELD
223
then deg ( A .rill ( A) = m;p6i. By the same argument as above there are exactly mi1 distinct maps A, and hence mi1 maps 7 . e uivalent to T~ . Thus XI mi1 = no , 3.4 where the summation is restricted to the isomorphisms T, which yield inequivalent composite extensions. Finally, since si ,< s, we have
i=l
This shows that the number of distinct maximal ideals in R is finite. If we take all the maximal ideals in our decomposition (and these are all the prime ideals of R), we have R / n p i = RIN
once more. This shows that A X, B has radical zero if and only if all the s, = s. I n particular A x B has radical zero if B is a separable extension.
direct sum of all composite extensions. 4. Extension of the Valuation of a Non-Complete Field
Hence we see that X mi = n if and only if N = (0). T o show that there may exist fields A and B for which A x, B has a non-zero radical, we consider the case where k = Ip(x) (I, is the field with p elements), and let A = k(a), B = k(p) where a p = /3p = X. Then in the Kronecker product a - ,5l is clearly nilpotent: (a - j?)" = a" - j?" = 0. Now let 2 be the algebraic closure of A, and restrict the isomorphisms T to be mappings of B into A. Certainly A contains &, the algebraic closure of k. Let B, be the separable part of B ; no = deg (B, I k); n = nopS where p is the characteristic of k. Then it is known from Galois Theory that there are precisely no distinct maps T of B into &. Hence there are no maps of B into A. If two of these maps T ~ T~ , yield equivalent co~npositeextensions A . riB and A . T ~ Bthen , A . T ~ is B isomorphic to A . r j B under an isomorphism X which leaves A fixed. Now A . T ~ B ,is in the B A. Since B j B, is purely inseparable separable part of A . T ~ over of degree pS, A . T ~ Bis purely inseparable over R T ~ B ,with degree psi <pa. Thus A . riBO is precisely the separable part of A riB over A; if
-
deg (A riBO1 A) = m; ,
Let k be a field with a valuation, not necessarily complete, and let E be a finite extension of k. Suppose first that we have obtained an extension of the valuation of k to E. The completion I? of E under this valuation will contain the completion h of k under the original valuation; I? also contains E, and hence the compositum E&.This field E&is a finite extension of &; hence (Theorem 2 of Chapter 2) it is complete. I t follows that E& contains I?; hence E& = I?. On the other hand, if an extension to E is not yet known, we may construct the algebraic closure C of & (any sufficiently high finite extension would suffice); we can extend the valuation of KC. Since C contains the algebraic closure of k, there exist isomorphic maps a of E I k into C I k. The valuation of C induces a valuation on aE which is an extension of the valuation of k. We may then define a valuation in E by writing I a I = I oa I for a E E. We notice that the completion of aE will be oE . &. There are several different maps of E into C, and we may ask when two of these maps give rise to the same extended valuation on E. So suppose a and T are maps of E into C which yield the same valuation. Then TO-I is a map of oE onto T E which preserves the valuation of aE; this map can be extended to the completions of
224
XI.
PREPARATIONS FOR THE GLOBAL THEORY
-
oE and TE, that is to the composita oE k and T E k. Since TU-1 is identity on k, this extension is identity on A. Hence if a and T yield the same valuation on E, then they are equivalent. Conversely, suppose a and T are equivalent; let h be the map of oE . 6 onto T E such ~ that h is identity on k and T = ha. Then o induces the valuation I a ,1 = I ua I and T induces the valuation I a l2 = 1 TCY1 = 1 how I. But ua and hoar are conjugates over a complete ground field 6; hence they have the same valuationIorl1=
Ia12.
CHAPTER TWELVE
Characterization of Fields by the Product Formula
Referring to our previous discussion, we see that there is exactly one extension of the valuation of k to E for each prime ideal in the Kronecker product 6 X, E. Furthermore if we call the degree of oE . k over k the local degree, we see that the sum of the local degrees for all extensions is not greater than the degree of E I k. We saw in Chapter 1, Section 5, that the field of rational numbers and fields of rational functions over arbitrary ground fields satisfy a product formula ITp( a , 1 = 1 for all non-zero elements a. We shall now study arbitrary fields with a product formula of this type. More precisely, let k be a field satisfying Axiom 1: There is a set 9-R of inequivalent non-trivial valuations of k such that for every non-zero element or of k, I a ,1 = 1 for all but a finite number of I ,I and I a ,1 = 1. We denote by p the equivalence class of valuations defined by I ;,I and we call p a prime of k. In this chapter I ,1 shall always denote the special valuation in the equivalence class p which occurs in the product formula of Axiom 1. For each non-archimedean prime p in 9-R we may form a ring o, of p-integers, consisting of those elements of k for which I or ,1 1. The elements a for which I a ,1 < 1 form a prime ideal of 0,; we shall denote this ideal by p-no confusion will arise from this notation. Finally we denote the residue classjeldat p by K p :
I ,1
n,
<
<
Consider the set k, of elements of k such that I or ,1 1 for all Then clearly either w = 0 and I or ,1 = 0 for all p E m; or else, because of the product formula, I a ,1 = 1 for all p ~ m I n the latter case we must distinguish two possibilities: p
E 9-R.
225
.
226
XII.
CHARACTERIZATION OF FIELDS
(I) rn contains no archimedean prime. Then k, is a field, for if j a l p = I p l p = l for all p ~ m ,then I a f j ? I , < l , I ap 1, = 1 and I a-l 1, = 1 for all p E %I. k, is the largest subfield ' R are trivial: we call k, the field of k on which all the primes of % of constants. It is easily seen that k, may be isomorphically mapped into every residue class field A,; under this map we may consider k, as a subfield of all K, . The degree of Kp I ko will be denoted by f(p). (2) rn contains an archimedean prime q. Then k, is not a field, for 1 ~ k , , b u t l 1 = 2 $ k o , s i n c e 121,>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
<
Consider the set S consiting of elements a such that I a 1, x, for a11 p E where the x, are fixed real numbers. We may think of S as a parallelotope in the space obtained by forming the Cartesian product of k with itself as many times as there are primes in 'D. We wish to find an upper bound for the order M of S. First we prove Theorem 1: Let q be a reasonable prime. Let S be a set of elements of k with order M > 1. If I a 1, x for all a E S, there exists a non-zero element 8 E k, which is either an element of S or a difference of two elements of S, such that I 8 1, A,X/M~(~); A, is a constant depending only on q.
<
<
Proof:
We must consider several cases:
Case 1: q is archimedean and the completion is real. In this case the order M is the number of elements of S. Since I a 1, ,< x, we have 11 a 11, ~ l l p (for ~ ) a11 a E S. Hence if we decompose the interval [- x l l ~ ( ~ )l l, p ( ~on) ]the real line into M - 1 equal parts, then, by the Pigeon-holing Principle, at least one of these parts must contain two elements of S. Hence there is a non-zero element 8 E k, the difference of two elements of S, such that
<
228
XII.
CHARACTERIZATION OF FIELDS
Since M 3 2, we have M - 1 2 M/2, and hence W/P(Q)
There are now two cases to distinguish: (a) Residue class field is finite. Then the ring oq/qr has (Nq)" elements. I t follows from the Pigeon-holing Principle that there is a non-zero element 8 in k such that Olor, is the difference of two elements of S' and
Il e ll. G 7 .
Finally we have
1 8 1,
< A , ~ / M P (where ~ ) A, = 4p(Q).
Case 2: q is archimedean, and the completion is complex. Here also M is the number of elements of S. If I or 1, X, we have the ordinary absolute value of a ~ ~ l ~ p ( ~ ) ; hence a lies in the square (in the complex plane) with center the origin and side 2 ~ l f ~ p If ' ~we ) . decompose this square into N2equal < N I, i.e. N2 < M ( N squares, where N <M then at least one of these smaller squares must contain two elements of S. Hence there is a non-zero element 8 E k, the difference of two elements of S such that the ordinary absolute value of
<
<
< +
B/a,
= 0 mod qT.
Hence ord 8/al 2 r, and so
< +
whence
and Now where A,
.
= (Nq)P(Q'
, of degree f.
hence ordinary absolute value of
(b) Residue class field Kq is a finite extension of k, Then if a is any element of D,
Therefore,
with ai E K,; thus there cannot be more than f~ elements linearly independent over k, modulo qr. But M = cairns, and (Nq)" = cfr; hence cfr < cairns', i.e. dim S' >fr. Thus there are more than fr elements of S' linearly independent with respect to k,; these cannot be linearly independent modulo qr. Hence there is a non-zero element 8 of S such that O/a, is a linear combination of more than fr linearly independent elements of S', such that this combination is non-trivial modulo qr and hence such that 8/al 0 mod qr. As in case (a) we deduce that 1 0 1, A,x/M~(~). This completes the proof of Theorem 1 in all cases. We can now give the desired bound for the order of the elements in a parallelotope:
where A, = 27~'"'.
Case 3: q is discrete. Let or, be an element with maximal I a, 1, in S. We replace the set S by S' = Stor,; then every element of S' satisfies I or' I 1, i.e. the elements of S' lie in the ring of q-integers oq. We can find an integer r such that
<
<
-
230
XII.
CHARACTERIZATION OF FIELDS
3.
<
Theorem 2: Let S consist of elements a such that I a ,1 x, for all p E a , where the x, are fixed real numbers. Let q be a fixed reasonable prime in W. Then if M is the order of S,
where D, is a constant depending only on q.
Proof:
By Theorem 1 there is an element 8 in k such that
Next we estimate / 8 1, for p # q. If p is archimedean, there is no constant field k,; hence 8 must be a difference a - b of elements of S. One easily finds a constant p such that 18 1, px,. If p is non-archimedean, whether 8 is a difference of two elements x, . of S or is itself an element of S , we may conclude that I 8 ,1 Using the product formula, we obtain
<
<
23 1
DESCRIPTION OF ALL PF-FIELDS
Case 2: When there is a field of constants k, , let z be a fixed element of k which does not lie in k,; define R = k,(z). This is a transcendental extension of k, , for k, is algebraically closed in k ; namely, all primes are trivial on k, , and hence on any algebraic extension k , , i.e. k, C k , . I n both cases we have already determined all the valuations of R (see Chapter 1, Section 5); we denote the equivalence classes of valuations of R by Latin letters, p , q, ... . Each prime p in W induces on R a valuation p ; we say that p divides p , and write p I p. We restrict our attention to these primes p in '%Tl which induce non-trivial valuations in R. A given prime p in R may have several divisors p in a ; but there can be only a finite number, for if I a ,1 > 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.
<
232 M
XII.
3.
CHARACTERIZATION OF FIELDS
We first obtain a lower bound for M, namely we prove that > 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
<
+
<
since ( A 1, ,< 1; this value, max I a, 1, , is 1 for all but a finite number of primes p. Let q be a fixed reasonable prime; then by Theorem 2 we have
where E is a constant, depending on q and the or, but not on I A Ipm. We now compare the upper and lower bounds for M, and obtain
233
DESCRIPTION OF ALL PF-FIELDS
Thus r is bounded; hence k is a finite extension of R. This completes the proof. Corollary 1: No prime in YJI can induce a trivial valuation on R.
Proof: Suppose p e W induces a trivial valuation on R. Then p also induces a trivial valuation on every finite extension of R, in particular on k itself. This contradicts Axiom I. Corollary 2:
Every prime in YJI is reasonable.
Proof: All the primes in R are reasonable. It follows from the local theory that an extension of a reasonable prime to a finite extension is reasonable. Let n be the degree of R. By raising the product formula to a suitable power, we may assume that E h(pw)= n: from now on we shall assume that the product formula of Axiom 1 already has this property. Now set r = n in (*); we see that ~ ( q ) 1 where q was a reasonable prime. By Corollary 2, p(p) 1 for all p E YJI. Let p be a non-archimedean prime of m, which induces a (nontrivial) prime p of R. Let k*(p), R*(p) denote the completions of k and R respectively. Let n, , e, , f, denote respectively the degree, ramification and residue class degree of the extension k*(p) I R*(p). Then we have the following statements:
<
<
and, if a is any element of R, For any element a E R we may write X(p,) is a certain constant; hence
I a J p m = 11 a @-')
where
ord, a
= ev
ord, a.
Now we have already defined the normal valuation equivalence class p of R (see Chapter 1, Section 5)Now keeping r and the elements tend to infinity, we obtain
a,
fixed, and letting
11 A I/,@ We have also
11 1,
in the
234
XII.
CHARACTERIZATION OF FIELDS
Hence
4.
FINITE EXTENSIONS OF PF-FIELDS
235
Let k be a PF-field, R its rational subfield. Then (a) the set W consists of all primes p in k which divide all the primes p of R ; (b) for each such prime p, XplPn, = n; (c) the valuations I , 1 which occur in the product formula (raised to a power if necessary to make C h(p,) = n) are precisely the normal valuation 11 1 , . Theorem 4:
The same result is easily verified when p is an archimedean prime. Hence for all primes p E Im and all a E R, we have
We now apply the product formula of Axiom 1 to the elements of R:
n I I,=n(n' I I,)=n(nl 11 a
a
where the accents mean that the product or sum is taken for all the primes p E 101 which divide p. We thus obtain a product formula
in the rational subfield R. In Chapter 1, Section 5, however, we saw that in such a formula (a) all primes p of R must appear, (b) the exponents C' n,p(p) are all equal. Let 8' n,p(p) = d; d is a constant. When p = p, we already have
1 a *,I
=
11 a 11nwm 9 P'pm'
4. Finite Extensions of PF-Fields
a~:vp(v))
=
We shall now show that every finite extension of a PF-field is again a PF-field. Since the rational number field and fields of rational functions are PF-fields, this will prove the converse of Theorem 3, namely that all algebraic number fields and algebraic function fields are PF-fields. Lemma 1 : Let k be a field which satisfies Axiom 1; let F be a subfield not consisting entirely of constants. Then Axiom 1 holds in F for the set 8 of primes p induced by those p in W which are non-trivial on F.
Proof: We define I a 1, = IT I a ,1 where a E F , p E X and the accent denotes that the product is taken over all p E W which divide p. We remark that this cannot be an infinite product, and hence I a , 1 is a well-defined valuation in the equivalence class p. Clearly
1 1 a llh(Pa).
Hence
<
Since p(p) 1, we have C,,, n, 2 n. But we have already seen that C n, ,< n, where this summation is extended over all primes in k (not only in Im) such that p I p. We conclude (1) 101 includes ) 1. We sum up our all primes p of k which divide p, and (2) ~ ( p = results in
Thus Axiom 1 holds in F. We may remark that if k I F is a finite extension, no prime of 9.R can induce a trivial valuation on F. In this case, therefore, 8 consists of all valuations induced on F by primes in W. Theorem 5:
Let k be a PF-field, E I k a finite extension. Then
E is also a PF-field.
236
XII.
CHARACTERIZATION OF FIELDS
4,
Proof: It follows at once from the Local Theory that an extension of a reasonable prime to a finite extension is also reasonable. Hence Axiom 2 holds in E. It will be sufficient to prove that Axiom 1 holds in E when E is an automorphic extension; for if Axiom 1 holds for an automorphic field containing E, then, by the Lemma, it holds in E also. So let E I k be automorphic, with defect pZ and automorphisms a, T, -... If a E E, then N (a)=
IT (Q"
= a E k.
a
Let 'p be some extension of p to E; then
thus Axiom 1 holds in E. This completes the proof. The following remark about the situation of Lemma 1 will be useful in the sequel: Lemma 2: If F, is the set of constants under %, then
F , = F nk,.
Proof: Clearly F n k, C F, . On the other hand, if I a ,1 1 for a E Fand all p ER, then clearly I a ,1 < 1 for all p E %X which divide primes of %. Since arelltrivial on F, we have I a 1, 1 for all the other primes in % p E %, and hence a E F n k, .
<
<
FINITE EXTENSIONS OF PF-FIELDS
237
Theorem 6: Let k be a PF-field with no archimedean primes. Let E be a finite extension of k . Then, if E, , k, are the constant fields of E, k respectively, Eo is the algebraic closure of k, in E.
Proof: Clearly every element of E which is algebraic over k, lies in E,. On the other hand, k I k , has transcendence degree 1, and since E is a finite extension of k , E I k, also has transcendence degree 1. Thus if c is an element of E transcendental over ko , E I ko(c) is an algebraic extension. Hence if c E E, , E I Eo is algebraic; and therefore, since all valuations of m ( E ) are trivial on E, , they are also trivial on E contrary to Axiom 1. Hence E, cannot contain any element transcendental over k, .
CHAPTER THIRTEEN
form a multiplicative subgroup I of V; the element of this subgroup are called idiles. I contains a subgroup isomorphic to the group of non-zero elements of k. We define a topology in the ring of valuation vectors by means of the iddes. If a is an idkle, we define the parallelotope 17, of V / a I,. These to consist of the vectors 5 in V for which 1 5 1, parallelotopes are taken to form a fundamental system of neighborhoods of zero in the additive group of V. This defines the socalled restricted direct product topology on V; this is not the same as the topology induced by the ordinary Cartesian product topology in P. We recall the definitions of some topological notions: Ajilter in V is a family 5 of sets such that:
<
Differentials in PF-fields 1. Valuation Vectors, Iddes, and Divisors
Let k be a PF-field. We form the completions k*(p) for all the primes p in the set XJt. Let P be the Cartesian product of all these completed fields; P consists of vectors
with one component Ep from each k*(p). P forms a ring under component-wise addition and multiplication. I t is easy to see that the mapping a+ (a,a, a, "') is an isomorphism of k into P. If 5 is any element of P, we define I 5 Ip to be I E p ,I We now define the subset V of P consisting of vectors 4 for which
I [ ,1 < 1
for almost all primes p.
