Lawrence C. Washington

Introduction to Cyclotomic Fields

% Springer-Verlag New York Heidelberg Berlin

Lawrence C. Washington Department of Mathematics University of Maryland College Park, MD 20742 U.S.A.

Editorial Board

P. R. Halmos

F. W. Gehring

C. C. Moore

Managing Editor Indiana University Department of Mathematics Bloomington, IN 47401 U.S.A.

University of Michigan Department of Mathematics Ann Arbor, MI 48104 U.S.A.

University of California at Berkeley Department of Mathematics Berkeley, CA 94720 U.S.A.

AMS Subject Classifications (1980): 12-01

Library of Congress Cataloging in Publication Data Washington, Lawrence C. Introduction to cyclotomic fields. (Graduate texts in mathematics; 83) Bibliography: p. Includes index. 1. Fields, Algebraic. 2. Cyclotonly. I. Title. 11. Series. 512'3 QA247.W35 82-755 AACR2 © 1982 by Springer-Verlag New York Inc. All rights reserved. No part of this book may be translated or reproduced in any form without written permission from Springer-Verlag, 175 Fifth Avenue, New York, New York 10010, U.S.A. Printed in the United States of America. 987654321

ISBN 0-387-90622-3 Springer-Verlag ISBN 3-540-90622-3 Springer-Verlag

New York Heidelberg Berlin Berlin Heidelberg New York

To My Parents

Preface

This book grew out of lectures given at the University of Maryland in 1979/1980. The purpose was to give a treatment of p-adic L-functions and cyclotomic fields, including Iwasawa's theory of 4-extensions, which was accessible to mathematicians of varying backgrounds. The reader is assumed to have had at least one semester of algebraic number theory (though one of my students took such a course concurrently). In particular, the following terms should be familiar: Dedekind domain, class number, discriminant, units, ramification, local field. Occasionally one needs the fact that ramification can be computed locally. However, one who has a good background in algebra should be able to survive by talking to the local algebraic number theorist. I have not assumed class field theory; the basic facts are summarized in an appendix. For most of the book, one only needs the fact that the Galois group of the maximal unramified abelian extension is isomorphic to the ideal class group, and variants of this statement. The chapters are intended to be read consecutively, but it should be possible to vary the order considerably. The first four chapters are basic. After that, the reader willing to believe occasional facts could probably read the remaining chapters randomly. For example, the reader might skip directly to Chapter 13 to learn about 4 -extensions. The last chapter, on the Kronecker—Weber theorem, can be read after Chapter 2. The notations used in the book are fairly standard; Z, Zp , and Up denote the integers, the rationals, the p-adic integers, and the p-adic ration als, respectively. If A is a ring (commutative with identity), then A denotes its group of units. At Serge Lang's urging I have let the first Bernoulli number be B 1 = — 4 rather than +I. This disagrees with Iwasawa [23] and several of my papers, but conforms to what is becoming standard usage. vii

Throughout the preparation of this book I have found Serge Lang's two volumes on cyclotomic fields very helpful. The reader is urged to look at them for different viewpoints on several of the topics discussed in the present volume and for a different selection of topics. The second half of his second volume gives a nice self-contained (independent of the remaining one and a half volumes) proof of the Gross-Koblitz relation between Gauss sums and the p-adic gamma function, and the related formula of Ferrero and Greenberg for the derivative of the p-adic L-function at 0, neither of which I have included here. I have also omitted a discussion of explicit reciprocity laws. For these the reader can consult Lang [4], Hasse [2], Henniart, Ireland-Rosen, Tate [3], or Wiles [1]. Perhaps it is worthwhile to give a very brief history of cyclotomic fields. The subject got its real start in the 1840s and 1850s with Kummer's work on Fermat's Last Theorem and reciprocity laws. The basic foundations laid by Kummer remained the main part of the theory for around a century. Then in 1958, Iwasawa introduced his theory of Zp -extensions, and a few years later Kubota and Leopoldt invented p-adic L-functions. In a major paper (Iwasawa [18]), Iwasaw a interpreted these p-adic L-functions in terms of-extensions. In 1979, Mazur and Wiles proved the Main Conjecture, showing that p-adic L-functions are essentially the characteristic power series of certain Galois actions arising in the theory of Zp-extensions. What remains? Most of the universally accepted conjectures, in particular those derived from analogy with function fields, have been proved, at least for abelian extensions of G. Many of the conjectures that remain are probably better classified as "open questions," since the evidence for them is not very overwhelming, and there do not seem to be any compelling reasons to believe or not to believe them. The most notable are Vandiver's conjecture, the weaker statement that the p-Sylow subgroup of the ideal class group of the pth cyclotomic field is cyclic over the group ring of the Galois group, and the question of whether or not A = 0 for totally real fields. In other words, we know a lot about imaginary things, but it is not clear what to expect in the real case. Whether or not there exists a fruitful theory remains to be seen. Other possible directions for future developments could be a theory of k-extensions = fi 7L,; some progress has recently been made by Friedman [1]), and the analogues of Iwasawa's theory in the elliptic case (Coates-Wiles

[4]). I would like to thank Gary Cornell for much help and many excellent suggestions during the writing of this book. I would also like to thank John Coates for many helpful conversations concerning Chapter 13. This chapter also profited greatly from the beautiful courses of my teacher, Kenkichi Iwasawa, at Princeton University. Finally, I would like to thank N.S.F. and the Sloan Foundation for their financial support and I.H.E.S. and the University of Maryland for their academic support during the writing of this book.

Contents

CHAPTER

Fermat's Last Theorem

1

CHAPTER 2

Basic Results

9

CHAPTER 3

Dirichlet Characters

19

CHAPTER 4

Dirichlet L-series and Class Number Formulas

29

CHAPTER 5

p-adic L-functions and Bernoulli Numbers 5.1. p-adic functions 5.2. p-adic L-functions 5.3. Congruences 5.4. The value at s = 1 5.5. The p-adic regulator 5.6. Applications of the class number formula

47 47 54 59 63 70 77

ix

CHAPTER 6

Stickelberger's Theorem

87

6.1. Gauss sums 6.2. Stickelberger's theorem 6.3. Herbrand's theorem 6.4. The index of the Stickelberger ideal 6.5. Fermat's Last Theorem

87 93 100 102 107

CHAPTER 7

Iwasawa's Construction of p-adic L-functions 7.1. Group rings and power series 7.2. p-adic L-functions 7.3. Applications 7.4. Function fields 7.5. p = 0

113 113 117 125 128 130

CHAPTER 8

Cyclotomic Units 8.1. Cyclotomic units 8.2. Proof of the p-adic class number formula 8.3. Units of0( p) and Vandiver's conjecture 8.4. p-adic expansions

143 143 151 153 160

CHAPTER 9

The Second Case of Fermat's Last Theorem 9.1. The basic argument 9.2. The theorems

167 167 173

CHAPTER 10

Galois Groups Acting on Ideal Class Groups 10.1. Some theorems on class groups 10.2. Reflection theorems 10.3. Consequences of Vandiver's conjecture

184 184 187 195

CHAPTER 11

Cyclotomic Fields of Class Number One 11.1. The estimate for even characters 11.2. The estimate for all characters 11.3. The estimate for h; 11.4. Odlyzko's bounds on discriminants 11 .5. Calculation of h„±,

204 205 210 217 221 228

CHAPTER 12

Measures and Distributions 12.1. Distributions 112. Measures 12.3. Universal distributions

231 231 236 251

CHAPTER 13

Iwasawa's Theory of 4-extensions 13.1_ Basic facts 13.2. The structure of A-modules 13.3. Iwasawa's theorem 13.4. Consequences 13.5. The maximal abelian p-extension unramified outside p 13.6. The main conjecture 13.7. Logarithmic derivatives 13.8. Local units modulo cyclotomic units

263 264 268 276 284 290 295 299 310

CHAPTER 14

The Kronecker-Weber Theorem

319

Appendix

331

I. Inverse limits

2. Infinite Galois theory and ramification theory I Class field theory

Tables 1. Bernoulli numbers 2. Irregular primes 3. Class numbers

331 332 336

347 347 350 352

Bibliography

361

List of Symbols

386

Index

388

Chapter I

Fermat's Last Theorem

We start with a special case of Fermat's Last Theorem, since not only was it the motivation for much work on cyclotomic fields but also it provides a sampling of the various topics we shall discuss later. Theorem 1.1. Suppose p is an odd prime and p does not divide the class number of the field Ggp), where is a primitive pth root of unity. Then 4 yP

zP,

(xyz, p) = 1

has no solutions in rational integers.

Remark. The ease where p does not divide x, y, and z is called the first case of Fermat's Last Theorem, and is in general easier to treat than the second case, where p divides one of x, y, z. We shall prove the above theorem in the second case later, again with the assumption on the class number. Factoring the above equation as p-

fl (x

+ Vp y) = zP,

i=

we find we are naturally led to consider the ring Z{(p]. We first need some basic results on this ring. Throughout the remainder of this chapter, we let = Proposition 1.2. ZM is the ring of algebraic integers in the field GO. Therefore 1{(] is a Dedekind domain (so we have unique factorization into prime ideals, etc.).

2

1 Fermat's Last Theorem

Let C denote the algebraic integers of Q(0. Clearly l[C] must show the reverse inclusion.

PROOF.

Lemma 1.3. Suppose r and s are integers with (p, rs) — 1) is a unit of[].

PROOF. Writing r (ir

C. We

1. Then Gr — 1)/

st (mod p) for some t, we have

1

(s — 1

(st

(s — 1

=1+

'

',.s(r

1)

Similarly,( — 1)/Gr — 1) e [C]. This completes the proof of the lemma. CI Remark. The units of Lemma 1.3 are called cyclotomic units and will be of great importance in later chapters. Lemma 1.4. The ideal (1 — ) is a prime ideal of and (1 — OP -1 = (p). Therefore p is totally ramified in Q(C).

PROOF. Since Xi' + XP - 2 + • • + X + 1 = Fr= 11- (X — Ç), we let X 1 to obtain p = - 0. From Lemma 1.3, we have the equality of ideals (1 — = (1 — C). Therefore (p) = (1 0". Since (p) can have at most p — 1 = deg(Q(0/Q) prime factors in Q(0, it follows that (1 — ) must be a prime ideal of C. Alternatively, if (1 —) = A B, then p = N(1 — — NA NB so either NA =1 or NB = 1. Therefore the ideal (1 — ) does not factor in C. —

We now return to the proof of Proposition 1.2. Let y denote the valuation corresponding to the ideal (1 — so v(1 —) = 1 and v(p) = p — 1, for example. Since QG) -= Q(1 — 0, we have that {1, 1 — (1 — ç) ..., )P - 2 } is a basis for Q ( ) as a vector space over Q. Let a E C. Then (1— ,

a = ao + a 1 (1 —) + • - • + ap _ 2 ( 1 —

-

with a; e Q. We want to show a; E Z. Since v(a) 0 (mod p — 1) for a E 0, the numbers v(a1(1 — 0 i p — 2, are distinct (mod p — 1), hence are distinct. Therefore, by standard facts on non-archimedean valuations, v(a) = min(v(41 — 0 i)). Since v(a) 0 and v((1 — 0 i) < p 1, we must have v(ai) > O. Therefore p is not in the denominator of any ai . Rearrange the expression for a to obtain

with b i E Q, but no bi has p in the denominator. The proof may now be completed by observing that the discriminant of the basis 11, (, , V - 2 } is a power of p. More explicitly, we have

= 1)0 + b 1 Ç + •• +

2g6)P

2

3

1 Fermat's Last Theorem

(Z/ pZ) X . Let oci =

where u runs through Gal(Q()/0) Then we have

e, where u:

/ 11 bp _ 2

But the determinant of the matrix is a Vandermonde determinant, so it is equal to

-

= (unit)(power of 1 —

1<j
Therefore b i = (algebraic integer)/(power of 1 —). Since b i has no p in the denominator, we must have b i = algebraic integer; therefore b i E Z, so we are done. Alternatively, we could finish the proof as follows. Since C - iot is an algebraic integer, its trace from Q ( ) to 0 is a rational integer: Tr(Ç) E Z. Now the minimal polynomial for V, (j, p) 1, is XP- 1 + xp- 2 + X + 1, so Tr(V) = —1. We obtain 2

pbi —

b

(p — 1)b 1 —

i =0

;*i

b. = TrG - e Z.

Using this equation for i = 0 and i = i and subtracting, we obtain p(bo — b i) E Z, therefore bo — b i e Z. It remains to show bo e Z. Write a=

+ + • + V -2 ) + [01 — b oK +•• + (bp_ 2 —

bO

)

P

By the above, the expression in brackets is an algebraic integer. Therefore —b 0 =

++••+

V -2 )

E

so b o eCnQ= Z. Therefore b i cZ for all i, so again we are done. This finishes the proof of Proposition 1.2. LI Before proceeding to the proof of Theorem 1.1, we need the following result, which will be discussed in more detail later.

Proposition 1.5. Let E be a unit of Z r

[

p].

Then there exist E 1 c QG + Ç 1 ) and

such that £ =

Remark. Take any embedding of O ( ) into the complex numbers. Complex conjugation acts as an automorphism sending to Ç 1 • The fixed field is QG + = Q(cos(27c/p)) and is called the maximal real subfield of Q (4 ). The proposition says that any unit of Z [ ] may be written as a root of unity times a real unit. This result is plausible since the field QG + Ç 1 ) has (p — 1)/2 real embeddings and no complex embeddings into C, while Q(c)

4

1 Fermat's Last Theorem

has no real embeddings and (p — 1)/2 pairs of complex embeddings. Therefore the 7L-rank of the unit groups of each field is (p — 3)/2, so the units of Q(C + 1 ) are of finite index in those of Q(0. However, it does not appear that Dirichlet's unit theorem can be used to prove the proposition. PROOF OF PROPOSITION 1.5. Let a = Then a is an algebraic integer since t is a unit. Also, all conjugates of a have absolute value 1 (this follows easily from the fact that complex conjugation commutes with the other elements of the Galois group). We now need a lemma. Lemma 1.6. If a is an algebraic integer all of whose conjugates have absolute value 1, then a is a root of unity. PROOF. The coefficients of the irreducible polynomials for all powers of a are rational integers which can be given bounds depending only on the degree of a over Q. It follows that there are only finitely many irreducible polynomials which can have a power of a as a root. Therefore there are only finitely many distinct powers of Œ. The lemma follows. LI Remark. The assumption that a is an algebraic integer is essential, as the example a = + i shows. Also we note that it is actually possible for an algebraic integer to have absolute value 1 while some of its conjugates do not. An example is a = N/ 2 —

+

— 1. One conjugate may be ob-

tained by mapping .\/i to — , which yields -\,/2 + ± + 1, neither of which have absolute value 1. However, if Q(a) is abelian over Q then all automorphisms commute with complex conjugation; so if ad = 1 then ea' = 1 for all u. Returning to the proof of Proposition 1.5, we find that c/Z is a root of unity, therefore c/ = + for some a (the only roots of unity in Q(() are of this form. This will follow from results in the next chapter). Write E = 1)0 + b 1 + - • - + bp-2CP-2. Suppose first that c/Z = Then E 1)0 + b1 + - + bp _ 2 (mod 1 — Also Z = bo + +•••= 1) 0 + b 1 + - • • + b_ 2 E = —c`Z Therefore 2Z 0 (mod 1 — But 2 0 (1 —). Since (1 —) is a prime ideal, Z e (1 — which is impossible since Z is a unit. Therefore = +Ca. Let 2r a (mod p), and let g l = Then E = LI = g i . This proves Proposition 1.5. CI. El, and

1.1. We first treat the case p = 3. If 3 x then x 3 + 1 (mod 9) and similarly for y and z. Therefore x 3 + y 3 —2, 0, or +2

PROOF OF THEOREM

(mod 9) but z 3 + 1. Therefore x 3 + y 3 0 z 3 . Similarly, we may treat the case p = 5 by considering congruences mod 25. However, we must stop at

5

I Fermat's Last Theorem

p = 7 since 1 7 + 3Q7 31 7 (mod 49). In fact there are still solutions if we consider congruences to higher powers of 7 (see the Exercises). So we need a new method. Assume p > 5 and suppose ,C P yP zP, p xyz. Suppose x # y # —z (mod p). Then — 2zP zP, which is impossible since p ,f/ 3z. Therefore we may rewrite the equation if necessary (as xP ( —z)P = (_ y)P) to obtain x # y (mod p). We shall need this assumption later on. Also we may assume x, y, and z are relatively prime, otherwise divide by the greatest common divisor.

Lemma 1.7. The ideals (x + y), i = 0, 1, . , p — 1, are pairwise relatively prime.

PROOF, Suppose g is a prime ideal with c?, I (x + Ciy) and g I (x + ..)3)), where i j. Then (CiY C3 ,1)) = (unit)(1 — Oy. Therefore g = (1 — or YIY. Similarly, g' divides Y(x + C iy) — C(x + = (unit)(1 so (1 — or x. If g' (1 —) then g' x and g' y, which is impossible since (x, y) = l. Therefore = (1 — 0. But then x + y _= x + C'y 0 mod g, the second congruence being by the choice of g. Since x + y e 1, we have x + y 0 (mod p). But zP = xP + yP x + y 0 (mod p), so p I z, contradiction. The lemma is proved. Lemma 1.8. Let a G l[ ] . Then ocP is congruent mod p to a rational integer (note this congruence is mod p, so it is much stronger than a congruence mod 1 — C). PROOF. Let a = /3 0 ± b + (bp _ 2 CP 2 = bt; + )"

+ • • • + bp _ 2 V -2 . Then aP

bt + (b 1 OP + • • •

+ • • • + 14 _ 2 (mod p), which proves the lemma.

D

Lemma 1.9. Suppose a = ao + ai Ç + • • • + op _ 1 CP -1 with ai least one ai = O. If n E 1 and n divides a then n divides each al .

El

and at

PROOF. Since 1 + + • + CP - 1 — 0, we may use any subset of 11, C, CP -1 1 with p — 2 elements as a basis of the l-module Z []. Since at least one ai = 0, the other aj's give the coefficients with respect to a basis. I=1 The result follows. We may now finish the proof of Theorem 1.1. Consider the equation p— 1

fl0 (x + V))) =

(z

as an equality of ideals. Since the ideals (x + C iy), 0 i p 1, are pairwise relatively prime by Lemma 1.7, each one must be the pth power of an ideal: —

(x + (;ry) = A. Note that Af is principal.

6

1 Fermat's Last Theorem

Now comes the big step: since the class number of CP( ) is assumed to be not divisible by p, the ideal A i must be principal, say A i — (oc i). Consequently (x + C iY) same as we could have obtained under the stronger assumption that Z[C] has unique factorization, rather than just class number prime to p. Let i =-- 1 and omit the subscripts, so x + Cy =-- eaP for some unit E. Proposition 1.5 says that e =-- Cr e l for some integer r and where i — Lemma 1.8 says that there is a rational integer a such that a" a a (mod p). Therefore x + 0) = je l aP a Cre i a (mod p). Also x + C- l y = C - re, a (mod 15) = c - rg l a (mod p) since 5 — a and p =-- 15. We obtain

- r(x + 07) ----= C r(x + C - 1 Y) (mod p) or x + cy ___ 2rx _ 2r-

'y

_.=

0 (mod p).

(*)

If 1, , c2r , v2r ç 1 are distinct, then (since p > 5) Lemma 1.9 says that p divides x and y, which is contrary to our original assumptions. Therefore, they are not distinct. Since 1 0 C and C2r o c 2r- 1, we have three cases:

(1) 1 =-- Cr. We have from (*) that x + cy — x — c - 'y 0 (mod p), so, cy V - 'y 0 (mod p ) . Lemma 1.9 implies that y=-- 0 (mod p ) , contradiction. (2) 1 ___, 2r- 1 or, equivalently, C = c, 2r . Equation (*) becomes —

Y

(x — y) Lemma 1.9 implies x — y---- 0 (mod p), which contradicts the choice of x and y made at the beginning of the proof. (3) c2r - 1 • Equation (*) becomes

x—

0 (mod p),

so x ==- 0 (mod p), contradiction. The proof of Theorem 1.1 is now complete. Ill

Remarks. (Proofs for the following statements will appear in later chapters). The obvious question now arises: How can one determine whether or not p divides the class number of Q(0? Kummer answered this question quite nicely. Define the Bernoulli numbers fin by the formula t

'

t" B

et — 1 =n=0 1 n

n!

—4, B2 = i, B 3 =

0 for k > 1, B4 = -310, B 6 = 412 , B 8 = — 310, B lo = 656, B12 — — 2679310)* Then p divides the class number of O() if and only if p divides the numerator of (for example, B o -- 1, B 1 =--

0 and in fact

B 2k+ 1 =

7

Exercises

some Bk, k = 2, 4, 6, .. . , p — 3. For example, 691 divides the numerator of B12 so 691 divides the class number of Q(C 691 ). If p does not divide the class number of Q( ) then p is called regular, otherwise p is called irregular, The first few irregular primes are 37, 59, 67, 101, 103, 131, 149, and 157 (which in fact divides two different Bernoulli numbers). The irregular primes up to 1 25000 have been calculated by Wagstaff. Ap39 % of primes are irregular and e — 1/2 = 61 % are proximately 1 — e - 1/2 regular. There are probability arguments which make these empirical results plausible. It is known there are infinitely many irregular primes, but it is an open problem to show there are infinitely many regular primes. Moreover, it is not even known whether or not Fermat's Last Theorem, even in the first case, holds for infinitely many p. One may also ask how often 2[C] has unique factorization, or equivalently when the class number is equal to one. It turns out that the class number grows quite rapidly as p increases, so there can only be finitely many p for which there is unique factorization. In fact, Montgomery and Uchida proved (independently) that the class number is one exactly when p < 19. To finish this chapter we shall show that O( 23) does not have class number one. It is known that Q(,/— 23) Q(C 23 ). For a proof, see the Exercises for the next chapter, or use Lemma 4.7 plus Lemma 4.8. The prime 2 splits in Q(\/ — 23) as AA, where h - (2, (1 + .\/— 23)/2) (see the Exercises). Let Y be a prime of CD(C 23 ) lying above h. We claim that Y is nonprincipal. The norm of .(93 from Q(C 23 ) to CP( \/ — 23) is hf, where f is the degree of the residue class field extension. In particular, f divides deg(Q(C 23)/Q(\/— 23)) = 11, so f 1 or 11 (actually, f = 1). Since h is nonprincipal and h 3 is principal, Att is nonprincipal. Therefore hf cannot be principal. But if P is principal, so is its norm. Therefore is nonprincipal, so 2 23] cannot have unique factorization. [

NOTES

The proof of Theorem 1.1 is due to Kummer [2]. At present, the first case has been proved for p < 6 x 10 9 (Lehmer [4]) using the Wieferich criterion: if 2P - I # 1 mod p2 then the first case is true. For more on Fermat's Last Theorem, see Vandiver [1] and Ribenboim [1]. EXERCISES

1.1. (a) Show that the irreducible polynomial for („ is X (P - )P" X 4P- 2)P" + • ' ' X P" + 1 (one way to prove irreducibility: evaluate the polynomial as geometric series to get a rational function, change X to X + 1, rewrite as a polynomial reduced mod p, then use Eisenstein). (b) Show the ring of integers of (13(Cr) is

1.2. Suppose p 1 (mod 3)_ Using the fact that 2, contains the cube roots of unity, show that xP + yP zP (mod p"), p,rxyz, has solutions for each n > 1.

8

I Fermat's Last Theorem

1.3. Using the fact that [ —5] has class number 2, show that x2 + 5 = y3 has no solutions in rational integers. 1.4. Show that the ideal 1 = (2, (1 + v/ —23)/2) is nonprincipal in l[(1 + but that its third power is principal. Also show that le le = (2). 1.5. Show that the class number of a2 (423) is divisible by 3 (in fact, it is exactly 3, but do not show this).

Chapter 2

Basic Results

In this chapter we prove some basic results on cyclotomic fields which will lay the groundwork for later chapters. We let denote a primitive nth root of unity. First we determine the ring of integers and discriminant of (LK„). We start with the prime power case.

Proposition

2.1.

The discriminant of (Ç) is p pn - l(p n _ n — 1 )7

where we have — if pn = 4 or if p

3 (mod 4), and we have + otherwise.

PROOF, From Exercise L1, the ring of integers is Z[4,2], so an integral basis is { vin)-1}. The square of the determinant of ()0
0<j<j)",j,j

gives the discriminant. But this determinant is Vandermonde, so it equals

11 (4.2 — Ckrn) =

(root of unity) • n (1 _ k<j ptik

0
Since (1 — C;„a) —(1 — 4,), we may include all pairs j, k with j get the discriminant det(.) 2 — (root of unity) • P

k to

11 (1 — 0<j,k
We immediately see that the discriminant, up to sign, must be a power of p. Let y denote the valuation corresponding to the prime ideal (1 — of As in the first chapter for the case n 1, we have (1 — Cp Otij "P" I = (p). It follows that v(p) = (p — 1)p n' and y(1 —) = "1 for 1 in n. 9

10

2 Basic Results

Consequently, if k j (mod pm) but k j (mod pm + 1 ), we have u( 1 — V:„7- j) = pm since 1;3,7 is a p'th root of unity. Fix j with p,/j. It is easy to see that there are (p — 2)p' of k with j k (mod p), and (p — 1)p n - values of k such that j k (mod pr) but j k (mod p i + Also, there are (p — 1)p"' possibilities for j. Therefore, the valuation of the discriminant is n—1

OP"- '[(P = (p

1 )pn

2)p I

[pn 1 (pn

E= (13 — n

OP'

• Pr]

_ 1)] .

Since v(p) = (p — 1)p" - 1 , we must have the discriminant = p P n-I {Pr1 - 11 — 1). To determine the sign, we use the following lemma. Lemma 2.2. Let k be a number field with r 2 pairs of complex embeddings. Then d(k) = discriminant of k has sign (-1)r2 .

PROOF. Let loc i ,

, ct,,} be a z-module basis for the ring of integers of k. Then d(k) = (det(4),,, i) 2 ,

where a runs through all embeddings of k into C. If a is a complex embedding, then Fr is another embedding, where the bar refers to complex conjugation. Therefore det(4) = ( —

det(aD,

since r 2 pairs of rows are interchanged. If r 2 is even then det(4) is real, so d(k) > 0. If r 2 is odd, then det(4) is purely imaginary, so d(k) < 0. This proves the lemma. Returning to the proof of Proposition 2.1, we note that r 2 = +(p 1)p which is even unless pn = 4 or p E2 3 (mod 4). This completes the proof. —

1,

Now let m = fl p7 be a positive integer. We shall always assume that m 0 2 (mod 4), since if ni is odd then CD(C 2m) = G(C„). Clearly O(Cm) is the compositum of the fields Proposition 2.3. p ramifies in G()<,› p divides

PROOF. If p divides ni then 0() g CE,n). Since p ramifies in 0(4), it ramifies in CD(Cm). Conversely, suppose p does not divide ni = 11 P7'. Then p is unramified in each (13() since p does not divide the discriminant. Therefore p does not ramify in the compositum, which is CD(C). This completes the proof. Note that the proposition implies that p divides the discriminant if and only if p divides in.

II

2 Basic Results

Proposition 2.4. If (m, n) = 1 then O ( ,i ) n OG.) = O.

Let K = Cl ( .) n C(J,i ). If K la then there is some prime, call it p, which ramifies in K (this follows from the fact that I d(K)1 > I. See Lemma 14.3). By the previous proposition, p I m and p n, which is impossible. Therefore K = PROOF.

Theorem 2.5. deg(CI(,)/a) = 0(n) and Gal(13(,)/0) (Z/nZ) X , with a mod n corresponding to the map fl —o

PROOF. Since (Km ) is normal over G, Proposition 2.4 implies that if (mn, n) = 1 then deg(O( 1 )/0) = deg(C(.)/0) • deg(O()/Q). It therefore suffices to evaluate the degree for prime powers, which we have already done (Exercise 1 and 0(mn) = 0(m)(19(n) for (m, n) = 1, we 1.1). Since çb(pn) = (p — obtain deg(O(,)/Q) q5(n). It is a standard exercise in Galois theory to show that Gal(egnyla) is a subgroup of (7Lin.7)x . Since they are of the same order, they must be equal. This completes the proof. LI

Theorem 2.6. ZE n] is the ring of algebraic integers of CP(). PROOF. We need the following result (for a proof see Lang [1], p. 68): Suppose K and E are two number fields which are linearly disjoint (<=> deg(KE/G) = deg(K/0) • deg(E/0)) and whose discriminants are relatively prime. Then KE = 6K6E, where 6F denotes the ring of algebraic integers in a field F. Also d(K E) = d(Kyieg (E10) d(E)deg(Km) . Applying this result to cyclotomic fields, using the fact that Theorem 2.6 is true in the prime power case, we obtain the theorem for all n. LI We now compute the discriminant of O ( ,i). The above-mentioned result may be written as logld(KE)I _ log AK) log d(E)1 deg(KE/0) deg(K/Q) deg(E/0) . Therefore if n = fl Ki we have

z - i(p i ai

log d(GGO) n

0()

(ai

=

ai — 1) (log p 1 )/0(Ki)

1 (log p i) = log n — p i — 1)

E (log p)/(p — 1). pin

We obtain the following (the sign is determined from Lemma 2.2).

12

2 Basic Results

Proposition 2.7.

n°(") p0(n)/(P - 1) .

d(o()) = ( — 1)°(")/2

One difference between the prime-power case and the case of general n is given in the following. Proposition 2.8. Suppose n has at least two distinct prime factors. Then 1 — is a unit of[] and n0<< (1 — C4) = 1. ci,n)= 1

• • + X + 1 = F17 :(1 (X — fli ), we may let Since X' + X n- 2 X = 1 to obtain n = ][174 (1 — If pa is the exact power of p dividing n then, letting j run through multiples of n/pa, we find that this product contains Fro' ( 1 — Vp ,)= pa. If we remove these factors for each prime dividing n, we obtain 1 = where the product is over those j such that is (1 not of prime power order. Since n is not a prime power, 1 — C„ appears as a factor in this product; hence is a unit. But Flu,n)=1 ( 1 — is the norm of (1 — C n) from (1-,1)( n) to 11, therefore equals a unit of Z, namely + 1. Since complex conjugation is in the Galois group, the norm of any element may be written in the form cer, which is positive. It follows that nu,n)= (1 — Vri) = + 1, which completes the proof. We remark that the proof works even if n 2 (mod 4). PROOF.

H

a

One might ask what the irreducible polynomial for the nth cyclotomic polynomial

On(X) =

(jn)= 1

looks like. We define

-

Since deg(Q()/0) = O(n) = deg (D„(X), it follows that lDn(X) is the irreducible polynomial for Cn . Also, (1)n(X) E Z[X] since the coefficients are rational and also are algebraic integers. In addition, it is easy to see that - 1 = H od(X)• din

The first few cyclotomic polynomials are (1) 1 (X) = X — 1,

(I) 3 (X) = X 2 + X + 1, (I) 2 (X) = X + 1, 04(X) = X 2 + 1.

All these have coefficients + 1 and 0; however, this is not true in general. By choosing n with many prime factors one can obtain arbitrarily large coefficients. One use of cyclotomic polynomials is to give an elementary proof of a special case of Dirichlet's theorem on primes in arithmetic progressions (Corollary 2.11).

13

2 Basic Results

Lemma 2.9. Suppose p,rn and a e Z. Then p lt.„(a) the multiplicative order of a mod p is n (Le., an 1 (mod p) and n is minimal). PROOF. Suppose pl(15n (a). Since X n — 1 =1"-- Mi n 4) d (X), we have a" -a- 1 (mod p). Let k be the order of a (mod p). Then kln. Suppose k < n. As above, we have 0 ak — 1 ndik (1),(a) (mod p). Consequently 04(a) 0 (mod p) for some d 0 . Therefore a" — 1 = 0 fl (a) d0(a) . (other factors) 0 (mod p 2 ). Since (I)„(a + p) (1)„(a) 0 (mod p), and similarly for Odo , we also have (a + p)" — 1 0 (mod 1;1 2 ). Therefore 0 (a + p)" — 1- -- a" + npa' — 1 npa" - 1 (mod p 2 ). Since pj/na, this is impossible. Therefore k = n. Conversely, suppose a" — 1 = 0 (mod p). Then (I),(a) 0 (mod p) for some d In. But if d < n then the order of a would be less than n since we would have ad — 1 0 (mod p). Therefore (I(a) 0 (mod p), and the proof is complete. Ill Proposition 2.10. Suppose p,rn• Then p divides (a) for some a EZ <=> p = 1 (mod n). PROOF. If pl(1)„(a) then a (mod p) has order n. Since the order of an element divides the order of the group, n divides p — 1. Conversely, if p 1 (mod n), then there is an element a (mod p) of order n, since (Z/pZ)x is cyclic. Therefore 0 Corollary 2.11. For any n

1 there are infinitely many primes p

1 (mod n).

PROOF. Suppose there are only finitely many, say p i , ... , pr . Let M = np i - — p r and let N e Z. Then (1), (NM) (1)„(0) _= + 1 (mod M), therefore mod p i and mod n (q) „(0) = +1 since it is a root of unity by the definition of (1),(X)). In particular 11),(NM) is not divisible by p i and none of its prime factors divides n. As N —)- co, (I)„ (NM) co, so for large N we have 0„(NM) 0 +1. Therefore there is a prime p dividing (13.„(NM). By the proposition, p 1 (mod n). From the above, p 0 pi , 1 s i r. Therefore we have obtained a new prime. This completes the proof. 0 We remark that Euclid's classical proof is just the above proof using 02 (X). Similarly, 04 (X) is used to obtain primes of the form 4n + 1. Now we turn our attention to the splitting of primes in cyclotomic fields. First we need the following useful result. Lemma 2.12. Suppose p,i'n and let Y be a prime of() lying above p. Then the nth roots of unity are distinct mod Y . PROOF. The result follows immediately from the equation n- 1

n=

n 0 - w. J-0

0

14

2 Basic Results

Note that this result is not true for pin. In CP(;) we have ;

1 mod

( 1 - Cd. Assume p,'n and let Y lie above p in CP(C„). The Frobenius automorphism of CP ( „) is defined by

up x

xP (mod Y)

for all

XEZ

[Ca

Since orp C, is an nth root of unity, Lemma 2.12 implies that u p C„ = U. The order of o-p is the degree of the residue class extension Z[ n] (mod Y)/ Z (mod p). Now orf, = 1 <:,› 0) = C, -4- = C, .4 ,› pf 1 (mod n). Since p is unramified in C ( „), it is a standard fact from algebraic number theory that the degree of the residue class extension multiplied by the number of primes above p equals the degree of the extension 0( ,)/ 0 . We have therefore proved the following. Theorem 2.13. Suppose p,rn and let f be the smallest positive integer such that pf 1 (mod n). Then p splits into g = 0(n)/ f distinct primes in CP ( „), each of which has residue class degree f. In particular, p splits completely <=> p 1 (mod n).

El

Remarks. The fact that p splits completely if and only if p 1 (mod n) means that CP(C„) is the ray class field modulo nco in the sense of class field theory. Since every abelian extension of the rationals is contained in some ray class field, this proves the celebrated Kronecker-Weber theorem: Every abelian extension of CP is contained in some CP ( ,T). Later we shall give a proof of this result without assuming class field theory. We also note that the splitting type of p depends only on its congruence class modulo n. This is a characteristic of abelian extensions which can be proved using class field theory. Theorem 2.13 is sometimes called the cyclotomic reciprocity law. One purpose of a reciprocity law is to give nice conditions for when a prime splits. For example, the quadratic reciprocity law (see Exercises) allows one to change a statement about g splitting in (13(j)), which depends on whether or not p is a square mod g, into a question of whether or not g is a square mod p. If one wished to make a list of the primes which split in CP( \/P), then checking whether or not p is a square mod g for each g would be rather laborious. However, after making an initial list of squares mod p. one would find checking each g mod p to be rather easy. The cyclotomic reciprocity law has the same advantages. As an example for the theorem, let p = 2 and n = 23. Then 2 11 1 (mod 23), so f = 11. Therefore 2 splits into two factors in 0 ( 23 ). But we already know that 2 splits as Y - in C1( \/ -23), where Y -= (2, (1 + \./- 23) 7 2) (see Exercise 1.4). Going from 0(\/ -23) to 0( 23 ), Y and must remain prime. Therefore (2) = (2, (1 + .,/ - 23)/2)(2, (1 -23)/2) is the explicit factorization of (2) in O(C23 ). As shown at the end of Chapter 1, neither of these ideals can be principal.

15

2 Basic Results

We can also use Theorem 2.13 to treat the case p n, since e() is the compositum of the linearly disjoint fields (Cn/p a) and CEO, where pa is the exact power of p dividing n. We determine f and g for CPG.) by Theorem 2.13 and then note that p is totally ramified in CP Gp „), with ramification index e = (p — Op' 1 . Therefore, the ramification index, residue class degree, and number of primes above p in (1::PG,i) must be at least e, f, and g, respectively. Since efg = deg(n)/Q), these e, f, g must give the correct answers for the full extension. It is now easy to see that C:P(,itp.) is the inertia field and that if we identify Gal(P()/l) with (Z/nZ)x (Z/paZ)x C) (2/(n/ pa)Z)x , then the inertia group for p is (/pa) x and the decomposition group is generated by (/pa)X and p (mod(n/pa)1). It is possible, in theory, to give explicit generators for the prime ideals lying above a rational prime. We need the following result (for a proof, see Lang [1 ], p. 27). Proposition 2.14. Let A be a Dedekind domain with quotient field K, let E/ K be a finite separable extension, and let B be the integral closure of A in E. Suppose B = A[a] for some a E E and let f (X) be the irreducible polynomial for a over K. Let be a prime ideal of A. Let f (X) denote reduction modulo Suppose

f(X) = P 1 (X)el

Pg(X)eg

is the factorization of f (X) mod into powers of distinct monic irreducible polynomials over (A/ M[X]. Let Pi(X) E A[X] be a monic polynomial which reduces mod to Pi(X). Let Pi be the ideal of B generated by and Pi(a). Then P is a prime ideal of B lying over 1 e 1 is the ramification index, the Pi's are distinct, and ,

=

•••

Pegg

LI

is the factorization of1 in B.

Applying the proposition to our case, we factor the cyclotomic polynomial mod p as (1)„(X) = 131(X)e'

P9(X)

(actually, e l = e2 = • -= eg since we are working with a Galois extension). Then Pi = (p, Pi()). In the special case p 1 (mod n), the ideals lying above p are of the form (p, a — where a (mod p) is of order n. When g is small, in particular g = 2, and p is unramified, then perhaps it is easier to determine the generators for the primes in the splitting field. Since these primes are then inert the rest of the way up to the full cyclotomic field, these generators work for WO also. This is the method we used for p = 2 in O ( 23 ) previously. Finally, we discuss subfields of O ( n ). The most important for our purposes is the maximal real subfield EK„ + „ 1 ), denoted EK,i) + . The extension 13G,i)/C:C„ + i: 1 ) is of degree 2 since the lower field is the fixed field of

16

2 Basic Results

complex conjugation. Alternatively, C„ is a root of X 2 — (C„ + C i; ' ) X + 1. One interesting fact is the following.

Proposition 2.15. (a) Ifn = pm then (1)(,2)/(lag,,) + is ramified at the prime above p and at the archimedean primes, and unramified at the other primes. (b) If n is not a prime power, EP(C„)/0(C„)+ is unramified except at the archimedean primes. PROOF. In both cases, the archimedean primes ramify since CP(C„ + C,; ') is totally real and CP(C„) is totally complex. Part (a) is true since p is totally ramified in CD(Cp-) and is the only ramified finite prime. Part (13) may be proved as follows. Let p and q be two different prime divisors of n (if p or q is 2, then use 4 instead of 2). Then 4 and q are in U(C,,), but not in (I)(C„ + Ç ,, 1 ), since the latter field is real. Since EP(C„)/Q(C,,)+ is of degree 2, we must have 1 , = Q(C„ + Adjoining 4 allows ramifica= CD(C„ tion only at primes above p and the archimedean primes (if L/K is unramified at a prime g, then LF/KF is unramified at primes above g. Here F = (1)(C, + L = Q(C p), K = CO. Similarly, adjoining Cci allows ramification only above q and at the infinite primes. Therefore there is no ramification at finite primes, so the proof is complete. Eli We now look at subfields of 0(4) in more detail. Let g be a primitive root mod p, let e be a fixed divisor of p — 1, and let f = (p — 1)/e (there is no relationship with ramification indices, etc.). Define f-1

E

i = 0, 1, .

e — 1.

=0

These numbers are called periods and their significance is as follows. Let a- be the automorphism of 0(C p) which maps cp to Ç. Since g is a primitive root, generates the Galois group. The subgroup of order f is H = {1, a',

,

which corresponds to {1, ge, . . . , g -1) } ç (Z/pZ)x . Consequently, {gei'10 <jf — 1} is a coset of this subgroup. It is easy to see that H fixes ?h. Also (Oh) = rj i+ for 0 e — 2, and o-(ne _ 1 ) = no . So ni has exactly e conjugates under Gal(Q(( p)/Q). It follows that ni , for any i, generates the subfield of 0(4) of degree e over Q. For example, if e = (p — 1)/2, f = 2, then = (f: + Ç, which in the case i = 0 gives us cp + Ç '• One may ask whether or not 7/[] is the ring of integers of Q(q 1). In general, the answer is no (see below), but for f = 2 the answer is yes. In fact, we have the following.

