goldstein l.j. analytic number theory - index - free

Analytic Number Theory


Larry Joel Goldstein Mathematics Department

University of Maryland

P r e n t i c e - H a l l , I n c . , E n g l e w o o d C l i f f s , N e w J e r s e y

0 1971 by Prentice-Hall, Inc. Englewood Cliffs, New Jersey

All rights reserved. No part of this book may be reproduced in any manner or by any means without permission in writing from the publisher.

Current printing (last digit): 10 9 8 7 6 5 4 3 2 1

13-034843-0

Library of Congress Catalog Card Number 77-140688

Printed in the United States of America

To Sandy, without whose love and encouragement

this book could never have been written.

PRENTICE-HALL INTERNATIONAL, INC., London PRENTICE-HALL OF AUSTRALIA, PTY. LTD., Sydney PRENTICE-HALL OF CANADA, LTD., Toronto PRENTICE-HALL OF INDIA PRIVATE LIMITED, New Delhi PRENTICE-HALL OF JAPAN, INC., Tokyo

Preface

The purpose of this book is to provide an introduction to algebraic number theory and class field theory that is accessible to most graduate students. Although the current trend in class field would dictate a cohomological approach, we have decided to let the theory of L-series play the central role in this development. Thus, the theory of Hecke's L-series is developed a la Tate, and the Artin reciprocity law is a simple statement about Hecke L-series. By following these lines of development, it is possible to include, in a connected theory, many beautiful and fruitful ideas which helped give class field theory its start, and which promise to help class field theory develop even more fully in the near future, as is indicated, for example, by the recent work of Langlands and Jacquet.

It is assumed that the reader has a background in algebra roughly equivalent to a one-year van der Waerden-level course. Moreover, it is assumed that the reader knows the basic facts about topological groups and Haar integration on a locally compact group. The latter chapters of the book also require a basic knowledge of complex analysis, say on the level of Ahlfor's Complex Analysis. Thus, in short, we assume that the reader has at his disposal the material covered in the standard first-year graduate courses taught in American universities.

This book grew out of a course that the author gave at Yale University in the spring of 1968. Over the last two years, the author has received asssitance and advice from many people who deserve his hearty thanks. The author wishes to express his sincere appreciation to Drs. Howard Garland, Murray Schacher, Paul Ponomarev, and Lawrence Corwin and to Mrs. Judy Sunley for pointing out errors in his lectures and in preliminary drafts of the manuscript; to Mrs. Madeline Delaney for typing the

vii

viii / PREFACE

class notes from which the book evolved; to Miss Dorothy Crouch and Miss Nancy Antizzo for their kind editorial assistance; to Mr. Robert Martin and Mr. Arthur Wester, both of Prentice-Hall, Inc., for their encouragement and kind cooperation during the preparation of the book.

LARRY JOEL GOLDSTEIN College Park, Maryland

Contents

1. Historical introduction, 1

PART 1

Algebraic Number Theory

2. The ring of integers of a number field, 9

2-1 Integers, 9 2-2 Ideal theory, 13

3. The locally compact groups of number theory,

3-1 Completions, 20 3-2 Adeles, 32 3-3 Ideles, 38

4. Local fields, 46

4-1 The topology of local fields, 46 4-2 Completions of number fields are local fields, 4-3 The arithmetic of local fields, 55 4-4 Local ramification theory, 62 4-5 Local Hilbert theory, 67

5. Global arithmetic, 72

5-1 Splitting of primes, 72 5-2 The valuations of an algebraic number field, 79 5-3 Global ramification theory, 80 5-4 Galois actions, 82 5-5 Global Hilbert theory, 84 5-6 An example-quadratic fields, 88

6. Applications, 93

6-1 Special functions, 93 6-2 Cyclotomic extensions, 96

PART 2

Harmonic Analysis

7. Hzrmonic analysis on locally compact abelian groups, 11 1

7-1 Duality theory, 11 1 7-2 Fourier transforms, 121 7-3 The Schwartz-Bruhat space, 127

PART 3

Hecke's L-Functions

8. Hecke's L-series, 133

8-1 Local zeta functions, 133 8-2 Global zeta functions, 145 8-3 Hecke L-series, 156

9. Applications of Hecke L-f unctions,

9-1 Splitting of primes, 162 9-2 Abelian L-functions, 164 9-3 Tchebotarev's density theorem, 9-4 Dirichlet's L-functions, 171 9-5 Class number formulas, 172

PART 4 I I Class Field Theory

10. Class field theory, 181

10-1 Abelian L-functions are Hecke L-functions, 182

11. The inequalities of class field theory, 188

11-1 The second inequality, 188 11-2 The first inequality, 197

12. Class field theory, 206

12-1 Proof of the reciprocity law, 206 12-2 Local class field theory, 218 12-3 Global class field theory, 226

13. Applications, 238

13-1 Kronecker's theorem, 238 13-2 The power reciprocity law, 241

PART 5

The Prime Number Theorem

14. The prime number theorem, 251

14-1 Natural densities, 25 1 14-2 Analytic preliminaries, 257 14-3 The prime number theorem, 260

Suggestions for further reading, 273

Appendix : algebraic preliminaries, 275

Bibliography, 277

Index, 279

Glossary of Notation

Symbol

B [R [R+ c Z 19 [C: HI or

(G: H ) N x )

Dejinition Page Reference

field of rational numbers field of real numbers positive real numbers field of complex numbers ring of rational integers order of the group G index of the subgroup H in the

group C number of rational primes 5 x

Riemann zeta function gamma function real part of the complex number s number of rational primes a(mod d)

and x ring of integers of the number

field K discriminant of the basis a l , . . . , a, discriminant of the number field K ideals of a number field primes of a number field inverse of the ideal W the power to which the prime P

appears in the canonical factorization of the ideal W

xiv 1 GLOSSARY OF NOTATION

Definition Page Reference

another notation for v, (91) the group of ideals of the number

field K greatest common divisor of the ideals

91 and 23 ith conjugate of x over Q normalized valuation corresponding

to the prime p ring of p-adic integers maximal ideal of 8, completion of K at the prime p the norm of the ideal BI adele ring of K set of infinite primes of K ring of S-adeles idele group of K group of S-ideles modulus of the automorphism a of

the group G volume of the idele a group of K-ideles of volume I group of S-units of K (also denoted

K S ) group of roots of unity of K ideal mapping S-ideal mapping field of p-adic numbers ring of p-adic integers value group of I 1, prime divisor of local field K unit group of local field K ring of integers of local field K ramification index of L over K residue class field of K residue class degree of L over K ideal norm of 9[ complementary set of z-module R relative different of the extension L/K relative discriminant of L/K decomposition group at p decomposition subfield at $1

inertia group at p inertia subfield at p

Frobenius automorphism

GLOSSARY OF NOTATION / xv

Symbol Definition Page Reference

Ti(i = 0, 1 , . . .) ramification subfields

Legendre-Jacobi symbol

primitive mth root of unity Euler's pfunction

Kronecker symbol

circle group orthogonal complement of H Fourier transform off self-dual measure on the local field K Schwartz-Bruhat space self-dual measure on AK local zeta function, global zeta

function factor in local functional equation regulator of K number of roots of unity in K class number of K ~ e c k e character conductor of x Hecke L-series associated to x Artin symbol

abelian L-series generalized conductor of conductor of LIK generalized conductor of LIK norm residue symbol ray modulo 111

admissible subgroup corresponding to L/K

conductor of H class group of H

power residue symbol

natural density of the set A

Historical introduction

Analytic number theory is devoted to solving problems of arithmetic by using the machinery of anlysis. The subject began about two centuries ago with Euler's proof of the infinitude of rational primes. Euler's proof is as follows: Suppose that there are finitely many primes p , , . . . , p,. Consider the product

for every positive integer r. By the unique factorization theorem for Z, every unit fraction Iln appears in the expansion (*), provided that we choose r large enough. In fact, given any positive integer M, we can choose r = r ( M ) such that (*) exceeds

If we let M tend to co, this last sum tends to oo, so we have a contradiction. This proof is very much in the spirit of analytic number theory. The result is deduced from an analytic fact-the divergence of the harmonic series.

Throughout the nineteenth century, much effort was expended in order to determine the law of distribution of the rational primes. Let ~ ( x ) - the number of rational primes I x . It is clear that

Euler's theorem asserts that

Z ( X ) - oo as x oo.

CHAP. 1 HISTORICAL INTRODUCTION / 3

Empirical evidence suggested to Gauss and Legendre the law

[We have used the Landau o-notation: We write f(x) = o(g(x)) as x -+ m to mean that f(x)/g(x) -* 0 as x -+ m.]

Gauss discovered that a much closer numerical approximation to w(x) is given by

Integrating once by parts shows that

X Li(x) = - " dy logx + Jz ii$j

Note that this estimate is consistent with equation (1). Therefore, Gauss conjectured that

which statement became known as the prime number theorem. The fundamental ideas involved in the first proof of the prime number

theorem were exposited by Riemann in a memoir in 1860. Riemann returned to Euler's proof of the infinitude of rational primes and was motivated by it to introduce the function

where Re (s) > 1 and the product runs over all primes. This product is absolutely convergent for Re (s) > 1 and represents an analytic function in this region. Furthermore, by expanding each factor in the product and multiplying the resulting series formally, we can formally deduce the relation -

C ( ~ ) = x n - ~ (Re(s )>l ) . n= 1

Euler's proof of the infinitude of primes shows that the behavior of C(s) as s approaches 1 through real values greater than 1 is intimately related to the distribution of primes. The function C(s) is called the Riemann zeta function.

Riemann showed that C(s) is analytically continuable to a meromorphic function on the whole s-plane, and has a simple pole with residue 1 at s = 1, this pole being the only singularity of the function. Moreover, Riemann showed that the zeta function satisfies a functional equation: If we define

where r ( s ) is the gamma function, then R(s) is an entire function, and

R(s) = R(l - s).

From the product representation for &s) (called the Euler product representation), it is clear that C(s) # 0 for Re (s) > 1. From the functional equation and the fact that T(s) has a simple pole at each negative integer, we see that C(s) has zeros at s = -2n (n a positive integer) [since R(s) is entire] and that C(s) # 0 for any other s in the half-plane Re (s) > 0. Thus, the only zeros of C(s), aside from the above "trivial" zeros, lie in the strip 0 < Re (s) < 1, the so-called critical strip. By the functional equation, the zeros are sym- - metric about the line Re (s) = +. The Riemann hypothesis asserts that all the nontrivial zeros of the zeta function actually lie on the line Re (s) = +. This conjecture, first enunciated in Riemann's 1860 memoir, has been the object of much research and currently remains unproven.

It was Riemann who first realized the connection between the distribution of primes and the distribution of the zeros of the zeta function. However, it was not until 1896 that Hadamard and de la Vallee Poussin independently succeeded in deducing the prime number theorem from the fact that [(l + it) f 0 for all real t. The connection between the distribution of the zeros and the prime number theorem runs even deeper. For if C(s) # 0 for Re (s) > 0 > 0 for some 8, then

n(x) = Li(x) + 0(9+c)

for every E > 0. Conversely, if this last equation holds, then ((s) # 0 for Re (s) > 8. Thus, the zeros of the zeta function control how many primes exist. The minimum possible value of 8 is 3, since a theorem of Hardy asserts that infinitely many zeros of C(s) actually lie on the line Re (s) = 3. So much for the classical theory of the distribution of primes.

In another outgrowth from Euler's work, Dirichlet considered the problem of the existence of primes in an arithmetic progression

a , a + d , a + 2 d ,..., ( a ,d )= 1.

In order to prove that there were infinitely many such primes, he was led to introduce the Dirichlet L-series: Consider the group of units (Z/(d))" of the finite ring Z/(d). This group consists of the cosets that contain an integer that is relatively prime to d. Let x be a homomorphism of (Z/(d))" into the circle group T. Let us define a function x on Z which is attached to x as follows: If (m,d) = 1, set

x h ) = x(m(m0d 4 ) . If (m, d) # 1, set

~ ( m ) = 0.

From the function X, we construct the series

(Re (s) > 1).

Such a series is called a Dirichlet L-series. It can be shown to have an Euler product representation of the form

L(s7 x) = IT P (1 - X(P)P-"-',

where the product is over all primesp. It is easily seen that both the series and product representations for L(s, X) converge absolutely for Re (s) > I . More- over, L(s, X) has a functional equation and analytic continuation similar to that of the Riemann zeta function. The behavior of L(s, X) as s - 1 + can be studied and knowledge of this behavior implies that there are infinitely many primes in the above arithmetic progression. Just as in the case of the Riemann zeta function, the zeros of the Dirichlet L-series are conjectured to lie on the line Re (s) = $ (except for some trivial zeros which are determined directly from the form of the functional equation). By proving that L(1 + it, X) f 0 for all real t, one can, just as in the case of the Riemann zeta function, derive an asymptotic formula for the number n(x, a, d ) of primes <x in an arithmetic progression with difference d and first term a, (a, d) = 1. The result states that

where p denotes Euler's function. A detailed account of the theory of Dirichlet L-series is to be found in Prachar's book.?

Just as the Riemann zeta function reflects the arithmetic structure of @, the Dirichlet L-series play an analogous role for cyclotomic fields-that is, fields of the form @([,), where c, is a primitive mth root of unity. The precise sense of this analogy will be one of the principal themes of this volume. For the moment, however, let us be content to speak of vague generalities.

Once one understands the connection between the Dirichlet L-series and the cyclotomic fields, generalizations of the L-series immediately suggest themselves. The main point is that the cyclotomic extensions are abelian extensions of @-that is, they are Galois extensions of @ with abelian Galois group. One can consider an arbitrary algebraic number field (that is, a finite, algebraic extension of @) K. To K we can associate a zeta function, whose role is exactly analogous to the role played by the Riemann zeta function for @.

Corresponding to abelian extensions of K, we can construct generalized Dirichlet L-series, which play the same role as the Dirichlet L-series play for abelian extensions of @. The generalized Dirichlet L-series are called abelian L-series. Their analytic properties were investigated in the latter part of the nineteenth and the beginning of the twentieth centuries. Their analytic continuation and functional equation were first established by Hecke.

Using the properties of L-series, number theorists at the turn of the century developed a complete theory of abelian extensions of algebraic number fields. At the hands of Hilbert, Weber, Takagi, and Artin this theory of abelian extensions resulted in several crucial theorems: First, it was shown how to classify all abelian extensions of a given algebraic number field. Second, it was shown how to relate the arithmetic of an abelian extension to the arithmetic of the ground field. Third, it was shown how canonically to realize the Galois group of an abelian extension in terms of the arithmetic of the ground field. This last result is the so-called Artin reciprocity law, and today it is used as the central result in the theory, from which all other crucial results are deduced. The reciprocity law of Artin generalizes the classical quadratic, cubic, and biquadratic reciprocity laws of Gauss and Eisenstein, and provides some deep insight into their meaning. Although the reciprocity law is a purely algebraic statement, it was first suggested to Artin by a problem in the theory of L-series. Even more significant is the fact that the reciprocity law is equivalent to a certain statement about abelian L-functions. Thus, the L-series do not casually intrude into the province of arithmetic, but are inseparable from it.

In the present volume, we shall study the arithmetic of algebraic number fields from the point of view of the L-series. Our point of view will be quite modern, based on p-adic methods, while most of our results will be quite classical, going back to the early part of the century. Our program will be to make clear to the reader the deep connection between the arithmetic and analytic theory of algebraic number fields.

In Part 1, we shall present an account of the elementary arithmetic theory of algebraic number fields. Part 2 will be devoted to the required tools from harmonic analysis that we shall use in order to introduce and study Hecke's zeta functions in Part 3. Part 4 is devoted to class field theory and Part 5 to the generalized prime number theorem.

t K. Prachar, Primzahluerteilung, Springer-Verlag, Berlin, 1957.

PART 1

Algebraic Number Theory

The first part of this book will be devoted to an exposition of the classical portions of algebraic number theory, namely, the theory of algebraic integers, ideal theory in algebraic number fields, valuation theory, and ramification theory. Also included is an introduction to adeles and ideles, which will lay the foundation for our discussion of Tate's thesis in Part 3. Our treatment has been motivated by two goals: compactness and clarity. For these reasons, we have avoided the usual valuation-theoretic approach, discussed in Artin [3] and Weiss [32]. The theory of Part 1 can also be completely developed by using the techniques of Haar integration. For this approach, the reader is referred to the recent treatise of Weil [28]. Our approach is a pragmatic mixture of the valuation theoretic and Haar integral approaches.

The ring of integers of a number field

2-1 INTEGERS

In what follows, all fields will be assumed commutative.

Definition 2-1-1: An algebraic number jield is a finite algebraic extension of the rational numbers @. An element of K is called an algebraic number.

Throughout Chapter 2, K will denote an algebraic number field of degree n over @.

Definition 2-1-2: An algebraic number x is integral (or is an algebraic integer) if x satisfies an equation of the form

xm + alxm-l + . . . $ a, = 0, a, E z. Such an equation is called an equation of integral dependence of x over Z.

Proposition 2-1-3: x E K is integral - Z[x] is ajinite h-module.*

Proof: If Z[x] is a finite Z-module, then for some positive integer m,

Z[X] E h + h~ + .. . + hxm-l, so that

xm = alxm- ' + + a,, a, E Z,

which proves that x is integral. Conversely, if x is integral,

Z[X] = h + ZX + + - + Zxm-1

if x has an equation of integral dependence of degree m. I1

* We shall always use "finite h-module" instead of "finitely generated h-module."

Corollary 2-1-4: I f x,, . . ., x, E K are integral, then Z[.Y,, . . . , x,] is a jinite Z-module.

Proof: It suffices to show that Z[x,, . . . , x,-,][x,] = Z[x,, . . . , x,] is a finite Z-module. But we may apply induction on r. The case r = 1 is the preceding proposition. Assuming the assertion for r - 1, we see that Z[x,, . . . , x,-,] is a finite Z-module. Also, Z[x,] is a finite Z-module. Thus Z[X,,. . . , x,] is a finite Z[x,, . . . , x,-,]-module, and therefore a finite Z-module. 11

Corollary 2-1-5: The integers of K form a ring, which we shall denote OK.

Proof: Let x, y E OK. Then Z[x, y] is a finite Z-module by Proposition 2-1-4. But x + y, xy are contained in Z[x, y] and the Z-submodules Z[x + y], Z[xy] are finite.* Thus, by Proposition 2-1-3, x f y, xy are in 0,. I1

EXERCISE: Op = z. If x E OK, then all conjugates of x in a fixed algebraic closure of K are

integral. (For they satisfy the same equation of integral dependence as x.) Thus, by Corollary 2-1-5, the minimal polynomial and the field polynomial of x over @ (that is, the unique power of the minimal polynomial of degree n) have rational integers as coefficients by the above exercise. Thus, the minimal polynomial and the field polynomial of x over @ give equations of integral dependence for x. Throughout Chapter 2, let x -+ ~ ( " ( 1 5 i < n) denote the distinct conjugation maps of K into C.

Proposition 2-1-6: I f x E K, there exists c E Z - {O] such that cx E 0,.

Proofi Let the minimal equation for x over O be

xm + + + a, = 0, a, E @.

Let c denote the least common multiple of the denominators of a , , . . . , a,. Then cx is integral, since

(cx), + c ~ , ( c x ) ~ - ~ + . . + cma, = 0,

and ca,, . . . , cma, E Z. 11

Proposition 2-1-7: OK is a free Z-module of rank n.

Proof: By Proposition 2-1-6, there exist bases {a,, . . . , a,} of K/@ such that all a, are integral. For each such basis of K/@, let us consider the determinant

Since a submodule of a finite Z-module is finite. See the Appendix.

A simple matrix computation shows that

A(u,, ..., a,,)=

W , , . . . ,fin) = det (cij)"(al, . . . , a,). Since det (c,) # 0, we see that if the discriminant of one basis is nonzero, then the same is true of all bases. But there exists 8 E K such that {1,8, . . . , & - I ) is a basis of K/@. It is well known that the discriminant of this basis, the square of the so-called "Vandermonde determinant," has the valuet

(- l p - 1)/2 (OW - OW), i# j

ail) ... a!l)2

- . ,

... up'

which is nonzero since K/Q is separable. Among all bases of K/Q that consist of integers, choose one, say {a,,

. . . , a,}, for which I A(a,, . . . , an I is minimal. We claim that {a,, . . . , a,) is a set of free Z-generators for 0,. Let x E OK. Then

which is called the discriminant of the basis. Let M = (a,,), a,, = ay) (1 5 i, j 5 n). Then A(a,, . . . , a,) = det (Mf M), where 'M denotes the trans- pose of M. A simple calculation shows that M f M = (bij), bij = TrK,,(a,aj), where TrK,, denotes the trace from K to Q. But bij E Q, SO that A(a,, . . . , a,) E Q. Moreover, the discriminant is integral, since the a:') are. Therefore, by the above exercise, A(a,, . . . , an) E Z.

Let (ply . . . , p,) be another basis of K/Q, related to the first by the system of equations

Claim that all ai are in Z. If not, suppose that a , 4 Z and set a , = dl + a:', 0 < a', < 1, d,' E Z. Define a new basis of K/Q by

a', = x - d,'a, a: = a . , (i + 1).

The matrix of transition between the two bases is

t See Exercise 2-1-9 (1).

The determinant of this matrix is a',. Therefore,

A(a:, . . . , a',) = (dl)2 A(al, . . . , a,),

which contradicts the fact that I A(al, . . . , an) I is minimal, since 0 < a', < 1. Thus, a , E Z, and the theorem is proved. /I

Definition 2-1-8: A set of free Z-generators for OK is called a minimal basis (or integral basis) of K. Any two minimal bases must be related by an invertible matrix with entries in Z. Such a matrix has determinant & I . Thus, the discriminants of any two integral bases are the same. This common value is called the discriminant of K, denoted dK.

EXAMPLE: Let K/@ be quadratic. Such an extension can always be written in the form K = @(n), where D E z is square-free. If D is positive, then K is called a real quadratic field; if D is negative, then K is called an imaginary quadratic field. Let us determine an integral basis for K and let us compute dK: If x E K, then x satisfies a quadratic equation over @ (its field equation):

x 2 + A x + B = 0 ,

Now x E OK A, B E z. Let x = n + b n , a, b E @. Then

TrKia(x) = 2a

NK,,@(x) = a2 - b2D.

Set a = a1/2, b = b'/2. Then b' E z since D is square-free, and

x E OK o a' E z, (a'2 - bI2D)/4 E z. Now a'2 and b'2 are congruent either to 0 or 1 mod 4; D, being square-free, is congruent to either 1,2,3 mod 4. Analyzing all the various cases, one quickly verifies that

D - l(mod 4) =. x E OK <- a' and b' are of the same parity, D z 2, 3(mod 4) =. x E OK C--* a' and b' are both even.

Thus, if D = l(mod 4), then 11, (1 + f l ) / 2 ) is an integral basis. If D = 2, 3 (mod 4), then 11, a ] is an integral basis. In these two cases, we have, respectively, dK = D and dK = 40.

EXERCISES 2-1-9: (1) Let X , , . . . , X, be indeterminates. Show that

the so-called Vandermonde determinant, is equal to

Therefore,

(2) (The Sign of the Discriminant) Let K be a number field of degree n over Q, and let {a,, . . . , an) be an integral basis of K. Let

Then det (T)2 = dK. Thus, det (T) is either real or purely imaginary. Show that det ( F ) = ( - l ) r a det (T), where r2 denotes the number of conjugate fields that are not contained in [k. Thus, show that d, > 0 - r2 is even.

2-2 IDEAL THEORY

In the last section, we demonstrated, in particular, that Z [ a ] is the ring of integers of the number field @(-). Moreover, it is well-known? that Z [ G ] is not a unique factorization domain. Thus, in order to explore the arithmetic of the ring of integers of an algebraic number field, it will be necessary to introduce a substitute for unique factorization. In the following discussion, we shall prove that every ideal of OK can be written uniquely as a product of prime ideals, where uniqueness is understood to mean uniqueness up to the order of the factors. Thus, in an algebraic number field, the prime ideals (and not the irreducible elements) play the same role as the prime numbers do in rational number theory.

From now on, we shall assume that all ideals are nonzero. Let 8 be an ideal of OK. First observe that 8 n Z # (0). For if x E (35 - {O}, and if the minimal equation for x over @ is xm + alxm-' + - + am = 0, then a, E 8 n (Z - (0)). There exist bases of K/@, all of whose elements belong to @. For let a E 8 - (01, (a,, . . . , a,) an integral basis of K. Then {aa,, . . . , aa,} is such a basis. By considering a basis of K/@, whose elements are in 8 and whose discriminant has minimal absolute value, we conclude, as in the preceding section, that 8 is a free 7-module of rank n. In particular, 8 is finitely generated, so that

Proposition 2-2-1 : OK is a Noetherian ring.

t For example, in Z [ n ] the numbers 3,7 , 1 + 2&3, and 1 - 2- are all irreducible elements, pairwise nonassociated with one another, but nevertheless

3.7 = (1 + 2 c 3 ) ( 1 - 2-1.

Definition 2-2-2: A fractional ideal of K is a nonzero OK-submodule 8 for which there exists c E Kx such that c 8 c OK.

I EXAMPLES: (1) Let 8 be any ideal of 0,. Such a fractional ideal is called integral.

(2) Let y E KX. Then 8 = yOK is a fractional ideal of K. Such a fractional ideal is called principal. (To verify that this is a fractional ideal, merely refer to Proposition 2-1-3.)

The usual operations of addition, multiplication,and intersection of ideals can be defined in an obvious manner for fractional ideals. A straight-

i forward verification shows that a sum (respectively, product, intersection) of i two fractional ideals is again a fractional ideal. Throughout the remainder of

Section 2.2, let 8 denote a fractional ideal of K. !

Definition 2-2-3: The inverse of 8, denoted is the Z-module

{x E K1x8 c OK).

It is clear that 8 - I is an OK-module. Moreover, there exists c E K x such that c 8 0,. Then c E @-I, which therefore contains a nonzero element. Let d E - (0). Then dB-' G 8 8 - I c OK. Therefore, 8 - I is a fractional ideal.

EXERCZSE: Let 8 be a fractional ideal of K. Then there exists x E OK - {0) such that x 8 E OK.

Proposition 2-2-4: Every prime ideal '$3 of OK is maximal.

Proof: Assume that '$3 is not maximal. Then it is contained in some maximal ideal la. Now '$3 n Z is a prime ideal of h. Reasoning as above, we see that '$3 n Z # (0) or Z. Thus, '$3 n Z = pZ for some rational prime p. But then la n Z = pZ, and we have found two prime ideals '$3 $ 0 , which when intersected with Z give pZ. We shall show that this is impossible. Let x E

0 - '$3 have minimal equation over Q

xm + alxm-I + . . . + a, = 0, ai E Z.

By transposing a,, we see that a, E la n Z = '$3 n Z * a, E '$3. Thus,

x(xm-I + a l ~ m - 2 + . - - + a,_,) E '$3.

But x $ '$3, so that

xm-l + a l ~ m - 2 + . . . + am-1 E '$3.

Repeating the above reasoning, we get that a,-, E '$3 and hence that

xm-I + ~ , x " - ~ + - . . + a,-, E '$3.

Proceeding in this way, we eventually arrive at the statement

x g ' $ 3 which is a contradiction. //

Definition 2-2-5: A fractional ideal 8 is said to be invertible if there exists a fractional ideal 23 such that 823 = OK.

EXERCZSE: If 8 is invertible, then 23 is unique and is precisely 8-l.

Proposition 2-2-6: Every prime ideal '$3 is invertible.

Proof: We clearly have

'$3 c '$3'$3- c OK.

Since '$3 is a maximal ideal (Proposition 2-2-4), either '$3'$3-l = OK or '$3'$3-1 =

'$3. Assume that the latter is true. Let {a,, . . . , a,} be a free Z-basis of '$3. Let x E $-I. Then ~ ' $ 3 c $3, so that

n

xu, = C aijaj (i = 1 . . . n ) a,, E Z. j = 1

Therefore, setting A = (a,,), I = the n x n identity matrix, we see that det (XI - A) = 0, which is an equation of integral dependence for x over Z. Thus, x E OK, which implies that '$3-l c OK. But it is clear that 2 OK, so that = OK. Let us show that this is impossible.

First claim that every integral ideal 8 contains a product of prime ideals. If 8 does not contain such a product, assume that 8 is maximal with respect to this property. Then 8 is not prime, so there exist a, b E OK - 8 such that ab E 8. The ideals (8, a) and (8, b) properly contain 8 and (8, a)(@, b) = 8. But these latter two ideals contain products of primes and so, therefore, does 8. Now assume that '$3-' = OK. Let a E '$3 - {O). Then there exist prime ideals '$3,, . . . ,'$3, such that

'$31 * - '$3, c aOK '$3 $ OK. Assume that g is minimal. Now '$3 3 '$3, for some i. Since '$3, is maximal, we have 8 = '$3,. If g = 1, then '$3 = aOK, so that a-I E '$3-' -OK. If g > 1, then '$3,. . . '$3, Q: aOK, since g is minimal. Let b E '$3, . . . '$3, - aOK. Then

b'$3 c '$3, . . . '$3, c a@,. Therefore, ba-I E '$3-I - OK. //

Corollary 2-2-7: The set of powers (positive and negative) of a prime ideal form an infinite cyclic group under ideal multiplication.

We now come to the main result of this section:

Theorem 2-2-8: (1) Every fractional ideal 8 of K can be expressed uniquely in the form

where the product runs over all prime ideals of OK. (It will turn out that all but a finite number of the exponents are zero, so that the product is actually well defined.)

(2) 8 is an integral ideal -> vs(8) 2 0 for all !$.

Proof: (1) Claim that 8 can be written in the desired form in at least one way. It suffices to consider the case of 8 integral. For if 8 is arbitrary, let c E OK be such that c 8 is integral. Let cO, = Dl . . . Oh, c 8 = !$, . . . 8,. Then

a=!$' . . . !$&;' ... ail.

Thus, assume that 8 is integral and is not a product of prime ideals. Since 0, is Noetherian, we can choose 8 maximal with respect to this property. Then 8 is not prime, so 8 $ !$ for some prime (and hence maximal) ideal !$. Since !$ is invertible,

8 c 89-l $0,. (*I Claim that 8 f @!$-I. If @!$-I = 8, let x E !@-I and let { z , , . . . , z,) be a Z-module basis of 8. Then 8 x G 8 =>

xzi = 2 a,,z,, a,, E Z (1 I i i n). j = 1

Reasoning as in the proof of Proposition 2-2-6, we see that x E OK. There- fore, !$-' G 0, * '$-I = OK, a contradiction. Thus, @!$-' # 8 , so that (*) implies that a!$-' is a product of primes by the maximality of 8. But then, if a!$-' = Dl . . . a,, 8 = !$al . . . Oh, which is a representation of 8 as a product of primes. Note that the above argument also shows that an integral ideal can be expressed as a product of nonnegative powers of prime ideals. This result, in conjunction with the uniqueness statement about to be proved establishes (2).

Uniqueness: Without loss of generality, let us restrict ourselves to integral ideals. Let

8 = !@, . . . '$, = 0, . . . El#, (**I be two factorizations of 8 into prime factors. Then !$, 3 8 => pl contains some Q, say El,. But 0 is prime and hence maximal, so that 3, = Ell. Multiplying both sides of (**) by !$;I, we arrive at

p 2 . . . $ , = D 2 . . . 0,.

The proof may now be completed by induction.

(3) Note that (1) implies that every K-ideal is invertible. Now 8 c 8 F, 88-l is integral - ~ ~ ( 8 8 - l ) 2 0 for all 9. But it is clear that ~ ~ ( 8 8 ) = ~ ~ ( 8 ) + ~ ~ ( 8 ) . Therefore, 8 c 8 - vs(8) i ~ ~ ( 8 ) for all !$. Since 8 + 8 (respectively, 8 n 8 ) is the smallest ideal containing both 8 and 8 (respectively, the largest ideal contained in 8 and 8 ) , the formulas of (3) follow. I/

If 8 is a K-ideal, then we shall denote by vp(8) or ordp (8) the power to which the prime p appears in the canonical factorization of 8. If x E K x we shall abbreviate vp((x)) (respectively, ordp ((x))) by v,(x) (respectively, ordp (x)). We shall formally set ord, (0) = +oo.

In algebraic number theory, we often have need of a notion of congruence that generalizes the usual notion of congruence modulo an ideal. Let 8 be an integral K-ideal, x, y E K. Then we write

x = y (mod 8 )

(read "x is congruent to y modulo 8") whenever

for all primes p. If x and y are integers, then V,(X - y) ) 0 for all p. There- fore, x = y(mod 8 ) - vp(x - y) ) ~ ~ ( 8 ) for all p - (x - y) E 8 - x - y E (3, by Theorem 2-2-8. Thus, in case we are dealing only with integers, our notion of congruence coincides with the usual one.

EXERCISE 2-2-9: ( 1 ) Congruence modulo an ideal is an equivalence relation.

(2) If a - b(mod 8) and c - d(mod a), then a + c r b + d(mod 8 ) and ac r bd(mod 8), provided that a, b, c, d are integers.

By Theorem 2-2-8, the K-ideals form a free abelian group with the prime ideals as free generators. We shall denote this group by I,.

Definition 2-2-10: Let 8 , 8 E I,. Then we say that 8 divides 8 (denoted @/B if 88- is an integral ideal.

EXERCZSE 1-2-11: @I8 -, 8 E @.

Theorem 2-2-8, (3) implies that 8 + 8 divides both 8 and 8 and that any K-ideal which divides both 8 and 8 divides 8 + 8. For this reason, we sometimes call 8 + 8 the greatest common divisor of 8 and 8 , and sometimes denote it by (8 , 8) . If ( 8 , 8 ) = OK, then we say that 8 and 8 are relatively prime and write (@,8) = 1. If 9Jl is an integral K-ideal, then the set of all K-ideals relatively prime to 9Jl forms a subgroup of I,, which we denote IK(rn).

EXERCZSE 2-2-12: Let 8 , 8 be a pair of relatively prime, integral K-ideals. Then, for x, y E K, x = y(mod 89) - x = y(mod 8) and x - y(mod '23).

One consequence of the preceding exercise is that in order to solve a congruence it suffices to solve simultaneously a system of congruences modulo powers of distinct primes. The following fundamental result asserts that such a system always has a solution.

Theorem 2-2-13 (Chinese Remainder Theorem): Let Qi(l 2 i ( r ) be distinct K-primes, a,(l ( i ( r ) nonnegative, rational integers, x,(l ( i ( r ) elements of K. Then there exists x E K such that

x =. x,(mod $9') (1 ( i ( r)

v s ( x ) 2 O if '$ E (8, . . - '$,I. In particular, if all xi are in OK, then x can be chosen in 0,.

We require two lemmas.

Lemma 1: Let a , , @, be two relatively prime, integral K-ideals. Let x , , x , E 0,. Then there exists x E OK such that

Proof: Since (a,, 8,) = 1, there exists a , E a,, a, E 8, such that a , + a, = I . Set x = x,a, + x,a,. Then x - x , = ( - x , + x,)a, E 8, and similarly, x - x , E 8,.

Lemma 2: Theorem 2-2-1 3 is true if all xi are in OK.

Proof: For 1 I i < r, set

Then 8':) and @y) are relatively prime, so that by Lemma 1, there exists y, E OK such that

y 1 (mod 8 , y, zz O(mod at)) (1 < i I r).

Since 8t' c '$41 for all j f i, we see that

yi F 1 (mod 87) yi -- O(mod '$?) ( j # i).

Set x = x , y , + . . . + x,y,. Then x E OK and

x - x j = x , y , + . . . x j (y , - I )+ +x , y , E 1 3 ~ ( 1 I j I r ) .

Proof of Theorem 2-2-13: Let a be a nonzero, rational integer such that ax, E: OK(l 5 i 5 r) . Let p = max {vs(a) 1 ' $ 1 (a)}. By Lemma 2, there exists w E OK such that

w E ax,(mod w E O(mod '$33 (13 1 (a ) '$ #'Pi) . Then

Since vs(w/a) 2 0 if !$ # '$3, or a divisor of (a), we may set x = w/a. // Remark: The conditions of Theorem 2-2-13 may be written

x r 0 (mod 91) where

The locally compact groups of number theory

3-1 COMPLETIONS

Throughout this section, let K be an algebraic number field.

Definition 3-1-1: An absolute value (or valuation) of K is a mapping v: K -4

R + u (0) having the following properties: (1) v(0) = O;v(x) = 0 - x = 0. (2) ~ ( x Y ) = v(x)v(Y). (3) There exists a constant C > 0 such that v(x + y) < Cmax (v(x), v(y))

for all x, y E K.

If (3) holds with C = 1, then v is said to be non-Archimedean or finite. Otherwise, v is said to be Archimedean or infinite. The most common way of satisfying (3) is by proving that v(x + y) < v(x) + v(y) (the "triangle inequality"), which implies (3) with C = 2. Conversely, if (3) holds with C = 2, then the triangle inequality holds. Let v be a valuation of K with constant C and let a > 0. Define

va:K+R+ u {O}

by V"(X)=Y(X)~ ( X E K).

Then v u is a valuation of K with constant Ca. Moreover, va is equivalent to v in the sense of Definition 3-1-4 below. Therefore, every valuation is equivalent to a valuation for which C = 2. For our purposes, we can always replace a given valuation by an equivalent one. Therefore, we will henceforth assume that all valuations considered satisfy the triangle inequality.

The valuation defined by v(0) = 0, v(x) = 1 for x # 0 is called the trivial valuation. Henceforth, we shall assume that all absolute values are nontrivial.

If K is a number field and p a prime of K, then we associate to p an absolute value vp as follows: Let x E K ".

(x) = prva, (a, p ) = 1. Then we define

1 x l p = v,(x) = c-.v, c > 1.

Proposition 3-1-2: v, is a non-Archimedean absolute value.

Proof: Immediate from Theorem 2-2-8. //

Archimedean absolute values of K can be constructed as follows: Let x -+ x(') (1 I i I n) be an embedding of K into C. Let

v i ( x ) = 1 x 1, x E K, where I I denotes the ordinary absolute value of C. In this way, we obtain n Archimedean absolute values of K. However, not all of these are distinct, since complex-conjugate embeddings give rise to identical absolute values. Thus, arrange the embeddings so that the first r, map K into R. Whereas the remaining ones map K strictly into C. The second group of embeddings consists of r, pairs of complex-conjugate embeddings, where r, + 2r, = n. Arrange the second group so that the jth and the ( j + r,)th constitute a conjugate pair. Then we shall show later in this section that the absolute values v,,,(l < i r , + r,) are distinct.

Let v be an absolute value on K. Associated to v there is a metric on K, namely d(x, y) = v(x - y).

Proposition 3-1-3: With respect to the topology induced by the metric d, K is a topologicalfield. That is, K is a Hausdorff topological space, in which the field operations are continuous.

Proof: The verification of this Proposition mimics the proof of the corresponding one for the real or complex numbers. Let us show that addition and subtraction are continuous. We leave the proof for multiplication and division to the reader. It suffices to show that the maps

x- -X

K-K and

(x, Y) - x + Y K x K - K

are continuous, for subtraction can be expressed as a composite of these two. Since v(x) = v(-x), the first map is an isometry and is a fortiori continuous. To show that the second map is continuous, we merely observe that v(x + y)

< - v(x) + v(y). This implies continuity at (0,O). Continuity at all other points of K x K is deduced by translation. /I

Definition 3-1-4: Two valuations v, and v, of K are said to be equivalent if they induce the same topology on K. (For example, if '$ is finite, different values of c in the definition of vp give rise to equivalent valuations.)

Later in this chapter, we shall prove the following result:

Theorem 3-1-5: Every valuation of an algebrazc numberjield K is equivalent to one of the valuations discussed above.

Let us now discuss the important notion of completions. Let K be a field with valuation v.

Definition 3-1-6: Consider K as a metric space with respect to the metric induced by v. A completion of K with respect to v is a completion of K as a metric space. More specifically, a completion of K with respect to v is an ordered pair (R, p) consisting of a metric space R and an isometry p:K -4 R such that

(1) R is a complete metric space (i.e., every Cauchy sequence is convergent).

(2) p(K) is dense in E. Two completions (R,, p,) and (K,, p,) are said to be equivalent if there exists an isometry q: Rl - R2 such that the diagram

'l Rl + R2

is commutative. It is proven in the study of metric spaces that a metric space has one and,

up to equivalence, only one completion. In the special situation that we are considering, the completion can be given the structure of a field with a valuation such that the valuation induces the metric. Let us briefly indicate how such a completion is constructed.

Let 9 denote the set of all v-Cauchy sequences. Under componentwise addition and multiplication of sequences, 9 has the structure of a commutative ring with identity (1, 1, . . . , 1, . . .). Let 9 denote the set of all null Cauchy sequences, i.e., the set of all Cauchy sequences convergent to 0.

Lemma: 9 is a maximal ideal of 9.

Proof: It is clear that 9 is closed under addition. Let x = (x,) E 9 , y =

(yn) E 9. Then v(y,) is a bounded sequence, so that v(x,y,) -+ 0 (n --. m). Thus, xy E 9. Thus 9 is an ideal of 9. Let x = (xJ E 9 - 9. By adding

an element of 9 to x, we can find a sequence y = (y,) E 9 - 9 such that y, # 0 for all n. Then y- ' = (y;') E 9 , for vb,) 2 C > 0 for some C since y $: 9. Then we deduce that y-I is a Cauchy sequence, since v(y;' - y i l ) < C-2v(y, - y,) + 0 as m, n -+ m. Then y- 'y is contained in the ideal - generated by x and 9. Thus, (x, 9 ) = 9 , and 9 is maximal. //

Since 9 is a maximal ideal in 9 , the quotient 919 is a field, which we shall denote K. Let p: K -4 R be the map

Then p is a field injection. Let us define v' on R. Let (x,)(mod 9 ) be an element of l?. Since (x,) is a Cauchy sequence, (v(x,)) is a Cauchy sequence of real numbers, which converges to a real number that is independent of the representative sequence (x,) in (x,)(mod 9 ) , since 9 consists of null Cauchy sequences. Therefore, we set

vt((xn) (mod 9 ) ) = lim v(x,). "-00

It is a trivial exercise to verify that v' is a valuation on R Claim that (K, p) is a completion of K with respect to v. First, let us

show that K is complete with respect to the metric induced by v'. Let (z,) be a Cauchy sequence in $ and let (x?)) be a representative sequence of z,. By the definition of v', we see that

vf(z, - z,) = lim V(X;) - x:)). m-00

Since (z,) is a Cauchy sequence,

lirn v(xi) - xk)) = 0. I , S, m--

Thus, for each natural number m, there exists a natural number t(m) >my such that if r, s 2 t(m), then v(xi) - xk)) < 2-"'. The sequence (x'Am)) is a Cauchy sequence and is thus in 9. Claim that (2,) converges to x =

(x'Am))(mod 9). Indeed,

lirn vl(zn - x) = lirn v(x?) - ~ 2 " ) ) = 0. ,400 m,n-so

Next claim that p(K) is dense in K. Let z = (x,)(mod 9 ) E E , 6 > 0. Since (x,) is a Cauchy sequence, there exists no = no(€) such that v(xn - x,) < 6

whenever n, m 2 no(€). Then

vf(z - ~(x.,)) = lim v(x, - x,,) < 6. n-m

Thus, p(K) is dense in R, and (I?, p) is a completion of K with respect to v. Throughout the remainder of the book, we shall denote by K, the com-

pletion of K with respect to the valuation v. We shall denote the canonical embedding of K into K, by p,. Whenever confusion will not arise, we shall identify K with p,(K) and consider K as a subfield of K,. This identification carries over to topological considerations also, for since p, is an isometry,

the topology induced on K by p, is the same as the topology induced on K as a subspace of K,.

EXERCISE 3-1-7 (Universal Mapping Property for Completions): Let L be a field that is complete with respect to an absolute value q. Let K be a field with an absolute value V. Assume that there exists an isometric field injection 1: K --t L. Then there exists a unique isometric field injection p : K, -4 L such that the diagram

L

commutes.

EXAMPLES OF COMPLETIONS: (1) Let K be an algebraic number field, van Archi- medean valuation of K corresponding to the embedding x --t x") of K in c, 1 5 i 5 r, + r2. Then K, is just the closure of K") in c. Two cases are possible. Either K") c [R or K( i ) Q [R. In the first case, corresponding to 1 i 5 r, , K. = [R.

(For K,, contains the closure of @ in [R, which is just [R.) In the second case, corresponding to r , + 1 5 i 5 r, + r,, K, = c. (For K, contains [R properly and K, c c.) In the first case, we call the valuation real; in the second case, complex.

(2) Let k be any field, x an indeterminate over k, K = k(x). Define a valuation v on K as follows. Set v(0) = 0. If a = f(x)/g(x) E K, with f, g polynomials in x, set ~ ( a ) = 2 - ( d e ~ ( f ) - ~ e g ( ~ ) ) . This quantity is independent of the_choice o f f and g. One can easily verify that this defines a valuation on K. Then K is just the field of formal power series k{x). That is, 2 consists of formal series

m

aixi, ai E k, r E Z, 1=r

where addition and multiplication are defined in the obvious way. (3) Let $ be a finite prime of the algebraic number field K.The remainder of this

section will be devoted to determining the structure of Ko.

Before we proceed with the study of the completions of algebraic number fields with respect to non-Archimedean valuations, let us fix some notation. From now on, we shall denote the valuation corresponding to the K-prime !$ by I 1% instead of vo. Let us choose c > 1 in the definition of ( arbitrarily for now. We shall normalize c a t the end of the discussion. We shall denote the completion of K with respect to !$ by Ko. We shall denote the extension of the valuation ( Is t o Ko also by I 1%.

Proposition 3-1-8: 1 Is is a non-Archimedean valuation of Kv.

Proof: Let x E Z. Then

I x l o = I I + . . . + l Is(max(I1la, . . . , I l l s )= 1. . x times

Thus, lxla < 1 for all x E 7. Let x, y E Ks be arbitrary, n E 7. Then

Ix -t YI; = I(x + YYIs

= lx"+(;)x"-' + . . . +ynlu

I I x l ; + l x l ~ - i ; ' l ~ l o + . . . + Iy l i by the triangle inequality and the fact that I z 1s I 1 for all z E 7. Assume that I x lo I l Y lo. Then the right-hand member of the last inequality is a t most (n + 1) ly 1;. Taking nth roots, we see that

Ix + Y lo < (n + I>','"ly Is Letting n -+ oo, we arrive at

I x + Y Is I IY Is = max {I x lo, IY 131. Thus, 1 Is is non-Archimedean. //

Proposition 3-1-9: The set of values assumed by I 1, on Ks coincides with the set of values taken b j ~ I Is on K.

Proof: Let x E KG. We shall find y E K such that = 1 ~ 1 ~ . Since K is dense in K3, there exists y E K such that

l x - yJs < 1 ~ 1 ~ . It is clear that

ly la = I (I? - x) + x lo I max {IY - x IT, I x Is]

= I x 1% which implies that 1 y 1% < I x 1s. Also,

I x I? = I (x - Y) + y 1s max {I x - y 13, IY IT) I max {I x lo, IY lo} = I X I$.

Thus, all of the inequalities in the above string become equalities, so that I ~ l u = m a x { l x - Y l s . I Y I s ) = I ~ I o . /I

Corollary 3-1-10: 1 1.11 assumes only values of the form cm, m E 7.

EXERCISE 3-1-11: Let K be a field with a non-Archimedean valuation v. Let x, Y E K be such that v(x) # v(y). Then v(x + y) = max {v(x), v(y)).

Set Q = { x E Kv I ] x IV<1)

$ = { x E K%IIxlg<1}. Using the fact that 1 IT is a non-Archimedean valuation on Kv, we see that 0% is a closed subring of Kt,, called the ring of !$-adic integers.? @ is a maxi-

? In case K = @, we shall write E , instead of 6,.

ma1 ideal of Oa. In fact, $ is the unique maximal ideal. For let 5352 be a maximal ideal, a E 5352. If 1 a lv < 1, then, a E $. If 1 a Is = 1, then 1 a-' Is =

1 s- a-I E 0% * a-la = 1 E (Dj, a contradiction. Thus, E01 = $. This same argument shows that the units Us of O0 are precisely the elements of absolute value 1.

Let us choose n E 0% such that Inlg = c-'. Clearly, n E $. Such an element exists-any element of $3 - q2 will do. The element n is called a local uniformizing parameter at $3. Any two local uniformizing parameters at

differ from one another by multiplication by an element of U$.

Proposition 3-1-12: Every ideal of Oe is of the form $' (r > 0). Moreover, $3. = (nr), so that Op is a principal ideal domain.

Proof: Let a € $ . Then l a l a < c - ' * l a n - ' I s < l * ~ = n ~ ( ~ ~ O s ) * $ c (n). But since $ is maximal, $ = (n). Thus, $' = (nr). Let BI be an ideal of Og, a E '$I an element of maximum absolute value. Such an element exists by Corollary 2-3-10. Let lala = c-.(r > 0). Then Ian-'I$ = 1 * a = u'nr(u' E U%) * 91 c $'. Let b E $' - 91. Then ba-' E Us. Thus, b E 91, a contradiction. //

Proposition 3-1-13: 0% (respectively, $) is the closure of OK (respectively, $3) in K$.

Proof: From the definition, we see that Oip is closed in Kg. Since

the same is true of $. It is clear that OK c O$, $3 c $. Let x E 03. Let f be given, 0 < E < 1 . To show that the closure of OK is 03, it suffices to show that there exists z E 0, such that I x - z < E . Since K is dense in K;@, there exists y E K such that I x - y Is < f. Then 1 y 5 1. By the Chinese remainder theorem, there exists z E K such that \ y - z \$ < f , I z la < 1 for all K-primes .C1 # $3. But

I z l v i m a x ~ l ~ l s ~ l ~ - z l s I I 1 * lzla 1 for all K-primes Q,

5 ordn (z) 2 0 for all K-primes, * z E 0,.

Moreover, I x - z 1% < max {I x - y I%, 1 y - z I$} < E. The proof of the analogous assertion for $ is left to the reader. //

Theorem 3-1-14: 8% is a compact, open subring of K*. All of its ideals @' (r > 0) are compact and open. In particular, K$ is a locally compact, totally disconnected topological field.

Lemma: Let r > 0 be an integer. Then the ring 6,/$3' isfinite.

Proof: We have proven in Sections 2-1 and 2-2, that OK and $3' are free Z-modules of rank n. By the elementary divisor theorem, O,/Qr is a direct sum of finite cyclic groups. //

Proof of Theorem 3-1-14: We have already remarked that O$ is a closed subring of Kv. Since

we see that 0% is also open. Similarly, $' is both open and closed. Since {$q is a fundamental system of neighborhoods of 0, it follows that KQ is totally disconnected.? In order to prove the remaining assertions, it suffices to show that 0% is compact. From this it will immediately follow that Qr is compact, since $' is the image of 0% under the continuous mapping x -+ nrx, n a local uniformizing parameter at $3. Therefore, it suffices to show that 0% is complete and totally bounded (i.e., 0% can be covered by a finite number of sets having diameter less than E , for any prescribed E > 0). Since 05 is a closed subspace of a complete metric space Kg, it follows that 0% is complete. Observe that qs (s > 0) has diameter c-". To show that OP can be covered by a finite number of sets of diameter less than 6, it suffices to show that Oa/$" is finite. For then we can choose s so that c-" < f; 0% is covered by the cosets of $"n @Ig, and each has diameter c-". By the Lemma, we are done if we show that o~/$% OK/ps. Let x E OQ. By Proposition 2-3-13, there exists 2 E OK such that Ix - 21 < c-". Such an x is unique up to addition of an element of Tps. Thus, we may define the mapping

x -+ 2 (mod '$I).

This mapping is well defined and has kernel qs. Thus, O%/$% OK/$: This completes the proof of the theorem. //

Theorem 3-1-15: The valuations introduced on an algebraic number field K are all inequivalent.

Proof: First, let us show that no two non-Archimedean valuations are equivalent. For if '$3, f: $3,, let x E $3, - $3,. Then 1 x < 1 * I xn (5, -+ 0 as n + oo xn 3 0 in the $3,-topology. But 1 XI%, 2 1 3 x" does not converge to 0 in the $3,-topology. Thus, the $3, and $3, topologies are not equal. Next, assume that vi and vj are two Archimedean valuations corresponding to the embeddings x -4 x(') and x - x"), respectively. Assume that v, is equivalent to vj. Then the conjugation mapping ~("5 K(j) is a homomor-

t A topological space is said to be totally disconnected if each point has a basis for its neighborhoods which consists of sets that are both open and closed. Equivalently, every component of the space consists of a single point.

phism. This conjugation mapping extends to a mapping from the closure K"' of K(') in C to the closure K T o f K(j) in C so that the diagram

is commutative. If K"' = IR, then K"' = IR and the map 0 is the identity, since IR has no nontrivial homomorphic field isomorphisms. (Why?) Thus, in this case i = j and the two valuations are identical. If X"' = C, then K"' =

C and the map 8 is either the identity or complex conjugation. But since only one embedding from each pair of complex conjugate embeddings was used in constructing the Archimedean valuations, this implies that i = j and again the two valuations are the same. Thus, the Archimedean valuations are inequivalent. Finally, suppose that an Archimedean valuation is equivalent to a non-Archimedean valuation. Let these two valuations be v and w, respectively. We can extend the identity map on K to an algebraic and topological isomorphism of K, onto K, such that the diagram

K-K

commutes. But K, is either IR or C, both of which are connected. By Theorem 3-1-14, K, is totally disconnected. This is a contradiction. //

At this point, let us make some conventions that will remain in force throughout the remainder of the book. We shall introduce r, + r, symbols, called infinite primes, denoted 1),, , , . . . , @,,,,+,,. These symbols will correspond to the Archimedean valuations of Kin the same way that the prime ideals of 0, correspond to the non-Archimedean valuations. From this point on, when we speak of a "prime," we shall mean either a prime in the previous sense (now referred to as a "finite prime") or an infinite prime. The symbol 1) will denote a typical K-prime. We may associate, in a uniform manner, to every K-prime '$3, a valuation and a completion, denoted I 18 and K$, respectively. An infinite prime is called real (respectively, complex) if the corresponding completion is the field of reals (respectively, the field of complexes). The set of infinite primes will be denoted S,.