This subset is a subring of P , and it contains the isomorphic replica of k. V is called the ring of valuation vectors: we shall sometimes write V(k) instead of V when it is necessary to emphasize that we are dealing with the valuation vectors associated with a particular field k. T h e vectors a of V, for which IaIp=1
for almost all primes p,
I a ,1
for all primes p, 238
#0
(1) Every set containing a set of 5 is itself a set of 5. (2) Every finite intersection of sets of 5 belongs to 5. (3) 5 does not contain the empty set. 8 is a Cauchy jilter in V if, in addition, given any IT, , there exists a vector ( , such that 5, $ I?, is a set of 5. A Cauchy filter 5 is said to be convergent, with limit (, if there is a vector 5 such that 5 17, is a set of 5 for all parallelotopes 17,. Finally, V is complete if every Cauchy filter converges.
+
Theorem 1:
V is complete in the restricted direct product
topology.
Proof: Let 5 be a Cauchy filter in V; let 5, be the filter induced (by projection) on the component space k*(p). Since each k*(p) is complete, each filter 5, is convergent, with limit a, . We show first that / a, 1, 1 for almost all p. Let 17, be a fixed parallelotope with ~rojectionN, on k*(p); let 5, be a vector of V is a set of 5.Then (5,) N , is a set of 5, , and such that 5, hence contains a, . By the definitions of parallelotopes and valuation 1, ,< 1 for vectors, N, is the unit circle for almost all p, and I almost all p. Hence, for almost all p, the unit circle is a set of 5, and therefore I a, ,1 < I for all but a finite number of p. T h u s a = ('-., ap , .") is a valuation vector. We claim now that a is the limit of 5. Let 17, be a fixed parallelobe so chosen that U b 17, C17,. Let 5 be a tope, and let
<
+
+
<,
+
240
XIII.
DIFFERENTIALS I N PF-FIELDS
+
+
vector such that 6 lIbis a set of 5 ; then (t), (lib), contains a, . It follows that a E 5 Db, and so 6 E a lib . Now we have
+
+
+
a set of 5.Hence a lI, is a set of 5 ; i.e. 5 converges. If a is an idkle, we may define its absolute value
2.
VALUATION VECTORS IN AN EXTENSION FIELD
A divisor (or ideal) a = n p r p is said to be integral if p, for all p. Let b = n p v v be any divisor (ideal), and define
241
>0
Then b = b1b;l = blib,; we naturally call bl the numerator and b, the denominator of b. Thus any divisor (ideal) can be expressed as a quotient of integral divisors (ideals).
Then I a ( may be regarded as the volume of the parallelotope 17, . If a is an idkle, a = a, , ...), then aa = ( - - - ,aa, , ...) is also an idkle, and we have
2. Valuation Vectors in an Extension Field
(..a,
by the product formula. Hence the parallelotopes l7, and 17,, have the same volume. We see that when we describe a parallelotope by means of an idde only the absolute values of the components play a role, not the actual components. This suggests that in order to describe parallelotopes we need merely prescribe an absolute value, or an ordinal number, for each prime p. T o this end we introduce formal symbols
Let k be a PF-field; let K I k be a finite extension of degree n. We adopt the following notation: V(k), V(K) shall denote the rings of valuation vectors in k, K. a, shall denote the idkles of k, K. 17, , 17,' shall denote the parallelotopes of V(k), V(K). V(k) can be mapped naturally into V(K) as follows: map 6 E V(k) onto the vector whose '$-component is 6, , whenever $!3 divides p . Since 5, E k*(p) C K*('$), and since there is only a finite number of primes dividing a fixed p in k, the image is indeed a valuation vector. It is easily seen that this mapping is continuous in the restricted direct product topologies of V(k) and V(K). Our aim is to give a description of the space V(K) in terms of V(k).For this purpose we introduce the space V n= V(k) x V(k) x
with v p = 0 for almost all primes p. In the case of algebraic function fields the product is extended over all primes p, and a is called a divisor. In the case of algebraic number fields the product extends only over the finite primes, and a is called an ideal. It is easy to see that each idkle a defines a unique divisor (or ideal) which we may also denote by a:
and its subspace kn = k
x k x ... x k.
We choose a fixed field basis for K I k: w 1 ,w , the mapping : Vn -+ V(K) by writing
+
+(ti
2
Similarly every element a E k defines a divisor (or ideal). If a is a fixed divisor (or ideal) of k, the set of all divisors (ideals) of the form aa is said to form a divisor class (ideal class).
x V(k)
52
...,5,)
=5
+
1 ~ 1
Then +(Vn) C V(K) and +(kn) = K ; tinuous mapping.
, .--, wn , and define
+ +
6 2 ~ 2
4 n ~ n
+ is easily seen to be a con-
Lemma 1: +(Vn) is everywhere dense in V(K).
242
XIII.
2.
DIFFERENTIALS IN PF-FIELDS
Proof: Let X be an element of V(K), ITa' any parallelotope; we have to prove that there is an element Y in q3(Vn) such that X- Y E 172r'To . this end we construct a non-empty set S' of primes p in k, containing at least (1) all archimedean primes, (2) all primes p which possess an extension '$ in K for which either l X I @ > lor lal,# 1. Let S be the set of all primes of K which are extensions of the primes in S'. S contains only a finite number of primes, so by the Approximation Theorem we can find an element a E K such that I-XIrpG
VALUATION VECTORS IN AN EXTENSION FIELD
243
We now show that if k is any field in which the lemma holds, then it holds also in any finite extension K of k. Suppose therefore that in k we have V(k) = k nb. Then
+
But q3(nbn)is contained in a parallelotope 17,' of K, for
where MIJ depends only on the w , . Since q3(Vn) is everywhere dense in V(K) we have V(K) = +(Vn) 17,and hence
+
IBI@
for all !l3 in S. Let then define vectors tl, t, , ..., tnin V(K) by writing (ti)@= ai for '$ E S, and (ti)$ = 0 for '$ 6 S. Then
where II,' 317,' Lemma 3:
Proof:
+ k = V(k).
There exists an idtle 6 such that l7,
Proof: We prove the theorem first for the rational subfield R. Let [ be a valuation vector; then tPE k*(p). We have seen that tPmay be written as a power series in p with integral coefficients (when p is non-archimedean); when R is the field of rational =m 7, where m is an integer and 0 17, < 1. numbers, tPm We define the principal part of $. at a prime p to be
+
The mapping
We have already remarked that
4
is continuous. Let
<
that X is contained in 17: where nais a parallelotope of V(k). , we can write X = Y Z where Z E kn Since V(k) = k Now +(Z) and Y E IIbn,so that +(Y) is in a fixed parallelotope nB1. lies in K, and
+
+
where n2'is a parallelotope of V(K) independent of X. We have seen, however, (Cf. Ch. 12, Thm. 2) that the totality of all such #(Z) is either finite, or else a finite dimensional vector space over and hence K O ,hence over k, . It follows that Z lies in a fixed nCn
( p non-archimedean),
X
=Y
+Z C l I ~ f l 7 ~ C s o m e L f ~ .
Thus if
81% Write a = 2, Prp(t). Then a E R. If = a lies in the parallelotope n, . Hence V(R) = R the lemma for the rational field.
+ is bicontinuous.
l7: be a fixed parallelotope in V(K) and let X be an element of q3-l(C fl +(Vn)); i.e. X E Vn and $(X) en,'. We have to show
is a vector of V(K) such that X - Y E &'. This completes the proof. Lemma 2:
+ 17,' . This completes the proof of the lemma.
+ 7 we see that + Ill. This proves r]
then
6, ,
a*.,
+ ... +
tn~ l . 7Finally ~
5,w.n
en;,
let c be any idhle; we see that if
244
XIII.
4.
DIFFERENTIALS IN PF-FIELDS
then tt1, "', ~ 5 E%c&. This shows that 4 is bicontinuous. From this it follows that is a (I, 1) mapping, and hence a homomorphism. For suppose
+
DIFFERENTIALS IN THE RATIONAL SUBFIELD
245
This is the required result. Let V be any vector space over an arbitrary field ko , of finite dimension n. Then it is well known that the linear KO-homogeneous mappings of V into KO form a vector space p, which is also of dimension n over KO; is called the dual space of V.
v
Let K be a finite extension field of ko . Then I? is a one dimensional K-space. Theorem 4:
Then, since 4 is bicontinuous, all the ti lie in every parallelotope, however small; hence, since V(k) is Hausdod, all the & are zero. Thus (b is (1, 1) and bicontinuous. Since V(k) is complete, Vn is complete and so +(Vn) is complete. Hence +(Vn) is closed. But #(Vn) is everywhere dense. It follows that 4(Vn) = V(K) and we have
x k K both topolog-
Theorem 2: V(K) is isomorphic to V(k) ically and algebraically.
Proof: I? consists of the linear KO-homogeneousmapping of K into k, . Let A, be an element of I? such that A,(t) is not identically zero. We shall show that every element of I? can be expressed as A(f) = hO(af) where a is an element of K. Clearly every such function A(() is an element of I?. On the other hand, we shall tnform a basis for K, then AO(tlt), ho(5&), show that if tl , t, , .--,A,((&) are linearly independent in K over k,, and hence form a KO basis for I? since dim I? = dim K = n. T o this end we notice that if ;\,(at) = 0 for all .$ E K then a = 0; for if or # 0 lies in K, and hence then
3. Some Results on Vector Spaces
Let X be a vector space over an arbitrary field KO. Let Y be a subspace of X. The KO-dimensionof the factor space X / Y will be denoted by (X: Y ) k o .When we may do so without causing confusion we shall omit the subscript k, . Theorem 3: Let A and C be subspaces of the same vector space. Let B be a subspace of A. Then
(A:B)=(AnC:BnC)+(A+C:B+C).
for all
5 contrary to hypothesis. Now we have
Hence the AO(&f)are linearly independent over KO. Thus every element h in I? can be written in the form
+
Proof: We map A onto (A C)/C by mapping or E A onto the coset a C . The kernel of this mapping is A n C. Hence
+
( A :B
+ ( A n C ) )= ( ( A + C)/C : ( B + C)/C)
=(A
+ C : B + C).
This completes the proof.
Further,
+
+ +
( A : B) = ( A : B ( A n C ) ) ( B ( A n C ) : B) =(A+C:B+C)+(AnC:Bn(AnC)) =(A+C:B+C)+(AnC:BnC).
4. Differentials in the Rational Subfield of a PF-Field
Let k be a PF-field. Then by Theorem 3 of Chapter 12, k is either:
246
XIII.
4.
DIFFERENTIALS IN PF-FIELDS
247
DIFFERENTIALS IN THE RATIONAL SUBFIELD
< an/2" <
(a) a finite extension of the field of rational numbers (briefly, a number jield); or (b) a finite extension of the field of rational functions in one variable x over a constant field k, (briefly, a functionjield). We introduce the following notation: R shall denote (a) for number fields, the field of rational numbers and (b) for function fields the field k,(x). (In both cases we shall call R the rational subJield as in Chapter 12.) P shall denote (a) for number fields, the field of real numbers; and (b) for function fields, the completion of ko(x) at p, . shall denote (a) for number fields, the additive group of real numbers modulo 1, with the natural topology; and (b) for function fields, the constant field ko with the discrete topology. N shall denote (a) for number fields, a fixed sufficiently small neighborhood of zero in r. For definiteness we take (- 3, 4). (b) for function fields, the zero element of k, . A differential of k is now defined to be a continuous linear map h of the ring V(k) into r such that h vanishes for elements in k. When k is a function field, h is further required to be ko-homogeneous. I n both the number field and function field cases this definition implies the existence of a parallelotope 17, in V(k) such that A(&) C N. I n the number field case the elements of a parallelotope & do not form an additive group. We therefore define the set
Proof: Let p(1/2") = an/2,, where - & is real. Since p is continuous,
where 5, is the valuation vector with component 1 at all nonarchimedean primes and component zero at all archimedean primes (thus for function fields 172 = K ) . The set 1 7 , O is an additive group. Its image h(naO)is also an additive group, and is contained in N. But N contains no subgroup of r other than the zero element, so h(17,0) = 0. We must now prove separately for the number fields and function fields two lemma concerning the linear maps of P into r.
Proof: The continuity of p implies that p(x4) vanishes for all large enough n, say n > N. Let
and a,
a, = 0. lim 27'
n+m
Now,
= p (0) =- 0 (mod 1).
Thus (a,,,
- an)/2, is an integer. But
r
Lemma A: In the number field case, let p(x) be a continuous linear map of P into Then p(x) E - ax (mod I), where a is a fixed real number.
r.
-
some fixed N. We write a, = - a Hence a,,, = a, for all n for n N. Then p(1/2n) = - a/2%, and by the additivity, p(r/2") r - ar/2" for all n 2 N. Hence we have , ~ ( 1 / 2 ~ )- (1112" for all n. It follows that p(x) E - ax for all rational dyadic fractions x, and hence by continuity
>
p(x)
- ax (mod 1)
for all real numbers x. Lemma 6: In the function field case, let p(5) be a continuous linear, KO-homogeneousfunction of P into r such that p(5) = 0 if 5 is an integer of R. Then p(5) = Residue at p, of (f(x) 5) where f(x) is a fixed polynomial.
Set p ( ~ - ~=) - ci . Then p(t)
= cl
Res, a (6)
= Re%,,,
+ c, Res,
( f ( 4 0,
a
(xt)
+ + c,
Res, a (xn-lt)
248
XIII.
DIFFERENTIALS IN PF-FIELDS
where f (x) = C1
+ c,x + + c,xN-l.
We investigate first of all the differentials in the rational field R. Let ( be a valuation vector of V(R), ( , its component at a finite prime p. We may write ,8 in the form
<
where 1 vp' Ip 1. Hp(5) = 0 for all but a finite number of primes p. Thus a = 2 H,(() is an element of R. We define for each finite prime p
and form the valuation vector v which has components rl, at the finite primes and components 0 at the infinite prime. We may now write E=a+v+5, with 5 component 0 at the finite primes and suitable component 5, at the infinite prime. Suppose ( may also be written in the form
4.
DIFFERENTIALS I N THE RATIONAL SUBFIELD
249
Now let p be a differential, and let a be an idde such that p(na) C N. We can find an element a E R such that ord, a = ord, a at all finite primes?. Hence ITa = IT,, where u is a valuation vector with component 1 at all finite primes and suitable component at infinite prime. We define p(() = p(a(). Then p is also a differential and
Clearly p(R;O) = 0 and hence p(v) = 0. Since a differential, p(a) = 0. Hence
ER
and p is a
since 5, is only unique modulo 1, and p(f) is unique. I t follows that p(5) is a function of 5, such that p(m) = 0 for all integers m. p(5) is linear and continuous since the map of ( -t 5 is linear and continuous. Hence we may apply Lemmas A and B. Lemma B gives us immediately for the function field case
In the number field case, Lemma A yields p(5) = - a5, (mod 1).
Then (a - a') has at every finite prime the same component as 9' - 7, i.e.
Ia-a'Ia=lq'-rllD
where m' has component m at every finite prime and component zero at the infinite prime, while m" has component m at the infinite prime and zero component elsewhere. I t follows that the mapping
5+ 5, is defined uniquely modulo 1.
Here we must apply the additional condition that p vanishes on integers. Putting 5, = 1 we see that a r 0 (mod I), i.e. a must be an integer. We now define the mapping A(() X(l) = - 5m (mod 1)
(for number fields),
h(l) = Res (5m)
(for function fields).
Then A is a differential. It is easily seen that A vanishes on field elements, and the continuity of A is proved as follows: (a) For the number field case, consider the parallelotopell,, , where E' is the idde with component 1 at all finite primes and component E at the infinite prime. For f E flE, we have
250
XIII.
5.
DIFFERENTIALS IN PF-FIELDS
Then A([) = h(5,), which can be made as small as we please by suitable choice of E . Thus h is continuous. (b) In the function field case, h clearly vanishes on the parallelotope 17,., where E' has component 1 at all finite primes and 2. ord,, E' We have therefore proved the existence of differentials for the rational field R. We now deduce two important properties:
>
Property I: The differentials of R form a 1-dimensional R-space. Let p be any differential; then we have
25 1
DIFFERENTIALS I N A PF-FIELD
tion to the statement that 6 E 17.: Thus we must have ord, a = 0 for all finite primes. Hence 17, C a O (0,0, E ) where E is chosen so that A(&) C N. In the number field case, I E 1 and in the function field case ord,, E 2 2. Thus in the function field case 17, C17,: . This proves Property 11.