Proposition 2.16. ZR„ + C,;-1 ] is the ring of integers of Q(C„ + C i; 1 ). PROOF. Suppose a = ao + a i gn + + + aN(Ç + Ci; 1)N is an algebraic integer, with N < 10(n) — 1 and with ai c Q. By removing those terms with ai c Z, we may assume aN 0 Z. Multiplying by C,,N and expanding the result

17

Exercises

as a polynomial in C„, we find that CnN = aN + • • ' a N UN is an algebraic integer in CD(C„), therefore in Z[C]. Since 2N (1)(n) — 2 0(n) — 1, •. , ,11v 1 forms a subset of a i-basis for the ring Z[C,J. Therefore aN E Z. This completes the proof. 0 For the case of i o 0, we may take Galois conjugates of everything to find that Z[Ili] is the ring of integers of OW ( = 0(1 0)) for f = 2. There are many counterexamples for f > 2. Several may be obtained in a way similar to the following. Let p = 31, f = 5. Since 2 5 1 (mod 31), 2 splits completely in the extension over Q of degree 6, which is 0(0. Suppose the ring of integers of QM has the form Z[a] for some a. Let f (X) be the irreducible polynomial for a over Q. By Proposition 2.14, f (X) must factor as a product of 6 distinct linear factors over Z/2,7. But X and X + 1 are the only linear polynomials mod 2, so this is impossible. Therefore the ring of integers cannot be Z[a]; in particular, it cannot be NOTES

Most of the results in this chapter are due to Kummer. For an application of the cyclotomic splitting laws to primality testing, see AdlemanPomerance-Rumely [1] and Lenstra [7]. EXERCISES 2.1. Show CP(Cp ) contains a quadratic subfield (p is an odd prime). Using the fact that only p can ramify, show that this field must be (1)(,/+p), where we have + if p 1 (mod 4) and — if p 3 (mod 4). 2.2. Show that 0((8 ) contains 3 quadratic subfields. Determine which ones they are. 2.3. Show that the only roots of unity in O ( ,) are of the form + 2.4. Let p be an odd prime. Show that there is a unique subfield K of CP(Cr z) of degree p over O. Show that 2P -1 1 (mod p 2 ) if and only if 2 splits completely in K (the primes 1093 and 351 1 are the only primes known which satisfy this relation. It can be shown that if 2' (mod p 2 ) then the first case of Fermat's Last Theorem holds.). 2.5 (a) Determine explicitly the factorizations of 2, 3, 5, 7, and 1 1 in O( ( 20 ), and show that all the prime ideals lying above these primes are principal (Hints:The quadratic subfields of O ( 20) are (1)( N/ 0(0, 0(0). The prime 2 may be treated via 0(4 For 3 and 7, observe that w + co; = 3 and oil' + w 7, where w 1 = (1 + N/75)/2 and w 2 = (1 — N.F5)/2. For 5, show that the norm from O ( 2 0) to O ( 5 ) of ( 5 + 1) + Ci• i is 1 — . For 11, first determine its prime factors in it)((5 )). (b) A theorem of Minkowski states that in every ideal class of a number field K, there exists an integral ideal A satisfying n! (4 2y

NA — — n" it

18

2 Basic Results where NA is the norm of A, n is the degree of K over 0, d(K) is the discriminant, and r 2 is the number of pairs of complex embeddings (see Lang [I] P. 119). Show that all ideals A satisfying this inequality are principal, so Q(C 20) has class number 1 (note however that it has a subfield C(1— 5) of class number 2).

2.6. Let p and q be distinct odd primes. Let (p/ q) = + 1 if x 2 p (mod q) has a solution, and (p/q) = I otherwise. The quadratic reciprocity law states that —

(2 ) = (9-)

if either p 1 (mod 4)

or

1 (mod 4)

q

P

and

3 (mod 4)

if both p

and

q

3 (mod 4).

Justify the steps in the following proof. (a) Assume p 1 (mod 4), q arbitrary. Then (2 ) = 1 <4. x 2 — x +

1 — p 4

0 (mod q) is solvable (I

q splits in 0( )

+2\4)

Uq (= Frobenius for q in C/(p)) fixes 0(fp) q a2 (mod p) is solvable

(-1-) = 1.

The same argument works if q 1 (mod 4) and p is arbitrary. (b) Assume both p 3 (mod 4) and q 3 (mod 4). Imitate part (a) to show (p/q) = — (q/ p). (Since — 1 is not a square modulo a prime congruent to 3 mod 4, we have x 2 p (mod q) has a solution -1r> x2 p (mod q) does not have a solution). —

2.7. Using the techniques of the previous exercise, show that x 2 solution if p + 1 (mod 8) and does not have a solution if p

2 (mod p) has a +3 (mod 8).

2.8. (Lenstra). This exercise gives another proof of Proposition 2.7. (a) Let 43(X) be the nth cyclotomic polynomial. Show that (D ( x)

1

-1

din

where it(d) is the M6bius function : ti(d) = 0 if d is not square-free; if d is square-free then ti(d) = (— 1)n where i is the number of prime factors of d. A useful fact: Edin p(d) = 1 if n = 1, =0 if n > 1. (X). Show that (1)(C„) = nt", n / (b) Let X" — 1 = (1)„(X)Since Cn generates the ring of integers of Q(C), (1(C„) generates the different, and its norm to gives the discriminant. (c) Show that 14-1, „) = (unit) • lipin ( n/P — 1). (d) Show that Norm( p — 1) = poo)/(P - 1 ). (e) Deduce Proposition 2.7. (

Chapter 3

Dirichlet Characters

In this chapter we introduce the basic facts about Dirichlet characters. We then show how they may be used to obtain information about the arithmetic of number fields. As a result, we show how to obtain ideal class groups containing prescribed subgroups. A Dirichlet character is basically a multiplicative homomorphism z:(ZInZ) —> C x . If n I in then x induces a homomorphism (Z/mZ)" —> C x by composition with the natural map VP-117/r —)- (Z/Z`Z)x . Therefore, we could regard x as being defined mod in or mod n, since both are essentially the same map. It is convenient to choose n minimal and call it the conductor of x, denoted f or h. Let X: ( 7/ /8 7/ ) ' —> C ' be defined by x(1) = 1, x(3) = — 1, z(5) — 1, x(7) = — 1. Since x(a -1- 4) = x(a), it is clear x may be defined mod 4 by x(1) = 1, x( — 3) — — 1. Since 4 is minimal, fx = 4-

EXAMPLES.

(1

)

(2) Let x; ( 7/ 16 7/ ) ' —> C ' be defined by x(1) = 1, x(5) = — 1. Then x is induced by the map ( 7/ /3 7/ )" C ' which sends 1 to 1 and 2 to —1. Therefore fx = 3.

It is convenient to classify characters into two types: if x( — 1) = 1 then x is called even; if — 1) = — 1 then x is called odd. Both of the above examples are odd characters. Many times we regard x as a map Z —> C by letting x(a) = 0 if (a, fx) # 1. It is therefore important to make a convention regarding the modulus of definition of x. We shall always regard x as being defined modulo its conductor. Such characters are called primitive. Essentially, this choice makes x(a) = 0 happen as little as possible. Also x is then periodic of period h. (

19

3

20

Dirichlet Characters

In Example 2, the fact that r defined mod 6 did not have period 3 can be explained by the fact that 6 contains the extraneous prime 2, so all even a had x(a) = 0. In the following, when we talk of the characters of (1/n7L)X, or of the characters mod n, we shall be including characters of conductor dividing n, for example the trivial character of conductor 1. The convention that all characters are primitive plays a part in the multiplication of characters. Let r and ip be Dirichlet characters of conductors fx and 40 . We define rt,P as follows. Consider the homomorphism

Y:(Zlicm(f x , foW) x —> defined by y(a) = x(a)tP(a). Then rtP is the primitive character associated to y.

EXAMPLES. (3) Define r mod 12 by r(1) = 1, r(5) — 1, z(7) = — 1, r(11) -- 1 and define ip mod 3 by tP(1) = 1, tP(2) = —1. Then rtP on (/12)X has the values xtP(1) 1, rt 1' (5) = r(5)tp(2) -- 1, rtp(7) = — 1, p(11) = -1. It is easy to see that rtP has conductor 4 and satisfies xtp(1) = 1, 20(3) = — 1. Note that r0(3) = — 1 x(3)0(3). (4) Let r be any character and let tp = (complex conjugate). Then ts(a) = Aar- if (a, fx) = I. It follows that rk is the trivial character: xX(a) — 1 for all a (including a = 0).

(5) If (f, 4) = 1 then /a, = fx fo (see Exercises). The advantage of using primitive characters become evident when one takes a product of several characters of various conductors, since otherwise the modulus of definition could grow quite rapidly. Also, with our convention, there is only one trivial character, rather than one for each modulus. It is sometimes advantageous to think of Dirichlet characters as being characters of Galois groups of cyclotomic fields. If we identify Gal(O(„)/Q) with (Z/nZ) X then a Dirichlet character mod n is a Galois character. The Examples 1 and 2 above may be interpreted as follows: (1) The kernel is 1 (mod 8) and 5 (mod 8). In the Galois group these form Gal(1(C 8 )/0(C4 )), so r is a character of the quotient of (Z/8Z )X by this subgroup. Consequently r is a character of Gal(G( 4 )/(13) (Z/4Z) X . (2) In this case EK6 ) = U(C 3 ) so a character mod 6 and a character mod 3 are characters of the same Galois group. In general, let r be a character mod n, hence a character of Gal(a( n)/(20). Let K be the fixed field of the kernel of r. Then K E e(C n ), and if n is minimal then n = fx . The field K depends only on r and is called the field belonging to r. More generally, let X be a finite group of Dirichlet characters. Let n be the least common multiple of the conductors of the characters in X, so X is a subgroup of the characters of Gal(e(C„)/0). Let H be the intersection of the

21

3 Dirichlet Characters

kernels of these characters and let K be the fixed field of H. Then X is precisely the set of homomorphisms Gal(K/Q) C X (see below). The field K is called the field belonging to X, and we have deg(K/Q) = order of X; in fact, X Gal(K/Q) (see below). If X is cyclic, generated by x, then K is precisely the same as the field belonging to x mentioned above.

(6) If X is the group of characters of (Z/nZ)" satisfying 4-1) = +1, then complex conjugation (C„ („- '} is in the kernel of each x E X. The field associated to X is Q(C,, (,; 1 ), which is the maximal real subfield of Q(C.). Similarly, if x is any character then the field belonging to x is real if and only if x(- 1) + 1. EXAMPLES.

(7) The character x of example (3) must correspond to a quadratic subextension of O (C i2) = G(C 3 )0(C4) G(‘..• — 3)U(i). The three quadratic subfields are (11(x.: — 3), Q(i), and Q( The first two choices would force y to have conductor 3 and 4, respectively, which is not the case. Therefore, 0(v 3) is the field belonging to x. Alternatively, since x( — 1) = 1,_the field belonging to x must be real. The fact that the discriminant of U(0) is 12 can be used to explain the fact that x has conductor 12 (see Theorem 3.11). The preceding notions can be put in the setting of characters of finite abelian groups, which we now review. Let G be a finite abelian group and let denote the group of multiplicative homomorphisms from G to C X . Lemma 3.1. If G is a finite abelian group, then G

(noneanonically).

G may be written a direct sum of groups of the form Z /in. Therefore is the product of groups of the form (i/mZ) - . But if y E (Z/MZ)— , then x(1) determines x (remember i/mZ is additive). Since x(1) can be any ,nth root of unity, the lemma is true for Z/mZ, hence for G. III PROOF.

Corollary 3.2.

6

G (canonically).

Let g E G. Then g: Ô —) Cx by g: x —) x(g). Suppose x(g) — 1 for all x E Ô. Let H be the subgroup of G generated by g. Then G.' acts as a set of distinct characters of G/H. But there are at most # (G/H) of these_ by the lemma. Therefore, H= 1, so g = 1. Consequently G injects into O. Since # (G) = # (G) = # (6), we are done. El PROOF.

Many times il is convenient to identify t

Gx (g,

=

G. We have a natural pairing

-> Cx

x(g).

This pairing is nondegenerate: if x(g) = 1 for all x c Ô then g = 1 by the above argument. If x(g) = 1 for all g e G then, of course, y. = 1.

22

3 Dirichlet Characters

Now let H be a subgroup of G. Let H L = tx e 61x(h) = 1, Vh

G

HI.

We clearly have a natural isomorphism H L -= (G/H). Proposition 3.3. il --' G/H - L.

By restriction we have a map 0 -. R. The kernel is H. It remains to show surjectivity. But #(H L ) = # (G/ H) = # (G/ H) = # (G)/ # (H). Therefore # ( R) = # (H) = # (G)/ # (H L ) = #( 6)/ #(H L ). The proposition follows. PROOF.

El Proposition 3.4. (HT = H (we equate 6 = G).

PROOF. As in the preceding proof, a straightforward calculation shows both groups have the same order. If h e H then h: x -, x(h) maps H L 1. Therefore H H11 . Therefore they are equal. III G, we may reverse the roles of G and 0 in all the above. The above results, with the exception of Lemma 3.1, hold for locally compact abelian groups. However, the proofs are more difficult since counting arguments cannot be used.

Remarks. Since

6=

We now return to Dirichlet characters. Let X be the group of Dirichlet characters associated to a field K. Then we have a pairing Gal(K/U) x X -) C x . Let L be a subfield of K and let Y = Ix e X I x(g) = 1, V g e Gal(K/L)).

Then Y = Gal(K/L)' = (Gal(K/C)/Gal(K/L)) -

= Gal(L/CPr.

Conversely, if we start with a subgroup Y c X and let L be the fixed field of 1/ -1- = tg E Gal(K/CP)1 AO = 1, Vx G Y},

then 11-1 = Gal(K/L), by Galois theory. Therefore Y - Y il = Gal(K/L) L = Gal(L/CP)". lt follows that we have a one-one correspondence between subgroups of X and subfields of K given by Gal(K/L) I 4- L Y-

fixed field of 17-1-.

23

3 Dirichlet Characters

This gives us a one-one correspondence between all groups of Dirichlet characters and subfields of cyclotomic fields, since any two groups may be regarded as subgroups of some larger group. Since Gal(L/0) is a finite abelian group we have Y = Gal(L/U) Gal(L/Q). This isomorphism, though useful, is noncanonical and is better expressed by the natural nondegenerate pairing

Gal(L/Q) x We leave the following statements as exercises: Let X i correspond to K. Then K2 , X 2 .4=> K i (1) X 1 (2) The group generated by X 1 and X2 corresponds to the compositum KiK2'

We now show how ramification indices may be computed in terms of characters. Let n = fl p. Corresponding to the decomposition (z/nz) X

Fi (

„z) x

we may write any character x defined mod n as

11 xp where xp is a character defined mod pa. If X is a group of Dirichlet characters, then we let Xp

{Xpl X E

In Example (3), x may be written as x = )( 2 Z3 where x2 is the character x6 of conductor 4 from that example and x3 = zit 1 = Theorem 3.5. Let X be a group of Dirichlet characters and K the associated field. Let p be a prime number with ramification index e in K. Then e =# (X p). Let n be the least common multiple of the conductors of the characters of X, so K c U(C,i). Let n = pa. m with p m. Form the field L = K(.) — K. Ug,n). (See diagram on p. 24). Then the group of characters of L is generated by X and the characters of (Z/n2)x of conductor prime to p (i.e., the characters mod m). Therefore it is the direct product of X p with the characters of U(Cni). Consequently L is the compositum of 1)(C,n) with the field F g 41Gp.) belonging to X. Since p is unramified in (LK.), the ramification index for p in K is the same as for p in L. Since L/F is unramified for p, this ramification index is the same as that for F, which is deg(F/CD) = This completes the proof. PROOF.

24

3

Dirichlet Characters

What happens in the proof is maybe best explained by an example. Consider the quadratic character z mod 12 corresponding to the field 0(0) = 0(.\/12). Then z = X2 X3 as above. The ramifications at 2 and 3 are occurring simultaneously, but we want to isolate, say, the prime 2. So we adjoin the character z3 and obtain the group generated by z2 z3 and z3 , which is also generated by z2 and z3 . We now have the picture CPO,

3)

CP(j3)

Q(\/-3)

X Z2

Z3

So we have " split " the field 0 (\./) so as to isolate the ramification at 2. Corollary 3.6. Let x be a Dirichlet character and K the associated field. Then p ramifies in K <=> x(p) = 0 (equivalently p I f). More generally, let L be the field associated with a group X of Dirichlet characters. Then p is unramified in L/0 <=> x(p) 0 0for all x e X. p ramifies in L/0 <=> X p 0 1 <=> iz e X with xp 0 1 <=>3z e X with LI Pi fx <=> 3 X e X with z(p) = O. PROOF.

Theorem 3.7. Let X be a group of Dirichlet characters, K the associated field. Let

Y= {X e X IX(P)

0},

Z = {X e

x I x(P) = 1}.

Then

e = [X: Y],

f = [Y : Z],

and

g = [Z: 1]

are the ramification index for p in K, the residue class degree, and the number of primes lying above p, respectively. In fact

X /Y -= the inertia group,

X /Z -= the decomposition group,

Y /Z is cyclic of order f.

25

3 Dirichlet Characters

PROOF. Let L be the subfield of K corresponding to Y. By Corollary 3.6, L is the maximal subfield of K in which p is unramified. It is a standard fact from algebraic number theory that L is then the fixed field of the inertia group, so the inertia group is Gal(K/L). Under the pairing Gal(K/Q) x X Cx we have Y = Gal(K/L) -I- by the correspondence between subgroups and subfields, so Y = Gal(K/0)7Gal(K/L) -1 = Gal(K/L) -Gal(K/L). Since the ramification index equals the order of the inertia group, we have e = [X: Y]. We now restrict our attention to the extension L/Q, which is unramified at p and has Y as its group of characters. Let n = icm fx (x E Y). Then p n and L ,j. The Galois group of (K„)/Q is (11n1) X and the Frobenius for p is p (mod n), which corresponds to the map C, The Galois group of L/0 is a quotient group of (11nl) X by Gal(QG„)/L). The Frobenius o-p for L/Q is just the coset of p in this quotient. But if Z E Y then 2( kills Gal((C,1 )/L). Therefore x(ap) = x(p), so x(ap) = 1 <,› x(p) = 1. Therefore Z = (u,> under the pairing Gal(L/Q) x Y C where denotes the cyclic group (of order f) generated by u p . Consequently

so f = [Y: Z]. Since the fixed field of the Frobenius is the splitting field for p,

it also follows that Z is the group of characters corresponding to the splitting field, which is of degree g; so g = [Z: Returning to the extension K/Q, we note that the splitting field is the fixed field of the decomposition group (which is generated by the inertia group and an extension of o-p to K). Therefore, as above, we obtain X/Z the decomposition group. This completes the proof. We now show how Theorem 3.5 may be used to construct unramified extensions. Proposition 3.8. Let G be any finite abelian group. Then there exist fields L and K such that G, and (a) Gal(L/K) (b) L1K is unramified at all primes (including the archimedean primes). We may also make L10 abelian and K10 cyclic.

26 PROOF.

3 Dirichlet Characters

By the structure theorem for finite abelian groups, we may write

G

0 1/nr 1

Z/n i

for some integers n 1 , , nr . Let p i , 'Pr be distinct primes satisfying pi 1 (mod 2n1). Since Gal(0((p ,)/Q) is cyclic of order pi — 1, there exists a character tPi of conductor p i and order p i — 1. Let xi -= "in'. Since (pi — 1)/n 1 is even, X( — 1) + 1. Let pr ., I be another odd prime and let X r + 1 be an odd character of conductor P r+ t (for example, (II r 1 ). Define X

X1 "

and let K be the corresponding field. Since Xi( — 1 ) • • • Xr( — 1 )Xr+ i( — 1 ) =

1,

K is complex, so every extension of K is unramified at the archimedean primes. Let X be the group generated by Ix 1 , . . . , L. + 1 } and let L be the corresponding field. Clearly X generated by x i for 1
X/<X> = <Xi, .., 1/n 1 1 C) . • • 0 1/nr i

G.

This completes the proof. Corollary 3.9. Given any finite abelian group G, there exists a cyclic extension K of ca such that the ideal class group ofK contains a subgroup isomorphic to G. We shall use one of the main results of class field theory : Let K be a number field and let H be the maximal unramified (at all primes, finite and infinite) abelian extension of K. Then Gal(H/K) is isomorphic to the ideal class group of K (the field H is called the Hilbert class field of K). By Proposition 3.8 there exists K, and a subextension of H/K with Galois group G. Therefore the ideal class group has a quotient group isomorphic to G. The corollary follows from the next lemma. PROOF.

Lemma 3.10. If A is a finite abelian group and B is a subgroup, then A contains a subgroup isomorphic to A/B. This result could be proved via the structure theory of finite abelian groups. Another proof is the following: PROOF.

A/B

(A/ Br

Â

A.

D

27

Exercises

It is unknown whether or not every finite abelian group occurs as the class group of some number field. We finish this chapter with a useful result which relates to the material of this chapter but which will be proved in the next chapter. Theorem 3.11 (Conductor—Discriminant Formula). Let K be the number field associated to the group X of Dirichlet characters. Then the discriminant of K is given by

d(K)

fx

.

x EX

This theorem can be very useful for computing discriminants of abelian number fields. For example, consider the real subfield of Q(Cp). The group of characters consists of the trivial character of conductor 1 and (p — 3)/2 other characters, all of conductor p. Since r2 = 0 we have d(O(C p + Cj 1 )) = (p- 3)/2

NOTES

The use Dirichlet characters to describe the arithmetic of an abelian field can be found in Leopoldt [9]. The above proofs of Proposition 3.8 and Corollary 3.9 are from Hasse [3]. For a generalization of Proposition 18 to non-abelian groups, see Friihlich [1]. It can be shown that any finite abelian group occurs as a subgroup of the class group of some cyclotomic field CaGn). See Cornell [1]. The techniques of Corollary 3.9 do not suffice for this, since the unramified extension constructed there is contained in a cyclotomic field ; hence it collapses when it is lifted (Proposition 4.11 fails). It is not known whether or not every finite abelian group occurs as the class group of some number field. However, every finite abelian e"-group is the (-Sylow subgroup of the class group for some number field (Yahagi [1]). The corresponding result for divisor classes of degree 0 is false for function fields over finite fields (Stichtenoth [1]). The techniques of Proposition 3.8 are part of what is known as genus theory. For more on this useful subject, see Ishida [1]. For more on the conductor—discriminant formula, see Hasse [2].

EXERCISES

3.1. Show that if (fx , fo,) = t then fx4, =

fo •

3.2. Suppose X i is a group of Dirichlet characters corresponding to the field K i ,i = 1, 2. Show that (a) X 1 X2 <=>K1 K2, and (b) the group generated by X 1 and X2 corresponds to the composition K1K2. 3.3. Let K be the field corresponding to the group X. Describe, in terms of X, the maximal abelian extension L of K which is abelian over Q and is unramified (over K)

28

3 Dirichlet Characters

at all primes. Do this for both the case where there is no ramifi cation at the infinite primes and the case where ramification at infinity is allowed. You may assume the Kronecker—Weber theorem, which says that all abelian extensions of Q lie inside cyclotomic fields.

3.4. Using the notation of Exercise 3.3, let K be a quadratic field. Give the field L explicitly in the form Q(1, (3, ...). Conclude that if the discriminant of K has s distinct prime factors then (a) divides the class number if K is imaginary, 22 (b) divides the class number if K is real. (This relates to the theory of genera. L is called the genus field.) 3.5. (a) Show that there are two quadratic characters of conductor exactly 8, one of which is even, the other odd. (b) Show that if f = 4 or an odd prime, then there is a quadratic character of conductor exactly f. (c) Let D be the discriminant of a quadratic field. Show there is a quadratic character of conductor 1 DI and show this character is unique unless 81D, in which case there are two such characters, one even and one odd. (d) Show that for any integer D there is at most one quadratic field whose discriminant is + D, unless 81 D, in which case there are can be two such fields, one real and the other imaginary. (e) Show that every quadratic field is contained in a cyclotomic field. If the discriminant is D, we may use Q(A); show that Q() with n < 1 DI does not contain the quadratic field. 3.6. (a) Let x be a nontrivial character. Show that

X x(a) =

O.

a=

(Hint: multiply by x(b), with x(b) 1.) (b) Suppose n is a positive integer and suppose a 1 (mod n) and (a, n) = 1. Show that there is a character x defined modulo n (possibly of smaller conductor) such that x(a) 1. Use this fact to show

X x(a) = O. z mod n

If we do not assume (a, n) = 1 then this is not necessarily true. Show

X

x mod 12

(4)2.

(This is one disadvantage of using primitive characters.)

3.7. Let x = xi, be the decomposition of a character x as in the discussion preceding Theorem 3.5. (a) Show that (x0),, = xp O p . (b) Show that if ( fx ,, fo) = 1 then x(a)0(a) iii(a) for all a. (c) Show that x(a)0(a) x Oa) unless x(a) tif(a) = 0 (x and tp arbitrary).

Chapter 4

Dirichlet L-series and Class Number Formulas

In this chapter we review some of the basic facts about L-series. Then their values at negative integers are given in terms of generalized Bernoulli numbers. Finally, we discuss the values at 1 and relations with class numbers. Let x be a Dirichlet character of conductor f. The L-series attached to x is defined by

x(n)

L(s, x) = E

„=

ns

'

Re(s) > 1.

For x = 1, this is the usual Riemann zeta function. It is well known that L(s, x) may be continued analytically to the whole complex place, except for a simple pole at s = 1 when z = 1. Let F(s) be the gamma function, TOO = 1 x (a)e2nia/f be a Gauss sum, and 6 = 0 if x( — 1) = 1, 6 = 1 if x( — 1) = — 1. Then

(f\s12 F (s + 6\Lt ) k s ' x)

w (f\" - s)12 F il — s + 6 \ L(1 — s, x n) ) 2

where

w

X

2(X)

Jff iô .

It will follow from Lemma 4.8 that I 1,11, I = 1. The functional equation may be rewritten as

F(s) cos(74s

6) )L(s, x) = T(Z) (27c YL(1 — s, 2 / f 29

30

4 Dirichlet L-series and Class Number Formulas

Also L(s, x) has the convergent Euler product expansion

L(s, x) = H (1 - x (p)p - sy 1,

Re(s) > 1.

P

It follows that L(s, x) 0 0 for Re(s) > 1. It is also true that L(1, x) o 0, but this is a deeper fact which will be proved later. From the functional equation we find that for n E Z, n > 1, we have

L(1 — n, x) O 0

if n 6 (mod 2)

and

L(1 — n, x) = 0 if n # 6 (mod 2), except for the case x = 1, n = 1, where we have L(0, 1) = ( 0) = —1. This exception is easily seen to result from the fact that the only pole for L-series occurs for x = 1 at s = 1. More generally, we may define the Hurwitz zeta function

. n= 0 (I)

1 + n)S '

Re(s) > 1,

0 < b < 1.

Then

a

f

L(s, x) =

f) .

The functions f - s(s, a/ f) = Z,,,„(f) in' are sometimes called partial zeta functions. They do not usually have Euler product expansions or nice functional equations, but they may be analytically continued to the whole complex plane, except for a pole at s = 1. We wish to give the numbers L(1 — n, x) explicitly. For this we need the generalized Bernoulli numbers. The ordinary Bernoulli numbers B„ are defined by co t t" et —

1

=

n= o

are defined by

The generalized Bernoulli numbers B

co

i'-, X(a)teal Ld a, 1

n!

t ef — 11 -=

in

Z Bn, X n -- r 0

n!

•

Note that when x = 1 we have cc

t"

tet

Z Bn, 1 n!— = et — 1 n=0

=

et

t —1

+

t,

so Bn 1 = Bn except for n = 1, when we have B 1 , 1 =12, B i -- —1. Also observe that if x o 1 then B o , x = 0, since Zuf =1 x(a) = 0.

31

4 Dirichlet L-series and Class Number Formulas

We shall also need the Bernoulli polynomials B(X) defined by tex t et — 1

— n=1

An easy calculation shows that

B,(1 —

X) = (-1)nB,(X).

Since the generating function is the product of

t" t = ZB el — 1 " n!

and

ex t

— Z X"

t" n!'

it follows easily that n

n B(X) ---= Z ( )13i Xn -i. i=o i Proposition 4.1. Let F be any multiple off Then F Bn, x = Fn— 1 Z

a x(a)B„(-- ). F a= 1

PROOF. co

Z Fn — 1 = n0

F

.

z x(a)Bn ( a\

tn

F 1 n!

a=1

--= Z Z(a) a1 =

te F (alF)Ft Ft 1 e — I '

Let g = Fl f and a = b + cf. Then we have (b+cf)t fg—1 te Z Z x(b) fg , e — 1 b = 1 c0

te bt = co tn = Z x(b) ft Z B n, x e—1 n! b= 1 n=0 f

The result follows.

El

In particular, since B i (X) = X — 1, we have If

B i , = 7 Z x(a)a,

x

1.

J a= 1

It is easy to see that the defining relation for the B t when x is even and odd when x is odd. Therefore

is an even function of

0 if n * o (mod 2), with the usual exception B 1 1 = 1 (or B 1 = —4 B

At this point the reader has probably conjectured that there is a relationship between L(1 — n, x) and B; so we prove the following result.

Theorem 4.2. L(1 — n, x) = — B,/n, n __ 1. More 1. —B(b)/n, 0 < b

generally,

(1 — n, b) =

32

4 Dirichlet L-series and Class Number Formulas

PROOF. Let F(t) =

t 13„(1 — b) — . = n! — 1 n=0

Define H(s) = J F(z)zs - 2 dz, where the integral is over the following path

which consists of the positive real axis (top side), a circle C, around 0 of radius e, and the positive real axis (bottom side). We interpret zs to mean exp(s log z), where we take log to be defined by log t on the top side of the real axis and log t + 27ti on the bottom side. It is easy to see that H(s) is defined and analytic for all s. We may write H(s) = (e2 "" — 1)

F(t)ts - 2 dt +

I F(z)z

2 dz.

CE

Assume first that Re(s) > I. Then J —> 0 as g H(s) =

( e 21T1S

ci,

so

F(t)i-S 2 dt

1)

o (e 2 ni s _ 1)

-(b+ m )t dt m=0 e co f co

E ts —

= (e 2nis _ 1)

= (e 2nis _ 1)

m=0 0 oo

1

1e

(1) + m)t

dt

F(s)

0 (m + = (e 2nis _ 1)F(s)C(s,b).

Therefore C(s, b) = H(s)/(e2 ' — 1)F(s), which by analytic continuation holds for all s 1. Incidentally, this gives the analytic continuation of b). We now assume that s = 1 — n, where n > 1 is an integer. Then e2"is = 1, SO

H(1 — n) = f F(z)z

1 dz =

(2i)

B(1

It is easy to show that lim (e 2 'is — 1)F(s) = 1-n

(n — 1)!

b)

33

4 Dirichlet L-series and Class Number Formulas

Therefore ,— n, b) = ( — 1y

B n (b )

1 B n (1 — b) n

n

Consequently

L(1 — n, x) =

x(a)fn

—

a=i 1f

n_ 1

a

-

n a=1

B n (f)

n•

This completes the proof.

LI

We now turn our attention to the value of L(1, x). It is well known that L(1, x) O. One proof uses the following. Theorem 4.3. Let X be a group of Dirichlet characters, K the associated field, and fc (s) the Dedekind zeta function of K. Then

cK(s) =

fl L(s, x) •

xeX

PROOF. I t suffices to consider the Euler factors corresponding to each prime p. Suppose (p)

=

is the prime factorization of p in K, and each = pd'. Then Ç K (s) contains the factor

has residue class degree f,

no — (N s) - ' = (1 — p - fs) - g. I us fix — x (p)p - syi. Those x with )

-

The L-series gives x(p) = 0 do not contribute so we ignore them. By Theorem 3.7, Y/Z is cyclic of order f, where Y is the group of those x E X with x(p) o 0 and Z consists of those with x(p) = 1. As x runs through a set of coset representatives for Y/Z, x(p) runs through all f th roots of unity. Each coset has g elements. Since f—1

fl (1

iffp - s)

=

—

P —fs ),

a=0

LI

the result follows. Corollary 4.4. L(1,

x)

O.

34

4 Dirichlet L-series and Class Number Formulas

PROOF. Let K be the field belonging to x. It is well known that the zeta function of K has a (simple) pole at s = 1. Let b be the order of x. Then b—1

„(s) =

b—1

fl Us, )(a) = C(s) • a=fl1 L(s, Z a).

a= 0

Since (s) has only a simple pole at s = 1, none of the factors L(s, X') can vanish at s = 1. This completes the proof. The classical application of Corollary 4.4 is the following. Theorem 4.5 (Dirichlet). Let (a, n) = 1. Then there are infinitely many primes p a (mod n). PROOF. Let x be a Dirichlet character. Then for Re(s) > 1 we have log L(s, x) = — log(1 —

Ap)p —s )

Aprp' =

p rn= 1

X(P) = p— g X( S ),

where trivial estimates show that g(s) is holomorphic for Re(s) > 1. Therefore, summing over all characters x mod n, we have ( a -1 ) log L(s, xmodn

pEa(n)

0(:/) P

g(s)

with g(s) holomorphic for Re(s) > (we have used Exercise 3.6). Now let s 1. Since L(s, 1) = C(s) has a pole at s = 1, log L(s, 1) log(s — 1) —» co. Since L(1, x) 0 0, oo for x 1, log L(s, x) remains bounded. Therefore the left-hand side —> oo, so the same is true for the right-hand side. Since g(s) is holomorphic at s = 1, we must have 0(n)

lim s —■

1

pa(n)

=oo

Ps

Therefore the sum cannot have finitely many terms. This completes the proof. We now use Theorem 4.3 to give a proof of Theorem 3.6, the ConductorDiscriminant Formula. Recall that X is a group of Dirichlet characters associated to a field K, so Theorem 4.3 applies. It is known (see, for example, Lang [1], p. 254) that 4(s) satisfies the functional equation

Asq_s y l roy24( s) _ 2

—

2

ri — sy24(1 — s),

35

4 Dirichlet L-series and Class Number Formulas

where A

2 - r2n -N / 2.\/Id(K)1.

Here N = deg(K/O) and r 1 and r2 have their usual meanings. Since K/CP is Galois, either r 1 = 0 or r2 = O. Suppose first that r2 = 0, so K is totally real and x( — 1) = 1 for all x. The functional equations for the L-series read (f) 6 /2 r (s) L ,s, x) = wx (f)(1- s)/2 F( 1 _ s)L(1 — s, )?). 2 2 Taking the product over all we find that we must have

A2 =

x

and comparing with the equation for CK(s),

(h)

and

n wz

Consequently I d(K)I = fx , as desired. If r i = 0 then r2 = N/2. In this case, half the characters are even and half are odd. Using the identity r (s\ r is

+ 1\ 2

2)

21 ,

)

nr(s)

(see, for example, Whittaker and Watson [1] p. 240), we again obtain the desired result. The sign of the discriminant is determined, as usual, by Lemma 2.2. This completes the proof of Theorem 3.6. Corollary 4.6. if K is totally real if K is complex.

Id(K)I, 11 T(z) { ideg(K/ CO 2 \,/ I AK) I, xEx

PROOF. It follows from the above proof that immediately.

[L E ,W = 1. The result follows 111

Note that this corollary contains the famous theorem on the sign of the Gaussian sum: if x is the unique quadratic character mod p then t(x) = if p 1 (mod 4) and z(x) = i.j) if p 3 (mod 4). We now evaluate L(1, x). For odd x this is easily accomplished via the functional equation:

L(1,

x)

=

z(x)

27c _ — L(0, x) f

niT(Z) B 1 , For even characters the argument is somewhat more difficult. We first need some lemmas.

36

4 Dirichlet L-series and Class Number Formulas

Lemma 4.7. For every integer b, f

L (a)e 27rtab, = a= 1

In particular,

t(x) = x( - 1).[(i). If (b, f) = 1, then change variables: c = ab (mod f). Since everything depends only on residue classes mod f, the result follows in this case. If (b,f) = d > 1 then the result is still true, since both sides vanish. The right-hand side is obviously O. For the left, observe that if x(y) = 1 for all y 1 (mod fld), (y, f) = 1, then z would be defined mod fld (note that (Z If Z) X maps onto (Z I ( fI d)Z) x ), hence could not have conductor f There1 (mod f/d), (y, f) = 1, such that x(Y) 0 1. Since fore there exists y dy d (mod f), so by b (mod f), we have PROOF.

E i(a)e f

= E f

2.abif

a z-- 1

A a)e 2rriaby 1 f

a= 1

= xy) E i(a)e 2n,abif . f

a=1

Since x(y) 0 1, the sum is O. For the second statement, use the first statement with b = —1.

D

Lemma 4.8. I T(X) I ' ": 1X PROOF. f

0 (i )1T( X ) 1 2 = / 1 ANT(X)1 2 (note only (f) terms are non-zero) b =1

= E E me ff

2iriab,

b = 1 a= 1

=

a

=

IJC(aV(e) e

x(a)i(a)f

i 2(c)e -21tibet

f

(by Lemma 43)

c= 1

b

e

2 Trib(a — off

(the sum over b is 0 unless a = c)

a

= fq5(f),

since x(a)i(a) = 1 if (a, f) = 1, and is 0 otherwise. This completes the proof. D

37

4 Dirichlet L-series and Class Number Formulas

We now evaluate L(1, x). We ignore questions of convergence, most of which may be treated by partial summation techniques. L(1,

" x(n)

=

" 11

=

1f v i(a) TOO a d1

Since t(i) =

i(a)e2 niani f

n= 1 n TO?) a=

n= 1 n

=

f

1 n1

—1 f

_

.

n

e2' f

Cf= e 2 nilf.

log( 1

T(X) a= 1

WOO = A — 1 )f tr(z), we obtain

A - 0'4Z) f

L(1, x) = —

log(1 —

CD.

a= 1

Now log(1 — Caf ) + log(1 — g. a) = 2 log I 1 — even, so x(a) = x(— a), we have T(X)

L(1, x) =

f

. Consequently, if

is

/ AO log I 1 — I.

a= 1

We have proved the following (odd characters were treated before Lemma 4.7). Theorem 4.9. L(1,

x)

= ni

TOO

f

1f ali z-(a)a

if z( — 1) = —1.

x

f

L(1, z) =

TOO

t(x) B 1 - = f

J a1

i(a)logl 1 —

Ca» if Z( — 1) = 1, Z 0 1.

Cl

Note that the theorem implies that /3 1 , x o 0 if x is odd. There is no elementary proof known for this fact. Later we shall give algebraic interpretations of these formulas. We now discuss class number formulas. The zeta function of a field K has a simple pole at s = 1 with residue

2r 1 (2n)r2hR w.N/ Id! where r 1 , r2 are as usual, h is the class number of K, R is the regulator (see below), w is the number of roots of unity in K, and d is the discriminant. Suppose K belongs to a group X of Dirichlet characters. Using the relation CK(s) = fl L(s, x), and the fact that (s) has a simple pole at s = 1 with residue 1, we obtain 2"(27r)r2hR

w\IIdI

= xex H L( 1, x)x*1

38

4 Dirichlet L-series and Class Number Formulas

Using Theorem 4.9, we now have in theory a method for calculating the class number of an abelian number field, as long as we can calculate the regulator, which involves finding a basis for the group of units. Usually this computation becomes too lengthy to be practical. So we need another method of obtaining information about the class number. For this, we shall factor the class number into two factors, one of which is relatively easy to work with. The next few results hold not only for abelian fields, but also for a wider class, namely CM-fields (also called J-fields). A field is called totally real if all its embeddings into C lie in R and totally imaginary if none of its embeddings lie in R. A CM-field is a totally imaginary quadratic extension of a totally real number field. Such a field may be obtained by starting with a totally real field and adjoining the square root of a number all of whose conjugates are negative. All of the fields 0(C,) are CM-fields. Their maximal real subfields are El((u + Cu- 1 ), and we obtain Q(Cu) by adjoining the square root of nc22 2 (the discriminant of X 2 — (C, C u-1 )X + 1), which is totally negative. One feature of a CM-field is that complex conjugation on C induces an automorphism on the field which is independent of the embedding into C. Namely, let K be CM, K + the real subfield. Let 4), qt: K C be two emfor all a e K. First note beddings. We claim that 4) -1 (0(a)) = 41 that 4)(K)/4)(K + ) is quadratic, hence normal, and complex conjugation fixes 4)(K + ). Therefore RK) = 0(K). In particular, 4) - V) is defined. Clearly both 4) - 1 ((T.) and i' are automorphisms of K and both fix K + since it is totally real. Since K is totally imaginary, neither automorphism can be the identity. Therefore they must be equal since Gal(K/K + ) has order 2. Consequently, when working with CM-fields we may talk about which is well-defined. Also, I oc1 2 = cci is independent of the embedding. This is useful when applying Lemma 1.6. For example, if y is a unit then elE is an algebraic integer of absolute value 1, hence a root of unity.

'No»

Theorem 4.10. Let K be a CM-field, K + its maximal real subfield, and let h and h + be the respective class numbers. Then h + divides h. The quotient h - is called the relative class number (some authors call h the first factor; a few call it the second factor).

PROOF. We need the following result from class field theory. Proposition 4.11. Let K/L be an extension of number fields such that there is no nontrivial unramified (at all primes, including archimedean ones) subextension F IL with Gal(F/L) abelian. Then the class number of L divides the class number of K. (Note• this proposition is usually used in the case that KIL is totally ramified at some prime. However it could also be used if K IL is normal with a non-abelian simple group as Galois group).

4 Dirichlet L-series and Class Number Formulas

39

Let H be the maximal unramified (at all primes) abelian extension of L. By class field theory, Gal(H/L) is isomorphic to the ideal class group of L. The assumptions on K/L imply that H n K = L. Therefore [KH:K] = [H: L]. But KH/K is unramified abelian, so is contained in the maximal unramified abelian extension of K. Therefore the class number of L = [H : L] = [KH : K] divides the class number of K. This proves the proposition. We remark that H is called the Hilbert class field of L. 0 PROOF.