By a K-modulus, we shall mean a formal product of primes of K (not necessarily distinct). A K-modulus can be canonically decomposed into a product of an integral K-ideal (its finite part) and the product of a certain collection of infinite primes (its infinite part). The main use of K-moduli is as moduli of generalized congruences. Let r be a nonnegative integer, x E K. Then we write x = 0 (mod $') to mean: (1) ord8 (x) 2 r if 9 is finite; (2) x(') > 0 if 1) = QmZi is a real prime and r > 0; (3) no condition in all other

cases. If 91 is a K-modulus, a = Qr8 the decomposition of % into powers of distinct primes, we shall write ;-- O(mod %) to mean that x = 0 (mod '$1) for all '$3. If there are no infinite primes in the decomposition of %, then this reduces to our old notion of congruence.

EXERCISE 3-1-16: Prove the Chinese remainder theorem, allowing the primes appearing to be infinite.

EXERCISE 3-1-17 (The Approximation Theorem): Let 6 > 0 be given. Let !$,, . . . , 9, be distinct K-primes, XI , . . . , x, E K. Then there exists x E K such that Ix - xils, < 6 (1 _( i r) and v ~ ( x ) 2 0 for !$ f 9,. (Hint: If all 9, are finite, then this is just the Chinese remainder theorem.)

EXERCISE 3-1-18: Let K be a field with valuation v. Then v is non-Archi- medean -a there exists a positive constant C such that

for all n E z. (Hint: Examine the proof of Proposition 3-1-8.)

If is a K-modulus, then write x - y (mod a ) to mean that x - y zz 0 (mod g). The usual rules for operating with congruences apply to this notion of congruence. We shall use the notion of congruence also for the elements of the completion K9. The definitions are analogous to those given above and are left to the reader to fill in.

The remainder of this section will be devoted to two tasks: First, we shall normalize the valuations corresponding to the various primes of K. For finite 9, this amounts to picking the constant c. The choice that we describe has a certain naturality. In fact, without the choice that we shall describe, the product formula of the next section would not hold. Our second task is to give an explicit description of the completion K$ for 1) a finite prime.

Let '8 be an integral K-ideal. By the elementary divisor theorem, the ring OK/% is finite. The number of elements in this ring is called the norm of a, denoted N%.

Proposition 3-1-19: Let $ = JJ Qrn be the decomposition of 1) into a product of prime powers. Then

Proof: Let us consider the mapping

This map is clearly a homomorphism with kernel '8. The Chinese remainder theorem asserts that the map is surjective. Thus,

It suffices to verify that N(qr) = (N'$3)', for all finite K-primes '$3, r a nonnegative integer. In the proof of Theorem 3-1-14, we showed that OK/Qr = OV/$'. Thus,

N(gr) = (O$:$)($:$~) . . . ($,-I : $ r ) . - - Now (o$:$) = N'$3, since Op/$ = OK!!@. Also, 0$/$ * a > 0, via the isomorphism x + $ -+ xna + '$Pfl. Therefore, ($":$"+ l) = N'$3. Finally, N(Pr) = N('$3)'. //

Let 9 be an arbitrary K-ideal. Write % = B-E- ' , with 23 and E integral K-ideals. Let us define N(9) = N(B)N(B)-I. Proposition 3-1-19 and the unique factorization of % into prime ideals guarantees that N(g) does not depend on the particular decomposition 23-E-I chosen. Moreover, we immediately see that

Corollary 3-1-20: Let 9 and 23 be two K-ideals. Then N(%B) = N(%)N(B).

Let us introduce normalized valuations I as follows: If '$3 is a real

prime, '$3 = Q,,, set

I X I$ = I x ( ~ ) 1. If '$3 is a complex prime, '$3 = Q,,, set

IXl$ = I X ( ' ) ~ ~ . If '$3 is a finite prime, set

I x I?, = (N'$3)-ord$(x).

As mentioned previously, the significance of this normalization will be apparent in the next section.

Let us now attack the problem of explicitly describing Kg. If '$3 is infinite, then there is no further analysis necessary-Kv is either R or C. Thus, let '$3 be a finite K-prime. The field 0K/'$3 is called the residue classfield at '$3. The residue class field at '$3 can be computed if one knows only the completion Kg, since 0,/'$3 * o~/$. We express this fact by saying that the residue class field is a local object. Let us fix a local uniformizing parameter n at '$3. Let a2 = (a,) be a system of representatives of the cosets of 0K/'$3 in OK.

Theorem 3-1-21: Let a E Kv. Then a can be written uniquely in the form

Conversely, every such series converges in K. Moreover, if b, # 0, then 1 a =

(Nw-r-

Proof: Let r = orda(a). Then an-' E Us. The image of a r r mod $ is nonzero. Thus, there exists b, $ $ such that an-' - b, E $. Then, a - brnr E $'+I. Applying the same reasoning to a - brnr as we applied to a, we find b,+, such that a - b,nr - h,+,nr+' E Q r + , , and so forth. We arrive at a sequence

But I cn - c, I 2 N'$3-"'" (m,n). Therefore, {c,) is a Cauchy sequence, which converges to an element of Kg since K$ is complete. Since cn E @, the sequence converges to 0. Thus we derive the desired representation of a. The other two assertions are trivial. //

Theorem 3-1-14 refers to the additive structure of X;p From it we can deduce the following multiplicative analogue:

Theorem 3-1-22: K$ is a locally compact group with respect to field multiplication and the subspace topology. I f @ is finite, then Uv is a compact, open subgroup. The subgroups of UQ of the form 1 + $r , r a positive integer, are all compact and open, and form a fundamental system of neighborhoods of 1 in K$. Thus, if$ is finite, KG is totally disconnected.

Summary: In the preceding section, we have introduced the completions of an algebraic number field. We have shown that the completions are all locally compact fields. These fields fall into two essentially distinct types-the Archimedean and the non-Archimedean. In what follows, we shall use various analytic techniques on the completions in order to deduce arithmetic information about the number field. From a topological point of view, the Archimedean completions are very different from the non-Archimedean valuations. For example, the completions at the infinite primes are connected; at the finite primes they are totally disconnected. At the finite primes, the completion contains a compact, open subring, called the ring of '$3-adic integers. No such ring exists in the infinite case, since the completions are connected. In some sense, number theorists do not really understand the role played by the infinite primes in the arithmetic of number fields. But both the finite and infinite primes seem to play an essential role.

EXERCISES 3-1-23: (1) Let I I be a valuation of K, a > 0. Then I la is a valuation of K which is equivalent to I 1.

(2) Suppose that v, and v, are equivalent valuations of K. Then vl(x) < 1 -- vz(x) < 1; vl(x) = 1 -- v,(x) = 1; vl(x) > 1 - v,(x) > 1. [Hint: vl(x) > 1 * > The sequence (xn) converges to 0 in the v,-topology.]

(3) Suppose that V, and v, are equivalent valuations of K. Then there exists a positive constant a such that v = v", [Hint: Choose x E K such that v,(x) > 1.

Then, by (2), v,(x) > 1. Therefore, we may set a = log vl(x)/log v,(x). Then a > 0. Set Ai = log v,(b)/log vi(a), b E KX. It suffices to show that A , = A,. Show that for every rational number r/s 2 Al, we have rls 2 A, by using (2).]

(4) If I I is a non-Archimedean valuation, then if I x 1 f 1 y 1, Ix + y l = maxIlxl,l~ll.

Show that this assertion is not necessarily true if ( x 1 = 1 y 1.

3-2 ADELES

In the preceding section, we introduced the completions of an algebraic number field with respect to its various absolute values. These completions are called localjelds, and the study of their arithmetic is called local number theory. Later in this chapter, we shall make an extensive study of local fields for their own sake. However, in the present section, we shall confine ourselves to global number theory, that is, the study of the arithmetic of K and 0,. In modern number theory, one tries to obtain global facts by studying a collection of corresponding local data. In the following section, we shall give a general procedure for "pasting together" local data into a global object. In particular, we shall show how to paste together all of the completions of an algebraic number field so that we still retain the arithmetic information possessed by the local fields but still have a locally compact object on which we can perform analysis.

Let the symbol '$3 run over a certain index set that we fix throughout the discussion. For each '$3 suppose that there is given a locally compact group Gq. Suppose that for all '$3 not in some finite set So there exists a compact, open subgroup Hrp of Grp. Let us define a locally compact group called the restricted direct product of the family {Gq) with respect to the family {Hrp}, denoted 9 (Gu:Hrp). Algebraically, the restricted direct product is the group

of all (av) E Grp such that arp E Hrp for all but a finite number of '$3. If we

were to topologize the restricted direct product as a subspace of the direct product, then the resulting group would not, in general, be locally compact. Thus, we must topologize the restricted direct product with some care.

Let S be any finite subset of '$3, including all '$3 for which Hq is not defined. Let us topologize 9 (Grp: Hrp) so that gS GO x #Js Hq = GS is an

open subgroup for all S considered. The weakest such topology makes the restricted direct product into a topological group (trivial verification). In this topology, each neighborhood of the identity contains a neighborhood of the identity of the form

where Nu c Gv is a neighborhood of the identity. (Use the fact that Hv is

open.) Moreover, with respect to this topology, the restricted direct product is a locally compact group. For let Scontain all '$3 for which Hrp is not defined. Then let Nu be a compact neighborhood of the identity in Grp, '$3 E S. A compact neighborhood of the identity in the restricted direct product is given by

WNv "8 Hrp. (Hrp is compact.)

Convention: We shall use the phrase "almost all" to mean "for all but a finite number of." Thus, we shall speak of the group Grp having a certain property for almost all '$3 to mean that Grp has the property for all but a finite number of '$3. This convention is similar to the "almost everywhere" convention of measure theory.

EXERCZSE 3-2-1: Let Grp be a locally compact topological ring for all 9 (i.e., a ring with a topology with respect to which addition and multiplication are continuous). For almost all 13 let there exist an open, compact subring Hrp. Then ll (Grp : Hrp) is a locally compact topological ring. v

EXAMPLES: (1) Let 13 run over all primes of an algebraic number field K. For '$ $ S,, 09 is an open, compact subring of Krp. The ring ll (Krp: Oq) is called the ring of adeles of K, denoted A,. rp

(2) Again, let fp run over all primes of the algebraic number field K. For8 $ S,, Urp is a compact, open subgroup of KG. The group ll (KG : Urp) is called the idele group of K, and is denoted Q,. rp

(3) Let S 3 S, be a finite set of K-primes. Then the group

is an open subgroup of A,, which we shall denote A;. Similarly, the open subgroup

of DK will be denoted &. The ring (respectively, group) A$ (respectively, 1);) will be called the ring of S-adeles (respectively, the group of S-ideles).

EXERCZSE 3-2-2: Algebraically, the ideles of K can be considered as a subset of the adeles under the obvious injection. When so considered, the ideles are precisely the units of the ring of adeles. The topology on 9, as a restricted direct product does not coincide with the induced topology on g, considered as a subspace of A,.

The arithmetic properties of K are intimately connected with the topological properties of the adeles and ideles. Therefore, we shall begin to study these topological properties. Let x E K. Then x E Krp for all '$3, and x E 0rp for almost all '$3. Thus, we may embed K i n AK via the diagonal map

X-(X,X,X ,..., X) E A,.

Theorem 3-2-3: K is a discrete subring of A,.

Proof: Let c > 0. Let N9(c) be the neighborhood of 0 in Ks defined by NB(c) = {x E Kg I I x ly < c]. Then

is an open neighborhood of 0 in A,. Moreover, if x E N(c) n K, then vlp(x) 2 0 for all finite $. Thus,

N(C) n K c 9,. (*I Let us show that OK, considered as a subspace of A,, is discrete. This will suffice to show that K is discrete in A,. For (*) would imply that there exists a neighborhood N of 0 is A, such that N n K = 103. Let {el, . . . ,On] be an integral basis for K. Let us consider the vector space

of dimension r, + 2r, = n over R. We can view R,,, as a closed subgroup of A, and we embed Kin R,,, via the map

We claim that the images of 8,, . . . ,8 , are a basis of R,,, over R. Indeed, the image of 8, can be represented by the vector

By elementary row operations on a matrix, we see that the determinant of this collection of n vectors is just det (8,")). But this is just the discriminant of the basis {8,, . . . ,On) and is therefore nonzero. Thus, the images of 8,, . . . ,On are linearly independent and therefore form a basis of R,,,. The Z-module generated by the images of d l , . . . On, is a discrete subgroup of R,,,. But this Z-module is precisely the image ofO, in R,,,. Thus, there exists a neighborhood Nq(c) of 0 in Kg such that

Then 9, n N(c) = (03. /I

Theorem 3-2-4: A,/K is compact.

In order to prove this important result, we require the following preliminary result, which will be useful in its own right.

Proposition 3-2-5: Let S be a finite set of K-primes containing S,. Then A$ + K = A,. Proof: Let a = (ao) E A,, T = {$ I a(p + Oa). Then T is finite. By the approximation theorem, there exists y E K such that vg(y - a%) 2 0 for

'$ E T, vdy) 2 0 for $ $ T U S,. If '$ E T, then y - a$ E Op by the choice of y. If $ + T U S,, the y - a$ E 0s since vs(y - as) 2 min (vv(y), vdao)) 2 0. Thus, y - (as) E A; a a E A$ + K. //

Proof of Theorem 3-2-4: It suffices to show that there exists a compact subset D of A: ( S any finite set of K-primes containing S,) such that

D + O , = A;. For then, by Proposition 3-2-5,

D + K = A ; + K = A , ,

and A,/K is the image of D under the continuous map

A, + A,/K.

As in the proof of Theorem 2-4-3, set R,,, = JJ Kg. Recall that we embedded qts,

K in R,,, via the map

We proved that the image of an integral basis ( a , , . . . , a ,} in R,,, is a basis of R,,,, considered as a vector space over R. For simplicity in notation, let us identify 0, and its image. Then every element a of R,,, can be written uniquely in the form

Thus, let R, be the set of elements of R,,, which are of the form

Since R, is compact, so is

Moreover, since R, + 0, = R,,,, we see that

D + O , = A i . // Largely from the proof of Theorem 3-2-4, we deduce

Proposition 3-2-6: Every adele can be expressed uniquely in theform k + s, where k E K, s E Aim, and where the infinite component of s is of the form

Proof: The proof of Theorem 3-2-4 showed that every adele could be so written. If k, + s, = k, + s,, then k, - k, = s, - s, a s, - s, E K.

36 / THE LOCALLY COMPACT GROUPS OF NUMBER THEORY CHAP. 3

Since s, - s, E A;-, we see that s , - s, E OK. But by the condition on the infinite component, this implies that s, = s,. Thus, k, = k,. //

The next result provides the motivation for the choice of the normalized valuations defined in the preceding paragraph.

Theorem 3-2-7 (Product Formula): Let x E K ". Then

n I ~ I , = 1, 13

where the product is over all primes (jinite and infinite) of K.

Before proceeding with the proof of this theorem, let us review some of the basic facts about Haar measure. If G is a locally compact group, then G possesses a nonzero measure p such that

(1) All continuous complex-valued functions with compact support are p-integrable.

(2) p is invariant under left translations; that is,

for all p-integrable functions f. This measure is unique up to a positive constant multiple. Such a measure is called a left Haar measure on G. If p denotes a left Haar measure on G, then for any p-measurable set A G G, x E G, we have

p(A) = p(xA).

Every Bore1 set of G is p-measurable. Moreover, if A G G is open, then p(A) > 0, whereas if A is compact, then p(A) < co.

Let G be a locally compact group, p a left Haar measure, and a an (algebraic and topological) isomorphism of G. If A c G is p-measurable, then aA is also. We may define a measure p' on G by

&(A) = p(oA), A p-measurable.

It is trivial to verify that p' is another left Haar measure which must differ from p by a positive constant factor. We shall call this positive factor the modulus of the automorphism a , denoted mod (a) [or mod, (a)]. Thus, the modulus is characterized by the equation

p' = mod (a)p.

It is clear that the modulus is independent of the original choice of Haar measure. If G is compact, then G is p-measurable and a(G) = G for every automorphism a. Thus, if G is compact mod (a) = 1 for every automorphism a . If G is discrete, then {e} is p-measurable, where e = the identity of G. Also, a(e) = e for every automorphism a. Thus, if G is discrete, then mod (a) = 1 for every automorphism a .

Suppose that H c G is a closed, normal subgroup and a is an automorphism such that a(H) = H. Then a, = a 1, is an automorphism of H. Also, a, = the map induced on GIH by a is an automorphism of G/H. A simple application of Fubini's theorem shows that

Proposition 3-2-8: mod, (a) = mod, (a,) mod,, (a,).

Proof of Theorem 3-2-7: Let x E K x , ox: A, --+ A, the automorphism defined by ax(u) = xu. Set H = K, G = A, in Proposition 3-2-8. It is clear that ax(H) = H, so that by the Proposition,

~ o ~ A ~ ( Q , ) = mod, ( a , I ) ~ o ~ A ~ / K (ax, 2 ) .

But K is a discrete subgroup of A, and A,/K is compact. Therefore, by the above discussion, the two factors on the right-hand side are 1. Thus, modAK(ax) = 1. Claim that

= n l x lw 13

This will suffice to complete the proof. Let S = the set of K-primes consisting of the infinite primes and all finite primes such that x $ U$. Then S is finite. Let H = g S K g x&OS. Then H is an open subgroup of A,. Then A,/H is discrete. Thus, by Proposition 3-2-8,

(ax) = mod, (ax, I ) .

Let HI = n K,, Hz a: = the restriction of a , , to H, (i = 1,2). 13ts

Then, directly from the definition of the modulus, we verify that

But Hz is compact, so the second factor on the right is 1. Thus,

where a, = the restriction of a', to K$. But as(y) = xy(y E K,). Let us now claim that mod,, (a,) = Ix 113. This will suffice to complete the proof, since for $ S, 1 x Is = 1. Without loss of generality, we may assume that I x 1, < 1. For if I x 1, > '1, then we may replace x by x-I and a, by oil. Then from the assertion for I x 1, < I , we see that

mod, (a,)-' = mod,, (oi l ) 3

= I X - ~ ~ , = Ixlil .

Thus, assume that 1x113 2 1. Since a, = multiplication by x,

where ,ug is a Haar measure on Kg. Now x8g = $", s = ords (x). We have proved that o~/$" is a finite ring with N(vs) = (N'$)" elements. Let

be a coset decomposition. Then

P $ ( ~ Q ) = 2 xi + QS) , = I

= r,u%(m

by the invariance of the Haar measure. Thus,

,u&" = = - 1,ud0?3) = (N'$)-ord~(X)p$(Orp).

This last equation, the remark that lxlv = (N'$)-o'd$(x), and equation (*) suffice to prove the claim. //

Using the same reasoning used at the end of the preceding proof, we can prove

Proposition 3-2-9: Let '$ be a K-prime, x E KG, a, = the automorphism of K defined by a X b ) = xy. Then modKg (a,) = 1 x 1%.

It will be convenient to introduce the language of congruences for adeles. Let a E A,, '532 = '$'% a K-modulus. We write

v a = 0 (mod 5332)

to mean that ag r 0 (mod '$%) for all K-primes 9. (We write au = 0 (mod qO), '$ finite, to mean that a% E 013.) As usual, we write a =- /3 (mod '532) to mean that a - /3 - 0 (mod '532). From Exercise 3-1-16, we immediately deduce

Proposition 3-2-10 (Additive Approximation Theorem): Let 9Jl be a K- modulus, a E A,. Then there exists x E K such that

x r a(mod 5332).

Tn the following discussion, we shall continue the program begun in the preceding section. Our main goal is to prove analogues of Theorems 3-2-3 and 3-2-4. As in the preceding section, the topological statements are equivalent to certain arithmetic facts. In this case, the analogue of Theorem 3-2-4 will be used to prove the finiteness of the class number and the Dirichlet unit theorem.

Theorem 3-3-1 : K x is a discrete subgroup of 4,.

Proof: Since K is a discrete subgroup of A,, there exists a neighborhood N of 1 E AK such that N n K = (I). Without loss of generality, we may assume that N is of the form

N13 x II 013, =g 139s

where Nv is an open neighborhood of 1 in Kg and S is a finite set of K-primes containing S,. Since U$ is a compact, open subset of Kg, Ug n Ng and U% n 013 = Ug are open subsets of KG so that

=g (N. n U d X gsu13 is a neighborhood of I E 4, and N' n K = (1). //

It is not quite true that 4,/KX is compact. However, we shall be able completely to describe the structure of this group. It will turn out that 4,/KX is the direct product of a compact group and R,. The compact group introduced will play an essential role in our study of Hecke's zeta functions.

If a = (av) E Q,, define

1.1 = IJ Iagb.

Since a E Ug for all but a finite number of 9, we see that I a9 = 1 for all but a finite number of '$. Thus, I a I is well defined.

Proposition 3-3-2: The mapping I I : JK --+ R+ is continuous.

Proof: Define the two subgroups

J, = KG gES-

J2 = II (KG: Ug), g 4 s m

considered as closed subgroups of 4, in the obvious way. Let qi (i = 1,2) be the restriction of I I to J,. It suffices to show that qi is continuous, since 'OK = J1 x J2 , 1 I = qlq2. But the mappings

x - 1x19 KG - R+

are continuous, so that ql is continuous. The homomorphism q2 contains the open subgroup Uq in its kernel, and is therefore continuous. //

139s-

Definition 3-3-3: 1 a ( is called the volume of the idele a.

Let a E 4,. Then a defines an associated automorphism & of A, via a(p) = US.

Proposition 3-3-4: modAK (6) = I a 1. Proof: Let G be the open subgroup of A, defined by

Kn x II 0x3, =a n$s

where S = S.. u {!$ 1 1 a% In # 1). Since G is open, AK/G is discrete, SO that by Proposition 3-2-8, and a remark from the preceding section,

mod,, (6) = mod,(&)

= gs (& 14. But & 1 (x) = a ~ x (X t Ka). Thus, mod, (6 = mod (04 = 1 an In, by

3 Propos~t~on 3-2-9. Thus,

mod,. (a) = II l an In = IJ l an In. I1 n c s

Let 9; = {a E JK ( 1 a ( = 1). Then we have the exact sequence

(1) + Jk + JK + R + + (1). (*>

By the product formula, K x c J;. We shall prove

Theorem 3-3-5: Jk/Kx is compact.

Once we have proved this theorem, we shall have established the desired decomposition of J,/K. For we can map R + into a, by the mapping

j : t + (tl'", . . . , tl'", 1, 1, . . .). -- components components

Then j(t) = I trlln+2rz'n I = t ( t E R+). (Recall that we normalized the absolute value at complex primes to be the square of the usual absolute value.) Thus, the composition of j followed by the volume mapping gives the identity on R,. Thus, the exact sequence (*) splits and

9, = J:, x R,. Since K x c Jk,

.JK/Kx = 4:/Kx x R,.

Before proving Theorem 3-3-5, we shall demonstrate a modern version of Minkowski's theorem on the existence of lattice points in convex regions of Euclidean space.

Theorem 3-3-6 (Minkowski-Chevalley-Weil): There exists 6 > 0 such that q E JK, I q I > 6 imply that there exists x E K x such that I xg 1% ( I pv 1%

for all primes q. Prooft Let N% = {x E Kql lx lv( l ) , M = %ES- Nv x nzs- JJ 0% E A,. Our

assertion is equivalent to the statement: lq 1 > 6 implies that there exists

x E K x such that xq-I E M - q M n K # {O). We will show that if q M n K = {0), then jq) _( 6 for some suitably chosen constant 6, depending only on K. Since M is a compact neighborhood of 0 in A,, there exists a compact neighborhood V of 0 such that V - V c M. Let v' be a Haar measure on the compact group A,/K such that vf(AK/K) = 1 ; let v" be a Haar measure on the discrete group K such that each singleton has measure I . By Fubini's theorem, there exists a Haar measure v on A, such that

Let q :AK -+ A,/K be the canonical map. Then y, is 1 - 1 on qV. For if q(q(v, - v,)) = 0, then q(v, - v,) E K n q M = {0} * v, = v,. Let f denote the characteristic function of qV, f the characteristic function of y,(qV). For fixed a t A,, there exists at most one x E K such that a + x E

q V. Therefore,

Thus, if qM n K = {O), then v(qV) 5 1. By Proposition 3-3-4, v(qV)=Iqlv(V), so that I q l l l/v(V). Thus, if l q l > l/v(V), then tlM n K f (0). I/

Remark: We can eKectively compute 6 by choosing a particular V and computing v(V).

Proof of Theorem 3-3-5: Let 6 > 0 be as in Theorem 3-3-6. Let to E J, be such that I lo I > 6, and let M be as in the proof of Theorem 3-3-6. First observe that r i M - M is compact, since it is the continuous image of a compact set in A, x A,. Since K x is discrete in J,, K x n r i M . M is discrete and compact and hence finite, say equal to {y,, . . . , y,}. Let a E Jk. Then Ia-'r, I > 6. By Theorem 3-3-6, a - ' t ,M n K # {O). Let x be contained in this intersection. Then xa{,M c t i M - M . By the product formula, I xu{, I > 6, so that by Theorem 3-3-6, there exists y E K x n xar,M. But then y is one of

they,, say y = y, . Therefore, (ax)-, E y;'(;,M. Let C = C,M U ( & y ; ' r , ~ ) ,

C' = {I t JK1 q or q-I E C ) . We have shown that if a E 11:,%ere exists x E K such that a - lx- ' E C a E C'K. But C and hence C' are compact. Therefore, since Jk/K is the continuous image of C' under the canonical projection, the assertion is established. //

Let us immediately apply Theorem 3-3-5 to the problem of showing the finiteness of the class number of K. Let S be a finite set of K-primes containing S,. Let I, [respectively I,(S)] denote the group of all K-ideals (respectively, the group of all K-ideals prime to the elements of S). Let PK denote the group of all principal K-ideals; let P,(S) - I,(S) n P,. It is clear that I, = I,(&), P K = P,(S,).

Definition 3-3-7: The order of the group IK(S)/PK(S) is called the S-class number of K. If S = S,, then the S-class number will be referred to as the class number of K, denoted h,.

The class number of K is an important arithmetic invariant of K whose computation has been the subject of many investigations. In the most elementary context, the class number enters. For instance,

Proposition 3-3-8: h, = 1 - OK is a unique factorization domain.

Proof: Exercise. (Hint: Use the decomposition of an ideal into a product of prime ideals.)

Theorem 3-39: The S-class number of K isjnite.

Before proceeding with the proof of this theorem, let us introduce some notions that will be useful in the later parts of this book. Let a E J,. We can associate to a an ideal is(a) belonging to I,(S) by the formula

ida) = Q Prd@"@, a = (ap).

- the The mapping is is clearly a surjective homomorphism with kernel fl, - group of S-ideles. When S = S,, we shall denote the mapping is simply by i, in which case we shall refer to i as the ideal mapping.

Proof of Theorem 3-3-9: By the above discussion,

J,/& 2 I,(S).

The inverse image of P,(S) under this isomorphism is Kxfl;/fl;. Thus,

Now 1); is an open neighborhood of 1 in 9,. Thus, Kxfl; is open 3 f l , / ~ ~ f l ; is discrete. On the other hand, it is clear that fl, = Jk.fl;C- =

f l i -K "fli. Thus, by one of the isomorphism theorems of group theory,

B,/K "fl; 2 flk/[Kx .(DL n fl;)]. The group on the right is the continuous image of fli/Kx under the canonical

(This map is continuous, since 1; n flk is open in 9:). Thus, by Theorem 3-3-5, fl:/[Kx-(fl; n Jk)] is compact. Therefore, JIK/Kxfl;C is both compact and discrete, and hence finite. /I

Next, we shall explore the structure of the group of units UK of OK. It will be useful to explore the structure of a more general group-the group of S-units. As above, let S be a finite set of K-primes containing S,.

Definition 3-3-10: An S-unit is an element x of K such that ordy (x) = 0 for all '$3 $ S.

If S = S,, then an S-unit is just a unit of OK. Let us denote the group of S-units of K by U,(S). Let WK denote the group of all units of OK that have finite order. It is clear that WK is precisely the group of all roots of unity in K, and that WK c UK(S) for all S.

Theorem 3-3-11 (Dirichlet-Minkowski-Hasse-Chevalley): Let S contain S elements. Then

(1) WK is a jn i t e group. (2) VK(S) = U,(S)/ WK is a free abelian group of rank s - 1. (3) UK(S) = WK x VK(S).

Proof: From the above remarks, we conclude that W, is precisely the subgroup of UK(S) consisting of all elements of finite order. Moreover, by (1) and (2), UK(S) is finitely generated. Thus, we can conclude (3) by the fundamental theorem of finitely generated abelian groups. Thus, it suffices to prove (1) and (2).

For all K-primes '$3, set Uy = {x E Ky / I x 1% = 1). Define two subgroups G and Go of J; by

Then

G n K x = U,(S)

Moreover, it is clear that G is an open subgroup and Go is compact. There- fore, since K x is discrete, Go n K x is both discrete and compact, hence finite 3 (1).

Now

First observe that G n K x is discrete, Go n K x is finite 3

is discrete => Go(G n Kx)/Go is discrete. Next, observe that since G is an open subgroup of J:, G is closed +- GKx/Kx = the image of G under the canonical map 4; - .lJ:/Kx is compact by Theorem 3-3-5. Thus, GKx/GoKx is compact, being the continuous image of GKx/Kx. Thus,

is compact. Thus, we have shown that Go(G n Kx)/Go is a discrete group, whose quotient with GIGo is compact.

Let us now study GIG,. Let us remove one prime from S, say an infinite prime for the sake of definiteness. Denote the resulting set of primes by S,. Let us construct a map

[It does not matter which prime is omitted from S, since (xa) E D: +- Cg log ( x s = 0 => log (xcp Is for 3 the missing prime can be computed from knowing the other terms.] The image of G is just [R at every infinite prime and an infinite, cyclic group at all finite primes in s,. Thus, Rs-'/im (G) is compact. The kernel of the mapping is Go. Thus, GIGo can be viewed as a subgroup of R r l with compact quotient. Similarly, Go(G n Kx)/Go can be viewed as a subgroup of GIGo with compact quotient. Thus, Go(G n Kx)/Go =r VK(S) can be viewed as a discrete subgroup of [ R S - I with compact quotient. (For if A 3 B 3 C are a sequence of topological groups, B a closed subgroup of A, C a closed subgroup of B, then A/B compact and B/C compact imply that A/C is compact. Proof: Exercise.) The proof will be complete if we prove the following

Lemma: Let r be a discrete subgroup of Rm such that [Rm/r is compact. Then l? is a free abelian group of rank m.

Proof: Let E = the R-vector space generated by r. Then we have the exact sequence of continuous maps

Thus, Rm/E is compact, so that E = Rm. Let (x,, . . . , x,} c r be a basis of Em. Set r, = the free Z-module generated by x,, . . . , x,. Then r, c and the group r/r, is a closed subgroup of the compact group Rm/rl. Thus, r/r, is compact. Since r, is open in r ( r is discrete), r/r, is discrete. Thus, r/r, is finite, being both compact and discrete. Therefore, r is finitely generated, since r, and r/r, are. By the fundamental theorem of abelian groups, r is a free abelian group of rank r, say. Since Rm/T is compact, r 2 m. Assume that r > m. Let (el, . . . , e,) be a set of free Z-generators of r and let there be an R-linear dependence of the form

Let E > 0 be given. Then there exists an integer N such that Nu,, . . . , Nu, are all within E of an integer. Then Ne, is the sum of a Z-linear combination of e,, . . . , e, and an R-linear combination of the same whose coefficients are 2 E in absolute value. By making E small enough, the discreteness of r will imply that the latter linear combination is zero, so that

which is a contradiction to Z-independence of the e,'s. //

EXAMPLES: (1) Let K = Q ( o ) , D < 0. Then r, = 0, r2 = 1. (The only possible embedding of K into a: is not contained in [R and is determined up to complex conjugation.) Thus, Theorem 3-3-1 1 implies that UK(S,) = the group of units of the ring of integers is equal to WK, since the free part is of rank r, + r , - 1 = 0.

(2) Let K = Q ( n ) , D > 0. Then r, = 2, r2 = 0. Thus, Theorem 3-3-11 implies that UK(S,) is the direct product of {f 1) and an infinite cyclic group. This

cyclic group possesses a unique generator which is > 1. This generator is called the fundamental unit of the real quadratic field K.

EXERCISES 3-3-12: (1) Show that in the case of an imaginary quadratic field, UK(S,) = {f 1, h i ) (D = -I), {f 1, & a , f a 2 } (D = -3, a = a primitive cube root of I), {*I} (D f -1, -3).

(2) Determine the fundamental units of Q(&), Q(&). [Hint for (I) and (2): x is a unit of the ring of integers implies that x&x) = f l , where 0 is the nontrivial automorphism of K. Thus, if x = (a + bfi) /2 , then a and b satisfy a2 - b2D = f 4. If x = a + 6 0 , then a and b satisfy a2 - b2D = f 1.1

(3) By Proposition 3-3-8, the class number of Q ( 6 5 ) is # I . Show that it equals 2.

(4) Show that for S sufficiently large, the S-class number is 1 and that J;KX = a,.

(5) (Multiplicative Approximation Theorem) Let 32,, . . . ,!$, be distinct K- primes, a , , . . . , a, nonnegative integers, xi E KG, (1 5 i 5 r). Then there exists x E K such that

xx,: z l(mod 3;1) (1 5 i ( r). [Hint: Use the additive version of the approximation theorem.]

(6) The x that is chosen in (5) cannot always be chosen so that x E Us for '$3 $ {PI, . . . , Qr} u S-. [Hint: There exist fields with class number greater than 1.1

EXERCISE 3-3-13: Let v be the measure on A, which was constructed in the proof of Theorem 3-3-6, and let N be a compact neighborhood of 0 in A,. If q E QK is such that I q I > l/v(N), then q(N - N ) n K # 10). [Hint: Mimic the proof of Theorem 3-3-6.1

Local fields

We have seen how the arithmetic properties of an algebraic number field K are reflected in the properties of the adele ring A, and the idele group JK. These latter two structures are constructed from the completions of K at its various primes. In this chapter, we shall make a detailed study of these completions for their own sakes. In the next chapter, the arithmetic theory of these completions will be applied to yield additional facts about algebraic number fields.

4-1 THE TOPOLOGY OF LOCAL FIELDS

Definition 4-1-1: A localjield is a finite, algebraic extension of @, for some @-prime p.

In the next subsection, we shall show that the completion of an algebraic number field at a prime is a local field. If this result is true, then we might sus- pect that a local field has a valuation, as do the completions of algebraic number fields. This assertion is true, and, in fact, we shall show that the valuation on a local field is unique up to equivalence. Our machinery, which will rely mainly on Hensel's Lemma, will allow us explicitly to write down a formula for the (almost unique) valuation on a local field. If K 2 @, is a local field, and if there exists a valuation I 1 on K which induces 1 1, on @,, then K becomes, in an obvious way, a topological vector space over the topological field @,-that is, K is a vector space over @,, on which there is defined a topology with respect to which addition and scalar multiplication are continuous operations. Therefore, let us begin by studying such objects.

A typical example of a topological vector space over @, is @;, given the product topology. Let V be an arbitrary topological vector space over @,, and let (vl, v,, . . . , v,} be a set of linearly independent vectors from V. Let us define the vector space homomorphism

\ f:@:,--. V

n

(al, . .. , a n ) - + Z a,vi i= 1

of @; onto f(@;).

1 Theorem 4-1-2: (1) f is an algebraic and topological isomorphism of @",

(2) f (@;) is closed in V.

Proofi Since V is a topological vector space over @,, it is clear that f is continuous. In order to prove that (1) holds, it therefore suffices to show that f is open. In turn, in order to prove that f is open, it suffices to show that f is open at (0, . . . , 0 ) E @;. Let A denote the unit cube in @;, i.e.,

t

A = {(a , an ) E @;I lailp< 1 , i z 1 ,..., n}. To prove that f is open at (0, . . . , 0), it suffices to show that f(A) is a neighborhood of 0 in f(@",. For then, if N is a neighborhood of (0, . . . ,0) in @;, there exists a E @, such that aA c N, and aA is a neighborhood of (0, . . . , 0). Thus,

and af(A) is a neighborhood of the origin in f(@;), since multiplication by a is a homeomorphism. Let B denote the boundary of A. Then

B = ( ( a l ,..., a n ) € Q", I a , \ , _ < l ( i = l , . . . , n); I a , l p = l forsomej} I B is a compact subset of @;, and 0 $ B. Therefore, f(B) is a compact subset

off(@;), and 0 $ f(B), since f is 1 - 1. Thus, f(B) is closed in f(@;); let C be a neighborhood of 0 in f ( Q 3 such that C n f(B) = 0. Since scalar multiplication in V is continuous, there exists 6 > 0 such that

(a E @,I la1 < 4 - C cf(@", -flf(B). (*) Let a, E @, be chosen so that I a, 1 , < 6. We claim that

Since Cis a neighborhood of 0 in f(@;), this will imply that f(A) is a neighborhood of 0, which will suffice to prove (I). Let v G a,C, v # 0. Then

with not all ai = 0. Suppose that

Then ai 'v E f(B), so that a i1a0C n f(B) # 0. Therefore, by (*), I ai 'a, 1, 2 6. Thus,

6 1 I a i l a o I, < 6 1 a i l I,,

by the choice of a,. Therefore,

I aio I p < 1

lal lp < l , . . . , l a n l p < 1. n

To summarize, we have shown that if v = C aiv, is contained in a,C, v # 0,

then 1 a, 1, < 1 for all i. Therefore, v t f(d)?his last statement is clearly true if v = 0, so we have shown that

a0C c f ( 4 , and (1) is proven.

To prove (2), let v E V - f(@;). We shall show that v is not contained in the closure off(@", in V. Embed @; in @",I via

(a,,. . . , an) + (a,,. . . , an, 0).

Then @; is a closed subspace of @",I, and (0, . . . ,0 , 1) 4 @;. Therefore, there exists a neighborhood U of (0, . . . , 0 , 1) in @",I such that

U n @;= 0.

Let f ' : @",I - V be the map associated with the linearly independent set (v,, . . . , v,,, v]. Since f ' is a homeomorphism of @;+I onto f '(@I), f '(U) is open, and f'(@;) is closed in f1(@;+:"); moreover,

f '(U> n f '(@3 = 0.

But f I(@", = f(@",. Also, there exists an open subset T of V such that

T 17 f '(@", I ) = f '(U). Therefore, T n f(@;) = 0 and T is a neighborhood of v E V, so that v is not in the closure off(@,) in V. //

Corollary 4-1-3: Let V be a finite-dimensional vector space over @,. Then there is a unique topology with respect to which V is a topological vector space over @,.

Proof: Choose a basis of V. Then the mappingf, constructed with respect to this basis, allows one to construct the topology on V from the topology of @", /I

Corollary 4-1-4: Let V be a finite-dimensional topological vector space over @,. Then the topology is metric and complete.

Proof: By Theorem 4-1-2 V is homeomorphic to @;, and the topology on @",is metric and complete. //

Corollary 41-5: Let L 3 @, be a localjield. Then there exists at most one valuation on L that coincides with I 1, on @,.

Proof: If 1 I is such a valuation, then the topology it induces turns L into a topological vector space over @,. Since the topology on such a topological vector space is unique, I I is determined up to equivalence. The set of all valuations equivalent to I ( is (1 la 1 a > 0). There is precisely one choice of a for which I la = I 1, on @,. Therefore, there is at most one valuation on L that coincides with I 1, on @,. //

In order to write down a formula for the unique valuation on L that coincides with I I, on a,, we require Hensel's Lemma. For the moment, we shall state a particular case of this important result. The proof we give will generalize immediately, once we know that a valuation exists on every local field. Let p be a finite @-prime, f E Z,[X]. Let J' E @,[XI denote the polynomial obtained from f by replacing each of its coefficients by its respective image under the canonical map h, -+ @, = Z,/pZ,. We say that f is primitive if some coefficient o f f is a p-adic unit, that is, is contained in Up. Iff =

a, Xn + . . . + a,, set 1 1 f 1 1 = max (1 a, I,). Let Zp,,[X] denote the finite- dimensional vector space of polynomials in Z,[X] of degree at most n. With respect to the norm 1 1 (I, hp,,[X] is a topological vector space over @,-that is, Z,,[X] becomes a topological vector space when given the topology derived from the metric associated with the norm 1 1 1 1 . It is clear that Corollary 4-1-4 implies that Z,,,[X] is complete with respect to the metric associated to I I 1 1 . Note that f is primitive u 1 1 f 1 1 = 1.

Theorem 4-1-6 (Hensel's Lemma): Let f E h,,[X] be primitive. Suppose that

f = gh, g, h E @,[XI, where g and h are relativelyprime. Then there exist G and H in Z,[X] such that

(1) f = GH. (2) G = g, I7 = h. (3) deg (G) = deg (g).

Proof: We first claim that we can construct two sequences (G,) and (H,) of elements of Z,[X] which satisfy the conditions

(1) G" - G,,, E B"+'[X], H, - H,,, E B"+l[X]. (2) G,H, - f E jj"+'[X]. (3) deg (Gn) = deg (g), deg (Hn) s deg ( f - deg (g). (4) (7, = g, I7" = h.

Choose Go and H, so that (3) and (4) hold. Then (1) holds vacuously for n = - 1, and (2) holds since f = gh. Thus, let us construct the two sequences inductively, using Go and H, as the initial terms. Let us tentatively set

where a is a local uniformizing parameter at p and K, and L, belong to Zp[X]. Let us choose K, and L, so that (1)-(4) are satisfied: By the definition of G,,, and H,,,, (1) holds. Also,

G,+, = (GI - Go) + (G, - GI) +, . . + (Gn+1 - Gn) + GO.

Thus, by the inductive hypothesis and the choice of G,, G,+, = g. Similarly, = h. Thus, (4) is satisfied. Since f - G,Hn E jj"+'[X], f - GnHn =

nn+ M,, M, E Zp[X]. NOW

f - G,+,H,+, = f - GnHn + an+l(GnLn + KnHn) + n2n+2KnLn = an+ l(Mn + G,L, + K,H,,) + a2n+2KnL,.

Thus, a necessary and sufficient condition that (2) hold is that the coefficients of the quantity in parentheses are all divisible by n. By (4), this is equivalent to

a" + gen + hif" = 0. (*)

Since g and h are relatively prime, we can find Kn and en satisfying (*). More- over, I?, is determined up to addition of a multiple of g. Thus, by the Euclidean algorithm in @,[XI, we see that we can choose I?, so that deg (En) < deg(g). By the induction hypothesis, deg (M,) ( deg (f ). Thus, by (*), deg (en) ( deg (f) - deg (g). Now choose K, and L, so that deg (K,) < deg (g), deg(L,) deg ( f ) - deg (g), and so that the images of K, and L, mod p I?, and e n , respectively. Then G,+, and H,+, satisfy (1) through (4). This completes the construction of the two sequences. Let n = deg (f). Then all elements of the sequences lie in ZP,,[X]. Moreover, (1) shows that both sequences are Cauchy sequences with respect to the 1 1 1 1 metric. Thus, by completeness, there exist G and H in Z,[X] such that

lim G, = G, lim H, = H. n-== n-m

BY (21, f = GH. BY (31, deg (GI = deg (g), deg (HI I deg ( f - deg (g). Finally, by (4), G = g, = h. //

Corollary 4-1-7: Let f = Xn + alXn-I + . . + an E @,[XI be irreducible. If a, E Z,. then all a, E Z,. Proof: Let a be a local uniformizing parameter at p. Assume that some a, $ Z,. By mulitplying f by a suitable power of a, we derive a polynomial f ' which is primitive and whose leading coefficient is a positive power of a. Then f' = f' 1 is a factorization of f', where deg (f') < deg (f). By Hensel's Lemma, there exist polynomials G and H such that deg (G) =

deg (f'), and such that f ' = GH. This leads to a proper factorization of S, which contradicts the irreducibility off. //

Theorem 4-1-8: Let L be a localfield, of degree n over @,. Then there exists a valuation I 1, on L which coincides with 1 1, on Qp. This valuation is unique and is given by

Proof: The uniqueness of I 1, was proved in Corollary 4-1-5. Suppose the I 1, is defined by the given formula. Then IxlL = 0 - x = 0, Also I 1, is multiplicative, since the function NLIQp is multiplicative. Thus, it suffices to show that I x + y 1, ( max (1 x I,, 1 y I,), which will also prove that I 1, is a non-Archimedean valuation. It suffices to show that 1 x 1, L, 1 => I 1 + x ILL 1.

I I For then, if x # 0, I x I , > 1 y I,, then

Ix +yI,= I x I L I ~ +yx-'I, I l x It = max Il x lL7 I Y 13.

Therefore, assume that I x L ( 1. Without loss of generality, assume that

I x # 0. Let f be the minimal polynomial of x over @,, f = Xm + a, Xm-l + - . - + a,. Then a i m = 5 NLrQD(x). Since I x IL ( 1, we see that 1 a, 1, ( 1. Since f is irreducible, Corollary 4-1-7 implies that all a, E Z,. Let g(X) =

f(X - 1). Then g is the minimal polynomial of x + 1 over @,. Moreover, f E Zp[X] => g E Zp[X]. In particular, I NLrQ,(l + x) I,( 1 * I 1 + x 1, I 1. Thus, I 1, is a valuation. If x E @,, then

Corollary 4-1-9: Let R be an algebraic closure of @,. Then there exists a unique valuation I I on R such that for every localfield K, the restriction of I 1 to K coincides with 1 I,.

1 Proofi The uniqueness is clear by Theorem 4-1-8. To prove existence, it 1 suffices to show that if x E L, then

I X IL = I X

For in this case, we may define i

1 X I = 1 IQAx).

Equation (*) guarantees that the restriction of I I to L coincides with I 1,. I t is also clear that I I defines a valuation on R. To prove (*), note that

I I Corollary 4-1-10: Let a be an aufomorphisrn of R that leaves gP pointwise

jixed. Then

1x1 = la(x>l. In particular, if x E L,

Proofi The formula

1x1' = Ia(x>I defines a valuation on S Z . For every local field K, the restriction of I 1' to K coincides with I 1, by the uniqueness portion of Theorem 4-1-8. There- fore, by Corollary 4-1-9,

Ixl'=Ia(x)lc(L). I1

Corollary 4-1-11: Let K and L be localfields, both containing @,. Let a :K -+ L be an algebraic isomorphism which coincides with the identity on Q,. Then a is continuous and open.

Proof: By Corollary 4-1-10, a is an isometry. /I

Theorem 4-1-8 can be used to yield important information about the local norm mapping. A trivial computation using Theorem 4-1-8 shows that if K c L are local fields, then

I x IL = I NL/K(x) Ii'dc8(L'K). Immediately from this formula, we deduce

In Chapter 6 , we shall supplement this result with the fact that N,,, is an open map.

EXERCISES 4-1-13: (1) Suppose that L/K is an extension of local fields. An element of L is said to be integral over K if it satisfies a monic, irreducible polynomial with coefficients in OK. Show that OL is the set of all elements of L that are integral over K.

(2) Let L/K be an extension of local fields. Then TrLIK is continuous. (Hint: Use Corollary 4-1-1 1 .)

(3) Let L/K be an extension of local fields. Then

Tr (OL) C OK.

(Hint: Use Corollary 3-1-7.)

4-2 COMPLETIONS OF NUMBER FIELDS ARE LOCAL FIELDS

In this section, we shall prove

Theorem 4-21: Let L be the completion of a number field K a t a prime p . Then L is a local field.

The main fact required for the proof is contained in

Theorem 42-2 (Weil): Let V be a locally compact vector space over Q,. Then dim (V) is$nite.

Proof: Let W be a finite-dimensional subspace of V. It suffices to show that dim (W) is bounded by an absolute constant. By Theorem 4-1-2, W is closed in V, so that V/ W is locally compact. Choose a E Q, such that 1 alp > 1. By Theorem 4-1-2, as an additive group,

1 w - Q p @ - - - @Q,, where the isomorphism is algebraic and topological. Therefore, if a" denotes the automorphism x -+ ax of W, then

mod, (a") = m~d,,(a")~'"(~) dim(,)

= lal, (1)

by Proposition 3-2-9. It is clear that a" maps W onto itself. Therefore, by Proposition 3-2-8,

mod, (3 = mod,,, (a" I,,,) mod,@ I,), (2)

I where a" I,,, and a" 1, denote the automorphisms of V/ W and W, respectively, induced by a". We claim that mod,,, (a" I,/,) > 1. This suffices to prove the

I Theorem. For then, by (1) and (2), we would have

mod, (a") > 1 a (3)

so that dim ( W) is bounded by an absolute constant. Let ,u be a Haar measure on V/ W and let A c B be two compact neighborhoods of 0 in V/ W such that

< ,u(B). Let C = {x E @,IxB G A].

I

i Then 0 E C. Moreover, C is a neighborhood of 0, since the mapping

@, x v--+ v is continuous. Therefore, since a-" -+ 0 as n 4 oo, we see that a-" E C for sufficiently large n. Thus, for n large,

a-"B & A c B

=. ,u(a-"B) < N

mod,,, (a-" I,,,) < 1

N =- mod,,, (a" I,,,) > 1,

since mod,,, (a-" I,,,) = modvlw (a" I,,,)-". /I

In Section 3-1, we promised that we would prove that every valuation of an algebraic number field K is equivalent to I 1, for some K-prime p . In order to prove Theorem 4-2-1, we will require the special case of this result for

K = @. This special case will be used as the starting point in our complete proof of the general case, which will be presented in Section 5-2.

Proposition 42-3: Every valuation of @ is equivalent to I 1, for some @- prime p.

Proof: Let us first show that if ( ( is a non-Archimedean valuation on @, then I I is equivalent to I 1, for some finite @prime p. Since I I is non- Archimedean, 1x1 5 1 for all x E Z. Since I I is nontrivial, there exists p E Z+ such that Ip I < 1. Choosep to be minimal. Thenp is a rational prime. For if p = p'p", 1 I p', p" < p, then either Ip' 1 < 1 or Ip" ( < 1, which contradicts the minimality ofp. Claim that ( ( is equivalent to 1 1,. It suffices to show that there exists a E R+ such that

1x1 = IxI;, X E 72,.

For the validity of this equation for x E Z+ implies its validity for all x E Q. Let x E Z+, x = p ~ ~ ~ p ( ~ ) x ' , (XI, p) = 1 . By the Euclidean algorithm, there ex is tq , r ,q>O,O<r <p ,suchtha tx l = q p + r.ThenIqpI< l , a n d I r l = 1, since 0 < r < p and p is minimal. Thus, by Exercise 3-1-22, (4),

Finally, I x 1 = IpOrdp(x) I = Ip l"'dp(x). But Ip 1 = Ip I;, a > 0. Then

Let I I be an arbitrary valuation on Q, and let m and n be integers greater than 1. Write m in the base n number system:

m = aons + a,nrl + - + a,, 0 < a, < n, a, # 0.

Then n" m m s < log mllog n. Since 0 < ai < n, I a, I I n. Thus,

Iml<n ln lS+ nlnls-I + .. . + n-1

I n(s + 1) max (I n Is, 13

< n ( k m log n + l)max(lnl, 1]"'"hn.

Replace m by mr, t E Z+ to get

log m ( log + 1) mar (1 n 1, lPog n. J m J ' l n t-

Taking tth roots and letting t -+ oo, we see that

I m I ( max (1 n 1, 1 ]'Og mllog " 3 I m 1'/'08 < max (1 n 1, 1)'/Io8 ". (*I

If In l< 1, then (*) shows that I m l < l = > I x l < 1 for all x E Z * I l is non-Archimedean (see Exercise 3-1-18) m I I is equivalent to I 1, for some

finite @prime p by the first part of the proof. Thus, we may assume that I n 1 > 1 for all n > 1. Then (*) implies that

I m l l / l o g m < l / logn - I I . Since this last equation is symmetric in m and n, this implies that

J m J l / l o e m = ) n / l / l o a n , m , n € Z + .

Choose an n and fix it. Set a = n1/'Ogn. Then we have shown that

l m l z a l o 8 m = lm1:8a.

Thus, 1 1 is equivalent to 1 I,. //

We now come to the main result of this discussion:

I Theorem 42-4: Let L be the completion of the number field K at the prime P . Then L is a localfield.

Proof: The result is clear if p is infinite, so assume that p is finite. By Theorem 4-2-3, the restriction of I I,, to Q is equivalent to I 1, for some @-prime p. Thus, by Exercise 3-1-22, (3),

Let us replace I 1, by the equivalent valuation I I & = I I;/". Then, assuming @ to have the p-adic absolute value, the embedding Q - L is an isometry. Therefore, by the universal mapping property for completions, there exists a unique isometry j such that the diagram

commutes. Identifying @, with j(@,), we see that L 2 @,. With respect to the topology on L induced by I I;, L is a topological vector space over @,. By Theorem 3-1-14, L is locally compact. Therefore, by Theorem 4-2-2, dim,, (L) is finite. //

4-3 THE ARITHMETIC OF LOCAL FIELDS

We have shown that a local field K has an essentially unique valuation, with respect to which K is a locally compact, topological field. The presence of a valuation on a field endows the field with much in the way of arithmetic structure. In this section, we shall develop this arithmetic. Much of what we say has been developed previously for fields that are completions of an algebraic number field at a prime. This is reasonable, since such fields are local fields by the results of the preceding section. However, our point of

56 1 LOCAL FIELDS CHAP. 4

view in the present discussion is that the arithmetic structure on a local field depends intrinsically on the field and not on the way in which the field was constructed as the completion of an algebraic number field at a particular prime. Throughout, let K be a local field with canonical valuation I I,. Suppose that K 2 @, for p a @prime. We say that K is a non-Archimedean or Archimedean field according as I 1, is Archimedean or non-Archimedean. From the proof of Theorem 4-1-8, 1 1, is non-Archimedean if p is finite. Conversely, it is clear that if p is infinite, then I 1, is Archimedean. Thus,

Proposition 4-3-1: 1 1, is non-Archimedean -. p is finite.

Most of the arithmetic theory with which we shall be concerned is valid only for non-Archimedean local fields. Therefore, unless explicitly mentioned, we shall always assume that K is non-Archimedean.

Let r, = (1x1, 1 x E Kx}. By Theorem 4-1-8, T, consists of certain powers of p. Therefore, T, is an infinite cyclic group. Moreover, if K G I,, then r, G r, and [T,:r,] is finite.

Definition 4-3-2: [r,:r,] is called the ramification index of the extension LIK, and is denoted e(L/K). If e(L/K) > 1, then the extension LIK is said to be ramified. If e(L/K) = 1, then the extension LIK is said to be unramiJed. If K is an Archimedean local field, then K = R or C. In this case, we set e(L/K) = 2 if L = C and K = R, and e(L/K) = 1 otherwise.