+
Corollary:
a*.,
<
There is a maximal parallelotope 17, such that
h(I7,O) = 0. 5. Differentials in a PF-Field
where m is an integer and a E R. Thus p([) = h(m/a [). This proves Property I. Property 11: There is an upper bound to the parallelotopes 17, for which p(l7,) C N, where p is any differential of R. By Property I, it is clearly sufficient to consider the special differential A. Suppose A(&) C N. Then, since h(L!lo) = 0, we G o ) C N. Now if [ E 17, 1&0, then ord, [ min have h(n, (0, ord, a) for all finite primes. Thus we may already assume that o r 4 0 for all finite primes. Suppose ord, a < 0 for a certain finite prime. Then clearly, if m is any integer of R, the vector
+
+
>
Now let k be a PF-field. We shall prove the existence of differentials in k, and shall show that the differentials in k also satisfy Properties I and 11. The proofs do not use the special properties of PF-fields, and hold for any finite extension k of a field R in which differentials exist and satisfy the two properties. So let k be a finite extension of R. In the function field case we assume that k and R have the same constant field. Let w, , w,, w, be a basis for k/R. Then if X E V(k) we can write .a-,
<
(with component zero at the infinite prime) lies in naO. But
where the tiare valuation vectors of V(R). Let p be a differential of k. Then
The functions p([pi) are continuous linear (and in the case of function fields KO-homogeneous) maps of V(R) into I'. Since awi E k, p(awi) = 0 and hence p(<wi) is a differential of R. Thus p([wi) = X(ai[) where ai E R. Finally we have
and hence A ( [ ) = m/p (mod 1) in the number field case, and Conversely for arbitrary ai in the function field case. In the first case we can certainly choose m such that m/p > 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.
<
Lemma 1: If a I b , then
(na:nb)= n(b) - n(a). Proof:
It is sufficient to consider the case where b
= ap.
Then
262
XIV.
2.
THE RIEMANN-ROCH THEOREM
where o, denotes the ring of integers, It follows that
n
a prime element in k*(p).
(17, :17,) =
(ord, b - ord, a)f(p) = n(b) - n(a).
We now define the function m(a) to be the dimension over k,, of the space 17,n k . It follows from Chapter 12 that m(a) is finite for all divisors a. We shall now show that m(a) is an invariant of the divisor class of a. We have
263
FIRST PROOF
We must first show that 1(b) is finite for all divisors b. T o this end we use the result (see Chapter 13, section 2) that there exists an idble a, such that no, k = V . We select a divisor a such that 17. = l7, . Then certainly a divides b and we have, by formula (*) n(b) - n(a) = m(a) - m(b) (V : nb k ) = m(a) - m(b) l(b).
+
+
+ +
+
Thus l(b) is finite, as we set out to prove. We may therefore rewrite the formula (*) as follows:
m(aa) = dim ( n u , n k ) = dim ( d o n ak) = dim (l7,n k ) = m(a), since multiplication by a. ( # 0) is an isomorphic map both of and of k as k , spaces. Thus m(a) is a function of the divisor classes. Lemma 2:
If n(a) > 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
ord,, a Let a = l/D, then
= s ord PODx
+ e,
ord, a = c p , where e and c, are suitably chosen constants (almost all c, = 0) and s is sufficiently large.
266
XIV.
THE RIEMANN-ROCH
THEOREM
Consider the set of elements or in k which have the form = A('> w l
+ A(x) + + +n(x) wn , ~2
where deg +,(x) ,< s. For finite primes p we have
I I,
G m y I wv lp ;
OL
3.
SECOND PROOF
It now follows that - %(a)- m(a)
< sn + e1 - n(s + 1) = e1 - n.
Thus r(a) is bounded from above. We can draw several consequences from this lemma. First we notice that for a fixed divisor b, the function +(a, b) = (If,
whence ordp OL 2 min (ordpw,)
= c,
,
Y
and almost all c, = 0. At the infinite primes pa we have
+ k :If, + k)
is bounded from above. Suppose this upper bound is attained when k = V; for if f is a vector of V not a = a,. Then clearly contained in k, f must lie in some parallelotope 17, hence in k which contains 17b k and has higher dimension k, contrary to the definition of a, . Thus we have over it than attained the result
+
+ a + K O+ +
+
+
Hence ordPa OL 2 s ord
pa
x
+ e,
= V. Corollary 1: There is a divisor a, such that k This may be interpreted geometrically. It means that the elements of k are distributed in V like the points of a lattice. Since
where we have the desired result: All such elements a are contained in the parallelotope l7, . Among these elements there are n(s 1) linearly independent over k, . Hence m(a) 2 n(s I). Next we must find an approximation for n(a). We have
+
+
where e' is a suitable constant. Since l/x is a prime at p, ordPa
1
= - ord,,
2 e(p,)
fpm
+
.(a)
+ m(a) - l(a) = 1 - g,
where g is the genus. We also use the method of the first proof to interpret l(a) as the dimension over k, of the space of differentials which vanish on 17,. It is clear that differentials exist, for if not, l(a) = 0 for all divisors a, and hence n(a) 1 - g which is certainly impossible. k) As in the first proof we see that every linear map of V/(IT, gives rise to a differential which vanishes on 17,. If A ( [ ) is a differential and or is any non zero element of k, then )((or[) is also a differential. T o show this let a be a divisor such that
<
x = e(pm).
Further f(pm) =fPa (the relative residue class degree) and hence n(3 2 - s
For every divisor b, l(b) is finite. We now use the same procedure as in the first proof to show that n(a) m(a) - l(a) is an invariant of the field: Corollary 2:
- e'
=
- sn - e'.
+
268
XIV.
3.
THE RIEMANN-ROCH THEOREM
Then
h(aa-'(17,
+ k ) ) = A(a(17a-la+ k ) ) = 0
SECOND PROOF
269
Hence the number of linearly independent a such that X(a 17,) = 0 is at least equal to m(l/aD). Similarly the number of linearly = 0 is at least equal to m(l/ac). independent a such that p(a na) We have therefore
and so h(a5) is a differential. It follows that the differentials form a vector space over k. Scalar multiplication is defined by setting a ( [ ) = A(&). We now prove Lemma 4:
The differentials form a 1-dimensional k-space.
Replacing a in turn by l / a 9 and ljac we obtain
Proof: Let p ( t ) be a differential which is not of the form A(&). Suppose (a1 , a2 , .--,a,) and (/3, , /3, , ..., ps) are sets of elements linearly independent over ko . We shall show that X(a,f), .-., )((aZ.$, p(&.$) ..-,p(/3&) are linearly independent over k, . A relation of linear dependence would be expressible in the form
+
We can write this X(af) p(/3f) = 0 where a = X cp* and /3 = Z dipi. Then if /3 # 0 we can replace .f by /3-?$ obtaining
Adding, we obtain
contrary to our assumption on p. Hence /3 = 0 . This implies also and hence all ci , d j = 0. Thus the A(ait), p(/3jt) are linearly independent. Suppose
This gives a contradiction since n(a) may be chosen so large that this inequality cannot hold. Thus every differential p(() can be written in the form X(a0 and our lemma is proved. We have now the inequality l(a) 2 m(l/aD). Replacing a by 1/aD we have l(l/aD) >, m(a). Hence
a =0
We now estimate how many linearly independent elements a have the property that A(&) vanishes on a given parallelotope l7,. Certainly X(a17,) = 0 if and only if h = 0, and we get
(n,,)
We do not yet have the reverse implication. However,
Adding, we have
270
XIV.
THE RIEMANN-ROCH
THEOREM
Hence there is a divisor 3 such that &-I is the maximal parallelotope on which h vanishes. For this D we have the missing implication in the proof of Lemma 4, namely that CHAPTER FIFTEEN
I t follows that l(a) = rn(lla3) where is the special divisor. As in the first proof we can show that D belongs to a fixed divisor class W. We have therefore proved again the Riemann-Roch Theorem:
Constant Field Extensions 1. The Effective Degree
Let K be a PF-field with a field of constants K O .Then K is a finite extension of Ko(x) where x is an element of K not in KO. We now adopt the following notation: rp shall denote the generic prime in K. k shall denote a subfield of K in which Axiom 2 holds. (By section 4 of Chapter 12, Axiom 1 also holds in k.) K/k is not necessarily finite. which is p shall denote the generic prime of k induced by a non trivial on k. ko shall denote the constant field under all p. By section 4 of Chapter 12, ko = KOn k. f(Q) shall denote the degree of the residue class field Rep over KO. f(p) shall denote the degree of the residue class field h, over k, . f9 shall denote the degree of Rep over whenever finite. ecpshall denote (ordcpa)/(ord, a) for a E: k. ncp= e9 f 9 = degree of K*(Q) over k*(p) whenever finite. We have already seen that Axiom 1 holds in k for the primes p if we use the valuation
z,
I a lv = rI I1 a llcp -
PIP
Since k is a PF-field, it follows that the valuation I 1, is either the normal valuation 11 11, or a constant power of it (the same for all p). Let us agree to use the same constant c for defining the normal valuations in both K and k. Then we may write
272
XV.
CONSTANT FIELD EXTENSIONS
1.
We call the exponent m ( K / k ) the ejfective degree of K / k . I t is necessarily finite, but need not be an integer. We shall see that in a certain sense it measures the size of the extension K / k , the constant field extension being disregarded. Theorem 1:
THE EFFECTIVE DEGREE
We shall need two lemmas about the rational case:
Proof: Let po be the valuation in ko(x)defined by the irreducible polynomial x. The residue class field RPo = ko and SO f(po) = 1. If 'Po 1 po in Ko(x), Po is also the valuation corresponding to x. Hence f ( g o ) = ello = 1. From Theorem 1 it follows that
for all primes p .
Proof: Since the same c is used for
11 11,
and
11 11%
we have Lemma 2:
If Ko/ko is finite, then
Proof: Since
Hence we have
T o prove the equality it will be sufficient to show: If w , , o,, .-.,w, are elements of K O ,linearly independent over ko , then they are linearly independent over ko(x). Suppose therefore Theorem 2:
If F 3 K 3 k , where F, K , and k are PF-fields,
then m(F/k) = m ( F / K )m(K/k).
Proof: We denote the generic prime in F by element of k . Then
and this proves the theorem.
9. Let
a be an
where each f,(x) E ko(x). Then, multiplying by the denominators of the f,(x) we obtain a relation with polynomials as coefficients. So we may suppose the fi are already polynomials. If a , is the coefficient of xV in f,(x), then
as coefficient of xu on the left side. Hence a,, = 0 for d i, v or h ( x ) = 0We are now in a position to determine m ( K / k ) if K is a finite extension of k . T o do this we first prove
274 Theorem 3:
XV.
1.
CONSTANT FIELD EXTENSIONS
If K / k is finite, then
2 n@= deg (Klk)
THE EFFECTIVE DEGREE
Further for all
n(Ko(x)lko(x))= m(Ko(x)/ko(x))deg (Kolko) by Lemma 1 = deg (Kolko) = deg (Ko(x)/ko(x) by Lemma 2.
p.
@lp
Proof: We evaluate the degree of the residue class field over ko in two ways. The first computation yields:
R
NOWwe compute n(K/ko(x))in two ways: n(Klko(x))= n(KIKo(x))n(Ko(x)lko(x)) = deg (KIKo(x))deg (Ko(x)lko(x) = n(Klk)n(klko(x))= 4 KIk) deg (klko(x)).
Hence we have:
ko
Hence n(K/k)= deg (KIA).
The definition of n ( K / k ) yields whence
Putting n(K/k) = m(K/k) deg (Ko/ko),we see that
Theorem 4:
If K / k is finite, then
If K / k is not necessarily finite we prove Theorem 5:
If we have three fields F 3 K 3 k with constant fields Fo , K O , ko , then
for any x E k, not in ko
.
Proof: Since K / K o ( x )and k/ko(x)are finite we have
By the results of Chapter 12, section 3, we know from We now compute m(K/ko(x))in two ways: we obtain that
Since m(K,(x)/ko(x))= 1, we have the desired result,
275
XV.
276
1.
CONSTANT FIELD EXTENSIONS
Corollary: If K O = k, then m ( K / k ) = deg (Klk). In general k C Kok C K and the constant field of Kok is K O. Therefore
m(K/K&) = deg (K/K,k)
and
m(K/k)= m(K,k/k)deg (K/K,k).
This reduces the question of the general case to that of pure constant field extensions.
Dejinition: K is called a constant field extension of k if = K,k. Such constant field extensions of k may be obtained in the following way. Let K l be any extension of k, and Klk a composite field such that Klk is of transcendence degree 1 over K l . The primes of Klk are to be taken as those valuations of Klk which are trivial on K l . They are then trivial on k, but not all of them are trivial on k (if x E k not in k, we may take any extension to K,k of a prime in K l ( x ) associated with the polynomial x). The field K = K l k and k are then in the relation that we have considered. This field K will have a constant field K O3 K , since the primes are trivial on K l . Therefore K = Klk = Kok is a constant field extension. One might expect K l = K O but this is not always the case. A part of our investigation will consist in deriving conditions under which it can be stated that K l = KO. K
THE EFFECTIVE DEGREE
277
Let a, , a 2 , a, be the elements of k whose linear dependence over k, and K Ois to be investigated. We can write each ai as linear combination of the w , with coefficients in k,(x). We can assume that all the coefficients are written with the same denominator f(x) E ko[x]. Since the linear dependence questions are the same for the at and the f(x) a, we may assume to begin with that all coefficients of the ol, are polynomials in k,[x]. This means that the a, are elements of a vector space over k, spanned by the elements xYw, ( p = 1,2, n; v = 1,2, -..,N, N sufficiently large). The fact that the w, form a basis means that the xYwpare linearly independent over k, as well as over K O . If the ai are independent, then r of the basis elements xvwp may be replaced by the a , . Going over to K Othe dimension of the whole space does not change, so the % must remain independent over K O. a*.,
a*.,
(3) If m ( K / k ) < 1, then w1 , w, , w, become linearly dependent over K,(x). Clearing denominators we obtain a relation (non trivial) a*.,
fl(4~1
+ ...
+fn(x) % =0
with coefficients in K,[x]. This means a non trivial relation between certain xVwpover K Owhereas these elements are linearly independent over k, .
Theorem 6:
Theorem 7: If for one prime p, f(p) = 1, then m(K/k)= 1 for any constant field extension K .
Proof: ( 1 ) We have
Proof: Theorem 1 shows that m ( K / k )is an integer # 0. Theorem 6 shows it is 1, whence m ( K / k ) = 1. In particular if k, is algebraically closed, f(p) = 1 for all p, hence m ( K / k ) = 1. Theorem 1 shows that there is only one q I p and that e~ = f ( p ) = 1 .
If K is a constant field extension of k, then m ( K / k ) 1, the equality holding if and only if K O and k are linearly disjoint over k, (i.e. if elements of k that are linearly independent over k, are also linearly independent over KO).
<
since elements of k linearly independent over k,(x) may become independent over K,(x). (2) Suppose
m(K/k)= 1,
deg (KIK,(x))= deg (klk,(x)).
If wl , w, , ..., w, is a basis for k/k,(x), then it is also a basis for
KIKo(x).
<
Theorem 8: If K = Kok = Klk, m(K/k) = 1, then K O = K l .
Proof: We have
where
K l C K O and
278
XV.
CONSTANT FIELD EXTENSIONS
Theorem 9:
For any divisor a in k we have
Proof: It is sufficient to show this for a = p. Then n,(p) =f(p). Therefore according to Theorem 1 2. Divisors in an Extension Field
Let K be a PF-field, k a subfield of K that is a PF-field. Suppose for an element x E k, considered as divisor of k, we have
x also gives rise to a divisor in K. Since p ranges over these valua-
tions induced by primes !$3 in k which are non trivial in k, only primes $! in K that divide the primes p of expression (1) can appear in
as read off from factorization (3). Corollary: If
KO= k,, then
Now let x E k but x $ ko . Consider the two fields ko(x) and k. In ko(x) we have the factorization
where p, is the prime in k,(x) corresponding to the polynomial x, p , the infinite prime. In k both po and p, will split up, no cancellation takes place so that the factorization of po and p, in k gives us numerator and denominator of x if written as divisor in k. The corollary gives us This suggests to map the primes p of k into the divisors of K by the correspondence
nk (numerator of x) = deg (k/ko(x))n,,(,,(po) = deg (klko(x))
and the same for the denominator. to extend this mapping in the obvious way to all divisors a of k and to identify a with its map. In this sense expression (1) already gives the factorization of x in K and instead of (2) we write
Theorem 10: If x E k, x $ k, and a is the numerator or the denominator of x written as divisor in k, then
I t is to be remarked that different primes p will have factorizations without common factor. If we denote by n,(a) and nK(a)the degrees of a divisor in k and K respectively, we contend
3. Finite Algebraic Constant Field Extensions
In the beginning of this section we prove algebraic theorems for more general fields.
280
XV.