Returning to the proof of the theorem, we observe that K/K + is totally ramified at the archimedean primes, so the proposition applies. This completes the proof of the theorem. 0 We now prove a result which generalizes Proposition 1.5. Theorem 4.12. Let K be a CM-field and let E be its unit group. Let E + be the unit group of K + and let W be the group of roots of unity in K. Then Q

t [E:WE + ] = 1 or 2.

PROOF. Define 0: E —> W by 0(e) = e/E. Since 7 = (Oa for all embeddings û(K is CM), we have 10(e)a I = 1 for all a. By Lemma 1.6, OW G W. Let tP: E—> WIW 2 be the map induced by (/). Suppose e = C i , where (E W and i G E + . Then 0(e) = C 2 E W 2 , so e E Ker(t/i). Conversely, suppose OW = C 2 G W 2. Then it is easy to see that 6 1 = C -- 's is real. It follows that Ker(0) = WE. Since I W/W 2 I = 2, we are done. Note that if 0(E) = W then Q = 2 ; if (1)(E) = W 2 then Q — 1. D Corollary 4.13. Let K = Q( ( n ). Then Q = 1 if n is a prime power and Q = 2 if n is not a prime power. The proof when n is an odd prime power is exactly the same as that given in Proposition 1.5. For p = 2 we must argue a little differently. Suppose e is a unit in Q( ( 2 ) such that E IE (t W 2 . Then E/g = c = a primitive 2mth root of unity. Let N denote the norm from G(( 2 »,) to OW. Then N(() = where PROOF.

2 ,,

_2_ 1

1 b = 1 (1 + 4j) , 2"1 -2 + 2'n-1 (2'n-2 + 1)

a =

j=0 b

1(4)

2-2 (mod 2m - 1 ) Therefore Ca is a primitive 4th root of 1: Ca = + i. It follows that N(e)IN(e) — +i. But N(e) is a unit of G(i), therefore + 1 or + i. None of these possibilities works, so we have a contradiction. So Q = 1 for G(„).

40

4 Dirichlet L-series and Class Number Formulas

Now assume n is not a prime power. By Proposition 2.8, 1 — is a unit. But (1 — — „) = — Suppose — e W 2 . Then — = (± Cr,,)2 = „2 r, so — 1 = Clearly n must be even, so n 0 (mod 4). Since — 1 = C1,112 , we have n/2 2r — 1 (mod n), therefore n/2 — 1 (mod 2), which is impossible. It follows that — 0 W 2, so Q = 2. This completes the proof.

When K = (EIgn), we may prove a result which is stronger than Theorem 4.10. Theorem 4.14. Let C be the ideal class group of(,,) and C + the ideal class group of the real subfield CK„) + . Then the natural map C+ C is an injection. PROOF. Suppose / is an ideal of OG„) + which becomes principal when lifted to OG„). We must show / was principal to begin with. Let / = (a) with a e (LIG,,). Then (i/a)=1/I = (1), since / is real. Therefore /a is a unit and has absolute value!. By Lemma 1.6, /a is a root of unity. If n is not a prime power, Q = 2; the proof of Theorem 4.12 shows that there is a unit e such that = 5/a. Then ca is real and / = (a) = (aE). It follows from unique factorization of ideals that / = (1E) in (LIG,)+, so / was originally principal. Now suppose n = pm. Let TE = çna — 1. We have n/Tr = —Ci,„„ which generates the roots of unity in 0(Cpm). Therefore 5/I = (7r/)' for some d. Since the it -adic valuation takes on only even values on G ( p „,)+ and since ITEd and / are real, d = v.(ctird) — v.( 1) = v.(curd ) — v(I) is even. Hence a/a = (—Ç„,)d E W 2 . In particular, ci/a = C/C for some root of unity C, and ct is real. As before, / = (1 , so / was originally principal. This completes the proof. 111 )

This theorem is not true for arbitrary CM-fields: Since (2, 10)2 = ( —2) in 43(.\/10), the nonprincipal ideal (2, 10) becomes principal in 4:1(\/10, — 2). In general, at most one nonprincipal class becomes principal (see Theorem 10.3). Theorem 4.12 may be used to give a relation between the regulator of K and that of K +. Recall that the regulator of a number field L is defined as follows. Let r = ri + r2 — 1 and let E l ,— , Er be a set of independent units of L. Write the embeddings of L into C as al, • • • , , ar, + C r.+ 19 6 1, 1+11 • • • (5- r +19 where ap 1 j r l , is real, and up d, r i +1 jr+ 1, is a pair of complex embeddings. Finally let 6, = 1 if a; is real and S i = 2 is a, is complex. The regulator is defined to be

R L(e i ,

, er) = absolute value of det(6 i log I E i)1/1
Note that we omit one a. Since the norm of each E is + 1, the sum over all ai , 1 < i < r + 1, of S i log I q t 1 is O. Since we take the absolute value of the determinant, the possible sign change from omitting a different a does not happen.

41

4 Dirichlet L-series and Class Number Formulas

If , Er is a basis for the group of units of L modulo roots of unity, then R L(E 1 , , Er) = R L is called the regulator of L. Again, the fact that we took the absolute value of the determinant makes R L independent of the choice of basis and ordering of the a's. Now let El, .. Er be a basis for the units of K + modulo { + O. Then El, , Er forms a basis for a subgroup of index Q (= 1 or 2) in the units of K modulo roots of unity. However each 6i = 1 for K + and each 5 = 2 for K. Therefore

R K(g 1 ,

TR K +(el ,

'gr)

, gr) = 2rRK , .

We now need the following result. Lemma 4.15. Let g 1 , , Er be independent units of a number field K which generate a subgroup A of the units of K modulo roots of unity, and let '1i, , generate a subgroup B. If A B is of finite index then [B: A] = RK(61 & (1il,. • • , 11r) • '

.

PROOF. We may write Ei =

(

Fii re"

- (root of unity), with ail E Z.

Therefore

S i log I E7-1 I =

a11 (5i log I

I.

Consequently RK(El, • '

'Er)

Rx(r/i, • • • , lb)

det(a 11 ) •

By the theory of elementary divisors, there exist integer matrices M and N of , dr); so det(a ii) = + fI di. determinant + 1 such that M(aii)N = diag(d 1 , But M and N correspond to changing bases of A and B, so we have bases i Z/d i Z , X,. of A and Y1• , yr of B with xi = di y i . Therefore B I A and [B: A] = I d i l. This completes the proof of the lemma.

ni

From the lemma, we see that R K(E., . . , Er) = QRK, in the above notation. We have proved the following.

Proposition 4.16. Let K be a CM-field and K + its maximal real subfield. Then

R

K RK+

=

2" where r = deg(K/O) — 1.

42

4 Dirichlet L-series and Class Number Formulas

We may now return to the class number formulas. Let X be a group of Dirichlet characters and K the associated field. We assume K is totally complex, so half of the characters in X are odd and half are even. Let n = deg(K/U). Then 2./2h(K-F)R K ÷

=H

2N/Id(K + )1

,Ex x even x#

L(1, x),

1

and (27q112 h(K)RK 142.\/ I Cl(K)

I

r-r — xlelx L O , x).

x# 1

Dividing, we obtain

70/2h - (102./2 12w.\/ I d(K)/d(K +) I

, H L(1, x). x odd

Now L(1, x) = (nit(x)/fx)B i , 2 for x odd, and by the conductor-discriminant formula N/ I d(K)/ d(K ± )1 = ( nxodd fx ) 1/ 2 . Also, by Corollary 4.6,

ii

-,( x )

=

ini V I d(K)/d(K + )1.

x odd Putting everything together, we obtain the following. Theorem 4.17. h - (K) = Qw 11 (-1B 1 ,x). x odd

111

Observe that this formula, as opposed to that obtained earlier, involves no transcendental quantities. It is therefore possible to use it to obtain divisibility properties of h - (K). Sometimes it is possible to obtain results about the full class number h(K) from those about h - (K). Later we shall show that p divides h(0()) <=> p divides h - (0((,)). The above formula will allow us to translate the statement about p dividing h - into one about p dividing certain Bernoulli numbers. We close this chapter by showing that the class number of 0((n) grows quite rapidly with n. For this we need the Brauer-Siegel theorem (see Lang [ 1 ]): Suppose K runs through a sequence of number fields normal over U such that

[K : 0] 0 logid(K)I —' .

43

4 Dirichlet L-series and Class Number Formulas

Then

log(h(lORK) logN/Id(K)1 Unfortunately this result involves the regulator, so we do not immediately obtain any information about h by itself. However we may apply the result to both G(C„) and its maximal real subfield, and then compare. For convenience, let d„ = Id(O(C.))1, h„ = h(G(C.)),R R".) , and let h, denote the corresponding objects for e(C„) + . We first estimate d„. Lemma 4.18. log d„ = 0(n) log n + o(0(n) log n). PROOF. From Proposition 2.7, we have log p

log d„ = 0(n) log n — 0(n) pn

P—

.

Let in = log n/log 2. Since 2' = n, it follows that n has at most in prime factors. Clearly log p < pin P

—

1 —

log pi i=1

Pi

1

log pi

<2 —

1=1

Pi

where the last two summations are over the first in primes. From the prime number theorem, it follows easily that there exists a constant C such that the mth prime is less than x = Cm log m, and the number of primes less than x is less than Dx/log x for some D. Therefore

p,„

log p log p < log x 2 E 2 E 1+ 2 E p—1 p,x P — p< -/ ....r,ri
_2

+ 2

log x Dx = 0(.1.--x-) log x

= O( N/m log m) = 0(\/log n log log n) = o(log n). This estimate gives the result. Lemma 4.19. If n is not a prime power then d„ = (d„+ )2 . If n = pa then d„ = p(d,',- )2 if p 0 2, d = 4(d„+ )2 if p = 2. In all cases we have

log d,;' = 10(n) log n + o(0(n) log n). PROOF. Recall the formula (see Lang [1], pp. 60, 66, or Long [1] p. 82)

d(L)I = (NgL/K)I

AK) ideg(L/K),

where L/K is any extension of number fields, g LiK is the relative different, and Nis the norm from L to G. If the ring of integers C L of L can be written in the form (°K[c] for some a E CL , then g LIK is the ideal of C L generated by f'(2),

44

4 Dirichlet L-series and Class Number Formulas

where f (X) is the irreducible polynomial for a over K. In the present case, we have

z n] = zEn + ,T 1 ] R„J [

and f (X) = X 2 — (Cn + C, 1 )X + 1, so f '() = C. — C, 1 = C, 1 2 — 1). If n is not a prime power then („2 — 1 is a unit, so gLiK = 1. Therefore

d„ = (d12 . If n = pa, p 0 2, then Ng = N(C„2 — 1) = p, so d„ = p(d:)2 . If n = 2, then C„2 is a 2" - 'st root of 1; so Ng = N((„2 — 1) = 4. Therefore d„ = 4(d) 2 . The final statement follows from Lemma 4.18 and the fact that log p = 0 (log n). This completes the proof. CI From Lemmas 4.18 and 4.19, we have

0(n) —> 0 and

log d„

F( n) log d:

O

,

so the Brauer-Siegel theorem applies. Therefore log h„R„ = 1 log d„ + o(log d„) and log h: R: = log d: + o(log do). By Proposition 4.16,

R = 0(0(n)). , log(R n") ± Therefore log h: = log(h„R„) — log(h:R) — log( R" )

R:

= 1 log d„ — i log d: + 0(0(n)) + o(log d„) = 10(n) log n + o(0(n) log n). We have proved the following result. Theorem 4.20. Let h, denote the relative class number for O ( .). Then log h: — 10(n) log n

as n —> oo.

Therefore h„ —> oo as n —> oo, so there are only finitely many Z[„] has unique factorization.

(Note: a — b means a/b —> 1).

n such that III

45

Exercises

Unfortunately the above result is not effective, in the sense that it does not allow us to compute a constant no such that hu > 1 if n > no . To do that we need other techniques. See Chapter 11. NOTES

For more on ordinary Bernoulli numbers, see Nielsen [1]. The generalized Bernoulli numbers were defined by Berger [1], by Ankeny-Artin-Chowla [1], and by Leopoldt [3], who used them extensively. The standard reference for class number formulas is Hasse [1]. See also Borevich-Shafarevich [1]. The book of Hasse also contains a detailed study of the unit index Q (warning: Satz 29 is incorrect). Another good reference for some of the topics of this chapter is Iwasawa [23]. EXERCISES 4.1. Show that for n, k > 0, Ika :;) an = [ 1 /(n + 1)](B+ 1(k) — Bn + JO)). 1 and z( — 1) = 1 then 4.2. (a) Show that if z (b) Show that if z( — 1) = 1 then 1

B2. 1 = (1/f)

Ija=1(a)a2.

f

B3 01 =

a ;Z( 0 ) 03

f a=1

therefore 1 B3 1

— r

Z(C0a 3

f a= 1

4.3. (a) Use the definition of Bn to show that lim sup n.00

( Bn.,1) 11

"

n.

f =— 271 .

(b) Use the functional equation for L(S, z) to show that lim n-0 n

\Ff (2i

I B n1 I n! 2 \f )

6(2)

4.4. Show that if In In then h(G(.)) divides h(Q( n)). 4.5. Let p> 3 and p = 3 (mod 4), and let h be the class number of G(.\/—p). Let z be the quadratic character mod p. (a) Show that hp -= —2 /0
46

4 Dirichlet L-series and Class Number Formulas

4.6. (a) Show that for p 1 (mod 4) the number of quadratic residues in the interval (0, p/2) equals the number of nonresidues. (b) Let p 1 (mod 4), and let h and z be the class number and character for the field 0(fp). Let E > 1 be the fundamental unit and let = e'IP. Show that p—1

-1(1 _

E 2h =_ 1

a= 1

and Eh =

(

b

sin

_)/n nb

P

TCC

(sin —), eP

where b runs through the quadratic nonresidues in the interval (0, p/2) and c runs through the residues in (0, p/2). Since sin x is monotone increasing in (0, Tr/2), this shows that the residues tend to cluster near the beginning and the nonresidues near the end of the interval (0, p/2).

Chapter 5

p-adic L-functions and Bernoulli

Numbers

In this chapter we shall construct p-adic analogues of Dirichlet L-functions. Since the usual series for these functions do not converge p-adically, we must resort to another procedure. The values of L(s, x) at negative integers are algebraic, hence may be regarded as lying in an extension of Q p . We therefore look for a p-adic function which agrees with L(s, x) at the negative integers. With a few minor modifications, this is possible. The resulting p-adic L-functions will be used to prove congruences for generalized Bernoulli numbers, from which we deduce Kummer's criterion for irregularity of primes. We shall also show there are infinitely many irregular primes. Finally we evaluate the p-adic L-functions at s = 1 and find a formula remarkably similar to the classical one. This yields a p-adic class number formula, from which we deduce Kummer's result " p1h; p 117; ," and also a congruence for class numbers of real quadratic fields due to AnkenyArtin—Chowla. Along the way, we define the p-adic regulator and prove that it does not vanish (Leopoldt's conjecture) for abelian number fields. There are several ways to construct p-adic L-functions. We have taken the quickest approach here. Later, we shall give other methods which give additional insights into relationships with cyclotomic fields.

§5.1 p-adic Functions First, we need some basic results on p-adic analysis. We start with the p-adic rationals Op Since we shall need to consider algebraic extensions (e.g., generated by values of Dirichlet characters) we extend to O p, the algebraic s

47

48

5 p-adic L-functions and Bernoulli Numbers

closure. The absolute value on (IA, extends uniquely to Op; we normalize by I p I = 1/p (throughout this chapter, I x I will be the p-adic absolute value; so we write Ix Ip only for emphasis).

Proposition 5.1. Op is not complete. PROOF. Let 00

n= 1

where n' = n if (n, p) = 1 and n' = 1 otherwise. If Op were complete, then the series would converge to a e Op* Therefore a would lie in a finite extension K of Op e Suppose e K for all n < tn. We may assume p,rm. Then en — 1

iq =P — m(ot

—

E

„,p')EK

n--=-0

and [3 C. (mod p). Therefore X' — 1 0 (mod p) has a solution in K. By Hensel's Lemma (since p)(m), K contains a solution of X' — 1 = 0 which is congruent to /3 mod p, hence to , mod p. Since the mth roots of unity are distinct mod p (recall in = n (1 - 0)

;ti it follows that C. e K. By induction, C. e K for all in with p,}1m. Since, as above, the roots of unity of order prime to p are distinct mod p, we have infinitely many residue classes mod p in the ring of integers of K. Since K/O p is a finite extension, this is a contradiction. Therefore a 1Ft Op and Op is not complete. 0 Since it is more convenient to do analysis in a complete field, we let Cp be the completion of Op with respect to the p-adic absolute value. The p-adic absolute value naturally extends to Cp and Op is dense in Cp •

Proposition 5.2. Cp is algebraically closed. PROOF. We need the following lemma, due to Krasner.

Lemma 5.3. Suppose K is a complete field with respect to a non-archimedean valuation. Let a, p e K, the algebraic closure of K, with a separable over K(fi ). Finally, suppose that for all conjugates at a of a we have hq — al < loti — a I. Then K(a) g K (f3 ) (1x1 denotes the unique extension of the absolute value on

K). In other words, if (3 is sufficiently close to a then a e K(f3).

49

§5.1 p-adic Functions

PROOF. Consider the extension K(oc, [3)/ K(t3) and let L/ K(t3) be the Galois closure. Let o- Gal(L/K(t3)). Then a(fl — a) = fl — a(a). Since l ax I = 1x 1 for all x (by the uniqueness of the extension of the absolute value), we have 1[3 — a(of)l = a. Therefore

for all ai

< I

— cx1

Max( l — 131 , 113 — a(1)1) < I ai — I' — a(a) I It follows that a(a) = a, so a e K([3), as desired.

I

Returning to the proof of the proposition, we let K = C. Suppose a is algebraic over Cp and let f (X) be its irreducible polynomial in Cp [X]. Since Op is dense in Cp , we may choose a polynomial g(X) e p [X] whose coefficients are close to those of f (X). Then g(a) = g(a) — f() is very small. Writing g(X) = fi (X — 13 i), we see that la — fil is small for some root [3 of g(X). In particular, we can choose g(X) and then t3 so that l fi — al < lai — l for all conjugates a i a. Therefore a E C p(#) = C p , since [3 G Op C. The proof is complete. D Sometimes, for technical reasons, it is convenient to embed Cp in C, or vice versa. In fact, the two fields are algebraically, but not topologically, isomorphic: Both fields have the same uncountable transcendence degree over ED, and both are obtained by starting with 0, adjoining a transcendence basis, and then taking the algebraic closure. From now on, unless otherwise stated, we shall be working in Cp , which may be regarded as the p-adic analogue of the complex numbers. We next introduce the p-adic exponential and logarithm functions. Define

exp(X) =

E

X" .-

n=0 Pi!

Since there are [n/pi] multiples of p i less than or equal to n, it is easy to see that the exponent of p in n! is

1+ <

[n] + [ P P

p—

If pa < n < pa + I then the sum is greater than

—+ • • +--a= — a= Pa

p—1

a—

np - a

p—1

n— p p —1

log n log p .

Therefore

n— p

log n

p— 1

logp

< v (n !) < P

co if It follows that IX7n! I —> 0 as n cc if 1 X1 < p-1 / (P -1) and Xn/n!I 'XI > p 1 l( P 1). Therefore exp(X) has radius of convergence p - 11(P - 1) < 1 (recall that a non-archimedean series converges <=> its nth term —> 0). Note

50

5 p-adic L-functions and Bernoulli Numbers

that e = exp(1) is undefined, but eP (e4 if p = 2) is defined. We could of course let e = (exp(p)) 1 /P but this would not be unique. We now define 10gp(1 +

—

=

1

Xn

n

n=1

Since the exponent of p in n is at most log n/log p, we find that the series has radius of convergence 1. However in this case we can extend the function. Note that since logp(X Y) = log(X) + logo( Y) is an identity for formal power series, it is true whenever the series converge. Proposition 5.4. There exists a unique extension of logy to all of C; such that log(p) = 0 and logo(xy) = logo x + logo y for all x, y e C. We need to investigate the multiplicative structure of C; . For each rational number r choose a power pr of p in such a way that pre = !I's (one way: let pr be the positive real rth power of p in 0, then embed in Cp). Denote by pu this set of fir, r e G. Let a e C; . If a l e O p is sufficiently close to a then I a I = I a i l. But I a i I = (p 1 /e)n for some n where e is the ramification index of Op(i)/Op. Therefore la 1 = p' for some r e U, SO l ap' I = 1. Now suppose fi e C,, 1 131 = 1. Choose S i e Go close to fi. Every unit of the finite extension Clp(fi,)/O p is congruent modulo A (the prime above p) to a root of unity of order prime to p (lift from (e / Ay via Hensel's lemma). It follows that I13 1 — wj < 1, hence 113 — wi < 1 and I fico - — fl < 1, for some root of unity co of order prime to p. Since such roots of unity are distinct modulo A, w is unique. Let W denote the group of all roots of unity of order prime to p in C,<. We have proved that PROOF.

C; = p`u xW x U t where U 1 = {x e Co l lx — l < 1 } . Now let a = prwx e C,. Define log o a = logo x. Since x e U 1 , logo x is defined by the power series. Clearly this extension satisfies the desired properties. Suppose f( a) gives another extension. If ctiy = 1 then

1 1 1 1 f(1) + f (x N ) ) + — f (prN f (0eN) = — f (a) = — N N N N

=o+o

+ — log (X N) = logo(x). N P

Therefore the extension is unique. This completes the proof of the proposition. EI If a e Gal(Cl o/Op) then since I ax 1 = I x 1 for all X E Op we may extend a to a continuous automorphism of C. By continuity,

logp(1 + (Tx) =

— 1)n ±

(6x)n/n = a

(—

l)n Xn/n =

a

lOgp(1 +

X).

51

§5.I p-adic Functions

By the uniqueness of logo , we therefore have a- log(ac) = logo a for a E , i.e., log(a) = o-(logo a). It follows that for a E Op , 1Ogp(0) E Go(a); this fact also follows from the power series expansion. For Up we may carry out the construction of the proposition more explicitly. For convenience we introduce the notation q

if p 0 2 if p = 2.

f p, 14,

Given a E Zp p)(a, there exists a unique 0(q)th (i.e., (p — 1)st if p 0 2) root of unity co(a)E ip such that a E2 co(a) (mod q). Let so =- 1 (mod q). Then logo a = logo. Alternatively, aP-1 so logo a = logp(aP-1 )/(p — 1).

1 (mod p),

Lemma 5.5. If 1 x 1 < p RP - 1) then I1og p (1 + x)I = 1x1 and if 1x1 < P 11(P 1) p(1 thenIlog x) I lx I. PROOF. If n < p then ni = 1, and in general I n I > 1/n. Therefore, if 1x1 < we have xn n

= lx r

<

Ix I if 2 < n < p

and

f < np(i -flop-1)1 x ' < 1 x 1

if n

since n • p(1 ')/ (P - 1) is decreasing for n > p. Therefore Ix — x 2/2 + • • •I = I x I as desired. The second part follows similarly. This completes the proof. Proposition 5.6. logo x = O <=> x is a rationalpower ofp times a root of unity (of arbtirary order).

PROOF. Clearly such x satisfy logo x O. Conversely, suppose log o x = O. Since C; = x W x U i , we may assume x = 1 + y with yl < 1. Let N be large enough that 1 yPN 1 < p 1 I(P 1) . Then ,A1 X PN = (1 + y)P N = 1 + pN y + - - • + ( F . yi + • - • + yPN.

J

All the middle terms have absolute value at most Ipyl < II < P-1 " - 1), and by the choice of N we have lyPN I < p - 1 /(P - 1) . Therefore I x - il < p - 1 413- 1) and by Lemma 5.5

0 = l logo(xPN) I = I X PN

-

i I.

Therefore x is a pNth root of unity. This completes the proof.

Ill

52

5 p-adic L-functions and Bernoulli Numbers

Proposition 5.7. If I xl < p-1 I(P - 1) then logo exp(x) = x and

exp logp(1 + x) = 1 +

X.

PROOF. Both are formal power series identities, so we need only check convergence. Since I xn/n!I < 1 for n > 1 (because v p(n!) < n/(p — 1)), we have 1 exp(x) — 1 1 < 1 for all x with 1x1 < p -11(" ), so exp(x) and logy exp(x) converge. Similarly, Lemma 5.5 may be used to treat the second identity. 0

Note that the first identity is true whenever exp(x) converges. But the second is not true for all x. Let x = 4 1. Then 0 = log(), so exp(logo(4)) = 1 0 4. This is true even though log() converges ( I 4 1 1 < 1). The point is that 1 logo x 1 is not less than p -11(P -1) for all x with 1x1 1 Cp — 1 1 = p - 11(P -1), so the formal rearrangement of the power series to get

exp logp(1 + x) = 1 + x does not work. Finally, let a e Z p , p ,j/ a. We may define

<0' = exp(x logp) = exp(x logo a). Since 1 logo 1 II = 1/q, this converges if 1 x 1 < qp- 1/(p— 1) > 1. If x = 1 then 1 = by Proposition 5.7. Similarly, if n e Z then " agrees with the usual definition. In particular, if n 0 (mod p — 1), or 0 (mod 2) if p = 2, then
(X ) = X(X — D. . . (X — n + 1) n

n!

Then (xn ) is a polynomial of degree n in X and if X is an integer we obtain a binomial coefficient. If X e lp then X is close to a rational integer, so (xn ) is close to an integer. It follows that (xn ) E Zp if X E Zpe However this is not true for extensions of Op . For example, ( P2)

e ZpOl <=> N/i is congruent mod p to a rational integer <=>

e Op <=> p

± 1 (mod 8).

A classical theorem of Mahler states that any continuous function f (X) from Zp to Op may be written uniquely in the form

' (X f (X) = Z an ) n=0

n

with

an 0 as n

00.

53

§5.1 p-adic Functions

Clearly any such function is continuous since it is a uniform limit of continuous functions. Since f(m) = 1,T= 0 ac), we may use the identity

0

(m i

( mi 1-1 )

to obtain m

E(-1)— (ni)f(i) =

i= 0

(;)

(—

i

0 j= 0 ni

m

j =13

(— )a. (m ji=i

m = j=0

(. 1a. (1 — j

= am

(note that 0 0 = 1 since it comes from (Zi)(— 1) ° = 1). Therefore f (X) determines am . The hard part is showing am -4 0. We shall not prove this here because we do not need it. See Lang [4], p. 99. If a„ 0 sufficiently rapidly, then f (X) is analytic; that is, f (X) may be expanded in a power series. Proposition 5.8. Suppose r < p -11( P -1) < 1 and f (X) = a„(X ) n=0

with lard Mr"for some M. Then f(X)may be expressed as a power series with radius of convergence at least R = (rp l l(P -1) ) -1 > 1. Lemma. Let Pi(X) = Enc°_ 0 an, i X, i = 0, 1, 2, ... be a sequence of power series which converge in a fixed subset D of Cp and suppose oo for each n, and (1) a„, a0, 0 as i (2) for each X e D and every E > 0 these exists an no = no (X, E) such that i X"I < e uniformly in i (=0, 1, 2, ...).

Then lim i , Pi(X) = Po(X) for all X e D. PROOF OF LEMMA.

Given e and X, choose no as above. Then

P(X) - Pi(X)I

Max{, I a

a„,i1 . 1)0 1} = E

n <no

for i sufficiently large. PROOF OF PROPOSITION

Pi(X) =

5.8. Let

X) a0 ( = la • X " n
i=

1, 2, 3, ... .

54

5 p-adic L-functions and Bernoulli Numbers

Then integer

an integer + an+ +..., (n + 1)! n!

an, = SO

Max

lan,il

a•

< MR'.

Also,

I

an, i+k I =

a"-

integer integer ai + k (i + 1)! + • • + k)! (1

0 as i

co.

Therefore {a n i } i is a Cauchy sequence. Let an, 0 = a„, i . Then MR Let Po(X) = En-= 0 an, o so Po converges in D = I an, o I {x E Cxl < R}. The polynomials P 1 , P2,... of course also converge in D. Finally, if X e D then

E an i X"

< Max{MR

I X In} —> 0 as no —> co,

nno

nno

uniformly in i. Therefore limi _ co Pi (X) = Po(X), so f (X) is analytic in D, as desired. As an application, let us reconsider the function 5. We may expand it as a binomial series (1 +

—

= E (- )( — 1) 11 . n

Since I — il < q- ,we may let r = .4-1 . We find that the series represents an analytic function with radius of convergence at least qp - 11 (p-1 ) , just as before. In fact

exp(s logp(a>) = E n-=0

( n )( — 1)"

since the functions are analytic in s and are equal when s is a positive integer. Since the positive integers have 0 as a (p-adic) accumulation point, the functions must be identically equal.

§5.2 p-adic L-functions We are now able to consider the main subject of this chapter. Let

H(s, a, F) =

E

mEa(F) m>

o

1 =F n= 0 (a + nF)s

m— = E

a

55

§5.2 p-adic L f unctions -

where s is a complex variable, a and F are integers with 0 < a < F, and b) is the Hurwitz zeta function. Then

11(1 — n, a, F) = —

F" - 1 B„(a1F)

e

n > 1,

and H has a simple pole at s = 1 with residue 1/F. Theorem 5.9. Suppose gIF and function H p(s, a, F) on G

Then there exists a p-adic meromorphic

CplISI < qP — "P-1)

> 1}

such that

lip(1 — n, a, F) = w(a)H(1 — n, a, F), In particular, when n

n > 1.

0 (mod p — 1), or (mod 2) if p = 2, then p (1

— n, a, F) = 11(1 — n, a, F).

The function H p is analytic except for a simple pole at s = 1 with residue

11F. PROOF. Let

H p(s, a, F) =

1

1 1—s )(B i)(—FY 1- s a s— 1 F =o

Assume convergence for the moment. Then

H p(1 — n, a, F) =

nF

)(B i) ( a

i =o

J

F" - l o) - "(a) B ( a ) = w(a)H(1 — n, a, F),

as desired.

At s = 1, we have residue

1 0

(0 \ (Bi) LF\i

1 F

It remains to prove convergence. We need the following well-known result. Theorem 5.10 (von Staudt-Clausen). Let n be even and positive. Then

B„ + (p —

l)In P

where the sum is over those primes p such that p — 1 divides n (in particular, 2 and 3 appear in the denominator of each Bernoulli number). Consequently, pB„ is p-integral for all n and all p.

56

5 p-adic L-functions and Bernoulli Numbers

PROOF. We shall show that for each prime p we have Bn — 1/p or 0 mod 4, depending on whether p — 1 does or does not divide n. Assume by induction that this is true for m < n. In particular, pB,„ e Zp for m < n. Since the cases in = 0, 1 are easily treated, we assume also that n > 2 is even. From Proposition 4.1 we have B„ = B, 1 = Pn

E B n0 a=1 P

p n—

p

n

E E

(-. ) (13,)

P

a=-1 j=0 J

, EP En

n (

a

(pB i)an-ipi - 2

a=-1 j=0 i

E (pB o anp - 2 + np13 l a" - 1 p -1

pB p pn - 2 ) (mod

Zp)*

a=1

Since B 1 = B 1 e Zp if p 2. Since n is even, nB e 7Z2. Therefore we may omit the term with B 1 . We obtain

(1 —

1 P 1 )B„ a" P a=1

E

P—1 P

9

if

1 )1

n

if (p — 1)n.

Since 1 — 1 (mod p), we have Bn — 1/p or 0 mod 17. Now consider Bn + E(p _ i)i„ 1/p. By the above, this is in Z p for every p, so there are no primes in the denominator. Therefore it must be an integer. This completes the proof of Theorem 5.10. El Returning to the proof of Theorem 5.9, we note that I(Bi)(F /a)-1 1 Therefore, by Proposition 5.8 with r = iqi =1/q, we find that

MO.

()(B)() a j=o I is analytic on D = {s e C p II si < qp - 11(P - 1) }. Since qp - 11(P - > 1, this is the same set as {s e CDII 1 si < qp-11(P - 1) }, so

E

(1 s) (Bi) (li

is analytic in D. Similarly
We are now ready to construct p-adic L-functions. Let x be a Dirichlet character. If we fix, once and for all, an embedding of into Cp , we may regard the values of x as lying in Cp Also, observe that w(a) is a p-adic Dirichlet

57

§5.2 p-adic L-functions

character of conductor q and order 0(q) (= 2 or p — 1). It may be regarded as coming from a complex character if desired, but the choice is noncanonical and depends on an embedding of C C!, 1 ) into Up It is better to regard a) as a p-adic object. Note that it generates the group of Dirichlet characters defined mod q. •

(

Theorem 5.11. Let x be a Dirichlet character of conductor f and let F be any multiple of q and f. Then there exists a p-adic meromorphic (analytic if x 0 1) function L p(s, x) on Is E C p Hs' < qp- 11(P - 1) } such that

L IP

n, x) = —(1 — zo) -n(P)Pn- 1 )

zo]

Bry

n

n

n > 1.

Ifx = 1 then L p(s, pis analytic except for a pole at s = 1 with residue (1 — 1/p). In fact, we have the formula F

=

Fs—1

ai =

c°

(a) '-s

1 Lp(S,

—S

j=0

)(B.)) (—FY. a

p ,t'a

Remarks. The factor (1 — xo) -- "(p)p" - ') is the Euler factor at p for L(s, zw - "). It is a general principle that to obtain p-adic analogues of complex functions, 1/n5 has p-adically arbitrarily the p-part must be removed (intuitively, large terms if p is allowed to divide n, while at least the terms are bounded if p)(n). The expression zw - "(p) is taken in the sense of multiplication of characters given in Chapter 3. In general, xo) - "(p) x(p)w - "(p). For example, if x = Oit # 1, then xo - "(p) = 1, while x(p) = w"(p) = O. Note that

Lp(1 — n, x) = (1 — x(p)p" - 1 )L(1 — n, x)

if n 0 (mod p — 1)

(mod 2 if p = 2). In general, L p(s, x) i s an intertwining o f the functions L(s, xo)j), j = 0, 1, . , p — 2. If x is an odd character then n and xw -" have different parities so Bn , = O. Therefore L p(s, x) is identically zero for odd x. If x is even then 0 so L p(s, x) is not the zero function. The nature of its zeros is not yet understood. PROOF OF THEOREM 5.11. We show that the formula gives the desired function. Since Lp(S,

z(a)Hp(s, a, F),

= a=1 pa

the analyticity properties follow at once. At s = 1, Lp(s, X) has residue L-- Aa)( 1 /F). If X = 1 then this sum equals 1 — 1/p. Hz # 1 then the sum is p)r. 1F

a=1

Flp

z(a)

—

z

oo.

58

5 p-adic L-functions and Bernoulli Numbers

The first sum is O. If pl f then z(pb) = 0 for all b. If p,i/ f then f (F/p), so again the second sum is O. Therefore L p(s, z) has no pole at s = 1 if z # 1. If n > 1 then F

E x(a)H p(1 — n, a, F)

L p (1 — n, x) =

a=1 Oa

F 1 a- ) zw'(a)B n (-F -= - - Fn - 1 n a=1

E

(cf. Exercise 3.7(c))

Oa

F

1

= - - Fn-1

n

E zw - n(a)B4a )

a =1

eFY -

+ —1 p.- l Pi P

zo.)-"(pb)B

1 b=1

b \ n(F/p) .

If plf-. then xw - n(pb) = O. Otherwise fx' .-.1(F/p). By Proposition 4.1 we have 1 _n(p)p"_ Lp (1 — n, z) — — (Bn . - n — n 'x 1 = —(1 — za)'(p)pn -1 )B n, x .-..

El

This completes the proof of Theorem 5.11.

What happens at the positive integers'? We shall treat the case s = 1 shortly. Let n > 1. It is classical that (

on+ 1

dn+ 1

OD

E0 (z + 1mr+ 1

log F(z) = m (dz) n + 1

n!

(1 ± n, z).

Recall Stirling's asymptotic series (see Whittaker and Watson, [1 ], p. 241) log

F(z)

— (z

' 1) log z z + E

B.+1

i

z

1)

The series does not converge for complex z, but log(F(z)/\/27r) equals the mth partial sum + 0( I z 1 - rn -1 ) as z —> oo. If we differentiate (n + 1)-times (this can be justified) we obtain (_ on+ 1

dn+ 1

n!

(dz) n + 1

log F(z) — 1 i

—. n (B .)z -(n+ i) .

Note that the right-hand side converges p-adically if I zi p > 1. We therefore regard

59

§5.3 Congruences

as the p-adic analogue of 1 n+

n, z).

=

.= 0 (z + m)

Letting z = a/F, we see that for n > 1 H p(1 + n, a, F) is the analogue of F a ) = con(a)H(1 + n, a, F). co - n(a)F" +1C(1 + n, — An easy calculation shows that L p(1 + n, x) is the analogue of (1 — of(p)p + ")L(1 + n, )(con). Note that L(1 + n, zw") gives the values for even characters at odd integers and for odd characters at even integers. Very little is known about these numbers, either in the complex case or the p-adic case.

§5.3 Congruences Theorem 5.12. Suppose x # 1 and pq fx . Then L p(s, x) = ao + ai (s — 1) + a2 (s — 1) 2 + • • •

with I ao I 1 and with plai for all i 1 (note that since L p(s, x) has radius of convergence greater than 1, ai —> 0 as i oo ; so we a priori have p I a ifor large i).

We may choose F as in Theorem 5.11 so that ql F but pq,f/ F. Also we may assume z is even since everything is 0 otherwise. If] > 6 then PROOF.

Bi j! ai

1

< pi/(P — 1) . n .

il —1

r

A check of the cases j = 3, 4, 5 shows that the inequality holds for j Therefore all coefficients in the power series expansion of 1 — F.> 3

1 — s) \

3.

(F)i

j

are divisible by p. Also, the terms for ] < 2 have possibly q, but not pq, in the denominator. Similarly, 1

exp((1 — s)logp ) =

E

1

7-, (1 -

j= 0 J

S) i (lOgp
60

5 p-adic L f unctions and Bernoulli Numbers -

has all coefficients in

Zp ,

and they are divisible by pq for j

2, since

q I logp .

Therefore we need only consider

E z(a)(1 + (1 — s) log p (a>)( F s — 1 a-- i

S

— + 2a

1—

—'

12a2

pq'a

F)

We find that F 1 1 F ao __=--- — Z Aa)( — logp — — ) (mod p). F 2a 12a2 a=-1 p)ra

Clearly (1/F) logp and F/12 are in

Zp .

Since a

co(a) (mod q),

1 A Z x (a) = A 1 zw - '(a) = 0 (mod lq) (we need the same reasoning as was used in the proof of Theorem 5.11 to handle the fact that the sum only includes a with p ,i' a). This shows that

iaol 1. Next, we have

logp(al , (mod p).

logp F Z(a) ( l2 13 •1"el

F

2a

12a -

Clearly F logp/12a 2 and logp /2a are divisible by p. If p __ 5 then F/12 E p2p , SO p 'a l . If p = 2 or 3 then F/12 E 4. But a 2 1 (mod p) if p )( a, so Zp 4-, x(a)a - 2 O. Again we have pia,. Zpta AO Finally, we have

a2

— Z z(a)(log p) P•ra

F

12a2

= 0 (mod p).

Since all the higher coefficients are already divisible by p, from the above, the theorem is proved. CI Most of the congruences for Bernoulli numbers and generalized Bernoulli numbers follow from this theorem. We give a few examples. For another approach, see the Exercises for Chapter VII.

Corollary 5.13. Suppose y 0 1, pq,1' f. Let m, n E Z. Then L p(m, y)

L p(n, y) (mod p),

and both numbers are p-integral.

PROOF. Both sides are congruent to (1 0 in the notation of the theorem.

El

61

§5.3 Congruences

Corollary 5.14 (Kummer's Congruences). Suppose m n $ 0 (mod p — 1) are positive even integers. Then B. B

—

n

in

" (mod p).

More generally, if m and n are positive even integers with m and n $ 0 (mod p 1), then

(1 — pm -1 ) Brn

(1 — p" --1 )

n (mod(p — 1)pa)

(mod pa + 1 ).

PROOF. Consider L p(s, co') = L p(s, con). Then 4(1 — m, ( f") = —(1 — pm - 1 )(B.Im) and similarly for n. Also

Lp(1 — m, co') = ao + a 1 ( —m) + a 2( m)2 ao + a 1 ( —n) + a 2 ( n)2 + (since pla,, i > 1) L p(1 — n, con).

(mod Pa+ ')

The result follows. Corollary 5.15. Suppose n is odd, n $ — 1 (mod p

1). Then

B„ + 1 (mod p) n+1

B I ,„,„ and both sides are p-integral.

PROOF. Since n $ — 1, co" +1 0 1. Also con(p) Corollary 5.13, B 1 ,. = (1 — co"(p))13 1 ,„,„ =

0 since co" 0 1. Therefore, by

L(0, con +1)

L(1 — (n + 1), con ' 1 ) = (1 — p )

B

+1

Bn+1 (mod

p).

The p-integrality also follows from Corollary 5.13. Theorem 5.16. Let p be an odd prime and let hp- be the relative class number of CD( p). Then pih; •=,› p divides the numerator of Bjfor some j = 2, 4, ... , p — 3. (Later we shall show pihp <-> pih; .)