Proposition 4 3 3 : Let K G L E M be a tower of local fields. Then

e(M/K) - e(L/K)e(M/L).

Proofi Exercise.

Let

0, = {x E KI I x l K I 11,

P, = {x E KI 1x1, < 1)

U, = {x E KI 1x1, = 1)t

Since I 1, is a non-Archimedean valuation, 0, is a subring of K. A simple computation shows that 0, is a local ring with P, as its unique maximal ideal and U, as its group of units. 0, is called the ring of integers of K, P, the prime divisor of K. P, is a principal ideal; in fact, if n E P, is chosen so that 1x1, is maximal, then P, = no,. Such an element n is called a local uniformizing element for K (or for P,). One trivially sees that every ideal of 0, is of the from P$ (a E Z-by ideal, one means "fractional ideal"). In

t UK will also be defined for K Archimedean.

particular, 0, is a principal ideal domain. Let x E K x , a a local uniformizing parameter for K. Then x can be written uniquely in the form

x = nau a E Z, u E U,).

The integer a is independent of the choice of n. We may, therefore, define

ord, (x) = a (x E K ')

ord, (0) = + w.

Then ord, ( ) obeys the rules

(1) ord, (XY) = ordK(x) + ord, (Y) (2) ord, (x + Y) 2 min (ord,(x), ord, (y)),

where the obvious rules for manipulating with + oo are observed. Moreover,

0, = {X I OrdK (x) 2 o}

PK = (x 1 ord, (x) > 0)

U, = {x 1 ord, (x) = O}

I X 1, = In kd~(x). (*I Let us set NPK = Inl;'. Then NPK is a positive power of p, independent of the choice of n. Thus, by (*),

( x 1, = Np;ord~(x) (X E a. Proposition 43-4: Let K 5 L be local fields. Then

ord, (x) = e(L/K) ord, (x) (x E K).

I Proofi Let n and ll denote local uniformizing parameters at P, and P,, respectively. Since 1 n lK generates the cyclic group r,, and 1 ll1, generates I?,, we see immediately that

Without loss of generality, we may assume that x # 0. Let

Applying I 1, to both sides of this last equation, we get the desired assertion. I /I

~ Let L/K be an extension of local fields, 2 a fixed algebraic closure of

L. Let {a,, . . . , a,) be the set of conjugacy mappings of L/K into E. If {a , , . . . ,on] is a basis of LIK, we define the discriminant A(ol, . . . , o n ) by

A(o,, . . . , on) = det (M)2,

where M = (coy)). A simple computation shows that

A(o, , . . . ,con) = det (N),

where N = M f M = (Tr,,,(o,o,)). As in Section 2-1, we verify that A(o,, . . . , a,) # 0 and that if {a:, . . . ,o:] is a second basis of LIK with

then A ( 4 , . . . ,a:) = det A(ol, . . . , on).

By Exercise 4-1-15, if all o, E 0,, then A(o,, . . . , on) E OK. (If n is a local uniformizing parameter for PLY then we can manufacture such a basis for L/K by multiplying the elements of an arbitrary basis by a suitable power of n.1

Proposition 4-3-5: There exists 8 E 0, such that

O, = 1 .o, + eoK + .. . + P-~o,.

Proof: Consider all bases of L/K of the form {1,0, . . . , 8"-'1 with 8 E 0,. Choose one for which ord, (A(1,8, . . . , P I ) ) is minimal. Then this basis is of the desired sort. (To see this, just imitate the proof of Proposition 2-1-7.)

I/

Corollary 43-6: Or, is a free OK-module of rank n.

Remark: A set of free generators for the OK-module 9, is called a relative integral basis ofL/K. If (o, , . . . , on} and (q,, . . . , q,} are two relative integral bases of LIK, and if (a,,) is the matrix of transition from the first basis to the second, then a,, E 0, and det (a,,) E UK (Proof: Exercise). Moreover, a basis is a relative integral basis if and only if ord, (A(o,, . . . , on)) is minimal.

Corollary 4-3-7: OK/PK isfinite.

Proof: Choose 8 as in Proposition 4-3-5, for the extension KlQ,. Since P, 2 P,,, if (a,, . . . , a,) is a complete system of representatives of Z,/p x Z/(p), then a system of representatives of OK/PK can be chosen from among

where b, E {a,, . . . , a,). //

Corollary 4-3-8: O,/P",a 2 1) is$nite.

Proof: [[O,:P", = [OK:PK][PK:P:] . . . [Pg-' :P$]. Moreover,

P:/Ppl 1 OK/P,

under the isomorphism induced by

x --+ x7rb (X E Pi) ,

with n a fixed local uniformizing parameter for P,. //

Corollary 43-9: 0, is compact.

Proof: It is clear that OK is closed in K and is therefore a complete metric space. 9, is totally bounded, since OK can be covered by finitely many cosets of @,/Pi for any b 2 1 by Corollary 4-3-8. //

Proposition 43-10: Let x E K x , and let i denote the automorphism

Y - XY of the additive group of K. Then mod, ( i ) = I x 1,. Proofi Without loss of generality, assume that x E 0,. Then

XOK = PF~K(x).

Therefore, reasoning as in the proof of Proposition 4-1-9, we see that

mod, (x) = [OK : Pgd~(")] - - [OK: P K Fd~(") (See proof of Corollary 4-3-8.) = N p i a O I ~ K ( X )

= I x t ,

where a = log [9,:PK]/log NPK. Let x E Q,. Since I x 1, = I x I,, Proposition 4-1-9 implies that a = 1. Therefore, the Proposition is proven. //

As an immediate corollary to the proof of the last result, we derive

Corollary 4-3-11: NPK = the number of elements in OK/PK.

Definition 43-12: OK/PK is called the residue classJield at the prime divisor P, (or at K) and is denoted I?.

If K G L, then G L via the injection induced by the mapping

OK ---t 0,.

Proposition 4-3-13: I? has characteristic p, where K 2 Q,.

Proof: g, = Z/(p) E z. //

60 / LOCAL FIELDS CHAP. 4

Definition 4-3-14: deg is called the residue class degree7 of L with respect to K, and is denoted f(L/K).

Proposition 4-3-15: If K 5 L G M is a tower of localfields, then

f (MIK) = f (LIK)f (MIL).

Proof: Exercise.

Proposition 43-16: NP, = pf, f = f(K/@,). (f is called the absolute residue class degree of K.)

Proof: NP, is the order of by Corollary 4-3-11. By the definition of the residue class degree, x is an extension of @, = Z/(p) of degreef. Since @, contains p elements R has pf elements. //

Theorem 4-3-17: Let LIK be an extension of' local fields of degree n. Then

n = e(L/K) f (LIK).

Proof: By Theorem 4-2-2, m = deg (L/@,) is finite. By Theorem 4-1-2, the additive group of L is algebraically and topologically isomorphic to

@, @ - . . @ @, (m copies).

Similarly, if k = deg (K/@,), the additive group of K is algebraically and topologically isomorphic to

@, @ . - . @ @, (k copies).

Therefore,

L = K @ - @ K (n copies), ( 1 )

where the isomorphism is algebraic and topological. Let x E K be such that 1 x 1, > 1. Then, by Proposition 4-3-10,

mod, (x) = I x 1,. Therefore, by (I),

mod, (x) = I x k. on the one hand, and

mod, ( 4 = l x I, on the other. But

t If LIK is an extension of Archimedean local fields, then we shall set f ( L / K ) = 1.

Therefore,

which completes the proof of the Theorem. //

This last result gives some motivation for the introduction of the ramification and residue class degrees. In the next section, we shall understand further the application of these concepts to global number theory.

EXERCISES 4-3-18: (1) (Structure of the additive group of a local field). Let K be a local field, n a local uniformizing parameter at PK, W a system of representatives of the cosets OK/PK. Then every x E K can be written uniquely in the form

where the convergence of the series means that the sequence of partial sums converges in the ) IK-metric.

(2) Let K be a local field, f E OKIX]. We say that f is primitive if some coefficient off is in UK. Let f E Rx] be the polynomial obtained from f by reducing the coefficients off modulo P,.

(a) (Hensel's Lemma-General Form) Let f E O,[X] be primitive. Suppose that f =gh, where g and h are relatively prime. Then there exist G, H E

OKIX] such that (i) f = GH. (ii) G = g, = h.

(iii) deg (G) = deg (g). (b) Let f E K[X] be irreducible. Iff = Xn + alXn-I + - - - + a,, with

a, E OK, then all ai E OK. (c) Let K c L be local fields. If x E L, then

I X l L = I NLIK(~)lkldeg(LIK).

(d) Let K c L be local fields. Then

[Hint: It suffices to verify the assertion for x = n , n a local uniformizing parameter at P,. Use the fact that if n is a local uniformizing parameter at PK, then a =

lle(LIK)u(u E UL), as well as Exercise (2c) above.] (3) Let LIK be a local extension, @ = aO, an L-ideal, a E L x . Let us define

the norm of 8 from L to K by

(a) NLIK is a K-ideal and does not depend on the choice of generator a. (b) NLIK is a homomorphism from the group of L-ideals into the group of

K-ideals.

(c) NL,PL = [Hint: Use Exercise (2d).] (4) (Decomposition of the Multiplicative Group of a Finite Field) Let K be a

local field. (a) K contains the (NPK - 1)st roots of 1. (Hint: Apply Hensel's Lemma

to XNPK-~ - 1, which splits into NPK - 1 distinct linear factors when reduced modulo PK.)

(b) 1 + PK is a subgroup of UK. (This subgroup is called the group of principal units of K.)

(c) Let UB denote the group of (NP, - l)st roots of 1. Then U$ E UK. (d) UK = UB x (1 + PK). (e) Let n be a local uniformizing parameter at P,. Then

where (n) denotes the group generated by a.

4-4 LOCAL RAMIFICATION THEORY

Let K and L be local fields, K 5 L. In this section, we shall give a simple criterion to determine when the extension L/K is ramified-that is, when e(L/K) > 1. Without loss of generality, assume that K is totally disconnected, since otherwise K = [R or C.

If R is an additive subgroup of L, the complementary set R' of R is defined by

R' = (x E L 1 TrL,K(xR) z OK].

It is clear that R' is an additive subgroup of L. Our first major result will state that if R is an L-ideal, then so is R'.

Consider the bilinear form

( , ) : L x L-+K

defined by

(x, Y> = T~LIK(XY).

Since LIK is separable, this bilinear form is nondegenerate-that is, if (x, y) = 0 for all y E L, then x = 0.

Proposition 4-4-1: Let K, L be fields, L = K(8) a separable extension of degree n. Let f be the minimal polynomial of 8 over K. Then

where f ' denotes the formal derivative off.

Proof: Let C be a separable algebraic closure of K. Let Bi(l ( i < n) be the conjugates of 8 = 8, in C . Define the polynomials Gj(X) E C[X] (1 _< i _< n) by

Gj(X) = r]: (X - ei) (1 I j n). i f j

Then

This implies that {GI, . . . , Gn) are linearly independent over c and therefore span the space of polynomials of degree < n - 1 over c. Thus, in particular, we may write

Substituting 8, in these equations gives

Set X = 0 in the equations (*) and insert the values of cij, cj, to derive

(-81) . . . (-en) = 0 j = l fl(ej) * + .

-0, missing

2 L (-8,) . . . (-en) = -f(o). j = ~ f1(8j) .

-01 missing

These last two equations imply that

which imply the assertion of Proposition 4-4-1. //

Corollary 4-4-2: Let the notation be as in Proposition 4-4-1, with K and L localfields, 8 E 8,. Then

for all i 2 0.

Proof: By induction on i, we immediately see that

n- I Let 8' = C cij8j, cij E OK. Then

j = O

CHAP. 4

by Proposition 4-4-1. /I

Corollary 4 4 3 : Let notation be as in Proposition 4-4-1 and CoroNary 4-4-2. Then f '(B)O', 0,.

Proof: Let a E O',. Since 1 &- 1 -

f'(e>' . - ' 3 fye)

is a basis of LIK, we may set

Since a E Oi, TrLiK (I .a) E OK. By Proposition 4-4-1, this implies that a,-, E OK. Also,

8" T~L,K (a@ = on-, + an- 1 T ~ L , K ( f'(e> ) E OK.

But since an-, E OK, the last term of this expression is contained in OK by Corollary 4-4-2. Therefore, a,,-, s OK. Continuing in this manner, we see that all ai E OK. Therefore, af'(8) E OL. I/

where 8 is as above, then R' = -

f ! ( e ) ~ -

Therefore, if 8 is chosen so that [l , 8, . . . , &-'I is a relative integral basis of LIK, then O', = (11 f '(8))OL.

Proof: By Corollary 4-4-3,

Let

By Proposition 4-4-1, Tr L!K (a&-j-l)= a, ( j = o , . . . , n-1) .

Therefore, a, E O,(j = 0, . . . , n - I), since u t R'. Thus, 1 7 R E R'. f (6)

Corollary 4-4-5: If@ is an L-ideal, then @' is an L-ideal, and 8' = @-'Oh.

Proof: a E @' - TrL,(a@) E OK

T~L,K(~@OL) E 0, -- a@ 5 0; a - ' 0 . //

Since O', is a fractional L-ideal, so is (@',)-I. Moreover, since 0, G O',, ( 03 -1 c 0,.

Definition 44-6: The integral L-ideal bLIK = (03-I is called the dzflerent of the extension LIK.

Proposition 4-4-7: Let 8 E 8, be such that L = K(8). Then f'(8) E bLIK, where f = the minimalpolynomial of 8 over K.

Proof: Corollary 4-4-3. //

Proposition 4-4-8: Let 8 be chosen so that {I, 8, . . . , P I ] is a relative integral basis for LIK. Then bLiK = f '(8) 8,.

Proof: Corollary 4-4-4. //

EXERCISE 4-4-9. Let K c L c M be a tower of local fields. Then

~ M I K = ~MIL~LIK, where on the right, we view bLIK as the M-ideal bLIKOM.

The connection between the different and the ramification index is given by

Theorem 4-4-10: e(L/K) > I - b,, + 0,.

Proof: Let L = K(B), 8 E O,, n = deg (LIK). Let 8 generate a relative integral basis.

-e Assume that e(L/K) = 1. Then f(L/K) = deg (LIK). But f(L/K) =

deg (Ell?), so that 6 has n conjugates over K. Let f = the minimal polynomial of 8 over K. Then f is a separable polynomial and, therefore, J'(8) # 0. Therefore, f'(8) q! P,. But this implies that f'(8) E UL. By Corollary 4-4-7, fye ) E bL/K bL,K = 0,.

Assume that e = e(L/K) > 1. Suppose that K 1 Q,. We want to show that

bL,K G PL -- Pi1 E 01, => TrL,K (Pi1) 5 OK.

It is immediately obvious that c PK. Therefore, by a trivial computation, we see that for a E O,,

The remaining assertions of Corollary 4-4-4 follow immediately. //

Let a E Pi-'. Since p(e - 1) 2 e, UP E pfk- 1 ) - c p e - .c - P K 0 L

3 Tr,, (UP) E PKO, n K = P,

3 T ~ L I K (a) E PK * TrL/K (Pi-') G PK => TrLIK (Pi1 Pi-') E 0, (since the trace is K-linear)

=> T r ( P i 0 . /I

In practice, the different of an extension is difficult to compute. However, there is a much more amenable invariant of a local extension that one can use to determine if e(L/K) > I-namely, the discriminant. In order to define the local discriminant d,,K, let us recall some facts from the last section. By the remarks following Corollary 4-3-4, we see that the discriminants of any two relative integral bases for LIK differ by the square of an element belonging to UK. Therefore, if (w,, . . . , on) is a relative integral basis of LIK, the K-ideal A(w,, . . . , o,)QK depends only on the extension LIK and not on the choice of relative integral basis. We call this K-ideal the relative discriminant of the extension LIK and denote it by dLIK.

Let LIK be a local extension, and let 0 E 0, generate a relative integral basis of LIK. We know that such 0 exists by Proposition 4-3-5. Let L" be a fixed algebraic closure of L and let (a, , . . . , a,) be the set of conjugacy maps of L into z, leaving K pointwise fixed. Since the determinant defining A(1,8, . . . , P1) is a Vandermonde determinant, we immediately deduce that

A(l ,e , . . . , P 1 ) = f fl i t j (ciiB - a,@ (*I

Let f E K[X] be the minimal polynomial of 8 over K. Then

f ( x ) = fi (X - ale) i= 1

is the factorization off over L. Therefore, a simple computation shows that

where we have assumed that the conjugation maps are numbered so that a, = the identity. Therefore,

For fixed j, aiaj runs through all conjugation maps except a j as i runs from 2 to n. Therefore,

NuKf '(0) = i # j fl (aje - a#>

by (*). Taking into the account the definition of the norm of an L-ideal, we see that

Theorem 4-4-11 : NLIKbLlK = d,,.

The connection between the discriminant and ramification is given by

Theorem 4 4 1 2 : e(L/K) > 1 - d,, c PK.

Proof: e(L/K) > I - bLIK G P, (Theorem 4-4-10)

NLIK~LIK C NLIKPL dLIK G Pi(L1K) [Theorem 4-4-1 1 and Exercise

4-3-18 (3~)] - dLIK c PK (since dL/K is the norm of an L-ideal). /I

4-5 LOCAL HILBERT THEORY

Throughout the following discussion, let L/K be an extension of local fields, with K non-Archimedean. Let us assume that LIK is Galois with Galois group G. We would like to consider the following problem: How does the arithmetic structure of L and K reflect itself in the Galois theory? It will turn out that there is an intimate connection. For example, it is possible to compute the ramification and residue class degrees of LIK in terms of G.

In Section 4-1, we showed that if a E G, then a is an isometry with respect to the metric induced by [ (,. In particular,

Therefore, we can define a mapping

G -+ Gal (QR)

via

6(x mod P,) = a(x)(mod P,). It is clear that this mapping is a homomorphism. We will presently see that it is surjective. The kernel I of the homomorphism is given by

I = (a E GI a(x) = x(mod P,) for all x E 0,j

Definition 4-5-1: I is called the inertia subgroup of G. Its fixed field is called the inertia subjield of L, and is customarily denoted T (after the German Tragheitskorper).

The fundamental theorem that connects the arithmetic theory with the Galois theory is

Theorem 4-52: The homomorphism a -+ d gives rise to the exact sequence

(1) - I - G + Gal (LIE) --+ (1).

Before proving this result, let us establish a Lemma. Since E is a finite field, Ex is a cyclic group. Let x E Ex be such that x(mod P,) generates Ex. Let y E L be such that L = K(y). Without loss of generality, we may assume that y E P,.

Lemma 4-5-3: There exists z E 0, such that z(mod P,) generates Ex, and such that L = K(z).

Proof: Let w -+ w(') (1 < i 5 n = deg (LIK)) denote the elements of G. Consider the polynomial

At) = n (X(C - X(j) + t(Y(i) - '+I

Y"')) E OK[tl.

Since f has only a finite number of zeros, there exists u E 0, such that f(u) f 0. Set z = x + uy. Then z E 0, and

0 f n (x(') - x(j) + u ( y ( i ) - y(j))) i#j

Therefore, all the conjugates of z are distinct, so that L = K(z). Moreover, z(mod P,) = x(mod P,) since y E P,. Thus, z(mod P,) generates Ex. //

Proof of Theorem 4-5-2: Let z be as in the Lemma. Let F be the field polynomial of z over K. Then

Claim that F(XNPK) - F(X)NP~ E PK[X].

Indeed, F(X)Np~ = (XNP~). + ayP~(XNPK)n-l + . - . + arPK + NPKG(X),

where G(X) E K[X]. Now ord, (NP,) = e(K/Q,)f(K/Q,) > 1. Thus, NP,G(X) E P,[X]. Moreover, since E is a field with NP, elements, rNPx = r(r E E), so that aYPK - a,(mod P,)(I 5 i n). Thus, the claim is established. Put X = z in the claim, and recall that F(z) = 0. Then we see that F(zNP~) E PKOL c P,. Thus,

]CI ( z N p ~ - az) E PL. atG

Since P, is a prime ideal, there exists a, E G such that z N P ~ - a l z E P,, so that ( Z " ) ~ ~ K - al(za) E P,, for all a E Z. Claim that

W N P ~ - 0 ,(w) E P L (W E 0,)' (*I This is true for w E P,, since o,(P,) = P,. If w $ P,, then w = za(mod P,) for some a E Z [since z(mod P,) generates Ex]. Thus, since a(w) = a(za) (mod P,), w N P ~ = zaNP~ = al(za) = a,(w)(mod P,). Thus, the claim is established. Therefore, d l is the Frobenius automorphismt and the mapping o 4 d is surjective. This completes the proof of the Theorem. //

Corollary 4-5-4: I has order e(L/K).

Proof: G has order n = e(L/K)f(L/K) (Theorem 4-3-17), and Gal (EIR) has order f(L/K) from the definition of the residue class degree. Therefore, the Corollary follows from the above exact sequence. //

Let us now assume that LIK is unramified. Then e(L/K) = 1, so I is trivial. Therefore, by Theorem 4-5-2, the mapping a -+ d induces an isomorphism of G onto Gal (LIE). But Gal (LIE) is cyclic, and is generated by the Frobenius automorphism x -+ xNP~ . Therefore,

Theorem 4-5-5: Suppose that LIK is unram$ed. Then Gal (LIK) is cyclic and is generated by an element a satisfying

a x = xNPK(mod P,) (X E 0,). The element a is uniquely dejined by this condition and is called the (local) Frobenius automorphism.

Let us now return to the situation of a general local extension LIK, which is Galois of degree n. Let T be the inertia subfield of LIK. Since I is a normal subgroup of G, TIK is Galois with Galois group GII. It is trivial to see that the inertia subgroup for the extension TIK is trivial, so that T/K is unramified of degree f(L/K). Thus, LIT is totally ram$ed-that is, e(L/T) = deg (LIT), f(L/T) = 1. Moreover, since deg (LIT) = e(L/K), we see that e(L/T) = e(L/K). Thus, we may roughly describe the tower K G T G L by saying that in the portion TIK only the residue class degree can grow, whereas in the portion LIT only the ramification index can grow. By Theorem 4-5-5, the portion TIK is a simple sort of extension4yclic with a canonical generator for the Galois group. Therefore, it remains for us to study the totally ramified extension LIT.

From the definition of T, Gal (LIT) = I. Let us define the sequence of subgroups of I:

t Let V E W be finite fields, with V containing q elements. Then W/V is Galois and Gal ( W / V ) is cyclic, generated by the Frobenius autornorphisrn x + f l .

Gi = ( a E I1 a x -- x(mod PF1)3 ( i 2 0).

We clearly have Go = I,

Go 2 GI 2 G 2 2 . - - . Corresponding to this decreasing sequence of subgroups, we have the increasing sequence of subfields

V o 5 V , E v2'z G L ,

where V o = T. The sequence of subgroups (G,) is called the sequence of ramification subgroups of the extension LIK. The fields Vi are called the ramification subfields.

Lemma 4-5-6: G,+, is a normal subgroup of G,(i 2 O), so Vi+, is a normal extension of V,, and Gal (K.+,/V,) = G,/G,+, . Proof: Obvious.

Let p denote the rational prime for which K 2 a,. Then I? has characteristic p.

Proposition 4-5-7: deg ( V , / T ) is relatively prime to p. In fact,

deg (V , /T ) I NP, - 1 .

Proof: Let n be a fixed local uniformizing parameter at P,. If a E Go, then anln E U,. Therefore, the mapping

is a homomorphism of Go into UK/(l + PK). The kernel of this homomorphism is clearly G,. Therefore, Go/Gl is isomorphic to a subgroup of UK/(l + P,). But by Exercise 4-3-18, UK/(l + P,) is isomorphic to the group of (NP, - 1)st roots of unity, and is therefore a cyclic group of order NP, - 1. Since NP, is a power of p, this completes the proof. //

Corollary 4-5-8 : V , / T is cyclic.

Proof: Go/G, is isomorphic to a subgroup of the cyclic group UK/(l + P,). /I

Proposition 4-5-9: Gal (K.+,/V,) is an elementary p-group ( i 2 1) .

Proof: Consider the homomorphism

The kernel is G,+, so that Gi/Gi+, is isomorphic to a subgroup of (1 + Ph)/ (1 + Pg1) . Consider the homomorphism of additive groups

1 + P:,+I?

The kernel is 1 + P p l , so that ( 1 + Ph)/(l + P p l ) is isomorphic to a subgroup of $ which is an elementary p-group, being a vector space over Z/(P). 11

EXERCISES 4-5-10: ( 1 ) Show that ( 1 + P;)/(l + P p l ) * E. (2) I f L/K is a Galois extension of local fields, show that Gal (L /K) is solvable.

The portion of the extension L/K corresponding to V, /T is called the tamely and totally ramified part. The portion of the extension L/K corresponding to L/V , is called the wildly (and totally) ramified part.

Global arithmetic

5-1 SPLITTING OF PRIMES

Let K be an algebraic number field, L/K a finite extension, p a finite K-prime. The L-ideal p0, is not in general prime. For example, let K = @, L = Q(fl), p = 5. Then

50, = (2 + i)(2 - i),

and neither ideal on the right is the unit ideal (since 2 + i are not units in 0,). In this section, we shall consider the following general problem: Given p, determine the factorization of p0, into prime factors. As stated, the problem has no known solution and none is expected at an early date. If LIK is an abelian extension, then fairly explicit answers to the problem are available. But this is part of class field theory and will be treated in Part 4. For the moment, let us be content with some crude results.

Let

p0, = ' $ 3 ~ . . . '$37, where '$3,, . . . , '$3, are distinct L-primes. Then the Pi are called the prime divisors of p in L. We denote the fact that '$3 is a prime divisor of p in L by writing '$1 p. If '$31 p, then p is called the restriction of '$3 to K, since

Let p now be an infinite K-prime, corresponding to the embedding a:K - C. An infinite L-prime !$ is said to be a prime divisor of p in L if '$ corresponds to an embedding z such that z 1, = a.

Theorem 5-1-1: Let p be a K-prime, !$ an L-prime such that '$3 1 p. Then K, E LI. More precisely, there exists a unique isontetricfield isomorphism

q: Kp ---t LI

such that the diagram

K d L

is commutative.

Proof: Since the restriction of 1 11 to K coincides with I I,, the mapping K -+ L + LI is an isometry. Therefore, the assertion follows from the universal mapping property for completions. //

Theorem 5-1-2: Let notation be as above. Then ei = e('$3,/p)(= e(LV,/Kp)).

Proof: ords, (x) = ei ord, (x)(x E K,). // Let f, = f('$3,/p) be the residue class degree of LI, with respect to Kp.

Let e, denote the corresponding ramification degree.

8

Theorem 5-1-3: C eiJl = n. i = 1

Corollary 5-1-4: p has at most n prime divisors in L.

Proof of Theorem 5-1-3: The case p infinite is clear from the definitions. For A = 1 always and ei = 1 or 2 according to whether the extension Lg,/K, is trivial or not. In the former case, the embedding of K into C has a unique extension to L, whereas in the latter case, the embedding of K into C has two extensions to L. Thus, we may assume that p is finite. Let p be the restriction of p to @. Let p = p,, . . . , p, denote the prime divisors of p in K. Choose -v E p - p2. Then ord, (y) = 1. By the Chinese remainder theorem, there exists x E OK such that

x - l(mod pi) (i = 2, . . . , r). Then i(x) is relatively prime to p,(i = 2, . . . , r ) and ord, ( x ) = 1. By the product formula,

However,

= f i l x ( i ) 1 i= l

= I N~lQ(x) I. (2)

On the other hand, by the choice of x, the precise power of p appearing in

is Np-l. Thus, by (I) and (2), the exact power of p dividing NK,,(x) is Np = pf (pip).

By the product formula in L, the same reasoning as above yields

The exact power of p appearing in the first factor is

by Proposition 4-3-4. However, by Proposition 4-3-16, N T ~ , = ~ p f ( ! p * l ~ ) .

Thus, the exact power of p dividing

on the one hand, and equals the exact power of p dividing (N,,,(x)-' I on the other. But

Thus, by the reasoning p dividing I NLtQ(x)-' I is

NL/Q(x) = NKIQNLIK(X) = NK/Q(x)"'

at the beginning of the proof, the exact power of Np-". Thus,

Np-n = Np-(e,f,+.~.+cofo) 9

which completes the proof. 11

EXAMPLE: Let L = Q ( O ) , D a square-free integer, p a finite @-prime. Theorem 5-1-3 implies that pOL has one of the three following forms:

(1) '$,'$,, qi distinct, f('$,/p) = l(i = 1, 2). In this case, we say that p is decomposed (or splits completely) in L/@.

(2) '$2, f($/p) = 1. In this case, p ramifies in L/@. (3) '$, f('$/p) = 2. In this case, we say that p is undecomposed (or inertial) in

L/@.

Later in this section, we shall show that there are only finitely many primes falling into case (2). But there are infinitely many falling into the

other two categories, with roughly half belonging to each type. This last fact will be proven in Chapter 9.

Generalizing the above example, we say that a K-prime p splits completely in an extension LIK of degree n if p has n distinct prime divisors in L. [Neces- sarily, e(Tp/p) = f(Tp/p) = 1 for every such prime divisor 9.1 We say that p is undecomposed in LIK if p has a unique prime divisor $3 in L for which f(@/p) = n. We say that an L-prime @ ramijies in LIK if e(Tp/p) > I , where p is the restriction of Tp to K. We say that p is ramified in LIK if some prime divisor of p in L is ramified in LIK. One of the important problems we shall settle below is that of giving an effective procedure for determining the ramified primes Tp. It will turn out that there is only a finite number of them and all of them divide a certain L-ideal, called the relative different of the extension LIK.

Let us now return to the general situation of an extension LIK of degree n and an arbitrary K-prime p. Let 0 E 0, be chosen so that L = K(@, and let f be the minimal polynomial for 8 over K. Let

f = f 1 . . . f , , J;.EKv[Xl.

where theJ;. are irreducible and distinct (f is separable). Let us fix an algebraic closure of K,, and let us choose a zero 8, ofJ from it. By Theorem 4-1-8, the field K,(8,) has a unique valuation which extends the p-adic valuation on Kp. Let Tp,, . . . , @, be the prime divisors of p in L.

Theorem 5-1-5: (1) h = g. (2) We can suitably renumber the J's so that there exists an isometric

field isomorphism Kv(i3,) -+ LO, such that the diagram

KV(0i) ---+ LB' f f

is commutative (1 I i I g), where j is the identity, and the mapping L --+

Kp(Oi) is the one induced by the condition 0 -+ 0,.

Proof: Let L, = K, @ L, considered as an algebra over K,. From the K

elementary theory of tensor products,

dimq (L,) = n. (1)

Since L = K[X]/fK[X], we immediately see that

LP = K,[xllfK,[xl. (2) An elementary exercise in algebra shows that (2) implies that

LP = K,[XlIflK,[XI 0 . 0 ~,tXlIf,Kp[XI, (3) where the summands on the right are considered as K,-algebras. From Galois

theory, each of the summands on the right side of (3) is a field, and with respect to an obvious identification

KPtlYllf;~KP[Xl = Kp(8i). (4)

Therefore, by (3),

LP = KP(O1) 0 . . . 0 KP(eg). (5)

Let {w,, . . . , o n ) be a basis of LIK. Then (1 0 w,, . . . , 1 @ o,) is a basis of L,/K,. From the theory of tensor products, there exists an algebra homomorphism L, L$, defined by

Since qrp,(Lp) is a Kp-vector subspace of Ls,, Theorem 4-1-2 implies that qs,(L,) is closed in Ls,. But clearly qs,(Lp) 2 L, which is dense in Ls,. There- fore,

By equation (5), we may regard Kp(Bj)(l _< j 5 g) as a closed subalgebra of L,. Since qp, is either injective or identically zero on KP(ej), we see that, upon renumbering thef;., m, induces an automorphism of Kp(ei) onto Lv,, with the automorphism leaving K, pointwise fixed. In particular,

g i h, (7) and

where ni = deg (LsJK,). However, by Theorem 4-3-17, ni = eif;., ei =

e(%/p),f;. =f(Pilp). Therefore, by (81, h

C eifi 5 n. i = 1

Therefore, g = h by Theorem 5-1-3. The desired isometries are the field isomorphisms qs,(l 5 i < g). (They are, in fact, isometries by Proposition 4-1-10.) The commutativity of the diagram is a matter of chasing through all of the identifications that we have made, and is left as an exercise for the reader. //

Corollary 5-1-6: Let notation be as in Theorem 5-1-5. Then

L O K ~ K = $ ,~n . .

Proof: See the statement of Theorem 5-1-5 and equation (5) in the proof. /I

The integer ns = deg (LV/Kp) is called the local degree of the extension LIK at the prime '$.

Corollary 5-1-7: The global degree is the sum of the local degrees-that is,

n = E n l p . $1,

Proof: The proof follows from Corollary 5-1-6 and equation (1) in the proof of Theorem 5-1-5. //

Corollary 5-1-8: Let x E L. Then, for any K-prime p.

Thus, the global trace is the sum of the local traces.

Proof: Let 2 denote the linear endomorphism of the K-vector space L defined by I(y) = xy. Then from the definition of the trace,

TrLIK(x) = Tr(I)

Let (y,, . . . , y,] be a basis of L/K. Then {I O y, , . . . , 1 O y,] is a basis of L, over K,. Let x* denote the linear endomorphism of L, defined by x*(a @ y) = a O (xy)(a E K,, y E L). Then the matrix of x with respect to the basis (y, , . . . , y,] is the same as the matrix of x* with respect to the basis {I 8 y,, . . . , 1 @ y,]. Therefore,

But x* restricted to Lv, is just the linear endomorphism x* defined by x:(y) = xy(y E L$,). Thus, by the definition of the trace,

8

Tr (x*) = C Tr (x*), i= 1

since x* maps Lg, into itself (1 i i 5 g). Assembling the data above implies the Corollary. //

Using essentially the same argument as used to prove Corollary 5-1-8, we can prove

Corollary 5-1-9: Let x E L, p an arbitrary K-prime. Then

Corollary 5-1-10: If L/K is normal and 1, is a K-prime, $ a prime divisor of p in L, then Ls/Kp is normal.


In Theorem 5-1-5, we saw how the way in which a prime y of K splits in L is determined by the way in which the polynomial f factors in K,[X].

In practice, it is often not possible to effect the factorization off , however, and one is forced to resort to a more practical device. Unfortunately, one must pay for viability with a limitation in applicability. This is precisely the state of affairs with the following result, due to Kummer:

Theorem 5-1-11: Let LIK be an extension of number jields, p a K-prime. Suppose that there exists 8 E 8, such that {I, 8, . . . , &-'3 is an integral basis of L. Let f be the minimal polynomial of 8 over K, f = the polynomial obtained from f by reducing its coeficients modulo p. Suppose that

where the pi E Kv[X], are irreducible and distinct. Then p splits into g prime divisors in L. I f these prime divisors are denoted q,, . . . , q,, then, after a suitable permutation,

Proof: By Theorem 5-1-5, if we write

where thef;. are distinct and irreducible, then the prime divisors q,, . . . , 3, are in 1-1 correspondence with thef;. and after a permutation of thef;., we may assume that deg (A) = niv,(l i I g). Let us reduce the equation (*) modulo p:

- - f = f , ...f,. (**)

Sincef;. is relatively prime toS,(i # j), f is relatively prime to f By Hensel's Lemma, fi = p~ for some pi E KJX], irreducible. For if fi is divisibleby two nontrivial, relatively prime polynomials, then this factorization can be hiked back to a factorization ofL, which contradicts the fact that A. is irreducible. Therefore, the factorization of f into powers of distinct, irreducible polynomials is given by

where g = the number of prime divisors of p in K. By Theorem 5-1-5, Lg, = KV(8,), where Oi is some root of f;.. Therefore, since L has an integral basis of the form 1,8, . . . , &-I , eg, = Rp(ei), where ei is a root of pi. Therefore, fv, = deg (Lg,/Kp) = deg (pi). Since nv, = es, fg,, on the one hand, and nrp, = deg (f;.) = ei deg (pi) on the other, we see that ei = eg,. /I

We have already shown that quadratic fields have an integral basis of the type required in Theorem 5-1-1 1, and in Chapter 6, we shall show that cyclotomic fields also have this sort of integral basis. However, not all fields have such a basis. For example, see [32, p. 1701. Moreover, if the condition that 8 be of the specified type is removed, then Theorem 5-1-1 1 is patently false. We leave to the reader the task of finding a counterexample.

5-2 THE VALUATIONS OF AN ALGEBRAIC NUMBER FIELD

Using Theorem 5-1-5, we can now prove the following result, which was anticipated in Paragraph 3- 1, namely,

Theorem 5-2-1 : Let L be an algebraic numberjield, and let v be a valuation on L. Then v is equivalent to 1 1rp for some L-prime 8. Proof: By Theorem 4-2-4, the restriction of v to @ is equivalent to I I, for some @-prime p. Without loss of generality, let us assume that vl, = I I,. By the universal mapping property for completions, we may identify @, with a subfield of L,. Moreover, by Theorem 4-2-3, L,/@, is finite. Let 8 E 0, be chosen so that L = @(8). Then @,(8) is a closed subspace of L, (Theorem 4-1-2), which contains the dense subspace L, so that

Moreover, the canonical valuation on @,(8) coincides with the extension of v to L,. Let h = the minimal polynomial of 8 over @,, f = the minimal polynomial of 8 over @, and let f = f, . . . f, be the factorization off into irreducible polynomials in @,[XI. Then h is one of the f;., say h = fl. By Theorem 5-1-5, there exists an L-prime '$3, a prime divisor of p in L, and an isometric field isomorphism q:@,(8) -+ Lg, such that the diagram

is commutative. In particular, v = I 1%. //

EXERCISE 5-2-2 (Ideal Norms): Let L/K be an extension of number fields, '$3 a finite L-prime, p the restriction of $3 to K. Define NK/L($3), the relative norm of 3 from L to K, by

NLIK($3) = pf(Bio). Extend NKIL to a mapping

NLIK : IL + ZK by multiplicativity.

(1) NLIK is a homomorphism. (2) NL,K(xQL) = NLIK(x)Q~, where NK/,(x) is the usual norm of an element

x E L. [Hint: Use Exercise 4-3-18, (2)dl. (3) NL/,(8) = N 8 . Z . (4) Let M be a number field such that MIK is normal and M 2 L 2 K. Let

Q be a prime divisor of '$3 in M. Then

5-3 GLOBAL RAMIFICATION THEORY

Let L/K be an extension of number fields. Let us consider the problem of determining which primes of L (or of K) ramify in LIK. Since this problem reduces to a purely local question, it can be settled by using the local machinery developed in Chapter 4. However, for the purposes of application, it is most convenient to provide some global criteria for determining ramification.

If A is an additive subgroup of L, let us define

A' = {X E L I TrWK ( ~ 0 , ) C OK).

As in the local theory, A' is an additive subgroup of L. By using Proposition 4-4-1, and by proceeding exactly as in the local case, we deduce

Proposition 5-3-1: (1) I fA is an L-ideal, then so is A'. Moreover, in this case, A' =O),A-'. (2) If0 E 0, is a generating element for the extension L/K, then f'(0)0), E O,, where f is the minimal polynomial for 0 over K.

It is clear that 0, E Oi, so that 0;-I is an integral L-ideal, called the relative dzrerent for the extension LIK, denoted bLrK. The integral K-ideal N,,,(b,,,) is called the relative discriminant of L/K, and is denoted d,,,. Let

dLrK = TI ~ " ( 0 ) . P

Theorem 5-3-2: Let p be a finite K-prime and '$3 a prime divisor of p in L.

Remark: Sometimes Theorem 5-3-2 is expressed in the form "The global different is the product of the local differents; the global discriminant is the product of the local discriminants." Of course, this is not precisely true, since the various local differents are fractional ideals of different fields. However, the imprecision will cause no harm if the reader interprets the statements as in the assertions (1) and (2) of Theorem 5-3-2.

Proof: Since NLOlKp('$3) = pf, f = f(@/p), the second assertion follows from the first. Suppose that b,,,,, = @",.

Let x E b~;,,,. Choose y E L very close to x at '$3, very close to zero at all other prime dlvisors of p in L, and of value at most 1 at all other finite L-primes. Then, by Corollary 5-1-8,

Tr,,,(yz) = TrLDIKq (yz) E Oq for all K-primes q 91q

for all z E 0,. Therefore,

y E O), E '$3-6'') * x E

3 '$P(') E bLqrKP.

Conversely, assume that x E 'fk6($). Choose y E L very close to x at '$3, very close to zero at all other prime divisors of p in L, and of value at most 1 at all other finite L-primes. Reasoning as above, we see that for all z E O,,

* TrLOrL, (xOs) E Op (by the continuity of the trace)

=. x E bi&Kp - * '$3'"@) 2 bLglKp. /I

Corollary 5-3-3 : '$3 ramijies in L/K o '$3 divides bL,, . Proojl Immediate from Theorem 5-3-2 and the local ramification theory.

/I

Corollary 5-3-4: p ramijies in LIK o p divides dLIK.

Theorem 5-3-5: Let L be an algebraic number jield. Then

dLIQ = dLZ.

Proof: From the definition of ideal norm,

dL/Q = NL,QbL,Q = NbL,,Z.

Now Nb,,, = the number of elements in O,/b,,, - the number of elements in O;/O,. Let l o l , . . . , a,) be an integral basis of L/@. Since the trace form Tr,,, (xy) is nondegenerate, there exists a dual basis for this integral basis- that is, there exists (o f , . . . , o,*) a basis of L/@ such that TrLIQ (wioT) =

a,,, where aij is the Kronecker delta function. Suppose that

Then the number of elements in O',/O, is I det (bij) I . However,

Therefore, I det (b,,) I = Id, 1, which completes the proof of Theorem 5-3-5. /I

5-4 GALOlS ACTIONS Proposition 5-4-2 : The mapping

Let LIK be a Galois extension of number fields, of degree n, with Galois group G. Fix a K-prime p, let $3,, . . . , $3, be the prime divisors of p in L, and let $3 = $3,. If $3 is finite, a E G, then ~ ( $ 3 ) is a finite L-prime. Moreover, if

then, since a leaves K pointwise fixed,

pOL = ( ~ $ 3 , ) ~ ~ . . . (a$,)% a'$ is one of the $3, (1 I i I g).

Therefore, the action of G permutes the set {$3,, . . . , $3,). If $3 is infinite, corresponding to the embedding z of L into C, let a@ be the infinite L-prime corresponding to the embedding zg. Again, it is clear that a$3 divides p, so that G acts as a transformation group on the set of L-primes dividing p.

Let Dg denote the isotropy group of $3-that is,

Dg is called the decomposition subgroup of G at $3.

Lemma 5-41 : Dog = a D@o-'.

Proof: Clear

Let L/K be a given normal extension, and let p be a K-prime, $3 a prime diviser of p in L. Suppose that L, = K(0), and let f be the monic, irreducible polynomial satisfied by 0 over K. Since LIK is normal, L contains all the zeros off. Let f = p, . . . p , be the factorization off overK,. Assume that 0 is chosen so that p,(@ = 0 and Lg = Kp(0). The conjugates of 0 over Kp are the zeros of p,. All of these zeros lie in L and hence in Lg. Therefore, Lg/K, is normal. If a E Gal (LglK,), then a is completely determined by its action on 0. Therefore, the restriction of a to L is contained in Gal (L/K). Claim that 6 = a 1, is contained in Dg. In any case, 6 is an isometry by Corollary 4-1-10. If '$ is finite, then a($) = @, so that 6($) = $3. If $3 is infinite, then, since 13 is an isometry, I x Ig = I bx 1g. Therefore, $3 = 6-'$3. Thus, 6 E Dg. Converse- ly, given 6 E Dg, the mapping L 5 L is an isometry, where L is given the $3-adic metric. Therefore, there exists a unique isometry Lg 5 Lg such that the diagram

a L - L

is commutative. It is trival to check that a E Gal (Lg/K,). Therefore, we have proven

Throughout the remaining discussion of this subsection, we shall assume that p is finite. Under the correspondence of Proposition 5-4-2, the inertia subgroup Ig of Gal (Ls/K,) corresponds to

This subgroup of Dg will be called the global inertia group at SQ, and will be denoted Ig also. (No confusion will arise from this abuse of notation.)

1 EXERCISE 5-4-3: I,$ = u Z ~ U - ~

I Under the identification of Proposition 5-4-2, the homomorphism

I Gal (LgIK,) - Gal (&/&)

of Theorem 4-5-2 corresponds to the homomorphism

Dg --+ Gal (OLl$3)I(QKl!J) a-B

where

d(x(mod $3)) = o(x)(mod $3).

(Note that 0L/$3 * &, OK/p = Z?,.) Therefore, by Theorem 4-5-2,

I Theorem 5-4-4: Let p be a jn i t e K-prime. Then we have the exact sequence

(I) ---t + D9 -+ Gal (Lip/$) --t (1). I

I Corollary 5-4-5: (1) Dg has order egj$ (2) Ig has order eg.

Proposition 5-4-6: Let S be the set of prime divisors of p in L. Then G operates transitively on S.

Remark: It is clear that Proposition 5-4-6 also holds if p is infinite.

Proof: Let S = {$3], . . . , $,I. Set $3 = $3,. Let us assume that 0$3 # qi for all a E G. Set 81 = 9,. . . $3,. Since (81, $3) = I , there exists a E '3, b E $3 such that a + b = I. Therefore, b E $3, and a(b) $ $3, for any a E G. Therefore,

on the one hand, whereas on the other hand,

NL,K(b) E '$ n K = p G %, which is a contradiction. /I

Proposition 5-4-7: Let LIK be normal, and let '$,, . . . , 9, be the prime divisors of D in L. Then

es, = - - - = efp,

f& = . - = f&.

Proof: If '$ = 3, for some i(l 5 i 5 g), then e,g = e~ for all a E G by Corollary 5-4-5 and Exercise 5-4-3. However, every '$, is of the form a'$ by Proposition 5-4-6, whence the first assertion. By Corollary 5-4-5 and Lemma 5-4- 1,

es,fs, = . . . = es,ls,-

Therefore, the second assertion follows from the first. //

In case LIK is normal, we shall denote the common value of the ep, (respectively, fp,) by e, (respectively, f,). If we write g, instead of g, then we may reformulate Theorem 5-1-3 as

Corollary 5-48 : e, f,g, = n.

Remark: In case LIK is abelian, Lemma 5-4-1 and Exercise 5-4-3 imply that Dv and Is depend only on +.I and not on the choice of 3. Therefore, in this case, we shall write D,(respectively, I,) instead of Ds (respectively, Is).

5-5 GLOBAL HILBERT THEORY

Suppose that LIK is a Galois extension of degree n, G = Gal (K). Let p be a finite K-prime, '$ a prime divisor of p in L. As we did in the local theory, let us connect the Galois theory of LIK with the ramification theory.

First, let us suppose that !@ is unramified in LIK. By Proposition 5-4-5, Is = {I], SO that by Theorem 5-4-4,

Dp = Gal (LQ/K,) (*)

where the isomorphism is gotten by "reduction modulo '$." By the local theory, Gal (Ls/K,) is a cyclic group of order f,, and is generated by the Frobenius automorphism x -4 xNP. Therefore, by the isomorphism (*), Ds is cyclic of order f, and possesses a canonical generator a s , called the Frobenius automorphism at '$, such that

a s x - xNp(mod '$) ( x E OL).

Recall that in Section 5-4, we identified Dg with Gal (Lv/K,). With respect

to this identification, as is identified with the local Frobenius automorphism (Ly). More precisely,

Whenever we wish to make the extension LIK explicit, we shall write (L/K)/'$ instead of as .

Proposition 5-5-1: Let z E G. Then 43 is unramified in LIK and

b,cg = ZO$Z-'.

Proof: Since e,s = es, T'$ is unramified in LIK. Moreover, a,p is uniquely characterized by the property

However, from the corresponding property for a ,

a 1 3 z - 1 ~ = ~ - 1 ( x N ~ ) ( m ~ d ' $ ) ( ~ € 0 , ) => zapz-lx r xNp(mod z'$) (X E OL) * 0,s = ~ f 7 ~ ~ - ' . //

Remark: By Proposition 5-5-1, if LIK is abelian, as depends only on p and

not on '$. In this case. we shall write a, [respectively, (?)I instead of 09

[respectively, (y)]. If PI = '$? . . . '$7 is an L-ideal, with (PI, dL,K) = 1,

we define the symbol

If LIK is abelian, then the symbol Lx for W a K-ideal, is defined analo-

gously. ( 8 )

Let K c M c L be a tower of number fields, p a finite K-prime, '$ a prime divisor of p in M, Q a prime divisor of '$3 in L. Assume that Q is unramified in LIK.

Proposition 5-5-2 :

Proof: Note that NQ = Npf, and that Q is unramified in LIM. Therefore,

(=#) is well defined and

which is the defining relation for ("$") - . I/

Proposition 5-5-3: Assume that MIK is Galois. Then

Proof: Exercise.

From Proposition 5-5-3, we immediately deduce

Corollary 5-5-4: Let K c M c L be a tower of number fields with LIK and MIK Galois. Let % be an L-ideal such that (9, dLIK) = 1. Then

MIK , (F) l M = (T) where 8 = % n M.

A similar statement can be made for the situation of Proposition 5-5-2. However, the statement in this case can best be put into an elegant form when LIK is abelian:

Corollary 5-5-5: Let K c M E L be a tower of abelian extensions. Let %be an M-ideal such that (go,, dLIK) = 1 . Then

Proof: If $ and p are as in Proposition 5-5-2, then the assertion is clear for % = 9, upon noting that NMIK($) = pf, f = f($/p). The general assertion now follows in general from the definition of the symbol. //

Having explored the properties of the Frobenius symbol, let us now turn to the real subject of this discussion-the global Hilbert theory. Let us fix a finite K-prime p and a prime diviser '$3 of p (possibly ramified) in L. For each nonnegative integer i, let us define

Gi = { a E G la(x ) r x(mod pi+'), x E O,].

It is clear that Gi is a subgroup of G, and is called the ith ramification subgroup of G at $. It is clear that Gi is a subgroup of DV, and that

Go = I, D B ? G o z G l 2 . . . .

It is also clear that under the identification of D8 with Gal(L@/K,), Gi cor-

responds to the ith ramification group for the local extension La/K,. There- fore, from the local theory, we have

Proposition 5-5-6: (1) Gi ( i 2 0 ) is a normal subgroup of D$. (2) Dp/Go has order f =f,. (3) Go/Gl has order = the largest factor of e, that is relatively prime to

Np. The order of Go/Gl divides Np - 1. (4) Gi/Gi+, ( i 2 1) is an elementary p-group, where p is the restriction

of P to Q.

Corresponding to the decreasing sequence of ramification subgroups, there exists an increasing series of subfields of L which contain K:

where Z is the fixed field of D@ and T, (i 2 0) is the fixed field of Gi. Z is called the decomposition subfield of L at p. To is called the inertia subfield of L at 9. Since Drp is not necessarily a normal subgroup of G, Z is not necessarily a normal extension of K. However, Gi is a normal subgroup of Dq, so T, is normal over Z. In particular, T o is normal over Z and T.+, is normal over T, ( i 2 0). Let g = g,, f = f,, e = e,. Let p = the restriction of $ to Q, e = eopr, (e,, p) = 1 . Then by Proposition 5-5-6,

Proposition 5-5-7: (1) deg (ZIK) = g. (2) deg (T, /Z) = f. (3) deg (TlITO) = eo. (4) deg (Ti+JTi) = a power of p.

Proof: The order of D@ is ef, and the order of Z* is e. //

Since Gal (LIZ) = Dv, the correspondence between the local and global theories given by Proposition 5-4-2, implies that there exists an isomorphism Drp =t Gal (L@/ZQ), where = the restriction of $ to Z . Since Z Q 2 K, and since Gal ( L r p / Z ~ ) Gal (L9/KD), we see that ZQ = K,. Therefore, the analysis of the tower of extensions

can be reduced to the analysis of the corresponding tower of local extensions. By the local theory and Proposition 5-5-7, T,/Z is unramified at the restriction of to To, and deg (T, /Z) =f. Therefore, the extension T,/Z multiplies the residue class degree by S. Similarly, LIT, is totally ramified at !Q, so that the ramification degrees are multiplied by e, and the residue class degrees are left unchanged. Thus, since efg = n, the only possibility left for the extension Z / K is for p to split into g factors in Z . To summarize: In ZIK, p does all its splitting. In T,,/Z, the residue class degree is increased, but the ramification degree remains constant. In the extensions r . + , / r . (i 2 O), the ramification

degree is increased, while the residue class degree remains constant. In Tl/To, the ramification degree is multiplied by an integer prime top. This type of ramification is called tame ramification. In Ti+,/Ti, the ramification is multiplied by a power of p.

Remark: Ds is defined even when '$3 is infinite. Let us define 1s to be equal to Drp if '$3 is infinite. Let Z and To denote, as above, the fixed fields of Dip and I$, respectively. Then Z = T. Moreover, the assertion that Ds has order e, f, and Zip has order e, holds even if '$3 is infinite. Moreover, the statement that a K-prime p does all its splitting in the portion ZIK of the extension also holds in general. This definition of D$ and Zip for $ infinite will prove useful when we treat class field theory in Part 4.

5-6 AN EXAMPLE-QUADRATIC FIELDS

Let us illustrate the theorems of the preceding discussion in the case of a quadratic field L = @(a), d a rational integer. Let us identify Gal (L/@) with {f 1). Let p be a positive, rational prime. Set

($) = 0 if p ramifies in L/@,

L1@ otherwise. = (7)

Proposition 5-6-1: (1) In case d = the discriminant of L, (d/p) = 0 o p divides d.

(2) (dip) = + 1 (respectively, - 1) o p splits completely in L/@ (respectively, is inertial in L/@).

Proof: (1) Theorem 5-3-5.

(2) (L+) = 1 o p is unramified in L/@ and f, = 1 9 e, = f, = 1

The symbol (dip) coincides with the Kronecker symbol of elementary number theory. In order to prove this, we require the following

Lemma: p splits completely in L/@ o X2 - d factors into linear factors in @,[XI.


We shall prove the coincidence of our symbol with the Kronecker symbol by separately evaluating the symbol for p = 2 and p an odd prime.

Proposition 5-6-2: Let p be an odd prime, p I / d. Then (dip) = 1 .c=. the congruence

has a solution in rational integers.