CONSTANT FIELD EXTENSIONS
Theorem 11: Let k be any field, k, a subfield algebraically closed in k. If a is an element of an extension field of k that is algebraic over k, , then
Proof: Let f(t) = Irr (a, k, , t). We must show that f(t) remains irreducible in k[t]. Suppose +(t) is a factor of f(t) in k[t] with highest coefficient 1. The other coefficients of +(t) are symmetric functions of the roots of +(t). These roots are also roots of f(t) and therefore algebraic over k, . It follows that the coefficients of +(t) are algebraic over k, . Since they lie in k, they must already lie in k, and +(t) E k,[t]. Since f(t) is irreducible in k,[t] it follows that +(t) = f(t) and the Theorem is proved. Let k, be algebraically closed in k, a as before and F a field between k, and k such that deg (k/F) is finite. Then Theorem 12:
Proof: k, is then algebraically closed in F as well as in k and Theorem 11 gives
But in spite of the fact that k,(a) is algebraically closed in F(a) it is not true in general that k,(a) will be algebraically closed in k(a). We now seek conditions under which this additional result will hold. Suppose /3 E k(a), /3 algebraic over k,(a). Then
Let us assume that k,(a, 8) can be generated by a single element: k,(% B) = ~O(Y).Then
and we see that F(a, B) = F(a). This means that /3 lies already in F(a). Since /3 is algebraic over k,(a) and k,(a) is algebraically closed in F(a), it follows that /3 lies in k,(a), i.e. k,(a) is algebraically closed in k(a). Now k,(a, /?) can be generated by a single element when either a or /3 is separable over k, (see van der Waerden, Modern Algebra, section 40). It follows that k,(a) is algebraically closed in k(a) in the following two cases: (I) a is separable over k, ; (2) k is separable over K (i.e. k is separably generated over k,).
and our theorem follows. We specialize now the field F. It is well known that k, is algebraically closed in a pure transcendental extension k,(x). By induction one shows that k, is algebraically closed in F = k,(x, , x, , x,) where the xi are algebraically independent over k, . For an a algebraic over k, it follows from the same fact that k,(a) is algebraically closed in F(a). Let now k be an algebraic extension of such an F for which it is known that k, is still algebraically closed in k. The preceding theorems show that
In the second case we conclude as follows: k/F separable 3 k(a)/F(a) separable => /3 separable over F(a). Now /3 was algebraic over k,(a). k,(a) is algebraically closed in F(a). +(t) = Irr (/3, kO(a),t) then +(t) remains irreducible in F(a) according to Theorem 11. So +(t) is separable, therefore /3 is separable over k,(a). A certain power jgpv will be separable over k, and still generate the field KO(%B) over k,(a). Summing up we have Theorem 13: With the notation as described, k,(a) is algebraically closed in k(a) if one of the following conditions is fulfilled:
(1) k,(a)/k, is separable ; (2) k is separably generated over k,
.
282
XV.
3.
CONSTANT FIELD EXTENSIONS
We specialize now to the case of transcendence degree r = 1, when k, is the constant field of a PF-field. Whenever k,(a) is algebraically closed in k(a) it follows from earlier discussions that k,(a) is the constant field KOof k(a) = K. Theorem 11 shows now that m(K/k) = 1. Let a be algebraic over k, , K = k(a). We have 1 and KO= k,(a) in either of the two cases:
FINITE ALGEBRAIC CONSTANT FIELD EXTENSIONS
283
If k e ) C kip), then
On the other hand
Theorem 14:
m(K/k)
=
(1) ko(a)/ko is separable; (2) k is separably generated over k,
<
.
If Ko/ko is finite and purely inseparable (characteristicp > O), then K/k is purely inseparable. Theorem 4 shows that m(K/k) is a power of p. If K,/k, is an arbitrary finite extension, let K, = k,(a) be the separable part of K O .We obtain from Theorem 14 m(K/k) = m(K/K,k) m(Klk/k) = m(K/Klk) = power of p. Theorem 15: If Ko/kois finite, K = Kok, then m(K/k) is 1 if the characteristic is 0 and of the form l/pv(v 2 0) if the characteristic is p > 0. In order to treat infinite constant field extensions we use the following Definition: A family K(,) of constant field extensions of k shall be called a C-family if to any two fields K(a), Kc@)a third field K(") of the family can be found that contains both of them. The union of the elements of all K(a) shall be denoted by K. It is again a constant field extension of k and and one shows easily KO= Ua KF). Theorem 16:
If m(Kfa)/k)= 1 for all a, then m(K/k)
Theorem 15 shows that the numbers m(K(a)/k) take on their For all K(a) 3 K(") we have minimum value for a certain PY). m(K(a)/k) m(K(")/k) and therefore m(K(a)/k)= m(K(Y)/k) or =( 1. ~Z(K(~)/K ~ )If) we apply Theorem 16 to all K(a)3 K("), we obtain m(&k(")) = 1 or m(R/k) = m(PY)/k)= a power of p. Let now KObe an arbitrary (not necessarily algebraic) extension of A,, K = Kok and KOthe algebraic closure of KO. We have
=
1.
Proof: K F ) and k are linearly disjoint over k,. Therefore KO= U, K F ) and k are linearly disjoint over k, . Let 6, be the algebraic closure of k, and A = &,k. Since A, is algebraically closed, it is the constant field of k. Let k p ) stand for all finite algebraic o Then K(a) is a C-family extension of k, and put K(") = k(a)k. with k as union.
Since &,k has an algebraically closed constant field, Theorem 7 shows that m(I?,k/h,k) = 1. So m(K/k) =
m(J&lk) m(%klK)
= a power of p.
<
Since m ( R , k / ~ ) 1 , m(K/k) 2 m(k$z/k)= m(K(")/k). Theorem 17: If the characteristic is 0, then m(K/k) = 1 for all constant field extensions. If it is p > 0, then m(K/k) = l/pv(v 2 0) and the minimum value is taken on by a finite algebraic extension K("). Theorem 18: Let K f a )by any C-family above k. There exists a K(") such that m(K/k) = m(K(Y)/k).For any K'a) containing K(Y) we have m(K(a)/K(~)) = m(K/K(y))= 1.
Proof: m(K/k) = m(K/K("))~z(K(~)/k) < m(K@)/k).
284
XV.
CONSTANT FIELD EXTENSIONS
Since m(K(a)/k)= l/pv,there exists a K(Y)for which the minimum ) has value is achieved. For K(")3 K ( Y one
4.
PURELY TRANSCENDENTAL CONSTANT FIELD EXTENSION
285
Suppose the strict inequality holds. Dividing g(t) by the x, with the maximal absolute value we obtain a new polynomial where Max, 1 x, ,1 = 1, and where (g(t) 1, < 1 or g(t) r 0 (mod 9). This congruence would mean that t (as element of the residue class field is algebraic over the residue class field K, . Since R, is algebraic over k, , t would be algebraic over k, . But the residue class field Kq contains an isomorphic replica of ko(t) where t is transcendental over k, . We have therefore
4)
whence m(K(u)/k)= m(K(Y)/k) and therefore = 1. m(K(")/K(Y))
This implies m(K/K(Y)) = 1.
This shows that there is only one 9 I p and that eq = 1, p = 9. An arbitrary element X E K can be written in the form
4. The Genus in a Purely Transcendental Constant Field Extension '
We consider the case K = k(t) where t is transcendental over k. The primes 9 of K are those that are trivial on k,(t). We have to distinguish two kinds of primes: (1) 9 is also trivial on k. They are then among the well-known valuations of the rational field k(t) which come from irreducible polynomials p(t) E k[t] (with highest coefficient 1. The infinite prime would not be trivial on k,(t)). But we have to single out those p(t) which give rise to the trivial valuation on k,(t). T o find them, suppose 9 induces a non trivial valuation on k6[t] that comes from the irreducible polynomialp,(t) E k,[t]. Since k, is algebraically closed in k we know from Theorem 11 that p,(t) remains irreducible in k. Since p(t) I po(t), we have p(t) = po(t). The primes 9 in question are therefore those irreducible polynomials p(t) E k[t] with highest coefficient 1 that have at least one non constant coefficient. (2) The primes that are non trivial on k, 9 1 p. Consider first a polynomial Since 9 is trivial on k,(t) we have
I t Iv
=
1 and
where x E k and g(t) and h(t) are relatively prime polynomials with highest coefficient 1. We first ask for those X that satisfy I X Iv 1 for all 9 that are trivial on k. If h(t) were divisible by an irreducible polynomial p(t) with non constant coefficients, our condition would be violated at that prime since g(t) is relatively prime to h(t). This shows that h(t) E k,[t]. If conversely h(t) E k,[t] it is a constant of K and our condition is obviously satisfied. This allows us first to determine K O . If namely I X lv = 1 for all 9 then both g(t) and h(t) have to be in k,[t]. Therefore , X 1, = I x ,1 = 1 which means that x E k, , so that X E k,(t). This means KO= k,(t). Let now x, , x2 , x, be elements of k linearly independent over k, . Suppose
<
..a,
Clearing denominators, we may assume that the fi(t) are polynomials. Collecting terms we obtain an equation for t. Since t is transcendental over k, all coefficients have to vanish. Since the xi are independent over k, this means all f,(t) = 0. k and KO are therefore linearly disjoint over k,, and this means m(K/k) = 1. Let now a be a divisor of k, lI, the parallelotope of V(k). a may
286
XV.
5.
CONSTANT FIELD EXTENSIONS
be viewed as divisor of K and let 17,' be the parallelotope of V ( K ) . We ask for the elements X E K such that X E 17,'.Then I X Iv 1 for all 'p that are trivial on k. Therefore X = g(t)/h(t)where h(t) is in KO. Let g(t) = xo x1t
<
+ +
So g(t) E 17,' if and only if xi E 11,. Hence
GENUS I N ARBITRARY CONSTANT FIELD EXTENSION
5. The Genus in an Arbitrary Constant Field Extension
We first consider the case where Ko/ko is a finite algebraic extension. Since K = Kok we can find a basis w l , w , , w, for K/k that consists of constants. These wi need however not form a basis for Ko/ko. According to Chapter 13, section 2, we have -a*,
We denote the genus of k by g(k), that of K by g(K). Then g(k) = l(1) = dimko( V ( k ) :I l l
is a linear combination of elements of & n k with coefficients in KO. This gives (because of the linear disjointness)
Theorem 9 shows nK(a) = nk(a). Let g(k) be the genus of k, g(K) that of K. Let a be a divisor of k such that nk(a) < 2 - 2g(k)
and
n,(a)
+k);
whence ng(k) = dim,, ( V n :n:
+ k"),
where the powers are meant in the sense of a Cartesian product. We now apply the mapping of Chapter 13, section 2. Then
< 2 - 2g(K).
Then by the Riemann part of the Riemann-Roch Theorem we have
+ mk(a) = 1 - g(k), + mK(a)= 1 - g ( K ) .
nk(a) nK(a)
Hence g(k) = g(K). Finally, let K = k(tl , t, , t,) where the ti are algebraically independent over k. Induction shows Theorem 19: Let K = k(t, , t, , -..,t,) be a purely transcendental constant field extension. Then
K O = ko(tl , t2 , ...,tn), mK(a) = mk(a)
287
n ~ ( a= ) nk(a)r m ( K / k ) = 1, and g ( K ) = g(k).
n K are linear combinations of the elements The elements of of Ilan k with coefficients from KO.
where
17' is not a parallelotope of V ( K ) but is a subspace of the unit parallelotope n1'of V ( K ) .On the other hand we have g ( K ) = dimKo( V ( K ) :l&'+K ) ;
whence deg (Ko/ko)g ( K ) = dimko( V ( K ) :R;'+ K ) .
A comparison with (1) yields
XV.
288
5.
CONSTANT FIELD EXTENSIONS
if we recall that n = m(K/k)deg (Ko/ko).An immediate consequence is the inequality
Thus the genus drops if m(K/k)< 1 (and g(k) # 0). But we shall show that the genus may drop even if m(K/k)= 1 and if 5 is separably generated. Suppose now m(K/k)= 1 .Then w, , w, , w, form a full basis for Ko/ko. We may then write nl'= nlF0 . Since both Dl' K and D,K0 K are KO-spaces the relat~on(2) takes on the simpler form me.,
+
+
f ( k ) = g(K)
+ dimKo(Ill + K :nlKo + K )
(4)
Now from our lemma on vector spaces, Chapter 13, section 3, we have (I&': nlKo)= (I&'+
K :nlKo
+ K) + (n;n K :nlKo n K).
Butn,' n K = KO and nlK0 n K 3 KO so that the second term is zero. We have therefore Theorem 20: If K/k is a finite algebraic constant field extension and m(K/k)= 1, then
GENUS I N ARBITRARY CONSTANT FIELD EXTENSION
Since X €&KO and the expression of an element of V ( K ) in terms of the basis oi is unique, we conclude that all x5 €17,. The converse is obvious. The linear disjointness of KO and k over ko gives now
for all divisors a of k. Consider next the more general case that KOis finitely generated (possibly by transcendental elements). If t, , t, , -..,t, is a basis of transcendency of Ko/koand we put Ki = ko(tl, t,), K' = K,'k, then KO/Ko1 is a finite algebraic extension. Theorem 19 shows that the inequality (3)holds in this case also. Should g(K) = g(k) then m(K/k)= 1, mK(a)= m,(a) and the elements of n,' n K are linear combinations of those of n,,n k with coefficients in K O . Let now K/k be an arbitrary constant field extension. Let K(a) be the C-family of fields obtained from k by adjuction of a finite number of elements of KOto k. The union of the Kc.) is K. We have 0 ,< g(K(cL)) - g(k). Let K ( Y )be a field with minimal genus. If K ( a )3 Q Y ) , then g(K(=)) ,< g(K(Y)on one hand, whence, since g(K(a))is the minimum, we have g(K(u))= g(K(Y)). An element X E nu'n K will be in some K ( u )3 K ( Y )Denoting . n by l7:) the parallelotope for K(a) we have X ~ n t ' K(a). Therefore it is a linear combination of elements in K ( Y n ) K(Y). with coefficients in K 2 ) C K O . Since m(K(a)/K(Y)) = 1 we have = 1. This linear disjointness shows also m(K/K(Y)) a*.,
a result that may also be written in the form
Using the Riemann-Roch Theorem on a divisor a with Suppose now that K/k is a finite algebraic constant field extension such that g(k) = g(K).( 3 )shows that m(K/k)= 1. Theorem 20 shows that 17,' = ITIKo. Let a be a divisor of k, and 5 E V(k) such that ord, 5 = ord, a for all p. Then 517; = n,,'and 517, = na. Therefore nu'= &KO . Let
289
we obtain
Therefore g(K) = g(WY)).
292
XV.
CONSTANT FIELD EXTENSIONS
would be highly desirable to find some criterion for conservative fields. We make in this direction the following remarks: Let KObe any extension of k , , h, and KOthe perfect closure of k, and K O . We have g ( x 0 k ) = g(hok) since R, is perfect. From g(K,k) 2 g(Kok) we get g(K,k) 2 g(kok). So the greatest change in genus is already obtained by going over to A&. According to our results, a finite subfield of &, will already achieve the same. We see that one need only consider finite algebraic extensions of k, that are purely inseparable. Theorem 20 shows now that k will be conservative if and only if for all primes alp=. D,5 (for all finite purely inseparable extensions KO). But a crlterlon should be found that expresses this behavior in k itself. We finally show by an example that the genus of a separably generated field k need not remain unaltered under a constant field extension. Let k, be the field of rational functions in one variable u over the prime field of p elements ( p odd). Then let k = k,(x, y ) where y2 = xp - U . Since p is odd, k is separable over k,(x). T h e genus of k is ( p - 1)/2 (see section 7 of the next chapter). Now let KO = k,(Qu), K = Kok. Then K = K,(x, y ) where y2 = ( X . Putting y = z ( x - $ ~ ) ( p - l ) 1 ~ we have z 2 = x - Qu. x E K o ( z ) s o y E Ko(x) whence K = K,(x), a rational field with genus 0 . I t shall be remarked that the mysterious amount ( p - 1)/2 of the genus drop becomes understandable from a result of J. Tate according to which the drop in genus is always a multiple of
Tu)~
( p - 1)/2. (Genus change in inseparable extensions of function jield theory, Proc. Amer. Math. Soc. 3, 1952, pp. 400-406.)
CHAPTER SIXTEEN
Applications of the Riemann-Roch Theorem 1. Places and Valuation Rings
Let {F +-a) be a system consisting of a field E and a single additional element co which satisfies the following formal properties:
p,
a f a3 = oo
for all
a
a -oo=oo
for all
a # 0.
E
Let k be any field. Following Dedekind we define a place of k to be a homomorphism of k into F +a.It is easily seen that the elements a of k for which +(a) E F form a ring o; o is called the ring of local integers. The elements which are mapped onto zero in P form an ideal p of o. It is easily seen that p is a maximal ideal, for a E 0, a $ p * +(a) # 0 , f co * +(a-l) # 0 , # co 3 a-l E D . Hence a is a unit of o, and (p, a) contains with a also a-la = 1 , and so ( p , a) = O. It follows that p consists of all non units of o, and hence p is the unique maximal ideal of o. We contend that o has the following characteristic property: If a E k at least one of the elements a, a-l lies in o. A ring which has this property is called a valuation ring. Thus every place of k defines a valuation ring. Conversely let o # k be a valuation ring in k. We shall show that o defines a place in k. T o this end we show that the non units of 293
+
294
XVI.
1.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
o form a proper ideal of o. Let a, #? E o, and let a f Either a/#?or #?/alies in o, say a/#?E o. Then
+
#? be a unit of o.
Hence, since (a #?)-I E o we have l/#?E 0, i.e. p is a unit. Hence, if a and #? are non units, then a #? is a non unit also. Similarly, if a, #? E o, and a#? is a unit, then a/a#?= 1/43 and #?/a#?= l/a lie in o. Hence a and #? are units. Hence if a is a non unit, #? any element of o, then a#?is a non unit. Thus the non units form an ideal p. Since all proper ideals are contained in the set of non units, p is the unique maximal ideal of o. We now map k onto o/p +a in the obvious way. Namely if a E o, we set +(a) = residue class of a modulo p, and if a # o, we set +(a) =a.This mapping is known to be homomorphic on the elements of o. We must verify that it is homomorphic on the whole of k. If a E o, b 4 o, then
+
+
b is not an element of o. If a #o, then +(a)=a.But since a a-l E o, and since a is a non unit of o a lies in p. Hence
Thus
+ is a place of k and we have proved.