PROOF. The odd characters corresponding to Therefore, by Theorem 4.16 = 2p fl •

2

=i

j odd

( - 01 ,

Q

(

p)

are co, w3 ,...,

62

5 p-adic L-functions and Bernoulli Numbers

(Q = 1 by Corollary 4.13; w = 2p). First, note that 1 P— =aor- 1 (a)

E

B 1 ,.„ -2 = 8 1

p—1

P a=1

Therefore (2p) (

mod

ZP•

1131 , - 2) -= 1 (mod p), so we have p-4

H

h;

—1/3 1, .;) (mod p).

r= 1 j odd

By Corollary 5.15, this may be rewritten as p-4

(

h; j

=11

1 Bi + 1 )

2j+1

(mod p).

j odd

The theorem follows immediately. As mentioned in Chapter 1, a prime is called irregular if p divides Bi for some j = 2, 4, ... , p — 3. Theorem 5.17. There are infinitely many irregular primes. , pr are all the irregular primes and let in = N(pi — 1) Suppose p i , • • • (A. — 1), where N will be chosen later. It follows from Exercise 4.3 that B „In j oo as n co, n even. If we choose N large enough, then 113.1m I > 1. There then exists a prime p which divides the numerator o f B.Im. Since pi is in the denominator of B. for i = 1, . . . , r by Theorem 5.10, we cannot have p = pi for any i. Also in # 0 (mod p — 1) for similar reasons. Let in' in (mod p — 1), 0 < m' < p — 1. Then PROOF.

,

m

(mod p),

so pIB.,. Therefore, p is irregular. It follows that there must be infinitely many irregular primes, as claimed. It is not known whether or not there are infinitely many regular primes. However, numerical evidence indicates that about 61 % of all primes are regular. More precisely, let i(p) be the number of Bi , j = 2, 4, .. . , p — 3, which are divisible by p. This number is usually called the index of irregularity. Assume that the Bernoulli numbers are random mod p in the sense that Bi is divisible by p with probability 1/p. There are (p — 3)/2 Bernoulli numbers in consideration for a prime p. The probability that i(p) = k is therefore

63

§5.4 The Value at s = 1

which approaches (1)k e -112/k! as p co (Poisson distribution with parameter 1). For i(p) = 0, we find that e - 112 60.65 % of all primes should be regular. The remaining 39.35 % should be irregular of various indices. This heuristic argument agrees closely with the numerical evidence. For the 11733 odd primes less than 125000, the computer calculations ofWagstaff [1] yielded the following data: i(p)

Fraction with i(p)

0 1 2 3 4 5 >6

0.6075 0.3033 0.0746 0.0130 0.0014 0.0002 0.0000

1 1

e- 1/2

— —

k! 0.6065 0.3033 0.0758 0.0126 0.0016 0.0002 0.0000

§5.4 The Value at s = 1 We now turn our attention to the evaluation of L p(1, x). The answer is the padic version of the classical formula with the Euler factor at p removed. Theorem 5.18. Let x be an even nontrivial Dirichlet character of conductorf, let = x 1 , let be a primitive fth root of unity, and let -r(x) = Eaf AciV be a Gauss sum. Then

L(1, x) = —

(1

AP)) TOO P

Ef x(a) log

f a=1

(1 — P

PROOF. (The proof is not especially enlightening. The reader could possibly omit it without seriously impairing the understanding of subsequent results.) We shall consider the cases f = p and f p separately. I. f = p (the argument in this case is essentially due to Kummer). Then x = d for some even k # 0 (mod p — 1), and p must be odd. Let 0(X) E 0[X] be a polynomial with p-integral coefficients such that 0(1) = 1. Then 0(X) = 1 + b,(X — 1) + b 2(X — 1)2 + • • • so we may formally expand (1 — (XXV log O(X) = — E = E -Ci(1 We claim the C's are p-integral. When

1 - ( —b 1 (X — 1) — b 2 (X — 1) 2 — • • — b k(X — 1 )ky

64

5 p-adic L-functions and Bernoulli Numbers

is expanded, we obtain terms of the form i

(p-integral coefficient)!

(X —

(X — Okak

where the expression in parentheses is a multinomial coefficient. Note that for any j with ai # 0,

is another multinomial coefficient, hence integral. Therefore 1 i

al,

, ak

)

E jai e Z.

But this expression, times the "p-integral coefficient" above, is the form of the contributions to Cn , with n =Ijai . This proves the claim. Returning to the above formula, we expand further and obtain Ci

log (/)(X) =

E )(.1

i =0

i=1

Now let X = et, so

log 4(e t) =

" Ci

E i=1

co

(i

E )(- i)i E j=0

in, tm.

J

Lemma 5.19.

4 L

(-

=o

= 0 for

j>

PROOF. The left-hand side is the coefficient of tm/m ! in the Taylor expansion of (1 — et)i = ti + higher terms. The result follows immediately. 1=1

The lemma shows that the coefficient of tin/m ! is a finite sum, in fact it is m Ci

E

j=0

i=1

(— Ojim.

Now let g be a primitive root modulo p (so and let

d)(X) =

gX—1

—

ga

-a 1 mod p p — 1 divides a)

1 (X 9- 1 + X9-2 + g

+ 1).

Then gel

(1- log (/)(et) =

dt

egt — 1

=g—1+

et

—1

1 gt t e.9t — 1

— 1)B,

ni

et

—1

+ g —1

65

§5.4 The Value at s = 1

It follows that the coefficient of tm/m ! (m so CB (g m 1) m =

2) for log 0(e1) is (gm —

(i

E :)(- oitn.

7 2.

V

Let m = kp". Then cokP" = W k and

B Lp(1, w k ) = lim Lp(1 — kp", (o k ) = lim —(1 — p k P") B kP " = lim — kP" kp" kp" n—■ OE) n—■ Since gk P" # co(g)" , we have

(co(g)k — 1)L p(1, wk) = lim — (gkPn — 1) n—■ op

ci

iim - E i=1

n—■ oo

Bkpn

kp"

( i) E . (- oitPn J

j =1

- 1)

iim - E ci E (Note:

—1

j= 1

i= 1

n—■ oo

(ii )

(i

i ))• —

Since each Ci is p-integral, we may evaluate lim ik limit is O. Otherwise we obtain wk(j)/j. Therefore

(0(9)k — 1)4 (1, wk) =

1=1

1

1

termwise. If p I j then the

1 Ywk(j).

j= 1 J rj

We now return to the original formula for log 0(X). Let = tive pth root of unity and let (a, p) = 1. Since I — 1 I may expand logo OGa) =

=E i=i

E

c.

o - ca»i

1)iikP"-

V c°

=

i=

4 be any primiCa — 1 I <

( 1 — CaDI

i (ji E .)(- ojcai, i=0

with the same Ci as above. Therefore, P—1

E

co ' ( a)

10gp «) Gcl ) = E

a= 1

Ci —

P —1

(i j

E .)(-

E w - kfrovi a= 1

i=1 i j= 0

" ci " = t(wk) E E (i) .j ( iywk(i) i=1 1 j=0 rj

since P—1

E co _ k(axai

a

=1

cok (j) ,((o _k ) if p /1,/

1, we

66

5 p-adic L-functions and Bernoulli Numbers

and equals 0 if pli. We now have p-1

— I )4(1, w k ) = —

logy (1)(Ca) a=1

p-1

Ei C0 — k (a)[ —logp g + logp(1 — Caa) — logp(1 — Ca)]

= — a

p— 1

w"(a) log p (1 —

= — (a)(g)k — a=

Finally, since r(cok)-r(w -k ) = cok( —1)p = p by Lemmas 4.7 and 4.8, we obtain (note wk(g) gk 1 mod p, so cok(g) # 1) T(w kl

p-

L p(1, w k ) =

w -k (a)logp(1

P

a= 1

as desired. Note that the Euler factor (1 — ok(p)/p) = 1 so it does not appear explicitly. II. f

p.

Lemma 5.20. Let x 0 1 be a Dirichlet character of conductor f and let be a primitive fth root of unity. Then for n > 1 Bn, x

Xa)

TO()

=

L

VI

- 1 )1 i =

=1 =1 » f . 1"-

( -1 )

..n

(we may also sum for 1 i < co by Lemma 5.19.)

PROOF. Let co

oo

1

fa(t) = a

n= 0 i=

i

.

)(—

iy —jin 17 !

l) .J =0 •

00

_ 1 ) i= 0

=

(ii) (

( et_ 1)

00 =

iz=1 i

Ca — 1

Then fa(0) = 0 and f a' (t) = et/(Ca — et). 1 x(b)X' and consider the partial fraction expansion Let g(X) = g(X) Xf — 1 ------

ra 1X—

Ca •

67

§5.4 The Value at s =

Computing residues at a we obtain ra

i(a)

g(Ca)

= f L )((b)ab =

f

(f)(C a)f-1

TOO.

Therefore

E Bn, x

tn — 1

f

,(b)ebt

E 4r‘

=

n= 1

TO)

1

pJ t

b= 1 -

T(X) f -1

f

E i(a) a= 1

t a e—

_

f —1

Therefore

B„, x t" = n n!

n=1

T

oo

— —

f-1

E

i(a)fa(t).

f a=1

If we equate the coefficients of tn/n!, we obtain the lemma.

[11

As in the case f = p, we have f - 1 (p - 1)p"

L P (1 '

T( X) Ern J

i

co a

=1

(_

E

EE

1)p" .

— 1 )1 i= j

i= 1

Postponing for the moment the justification of the termwise evaluation of limP -1)" = l if p,/j, we obtain

TOO

fE - 1 lim(P-1)P E " a=1

2(a)

i= 1

1 ) I j= 1 • P

ti

But

E=

j=0

(

1— = 0,

J

SO

P

J

1

—E (.)(— J

— n

EE

r aP=1 j=0

Pli

1

= — P—.v,E=1 (cx—

1 )1•

We now have

Pi

xa ; a = 1 ŒP=1

1=1

, —

I

68

5 p-adic L-functions and Bernoulli Numbers

If is not a p-power root of unity then i — 1 i,, = 1 SO 1(a — 1 )/(V — l) l p < 1. If is a p"th root of unity then n 2 (f p). Therefore again we have — 1 )1 G 1. In both cases we have convergence, so we get I — T( 13

; )

Ea Ea i(a) logp (1 =

t(

11

c°ta

)

x) E E i(a)10gp (°; _ a )

Pf

aa

xjaglogp(1 — ciP ) — p logp(1 — c1 )1

T(X

)

PJ a= 1

(1

X(P) ) T(X) P

i(a) logp(1

fa=1

as desired (the logp(1 — ciP) is treated by a change of variables if p,}' f. If pI f then use the same technique as in the proof of Lemma 4.7). We now justify the termwise evaluation of lim j(P - "P" as promised above (yes, even in the p-adics things like this need to be checked once in a while). We know that j(Ii "P" = J + small, where J = 0 or 1. Consider the inner sums over i and j. We have

E E (coefficient)(J +

small) =

E E (coeff.)(J) + E E (coeff.)(small).

When the coefficients are p-integral, the second term is small. The problem is that the coefficients have ( — 1) i in the denominator, so sometimes they are large. Therefore we must show that (large)(small) = small. Lemma 5.21.

(i)( i — 1) i- if"

and

i= 1

7 (.(— 1 )` - ' i)

i= 1 13 j

are both divisible by i! for ni 1, the first divisibility being in Z, the second in Zp (note that we do not get /-divisibility for the second expression: let p = 3, i = 4. Then 24 divides 3 in /3 but not in 4 PROOF.

Write the monomial X'" as

xm =

at

(xi ).

1= 0

Then the ai are uniquely determined and (i ai = i= J (see the discussion preceding Proposition 5.8). Since we may obviously write any polynomial in Z[X], in particular Xi", as a /-linear combination

69

§5.4 The Value at s = I

of polynomials of the form (X)(X — 1) • (X — j + 1) = Xj + lower terms, we must have a1/i! E Z. This proves the first half of the lemma. But for any i we may let in = (p — 1)p" oo to obtain the second expression, so the lemma is proved. If is not of p-power order we have I — 1 I = 1, so we may proceed as in the case f = p: write 1 (i) i(p _ op „ = (i — 1) 119- 1) ip-1)pn- 1 — 1

Everything else inside the limit is p-integral so we may take the limit termwise. But if is a prnth root of unity (in 2), then i(Ça — 1) i is very small padically for large i, so we must proceed more carefully. Fix n and first consider i < n. Then log i i log n + v p(i ( a — 1) i) < log p 4(pm) log p (p — 1)p If plj then v p (fP - "P") > (p — 1)pn which grows faster than v p(iGa — So omitting the terms with plj does not change the limit. If p ,Kj then v(j" 1)P" — 1) > n + 1. Therefore v p(j(P - "P" — 1) — v p(i(V — 1)i) oo as n oo uniformly for i < n. It follows that we may replace j( P - "P" by 1 for all terms with i < n. Now consider i> n. By the above lemma 1 =1 ( rpvv _ y iL

0

10

./

/ (P-

'A( — 1) i)

i—1—p

P

1

log(i — I) log p

(p — 1)p

> ci > cn

for some c > 0 (for the estimate on (i — I)! see the discussion preceding Proposition 5.4). Therefore the terms with i > n do not affect the limit. We obtain the same result when j" "P" is replaced by 1 (p j) or 0 (p 1 j). To summarize, if i < n then the denominator is not small enough to cause problems, so we may take the limit termwise. The terms for i > n become 0 in the limit, so may be ignored. This completes the justification. The proof of Theorem 5.18 is now complete. The above reasoning also yields the following result, which shows that the p-adic L-functions are Iwasawa functions (see Exercise 12.3 and Theorem 7.10). Proposition 5.22. Suppose x 0 1,f p, = f . Then for s i • -r(x) f OE) 1 L p(s, x) = f a=i

- 1 )I j=

E Zp

we have

70

5 p-adic L-functions and Bernoulli Numbers

PROOF. L p(s, x) = lim L p(1 — n, x), where n = (p — 1)m —> co, and m (1 s)/(p — 1) p-adically. So —

L p(s, x) = lirn — (1 — x(p)pn -1 ) Bn, Now use Lemma 5.20. Since lim jn = lim(co( j>)" = lim< j>" = <j> 1 if we may use the above reasoning to justify the termwise evaluation of the limit and obtain the result. The details are left to the reader.

§5.5 The p-adic Regulator The question now arises regarding whether or not Lp(1, x) is nonzero. As in the complex case, we have Lp(1, z) 0, but it is a rather deep fact. However, we may quickly dispose of a special case.

Proposition 5.23. If p is a regular prime and k is an even integer with k (mod p — 1), then Lp(1, # 0 (mod p). In particular, L p(l, 0) 0 O. PROOF. We know from Corollary 5.13 that L p(1, = —(1 — p k-1 )(B k/k) 0 0 (mod p), since p,1' B k .

0

Lp(1 — k, cok)

To treat the general case, we introduce the p-adic regulator. Let K be a number field. If we fix an embedding of Cp into C, then any embedding of K into Cp becomes an embedding into C, hence may be considered as real or complex, depending on the image of K (this classification possibly depends on the choice of the embedding of Cp into C. We therefore sometimes obtain an ambiguity in the definition of the p-adic regulator. See Exercises 5.12 and 5.13). Let r = r1 + r2 — 1, with r 1 , r 2 defined as usual for K. The embeddings of K into Cp may be listed as a 1 ,..., o r i , 0 1 1 , d ri + 1 • • • + r2 6ri A- r2 1 where the ai , 1 < i < r 1 are real in the above sense, and the other embeddings , Er are complex. Let 6i = 1 if a, is real, Si = 2 if ai is complex. Let e l , be independent units of K. Then RK , p(Ei, • • ,

Er) =

det(6i logp(o-i )) <


Note that this regulator is only defined up to a change in sign, since changing the order of the o-i's could introduce a factor of — 1. We are mostly interested in p-divisibility properties, so this will not present a problem. Since there are additional ambiguities unless K is real, or CM (see Exercise 5.13), we shall usually only discuss p-adic regulators in these cases. If { e 1 Er } is a basis for the units of K modulo roots of unity, then R p(K) = RK , p ( C Er) is called the p-adic regulator of K. In Chapter 8 we shall prove the following result. The proof will rely heavily on the above formula for Lp(1, z). ,

,

71

§5.5 The p-adic Regulator

Theorem 5.24. Let K be a totally real abelian number field of degree n corresponding to a group X of Dirichlet characters. Then

n (1

2" - 'h(K)R (K)

x(p))

x Ex xo 1

\ d(K)

1 4(1 , x).

P

(Since both R(K) and \./d(K) are only determined up to sign, the above equality actually means that we can choose signs so as to obtain equality.) If we define the p-adic zeta function of K to be p (S)

= x Ex

L p(s, x)

then we obtain lim (s 04,p(s) 2" - hR p H ( 1

s-1

,E

P)

so if R p 0 0 then K p (S) has a simple pole at s = 1 with a residue which is the p-adic analogue of the residue for the complex case. We shall prove that R(K) 0 0 when K is abelian over G, so that Lp(1, O. From Proposition 5.23 combined with Theorem 5.24, we already have R p(K) 0 0 when K = G(Cp)± and p is regular. In general, there is the following. Leopoldt's Conjecture (Preliminary Form). R p(K) 0 0 for all number fields K. At present, there is no general proof of this result, although it has been verified in several cases.

Theorem 5.25. If KIO is abelian then R(K) 0 O.

PROOF'. We shall need several preparatory results. Lemma 5.26. Let G be a finite abelian group and let f be a function on G with values in some field of characteristic O. Then (a)

det(f

= fl E x(a)f(a), xEG crEG

(b)

det(f (at -1 )

f(a)) ,* = x#1 crEG

(c)

f (a) = 0 then det(f(a -1 )), t , 1 = GI 1 1T E x(a)f(0x*1 aEG

PROOF. (a) Consider the finite-dimensional vector space V of all functions h(X) on G. Then G acts on V by translation: ah(X) = h(crX). Define the

72

5 p-adic L-functions and Bernoulli Numbers

linear transformation T = L f(a)a. Let Or(X) be the characteristic function of {T} c G, so (/),(a) = 1 if a = T and 0 if a T. Then {0,}„ G forms a basis for V. Since

Tor(x) =E f (c)(/),(a X)

=

E f (04) c, _ ,(X)

E f(ra - ')C(X), the matrix ([ (at G is the matrix for T with respect to this basis. Since the characters E are linearly independent, they also form a basis for V 1X), they span V, hence form a basis). (alternatively, since 0,(X) = But Tx(X) = f (u)x(u)x(X), so the character is an eigenvector with eigenvalue Au) f (a). Consequently, T is diagonal with respect to this basis. The determinant is the product of these eigenvalues, so the first part of the lemma is proved. (b) Let W be the subspace consisting of functions h(X) with h(o) = O. Let 0,(X) = 0,(X) — 1/1 G I. Then {ll/ r(X)IT 0 1} forms a basis for W. Using the fact that tp i (X) = O (X), we easily find that (f (ar 1 ) — f is the matrix of T restricted to W for this basis. As before, the nontrivial characters diagonalize T restricted to W, so part (b) follows. (c) Adjoin a row and column to (f(at 1 ) — f (a)), 1 to obtain the following (index the rows by cr, the columns by T): „

x

E, x(

E

x

E,

(o 1 f()

f (at') — f(a) . •:i

Now add the first column to each of the other columns, then add each of the columns of the resulting matrix onto the first column. The final result is 1 GI 1 • • • 0 1. 07 - 1 ) • • (

E,

We have used the fact that f(a) = 0 to obtain the zeroes in the first column. Using the result of part (b), we obtain the result. This completes the proof of Lemma 5.26. 0

Lemma 5.27. Let K/0 be a finite Galois extension. If K is real then let a 1, • • - ar+ 1 be the elements ofGal(K/0). If K is complex then let 0. 1, • • • , + 11 di, • • • 1 6r+ 1 be the elements of Gal(K/0) (we regard K as a subfield of C). There exists a unit e of K such that the set of units {ecrIl < i r} is multiplicatively independent, hence generates a subgroup of finite index in the full group of units (such a unit is called a Minkowski unit).

We shall find a unit E such that I Ea I > 1 but I Eai I < 1 for i 1 (the absolute value is the complex absolute value corresponding to a fixed

PROOF.

73

15.5 The p-adic Regulator

embedding of K into C). The existence of such a unit is usually implicitly proved during the proof of Dirichlers Unit Theorem. However, since it is rather difficult to isolate this step from many treatments of the subject, we shall reverse the steps and derive the existence of e from the Unit Theorem. Let E be the group of units of K and consider the mapping L: E W defined by L(n) = (log 117' 2 1, ..., loglr(r+ 1 1). Note that logl if" I = I . The kernel of L is exactly the roots of lOg unity in K by Lemma 1.6. By the Unit Theorem, the image must be a free abelian group of rank r. A bound on L(ri) gives a bound on the conjugates of hence on the coefficients of the irreducible polynomial for q. It follows that there are only finitely many images L(q) is any bounded region of W, so the image of L is discrete. Therefore it is a lattice M of maximal rank. Consider the "quadrant" Q = {(x 2 , , x r , e prIx. < 0 for 2 < i < r + 1). Then M Q # 0. Let E e E satisfy L(E) e M Q. Then logl Err' 1 < O for 2 i r + 1 and log 1 Ea 1 1 = —Eri log 1 > O. It follows that 1E 6'1 > 1 but < 1 for j# 1, as desired. We claim that e is a unit of the type asserted in the lemma. For the proof we need the following.

Lemma 5.28. Let (a ii) be areal square matrix with aii > 0, ao 0 for i j, and such that E, a > 0 for all]. Then det(a ii) 0 O. PROOF. If det(a ii) = 0, there exists a non-zero vector (x i) such that E au x, = 0 for each j. Let Ix, be maximal among the entries of the vector. By changing signs if necessary, we may assume xk > 0, hence x k > x i for all i. Then 0=

E ax , E aik xk (since aik

0 for j# k)

= (E aik )x k > O. contradiction. Returning to the proof of Lemma 5.27, we may assume a i = id and let au = bi log1 E6

.

Then a S i log e I > 0 and aii < 0 for i j. Since a 0, we have Iri=1 au = —ar+l j > 0 for ] r + 1. Lemma 5.28 implies that R K (e' '

, ecr r) = Idet(a ii)1 # O.

Therefore dri , . . , e must be multiplicatively independent, otherwise there would be a linear relation among rows of the determinant. This completes the proof of Lemma 5.27. El We now prove Theorem 5.25. We may assume that K is totally real, since if K is imaginary then R(K) = (11Q)2rR p(K+) (use the same proof as for Proposition 4.15, changing log to logo). Fix an embedding of K into Co,

74

5 p-adic L-functions and Bernoulli Numbers

let {1 = al, ar+ = Gal(K/0), and let E be as in Lemma 5.27. By Lemma 5.26(c) (this is where we need G = Gal(K /0) to be abelian) we have ,

Ro(e2,

det(logo(e"J )) 2 j, j r + 1

i)

1

IGI

H z( a) log(). xE

We now need the following deep result, which is the p-adic analogue of a theorem of Baker.

Theorem 5.29. Let ce i ,

, an be algebraic over 0 and suppose log o ai,. logo ; are linearly independent over Q. Then they are linearly independent over CD = the algebraic closure of CI in Cp ( for a proof; see Brumer [1]). 0

Since el, . , e'r are multiplicatively independent, it follows easily that 1ogo(0), , 1ogo(0) are linearly independent over 0 (we need Proposition 5.6). Since log er + 1 ) = — (

i=1

we have

x (0-) log

(x(0-) — Aar+

() =

log (').

i=

If z 0 1 then z(o- i) z(o-, i ) for some i, so not all coefficients are zero. By the above theorem, the sum does not vanish. Therefore R o(E'2,

,

O.

But Ro(e62, , err+ = [E: ElRp(K), where E is the full group of units of K and E' is the subgroup generated by + 1 and e2, , e'r+', which is the same as the subgroup generated by + 1 and {ell < i < r} . This completes the proof of Theorem 5.25. Li Corollary 5.30. Let x

1 be an even Dirichlet character. Then / ( 1, z)

O. 111

By Theorem 5.18, we know that L o(1, z) is essentially a linear form in logarithms. So why did we not apply Theorem 5.29 directly'? The problem is that the logarithms in question are generally not independent over Q. In certain cases we know all relations. For example, the only relation among llogo(1 — 43)1 1 a < (p — 1)/2) is that the sum is O. So we may use the argument used above to obtain a situation where Theorem 5.29 is applicable. But in the general situation, there can be many more relations and the analysis becomes much more complicated. We shall discuss this matter more fully in Chapter 8.

1 5.5 The p-adic Regulator

75

For later reference, we now give another version of Leopoldt's Conjecture. Let K be a number field. For each prime A lying above p, let U, denote the local units of Ko and U 1 , 0 denote the principal units, that is, the units congruent to 1 modulo A. Let U=

fl

U0 and

U 1 = T1 U 11,A. .

AI P

We may embed the global units E in U: Ec+U

Let E1 denote those e. whose images are in U 1 . Then E l is a subgroup of E of finite index (since r."A E U 1 0 ), so E l is an abelian group of rank r = r1 + r2 — 1. Let E 1 denote the closure of E 1 in the topology of U 1 . Since U 1 is a Zn module (s:uF-- us), E 1 is also a 4-module. What is its rank? Leopoldt's Conjecture. The 7L-rank of E 1 is r1 + r2 — 1. Theorem 5.31. Let K be totally real. Then R p(K) 0 0 -4=> the 7L-rank of E l ri — 1. is Remarks. One may be tempted to think that E 1 must have rank r 1 + r2 — 1 since E l has that for its 7/-rank. But consider the following. The group generated by 7 and 13 in 0 3x has 7/-rank 2. But 7 1°g3 13 = 131°g37, and log3 13/ log3 7 E . Therefore 7 generates the closure of the group, so the 7/ 3 -rank of the closure is 1. If there is only one A above pin K, then the theorem says that R p(K) 0 -4.> units which are independent over Z are independent over Zn . This is not necessarily true for nonunits, as the above example with 7 and 13 shows. Also, if there are several primes above p, it is not sufficient to consider only one U 1 , A . For example, if p splits completely then each U 1 , 0 is a Znmodule of rank 1; so if r1 + r2 + 1 > 1 then the units must be 4-dependent in each U 1 A . But the relations are different for different A, so it is still possible for the units to be I n-independent in U 1 . To be more precise, suppose e. 1 , , Er are Z n-dependent in U l . Then there exist al , , a,. e Zi, such that Eli • • • = 1 in K o for all A. This means that if ai are rational integers with ai —> ai p-adically ,then - ••

A. Since the h-adic valuations are different for different A, the fact that the limit is 1 for one K0 does not imply anything about the limit for other A. However, Erarn —>

1

in KA for each

if the units are 4-dependent then it is possible to get 1 as a limit for all K0 simultaneously.

76

5 p-adic L-f unctions and Bernoulli Numbers

5.31. Suppose the Zp- rank of E l is less than r = r1 — 1. Let c i , , E r be a Z-basis for E l modulo roots of unity. Then must be Zp- dependent in U 1 , say

PROOF OF THEOREM

ra

=

(a E p , some ai o 0).

Let L be the Galois closure of K/U. Suppose that Ix I A , and IxI 2 are the absolute values corresponding to primes h i and h 2 of K lying above p. When these absolute values are extended to L they are related by I x1 1, 1 = I axi o2 for some a E Gal(L/U) (in fact, a = Y 2 , where Y i lies above hi). Therefore, if in K ft , (hence in (4 1 )

eV. • • • crar- ---÷ 1

then (1)a 1n • • •

(s7.)arn

1

in 42 .

Fix a prime ho lying above p in K and a prime Y o lying above • • era' = 1

in L. Then

in Ko for all ,h

if and only if ()'" • • • Kr = 1

in L,0 for all a e Gal(L/0)

(does (Ea) = ( ) a? No, since a is not always an automorphism of 40/Up ; it does not necessarily fix Up if p splits, so a does not even make sense as a p-adic map. It is defined only before embedding in 40). Taking logarithms (we may assume L y . c Cp), we have ai 1ogp(c7) = 0

for all a-.

Clearly this implies that R p(E l , , E,.) = 0, hence R p(K) = O. Conversely, suppose R(K) = O. Then there exist a, E L , o such that ai 1ogp(e7) = 0

for all a;

but we want ai E Zp a We may assume that one of the ai's equals 1. Let Gal(4 0/0p). Then

TE

Ea since

T

log,,() = 0

for all a;

permutes the a's, we have aT 1ogp(c7) = 0

for all a.

Letting T denote the trace from Lyo to Up we obtain T(ai) log() = 0

for all a.

77

§5.6 Applications of the Class Number Formula

Since one ai = 1, at least one of the T(ai) O. Upon clearing denominators we obtain a relation with coefficients in ZpO Reversing the steps from the first half, we find that (we now may assume ai c Zp for all 0 (gq)ai • • • (g ro- yir =

in 40

(root of unity)

for all a. If we multiply each ai by the same integer, chosen suitably, we may assume the root of unity is 1. Then, continuing backwards through the above, we have ell • • • era' = 1 in KA for all h, so Ei, , Er are Zp - dependent in U 1 . Since E l is generated (modulo torsion) over Zp by e1,. , Cr ' we must have the Zp-rank of R I < r = ri — 1. This completes the proof. 0

Corollary 5.32. If K/CP is abelian then the 7Z-rank of E i is ri + r2 — 1. PROOF. If K is real, use Theorems 5.25 and 5.31. If K is complex, then the corollary is true for K + Since 1.1 + r2 — 1 is the same for both fields, the result follows easily. 111

§5.6 Applications of the Class Number Formula We now use the p-adic class number formula (Theorem 5.24) to deduce results on class numbers. Proposition 5.33. Suppose K is a totally real abelian number field. If p does not divide deg(K/0), if there is only one prime of K above p, and if the ramification index of p is at most p — 1, then I R(K) I .\/d(K)

<1.

PROOF. Let K p denote the completion of K at the prime above p and let (9 p be the ring of integers of K. By the assumptions on p,deg(K p/Op) = deg(K/0) and also the Galois groups may be identified. If x E K p and 1x1 < 1 then 1x1 < p -11(P - 1). Therefore (cf. Lemma 5.5) logp(1 ± X)

(— 1) n+ 1

E

n=1

since all terms in the sum are in (Op . It follows easily from the definition of the extension of logy that log y E E &, for all E E K. Let El, , E-n- 1 (n = deg(K/0)) be a basis for the units of K modulo , fin } 1. Let 13„ = 1. Then {f3, 1+11, and let fii = logy ci , 1 < i n generates a 7L,,-submodule of Op . Let , an } be a basis for (9,, as a 7/,,module. Then we can write =

au cx; with j= 1

aii E Z p

78

5 p-adic L-functions and Bernoulli Numbers

Let g e Gal(K p/CP p). Then f3 = E aij oc7 , so det(r) i

=

det(api,,.

The p-part of the discriminant of K is the discriminant of K p/Up (there is only one prime above p), so we have I N/ d(K)I p

I N/ d(1
Also log p(q) • • 1

det(K) = det( • • • Since obtain

log() = 0, we may add all the columns onto the last one to det(fi) = deg(K/O)R p(K).

Therefore, since p,F deg(K/0), we have

R(K)

det(K) det(a)

/d(K)

\

because aii c

Zp

= Idet(a o) l p

for all i, j. This completes the proof.

1=1

Remark. Actually, the proposition is true in much more generality. Let K be totally real of degree n and for each prime A above p let NA denote its norm and vi, the number of p-power roots of unity in Kit . Then npR p

<1. (NA)(v,)

In particular, nR p N

<1. /d p

The proof involves an extension of the above ideas (see Coates [7]). Theorem 5.34. If plh+(0(4)) then plh - (0(p)). Therefore plh(O(p))<=>/) divides B i for some j = 2, 4, ... , p — 3. Remark. At present, there are no known examples where p1h+(O( P))• It is a conjecture of Vandiver that this never happens. PROOF OF THEOREM 5.34. The characters corresponding to 1, w2 , , (DP'. Let n = — 3). Then

h+R; = P— 3 II / N/d+ j= 2 i

j even

(

1,

a).

Q(4) 1 are

79

§5,6 Applications of the Class Number Formula

Since

satisfies the hypotheses of Proposition 5.33, we have 1. If plh÷ then pl L( 1, co') for some j = 2, 4, ... , p — 3. By IR; /,,/d + Corollary 5.13, 4:DGp) +

0

L o(1, 0)

L(0, (o')

—(t — w - (p))B Œ„ _L = — B 1 _ (mod p).

Since p—4

h-

H (-1/3 1 , w ) (mod p)

1 i odd

(see the proof of Theorem 5.16), and since all these B" are p-integral (Corollary 5.13), we have p lh - , as desired. This completes the proof. Later, we shall give another proof of Theorem 5.34 which depends on class field theory but not on p-adic L-functions. Before giving more applications, we need to know about logarithms of units. Lemma 5.35. Let KILD be an extension of degree n, with n < p — 1. Assume that p is totally ramified: (p) = An . Suppose g is a unit of K which is congruent to a rational integer modulo Ac 0). Then log o e 0 mod Ac . PROOF . Let 71 generate in (90 , so e = a + + ••• with a, b, . . e Z. Since = p cin < p - 1,1(p 1) ,

a(1 + (b I a)n( + .)

we have log o(1 + (bla)n` + . • ) 0 mod Ac by Lemma 5.5. So log o e logo a. Let N denote the norm from K0 to 13 p (which may be identified with the norm from K to 0). Then + 1 = Ne Na = a" mod Ac Therefore n logy a = log y a" = logo(+ a") is complete.

0 mod Ac. Since p n, the proof

We are now able to prove a famous result of Kummer which will be useful for treating the second case of Fermat's Last Theorem. Theorem 5.36. Assume p is a regular prime and let g be a unit of 43G o If e is congruent to a rational integer mod p then g is the pth power of a unit of (Ç). ).

(Note that the congruence is mod p, which is much stronger than mod(1 — C), which always holds. Also, the converse of the theorem is true (Lemma 1.8). See also Exercise 8.1.) PROOF. We may write e = Cag i with e t real, by Proposition 1.5. Every element of ZE + C -1 ] is congruent mod(1 — C)(1 — c -1 ) = 2 — (( + --1 ) to a

80

5 p-adic L-functions and Bernoulli Numbers

rational integer (simply replace C + by 2). Also Ca= (1 + (C — 1))" ça 1 + a(C — 1) mod( C — 1) 2 . If congruent to a rational integer mod( C — 1) 2 , we must have pia. Therefore E = e i is real. From now on we work with K = Q(Cp)+. Let A = — c)(1 — 1)/ the prime above p. Then A(P- 2 (p). By Lemma 5.35, log,, e —= 0 (mod p), so (1/p) log p Suppose E is not a pth power. Then we may find (real) units e2 , • • • 9E. (r = (p — 3)/2) such that the group E' generated by + 1, e, e2 er is a subgroup of index prime to p in the full group of units E (Proof: Let ri 1 , , rh. be a basis for E, so e= +H ri`,4 with some ai 0 0 mod p, say i = 1. Let = ?if for j > 2. Then E' has index +al 0 0 (mod p).) Therefore ,

,

iRp(E)1 = I[E : E]R,,(K)I = I R(K)I. i Let J3 = (1/p) log,, 8 1 , = log,, ei , 2 r, fir + = 1. Then ,B1, • • 9 fir + 1 generate a 7L,,-submodule of Co . As in the proof of Proposition 5.33, we have

det([37) /c/(K)

<1

1 and det(K) — • deg(1(C7, )/Q) • R p(E').

\

Therefore R(K)

II < 1 .

But 2r 11

P-3

+R P

= H L,,(1, w),

jc-1

j= 2 jeven

so p must divide L,,(1, wi) for some j. This contradicts Proposition 5.23. The proof is complete. We now give another proof using class field theory. As above, we may assume e is real. Raising e to the (p — 1)st power if necessary, we may assume that e 1 (mod p) (eP - is a pth power is a pth power). Let n = 4 - 1. Note that nP - Vp is a unit of Z[C] and that every element of Z[C] is congruent to a rational integer modulo E. We may write

e = 1 + pa + pity with

y

E Z[]

and

a

E

Z.

Then 1 = N41,(0/Q(e) (1 + pa) P - —= 1 + (p — 1)pa 0 1 — pa (mod pm), so nia and e 0 1 mod pm. Since E — I is real, v p(e — 1) is a multiple of 2/(p — 1)(= 1/e for Q(C p)+ ). Therefore vp(e. — 1) > 1 + 2 P — 1'

or

e 1 mod pn 2 (or mod nP').

81

0.6 Applications of the Class Number Formula

Consider the polynomial f (X)

= (7rX — 1)" TtP

E

Clearlyf(X) is manic and since E 1 mod nil + 1 , the constant term is in But pl(lj) for 1 p — 1, so the other coefficients are also in Z[C]. Since f(0) = ( — 1 + )17cP 0 (mod 7r) and f ' (0) = pI 7tP 1 0 (mod it), Hensel's Lemma implies that f(X) = 0 has a solution in the completion Z[C]. It follows that E l /17 E p[(] (alternatively, since E 1 mod 7tP+ 1 , we find that exp((l/p) log p E) converges, and its pth power is 8). Suppose now that E is not a pth power. Then C)(C p , E l /P)0(C p) is a nontrivial abelian extension of degree p. Since E llp E Op((p), the prime (n) splits completely ; in particular, it does not ramify. The archimedean primes are all complex, so cannot ramify. Let g(X) = X P — E. The relative discriminant divides N(08 1 /P)) = N(pE (P - 1 ") = (n) (P - 1)p, where N is the relative ideal norm. Therefore the primes other than (g) are also unramified. The extension is therefore unramified everywhere. By class field theory, the degree of the maximal unramified abelian extension equals the class number. Consequently, p divides h(CP(C p)), which contradicts the assumption that p is regular. Therefore E is a pth power, and the proof is complete. We conclude this chapter with two results on quadratic fields. Theorem 5.37 (Ankeny-Artin-Chowla). Let p 1 (mod 4) and let h and e = (t + u p)/2 > 1 be the class number and fundamental unit for C)(\/P).

Then U —

B (p -

1 )/2

(mod p)•

Until now we have been able to ignore the ambiguity in sign for the p-adic regulator. But now we are forced to choose signs. From the classical class number formula for 0( p), we have (Exercise 4.6) PROOF.

p- 1

E — 2h =

H (1—

a=1

where = e 2'71/ and x is the character for CQ(\/P). We also have the Gauss sum r(x) = E z(a)cap = Note that if we had chosen a different pth root •), and also e+ 211 could have of unity for IV we could have had r(x) = appeared on the left-hand side of the above formula. We also made the choice e > 1. However, in the p-adics, there is no canonical way to choose p p, and E. But we can choose them so that the above relation holds and also T(x) = p: Fix an embedding of C11 into Cp(note that since x(a) = + 1 or 0,

82

5 p-adic L-functions and Bernoulli Numbers

everything is algebraic). Since the above is an equality in Now take p-adic logarithms:

a, it holds in Cp .

13-1

2h logo e = —

E z(a) iogpo — 4)

a=1

P L Z(a) 1og p (1 — 4)

=

IT 1

= j)(1

L ,(1, y)

=/L(1, y).

Therefore 2h logp E

= Lp(1, y),

NA" which is the class number formula with no ambiguity of sign. Clearly z = di 1)/2 since y is quadratic of conductor p. By Corollary 5.13, Lp(1, y) _=. Lp(1 —

p

— 2

1 ) ,x

= _(1 _ pp- 0/2) B = 2B( p _ 1)/2

(

_

1)/2

(mod p).

But (cf. Exercise 5.15) t logo e = 1ogp (i) + logp (1 + li)

u 0+— t

p (mod p).

Therefore

hu t

B (p_ 1)/2 (mod ji),

but since both sides are rational, the congruence actually holds (mod p). This completes the proof. 171 Since it can be shown that h < p, this congruence actually determines h if u * 0 (mod p), or equivalently if p does not divide B (p _ 1)/2 (note p,}/ t since N 5 , 1 / E) . For p < 6,270,713, no examples of u 0 are known (Beach, Williams and Zarnke [1]). However, if we assume B (p _ 1 ) / 2 to be random

§5.6 Applications of the Class Number Formula

mod p (but see Exercise 5.9), then the number of p p -- 1 mod 4) should be

83 x with pIB (p _ 1 )1 2 (and

I log log x; so up to 6,270,713 one would expect only around one or two examples. Therefore the fact that none exist should not be considered decisive. Proposition 5.38. Let in 1 be squarefree and assume 3 does not split completely in 12)(i— m). If 3 divides the class number of cU( \,/3m) then 3 divides the class number of 0( \/— in) (we allow 31 m, in which case 1:)(0m) = CI( \/m/3)). Remark. The result is also true if 3 splits, but our proof does not work. Later we shall prove the following more precise result, due to Scholz: Let r and s be the 3-ranks of the ideal class groups of (1:1, /3m) and CP(/— m), respectively. Then r + 1 > s > r. Whether s = r or r + 1 depends partly on the units of 0(3m). That the units could have an effect can be seen in the

present proof. If m = 3387 then the class number of GX.../3m) = (14\/1129) is 9, but the class number of 11(.\/— 3387) is 12. Therefore we cannot replace 3 by 9 in the proposition. PROOF OF PROPOSITION 5.38. We may assume M > 3 since the proposition is vacuously true for in < 3. Let x be the character for ED( \/— m). Then xa) = xo) 3 is the character for 0(0m). Let E, h, and D be the fundamental unit, class number, and discriminant for (1/3m). As in the proof of Theorem 5.37, or by the class number formula since we need not worry about signs here, we obtain (

1

xo)(3)) 2h log 3 E

= L 3 (1, xw)

L3(0, x(0) = —(1 — Z( 3))13 1, x (mod 3).