Proof: Assume that (dlp) = 1. Then by the Lemma, X2 - d factors into linear factors in @,[XI, so that X2 - d factors into linear factors in @,[XI = Z/(p)[X]. Thus, X2 - d(modp) has two distinct zeros in Z/(p). Thus, the congruence has a solution in rational integers. Conversely, assume that the congruence has a solution in rational integers. Then X2 - d(modp) factors into linear factors. Since p is odd, these factors are distinct (2X $O(mod p)). Therefore, by Hensel's Lemma, X 2 - d factors nontrivially in @,[XI, so that (dip) = 1 by the Lemma. The corresponding assertion for (dip) = - 1 follows since (dlp) # 0, since p is odd and p ) d. //

Remark: Let p be an odd prime, d any rational integer (not necessarily a discriminant). We shall define the symbol (dip) to be + I if the congruence x2 3 d(modp) is solvable in rational integers and - 1 otherwise. By Pro- position 5-6-2, this symbol extends the definition of the symbol (dlp).

Proposition 5-6-3: Let d be odd. Then

Pro05 By the Lemma, it suffices to show that X2 - d factors into linear factors in @,[XI o d r I or 5(mod 8). This is equivalent to showing that d is a square in @, o d = 1 or 5(mod 8). If d is a square in @,, then d is a square in Z,, and conversely. Assume that d = a2 with

m

a = C ai2' (a, = 0 or 1). i = O

Then a 2 r a: + aoa122(mod 23)

= 1 + aoa122(mod 23),

since dis odd. Therefore, d r 1 or 5(mod 8). Let the binary expansion of d be

Assume that d E 1 or 5(mod 8). Then bo = 1, b, = 0. Then one can define a, E {O, 1) (1 < i co) recursively, so that for each positive integer m,

The details of this routine calculation are left to the reader. Then

ca

a = C ai2' E @,, i = O

a n d d = a2. /I

From Propositions 5-6-2 and 5-6-3, it follows at once that if d is the discriminant of a quadratic field, then (dip) coincides with the Kronecker symbol.

Proposition 5-6-4: Let dl and d2 be rational integers, p a positive rational prime. Then

Proof: Without loss of generality, assume that p does not divide dld2. As- sume that p = 2. Then ((d1d2)/2) = + 1 o dld2 - 1 or 5(mod 8) 0 dl =. 1 or 5(mod 8) and d2 = 1 or 5(mod 8), or dl - 3 or 7(mod 8) and d2 = 3 or 7(mod 8) o (d,/2) = (d2/2) by Proposition 5-6-3. Assume that p is odd. By Proposition 5-6-2, (dld2/p) = 1 o the congruence x2 r dld2(modp) is solvable in rational integers. If this is so, then either the congruences

x 2 = dl(modp), x2 - d2(modp)

are both soluble or are both insoluble. Conversely, if the two congruences are both solvable or are both insoluble, then the congruence

x2 - dld2(modp) (*I is solvable. Indeed, the reduced residue classes modulo p which are represented by squares are (p - 1)/2 in number, since they are represented by the residues modulo p of 12, 22, . . . , ((p - 1)/2)2. Therefore, the reduced residue classes modulo p which are represented by squares form a subgroup of index 2 in (Z/(p))". Thus, if d, and d2 are both squares modulo p or are both nonsquares modulo p, then dld2 is a square modulo p. We have, therefore, demonstrated that ((dld2)/p) = 1 o both (dl/p) and (d,/p) are 1 or -1, whence the assertion of Proposition 5-6-4 for p odd. //

It is customary to extend the definition of the Kronecker symbol to denominators that are arbitrary nonzero rational integers. If n is such, with

n = (sign (n))pyl . . . P;~,

then we set

I f p is odd and unramified in L/@, then

where (n) denotes the principal @ideal generated by n.

EXERCISES 5-6-5: (1) (The Frobenius Under Composites). Suppose that LIK and L'IK are Galois extensions with Galois groups G and G', respectively.

(a) LL' is Galois over K and Gal (LL'IK) is isomorphic to a subgroup of G x G' under the mapping a -+ (a I,,, a I,.).

(b) Suppose that Q is a finite LL'-prime, '$ = Q I,, '$3' = Q I,,. Suppose that '$ is unramified in L/K and that 9' is unramified in L'IK. Then Q is unramified in LL'IK.

(c) Using the same notation as in (a) and (b), assuming that '$and '$' are unramified in LIK and L'IK, respectively, show that

(2) (Linearly Independent Extensions) Let L/K and L'/K be two extensions of number fields. We say that these two extensions are linearly independent (or that L and L' are linearly independent over K) if deg (LL'IK) = deg (LIK) deg (L'IK). Throughout this exercise, let us assume that L and L' are linearly independent over K.

(a) If {xl, . . . , x,} is a basis of LIK and {yl, . . . , y,} is a basis of L'IK, then {xiyj) is a basis of LL'IK.

(b) If LIK and L'IK are both Galois with Galois groups G and G', respectively, then the mapping

Gal (LL'IK) -+ G x G'

defined in Exercise (la) is an isomorphism. Given a E G, z E G', there exists an element a z E Gal (LL'IK) such that a z IL = a , a z IL. = z.

(c) OLL, = OL.OLr (= the smallest subring of LL' containing both 9, and 9 , ) .

(d) If '$, Q, and $3' are as in Exercise (2b) above, then

(e) Let all notations be as in (a) through (d) above. Then

(3) (Idele Norms) Assume that L/K is a Galois extension of number fields. Let us consider QK as embedded in 9, in a locally diagonal fashion-that is, a E QK is identified with the L-idele with component up for all L-primes '$ which divide p. Let us define an idele norm mapping

(a) N L I ~ is a homomorphism. (b) NLIK is continuous. [Hint: NL?31Kp(Ug) E UP.] (c) In Section 11-2, we shall show that NLgIKp is an open map and that

NLgIKp(Ug) = UP if LglKp is unramified. Use these facts to show that NLfK is open.

(d) The diagram

[where the lower norm is the ideal norm of Exercise (I)] is commutative. (e) The diagram

Applications

(where the lower norm is the usual norm from L to K) is commutative.

6-1 SPECIAL FUNCTIONS

Let K be a local field. If K is non-Archimedean, then we can introduce special functions on Kin much the same way as we introduce such functions on R or C. In this section, we shall illustrate this process by introducing the root, logarithm, and exponential functions on K. For t E K, n E Z, we write formally

" t j log (1 + t) = C (-I)/ - j = o j! (2)

" t' exp (t) = C T.

j = o J! (3)

The main questions we wish to answer are: 1. For what t do the series (1)-(3) converge? 2. Is (1 + t)'In the inverse function to t" - 1, and conversely? 3. Is log (1 + t ) the inverse function to exp (1 + t), and conversely?

The techniques are the same for dealing with all three functions, so we will give details only for the root function.

rn

Proposition 6-1-1: A series C a, in K converges - a, -+ 0 as n --, cm. n = 1

Proof: => Clear.

Let s, = C a,. Then for m 5 n, m= 1

I max l akl, m + l g k l n

Thus, s, - s, - 0 as m, n - co. Therefore, the sequence of partial sums is a Cauchy sequence. Since K is complete, the assertion follows. //

Proposition 6-1-2: The exact power of the prime p dividing n ! , n E Z+, is

where [x] = the greatest integer less than or equal to x.

Proof: Among the integers 1 , 2, . . . , n, there are [nip] multiples of p, [n/p2] multiples ofp2, etc. 11

Suppose that K 3 @,, e = e(K/@,), m = ordK (n).

Theorem 6-1-3: The binomial series ( 1 ) converges for

ord, ( t ) > rn 4- e/(p - 1 ) .

Proof: By Proposition 6-1-1, it suffices to show that the value of the jth term of the series (1) tends to 0 as j + m. Since

(17) = (1 /4 (1 /n - 1 ) . . . ( l / n - j + 1 1 , j!

we see that

ord, ( ( I F ) ) 2 o r d , (nj) - ord, ( j ! )

= -jm - ord, ( j ! ) .

By Proposition 6-1-2, we see that

ordK ( j ! ) 5 e([$] + [f] + . . .)

+ ) (where pk < j < pk+l)

Thus,

so that

ord, (('?)tj) 2 (ord, ( t ) - m - e e

-)j+--- p - I p - 1 (*I

Corollary 6-1-4: Vord, ( t ) > nz + e/(p - l ) , then ( 1 + t)'ln exists and

ord, ( ( 1 + t)lin - 1 ) = ord, ( t ) - m.

Proof: Formal properties of the binomial coefficients guarantee that the series ( 1 ) represents an nth root of 1 + t, as soon as the series converges. The ordinal of the term j = 1 is ord, ( t ) - m. If j 2 2, then the ordinal of the jth term is at least

e 2 ord, ( t ) - 2rn - - e > ord, (2) - m $- ord, ( t ) - m - - P - 1 P - 1

2 ord, ( t ) - m.

Therefore, the ordinal of the quantity ( 1 + t)I1" - 1 is the ordinal of the term j=1. //

Corollary 6-1-5: (1 + Pi)" = 1 + Pi+", if i > max {m, e / (p - I ) ) , P = P,.

Proof: Theorem 6-1-4 shows that 1 + Pi+" G ( 1 + Pi)" if i > e/(p - 1). Let us prove the reverse inclusion. It suffices to show that the mapping x - xn of 1 + Pi into U, is injective if i > m. The kernel of this mapping is contained in the group of nth roots of unity in 1 + Pi. Let us show that I is the only such. If x is a nontrivial nth root of unity in 1 + Pi, then

1 + x + . . . + x n - ' = 0 .

Moreover, xO = l(mod Pi) (a = 0, 1, . . . , n - 1) . Therefore,

1 + x + . . - + xn-I = n(mod Pi),

so that n = O(mod Pi) * m 2 i, a contradiction to the assumption that i > m. /I

Corollary 6-1-6: Let i > max {m, e/(p - 1)). Then the functions tn and tlln, defined, respectively, on 1 + Pi and 1 + Pi+", are inverse functions of one another.

We shall just state the properties of the logarithm and exponential. The proofs of the statements are left to the reader. The proofs required closely parallel the arguments given above.

Theorem 6-1-7: The series for the logarithm converges for ord, ( t ) > 0. In this region,

ord, (log ( 1 + t ) ) = ord, (t) .

Moreover, in the region of convergence,

1% ((1 + t)(l + 4) = log ( 1 + t ) + 1% ( 1 + u).

Therefore, log is continuous for ord, ( t ) > 0.

Theorem 6-1-8: The series for exp converges for ord, ( t ) > e/(p - 1). In this region,

ord, (exp (t) - 1) = ord, (t).

Moreover, in the region of convergence,

exp (t + u) = exp (t) exp (u).

Therefore, exp is continuous for ord, (t) > e/(p - I).

Theorem 6-1-9: Let r > e/(p - 1). Then log and exp are inverse isomorphisms (algebraic and topological) between 1 + Pr and Pr.

As an application of the results on special functions, let us prove

Proposition 6-1-10: Let LIK be an extension of local fields. Then NLl, is an open mapping of L into K.

Proof: Since N,,, is a homomorphism, it suffices to show that it is open at 1. Without loss of generality, it suffices to show that NLl,(l + P3 is a neighborhood of 1 in K for all sufficiently large i. Since 1 + Pf, ? 1 + Pi, NL,,(l + P i ) 2 (1 + Pi)" = 1 + Pgm for all sufficiently large i, where n = deg (LJK), m = ord, (n). Thus NL,, is open at 1.

6-2 CYCLOTOMIC EXTENSIONS

Let m be a positive integer. The roots of the polynomial Xm - 1 in C are called the rnth roots of unity. The rnth roots of unity form a multiplicative group, as is easily seen. Since a finite subgroup of C is cyclic, the group of the mth roots of unity is cyclic. A generator of this group is called a primitive rnth root of unity. Throughout this section, let 5, denote a fixed primitive rnth root of unity. Then all primitive rnth roots of unity are of the form k, (a, m) = 1, and conversely every such quantity is a primitive mth root of unity.

If K is an algebraic number field, an extension field L of K of the form L = K([,) is called a cyclotomicfield. The arithmetic of cyclotomic fields gives us a chance to give an impressive demonstration of the machinery we have built up. In addition to providing us with an example of our general theory, cyclotomic fields are important in their own right. They play a central role in number theory, the full extent of which is not understood at the present time. For example, we shall prove the following theorem in Chapter 13.

Theorem (Kronecker): Every abelian extension of Q is a subjield of a cyclo- tomicjield.

Moreover, the cyclotomic fields will provide the means of proving Artin's reciprocity law.

Since K([,) contains all the zeros of Xm - 1, K([,)/K is a Galois extension. Let Gm = Gal (K(C,)/K). If a E G,, 4 , ) is a primitive rnth root of unity; indeed, 0([,)" = a([$ = I + a([,) is an rnth root of unity. If

a([,) is not primitive, a([,). = 1 for some a, 0 I a < m - 1 a(&) = 1 => k = 1, which is a contradiction to the fact that 5, is primitive. Thus,

Let Z; denote the group of units of the finite ring Z/(m). We have shown that there exists a mapping

a - a,(mod m).

It is trivial to check that the mapping is a homomorphism. Moreover, since the action of a on Cm determines a completely, this homomorphism is injective. Therefore,

Proposition 6-2-1 : A cyclotomic extension K([,)IK is abelian-that is, it is Galois with an abelian Galois group. The Galois group is isomorphic to a subgroup of Z;. In particular, deg (K(5,)JK) ( p(m), where p is Euler's function.

Proposition 6-2-2: TI (em - c) = (-l)m-lmm O<i, j S m - 1

i t j

Proof: Note that m - l xm- 1 = TI (X-Pm), i= 0

Equating the constant terms, we see that

Differentiating (I), and setting X = CA, we derive

i t ?

We now take the product of equations (3) for j = 0, . . . , m

By (2), the left-hand member of (4) is (-l)m-lmm. //

(3)

1, and get

(4)

Corollary 6-2-3: Let p be ajinite Q-prime. Ifp ramijies in Q([,)/Q, p divides m.

Proof: Let Gal (Q([,)/Q) = (a , , . . . , a,]. Then {I, [, . . . , Cf - I ] is a basis of Q(Cm)/Q, where, for simplicity of notation, we have written [ instead of [,.

The discriminant of this basis is easily calculated (it is given by a certain Vandermonde determinant) and equals

This product is a subproduct of the product of Proposition 6-2-2. Therefore, the discriminant of the basis divides mm. Since the discriminant of the field @([,) divides the discriminant of every basis, we see that the discriminant of @([,) divides mm. Therefore, i fp is a finite @-prime that ramifies in @(cm)/@, p divides mm, which implies that p divides m. //

Corollary 6-2-4: Let p be a finite K-prime. I f p ram$es in K(Cm)/K, then p divides m.

Proof: Letp be the rational prime which p divides. Recall that Gal (K(cm)/K) can be identified with a subgroup of Gal (@([,)/@). If p ramifies, then I, is nontrivial, since its order equals the ramification index. Directly from the definition of I,, we see that I, G I,. Therefore, Ip is nontrivial, s o p ramifies in @([,)/@. By Corollary 6-2-4, p divides m. Therefore, p divides m. //

Henceforth, we shall be concerned only with cyclotomic extensions of @.

Theorem 6-2-5: deg (@([,)/@) = ~ ( m ) .

Proof: For the sake of simplicity of notation, let us write simply 5 instead of 5,. Let f denote the minimal polynomial of 5 over @. Then f 1 Xm - 1. It suffices to show that f ( e ) = 0 if (a, m) = 1. To prove this last statement, it suffices to show that f(5p) = 0 i fp is a prime, p ,/' m. An elementary computation shows that f(X)P - f(XP) E pZ[X]. Since f(5) = 0, this implies that f((p) is divisible by p in OQco. Over C, f can be written in the form

where I is a certain set of integers i which satisfy 0 ( i ~ ( m ) - 1, (i, m) = 1. Assume that f(C;P) # 0. Then f(5p) can be written as a product of nonzero factors of the form 5~ - ri (i E I). Therefore, by Proposition 6-2-2, f(5p) divides mm. Thus, p divides mm, which implies that p divides m. This is a contradiction. //

Our next job will be to calculate the discriminant of the extension @(cm)/@ and to determine the ring of integers of @(cm). It is most efficient to tackle both of these problems at the same time.

Proposition 6-2-6: Assume that m = pa, p a prime, p a prime divisor of p in a r m > . Then

(1) 1 - 5, is a local uniformizing parameter at p. (2) (1 - Cm>""' = (p), where (x ) = the principal @(5,)-ideaI generated

by x. (3) PIP) = dm), PIP) = 1, PIP) = 1.

Proofi Note that Xm'p - I divides Xm - I. Moreover, the zeros of

are the primitive mth roots of 1. Therefore,

From equation (I), we see that

xm - 1 p = lim = lim IT (X- C) X-1 Xmlp - 1 X-1 O S b S m

( b , m ) = l

= (1 - [)p(m) 1-C It- O S b S m 1 - cm ( b , m ) = l

In order to prove (2), it suffices to show that (1 - Ck)/(l - Cm) is a unit in OQ((,), if (b, m) = 1. If (b, m) = 1, there exists a E Z such that ab E l(mod m). Then

are both integers, so that (1 - imb)/(l - im) is a unit. From equation (2) we deduce that e(p/p) 2 q(m) = deg (@(im)/@). Therefore, since efg = p(m), we derive (3). Since ord, (p) = e(p/p), equation (2) implies that

Corollary 6-2-7: p ramiJies in @(im) -- p divides m.

Proof: => Corollary 6-2-3. + Let pa(a > 0) divide m. Observe that @([,.) c @(Cm). Let p be a prime divisor of p in @([,.), El a prime divisor of p in @(im). By Proposition 6-2-6, e(p/p) > 1 . Therefore, e(D/p) = e(D/p)e(p/p) > 1. Thus, p ramifies in @ / @ 11

Proposition 6-2-8: Assume that m = pa, p a prime. The discriminant of the basis (1, c, . . . , p(m)-') is (-

Proof: As usual, the discriminant is given in terms of a Vandermonde determinant, and up to sign, equals

which equals NQcolQ(y), where

O<b<m ( b , m ) = 1

A trivial computation shows that y = fl(l;), where f = the minimal polynomial of C over @. Therefore, the discriminant of the basis considered is NQcolQ( f '(c)). Since m is a prime power,

Therefore,

From this follows

In order to compute NQ(q)lQ(q - 1)' observe that

= p.

Assembling all of the information tabulated above, we see that

The sign of the discriminant is (- l)v(m)f2 by Exercise 2-1-9, (2). //

Proposition 6-2-9: Assume that m = pa, p a prime. Then { 1 , c, . . . , i.(")} is an integral basis for @([).

Proof: It is clear that Z[c] G OQ,,,. It suffices to prove the reverse inclusion. Let iZ = 1 - c. Since

n j = (1 - 5)' E z[cl,

p = ( 1 - A)' E Z[A],

we see that Z[c] = Z[L]. Thus, it suffices to show that z[L] 2 OQ(,,. Let

be contained in OQ(C). It suffices to show that a, E Z ( j = 0, . . . , q(m) - 1). Since

Cramer's rule and Proposition 6-2-8 show that

Therefore, a, has at most a power of p in its denominator. Let n be the smallest nonnegative integer such that pnaj t Z ( j = 0, . . . , q(m) - I). Assume that n > 0. Let k be the smallest nonnegative integer such that p does not divide pnak. Such a k exists by the minimality of n. Then

a,Ar E PO@(,) = A P ( ~ ) O ~ ( , ) (by Proposition 6-2-6)

for r = 0, . . . , k - 1. Therefore, since u is an integer, pna E pO,,,, = Av(m)O

Q (0

* pnak E IZO,,,, n Z = pZ * pnak is divisible by p, a contradiction.

Therefore, n = 0 and all a j t Z. //

Corollary 6-2-10: I f m = pa, p a prime, dQ,,m) = (- l)9(m)/2mv(m)/pmlp.

Proof: Propositions 6-2-8 and 6-2-9.

Theorem 6-2-11: Let m be arbitrary. Then

Proof: Let pa be the exact power ofp dividing m. Since the global discriminant is the product of the local discriminants, it suffices to show that

since the sign of the discriminant is (-l)c(m)/2 by Exercise 2-1-9, (2). By Corollary 6-2-10,

since @(em) = @(cm/,=, (,a), the local extension @p(Cm)/@p(cp.) is unramified by Corollary 6-2-4. Therefore,

Theorem 6-2-12: Let m be arbitrary. Then (I,[, . . . , 59(m)-1) is an integral basis of @(5,).

Proof: It suffices to show that the discriminant of this basis is equal to the discriminant of the field, which is given by Theorem 6-2-1 1. This reduces to a routine calculation of a determinant, which we shall leave to the reader. //

Now that we have explicitly computed the discriminant and the ring of integers of a cyclotomic field, let us describe the law of factorization of primes in such a field. Let us first compute the Frobenius automorphism, which is really the heart of the matter.

Proposition 6-2-13: Suppose that p 4 m. Then

k = 56,(mod P)

i f and only i f & = 56,. Proof: It suffices to show that

5", - l(modp) (*>

if and only if r", = 1. e is clear. Thus, suppose that (*) holds. Assume that pm # 1. Note that

Differentiating and setting X = 1, we derive m- 1

m = n (1 - Pi). j = 1

In particular, 1 - C". divides m, which is a contradiction to (*), since p t m . /I

Theorem 6-2-14: Suppose that p I( m. Then

Proof: The Frobenius symbol is well defined by Proposition 6-2-7. The Frobenius symbol is characterized by the property

( Y ) x ~ ~ ( m o d p ) , x E oQIC,,,).

In particular,

But the left-hand side is of the form e. By Proposition 6-2-13, e = c. /I

Theorem 6-2-15: Suppose that p 4 m. Let f be the smallest positive integer such that pf = I(mod m). Then

(1) f = f,, the residue class degree at p. (2) p splits into a product of q(m)/f distinct primes, each having residue

class degree f. (3) p splits completely in Q(5,) (i.e.,p splits into a product of distinct

primes of degree 1) i f and only ifp = l(mod m). (4) Suppose that (a, m) = 1. Then all primes p such that p r a(mod m)

split in the same way-that is, they have the same e, f, and g.

Proof: (1) Since p 4 m, p is unramified, so e, = 1. We have shown that

generates the decomposition group D, of order ep fp = fp.

heref fore, the Frobenius at p has order f,. Thus, by Theorem 6-2-14, fp

is the smallest positive integer f such that pf = l(mod m). (2) epfpgp = deg (Q(5,YQ) = dm), and e, = 1. (3) Follows from (2). (4) By (I), the residue class degree of a prime p, p 4 m, depends only on

the residue class ofp modulo m. From above, ep = 1 and gp can be computed from fp via (2). //

In the above result, we have demonstrated the connection between arithmetic progressions and the splitting of primes in cyclotomic extensions of Q. Roughly speaking, all primes in the same arithmetic progression modulo m belong to the same splitting type in the field of the mth roots of unity. From the properties of the Frobenius automorphism, we can prove a similar result for quadratic extensions-namely,

Theorem 6-2-16: Let Q ( 0 ) be a quadratic extension of discriminant d. There exist q(d)/2 integers a , , . . . , a,,,,,,, (a,, d) = 1, such that

(1) I f p J d is a prime and f is the smallest positive integer such that pf = ai(mod d) for some i, then f = f,, the residue class degree at p.

(2) p splits completely in @ ( a if and only ifp = ai (mod d ) for some i. In particular, if two primes belong to the some arithmetric progression

moduro d, then they are of the same splitting type.

We shall deduce Theorem 6-2-16 from the corresponding theorem in the cyclotomic case. Let m and n be positive integers, (m, n) = 1. From Theorem 6-2-14, we see that

Thus, the left-hand side depends only on the residue class of n modulo m. The Frobenius automorphism therefore induces an isomorphism

defined by (@(i,)/@) a (mod m) ---t ------- .

72) @(dm G i2&i (3) Ifp is an odd prime, then @(cp) contains a unique quadratic subfrld,

namely, @(d(- 1) '~ - 1)12p).

Also, @(.J--8) = @(C,.\/8) @(5,).

(3) Since p is an odd prime, Gal (@(cp)/Q) - (Z/(p))" is a cyclic group of even order p - 1. Therefore, (Z/(p))* contains a unique subgroup of index two. Call this subgroup G. The fixed field of G is a quadratic extension of @, contained in @(cp), and is, in fact, the unique such. Suppose that this field is given by @ ( o ) , where d is its discriminant. Since p is the only prime that ramifies in @(cp)/Q, p is the only prime that divides d. Since 2 cannot divide d, d -- l(mod 4). Therefore, d = (- I)(P-l)/'p, and we have exhibited the unique quadratic subfield of @(cp). //

The discriminant of a quadratic field is called a prime discriminant if it is divisible by only one prime.

EXERCISES 6-2-18: ( 1 ) The prime discriminants are -4, 1 8 , (- 1)'~-l)I2p, p an odd prime, p > 0.

(2) Every discriminant of a quadratic field can be expressed as a product of prime discriminants.

The next result is a special case of Kronecker's theorem, which was cited at the beginning of this section.

Theorem 6-2-19: Let Q ( a ) be a quadratic field of discriminant d. Then @ ( a ) @(Cldl).

Proof: In the proof of Theorem 6-2-17, we showed that a(-) = @((,), @(a) S @(I;,), @(d(-l)(p-1)/2p) c @(Cp). It is easily checked that @(a) c @(C8). Thus, Theorem 6-2-19 is true for prime discriminants. By Exercise 6-2-18, (2), an arbitrary discriminant d can be written as a product p,, . . . p, of prime discriminants pi. Then

Let (n, d) = 1, n > 0. By Theorem 6-2-14,

We have observed above that the left-hand side depends only on the residue class of n modulo d. Therefore,

@(cd)/@ (7) lQ(6) = ($)

depends only on the residue class of n modulo d. Moreover, since

runs over Gal (@(Cd)/@) as n runs over (Z/(d))", (d/n) runs over Gal ( @ ( a ) / @ ) as n runs over (Z/(d))"; that is, (din) assumes values f I. Therefore, the mapping

is a surjective homomorphism. Its kernel is a subgroup of (Z/(d))" of order p(d)/2. Let the cosets of the subgroup be represented by a,, . . . , apcd,,2. Then

p is decomposed in @(G) )o (A) n = + I - p == ai(mod d) for some i.

If p is inertial in @ ( a ) , then p2 is contained in the kernel of the above mapping, whereas p is not. Therefore, 2 is the smallest positive integer such that p2 = a,(mod d) for some i. Conversely, if 2 is the smallest such integer, p is not contained in the kernel and p is inertial in @ ( a ) . This completes the proof of Theorem 6-2-16, (1). Part (2) of this theorem follows immediately from part (1). //

We should remark at this point that the computation of the Frobenius automorphism for a cyclotomic extension is the key step in determining the law of decomposition of primes in both the cyclotomic extension and in its

subfields. What we have shown above is that all quadratic fields are contained in cyclotomic fields. The same is true of normal cubic extensions, although this is not so easy to see. We shall return to the question of the law of decomposition of primes in Part 4, where the results of this section will appear in a much more intelligible light.

Theorems 6-2-14 and 6-2-19 contain the classical law of quadratic reciprocity as a simple consequence. Let us therefore close this chapter with a proof of this important result. In Chapter 13, this result also will appear in a much more intelligible framework, as a special case of the Artin reciprocity law.

Theorem 6-2-20 (Law of Quadratic Reciprocity-First Form): Let p and q

be distinct, odd, positive primes. Then

9 = (-~)'P-'"P- lV4-

(3) (%p) Proof: ( I ) By Proposition 6-2-17, (I), @((,) = @(s"i). Therefore,

(G) = 1 o p is decomposed in @(n)

(2) Observe that @((,)I@ has degree 4. The Galois group of this extension consists of the automorphisms (, + ( g a = 1, 3, 5, 7). The elements of the Galois group that fix the subfield @(n) [Proposition 6-2-17, (2)] are c, - cb8 (b = 1, 7). Therefore,

($) = 1 o p is decomposed in @(a)

(3) Let d = (- l)(p-l)"p. By Proposition 6-2-17, (3), @(d) is the unique quadratic subfield of @(Cp). Therefore, Gal (@((,)/(@JZ) consists of the squares of elements in Gal (@(Cp)/@). Using the same reasoning as in (1) and (2), we deduce that

o (; = (f

o q r a2(mod p) (Proposition 6-2-1 3)

Thus, from the definition of d, and the multiplicativity of the Legendre symbol, we derive

Utilizing (I), we find that the assertion (3) follows. //

The quadratic reciprocity law has a natural generalization.

Theorem 6-2-21 (Law of Quadratic Reciprocity-Second Form): Let M and N be nonzero integers, both odd, (My N) = 1. Then

(1) (2) I (- 1)W- l)/Z+(sim(M)- 1)/2.

1)W- 1)/8.

EXERCISES 6-2-22: Let K be a number field, L = K((,), p a K-prime not LIK dividing rn. Then (T)(. = Cv.

PART 2

Harmonic Analysis

Harmonic analysis on locally compact abelian groups

In Chapter 3, we saw how we could associate various locally compact groups to an algebraic number field, and we saw how the topological properties of these groups translate into arithmetic properties of the field. In the next two chapters, we shall continue our previous line of inquiry by applying the theory of harmonic analysis to the locally compact groups arising in number theory. In the present chapter, we shall review the basic facts about harmonic analysis on a general locally compact abelian group. We shall state only the main facts we require. An accessible reference for the proofs is Elements of Abstract Harmonic Analysis by G. Bachman (Academic Press, 1964). We shall illustrate the general theorems in the case of p-adic fields and adele rings. These illustrations, which are due to Tate [25], will form the basis for our introduction to the Hecke zeta functions in Chapter 8. Throughout Chapter 7, let G be a locally compact abelian group with Haar measure p.

7-1 DUALITY THEORY

A. General Theory

A character x of G is a continuous homomorphism of G into the circle group U = {z E C I l z 1 = 1). The set of characters of G can be made into an abstract abelian group by defining the product X,X, via

x,xz(x) = x1Wxz(x) (x E GI.

This abelian group is called the character group of G , and is denoted G*.

A generic element of G* will be denoted x*. We shall denote the value of x* at x by either x*(x) or (x, x*).

G* can be given a topology that turns it into a topological group as follows: Let K be a compact subset of G, and let N be an open neighborhood of 1 E 8. Then a typical basic open neighborhood of x* E G* is

N(x*: N, K) = {y* E G* I y*(x)x*(x)-I E N for x E K}.

This topology is the topolgy of uniform convergence on compact subsets of G. With respect to this topology, G* is a locally compact abelian group. (Not an obvious assertion! See Bachman, p. 202.) One of the most fundamental theorems about G* is given by

Theorem 7-1-1 (Pontryagin Duality Theorem): The algebraic homomorphism of G into (G*)* given by

X - - + x**

x**(x*) = (x, x*) (X E G, x* E G*)

is an algebraic and topological isomorphism of G onto (G*)*.

Proof: See Bachman, p. 241.

Directly from the definition of the topology on G*, it is easy to prove

Theorem 7-1-2: If G is compact, G* is discrete. If G is discrete, then G* is compact.

From the definition of the topology in G*, we see that if H is a closed subgroup of G, x 4 H, then there exists x* E G* such that x*(H) = 1, x*(x) # 1. Let

HL = {x* E G* 1 (H, x*) = 1).

Under the Pontryagin duality, we may identify (HI)L with a subgroup of G.

Theorem 7-1-3: H I is a closed subgroup of G* and (HI)L = H.

If x* E G*, then x* 1, E H*. It is clear that the kernel of the homomorphism x* -+ x* 1 , is HI. Therefore, there exists a homomorphism rH of G*/HL into H*.

Theorem 7-1-4: rH is an algebraic and topological isomorphism of G*/HL onto H*.

Let x* E (G/H)*. We can define i,(x*) E HL via i,(x*)(x) = x*(xH) (x E G*). It is easy to check that iAx*) E HL. We therefore have a homomorphism

iH: (G/H)* -+ HL.

Theorem 7-1-5: iH is an algebraic and topological isomorphism of (G/H)* onto HL.

Remark: Theorems 7-1-4 and 7-1-5 are dual statements to one another under the Pontryagin duality.

6. Some Examples

Example (I): Let G be a finite, abelian group. If G is given the discrete topology, then G becomes a locally compact abelian group. A character of G is a homomorphism of G into U. (Continuity is guaranteed by the discreteness.)

Proposition 7-1-6: Let x* E G*. Then

C x*(x) = g, if x* is trivial X E G

= 0, otherwise,

where g is the order of G.

Proof: If x* is trivial, then the assertion is obvious. If x* is nontrivial, choose x, E G such that x*(x,) # 1. If x runs over G, so does xx,. Therefore,

C x*(x) = C x*(xx0) X E G x € G

= x*(x,) C x*(x). X t G

Thus,

(1 - x*(x,)) C x*(x) = 0 x E G

=> C x*(x) = 0 X E G

since x*(x,) # 1 . //

Proposition 7-1-7: Let x E G. Then

C x*(x) = g, i fx = the identity of G X'EG'

= 0, otherwise.

Proof: Let us identify G with G** under the identification given in Theorem 7-1-1. Then x is identified with x** E G**, and

C x*(x) = C x**(x*). x.tO' * .€Go

The result now follows from Proposition 7-1-6. I/

Corollary 7-1-8: Given x E G, x f the identity, there exists x* E G* such that x*(x) # 1.

Proof: If x*(x) = 1 for all x* E G*, then C x*(x) = g + x is the identity

~ ' € 0 '

by Proposition 7-1-7. /I

Corollary 7-1-9: Given x, y E G, suppose that x*(x) = x*(y), for all x* E G*. Then x = y.

Proof: Immediate from Corollary 7-1-8. //

Example (2): G = K, a local field. Let us explicitly describe G*. Let x* be a nontrivial character of K. Given x E K, let us associate the function

I: K--+ U

Since y + xy is a continuous mapping of K into itself, I is continuous. Moreover, since

I(Y 1 + Y2) = MY 1 + Y2)7 x*) = ( X Y ~ + xy27 x*) = (xYI, x*)(xY~, x*) = %Y l)I02),

2 is a character of K.

Theorem 7-1-10: The mapping x -+ I is an algebraic and topological isomorphism of K onto K*.

Proof: Since

= %(Y)22(~)7

we see that x -+ I is a homomorphism of K into K*. If x is contained in the kernel of this mapping, then (xy, x*) = 1 for all y E K =+ x = 0 since x* is nontrivial. Thus, x -+ I is injective. Next, claim that the map is bicontinuous. For large, positive A, set

B = ( x E KI Ixl,<A}.

Then B is a compact subset of K. If N is a small neighborhood of 1 E T, then N(l: N, B) is a small neighborhood of 1 E K*. Moreover,

x close to 0 in K s- x*(xB) is contained in a small neighborhood of 1 E T

+ I is close to 1 E K*,

so x --t I is continuous. Fix yo E K so that x*(yo) # 1. Then, if x is so close to the identity character that x*(yo) 4 x*(xB), then yo 4 xB, and thus IxlK < 1 yo IK/A, which implies that x is close to 0 in K. Thus, x -- I is bicontinuous.

In order to complete the proof of Theorem 7-1-10 it suffices to show that R = K*. Let H denote the closure of R in K*. By Theorem 7-1-5,

(K*/H)* = H I = (x E K 1 (x, y) = 1 for all y E H)

~ ( x E Kl(x,jj)= 1 forally E K)

However, if (x, jj) = x*(xy) = 1 for all y E K, then x = 0 since x* is nontrivial. Thus, (K*/H)* = (0) => H I = (0) => H = (HL)l = K*. Thus R is dense in K*.

Let us think of K* as an additive group. We can view K* as a vector space over Q, by defining ax*(x) = x*(ax) (a E @,, x* E K*, x E K). It is trivial to verify that K* is a topological vector space over Q,. Moreover, K* is locally compact. Therefore, by Theorem 4-2-2, K* is finite-dimensional over Q,. l? is a vector subspace of K* and is therefore closed in K* (Theorem 4-1-10). Thus, since R is dense in K*, R = K*. //

By Theorem 7-1-10, K can be identified with its dual via any nontrivial character x*. For use in Chapter 8, let us normalize our choice of x*. To begin with, let us assume that K = Q,. What is required is an additive, continuous mapping A,: Q, --t R/Z, not identically 0. For then we may set x* = exp A,, where exp ( ) = e2"'( ).

Case a-p = oo : Set

A,(x) = -x(mod Z) (x E [R). (1)

Case b-pfinite: Let x E @,. Choose v E Z so that pvx E Z,. Since Z is dense in Z,, there exists c E Z such that

I pUx - c 1, < p-".

Set A,(x) = c/pv(mod Z). Then A,(x) - x E Z,, A,(x) E Q, and the denominator of A,(x) is a power of p. If a different pw and c' are chosen, then

so that

since only powers of p appear in the denominators. Therefore, A,(x) is well defined. It is clear that A, is additive and continuous.

In both cases,

is a nontrivial character of @,. Let K be any local field. Then K contains @, for some @-prime p.

Define the mapping A,: K-+ R / Z

by A A x ) = J p ( T r K ~ d x ) ) .

This mapping is clearly additive and continuous, since Tr,,,, is additive and continuous. Therefore,

X*(X) = e 2 n i A x ( x )

is a nontrivial character of K. By Theorem 7-1-10, we immediately deduce

Theorem 7-1-11: The mapping x -4 eZniA~cx) is an algebraic and topological isomorphism of K with its dual, where by e 2 K i A ~ ( x ) we mean the character y , e 2 n i A x ( x ~ ) .

Remark: When K is realized as the completion of a number field at a prime p, we shall write A , instead of A,.

Example (3): Let G = K x , K a local field. Let c be a character of K x . Before considering the structure of a general c, let us discuss two special cases :

Case a-c(U,) = {I): In this case, c(x) depends only on I x I,, so that the mapping I x 1 , - c(x)(x E K x ) is a character of T,. If K is Archimedean, then T , = R+, so that

c ( x ) = I x I " ( t E R ) , (*I since every character of [R+ is of the form y -4 yia(a E R). If K is non- Archimedean, then T , is the infinite cyclic group generated by NP,. There- fore, an easy computation shows that again c has the form (*). If K is Archimedean, then t is uniquely determined by c. If K is non-Archimedean, then t is determined up to addition of a multiple of 2a/log NP,.

Case b: Suppose that c' is a character of U,. Then c' determines a character c of K as follows. If K is Archimedean, x E K x can be written uniquely in the form

x = ru (r > 0, u E U,).

In this case, let us define c(x) = cl(u).

If K is non-Archimedan, let a be a local uniformizing parameter at P,. Every x E K x can be written uniquely in the form

In this case, we define

It is trivial to check that c is a character of K and that c I , , = c'. We say that c is induced by the character c' from the subgroup U,, and we denote this situation

c = ind (c').

Let c be an arbitrary character of K. Then c' = cl,, is a character of U,. Let d = ind (c'). Then d is a character of K and dl,, = c. Therefore,

e = cd-I is a character of K such that e(U,) = (11. Therefore, e is of the type considered in Case a. Therefore, there exists t E [R such that

c = ind (c') I 1%. (2) In order to get a firm handle on the characters of K x , it remains to

investigate the characters of the form ind (c'), where c' is a character of U,. There are three cases to consider.

Case b,: K real. Then U , = (f 1). Therefore, there are two possible c':

c: r 1

c:(x) = sign (x ) ( x E U,), where

sign (x ) = + 1 if x > 0 - - -1 i f x < O .

The corresponding induced characters are

C J X ) = sign (x ) ( x E K x ) .

Case b,: K complex. Then U, = U. It is well known that the characters of T are the mappings of the form eie -3 eine (n E Z ) . The corresponding induced characters are

where I I denotes the usual absolute value on C.

Case b,: K non-Archimedean. The set of subgroups of the form 1 + P ; (r > 0) of UK form a fundamental system of neighborhoods of 1 E U,.

Since a small neighborhood of 1 E U cannot contain a subgroup other than (11, and since c' is continuous, c'(1 + P i ) = (1) for all sufficiently large i. Suppose that i is sufficiently large for this to occur. Then c' is constant on the cosets UJ(1 + Pi) . But U, is compact, and 1 + P i is open in U, 3

U,/(1 + Ph) is compact and discrete, hence finite. Thus, c' is essentially a character of the finite group UK/(l + P3. Conversely, given a character d of this finite group, we can construct a c' by defining

Remark: Suppose that c is a character of K, with K non-Archimedean. We associate to c a K-ideal f which is called the conductor of c as follows: Set f = Pi, where i is the smallest nonnegative integer such that c(l + P i ) = (1). (By convention, 1 + P; = U,, P i = OK). Such an i exists by the reasoning used in Case b, and equation (2).

Example (4): As the reader might guess, our next examples are the global counterparts of Examples (2) and (3)-namely, the calculation of the dual groups of the adeles and ideles. Since both the adeles and ideles are constructed as restricted direct products, let us consider the general problem of calculating the dual of a restricted direct product G = J'J (G,: H,). For each

v p, G, is contained in G as a closed subgroup via the identification

X, -+ (1, 1, . . . , x,, . . . , 1, . . . ). Let x* E G*. Then the restriction of x* to G, is a character x,* of G,. More- over,

Proposition 7-1-12: For all but a finite number of p, x*(H,) = 1 and for x = (xP) E G,

x*(x) = n xp*(x,). P

Proof: Let U E U be a neighborhood of 1 that contains no nontrivial subgroup of U and let N be a neighborhood of the identity in G such that x*(N) c U and such that

N = n Np x IT Hp, P E S P f S

where S is a finite set of p and N, is a neighborhood of the identity in G,. But then

Gs = n (1) X HP G N P E S P'+S

and G, is a closed subgroup of G; thus x*(Gs) E x*(N) E U, so x*(G,) = 1. Therefore,

xp*(Hp) = 1, P $ S.

If x = (x,) E G and S' is a finite set of p containing all p for which (a) X, 4 H,, (b) p E S, or (c) H, undefined, then

x = (x,; 1, . . ., l)(l, . . . , I ; x,) -- ,$S' VES'

and

X*(X) = IT, xp*(x,). x*(l, . . . , 1 ; x,) P t s

The converse of Proposition 7-1-12 is true, namely,

Proposition 7-1-13: If x,* E G: and x,h(H,) = 1 for all but a finite number of p, then

dejines a character of G.

Proof: Since x, E H, for all but a finite number of p, the product is well defined. It suffices to show that x* is continuous. If x is close to the identity

in G, then threre exists a small neighborhood of the identity N = N, x P E S H,, N, a neighborhood of the identity in Gpy such that x E N. Without loss

P f S

of generality, assume that S contains all p for which x,* is nontrivial on H,. Then for the NP sufficiently small,

is close to the identity in U. I/

For each p we may construct

H: = (x,* E Gp* I (H,, xp*) = 13.

Since H, is open in G,, G , /H~ is discrete * (Gp/Hp)* is compact * H: is

compact. Also, H, is compact 3 H,* is discrete * GC,*/H, is discrete H i open in G,*. Thus for almost all p we have the compact, open subgroup H:. Thus we can form

Theorem 7-1-14: G* z IT (G:: H:) under the mapping x* -+ (x,*). P

Proof: We have already showed that the mapping

is an algebraic isomorphism. Let us show that it is also a topological isomorphism. Now x* E G* is close to the identity character o x*(B) is contained

in a small neighborhood of 1 E U for B G G some large compact set. Without loss of generality, assume that B is of the form

NP G G, compact, S a finite set of p. Assume that S is so large that if x,*(Hp) =& {I), then p E S. Then x* is close to the identity character o x*(B) is close to 1 o x,*(Nv) is close to 1 for p E S , x,*(H,) = 1 for p 6 St;. x,* close to the identity character in G,*(p E S ) , x,* E ~ ; ( p 6 S ) o x* = (x,*) is close to the identity in IT (G,*: H:). //

P

Example (5): As a particular case of Theorem 7-1-14, let G = A,, the adele ring of a number field K. Then

G = IT (K,: 0,). P

Therefore, by Theorem 7-1-14, and the fact that Kp is self-dual,

A: = ( K p : 0;). P

Proposition 7-1-15: Let p be Jinite. Then 0; = b-', where b = bKpl,, and KP 2 @,, where K,* has been iden t$ed with Kp as in Theorem 7- 1 - 1 1.

Proof: x x 0; o ( y , x ) = 1 for all y E OP t;. e2ziA~v(xy) = 1 for all y E 0, + A,$xy) E Z for all y c 0,- TrK,l,p(xy) E Zp for all y E 0, e x ~ b - l . /I

Since there is only a finite number of primes that ramify in the extension K/@, b-' = 0, for almost all p . Therefore, by Proposition 7-1-15, 0: = Op for almost all p. Thus,

A: = TI ( K P : OP) = AK. P

The isomorphism between A, and A: can be explicitly realized as follows: Define the continuous, additive mapping

A(x> = C M x P ) ( X = (xv)). P

This sum is well defined, since xP E b-I for almost all p, so Ap(x,) = 0 for almost all p. To x E A,, let us associate the character

of A,. Then

Theorem 7-1-16: A, is isomorphic to its own dual under the isomorphism - e2z iA(x~) .

EXERCISE 7-1-17: Let c E Q:. Set cp = the restriction of c to K;. Then ( 1 ) cp(Up) = ( 1 ) for almost all p. (2) c(a) = lIcv(a,).

P (3) Given a collection (c,], cp E (KpX)* such that cp is trivial on Up for almost

all p ; then = n cp E Q:.

P

(4) Use Exercise (3) and Example (3) to find the general form of c E g:.

7-2 FOURIER TRANSFORMS

Let f be a continuous function on G that is integrable with respect to the Haar measure p. The Fourier transform f * off is a function on G* defined by

One can show that f * is continuous. If, in addition, f * is integrable, then the Fourier inversion formula holds:

for a suitable choice of Haar measure p* on G*. Such a Haar measure p* is said to be dual to p. ( M . A. Naimark, Normed Rings, Noordhoff, 1964, p. 414.)

In the remainder of this section,.we shall explicitly determine a pair of dual measures for a local field and for an adele ring. As usual, let us begin with local theory.

Let K be a local field. Let us choose a Haar measure dx, on Kas follows: 1. If K is non-Archimedean, K 2 Q,, set dx, equal to the Haar measure

for which

2. If K = IR, set dx, equal to the usual Haar measure. 3. If K = C, set dx, equal to twice the usual Haar measure.

Then

Theorem 7-2-1 : The normalized measure dx, is self-dual. That is, on identijjy- ing K with its dual as in Theorem 7-1-1 1, the Fourier inversion formula holds for dx* = dx,.

Proof: It suffices to verify that the Fourier inversion formula holds for a singlef. Let us consider the above three cases separately.

Case 1: K non-Archimedean. Let f = the characteristic function of OK. Let y E K be identified with the character

eZn iA~(xy)

of K. Then

f*(y) = J f(x)e2niA~(x~) dx K K

If y E b-', b = bKIBD, then AK(xy) = 0 for all x E OK * f *(y) = Nb-'I2. If y $ b-l, then there exists x, E OK such that AK(x,y) $ Z, so that the character

e 2 z i A ~ ( x ~ )

is nontrivial. Therefore,

= e z n i ~ ~ ( ~ ( x + x ~ ) ) dxK (by translation-invariance of dx,) OK

+ (1 - e i ~ ~ t x ~ ~ J e z + ~ ~ ( x ~ ) dxK = 0 QK

+ f *(y) = 0 (since AK(xoy) $ Z).

We have, therefore, shown that

f * = Nb-'I2 . (characteristic function of b-I).

Using the same reasoning as that used to calculate f *, we see that for x E OK,

[n a local parameter at PK, = ~ a - l ~ ~ J zroK dyK r = ord, (b-')I

= Nb-If2 mod, (nr) ) QK dy,

- - Nb-l/2.NpirNb-112

= 1.

Similarly, if x 4 OK, then,f**(x) = 0. Therefore,

f **(XI = f(-x),

and the Fourier inversion formula holds.

Case 2: K = R. Set f(x) = e-""'. Then

By using Cauchy's theorem, we may shift the line of integration to see that

But this last integral is 1. (Multiply it by itself and transform to polar coordinates.) Therefore,

f *(XI =fW and f **(x) = f(x) = f(-x), so again the Fourier inversion theorem holds.

Case 3: K = C. Set f(z) = e-zlzl~ and parametrize z as a point x + iy of the complex plane. Then dz = 2 dx dy and 1 z 1, = x2 + y2. (Recall that in the complex case I 1, was normalized to be the square of the usual valuation.) Therefore,

- - 2e-4n(xq2+y*2)

= 2 f (2x* + 2iy*)

Therefore,

Theorem 7-2-1 together with Theorem 7-1-1 1 implies

Theorem 7-2-2: Let f be a complex-valued function on the local field K such that both f and f * are continuous and integrable. Then

Remark: When the local field K is explicitly realized as the completion of a number field at a prime p , we shall write dx, instead of dx,.

Let us now turn to the problem of constructing a self-dual measure on A,, the adele ring of an algebraic number field K. It is just as easy to consider

the problem of constructing a pair of dual measures on a restricted direct product

G = IT (Go : HP) P

of a family of locally compact groups {G,) with respect to the family {Hp) of open compact subgroups.

Suppose that for each p, there is given a left Haar measure dx, on G,. Suppose that

jnPdxP = 1

for almost all p. Let S be a finite set of p which includes all p for which Hp is undefined and all p for which Jq dx + 1. Let

Then G, is an open subgroup of G and GS is compact. sure dxS on GS so that GS is assigned the measure 1. measure

dx, = IT dxP x dxS PC,

Choose a Haar mea- Give G, the product

G is uniquely defined Since G, is an open subgroup of G, a Haar measure on by the requirement that its restriction to G, is dx,. It is easy to see that this Haar measure on G is independent of the choice of S. We shall denote this measure by dx or dx,.

P

EXAMPLE: Let G = AK. For each K-prime p, let dxp be the normalized Haar measure on KO. Then

J,P dxp = (Nhp)-'I2, hp = bKplBp.

= 1

whenever p is unramified in K/@. Thus, since only finitely many primes ramify,

j,, d"P = 1

for almost all p. Therefore, the measure dx makes sense.

Proposition 7-2-3: For each p , let

(a) f,(HP) = (1) for almost all

there be given f, E L1(GV). Suppose that

P .

Proof: In order to prove ( I ) , it suffices to show that

where B ranges over all compact subsets of G, by the Lebesgue dominated convergence theorem. But every compact subset of B is contained in G, for some finite subset S of p as above. Assume that S is so large that p $ S * fP(HP) = (1). By the definition of dx,

This proves (*), so that f E L1(G). By the Lebesgue dominated convergence theorem

where the limit is taken over the collection of compact subsets of G, partially ordered by inclusion. As above, we may assume that B runs only over sets of the form

Np c Gp a compact subset. Using the same reasoning as above, we have, for such B,

Therefore,

We can now use Proposition 7-2-3 to construct a pair of dual measures on G: Let dx,* be the Haar measure on G,* which is dual to dx,. Let Set

f , E L1(GP) be continuous for all p, f , = the characteristic function of H, for almost all V . Then

f ( 4 = n f,(x,) P

is continuous on G, for every S so that f is continuous on G, since the topology on G is the weakest one in which every G, is an open subgroup. More- over, by Proposition 7-2-3, f E L1(G). //

Proposition 7-2-4: (1) f,* = the characteristic function of H t for almost all p. (More precisely, this is true if J,dx = 1).

(2) f *(x*) = n f ,*(x,*) (x* = (x,*)). P

Proof: ( 1 ) Iff, = the characteristic function of H,, then

If x,* E H:, then the integrand is identically 1 3

for almost all p. If x,* 4 H:, then x , --+ (x,, x:) is a nontrivial character of the compact group H,. Therefore, by the trick used in the proof of Theorem 7-2-1,

(2) Let x* = (x:) E G* be given. For almost all p, x,* t H:. Thus for almost all p, f,(xP)(xP, x,*) is the characteristic function of H,. Moreover,

Since almost all factors of the product are 1 and f , E L1(GP). Therefore, by Proposition 7-2-3,

Proof: Apply the Fourier inversion formula and part ( I ) of Proposition 7-2-4. //

By Corollary 7-2-5, it makes sense to construct the measure dx:. P

Proposition 7-2-6: ]1I dx,* is the measure dual to n dx,. P lJ

Proof: Applying Proposition 7-2-4, (2), twice and using the fact that dx: is dual to dx,, we get

f **(XI = I71 f t*(x,) P

= n f,( -.,I P

= f(-47 for any f of the form considered in Proposition 7-2-4. //

Theorem 7-2-7: Let K be an algebraic number$eld, and for each K-prime p , let dx, denote the self-dual measure on K,. Then JJ dx, is a self-dual measure on A,. P

Proof: Immediate from Proposition 7-2-6. //

EXERCISE 7-2-8: Let G be a compact abelian group, c a character of G. Then

c(x) dx = dx if c is trivial

= 0 otherwise, where dx is a Haar measure on C. (Hint: Consult the proof of Theorem 7-2-1, Case 1.)

7-3 THE SCHWARTZ-BRUHAT SPACE

In the preceding section, we had occasion to study functions f for which both f and f * are continuous and integrable. In this section, we shall introduce a class of functions for which this is true-the Schwartz-Bruhat functions. It is possible to define the space of Schwartz-Bruhat functions for an arbitrary locally compact abelian group. However, in order not to get involved in a technical digression, we shall define this class of functions for three classes of locally compact groups G-namely, G = IR", G = K a non-Archimedean

local field, and G = A, the adele ring of a number field. These special cases will suffice for our purposes.

Definition 7-3-1: A Schwartz-Bruhat function f on P is a complex-valued C" function on P such that for every polynomial function p on P and every invariant differential operatort D on Rn, the function p Df is bounded.

Roughly speaking, a Schwartz-Bruhat function on Rn is a function which, along with all of its derivatives, is rapidly decreasing "at co." The collection of all Schwartz-Bruhat functions on P is clearly a complex vector space, which we shall denote Y ( P ) .

Let f E Y ( P ) . Since f is C" and rapidly decreasing, we see immediately that f is continuous and integrable with respect to Haar measure on P. The main fact about Y ( P ) which we would like to prove is

Theorem 7-3-2: Iff E Y ( P ) , then f * E Y(Rn).

As immediate consequences of this result, we get

Corollary 7-3-3: Iff E Y(lRn), then f and f * are continuous and integrable.

Corollary 7-3-4: The Fourier inversion theorem is valid for Schwartz-Bruhat functions. Moreover, the mapping f -+ f * is an isomorphism of Y ( P ) onto itself.