There is a 1 - 1 correspondence between the places and valuation rings of a field. Now let o be any ring contained in a field k. Let be any homomorphism of o into any field E, with kernel p. We shall prove that may be extended to a valuation ring of k. Should it happen that p = 0, then is an isomorphism of o into E and can be extended to an isomorphism of k into an extension field of P that contains an isomorphic replica of k. Henceforth we exclude this trivial case. We now prove Theorem 1:
+
+
+
Theorem 2:
+ can be extended to a valuation ring of k.
Proof: The set S of elements of o, which have non zero images under
+, forms
a semigroup. The set o' of quotients a/s with
PLACES AND VALUATION RINGS
295
a E D , s E S therefore forms a ring, the quotient ring of o with respect to p. It is easily verified that can be extended to o' by defining +(a/s) = +(a)/+(s). This extended map carries o' onto a subfield F of P. Since F is subject only to the condition that it contain F, we may replace it by any field extension of F, in particular by the algebraic closure of F. From now on we assume P to be algebraically closed. Now let x # 0 be an arbitrary element of k. We shall attempt to extend to the ring of o[x], consisting of polynomials in x with coefficients in o. Let
+
+
* ( t ) = a,,
+q t +
"'
antn
be a polynomial in o[t] and write
+
In order to extend to o[x] we must define +(x) = x, EEand then write +($(x)) = fix,). The element x, must be selected so that the extended mapping is well defined, i.e. so that if $(x) = 0 then $(x0) = 0. The set of polynomials $(t) such that $(x) = 0 is an ideal in o[t]. Hence the images $(t) form an ideal in Fit], and since F is a field this is a principal ideal ($,(t)). In order, therefore, that our extended mapping be well defined, xo must be a zero of $,(t). It follows that we can extend 4 to o[x] provided $,(t) is not the polynomial 1. Suppose it is impossible to extend to o[x]. We shall now show that it is possible to extend4 to o[y] where y = Ilx. Since $,(t) = 1, x is a zero of a polynomial
+
where the p, lie in p. Thus
Suppose it is impossible to extend similar equation for y:
+ to o[x].
Then we have a
296
XVI.
2.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
with pif in p. Suppose these equations are of minimal degree, n. The element (1 $- pof)has an inverse in o, so (2) and let m yields
2. Algebraic Curves
<
and hence
This relation may now be used to shorten (I), contrary to the minimal nature of n. Hence m = n = 0. But this is impossible, since 1 p, # 0. It follows that we must be able to extend 4 to one at least of o[x], o[x-l]. We now consider the set of extensions of 4 to rings containing o. If we define an ordering in this set by letting 4, > 4, whenever 4, is an extension of 4, we see that the set is inductively ordered. By Zorn's Lemma there exists a minimal extension 4, of 4 to a ring o, containing o. Since 0, is maximal, o, = olf and o, contains at least one of the elements x, x-l for every x E k. Thus o, is a valuation ring. This concludes the proof of the theorem. We now explain the application of this theorem to algebraic x,] the ring of geometry. Let k, be any field, R = k,[x, , x, , polynomials in n variables over k, . Let p be a prime ideal in R. p is finitely generated, say
+
..a,
where we write f,(x) as an abbreviation for f,(xl, x 2 , ..., x,). The a,) is said ideal p defines an algebraic variety. A point (a, , a, , to be an algebraic point on the variety if the a, lie in k, or some algebraic extension, and f,(al , a, , -..,a,) = 0 (i = 1, ..., r). The quotient field k of R/p is called the function jield of the variety. Suppose (a, , a, , ..., a,) is an algebraic point of the variety. Then the map 4 : xi -+ a, carries R into an algebraic extension of k,. I t may be extended naturally to a map 4' of R/p into this extension field. Then by the preceding theorem, 4' may be extended to a valuation ring of k. T o this valuation ring corresponds a place and so we have
297
ALGEBRAIC CURVES
An algebraic variety whose function field is of transcendence degree 1 over k, is called an algebraic curve. We shall see that in this case there is an intimate connection between the places of the function field and the valuations. First let k be any field, o a valuation ring. We shall define an equivalence relation in the group of non zero elements of k. We say that a is equivalent to b (a b) if both a/b and b/a lie in o, i.e. if a o = bo. This relation is easily seen to be reflexive, symmetric and transitive. We denote the equivalence class of a by j a 1, and define a multiplication between the equivalence classes by writing I a / I b 1 = I ab I. To justify this definition we must show a' and b b' then ab a'b'. This is easily verified. that if a Under this multiplication the equivalence classes form a group in which the identity element is the equivalence class consisting of units of o. We may define a total ordering of the equivalence classes. We write
-
-
-
-
a
I a I < I b l + -b~ p + $
(3 -
=0,
where 4 is the place corresponding to o. It is easily shown that this relation is independent of the choice of representatives for the equivalence classes. We can verify that
em.,
Theorem 3: T o every point on an algebraic variety corresponds at least one place of the function field.
If we now define I 0 I to be 0, we have a function I / defined for all elements of the field k. Instead of defining I a I to be the equivalence class of a, we may take it to be an isomorphic image in any isomorphic multiplicative group. We see that I I satisfies the axioms for a valuation in k, apart from the requirement that the values be real numbers.
298
XVI.
2.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
Now let k be a PF-field, that is k is either an algebraic number field or an algebraic function field of transcendence degree 1 over its constant field. Thus we include as a special case the function field of an algebraic curve. Let # be a non trivial place of k. In the case of a function field we restrict ourselves to those places # which set like the identity map on the constant field k, . I n this case we select an element x E k, x 6ko such that #(x) #a. (This can always be done, for if $(x) =CO, then #(l/x) = 0, # a . ) Let R be the rational subfield. R is the field of rational numbers if k is an algebraic number field and R = k,(x) where #(x) #a in the case of a function field. Then #(a) #a where a is any integer of R. I n the number field case this follows from the fact that #(I) = 1. I n the function field case it follows from our choice of x. We now show that # is not the identity on the integers of R. Since # is non trivial there is an element n- # 0 such that #(T) = 0. This T satisfies an irreducible equation with integral coefficients in R. Applying the homomorphism 4 we obtain #(a,) = 0, since #(T) = 0. But a, # 0 since the equation is assumed to be irreducible. Hence # is not the identity on the integers of R. It follows that # maps the integers of R into a field. The kernel is a non zero prime ideal (p) of R. If a is any element of R, and a = pvb,wherep does not divide the denominator, or the numerator of b, then I a I = I p Iv. Hence the value group v, of I I over the rational field R is cyclic. Now let or be any integer of k. or satisfies an equation Since I f(or) I = 0, two of the terms must have maximal absolute value, say I a,& I = I a& I. Then I or$-j I = I aj/a, I = I p Im since the a* lie in R. Since i - j may vary as or varies, we introduce a uniform exponent by writing Hence if vk is the value group of I I over k, we see that v;' is a subgroup of the cyclic group v, . But since vk is an ordered group,
ALGEBRAIC CURVES
299
the mapping I or I -+ I a! In' is 1 - 1 and order preserving, and so = v , . Thus vk is a cyclic group of positive real numbers. Hence we may regard the mapping or + I or I as a valuation of k. In particular, we have
$1
Theorem 4: To every point on an algebraic curve corresponds a valuation of its function field. Let k = k,(x, y) where x and y are connected by the polynomial relation F(x, y) = 0 (not necessarily irreducible). k is then the function field of the curve defined by F(x, y). Let (x, , yo) be an algebraic polnt on an irreducible constituent of this curve, i.e. an algebraic zero of an irreducible factor of F(x, y). Then (x, ,yo) gives rise to a valuation of k. Theorem 5: If aF/ax and aF/ay are not both zero at (x,, yo), then the valuation induced by (x, ,yo) is unique and the residue class field under this valuation is ko(xo,yo).
Proof: Suppose aF/ay (x, ,yo) # 0. Let p(x) = Irr (x, , k, , x). If # denotes the mapping defined by (x,, yo), i.e. the mapping described by $(a) = a!
for
a E KO,
then #Cp(x)) = p(x,) = 0. Hence if p denotes any valuation induced by #, we see that I p(x) ,1 < 1. Thus p induces on R = k,(x) the valuation defined by p = p(x). The map # can be extended to each completion k*(p) and hence also to R*(p). Let F,(x, t) be the irreducible factor of F(x, t) in R of which y is a root. Then the different inequivalent completions k*(p) are obtained by adjoining to R*(p) roots of the distinct irreducible factors of Fl(x, t) in R*(p). Suppose f,(t) is Irr (y, R*(p), t). Then 1, fl(t) is one of these factors. Since y -+ yo f a , we have I y 1 , so y is an integer and hence all the coefficients of fl(t) are integers (cf. Chapter 2, section 5). Since y is a root of F(x, t) = 0 we have
<
300
XVI.
3.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
and g ( t ) has integer coefficients. Then
301
LINEAR SERIES
are divisors of f we have a,/ao = a, where We define the expressions
ai
is an element of k .
(with c, in the constant field KO) to mean since f l ( y ) = 0. Now apply the map C$ which has been extended to the completions R * ( p ) and k*(p); we obtain
But if there were any irreducible factors of Fl(x, t ) distinct from f l ( t ) these would be contained in g ( t ) and we should obtain +(g(yo)) = 0. Thus the valuation p induced by is unique. Let fo(t) = 4( fl(t)). fo(t) is a polynomial in the residue class = ko(ko(xo). Since f,(t) is irreducible over R it follows that field fo(t) is either Irreducible or a power of an irreducible polynomial (cf. Chapter 3, Theorem 8). Since fo1(y0)= $( fl1(y0)) # 0, yo is not a multiple root of fo(t) and hence fo(t) is irreducible in 4 . Hence deg (ko(x0 ,Yo)lko(xo))= deg (k*(p)lR*(p)).
+
Now the residue class field 6, of k certainly contains ko(xo)and yo . Hence we have
Thus we have the required result that 6, = ko(xo, y o ) . We notice also that the residue class field is separable over k,(x,). The extension ko(xo)/komay however be inseparable.
The totality of all such expressions is called a linear series. The maximum number of linearly independent functions or, is called the dimension of the linear series. We see that the linear series does not depend on the choice of a, . For if we choose a,' as a new fixed divisor in f, we have
Hence
Thus the totality of all expressions (*) is unaltered by the change. Now let the ai be integral divisors. We have at ai integral c- ai= El 7 -,9 a,ao integral. a, a,
Hence
The set of expressions (*) constructed using all the integral divisors of f is called a complete linear series. The dimension of f is defined to be the dimension of the integral divisors of f, i.e. dim f = m(ail).
3. Linear Series
Let k be an algebraic function field in one variable. Let f be a fixed divisor class of k , a, a fixed divisor in f. Then if a,, a , , ... or
The Riemann-Roch Theorem,
302
XVI.
4.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
303
FIELDS OF GENUS ZERO
+
ITa contains all polynomials of the form $(x) yt,h(x) where s and deg t,h(x) s - 1. Hence m(a) 2s 1. T o deg +(x) show this we have only to remark that for finite primes p, I y ,1 - 1, and that I y l V co I x lVm . These results follow from an exarnination of the dominant terms in the equation of the conic. We now have 1 l(a) 1 - g as in Part 1, and the result follows as before.
can now be written in the form
<
<
< +
<
where W is the canonical class.
<
4. Fields of Genus Zero
In our first theorem we give a complete description of the fields of genus zero. Theorem 6:
A field has genus zero if and only if it is of one
of the following types: (a) a field of rational functions, (b) the function field of a conic, i.e. a field k,(x, y), where x and y are related by an equation ay2
+ (b + cx) y + (d + ex +fx2) = 0.
Proof: We divide the proof into three parts. Part I . k = k,(x) is a rational field. We show it has genus zero. Consider the parallelotope na where a = p;%(a) = - s. T o find m(a) we notice that a = $(x) €Ilaif and only if $(x) is a s. Hence m(a) s 1. It follows that polynomial of degree 1 l(a) 1 - g, or g l(a). But l(a) is zero if - s < 2 - 2g (by the Riemann part of the Riemann-Roch Theorem). Hence g = 0.
<
< <
+
+
Part 2. k = k,(x, y), where ay2
+ (b + cx) y + (d + ex +fx2) = 0.
We show that k has genus zero. If this equation is of the first degree, or is reducible or yields a constant field extension, then k is a rational field and g = 0 by Part 1. We may suppose none of these is the case and that a # 0, say a = 1. Consider again 17, where a = p;" In this case n(a) = - 2s since the degree of k over k,(x) is 2. We contend that
+
Part 3. k is a field of genus zero. We show it is one of the types described above. If b is a divisor of the canonical class,
Therefore m(b) = 1 - n(b) = 3. Hence the parallelotope & contains three linearly independent elements of k, say a, 8, y. If we set x = B/a, y = y/a, and a-l= a-la we see that contains, 1, x, y. Since 1 E , a is an integral divisor. Since x E I ~ , - ~ax, is an integral divisor, b say, so that x = b/a. Then since deg (k/k,(x)) = n (denominator of x), we have that deg (k/k,(x))
< n(a) = n(b-l) = 2.
(The inequality is written because there may be cancellation between a and b, leaving a single prime in the denominator of x.) If deg(k/k,(x)) = 1, then k is rational. If deg (k/k,(x)) = 2, then the denominator of x is a. Since y EIT,-~we may write y = c/a. It follows that y is integral at all finite primes dividing x. Hence if y can be expressed as a rational function of x, this function must be a polynomial. The only polynomials in x which can lie in are linear polynomials. But y is not a linear polynomial in x since it is linearly independent of x. It follows that k,(x, y) is a proper extension of k,(x). Hence k = k&x, y). In order to find the defining equation of this extension (which must be an irreducible quadratic) we consider the parallelotope ITa-= . n(a-2) = 4, and consequently the following elements of k
304
XVI.
4.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
lie in 1, x, y, x2, xy, and y2. But m(a-') = 5; hence there must be a relation of linear dependence between these six elements: ay2
+ (b + cx) y + (d + ex +fx2) = 0.
This completes the proof. In order to distinguish between the two types of field of genus zero, we prove the following result: Theorem 7: A function field of genus zero is rational if and only if there exists a divisor of odd degree.
Proof: If k = k,(x), the denominator of x is a prime of degree 1. In addition, the numerator of x and the numerator of x 1 are also primes of degree 1. Conversely, suppose k contains a divisor of odd degree. Since k contains a divisor of degree - 2 (any divisor of the canonical class), we can construct a divisor a such that n(a) = 1 and so m(a-l) = 2. Then 17,-1 contains two linearly independent functions a, p. If we set x = p/a we have 1 and x in 17,-1,-1. Hence aa is integral, and n(aa) = n(a) = 1. Thus aa is a prime p. But xp is integral, so x = b/p. It follows that if k, is the constant field of k, then
+
deg (k/kO(x))= n(p) = 1.
FIELDS OF GENUS ZERO
305
Suppose the function field k is rational, with constant field k, . Then there are three primes of degree one (cf. Theorem 7). Let them be pl , p2 , p, . If a is the denominator of y, n(a) = 2. Hence at least one of the pi, say p1 is not in the denominator of y. Hence ordvly 0. From the equation of the conic ordPl x 2 0 also. I t follows that in the homomorphic map of k into k, +a defined by pl (see section I), both x and y have images in k, . This the conic contains a rational point. Conversely, suppose (x, , yo) is a rational point on the conic. If we write x = x, f u, y = yo f v we obtain a new equation of the form au2 buv cv2 du ev = 0.
+
+ + +
This may be written in the form
Thus llv is a rational function of ulv (unless d = e = 0, in which case the equation yields a constant field extension). Hence v , and consequently u are rational functions of ulv. It follows that the function field k = k,(u/v), i.e. k is rational. Theorem 9: In a field of genus zero, every divisor of degree zero is a principal divisor.
Hence k = k,(x). Consider the algebraic curve C defined by an ideal a in the polynomial ring k,[x, y]. The algebraic points of C are the sets (xo,yo) where x,, yo lie in ko or some algebraic expression of k,, and are such that a is annihilated by (x, ,yo). A point (x, ,yo) on C with coordinates in k, is called a rationalpoint.
Proof: Let a be a divisor such that n(a) = 0. Then m(a) = 1, so & contains a single function a. Hence contains 1. It follows that aa-l is integral. But n(a-la) = 0, and so a-la = 1. Hence a = a. Finally we prove the well-known
Theorem 8: The function field of a conic is rational if and only if the conic contains a rational point.
Theorem 10: The only admissible generators of the field k,(x) bx)/(c dx) where a, b, c, d lie in k, . are elements of the form (a
Proof:
Let the conic be
If a = b = c = 0, then the conic is a straight line. The function field is rational, and the conic contains rational points. If one of a, b, c is non zero, we can make a change in the generators x, y if necessary, so that a # 0, say a = 1.
+
+
Proof: Let p1 , p2 , be divisors of degree 1. Then x = p,/p, say. If u = p3/p1 , then u is integral at all finite primes dividing x. Thus u is a polynomial in x. Since ordvl u = - 1, thus polynomial is linear: u = a bx. Similarly if v = p,/pl, then v = c f dx. Hence a*.,
+
306
XVI.
5.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
Since n(p,) = 1, w is a generator of ko(x), and conversely, each generator is of this form. 5. Elliptic Fields
Fields of genus 1 are called elliptic jields. For such fields the Riemann-Roch Theorem has the form
where b belongs to the canonical class, n(b) = 2g - 2 = 0, m(b-l) = g = 1. There are several possibilities for the values of m(a) : (I) If (2) If (3a) If (3b) If
n(a) > 0, then m(a) = 0. n(a) < 0, then m(a) = - n(a). n(a) = 0, and a is a principal divisor, then m(a) n(a) = 0, and a is not principal, then m(a) = 0.