If 3)(m then 31D. As in the previous proof, or by Proposition 5.33, we have llog3 C/.\ /DI 1. Also in this case x0)(3) = 0, so the Euler factor disappears. If 31m then 3,I'D, so xo)(3) 0 0. Therefore the Euler factor contributes a 3 to the denominator. But llog 3 E I < 1 so log 3 e 0 mod 3 (since 0 3 (e)/(1) 3 is unramified, 3 generates the maximal ideal). This cancels the denominator. Consequently, in both cases the right-hand side is h times something integral. If 31h, we therefore find that 3 divides (1 — x(3))B 1 , x (note that if log 3 E 0 (mod 9) then we do not need 31h). Since we have assumed 3 does not split in 12)(/— m), x(3) 0 1. Therefore 3 divides —B1 x = h(G m)). This completes the proof. LI

84

5 p-adic L-functions and Bernoulli Numbers

The general philosophy to be learned from the proofs of Theorem 5.34 and Proposition 5.38 is that the p-adic L-functions at s = 1 contain information about units and class numbers for real fields, while at s = 0 they contain information about relative class numbers. Since we have congruences between these values, we can obtain results as above. However, the character w appears, so it is helpful to have Q(Cp) nearby, either explicitly, as in Theorem 5.34, or implicitly, as in Proposition 5.38. All this will be made more precise later, when we discuss reflection theorems.

NOTES

For more on p-adic analysis, see Amice [1], Iwasawa [23], Koblitz [1], or Mahler [I]. A simple proof of Mahler's theorem is in Lang [4]. A version of the p-adic logarithm and exponential appeared in the work of Eisenstein [1] and of Kummer [3]. The construction of p-adic L-functions given above and the analogy with the values at positive integers is from Washington [2]. Other constructions can be found, for example, in Kubota-Leopoldt [1] ( = the original construction), Amice-Fresnel [1], Coates [7], Fresnel [1], Iwasawa [18], [23], Serre [2], and Lang [4]. For other treatments of the positive integers, see Diamond [3], Hatada [I], Shiratani [5], and Koblitz [3]. p-adic L-functions have been constructed for all totally real fields by Barsky [4], Cassou-Noguès [4], and Deligne Ribet [1]. See also Katz [7]. For p-adic L-functions in other settings, see. Amice-Vélu [1], CassouNoguès [6], Coates-Wiles [4], Lichtenbaum [4], Manin [2], [3], [4], Manin-Vigik [1], Via [2], Mazur-Swinnerton-Dyer [1], and several of the papers of Katz. For work on the zeros of p-adic L-functions, see Barsky [6], Sunseri [1], Wagstaff [2], and Washington [12]. For information about the behavior of p-adic L-functions at s = 0, see Federer-Gross [1], Gross [2], Ferrero-Greenberg [1], Lang [5], and Koblitz [4]. The last three give the relationship with the p-adic F.-function

(Morita [1]). Theorem 5.16 is due to Kummer. For generalizations, see Adachi [I], R. Greenberg [3], and Kudo [3]. Theorem 5.17 has been generalized: for any N > 2 and for any proper subgroup H of (Z/NZ) , there are infinitely many irregular primes not in H. See Metsankyla [9]. The probability arguments seem to have originated with Lehmer [1] and Siegel [1]. The calculation of L p(1, x) given above is partly from Washington [6], which is based on ideas of Kummer, and partly from a modification of Washington [4], which is based on the proof of Leopoldt [10] (see also Iwasawa [23]). Other methods may be found in Amice-Fresnel [1], Koblitz [3], [4], and Shiratani [3].

85

Exercises

It is possible to determine the sign of R/ /d canonically: choose orderings in the determinants for R and \./d so that the archimedean RI \/(-1 is positive. Then use the same orderings in the p-adic case. This gives the correct sign in the class number formula. See Amice-Fresnel [I]. Leopoldes conjecture is from Leopoldt [9]. The reduction of the conjecture to the p-adic version of Baker's theorem (proved by Brumer) is due to Ax. For other work on the conjecture, see Bertrandias-Payan [1], Gillard [3], G. Gras [1 ], Serre [3], and Waldschmidt [I]. The last paper shows that the l p-rank of the units is always at least half of the i-rank. There have been occasional papers claiming to prove Leopoldt's conjecture, but they all appear to be incorrect. Some of the congruences of Ankeny-Artin-Chowla [1] were also discovered by Kiselev [I]. EXERCISES 5.1.

Let K be a finite extension of Up of degree n. Show that there is a constant C depending on n, but not on K, such that I log o x I C for all x E K.

5.2. Show that log o : Co

Cp

is surjective.

5.3.

Let K be a number field and let S be a finite set of places of K including the archimedean places. An S-unit a c K is an element satisfying It = 1 for all places çt S. Show that if S is sufficiently large then Leopoldt's Conjecture is not true for S-units; namely in-rank < E-rank = # (S) - 1.

5.4.

Show that

L p(1, x)

1' B i -F x(a)(-log p + a F a=1 j=1

E

(there does not appear to be an easy way to transform this expression into that of Theorem 5.18).

5.5.

Let K be an abelian field with X its group of Dirichlet characters. Let K,

p(S) =

Lp(s, x).

•

n) 11, (1 - x(p)p' - 1 ) if n > 0, n Show that 411 - n) = 4(1 Show that K AO vanishes identically if K is complex.

0 (mod p - 1).

5.6. Let i be even, 0 < t < p - 1. Let tri be the smallest integer u > 0 such that B, p,, (mod p2 m + 1). Show that tti v p (L p(1, w i)). (Hint: Theorem 5.12.)

0

5.7. Let e be a unit of (11((;). Show that if c is congruent to a rational integer modulo a sufficiently large power of p then e is a pth power (this result will be refined in Chapter 8). 5.8. (Ankeny-Artin-Chowla). Let m > 1 be square-free, m 1 (mod 3). Let (t u\/3m)/2 he the fundamental unit for LP( \/3m). Show that h(CP( \/ - m)) - (to It)h(C.P(137n)) (mod 3). (Be carefirl: it is necessary to expand log 3 to the third term.)

86

5

5.9.

Let p 3 (mod 4). Use the Brauer Siegel theorem to show that log h(0(y — p)) — logy p. Conclude that p,1/13(1, ± 1)/2 for all sufficiently large p. Also, show that if b 13(,, iv 2 (mod p), O < b < p, then blp 0 as p or.). Therefore B(pi 1 )12 is not "random" mod p. (Actually, h < \,/p log p, hence p 13( , 1) / 2 , for all p 3 (mod 4)).

5.10. (J. C. Adams) Show that if (p

p-adic L-functions and Bernoulli Numbers

OA' i but p9Ii,then pg I B,

5.11. (a) Show that L v(s, 1) = (1 (11p))(s 1) - + ao + a ,(s af E Ep for i O. (b) (Carlitz) Show that if pg(p — 1)11, then p9 1(B 1 4 1 lp — 1).

1) + - - • where

Cp, 5.12. Let K = CP(a), where a 3 = 2. The fundamental unit is a - 1. Let 6 i : K i = I, 2, 3, be the ernbeddings of K into C p . Show that for any i we may choose the embedding Cp C so that (A is real and the other two ernbeddings are complex. We therefore have three possible regulators: R 1 , R2, R3. Show that if i j then R i/R i is transcendental (use Theorem 5.29). 5.13. Use Theorem 4.12 to show that if K is a CM-field then the p-adic regulator is independent of the choice of labelings of the ernbeddings of K in Cp . 5.14. Let it = (1, 1. Suppose c is a unit of Z[(p] such that e a t- bit' mod It' with a, b E p,1/ ab, and c 2. Show that if cl(p — 1) kt E then ep(logp e) = cl(p — 1). In fact, show that log!, e (bia)n` mod Tr" 1 - (Hint: look at the proof of Lemma 5.35.) 5.15. (a) Show that if r E(LP then low Umod p. (b) Let It = p - 1, and let a E C(Cp). Show that a' 1 = rst, where r c (LP (possibly divisible by p), s is a root of unity, and t 1 mod n2 . (c) Let a E Cg,). Show that log p a 0 mod n2 . 5.16. (a) Let x be an even Dirichlet character. Show that 4(s, x) = 0 if and only if XaC i (P) = 1 . (b) Show that there exist quadratic odd characters x l and y 2 with x i (p) = 1 and L2(P) = — 1. (Hint: Quadratic reciprocity plus Dirichlet's theorem). (c) The classical complex L-functions for even characters satisfy a functional equation of the form f (s)L(s, y) = hx(s)g(s)L(1 s, where f and g are analytic and independent of x, while hi(s) may depend on x but is nonvanishing. Show that there is no such functional equation for p-adic L-functions (of course. we do not allow f and g to be identically zero).

Chapter 6

Stickelberger's Theorem

The aim of this chapter is to give, for any abelian number field, elements of the group ring of the Galois group which annihilate the ideal class group. They will form the Stickelberger ideal. The proof involves factoring Gauss sums as products of prime ideals, and since Gauss sums generate principal ideals, we obtain relations in the ideal class group. As an application, we prove Herbrand's theorem which relates the nontriviality of certain parts of the ideal class group of U(Ç) to p dividing corresponding Bernoulli numbers. Then we calculate the index of the Stickelberger ideal in the group ring for 41( ptl) and find it equals the relative class number. Finally, we prove a result, essentially due to Eichler, on the first case of Fermat's Last theorem. In the next chapter we shall use Stickelberger elements to give Iwasawa's construction of p-adic L-functions.

§6.1 Gauss Sums In order to prove Stickelberger's theorem, we need to study Gauss sums, which are also interesting in their own right. The Gauss sums used here are not the same as those used earlier, but there are many similarities. Let F = Fq be the finite field with q elements, q being a power of the prime p. Let Ç be a fixed primitive pth root of unity and let T be the trace from No 21 pZ. Define tif : F —, C x ,

if/(x) = V,; (x),

which is easily seen to be a well-defined, nontrivial (T is surjective) character of the additive group of F. Let X: IF' ---■ C' 87

88

6 Stickelberger's Theorem

be a multiplicative character of F . We extend z to all of F by setting z(0) = 0 (even if z is the trivial character). Note that zq - is the trivial character, so the order of z is prime to p. If q p, such characters are not the Dirichlet characters studied earlier. The concept of conductor will not enter into the present discussion. Define the Gauss sum g (X)

=—

ACI)IP(a) tie [F

If z has order m, then g(Z) E

A quick calculation shows that g(1) = 1.

0( ,p)*

Lemma 6.1. (a) g(x) = — 1 )g(x) (b) if z (c) fz

1, q(z)g(i) = z(-1)q; 1, g(z)g(z) = q.

PROOF. (a) is straightforward. (b) follows from (a) and (c). For (c), z (ab -1 )tp(a — b)

g(z)g(z) = cs,13*0 =

b)

Z(C)110C —

(let c = ab 1)

b,c *0

= E ;((1)17fj(0) + E x(c) E tfr(b(c — 1)) b*0 ct0,1 1;00 = (q — 1) +

E z(c)(— 1) = q. c# 0, 1

This completes the proof. If x i , z 2 are two multiplicative characters, we define the Jacobi sum J(Zi, X2) = — E xi(a)z2(l — a)aeF

More generally, we have

E

J(Xi, • • • , r.) = a1-1-

Xi(a1) • • •

+a„ 1

but we do not need this for n > 2. Note that if x i and z 2 have orders dividing in then J(x 1 , z 2 ) is an algebraic integer in CDG.). Lemma 6.2. (a) J(1, 1) = 2 — q; (b) J(1, z) = 1) = 1 if x A 1; (c) AZ, = A -1 ) if z 1 ; (d) J ( (1, Z2) = 9(x1)9(x2)/9(z' X2) if xi o 1, z 2

1, and Z1Z2 0 1.

89

§6.1 Gauss Sums

PROOF. (a) and (b) are easy. To prove (c) and (d), we compute Z1(a)Z2(b)0(a + b)

g(X1)9(X2) = a, b

=

>

1(CI)

—

(b)

a, b

=

a, b

Xi (0)(2(b — a)0(b) + z 1(0)0( — a

b*0

11. X1X2 1, then the second sum vanishes. If x i x2 = 1, then it equals Xi( — 1)(g — 1). The first sum equals (let a = bc) b,c b 0

x i (b ) 2 (b) i (c) 2 (1 — c)0( b ) = g ( x x 2) A xi, z2)

•

1, we obtain (d). If x i X2 = 1, use Lemma 6.1(b), along with g(1) = 1, If Xi Z2 to obtain (c). This completes the proof. LII Corollary 6.3. If x i , x 2 are characters of orders dividing

in,

then

g(X1)9(X2) g(XiX2)

is an algebraic integer in O(C,b). If x i , , and x 1 x2 are nontrivial, use the above result. The remaining cases are quickly checked individually. LI PROOF.

The significance of this result is twofold: not only is the expression integral, it also eliminates Cp. This will be useful later. Let in be an integer with (m, p) = 1. Then the fields O(C„,) and O(Cp) are disjoint. Let (b, m) = 1. We may define ab e Gal(Q(C„„ 4)/0) by ab: CpH± Cp, CmH ± Cbm•

(perhaps it would be better to use double indices and call this cr,, b , but usually Cp will drop out early, leaving only Cm). Lemma 6.4. Assume xm is trivial. Then

g (x )b g(grb

and g(x)m E

CP(C,n).

= goo b-cr b e

90

6 Stickelberger's Theorem

PROOF. The second follows from the first if we let b = 1 + m. For the first, we have

Z(a) btfr(a) = 9((b). Cm , Cp Cep for some c, (c, p) = 1. Then

90(Yrb = Let -c E Gal(11(C,)/(1(C„,)), so r: 9(ZY = =

Z(a)thca) AO - E z(a)0(a) = z (c) - g(z),

and similarly

y(?)T = Therefore r fixes g(z)b- crb. The result follows. Lemma 6.5.

g(e)

=

PROOF. Since a i—aP is an automorphism over Z/pZ, T (a) = T(aP), and aP yields a permutation of F. Therefore 9(e) = —

= g(x).

=

-

111

This completes our list of basic properties of Gauss sums. We now digress to give an application to the Fermat curve. We wish to count the number of solutions of xd + yd

= 1, with X, Y c Fq .

As is usually the case, it is more natural to count points in projective space. That is, we consider solutions, except (0, 0, 0), of xd + yd = zd,

and identify two solutions if they differ by a scalar multiple. If Z 0, we may identify (X, Y, Z) with (X/Z, Y/Z, 1) and obtain a solution of the original equation. But if Z = 0, we obtain the "points at infinity" (X/O = cc), which correspond to solutions of X" + Yd = 0. Since we do not count (0, 0, 0), we must have Y 0 0, so any solution (X, Y, 0) may be put in the form (X /Y, 1, 0). The number of points at infinity is exactly the number of solutions in Fq of X d = —1. Despite all this, we shall start by counting the solutions of X d + Yd = 1, and make the correction later. We first assume d divides q — 1. Since F: is cyclic of order q — 1, there exists a character z of F: of order exactly d. The cyclicity implies that z(u) = 1 if and only if u e F: is a dth power. For u E F:, let N d(u) be the number of solutions in Fq of X" = u, so

1, u 0 Nd(u) = {0, u 0 0, u 0 dth power d, u 0, u = dth power (since d I q — 1).

91

§6.1 Gauss Sums

It follows easily that d

Nd(u) = 1

a=1

xa(u)

if u 0 0.

Therefore, the number of solutions of X d + Y d = 1 is

1 N d(u)N d(v) + 2d a+v=1 ay*o (the second term corresponds to X = 0 or Y = 0) d

d

= 2d + 1 1 1 za(u)x b(1 — u) u*0,1 a=1 b=1 d

d

= 2d —

Zb)* a=1 b=1

From Lemma 6.2 we see that the term a = b = d contributes 2 — g; the terms with either a = d or b = d, but not both, contribute a total of 2(d — I); those with a±b=d yield Ida : 1 xa( — 1) = N d(-1)-1; and the remaining terms can be expressed in terms of Gauss sums via Lemma 6.2(d). Using the fact that N d( — 1) is the number of points at infinity, we find that the number of solutions of X d + Yd = Zd in projective space is

q±1

d- 1

1 Aza,

X.

a,b=1 a-4-b#d

Since I g(z) I = ji i f x 0 I, w e have, using Lemma 6.2(d), that

I / AZ", Zb ) I

(d — 1)(d — 2)\/4.

If N denotes the number of solutions,

IN — (q + 1)1

(d — 1)(d — 2)\/-4.

This is a special case of a more general result which states that for a curve of genus g we have

IN — (q + 1)1

2W -4.

Note that g + 1 is the number of points on a line aX + bY = cZ, so the number of points on a curve is approximately the same as for a line, the possible error being bounded in terms of the genus. Now assume d is arbitrary, so we do not necessarily have dIg — 1. Let e = (d, g — 1). Then N d(u) = N e(u), so

N = # {Xd ±

=E

u+v,

yd _ zdi(x ,

Y, Z) * (0, 0, 0))/(g — 1)

N d(u)N d(v)N d(w)/(g — 1)

= # {Xe ± Ye = Ze I(X, Y, Z) 0 (0, 0, 0)) /(g — 1).

92

6 Stickelberger's Theorem

Since e divides q — 1, we have

N (q + 1)1 (c — 1)(e — 2)\,/q so we have proved the following.

(d — 1)(d — 2)".4,

Proposition 6.6. Let N denote the number of projective space solutions of xd + Yd = Z in Fl . Then N — (g + 1)1 (d — 1)(d — 2) \/Corollary 6.7. For any given d, X d + Y d X Y # 0 (mod p), for all sufficiently large p.

1 (mod p) has solutions with

PROOF. From the above, the number of points at infinity is N d( — 1) (or N e(— 1)), which is at most d. The number of solutions with X 0 or Y 0 is at most 2d. Therefore we have a nontrivial solution as soon as N > 3d. Since N — p = 0 (\/ ), the result follows. This corollary shows that it would be difficult to prove Fermat's Last Theorem using only congruences. To finish this digression we show that Proposition 6.6 is essentially the Riemann hypothesis for the Fermat curve. Fix d and p and let N„ be the number of solutions of Xd + yd = zd projective space over IF,— The zeta function Ç(S) of the curve may be defined as follows: Define Z(T) by

Z'(T) Z(T)

N „T" - 1 ,

Z(0) = 1.

Then

= Z(P s). This function satisfies many properties similar to those for Dedekind zeta functions: for example, there is an Euler product, and also there is a functional equation relating the values at s and 1 — s. It can be shown that Z(T) is a rational function of the form

Z(T) =

P(T) (1 — T)(1 — pT)'

where P(T) E Z[T], P(0) = 1.

(for further properties and proofs, see Weil [6] or Eichler [3 ]). Writing P(T) = fl (1 — c, T), we see that

Z'(T) = Z( T) n=1

+ p" — oc7)Tn 1 ,

SO

N„ = 1 + p" —

.

93

§6.2 Stickelberger's Theorem

To answer the question that arises when one compares this with a previous formula, yes, the a i's are Jacobi sums, but we shall not prove this here. It follows easily from the Davenport-Hasse relations (see the Exercises). From Proposition 6.6 we have a7 < (d — 1)(d — 2)11 2 . Lemma 6.8. 1/n

lim sup n—■ oo

EL

= Maxlai l.

PROOF. The only problem arises when two a's have the same absolute value, in which case a straightforward proof would have to show that "cancellation" does not decrease the lim sup. However, there is the following classical trick. Consider the complex function f (z) =

E1 =E 1 — oci z

n=o

(E a7)e.

The radius of convergence of the power series is the distance to the nearest singularity, namely 1/Max I ai I. But it is also the reciprocal of lim sup I E oc7 I' In. The result follows. I=1 We now have I ai p for each i. Returning to Us), we see that Us) = = ai for some i. Therefore Re(s) < I. But the functional equation for Us) implies that if Us) = 0 then Cd(1 — s) = O. Therefore Re(s) = for each zero s. This is the Riemann hypothesis for the Fermat curve. All of the above is part of a much more general situation, which applies not only to cuves but also to higher dimensional varieties (the Weil conjectures, now Deligne's theorem). The reader is strongly urged to read the classic papers of Weil ([1], [2]), where this is discussed and where additional results on Gauss and Jacobi sums are proved.

§6.2 Stickelberger's Theorem Let M/0 be a finite abelian extension, so M g 0() for some m (by the Kronecker-Weber theorem, proved in Chapter 14). We assume in is minimal. G = Gal(M/0) may be regarded as a quotient of (Z/InZ ) X . We let o-a , (a, m) = 1, denote both the element of Gal(0(C„,)/(1) and its restriction to M. Let {x} denote the fractional part of the real number x; so x — E Z and 0 < {x} < 1. Define the Stickelberger element

a

= 0(M)

—} 0-7 1 E Q[G]. a(mod m) (a, m)= 1

M

94

6 Stickelberger's Theorem

The Stickelberger ideal l(M) is defined to be Z[G] n OZ[G], in other words, those Z[G].-multiples of 0 which have integral coefficients. Lemma 6.9. Suppose M = OG„,). Let l' be the ideal of Z[G] generated by elements of the form c — c c., with (c, ni) = 1. Let fi eZ[G]. Then 130E Z[G] .4.> fie T. Therefore I =

PROOE Since (c coo

=

Oa 1

jacl\ca 1 e

tali)

a

we have" To prove the converse, first note that ni = (1 + nt) — a l c Suppose (Ia xa aa)0 E Z [G], with xa e Z. A short calculation shows that

ft

x a{ amb }) a I; 1

x a a a)

Looking at the coefficient of a 1 , we find that ni divides > x a a. Since ni e I', so is x a a. Therefore

xa aa

x a(aa — a) +

x a a e I'.

LI

This completes the proof.

This result is not true in general if M is a proper subfield of O ( ,,,). The problem is that ab = 1 for several b, so the "coefficient of ai-1 " involves a sum over various values of b. For example, let M = 0(.\/12) = O ( 12 + 11) OE (LK 12 ). Then a l = a 11 = 1, while 0-5 = a7 = a, say. We have 0(M) = 1+ [G], so 1 . E Z[G]. But I' is generated by {5 — a, 7 — a 11 — 1 } , therefore by {2, 1 + a}. In particular, 1 0 I'. If x = x, Z[G] then x acts on ideals and ideal classes in the natural way: A' =fl (A')x,

Theorem 6.10 (Stickelberger's Theorem). Let A be a fractional ideal of M, let e Z[G], and suppose 130 e Z[G]. Then A" is principal. Therefore, the Stickelberger ideal annihilates the ideal class group of M. Before starting the proof, we give two examples. (a) Suppose M is real. Then aa = a_ a and + 0(M) =

1 L a(mod m)

aa =

alm> = 1, so

0(m) Normm,Q.

2 deg M

95

§6.2 Stickelberger's Theorem

In this case we find that the norm, or some multiple of it, annihilates the ideal class group. This of course is already obvious, since 0 has class number one. We therefore can obtain nontrivial results only if we look at imaginary fields. (b) Suppose M = Q( ‘/— m) is imaginary quadratic. Gal(M/0) = {1, a}, where a is complex conjugation. Since A 1+6 is an ideal of 0, hence principal, a acts by inversion on the ideal class group. Let t3 = ni. Then (30 = ao-a- 1 e Z[G], and in the ideal class group flO acts as where x is the quadratic character for M. We find that E a(a) = m13 1 , x annihilates the ideal class group. Of course, the class number formula implies that this number is just — mh (if m > 4), so what we have is a weak form of the analytic class number formula; however, it will be proved algebraically. More generally, let F be any totally real number field and M/F a finite abelian extension. Let G = Gal(M/F). Via the Artin map A a-A e G, one can define partial zeta functions for a e G:

1 4(a '

= 6 E 6 NA'

(Re(s) > 1).

These may be meromorphically continued to the whole complex plane, and the values F(o-, — n) are rational numbers for n > O. Define

On(M/F) =

E

— n)o--

aeG

and let I n (M/F) be the ideal generated by elements o f the form

(N

— aA )0„(M /F),

where A ranges over a set of integral ideals of F not divisible by a certain finite set of prime ideals. Then In(M/F) should annihilate some natural object, perhaps a K-group. For example, / 1 (M/0) annihilates K2em, except possibly for the 2-part, where Cm is the ring of integers of M. When M = 0(Ç,n) and F = 0, we have a,

)=

b a(m) b>0

bs

which is essentially a Hurwitz zeta function. If n = 0, we have

1 — {17?1 })cr,-1 , 00 ( 0 ) = conE ii d which differs from 0 by half the norm (in fact, we shall need 00 later). Since Ko (9„, is essentially the ideal class group of M, the above may be regarded as an appropriate generalization of Stickelberger's Theorem. For details, see Coates [7].

96

6 Stickelberger's Theorem

We are now ready to start the proof of Stickelberger's theorem. The major step will be the factorization of certain Gauss sums. Let p be a prime and let q = pf be a power of p. Let A be a prime ideal of et) ( 1 ) lying above p. Since Z[Cq ,] mod A is the finite field with q elements (.1 = residue class degree by Theorem 2.13), and since the (q — 1)st roots of unity are distinct mod A, there is an isomorphism

=

(q — 1)st roots of 1

IFqx

satisfying co(a) mod

A =a c .

This a) is essentially a generalization of the a) of the previous chapter. Let be the prime of CP(Cq - 1 , lying above A. For a c Z, let s(a) = where v '0 is the valuation. Clearly s(a) depends only on a (mod q — 1),

Lemma 6.11. (a) s(0) = 0; (b) 0 < s(c + fl) s(a) + s(fl); (c) s(a + fi) s(a) + s(6) (mod p — 1); (d) s(pa) = (e) D: 21. s(a) = (q — 2)(f)(p

1)/2.

PROOF. (a) is obvious; (b) and (d) follow from Corollary 6.3 and Lemma 6.5, respectively. Since L = A, the values of vs, on et)( 47 _ 1 ) are divisible by p — 1. Therefore (c) also follows from Corollary 6.3. Since g(co - ')g(co') = ±q = + p f , we have s(a) + s(q — 1 — = vo(pf) = (p — 1)f. Pairing up the terms in the sum (the term for a = (q — 1)/2 pairs with itself), we obtain (e). This completes the proof. LI

Lemma 6.12. s(a) > 0 if a PROOF. Since rt =

0 (mod q — 1), and s(1) = 1.

1e

g(w - Œ) =

,

—E —(04(a) —E co —(a)

0 (mod

Therefore s(a) > O. Also,

g(coi) = =

i (a)cp•(.) co - (a)(1 It )T(a) —

_L co— 1

- 2). nT (a )) (mod .0

w 1 (a)T(a).

Regarding IF as ZUg _ J mod

A, we have

T (a) = a + aP + • • +

(since a

(0)(1

(mod A)

aP generates the Galois group mod A), and

E co - 1 (a)T (a)

> a - 1 (a + aP + • • +

U “)

a mod y4

') (mod A).

97

§6.2 Stickelberger's Theorem

Ea 1 =

If° < b < f, then La o aPb-1 =. 0 (mod A), so the sum reduces to q—1 —1. Therefore (mod ;) 2 ),

hence s(1) = v-4(n) = 1 (since 03(Cq _ 1 , Cp)/Q(Cp) is unramified at p). This completes the proof. 0 Proposition 6.13. Let 0 5 a < q — 1 and let a = ao + aiP + • • • + ai p — 1, be the standard p-adic expansion of a. Then

af-lPf 1 7

s(a) = ao + a 1 + • • • + aj_i.

From Lemma 6.11(a), (b), (c) and Lemma 6.12 we immediately have s(a) = a for 0 p — 2. If q = p, we are done. Otherwise, s(p — 1) > 0 and we similarly obtain s(p — 1) = p — 1. Lemma 6.11(b) and (d) imply that s(a) .a,:;,+•••+af _ 1 . When a runs through the integers from 0 to q — 1, inclusive, each coefficient of the p-adic expansion takes on each of the values from 0 to p — 1 exactly pf -1 times, so PROOF.

q—

1 (a0 + • • + a1 - 1 )

p(p — 1) =

L.

a=0

(f)p 11

=

fq.

L.

If we omit a = q — 1 = (p — 1) + - • + (p — 1)pf -1 we obtain q-

2

(ao + • • • + a f _ 1 ) = a=0

1

P

2

q-

fq — (p —

2

s(a),

= =0

LI

by Lemma 6.11(e). The result follows. Remark. Let

Cp - 1 and 0 < a < q — 1, as above. Then 700 4- ••• + af _ +a 1 + 1 ). 9(€0-2) (a0 !) • (a f _ 1 !) (mod Ira°

TE =

(see Lang [4], [5]). In Lemma 6.12 we verified a special case of this formula. The general argument follows a similar line, but involves a rather delicate analysis of binomial coefficients.

Now fix a positive integer m. Let p be a prime, (p, m) = 1, and let f be the order of p (mod m), so in divides pf — 1 = q — 1. Fix a prime ho of U(C„,) lying above p; let ho above h o , so Ag 1 = ho; - be the prime of (m' let („90 be a prime of (13(ç _ 1 ) lying above h o ; and let gs o be the prime of Cp) lying above (and fio). Let co = (24, 0 be as above and let = d , where d = (q — 1)/m. Then xm = 1, so g(x) e O(Cm Cp). Since g(x)g(x) = q = pf the factorization of g(x) involves only primes of O(Cm, Cp) above p, that is, the conjugates over U of j0. Let (a, in) = 1 and let 6a E Gal(0(Cm)/0) be the corresponding element of the Galois group. For each such a, fix an extension of at, to U(C,. 1 , Cp) such that C'T,“ = Cp

°

CO

98

6 Stickelberger's Theorem

Up

The decomposition group for p in (Z/mZ) X is generated by p (mod m) (see the discussion after Theorem 2.13). Let R denote a set of representatives for (Z/m2) modulo this decomposition group. Then {h' la e RI is the set of conjugates of ho . Since io is the unique prime above ho , all conjugates of A have the form Ag" 1 . Let = ' be one of them. Then v(g(x)) = v A-0(g(x) 6°) = vi,-0(g(xa)) = v41 0(g(x")) = s(ad) (140 = v4,0 since

is unramified). Therefore (g(z))

ARs(ad)a.-i .

Lemma 6.14. Let 0 < h < q — 1. Then s(h) = (P

0

i 1} q Ph

PROOF. Let h = ao + ai p + • • • + af _ i pf - 1 . Then (mod q — 1).

ph ao + ai pi+1 + • • • + a1 _

It follows that 1 ph 1 = 1

lf q— 1

+ • + af - 1Pi-1 )*

(ao pi

111

Summing over i, we obtain the result. We now have s(ad) = (p —

{pa/m}, so f -1

= (p — 1)

f nin

r-}oi=0 R M

99

§6.2 Stickelberger's Theorem

Since group),

= A o and since api(A o) = A o (definition of decomposition (g (x )m) = /qm Eli Pfll

where 0 = {b/tn}a,-1 is the Stickelberger element (we raise to the mth power to avoid denominators). We now have a partial result: If Ao is a prime of (I:( m) with AO' m, then Arno ' is principal in U(C, cp). The main problem is now to get down to a(Cm), then to M. Suppose that A is an ideal of M Ia ( m ) with (A, m) = 1. Let A = be its factorization into (not necessarily distinct) prime ideals in aG.). Then A m° = ( 11 9(4)m ), where we write xo , to indicate that x depends on A,. Suppose 13 EZ[G] (G = Gal(M/Q)) and 130 E Z[G]. Extending the elements of G, we may regard 130 as an element of Z[Gal(CK,,,,,)/CD)]. Then Am" = ("m),

where y =

OGpm),

where P fl p, is the product of the rational primes divisible by the A's. Since 7"7 E CD( m) by Lemma 6.4, and it is the mth power of an ideal of Q(Cm), namely A", it follows that the extension 1.)(4„ y13)/(1:Wm) can be ramified only at primes dividing m (proof: locally, A" is principal, so we are adjoining the mth root of a local unit). But P( 'm' 711) g U ( m) Therefore, ramification can occur only at pi's. Since (P, in) = 1, the extension must be unramified. Lemma 6.15. If 43 ( m ) then K = OGm).

K

1a

(

n

) and K/( m ) is unramified at all primes,

Suppose K (1:C - ). Then there is a character x for K of conductor not dividing m. By Theorem 3.5, K/CP(,n) must be ramified at some prime. Contradiction. PROOF.

We find that ya E la ). Therefore A" = (y13) is principal as an ideal of CP(Cm). But this does not necessarily mean that it is principal as an ideal of M. So we show that 713 E M, which suffices, since if two ideals of M are equal in CP(Cm ) they must have been equal originally because of unique factorization. Let g be a prime of ( q . i) lying over one of the prime factors Ai of A. Then x o, a priori depends on the choice of Y, so we temporarily let x o, = Let a E Gal(a( q _ 1 )/M). Then (

m

:Z['q _ 1 ] mod Y i[( q _ 1 ] mod ga and correspondingly if x,9(a) then x,,,,r(a) = V. Therefore x°,9', x But y';= 1, so xj, = xj. for a E Gal(aGq _ VaGm)). Therefore depends

100

6 Stickelberger's Theorem

only on A, so we may return to the notation x o,. The above reasoning shows that 4, = for a e Gal(O(„,)/M). If we extend a by letting a( , p) = then g(x i„)' = g(4) = Since A a = A for a e Gal(CD(L)/M), a permutes the A;s. Therefore /la

g(XA,) ficr =

=

g(zgr)fl = Tfl.

But we already have yfl c (LIG„,); hence yfl c M. So A° is principal in M. Finally, if A is an arbitrary ideal of M, we may write A = (a)A 1 , with a e M and (A 1 , rn) = 1. Then Am) = (a")Ar, which is principal. This completes the proof of Stickelberger's theorem. 0

§6.3 Herbrand's Theorem Let G be a finite abelian group and Ex

=

its character group. Let x

E

and define

1

1G1aeG

X(a)a

1E

where C-P is the algebraic closure of O. One may easily verify the following relations: (a) = Ex ; (b) r = 0 if x i/'; (c) 1 = Ex ;

Lze6

(d) E x a =

The Ex 'S are called the orthogonal idempotents of the group ring 0[G]. If M is a module over C-P[G] then we may write M = e M, where Mx =

(use (c) to get the sum; if 0 = Ex a x , then use (b) and (a) to show ex = 0 for all x). Each a e G acts on M, and M, is the eigenspace with eigenvalue x(a), by (d). Of course, all the above works if EP is replaced by any (commutative) ring which contains the values of all x e and in which G1 is invertible. In particular, let p be an odd prime and let G = Gal(0( p)/0) (Z/pZ) X . Then = Ico 1 10 < i < p — 21. We shall work in the group ring Zp[G]. The idempotents are p—1

E=

p — 1 a= 1

al(a)a-1 a

0 < i < p — 2.

101

§6.3 Herbrand's Theorem

Later, we shall also need

=

1 - a_ , = 1 E i and 2 i odd

r, ± =

1 + a_ , = 2 i even

There is a decomposition A = A - G A + for any Zp [G]-module, for example the p-Sylow subgroup of the ideal class group. Let 0 = ( 1 /p) be the Stickelberger element. Using (d), we find that 1

i

0 P

P—'

B1,

aw a=1

and 3i(e -

( )0 = (c -

(c))B 1 ,

Let A be the p-Sylow subgroup of the ideal class group of WO. Since p" A = 0 for sufficiently large n, we may make A into a Zp- module by defining

E bi pi)a = =o

=o

since the latter sum is finite. G also acts on A, so A is a

Zp [G]-module.

Let

p-2

A =A i i =0

be the decomposition as above. Stickelberger's theorem implies that (e - a( )0 annihilates A, hence each A i . Therefore we have proved the following: Let annihilates A 1 . c E Z, (c, p) = 1. Then (c - w i(c))B

Remark. Since p0

(p - 1)E, i (mod p), it is not very surprising that p0 annihilates A 1 for i 1. The fact that it annihilates A,, however, requires Stickelberger's theorem.

Now, suppose i 0 is even. Then B 1( I = 0 so the above says nothing. If i = 0 then (c - 1)/2 annihilates A o , so Ao = 0. But this is already obvious since r, = (Norm)/(p - 1). Let i be odd. Consider first the case i = 1. Let c = 1 + p, so we have 1 (C —

(o(c))B 1 , -1 = pB

=

aw - (a) a=1

p - 1 0 0 (mod p).

Since A is a p-group, we must have A = 0. (It is easily seen that A 1 = 0 is related to von Staudt-Clausen, which is related to the fact that the p-adic zeta function has a pole. Perhaps this explains why A 1 is a special case). If i o 1, we may choose an integer c, for example a primitive root (mod p), such that e e i WV) (mod p). We may consequently ignore the factor c - (e), so we obtain the following.

102

6 Stickelberger's Theorem

Proposition 6.16. Ao = A 1 = O. For i = 3, 5,

, p — 2, B 1 ,„_ annihilates

Ai.

Suppose A, O. Then we must have B 0 (mod p). But B 1 (,, — i) (mod p) by Corollary 5.15. We have proved the following'. Theorem 6.17 (Herbrand). Let i be odd, 3

p — 2. If A,

0 then plB p _,.

This theorem is much stronger than the theorem "plh = p divides some Bernoulli number" since it gives a "piece-by-piece" description of the criterion. Even better, the following is true. Theorem 6.18 (Ribet). Let i be odd, 3

i

p

—

2. If pl Bp-, then A,

O.

We shall not give the proof, which is very deep. It uses delicate techniques from algebraic geometry to construct an abelian unramified extension of degree p which corresponds by class field theory to A,. See Ribet [2]. One corollary of Ribet's theorem is that the p-rank of the ideal class group of (3(4) is at least the index of irregularity (p-rank = number of summands when A is decomposed as a direct sum of cyclic groups of p-power order). However, it is not known whether or not there is equality, since the rank of some A i could possibly be two or larger. If p)(17(E)(4)) then we do have equality, as we shall prove in Chapter 10.

0.4 The Index of the Stickelberger Ideal Let p be an odd prime, n

1, G = Gal(0(4„)/11), and R = Z[G]. As before, Pn

=

CUT a

Pn

a=1 (a, 11= 1

is the Stickelberger element and I = RO n R is the Stickelberger ideal. Let J = o- denote complex conjugation. Then R - = {x E

= — = (1 — J)R,

the first equality being the definition, the second following from a short calculation. We define In R = RO n R -

Note that we have to be careful about using the idempotent (1 — J)/2 since it has a denominator. However, observe that x c R - <,› [(1 — J)/2]x = x.

103

§6.4 The Index of the Stickelberger Ideal

Theorem 6.19 (lwasawa). [R - : I -] = Remark. The above definitions hold for arbitrary a(C.). Sinnott has shown

in this case that [R - : I - ] = 2ah - (0(C„)), where a is defined as follows: Let g be the number of distinct primes dividing m. Then a = 0 if g = 1, and a = 21 ' — 1 if g > 1. PROOF OF THEOREM 6.19. The proof will proceed by considering completions, since we can then work with one prime at a time, which is slightly easier. Let q be a prime, R 4 14 [ G], = ;I. Clearly / is dense in 14 in the natural q-adic topology. Also, R q-- = (1 — J)R 4 and /4- = 14 n R 4- . Lemma 6.20. (a) 14 = R 4 0 n R 4 , (b) Ç = R 4 0 n R47, (c) /; = /- • Z g , (d) lip q then I q R q 0.

The proof of Lemma 6.9 works for both R and R q , with I:, = I'Z q , so we obtain R4 0 n R q = F4 0 = I7L 1 El = PO7L4 = I7L4 = 14 . PROOF.

This proves (a). Part (b) follows easily from (a). If p q then 0 e R 4 , so (d) also follows from (a). We now prove (c). Since {z} + { — z} = 1 for z Z, we have (1 + J)0 = N, where N =

G. creG

Let x E I. Then x0 e 1, and we have x0 E /- (1 + J)x0 = 0 xN = 0. Similarly, suppose y e 14 . Then y0 E 1'4 0 = /4 , from the above, and y0 c Ç <,> yN = 0. Clearly /- Z

Ç. Suppose now that y0 c Ç, with y e I. We may write y

=

d„c(c — cc),

a, c ,E4 .

C

The condition yN 0 becomes Lc da(c — 1) — 0. We want to approximate y by an element x E T such that xN 0. This then will give us an element x0 of I - near yO, which will show that /; closure of I - = I - 74, as desired. The approximation will reduce to the following. Fact. Suppose bi c Z, s i E 74 , and suppose sequence (t in) , . , t(in }) E Et' whose limit is (s t ,

IT_

b i si = 0. Then there is a , sm ) and such that b i d') = 0.

PROOF. We may assume (q, b 1 ) = 1. For 2 < i < a-t, choose 11n) 0 mod b 1 tr near si (this is where (q, b 1 ) = 1 is needed). Then bi t" 0 mod b 1 ,with so we can choose t (in ) E Z. Since tr) is near s, for i o 1, we must also have t(r) near ,s 1 This completes the proof of the fact.

104

6 Stickelberger's Theorem

If we let a`, = s(=si), c - 1 = b,,„ and x = ta(c - ac) c l', then x is near y and xN = 0, as desired. This completes the proof of Lemma 6.20. D We have R q R 0 Zq , and under this isomorphism Rrq- R - 0 Zq and / 0 Zq , by (c). It follows easily that R:7 - /Ç (R - /1 - ) 0 Z q , which is isomorphic to the q-part of R - /I - . Therefore it suffices to prove the following. Theorem 6.21. [R t-i - : /,;-] = q-part of PROOF. We first consider q 2, p. Then (1 + J)/2 E R q ; and we get R q = Rq+ C) IÇ and /q = / (I) /(i from the relation 1 = (1 + J)/2 + (1 - J)/2. Therefore

1-J

=

Iq

2

=

1-J 2

R 0 = R - O. q

Consider the linear map A: R q-

R q- ,

x 1-* x9.