For the proof of Theorem 7-3-2, let us adopt the following notation: Let x = (x,, . . . , x,) E P , dldx = (dldx,, . . . , dldx,). We require several Lemmas :

Lemma 1: I f f E Y ( P ) , and p E C[X,, . . . , X,], then pf E Y ( P ) .

Proof: Obvious.

Lemma 2: If f E Y ( P ) , and p E C[X,, . . . , X,,], then pf* is bounded.

Proof: It suffices to consider the case p = x; (r > 0). Observe that if D, = dj/dx/, (1 < j < oo), then Djf E Y ( P ) . Also,

where x. y = x, y, + . . - + x,y,, dy = dy, . . . dy,. Integrating this formula r times by parts with respect toy , , and using the fact that f and all its derivatives are rapidly decreasing, we see that

t An invariant differential operator on Rn is a differential operator of the form q(d/dxl, . . . , dldx,) for some q E C[XI, . . . , X.].

(- l j r f x = D. f (y)e2""-y dy. (2azx,) R-

The integral is a bounded function of x, since D, f is in Y ( P ) , and therefore integrable. Thus, x; f * is bounded. //

Lemma 3: Iff E Y(lRn), p E CIX1, . . . , Xnl, then

Proof: The statement is linear, so it suffices to prove it whenp is of the form x; (r > 0). However, by induction, it suffices to verify the assertion when r = 1. Therefore, we are reduced to showing that

Let h E R, x + h = (x, + h, x2, . . . , x,). Assume that h + 0. Then

Note that e i x ~ h - 1

lim f (x) ------ = ix, f (x), h-0 h

and x, f is integrable on R". Therefore, by the Lebesgue dominated convergence theorem,

Proof of Theorem 7-3-2: Let p, q E C[X,, . . . , X,,] be arbitrary. By Lemma 1, p(ix) f (x) E Y ( P ) . Therefore, by Lemma 2, q(p(ix) f (x))* is bounded. By Lemma 3, this implies that q(x)p(d/dx) f (x) is bounded. Lemma 3 also implies that f * is a C" function on P. //

Next, let K be a non-Archimedean local field.

Definition 7-3-5: A Schwartz-Bruhat function f on K is a locally constantt function with compact support.

A typical element of the space of Schwartz-Bruhat functions Y(K) is the characteristic function of the set x + P;, (r E Z). We leave to the reader the proof of the elementary fact that every element of Y(K) is a linear combination of such functions. Now we are faced with proving the analogue of Theorem 7-3-2 for Y(K). Fortunately, our task is much easier in this case.

t A locally constant function is one that is constant in some neighborhood of each point.

Theorem 7-3-6: Let f E Y(K). Then f is continuous and integrable, and f * E 9'(K), SO the same is true off *. Proof: The only nontrivial assertion is that f * E 9'(K). By our remark above, it suffices to consider the case where f = the characteristic function of a + P;. Let x E K. Then

It suffices to show that the first factor is locally constant, whereas the second factor is locally constant with compact support. We claim that the first factor is constant on the neighborhood x + a-'bil of x. indeed, if y = x + a-'z (z E bi l ) , then

e2n iA~(ay) = e 2 n i A ~ ( o x ) e 2 n i A ~ ( z )

The integral is locally constant, since if w = x + z (z E bi1P;*), then e2n iA~(wy) = e2niAp(Xy) (Y E Pk).

The integral vanishes if x 4 &'Pir , since in this case, the function y --+

ezn'A~(xy) is a nontrivial character of the compact group P;. //

It now remains to define the space Y(AK) of Schwartz-Bruhat functions on the adele ring A, of a number field K. Let

Then

A Schwartz-Bruhat function on A, is a linear combination of functions of the form

f = fm IT fp, f m E Y(A=-)) f p E ~ ' (KP) , pes-

where f, = the characteristic function of 9, for almost all p. It follows from the definition of the topology on A, that such a function is continuous. Moreover, by Theorem 7-2-3, f is integrable.

Theorem 7-3-7: Iff E Y(AK), then f * E Y(AK). Jn particular, f * is continuous and integrable.

Proof: Without loss of generality, assume that f is product of local functions. By Proposition 7-2-4, f ,* = the characteristic function of 0, for almost all p. Moreover, by Theorems 7-3-2 and 7-3-6, f 2 E Y(A,), f,* E Y(KJ (p 4 S,). Therefore, by Proposition 7-2-4,

f * = f5IIf,* P

PART 3

Hecke's L-Func tions

is contained in Y(A,). //

Hecke's L-series

In this chapter, we shall introduce the Hecke L-series, which generalize the various Dirichlet series introduced in Chapter 1. Our main objective will be to show that a Hecke L-series has an analytic continuation and a functional equation. The main idea of the proof is to break a Hecke L-series up into an Euler product and to interpret the factors as "local zeta functions." By adding local zeta function for the infinite primes, we get a "global zeta function," which has an analytic continuation by the Poisson summation formula. Our theory, then, is divided into three parts. In Section 8-1 we shall study local zeta functions; in Section 8-2, we shall study global zeta functions; in Section 8-3, we shall define and analytically continue the Hecke L-series. This chapter is, essentially, an exposition of Tate's thesis 1251. The main change has been to introuduce the local and global Schwartz-Bruhat functions as a means of realizing the axioms in Tate's thesis.

8-1 LOCAL ZETA FUNCTIONS

Through Section 8-1, let K be a local field, K 2 @,. If K is non-Archi- medean, let P, and b, (or just b if no confusion can result) denote, respectively, the prime divisor of K and the different of K/@,.

Before defining the local zeta functions, let us explicitly construct a Haar measure on K x : Let Y ( K x ) denote the space of continuous, complex- valued functions on Kx which have compact support (i.e., the closure of {x E K x I f ( x ) # 0) is compact). Consider the linear mapping of X ( K x ) into C defined by

where dx, is the normalized Haar measure on K (Section 7-2). For any a E K X ,

d(a-lx), = mod, (a-l) dx,

= 1 a 1;' dx, (Proposition 2-6-26).

Therefore,

Thus, the linear functional defined by (*) is invariant under translation by K. Iff 2 0, then the value of the functional on f is nonnegative, so the linear functional is positive. By a well-known result in measure theory, this linear functional can be extended to a Haar integral on Kx. Throughout this section, let a denote a variable ranging over Kx , x a variable ranging over K. Let d*a, denote the Haar measure induced by the above linear functional. Finally, let us normalize the Haar measure on K as follows:

d "a, = d *a, (K Archimedean)

d X a K = NPK(NPK - I)-' d*aK (K non-Archimedean).

Proposition 8-1-1 : 1 d a, = (N b)-11' (K non-Archimedean). U K

Proof: From the definition of dxaK,

I d x a K = NP,(NP, - I)-' I dx,. U K U K

Break 0, up into cosets modulo P,, and notice that U, is the union of all cosets except P,. This implies that

dx, = (NP, - 1) 1 dx, (translation-invariance of dx,) U K PK

= (NP, - 1) I d(nx), (n a local uniformizing parameter) OK

= (NP, - 1) mod, (n) j dx, OK

= (NP, - 1)NP;'Nb-'I2 (Proposition 2-6-26 and the definition of dx,). //

EXERCISE: Let f be a complex-valued function on K ". Then f(x) is integrable with respect to d X a K + f(x) 1x1~' is integrable on Kx with respect to dxK. Moreover, if one of these conditions holds, then

Let c be a character of K x , f E Y(K).

Definition 8-1-2: A local zeta function on K is a function of a complex variable s, defined by an integral of the form

Proposition 8-1-3: The above integral converges for Re (s) > 0.

Proofi If K is Archimedean, then f is rapidly decreasing, and the convergence of the integral is an exercise in advanced calculus, which we leave to the reader. If K is non-Archimedean, it suffices to consider the case f = the characteristic function of PF. In this case, the absolute value of the integral is at most

l a l;d "a,, I,,-,,, where a = Re (s). Let

A v = { a E K I I a l K = N P ; V ) ( v E Z ) . Then

m

= Nb-'I2 NPiva (Proposition 8-1-l), v=r

where n is a local uniformizing parameter at P,. However, it is clear that the last series converges for a > 0. //

Next, we would like to investigate the analytic properties of local zeta functions. In order to do this, we must prove a criterion which implies that a function defined by an integral on a locally compact group is analytic. We shall phrase our results in a general enough way that they suffice for use in both the local and global theory.

Let G be a locally compact group with left Haar measure dx. Let D E C be a region and let f:G x D + C be a continuous function such that:

(1) For fixed x E G, f(x, s) is an analytic function of s E D; (2) For fixed s E D, f(x, s) and f,(x, s) are integrable with respect to

dx, where the latter means the derivative of f(x, s) with respect to s. We may then define the function

and ask the auestions:

( 1 ) When is g analytic in D? (2) When can the derivative of g be calculated by "differentiating under

the integral sign"; that is, when can we compute the derivative using the formula

Definition 8-1-4: Let G, D, and f be as above. If E E D is any subset, then we say that

converges uniformly on E if, given 6 > 0, there exists a compact subset C = C(6, E) such that

for all s E E. We say that the integral ( 1 ) converges uniformly on compact subsets of D if it converges uniformly on every compact subset of D.

Theorem 8-1-5: Let f, g, G, and D be as above. Assume that

converges uniformly on compact subsets of D. Then g(s) is analytic on D and

Lemma: Let C be a compact subset of G. Then

is analytic on D and

Proof: By hypothesis, we may expand f (x , s ) in a power series about s E D:

Note that

a d 4 = f ( x , 4 (1)

a m = f,(x, 8). (2)

Fix x E D. Let 6 > 0 be chosen so that the disc of radius 6 about s is

! contained in D. Since f is continuous on C x {z I 1 z - s I 61, there exists a positive constant A such that

I f (x ,z)I_<A ( ~ E C , I Z - S J I ~ ) .

By Cauchy's inequality,

I a,(x) I I A&" ( x E C). I

(3) From equations ( 1 ) through (3), we readily deduce that for x E C, l z - sl

I

1 612,

Therefore,

I g c ( ' ) - gc(S) - fs(x, s )dx l_< 2A6-' z - sl J dx. z - S C

Letting z approach s, we derive the assertion of the Lemma. //

Proof of Theorem 8-1-5: Consider the (generalized) sequence {gc} of functions on D, where C ranges over all compact subsets of G, ordered by inclusion, and

Since

converges uniformly on compact subsets of D, the same is true of the sequence {g,}. But since a uniform limit of analytic functions is analytic,

I lim gc = g C

is analytic. Moreover, by a well-known theorem in complex analysis,

g m = l i ~ gc, m

Proposition 8-1-6: [(f, c, s ) is analytic for Re ( s ) > 0.

Proof: Assume that K = C. Then

I I f (a)c(a) 1 a 1: 1 d X a r = 2J2= [ 1 f(reie) 1 r2--I dr d o (0 = Re (s)). C X 0 0

Let D be a compact subset of the upper half-plane, and let 6 > 0 be such that

s E D 3 Re (s) 2 6. Since f is rapidly decreasing, there exists a constant A > 0 such that

If(reie)(<Ar-36 ( r 2 1 ) .

There also exists a constant B > 0 such that

If(rei"><B ( r < l ) .

Therefore, if 0 < R < 1 < S,

Thus, the integral

converges uniformly on D. Therefore, Proposition 8-1-6 follows from Pro- position 8-1-5 in case K = c.

The case K = R is similar to K = and is left to the reader. Finally, assume that K is non-Archimedean. It suffices to show that

converges uniformly for Re (s) 2 6. Let n be a large positive integer. With- out loss of generality, we may assume that f is the characteristic function of Pr, for some integer n 2 r. Then

Proceeding exactly as in the proof of Proposition 8-1-3, we see that this latter integral has the value

Nb-'IZNPir"(l - NPi")-' < Nb-' /2NPp(1 - NPi6)-I, -

which completes the proof of the uniformity of convergence on compact subsets. Therefore, by Proposition 8-1-5, we see that((f, c, s) is analytic for Re (s) > 0. /I

The next result will be the main step required in showing that a local zeta function has an analytic continuation and a functional equation:

Proposition 8-1-7: Let f l y f2 E Y(K), and let c be a character of K. Then, forO< Re(s) < 1,

&fly c, s)5(f $3 c-', 1 - s) = C(f T, c-'9 1 - s)C(f,, c, s). (*> Proofi By Proposition 8-1-3, both members are defined for 0 < Re (s) < 1. Write the right-hand side as an iterated integral and apply Fubini's theorem to get that the right-hand side of (*) is equal to

Here d "aK d "/IK denotes the product measure on K x x K ". Replace (a, /I) by (a, aS) in the last equation and use the invariance of Haar measure under translation to get

C(fT, c-'5 1 - s)5(f27 CY s)

= J J f Xa)fz(aP)c(P) lalK IS I: dxaK dxSK KX

It suffices to show that

is symmetric in f l and f2 for y # 0. Replacing x by xy-I, and recalling that d(xy-I), = I y l i l dx,, we see that the integral equals

which is symmetric in fl and f,. //

We now come to the main result on local zeta functions.

Theorem 8-1-8: A local zeta function c ( f , c, s) possesses an analytic continuation to the whole s-plane as a meromorphic function of s. Moreover, there exists a meromorphic function p(c, s) such that

Rf , c, = p(c, sX(f *, c- , 1 - 3).

Proof: We have shown above that c = c'l Iit(t E R), where c' is induced from UK. Since

R f , c, $1 = R f , c', s + it), it suffices to assume that c is induced. For each induced character c, we shall choose a function f, E 9 ( K ) such that ((f,*, c-l, 1 - s) + 0. We shall set

We shall observe that p(c, s ) is meromorphic in the whole s-plane. Moreover, by Proposition 8-1-7,

C ( f , c , s ) = P ( c , s ~ ( ~ * , c- l , 1 - S ) (*I for 0 < Re ( s ) < 1, where the equality holds whenever both members are defined. The left member is analytic for Re ( s ) > 0 , while the right member is meromorphic for Re ( s ) < 1. Therefore, the functional equation (*) provides an analytic continuation of c( f , c, s ) as a meromorphic function in the entire s-plane, and the functional equation is valid for all s. Thus, it suffices to exhibit f , and to compute p(c, s). We break our computations up into three cases: K = R, K = C , K non-Archimedean.

Case 1: K = R. Let x , a be additive and multiplicative real variables, respectively. Then

A,(X)=--x(mod7) ( X E R), dx , = d x = ordinary Lebesgue measure on R,

da d x a - -, I I = the ordinary absolute value on R. ,- l a l Define

f+(x ) = e-..', f L ( x ) = ~ e - ~ ~ ' .

Both of these functions belong to Y ( R ) . We have previously shown that

f T(x) =f+(x>. Therefore,

/+(XI = Jm f+(y)e-2Kixy dy. -- Differentiating this equation, using Theorem 8-1-5, we derive

= -2ni f T(x) + f T(x) = - f i ( ~ ) .

Computing the zeta functions corresponding to ( f , , c+) and ( f , , c,), we see that

Case 2: K = C. Let z = x + iy be an additive complex variable, a = u + iv a multiplicative complex variable. It will be convenient to consider a in polar coordinates, a = rei8, r > 0. Then

dz, = 2 dx dy, da, - 2 dr dB d x a K = ----, - _, l alK

AK(z) = L ( 2 x ) = -2x(mod 7).

For n E 7, define

fn(z) = ( X - iy)lnle-2nlzl' ( n 2 01, = ( X + iy)lnle-21z12 ( n 1 O),

where I I is the usual absolute value on C. First we claim that

f :(z) = il"lf-,,(z). (10)

We have previouslyt verified the case n = 0. Let us proceed by induction on n. Assume (10) for n > 0 ; we have

Apply the differential operator

t See proof of Theorem 7-2-1.

to both sides of the last equation. We may differentiate under the integral sign by Theorem 8-1-5. Also, Dzn = 0 by the Cauchy-Riemann equations. Therefore, we see that

= 21:- J- (S A j t ~ + ~ e - ~ n ~ s z + t ~ ~ - 4 ~ i ( s x - t y ) dS dt - -

= f f + , ( x + ~ Y I , which is the inductive step. Thus, (10) is proven for n > 0. If n < 0, then by (10) for a positive integer,

by the Fourier inversion formula. Therefore,

which proves (10) in general. In order to compute the zeta functions, it is best to transform to polar

coordinates:

Therefore,

Case 3: K non-Archimedean. Let us denote the prime divisor of K by P instead of PK. Let b be the different of K. Let cn be a character of K having conductor Pn = f. Set

fn(x) = e 2 ~ i l \ ~ ( ~ ) , x E b-lp-"

= 0, otherwise.

Lemma: f f = Nb1I2 (characteristic function of OK) (n = O), = Nb'/2NPn.(characteristic function of 1 f Pn) (n > 0).

Remark: In particular, the Lemma implies that f f is locally constant with compact support* f .* E Y ( K ) 3 fn E Y ( K ) .

I Proof: Let us give the argument for the case n > 0. The argument for n = 0 is similar and is left to the reader. We clearly have

If x 4 1 + Pn, the mapping y -4 e2niA~((x-1)y) is a nontrivial character of the compact group b-'P-". In this case, f,*(x) = 0 by Exercise 7-2-8. //

Let us now compute C( f f , c i l , 1 - s). Let n be a local parameter at P. If n = 0, then the Lemma implies that

The zeta function C( f n , cn, s ) is somewhat more tricky to compute: Suppose that b = Pa. Then

, c s = 2 NP-" J e2nm~(5)c u r=-a-n pr-pr+~

"( > dXUK. (16)

Note that pr - pr+l - - nr(OK - P) = nrUK.

Therefore,

J e2ziA~(5)c,,(a) d X a K = J e 2 K i A ~ ( a ) c n ( ~ ) dXuK. pr-p-I n'UK

Let us denote this quantity by I,.

If r 2 -a, then nrUK G b-I =. AK(nrUK) G Z =>

where

If -a -n < r < -a, then -a - r > 0, so that 1 + P-"-'is a subgroup of U,. Since -a - r < n, 1 + Pn is a nontrivial subgroup of 1 + P-"-' =- c is a nontrivial character of the compact group 1 + P-"-', since Pn is the conductor of c. Represent nrUK by a disjoint union of the form

nru, = U /3(l + P-"-'). B

If y E /3(1 + P-"-'), then y = /3 + 6, 6 E b-l eZniAK(y)= e2niA~(B) .

Therefore, if -a - n < r < -a,

= c,,(/3)eZniA~(fl) 1 en(.) d "a, B 1 +P-'-'

= 0.

Combining (16), (17), and 18), we see that

Let us refine formula (20) further. Let

UK = U 8(1 + P"> B

be a disjoint union. Then

n-"-"UK = U n-"-"/3(1 + P") B

is a disjoint union. If y E ra-"B(1 + Pn), then y - n-"-"/3 E b-I * AK(y) = AK(/3/na+"). Moreover, since c, has conductor Pn, c,(y) = c,(/3/na+"). Therefore, by equation (20), we see that

By the multiplicative invariance of dxaK,

d x a K = j dxa,. 1 +P"

Therefore,

where p*(c,) is a generalized Gaussian sum given by

p*(cn) = ~ f - 1 1 2 cn (A) e2niAK(fllna+n).

I B

i Collecting the data in equations (14), (15), (19), and (21), we finally derive

1 p*(c,, s) = ~ ~ ( a ) - " N b ~ - ~ ~ ~ ( l - co(n)NPs-')(l - c0(n)NP-")-l (n = 0) = N(f b)s-'12p*(c,) (n > 0).

This completes the proof of Theorem 8-1-8. //

If the reader is wondering about the motivation for introducing the local zeta functions, let us consider the case where K is the completion of an algebraic number field at a finite prime p. Assume that p is unramified (from @), and that c is a character of K x with conducter pO, f = the characteristic function of 0,. Then equation (19) implies that

I &f, C, S) = (1 - c(~)Np-~)- ' .

This is a typical factor of an Euler product for a "zeta function" such as the Riemann zeta function or the Dirichlet L-series.

EXAMPLE: Let K = @,, and let x ---t Z denote reduction mod p. For x E UK, define c(x) = (nlp), where (RIP) denotes the Kronecker symbol. Set c(p) = 1, and extend c to K ? by multiplicativity. Then c has conductorp. The cosets UK/(l + j) are represented by (1,2, . . . , p - 1). The Gaussian sum p*(c) is given by

This is the classical Gaussian sum.

EXERCISE 8-1-9: Show that I p*(cn) I = 1. (Hint: Use the functional equation twice to relate [(f, c, s) to c( f **, c, s), and set s = i. This gives onerelation between p*(c,), p*(cil) and c,(-1). Next, conjugate the equation connecting c(f, c, s) to 5( f **, c, s) to get a second relation.)

8-2 GLOBAL ZETA FUNCTIONS

Let K be an algebraic number field.

Definition 8-2-1: An idele class character (or grossencharacter) of K is a character of flK which is trivial on K x . Equivalently, an idele class character of K is a character of flK/Kx.

Throughout this section, let c be an idele class character. As shown in Exercise 7-1-17, (3), c can be written in a canonical way as a product of local characters :

146 / HECKE'S L-SERIES CHAP. 8

where p runs over all K-primes, cp is a character of K," for all p,- and cp(Up) = {l) for almost all p.

Throughout this section, let f E Y(AK). Then f is a linear combination of elementary functions of the form

where f, E: y(K,), fp E y(KP),fp = the characteristic function of 0, for almost all p.

Let d x a = n, d x a p denote the canonical restricted direct product measure on -OK.

Definition 8-2-2: The global zeta function C(f, c, s) is a function of the complex variable s defined by

Remark: For simplicity in exposition, we shall assume that f is product a of local factors

f = JJ fP, f p E 9(KP), f p = the characteristic function of 0, for P almost all p,

instead of allowing the more general form (2). All our theorems will hold for the more general functions; we leave the general case to the reader.

Proposition 8-2-3: C(f, c, s) = a, ((fp, c,, s) for Re (s) > 1.

Proof: The integrand in (3) is a product of the local functions

f& I 1;- Let us show that for a = Re (s) > 1,

Proposition 8-2-3 will then follow from Proposition 7-2-3. Since all the factors in (4) are finite by the local theory, it suffices to consider the product over all primes 6 S, where S is some finite set of K-primes. Let S be chosen to include all infinite primes, all p for which fP # the characteristic function of@lp, and all p that ramify in K/Q. Then by the local theory, if c, is the trivial character of K,",

= (1 - Np-")-' (P f$ 9. In order to verify (4), it suffices to show that

n (1 - Np-")-l < PeS

for a > 1. Let p denote the @prime that p divides. Then

Therefore, since there are at most n primes p that divide a fixed p, where I n = deg (K/@), we see that

where Tis a finite set of @primes. However, it is clear that the product on the right converges for a > 1. //

Proposition 8-2-4: [(f, c, s) is analytic for Re (s) > 1 . 1

Pro08 By Proposition 8-2-3, C(f, c, s) is a product of local zeta functions, each of which is analytic for Re (s) > 0. Therefore, it suffices to show that

11 C(fp9 CP, s) P c s

1 converges uniformly for Re (s) 2 1 + S , 6 > 0, where S is as in the preceding proof. By the estimate (9,

so that by the analogue of the Weierstrass M-test for infinite products, the uniform convergence is proven. //

The main theorem of the global theory asserts that C(f, c, s) can be analytically continued to a meromorphic function on the whole s-plane, and that C(f, c, s) satisfies the functional equation

This theorom is much more delicate to prove than the corresponding result in the local theory. The proof will be based on the Poisson summation formula. Before proving this important formula, let us prove a preliminary result :

Theorem 8-2-5: K l = K relative to our ident$cation of A, with its dual.

Proof: Assume that x E K. Then x E KL o e2"iA(xy) = 1 for all y E K. Without loss of generality, assume that x # 0. Then, to show that x E KL, it suffices to show that e2"'*(y) = 1 for all y E K. Note that

= p ( Q ( ) ) (Theorem 2-7-8). P

Therefore, it suffices to show that

C P LP(4 E Z (2 E m Recall that Lp(z) is a rational number whoa denominator is a power of p for p # co. Also, 1,(z) - z(mod Z)(P # co) and L(x) = -z(mod Z). Therefore, if q is a finite @prime,

c n P ( ~ ) = c n P w + ~ , ( z ) + U ) I P Pfq,-

= C Lp(z) (mod Z). Pf q,

But the quantity on the right is rational, and its denominator is not divisible by q. Since q is an arbitrary, finite @-prime, C, I,(z) EZ O(mod a. This completes the proof of the inclusion K G KL.

By Theorem 7-1-5,KL=(AK/K)*. But AK/Kis compact, so KLis discrete by Theorem 7-1-2. By what we have shown above, K E KL, SO KL/K E AK/K is compact. Therefore, KL/K is compact and discrete, hence finite. Therefore, given x E KL, there exists a positive integer n such that nx E K. Therefore, x E K, and KL = K. //

In what follows, let us normalize the Haar measure on the compact group A,/K, so that the whole group is assigned the volume 1.

Theorem 8-2-6 (Poisson Summation Formula): Let f E Y(AK), a E 4,. Then

In the course of the proof, we shall show that both sums in this equation are absolutely convergent. We shall prove the result in a series of Lemmas.

Lemma 1: Let f E Y(AK). Then

g(P) = C f(x + P) X E K

converges uniformly for /3 in a compact subset of A,. Thus, g is continuous.

Proof: Let B G A, be a compact subset. Then B is contained in a compact set of the form

where S is a finite set of K-primes and N, is a compact subset of Kp. For purposes of proving that the series converges uniformly, we can assume that /I varies over a set of the form (*). Without loss of generality, assume that

Choose S so large that p 4 S *f, = the characteristic function of 8,.

Assume, without loss of generality, that for p E S - S,, f , = the characteristic function of fh. If f(x + /I) # 0 (X E K, P E C), then x + j3 E jjsp

(p E S - S-), ord, (x) 2 0 (p $ S). Therefore,

Let PI denote the K-ideal

Then PI depends only on f and C, and f(x + 8 ) # 0 (x E K, jl E C) 3

x E 91. Therefore,

Let

Then

As we have shown in Chapter 3, PI is a discrete subgroup of R,,, =

IlpEs,Kp. Using the fact that f, E 9(Rm,,), we see that f,(x + B,) can be uniformly estimated as P, varies over the compact set

where the estimate is rapidly decreasing in x. Using this estimate, we see that

C f J x + P-) xtpl

converges uniformly on C,. Therefore, the Lemma is proved. //

The function g defined in Lemma 1 is periodic-that is, g(/3 + x) = g(B) (/3 E AK, x E K). By Lemma 1 , g is continuous, being the uniform limit of continuous functions. Therefore, g induces a continuous function on the compact group A,/K. Conversely, every continuous function of A,/K is induced by a continuous, periodic function on A,. The dual of AK/K is KL (Theorem 7-1-5), and KL = K (Theorem 8-2-5). Therefore, we may view g* as a function on K.

Let D, be the additive fundamental domain for A,/K which was constructed in Proposition 3-2-6. Then

A,= U(DKf X) x E K

is a disjoint union. Throughout the remainder of the proof let ,u be the normalized Haar

measure on A,.

Proof: The functional

defined for continuous functions on A,/K, has the properties of a Haar integral on AJK, and assumes the value 1 for g -- 1. Therefore, the functional defines the Haar measure which we have associated to AK/K. The characters of AK/K are yx(a) = e2ziA(ax)(x E K). Thus the Lemma is proved. /I

Lemma 3: gS(x)=p(D,)-lf*(x) ( x E K ) .

Proof: By Lemma 2,

(The interchange of summation and integration is justified by Lemma 1 and the relative compactness of D,.)

Lemma 4: g* is integrable with respect to Haar measure on K and

g(a) = C g * ( ~ ) e - ~ ~ ~ " ( ~ ~ ) , a E AK. x E K

Proof: f E Y(AK) 1 f * E Y(A:) = Y(AK). Therefore, by Lemma 3, g* is continuous and integrable with respect to Haar measure on K. There- fore, the Fourier inversion theorem can be applied to g*:

(Here dx is the measure which assigns to each singleton of K the measure 1. This measure is the dual measure to the measure on A,/K which gives the whole group volume 1.) The Lemma now follows immediately. //

Lemma 5: For f E Y(AK),

Proof: Set a = 0 in Lemma 4 to get

Lemma 3 completes the proof. // Proof of Theorem 8-2-6: Let a E YK and let

By Lemma 5, we get

Applying Lemma 5 to the sum on the right, we see that

by the Fourier inversion formula. Therefore, p(DK) = 1 and the proof is complete. //

Corollary .8-2-7: p(DK) = 1. Before we turn to the problem of analytically continuing [(f, c, s),

it is necessary to construct a fundamental domain for 4:/Kx and to compute the volume of this fundamental domain with respect to dXBK.

Recall that we have defined a homomorphism

i:JK - IK, where i is the "ideal mapping" with kernel .Jim. Let ai (1 ( i ( h) be chosen from 4; so that i(ai)(l ( i ( h) represent the distinct ideal classes of K. Let us fix a E 4:. Then there exists j(l < j ( h) such that

Therefore, there exists x E K x , such that

. . The choice of j is uniquely determined by a, but x is determined only up to multiplication by an element of UK. By the Dirichlet unit theorem,

UK = W x v, where W is the finite group of roots of unity in K and V is a free abelian group of rank r = r, + r, - 1. Let w be a generator of the cyclic group W, and let w = the order of W. Let ( E , , . . . ,€,I be a basis of V. Let us construct a fundamental domain for ( J im n D;)/UK. This set will turn out to be a fundamental domain for 4:/Kx.

152 / HECKE'S L-SERIES (=HAP. 8

In the proof of the Dirichlet unit theorem, we considered the homomorphism

q:&- n 9; --+ IRr v(a) = (1% I a, IP),ES,b

where S; denotes S, with some prime deleted. In the proof of the unit theorem, we showed that (q(ei))lrisr is a basis of IRr. Let

~1~ q-l({A , = I aiq(ei) 1 0 < ai < 11)

E, = a E El ( arg (a,,) < . ( 2n1 From equation (*), it is clear that there exists e E V such that

aa-lx-lE-l 1 E El.

Thus, there exists w, E W such that

a ~ ; ~ x - ~ e - ~ w < ~ E E 0 -

Since x, c, w, all belong to K x , we see that

Theorem 8-2-8: EK = 6 aiEo is a fundamental domain for i = 1

Proof: We have shown that = EK.Kx. Let a , a' E EK be such that

a = xu', x E Kx. Let a = elat, a' = e,a,, where el , e, E E,. Then el&, =

e,a,x 3- i ( e 1 3 = i(e,a,x) 3 i(ai) and i(a,) belong to the same ideal class + a, = a,, el = e,x. Since i(e,) = BK(k = 1,2), we see that i(x) = 0, +- x t U,. But UK n EK = (1} by construction 3 x = I + a = a'. //

Let the notations be as above. The regulator R = RK of K is defined by

RK = I det (a,,) 1, a,, = log I E , I,, (1 5 i, j < r). Since {q(~,)},,,,~ is a basis of IRr, RK # 0.

Remarks: (1) R is independent of the ordering of the ei. (2) The definition of R makes no use of the (r + 1)st conjugates of the

6,. If one computes the analogous determinant, with the jth conjugates omitted, then the result is unchanged, since

r + l

[since NK,.(ri) = , = I n!l ei 1 , = 11.

Let us now do some "fussing" with normalizations of measures. Choose p , t S,. Let us embed R+ as a closed subgroup of JK via

t ( t , l , l , . @,real) t P O

t + ( t 2 , 1 1 . . . (p, complex) t

P O

With IR, so embedded, we have the isomorphism

a, = a:, x R,. Choose a Haar measure dXj?, on .Ilk so that

Theorem 8-2-9 : I,, d x P ~ = 2'1(2n)'2h R w n l .

Proof: For convenience, set p, = p,+ , , where p, is the prime used to embed [R, in a,. Then

Let

However,

and 8(F) is the unit cube M spanned by the vectors

{6(c1), . . . , e(c,), (o,o, . . . , 1)). Since 8 is a homomorphism,

volume ( F ) volume (M) volume (8-'(N)) = volume (N)

where N is the unit cube in R ' + I spanned by the vectors

((1, 0, - . . 1 O), . - * 1 (0, 0, . . - 1 1))l

and volumes in 9;- are taken with respect to dxaK. (Since 4;- is an open subgroup of J,, the restriction of dxaK to 4;- is a Haar measure on Js-.) But volume (N) = 1 and volume (M) = R. Therefore,

Also,

d x a , = 2 (preal)

= 2w (p complex),

5 dxa, = Nb;Il2 (p finite). " 9

Thus,

Finally, we get the desired formula for the volume of EK. //

We now come to the main theorem of the global theory:

Theorem 8-2-10: c(f, c, s) possesses an analytic continuation as a meromorphic function of s and satisfies the functional equation

&f, c, 3) = C(f *, c-I, 1 - $1. lf c is nontrivial, then ((f, c, s) is entire. I f c is trivial, then C(f, c, s) can have poles only at s = 0 and s -- 1. These poles are simple, and their respective residues are -u f(0) and rc f *(0), where

Proof: For Re (s) > 1, Fubini's theorem implies that

c(f, c, S) = 1- J f c t m m I t s is d x s K $ . O Jk

Consider the inner integral, which is defined for almost all t by Fubini's theorem. This integral is equal to

by the multiplicative invariance of dXBK, the product formula, and the condition c(x) = 1, x E Kx. This last sum equals

I 4tB) d " h = c(O 1 4B) d '8, EK EK

= 0, if c is nontrivial = u, if c is trivial (Theorem 8-2-9)

since EK is a fundamental domain for J:/Kx and c can be viewed as a character of the compact group J:/Kx. Set

E(c) = 0, if c is nontrivial = 1, otherwise.

Then

= lS- l I [C f *($)I IBI ' - ' ~ (~s ) ~ ' B K EK uEK

- E(c)t"f(O)u (Theorem 4-2-6)

I - E(c)tff(O)u + E(c)ts- If *(O)K, (*)

where these equations are valid for almost all t. Assume that Re (s) > 1. Then

+ E(c) [--f(0)s-' + f *(O)(s - 1)-'1. (**I Consider the integral

It is a simple exercise to verify that this integral converges absolutely and uniformly on compact subsets of C, so that the integral is an entire function of s. The same is true of the corresponding integral with f replaced by f*, c by c-l. Therefore, by equation (**), the analytic continuation of [(f, c, s) is established. The statement about poles is obvious. Since the equation (**) remains invariant under the transformation (f, c, s) -+ (f *, c-l, 1 - s), the functional equation is established. //

8-3 HECKE L-SERIES

Let us now introduce the main objects of our investigation-the Hecke L-series. These Dirichlet series are the ones that seem to occur "naturally" in arithmetic. The main fact about the Hecke L-series which we seek to establish is that they are analytically continuable to meromorphic functions in the entire plane. The analytic continuation will be carried out by means of a functional equation that connects the behavior of the function at s with its behavior at 1 - s. The main tool required to establish this result is the local and global theory of zeta functions developed in Sections 8-1 and 8-2 above.

Throughout this section, let K be a fixed number field, S a finite set of K-primes containing all the infinite primes. Let I,(S) denote the set of all K-ideals that are relatively prime to the finite primes of S.

Let c = n c, be a grossencharacter of K, and let f, be the conductort P

of c,. The product f = p + S , I I f P

t If p is infinite, we formally define f p to be p if cp(Up) f ( 1 ) and p O otherwise.

(interpreted in the obvious way as an integral K-ideal) is called the conductor of c. Similarly, the K-modulus

is called the generalized conductor of c. Let p be a finite K-prime that is relatively prime to f. Let us define

where n is a local uniformizing parameter at p. This definition does not depend on the choice of n, since p does not divide f, so c,(U,) = (1). Extend x to IK(f) by multiplicativity. Then x is a character of IK(f) (considered as a discrete group). A character constructed in this way is called a Hecke character. Let J,,, = ((a,) I a, = 1 for all p E s ) .

Let % be a K-ideal such that f divides a. Then I,(%) s I,(f), so that we can obtain a character X' on I,(%) by restricting X. We will also call characters of the form X' Hecke characters. If a Hecke character x is defined on I,(%), then we will say that x is defined modulo a.

From the definition of a Hecke character, we see that to each Hecke character there exists a grossencharacter c which can be used to define the Hecke character as above. This griissencharacter is unique. For if x is a Hecke character defined modulo a , and if S = S, u {p I p 1 a), the gr'ossencharacter c is uniquely determined by x on fl,,,. Moreover, c is uniquely determined on K x , since c(Kx) = {I). Therefore, c is uniquely determined by x on KxflK,,. But by the multiplicative approximation theorem, this set is a dense subset of 9,. Therefore, c is uniquely determined on all of 9, by X.

If x is a Hecke character associated to the grossencharacter c, then the conductor (respectively, generalized conductor) of x is taken to be the conductor (respectively, generalized conductor) of c. If x is defined modulo its conductor, then we say that x is a primitive Hecke character.?

EXERCISE 8-3-1: Let x be a Hecke character defined modulo %, and let a E QK,, be such that a = l(mod 3). Then, if c is the grossencharacter associated to X,

c(a) = x(W).

Let us fix a Hecke character x defined modulo f. For the time being, we need not assume that f is the conductor of X. Let us consider the Hecke L-series attached to X, defined by the Dirichlet series

where the sum is taken over integral K-ideals %. Let us define ~ ( 8 ) = 0 if 8 is an integral K-ideal such that (a , f) # 1. Then

7 When we speak of "the" ideal character associated to a gfdssencharacter, we shall mean the unique primitive character.

1

EXAMPLE: Let x be the trivial character defined modulo OK. This character will be denoted x,. The corresponding Hecke L-series

L(s, xo) = N'U-I

is called the Dedekind zeta function of K, and will usually be denoted CK(s). CB(s) is just the Riemann zeta function.

The reader is referred to Section 9-4 Hecke L-series-the Dirichlet L-functions.

Proposition 8-3-2: (1) The sum

for some nontrivial examples of

over all K-primes p , converges uniformly.for a 2 1 + 6, 6 > 0. (2) The product

1 - ( p ) N p S ) ( s = a + it) P

converges absolutely and uniformly for a 2 1 + 6, 6 > 0. It therefore represents an analytic function of s for Re ( s ) > 1.

Proof: (2) clearly follows from ( 1 ) by an elementary theorem on infinite products. In order to prove (I), recall that a rational prime p splits into at most n = deg (K/@ K-primes. If p is one such, then

Np = p f ( P I P ) > - p.

Therefore,

so that ( 1 ) follows. /I

Theorem 8-3-3: ( 1 ) The series

C x(WN%-" PI

converges absolutely and uniformly for Re ( s ) 2 1 + 6, 6 > 0. (2) L(s, X ) is analytic for Re ( s ) > 1. (3) For Re (s) > 1,

L(s, x ) = l l ( 1 - x(p)Np-"-'. P

This formula for L(s, X ) is called the Euler product representation.

Proofi To prove ( I ) , it suffices to show that the series

C m-" PI

converges uniformly for a 2 1 + 6, 6 > 0 Let x > 0. By the unique factorization of K-ideals into powers of primes,

where C' is a sum over certain ideals % for which NPI I x. Assertion (I) NPISx

now follows from (2) of Proposition 8-3-2 with x = x,. Assertion (2) follows immediately from (1). In order to prove (3), let x > 0. As above, let us note that

Therefore,

By ( I ) , the series on the right tends to 0 as x -+ co. Therefore, letting x +

m, we get (3). I1

Let us now consider the problem of analytically continuing a Hecke L-series. Henceforth, we shall be concerned only with primitive Hecke characters. If x is a nonprimitive Hecke character associated to the grossencharacter c, and if X* is the primitive character associated with c, then

where f * is the conductor of x*. Therefore, the problem of analytically continuing L(s, X ) is the same as that of continuing L(s, x*). We shall, therefore, consider only the case of primitive characters, since in this case, the functional equation will assume a nice shape.

Theorem 8-3-4: Let x be the Hecke character associated to the idele class character c. Then L(s, X ) has an analytic continuation as a meromorphic function on the whole s-plane. I f x + x,, then L(s, X ) is entire; i f x = x,, then L(s, X ) has a simple pole at s = 1 as its only singularity. Moreover, L(s, X ) satisfies a functional equation: There exist A E C, T(s, X ) a product of gamma functions, such that i f

R(s, x ) = 4 s - l)AST(s, x)L(s, x), then R(s, X ) is entire and

R(1 - s, X - ' 1 = W(x)R(s , X I , where W ( X ) is a constant of absolute value 1.

Proof: Let

C = n c p ; P

let So = {p I cp is ramified}. For p 6 S,, let f p = pep be the conductor of c,, f p = fjep. For p c# S,, let bp = b,,,,,; let b = b,/,. For p E S,, let n, E Z and t, = [ I be defined as follows:

(1) p real: Let n, = 0 or 1 be chosen so that cp = c+"p I 1:~. (2) p complex: Choose n, E Z such that cp = c,, 1 I:,.

In each case, n, and tp are uniquely determined by c,. Let f, E Y(Kp) be defined as follows: Let S = So U S,. Set

f, = the characteristic function of O,(p 6 S), f, = the test function used in Section 8-1 to evaluate p(cp, s)(p E S).

Then f = nf, E Y(A,). Moreover, by Theorem 8-2-3, P

Let us evaluate the right-hand side of (2). For p 4_ S,

[(f,, c,, S) = Nbi-1/2~(bp)-1(l - x(p)Np-')-'.

Therefore,

n w p , Cp, s) = L(s, x) n Nbs- 1/2~(b,)- l . Pf S P e s

By the fundamental theorem of the local theory, each local zeta function appearing in the product (2) is meromorphic in the entire s-plane. Moreover, [(f, c, s) has an analytic continuation as a meromorphic function on the whole s-plane. Therefore, by (3), the same is true for L(s, x). If x # xo, then [(f, c, s) is an entire function. From the formulas of Section 8-1, we see that for p E S, [(A, c,, s)-I is an entire function. [The main point is that r(s)-l is entire.] Therefore, by equations (2) and (3), L(s, X) is entire. If x = xo, [(f, c, S) has poles at most at s = 1 and s = 0, these poles being simple, with residues K f *(O) and -K f(O), respectively. A simple computation shows that f(0) f 0 and f *(0) # 0, so that, in fact, the poles exist. A simple computation shows that

n RfP, C,? $1 P C S

(4)

has a pole at s = 0 of order r , + r, [since r(s) has a simple pole at s = 0). Therefore, since [(f, c, s) has a simple pole at s = 0, L(s, X) must be regular at s = 0. The product (4) is easily seen to be regular at s = 1. Therefore, L(s, X) must have a pole at s = 1, this pole being simple. It remains to establish the functional equation. By the functional equation for [(f, c, s) and the functional equations of the local zeta functions,

However, from our local computations,

Therefore,

n [(f ;, c,', 1 - s) = L(1 - s, x-'1. PfS

Assembling the above information, we see that

L(l - s, X-') = L(s, X) n Nb:-'/2~(bp)-1 JJ p(cp, s). P'+S ws ( 5 )

By the computations of the local theory,

= (-i)InPI(2,)1 - 2 ( r+ i tp) r((2(s + itp) + np)/2) r((2(1 - s - it,) i- np)/2)

(p complex).

Set

where p, = 1 if p is real and = 2 if p is complex. Then

where

U = C tP, V = C tP, p complex peS,

N = C I n , l , M = E n , . p€S, p real

then

( 6 ) Note that N(b) = Id 1 by Theorem 2-7-17. Therefore, if we set

A = (I dl N ( f ) n ~ 4 - ~ ~ ) ' / ~ ,

then equation (6) immediately yields the desired functional equation. //

Applications of Hecke L-functions

The following chapter will present a few applications of the analytic properties of Hecke L-functions. These applications can be taken as the "raison d'etre" for introducing the L-functions.

9-1 SPLITTING OF PRIMES

Let K be a number field, L a normal extension of K. We have shown that in L, a prime p splits into g distinct primes, each having ramification degree e and residue class degreef, such that

efg = n = deg (LIK).

Moreover, we showed that e > 1 occurs for only finitely many p. Thus, except for finitely many primes, we have

fg = n.

The numbers f, g determine the splitting type of an unramified prime p. If f = 1, g = n, then p is said to split completely in L. Iff = n, g = 1, then p is said to be undecomposed or inertial. One can ask whether there are infinitely many primes belonging to each splitting type. In general, this is not true. For if p is inertial, then pOL is a prime of L (which we again denote p),

and (y) is of order f = n. Therefore, (Lf) generates G = Gal (LIK).

Therefore, an inertial p can exist only if G is cyclic. The converse is true, and will be proven in our discussion of Tchebotarev's density theorem. For the moment, we shall content ourselves with the following

Theorem 9-1-1: There cxist infinitely many primes that split completely in L.

Actually, we shall show much more: Let A be a set of primes of K. Then we say that A has a Dirichlet density d(A) if

exists and has the value d(A).* It is clear that if A has a Dirichlet density, then

0 <d(A) 2 1 .

If d(A) is positive, then A has infinitely many elements, since CK(s) has a simple pole at s = 1, which implies that

1% CK(4 = -log (s - 1) + 0(1), s --t 1.

We shall prove

Theorem 9-1-2: Let A = (p I p splits completely in L}. Then A has a Dirichlet density d(A) = lln.

Proof: Since CL(s) has a simple pole at s = 1, we have

CL(s) = -log (s - 1) + 0(1), s -+ 1. But

CL(4 = IJ (1 - rn-9-l

Now N q = Npf, f = f(q/p), 91 p . Therefore, if p splits completely in L,

B (1 - m - " ) - ' = (1 - Np-q-".

But if p does not split completely in L, and 9 1 b, then either 1 - N9-" =

1 - ~ p - f ~ , f 2 2 or p ramifies in LIK. Therefore,

n p does not split completely in , K 8 (1 - N 9 - y

is bounded in a neighborhood of s = 1. Thus,

= lim -(lln> log (s - 1) + O(1) 1 + -log (s - 1) + O(1)

* The limit is understood to be taken with respect to real s.

Thus, there is an abundance of primes that split completely. If LIK is normal, then the primes that split completely actually characterize the extension uniquely. More precisely:

Theorem 9-1-3 (Bauer): Let L and L' be two normal extensions of K, and suppose that a prime p splits completely in L o p splits completely in L', with the possible exception ofjinitely many p. Then L = L'. Proofi Let L" = LL'. Set

A = {p I p splits completely in L),

with A' and A" defined similarly. Then, by assumption, A, A', and A" coincide except for finitely many p. By Theorem 9-1-2,

1 d(A) = -9 n = deg (LIK) n 1 d(Af) = ,, n' = deg (Lf/K) n 1

d(A") = 7, n n" = deg (L"/K).

9-2 ABELIAN L-FUNCTIONS

In the following discussion, we shall introduce a class of Dirichlet series that are closely connected with the decomposition of primes in an abelian extension. These abelian L-functions actually are special cases of the Hecke L-functions introduced in the preceding chapter. This last statement is essentially the Artin reciprocity law, a special case of which will be proved in the following chapter.

Let L/K be an abelian extension, G = Gal (LIK). Let p be a finite prime of K, !@ any prime of L that divides p. Recall that we have the exact sequence

(1) - I(!@) ---+ ---+ Gal ( W d -+ (11,

where D(Q) is the decomposition group at !@, I(!@) is the inertia group at !@, and Lcp (respectively, RP) is the residue class field of L (respectively, K) at !@ (respectively, p). If p is unramified in L/K, then D($) is isomorphic to

G a l ( , ) and D(!@) is generated by (q). Moreover, if a t G,

since G is abelian, D(!@), I(!@), and (y) therefore depend only on p. Let \ V '

us denote these three objects by D(p), I(#), and

Let x E G*. Associated to X, we define a character (also denoted X) of IK(dK) as follows: Let p be a finite prime of K, '$3 a fixed prime of L which divides p. Let a, E D(p) be any element such that its image in Gal is the Frobenius automorphism x-+xNV. Any other such element is ofthe form no,, a E I(p). Define

Then ~ ( p ) depends only on p and not on the choice of a,. Moreover, ~ ( p ) = 0 if I(p) + ker (x). (For then x is a nontrivial character of the compact group I@).) If I ( p ) c ker (x), then ~ ( p ) = ~(a , ) . In particular, if p is unramified in LIK, then

Define x on IK by extending the definition (1) by multiplicativity. Then x is a multiplicative mapping of I, into U u { O ) , and the restriction of x to IK(dK) is a character.

Associated to the above character X, let us define the abelian L-jiunction L(s, x; LIK) by

L(s, X ; L/K) = C x(%)N%-" (Re (s) > I), 81

where the sum is over all integral K-ideals 9. As in the case of Hecke L- functions, we may establish the following:

Theorem 9-2-1: (1) L(s, X; LIK) converges for Re (s) > 1 and represents an analytic function of s in this half-plane.

(2) L(s, X; LIK) = n (1 - x(p)NpWs)-I (Re (s) > l), where the product is extended over alljinite primes of K.

The abelian L-functions for the various x E G* are related to one another:

Before proceeding with the proof of Theorem 9-2-2, let us prove the following

Lemma: Let G be ajinite, abelian group of order n (considered as a discrete group), z t G of orderf, t an indeterminate. Then

Proof: Let G, be the subgroup of G generated by z. Then GI is cyclic of order f. By Theorem 7-1-4, GT is isomorphic to G*/Gf and has order f. Therefore,

166 / APPLICATIONS OF HECKE L-FUNCTIONS CHAP. 9

But since GI is cyclic of orderf, A(z) runs over all fth roots of 1. Let Cf denote a primitive fth root of 1. Then

f = 1

J-J (1 - A(z)t)"lf = n (1 - 5jt)"lf AEG: k=O

= (1 - tf)"/f. //

Proof of Theorem 9-2-2: Let p be a finite prime of K. It suffices to show that

Fix z, E G so that the image of z, in Gal (I$/%) (for some choice of 13 1 p) is the Frobenius automorphism. Then

Let us consider G and I(p) to be topological groups with the discrete topology. Then Theorem 3-1-2 asserts that

The order of z, is f = f(p), since z, generates Gal (Ev/Ko). Moreover, G/I(p) is of order nle. Therefore, applying the Lemma to (*), we obtain

11 (1 - x(@)Np-") = (1 - Np-fs)"Ief xEG'

Let us consider the following statement:

Theorem: L(s, X; LIK) is a Hecke L-function.

This Theorem is one version of Artin's reciprocity law, which lies at the root of class field theory. Chapter 12 will be devoted to proving the assertion. In the special case LIK quadratic, it will turn out that the assertion is equivalent to the classical quadratic reciprocity law in the case K = @. We can easily draw important conclusions from the reciprocity law as stated above :

Corollary 9-2-3: L(s, X; LIK) has an analytic continuation as a meromorphic function of s. I f x = x,, the trivial character, then L(s, x,; LIK) has a simple pole at s = 1, and is regular at all other points. I f x # xo, then L(s, X; L/K) is an entire function of s.

Corollary 9-2-4: IfLIK is abelian, then cL(s)/cK(s) is an entire function.

Proof: Corollary 9-2-4 follows immediately from Theorem 9-2-2, Corollary 9-2-3, and the fact that L(s, x,; L/K) = C,(s). //

It is possible to weaken the hypotheses of the last result, proving the analogous fact for L/K normal.

Corollary 9-2-5: If x # xo, then L(1, X; LIK) # 0, oo.

Proof: L(1, X; LIK) # oo by Corollary 9-2-3. Also, L(1, X; LIK) # 0, since the vanishing of any L-function at 1 would contradict the existence of a pole of cL(s) at s = 1, upon the application of Theorem 9-2-2. //

Theorem 9-2-6 (Dirichlet's Theorem): Let o E G, and define

Then A(o) has a Dirichlet density = lln, n = deg (LIK).

Proof: It suffices to show that

The expression on the left is equal to

Consider the sum T(s) = n- ' ~ ( 6 ' ) log L (s, X; LIK).

xEG'

On the one hand,

T(s) - - Cx(o-')logL(s,x;LIK) 1 lim - - 1 + lim X*XO - - - S-I+ log C,(s) n s-l+ log C K W n (2)

by Corollary 9-2-5 of the preceding theorem. On the other hand,

For all but a finite number of p, y(p) = X(Lx). Remark that P

= 0, otherwise.

168 / APPLICATIONS OF HECKE L-FUNCTIONS CHAP. 9

o or x - x(a-l(y)) is a character of the compact group G*. Therefore, I

which together with (1) and (2) proves Theorem 9-2-6. I/

Corollary 9-2-7: Let LIK be abelian, (f, g) a splitting type for LIK. Then a necessary and suficient condition that there exist infinitely many primes of splitting type (f, g) is that G have an element of order$ Let n, = the number of elements of G of order $ Then the set of primes of splitting type (f, g) has a Dirichlet density equal to nf/n.

Proof: If p is a prime that is unramified and of residue class degree f in LIK LIK, then G certainly contains an element of order f, namely (T). Con-

versely, if G contains n, elements of orderf, then Dirichlet's theorem asserts

that the set of primes p, unramified in LIK and such that Lx has order/, ( b )

has a Dirichlet density equal to n,/n. But - of order f implies that p has (LF) residue class degree f in LIK. I/

Corollary 9-2-8: (Dirichlet's Theorem on Primes in Arithmetic Progressions): Let n, m E h, (n, m) = 1. Then (p lp is a rational prime, p = m(mod n)} has Dirichlet density equal to l/p(n). In particular, there are infinitely many primes congruent to m(mod n).

Proof: Let L = Q([,). If a E Gal (LIQ) is defined by a([) = [;, then

Theorem 6-2-14 implies that = a e p = m(mod n). Corollary 9-2-8

now follows from Theorem 9-2-6. //

9-3 TCHEBOTAREV'S DENSITY THEOREM

In this section, we shall use Dirichlet's theorem (Theorem 9-2-6) to prove Tchebotarev's density theorem. Actually, Tchebotarev's theorem is a generalization of Theorem 9-2-6 to nonabelian extensions.

Throughout, let LIK be a Galois extension of degree n, G = Gal(L/K). If 9 is a finite L-prime that is unramified in LIK, then we have proven

that for r E G.

Therefore, as 9 runs through all prime divisors of some finite K-prime p that

is unramified in LIK, - runs over a certain lsonjugacy class of G. This (Lk3 LIK conjugacy class is called the Artin symbol at p, and is denoted (--). If LIK

is abelian, then the Artin symbol at p consists of a single element-namely, the Frobenius at p. By abuse of language (and notation), we identify the Artin symbol and the Frobenius symbol in the abelian case.

Let C be a conjugacy class of G that contains c elements.