=
1.
(I) has already been shown for all fields (Lemma 2 to the Riemann-Roch Theorem). T o prove (2) we notice that
Hence m(a) = - n(a). In case (3), we see that if n(a) = 0 and m(a) > 0, then the parallelotope 17, contains at least one function a. Hence 1 lies in 17+, . It follows that aa-l is integral. But n(acl) = 0. Hence aa-l = 1, a = a. Since m(b-l) = 1 and n(b-l) = 0, it follows that b-l, and hence b, is a principal divisor. Thus the canonical class is the unit class. A field of genus g # 1 has a divisor of degree 2g - 2. For fields of genus 1, the minimal positive degree of a divisor can be arbitrarily large, (S. Lang, J. Tate, Principal homogeneous spaces over abelian varieties, Amer. J. Math., July, 1958, remark at end of introduction). Case I : There exists a divisor a with degree 1. This is certainly true for function fields over algebraically
ELLIPTIC FIELDS
307
closed fields, because then all primes are of degree 1. We shall show later that it is true for function fields over finite fields. Since n(a) = 1, we have m(a-l) = 1 so 1 7 , - 1 contains a function a. Hence 17,-1,-1 contains 1; thus aa is integral and n(aa) = I. I t follows that we may assume our original a to be integral. Now m ( r 2 ) = 2 so 1 7 , - 2 contains two linearly independent functions I and x. Now xa2 is integral; hence if ko is the constant field of k, deg (k/ko(x)) = n denominator of x) = 2. We see that a2 = p, the infinite prime in ko(x). Since m ( r 3 ) = 3, Ra-3 contains three linearly independent functions 1, x, y. Now y is integral at all finite primes dividing x. Hence if y is a rational function of x, this function is a polynomial. But 17,-s contains no power of x higher than x itself, and y is not a linear polynomial in x since it is linearly independent of I and x. It follows that y does not lie in ko(x). Hence k = ko(x,y). T o find the equation which defines this extension, we consider the parallelotope 4 - 6 . This parallelotope contains the functions 1, x, x2, x3, xy, y, y2. But m(c6) = 6, so there must be a relation of linear dependence between these 7 functions. Thus
If the characteristic of k is unequal to 2, 3, we can reduce this equation to the familiar Weierstrass form
We consider the following situation now: Let k be the field ko(x,y) where y2 =f(x), deg f(x) = 3, the characteristic of k is unequal to 2, and f(x) is separable. Then k is the function field of the curve C F(x, y) =y2 -f(x) = 0. aF/ay = 231, hence aF/ay (x, ,yo) # 0 provided yo # 0, if yo = 0 then
since f(x) is assumed to be separable. It follows that aF/ax and aF/ay never vanish together on the curve. Hence by Theorem 5,
308
XVI.
5.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
each algebraic zero (x, ,yo) of F(x, y) gives rise to a unique place. The residue class field under the corresponding valuation p is ko(xo,yo), and f(p) = deg (ko(xo, yo)lko). In particular if (xO, yo) is a rational point on C, f(p) = 1 = n(p). Conversely let p be any prime of k of degree 1. Let 4 be the homomorphism defined by p. Then if
(xo ,yo) is a rational point of C. We saw in the preceding discussion that there exists an integral divisor a such that n(a) = 1. a has the property that a2 is the denominator of x in k. Hence a2 = p, , and a = p, . Hence n(p,) = 1. If 4, is the homomorphism defined by p, , we have &(x) = &(y) =co. We may call the pair (co,co) the point at injnity on C. We have the result: Lemma: There is a 1 - 1 correspondence between the primes of k of degree 1 and the rational points of C (including the point at infinity). Now let a be a divisor such that n(a) = 1. Then m(a-1) = I. Thus contains a single function a, unique up to constant factors. Hence aa is integral and n(aa) = n(a) = 1. That is to say, cia is a prime p. Thus every divisor class of degree 1 contains a prime. This prime is clearly unique, since m(a-l) = 1. Further, if a defines a divisor class of degree 1, then alp, defines a divisor class of degree 0.If a defines a divisor class of degree 0, then ap, defines a divisor class of degree 1. Hence we have: Theorem 11: There are 1 - 1 correspondences between the elements of the following sets: The rational points on C and the point at infinity. The primes of degree 1. The divisor classes of degree 1. The classes of degree 0. Corollary: The class number (= number of divisor classes of degree 1) is one more than the number of rational points on C. Clearly the point at infinity corresponds to the unit class.
ELLIPTIC FIELDS
309
We now examine the multiplication of divisor classes of degree zero. Let (x, ,y,), (x, ,y,) be rational points corresponding to the classes defined by divisors p,/p, ,p,/p, respectively. Let the inverse of the product of p,/p, and p,/p, be represented by p,/p, . Thus
must be a divisor of the unit class, i.e. must be a function a E k. Now ci lies in the parallelotope ITP;3. Hence ci has the form or = a
+ bx + cy.
Further a has zeros at p, ,p, and hence
Thus we may interpret a geometrically as the straight line joining (x, ,yl) and (x, ,y,). Unless (x, ,y,) and (x, ,y,) coincide, these two equations are sufficient to determine a up to a constant factor. When they do coincide, the argument must be modified slightly: a becomes the tangent to the curve C at (x, ,y,). Clearly p, corresponds to the point (x, ,y,) in which the straight line or meets the curve again.
Case 2: There exists a divisor a of degree 2. Since n(a) = 2, m(a-l) = 2, so 17,-1 contains two linearly independent functions a, p. Hence, writing x = p/ci we see that 11,-1,-1 contains 1 and x. I t follows that cia is integral. So, since n(aa) = 2, we may assume our original a is integral, and that contains 1, x. Since xa is integral, we see that
n,-,
deg (k/ko(x))= n(a) = 2,
where ko is the constant field. Now m ( r 2 ) = 4, so lIa-2 contains 1, x, x2 and a function y. By the argument of Case 1 (with the obvious modifications), we see that y does not lie in k,(x), and hence k = ko(x,y). To find the defining equation, we must consider a parallelotope containg y2 so we examine l7,-4 . This parallelotope contains the functions 1,
310
XVI.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
6. The Curve of Degree n
Let k be the function field of a curve of degree n, that is to say, let k = k,(x, y) where x and y are related by a polynomial equation F(x, y) = 0 of total degree n. We assume that F(x, y) is irreducible, and that it does not give an extension of the constant field. (If F does give a constant field extension we may make this extension first; then F is reducible over the new constant field.) We also require that the coefficient of yn in F(x, y) be non zero. If this is not already the case, we make a change of variables as follows: Let #(x, y) be the homogeneous part of F(x, y) of degree n. Make the transformation x = u av, y = v . Then # becomes +(u av, v). The coefficient of vn is #(a, 1). We choose a from k, (or from a separable extension, which does not change the genus) such that #(a, 1) # 0. The equation F = 0 is now in the form we require for the proof of
+
+
Theorem 12:
x2, x3, x4, y, xy, x2y,y2. But rn(c4) = 8, SO these 9 functions are connected by a relation of linear dependence:
X,
Case 3: There exists a divisor a of degree 3. We can show as in the previous cases that a may be assumed to be integral. Hence, since rn(cl) = 3, 17,-1 contains linearly independent functions 1, x, y. ax is integral. We may assume that a is the full denominator of x, for otherwise we recover either Case 1 or Case 2. Hence if k, is the constant field, deg (k/k,(x)) = 3. By the obvious adaptation of the method of Case 1, we see that y is not contained in k,(x), and so k = k,(x, y). T o find the defining relation we examine the parallelotope Do-, . This contains the functions 1, x, y, x2, xy, y2, x3, x2y, xy2, y3. But rn(c3) = 9 so there is a relation of the form
y3 +(a +bx)y2
+ (C+ dx +ex2)y + (f +gx +hx2+mx2) =0.
g(k) < *(n - 1) (n - 2).
Proof: Let p, be the infinite prime in k,(x). By examining the dominant terms in the equation F(x, y) = 0 we deduce that I y lvm I x l o rn for all prn dividing p, . Further I y 1 , 1 for all finite primes p. Now consider the parallelotope 17, where a = p;" . Since the effective degree is equal to the degree (there is no constant field extension) we have nk(a) = - ns. The parallelotope contains all functions of the form
5
<
where &(x) denotes a polynomial of degree i. I t follows that
We may choose s so large that m(a)
+ n(a) = 1 - g.
Hence 1 - g b 1 - g ( n - 1)(n-2) and so we have the required result g
< & (n - 1) (n - 2).
312
XVI.
APPLICATIONS OF THE RIEMANN-ROCH
7.
THEOREM
<
7. Hyperelliptic Fields
Let k be any function field, and let l7, be a parallelotope, with m(a) > 0. Lemma 1: If / 3 ~ kand / 3 a ~ l 7 ,for all a ~ 1 7 , , n k , then lies in the constant field k, .
/?
First Proof: Suppose a,, a,, a n form a KO-basis for the elements of k in 17,. Then /3af = 2 ci,a, (ci, E k,). It follows that det (C - PI) = 0 (C denotes the matrix [cij] and I the unit matrix). Hence /3 is algebraic over k, . But k, is algebraically closed in k, and therefore /3 E k, . a*.,
Second Proof: Let IIbbe the smallest parallelotope containing .--, a n . Then /3nbis also the smallest. It follows that /3b = b. Hence, considered as a divisor, /3 = 1, i.e. /3 is a constant. Suppose now that 17, C 17,.Then a ( b and b = at where t is an integral divisor. If a and b have the property that m(a) = m(b) # 0, i.e. if every field element which lies in 17,already lies in IIbwe may say that a can be shrunk to b. al,
Lemma 2:
If a can be shrunk to b, and b = ac, then m(c-l) = 1.
Proof: Since t is an integral divisor, 17,-1 contains the elements of k, . We must show it contains no other elements of k. Suppose /? E n k. Then /3c is integral. Let a be any element of II, A k. Then ab-l is integral. It follows that a/3b-lc is integral and so = 17, . By hypothesis, ajl E 176 , and hence, by Lemma 1, a/3 E 17bc-1 /3 E ko . This completes the proof. We apply the Riemann-Roch Theorem to this divisor c-1 obtaining
Thus
313
HYPERELLIPTIC FIELDS
<
It follows that 0 n(c) g. These results enable us to give simple examples of parallelotopes which can be shrunk, and other which cannot. First we show that if b is a divisor of the canonical class, then a = b-l cannot be shrunk. Suppose b = cb-l and m(a) = m(b). Thus
Hence n(c) = 0. Since t is integral, it follows that c = 1. Thus b-l cannot be shrunk.
On the other hand let p be a prime of degree 1. We shall show that b-lp-l can be shrunk to b-l. We have
since n(p)
> 0 implies m(p) = 0. Hence
Thus m(b-lp-l)
= g = m(b-l).
This shows that b-lp-l can be shrunk. It is easily seen that if a is an integral divisor of degree > 1, then b-lp-l cannot be shrunk. We shall now use these results to describe hyperelliptic fields: A function field of genus 2 2 is called hyperelliptic if it is a quadratic extension of a field of genus zero. Let k be a function field with genus g 2 2. If b is a canonical divisor, IIb-lcontains linearly independent elements x, , x, , xg contains ax,, ax2, axg. of k. If or is any element of k, norb-l Since the divisor class of b is an invariant of k, it follows that the subfield ..a,
.a-,
is an invariant subfield of k. Theorem 13:
If k # k', then k is hyperelliptic.
314
XVI.
7.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
Proof: Let deg (klk') = p 2. The effective degree m(k/k') = p also since k and k' have the same constant field. By choosing cu = xyl we may assume 1, x2, xg E 17D-land so
k'
= kO(x2, ---, x,).
Lemma: If k is any field of genusg 2 2, and k' is a subfield of genus zero, then k'x, ... + k'x, # k. k'x,
+
Proof:
Now b-l is the greatest common divisor of 1, x, , xg in k, i.e. ord, b-l = mini ord, xi; otherwise we could shrink b-l, which is impossible. It follows that b-l is also a divisor in k'. Now let ml(a), nl(a), g', b' have the obvious meanings for kt. We see that
Since
we see that V' = 17,'
Hence
n (b-l)
+
dim,, (V' :l7;
em.,
this last equality holding because 1, x, , ..., xg E k', and so We apply the Riemann-Roch Theorem to b-l in k' and get
315
HYPERELLIPTIC FIELDS
+ k') = Z'(1) =g' = 0,
+ k. Now
q x1 + -..+ 17; x, cn,-, .
&I.
b + ml(b-l) = m' (T) + 1 -g',
Hence xlV'
From this we obtain But if
xlk' Hence
But l/b' divides b/bl, hence
f
v.
+ x&' + .--+ xgkl= k,
some of the x, , say x, , x, , by Chapter 13, Theorem 2
-a-,
x, form a basis for klk', and hence
This contradiction completes the proof of the lemma. Theorem 14:
It follows that ml(b/b') = m'(l/bf). If g' = 0 this means that we can shrink the parallelotope II(bl,-i. But we have seen that this is impossible. Hence g' = 0, and m1(b/b') = g' = 0. It follows that (p - 2) (g - 1) = 0. Hence p = 2 and k is therefore a quadratic extension of k', i.e. k is a hyperelliptic field. For the converse of the theorem we need the following
+ x,V' + ... + x,V'
If k is hyperelliptic, then
is the only quadratic subfield of genus zero.*
* For the sake of brevity we say that k' is a quadratic subfield of k if k is a quadratic extension of k'.
316
XVI.
Proof:
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
Let k' be any quadratic subfield of genus zero. Then
+
d+l,y. By have m(a,) = g 3 and so 17,2contains 1, x, x2, familiar arguments we can show that y does not lie in k,(x) and hence k = k,(x, y). We have then ordPmy2 2g 2 for at least one p , dividing p, . T o find the defining equation of the extension k/kl, which is quadratic, we must find a parallelotope containing y2. Hence we choose i such that - g 1 - i = - 2g - 2, i.e. i = g 3. The parallelotope contains 1, x, x2, -.-,~ ~ gY+9 ~ , xy, ---, d+ly, y2. But m(ag+,) = 3g 5. Hence between these 3g 6 functions there is a relation of linear dependence, which we may write - a * ,
+
since deg (k/kf)= 2. If we adjoin to k' any element of k, not already in k', then we obtain all of k. It follows that xJx1 E k' for all i. Hence
+ +
+
+
By the preceding theorem, ko(l, x2/xl, subfield of k. Hence
.-a,
xg/xl) is a quadratic
<
where deg &(x) i. Now y determined only up to a polynomial of degree g 1 in x, so if the characteristic is not 2, we may assume y 2 = f(x) where 2g 2. Since y appears for the first time in the paraldeg f(x) lelotope n,;g-l,we must have deg f(x) > 2g. Hence the degree 1 or 2g 2. We notice also that f(x) must be of f(x) is either 2g square-free, otherwise we would obtain another y in an earlier parallelotope.
+
< +
Now let k be a hyperelliptic field. Let k' be its unique quadratic subfield of genus zero. If b is a divisor of the canonical class of k, b is also a divisor of k', and we have nf(b) = g - 1. Since k' is of genus zero every divisor of degree zero is principal. Hence all divisors of k' of the same degree belong to the same divisor class. It follows that b can be taken to be any divisor of degree g - 1 in k'. We notice that if k' is the function field of a conic without rational point, an hence has no divisors of odd degree, then the genus g of k must be odd. Case I: k' is a rational field, k' = k,(x). Here we may take b = pg-l, where p, is the infinite prime in k,(x). Then Ub-l contains the linearly independent elements 1, X, d-l. We now examine the parallelotopes ITfli where % =p-(g-l)-,. When i > 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,
+ +
According to Lemma 1 this can be accomplished if k , contains enough elements. Theorem 15:
If a and b are integral divisors, then
We may therefore assume at the outset that 17,-1 and Ub-1are the smallest parallelotopes containing &-I n k and nb-1 n k. If k , is an infinite field we have already enough constants for Lemma 2. If k, is a finite field, a sufficiently high separable extension will provide us with enough constants. Since m(a-l), m(b-l) do not change under a separable constant field extension, (see Theorems 21, 22, Chapter 15) we may assume that k , already contains enough elements. According to Lemma 2 we can find an element a ~ 1 7 ,n- ~ k such that ord, a = ord, a-l for a11 primes p dividing a and for all primes p dividing b. If cu. = al/a we see that a, is relatively prime to a and b. We also see that
and
Hence we may assume that a and b are relatively prime, i.e
320
XVI.
APPLICATIONS OF THE RIEMANN-ROCH THEOREM
Finally we have
+
Hence if a ED,-, f3 k, /3 ED,-, n k, and a /3 = 0, then a and fi are constants. The dimension of the set(&-I n k) (Ub-1 n k) is therefore m(a-l) m(bW1) - 1. But this set is contained in 17,-16-1 n k. Hence its dimension is at most m(a-lb-l). This gives the required result. If f is a divisor class, a any divisor in it, m(a-l) is the dimension of f. In the cases where m(a-l) = 0 or m(aib) = 0, the RiemannRoch Theorem gives a complete description of the dimensions of f and W/f . If m(a-l) # 0 then f contains integral divisors and we may assume that a is integral. If Wif contains an integral divisor 6 then W contains b = ab. Since m(b-l) = g we obtain the following complementary result to the Theorem of Riemann-Roch:
+-
+
Theorem 16: (Cliflord's Themem). If b is an integral divisor of the canonical class W, and b = ab where a and b are also integral,
then
CHAPTER SEVENTEEN
Differentials in Function Fields 1. Preparations
In this chapter we shall be dealing with function fields in the sense of Chapter 13, i.e. with fields of algebraic functions in one variable. In the present section we prove some preliminary results which will be used in the sequel. As usual, k shall denote a function field of our type; k, its field of constants. R = k,(x) shall denote a rational subfield. Theorem 1: There are infinitely many separable irreducible polynomials in k,[t].