By the theory of elementary divisors, as in the proof of Lemma 4.14, we have [R q-- : = q-part of det(A). But det(A) may be computed by working in aq[G] -, which has the advantage of being a vector space over an algebraically closed field. We have

y Qq [G]

I. q [G] = x odd

where Ex = (11 p") 1 (a, p)= 1 A* a 1 and each direct summand is onedimensional. As in the Previous section,

Ez 0 = 131 , z u,z , so A becomes a diagonal matrix. Therefore det(A) =

H z odd B 1 , 2 , hence

= q-part of fl B 1 z

[Rq- :

= q-part of 2p" n (-1--B 1 , = q-part of h - ( l:P( p.)). For q = 2, the argument must change slightly since (1 + .42 R2, so R 2 R R . Also, there is a power of 2 in the class number formula which must be accounted for. Since we are restricted in our use of (1 - J)/2, we modify 0 to obtain an element already in 02[G] . Let

= (41.

a 1 — - -)o - = 0 - , = Pn 2 a( Pn

P)= 1

105

§6.4 The Index of the Stickelberger Ideal

where N is the norm (one could also call it the trace). Clearly 0 qt R2, but a short calculation shows that 1 J = 2 so 0 is in the " — " component. We recall that 0 is perhaps a better Stickelberger element than 0, since it is the one that generalizes most readily (see the discussion after the statement of Theorem 6.10). Lemma 6.22. (a) I 27

R20;

(b) [R 2 0 : /2- ] = 2. PROOF. For the first statement, suppose x e R2 and x0 e I = R2 0 n R. Then x0 = [(1 — J)/2]x0 = x[(1 — J)/2 ]( 0 +1N) = x0 E R 2 0. For (b), we claim that if x e R2 then either x0 e R2 or x0 — e R2. To prove this, we note that x0 = x0 — XN e R2 lxN e R2 and similarly for (x — 1). Let x = E x,a. Then xN = (E x,)N and (x — 1)N = (-1 + E x,)N. Since either (1 x,) or (— 1 + E )c,) is even, the claim is established. The claim implies that [R 2 0 : R2 0 n R2] = 2 (the index is not 1 since çt R 2 ). The proof of the lemma will be complete if we can show that R20 n R2 = R20 n R 2 = I. We have already shown that U. s R 2 0 n R2. Now, let x0 E R 2 0 n R2, where x = E x, E R2. Then, as above, x0 C R2 +XN E R2 X 0 (mod 2). Let y, = x, for a1, J, and let Yi = — x, and y j = xj x a . Then E y, = 0, so y = E ya E R 2 satisfies yN = 0. Also, x — y =( x„)(1 + J), so (x — y)0 = 0. Hence x0 =, y() = y0 — vN = y0 E R 2 0. Since x0 E R2 and satisfies [(1 — J)/2]x0 = x0, we have x0 e R20 n R2 = 1. 2 ,

SO

R2 0 n R2 = 1 2 •

This completes the proof of Lemma 6.22. 0 Just as for the other primes, we have a linear map

A: R 2.- — 0 R 27,

X 1—* 0X

(since x E R2- , XN = 0, so there is no 2 in the denominator), and [R2- : R2- 0] = 2-part of det(A) = 2-part of

ri B 1 , 2

(x odd

Ex

= Ex 0)

xodd

2 1IG I .1 • (2-part of h - ).

Since this index is finite we must have = Z 2 -rank of R2- = Z2-rank of KO. Observe that R 2.70 = (1 — J)R 2 0 = R2 (20) = 2R 2 Û. Therefore [R 2 0: R 2- 0] = 211GI.

106

6 SLickelberger's Theorem

But

[R 2 0: /i] = 2, from Lemma 6.22. Putting everything together, we obtain : I] = 2 part of h - , -

as desired. Finally, we consider q = p. The main problem is that 0 has p" in its denominator. Let Ô = 0 — IN be as above. Suppose x = Lb , p)=1 XJ E R. Then x0 c RI; <=> x0 E R. We have 1 x0 = — axa; Pn c a hence x19 E R

a

axa,

axa, a

o (mod p") for c -1

all c with (c, p) = 1. But

ac xac.

axa (mod p"),

a

a

axa 0 (mod p"). It follows easily that (x — b)0 c R p so we only need for exactly one integer b (mod p"). Therefore

[Rp O : R p O n R;] = 13". But

R p O = R;,- 0 = R; (8 —

= R; 0,

since

1—J N = 0, 2

Therefore

R p O n R; = ÇO n R; RO n R; = Ç. If x0 c 1r„ then x0 = [(1 — J)/2]x0 e R; 0; hence RO n R; R;0 n Ç. Therefore /;

= R p O n R;.

From the above,

[R; 0 : I;] =p"• Let

A: R; —> R; ,

x p"Ox.

Then

[R; : p"R; 0] = p-part of det(A) p-part of p (n12)IGI fl B1, 2 p(n12)1G 1 (p-part of h - ). pn But [R; O : p'1 R 1 0] = p (n/ 2)IGI, so : 11-,] = p part of h - . -

This completes the proof of Theorems 6.19 and 6.21.

00

107

§6.5 Fermat's Last Theorem

The formula [R - : I - ] = h- may be regarded as an algebraic interpretation of the class number formula. It should be considered as being of a similar nature to the formula [E+ :C±] = h+,which will be proved in Chapter 8. The natural question arises: is there an isomorphism of G-modules R - /I - C/C + ? (C = class group, C + = class group of the real subfield). After all, both sides have the same order. In general, there is not such an isomorphism. Let p = 4027. The class group C, of Q(/- 4027) is Z/3Z (D Z/3Z. Note that since 1 + J is the norm to 0, J acts by inversion on the ideal class group. Therefore C = C, and Ct = 1. Since (111(C p)/(11— p) is totally ramified (at p), it follows from class field theory that the norm map on the ideal class groups is surjective. Suppose R - /I - C/C+. Since R - /I - is cyclic over R, generated by 1 — J, a similar statement holds for C/C+ ; so there exists CE C such that C/C + = cR mod C +. Therefore C, = Norm(C) = Norm(C/C + ) (since Norm(C) Ct = 1) is generated by c l = Norm(c) over R. If a E Ga1(C( p)/0) then a l (1)(\/— p) = 1 or J. Therefore el= It follows that c i R is the subgroup generated by c l , hence c i R C 1 Z/3Z Therefore R - /1 - is not isomorphic to C/C + . However, we may hope for less. Let A be the p-Sylow subgroup of the ideal class group of Q(C p.), so A = A ± (D A . Then is there a G-isomorphism R;/I;

A-?

Again, both sides have the same order. We shall show in Chapter 10 that if p,r11.+(0( p)) then the above holds, so A is cyclic (i.e., generated by one element) as a module over the group ring in that case.

§6.5 Fermat's Last Theorem Theorem 6.23. Suppose p is prime and suppose the index of irregularity of p (=the number of Bernoulli numbers divisible by p) satisfies i(p) < p — 2. Then

XP + YP = ZP,

(XYZ, p) = 1,

has no integer solutions.

Remark. This theorem was proved by Eichler under the assumption that the p-rank of the minus part of the ideal class group is less than fr; — 2. It was noticed (independently) by Bruckner, Iwasawa, and Skula that it is possible to use i(p) instead. This yields a stronger theorem. Ribet's theorem shows

108

6 Stickelberger's Theorem

that i(p) rank, but since possibly some component A i could have rank greater than 1, we could have strict inequality. Also, it is easier to compute i(P). Up to 125000, the largest value of i(p) is 5, and probability arguments indicate that we should have i(p) = 0(log p/log log p). Therefore, the theorem gives perhaps the best evidence yet for the first case of Fermat's Last Theorem. In fact, the probability that i(p) > — 2 is e - 1/2 k

2

(1)k k!

(see the discussion following Theorem 5.17), which is easily seen to be at most (1/2 k)(1/k!), with k =[IF)] — 1. Since the first case of Fermat's Last Theorem is known at present for all p < 6 x 109, we only need to consider larger p. Therefore, the total number of expected exceptions to the first case of Fermat's Last Theorem is at most 1

([] - 1)!

p >6 x10 9

which is less than 10-300000. 6.23. Let

PROOF OF THEOREM

P. As in Chapter 1, we assume we have a

solution and obtain

(x + C i y) = Cr,

i = 0, . , p — 1,

where C 1 isan ideal of WO. Let C be the subgroup of the ideal class group generated by C 1 , , Cp _ i . Then C is an elementary p-group, so Zp[G] acts on C, where G = Gal(G(Ç)/0). Since Cla = Ca , C1 generates C over the group ring, and = C

(<x> denotes the cyclic subgroup generated by x). Also, e<Ei C I > = C. i odd

Therefore p-rank C - = # {Ei C 1 0, i odd} < # {A i 0, i odd} < i(p) (by Herbrand's theorem) <

—2

(by assumption).

This type of inequality can prove useful whenever one is working with a group, such as C, which is cyclic as a module over the group ring, since in essence we have reversed the inequality "rank i(p)."

109

§6.5 Fermat's Last Theorem

We now proceed with the proof of the theorem, following Eichler's argument. We may assume p> 3. Let r = R/P] — 1. Consider the set of all products • • • C riir with 0 < bi < p. The number of such products is pr > prank(C) = I C . Therefore, two of them must agree in their C - components, so we may divide and obtain

H C E C+,

with — p < ai < p,

i=

and some ai is nonzero. Therefore

= (p)s, with p c CP(C p) and S an ideal with S = S, so

(x + i =1

c iy )a) =

(p

)S.

Since all the Ci's are prime to p, we may assume p and S are prime to p. The above implies that SP is principal in 0( Cp). Since the ideal class group of 0(Ç) + injects into that of 4:Ep), by Theorem 4.14, S P is principal in 4:1( p)+ : SP = (oc), with c = oc. Since any unit of CKp) is a root of unity times a real unit, we obtain

fl (x +

Ci y)a‘ = Cs'EapP, with p E Z and E a real unit.

j= 1

Therefore iy)a =

11 (x + i=1

By Lemma 1.8, pi'

p"

PgapP.

rational integer (mod p),

x + Ciy \

H x+c

ly

so

C 211 (mod p),

and

+ y cix

with y

iai(mod p), y

2p. — x;

{ y, x,

C" (mod p),

O. Let

if ai 0,

F(T) = fl (xi + T iy i)la I,

Yi =

{x,

if ai < 0 if ai > 0,

G(T) = fl (y i + Tix i)lail.

110

6 Stickelberger's Theorem

Then F yields the numerator and G yields the denominator of the above expression, so (mod p F(C) ),

SO

F(C) = CVG() + p K(0, for some K(T)

E

2[7].

It follows that F(T) = Tv G(T) + pK(T) + (1 + T + • • • + TP- 1 )H (T) for some polynomial H (T ) E 0(T), but since everything else has integral coefficients, H(T) e 2[T]. Multiply by 1 T, differentiate with respect to T, set X = C, and reduce mod p. Then —

(1

—

) F'(C)

—

F(C) = (1 -

CvG(C) + v(1

—

—

—

XC' -1 G(C) (mod p).

Dividing by F(C) CvG(C) (mod p), we find that (1

F'(C) C) F

C)

= (1

G'(C) G(C)

1 + y(1 — C)C - (mod p).

This is essentially the same as what we would have obtained if we could have CVG() with respect to C. The above taken the logarithmic derivative of F(C) may be rewritten as (1 —

xi Yt r -

i=1

- E

yi +

+ v(1 -

which is the same as (1 — C)

iai C1 (

Y . Cz y

y + C1x)

v(1

—

C) (mod p).

Multiply by firi = i (x + Ci y)(y + C x), which is a polynomial of degree r(r + 1) in C. Let i o be the index of the first nonzero ai . On the left we obtain a polynomial in C of degree 1 + io + r(r + 1) — 2io = 1 + r(r + 1) — io with leading coefficient i o a,o(x 2 — y 2)xryr. On the right we have a polynomial in C of degree 1 + r(r + 1) with leading coefficient — xryry. But

1 + r(r + 1) < 1 + (j) — 1)(j)) < p — 1, so we have a polynomial in C of degree less than p — 1. By Lemma 1.9, corresponding coefficients are congruent mod p. It follows that y 0 (mod p), so the right-hand side vanishes. The leading coefficient on the left now must vanish (mod p), so x 2 y 2, hence x + y (mod p). Interchanging y and z, we may also obtain x +z (mod p), so + xP + xP = ±x", which is impossible for p> 3. This completes the proof.

111

Exercises

NOTES

A good reference for much of this chapter is Coates [7]. See also the new edition of Ireland-Rosen [1]. For more on Gauss sums, see Ireland-Rosen [1] and Weil [1], [2], [3], [4]. Stickelberger's theorem was proved for 1:PG) by Kummer [1], and in general by Stickelberger [1]. For a proof which does not use Gauss sums, see Fralich [4]. There is a beautiful relation between Gauss sums and the p-adic F-function. See Gross-Koblitz [1], Lang [5], and Koblitz [4]. There are also analogues of Stickelberger's theorem for totally real fields (G. Gras [10], Oriat [3]), for group rings (McCulloh [1], [2]), and K-groups (Coates-Sinnott [1], [3]). For another extension, using Hecke characters, see Iwasawa [28]. Theorem 6.19 is due to Iwasawa [11]. The analogous formula for (WO and some other abelian fields has been proved by Sinnott [1], [2], [3]. For the general theory of Stickelberger ideals, see the papers of Kubert-Lang. Theorem 6.23 has been improved slightly by Uehara [2].

EXERCISES 6.1. Let F be a finite field. Show that the characters ip c(x) = 0(ex) for Conclude that all additive characters of IF are of this form. 6.2. Suppose the characters zt are multiplicative characters of F all i. Let ei c Z. Show that

H g(xi c u(.) 4-> 11 e(a) = 1 )"

C G

F are distinct.

and that zi-", = 1 for

for all a c (ZIp ) x

6.3. Let p 3 (mod 4) be prime, let R denote the number of quadratic residues mod pin the interval (0, p/2), and let N denote the number of nonresidues in this interval. Use Stickelberger's theorem to show that R — N annihilates the ideal class group of p). (Historically, this was known before the class number formula) (Hint: Exercise 4.5). 6.4. Let E/IF be an extension of finite fields, [E: F] = n, and let N be the norm for this extension. Let hand th be the additive characters. Let z E be a multiplicative character of IF and let z E = N. a multiplicative character of E. Let R

g(ZE)

g(ZO n (a) If = 1, show that R c (b) Show that R is a unit. (c) Show that R has absolute value 1, hence is a root of 1. (d) Use the Remark following Proposition 6.13 to show that R is congruent to 1 modulo primes above p (= characteristic of F).

112

6 Stickelberger's Theorem

(e) (Davenport Hasse) Conclude that for p > 2, g(h) = g(x ur (this also works for p --.= 2; see the original paper by Davenport and Hasse. For a different proof, see Weil [11). (f) Show that if y r has order exactly d then yl has order exactly d. (d) Suppose F has g elements and dl .9 — 1. Show that the number of solutions in projective space over IE of X d + Yd = Zd is 1 + gn — Idu.-fi l._ i. „ +b ,„ ../(74,x1r. (h) Let a be a root of the polynomial in the numerator of the zeta function for X d + Yd = Zd over Z/pZ. Suppose [0= : Z/ pi] = e. Show that au = J(1, x') for some a, h. 6.5. Show that for each positive integer d, the equation Xd + lid = Zd has nontrivial p-adic solutions for all p. 6.6. (a) Use the Brauer-Siegel theorem (or Theorem 4.19) to show that the index of irregularity satisfies 1(p) p/4. + o(p). (b) The probability that i(p) = k is e -1 / 2 (-1)&/k ! (see the discussion after Theorem 5.17). Let x be such that the expected number of p _._ x with i(p) = k is 1. Show that log x is approximately k log k. Assuming that x is approximately equal to the first p with i(p) = k, conclude that we should have i(p) = 0(log p/log log p). This gives a fairly accurate estimate. The first p with i(p) = 5 is 78233, and log p/log log p = 4.65.

Chapter 7

Iwasawa's Construction of p-adic L-functions

Following Iwasawa, we show how Stickelberger elements may be used to construct p-adic L-functions. The result yields a very useful representation of these functions in terms of a power series. As an application, we obtain information about the behavior of the p-part of the class number in a cyclotomic Zn-extension and prove that the Iwasawa p-invariant vanishes for abelian number fields. Also, we show how many of the formulas we obtain have analogues in the theory of function fields over finite fields.

§7.1 Group Rings and Power Series Let el be the ring of integral elements in a finite extension of ep• For example, be the e = di x = 7/ p [x(1), X 2),. .1 for some Dirichlet character x. Let maximal ideal of 0 and let TE be a generator of A, so (n) = A. Let F be a multiplicative topological group isomorphic to the additive group l p . Let y be a fixed topological generator of T; i.e., the cyclic subgroup generated by y is dense in T. For example, we may let y correspond to 1 G Zp under the above isomorphism, since 1 generates Z, which is dense in Zp . Since the closed subgroups of Zp are of the form pn 1 p , the closed subgroups of F are of the form FP'. Let rn . rirPn, so F„ is cyclic of order pn , generated by the coset of y. Consider the group ring VIF,,]. If in > n-_. 0 there is a natural map 0„,.„: c!2[Fm] aim] induced by the map F„, —> rn. Clearly ei[T]/((i + Tr" — 1).

where the isomorphism is defined by y mod FP" i—, 1 + T mod((1 + Tr — 1). 113

114

7 Iwasawa's Construction of p-adic L f unctions -

Since (1 + T) P" — I divides (1 + T)Pm - I when in > n > 0, there is a natural map in the polynomial rings corresponding to Om,„• If we take the inverse limit of the group rings (9[Fn] with respect to the maps 0,n,„ we get &UFA, the so-called profinite group ring of F. Clearly (9[F] Mill since an element a e (9[F] gives a sequence of elements a n E (9[Fn] such that 0,n,„(a,n) = oc n • However, as we shall see, (9[[F]] contains more elements. In effect, it is the compactification of (9[F] and contains certain "infinite sums" of elements of F. To understand (9[[F]] better, let us look at polynomial rings, since clearly 0[[11] lim 0[1]/ ((1 + Ty" — 1). Theorem 7.1. (all]

(9[[1 ] ], the isomorphism being induced by y

1 + T.

Before proceeding with the proof, we shall prove two preliminary results which are useful in their own right. Proposition 7.2. Let f, g E (9[[7 ] ] and assume f = ao + a i T + • • • , with • Then we may uniquely write ai E for 0 < i < n — 1, but a„ E g = q + r, where q E (0[[T]] and where r E C[T] is a polynomial of degree at most

n — 1.

PROOF. We first prove uniqueness, which reduces to considering qf + r = O. If q, r 0, we may assume that either 7,'r or n)(q. Reduction mod n shows that n Ir, so n I q f. An easy argument shows that since n f we must have n Iq, which gives a contradiction. So q = r = O. The existence is a little more difficult. Define an operator T = (9[M] O[[T]] by co

T(1 b i T) =

In essence, T is a "shift operator." Clearly T is (9-linear and satisfies (i) T(T"h(T)) = h(T) for all h(T) E (9[[7 ] ]; (ii) i-(h(T)) = 0 <=> h(T) E (9[T] with deg h(T)

n — I.

We may write f (T) = nP(T) + TU(T),

where P(T) is a polynomial of degree less than n and U(T) = an + an+ i T + T( f (T)). Since a„ c , U(T) is a unit of the power series ring. Let q(T) =

1

U(T) jto

(_

.(

P)i

T0 -

-c(g).

115

§7.1 Group Rings and Power Series

Here, for example, p• )2

c-j

t(g) =

T 0-

P T(— T —

UU

T(g))).

Note that possibly each summand contributes, say, to the constant term. But the factor it makes the sum of these contributions converge. So q(T) is well-defined power series in (9[[7 ] ]. Since qf = nqP + TnqU, we have

T(q1) = n-c(qP) + -c(r'qU) = nT(qP) + qU. But nt(qP) = it(T 0 1 1 0 (1Y7C(T — ° \ i =0

UI

T(g) — qU.

Therefore

-c(ql) = By (ii) above, g ----- qf + r, where deg r Proposition 7.2.

n — 1. This completes the proof of

III

Definition. P(T) E (9[T] is called distinguished if P(T) = r + a„_ 1 T" - 1 + • • + ac, with ai E A for 0 < i < n — 1. (Note that P(T) is almost an Eisenstein polynomial. But we allow 7E 2 la o , so we do not necessarily have irreducibility). Theorem 7.3 (p-adic Weierstrass Preparation Theorem). Let CO

f(T) =

ai r E

i=0

and assume for some n we have ai E A, O < i < n — 1, but a„ A (so a„ E x ). Then f may be uniquely written in the form f(T) = P(T)U(T), where U(T) E 0[M] is a unit and P(T) is a distinguished polynomial of degree n. More generally, iff (T) c O[[T]] is nonzero, then we may uniquely write

f(T) = nmP(T)U(T) with P and U as above and it a nonnegative integer.

PROOF. The second part clearly follows from the first part if we factor as large a power of n as possible from the coefficients off (T).

116

7 Iwasawa's Construction of p-adic L-functions

To prove the first part, let g(T) = T" in Proposition 7.2. Then T" = g(T) f (T) + r(T),

with deg r

n — 1.

Since g(T) f (T) g(T)(a n T" + higher terms) (mod 7r), we must have r(T) =-- 0 (mod 7E). Therefore P(T) = T" — r(T) is a distinguished polynomial of degree n. Let (1 0 be the constant term of g(T). Comparing coefficients of T", we have 1 g o a (mod n). Therefore (1 0 E C x , so g(T) is a unit. Let U(T) = 11g(T). Then [(T) = P(T)U(T), as desired. Since any distinguished polynomial of degree n can be written as P(T) = T" — r(T), we may transform the equation f (T) = P(T)U(T) back to T" = U(T) - if (T) + r(T). The uniqueness statement of Proposition 7.2 now implies the uniqueness of P and U. This completes the proof of Theorem 7.3. 0 Corollary 7.4. Let [(T) E CUTE be nonzero. Then there are only finitely many x E C p , x I < 1, with f(x) = O. PROOF. Assume f(x) = O. Write f(T) = nmP(T)U(T), as above. Since U(T) is invertible, U(x) 0 O. Therefore P(x) = O. The result follows. 0 Lemma 7.5. Suppose P(T) E (.9[T] is a distinguished polynomial, and let g(T) E e[T] be arbitrary. Ifg(T)1P(T) E MT]] then g(T)1P(T) E e [T]. Suppose g(T) = f (T)P(T) for some f (T) e C[M]. Let x e a zero of P(T). Then PROOF.

Cp

be

0 = P(x) = x" + (multiple of Tr),

so 1x1 < 1. Hence f (x) converges, so g(x) = O. Dividing by T — x, and working in a larger ring if necessary, we continue this process and find that P(T) divides g(T) as polynomials, therefore in e[T]. This completes the proof of Lemma 7.5. El We now prove Theorem 7.1. It suffices to show that C[[T]] ''-' lirn e[7 ]1((1 + T)P" — 1).

Note that P(T) = (1 + T)P" — 1 is a distinguished polynomial. In fact, we can say more. The ideal (rr, T) Q (p, T) is a maximal ideal of C[T] and also gives the maximal ideal of (9[[7 ] ]. Clearly P 0(T) e (p, T). Since Pn+1(T)

P „(T)

— (1 + T)P 9P -1) + (1 + T

)

"(P- 2)

+ • • - + 1 E (p, T)

(i.e., p divides the constant term), induction implies that 13,,(T)

E

(p, T)"+ 1.

117

§7.2 p-adic L-functions

By Proposition 7.2, there is a natural map from (9[[1]] to (9[T] mod P(T) for each n. Namely, f (T) f„(T), where f (T) = q(T)P(T) f n(T), with deg fn < pn. If ni > n > 0, then Pn qm)Pn•

MT) (q.

By Lemma 7.5, f,„ fn (mod PO, as polynomials. Therefore (fo,

.) E liM

e[T]/(P(T)).

This gives us the map from the power series ring to the inverse limit. Iffn = 0 for all n then P„ divides f for all n. Thereforef G n (pl TY 1+1 = 0, so the map is injective. We now show it is surjective. Suppose (f0 , f1 , . . .) is in the inverse limit. Then, for ni n 0, f,„ fn (mod Pn), therefore (mod(p, T)" + 1). Therefore, the constant terms are congruent mod p' 1, the linear terms mod p", etc. So the coefficients of the terms form Cauchy sequence (Alternatively, f = lim fn exists since (P[[T]] is complete in the (p, T)-adic topology). Let f(T) = lim f„(T) E (9[[T ]]. We must show f (f o , f1. .). If ni n 0 thenfni — f„ = q ,,P,, for some q„,, n E (9[7]. Let in x. Then ,,

q.,„ =

— fn f — f;,

P,

Pn

Since q„,,„ E (9[T], the limit must be in C[[T]] (i.e., no denominators), so f = (P„)(lim q mn ± fn . )

Therefore f

(fo , f1 , . .). This completes the proof of Theorem 7.1.

III

§7.2 p-adic L-functions We can now construct p-adic L-functions. The strategy is as follows. First, we use Stickelberger elements to obtain an element of (9[F, ] for each n, where (9 is an approximate ring. These elements are "compatible," so they give an element of MI-1], therefore a power series in C[[T]]. This power series will give us the p-adic L-functions. Let q = p if p 2, q = 4 if p = 2. The Galois group of CD(C,i,n)/0 is (Z/qpnZ) X . If we let

=

U 0('

„

then it follows from infinite Galois theory that Gal(O( gp .,)/0) = lim(Z/qpni)x =

118

7 Iwasawa's Construction of p-adic L-functions

More explicitly, let a = ap1 e 4 , and let

aa(C) =

=

=

for some n. Then

cad'',

which is a finite product since CP. = 1 for i > n. Clearly o-a gives an automorphism of (Eno .), and a moment's reflection shows that every automorphism must be of this form, since we know what happens at each finite level. Now, Z;

(Z/qZ)X x (1 + qZ p) (Z/qZ)X x Z p ,

the isomorphism being given by

a 1-4 (co(a) mod q, )1-4 (w(a) mod q,

log logp(1 + q)) .

Also, observe that 1 + q is a topological generator for 1 + qZ p ; i.e., (1 + q)'P = 1 + qZp . Let d be a positive integer with (p, d) = 1. We assume d 2 (mod 4) (hence qpnd 2 (mod 4) for 0). Let qn = qpnd, K n = CP(Cq), and K „ = U n > 0 CP(C q„). Then Kn = Ko ( qpn) and K.. K o (Co .). It follows easily that Gal(K 00/0) A x F. where A = Gal(K o/CP)

and

F = Gal(K „/K 0)

Zp •

More explicitly, F = 1 + qo Zp =7 (1 ± q0)4, so 1 + qo gives a topological generator. The elements of F which fix K n are clearly those in 1 + qnZp = ( 1 + q0) Pn2P = Fe". Therefore Gal(Kn/K 0) = F/FP" = F. Finally,

Gal(Kn/O)

A x F.

Corresponding to this decomposition, we write

= 6(a)y„(a), with 6(a) E A, y(a)

E

F.

Let x be a Dirichlet character whose conductor is of the form dpi for some j > O. Regarding x as a character of Gal(Kn/11), we see that we may uniquely write x = where 0

E

G

fn . Then

0 is a character with conductor d or qd (hence

pq fo), while 0 is a character of Fn , SO i/J has p-power order and is either trivial or has conductor of the form qpi with] > 1. We call 0 a character of the first kind and 0 a character of the second kind. Note that the characters of the first kind are associated with K 0 , while those of the second kind are associated with the subfield of Won) of degree pn over a Therefore the

119

§7.2 p-adic L-functions

characters of the first kind correspond to tame ramification at p (i.e., p does not divide the ramification index of p), if p 2, while those of the second kind correspond to wild ramification. Observe that 1// is an even character since it corresponds to a real field. Therefore, if x is even then 0 is even. We now consider Stickelberger elements. Assume x = Ot# is an even character, and let 0* = ail- 1 , so 0* is odd. Let n

1 = — — 1 a6(a) l yn(a) - 1 qn 0
(=(— 1) x Stickelberger), and let tin = (1— ( 1 + 410)Y.( 1 + q 0 ) -1 Kn y, aa(1 + q0 ) ( 1 + go) { 62 }) 6(a) -1 Yn(a) -1 Yn( 1 + q0) - 1 . t; 1,— ( qn 1 (A Note that ri„ E Z p [A x Fn]. Let 1

i (-II

Se A

be the idempotent for 0*. Then Eo.„ = „(0)E0. and

£0* tin

= 119(0)E0.„ where

1

„(0) = — — 1 a0c0 -1 (a)Y.(0 -1 E KO[Fn] qn a and

119(0) = (1 — (1 + q0)y„(1 + q 0) -1 g„(0) = 1 (0 ± qcof al ig ni

\ E 00 [F, ] . a

f a(1 + q0 1) x Oar l (a)yn(a) 1 yn(1 + q0)1 1 q.

(K 9 = Qp(0(1), 0(2), . .

=

Zp [OM, 0(2), . . .] ).

Proposition 7.6. (a) Itin( 0) E (b) VO 0 1 then Rn( 0) E (9 9[Fni; (c) if m _.. n _.. 0 then 17 9,(0)1— ri n(0) and 9,(0)i—, „(0) under the natural map from K 0[1"9,] to Ko [r]. PROOF. (a) is obvious except when p = 2. To take care of this case, note that 7„(a) = yn(q„ — a) but 0o) -1 (q„ — a) = —Ow - '(a). The terms for a and q„ — a in 1-17„(0) combine to give

1 (r1 + 610)1 a 1

2 k

fa(1 +

lqnj 1

qn

q0)1 (1 + q0) ( 1 f

\ =(1 + q0)

This proves (a).

a -0 + (1 f a(1 +

tqnJ)

jai

1

fa(1 + go)1 kni [ q_n j

q 0)1\) f)

qn

q° 2

120

7 Iwasaw a's Construction of p-adic L f unctions -

rn ,

We now prove (c). Under the map F„, Therefore 1

f -

we have y,„(a) 1—* y„(a).

E

afko - 1 (a)y„(a)qmo
(a,0)=1

The set of y„,(a) which map to y(b) is fy,„(b + iq„)I0

i<

Since (b + iq„)0co- 1 (b + iq„) = Ow - '(b)Z (b + iqn) =

0(.0 - 1 (b)(b + Pm- n — 2

qn

1 qn),

we have (p

fl( 190

m- fl —

1)

2

o <6
Ow - i (b)y(b) -

(6,0)=1

Since y(b) = yn(q„ — b), but Ow - (q„ — b) — Ow - 1 (b), the second sum „(0). vanishes. Therefore (8) — (1 ± q0 )7„(1 + goy so i 1 „,(0)1— Also, 1 — (1 + q 0 )y„,(1 + q 0) 1 IL(0). This proves (c). For (b), we use a slightly longer proof than is necessary, since it gives additional information which will be useful later.

Lemma 7.7. y(a) = y(b) <=>

(b> mod qp"•

The decomposition cra = (5(a)y„(a) corresponds to CP(Cqn) = ( go) • where 1:, „ is the subfield of 1:1(Cap„) which is cyclic of degree p" over CP. Since o-a restricted to IL„ depends only on (mod qp") (the w(a)-part gives LII the action on (13(Cq)), the lemma follows.

PROOF.

Let R denote the set of (p — 1)st roots of unity (2nd roots of 1 if p = 2) in 7L. ba yn(a) = y „(b) <=> EE 1 (mod qp") .4=> a/b w(a/b)<=>a (mod qp") for some a E R. If a E Zp , let s„(a) be the unique integer satisfying s„(a)

a (mod qp"),

0

s„(a) < qp".

Actually, s„(a) is a partial sum of the p-adic expansion of a. The above may be rephrased as s„(a) = s„(ba).The set of numbers a with 0 < a < q„ such that s„(a) = s„(ba)is {s„(ba) + iqpn 10

i

d

—

1}.

121

§7.2 p-adic L-f unctions

(Recall qc, = dq). Finally, let T denote a set of representatives of elements of (Z/q„Z)X such that Fn = {yn(b) I b E T } . We have R„(69 = —

a0(o'(a)y„(b)- 1 L qn

=— i

l

be T y(a)= y(b) d-1

Lqn

EEE

(sn(boc) + iqpn)0(o'(sn(ba) + iqp")yn(b) - 1 .

beT aeR i--=0

Ha E R then — a E R, so we let R' be a set of representatives for R mod{ + 1 } . Clearly sn( — ba) = qpn — sn(ba) (if ba * 0),

so No - 1 (s„(— ba) + (d — 1 — i)qp") = Ow - 1 ( — s„(ba) — iqp" + dqp") = --0(o'(s„(ba) + iqp"). (Recall fo,, - i divides q„ = dqp"). We now assume d> 1. Combining the term (oc, i) with (—a, d — 1 — 0, we obtain 1 v, ',„ '—'-ln

d-1

(s„(bce) ± iqp" — qp" + s(b)

beT cceR' i=0

—(d — 1 — i)qp")8w -1 (s„(ba) + iqp")y„(b)- 1 d-1

(s (b)+ iqpn — 3' 2-')oco - i(sn (bo) + iqPn)Y n (b) - 1 .

2

qn beT leR' 1=0

Lemma 7.8. Suppose s, t e Z, (t, d) = 1. Then d-1

1

0 W -1 (S +

itq) = O.

i=--0

PROOF. If p,' fk,-1, then fo - i = d. Since s + itq runs through a complete set of residue classes mod d, the result follows. Now suppose pi f0,,,,, so the conductor is qd. If pis then all terms in the sum are O. If p , f' s then s + itq runs through all residue classes mod d, but is fixed mod p. Since fo- i > q, there is an integer u 1 (mod q) with Ow '(u) 0 1, O. Multiplication by Ow 1 (u) permutes the sum, which must therefore be 0. This proves the lemma. 0 We now have d-1 i 1 „(0)

= — E E E —, ow - i(sn(bot) + beT aeR' 1=0

a

Since p,' d, it follows that Rn( 0) G Ce[F ] .

iqp")y„(b) -1 .

122

7 lwasawa's Construction of p-adic L f unctions -

If d = 1, then O is a power of w and q„ = qp". Since 0 0 1 and O is even, we cannot have p = 2 (or 3). Therefore we may ignore the 2 in the denominator. We find from the above that 1

R„(0) =

sn(ba)Oco'(s(ba))),(b)- 1 •

LqPn beT aeR

Since sn(ba) -= ba (mod qp"), we have Ow - 1 (s„(ba)) = Ow - 1 (b)Ow'(a) = Ow - 1 (b)0(a)a- 1 .

Consequently

1

E E botOw - 1 (b)0(a)a - 1 7„(b)- 1

LqPn beT aeR

0 mod Ce ,

since 10(1) = O.

This completes the proof of (b), hence of Proposition 7.6.

E

For future reference, we record part of what we just proved. Proposition 7.9. Assume 0 0 1. (a) Iffy = q, then

E E sn(ba)Oco"(ba)y„(b) - 1 ;

1

Rn( 0 (b) iffo

n ) —

LqP

n

beT aeR

q, then

1,(0) = —

1

d-1

i0co - 1 (s(ba) + iqpn)y„(b)- 1 .

0

Li beT aeR i=0

Combining Theorem 7.1 and Proposition 7.6, we find that there are power series f, g, h E 60 [M] such that (if 0

lim (9) 4- f (T, 0)

1)

lim 11, (9 ) 4- g(T, 0) lim 1 — (1 + q0)7„(1 ± q 0

) - 1

4- h(T, 0).

It is easy to see that h(T, 0) = 1

1 + qo 1+T.

Also, f (T, 0) = g(T, 0)

If 0 = 1, we take this as the definition off (T, 0).

123

§7.2 p-adic L-functions

Theorem 7.10. Let x = 6,li be an even Dirichlet character (0 = first kind, = second kind), and let Co = 0(1 + g0) -1 = X(1 + go ) - 1 (=a root of unity of p-powder order). Then p(s, X) = f((1 + q0)5 - 1, 0). Remark. This is a very useful result. Note the essential difference between the

contributions from the characters of the first and second kinds. This will be used in the proof of Theorem 7.14, and it is what one expects from analogy with function fields (Section 7.4). PROOF. Observe first that if si < gp -

1) then

1( 1 + 010)5 — 1 1 = exp(s log(t + g o ))

-- 1

I

<

1,

and since C o is of p-power order, (1 + q 0 )5 il < 1. Therefore the righthand side converges and is an analytic function of s. Consequently, we only need to prove the above equality for s = 1 — m, where ni is a positive integer. We shall work with ;i(0) and g(T, 0) since, in all cases, they have integral coefficients. Let i(a) = logp llogp(1 + g o). Since y„(1 + q 0) corresponds to 1 + T, it follows that y(a) = y „(1 + q 0 ) ) corresponds to (1 + T)' ) (mod(1 + T) 1). From the definition of IMO), we have —

"

E

g(T, 0)

+

0
a} q0){—

{ (1 + g0)a}) qn

qn

x Oar 1 (a)(1 + T

)

")-1

(mod(1 + T)P" — 1). Let (1 + g o )a = al + a2 g„, with 0 i(ai ) mod p", and g(T, 0)

al < g,. Then i(a) + 1 = i((1 + go )a)

E a2 0w - 1 (a 0(1 + T)- " 1) (mod (1 + Ty - 1). a

If rn is a positive integer and n is sufficiently large, then

g ( 0( 1 + .40) 1-m — I, 0)

E az

1 (ai)(4 1 (1 + Ro)mm i V ai)

a

od g„,

since

(1 + T)P" — 1 = (C,,(1 + q 0 ) 1 ')P n — 1 0 (mod g,). = (1 + q 0 ) (1- m)P" — 1 Therefore + q0Y (al) = tii(ai), and (1 + 0101(a1) = .

But (0-4° 1)

g(4( 1

qa) l-m — 1, 0)

a2 00) - 1 (a1)0(a1)
E az Xar m(a i )dr - 1 (mod g,). a

124

7 Iwasawa's Construction of p-adic L f unctions -

If n is large enough that fx l qn , then zco'n((1 + Ro)a) = Zco m(a 1 ). Also,

(( 1 + Ro)a) m aT + maT'q„a2 (mod qn2), SO

zw - m(1 + q0 )(1 + q o)m

ZW -m (a)ani a

ZCO -m (ai)dil

mq„

a 2 zco'n(a Oar (mod q).

Note that this last term is the one we need to evaluate in the above. As a runs from 1 to qn , so does a l , so the first two terms are the same. Also, zw -m(1 + q0 ) = z(1 + q0 ). We obtain g (C0(1 + 470)1—m 1, 9)

1 = ((1 + qoriz(1 + q 0) - 1) - lim 1—

m

= -h((1 +

1 1 — 1)— lim —

m

n-œ

z w - -(a)an

qn 0
z w - -(a)am.

qn a

The following lemma completes the proof of the theorem. Lemma 7.11.

1 1im—

zw - m(a)am = (1 - xw - m(p)pm

)13„,,

m.

n-o c1 qn 0
From Proposition 4.1 (recall LUX) =

Bm,,--

1 qn m (i — x co -- -(i)qnBm — qn qn 1

— zw—u)(7. — — fn - lqn)(mod - qn 2 qn

)

(1/p takes care of the 6 in the denominator of B 2). Since Zw -m (qn

./) • (qn - Am-1

- Zco -m(him-1 (mod qn),

we may pair terms to obtain

z w—wr - 1

o (mod q„).

Therefore Bm, ?cm —

1 tb, lim — zw - m(j)i m . n qn J-1 -. 00

125

§7.3 Applications

Finally, we obtain qn

(1 —

qn

-i

_

= Ern — zco — muy'n — lim — E zco m(PiXpir qn j=1 qn j=1 qn

= liM

ZW qn j= 1 P 9n

1 = liM qn

ZW =1 (i, go)= 1

This completes the proof of the lemma, and also of Theorem 7.10.

LI CI

§7.3 Applications Theorem 7.10 has many applications. For example, the congruences of Chapter 5 may be generalized (see the Exercises). In the following, we shall give an application to class numbers, but first we need a result about g(T, 0).

Lemma 7.12. If O = 1, then g(T, 0) is a unit of 7/ [[T]]. PROOF.

By Theorem 7.10, f(0, 1) =

1 B 1 , 0,-1 = - -

q =1

1 co - 1 (a)a - mod Z p ,

n

since w(a)

a (mod q). Also

h(0, 1) =

q.

Therefore Ig(0, 1) =

(0, 1)h(0, 1)

q mod g 2p 2 P.

It follows that 1-g(0, 1) A 0 (mod p), so the constant term of -3-g is a unit. This completes the proof.

Theorem 7.13. Let (d, p) = 1, qn = qdp", and h,; = d 0 2 (mod 4). Then

0

H

0#1

felo

o even

(Q(Cq.)). We assume

If(C - 1, 0) x (p-adic unit). =1 *1

126

7 I wasawa's Construction of p-adic L-functions

PROOF.

From the analytic class number formula we have

V = 2q0 Q n ( - 4B 11 A #1

1)

.4+ Igo (le yen

and hp;

= 2q,Q n

(--1.B,,,_,).

x# 1 fx I qn x e vert

The number Q equals 1 or 2, but is the same for all n > 0 by Corollary 4.13. Writing x = 04/, where 0 is of the first kind and ifr is of the second kind, we obtain

11 (---113 1 ,-L) =

x*,

n ( -1131, 0„_ 0 n ( _ 113 1 , 0 ,o _ 0 n ( -01,0 0 _ 0. O*1

0#1

61 ilil

The first product is the same as that for h o- . To treat the second product, note that —B 1

g(Co —1, 1) 1 = L(0, ip) = h(co _ i, 0

(tko'(p) = 0, so the Euler factor disappears. It is because of the Euler factors for the other characters that we must take the ratio hn- IhcT and require C 0 1; otherwise the formulas could reduce to 0 = 0). From Lemma 7.12, lggv, — 1, 1) is a unit; and — 1, 1) = 1 —

1 + q y - = 1 — Co; '(mod q).