Theorem 9-3-1 (Tchebotarev) : Let

p I p is finite, unramified in LIK,

Then A has Dirichlet density cln.

We shall follow a recent proof of McCluer (Acta Arithmetica, XV, pp. 45-48, 1969).

Lemma 9-3-2: Let $3 be ajinite L-prime that is unram$ed in LIK, p = the restriction of to K. Let M be afield such that K E M E L and such that

every prime divisor O of p in M is undecomposed in L. Let

there exist [Z,(a):H] prime divisors of p in M for which (L#) = a ,

where ZG(a) is the centralizer of a in G and H is the subgroup of G generated by a.

Proof: Since each prime divisor O of p in M is undecomposed in L, it suffices to show that there are [Z,(a):H] prime divisors $3 of p in L for which

($K) = a . Any such prime divisor is of the form r$3 for some r E G. By

equation (I), (@) = a o rar-I = a o r E Z,(O). Two such prime r@

divisors rl$3 and r,$3 are equal e rlr;'9 = $3 o r1r;l E Drp. But Drp is generated by a. Therefore, Drp = H, and there are [Z,(a):H] L-primes ~ $ 3

for which - = a. // ( 5 9 Lemma 9-3-3: Let A be a set of jinite K-primes. Then A has a Dirichlet density a o

CNp-"=-a log( s - l )$o ( log ( s -1 ) ) ( s+ l+ ) . PEA

Proof: Exercise.

Lemma 9-3-4: C Np-" =(log (s - 1)) (s --+ I f ) , where f , = f(p/p), v

f p > l

p = the rational prime which p divides.

170 1 APPLICATIONS OF HECKE L-FUNCIIONS CHAP. 9

Proof: If f , > 1, then Np-" = p-fpS <P-~" . Since at most n K-primes p divide a given @-prime p,

Proof of Theorem 9-3-1: By Lemma 9-3-3, it suffices to show that

Choose o E C, and let H be the subgroup of G generated by a, M = the fixed field of H. Then L/M is cyclic with Galois group H. Therefore, by Theorem 9-2-6 and Lemma 9-3-3,

where Q runs over M-primes. Note that if f(Q/p) = 1, then NQ = Np, where p is the restriction of Q to K. Therefore, by Lemma 9-3-4,

C Np-" = -[H: 11-I log (s - 1) + o(1og (s - 1)) (s - I+). (*) = ,,

We claim that M satisfies the hypotheses of Lemma 9-3-2 for each M-prime . -

Q that appears in the last sum. Indeed, if = o , then Do = H, so

that there can be only one prime divisor '$3 of Q in L, since if 9' is any one such, e('$3'/Q)f(!@'/Q) = deg (L/M) => g('$'/D) = 1. Therefore, by Lemma 9-3-2,

[ZG(a): HI C Np-" = -[H: I]-' log (s - 1) + o(log (s - 1)) (&+)=c

LIM since (r) = o + (y) = C. Therefore, by Lemma 9-3-4,

C Np-" = - 1 log (s - 1) + o(l0g (s - 1)) (s -+ l+). (Y) = [Z,(o) : HI

But 1 1 [H:l] 1

[Z,(o) : HI [H:] = [ZG(o) : 1]

- - 1 [ZG(@ : 1 I

- - C -. n

This completes the proof of Theorem 9-3-1. I/

9-4 DIRICHLET'S L-FUNCTIONS

Let K/@ be an abelian extension, G = Gal (KIQ), x E G*. Throughout this section, let us assume

Artin's Reciprocity Law for @: x is the (primitive) Hecke character associated to some grossencharacter c of @.

The proof of the reciprocity law will be the main task of Chapter 12. For the time being we shall just assume it and show how we can evaluate some abelian L-series and come up with some very classical objects.

Definition 9 4 1 : An abelian L-series L(s, X; K/@) is called a Dirichlet L-series.

Let us show that the Dirichlet L-series coincide with the classical Di- richlet L-series introduced in Chapter 1. Let f = f h( f > 0) be the conductor of c. Then x is defined on I&. Let Zf denote the multiplicative semigroup of rational integers relatively prime tof. For n E hf , define

P b ) = q c,(n)

= x((n)>c-(n>7 where (n) denotes the @-ideal generated by n. It is clear that is a multiplicative function.

Proposition 9-4-2 : Either c, is trivial or

c ~ ( x ) = sign (x ) (x E Rx). Proof: Since x E G*, x is of finite order. Therefore, it assumes only a finite number of values. Let S be the set of @-primes consisting of the infinite primes and all primes dividing f. Since the values of c on D,,, are values of x7 c assumes only a finite number of values on D,,. Therefore, the same is true on KXDK,,. But KxJK,, is dense in 8, by the multiplicative approximation theorem. Therefore, c assumes only a finite number of values on DK, so that c is of finite order. In particular, c, is of finite order. Thus, c, is constant on

the connected components of Rx . The only possibilities for c, are then the ones mentioned. /I

Corollary 9-43: If n E E f , n > 0, then

An) = x((n>).

By Corollary 9-4-3, we see that

u s , x ; Kl@) = 2 ~ ( n ) n - ~ - n= 1 nEZf

Theorem 9 4 4 : p(n + f ) = p(n) (n E Zf).

Proof: Note that 1 + f/n .= l(mod f ) so that 1 + f/n = l(mod q) for all rational primes q dividing f. Therefore, cp(l + f/n) = 1 for all q If. Thus,

~ ( n + f > = n cp(n + f ) P Y ~

= An). 11 By Theorem 9-4-4, p is essentially a character of (E/(f))" = the group

of residue classes modulo f that are relatively prime to f. Therefore, L(s, X; K/@) is a Dirichlet L-series of the type introduced in Chapter 1.

As a particular case of the above situation, consider the case K/@ a quadratic extension of discriminant d. Then there is exactly one nontrivial character of G, and the corresponding abelian L-series is

where (din) denotes the Legendre symbol. The conductor-discriminant formula (See Exercise 6-1-7) implies that the conductor of the Kronecker symbol (dl.) is dZ. Therefore, by Theorem 9-4-4, the Legendre symbol is periodic with period d. This fact may also be verified directly from the quadratic reciprocity law.

9-5 CLASS NUMBER FORMULAS

Let K/@ be an abelian extension. It is possible, at least in principle, to write down an explicit formula for the class number h of K. The basic idea comes from two results we have already established: On the one hand, by Theorem 9-2-2,

= C W II u s , X ; Kl@) xEG* X + X 0

(1)

since L(s, x,; K/@) = ((s). Multiply equation (1) by s - 1 and let s tend to 1. Recall that L(s, X; K/@) (X # xo) is a Hecke L-series corresponding to a nontrivial grossencharacter and is therefore an entire function of s. On the other hand, C,(s) and C(s) both have simple poles at s = 1, the latter having residue 1 and the former having residue

Therefore,

Provided that we can obtain explicit expressions for L(l, X; K/@), equation (3) gives a formula for the class number. The main difficulty with this formula is that it involves the regulator, which is usually difficult to calculate. Any improvement of the formula (3) that avoids the use of the regulator would rank as one of the major achievements of number theory in this century. In order to complete the development of equation (3), let us consider the following

EXERCISE: Evaluate L(1, X ; K/Q) for x an Artin character of K/@.

For simplicity, we shall write L(s, X) instead of L(s, X ; K/@). Let f be the conductor of X. Then f is an integral @-ideal. Let f be a positive generator for this ideal. By the result of Section 9-4, L(s, X) coincides with the Dirichlet L-series

where

An) = x((n)) (n > 0). Moreover, p is periodic with period f.

Lemma 1 : The sums

where C # 1 is a root of unity, are bounded in x.

Proof: Suppose that C is a primitive mth root of unity. It suffices to show that for every nonnegative integer k,

For then if K is chosen so that Km 5 x < (K + l)m,

By the periodicity of the function e2ziz1m,

However, it is clear that the right-hand side is zero, since c is a primitive mth root of unity, m > 1.

Lemma 2: The series

2 l"n-"(r = a primitive mth root of unity, m > 1) n = 1

converges uniformly for s 2 e, E > 0, and therefore represents a continuous function for s > 0.

Proof: Assume that s 2 e, E > 0. By Abel partial summation,

Lemma 3: Let c = eZzilf. Then f -1 C C(d-n)r = 0 if n + d(mod f ) r = O

= f if n=d(mod f ) .

(By Lemma 1)

Proof: The second assertion is clear. If n $ d(mod f), then c' = Cd-" # 1. Therefore,

We are now ready to complete the derivation of equation (3). Assume, to start with, that s > 1. Then by the periodicity of p ,

where the inner sum is convergent, since s > 1. By Lemma 3,

Set

I Gr(p) is called a Gaussian sum. With this notation, we see that

By Lemma 3, each of the inner sums on the right is a continuous function for s > 0. Therefore, we may let s tend to 1 in this last equation to get

In order to get the formula (4) in a better shape, let us recall that the c.3

I power series C zn/n converges for 121 < 1 and represents the function 1 n= 1

--log (1 - z), where the principal branch of the logarithm is understood. Since the series obtained by substituting z = c-' converges, Abel's theorem

I on power series [I, 2nd ed., p. 421 implies that

Therefore, by equation (4), f-1

L(1, X) = -f -' r=O C Gr(p) log (1 - Cr). ( 5 )

We shall simplify equation (5) using the following two Lemmas:

Lemma 4: I f (r, f ) # 1, then G,(p) = 0.

Proofi Let p be a prime dividing (r, f). We shall prove that there exists a rational integer y such that (y, f ) = 1, y = I(modflp), and p(y) # 1. Assume for a moment that such a y exists. Since cr is a primitivefl(r, f)th root of unity, P = c. Therefore, since (y, f ) = 1,

GAP) = C p(y4Cryd d h o d f )

= P(Y> C ~ ( d ) r ' ~ d h o d f )

= P(Y)G,(P). Therefore, G,(p) = 0, since p(y) # I . In order to prove the existence of y, let us consider the grossencharacter c = rl[ c,, which is associated with

P x via the reciprocity law. Let f, be the conductor of c,. Then

and f + 1 unless K = @ (since the primes that ramify in K/@ divide f and there exists at least one such prime by Corollary 13-1-6). Let us assume, without loss of generality, that K # @. Let us fix a prime p that dividesf.

,-l/

There exists x, E 1 + fp/p such that c,(x) # 1. Let us choose x, E 1 + 7, for each @-prime q # p, qlf. By the approximation theorem, there exists y E Z such that y > 0, and

ordr (y - x,) 2 ord, (f), r a @-prime, r If. Then y r l(modflp). Moreover, (y, f ) = 1. Finally, note that by construction, c,(y) = 1 for r # q, r If, and c,(y) # 1. Therefore,

Thus, p(y) + 1. /I

Lemma 5: If(r, f ) = 1, then p(r)G,(p) = Gl(p).

Proof: Since (r, f ) = 1, as d runs through a complete system of residues modulo f, so does rd. Therefore,

Applying Lemmas 4 and 5 to equation ( 9 , we see that

where G(p) = Gl(p). This is the penultimate form of the class number formula.

Let c be the grossencharacter that corresponds to x via the reciprocity law. In order to put equation (6) into final form, it is necessary to consider separately two cases:

Case 1: c is ramified at oo. In this case, p(- 1) = - 1. Therefore, replacing r by -r in the sum in equation (6), and adding the result to equation (6), we get

L(1, X ) = --G(p)(2 f)- ' C p(r)-'[log (1 - C?) - 1ogU - PI1 ( r , f ) = l r(mod f )

= -G(p)(2 f )-I p(r)-I2i arg log (1 - c-r). ( r , f ) = l ,(mod f )

But, arg log (1 - C-') = arg (1 - C-') = arg (C+" + arg ( P 2 - P 2 ) = -nrlf + 742. Therefore,

= niG(p) f -2 p(r)-lr. ( r , f )= l

(7) r b o d f )

Case 2: c is unramified at w. In this case, p(-1) = +1, and changing r into -r in the sum in equation (6) and adding the result to equation (6), we get

L(1,x) = -G(p)(2f)-' ,U(r)-l [log(l - i? + log (1 - C-r)l r = l

( r , f )= l

f-1 = -G(p)(2 f)-' C p(r)-I log

r= 1 (r.f)=l

since 0 < arlf < n. Equations (7) and (8) are the final form that we shall derive for the value

of L(l, x). Inserting the appropriate values in equation (3), we derive a formula for the class number. The most important feature of this formula is that all transcendental operations are missing, so that the class number can be computed in a finite number of steps, in terms of the invariants d, w, R, etc. It is possible to refine the class number formula yet further by studying the behavior of the Gaussian sum. However, we shall not pursue this subject any further here. The interested reader is referred to [8, Chapter 51 for a complete exposition of the subject of class number formulas and their application to quadratic and cyclotomic fields.

PART 4

Class I Field Theory

Class field theory

In the preceding chapter, we presented some applications of the theory of Hecke L-functions to the arithmetic of algebraic number fields. Let us now continue in this vein by considering one of the most striking developments ever to arise out of the theory of L-series-namely, class field theory. Broadly speaking, class field theory is the study of abelian extensions of algebraic number fields. From class field theory we obtain (1) a complete classification of all abelian extensions L of a given algebraic number field K, (2) a reciprocity law that describes the Galois group of such an extension in terms of the ground field K, (3) a law that describes the way in which a K-prime p splits in L. All of these facts come from essentially the same place-Artin's reciprocity law. We shall demonstrate below that Artin's reciprocity law is equivalent to the statement that all abelian L-functions are Hecke L-functions. Historically, Artin was first led to the reciprocity law when he attempted to prove that every abelian L-function has an analytic continuation and a functional equation. Over the years, however, the L-functions have been expunged from class field theory and have been replaced by an algebraic edifice based on the cohomology of groups. This was done in order to establish results of an algebraic nature in a purely algebraic way-without the aid of transcendental reasoning. Indeed, this point of view has much merit, since class field theory, although motivated and given an initial impetus by analytic techniques, has never been capable of development through the use of purely analytic machinery. However, since the L-functions play the central role in our point of view, one should not lament the use of analysis to obtain a purely algebraic result, but rather the fact that one must establish a purely analytic fact (that abelian L-functions are Hecke L-functions) by algebraic means.

10-1 ABELIAN L-FUNCTIONS ARE HECKE L-FUNCTIONS

In the first part of this section, we shall reduce the proof of the fact that every abelian L-function is a Hecke L-function to an algebraic statement known as Artin's reciprocity law. We shall then assume the reciprocity law and demonstrate some of its arithmetic consequences. The next two para- graphs will be devoted to a proof of the reciprocity law.

Throughout Section 10-1, let LIK be an abelian extension, G = Gal (LIK), x E G*, S, = S, u { p I ker (x) + I,). As in Section 9-2, we shall view x as a character on I,(S,). A character obtained in this way is called an Artin character. Let us now explore the consequences of the following assertion:

Theorem 10-1-1 (Reciprocity Law, First Form): Every Artin character x is the Hecke character associated to some grossencharacter c(X) of K.

Immediately, we deduce

Corollary 10-1-2: L(s, x ; LIK) is a Hecke L-function for all x E G*.

From the definition of x given in Section 9-2, we see that for p 4 S,,

In particular, (1) implies that

Proposition 10-1-3: cp) is unramiJied if p 4 S,y.

Since P ( x ) = 1 (x E KX), we see that

Thus, c,(x)(x E Kx) is one of the n possible values of X. By the multiplicative approximation theorem, K x is dense in R,,, = rl[ K,". Therefore, c, assumes

,€S,

only finitely many values on R,,, * c, is of finite order +- cf) is of finite order for p E S,. We claim much more, however:

Proposition 10-1-4: If p is complex, cp) is trivial. I f p is real, then either cp) = 1 or cp)(x) = sign (x) (x E IRx).

Proof: cp) I R + = C' is a character of IR, and is therefore of the form cl(x) = xa(a E R). Since cl(x) is of absolute value 1, a = 0 and c' is trivial. Thus, we deduce the assertion for p real. Let p be complex, U = the group of complex numbers of absolute value 1, c" = cp)I,. Then c" is a character of U and is thus of the form c"(x) = xn. But by the discussion preceding the statement of this Proposition, c" is of finite order, so that n = 0 and c" = 1. Thus, let x = reie E C, r > 0. Then cF)(x) = c'(r)c"(eie) = 1. //

Let f, denote the generalized conductor of d x ) .

Definition 10-1-5: The least common multiple of all f,(x E G*) is called the conductor of LIK, denoted fL,,. The exact power of p dividing f,, will be denoted f,.

Proposition 10-1-6: d,,, and f,,, are divisible by the same finite primes.

Proof: Proposition 10-1-3 shows that if a finite K-prime p divides f,,, then p E S, - S, for some x E G*. Thus, if p divides f,,,, p is ramified in LIK => p divides d,,. Conversely, suppose that p divides d,,,. Then p ramifies in LIK and I, is nontrivial. Let x E G* be such that x is nontrivial on I,. Such a character exists by duality theory. Then p E S, * p divides f, * p

I divides f,,,. //

1 Remarks: (1) In Section 12-2 we shall improve Proposition 10-1-6 to read " p ramifies in LIK o p divides fL,,." This version treats the infinite primes in the same way as the finite primes.

(2) Define the generalized discriminant d,,, of LIK to be the product of d,, and all infinite K-primes that ramify in LIK. Then Proposition 10-1-6 can be improved to read

Proposition 10-1-6': dLIK and f,,, are divisible by the same primes.

(3) Proposition 10-1-6 can be generalized to read

Theorem (Conductor-Discriminant Formula): f, = d,,,. x€G*

EXERCISE: Assume that K = @. Prove the conductor-discriminant formula by comparing factors in the functional equations for the Dirichlet series on both sides of the equation in Theorem 9-2-2.

Proposition 10-1-7: Let x E K x , x = 1 (mod f,). Then x(i(x)) = 1.

Proof: Let So denote the set of finite primes dividing the generalized conductor of X. Since x is a Hecke character,

But since x - 1 (mod f,), cp)(x) = 1 for all p dividing f,. Thus, by Pro- position 10-1-4, all the factors in (*) are 1, so Proposition 10-1-7 is proved.

//

Theorem 10-1-8 (Reciprocity Law, Second Form): Let x E K x , x = I (mod I

i f u x ) . Then

--

Proof: By Proposition 10-1-6, i(x) is relatively prime to dLIK, so that

is defined. Moreover, by Proposition 10-1-7,

for all x E G*. Thus, the assertion is proved. I/

Let p be a K-prime. The local norm residue symbol [ , LIK], is a mapping

[ LIKI, : Kp --+ G

x ---t [x, LIKI,

defined as follows: By the multiplicative approximation theorem, there exists y E K x such that

yx-' -- 1 (mod f,), y = 1 (mod fb), (1)

where fl, = fL,,fil. (If f, = po, then the condition z 1 (mod f,) will mean that z E Up for p finite; for p infinite, the condition will be vacuous.) Write i(y) in the formt

i(y) = pa%, (p, %) = 1, (2)

and set

Remarks: (1) The Frobenius symbol in equation (3) is defined, since (%, f,,,) = 1 => (%, d,,) = 1 by Proposition 10-1-6.

(2) The definition of [x, LIK], is independent of the choice of the auxiliary variable y: For if y' also satisfies equation (I), then yy'-' r l(mod f,,,) and i(yyf-l) is relatively prime to p. Thus, i(yl) = pa%', where a

is given by equation (2). By Theorem 10-1-8,

Since the conditions expressed by equations (1) and (2) are multiplicative, we see that

The main properties of the norm residue symbol are summarized in the following

Proposition 10-1-10: Let L/K be abelian, p a K-prime, x E K,". (1) If K c M c L, then

[ 7 L/Kl~ IM = [ 9

(2) Let MIK be an abelian extension, with M linearly disjoint from L over K. Then

[x, LMIKI, = [x, LIKl,[x, MIKIP

(3) Let MIK be an arbitrary extension. View Gal (LMIM) as a subgroup of Gal (LIK). Let !@,, . . . , !@, denote the prime divisors of p in M. Then, if Y E M X ,

fi [Y, LMIMIv, = [NM/K(Y), LIKI,. i = 1

(4) I f p does not divide f,,,, and x E U,, then [x, LIK], = 1. (5) I f p isjinite and is unramified in LIK, then for x E K,",

(6) If x E K X , then [x, LIK], = 1 for almost all p.

Proof: (1) Let the K-modulus f denote a common multiple of f,,, and fM/,.

Let f, = the exact power of p that divides f, f,, = ff;'. Let y E K x satisfy equation (1). Then this y can be used to define [x, LIK], and [x, MIK],. Therefore, using the notations of equations (2) and (3), we see that

from which the assertion follows. (2) Let f denote the least common multiple of the K-moduli f,,,,

f M I K , and I,,,; let f,h = the exact power of p dividing f. Choose y E K x such that

yx-I r 1 (mod f;), y =. 1 (mod ff,h-I).

Then y satisfies condition (1) for extensions LIK, M/K, and LMIK. There- fore, if we set i(y) = p*, (%, p) = 1, we have

Proposition 10-1-9: [ , LIK], is a homomorphism.

t If p is infinite, we always set a = 0.

However, since L and M are linearly disjoint over K,

by Exercise 5-6-5, (1). This suffices to prove (2). (3) Exercise. (4) Since p does not divide fP,-we see that f, = pO. Let y E K x satisfy

condition (1). Then y =l l(mod f,,,), since f b = yL,K. Moreover, since x E Up, ordP (y) = 0. Thus,

by Theorem 10-1-8. (5) By (4) and Proposition 10-1-9, it suffices to show that

where n is a local uniformizing parameter at p. Without loss of generality, choose w so that n E K and n - 1 (mod f,,,). (This is possible by the approximation theorem and the fact that p is unramified so that p does not divides fL,K.) Therefore, since i(n) = p%, (8, p) = 1,

by Theorem 10-1-8. (6) Follows directly from (4). //

We are now in a position to put the reciprocity law in its classical form, which is due to Hilbert.

Theorem 10-1-11 (Reciprocity Law, Third Form): Let x E Kx. Then

n [x, LIKlp = 1. P

Proof: The left-hand side is well defined by (6) of Proposition 10-1-10. Let S = {p,, . . . , p,] denote the set of K-primes that either ramify in LIK or divide i(x). By Proposition 10-1-6, we see that

Choose yi E K x so that

xy; r 1 (mod f,,), yi = 1 (mod f;,). Set i(yi) = pa121i, (8,, pi) = 1. Then, a, = ord,, (x), and

Moreover, since S contains all prime divisors of i(x),

Therefore, by equations (2) and (3), we see that

1 By equation (3),

But from the choice of the yi's,

- 1 (mod f,,), ( j = 1, . . . , t). Thus, by equation (I), we see that

X-l fi yi -- 1 (mod f,,K). i = 1

Thus, by Theorem 10-1 -8 and equation (4), Theorem 10-1-1 1 is proved. //

We have shown that Theorem 10-1-1 (Reciprocity Law, First Form) implies Theorem 10-1-8 (Reciprocity Law, Second Form), which, in turn, implies Theorem 10-1-11 (Reciprocity Law, Third Form). In Section 12-2, we shall prove that Theorem 10-1-11 implies Theorem 10-1-1, so that all three forms of the reciprocity are equivalent. At this moment, however, note that we need Theorem 10-1-8 in order to prove that the norm residue symbol is well defined. Thus, one of our principal tasks in Section 12-2 will be to give a definition of the norm residue symbol that is independent of the reciprocity law.

By Proposition 10-1-10, [x, LIK], = 1 if p 4 S. Therefore,

The inequalities of class field theory

11-1 THE SECOND INEQUALITY

As a preliminary to our proof of the Artin reciprocity law, we shall start with a modest result, the so-called "second inequality" of class field theory-namely,

Theorem 11-1-1 : Let LIK be normal. Then [J, : K x NLIKJL] 5 n.

Our proof of Theorem 11-1-1 will be derived from several propositions concerning admissible subgroups of 4, as well as some previously proven results on Hecke L-series, especially concerning their behavior near s = 1.

Proposition 11-1-2 (Hibert's Satz 90): Let L/K be cyclic of degree n, Gal(L/K) = (o). If x E Lx is such that NLl,(x) = 1, then there exists y E Lx such that x = yo(y)-I.

Proof: For 0 i < n - 1, let a, = x-o(x) . . . oi(x). Then

o(a i )=x- 'a i+ , ( O < i < n - 2 )

o(a,-,) = x-'.

Consider the endomorphism of L defined by

By the theorem on the linear independence of characters, there exists y E Lx such that the image w of y under the endomorphism is nonzero. Then

n- l o w = C o(ai)oi+'(y)

i = O

n- 2

= X-' C a,+, oi+'(y) + x-'y i = O

= x-'IV.

Therefore, x = wo(w)-'. //

Proposition 11-1-3 (Herbrand's Lemma): Let G be an abelian group, H a subgroup of finite index. Let T, and T, be endomorphisms of G such that

(1) TI T, = T, TI = 0. (2) Ti(H) E H (i = 1, 2).

Set Si = Ti 1 H. Then

[ker (TI) : im (T,)] [ker (S,) : im (S,)] , [ker (T,) : im (T,)] = [ker (S,) : im (S,)]

I where all the indices are finite.

Proof: Note that all of the indices mentioned in the assertion are well defined (e.g., ker (TI) 2 im (T,)) by (1) and (2). We clearly have

I [G: H ] = [G: H - ker (T,)][H. ker (T,): HI.

Since G/H-ker (T,) * im (T,)/im (S,), we see that

[G: H. ker (T,)] = [im (T,): im (S,)], (1) and this latter index is finite. By one of the fundamental homomorphism theorems,

H . ker (T,)/H * ker (T,)/ker (S,) so that

[H. ker (T,) : HI = [ker (T,) : ker (S,)],

with both indices finite. Since im (S,) E ker (S,), we see that

[ker (T,) : ker (S,)] = [ker (T,): im (S,)] [ker (S,) : im (S, )] '

From equations ( I ) , (2) , and (3),

[G:H] = [im (T,): im (S,)][ker (T,): im (S,)]

[ker (S,): im (S,)]

- - [im (T,): im (S,)][im (T,): im (S,)][ker (T,): im (T,)] [ker (S,) : im (S,)]

Since our reasoning was symmetric in the indices 1 and 2, we may interchange 1 and 2 in the last equation and then equate the two different expressions for [G:H]. This yields the desired identity. Since all of the indices in the course of the proof are finite, the finiteness assertion is also proved. //

1 -

Theorem 11-1-4: Let LIK be a cyclic extension of local fields, e = e(L/K). Then

[UK : NLIK U J = e,

[Kx : NLIK(Lx)] = deg (LIK).

Remark: In particular, Theorem 11-1-4 is applicable if L/K is an unramified normal extension, since in this case the Galois group is generated by the Frobenius automorphism. Thus, if LIK is a normal, unramified extension,

UK = NL,KuL-

Moreover, it is easy to show (Exercise) that if L/K is normal but with e > 1, then

u, ;t NL;, UL.

Proof: The case where K is Archimedean is trivial and is left to the reader. Thus, let us assume that K is non-Archimedean, n = deg (LIK), G = (a). Let a and p be the endomorphisms of UL defined by

a(x) = xa(x)-' (1)

P(x) = NL,K(x)- (2)

By straightforward computations, we see that

ker (a) = UK

ker (/I) = a(Lx) (Hilbert's Satz 90) (4)

[ker (P): im (a)] = e. ( 5 )

Moreover, since P(x) = x-a(x) . . . - . 8"-'x, we see that

u p = pa = 0, (6)

wheie 0 is the trivial endomorphism of U,. Choose a positive integer r so large that log provides an isomorphism between 1 + P: and P:. Since a and p both stabilize 1 + P;, and (U,: 1 + P:) < oo, Herbrand's Lemma may be applied to yield

[ker (a') : im (p')] e-' iUK: N L ~ ~ uLl = [ker (PI): im (a!)]'

where a' and p' are the restrictions of a and p, respectively, to 1 + P;, and where we have applied equations (3) through (5). Let us apply the isomorphism log to 1 + P:. To the endomorphisms a' and /I' correspond the endomorphisms y and 6 of Pi, where

[ker (y) : im (a)] [ker (6) : im (y)]

By the normal basis theorem of Galois theory, there exists x E L, so that (1, ax, . . . , an-'x) is a basis of L/K. Without loss of generality, we may suppose that x E P i , so that a f x E P: (0 5 j 5 n - 1). Thus,

A = xOK + axO, + - - + an-IxO, s P i ,

Since the left-hand side is both compact and open in P:, [P;:A] < oo, we may apply Herbrand's Lemma to the pair of endomorphisms y, 6 of P: and the additive subgroup A, finding that (8) is equal to

[ker (y') : im (6')] , [ker (6') : im (y')]

where y' (respectively 6') is the restriction of y (respectively, 6) to A. Note that every element of A can be expressed uniquely in the form r(o)x, where r(X) E OKIX] is of degree < n. Thus, if s(X) E K[X] is such that s(a)x = 0, then s(X) is divisible by 1 - P (just apply the Euclidean algorithm). Let

n- 1

z = C aVaVx, a, E OK. v-0

Then n- 1

z E ker (y') - (1 - a ) C avovx = 0 u= 0

n- 1 e+ (1 - X) C avXv = a(l - X"), a E OK

v = o

- z = a Tr, , (x) E TrLIK (A) e=+ z E im (6').

Therefore,

[ker (y ') : im (S')] = 1. Similarly,

n- 1 z E ker (6') - (1 + a $- - - + an-') C avavx = 0

v = o

Thus, ker (6') = im (y') and [ker (6'): im (y')] = 1. Thus, the expression in equation (9) is 1. /I And the right-hand side of equation (7) equals

Let 1132 be a K-modulus. Let

Vm = {a = (a,) E JK 1 a - 1 (mod 1132), up E Up for pt1132).

Then Vn is an open subgroup of 11,.

Definition 11-1-5: A subgroup H of JI, is said to be admissible if H 2 K x Vm for some K-modulus.

Proposition 11-1-6: Let LIK be normal. Then KxNL,DL is an admissible subgroup of Q,. Proof: Let S denote the set of primes (finite or infinite) that ramify in LIK. By Proposition 11-1-2, NLrplKvL$ 2 U, for p q! S, 9 a prime divisor of p in L. Let p E S - S,, and choose r, E Z+ so that NLrpIKv(Urp) 2 1 + prv. Such an r, exists, since the local norm is an open mapping. Set

Definition 11-1-7: Let LIK be normal. Then KXNL,,DL is called the admissible subgroup associated to the extension LIK. We shall denote this admissible subgroup by HLK.

Remark: One consequence of the reciprocity law will assert that every admissible subgroup of 9, is of the form HLIK for a uniquely determined abelian extension LIK.

Definition 11-1-8: Let H be an admissible subgroup of J,. If KxVm c H for a K-modulus '332, then we say that H i s dejined modulo '332 and we say that 1132 is a dejining modulus for H.

EXERCISES 11-1-9: ( 1 ) If H is defined modulo Fm and if Fm divides 8, then H is defined modulo 8.

(2) If H is defined modulo both % and 8, then H is defined modulo (Fm, %) = the greatest common denominator (g.c.d.) of TJl and 8.

(3) There exists a "minimal" defining modulus f H of H. (I.e., H is defined modulo f H , and if H is defined modulo Fm, then f H divides %.) Such a defining modulus is unique and is called the conductor of H.

Proposition 11-1-10: Let HLIK have conductor fLIK and let p be a K-prime. Then p divides fL,, o p ramiJies in LIK.

Proof: See Propositions 1 1-1-2 and 11-1-4.

Definition 11-1-11: Let H be an admissible subgroup of 4,. Then C, = S,/H is called the class group associated with H.

Let us now show that CH is isomorphic to a certain group defined in terms of ideals. It is via the isomorphic copy of CH that the properties of the class group will be explored. Let us assume throughout the present discussion that H is an admissible subgroup defined modulo 1132; let us set

PK(1132) = {i(a) 1 a E Kx, a E 1 (mod 1132))

J , , = (a E JK 1 a, = 1 for all p E S),

where i(a) denotes the ideal associated with a , and S = the set of primes dividing 1132. It is clear that H m = i(D,,, n H ) c IK(1132) = the group of all K-ideals prime to 1132.

Proposition 11-1-12: CH = IK(1132)/Hm.

Proof: Let us prove the desired assertion by constructing the isomorphism in two steps-namely,

c, -- [JK,sl(~K,s n = [1,('332)I~d. (1) Let a E 9,. Since K x Vm c H, there exists jl E H such that u p E D , , (multiplicative approximation). The product u p is determined only modulo D,,, n H. Thus, there exists a homomorphism

JK + JK, s / JK,s n H* The kernel of this map is H, whence the first isomorphism of equation (1). Let us now consider the surjective mapping

i: j K , S IK(a), gotten by restricting the ideal map to J, , . It suffices to show that ker (i) c H. For then, the kernel of the mapping

ker (i) = (1) x 11 Up x rJ K," VES P $ S U S , P E S m - S

Proposition 11-1-13: Hm 3 P,(m).

Proof: Let a E K x , a r 1 (mod m). We claim that i(a) E Hm. It suffices to show that there exists p E DK,s n H such that i(P) = i(a). Let y E DK be defined by

(since a = l(mod 1132)). Then ay E 4 , s and ay E KxVm c H. Moreover, since a is a unit at all p lm , i(ay) = i(a). Thus, we may set p = ay. //

Proposition 11-1-14: C, is finite.

Proof: By Propositions 11-1-12 and 11-1-13, it suffices to show that IK(5332)/PK(5332) is finite. Let P$(5332) = the group of all principal K-ideals that are prime to 5332. It suffices to show that both IK(5332)/Pi(5332) and Pi(5332)/PK(5332) are finite. Since every K-ideal class contains an ideal relatively prime to 5332, IK(5332)/Pi(5332) is isomorphic to the ideal class group of K, and is thus finite. On the other hand, let R0(5332) [respectively, R(1132)I denote the group of all a E K X that are relatively prime to 5332 [respectively, the group of all a E K x such that a r 1 (mod % I ) ] . Then there is a surjective homomorphism

Thus, it suffices to prove that the group on the left is finite. Let 5332, denote the finite part of 5332, Um = (OK/5332,)" = the group of units of the finite ring 9,/!JJlO. Then there is a homomorphism

where (a, 5332) = 1. By the multiplicative approximation theorem, this map is surjective. Since Um is finite, this proves the assertion. 11

Corollary 11-1-15: H is an open subgroup of ,OK.

Proof: JK/H = CH is finite and hence discrete. //

Let c be a character of C,. Then we shall view c as a character of J K such that c(H) = 1. [c is necessarily continuous, since ker (c) 2 H and H is open.] Since K x c H, c ( K x ) = 1, so that c is a grossencharacter of K.

Definition 11-1-16: A grossencharacter arising from a character of C, for some admissible subgroup H is called a grossencharacter of classical type. Equivalently, c is a grossencharacter of classical type o ker (c) is an admissible subgroup of J,.

EXERCISE 11-1-17: Let c be a grossencharacter of classical type, with generalized conductor f . Then ker (c) is an admissible subgroup of K having conductor f.

Let c be a character of C,, where H is an admissible subgroup of D K defined modulo 5332. Let xC denote the ideal character associated with c. Since c(H) = 1 and H 2 Vn, we see the c,(Uv) = 1 for ~ $ 5 3 3 2 . Thus, X , is defined on IK(5332). Moreover, x,(H=) = 1 (direct computation). Therefore, X , can be viewed as a character of IK((Jn)/H,. Thus, we have constructed a homomorphism

Claim that is an isomorphism. By Proposition 11-1-12, we know that there exists an isomorphism A: CH --t IK(5332)/Hm. Let A* denote the dual isomorphism. It suffices to show that the diagram

c; + (IK(5332)lHm)*

1 identity 1 identity

c; -+ ([K@W/Hm)* is commutative. We leave this routine calculation as an exercise. The net result of this discussion is

Proposition 11-1-18: Let H be defined modulo 5332. Then the mapping

Corollary 11-1-19: Each character of IK(n)/Hm can be viewed as the restriction of some Hecke character to IK(YJl).

Let x be a Hecke character of K. We shall require the following result on Hecke L-series:

Theorem 11-1-20: Let Re ( s ) > 1. Then

log L(s, x ) = C x(P)NP-" + g(s, x), P

where g(s, X ) is deJined and continuous for Re (s) > 3. Proof: From the Euler-product representation of L(s, x) , we see that for Re ( s ) > 1,

where the principal branch of the logarithm is always understood. It suffices to show that the double series on the right converges absolutely and uniformly for Re ( s ) > 5 + 6, 6 > 0. But this is immediate, since

< 2n C p - ( 1 + 2 e ) , - P

and this last sum converges. I/

Proof of Theorem 11-1-1: Throughout the proof, functions denoted g(s), g,(s), etc., will denote functions that are defined and bounded in some neighborhood of s = 1. Let H = HKIL be defined modulo 9ll. By Prop- osition 11-1-12, it suffices to show that (IK(911):Hm) < n. Let us denote IK(%R)/Hm by Gm. Let x E G g . By Proposition 11-1-19, we can view x as the restriction of some Hecke character (also denoted X) to IK(911). Let s > 1. Then by Theorem 1 1-1 -20,

1% L(s, x ) = C X(P)NP-~ + g(s, x) (P, a)= 1

(1)

Summing (1) over all X, and recalling that

= o , P $ we see that

We have shown that if x is a nontrivial character, then L(s, X) is an entire function that does not vanish at s = 1. Thus, for x f xo = the trivial character, log L(s, X) is bounded in a neighborhood of 1. On the other hand, L(s, xo) has a pole of order 1 at s = 1. Therefore,

log LO, xo) = -1% (s - 1) + g h ) (3)

Therefore, (2) and (3) imply that

-[Gm]-' log (S - 1) = C Np-" + g,(s). (P, m,= 1

(4) P%JJl

Now claim that if p splits completely in LIK, (p, 9ll) = 1, then p E Hm. Let 9 be a prime divisor of p in L. Let Kt (respectively, a) be a local uniformizing parameter at 9 (respectively, p). Set

a = ( l , l , . . . , l , I I , l ,... ) € J p d

Then NL,(a) E .DK,S n H and i(NLI,(a)) = pf(qI~) = p since p splits completely in LIK. Thus, p E Hm.

By (4) and the above result, we see that

-[Gal-' log (S - 1) 2 C NQ-' + g4(s) p splits completely

in LIK

Since l,,(s) has a pole of the first order at s = 1,

log l,,(s) = -1% (3 - 1) + gds) = C N V S + g,(s),

v by Theorem 11-1-20. Next, we claim that

C Nq-" = g7(s). '$ does not spl i t completely

i n LIK

Indeed, if 9 does not split completely in LIK, N$-" Np- f (~ /~ )= < NP-~". Thus, by (61,

n C ~ p - J = p splits completely i n L/K

C @ splits completely i n LIK

N$-"

= -log (s - 1) + g7(s).

BY ( 5 1 7

-[Gm]-' log (s - 1) 2 -n-I log (s - 1) + g,(s).

Dividing by log (s - 1) and letting s 4 1 + yields

EXERCISE 11-1-21: Let (332 be a K-modulus. The cosets in ZK((332)/PK((332) are called arithmetic progressions defined modulo Cm. ZK((332)/PK((332) is called the group of arithmetic progressions defined modulo Cm.

(1) Let K = Q, m > 0. Then a typical arithmetic progression defined modulo m is

{(a + bm) I b E Q, ord, (b) 2 0 for all p 1 m3,

where a E @ is such that ord, (a) = 0 for all p 1 m. There are ~ ( m ) such progressions. A typical arithmetic progression defined modulo mp, is

{(a + bm)Ib E @,ord,(b)~Oforal lpIm, b > 01,

where a is as before. There are 29(m) such progressions. (2) If 8 , % E IK(Cm), we write 8 - %(mod (332) to mean that 8 and 8 are in

the same arithmetic progression defined modulo Cm. Then 8 r 23 (mod (332)- 8 % - I

E PK((332). If K = Q, then (a) r (b)(mod m) o a = b(mod m) in our previously defined sense of congruence. If K = @, (a) - (b)(mod mp,) - a r b(mod m) and a and b are of the same sign.

(3) If H is an admissible subgroup of QK, then each element of C, is a union of arithmetic progressions.

11-2 THE FIRST INEQUALITY

In the preceding paragraph, we obtained an upper bound for the index [a,: H,,,]. Our computations were valid for LIK normal. Let us now study the case where LIK is cyclic in some detail. Our main result will be

Theorem 11-2-1 (The First Inequality): Let LIK be cyclic of degree n. Then

[JK : H L / K I 2 n.

An immediate consequence of this result is

L

Corollary 11-2-2: Let L/K be cyclic of degree n. Then

[J), : HLIK] = n.

Our proof of Theorem 11-2-1 will actually yield somewhat more than just the statement of the theorem. Let So be a set of K-primes containing all infinite primes and all primes that ramify in L/K. By Exercise 3-3-12, (4), we can choose So sufficiently large and guarantee that 4, = KX&. Let us assume throughout that So is so chosen. Let S denote the set of all L-primes that are divisors of elements of So. The statement we shall actually prove is

Theorem 11-2-1': Let L/K be cyclic of degree n, and let J); and LS denote the S-ideles and the S-units of L, respectively. Then

[J), : H,,,] = n[Kx n NL,J);: NL,LS].

A. Herbrand's unit theorem

As a prelude to the proof of the first inequality, we shall now prove a refinement of the Dirichlet unit theorem. This refined version is due to Herbrand.

Let all notations be as above and let Gal (L/K) = (a). Suppose that

SAK) = {p,, . . . , P,},

S O = {PI, - . ., ~ r ) ,

and let S have s + 1 elements. Then there exist s independent S-units. For 1 5 i< t, let n, = the local degree of L/K at pi, ai = n/n,. Then

D(pi) = (aa3 (1 5 i 5 t). (1)

Let 3, be a fixed prime divisor of pi in L. Then the set of all prime divisors of pi in L consists precisely of the primes

a"$, (0 5 v < ai - 1). Therefore,

r s + 1 = C a , .

i= 1 (3)

By the Dirichlet unit theorem, the group of S-units of L is the direct product of the group of roots of unity in L and a free abelian group of rank s. The Herbrand unit theorem asserts the existence of s + 1 S-units which are acted upon by Gal (LIK) in a particularly simple way, which satisfy a simple relation, and which contain among them s independent S-units. More precisely,

Theorem 11-2-4 (Herbrand): There exist S-units el, . . . ct, such that (1) aal(ci) = ci (1 < i j t). (2) The group generated by av(ei) (1 j i 2 t, 0 j v < a, - 1) is of

finite index in UL(S). (3) The units satisfy the relation

where Ai(x) = x.a(x) . . . aal-'(x). (4) All relations among the units av(e,) are generated by the single re-

lation of (3).

To prove Herbrand's unit theorem, we need two preliminary results:

Proposition 11-2-5: Let (a,,) be an n x n real matrix such that aij < 0 for i # j, and such that

Then (aij) is nonsingular.

Proof: Assume that the assertion is false. Then det (aij) = 0 and there exist real numbers x,, . . . , x,,, not all 0, such that

Without loss of generality, we may assume that I xi ( < 1 (1 2 i 5 n), x, = 1. Then

n

2 a , , + C a, (since for i 2 2, aij < 0) j = z

> 0. This yields a contradiction to (*). I/

Proposition 11-2-6: Let M be a number jield, SJM) = {p,, . . . , p r + , ) . Let E , , . . . , e,,, be elements of UM such that for 1 < (i, j) < (r + I),

Then e l , . . . , E,,, generate a subgroup of jinite index in UM. Moreover, there always exists such a system of units el , . . . , cr+,. Such a collection is called a system of Minkowski units for M.

rlr,

Since

200 / THE INEQUALITIES OF CLASS FIELD THEORY CHAP. 11

Proof: In order to prove the first assertion, Dirichlet's unit theorem implies that it suffices to show that E,, . . . , 6, are independent units. To prove this fact, it suffices to show that the matrix

(1% I f j b,)lSi,j<r

is nonsingular. But, by assumption, for i f j, the (i, j)th entry of this matrix is negative. Moreover, since

I Nw/dcj) I = I f j (PI . . I f j IP,+, = 19 r

C 1% I E j I = -1% I E j I,,+, > 0. i = 1

Therefore, the nonsingularity follows from Proposition 11-2-5. The existence of a system of Minkowski units follows directly from Theorem 3-3-6. 11

Proof of Herbrand's Unit Theorem: By Theorem 3-3-6, there exist elements . , qr+, of UL such that for 1 < i < r + 1,

I Vi IS, > 1

I ri IS < 1 (13 E U L ) , P f '$i)

I oVqi I,.S, = I q, IS,, upon taking account of equation (2), we see that

( o v q i I l < i < r + l , O < v < a , - 1 )

is a system of Minkowski units for L. For 1 i < r + 1, let us define

l i = Vi if ni = 1

= qianI2(qi) if n, = 2.

It is then trivial to show that

( o v l i I l < i < r + l , O < v < a , - - 1)

is also a system of Minkowski units for L and that oai(li) = l i . In particular, li is contained in the decomposition subfield of L at 13,. Set

C = A i ( l i ) ( i = l , . . . , r + l ) .

Then ri E K n UL = UK. Moreover, for i f j,

I ovAi IS*

I Ai I ,-"%,

I ri lP> = ",=o at- 1

= rI v = 0

Also, since Ci E UK, r + l

1 Ci 1, = 1 (Product Formula), j= 1

so that

I Ci I P ~ > 1.

Therefore, C', . . . , Cr+, is a system of Minkowski units for K and contains r independent units. Therefore, C,, . . . , Cr+, satisfy a relation

But by the definition of Ci, this implies that

Thus, let us set ei = lfl(1 < i < r + I). Then

( o V ~ , ( 1 _ ( i _ ( r + 1, O < v _ ( a i - 1)

is a system of Minkowski units for L such that aai(ei) = ei (1 5 i 5 r + I), and

In particular, the units oV(~, ) generate a subgroup of finite index in UL, and the one relation among them generates all others.

Let us now construct E , + ~ , . . . , 6,. For r + 2 ( i ( t , let Zi denote the decomposition subfield of L at '$,, let Qi denote the restriction of 9, to Zi, and let hi denote the class number of Zi. Then lap is a principal ideal. Let E, E Zi be a generator for this ideal. Then

EiOL = Q?OL = ' $ ; lh l ,

where ei = e(Pi/pi). Moreover, since D('$,) leaves Zj pointwise fixed,

oLlf(ci) = ci. r

Let c = eihi and let x be an S-unit. Then i = r + 2

xOL = fi aG ov('$P) (b, E Z), i = r + 2 v=O

where r

civ = biv J-J ejhj. j = r + 2 j t i

Therefore, xc is contained in the group generated by U, and oV(ei) (r + 2 2 i ( t, 0 < v I ai - 1). Thus, finally, we see that the group generated by oV(ci)(l _< i _< t, 0 5 11 ai - 1) is of finite index in LS. This is assertion (2). Assertion (1) is obvious from the construction of the units. Assertion (3) has also been established.

Let us now prove (4). Suppose that there exists a relation

Forming the ideal generated by the elements on both sides, we see that

* O j j av('$3iy,h'*,."". = 0, i = r + 2 u=

*biv=O for r $ 2 < i < t .

Therefore, the given relation involves only e l , . . . , e,+,, so that the relation is a consequence of the relation of (3). This is assertion (4). 11

B. An index computation

As our next step in the proof of the first inequality, we shall prove an index relation that follows from Herbrand's unit theorem. Throughout this discussion, let us keep all of the notations of the preceding subsection. More- over, let us assume that S has at least two elements. (This is always the case, but we do not need this fact.)

The main result of this subsection is f

Theorem 11-2-7: [KS: NL,,(LS)] = n ni/n, where KS denotes the So-units i= 1

of K.

Before turning to the proof of Theorem 11-2-7, let us prove a strength- ened version of Hilbert's Satz 90.

Proposition 11-2-8: Let x E LS be such that NLIK(x) = 1. Then there exists y E LS such that x = ya(y)-I.

Proof: By Hilbert's Satz 90, there exists z E L x such that x = za(z)-l. It suffices to show that z can be written in the form wy, where w E K and y E LS. Write

z0 , = AB,

where A is divisible only by primes in S and B is divisible only by primes not in S. Suppose that '$3 6 S. Let p denote the restriction of '$3 to K, and let D(p) = (aa(v)). Since xOL = zO,-t~(z)-~O, and since x is an S-unit, each conjugate of '$3 appears to the same power in the prime ideal decomposition of z0,. Therefore,

Since So contains all ramified primes, e('$3/p) = 1 for p 6 So and thus

But since 4, = Kx42 , the S, class number of K is 1. Therefore, A is a principal ideal, generated by an element of K. But then B is principal, and we may write z = wy for some w E K, y E LS. // Proof of Theorem 11-2-7: Define the endomorphisms A and B of LS by

A(x) = xa(x)-I

B(x> = NL&). It is trivial to check that AB = BA = the trivial endomorphism, and that ker (B ) = im (A) by Proposition 11-2-8. Therefore,

[KS: NL,(LS)] = [ker ( A ) : im (B)] [ker (B ) : im (A)] '

Let M be the subgroup of LS generated by the elements

ov(ci) ( 2 ( i < t , 0 v a, - 1 )

au(c l )o""(~ , ) - l (0 < v < a , - I). It is clear that M is stabilized by a and therefore by A and B. Moreover, M is of finite index in LS. For indeed,

* E,o(E,) . . . aal-'(6,) E M * E? E M* ~ " ( c : ~ ) E M (v = O , . . . , al - 1 ) => M is of finite index in LS [by (2) of Theorem 11-2-41,

Thus, we may apply Herbrand's Lemma to derive

[KS : NL,,(LS)] = [ker (A ' ) : im (B')] , [ker (B') : im (A')]

where A' (respectively, B') is the restriction of A (respectively, B) to M. Let y E M ,

Then y E ker (A') c> a ( y ) = y. Using the fact that the av(ei) satisfy an essentially unique relation, we see that

y ~ k e r ( A ' ) e b , , = O fora l lv

biu = bi, (i 2 2). Thus, ker (A') is generated by the independent units

q,. = . . . 0.1-'(e,.) (i = 2, . . . , t).

However, im (B') = the group generated by NLIK(ci) (i = 2, . . . , t ) and N L I K ( ~ l ~ ( ~ l ) - l ) = 1 . But

1

[ker (A') : im (B')] = JJ ni. i = 2

(2)

In order to compute the other index, let us adopt the following notation: Iff E Z[X], set

f (a)(x) = xa~o(x>.l . . . ar(x>.. (x E L),

where f(X) = a, + a , X + . . . + arX.

Then fg(u)(x) = f (cr)(g(u)(x)). Moreover, iff (a)€, = ei, then 1 - P I f (X). This fact follows from the Euclidean algorithm and the fact that the units aV(ei) (u = 0, . . . , ai - 1) are independent.

It is a simple computation to show that

ker(Bf) = {fl(c)(l - o)(€,), . ,fr(o)(l - a)(€,) Ifi Z[XII

ker (B') - {f(o)(l - 4( f l ) I . -- im (A') { g ( W - G I ~ (el)I

It is clear that every coset on the right is represented by an element of the form a(l - o)(e,) (a E Z). Two such elements a(l - a)(€,), a'(1 - a)(€,) belong to the same coset o

(a - 1 - ) ( , ) = ( 1 - u)2 ( , ) for some g E Z[X].

Without loss of generality, assume that deg (g) ( a, - 1. Then the two elements are in the same coset o

{(a - af)(1 - u) - g(u)(l - CJ)~](E,) = el for some g E Z[X]

o 1 - Xal 1 (a - af)(l - X) - g(X)(l - X)' for some g E Z[X]

o 1 + X + . . - + PI-' I(a - a') - g(X)(l - X) for some g E Z[X]

o a a'(mod a,). (For the => implication, set X = 1. To get the converse, note that there exists g E Z[X] such that 1 + X + . + Xal-l = g(X)(l - X) + a,.)

Therefore, there are a , = n/n, cosets, and

n [ker (B') : im (A')] = - "1

Thus, by equations (1) and (2), the assertion is proved. /I

C. Proof of the first inequality

With the aid of the index relation of subsection B, let us now prove Theorem 11-2-l', which implies the first inequality in a trivial way. Let all notations used in subsections A and B remain in effect. Then

[fl, : HLIK] = [K " 5; : K " N,,,(K " a;)]

= [Kx.J;/Kx : K " N ~ ~ ( K " ~ ~ ~ ) / K " ] ,

[@ : N,,,D;] = [Kxfli/Kx : K" N~,~~~;/K"][K"& : K " N ~ , J J ~ ]

= [fl,: H,,] [KXD$ : KXNL,,fli].

Therefore,

Let p be any K-prime and '$3 and '$3' be prime divisors of p in L. Then there exists an isomorphism of Lrp onto Lv which leaves Kp pointwise fixed. Therefore, NLgIKv (4) = NL ,, (Li;,). For each prime p E S, let us choose

8 v a prime divisor Tp of p in L. Then

for S contains all primes that ramify in L/K, and if p is unramified, then NL8,,,(UP) = Up by the remarks following the statement of Theorem 11-1-4. Therefore,

by Theorem 11-1-4. Thus, by equation (*),

= n[Kx n NL,,Jf: NLIKLS] (by Theorem 1 1-2-7). This is precisely the assertion of Theorem 11-2-1'. //

Class field theory

12-1 PROOF OF THE RECIPROCITY LAW

The present section will be devoted to a proof of the reciprocity law. Our proof will be a contemporary version of the proof contained in Cheval- ley's thesis, with improvements due to Artin and Herbrand. Our first job will be to give a definition of the conductor that does not use the reciprocity law, as was the case with the definition given in Section 10-1. We shall then prove a weak form of the reciprocity law, which will allow definition of the norm residue symbol. Then we shall prove Artin's reciprocity law. Finally, we shall prove (but not until Section 12-3) that our new definitions of the conductor and the norm residue symbol coincide with those introduced in Section 10-1. Throughout this section, let LIK denote an abelian extension of K with Galois group G. Let H,,, denote the admissible subgroup of JK associated with the extension LIK.

Definition 12-1-1: By the conductor f,,, (=f) of the extension LIK, we mean the conductor of HE,,.

From Proposition 1 1-1-10, we derive

Proposition 12-1-2: Let p be a (jinite or injinite) K-prime. Then p divides f L I K o p ramifies in LIK.

Let us define D L I K = {a E D K ( a -- 1 (mod f ) ) .

Then DL,, is an open subgroup of D,. By the multiplicative approximation theorem, we see that

KXDLIK = f l K .