Since by the Riemann-Roch Theorem
we obtain
First Proof: We may prove the theorem directly by considering two cases. If k, is an infinite field, all the linear polynomials t - a (a E k,,) are separable. If k, is finite, the existence of such polynomials is well known. Second Proof: It is possible to give a unified proof for both cases by adapting Euclid's proof that the number of primes is infinite. Suppose p, , p, , ..., p, are separable polynomials. Let
An equivalent statement to Theorem 16 is therefore Corollary:
If m(a-l)
> 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
<
1 for p # p, p, and hence Ag(Hp([)) since I Hp(0 1, q # p, p, . We have therefore
=0
for
Thus if we write
(the development of H,(S) at p,) where a is the coefficient of xmr-I in $(x) (r = degp(x)), we have
In Chapter 13 we defined for a rational subfield R the normed differential A([), describing A([) by its infinite component Let k1 be a finite extension of k, in which p(x) splits completely: The maximal parallelotope on which X vanishes is n,;. We now wish to examine the components of A([) at the finite primes p (i.e. irreducible polynomials p(x)). Thus we wish to find A,([) = A([,). Clearly A([,) = 0 if I t9 I 1. Any tP E R*(p) can be written in the form
<
Then R, = kl(x) has the same infinite prime as R in an obvious notation, we have
Further, where deg rj$(x)
< deg p(x).
We consider the principal part
For each or, we can write where
= k,(x),
and so,
XVII.
324
DIFFERENTIALS IN FUNCTION FIELDS
coefficients are polynomials in of a,. Thus if H,([) =
a,,
with coefficients independent
Similarly
~ ( a+n =
A!? +-'- + ..., x - a,,
all the A?! are conjugates over k, discussion &")(~,(0)
.
=JR
wRx.
The local component is written
According to the preceding and similarly for CL,(~). We have the relation
= A?;
':.
Hence Let k be a finite extension of R = k,(x). We define the differential dkx of k by giving its value When p(x) is a separable polynomial and a, one of its roots, we see that for each valuation vector X: Hence if p(x) is separable we have
3. Differentials and Derivatives in Function Fields
We now change our notation, and denote the normed differential in R, hitherto called A, by d,x. The differential p such that p ( [ ) = A(&) is then denoted by adRx. The value of the differential X for the valuation vector 5 is now written
where p is a prime in k dividingp in R and S, denotes Sk*(p)lR*(p) . Clearly dkx = 0 if k is inseparable over R. Suppose now k is separably generated, and x is selected so that k/k,(x) is separable. According to Theorem 8 of Chapter 13, two differentials are equal if they are equal at one local component. We make use of this fact and study differentials locally at a conveniently selected prime p(x). We choose an irreducible polynomial p(x) in R which is separable and unramified in k. This choice is always possible since the number of separable primes is infinite and the number of ramified primes is finite. In the notation of section 1 of this chapter, if p is a prime dividing p(x), we have k*(p)
= k,(x
- x&,
.
326
XVII.
DIFFERENTIALS IN FUNCTION FIELDS
where k2/kl is separable, and hence k,/k, is separable. We may write
3.
DIFFERENTAILS AND DERIVATIVES IN FUNCTION FIELDS
327
Hence p induces on R' = ko(y) the valuation defined by q(y). We can define a normed differential dRty in k,(y) as we did before for k,(x), and then we define dky by the relation
Then The differential dky is non trivial since AIR' is separable and dRly is non trivial. For our chosen prime p we can argue as before and obtain
Hence we have
f
Xpdk~
= Skalko (Rest
P
(x, $))
We compare this result with the previous one:
jPXPdkx= SkzIko (Res4X Now (x - x,) is a local uniformizing parameter for k*(p). In Chapter 10, section 3 we saw that if t is any other local uniformizing parameter then dx d(x - x ) Res, (X,) = Rest (xP dt ' = Reat (XP
)
=)
P
"3).
-
In Chapter 10 we defined the derivative dxldy and obtained the result
Hence we have the local result
The derivative dxldt was defined for power series fields in Chapter 10. Hence if t is any local uniformizing parameter in k*(p),
Now let k,(y) be another rational subfield of k, with the property that k is separable over k,(y). Let p be chosen in k to satisfy the additional conditions: does not divide the denominator of y, p is unramified over k,(y). p
Let k, be the residue class field of k at p, and let y = yo (mod p), . Then since k, is separable over k, ,yo satisfies a separable irreducible polynomial q(t) over k, . Then
yo E k,
Now let k be a finite separable extension of k,(y), x any element of k. We wish to define the derivative dxldy in k. The global field k is contained in each of the specially selected local fields k*(p), and according to Chapter 10 we can define dxldy in these local fields. We have to show (I) that dxldy as defined in k*(p) has a value in k, and (2) that this value is independent of p. Let F ( X , Y) = 0 be the relation satisfied by x and y in k. This relation is also satisfied in every k*(p). In every k*(p) we have
Hence
328
XVII.
DIFFERENTIALS IN FUNCTION FIELDS
which is an element of k, and is clearly independent of p. dxldy is well defined since F,(x, y) # 0, because x is separable over k,(y). Thus we have defined dxldy globally provided x is separable over ~O(Y). All the usual properties for derivatives then hold in k because they are already-satisfied in the local fields. Now since dxldy is an element of k, dxldy d,y is a differential of k. We have already seen that the local components of dxldy dky and dkx are equal at the specially selected prime p. Hence by Theorem 8 of Chapter 13,
4.
DIFFERENTIALS OF THE FIRST KIND
329
4. Differentials of the First Kind
Let k be a separably generated function field. Let x be a separating element of k, y any element. We shall study the differential ydx. First we require a Lemma: If k is separably generated and p is any prime, there
exists a local uniformizing parameter t in k such that t is a separating element. Proof: Let x be a separating element. If we replace x (if necessary) by llx, x 1, or llx + 1, we can ensure that p is neither a pole nor zero of x. Let T be any local uniformizing parameter in k. Suppose T is not a separating element. Then dr/dx = 0. Set t = TX. Then t is also a local uniformizing parameter. We have
+
We drop the subscript and denote the differentials simply by . Then Xdx denotes the value of the differentials dx dx, dy, at the valuation vector X. The formula (*) now yields
Sk
Theorem 2:
If k is separably generated and y is a separating
element, then
Evidently this result justifies both the name 'differential' and the notation we have introduced. Corollary:
In the situation of Theorem 2, we have for all
primes p
The formula (*) is also true when dx = 0, i.e. when k/k,(x) is not separable. We consider
Then F,(x, y) # 0 since x is separable over k,(y) by hypothesis. Suppose also that FJx, y) f 0. Then y is separable over k,(x) and hence k is separable over k,(x) contrary to our assumption. Hence dxldy = 0, and (*) is satisfied.
Hence t is a separating element. A prime p of k is said to be regular if the residue class field of k at p is separable over k, . Clearly if k, is perfect every prime of k is regular. Let p be a regular prime. Let t be a separating local uniformizing parameter. Let R = k,(t). Then p induces on R the valuation defined by the irreducible polynomial p(t) = t. Further p is unramified over k,(t) since t does not split and the residue class field i4p = k, is separable over k, . Hence, according to our previous discussion, we have
In Chapter 10 we defined
and we saw that this was independent of the choice of local uniformizing parameter t. We have therefore for regular primes p
330
XVII.
4.
DIFFERENTIALS IN FUNCTION FIELDS
If p is a prime of degree 1 (in particular, if k, is algebraically closed),
Let Y
dx
=
2 aJV,
where a, # 0. Then we see that
>
Hence, since dx dyldx = dy, we have
Let b be a divisor of the canonical class. Let A([) be a fixed differential which vanishes on Ub-l (maximal). Any other differential may be written in the form A(ol[) with a E k. The maximal parallelotope on which A(aQ vanishes is 17,-lb-1. A differential is said to be of the Frst kind if it vanishes on the unit paralleotope Ill in k. Hence A(a[) is a differential of the first kind if and only if Ul C 17,-lb-1 , which is equivalent to or E Ub-l. Since m (b-l) =g, we have
v-m
Hence if ord, 5, - m we shall have other hand, if ord, [,= - m - 1, say
DIFFERENTIALS OF THE FIRST KIND
$, 5,ydx
5P
= 0.
On the
= crrn-l, we have
which is not always zero. Hence if 17,is the maximal parallelotope on which ydx vanishes, we see that ord, a = - m. If m 2 0 we say that ydx is regular at p. If m = - r < 0 we say that ydx has a pole of order r at p. If the maximal parallelotope on which ydx vanishes is 17, where a = l7pVp,then we shall sometimes say that ydx = a-' = np-'p.
Let dx be defined as in section 3. Then the maximal parallelotope in k,(x) on which dx vanishes isup;. By Theorem 10 of Chapter 13, the maximum parallelotope in k on which dx vanishes is I I P 2 g where D, is the different. Writingp, = u, , to denote the divisor in k which is the denominator of x, we see that we have the factorization If
with a, # 0, at a regular prime p, then pm is the p-contribution to %,/ux2. We have a similar factorization dy = DY/uY2.
Theorem 3: There are precisely g linearly independent differentials of the first kind in a field of genus g. As an example we shall find the differentials of first kind in the field k = k,(x, y) where y2 = f(x), f(x) being assumed squarefree. We assume also that k has genus g >, 1 and that the characteristic of k is not 2. Let R = k,(x). We must find the different D, of k/R. Let p(x) be an irreducible polynomial in R. Clearly p(x) is unramified and makes no contribution to 2), if p(x) does not divide f(x). On the other hand, if p(x) divides f(x) then p(x) is ramified and the ramification number is 2. Hence the ramification is tame. If p in k dividesp(x) thenp(x) = p2, and the p-contribution to D, is pe-l = p . Thus the contribution made to D, by the finite primes is exactly equal to that of y = T o find the contribution made by the infinite prime, we replace x by I/x and obtain y2 = f(l/u); multiplying by ~ ~ we g obtain + ~ ( y ~ g + l= ) ~f(l/u) ~ ~ g pm + ~ will . make a contribution if and only if u is a divisor of the polynomial
m).
It follows that the infinite prime makes the same contribution as p:+ly. Hence we have D, = yp:+l. We have now
332
XVII.
4.
DIFFERENTIALS IN FUNCTION FIELDS
< <
Now let x = p,/p, . Then for 0 v g - 1, xVdx/y= ~,,"pg;l-~ is a differential of first kind. But since the xvdx/y are g in number, it follows that a basis for the differentials of first kind is given by
dx xdx ... xg-ldx dfrn' df(X)' ' d f i ) '
Hence
Since y
Let a be an integral divisor, a # 1. We wish to construct, if possible, differentials ydx having no poles of higher order than indicated by a, i.e. such that aydx is integral. Clearly if X(ci6) is a differential of this type, A(&) must vanish on IT,, i.e. 17, C17,-ib-1. Hence X(ci4) is a differential of this By the Riemann-Roch Theorem type if and only if ci ~17,-l~-i.
DIFFERENTIALS OF THE FIRST KIND
E k,
we have
Thus
2 R a p (ydx) = 0. P
From this result we see again that a differential cannot have a single pole of order 1. We also see that if a differential ydx has poles of order 1 at two distinct primes p1 and p, and no other poles, then since n(a)
> 0 implies m(a) = 0. Hence
Rap, (ydx) = - Resp, (ydx). Altering ydx if necessary by a constant factor, we can ensure that
We consider the special cases:
Case I : n(a) = 1, a = p . In this case there are g linearly independent differentials with the required property. But the differentials of first kind have this property (they have no poles at all). Hence only the differentials of first kind have the required property. Thus there are no differentials which have exactly one pole of first order.
+
1 linearly Case 2: n(a) = 2, a = p,p, . Here there are g independent differentials with the required property. Among these are theg differentials of first kind. It follows that there is a differential which has poles of order 1 at p, and p , . Suppose now that k, is algebraically closed, and hence perfect. Then k is separably generated. Every prime is of first degree, and if t is a separating local uniformizing parameter for p , we have
Now let p1 , p, , p , be given distinct primes and c, , c, , ..., c, given elements of k, such that C c, = 0. We shall show how to construct a differential which has simple poles at the given primes with residue ci at pi. T o this end we construct differentials yidx (i = 2, r ) such that em.,
..a,
Res,, ( y d x ) =
+1
and
Res,, (y,dx) = - 1 .
Then clearly
ydx = c0,dx
+ -.-+ c,y,dx
is a differential with the required property. This differential is not unique, but obviously two differentials with this property differ only by a differential of the first kind.
2.
APPENDIX
Theorems on p-Groups and Sylow Groups
If C is a subset of a group G , then the set of all x E G satisfying xC = Cx (or xCx-l = C ) is called the normalizer Nc of C in G. One sees immediately that Nc is a subgroup of G. Should C be a subgroup of G then C is contained in Nc and is a normal subgroup of Nc . N, is then the largest subgroup of G having C as normal subgroup. Definition 2: Let S be a subgroup of G , Cl and C, subsets of G . We say that Ci is S-equivalent to C, if C, = xClx-l for some x E S. It is easy to see that this is an equivalence relation. Let C be a given subset of G. In order to find the number of sets S-equivalent to C we ask when xCx-l = yCy-l for two elements x, y E S. This equivalent to y-lxC = Cy-lx, hence to y-lx E Nc . Since y-lx E S this means y-lx E S n Nc or x ~y - ( S n N,). Hence x and y have to be in the same left coset of S modulo S n Nc . This shows: Lemma 1: The number of sets in the S-equivalence class of C
is the index:
(s/sn NC) Should S = G this simplifies to (GIN,). In this case we call the G-equivalent sets also conjugate sets. 334
335
If C consists of one element only, then the conjugates of C have also only one element. This gives a distribution of the elements of G into classes of conjugate elements. The number of elements in the class of a is (GIN,). This number is 1 if N, = G, or if ax = xa for all x E G . The set Z of all these a forms an abelian normal subgroup of G , the center of G. Let G be a finite group of order n, Z the center of G, and x the order of Z. G is covered by its classes of conjugate elements. An element a outside Z lies in a class with (GIN,) elements and (GIN,) will be greater than 1. Counting the number of elements in G we obtain a formula of the type
1. S-Equivalence Classes Definition 1:
THEOREMS ABOUT P-GROUPS
n =z
+ 2 (GIN,)
where each term in the sum is greater than 1 2. Theorems About p-Groups
A group whose order n is a power pr of a prime is called a p-group. I n formula (2) for n = p7 each index of the sum is a power p8 > 1. Therefore z must be divisible by p. This shows: Theorem 1: A p-group G # 1 has a center Z of an order
> 1.
Corollary 1 : A p-group G # 1 contains an invariant subgroup
of order p. Corollary 2:
One can find a chain 1 = G,CGICG,C
CG, = G
of invariant subgroups Gi of G such that each index (G*+l/G*)=P.
Proof: Corollary 1 shows the existence of an invariant subgroup GI of order p. By induction one may assume the existence of a chain in GIG,:
4. The chain 1 =GoCGlCG2C..-CG,
has the desired properties. We prove: Theorem 2:
Let G be a p-group, S # G a subgroup. Then
N s # S.
Proof: Let Z be the center of G. We prove the theorem by induction. Case 1: If Z that N, # S.
i#
S, any element a of Z that is not in S shows
Case 2: Let Z C S . Consider the natural homomorphism of G onto G/Z. S is mapped onto S/Z. Since Z C S C N s , Ns is mapped onto N,/Z and Ns/Z is the largest subgroup of G/Z that has S / Z as normal subgroup. We have to show Ns/Z # S/Z. But this is our theorem in the group G/Z of smaller order. Corollary 1: If S is of index p in G, then S is normal in G. Indeed Ns can only be G itself. Corollary 2: If S C G, S # G, then there exists a group Sl between S and G such that S is invariant subgroup of S, of indexp.
Proof: S C Ns and S # Ns . Let Sl/Sbe a subgroup of Ns/S of order p. S1has the desired properties. Corollary 3:
There always exists a chain
Let G be a finite group of order n, p a prime and pr the exact power of p dividing n. A subgroup P of G of order pr is called a
337
Sylow subgroup of G. We prove that it exists always. If r = 0 the statement is trivial so we are concerned only with the case r > 0. Lemma 2: Let G be abelian and p I n. Then G contains an element of order p.
Proof: By induction. Select an element a # 1 of G and let d be the period of a. If p I d then adlp is the required element. Let p 1. d and call S the cyclic subgroup generated by a. The factor group G/S is of order nld < n and p I nld. So there is an element bS in G / S of exact period p. Let e be the period of b: b2 = 1. Then (bS)e = S. Therefore p I e and help is the required element. Theorem 3:
Every group G has a Sylow subgroup.