Since 0 equals ti/ evaluated at a generator of F„, Co determines ifr. Since there are p" elements of f'„, it follows that as ii/ runs through the characters of the second kind, C,I, runs through all pnth roots of unity. Putting everything together, we find that

4

H (-1-131,0.-t)) = vp ( H (1 - ç -1 ) -1 =

vp(p - ")= v P

40

(1

qPI

kil*i ;*1

For the third product, we proceed as above (again, since 0 0 1, the Euler factor disappears): — 413 i , 0 0, - 1 —

if (Co — 1, 0 ),

SO

11(-4/31,00.-i) = 9*1

El

11

0# 1 O' n = 1

If ( - 1, 0).

Combining all the above, we obtain the theorem.

D

127

§7.3 Applications

Theorem 7.14. Let pen be the exact power of p dividing hn- , in the notation of the previous theorem. There exist integers A, p, and y, independent of n, with A > 0, p > 0, such that en = An + pp" + for all n sufficiently large. PROOF. In the notation of the previous theorem, let A(T) = fl f (T, 0) e 61* 1

Then h:

H

n o V.

AG — 1) x (p-adic unit).

*1

By the Weierstrass Preparation Theorem, A(T) =-- pmP(T)U(T), where p 0, P(T) is a distinguished polynomial, and U(T) is a unit of Z pUTE. Therefore v(h) --- v(h) + (Pn — Oft + vp (

fl PG — 1)). *

Let A =-- deg P(T), so P(T) = TÀ + 0 < i < A — 1. If n is large enough and if then v(( — 1) A )

=

+ • • + ac, with plai for is a primitive path root of unity,

oA < (fl)

V p(p).

Hence v p(P(C — 1)) =-- vp( ( — 1)'). It follows that for n sufficiently large,

11 PG - 1) = v11 p( G - 1)A) + c =

vp

v(p)

+ c=

An + C,

r,"= 1 (

*1

where C is independent of n (it absorbs the effect of low-order roots of unity). LI The theorem follows easily. The above is part of a much more general theory of Iwasawa, which we shall consider in a later chapter. Suppose we have a sequence of number fields U K r„ K o c K 1 c • • • c K„ c • • • c K. with Gal(Kp/K 0) ZIpnZ. Then Gal(K co/K 0) = lim(Z/pnZ) =Z p , so the extension K cjKo is called a 4-extension (or F-extension). Let pen be the

128

7

lwasawa's Construction of p-adic L-functions

exact power of p dividing the class number of K„. Then there exist integers A, [4 y, as above, such that e„ = An + pp" + v for all sufficiently large n. If the fields K„ are CM-fields, then hn = hn+ h„- ,

en = en' + e,.

p = p." + ,u - ,

etc.

So what we have proved above is the existence of A - , it - , 1, - . We shall show later (Chapter 13) that for a given base field Ko there are at least (exactly if Leopoldt's Conjecture is true) r 2 + 1 independent 7extensions, so, for example, real fields should have only one la-extension, while imaginary quadratic fields have two independent l a-extensions. For the moment, we content ourselves with showing that every number field has at least one Z a-extension. Let B„ be the unique (unless p = 2 and n = 1) subfield of W on) which is cyclic of degree p" over (L) (use the isomorphism (11qp"Z) X '-' (21q2) x x (cyclic of order pn); let B„ be the fixed field of (21q2) x ). Then 0 = m o and 03 /0 is a Z p -extension. It corresponds to the group of all characters of the second kind. Now let K be any number field and let K, = KIL . We claim that K oolK is a Z p -extension. Let Be = K n R. „,. Then Gal(K,o/K) --, Gal( 00 /11„, n K)--, pe2 p - Z p , as desired. The extension KIK is called the cyclotomic 2 p-extension of K. If K contains 0 ( q) then the extension is obtained by simply adjoining all pnth roots of unity for all n. This is what happened in Theorems 7.13 and 7.14.

§7.4 Function Fields The theory of cyclotomic l a-extensions has a strong analogue in the theory of function fields over finite fields. Let Fq be the finite field with q elements (no relation to the previous q; but this is the standard notation). Let X, Y be indeterminates related by a polynomial equation over Fq , so k = IF q (X, Y) has transcendence degree one. We assume k n IF, = Fq . The field k is called a function field (of one variable) over Fq . It is well known that there are close connections between the arithmetic behavior of number fields and that of function fields; for example, both have zeta functions, satisfy class field theory, and have finite residue class fields at all (nonarchimedean) places. Let Ck(s) be the zeta function of k. Then

= (1

R(q — 1) _

—

q S)(

ql - S)

where R(T) E 2[T] (we have used a nonstandard normalization of the numerator; usually P(T) = R(T — 1) is used). The zeta function of the field 4 (X), the analogue of 0, is simply 1 (1 — q -5)(1 — ql-s)

129

§7.4 Function Fields

so the numerator is a product of L-series (at least when k is abelian over Returning temporarily to number fields, we assume, for simplicity, that K = U(C„) + and let K. p(S) =

{

1 p (S,

O even felP

Then

K ,p(S)

is the p-adic zeta function of K. Let A(T) = g(T, 1)

fl

O*

f (T, 0) c

1

Then

A((1 + p)' — 1) ho p)s 1)

which is a formula remarkably similar to the one above, except that A is a power series instead of a polynomial. But the Weierstrass Preparation Theorem says that a power series is " almost" a polynomial. Note in addition that h has a relatively simple form, and it may be traced to T:; , p(S) (i.e., 0 = 1), again in analogy with the function field case. Of course, this may also be done for fields other than U(C p) + , but then we must use q o in place of p. However, qo depends on the character 0; so either the above formula becomes a little more complicated, or we change variables in the f(T, 0) so that we may still use p (change T to (1 + Tr - 1, where a = logp(1 q 0 )/log p(1 We now return to function fields. The polynomial R(T) satisfies the following properties : (1) R(T) =

[p i (1 — ce,(T

+ 1)) where the of's are algebraic integers of

absolute value q 112 (the Riemann Hypothesis) and g > 0 is an integer called the genus of k. (2) R(0) = — = h(k) = the number of divisor classes of degree 0 for k. This number is the analogue of the class number. (3) If IF/F q is an extension of degree m then k' = kIF is a function field over F. The numerator of the zeta function of k' is 2g

Rk , (T — 1) =

fl (1

—

(V')

R k(cT — 1).

= m

J .,

1

From the theory of finite fields, there is a unique sequence of fields Fq

F q , c • • c If q ,•• c • • • c

=

UF

q P"

fl

which is clearly a 7 p - extension. Therefore, if k, = klFqp, then k = ko

c•-•c

c • • c=

/c o,

is also a 7 p - extension . Since everything except 0 in a finite field is a root of unity, we have obtained this extension by adjoining roots of unity; and if Fq contains the pth roots of unity (remember, p and q are not related) then

130

7 Iwasawa's Construction of p-adic L-functions

the extension k obtained by adjoining the p-power roots of 1 (since reducing the rings Z[C p„] modulo appropriate primes gives a 4-extension of finite fields). Therefore Ic,,lk is analogous to the cyclotomic Z a-extension of a number field. Combining (2) and (3) above, we find that h(k n) = fl R ( _ 1 )

k h.o ,

c pp7,, C#1

(the analogue of Theorem 7.13). The polynomial R(T) is not necessarily distinguished, so we write R(T) = p"P(T)U(T), where p 0, P(T) e Z p[T] is a distinguished polynomial (of degree < 2g), and U(T) is a unit of l[M] (actually, U(T) is a polynomial by Lemma 7.5). Since R(— I) = 1, by (1), 1.1 = 0. Alternatively, when R(T) is expanded as a polynomial in I + T, one of the coefficients, in this case the constant term, is not divisible by p. This is essentially what we shall do to prove p = 0 in the number field case. Now let pe" be the exact power of p dividing h(lc„). As in the proof of Theorem 7.14, we find that en = An + y

for n sufficiently large,

where /1 < 2g is the degree of P(T). Because of the strong analogy between cyclotomic l a-extensions and the above situation for function fields, plus some numerical evidence, Iwasawa was led to conjugate that p = 0 for cyclotomic Z r-extensions of number fields. For the special (and most important) case of 0(4.)/(1)(4), Iwasawa and Sims showed that p = 0 for p < 4001. Subsequent calculations (on a computer, of course) by Johnson and then Wagstaff extended the result to p < 125000. In the next section we shall extend the result up to p < ao .

§7.5 ti =0 Theorem 7.15. Let K be an abelian extension of G, let p be any prime, and let K 09IK be the cyclotomic Zp - extension of K. Then p = 0. Remark. Iwasawa has constructed examples of noncyclotomic la-extensions with p > 0. See (Iwasawa [24]). PROOF OF THEOREM 7.15. Before starting the main part of the proof, we state some facts, some of which will be proved in later chapters but which are needed now. I. K is contained in a cyclotomic field (Kronecker-Weber theorem). II. If K

Kr then p

p'.

131

PROOF. Lift the p-part of the Hilbert class field H„ of K„ up to K. The compositum H K is contained in the class field of K. Since H„ and K;, might not be disjoint, we could lose a factor of at most [K;T : K u] < [K' : K]. Therefore e;, + 0(1) > e„.The result follows easily. LI III. Assume the pth roots of unity are in K. Then p = p+ + p- , and if = 0 then p+ = 0 (proof in Chapter 13). IV. Suppose e„ = An + pp" + y for K Go/K o . Let K' = K. Then K „D/K 1 is a 4-extension and

A ' = A, PROOF.

p' = ,upm,

e'„ = en , = An + (ppm)p" + (y

y' = y + Am. Fil

Am).

We now claim that it suffices to prove p - = 0 for K of the form 0(0), where q = p or 4 and (d, p) = 1. (These are exactly those to which Theorems 7.13 and 7.14 apply). For if K is arbitrary, K O(, n,,,d) for some n and d. If p - = 0 for 4:1( 0) then p + = p = 0 by III. By IV, p = 0 for O( u nd) and by II, p = 0 for K. If 0 0 1 is an even character of O( qd) then if (T, 0) has p-integral coefficients. If we can show that for each 0,1f (T, 0) has at least one coefficient relatively prime to p (" pe = 0"), then it follows easily that p - = 0 (see the proof of Theorem 7.14). This is what we shall do. As in the function field case, it will be more convenient to work with 1 + T than with T. Recall that if a E Z p then s(x) is the unique integer satisfying 0 s(a) < qp" and s(u) a (mod qp"). For p 0 2, let CO

a =

ti(a)p , 0 < (0) p — 1, -0 be the standard p-adic expansion. Then s(a) = 0 ti(a)pl (we do not need ti(a) for p = 2). Proposition 7.16. Let R be the set of (p — 1)st roots of unity in 4 (R = 1+1) if p = 2) and let R' be a set of representatives for R modulo 1+1). (a) Suppose 0 = co", k # 0 (mod p — 1), k even. Then tie = O there exists )6' E Z px and n > 1 such that

tn(floc)oc k- # 0 (mod p). Œ€R

(b) Suppose 0 is not a power of co. Then p c, = 0 .4=> there exists 13 n > 0 such that d-1

i0o) -1 (s, (f3 1) + iqp")

0 (mod

A),

aeR' i=0

where

A

is the prime of (99 above p, and d or dq is the conductor of 0.

E

Z; and

132

7 Iwasawa's Construction of p-adic L-functions

PROOF. (a) Since 0 * 1 is even, we must have p 5, so we have the luxury of ignoring 2 and letting p = g. In the notation of Proposition 7.9, we see that if R„(0) is expressed as a polynomial in 1 + T, modulo (1 + T)P" — 1, then each 7„(b) - 1 corresponds to a different power of 1 + T. Let 7„(b) - 1 4(1 + T)'", 0 < ah < p". Then (T = set T = variable) If (T, 0) -- —

1 L L 40000w-1000o + 2p" +1 beT aeR

Tr mod(1 + T)P" — 1.

Since (1 + T)P" — 1 TP" (mod p), the above congruence determines the coefficients of (T, 0) modulo p, up to TP" - 1 . Suppose [to > 0, so p divides all the coefficients of If. Then p divides all coefficients of the above polynomial when it is expressed as a polynomial in T, or in 1 + T. Consequently, pe 0 0 <=> L s„(ba)Ow'(ba)

0 (mod pn+ 2 ),

aeR

for all n and all b E T. But we can arbitrarily change the choice of the set of representatives T (this does not change the sum since we sum over R). Therefore we can consider all b e Z, (b, p) = 1. It is more convenient to consider all b e Z; ; this does not affect anything since we are only looking at the beginning of the p-adic expansions. Since 0 = oik and w(a) = a, Ow 1 (c) = a' '• Also, we may factor off and ignore 00) -1 (b). Writing sn(ba) = sn + 1 (ba) — t„ + i(ba)p"± 1

ba — t n + i (ba)p"± 1 (mod pn+ 2 ),

we have iie 0 0 <=>

L (ba — t. + i(bo)p n ± 1 )ak- 1 --- 0 (mod P n +2) aeR

<=> L tn±i(bc)/k - ' -_ o (mod 13), aeR

for all n 0 and all b e Z;. This proves (a). (b) This part follows immediately from Proposition 7.9, in a manner similar to part (a). I=1

Proposition 7.17. Let ni, d be positive integers with (p, d) = 1. For all n sufficiently large, there exist P i , )32 e Zp , both congruent to 1 mod pm, and there exists ao G R' such that

sn+ ,,([31 a) = s„([31 a) _- 0 (mod d)

for all a e R',

4+,4)6'2 a) = s,,(13 2 ci)

for all a

0 (mod d)

ao , a e R',

sr, ± „,(02 a o ) = 4(3 2 ao) + gpn _- 0 (mod d).

Remark. What this means is that the p-adic expansion of each # l a and 132 a (a 10 ) has ni consecutive O's starting with the (n + 1)st place, while 132 ao has a 1 followed by ni — 1 O's. If a is a normal number (i.e., all possible combinations of digits occur with the expected frequency) then there are

133

§7.5 p -= 0

arbitrarily long sequences of O's. But we do not know whether or not any a is normal (even though almost all numbers are normal), so we must use )6' 1 and )6' 2 to help. Also, we are requiring the desired patterns to occur for all a simultaneously, which causes additional problems, especially since there are usually dependence relations among the a's. We shall postpone the proof of Proposition 7.17 in order to complete the proof of Theorem 7.15. We first treat criterion (a) of Proposition 7.16. Since

1 )Pm+ + — Or+ 2 + • • •

= P • Pm + (I) we have

if n > vp(y).

t„(— y) = t„(0 — y) = p — 1 — t(y)

Since k — 1 is odd, we may combine the terms for a and — a to obtain <=> 2 1 t (/61)1' —= (p — 1) Œ cceR' n

p)

aeR'

z;

for all )6' c and all n > 1. Note that the right side is independent of )6' and n. In Proposition 7.17 let in ---- d = 1. Then for n sufficiently large we have 13 1, )6'2 such that

tn+ a l a) = 0

for all a c R',

tn+ 02 a)

for all a

0

o' a

E

R',

tn + 1(fi2 ao) = 1 -

Therefore if [to 0 0, the above criterion (with n + 1 instead of n) yields 0

(p — 1)

ad-1

(fl =

and

24' (p —

(fl = 1% ).

This is impossible (recall p 2 in this case), so [to = O. We now consider part (b) of Proposition 7.16. Assume tte 0, so d1

6 '(s(0) + iqpn)

0 (mod

A)

aeR i=0

0 and all )6' c 4 . In Proposition 7.17, let in = 2, d = d. If n is sufficiently large, there exist fi 1' )62 1 (mod p2 ), in particular )6' 1 , )6' 2 = 1 (mod q), satisfying the criteria of Proposition 7.17. Therefore

for all n

sal

fi1»x= a

fl2

s(fi2 a) (mod q),

and

s„(fl i a) —= O

s.( 62 a) (mod d)

for a

oto •

134

7 lwasawa's Construction of p-adic L-functions

Hence s(/3 2 Œ) (mod dq),

s(fl i c)

2

0 Oto,

and Ow - 1 (4(fi 1 7) + iqp") = Oco'(sn(fi 2 cx) + iqp")

for all j and all a

ao . Similarly, sn(P2(x0) —= sn(/62(x0) + qp" (mod q)

sn(flicxo) and

-= s(/3 1 Œ0)

s(P2 Œ0 ) + qpn (mod d),

SO

19(0'(sn(fl i ao) + iqpn) = eco -1 (sn(fi2l0) + (i + 1)qpn).

Let a c, = s(/3 2 c), for convenience. Comparing terms above for fl = /3 1 and fl = #2 , we find that we must have d—1

d—1

'(a0 + iqp") i=0

i=0

i00) - 1 (4(/3 icto) +

d— 1

i0w - '(ac, + (i + 1)qpn) i=0 d

d

jOw - '(a0 + jqp") — j= 1

Ow - 1 (a0 + jqp") i=1

d—1

d

6w '(a0 + fqp")•

iow - '(a0 + iqp") + dOw - 1 (a0 + dqp") — i=0

j=1

The last sum vanishes by Lemma 7.8. Consequently, dew - '(a0 + dqp")

0 (mod A).

But p 4/d and Ow - 1 (a0 + dqp") = Ow - 1 (a0), so Ow - 1 (a0) = O. But ao = s„(fi2g0) — qp" (mod d), so (a0 , d) = 1. Also, 4((12 go) fi2cxo czo (mod q), so (a0 , q) = 1. Therefore (ao, fo.- i) = 1, hence Ow - '(ac,) 0 O. This contradiction completes the proof of Theorem 7.15.

PROOF OF PROPOSITION 7.17. Let r be a positive integer. A sequence of vectors x„ E [0, Or is called uniformly distributed mod 1 if for every open set U g [0, 1)r we have •

lim N

co

#{n Nix„E U)

— meas(U),

§7.5

p

135

=0

where we take the usual measure, normalized by meas([0, ly) = 1. There is the following well-known criterion of Weyl: The sequence { x„) is uniformly distributed mod 1 <=> for every z z A 0, we have

E

Zr,

N

e2Itix" . z = 0

lirn — N -■ oo N n=1

(x • z is the usual dot product; note that if z = 0, the limit is 1).

The proof of the criterion may be sketched as follows: the trigonometric polynomials are dense in the set of continuous functions on (R/Z)r, and j. e2lux-z dx = 0 if z 0, so the above is equivalent to

1 — N

f (x„) f f (x) dx

for all continuous f.

n=1

Now, if f approximates the characteristic function of an open set U then f (x„) is approximately # {n NI xn E U}, while f (x) dx is approximately meas(U). Since step functions can be used to approximate continuous functions, the argument also works in reverse. For fuller details, see (Kuipers and Niederreiter [1]).

E

Lemma 7.18. For 13 e Z p , let x„(13) = q - 1 p'(fi). For almost all fi (i.e., except for a set of measure 0 for the usual Haar measure on Z p), the sequence of numbers x(f3) e [0, 1) is uniformly distributed mod 1 PROOF. Let S(N, #) = (1/ N)L„N= e(x„(M), where e(x) = e2 '. Then 1 1_ 1 v f L N N 2 m#n

I S(N, #)1 2 d# = —

e(x n(#) — x m(#)) dfl.

'

Suppose n > m. We claim that the map 1/qp9/ 7L/qp9/,

s(g) Pn-ms.(g)

is a bijection. For suppose

s(c) — P n-ms.(g)

s(f3) Pn-msa) (mod qp").

Then —

s(c) — s„(#) — s, n(fi))

It follows that a — fi

p" - m(oc — )3) (mod qp").

0, so the map is injective, hence bijective. Since

x„([3) — x,n()(3) = q - 1 p'(s„()3) — p's,n(fi)),

136

7 Iwasawa's Construction of p-adic L-functions

it follows that if # runs through the congruence classes mod qp" in zp , e(x„(fi) — x„,(#)) runs through all qp"th roots of 1. Since each congruence class has the same measure, namely q -1 p - ", it follows that the integral above vanishes. Similarly, the integral is 0 if n < m. Therefore 1 f 1S(N, fi)1 2 dfi = KT , SO

i f 'SOW, I mi=11S(m 2, 13)1 2 dfi = m.1

' 1 fi)1 2 dfi = 1 2 < X— m=1 M

Consequently 1 1 S(M2, 13)1 2 c Ll(z p,) , hence the sum must converge for almost all #. Therefore lim1S(m2 , #)1 2 = 0 for almost all #. For arbitrary N, choose m such that m 2 < N < (m + 1)2 . Trivial estimates yield 1 S(N ,

13)1

1 S(m 2 ,

p)1 +

2m

N

, 0 as N oc

for almost all # E zp. Now let z E Z, Z 0 O. Since

zsf,(#) _-,-_- z# ---- s r,(z#) (mod qp"), we have e(zx„(#)) = e(x„(z fi)). By the above,

1N 1N le(ZX„(P)) nl =n - iv- E e(x n(z[1)) = S(N, zfl) --, 0

-ir -

for almost all #. Each z excludes a set of measure O. Since Z is countable, we exclude altogether only a set of measure O. For the remaining Ps, we may apply the Weyl criterion (r = 1). This completes the proof of Lemma 7.18.

0 Remark. A p-adic number # is called normal if for every k > 1 and every string of integers of length k, consisting of integers in 10, 1, . .. , p — 11, the standard p-adic expansion of # contains this string infinitely often, with asymptotic frequency p -k . It is not hard to see that # is normal if and only if x„(#) is uniformly distributed mod 1 (see Exercises). Therefore, almost all # e Zp are normal. Since the digits of the p-adic expansion can be regarded as independent identically distributed random variables, this fact may also be approached via theorems of probability theory.

§7.5 p = 0

137

Lemma 7.19. Suppose y„ ,Y r C Zp are linearly independent over O. For almost all fl C Zp the sequence of vectors X, = X n(fi) = (xn(1371), • • • xn(AY,))E [O, is uniformly distributed mod I.

PROOF. Let z = (z 1 , ...., z,.) E Zr, z q -. 1r n zi sn(S 7i )• . Since

E z i s,(py ,)

0, let 13 E Zp , and let y, = X, z =

E zi flyi

sn(fl

E z i y i) (mod qp"),

we have

y„

q -1 p - "s(#y) (mod 1),

where y =

y, z i y i .

Note that y 0 since the y i's are linearly independent. By Lemma 7.19 and the Weyl criterion (with r = I), 1 —

E

1

e(X „. z) = —

a°

E e(x(/3 y)) o

for almost all fl E Zp • As in the previous lemma, each z excludes only a set of measure 0, and E is countable, hence the Weyl criterion (with r = r) completes the proof of Lemma 7.19. LI Lemma 7.20. Suppose y l , , yr c Zp are linearly independent over Q. Let = , c (0, 1)% let y > 0, and let m and d be positive integers with (d, p) = 1. For each n sufficiently large, there exists # C Zp such that

(1) iY 1 (mod pm), (2) Ixr(fiY) < e for 1 < j < r, (3) s(#yi ) =- 0 (mod d) for 1 j r.

PROOF. For t c R, or H/Z, let lit denote the distance from t to the nearest integer, and let (ti, • • • , t i.) = max 'ltd. Let E = E/2d and x' We assume e is small enough that ; + < I and ; > 0 for all]. Since [0, l ] is compact, there exist y l , , y D E (0, 1)r, for some D, such that for each y E [0, 1] we have iLr - vJ < E' for some i. By Lemma 7.19, for each yi there exists /3i E 2p and n i E Z such that X(fli) - y i < v Let no = m + max n i . We claim that if n no we can satisfy (1), (2), (3). Choose yi such that 1 (note E ). - 4d) - v < 81 d P .

Let fl' = 1/d + pn - "13i . Then X(r) < x' -

d

- Yi

+ iiYi - X n,(fii)11 1 + iiX n( -) + X(fi) - X n(fi')1 d

138

7 lwasawa's Construction of p-adic L-functions

But

sn(1610

1 sn(-d

+ sn(13" -"Va j)

1 sn(-

+ PH n 's„,(fliyi) (mod qpn)

d

for 1 < j < r. Therefore

1 d

X() X n ( -) + X,„(fli) (mod 1),

so the last term in the above sum vanishes. Hence X'

- )4,6') < e' + e' +0 = - , d

SO

1)7

dX„(r )11 < e, and

for each j. Since i qpn. It follows that

< 1 and

sn(dry i) = ds,(ryi)

1)7; -

<e

- e > 0 by assumption, 0 < ds„(ryi) <

0 (mod d),

and

dX„()3') =

Since n - n i > ni we have dr 1 (mod pm). It follows that )3 = dr satisfies the conditions of the lemma. This completes the proof of Lemma 7.20. LI .

Remark. The location of the vector X n()3) depends on the coefficients of the

p-adic expansions of the fiyi's near the nth digit. By the choice of the vectors X„,()3)) are distributed throughout all of (0, 1)". Hence they give us a wealth of possible patterns of coefficients. We can add these onto existing patterns to obtain any desired pattern. In effect, this is accomplished by the term p" - "A in the definition of fi'. This is what allows us to get close to )T and also obtain the congruence mod d (cf. Ferrero-Washington [1]). We can now prove Proposition 7.17. We cannot apply Lemma 7.20 directly since the elements of R' are not necessarily linearly independent. If R' is linearly independent (4.>p is a Fermat prime) then the following argument can be simplified to yield the result. Therefore assume R' has dependence relations. If a is a primitive (p - 1)st root of unity and r = (/)(p - 1), then 1, a, , a' forms an integral basis for Z[a]. Consequently we may choose a i , , a, R' (let ar+ 1 , • , a„ t = (p - 1)/2, be the other elements) such that

ai =

1=1

j = r + 1, . . . , t.

§7.5 p

139

0

We may assume ail 0 0 for some j. Order a„ 1 , , a, lexicographically according to I aj,I, 1 i r. That is, let ai > al if for some j 0 we have la » I = Let be a maximal element for this ordering I a l, I for i < io and I a I > I (we do not care whether or not a unique). If necessary, change the signs of a 1 , , ar so that a 0 for 1 < i < r (this changes K). Note that aio > 1 (since a i 0 0 for some]) and ajo , > 0 for some other i (since ajo/oe, Z). Now change the signs of ai , r + 1 < j t, if necessary, so that the first nonzero a,. 1 — < i— < r3 is positive for each such]. Let x„ , E 1) be such that x, is much larger than x,, for each L Define

=

j t.

r+ i=

Since the first nonzero coefficient a, is positive and since x, is much larger than x i x i+ 2, etc., we must have xi >0 for each ]. By the choice of f 0 , xio > xi for r + 1 j < t, j Jo (even if ajo is not the unique maximal element, we have ("Jo, > 0 for all i; so any other maximal element must have a negative coefficient, hence a smaller xi). Also, x i, > aio x, x1 (since ajo, > 0 for some i # 1), so x > x i for 1 i r. Replacing x, by cx, for a suitable constant c, we may arrange that 0<x
for 1

td

jo,

and < 2p'.

13' <

By Lemma 7.20, for all n sufficiently large there exists )6 that q p's,,(POE,) — x 1 <

1 (mod p') such

(8 very small)

8

and sn(flce,)

0 (mod d)

foréi = 1,

,r.

If E is small enough,

E ajog - p's„(13a,) < p'

for r +

j t,j jo ,

and <

E ajoi q'p's„02,) < 2p'. i=1

Also,

E a 11 sn(fice,)

ai,13ce, (mod qpn)

and satisfies the appropriate inequality, so Woe.) = s r,(13 aj,a,) =

140

7 Iwasawa's Construction of p-adic L-f unctions

Therefore sn(ficxj)

0 (mod d)

for 1

j < t,

and

0< pm

For j

q -1 p - " sr,(ficci)

< pm,

1

jo ,

< q-1 p - " sr,(igajo ) < 2 p'.

Jo we have

0 < sn(ficz j) < qp"' and s(/3ccj)

igaj (mod qp""),

hence sn(/3cx,) =

Similarly,

0 < sn(flajo ) — qpn" < qp" - `", and

sn(flocio ) — qpn"

flock, (mod

qp" --),

SO

sn (igocio ) — qpn" = sn ,(Pajo). Therefore )3 gives us /32 in the statement of the proposition (change n to n + m and change n — ni to n). To get fi a , let c be small enough that 0 < xj < p - m for all j, including Jo, then proceed as above. This completes the proof of Proposition 7.17.

NOTES The construction given here is due to Iwasawa [18], [23]. For another approach, see Coates [7]. The proof that p = 0 had its origins in the work of Gold [1], who considered certain quadratic fields. Later progress appears in Ferrero [2]. The proof given above follows Oesterlé [1] (see also Gillard [5]). For a different, but equivalent, approach see the original paper by Ferrero-Washington [1], which also treats the simpler case of Q(C r) separately. The idea of the proof is summarized in Washington [10]. Iwasawa [24] has constructed examples of noncyclotomic 2r-extensions with p> 0. For the non-p-part of the class number, see Washington [7]. For composites of cyclotomic Z r-extensions (p = p i , , ps), see Friedman [1]. For values of the )-invariant, see Ernvall-Metsdnkyld [1], Ferrero [3], Kida [1], and several of the papers of Gold. For a heuristic estimate of A for Q(C p), see the appendix to Chapter 10 of Lang [5]. For upper bounds for A, see Ferrero [1], Metsânkyld [12], and the end of Ferrero-Washington [1].

141

Exercises

EXERCISES

7.1.

Let Cr; be a ring. Show that f (T) e C[[ll] is a unit -4=> f (0) e

7.2.

Using the fact that every finite abelian extension of U is contained in Q(c:,,) for some n (Kronecker-Weber theorem), show that 1;;/cLP is the only Z- p extension of (113 0,, is defined after Theorem 7 14)

7.3.

(a) Show that Theorems 7.13 and 7.14 can be generalized to any imaginary abelian field all of whose Dirichlet characters are of the first kind. (b) Let K be an arbitrary abelian number field and let K ,o1K be the cyclotomic Z p -extension. Show that there is a field F, all of whose characters are of the first kind, such that for some e > 0 and all n sufficiently large F = K„ (Hint: Let x = 00 run through the characters of K. Let F correspond to the group of O's. Note that O's correspond to It's). (c) Show that for arbitrary imaginary abelian K a modified version of Theorem 7.13 holds, and deduce Theorem 7.14 for K.

7.4.

(a) Let x L 1 be an even character of the first kind. Show that the constant term off (T, x) is —(1 — (b) Suppose p is regular. Show that p(Q)G p .)) for all n (note that for p = 2 we have an empty product in Theorem 7.13, so the result is trivially true!). (c) Let K be an imaginary abelian field. Show that /1. - > # {x Ix is odd and x(P) = (d) The class number of Q( \,/ — 5) is 2, and 3 splits in Q( \f— 5)/Q. Show that although 3 ,k`h(K 0 ), we have 31 h(K„) for all n > 1, where K ./K 0 is the Z 3 -extension of K 0 =

7.5.

Suppose 1 is not a character of the second kind (but also not necessarily of the first kind). Show that for n 1, (11n)B„, is p-integral, and if in n (mod pu) then ( 1 — Zw -m (P)Pm- I)

Rm. '

( 1 — Xal n(P)Pn+ `)

in

.

„ -(mod Pa+ 1 ) n

(this of course contains the Kummer congruences). 7.6.

(a) Suppose x = 1. Show that 1 f(0, 1) = - (mod 4),

hence that g(0, 1) =

—

1 (mod p)

if p

2,

g(0, 1) = 2 (mod 4)

if p = 2.

(b) Show that 1 — (1 + q ) — nq (mod nqp4). (c) (von Staudt-Clausen) Show that if p — 1 In then B r, — (1/p) (mod (d) More generally, show that for n > 1, —(1/p) (mod Z p). 7.7.

Suppose x

Zp).

1 is of the second kind and of conductor qtr. Show that for n > 1, B„.

where C; = x(1 + g).

q p1

1 —

(mod -q 4),

142

7 Iwasawa's Construction of p-adic L-functions

7.8.

Let R' be as defined in this chapter. Show that R' is linearly independent over (p — 1)/2 = q5(p — 1) •4=> p is a Fermat prime.

7.9.

(a) Show that 13 E Z r, is normal <=> the sequence cr 'p - "s„(0) is uniformly distributed mod 1. (b) Let , y, E 1 g,. Figure out a suitable definition of "joint normality" and show that it is equivalent to the sequence of vectors (61 -

nsik ). • • ,

-ns.(70)

being uniformly distributed mod 1. (c) Show that if y 1 , , y,. are linearly dependent over CP then they cannot be jointly normal. 7.10. Let 1( 0 be a function field over a finite field and let Ic/k 0 be the " cyclotomic" 4-extension. Let 1 p be another prime and let r- be the exact power of 1 dividing h(1 ) . Show that en is bounded as n oo. (The analogous result has been proved for cyclotomic 7L-extensions of abelian number fields. The proof involves uniform distribution mod 1, but works directly with the class number formula, rather than with lwasawa's power series. See Washington [7]).

Chapter 8

Cyclotomic Units

The determination of the unit group of an algebraic number field is rather difficult in general. However, for cyclotomic fields, it is possible to give explicitly a group of units, namely the cyclotomic units, which is of finite index in the full unit group. Moreover, this index is closely related to the class number, a fact which allows us to prove Leopoldt's p-adic class number formula. Finally, we study more closely the units of the pth cyclotomic field, and give relations with p-adic L-functions and with Vandiver's conjecture.

§8.1 Cyclotomic Units Let n * 2 mod 4 and let 1/„ be the multiplicative group generated by

{ -1- C„, 1 — i1 < a

n

—

1 }.

Let En be the group of units of CD( ( „) and define C = C„ = 17,, n En . C is called the group of cyclotomic units of cla (C n ). More generally, if K is an abelian number field, we can define the cyclotomic units of K by letting Kg_ Q(C„) with n minimal and defining CK = ER n C„. This works well for For other K, it is perhaps better to take norms, from CD((„) to K, of cyclotomic units. See (Sinnott [2]). Since the real units multiplied by roots of unity are of index 1 or 2 in the full group of units (Theorem 4.12), it will usually be sufficient to work with

143

8 Cyclotomic Units

144 real units. The following observation will be useful: 'a

—

a)/2

Sin(na/n) sin(z/n)

1 ___ 1 — C,

is real, and if a is changed to —a then we obtain the same unit multiplied by —1. Lemma 8.1. Let p be prime and in > 1. (a) The cyclotomic units of O(Cp..) + are generated by —1 and the units

=

r(1 — a)/ 2

1 — 4.„

1
1—

(b) The cyclotomic units of units of Q(Cpm) +

are generated by Cp,„ and the cyclotomic

Let C = Ç. The definition of the cyclotomic units involves 1 — Ca for all a 0 mod pm. If k < in and (b, p) = 1, then, using the relation 1 — XP k = — CY" k X ), we obtain

PROOF.

pk 1 1

_

pk = :1=0

Since (p, b + jp') = 1, we are reduced to considering only those a with (a, p) = 1. Also, 1 — and 1 — Ça differ only by a power of C, 'so we need only consider 1 < a < +pm. Suppose now that F 1<

a

< +pm

(a,p)=1

is a unit of Q(C). Since the ideals (1 — Ca) are all the same,

▪ cd 11(

_r

aya

Ca

= O. Therefore

i _fc

± Ce Ma) a*1 where e = d + ca(a — 1). Note that if p = 2 then (a, p) = 1 requires a to be odd, so Ce is in Q(C) in all cases. This completes the proof of (b). If E OGYE then since each factor in the above product is real, + Ce must be real, hence +1. This completes the proof. LI Remark. If n is not a prime power, not every cyclotomic unit is a product of roots of unity and numbers of the form (1 — C„a)/(1 — L) with (a, n) = 1. Namely, each such product is a real unit times a root of unity, while the cyclotomic unit 1 — Cn is not of this form (see the proof of Corollary 4.13).

145

§8.1 Cyclotomic Units

Our goal is to show that the cyclotomic units are of finite index in the full group of units. It suffices to work in the real subfield. We start with the important, and easier, case of prime powers.

Theorem 8.2. Let p be a prime and m 1. The cyclotomic units C p+m of CD(C pmr are of finite index in the full unit group E.", and hp+m = [E +m p :Cp+ m],

where hp+m is the class number of Q(Cpm) + . PROOF. We shall show that the regulator of the units of Lemma 8.1 is nonzero. Let C = Ç n. As usual, let aa : C Ca be in Gal(Q(C)/Q). The elements
c— 1 /20

(if p = 2, extend a a to Q(C 2 m+ i). Everything below works, since l(C -112 ( 1 — is all that matters, and it is independent of the choice of the extension). We now apply Lemma 5.26. Let f(a) = logl(C -1/2 (1 — Ma l = log I ( 1 — Oak

E G.

Then the regulator is R({ a}) = + det[log I a. r l]a, =

det [f(at) —

= +det[f (T0 -1 ) —

= ± 11 X

*1

X

E

aE

G

x(0.) log I ( 1

-

FI 1<- a< 4pmz(a) log 1 — ca

fl

1 Pm al

( a)

log I 1 —

I.

Iffy = pk with 1 < k m, then, using the relation

n (1 — c7,m) = 1 — 1
we obtain pk

Pm

a= 1

Aco log 1 —

AM logl 1

Ca l = b= 1

TOO

L(1, x) = --c(x)L(1,

j.

146

8 Cyclotomic Units

Therefore,

R(gal) = +

H - 1-rooLo, =

r' 0,

z*

where R + is the regulator of 1:1(Cp.)+ (we have used Corollary 4.6 to handle t(x)). Therefore the + ci's generate a subgroup, namely C+, of finite index in the full group of units, and

_ h+ = R({U) R+ — by Lemma 4.14. This completes the proof. Remark. This result should be regarded as the analogue for h + of Theorem 6.19. A similar question arises: is E + /C + isomorphic to the ideal class group of CP(Cp.) + as modules over 2[Gal(G(Cpm) + /Q)]? Let p 1 mod 4, so CPO)) g CP(Cp) + • Since Q(Cp) + /CP (\ p ) is totally ramified, the norm map on the ideal class groups is surjective (see the appendix on class field theory). Also, tin, norm of E ± is contained in the units of G( p), hence N(E ± )/N(C ±) is either cyclic or (2/22) x (cyclic). Therefore, if E ± /C + is isomorphic to the ideal class group as modules over the Galois group, then the ideal class group of G( ) must be cyclic (since p 1 mod 4, the 2-part is trivial). For p = 62501, Schaffstein found that the class group is 2/32 x /02. Therefore, the isomorphism does not always hold. Whether or not there is an isomorphism as abelian groups appears to be an open question. Finding nontrivial examples appears to be difficult since h: = 1 for small n and for large n the calculations required to determine E ± or the class group are extremely lengthy. The next question is whether or not the p-part of E ±/C+ is isomorphic to the p-Sylow subgroup of the class group of 1:1(4..)+. In this case, it is hard to know what to expect. Vandiver's conjecture predicts that the p-Sylow subgroup of the class group, hence of both groups, is trivial. This is true for p < 125000, but it is not clear that it should be true in general. Recent work of Mazur and Wiles on the "Main Conjecture" has yielded the following: Decompose the p-Sylow (E + /C + )p of E + /C + and the p-Sylow A of the ideal class group of O(;) via the idempotents ei of Chapter 6. Then lEi(E +/C +)pl = 101.

We shall return to O(Cp) later, but now we treat the case of CD(C,) for general n. We do not give a set of independent generators for the full group of cyclotomic units. However, we exhibit a set of independent units that generate a subgroup of finite index, which suffices to show that the cyclotomic units have finite index. One's first guess for a set of independent units would probably be

1 < a < In, (a, n) = 1. 1 — Cri This set has 10(n) — 1 elements and is the obvious generalization of Lemma 8.1. Unfortunately, this set does not always work. We shall show below 511 — a)/ 2

fr:

147

§8.1 Cyclotomic Units

(Corollary 8.8) that there are sometimes multiplicative dependence relations. Therefore, we use a set of units discovered by Ramachandra. Theorem 8.3. Let n

2 mod 4, and let n = fl7 p ei .' be its prime factorization. Let I run through all subsets of {1, , s), except {1, ,$), and let n1 = FL ' p. For 1 < a < t'z, (a, n) = 1, define da

Cn

11

1—

da = y1

- a) n1 .

Then ga) forms a set of multiplicatively independent units for CK„) + . If C„ denotes the group generated by —1 and the a's, and E„ denotes the group of units of C(„)+ , then

11 11 (0(Ki)

[E,;- C;i] =h,

x*1

Pal

+ I — AP,))

o 0,

fx

where hn+ is the class number of (n) characters of (Z1nZ)X .

and

runs through the nontrivial even

Remarks. The difference between the units (1 — Ca)/(1 — ) and the present units is that these new ones contain contributions from the units of proper subfields. We have not obtained generators for the full group of cyclotomic units of OG„)±. Sinnott has calculated the index of the full group of cyclotomic units to be [Er; ;C:] =

where b = 0 if g = 1 and b = 2g-2 + 1 — g if g > 2, and g is the number of distinct prime factors of n. See (Sinnott [1]). 8.3. The proof will be similar in many ways to that of Theorem 8.2, but will be more technical. As in the proof of that theorem, we have PROOF OF THEOREM

R({a}) = ± ,*1 -.,

a=1

x(a)

log I

—

1 where )( runs through the nontrivial even characters mod n. Clearly, this product should reduce to an expression involving ri L(1, )(), but there are a few problems : n might not befx , the restriction (a, n) = 1 may leave out some terms with (a, n) 0 1 but (a, f x )= 1, and 71 is not necessarily . The following lemmas treat these difficulties. (a, n) =

Lemma 8.4. Iff, (n1m) then Z(a) log I 1 — a= 1 (a, t ) = 1

148

8 Cyclotomic Units

PROOF. We claim that there exists b 1 mod (n/m) such that (b, n) = 1 and Cx may be factored through (Z / (n / m)Z)X , 1. If not, then z: (Z/ nZ) X ( b) so f I (n/m), contradiction. Since arn = C cril bm,

z(a) log

= z x(a) logii -

—

=

xor

abnhI

X(a) logl 1 — m

l

so the sum vanishes.