As usual, let i:DK -+ ZK denote the ideal map. Let us define the reciprocity map

This homomorphism is well defined, since (i(a), f) = 1 3 (i(a), d,,,) = 1 by Proposition 12-1-2.

Theorem 12-1-2 (Weak Reciprocity): r,, induces an isomorphism

DLIK/(HL/K n flLIK) * G .

In order to establish the weak reciprocity law, it suffices to prove the following two propositions :

Proposition 12-1-4: rL, is surjective.

Proposition 12-14: ker (r,,) = H L , n J,,,.

In order to prove Proposition 12-1-4, we require three lemmas.

Lemma 1 : It sufices to verify Proposition 12-1 -4 when L/K is cyclic of prime power degree.

Proof: By the fundamental theorem of abelian groups, we can write G = C, @ . - . @ C,, where Ci ( I < i < r ) is a cyclic subgroup of prime power order. Let Li(l < i < r ) be the fixed field of C, @ . . . @ ei @ . . .. @ C,, where the caret denotes that Ci is omitted. Then Li/K is cyclic with Galois group isomorphic to Ci. Moreover, the fields Li (I i < r ) are linearly disjoint over K. Let a E G. Write a = at1 . . . a:, where Ci = (ai) (I < i ( r). If we assume Proposition 12-1-4 in the cyclic case, there exists

ai t DK ( 1 I i < r ) such that

Lemma 2: Let LIK be cyclic of degree p = a prime. Then there exist infinitely many primes that are undecomposed in L.

Proof: Since deg (LIK) = a prime, a K-prime p either splits completely in L or is undecomposed. Assume that p splits completely in L for almost

208 1 CLASS FIELD THEORY CHAP. 12

all K-primes p . Then L = K by Bauer's Theorem (Theorem 9-1-3). Thus, a contradiction is reached. //

Lemma 3: Let LIK be cyclic of degree pr, p a prime. Then there exist infinitely many primes that are undecomposed in L.

Proof: Assume that the assertion is false. By Lemma 2, r > 1. Let S = the set of primes that are undecomposed in LIK. Then S is a finite set. Let G' be the normal subgroup of G of order pr-', L' = the fixed field of G'. Then L'IK has degree p. If p $ S, p splits in LIK => deg (Z,IK) > 1 for almost all p => Zp > L' for almost all p. Thus, almost all K-primes split in L', which is a contradiction to Lemma 2. //

Proof of Proposition 12-1-4: By Lemma 1, it suffices to assume that L/K is cyclic of degree pr, p a prime. Let G = (a). By Lemma 3, there exists a K-prime p such that p is undecomposed in L/K and (p, dL1J = 1. Then

LIK (T) generates G, since it is of order ef = f = deg(L/K). It is clear that

LIK is contained in im (r,,). // (7) The basic idea of our proof of Proposition 12-1-5 is to reduce it to the

case of a cyclotomic extension, where the result may be verified by an easy computation. In order to make this reduction, we require the following construction theorem:

Theorem 12-1-6: Let LIK be cyclic, S a finite set of K-primes containing S, and all primes that ramify in LIK. Let p be a K-prime not in S. Then there exists an abelian extension MIK such that

(1) LMIM is cyclotomic; i.e., L M M(5,) for some m. (2) p splits completely in MIK. (3) L n M = K.

In order to prove this crucial result, we require two lemmas about cyclotomic extensions. In the following discussion, let m and n be two positive integers. We shall always assume that m is relatively prime to p

and to all primes in S. Set Lm = K([J, rm = (LLK), P Gm = Gal (Lm/K),

dm = the order of z,.

Lemma 1: There exist infinitely many m for which n Idm.

Proof: By Exercise 6-2-22, rmcm = [zv. Thus we are reduced to proving

Reduction I: Let a > 1, n > 1 be integers. Then there exist infinitely many positive integers m such that the order of a mod m is divisible by n, and m

SEC. 12-1 PROOF OF THE RECIPROCI~ LAW 1 209

is relatively prime to p and S. For we need only take a = Np to establish the assertion of Lemma 1.

Reduction 2: In Reduction 1, it suffices to assume that n = pr, p a prime. For suppose that n = p;l . . . pp. Choose mi (1 j i 5 k) inductively so that (1) mi is relatively prime to p and S, and also to mj (1 5 j < i - I), and (2) the order of a(mod mi) is divisible by p?. It suffices to set m = m, . . . m,. Since the mi are pairwise relatively prime, the order of a(mod m) is divisible by n. It is clear that this process can be used to construct infinitely many such m.

Reduction 3: It suffices to show that for all sufficiently large integers r, there exists a prime q, such that the order of a(mod q,) is exactly pr.

It is clear that the q, are all distinct. Therefore, infinitely many can be selected that are relatively prime to p and S. This set of primes satisfies the conditions of Reduction 2.

Proof of Reduction 3: Let A = (UP' - 1)/(aPr-' - I). Choose r so large that A > p. Claim that there exists a prime q, dividing A, q, f p. If not, then A is a power o fp and is divisible by p2. Let B = UP'-' - 1. Then

where all coefficients in the sum of the right are divisible by p. Since p divides A, p divides B; provided that p > 2, p2 divides all terms of the above sum, except for the last. Therefore, A =p(modp2), which is a contradiction to the fact that p2 divides A. If p = 2, then A = B + 2, B = a2'-' - 1. If 4 divides A, then B r 2(mod 4), so that a2'-' G 3(mod 4), which is an im- possibility, since 3 is not a quadratic residue modulo 4. Thus, there exists a prime qr # p, q , I A. Then

so that the order of a(mod q,) is pr. //

Lemma 2: There exist infinitely many m such that there exists am E Gm such that

(I) 2, and a, are independent (i.e., a;r& = 1 a; = 1, rk = 1). (2) The orders of 7, and a, are both divisible by n.

Proof: By Lemma 1, there exist infinitely many m, such that n 1 dm,; there exist infinitely many m, such that dm, Idm,, (m,, m,) = 1. Let m = m,m,,

urn = (Ky) = rmi. Then (2) is clearly satisfied. Assume that

Since z, = z,,z,,, z;;l:"z;, = 1.

This implies that zCb = 1, z;, = 1, since L,, and L,, are linearly disjoint over K*d,,I(a+ b),d,,Ib*d,,Ia, d,,lb*a;= 1, z;= 1 *a, and 7, are independent.

Proof of Theorem 12-1-6: Let m be chosen so that the conclusions of Lemma 2 are satisfied. Without loss of generality, assume that L, and L are linearly disjoint over K. This can be accomplished since L contains only finitely many roots of 1, say all rth roots of 1. Choose m so that (m, r) = 1. Then

L, n L = K. (1)

Let G = Gal (LIK). By (1).

Gal (LL,/K) = GG,. (2)

Let G = (o), I = (g), H, = {aa,, IT,), M = the fixed field of H,. P

Since p is unramified in LIK and in L,/K, p is unramified in LL,/K. There-

fore, (Ly) is well defined. Moreover,

by (1). Thus,

p splits completely in MIK. Next, claim that H, n G = {I). [Here we view G as contained in

Gal (LLJK) by extending the elements of G to be the identity on L,, which is possible by (I).] Let q E H, n G. Then

But the left-hand side is in G,; the right-hand side is in G. And G, n G = {I) [viewing G, as a subgroup of Gal (LLm/K) in the same way as we viewed GI. Thus, ab, = 1, 7; = 1, since a, and 7, are independent * n I b, n I c, since the orders of a, and 7, are divisible by n => ab = 1, Ic = 1. But since ab--'p = 1, this implies that aa = 1 =. nla * q = a" = 1.

The fixed field of H, is M ; the fixed field of G is L,. Thus, the fixed field of H, n G is ML,. Since H, n G = {I),

LL, = ML,.

SEC. 12-1 PROOF OF THE RECIPROCITY LAW 1 211

Thus,

LM G LL, = ML, = M(c,).

Note that Gal (LLJK) = GG,, where the product is direct. Since L is the fixed field of G, and M is the fixed field of H,, L n M is the fixed field of G,H,. But aa, E H,, a, E G, * a, E H,G,. Thus, G c H,G,. Therefore, GG, = H,G,, and L n M = K. //

Remark: Suppose that we are given a finite set T = ( p , , . . . , p,) of K-primes such that T n S = 0. By Theorem 12-1-6, there exist abelian extensions MJK, . . . , M,/K such that

(1) LMi/M, is cyclotomic (i = 1, . . . , t). (2) p, splits completely in M, (i = 1, . . . , t). (3) L n Mi = K.

By choosing the values mi used to define M, so that

L,, n (LL,, . . . L ,,.,) = K (i = 2,. . . , t), we can replace (3) by

(3') L n M, . . . M, = K.

[To verify this, follow the same reasoning as used to prove (3) of Theorem 12-1-6.1

Proof of the Weak Reciprocity Law.

Reduction I: It suffices to prove that ker (r,,,) =I H,,, n DL,,. Indeed, it suffices to verify Proposition 12-1-5. If we knew that

ker (~L/K) 3 HLIK n DL,,, then

(-DL/, : ker ( ~ L I K ) ) < (DL,, : H L r K n DL,).

By Proposition 1 1-2-1, (DL,,: ker (r,,)) 2 n. Also, JLrK/(HLrK n DL,,) = J,/H,,, by the multiplicative approximation theorem. Therefore, by the second inequality, (DL,,: HLK n DL,) I n. Collecting all the information together, we see that (DL,,: ker (r,,,)) = (DL,,: H L r K n DL,) = n. Thus, ker (r,,) = H L / K n DL,,, which is Proposition 12-1-5.

Reduction 2: It suffices to consider LIK cyclic of prime power degree.

Every abelian extension of K can be written as a composite of a collection of pairwise linearly disjoint cyclic extensions of prime power degree. Thus, it suffices to show that if LIK and MIK are two linearly disjoint, abelian extensions such that the weak reciprocity holds for LIK and M/K, then weak reciprocity holds for LMIK. If a E JLMB, then

since L and M are linearly disjoint over K. Therefore,

Reduction 3: It suffices to verify the weak reciprocity law in the case L c K((,), for some positive integer m.

By Reduction 2, we may assume that LIK is cyclic of prime power degree. Let G = Gal (LIK). Let a = (a,) E ker (r,,,). Let T = the finite set of K-primes for which a, 4 Up. Suppose that

T = (PI, - . ,pr).

By Theorem 12-1-6 and the Remark following the proof, there exist abelian extensions Mi/K (1 < i 5 t ) such that

(1) pi splits completely in Mi/K (1 i j t). (2) LM, c Mi([,,) (1 2 i t ) , mi a positive integer. (3) L n (MI . . . M,) = K.

Let M = M, . . . Mr. By (3), LMIM is a cyclic extension with Galois group G. (The identification consists in extending the elements of G to be the identity on M.) By Lemma 3 to Proposition 12-1-4, there exist infinitely many primes fp of M such that fp is undecomposed in LMIM. Let us choose one such prime, subject to the condition that fp not divide any mi, any prime of S nor of T. Let ll be a local uniformizing parameter at p and set

generates G. Call this generator a. Then

Set a = ( 1 1 . . a , . . . ) (1 l i l t).

Then

Suppose that aK = am. Then ( i(ail )

Moreover, since pi splits completely in MilK7 ai = N,w,,K(yJ, Ti E JLM,IM,. Therefore,

Since L M c Mi([,,), the weak reciprocity law implies that ai E H,,,,,. Thus,

Finally, a E H L / K n JLIK. Thus, ker (rLIK) c HLIK n DL,, and the equality follows from the first inequality (Theorem 11-2-2).

Before we reduce the proof of the weak reciprocity law to a provable statement, let us make a short digression. Let E > 0, T = any finite set of primes containing all infinite and all ramified K-primes for the extension LIK. A T-neighborhood A, of 1 in JK is a neighborhood of the form

A,={a=(a , ) I I a , - l I ,<c ,p E T).

If E < 1, then a E A, => a, E U, for p E T - S, 3 i(a) is not divisible

by any prime that ramifies in LIK =>

Lemma: Weak reciprocity holds for LIK o There exists a T-neighborhood

A, of 1 in JK such that = 1 for a~ a E A..

I Proof: e. Let p E HLIK n JLIK. Then = xy, x E K x , y E NLIKJL. Since I pplp = 1 for p E T, x is a local norm at p for all p E T. Let p be a typical element of T and let $3 be a fixed prime divisor of p in L. Let yq E Ki; be such that

I x = N L ~ , ( Y ~ ) .

6 Let T be the set of L-primes '$3 that we have chosen. By the approximation theorem, there exists y E LX such that y is very close to y$'(fp E 7) and y

is very close to 1 at all other L-primes dividing p E T. For p E T, the p- component of the idele xNL,(y) is

and is therefore very close to 1. Thus, we may assume that xNLl,(y) E A,. Then

by the reduction assumption. But yNL,(y-l) = N,,(6), 6 E J,. Thus,

- - 1.

Thus, /3 E ker (r,,), so that by Reduction 1, the first implication is proved.

*. By the weak reciprocity law, = 1 for all a E HLIK n DuK.

Let J' = {a = (a,) I a, E Up for p E T - S,). Then J' is an open subgroup of J,, and so is HLIK. Thus, J' n HL, is an open subgroup of J, and contains

a T-neighborhood of 1 in 4,. Moreover, if a E J' n H,,,, then

Reduction 4: It suffices to prove the weak reciprocity law for L = K([,), m some positive integer.

By Reduction 3, it suffices to show that Reduction 4 implies that weak reciprocity holds for L c K([,). By the reduction assumption and the Lem- (4,:'") = 1 ma, there exists a T-neighborhood of 1 in flK, say A,, such that

for all a E A,, where T contains all infinite K-primes and all primi divisors of m. But A, is clearly a T-neighborhood for LIK. Therefore, by the Lemma and the fact that

we see that weak reciprocity holds for L/K. At last, we have reduced the weak reciprocity law into a simple cal-

culation. Let L = K([,). Let T = the finite set of K-primes containing all infinite primes and all prime divisors of m. Then T contains all primes that ramify in LIK. Let A be a T-neighborhood of 1 in JK such that

A c {a E DKIa = l(modm)].

SEC. 12-1 PROOF OF THE RECIPROCITY LAW / 215

By the Lemma, it suffices to show that

for all a E A. But, by Proposition 6-2-22,

= 5,Y since a E A * N(i(a)) - 1 (mod m). This completes the proof of the weak reciprocity law. //

Corollary 12-1-7: Let L/K be abelian with conductor f. Let x E Kx, x = l(modf). Then (3 - = 1.

Proof: Since x = 1 (mod f), for each prime p which divides f , there exists x, E LP such that x = NLPIKP(xP), where 9 is a fixed prime divisor of p in L. Let a E 8, be defined by

a = (x, . . . , 1, . . . , I, 1, . . .). - ,If

Then a E NLIKJL and xu-' E KxNLIKDL n DL,,. Thus, xu-' E ker (r,,,) by the weak reciprocity law. But since x is a unit at all primes dividing f, i(xa- l) = i(x). //

Let us now define the local norm residue symbol ( , LIK), just as in Section 10-1, except now using the conductor of Definition 12-1-1. Corol- lary 12-1-7 guarantees that the norm residue symbol is well defined, using the same sort of arguments used in Section 10-1. At the end of Section 12-2, we shall show that the conductor of the present section and the conductor of Section 10-1 coincide, so that the norm residue symbol we are presently using coincides with the corresponding symbol introduced in Section 10-1.

Theorem 12-1-8 (Hilbert's Reciprocity Law): Let LIK be abelian, x E Kx. Then

n (x, LlK)v = 1. P

Proof: First note that we can prove all of the assertions of Proposition 10-1-10 for the symbol ( , LIK),, using the same arguments as we used for [ , LIK],. The proof of Theorem 12-1-8 then proceeds exactly as does the proof of Theorem 10-1-1 1. //

In order to prove that ( , LIK), = [ , LIK],, let us prove a number of preliminary results about the kernel and the image of ( , LIK),.

SEC. 12-1 PROOF OF THE RECIPROC~TY LAW / 217

Proposition 12-1-9: Let p be a K-prime, Ip a prime divisor of p in L. Suppose that x = NLrplKP(z), z E L;. Then

(x, L/K)p = 1.

Pro08 We first claim that there exists y E L X such that

where f, = the exact power of p dividing the conductor f of LIK, f k = ff;'. Recall that we identified Gal (L%/KP) with Dp. Therefore, we may write

Let Ip = Ip,, . . . ,!Q, denote the distinct prime divisors of p in L, m a large positive integer. By the approximation theorem, there exists y E Lx such that

yz-I G 1 (mod Ipm), y = 1 (mod Ipr) (i = 2, . . . , g), y = l(mod fb).

Let a E Gal (LIK). If a E D,, then a(y)a(z-l) = l(mod 9"). If a 4 Dp, then a(y) = l(mod Ipm). Therefore,

E l(mod pm),

NLIK(Y) - l(mod fb).

Thus, there exists y E L x which satisfies (*). By the definition of ( , LIK),,

where i(NLIK(y)) = pa@, (@, p) = 1. But from the properties of ideal norms, we see that 8 = NLIK(8), where 8 is an L-ideal. Therefore,

by Proposition 5-5-5. //

Let a E D,. By Proposition 11-1-4, Remark, we see that (a,, LIK), 7 1 for almost all K-primes p. Thus, we may define the reciprocity mapplng

r:DK -+ G

r(a> = IT (up7 LIK)P. P

Theorem 12-1-10 (Strong Reciprocity Law): The reciprocity mapping r

induces an isomorphism

DKIHLIK * G7

where H,,, = KXNLlKD, is the admissible subgroup of DK associated to the extension LIK.

Proof: It is clear that r is a homomorphism. First we claim that r is surjective. Let 6: G --, G be the isomorphism defined by 6(x) = x - l . It suffices to show that rlJLIK = 80rLlK. For we have shown (Proposition 12-1-4) that rLIK is surjective. Let a E DL/,. Since a r l(mod f), (a,, L/K), = 1 for all K-primes p that divide f. Thus,

by Proposition 12-1-2. By the analogue of Proposition 10-1-10, ( 9 , for the symbol ( , LIK),, we see that

Therefore,

since i(a) is relatively prime to dLIK. But -, = rL,(a). This completes (;%) proof of the assertion that r is surjective. In bider to complete the proof of Theorem 12-1-10, it suffices to show that

ker (r) = KxNL,(DJ.

As a first step, we assert that NL,,(DL) c ker (r). Indeed, let a = NLm(b), p E 9,. Then

by Proposition 12-1-9. By Theorem 12-1-8, we know that K x c ker (r). Thus, we have shown that KxNLIK(DL) c ker (r). Thus, since r is known to be surjective, KxNLIK(DJ is of index at least n, n = deg (LIK). But, by the Second Inequality (Theorem 11-I-]), KxNL,(DL) is of index at most n. Thus, KxNLIK(flL) is of index exactly n. Thus, since KxNLIK(DL) c ker (r) and r is surjective, K x NLIK(DL) = ker (r). //

Immediately from the Strong Reciprocity Law and the fundamental theorem of Galois theory, we deduce the following Galois correspondence:

Theorem 12-1-11 : Let L/K be an abelian extension. Let r denote the isomorphism of DK/HL,, and G = Gal (LIK) described in the Strong Reciprocity Law. To each subgroup G' of G, we associate the admissible subgroup H(Gf )

of flK, which is defined by H(Gf)/HLIK = r-'(G'). To each admissible subgroup H such that H 2 HLIK, we associate a subgroup G1(H) of G, defined by G1(H) = r(H).

(1) The correspondences G' t?r H(G') and H H Gf(H) are inverse bijections between the set of admissible subgroups containing H,,, and the set of subgroups of G.

(2) Coupling the correspondence of (1) with the usual Galois correspondence, we obtain a bijection between the set of admissible subgroups containing HLIK and the set of intermediatejields M, K c M c L.

(3) I f H t.t M under the correspondence of (2), then H = H,,,. (4) I f H, t, Ml and Hz ++ M, under the correspondence of (2), then

H1 c Hz- M, c MI, H ,H2t+M, n M,, H, n H,t+MIM,.

12-2 LOCAL CLASS FIELD THEORY

In this section, we shall study the local norm residue symbol more closely than we did in the preceding one. In particular, we shall identify the image and kernel of the symbol and we shall show that the local norm residue symbol is really a local object-that is, it only depends on the local extension Lrp/K,, and not on the global extension LIK used to realize the local extension.

Proposition 123-1: (x, LIK), E D, (x E K,").

Proof: Let Z denote the decomposition subfield of L at p. Recall that in

the extension ZIK, e, =f, = 1 . In particular, is well defined, and,

having order f,, is trivial. Thus, by the analogue of Proposition 10-1-10, (1) and (9, for our norm residue symbol,

Thus, (x, LIK), E Gal (LIZ) = D,. //

Theorem 12-2-2 : The local norm residue mapping

( L/K)p: Kp --+ Dp

is surjective.

Proof: Let Z = the decomposition subfield of L at p, T = the inertia subfield of L at p. Then K c Z c T c L. Let Hz = Hz,,, H, = HTIK, H, = HuK. Then

Hz =, H, 3 H,.

Since T is the maximal subfield of L in which p is unramified, the Galois correspondence of the preceding section shows that H, is the minimal admissible subgroup H of fl, such that H 3 HL and such that ptf,. Let

A = {x E K X I x r l(modfb)}.

Then A c HT, so that AH, c HT. Moreover, H = AHL is an admissible subgroup of 4, that contains HL and is such that ptf,. Then, by the minimality of HT, AHL 2 Hp Therefore, we have shown that

AHL = HT. (1) Let a E D, = Gal (LIZ). We shall show that there exists x E K x

such that (x, L/K), = a. By the reciprocity law, there exists a E J,,

(i(a), ~L,K) = 1, such that

Assume that p is finite. Then T/Z is cyclic with Galois group generated

, !Q a prime divisor of p in Z. Let w E K x be a local uniformizing

parameter at p. Then w is also a local uniformizing parameter at 9. Let

P = ( l , l , ..., w ,..., I ) € JK.

Under the Galois correspondence, maps into (g). Thus, by the

Galois correspondence, Hz/HT is cyclic and is generated'by B = P(mod H,.) In particular, there exists a positive integer a such that

p a E H,. By (I), we can write p a in the form

By the multiplicative approximation theorem, we may assume that (y, pd,,,) = 1. Then x = 1 (mod fk) and

i(x) = pai(ay-I), (i(ay-I), dL,K) = 1. By the definition of the norm residue symbol,

If p is infinite, then I, = D,, so that by the Galois correspondence, H,/H, is trivial. Thus, (3) holds with a = 0. From here on, the proof proceeds as in the first case. //

The preceding two results completely specify the image of the local norm residue symbol. Theorem 12-2-4 below will calculate the kernel. First, however, we will need a local analogue of the Second Inequality.

Let p be a K-prime. We say that x E K, is a local norm at p if x can be written in the form x = N%iKp(~) , z E LG, for some prime divisor '$ of p in L. If x has such a representation for the prime divisor '$,, and if '$3, is any other prime divisor of p in L, then x has the same type of representation with respect to '$,. Indeed, if a E G is such that a('$,) = '$,, then x = a(x) =

N%2rK,(a(z)). The index of the group of local norms at p in Kp is called the local norm index at p, and is denoted (Lg: Kp). Let n, = deg (GlK, ) .

Theorem 12-2-3 : (Lrp : K,) I n,.

Proof: First we claim that Theorem 12-2-3 is true if &/K, is cyclic of prime order. Let Z = the decomposition subfield of L at p, 58 = the 2-prime that

divides. Since p splits completely in ZIK, Corollary 11-1-3 asserts that

NZn/Kp(Z6) = KPX'

Therefore, N,,,,, induces an isomorphism

We are thus reduced to proving that

(Lrp : Z d 5 n,. (*>

Recall that Gal (LIZ) = D, and that D, has order n,. By the Strong Rec- iprocity Law, there exists an isomorphism r of CLiz = ~ z / Z x N , l z ( ~ L ) onto D,. The natural embedding Z8 - 4 , induces a homomorphism

Therefore, we have the following commutative diagram

In order to prove (*), it suffices to show that j is injective, since CLiz has order n,. Without loss of generality, assume that D, is nontrivial. Then D, has prime order. Therefore, either j is trivial or injective. If j is trivial, the image of ZB in D, is trivial, which contradicts the fact that ( , LIK), maps onto D, (Theorem 12-2-2). Thus, j is injective and Theorem 12-2-3 is proved if &/KO is cyclic of prime order.

Let us now prove Theorem 12-2-3 under the general hypotheses. Let us proceed by induction on deg (Ls/KJ. Without loss of generality, we may assume that deg (LglK,) is not prime, since in this case, we are done by the above reasoning. Thus, there exists a field M # Lrp such that K, c M c &, LgIM cyclic of prime order. Since M f Lg, deg (MIK,) < deg (LglK,). Thus, by the induction hypothesis, we know that

By the special case of Theorem 12-2-3 proved above, we see that

which suffices to establish the general case of Theorem 12-2-3. //

Theorem 12-2-4: (x , LIK), = 1 o x is a local norm at p.

Proof: e. This is precisely Proposition 12-1-9. 3. Let 1 = ( , LIK),. Let us fix a prime divisor '$ of p in L. By Prop-

osition 12-1-9, we know that

Thus, the index of ker (1) in K, is at most equal to (Lyr : K,). On the other hand, by Theorem 12-2-2, the index of ker (1) is at least n, = deg (&/K,). But by Theorem 12-2-3, (Lg :K,) 5 n,. Thus, ker (1) has index exactly n,. By (*) and Theorem 12-2-3, NLglK,(L$) = ker (1). //

We constructed the local norm residue symbol at p using the arithmetic of the global extension LIK. We would now like to prove that the local norm residue symbol is actually a local object-that is, it depends only on the local extension Lp/Kp. This is certainly plausible, since by Proposition 12-2-1 and Theorem 12-2-2, we see that ( , LIK), is a homomorphism from K," onto D, = Gal (LglK,). But it is not obvious from the way in which we defined the local norm residue symbol that it does not depend in some way on the global extension LIK. In order to prove this, we require two preliminary results.

Proposition 12-2-5: Let LIK be an abelian extension, MIK an arbitrary extension. Let p be a K-prime, '$ a prime divisor of p in M. Let x E Mil. Then

Proof: Let f = JJ fa denote an M-modulus that is a common multiple of

fLMIM and fLIK. Let '$ = '$,, . . . , 9, denote the distinct prime divisors of p in M, m a large positive integer. By the multiplicative approximation theorem, there exists y E M such that

yx-I = 1 (mod '$"), y E I (mod '$7) (i = 2, . . . , g), y - 1 (mod f'),

where f f = f (6 fr,)-'. Using the same reasoning as that we used in the proof

of ~ro~ositiod-b-1-9, we see that

By the definition of the norm residue symbol,

Moreover,

(y, LM/M)@, = 1 (i = 2, . . . , g),

since for m sufficiently large, y is a local norm at '$,. Thus, by the analogue of Proposition 10-1-10, (3), for the symbol ( , LIK),, we have

Proposition 12-2-6: Let K c K' c L' c L be a tower of number jields. Let '$ be an L-prime and let '$", p', p denote the primes of L', K' and K, respectively, which '$ divides. Assume that L$ = Lh,, K, = Kb,. Then, if x E K,"

Remark: Throughout the remainder of this discussion, we shall always interpret the norm residue symbol as belonging to the local Galois group. By the assumptions of Proposition 12-2-6, the local extensions Lo/K, and Lw/K,, are the same. We mean the above equality with both sides viewed as lying in the common Galois group of these two extensions.

For since Kp = Kb,, e(pl/p) = 1, f(pr/p) = 1. Therefore, NK,p,lK,(~) = X.

Thus, the claim follows by Proposition 12-2-5. We now claim that

If we consider the left-hand side as an element of Gal (LIK') and the right- hand side as an element of Gal (Lf/K'), Proposition 10-1-10, (I), implies

that the two automorphisms agree on L'. By continuity, they agree on Lk = L,, whence the second claim. The two clainis suffice to prove the assertion. /I

Theorem 12-2-7 (Local Character of the Norm Residue Symbol): Let L/K and L'/K1 be two global extensions. Let p and p' be primes of K and K', respectively. Let Q (respectively, 0 ' ) be a prime divisor of p in L (respectively, a prime divisor of p' in L'), and suppose that Kp = Kk, La = L$t. If x E

K,", then

(x, L/K), = (x, L'/K')p,.

Proof: Let be a prime divisor of p in KK', and let '$' be a prime divisor of '$3 in LK'. By our assumptions, KKj3 = K,, LKj3, = 4,. Moreover,

NKK$/K,(x) = X.

Therefore, by Proposition 10-1-10, (3),

(x, LK'/KK1)$ = (x, LIK),.

Applying Proposition 10-1-6 to the tower KK' c LK' c LL', we see that

(x, LK'/KKf)$ = (x, LL'/KK')y.

Therefore,

The right-hand side of this last equation is symmetric in primed and un- primed fields. Therefore,

We now can state a local reciprocity law that is analogous to the strong (global) reciprocity law of Theorem 12-1 -1 0.

Theorem 12-2-8: The local norm residue symbol ( , LIK), induces an isomorphism

I K,"INL~,K,(Li> * 4. Proof: Theorems 12-2-2 and 12-2-4.

Corollary 12-2-9: Ler L/K be an abelian extension. Then

Corollary 12-2-10: Let LIK be an abelian extension. Then

NL~/K,(UP)I = ~(WP) .

Proof: Let TZ be a local uniformizing element at p. Decompose K," into

K," = <z) X U,.

where f = f((P/p). But the first factor on the right i s 5 By Corollary 12-2-9, the left-hand side is n, = e((P/p)f((P/p). Thus,

WP:NL~IK,(~V)I = PIP). 11

Proposition 12-2-11 : (Up, LIK), = I,. Proof: We first claim that (U,, LIK), c I,. For let T = the inertia subfield of L at p. Then p is unramified in TIK, so that if (P is a prime divisor of p in T, NTV,K,(U,) = U, (Corollary 11-1-3). Therefore, if x E U,, x = NTglK,(y), y E T;, and

(x, L / K ) ~ I T = (x, T / K ) ~ [Proposition 10-1-10, (I)]

(NT%IK,(Y)~ TIKh - - 1 (Proposition 12-2-5).

Thus, (x,L/K), E Gal (LIT) = I,, and the claim is proved. Let A = ( , LIK), I,,. Then

ker (A) = NLVIKv(L$) n Up (Theorem 12-2-4)

= NLcplKp(UV)'

By Corollary 12-2-10, ker (A) is of index = e((P/p). Thus, the image of A has e(@/p) elements. But I, has e((P/p) elements. Thus, the assertion of Proposi- tion12-2-11. 11

Remark: The preceding results are valid only for local abelian extensions LQ/Kp which can be realized from a global extension LIK. In the exercises at the end of this chapter, we shall sketch a proof of the fact that every local extension can be realized from a global extension and every abelian local extension can be realized from an abelian global extension. Thus, all of the results of this section are really results about arbitrary abelian extensions of local fields. Since the local norm residue symbol is a local object, all of our results depend only on the local extensions and not on the intermediate global extensions. There is a method of defining the local norm residue symbol directly in terms of the local extension. However, we have not presented it, since such a discussion would have involved introducing Galois cohomology. The interested reader can consult Cassels-Frohlich [9].

Henceforth, in order to indicate the local nature of the norm residue symbol, we shall sometimes change our notation from ( , LIK), to ( , Ls/K,). Also, we shall assume the results contained in the exercises as known.

Let us now use the local norm residue symbol to construct a corresponding symbol for the maximal abelian extension of a local field. This

construction will prove important in proving the existence theorem of the global theory. In order to construct the symbol, let us first review the essentials of infinite-dimensional Galois theory. (For details, the reader may refer to Jacobson, Lectures in Abstract Algebra, Vol. 111, pp. 147-151.)

Let LIK be an arbitrary (not necessarily finite) field extension, G a group of automorphisms that leave K pointwise fixed. We say that LIK is a Galois extension with Galois group G, if K is the set of elements left fixed by G. It is not true that there is a 1-1 correspondence between subgroups of G and intermediate fields K G M G L, which analogizes the usual Galois correspondence. It is necessary to topologize G, using the so-called Krull topology, and to consider only closed subgroups of G in order to develop a Galois theory analogous to the finite-dimensional case.

The Krull topology is defined as follows: Take as a fundamental system of neighborhoods of the identity of G, the collection of all subgroups of finite index. It is easily verified that this collection defines a Hausdorff topology on G, with respect to which G is a topological group. Moreover, we have the following

Proposition: G is a compact, totally disconnected group.

The analogue of the classical Galois correspondence is given by

Theorem: Let LIK be a Galois extension with Galois group G. There exists a 1-1 correspondence between closed subgroups H of G and intermediate subfields K c M c L given by

M(H) = {x E Lla(x) = x for all a E H},

H(M) = {a E Gla(x) = x for all x E M).

If Hi u Mi(i = 1, 2), then

HI G H2 - MI 2 M2,

HI n Hz +-+ M1M2,

[H,, Hz1 +-+ MI n M2.

If H is normal and H - M, then MIK is Galois with Galois group GIH.

Let K, be a local field. Let A, denote the maximal abelian extension of K,, that is, the composition of all finite, abelian extensions of K,. Then A, is a Galois extension of K,, in the sense of the above discussion. Let us denote the Galois group of A, over K, by G,. Let us define a homomorphism

Kv -Gv

x --, (x, Kv), where (x, Kp) is defined as follows: Let LV/Kv be a finite abelian extension. We set

This defines (x, K,) on A,, since A, is the composite of fields of type Lv. More-

so that

Thus, the definition is consistent.

Proposition 12-2-12: The mapping x + (x, K,) is continuous.

Proof: This is obvious if p is infinite, since in this case A, = c, so that G, is discrete. Assume that p is finite. Let H be a small neighborhood of the identity in G,. It is enough to show that there exists a neighborhood of 1 in K t , say N, such that (N, K,) E H. Without loss of generality, we may assume that H is a subgroup of finite index in G,. Let L correspond to H under the Galois correspondence. Let N be contained in the group of local norms at p for LU/K,, N a neighborhood of 1. Then (N, K,) IL1, = (N, L$/K,) = 1, which implies that (N, K,) c H, by the Galois correspondence. //

Proposition 12-2-13: Let Ls/K, be a Galois extension of local fields. Then

(x, Lu) IAp = (NL$,KV(x), KP).

Proofi Exercise.

EXERCISE 12-2-14: Let L be a non-Archimedean local field, L 7, @,. Suppose that L = @,(x), x E 0,; let f be the minimal polynomial of x over @,.

(1) f E Z,[XI. (2) Let g E Z[X] be chosen so that each coefficient of g is very close to the

corresponding coefficient off (with respect to the p-adic metric). Let y be a root of g. Then @,(Y) = @,(x).

(3) Let K = @(y). Then K is a number field. Define a valuation on K by using the embedding K - @,(y) which sends y into y. This valuation defines a K-prime

such that Ks = @,(y). (4) L is the completion of K at 9.

12-3 GLOBAL CLASS FIELD THEORY

In this section, we continue our investigation of the consequences of the reciprocity law. Our first task will be to prove the assertion that initiated our discussion of class field theory-namely, the statement that every Artin character is the Hecke character associated to some grossencharacter of K. We

shall then show that the two different definitions of the conductor of an abelian extension are equivalent. This will, in turn, lead to a simple proof that the three forms of the reciprocity law are equivalent. We shall then proceed to introduce the concept of class fields and to show how to classify all abelian extensions of an algebraic number field as class fields.

Theorem 12-3-1: Every Artin character x is the Hecke character associated to some grossencharacter C'X' of K.

Proof: For each K-prime p, let us define the local character c;) by

C;)(X) = x-'((x, L/K)p), x E KpX.

Now c:) is unramified cz cp)(U,) = 1

* (UP, LIK,) ker (XI o ker (x) =I I, (Proposition 12-2-1 1).

In particular, c:' is unramified for almost all p. Let us now define

Since cf) is unramified for almost all p, C(X) is a character of DK. Moreover, i f x E Kx,

c'"'(x) = JJ x-'((x, LIK)p) P

= 1,

by the Hilbert Reciprocity Law. Thus, c(x) is a grossencharacter of K. The corresponding ideal character IZ is defined on IK(Sx), S, = S, u {p I ker (x) + 1,). We now claim that L = X. It suffices to show that IZ(p) = ~ ( p ) if p 4 S,. Let w be a local uniformizing parameter at p . Then

= x- ' ( (~9 LIK),).

Let T = the inertia subfield of L at p. Let '$3 be a prime divisor of p in T, '$3 = a prime divisor of !$ in L. From the Hilbert theory, L% = T$. Therefore, by the local character of the norm residue symbol,

(as elements of the local Galois group). But p is unramified in T/K, so that by Proposition 10-1-10, (5 ) ,

In particular, the reduction of (w, L/K),(rnod p) is a;', where a, is the Frobenius automorphism of Gal (Ls/K,). Thus,

Theorem 12-3-2: Let LIK be an abelian extension. Let f denote the conductor of LIK given by De$nition 10-1-5; let f* denote the conductor of LIK given by Definition 12-1-2. Then f = f*.

Proof: Let p be a K-prime. Let f, (respectively, f,*) denote the exact power of p that divides f (respectively, f*). It suffices to show that f, = f,*. If p is infinite, we know that both f, and f,* are t, if p ramifies in LIK and are p0 otherwise. Thus, we may assume that p is finite. Then

pn divides f, o c,(x)(l + pa) = 1 for all x E G*

o ~ ( ( x , LIK),) = 1 for all x E G*, x E 1 + pa

o (x, LIK), = 1, x E l + p a

o 1 + pn c ker (( , LIKM o 1 + p i s contained in the group of local norms at t, o pn divides f,*.

(Note that by convention, 1 + Go = Up) . //

Corollary 12-3-3: ( , LIK), = [ , LIK],.

Corollary 12-3-4: The three forms of the reciprocity law are equivalent.

Proof: In Section 10-1, w.: showed that the first form of the reciprocity law implied the second form and that the second form implied the third form. By Corollary 12-3-3, the third form is equivalent to the Hilbert Reciprocity Law. But Theorem 12-3-1 shows that the Hilbert Reciprocity Law implies the first form of the reciprocity law. //

Let LIK be an abelian extension with Galois group G. By analogy with the local situation, let us introduce the global norm residue symbol

( , L/K):D,--+G

(a, LIK) = (a,, LIK),. P

The global symbol coincides with the reciprocity mapping r of Theorem 12-3- 10. Directly from the properties of the local symbol, we deduce

Proposition 12-3-5: Let K G L G M, M/K abelian. Then

( 9 M I K ) IL = ( , LIK).

Let A, denote the maximal abelian extension of K. As in the local theory, we conclude that AK/K is an (infinite) Galois extension with compact Galois group 8. Still following the analogy of the local theory, we define a global symbol ( , K)

( , K):JK+@

as follows: Let K c L r A,, L/K finite. Define ( , K) 1, by

( 9 K)IL = ( 7 UK). Proposition 12-3-5 guarantees that this definition is consistent.

Proposition 12-3-6: ( , K) is continuous.

Proof: Let @ be a small neighborhood of the identity in 8. It suffices to show that there exists a neighborhood N of the identity in 9, such that (N, K) c @. Without loss of generality, let @ be a subgroup of finite index and let L be the subfield of A, that corresponds to @. Then L/K is finite, since @ is of finite index. Let N be a neighborhood of 1 E 9, that is contained in the admissible subgroup HL/,. Then

( N , K) I, = (N, LIK) c ( H ~ K , LIK) = 113 by the reciprocity law. Thus,

( N , K) c Gal (A,IL) E @. /I

Let A, = ( , K). By the reciprocity Iaw,

K x 5 ker (A,).

Recall that we embedded R+ in 11, via

t + (PIn, P, . . . , t l / n , t2In, . . . , PIn, 1, 1, . . .). - - real prlmes complex prlmes

With respect to this embedding, we had the direct product decomposition

JJ, = R+ x s:. (2) Since [R, E NLi,(JL) for any finite extension LIK, we see that R+ E ker (A,). Thus, by (I), A, induces a homomorphism

such that the diagram

is commutative.

Proposition 12-3-7: A, is surjective.

Proof: From the above discussion,

A,(s,) = X,(~:/K~).

Therefore, it suffices to show that ~,(S:/K) = 8. Let @ be a subgroup of finite index, L = the subfield of A, that corresponds to @. Let x E HLIK. Then (x, K) 1, = (x, LIK) = 1 => (x, K) E @. Thus, since the set of subgroups of finite index in 8 is a fundamental system of neighborhoods of the identity, im (1,) is dense in 8. Since flh/Kx is compact (Theorem 3-3-9, X,(J:/&") is compact and hence closed (Proposition 12-3-6). Thus, im (A,) = im (A,) is both closed and dense. Thus, im (A,) = 8. I/

At the beginning of this chapter, we mentioned that the main problem of class field theory is to settle the problem of classifying abelian extensions of an algebraic number field. We can now easily accomplish this task, using the Propositions proved above.

Definition 2-5-8: Let H be an admissible subgroup of D,, LIK a Galois extension. We say that L is classfieldt to H if

The two fundamental assertions about class fields that we shall prove are

Theorem 12-3-9 (Existence Theorem): Let H be an admissible subgroup of 9,. Then there exists an abelian extension LIK, with L classfield to H.

Theorem 12-3-10 (Uniqueness Theorem): Let LIK and M/K be abelian extensions, both class field to H. Then L = M.

To prove Theorems 12-3-9 and 12-3-10, we need some information about the kernel of the global symbol A,. From the definition of the symbol and the reciprocity law,

where L runs over all finite, abelian extensions of K. By the reciprocity law, K x c ker (AK). Thus, let j denote the canonical homomorphism

J), ---!+ J,/K ',

and let i denote the homomorphism

DK/Kx 2 J:/Kx,

which was constructed. Then we have the following commutative diagram:

t It is possible to show that every class field is an abelian extension.

Theorem 12-3-11: ker (1,) is a divisible group. That is, given x E ker (A,) and a positive integer n, there exists y E ker (1,) such that yn = x.

Before proving Theorem 12-3-11, let us show how the existence and uniqueness theorems follow from it.

Proof of Theorem 12-3-9: Since H is admissible, H i s an open subgroup of finite index, say n. Claim that ker (A,) E H. For if x E ker (A,), Theorem 12-3-11 implies that there exists y E ker (A,) such that j(yn) = j(x). Then x = zyn, z E K X * x E H (since K x G H and H is of index n). Thus, ker (A,) E H. Since both j and i are open maps, io j(H) is an open and hence closed subgroup of the compact group .J:/Kx => io j(H) is compact 3 A,(H) is compact 3 A,(H) = @ is a closed subgroup of 8. Since ker (A,) c H, A, induces an isomorphism

D,/H = a/@. (*I In particular, $5 is of finite index in 8. Let L/K denote the extension corresponding to the group @. Then LIK is a finite abelian extension and Gal (LIK) = a/@. We claim that L is class field to H. Since A,(H) = @ and L is the fixed field of @, we see that A,(H) 1, = (1) * (H, L/K) = {I) 3 H E HLIK. By the reciprocity law,

deg (LIK) = [J,: HL/K] I [J,: HI.

On the other hand,

deg (LIK) = [ 8 : @] = [S,: HI

by (*). Therefore,

and L is class field to H. //

Proof of Theorem 12-3-10: In the proof of Theorem 12-3-9, we showed that AK(H) = @ is a closed subgroup of finite index in 8 , and that

[J,: HI = [a:@].

If L is class field to H, A,(H) 1, = (H,,,, L/K) = (1) 3 @ E Gal (AL/L).

But since deg (LIK) = [D,: HI, deg (LIK) = [@:a] =. Gal (AJL) = 6. Therefore, L is the subfield of A, corresponding to @, and L is uniquely determined by H. //

In order to prove Theorem 12-3-11, which is the heart of the matter in proving the Existence and Uniqueness Theorems, we shall require quite a bit of preliminary work.

Lemma 1 (Herbrand's Lemma): Let G be an abelian group, H a subgroup offinite index. Let TI and T, be endomorphisms of G such that

(1) TIT, = T,T, = 0. (2) Ti(H) E H (i = 1, 2).

Set Si = Ti I ,. Then

[ker (TI) : im (T,)] - [ker (S,) : im (S,)] , [ker (T,) : im (T,)] - [ker (S,) : im (S,)]

where all the indices are finite.

This result was proved as Proposition 11-1-3.

Lemma 2 : Let G be an abelian group, K, H, and Jsubgroups of G. Assume that K is offinite index in H. Then

[H: K] = [HJ: KJ][H n J: K n J].

Proof: Consider the canonical homomorphism

j : H/K -+ HJIKJ.

Clearly, j is surjective, and ker ( j ) = (KJ n H)/K = (J n H)K/K =

( J n H)/(K n H). This completes the proof. /I

Throughout the next three lemmas, let n be a fixed positive integer and let K be a fixed number field that contains all nth roots of 1.

Lemma 3: [K," : K,""] = n2/l n 1,.

Proof: If p is real, then n = 2 and the assertion is obvious. If p is complex, then In 1, = n2, so that again the assertion is obvious. Assume that p is finite. Since [Kt :K,""] = n[U,:U$ it suffices to show that

Let us apply Herbrand's Lemma with T,(x) = 1, T,(x) = xn, H = 1 + p , where i is sufficiently large. A direct application of the Herbrand Lemma shows that

[U,:U;] = n[l + $':(I + p)"].

It therefore suffices to show that

Let m = ord,(n). We claim that for i sufficiently large,

But this is immediate from Corollary 6-1-5. In order to complete the proof of the Lemma, (*) and (**) imply that it

suffices to prove that

But the various cosets of (1 + p)/(l + p+") are represented by

where n is a local parameter at p , and the a, range over a set of representatives of the various residue classes mod p. There are N p such representatives, so that [l + p: 1 + = N p m = Illnl,. //

Lemma 4: Let x E Kx, L = K(xl/"). Then (1) p splits completely in LIK o x E Ktn. (2) Assume that p isjinite and does not divide n. Then p is unramified in

L I K o x E UvKxn.

Proof: (1) p splits completely in LIK o X" - x splits completely into distinct linear factors in K,[X] (Hensel's Lemma) o x E Kin.

(2) G- let 13 be a prime divisor of p in L. By Proposition 4-1-5, ord$ (x) =

ord, (x) s ord, (x) is divisible by n, say ord, (x) = an. Let n E K x be a local uniformizing parameter at p. Then xn-On E Up =. x E U,Kxn. + Without loss of generality, assume that x E Up. Since L is generated by a root of Xn - x, the different of the extension LIK divides nxn-I, and is therefore prime to p , since x E U, and p is prime to n. //

Let us choose a finite set S of K-primes containing S, and all prime divisors of n. Choose S so large that D, = KxflS, [Exercise 3-3-12, (4)]. Let

L, = the maximal abelian extension of K of exponent n such that p 4 S G- p is unramified in L,/K,

L, = the maximal abelian extension of K of exponent n such that p $ S =- p is unramified in LJK, p E S =. p splits completely in

I L,/K. Set

By Lemma 4,

Note that A, = J;, B, = U,(S) = the group of S-units (Definition 3-3-10). Since L,/K is a Kummer extension,

deg (L, /K) = [B, Kn : K"] = [UK(S) : U,(S)"].

By Theorem 3-3-1 1, U,(S) is the direct product of the group of roots of unity in K and a free abelian group of rank s - 1, where s = the number of elements in S. Since K contains the nth roots of unity, a simple calculation yields

deg (L,/K) = ns.

The key step in proving Theorem 12-3-11 consists of calculating the admissible subgroup H,,,,.

Lemma 5: L, is class jield to K A,.

Proof: We first claim that

It suffices to show that A, c H,,,,. Let a = (a,) E A,. If p E S, p splits completely in L,/K => a, is a local norm at p, since the group of local norms is K,. If p $ S, a, E Up 3 a, is a local norm at p by Theorem 11-1-4. Thus, a E NL~IK(JLJ G HL,IK.

By the choice of S, KxAl = KxJ$ = J,. Thus, by ( I ) ,

HL*IK = JK,

so that by the reciprocity law, L, = K. We now claim that KxA, E H,,,,. As above, it suffices to show that

A, E HL ,,,. Let a E A,. Then we can write a in the form a = p y , p , y E J,, /IP = 1 for p + S, yp = 1 for p E S. Now Ll/K is of exponent n * J,/HL,,, is of exponent n (reciprocity law) 3 E H,,,,. As above, we see that y E NLIIK(JL,). Thus, a E HLIIK, and we have shown that

In order to complete the proof of the Lemma, it suffices to show that equality holds in (2). In the discussion prior to the statement of the Lemma, we showed that deg (L,/K) = ns. By the reciprocity law, this implies that [DK:HL,,,] = n\ Therefore, by (2), we are reduced to showing that

[J,: K ~ A , ] = nS.

But

[J, : K x A,] = [K "J; : K x A,] (by the choice of S )

- - us: 4 1 (by Lemma 2) [(Jg n Kx):(A, n Kx)]

= n-s deg (L,/K) n" (by Lemma 3 and Kummer Theoryt) P E S I nl,

- - n-sn2s (by the product formula) = ns.

This proves (3), and thus completes the proof of Lemma 5. /I Proof of Theorem 12-3-11: Let all notations be as in the proof of Lemma 5. Recall that L, = K. Since L, = K(Biln), this implies that

K" n A, c K ~ ~ . (1) Note that (1) holds as soon as S is sufficiently large. Let

J, -+ JklK

be the canonical projection, X E ker (A,), n E Z,. It suffices to show that x = ynz, y E J,, z E K X . For then, 3 = y". But x E ker (I,) a x E ker (A,) * x E ker(( , K)IL,) * (x ,L , /K)= 1 * x E HL,IK * x = wz, w E A,, z E K x (by Lemma 5). In order to complete the proof, it suffices to show that w is the nth power of some K-idele. To prove this fact, it suffices to show that w, is an nth power for all primes p. Since w E A,, this is true if p E S. Suppose that p $ S and set S' = S u (p). Then x E ker (A,), z E ker (A,) * w E ker (A,). Reasoning as above, we deduce that w = w'z', w E A', =

A,(Sf), z' E Kx. Since A, c A',,

ww'-I E K X n A', E Kxn

by (1). But w' is an nth power at p, since p E S'. Therefore, w is an nth power at P. /I

The following result gives the law of decomposition of primes in an abelian extension, as promised in the beginning of this chapter.

Theorem 12-3-12 (Decomposition Theorem): Suppose that LIK is abelian, p a K-prime, HLIK the admissible subgroup of D, associated to LIK. Let H, denote the smallest subgroup of 11, such that H, 2 H,,, and such that p does not divide f,,. Let T = the classJield associated to H,. Then

(1) T is the inertia subjield of L at p.

t See Appendix.

(2) e, = [H,: HL,A (3) Let p be finite and let n be a local uniformizing parameter at p. Let

f be the smallest positive integer such that t nf E H,. Then f = f,, the residue class degree at p of LIK.

(4) Let p be finite. Then p splits completely in L/K o n E H, and Hp =

H L I P

Proof: (1) T is the largest subfield of L in which p does not ramify. (2) By the reciprocity law,

deg (LIK) = [J,: HLlK1,

deg (T/K) = [J, : H,]. =. deg (LIT) = [H,: H,,,].

But from Hilbert theory and (I), deg (LIT) = e,. (3) By the reciprocity law, f is the smallest positive integer such that

[nf, TIK] = 1

since T/K is unramified at p [Proposition 10-1-10, (5)]. Therefore, f =f, (Proposition 5-5-2).

(4) Immediate from (2) and (3).

EXAMPLE 12-3-13: If K = @, L = @(c,), we may set n = p in (3). Then, if p 4 m, Hp = HLIK since @([,)/@ does not ramify atp. A simple computation shows that pf E Hp o p f - 1 (mod mp,). This is equivalent to the splitting law for @(c,) which we found in Chapter 6.

The next result generalizes the result which stated that if two rational primes are in the same arithmetic progression modulo m, then they split in the same way in Q([,).

In the exercises for Section 11-1, we introduced the notion of an arithmetic progression defined modulo a K-modulus %V. We refer the reader to this set of exercises for the notations and definitions of terms appearing in the following theorem.

Theorem 12-3-14: Let p and p' be K-primes, relatively prime to f,,,. Assume that p r p'(mod f,,,). Then p and p' belong to the same splitting type for LIK.

Proof: Since p and p' are relatively prime to the conductor, they are unramified in LIK. Therefore, their splitting type is completely determined by their respective Frobenius automorphisms. It thus suffices to show that

t We identify x with the idele (1,. . . , 1, n, 1, . . .). t D

Since p = pl(mod f L / K ) , pp1-I E PK(fLIK). Therefore, there exist a , a' E 4, such that i(a) = p, i(af) = p', am'-I E K X , act'-' = 1 (mod f,,,). From the reciprocity law,

[aar-I, LIK] = 1.

Moreover,

[a8 ah-', L/KIFp = 1

for all K-primes !$ that divides f,,,. Therefore,

By Proposition 10-1-10, ( 9 , this implies that

which implies Theorem 12-3-14. //

Applications

13-1 KRONECKER'S THEOREM

In this section, we shall prove the famous result of Kronecker cited in Section 6-2. The first few results of this section are aimed at calculating the admissible subgroup of 4, to which @(Cm) is class field. Throughout this section, let m be a positive integer greater than 2. We shall write Km instead of @(Cm).

Proposition 13-1-1: Let p be a Jinite @-prime such that p 1 m. Suppose that m =prn, (n ,p) = 1. Then the group of local norms at p for Km/@ is pf" x ( 1 + pr). Thus, the local conductor f , for K,/@ is given by f , = p"'.

Proof: Since

Km = K,. K,,,

the extension Km/Kpr is unramified at all prime divisors p o f p in K,,. Fix a prime divisor p of p in Kpr and a prime divisor !J3 of p in K,. Since Km/Kpr is unramified,

NKm,v/Kp' , p(Um,$) = p'

Therefore,

NKm,'P~~p(KmX,d = jfp- l N ~ p ~ ~ a p ( K ~ ~ , p).

By Theorem 12-3-3, the local norm index at p of Kp./@ is at most deg (@,(CPr)/@,). By Proposition 6-2-6, this degree is pr- ' (p - I), so that the local norm index of KpT/@ at p is at most pr-'(p - 1).

Note that

where U: is the group of ( p - ])st roots of unity. Since

[ I + p: 1 + B'] = pr-I,

[@; : p x ( 1 + B')] = ( p - l)pr-1.