Proof: We proceed by induction on the order n. Case 1: G has a subgroup S # G of an index prime to p. The order of S is then divisible by exactly pr and is less than n. Therefore a Sylow subgroup P of S exists and is Sylow subgroup of G. Case 2 (actually only possible for p-groups). Every subgroup S # G has an index divisible by p. We return to formula (2) and see that n as well as each term in the sum is divisible by p. Hence p I z the order of the center Z of G. Let S be the cyclic group of order p generated by an element a of order p of Z (use Lemma-2). S is normal subgroup of G. Let PIS be a Sylow subgroup (of order pr-1) of G/S. Then P is of order pr whence a Sylow subgroup of G.
such that G,-I is normal in G, of index p. The proof follows from the preceding corollary. 3. The Existence of Sylow Subgroups
THEOREMS ABOUT SYLOW SUBGROUPS
4. Theorems About Sylow Subgroups
Let G have order n, divisible by exactly pr, P a Sylow subgroup of G. Let S be any p-subgroup of G (not necessarily itself a Sylow subgroup of G). Lemma 3:
SnNp=SnP.
Proof: That S n P C S n N, is trivial. We show the converse. S n N, C S, hence S n N, = S, is a p-group. Since S, C N p we have x P = P x for each x E S1. Therefore SIPis a group and P is normal in it. We have for the index (S,P/P) = (Sl/Sl n P ) = power of p.
Index of Symbols
So SIP is itself a p-group containing P. Since P is a maximal p-subgroup of G we get SIP = P whence S, C P. S1 C S shows S
deg 6 , 60 dim f, 301 dx, 328 d K x , 325, 327 d R x , 324
n Np = Sl C ( S n P).
Consider all transforms of P: xPx-l, x E G. Their number is (GIN,). Since P C N, , this index is prime top. Distribute these transforms Pi into S-equivalence classes. The number of transforms in the S-equivalence class of Pi is given by the index (SIS n N p i ) so in view of Lemma 3 by (SIS n Pi). We obtain a formula of the type
Each term on the right side is a power of p. Since the left side is prime to p it must happen for some Pi that ( S / S n Pi) = I or SnP,=SorSCP,. Hence S is contained in some transform of P. Let for a moment S be itself a Sylow subgroup. Then S = Pi . So the transforms of P are all Sylow subgroup of G. Let S = P. Then
A, 135 A H , 116, 119 [A],92 ( A :B), 261 ( A ,HI, 115
e . 53 e p , 233 e v , 271
E, 135 [E : k], 127 (as used in Parts (11, III)*
( E : R )= f, 54 ( E : K), , 133
( S l s n Pi) = (PIP n Pi) = 1
happens exactly one, namely for Pi = P. All other terms in (3) are divisible by p. (3) shows (GIN,) E 1 (modp). All Sylow subgroups P of a group G are conjugate. Their number, the index (GIN,) is r 1 (modp). Everyp-subgroup S of G is contained in a Sylow subgroup of G. Theorem 4:
deg a, 13 deg a, 60 *Note: [E: K ] , as used on pages 36 and 60 denotes the degree of a field extension.
3 39
INDEX OF SYMBOLS
INDEX OF SYMBOLS
Pm , 13,231 P,238, 246 p, 54, 86, 136, 225, 271, 293 8, 54 pm , 231 PIP, 231 m(a), 262 m ( K / k), 272 !Dl, 225 n~ , 271 np , 233 n(a), 261 nda), 278 n(K I k), 274
k (a field) k", 241 kl , 322 kz , 322, 326 k', 313 k, 54 A, 17, 171 k*, 3 ( k ) i , 10 K D , 10
= deg ( K 1 k) = T h e degree o f a field extension* K, 282 K M , 181 KO, 271
(K: k)
N , 215, 216, 246 N , , 335 N , , 334 Na , 226 N ( 4 , 22 N(c), 113 N A , 130, 145 N K I ~145 , N K / L ,151, 174 Nil-, , 174
r(a), 265 R, 230, 246, 321 R, , 5 8 Resko(c),148 Res, x, 198 Rest y , 201 Res(ydz), 202
*Note: In Part I this is also referred to as ( K : k).
1
Xdx, 328
s, 226
S
=
S E I K ,86
SF^ , 257
S(EIK), 127 S'(E/K), 133 S ( X ) , 258
T,66 T'. 86 T,,153 T,,155 Tm , 206 5,72
ord a, 13 ord or, 37, 82, 172 ord, a, 13, 233 ord, a, 13 f (4,14 ordm f ( x ) , 14 OM 1, 98 ord 7 , 160, 178 ordp a, 226-7 ordp a, 233
f uxdy, 194
u,, 108 U E , 170 u , , 170 u , , 169
x dy, 194 X v , 255
co, co), 308 a1 < l b l , 2 9 7 a 1, 297 a 1, 37, 43, 223 a 1, 240,261 x I, 3, 11 @ ( x )1, 28 f ( x ) 1, 12, 13
INDEX OF SYMBOLS
Subject Index
Abelian group(s) - character of a finite, 71 - dual of an, 71 - pairing operations on, 71 Absolute value - of a divisor, 261 - of an idele, 240 - 'ordinary', 12 Algebraic - closure of a complete field, 43 - constant algebraic field extensions, 279 - curve, 297 - of degree n, 31 1 - rational point of a, 304 - point, 296 - variety, 296 Approximation Theorem, 10 Archimedean valuation, definition, 6 Axioms for PF-fields, 225-6 Basis, complementary, definition, 89 Brauer factor set relations, 124 Canonical - divisor class, 264 - generator (of a Galois Group), 128, 153 - homomorphism (from the second cohomology group), 115 Cauchy - criterion, 47 - filter - convergent, 239 - definition, 239 - sequence (with respect to a valuation), 17
Center of a group, definition, 335 C-family (of field extensions), definition, 282 Character (of a finite abelian group), definition, 71 Chinese Remainder Theorem, 167 Class - number, 308 - of differentials, 264 Clifford's Theorem, 320 Coboundary operator, definition, 114 Cochain - continuous, 117 - definitions, 114 Cocycle - continuous, 117 - definition, 114 - 'split', 116 Cohomology group, definition, 114 - second, 116 Complementary - basis, 89 - set, 86 Complete - fields, 17, 19 - linear series, 301 - ring of valuation vectors, 239 Completion - at p, 18 - at the infinite prime, 18 - of a field, 17 Composite extensions, definition. 218 - equivalent, 219 Conductor of an extension, definition, 143 - for cyclic p-Extensions, 203
343
344
SUBJECT INDEX
Conjugate sets (in a group), definition 334 Conservative field, definition, 291 Constant field extensions, 271 - c-family of, 282 - definition, 276 - finite algebraic, 279 Constants, field of, definition, 226 Continuous - cochain, 117 - cocycle, 117 Convergence - of a Cauchy filter, 239 of a power series, 48 - of a series. 47 Curve - see Algebraic curve
-
Dedekind, 293 Defect of an extension, definition, 63 Degree - effective, 272 - local, 224 - of a divisor (class), 261 of an algebraic curve, 311 - of an element of an extension field, 60 - of a residue class, 60 - residue class degree, 54 Denominator of a divisor (ideal), definition, 241 Derivative of a power series, 198, 200 - in function fields, 324 Different, definition, 87 - for cyclic p-Extensions, 203 - inverse, 86 - inverse 1-13, 256 - 1-9,257 - 1,257 - S-different, 259 Differential(s) - class of, 264 - definition, 194, 202 - equality of, 194 - in a power series field, 200 - in function fields, 321, 324 - in PF-fields, 245, 251 - local components of, 322
-
- of a field, 246, 325
- of the first kind, 331 - residue of a, 202 Dimension - of a divisor (class), 262 - of a Kronecker product, 218 - of a linear series, 301 Discrete valuations, definition, 56 - ramification theory for, 82 'Divides' - a divides a, 260 - a divides b, 260 - p divides p, 23 1 Divisor(s) - absolute value of a, 261 - canonical divisor class, 264 - class, 240 - definition, 240 - degree of a, 261 - denominator of a, 241 - dimension of a, 262 - in an extension field, 278 - integral, 241 - numerator of a, 241 Dual - group, 71 - space, 245 Dyadic square, definition, 35 e, definition, 53 Effective degree, definition, 272 Eisenstein - criterion, 93 - extension, 93 - polynomial, 93 Element - %(a, E), 95 - of EIF, 84 Elliptic field, definition, 306 Equivalent - composite field extensions, 219 - elements of k ,297 - extensions of a group, 112 - subsets of a group (s-equivalence), 334 valuations, 3 Euclid, 321
-
SUBJECT INDEX
Euler, 90 Existence theorem, 181 Extension - see Field or Group Extension Extension of a valuation, 21 - archimedean case, 24 - of a non-archimedian field, 28, 37, 43 - of a non-complete field, 223 - ramification of an extended valuation, 53 - t o k(i), 23 - to the completion of a field, 17
f, definition, 54 Factor set, definition, 110 - relations (Brauer), 124 Field - complete, 17 - completion of a, 17 - conservative, 291 - differential of a, 246, 325 - elliptic, 306 - function field - see Function field - genus of a, 263 - hyperelliptic, 312 - inertia, 66 - normed, 24 - number, 246 - of constants, 226 - of formal power series, 47 - of genus zero, 302 - of p-adic numbers, 18 - place of a, 293 - product formula - see Product formula fields - ramification, 70 - rational p-adic number, 206 - rational subfield, 230 - residue class field, 54 - splitting field (of a cocycle), 119 - valuation of a, 3 - valuation ring in a, 293 - value group of a, 53 - with perfect residue class field, 190 - withseparable residueclass field, 93
Field extension(s) - c-family of, 282 - composite, 218 - conductor of a, 143 - constant, 271,276 - defect of a, 63 - degree of an element of a, 60 - divisors in a, 278 - Eisenstein, 93 - finite algebraic constant, 279 - Galois Group of a, 103 - genus in a, 284, 287 - infinite separable, 291 - normal, 103 - of PF-fields, 235 - purely ramified, 83 - tamely ramified, 67 - unramified, 65, 127 - valuation vectors in a, 241 - with degree equal to the characteristic, 180 Filter - Cauchy, 239 - definition, 239 - convergent, 239 Finite intersection property, definition, 106 First countability axiom, 108 f-neighborhood, definition, 166-7 First inequality, 132 Fractional ideal, definition, 86 (see ideal) Function Field, definition, 246 - derivatives in a, 324 - differentials in a, 321, 324 - hyperelliptic, 313 - of a conic, 302 - of a variety, 296 - paralletopes in a, 262 Fundamental theorem of Galois theory (Infinite extensions), 105 Galois - cohomology theory, 114 - group, definition for infinite extensions, 103 - theory for infinite extensions, 103 - t o ~ o l-.o mon a. 104
.
346
SUBJECT INDEX
Genus of a field, definition, 263 - in field extensions, 284, 287 Group(s) - abelian - see abelian group - center of a, 335 - character of a, 71 - cohomology, 114 - dual, 71 - extensions, definition, 108 - equivalent, 112 - Galois, 103 - inertia, 72 - norma, definition, 127 - for infinite extensions, 174 - normalizer, 334 - pairing operations on, 71 - p-group, 335 - ramifications, definition, 72 - higher ramification groups, 77, 82, 85 - splitting, 116, 119 - Sylow subgroup, 337 - value, 53 Hensel's lemma, 29 Herbrand, 83 Higher ramification groups, 77 Hilbert, 84 Homomorphism, canonical, of b,(A, ) b&4, H), 112 Hyperelliptic function field, definition, 313 -+
Ideal, definition, 240 - denominator of an, 241 - fractional, 86 - ideal class, 240 - integral, 241 - numerator of an, 241 - period, 170 Idkle - absolute value, 240 - definition, 239 - Index of (a : an), 209 Inequality - First, 132 - Second, 133
Inertia - field, 66 - group, 72 Infinite prime - completion at the, 18 - definition, 13 Infinite product space, definition, I65 Integral - properties of integrals, 196 divisor (ideal), 194 Inverse - different, 86 - I-SP different, 256
-
Kernel (G and H), definition, 71 Krasner, 44 Kronecker product - dimension, 21 8 - of rings, 218 - of vector spaces, 216 Lang, S. 306
I-q different, definition, 257
- inverse, 256 I-different, definition, 259 Limit - of a cauchy filter, 239 -of a sequence with respect to a valuation, 8 Linear series, 300 - complete, 301 - definition, 301 - dimension, 301 Linear space(s) - see Vector space - dual, 245 - Kronecker product of, 216 - normed, 19 Local - class field theory (aim of), 127 - components of differentials, 322 - degree, 224 'near zero', 166 Newton('s) - diagram, 37 - polygon, 38
SUBJECT INDEX
Nilpotent element of a ring, definition, 215 Noether's equations, 118, 119 Non-archimedianvaluation,definition, 6 - triangular inequality, 7 Non-critical part (of a value group), definition, 76 Norm - for reasonable non-archimedian primes, 226 - group, definition, 127 - for infinite extensions, 174 - on a linear space, 19 - on E, 174 - residue, 13G - residue symbol, 158, 159, 175 - for power series fields, 193 - uniqueness, 187 - topology induced by a, 19 Normal - extension of a field, 103 - forms (for the valuation in a field), 209 - valuation - at p, 14 - for reasonable primes, 226 Normalizer of a group, definition, 334 Normed - field, 24 - linear space, 19 Null sequence with respect to a valuation, definition, 17 Number field, definition, 246 Numerator of a divisor (ideal), definition, 201 order (of S),226 ordering of classes in k*, 297 ordinal (see 'ord') - of a at p, 13 - of ar, 172 - of 7 , 160 p-adic numbers, definition, 18 - rational p-adic field, 206 Pairing operation, definition, 71 Parallelotope, definition, 227 - in a function field, 260
347
- in the ring of valuation vectors, 239 - upper bound for the order of a, 227 Period, definition, 168 - ideal, 170 PF-field - see Product Formula field p-group, definition, 335 Pigeon-holing principle, 227, 229 p-integers, definition, 225 Place of a field, definition, 293 Point at infinity, definition, 308 Pole at p, definition, 330 Polygon, Newton's, definition, 38 Polynomials - Eisenstein, 93 - universal, 51 Power series - convergnece of, 48 - derivative of, 198, 200 - field of formal, 47 Prime, definition, 225 - reasonable, 226 - regular, 329 Principal part (of a valuation vector at P), 242 Product Formula field(s) - axioms for a, 225-6 - definition, 226 - description of all, 230 - differentiation in, 245, 251 - finite extensions of, 235 Prolongation, definition, 95 Purely ramified field extension, 83 Radical of a ring, definition, 215 Ramification, definition, 53 - field, 70 - group, 75 - higher ramification groups, 77, 82, 85 - of a subfield, 95 - theory, 64 Rational-p-adic number field, definition, 206 - point (on an algebraic curve), 304 - subfield, 230, 245 Reasonable prime, definition, 226
SUBJECT INDEX
Reducibility Criterion, 36 Regular - prime, 329 - ydx is, 330 Residue, definition, 198, 201 - of a differential, 202 Residue class - degree (f), 54 - field, 54 - at p, 225 - ring, 54 Restricted direct product topology (on V), 239 Riemann part of a Riemann-Roch theorem, 265 Riemann-Roch Theorem, 264 - applications, 293 - first proof, 262 - for elliptic fields, 306 - second proof, 265 Ring(s) - Kronecker product of, 218 - nilpotent element of a, 215 - of integers, 53 - of p-integers, 225 - of valuation vectors, 239 - complete, 239 - radical of a, 215 residue class, 54 - valuation, 293
-
S-different, definition, 259 Second-Inequality, 133 - cohomology group, 116 Sequence - Cauchy, with respect to a valuation, 17 - limit, with respect to a valuation, 8 - null, with respect to a valuation, 17 S-equivalent (subsets of a group), 324 Series - convergence of a, 47, 48 derivative of power series, 198,200 linear, 300 - power, 47 'Shrunk' (a can be shrunk to b), 312 splitting-field (of a cocycle), 119
-
- group (of a cocycle), 116, 119 Sylow subgroup, definition, 337 - theorems about, 337 Tamely ramified field extension, definition, 67 Tate, 292, 306 Tempory symbol (c, K/k/r),144 Topology - in a Galois group, 104 - induced by a norm, 19 - induced by a valuation, 5 - in the ring of valuation vectors, 239 - new topology in K*, 170 Trace, 86 Tragheitslrorper, definition, 66 Transfer, definition, 149 Transform of a subgroup, 338 Triangular Inequality, 7, 44, 297 Trivial valuation, definition, 3 Tychonoff's theorem, 166 Ultrametric spaces, definition, 44 Unit class, 309 Universal polynomial, definition, 51 Unramified field extension, d e f i ~ t i o n , 65. 127 Valuation(s) - archimedian, 6 - cauchy sequence with respect to a, 17 - classification of, 6 - definition, 3 - discrete, 56 - equivalent, 3 - extension of a - see Extension of a valuation - limit of a sequence with respect to a, 8 - non-archimedean, 6 - normal, 14,209,226 - null sequence with respect to a, 17 - ramification of an extended, 53 - ring, 293 - topology induced by a, 5
SUBJECT INDEX
- trivial, 3 - vectors, 238 - in an extension field, 241 Value group, definition, 53 - non-critical part, 76 Van der Waerden, 281 Variety, algebraic, defined by an ideal, 296
349
- function field of a, 296 Vector space - 244 - see Linear space (valuation) Vector, definition, 238 Verzweigungskorper, definition, 70 Vorlagerung, definition, 149 Witt, 192