Therefore, if we can show that p' x ( 1 + p) is contained in the group of local norms a t p for Kpr/@, we immediately know that p' x ( 1 + p") is exactly the group of local norms and its index in @; is q(pr).

The minimal polynomial of 5, over @, is

Thus, the minimal polynomial of 1 - Cpr over @, is XP'P" + . . . + ( - 1 ) d ~ ' ) ~ .

Thus, p is the norm of 1 - C , . In particular, p is contained in the local norm group at p.

If p is odd, r > ord, (q (pr ) ) + I / (p - l), so that the binomial theorem (Theorem 6-1-3) implies that

( 1 + p"') ' /9 '~" g 1 + p' * 1 + B' z (1 + p ' y .

The right-hand side consists of local norms at p, so that 1 + j7 consists of local norms. This completes the proof of Proposition 13-1-1 in the case p odd.

Assume that p = 2. It suffices to show that

For r = 1 , this assertion is clear. Let us proceed by induction on r. Assume that

1 + 2.-1 = ( 1 + 2 y 3 ( r > 1) . Then

( 1 + 2)zr-' = ( 1 + 21-1)~.

By direct inspection of 2-adic decimal expansions, we see that the right-hand side is contained in 1 + 2r. A simple computation shows that ( 1 + 2 ' - ' ) 2

# 1 + 2r-1. Since [l + j r - l : 1 + 2.1 = 2, this implies that ( 1 + T r - l ) ~ =

1 + 2.. //

Proposition 13-1-2: If p, denotes the inJinite @-prime, then the group of local norms at p, for K,/@ is R,. Therefore, the local conductor f,, for K,/@ is given by f p _ = p,.

Proof: Let !$ be a prime divisor ofp, in K,. Since m > 2, K, $ R, so that

$ is complex. Therefore, Kmr9 = C, and the assertions of Proposition 13-1-2 follow. //

Theorem 13-1-3: The conductor of Km/@ is mp,.

Proof: By Propositions 13-1-1 and 13-1-2, the conductor of Km/@ is divisible by mp,, and the quotient of the conductor and mp, is divisible only by primes p that do not divide mp,. Let p be such a prime. By Corollary 6-2-3, p does not ramify in Km/@. Therefore, by Proposition 12-1-l,p does not divide fKmI,. Thus, a contradiction is reached and no such p exists. //

Theorem 13-1-4: Km is classjeld to @ " Vmp,.

Proof: From Propositions 13-1-1 and 13-1-2, we see that Vmp, G NKm/QDKm, so that @" Vmp_ G HKmIQ. In order to prove equality, let us show that the two groups have the same index in 4,. By the reciprocity law,

14Q: HK,lQl = deg (Km/@) = ~ ( ~ 1 '

Therefore, it suffices to show that

[J, : @" vmp,l = 444. Let U = 4,S-. We showed in Chapter 12 that JQ/@"U - C,, the ideal class group of @. Since Z is a unique factorization domain, C, is trivial (Exercise 3-34), so that 4, = @* U. Since

@XuI@xVmp, * @Xu~@X/@X~mp, l@x,

it suffices to show that

[@xu /@x : Q X vmp,/@"l = dm).

By Lemma 2 to Theorem 12-3-1 1,

[U: Vmp,] = [@"U/@": @"VmP,I@"I[U n @": Vmp_ n @"I.

Now U n @" = {f 1)

v m p , n @" = 111,

so that [U n @"VmP_ n @"I = 2.

Let m = rl[prD be the decomposition of m into a product of prime powers. P

Then

[U: Vmp,] = 2 p [Up: 1 + prq P m

= 2 p [u;: {l,][l + p: 1 + prq P m

= 2 p ( p - 1)p"-' P m

= 2q(m).

Theorem 13-1-5 (Kronecker): Every abelian extension K of @ is contained in a cyclotomic field.

Proof: Let HKIQ be the admissible subgroup of flQ associated to K/@. Since H,,, is admissible, it contains Vmp, for some positive integer m. Without loss of generality, assume that m > 2. Then

By Theorem 13-1-4 @(Cm) is class field to @"VmP,. By the Galois correspondence (Theorem 12-1-1 I), there exists a subfield K' of @(Cm) that corresponds to the admissible subgroup H,,,. By the uniqueness theorem (Theorem 12-3-10), K = K'. //

Remark: By the above proof, we see that if the conductor of the abelian extension K of @ is m or mp,, then K is contained in the field of mth roots of unity. Therefore, as a special case, we can assert

Theorem 13-1-6: If K is an abelian extension of @, then K ram$es over @ for at least one prime.

Remark: By using other reasoning (actually more elementary), it is possible to show that Theorem 13-1-6 holds for arbitrary K. (One need not even assume that K/@ is normal.)

13-2 THE POWER RECIPROCITY LAW

Let us now make explicit the precise sense in which the Artin reciprocity law is a generalization of the quadratic reciprocity law. We shall do this by showing that the Artin reciprocity law implies the so-called power reciprocity law, which is an analogue of the quadratic reciprocity law, except for the general case of nth power residues. This power reciprocity law contains the quadratic reciprocity law as a special case.

Throughout this section, let K be a number field that contains a primitive nth root of unity C. A typical example of a Kummer extension of K is L, =

K(al/") (a E K). Indeed, L, is an abelian extension of K with exponent n. For if a E Gal (LJK), then a(a1ln) = Cc'")a'~"(c(a) E Z) and the mapping

G - Zl(4

is an isomorphism. One can prove that every Kummer extension of K can be obtained as a composite of fields of the form La. However, we shall not require this fact, so its proof will be omitted.

Proposition 13-2-1: Let p be a Jinite K-prime. If p is ramified in LJK, then p divides anOK.

Proof: Let p(X) be the monic, irreducible polynomial satisfied by all" over K. By Proposition 5-3-1, (2), p'(alln) E b,,,,, so that p divides p'(alln). But p(X) divides X" - a, so pl(X) divides nX"-l. Therefore, p'(a1ln) divides n(alln)"-I in La. Finally, p divides naO,. //

Our next job is to write down the Hilbert form of the reciprocity law for La/K. Let a, b E K x , p a K-prime. The norm residue symbol (a, b), at p is defined by the relation

(a, Lb/K),bll" = (a, b),bll".

It is clear that (a, b), is an nth root of unity. It is independent of the choice of b1l". For if blln is a fixed nth root of b, then every other nth root of b is of the form cblln, and

(a, Lb/K), = cc(a, Lb/K)bl'" = (a, b),(CCb1l"),

since C E K. It is clear that (a, b), = 1 for almost all p, since the same is true of

(a, Lb/K),. Thus, the Hilbert reciprocity law implies

Theorem 13-2-2: JJ (a, b), = 1. P

Since (a, Lb/K), is multiplicative in a, the symbol (a, b), is multiplicative in the first variable. We claim that (a, b), is multiplicative in the second variable: Let b, b' E Kx, and fix nth roots of b, b', bb' so that bllnb"l" =

(bb')'ln. Let L, , , = LbLb,. Then L b , 5 L,,. and

(a, Lbb'/K)P(bb')l,'n = (a, Lb, b./L)p(blln) (a, L , b,/K)p(b'l'n) = (a, Lb/K),(b1l"). (a, Lb,/K),(bflI") = (a, b),(a, bl),(bb')' '",

so that (a, bb'), = (a, b),(a, b'),. By the multiplicativity in both variables, the symbol (a, b), is trivial if

either a o rb is an nth power. In particular, the symbol is trivial on K x n x Kx". Let us extend the symbol (a, b), to K," x K," as follows: If a, b E K t , choose c, d E K x so that ac-I and bd-I are in K,"", and set

This definition does not depend on the choice of c and d, since the original

symbol is trivial on Kxn x Kx". It is clear that the extended symbol is multiplicative in both variables.

Proposition 13-23: (a, b), = 1 $a is a local norm at p . Proof: If a, b E K x , then (a, LJK), is trivial by Theorem 12-2-4. In the general case, if a is a local norm at p, and ac-' E Kf", then c is a local norm at p, and the assertion follows. //

Proposition 13-2-4: (1) (-a, a), = 1 (a E K,"). (2) (1 - a, a), = 1 (a E K,", a f 0, 1).

Proof: In the light of Proposition 13-2-3, it suffices to show that -a and 1 - a are (global) norms from La. By remarks made above, Gal (La/K) is isomorphic to a subgroup of h/(n). Therefore, Gal (La/K) is cyclic. Let a be a generator, and suppose that a(al/") = rdal/". Let x E K. Then

xn - a = JJ (x - bCr) OSr<n

d- 1 "Id- 1 = n JJ (. - all"ckd+l)

1=0 k=O

d- 1 n/d- 1

= n JJ ak(x - all"C) I=0 k=O

d- 1

= n NL J K ( ~ - al/"cl). 1=0

Setting x = 0 and x = 1 in the above formula gives the two desired assertions. //

Proposition 13-2-5: (a, b),(b, a), = 1 (a, b E K,").

Proof: By Proposition 13-2-4, (I),

1 = (-ab, ab),

= (-a, a>,(b, a),@, b),(-b, b), = (a, b)p(b, a),. /I

Let us now introduce the power residue symbol. Let a E K X , 8 an integral K-ideal relatively prime to n and a. Then 8 is not divisible by any

K-prime that ramifies in La/K. Therefore, the Frobenius symbol (LiK) - is

defined. The power residue symbol is defined by the equation

It is clear that - is an nth root of unity and does not depend on the choice ($1 of a'". (The proof of this last fact follows exactly the analogous argument

for the norm residue symbol.) It is also clear that the power residue symbol is multiplicative in both variables. The multiplicativity in (3 is obvious; the multiplicativity in a follows using the same argument used to prove that the norm residue symbol is multiplicative in the second variable.

The reason for use of the name power residue symbol is contained in

Proposition 13-2-6: Let p be a jn i t e K-prime, a E Kx, p relatively prime to n and a. The following statements are equivalent:

The equation X" - a has n solutions in 0,. The congruence X" =2 a(mod p) has n solutions in &. The congruence Xn r a(mod p) has a solution in 4. First note that since p is unramified in La/K, e, = 1. Now

o f p = 1 o n, = 1 (since n, = e, f,) e X" - a splits into linear factors in K, (Proposition

5-1-1) o X" - a splits into linear factors in 0, (since a E Up)

o Xn = a(mod p) has n solutions in Ep [=> is clear. Conversely, if the congruence has n solutions, these solutions are distinct, since nX"-l + O(modp). Therefore, the factorization of X" - a follows by Hensel's Lemma.]

o The congruence Xn = a(mod p) has a solution in E, (3 is clear. But if one solution exists, then n solutions exist, since I?, contains the nth roots of unity.) //

The next result shows how to evaluate the norm residue symbol in terms of the power residue symbol.

Proposition 13-2-7: Let p be a jn i te K-prime, not dividing n. Let n be a local uniformizing parameter a t p, and let u, v E Up. Then

Remark: (1) through (3) completely determine ( , ), because of the multiplicativity in both variables.

Proof: Without loss of generality, we may assume that n, u, v are all in K. Then the extensions Lu/K and LJK are both unramified at p. Since LJK is unramified at p, u is a local norm at p by Theorem 11-1-4. Therefore, (1) holds by Proposition 13-2-3. By Theorem 10-1-10, ( 3 ,

from which (2) follows, when Proposition 13-2-5 is used. By Proposition 13- 2-4 and (2),

Let a, b E KX, and assume that b is relatively prime to n and a. If (b) denotes the K-ideal generated by b, let us define

Then

Lemma: Suppose that a and b are relatively prime, and let p be a j n i t e K- prime not dividing n. Then

Proof: Suppose that p 4 a. Then the left-hand side equals

On the other hand, (:r'rdv(b) *

(a, b), = (b, 4;' = (n, a);ordp(b)(~, a);

(b = norddb)u, n a local parameter at p)

(Proposition 13-2-7).

If p I a, then p 4 b, and the same reasoning as used before can be applied. 11 Theorem 13-2-8 (Power Reciprocity Law): Let a, b E K x be relatively prime and prime to n. Let S be the set of all infinite primes and all K-primes dividing n. Then

Proof: By the Lemma,

= (a, b);' (Theorem 13-2-2) PES

= rj (b, a), (Proposition 13-2-5). // 11tS

EXAMPLE: Let K = @, n = 2, S = (a, 2). Let p be an odd rational prime, a a nonzero integer, p j' a. By Proposition 13-2-6, (alp) coincides with the Kronecker symbol. Therefore, if a is the discriminant of a quadratic field, b an odd integer + 0, then (alb) coincides with the Kronecker symbol because of the multiplicativity in the lower variable. Moreover, if a, b are nonzero odd integers, relatively prime, then the power reciprocity law asserts that

Lemma 1: (b, a), = (- ~)("'Bu(u)- l ) ( " ' ~ d b ) - l ) .

Proof: (b, a), = + 1 o b is a local norm from &(alJ2). If a > 0, then @,(alJ2) = R and (a, b), = 1. If a < 0, then = C, and (a, b), = 1 o b > 0. /I

By Lemma 1 and the power reciprocity law, we see that

Therefore, by the quadratic reciprocity law, (a, b), = (-l)(b-l)(u-1)/4.

Thus, we see that the power reciprocity law and the quadratic reciprocity law say the same thing.

To close this chapter, let us make one comment about the generalization of the two supplementary reciprocity laws to the law of quadratic reciprocity

-that is, the two formulas for for a an odd integer.

Let S be as in Theorem 13-2-8, and suppose that a is divisible only by K-primes in S. Let b E K x be prime to n (and hence prime to a). As in the proof of the Lemma, we can prove that

where n is a local parameter at p . Reasoning exactly as in the proof of the power reciprocity law, we deduce

Theorem 13-2-9 (Supplementary Reciprocity Law): Let a and b be relatively prime integers, b relatively prime to n, a divisible only by primes in S. Then

If one specializes this theorem to the situation of the example, upon setting a = - 1 and a = 2, we see that the supplementary reciprocity law implies that

Thus, the supplementary reciprocity law in this case is equivalent to the evaluation of (-1, b), and (2, b),. The formulas for these are given in Theorem 6-2-21.

PART 5

The Prime Number Theorem

The prime number theorem

In this chapter, we shall consider various classical questions on the distribution of prime numbers. In particular, we shall prove the prime number theorem of Hadamard-de la VallCe Poisson and the prime ideal theorem of Landau. Our viewpoint will continue to focus on the theory of Hecke L-functions. Since we are interested in presenting as general a theory as possible, it has been necessary to omit some of the refinements in various theorems, which have been effected in particular cases. A noteworthy example of such an omission is the method of trigonometrical sums, due to Vinogradov.

The analytical requirements of this chapter are somewhat greater than has been the case in the preceding portion of the book. We have assumed that the reader is familiar with Ahlfors' Complex Analysis. All nonelementary theorems we require (in addition to being proved in Ahlfors) are stated in Section 14-2.

Throughout, we shall use the 0- and o-notations: Suppose that a is real and that f(x) and g(x) are defined in some neighborhood of a. Suppose that g(x) # 0 in this neighborhood. Then we say thatf(x) = O(g(x)) (respectively, f (x) = o(g(x))) as x - a if f(x)/g(x) remains bounded in this neighborhood (respectively, f(x)/g(x) --t 0 as x --t a). We make the obvious change in the definition to define f(x) = O(g(x)) as x - oo, f(x) = o(g(x)) as x --, co.

14-1 NATURAL DENSITIES

In Chapter 5, we defined the concept of Dirichlet density. This definition I seems somewhat artificial, but we saw how useful the concept is. The utility I of the concept makes up, to some extent, for the artificiality of the definition.

In this section, we shall introduce a new concept of density-natural density. This concept will seem to be the "natural" one. Nevertheless, theorems about natural densities of sets of primes are much deeper than the corresponding results for Dirichlet densities. In fact, we shall show that natural densities are intimately connected with the Riemann hypothesis for Hecke L-functions.

Throughout this section, let K be an algebraic number field, A a set of finite K-primes. For x a positive real number, let n(x, A) = the number of primes p E A such that Np x.

Definition 14-1-1: We say that A has a natural density if the limit

n(x, A) d(A) = lim -- x-- X

log x

exists. The number d(A) is called the natural density of A.

Remark: From the definition, it is not at all obvious that the density of a set of K-primes is at most 1. This will be one of our major results. For we shall show that the set of all finite K-primes has natural density 1. In case K = Q , this is the prime number theorem discussed at the beginning of this book. In the case K is an arbitrary algebraic number field, the statement that the set of all finite K-primes has natural density 1 is called the prime ideal theorem.

We can reformulate the definition of natural density in terms of the o-notation as follows. A has natural density d(A) if and only if

The relationship between natural and Dirichlet densities is given by

Theorem 14-1-2: I f A has a natural density d(A), then A has Dirichlet density d(A).

Proof: If suffices to show that

Expanding the logarithm on the left into a sum, then expanding the summands into series, and retaining only the first term of each series (the remaining terms adding up to a function of s that is bounded in a neighborhood of 1, and thus can be added into the o-term), we see that it suffices to prove that

z Np-" -d(A) log (s - 1) + log (s - I)), (s -+ I+). P E A

Note that we can omit a finite number of terms from the left side. We shall omit all those terms corresponding to Np 5 M, where M is to be chosen. Let 6 > 0 be given. We shall show that there exists M = M(6) such that

for all s > 1 that are sufficiently close to 1. This will suffice to prove Theorem 14-1-2. Choose M so that

Then, without loss of generality, we may choose M so that M # Np for any K-prime p. Fixing M, we see that

Denote the second sum in the last equation by T(s). Applying partial summation to T(s), and estimating n(x, A) by d(A)(x/log x), we immediately derive

--

U(s) = 2 log x log (x - 1) x = M xS

- - 2 x log (I - I/x) + log x x = M XS log x log (X + 1)

Therefore, in order to prove Theorem 14-1-2, it suffices to show that

a s s - l+ .

For then, if s is sufficiently close to 1,

I U(s) + log (s - 1) I < -5 log (s - 1). 2

Therefore,

I T(s) + d(A) log (s - 1) 1 1 26 log (s - 1)

for s sufficiently close to 1. In order to prove (*), assume that I < s < 2. The series

2 x - ~ x = 1

converges uniformly for s < t ( 2. Therefore, we may integrate the series term by term to get that

1 " 1 +2- J: (2 .-I) dt = - c --- x= 1 x= 1 x2 log x .=I x" log x

= C p +0(1 ) a s s - + I + . x-1 xS log x

On the other hand, since the Riemann zeta function has a simple pole with residue 1 at s = 1,

where B(t) is continuous in a neighborhood of 1. //

Let x be a Hecke character of K, and let X , denote the trivial Hecke character. For x a positive real number, let us define

where the sum is over K-primes p . The main object of this chapter is to prove the following generalization of the classical prime number theorem:

Theorem 14-1-3 (Generalized Prime Number Theorem): As x 3 oo, we have

= o ( ~ ) log x

The proof of Theorem 14-1-3 will be given in Section 14-3. For the moment, let us be content with remarking on the result and with providing some examples.

Remarks: (1) The o-terms in Theorem 14-1-3 will, in general, depend on the field K and the parameters of the Hecke character X. There exist some crude results on the nature of this dependence, but much is yet to be learned in this regard.

(2) The o-terms do not provide the best possible estimates that one is entitled to expect. For reasons that will become apparent later in this chapter, the positions of the zeros of L(s, X) determine the size of the error terms in Theorem 14-1-3. In order to explain the meaning of this last statement, let us introduce the function

Integrating once by parts shows that

Note that

=o(") log x (x-00).

Therefore, the.prime number theorem of Theorem 14-1-3 can be written in the form

where

E(x) = 1 if x is trivial = 0 otherwise.

Estimating Li(x) by a sum, we can restate (*) in the form

The precise connection between the prime number theorem and the zeros of L(s, X) is this: Let 8 = inf (a I a + it is a zero of L(s, x), a > 0). Then, the error term in the formula for n(x, X) given by (*) or (**) can be taken to be O(P+<) for all E > 0. This error term is smaller than the main term of (*) or (**) if and only if 8 < 1. There is reason to believe

The Riemann Hypothesis: All nontrivial zeros of L(s, X) lie on the line Re (s) = +. (By the functional equation, all such zeros lie either on the line or exist in pairs symmetric about this line.) If the Riemann hypothesis is true, then the error term in (*) and (**) can be taken to be O(X('/~)+') for all e > 0. Of course, the error term stated in (*) and (**) is much larger than predicted by the Riemann hypothesis. In fact, it can be shown that (*) and (**) are equivalent to the statement that L(s, X) does not vanish on the line Re(s) = 1, a statement that is far weaker than the Riemann hypothesis. A proof of the Riemann hypothesis, even for a single L-series, seems to be far beyond the grasp of contemporary mathematics.

Remark: In the above discussion, it is important to deal with the prime number theorem in the form (*) or (**), since

and the integral on the right is not O(2) for any 8 < 1. (3) Theorem 14-1-3 remains true if x is allowed to be a nonprimitive

ideal character obtained by restriction of a Hecke character, since the cor-

responding n-function differs from the corresponding function for a primitive character by a bounded function of x.

EXAMPLES: (1) Let n(x, K) denote the number of K-primes p such that Np 5 x. Then Theorem 14-1-3 [version (*)I for x = xo implies that

In particular, the set of all K-primes has natural density 1. In this case, the prime number theorem is called the prime ideal theorem and is due to Landau.

(2) Let L/K be an abelian extension, with Gal (L/K) = G of order n. Consider G* as consisting of certain Hecke characters (see Section 12-3). Let a E G, and set

n(x, a ) = C 1. N p S x

p unramificd in LIK

Recall that

= 0 otherwise

(Proposition 7-1-7). Therefore,

= C XEG' N p l x 2 x(++))

p unramified in LIK

On the other hand, by the prime number theorem,

(x 3 w).

Therefore, we have proved that

1 X n(x, a ) = _Li(x) + o(Ex) (x - m).

(3) Let '332 be a K-modulus. Let 8 be a K-ideal that is relatively prime to 9.2. Let

Let us use the results of (2) to show that

n(x, 8, '332) = n-lLi(x) + o(+) (X - m), og x

for some positive integer n that we shall define below. (See exercises at end of Section 11-1 for notation.) Let L be the class field corresponding to the admissible subgroup K Vm, where

Let G = Gal (LIK), n = deg (LIK), S = the set of all primes that either divide '332 or are infinite. By the reciprocity law, the global reciprocity mapping sets up an isomorphism

D K / ~ x VE G.

By Proposition 11-1-12, there exists a canonical isomorphism

JK/Kx Kn = I~(Fm)/i(.l),,s n K x Vm).

By Proposition 11-1-13, i(JK,S n K x Vm) 2 PK('332). However, the converse inclusion is clear. Thus, i(QK,S n K x Vm) = PK('332). Thus, putting together the two isomorphisms described above, we find that there exists an isomorphism

~K('332)PK('332) * G. (*)

A simple calculation shows that this isomorphism is induced by the mapping

n - (y) (8 t z,(n)).

Therefore, by (*), if p is a finite K-prime not dividing n , then

LIK (T) = (L*) e p -- @(mod 9.2).

Therefore, by Example (2),

(4) Let '332 = the trivial K-modulus, the other notations being the same as in Example (3). In (*), the left-hand side is now isomorphic to C,, the ideal class group of K. Moreover, p = @(mod D ) + p and 8 belong to the same ideal class, and n = hK by (*). Thus, let $) be an ideal class of K, and set

Then, Example (3) implies that

n(x, 8) = h i l Li(x) + o ( e X ) (x -- m).

In particular, every ideal class contains an equal proportion of prime ideals.

14-2 ANALYTIC PRELIMINARIES

In this section, we summarize the facts we require from the theory of complex variables. For proofs, the reader is referred to Ahlfors' Complex Analysis and Titchmarsh's Theory of Functions. We shall denote references to these works by [A] and [TI, respectively.

A. The gamma function [A, pp. 197-2041

The gamma function T(s ) is a meromorphic function in the entire complex plane, defined by

where y denotes Euler's constant, defined as

For Re ( s ) > 0, the gamma function may defined by

The gamma function generalizes the notion of a factorial, in the sense that

T(n) = (n - I ) ! (n E Z,). (4)

Equation (4) is deduced from the fact that r(1) = 1 and from the functional equation

T ( S + 1 ) = ST(S). ( 5 )

For 1s 1 large, I@) may be estimated by Stirling's formula:

1 0 g T ( ~ ) = ( ~ - ; ) l 0 g s - ~ + ; l o g ( 2 n ) + O ( I s I - ~ ) ( I s l - -m) (6)

uniformly in the angle -a + 6 < arg ( s ) < a - E , E > 0. [log r ( s ) can be defined as a single-valued function in this region, since r ( s ) has no zeros by equation (I).] Setting s = o + it in (6), we see that

I T ( o + it) 1 = e-n"lr2 1 t p-(1'2'(1 + 4 1 ) ) ( I tI A CCJ) (7)

uniformly for a < o < b. Differentiating equation (1) logarithmically, we see that

Applying the definition of Euler's constant to equation (8), we find, by a simple computation, that

B. The Phragmen-Lindelof theorem [T, p. 1801

The Phragmen-Lindelof theorem is a generalization of the maximum modulus principle to unbounded regions of simple topological type, such as angular regions or strips. We require only a relatively simple form of the theorem, which asserts, roughly, that if a function is analytic in a closed strip,

and if the function satisfies certain growth conditions on the boundary of the strip, then one can deduce information about the growth of the function within the strip.

Theorem 14-2-1 (Phragmen-Lindelof): Let f (s) be analytic in the closed strip D = ( s = o + it I a 5 a < b, It 1 2 c > 0). Further, suppose

( 1 ) f ( a + it) = O(exp (exp (On I t I/(b - a)))), It 1 A CCJ, uniformly in D.

(2) f(a + i t) = O(I t la), f(b + i t) = O(I t Is), I t 1 -. m, for some a, /.I > 0. -

Then f (a + it) = O(I t l y ) , I t 1 --* oo, uniformly for a I a I b, where y = p a + q is the linear function that equals a for a = a and /.I for a = b.

C. Entire functions of finite order [A, pp. 206-2101

Let f(s) be an entire function. For r 2 0, set

M(r) = max 1 f(s) I. I s I = r

By the maximum modulus theorem, M(r) is monotone increasing. We say that f (s) is an entire function offinite order if

f M = O(exp (Is Dl, Is l -+ Oo

for some t > 0. The infimum p of all possible t is called the order of f(s). Assume that f (s) is of finite order p. Let the nonzero zeros z,, . . . , z,, . . . o f f ( s ) be numbered so that each zero is repeated according to its proper multiplicity and so that

Ce

Theorem 142-2: The series C ( z, I-(p+f) is convergent for E > 0. n= 1

From Theorem 14-2-2, the theory of Weirstrass canonical products guarantees that f(s) has the following Weirstrass canonical product:

Theorem 14-2-3: Let p = [p]. Then

where m = the order of the zero of f(s) at s = 0, g(s) is a polynomial of degree at most p, and

We shall require Theorem 14-2-3 in the case p = 1:

Corollary 14-2-4: Let f(s) be an entire function of order 1. Then f(s) has a Weirstrass factorization of the form

where m = the order of the zero of f(s) a t s = 0, and C and a are complex constants.

14-3 THE PRIME NUMBER THEOREM

In this section, we shall prove the prime number theorem (Theorem 14-1-3). Throughout this section, let K be an algebraic number field, x a Hecke character of K. Then x is a character of IK(S,) (see Paragraph 9-1). Let xo be the trivial character of ZK(S,), and let x2 denote the square of x on IK(S,). We shall extend x0 and x2 to be zero on any K-ideal not relatively prime to S,. Note that X, and x2 are not necessarily primitive. Let 2, and 22 be the associated primitive characters. Then 2, and f 2 are Hecke characters, agreeing with x0 and x2, respectively, on IK(S,). Let L(s, x,) and L(s, x2) be defined by

These latter two Dirichlet series agree with L(s, 2,) and L(s, f2), except for multiplication by a finite number of elementary factors.

Let us define the following symbols:

E(x) = 1 if x is trivial, = 0 otherwise.

Proposition 14-3-1 : The following statements are equivalent:

(1) n(x, X) = E(X) 2 (log n)-1 + o(x/log x) (x - m). n = 2

(2) e(x, XI = E(X)X + (x - m). (3) y(x, x) = E(x)x + o(x) (x - 00).

Proof: (1) =. (2). By partial summation,

(2) =. (1). Again by partial summation,

Note that

< (log 2)-' "' dy + (log (x 112))-2 dy J: log2 y - I / x

= o(i&). Therefore,

which is equivalent to (1).

(2) 0 (3). First note that i f p is a finite @prime, then

log x 5 P<x'/' C C 1 0 g m m plP

( nxl J2 log x = o(x) (x ---+ 00).

Therefore,

Y(x, x) - e(x, x) = o(x) (x --+ m),

which shows the equivalence of (2) and (3). //

The form of the prime number theorem that we shall prove is (3). The basic idea of the praof is to apply Cauchy's theorem to

By logarithmically differentiating the Euler product for L(s, x), we see that

A simple exercise in residues shows that for x > 0,

= O i f x > l , where the integral is defined as

Therefore, assuming that we may integrate (1) term by term, we see that

A

2 + i m Xs LI 2.- [rn) -(2d)-I 5 --- (s, X) ds = C 2 x(pm) log Np I -

2-im S L 2-im S ds

p m = l

= w(x, x). (2)

On the other hand, let us calculate the integral on the left side of (2) by using Cauchy's theorem to shift the line of integration from Re (s) = 2 to Re (s) = c, 0 < c < 1, where c will be chosen. In order to accomplish such a shift in the line of integration, we must add in compensating terms to account for the poles of the integrand in the strip c < Re (s) < 2. Let us assume that c > 3. The only poles of the integrand are the points where L(s, X) has a zero or a pole. We know that s = 1 is such a point if and only if x is trivial. The residue of the integrand at s = 1 is, therefore, E(x)2nix. Let us assume that the Riemann hypothesis is true. Then s = 1 is the only pole of the integrand in the indicated strip. Thus, by Cauchy's theorem (modulo some estimation on horizontal line segments),

2+i- c+im X~ L1 I : (3, X) ds = E(x)2nix + - - (s, X) ds. 2-i- s L c-i- s L (3)

The remainder of the proof would then consist of showing that the integral on the right side of (3) is o(x) as x --. oo.

There are several serious difficulties with the above "proof" of the prime number theorem, the most serious of which is that we have assumed the

Riemann hypothesis. Further, we have assumed that the integral in (3) can be shown to be o(x). Finally, we have ignored some integrals over horizontal line segments. The latter two difficulties can be gotten around by using - various estimates for (s, x). These estimates will be proven below. The

big problem is to avoid the use of the Riemann hypothesis. The basic idea of our circumvention of this point will be to carve out a small zero-free region of L(s, X) to the left of the line Re (s) = 1. We shall use this zero-free region in the same role as the strip we used above. The next few results will be aimed at constructing the zero-free region.

Proposition 143-2 (Hadamard): Let t be real, a > 1. Then

Proofi The Euler products for L(s, x,), L(s, X) and L(s, x2) converge uniformly on compact subsets of the half-plane Re (s) > 1, and none of these functions vanish in this half-plane. Therefore, for Re (s) > 1,

1% L(s, x2) = - C 1% (1 - x(~)~Np-") , v + s x

where all log's are taken with respect to the principal branch. Moreover, the series on the right converge uniformly on compact subsets of the half- plane Re (s) > 1. Therefore, they can be differentiated termwise to give

L ' L ' L ' -3 - (a, x,) - 4 Re - (a + it, X) - Re - (a + 2it, x2) L L L

= f: Wma log Np(3 + 4 cos (ma,) + cos (ZmaJ), p f S x m = l

where a, = arg (x(p)Np-"). But

3 + 4 cos (a) + cos (2a) = 2(1 + cos (a))2 2 0,

which implies Proposition 14-3-2. /I

Theorem 143-3: Let x be nontrivial. Then L(1, X) # 0.

Proof: Case (1)-x2 # x,. Assume that L(1, X) = 0. Since L(s, x,) has a pole of order 1 at s = 1 , we see that

L ' Re - (a, X) = - a + 4 1 ) (a -+ 1 +, for some a 2 1)

L a - 1

L ' Re (a, x2) = - + 4 1 ) (a - 1+, for some b 2 0). a - 1

Setting t = 0 in Theorem 14-3-2, we see that these last three equations imply that

3 - 4 a - b a - 1 + o(1) 2 0 (a > 1).

But 3 - 4a - b < 0, which gives a contradiction for a sufficiently close to 1.

Case (2)-x2 = x,,: Let c denote the grossencharacter associated to x and let H = ker (x). Then H i s a subgroup of index 2 in J, which contains K x . In particular, H is open and is therefore admissible. Let L be the class field to H. Then L is a quadratic extension of K, and the reciprocity map r induces an isomorphism

J,/H .t: Gal (LIK).

Since J,/H has order 2, c induces the unique nontrivial character of J,/H. Let IZ be the unique nontrivial character of Gal (LIK). Then

IZ(r(a-I)) = c(u).

Therefore, the Artin character derived from IZ coincides with the Hecke character X, and

L(s, X) = L(s, 1; LIK). By Theorem 9-2-2,

CLs) = C,(s)L(s, 1 ; LIK) = C,(s)L(s, x).

If L(l, X) = 0, then C,(s) is regular at s = 1, a contradiction. //

Proposition 143-4: There exists c > 0 such that

L(a + it, X) = O(l t I"), t -+ m ,

uniformlyfbr -E 4 a < 1 + E, E > 0.

Proof: All that is required is a simple application of the Phragmen-Lindelof theorem. Observe that

I L(l + E + it, x)I < (1 - Np-('+")-I P

=0(1 ) a s l t l -+m.

By the functional equation for L(s, x),

By Stirling's formula and (I), we see from (2) that (2)

L(-E + it, X) = O(l t I") (It l - m) (3) for some c > 0. By (I),

L(1 + E + it, x ) = O(l t I") (1 t I --+ m). (4) Therefore, from the Phragmen-Lindelof theorem applied to L(s, X) in the strip -6 < Re (s) < 1 + 6, I Im (s) 1 2 a > 0, we deduce Proposition 14-3-4. //

Remarks: (1) The constant implied by the 0-term depends, in an unde- termined way, on K and X. The determination of this dependence is a difficult problem.

(2) A more careful application of the same argument as above suffices to determine the constant c explicitly.

Theorem 14-3-5: R(s, X) is an entire function of order 1

Proof: First claim that R(s, X) has order at most 1. It suffices to show that

R(s, X) = O(el"ll+') (I s 1 + oo) (1)

for all E > 0. By the functional equation, it suffices to prove (1) for Re (s) 2 4. Since

R(s, x) = 4s - l)r(s, xMs, x)L(s, XI, in order to prove (I), it suffices to show that

L(s, X) = O(el"l' +') (I s 1 -4 oo, Re (s) 2 t) . (4) Stirling's formula implies that

r ( ~ , x ) = O(elSll0glsl) (I s l -- m , Re ( 4 2 6), ( 5 ) which implies (2). Estimate (3) is obvious. Let 6 > 0. By Theorem 14-3-4, the estimate (4) holds in the strip + < Re (s) < 1 + 6, I Im (s) 1 2 a > 0. From the proof of Theorem 14-3-3,

L ( s ,x )=O( l ) (Isl-00) uniformly in the region Re (s) 2 1 + 6. Thus, (4) holds and R(s, X) is of order at most 1. In order to show that R(s, X) has order 1, we must show that the estimates we have made are, essentially, the best possible. Allow s to tend

to oo through real values. Then L(s, X) - 1. Moreover, by Stirling's formula, there exists A > 0 such that

log r(s , X) = AS log + O(S) (S --. oo).

Also, there exists B E R such that

Thus, as s -+ oo through real values,

R(s, X) = eAdog "1 + 0(1))7

which shows that R(s, X) has order a t least 1. //

Corollary 143-6: R(s, X) has a Weirstrass factorization of the form

for suitable constants C and a, where z,, z,, . . . are the nonzero zeros of R(s, X) and m = the order of the zero of R(s, X) a t s = 0.

Remark 14-3-7: From the Euler product representation of L(s, x), it is clear that L(s, X) # 0 for Re (s) > 1. Therefore, R(s, X) does not vanish for Re (s) > 1, since e(s, X) and r(s , X) have no zeros. By the functional equation, we then conclude that R(s, X) # 0 for Re (s) < 0. Therefore, 0 ( Re (z,) 1 for all m. Also, any zero of R(s, X) is a nontrivial zero of L(s, x). We shall later show that L(s, X) does not vanish in a small region inside the strip 0 ( Re (s) ( 1, so that the same is true of R(s, x). Since L(1, X) # 0, R(1, X) f 0, so that by the functional equation, R(0, X) # 0, so that m = 0 in Corollary 14-3-6.

The strip 0 < Re (s) ( I will be called the critical strip. By our discussion, the zeros of R(s, X) are precisely the zeros of L(s, X) in the critical strip.

Corollary 143-8: For all E > 0,

converges.

We now come to the construction of the zero-free region for L(s, x):

Theorem 14-3-9: There exists a constant c > 0 such that L(o + it, X) f 0 for

C - log 2 '

Proof: By Theorem 14-3-3, if x is nontrivial, the second condition is verified for all sufficiently small c. If x is trivial, then L(s, X) has a pole at s = 1, so that the second condition again holds for all sufficiently small c. It therefore suffices to show that the first condition is verified for some c. Let b + it be a zero of L(s, X) such that I t 1 2 2, and let s = a + it, a > 1. Logarithmi- cally differentiate the Weirstrass product for R(s, X) (Corollary 14-3-6) and take real parts to get

Note that L ( s , X) is a constant. By equation (9) of Section 14-2, e

for It 1 2 2, uniformly for 1 a ( 2, where A is a positive constant. There- fore, by equation (I),

for It 1 2 2, uniformly for 1 ( a < 2. Each term of the summation in (2) is nonnegative, since

Therefore, there exists C > 0 such that

- ~ e s ( a + it, x ) i Clog It

for It 1 2 2, uniformly for 1 < a 2. Since L(s, x,) has a simple pole at s = 1, there exists D > 0 such that

for 1 I a I 2. Therefore, by Proposition 14-3-2 and equations (3) through (9, there exists E > 0 such that

4(a - p)-' < 3(a - I)-' + E log t. (6)

Set a = 1 + €/log It 1, E > 0. Then (6) implies that

Choose E so that EE < 1 and set c = ~ ( 1 - eE)/3 + EE. Then Theorem 14-3-9 is satisfied for this value of c. //

We shall be able to prove the prime number theorem after establishing the following five lemmas:

Lemma 1: Let y > 0, a > 1, T > 0. Then

where the 0-term constants are independent of y, b, and T.

Proofi Assume that y < 1. Let U > 0. By the residue theorem,

since ys/s has a simple pole with residue 1 at s = 0. Note that

Similarly,

Equations (1) through (4) imply the desired result in case y < 1. The calculations for y > 1 are similar and are left to the reader. //

If m is a nonnegative integer, define

A h , x) = C x(Pr> 1% NP. Npr=m

( 5 )

and L - - (s, X) = C A(m, xM-" (Re (s) > 1). I/ (7) L m= l

Lemma 2: A(m, X) = O(1og m) (m -+ 00).

Proof: If m is not a power of a rational prime, then A(m, X) = 0. If m is of the form pK for p a rational prime, then

A(pk, X) Np'=pr C 1% NP = ~ ( P / P ) log P Npzpk I

I f (PIP) ~(P/P> 1% (pk) I deg (KlQ) log m. I /

Lemma 3: Let a > 1, T > 0, x = a positive integer + +. Then

-- - - (s, x) ds = ~ ( x , x) + 0(-) I

where the 0-term constant does not depend on a, x, or T.

Proof: The series (7) converges uniformly in each half-plane of the form Re (s) 2 1 + 6, 6 > 0. Therefore, we may multiply (7) by xys and integrate termwise over the interval [a - iT, a + iT] to get

1 " a+iT (f)' d~

by equation (6) and Lemma 2. Now

C log n = O( c $1 n>2x

2 na 1 1 0 ( ) 1 0 , n i ~ i i

= O(x-"+I log x),

with the 0-term constant independent of a, x , or T. If x < n < 2x, then

1 log (f) 1 = / log (:) 1 = / log (1 - 7 n - x ) I

n - x > A - 7

X

for some positive constant A. Therefore,

272 / THE PIUME NUMBER THEOREM CHAP. 14

pend on x. Therefore, the integrand over C, is O(1) as x -+ w. This completes the proof of the generalized prime number theorem. I/

EXERCISE 14-3-10: Instead of using Theorem 14-3-9, assume the Riemann hypothesis in the proof above. Show how the above calculations imply that

~ ( x , X) = E(x)x + O ( X * / ~ log x) (x -+ 00).

Use partial summation as in Theorem 14-3-1 to show that in this case

n(x, X) = E(x)Li(x) + O(X'/~ log2 x) (x -4 m). Suggestions for further reading

In the course of this book, we have tried to convey the spirit of contemporary analytic number theory. Of course, in anything smaller than an encyclopedic treatise, one is forced to omit many subjects whose importance to the subject is great. That is certainly the case with the present volume. Therefore, it seems in order to suggest some further reading in the field. The following suggestions are by no means a complete sampling of current work in the very active field of analytic number theory, but, rather, represent, in the author's opinion, beautiful and accessible points of departure for learning more about the subject.

We have not mentioned anything about nonabelian L-functions. The student can do no better than to start to learn about them by reading Artin's original papers [2, pp. 105-124, 165-1801. The best progress made so far in proving Artin's conjec- tures on the nonabelian L-series has been made by Brauer [6,7]. In these references, the reader will also find a proof of the so-called Brauer-Siege1 theorem, which tells how the class number grows with the discriminant. A summary of the theory of nonabelian L-series is given by Heilbronn in [9]. A proof of the Brauer-Siege1 theorem is to be found in [20].

The literature on the subject of class numbers is immense. Starting from the class number formulas developed in this book, one can prove Fermat's last theorem in the case of regular primes. For an excellent discussion of the application of the class number formulas, the reader can turn to [8]. After mastering this material, the interested reader can turn to the books of Hasse [16] and Meyer [21]. With little or no additional background other than that supplied by this book, the reader can study Stark's paper [24], which proves that there are exactly nine imaginary quadratic fields with class number 1. This conjecture goes back to Gauss and was just recently settled.

We have completely bypassed the theory of arithmetic in function fields. The interested reader may consult Chevalley [lo] or Eichler [12]. The most interesting feature of the function field situation is that here a form of the Riemann hypothesis is known. The proof is given, for example, in the cited book of Eichler.

The connection between analytic number theory and the theory of automorphic functions runs deep and its importance cannot be overestimated. An introduction to automorphic functions is given by Gunning [14]. But since the subject's contemporary interest stems from the work of Hecke, the reader is strongly advised at least to look at Hecke's works [17], as well as Godement's summary of their contents [13]. A very readable account of some of Hecke's work is given by Ogg[22]. The modem aspect of the number-theoretic applications of automorphic functions is given by Shimura [23].

One can read about explicit forms of the reciprocity law in Hasse's classical treatise [15]. Two other beautiful classical texts that are worthy of attention are by Hecke [17] and Landau [19]. Although these last two are rather old, they make inspiring reading even now.

The reader can learn about the cohomological development of class field theory in Cassels-Frohlich [9] and Artin-Tate [S]. Another approach to the reciprocity law, as well as a development of number theory solely from the point of view of analysis on locally compact groups, is found in Weil[28]. A reinterpretation of the functional equations of the Hecke L-series is given by Weil in [31]. A proof of the quadratic reciprocity law, as well as an adelic development of the Siege1 analytic theory of quadratic forms, is given by Weil in the germinal pair of papers [29, 301.

APPENDIX

Algebraic preliminaries

In this appendix, we summarize those algebraic facts that we require which are not always covered in elementary courses in abstract algebra. For proofs, the reader is referred to any standard treatise on algebra (e.g., van der Waerden or Lang).

1. MODULES

Let M be a module over Z, N c M a submodule. Then

Proposition A-1-1 : If M isfinitely generated, so is N.

Proposition A-1-2 (Fundamental Theorem of Abelian Groups): Assume that M is finitely generated. Then there exist nonnegative integers r, s, and positive integers dl , . . . , ds such that

and such that

dl 141 . - - 14. The integers r, s, dl, . . . , ds are uniquely determined by M.

The module M is said to be free if there exist elements w,, . . . , w, in M such that if x E M, then x can be written uniquely in the form

A free module is obviously finitely generated. In case M is free, the set consisting of w,, . . . , wn is said to be a basis of M. Every basis of M contains the same number of elements, and this number is called the rank of M.

Proposition A-1-4 (Elementary Divisor Theorem): Let M be a free module of rank m, N a submodule of M of rank n. There exist bases (w,, . . . , w,}, (v,, . . . , v,} of M and N, respectively, such that

vi = a,w, ( 1 5 i ( n),

where the ai's are positive integers satisfying

2. KUMMER THEORY

Let K be a field containing the nth roots of 1, and let a,, . . . , a, E K x . The extension

is said to be a Kummer extension of K. It is easy to show that LIK is a Galois extension and that Gal (LIK) is an abelian group with exponent n (i.e., if a E Gal (LIK), then an = 1). Moreover, if A, = { y E K 1 y = xn, x E Lx}, then

deg (LIK) = [A,: Kx"]. (*) The fundamental theorem of Kummer theory asserts that if LIK is a Galois extension, with Gal (LIK) of exponent n, then LIK is a Kummer extension. Moreover, (*) holds, and if {a,, . . . , a,} is a system of coset representatives of A,/Kn, then

Bibliography

1. Ahlfors, L. V., Complex Ana(vsis. McGraw-Hill Book Co., Inc., New York, 1953.

2. Artin, E., The Collected Works of Emil Artin, S. Lang and J. Tate, eds. Addison- Wesley Publishing Company, Reading, Mass., 1965.

3. - , Algebraic Numbers and Algebraic Functions. Gordon & Breach, New York, 1967.

4. - , Theory of Algebraic Numbers. Lecture notes, Gottingen, 1956. 5. - and J. Tate, Class Field Theory. W. A. Benjamin, New York, 1968. 6. Brauer, R., "On the Zeta Functions of Algebraic Number Fields, I, 11," Amer.

J. Math., 69 (1947), 243-250; 72 (1950), 739-746. 7. - , "On Artin's L-Series with General Group Character," Ann. Math.,

48 (1 947), 502-5 14. 8. Borevich, Z. I. and I. R. Shafarevich, Number Theory. Academic Press, New

York, 1965. 9. Cassels, J. W. S. and A. Frohlich, eds., Algebraic Number Theory. Thompson

Publishing Co., Washington, D. C., 1967. 10. Chevalley, C., Introduction to the Theory of Algebraic Functions of One Variable.

Amer. Math. Soc., Providence, 1951. 11. Davenport, H., Multiplicative Number Theory. Markham Publishing Co.,

Chicago, 1967. 12. Eichler, M., Introduction to the Theory of Algebraic Numbers and Functions.

Academic Press, New York, 1966. 13. Godement, R., "Sur les Travaux de Hecke I, 11,111, IVY" SCm. Bourbaki,

EXPOSCS 51,59,74, 80 (1951-52). 14. Gunning, R., Lectures on Modular Forms. Princeton University Press, Princeton,

N. J., 1962.

15. Hasse, H., Zahlbericht, Jahr. Deutsch. Math. Verein. (1928, 1931). 16. - , Ober die Klassenzahl Abelscher Korper. Akademie-Verlag, Berlin, 1952. 17. Hecke, E., Mathematische Werke. Vandenhoeck and Ruprecht, Gottingen,

1959.

18. - , Vorlesungen iiber die Theorie der Algebraischen Zahlen. Chelsea Pub- lishing Co., New York, 1948.

19. Landau, E., Einfiihrung in die Elementare und Analytische meorie der Alge- braischen Zahlen und der Zdeale. Chelsea Publishing Co., New York, 1948.

20. Lang, S., Algebraic Numbers. Addison-Wesley Publishing Company, Read- ing, Mass., 1964.

21. Meyer, K., Berechnung der Klassenzahl Abelscher Kcrper. Akademie-Verlag, Berlin, 1957.

22. Ogg, A., Modular Forms and Dirichlet Series. W. A. Benjamin, New York, 1969. 23. Shimura, G., Automorphic Functions and Number neory . Springer-Verlag,

Berlin, 1968.

24. Stark, H., "A Complete Determination of the Complex Quadratic Fields of Class Number One," Mich. Math. J. 14 (1967), 1-27.

25. Tate, J., "On Fourier Analysis in Number Fields and Hecke's Zeta Functions," Dissertation, Princeton University, 1950; reproduced in [9], pp. 305-347.

26. Titchmarsh, E. C., The Theory of Functions. Oxford University Press, London, 1939.

27. Weil, A., Sur l'lntkgration dans les Groupes Topologiques et ses Applications, Hermann, Paris, 1938.

28. - , Basic Number Theory. Springer-Verlag, New York, 1967. 29. - , "Sur Certaines Groupes dYOptrateurs Unitaires," Acta Math., 111

(1964), 143-21 1.

30. - , "Sur la Formule de Siege1 dans la ThCorie des Groupes Classiques," Acta Math., 113 (1965), 1-87.

31. - , "Fonctions Zeta et Distributions," Stm. Bourbaki, ExposCs 312, 1965-66.

32. Weiss, E., Algebraic Number Theory. McGraw-Hill Book Co., Inc., New York, 1963.

Abelian L-function, 5, 164ff, 182 Absolute value, 20ff Additive approximation theorem, 38 Adeles, 32ff Admissible subgroup, 192 Algebraic number field, 9 Analytic continuation, 139, 154, 159 Approximation theorem, 29, 38,45 Archimedian valuation, 20 Arithmetic progression, 197, 236 Artin character, 182, 227 Artin reciprocity law, 5, 166, 171, 182,

183, 186 Artin symbol, 169

Bauer's theorem, 164, 208

Character, 11 1 Chinese Remainder Theorem, 18 Class field, 230, 234

Class field theory, 181ff Class group, 192 Class number, 42, 172 Class number formula, 172 Completion, 22, 55 Conductor, 118, 156, 206, 215 Conductor-discriminant formula, 183 Congruences, 17, 38 Critical strip, 34 Cyclic extension, 190 Cyclotomic field, 4, 96K, 238

Decomposed prime, 74 Decomposition subfield, 87 Decomposition theorem, 235 Dedekind zeta function, 158 Defining modulus, 192 Different, 65, 80 Dirichlet density, 163 Dirichlet L-series, 3, 4, 171 Dirichlet's theorem, 167 Discriminant, 12, 67, 80, 81, 100, 101 Discriminant, of basis, 11, 58 Dual group, 11 1

of additive group of local field, 114

Dual group (con?.) : Hecke L-series, 133, 156ff of adele ring, 120 Hensel's lemma, 49 of finite group, 11 3 Herbrand's lemma, 189, 232 of multiplicative group of local Herbrand's unit theorem, 198ff

field, 116 Hilbert's reciprocity law, 21 5 of restricted direct product, 11 8 Hilbert's Satz 90, 188, 202

Duality, 11 1 Hilbert theory, 67, 84

Entire function, 259, 265 Euler product, 3, 145, 158 Existence theorem, 230

Field polynomial, 10 First inequality, 197 Fourier inversion theorem, 121, 128 Fourier transform, 121 Fractional ideal, 14 Frobenius automorphism, 69, 84ff,

102, 164 Functional equation, 139, 154, 159 Fundamental domain, 152 Fundamental unit, 45

Galois correspondence, 217, 225 Gamma function, 258 Gaussian sum, 145,175 Generalized conductor, 157, 183, 192 Generalized discriminant, 183 Global class field theory, 226ff Global zeta function, 14% Greatest common divisor, 17 Grossencharacter, 145, 175, 194, 227 Grossencharacter of classical type, 194

Ideal, 12ff Ideal character, 157 Ideal class, 257 Ideal mapping, 151 Idele class character, 145 Ideles, 33, 38ff Imaginary quadratic field, 12 Inertia group, 68 Inertia subfield, 87 Inert prime, 74 Integer, 9ff Integral basis, 12 Invariant differential operator, 128

Kronecker's theorem, 96, 238ff Kronecker symbol, 88 Kummer's theorem, 78

Legendre symbol, 172 Local field, 32, 46ff Locally constant function, 129 Local norm, 220 Local norm index, 220 Local uniformizing parameter, 26, 56 Local zeta function, 133ff

Hadamard's lemma, 263 Hecke character, 157, 195

Minimal basis, 12 Minimal polynomial, 10

Modulus, 28 Modulus of automorphism, 36, 37 Multiplicative approximation theorem,

45

Natural density, 252 Norm, 52, 77 Norm of ideal, 29, 72 Norm residue symbol, 184, 215, 218,

223, 227, 242

ord, , 17, 57

Phragmen-Lindelof theorem, 258 Poisson summation formula, 148 Pontryagin duality, 112 Power reciprocity law, 241ff Power residue symbol, 243 Prime, 28, 56

complex, 28 finite, 28 infinite, 28 real, 28

Prime ideal theorem, 252, 256 Prime number theorem, 2, 254, 271 Primes in arithmetic progressions, 3,

4, 168 Primitive character, 157 Product formula, 36

Ramification groups, 70, 86 Ramification index, 56, 62ff Ramified extension, 56, 62ff Ramified prime, 81 Real quadratic field, 12 Reciprocity law, 5, 106-7, 228 Regulator, 152 Relatively prime ideals, 17 Residue class degree, 60, 73 Residue class field, 30 Restricted direct product, 32 Riemann hypothesis, 3, 255, 272, 282 Riemann zeta function, 2

Schwartz-Bruhat function, 146 Schwartz-Bruhat space, 127ff Second inequality, 188 Self-dual measure, 121ff Splitting type, 162, 236 Strong reciprocity law, 216-17 Supplementary reciprocity law, 247

Tame ramification, 71, 88 Tate's thesis, 133 Tchebotarev density theorem, 168ff Topological field, 21 Totally disconnected, 26, 27 Trace, 52, 77

Quadratic field, 88 Quadratic reciprocity law, 1 0 6 7

Undecomposed prime, 162 Uniqueness theorem, 230 Unit theorem, 43

Valuation, 20ff, 54, 79 Archimedean, 20, 49, 50, 51 complex, 24 equivalence of, 22 finite, 20 infinite, 20 non-Archimedean, 20, 21, 24 real, 24

Vandermonde determinant, 11, 66, 100 Volume of an idele, 39

Weak reciprocity law, 207ff Weierstrass canonical product, 259, 266 Wild ramification, 71

goldstein l.j. analytic number theory - index - free

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