El

Lemma 8.5. Let n = mm' with (m, m') = 1, and suppose fx lm. Then n

ni

E x(a) log I 1 —

I = 0(m)

z(b) logll —

I

b= 1 (b,m)= 1

a =1 (a, n)= 1

PROOF. Write a = b + cm with 1 < b < m, 0 < c < in'. If (a, n) = 1 then (b, m) = 1. Conversely, for each b with (b, m) = 1 there are 0(m) choices of c such that (b + cm, m') = 1, hence (b + cm, n) = 1 (since (m, m') = 1). El Since z(a) and Cri" depend only on b, the lemma follows.

Lemma 8.6. Suppose F, g, t are positive integers with fx l F and g I F. Then Ft

z(a) log 1 —

Z(b) logl 1 — C bF I -

= b= 1 (b, g)= 1

a=1 (a, 9)= 1

PROOF. Write a = b + cF,1 s b < F,0 c < t.Then (a, g) = 1 <=> (b, g)= 1. Since t—1

11(1 —

c=0

c,t'cF) = —

and since z(a) depends only on b, the lemma follows easily.

El

Lemma 8.7. Assume LI m. Then m

mn

Ab) logl 1 —

=

b= 1

[

II (1 — AO] PI

x(b) iogi 1 — b= 1

al.

(b, m)= 1

PROOF. Let p, q, . . . represent the primes dividing m. We only need to consider those which do not divide f. The right-hand side equals

b=1

x(b) log — 1 — E ;0) E b=1

log

—

I

p

ni

+

—

AN log

x(Pq) p, q

b=1

—•••

p#q ni/p

ni

=

b=1

x(b) log II —al—E AP) p

Z(b) logl 1 — b=1

alpI ± • • •

149

§8.1 Cyclotomic Units

(by Lemma 8.6 with g = 1. Since p

we have fi l m/p)

= E AO log 11 E

D(b) log I — I + •

p

b=1

b=1 Pi b

logl 1 —

b=1 (b, m)= 1

(e.g., if pq l b then Ep subtracts the term for b twice but Ep # , adds it back once; this is essentially the relation E';_ o (— l)'() = 0 if n > 0). This completes the proof of Lemma 8.7. We may now finish the proof of Theorem 8.3. If m' = n1 for some / then

n = mm' with (m, m') = 1. If fx I m then

E x (a) log I — aiim" I = 0(nf) E

a=1 (a, n)= 1

x(b) log —

I

b=1 (b, m)= 1

m = O ( iv) 11(1 — AO] E x(b) log I 1 — _pjm

fx

1

= 004 11( 1 — im

=

I

b

AO]

E x(a) logl — C7f.„I

a= 1

y (1 -

0(Pn' ) r(f;) L( 1 1

x(p»

= -0(170,r(z)L(1, 2) no Pim

Therefore (let n = n1 n, so n1 = m' and

= m)

E 45(n r) n (1

E x(a) Elog 1 — arr i = — x(x)L(1,

a= 1 (a, n)-= 1

I

I

Pi WI

fxlni

Consequently R({a})

=

ItooL(1,

(n )

E Ix I WI

= hR

nf x*

E 5(n) fl

I fx i

II

— x(p))

PI

— x(p))),

pint

WI

where hn+ and R n+ are the class number and regulator of O ( n ) + , respectively. Recall n = n7 = p7% We claim that

n

E 0(ni) fl (1 — x(P)) = Pi WI

fxlni

Pt

f

fx

o(pi)

+1-

x(pi)).

150

8 Cyclotomic Units

If the right-hand side is expanded out, we obtain

E 0( E p)11iJ(1 — J

ieJ

where J runs through all subsets of fi I pi Ai fx if pli), then 1 — z(p i) = 1. Therefore we enlarge the set of i J to include all i J with 1 < i < s. If we let 'I, = fli E Ki and 'I', = nio , pP then we obtain }.

E 0(11.1) fl ( 1 — Since J is included in the sum 1 n, 1 J

AP».

fx ) = 1 <=>fx in,, the claim is proved. {1, . . , s} , as required.

Note that fx

Finally, since the real part of 0(Ki) + 1 — z(p i) is positive, the above product is nonzero. Since the index [E: : C] is the ratio of regulators R({.})/R„ + , the proof of Theorem 8.3 is complete. Corollary

8.8.

Let

Cn" be the group generated by —1 and the units of the form

r(1 — 012 1

1 < a < In, (a, n) = 1.

—

1— Then

[E: : C] =

—

z

om,

x*1 Pin

where z runs through the nontrivial even characters mod infinite if the right-hand side is 0.

PROOF. The regulator of

is n

+ —

n, and the index is

x

-2 E z(a) log I 1 —

.

a= 1 (a, n)= 1

By the above calculations (plus Lemmas 8.7 and 8.6), we find that this expression equals

±

kr( z )L(1,

x# 1L

(1. pi n

- x(P))] = h R,fl fl (1 — z(p)). ,

X*1

Pin

This completes the proof.

It is easy to see that there are many examples where C n" is not of maximal rank. For example, if n = 55, then 11 splits in 0 (\/), so x(11) = 1 for the quadratic character of conductor 5. Therefore [E 5 : C'5' 5] is infinite. If n has 4 distinct prime factors then the index is automatically infinite (see Exercises).

151

§8.2 Proof of the p-adic Class Number Formula

In the above, we have used only two basic relations, namely

—

a

=

a( 1

and (n/m) — 1

—

=

fl

j=0

if min.

(1 — a + rnJ)

The following theorem of Bass shows that these generate almost all relations.

Theorem 8.9. Let n # 2 mod 4 and let (A,?) + be the additive abelian group with generators

1

a n

11

a n

}

Z/Z, —

E

and relations

g (—a) n

g ( na)

and a

(n/m)

-

(a ± mi

1

g

g(w1) = 1=0

n

n (and alm 0 0).

en be the group generated by {1 —

Il a < n). contains some nonunits). Then, for some c, there is an exact sequence

Let

(

0 —> (Z/2ZY (A,?) + —> en/< ±> —> 0, where

4

a

—n )

1—

mod< -F

Cn>.

PROOF. The proof uses distributions, hence will be postponed until Chapter 12.

§8.2 Proof of the p-adic Class Number Formula In order to prove the p-adic class number formula (Theorem 5.24), we study the units of an arbitrary totally real abelian number field K of degree r over WO + and let N be the norm from 13(Cn) + to K. Let En+ and EK OE Let K be the respective unit groups, and Cn+ and CK the cyclotomic units (we can take CK = EK r Cn+ or N(C); either definition works here). If E E EK then NE = Ed where d = deg(Q(C„) + 1K). Therefore N(E) contains Eck , hence is of finite index in EK. Since [En+ : Cn+ ] is finite, and N(C) CK, it follows that [EK : CO is finite. But we need to be more explicit. ,

152

8 Cyclotomic Units

Let G = Gal(OGO -70) = {cr a ll a < In, (a, n) = 1 } , and let H = Gal(QQ + /K). Then G/H = Gal(K/0). Let R G be a set of coset representatives for G/H and R' R a set of representatives for G/H — {H}. In Theorem 8.3, a

clna Œa a

,

where a = fl (1 -

Ot

Letting fi = N (a), we have N() = C;',(f31 a / fl) with d'a E Z. Clearly f3/f3 = fiab/fi if o-a o-Ï, E H. So we only need to consider

{N a IJa E R' } . We have R({N a }) = +det(logl NnaeR' ae G/H,G # 1

= + det(log I fl"a I — log I

I)

= ± det(log I Par 1 — log I fia Iita, 1 -reG/H a,T#1

±

FI

(Lemma 5.26)

x(a) log I /3' I

Xe (G ill ) x# 1

E

G/H

E log It I. ± x#FI1 2a Au) reH Extending

x

to G by letting x(H) = 1, we obtain

± H aE Au) log I or' I )(# 1

=±

eG

H

x#1 L

E

x(a) log 'Cal'

a=1 (a, n)= 1

But these factors are exactly the ones that were evaluated in Theorem 8.3. Therefore, as before,

R({I\ T}) = ± H [frooL(1, X

*1

= hKRK

Hf x K') ((

X*1 Pi

so the group generated by {N iK = hK

H

((pt)

+ 1 - x(pid

Pit f x

+ 1 — APO)

CI,

} and — 1 has index

H FI (0(K.) + 1 -

x*1

pi

4. f x

in the full group of units. Observe that the preceding calculation of R({/V a}) would have worked just as well with log o in place of log, except for the use of the relation

153

§8.3 Units of CD(Cp) and Vandiver's Conjecture

Also note that (Theorem 5.18) fx

E z(a)logp(i - vfx ) = -(1

Z(P) )

a= 1

1 T(Z)Lp( 1 ,

so an Euler factor appears in the calculations. Therefore

Ri,({Na}) = ±

Efroo(i - x(P) ) 1 Lp(1,

x*1

((pi)

11

+ 1—

i))]

fx

2 1- VC1K

hK

n (1

x(p))1

x*

I, p (1,

But Rp(fA r cil)

1K

D

(p-adic version of Lemma 4.14),

p

SO

R K,p

+

= n (1

-.1c1K

x

x(p) p

*

-

1

Lp(1, z).

Since RK , p is only determined up to sign, we may choose RK , p so as to eliminate the" + ". This completes the proof of Theorem 5.24. 0

§8.3 Units of

(

ig)

and Vandiver's Conjecture

We now study more closely the units of WO for p an odd prime. Let = 4 and let G = Ga1(CP()/0) CE/p.EY . The characters of G are of the form wi, 0 < i < p 2, where w is the Teichmtiller character. Correspondingly, we have the idempotents —

1 Ei =

1

p-1

E

P — 1 a= 1

covoc a

p

p— 1

1

. 1 0'(a)o-c, e

It is easy to see that p-2

and i=0

Let E be the units of CI( 4). For N > 0, let EN

= ElEPN

0,

154

8 Cyclotomic Units

Usually we shall take N sufficiently large; if we wanted to, we could take the inverse limit for N co, but this is not necessary. Since E --= WE, where W is the group of roots of unity, we have E/EP N Z/pZ x (i/pNiyp — 3)/2 9

as groups.

We wish to study the action of G, and Z p[G], on EN. If ri E EN and a E then /la is defined in the natural way: let a ao mod pN with ao e Z; then ri ao . Consequently g i acts on EN for each i, so p -2

EN =

E pN• i= 0

We now analyze each summand. First, suppose i multiple of the norm, hence

O. Then go is just a

go Ep iv Norm(EN) g_ 1 mod EPN = 1. Next, let i be arbitrary. Let ri e E, so q = Crri o , where r e Z and ijo = Since c a(C) = Cw (a), and since this equation characterizes e l E p lY the subgroup generated by lies in g [ EN. Now consider the real unit /1 0 : p—1

rra-1 (rio

E i(110)P— 1

)w"

a= 1

) mod EPN

If i is odd, w i(a) — (A— a) while cra-1 (/ 0) = cri,l(ri o). The factors for a and — a cancel, so e,(1/ 0) 1 mod EP N. We have proved the following.

Proposition 8.10. EN

=

e

e

p- 3

p- 3

8,E,,v

and

EN =

/(E +

E p-N l i2

i=2

even

even

(
Note that EN is a direct sum of (p — 3)/2 cyclic groups of order pN (by the Dirichlet Unit Theorem) and that there are (p — 3)/2 summands in the above formula. One therefore might expect that each summand is a cyclic group. We shall show that this is the case, using the cyclotomic units.

Proposition 8.11. Let g be a primitive root mod pn. Then 1— 1—

generates C pi., as a module over

p fl

155

§8.3 Units of EI(, p) and Vandiver's Conjecture

PROOF. During this proof, let for some r> 0, so

Let (a, p) = 1. Then a

=

H

r- 1

rgr

1 — Ca

1—

1—

gl•

gr mod p"

2

,

1—

i=o

(c(1 -g)/2 =0

III

The result follows from Lemma 8.1. Remark. The above works for p = 2 if we let g either a gr or —a gr mod 2".

5 mod 8 and note that

Actually, Proposition 8.11 is a more explicit version of Lemma 5.27. Fix the primitive root g mod p and let i be even, 2 < i < p — 3. Let w N (a) w(a) mod p",

0)40

E

Z.

Define p-1

1

W)

corv(a) i a; 1

c( 1 - g)/ 2

5

1—

u=1

SO

1 _ vVp - 1)r, ( C (1- g)/2

1— ( )

mod(E + )P N.

and

Er) e In particular, define Ei

aLf

=

p-1 c(1 g)/2

H

Cg giga- 1

1—

a= 1 p-1(

1—

ca(1 - g)/ 2 1— CaggP

mod(E

a= 1

(change a to Since w N (a)

a' and note

G-a C

Ca).

a mod p it follows that

Er) is a pth power <=> E. is a pth power.

)P

156

8 Cyclotomic Units

This fact will prove useful in the following. For technical reasons it will often be convenient to let N be large. But we still obtain information about the case N 1. Since log o = 0, we have (change a to a-1 ) p- 1

logy Er) =

E coN(a - 1 )1 logp

a= 1

a—

I

El co(a)-` logp ( 1

P

1—

a= 1

1

1

Ca C'ag)(mod Ca

4Np] ; see Exercise 5.15(c))

(since logo la ( o) c

—(o1(g) — 1)T(o..) -1)L p(1, w i) (mod ply). From Proposition 6.13, we know that vo(t(w -i)) = il(p — 1) (ser. also Exercise 8.12). Since a(g) — 1 0 — 1 0 0 mod p, we have proved the following. Proposition 8.12.

If N > 1 + vp(L p(1, coL)), then vp(logp Er)) =

1 + v (L (1, p — "

Proposition 8.13. Let N > 1 and let i be even, 2 < i < p — 3.

Then

ZIP NZ.

E l Ep+N

Since / ( 1, col) 0 0, EN) 0 1, hence EiE p+N 0 0, for large N. Since Ep+,, (Ile2)(P -3)12 , and EiE p+N is a direct summand, E i Ep+, (Ile2)ai, for some ai 1. But ai = (p — 3)/2, so each ai = 1. This proves the proposition for large N. If N is arbitrary (i.e., smaller), we may choose M N large and take a quotient. Then PROOF.

E

E i E p+N

E t(E p+m/(E p+m) PN )

Z/PN Z,

as desired. Recall that h + = [E+ : C + ]. Let Co+N be the group generated by C + mod(E + )P N. If p" > h + then

(E +/C + )p

Ep+N/C '+ o N,

where, for a finite abelian group A, we let (A),, denote its p-Sylow subgroup. From Proposition 8.11, EiCp+N is generated by the unit Er), so P- 3

(E + IC + )p

EiEp+Argas1) ›. i= 2 i even

157

§8.3 Units of EI(, p) and Vandiver's Conjecture

Since Ei Ep+. is a cyclic group of p-power order, the ith summand is nontrivial if and only if Er) is a pth power. Since Er) is a pth power if and only if E. is a pth power, we obtain the following important result. Theorem 8.14. p1h + (0(4)) of a unit of (11(4) + •

<:>

some Ei (i even, 2

i p — 3) is a pth power

Corollary 8.15. If p ,117+(41(4)) then the Ei's generate E+ /(E+)P (this is also obvious from Theorem 8.2). Theorem 8.16. Ei is a pth power

pI131 .

PROOF. We may replace E. with Er ) for N sufficiently large. If Er), then logo Er ) = p logo q. Since log o q e Zp[(p] (cf. Exercise 5.15(c)), 1

+ v (L (1, (CI)) , P P

vo(logo Er )) =

SO

v p(L p(1, col)) > 0.

Since L p(1, co') EE L p(1 — j , co')

—

(mod p)

by Corollary 5.13, we have pi B i . This completes the proof. Remark. The converse is not true. In fact, for p < 125000, p Jr/ h + , so E. is

not a pth power. Corollary 8.17. p I h + (Q( p))

plh - (Q(4)) (this is the same as Theorem

5.33). 111

PROOF. Theorems 8.14, 8.16, and 5.16.

We have mentioned that p)(h'(0(4)) for p < 125000. The way this is verified (on a computer) is via the corollary of the following result. Its advantage is that it uses only rational arithmetic, hence is suitable for computer calculations.

Proposition 8.18. Let i be even, 2 < i < p — 3. Let I be a prime with 1 mod p, say 1 = kp + 1, and let t be an integer satisfying (t, 1) = 1 and tk mod 1. Define d = di = 1P- ` +

+•••+

and (p—

Q.

1)/2 (t kb _

= t_ k/2

b= 1

1 1

158

8 Cyclotomic Units

Let 1 be the prime of Q(Cp) above 1 such that t" following Proposition 2.14). Then

Cp

mod 7 (see the discussion

1 mod / <=> E, is a pth power mod!.

_c

—a l 2NaPPROOF. Let Ri = flI G ./2 ) I mod p. Changing a to ag, we find that

Recall that g is a primitive root

P-

R, =

(c.9/2 —

c - . 9/2)(Q9)- - (pth power),

=

SO p— 1

ag,/2

—agl 2

ra/2

al2

a= 1

'z

n

R'

aP -1- i

(pth power)

= Ei AP for some A E (11(0 Note that the only prime ideal which can divide R i is (1 — ); therefore R i 0 mod 1 will never happen. Since (gi — 1, p) = 1, Ei is a pth power mod 1 if and only if R. a pth power mod 1. The terms for a and p — a in the definition of R. by a pth power (note —1 = ( —1)P), so we may combine terms and find that R. a pth power mod 1 if and only if the same holds for (p— 1)12

FI (c

= c _d / 2

b/2

b=1

(p— 1)/2 n (c

, _ o bp -

b.-- 1

Since tic # 1 mod 1 but t" = 1 mod 1, tk is a pth root of unity mod 1. Proposition 2.14 yields the prime ideal 1 of 41(0 lying above 1 such that tk mod 1. Since (2[1]d)x is cyclic of order 1 — 1 = kp, it follows that 12 =- 1 mod 1, or mod 1, if and only if Qi is a pth power mod 1. Since (p— 1)/ 2 Qi

c—d/2

n (c

b _ 1 ),bp-

mod 1,

19=1

the proof is complete.

Corollary 8.19. Let p be an irregular prime and let , is be the even indices 2 < i < p — 3 such that pIA. Suppose there exists a prime I 1 mod p and an integer t, as in Proposition 8.18, such that OE # 1 mod 1 for all i E , i 5 1. Then p Jr' h + (0(Ç))* PROOF. Theorems 8.14 and 8.16, and Proposition 8.18.

Remark. Of course, we could have used a different prime 1 for each index i, but the above form is what will be needed in Chapter 9 when we treat Fermat's Last Theorem. The converse of Corollary 8.19 is also true. Suppose . Then none of the Ei's are pth powers. The density of the prime ideals 1 such that a given E. is a pth power mod 1 is 1/p by the Tchebotarev Density Theorem. Since there are less than p units Ei , there must be infinitely many 1

159

§8.3 Units of CP((;) and Vandiver's Conjecture

such that none of the E's are pth powers mod I. As usual, only primes I with residue class degree 1 over CI need to be considered, but these are precisely the primes that lie over rational primes 1 L mod p. Proposition 8.18 now applies and we have Q:` 1 mod I for all i, for an appropriate choice of t. Remark. Vandiver's conjecture states that p,1/ h +(()(C p)) for all p (Serge Lang has pointed out that the conjecture actually originated with Kummer: in a letter to Kronecker, Kummer refers to p h + as a "noch zu beweisenden Satz" (see Kummer's Collected Works, vol. I, p. 85)). The conjecture has been verified for all p < 125000. What are the chances that it is true in general? First, consider a probability argument similar to that used for h - in Section 5.3. Suppose each Ei is a pth power with probability 1/p. There are (p - 3)/2 indices i, so the probability that p h + would be t (P - 3)/ 2 (i - P -) e - f 2 = 0.6065.... This does not agree at all with the numerical evidence, Perhaps, then, is there something yet undiscovered which causes the E's to be non-pth powers? If so, could this force be strong enough to make Vandiver's conjecture true for all p? This question remains open. Another probability argument that could be used is a refinement of the above. Since the only E d's which can possibly be pth powers occur when p 13i , we consider only such indices i and suppose that the probability of being a pth power is again 1/p. The index of irregularity should take on the value k with probability e - 1/272kk (see the discussion following Theorem 5.17), so the number of exceptions to Vandiver's conjecture for p < x should be approximately

E E (Prob. i(p) = k)(Prob. some Ei is a pth power) p<x k=0

i e -112\

= E k=0 E ( -29„

1\k\

- (1 -)

(1 - e - 1/2P ) Px

p

1 E2p -

log log x.

Since log log(125000) = 1.23 ... , it is not surprising that no exceptions have been found. Moreover, most of the contributions to this sum come from the first few primes. If we started the sum at the first irregular prime p = 37, we would obtain a much smaller number. Finally, we could use a more naive approach. Suppose pill+ with probability 1/p. Then the number of exceptions to Vandiver's conjecture for p x should be

E-P

log log x.

160

8 Cyclotomic Units

The comments for the previous approach apply here also. However, another point arises. For h - we used Bernoulli numbers which were much larger than p. Hence it was reasonable to expect that they were random mod p. But it is possible that h+ is often small. In fact, very little is known about h + . For small values of p, h+ = 1. For some p we know h+ > 1. For example 3117 57 since h(11( \/257)) = 3. But, at present, for each p with h+ > 1 we do not know the exact value. The computer calculations would be too lengthy. In any case, assuming h+ is random mod p is rather dangerous. Whether or not Vandiver's conjecture is always true, the above arguments indicate that it should hold for most primes. In Chapter 10, several important consequences of the conjecture will be given.

§8.4 p-adic Expansions We now examine the p-adic expansions of units. Our main goal is to prove Theorem 8.22. Let = and let 7r = — 1

and

A = (C — 1)( -1 — 1) = 2 — (C + V-1 ),

so (7r) and (A) are the prime ideals lying above p in Z[] and ZR + respectively. Note that any element of Z[C + ( - 1 Z [A] is congruent to a rational integer mod A. Since (A) (P - 1)12 = (p), we have = gp + pp + Let i

with a, fl E.

+ 1 be a real unit. We may write = a + bAC + clAc+ + • • •

where p ab and c 0 mod(p — 1)/2 (if plb, add (p — 1)/2 to c. If c 0, use the formula for P -1)/ 2 above, and then modify a. If pla, then ri is not a unit). Note that c is the largest integer n such that ti is congruent to a rational integer mod A. Hence c is uniquely determined by tI. The integers a and b are unique mod A' +1 and mod p, respectively. With the above notation, we may use Exercise 5.14 to conclude that v p(logp ri) = p

2c

1'

so this gives us a method for determining c. For example, if N is large enough, Proposition 8.12 imples that Er) ai + bi Aci mod Aci with p—1 2 + 2 v P(L P(1 ' wi)). i

161

§8.4 p-adic Expansions

i p — 3. If N > 2c/(p — 1) and

Proposition 8.20. Let i be even, 2

=a+

+ • c ei Ep+N

then i c — mod

2

PROOF. Let (a, p) = 1. Since ir follows that

=

I)

2 — 1 = (Tr + 1)2 — 1 = air + • • • , it

OE 2c2c mod A C+ 1 .

„(A) -== a2A mod /12 and 0-(2,) Therefore

a + ha' A' mod 2' 1 .

0- 201)

From Exercise 5.14, log,

— A' mod A

1,

log p

ba2c A' mod 2` +1 . a

logp if

w i(a)logp q mod pN .

Since pi e ei EN,

Since N > 2c/(p — 1), we obtain . b ol (a) — A' a

ba2 ` A' mod Ac+ 1 , a

hence Oa)

Therefore i

a 2` mod p, for all (a, p) = 1.

LII

2c mod(p — 1), as desired.

Lemma 8.21. Let i be even, 2 < i < p — 3, and let ij be a generator for gi Ep+N. If N > 1 + vp(L,(1, w i)) then a; + b:/1» mod .1» + 1 (p )( a;b;, c; # 0 mod P

2

1

with

c'i

i p—1 2 + 2 vp(L,(1, W i)).

PROOF. We have Er) = iVi7P N for some di e

7 e E + . Therefore

log, Er) di log, q mod pN .

)

162

8 Cyclotomic Units

By Proposition 8.12, vp(logp EIN)) =

p—1

+ v (L (1, w i)) < N. PP

Therefore vp(logp Er)) = vp(d i logo 'Ili)

vpoog p

SO —

c; =

2

i

vp(logp it)

2+

1

p— 1 2 vp(Lp(1, (pi)), as desired.

Remark. Since v(d 1) = vp(logp Er) — v p(logp VI) = have from the discussion preceding Theorem 8.14, 3

2

vp(h + (0(4)))

— 1)](c 1 — c;), we

E (ci —

P — 1 i=2

c).

i even

We may now generalize Theorem 5.36 (see also Exercise 5.7).

Theorem 8.22. Let M = max i v p(L p(1, w i)), where i is even, 2 i p — 3. If II is a unit of which is congruent to a rational integer mod pl1+1 then is a pth power of a unit. []

PROOF.

As in the proof of Theorem 5.35, we may assume j is real. Write = a + bAC + • • • .

Then, by hypothesis, C>

p—1 (M + 1), 2

Let N M + 1 and let i-1 2 ,

2c so vp(logp ri) = — > M + 1. p 1 ,

3

be as in Lemma 8.21. We may write

= 'Y PN FI jg with gi e Z, y e Et We shall show that pig i for all i. Since 2c = vp(logp p—1

p

1

mod 1

by Proposition 8.20, it follows that the numbers vp(gi logp it) are distinct mod 1, hence distinct. Therefore v p(E g i logp

= min v p(g i log,,

Also, vp(E gi log,, it) = vp(logp — pN log,, y) vp(p" log,, y)) min(vp(logp > min(M + 1, N) = M + 1.

163

§8.4 p-adic Expansions

Therefore, for each i, the above, plus Lemma 8.21, yields M 1 _< vp(g i logp iii)

v(g) + p _ 1 + vp(Lp (1, (d)) < vp(gt) + 1 + M.

Consequently v(g) > 0 for each i and proof.

is a pth power. This completes the

Corollary 8.23. Suppose p 3 /1/ B pi for all even i, 2 < i < p — 3. If i is a unit [] which is congruent to a rational integermod p 2 then 1 is a pth power.

of

PROOF.

By Theorem 5.12, = ao + ai (s — 1) + • • •

L p(s,

with ai E Zp for all i, and p Iai for i > 1. Therefore —

B pi

.

B.

1 ) PI = Lp(1 — pi, coi)

— (1 —

pi

ao = L p(1,

oi) mod p 2 .

If p 3 Bpi then v p(Lp(1, w i)) < 1, so M 1. The result now follows from the theorem. We have been considering local properties of global units. As a final result, we consider the global units as a subgroup of the local units. We continue to assume p is an odd prime. Let U 1 = {x EZ p[2] Ix 1 mod 2}. Then U 1 is a Z p[Galel:KX/CP)]-module, so p- 3

U1

= fl ei U t . i=0

even

Let p- 3

(Y1 =

H e,U 1 . =2

I even

Since the norm to (U p is (p — = {x E Ul I Norm(x) = 1}. Lemma 8.24. Let i be even, 2 < i < p — 3. Then &U 1 is a cyclic Z p-module with

= (1 + /1") (P - n'i as a generator.

164

8 Cyclotomic Units

PROOF. As in the proof of Proposition 8.20, ovA i/2 )

ociAi/2 mod ) i/2-, 1

Therefore (1 ± Ai/2)(p— 1)Ei „____ 1

± .PE— I ar iorocx i)2 (

± ...

a= 1

1 — A" mod A i/2+1 .

It is easy to see that any set of elements whose 2-adic expansions start with 1

plus the element = 1 + p, may be used as a set of Zy -generators of U l . Since E e i L/ 1 for all i, including 0, we must have e i U i generated by as desired. (if 17 E C + n U 1 then co(q) is a power Let CP = C + n U t = c+ n of Norm(q), hence equals 1. Therefore C + n U l = C n U'i ). If n E C + then re — 1 E Cj. If N is chosen large enough that C+ n (E + )PN (C + )", then E 3 generate C +AC + )P, hence (EP 3)P — 1 generate CP AC P )P.

(EP)P-

The standard recursive procedure shows that Ct is contained in the Zy of U l generated by these elements; therefore the closure et-submodule of ct in U l is exactly the 7Ly -submodule generated by the (V ))P - i = 2, 4, ... , p — 3. Theorem 8.25. Let i be even, 2 < i < p — 3. Then [eityi

81

=

PROOF. Let d be the index, which must be a power of p since we are working with 7Zy -modules. Let be as in Lemma 8.24 and let ( Er) "- I be as above. Then = (Er p— 1)u, )

where u is a p-adic unit. Consequently (with N sufficiently large), v y(d logo

But logy = log y(1 —

= vy(logy Er)) — 2

p— 1

± • • -) = 2 i/2

vp(logp

=p

+ vy(L y(1, ol)).

mod ,%i/2+ (cf. Lemma 5.5), so 1.

Therefore v(d) = v p(L p(1, w i)). The proof is complete.

165

Exercises

Corollary 8.26. Res, i

p(s) = (1 - 1/p)[U; : C if ] • u, where u is a

p adic unit. -

PROOF.

We know from Chapter 5 that the residue is

(1 - -1 )

n Lp(1, (0`).

The result now follows easily from the theorem.

El

The above results will be generalized in Chapter 13. NOTES Theorem 8.2 is due to Kummer [4]. Many of the results in this chapter had their origins in his work, and also in that of Vandiver. The index of the cyclotomic units in the general case of (LPGri) has been determined by Sinnott [1], [2], [3]. Other results have been obtained by Leopoldt [2] and C.-G. Schmidt [2]. For the case of function fields, see Galovich-Rosen [1]. The cyclotomic units can be used to obtain information about class numbers, especially their parity. See D. Davis [1], Garbanati [1], Schertz [1 ], and several papers of G. Gras and M.-N. Gras. For elliptic analogues of cyclotomic units, see the papers of Robert, Gillard, and Kersey. For a nonnumber-theoretic application, see Plymen [1]. Vandiver states that he conjectured p)( hp+ in Vandiver [1]. Kummer tried but was unable to prove the conjecture (Letter to Kronecker, April 24, 1853; Collected Papers, vol. I, 123-124). The last section is from Washington [8], which is based on ideas of Dénes [2 ], [3 ]. EXERCISES 8.1. Suppose p h + (G( p)) but plh - (C(c:p)) (such p are called "properly irregular"). Show that there exists a unit in CP(;), in fact one of the Ei 's, which is congruent to a rational integer mod p but which is not a pth power. 8.2. Let a be as in Lemma 8.1. Show that (1 — 0(1 — (;') (1 — C)(1 — 8.3.

Let p be odd and let N be the norm from CPGpY to CP. (a) Show that a = '1 ' 2 (4 — 1)/ ( p — 1) a mod( p — 1). (b) Show that N( a ) a( P - 1 " mod p. (c) Show that if the Legendre symbol (a/p) = — 1 then N(„) = — 1. (d) Let p 1 mod 4 and let g be the fundamental unit of 1:1(ip). Show that Norm(g) = —1, where the norm is from CP(N/P) to G.

8 Cyclotomic Units

166

8.4.

Let N be the • norm from a(..) to 0((,,). Show (a) N((;.. L — 1) = — 1, (a, p) = 1; (b) N(,L) = (c) N: C Cp. and N --+ Cp+. are surjective.

8.5.

Show that [E:

8.6.

Let On be as in Theorem 8.9. Show + e C.

8.7.

Corollary 8.8 implies that there is a relation in

8.8.

Let n # 2 mod 4 have at least 4 distinct prime factors, say p < q < r < s. Show that at least one of the quadratic characters x of (Z qrs1) is even and satisfies x(p) = 1. Conclude that [E,' q] is infinite in the notation of Corollary 8.8. Show, however, that if n = 3 • 7 .11 then the index is finite.

8.9.

Suppose p 01 + (0(C0). Let tension

= [E + : C ] for (13(Cn).

C39.

Find one.

, is be the irregular indices. Show that the ex-

cl(Cp, 41P,

Ei1.1910(Cp)

is unramified (see Exercise 9.3) and has Galois group isomorphic to (Z/pZ) .s. Under the assumption p' , s is the rank of the ideal class group (Corollary 10.14), so this shows how to generate the "p-elementary" Hilbert class field of (13(Cp). In fact, for p < 125000 the ideal class group is (Z I pZ)s , so we get the entire Hilbert class field. 8.10. Suppose we have units ri2, r14, • • • qp- 3 of (L:P(Ç) f such that

ai + bi 2." mod /lc.'" I

p

and suppose the c i are distinct mod(p — 1)/2 (for example, c ; g 2 , ...,gp _ 3 be integers. Show that P-

1 /2). Let

3

fl

a + N.' mod /1' 1

I=2

with p,ab and with c = min i (ci + [(p — 1)/2]v e(gi)). 8.11. For i = 2, 4, . , p — 3, let Ei = bi A'' (mod ) C+ 1), with p 'ai b i and c i 0 0 (mod p — 1). Suppose p,}/.13. Show that c / = 112. 8.12. (a) Show that 0 < vp(r(co -f)) < 1 (i 0 (mod p — 1)). (b) Use Proposition 8.20 plus the proof of Proposition 8.12 (without Proposition 6.13) to show that vp(t(u) -1)) il(p — 1)(mod 1). (c) Conclude that vp(T(w - r)) il(p — 1).

Chapter 9

The Second Case of Fermat's Last Theorem

In Chapters 1 and 6 we treated the first case of Fermat's Last Theorem, showing that there are no solutions provided certain conditions are satisfied by the class number. We now study the second case, namely 0. xP + yP = zP, p,i/ xy, plz, z Again, the class number plays a role, but the units are also very important, which makes things much more difficult than in the first case. In fact, the second case is only known to hold for p < 125000, while the first case is known at present for p < 6 x 10 9 . In the following, we first give the basic argument which underlies all of the theorems we shall prove. Then we show how various assumptions make the argument work. A basic component of all the proofs is Vandiver's conjecture that pVt + (CD(Cp)). In fact, if a prime is ever found for which Vandiver's conjecture fails, it is not clear that we could prove the second case of Fermat's Last Theorem for that prime. However, the first case is probably safe, since Theorem 6.23 and also some other independent criteria all have a very low chance of failing, either simultaneously or even individually.

§9.1 The Basic Argument Consider the equation WP

+

OP

. rvIm 'P

where

P 3; A = (1 — C)( 1 —

-1 ),

= 4);

A, co, 0, e Z [A] are pairwise relatively prime ; ii is a (real) unit of 1[2]; and ni > p(p — 1)/2. 167

168

9 The Second Case of Fermat's Last Theorem

We shall show that under certain conditions this equation has no solutions. If this is the case then xP + yP = zP,

134 xy, plz, z 0 O.

has no solutions. Otherwise we could assume (x, y, z) = 1 and let w = x, 0 = y, and = z/pa, where a = vp(z). Then ni = pa(p — 1)/2 p(p — 1)/2, and pi = paP/Am is a unit. Suppose we have a solution. Then p—1

11= 0 (a) ± Ca

a

Suppose

A is a prime ideal of Z[C] such that Aiw + ç 6 and Alw + Cb 0, where a

b mod p.

Then = (uni0(1 —

b )(9

and

Ai cb — a( 0)

ca

w

b 0)

= (unit)(1 —

If A

(1 —) then Ale and fi 1w, contradiction. Therefore A = (1 — 4 Since A and 0 are relatively prime, A4/0; hence A2 4/(a) + Ca0) — (0) ± (60). Therefore A 2 = (.1) divides at most one of the factors (co + CO). Since w + 0 co" + OP 0 mod A., and since similarly w + Ca0 0 mod(1 — we may write (CO

l

P—1 0) n

w Ca a =l 1 — C a

= (unit)e

(P

1)/2p

where the factors on the left are pairwise relatively prime algebraic integers and

cao

(1 —

1

a < p — 1.

It follows that there are ideals Ba , 0 < a < p — 1, in Z[C] such that a 0) 1 < a < p — 1, (w

and

(co

= (A)tm - (P - 1)1213g.

+

Note that the ideals Ba are pairwise relatively prime. For later reference, we also observe that

= Bo B i • • B

and

(1 — C),I/Ba for 0 < a < p — 1.

169

§9.1 The Basic Argument

It is easy to see that B p _ a is the complex conjugate of B a . We shall write B _ a instead of Bp— a' We now need our first assumption. Assumption I. p,l'h±(0(( p)) (Vandiver's conjecture). Assuming this, we claim that B o is principal in Z[A]. Note that B o = Bo and (1 — so Bo arises from .7[4 Since Bg is principal in ZP1, because co + 0 and A are real, Assumption I implies that Bo = (p o),

with po real,

as claimed. Consequently, (t) + 0 = 110e -(P-1)/2 Pg, where ri o is a unit which must be real since everything else is real. Now let a 0 mod p and let

ico + (aO\ (co + (-1-1 = a =

y— a

w CO

_ ___ y - a

COG

— 61) ±

(

CO ± O) a

+ (a) +

± i0

± — a0

= 1 mod(1 — () 2 m - P,

since co + 0 -.= 0 mod )L m - ( P -1)/ 2 therefore mod(1 — AP — 1 ) — P = P(P — 2) p, we have

D2 m - P + 1 .

Since 2m — 1)

a EE 1 mod(1 — ()P. Lemma 9.1. If cx

E

Z[ p] satisfies a _= 1 mod(1 — ()P then U(Cp, a l/P)/Q(Cp)

is unramified at (1

— 0.

PROOF. Let

f (X) =

((1 — OX + 1)P — a ( 1 — CY

Clearly f is monic, and since p divides the binomial coefficients ( .'1), 1 < j < p — 1, it follows that f (X) has coefficients in Z[0. A root fi off generates the same extension as 1 11P .The different of this extension divides

f '(fi) _

—

(1—

P

OP-

1 (( 1 — ) L3 + 1 ) P-1

P 1 mod(1 — (). OP- (1—

Since p/(1 — DP -1 is a unit, the different is relatively prime to (1 — (), so O (1 — () is unramified. This completes the proof of the lemma.

170

9 The Second Case of Fermat's Last Theorem

Since a = (Ba/B_OP, the extension in Lemma 9.1 is also unramified at all other primes (Exercise 9.1).

Lemma 9.2. Assume p,l'h ± (0(4)). Suppose a G Q(Ç) satisfies (Yc = a -1 and suppose the extension Q(Cp , a l iP)/Q(C p) is unramified. Then a is a pth power in PROOF.

Assume the extension is nontrivial, hence of degree p. Let 0. :allp,,y,„1/p

generate the Galois group and let J denote complex conjugation, extended so that J(Œ) = (Ja)1 /P. Since Ja = a -1 , jo.(ot i/p)

....= j((oti/p) ..,

1

coe ll p '

and 0. j ( c i/p) = 0.(cx - 1/p) = c- l a -

11p .

Therefore Jo- = o-J, so o-J has order 2p and fixes Q(Cp ) . Since 1/P) : G(C p) ± ] = 2 P, [0(4, o-J generates the Galois group. Let K be the fixed field of J, so K/Q(C p ) s abelian of degree p. If a prime ideal A of Q(Cp)+ were to ramify in this extension, it would have ramification index p > 2, so the ramification could not be absorbed by Q(Cp)/Q(Cp)t Hence Q(Cp , a l iP)/Q(C p) would be ramified, contrary to hypothesis. Therefore K/Q(C p)± is unramified and abelian of degree p, so pIh + (Q(C p)), which is a contradiction. This proves the lemma. El

Remark. Note that the fact that ci = a -1 , hence a is in the " —" component, caused a l /P to yield an extension of the real subfield, which is the " + " component of 0(4). This phenomenon will occur again in Chapter 10. By the lemma, we have a=

(0

p

± Cao\( 0 ) ± C-6161- 1

= al

for some a l c Q( ) . But, as ideals,

(

a) ± Ca 0 (0 ± C la 0) _ i_

By the same reasoning as was used above for Bo , we have co + Ca@ co + C - at9

i— Ca

1 — C-a

— 111(r)P,

171

§9.1 The Basic Argument

where is a real unit and p' is real. Therefore ic°

+ caey

—

= 11(Œf3 ).

Raising both sides of the (p + 1)/2th power, we obtain w +

0

1—

—1 aPP,

where ria is a real unit (so ria = a) and pa e ZR]. Changing a to —a, we we find that (15a)" = pP_ a , so we may change p _ a by a power of and assume

= PWe have two equations:

+

19 = ( 1 —

+ C-a9 = ( 1 — - `1)11,d5Pa. Multiplying, we obtain (0 2

+ 02 + (ca C –

ARP a

where ÇI)(1 —

a)

Ca —

a.

Also, from a previous formula for or + 0, we find (0 2 + 0 2 +

2o0 =

2 A2m–p+ 1 2p

P0•

Subtract and divide by Aa :

= 11.2 (PaPar

114 )-2m-P+1 PZPA.-1 . Now let b # 0 mod p be another index and assume a # ± b mod p. This is possible if p > 3. For the case p = 3, see the Exercises. We have n 4A2m-p+ 1 2p — wO = Td(PbPb)P —(00

Po Ab 1 .

Subtract and rearrange: rd(Pa

— rd(Pb b) p = rlô A2m- p+ P o2P (Aa– 1 — Ab-1 ).

An easy calculation shows that (c–b

Aa 1

c -a)(ca+b _ 1\

6, A

where 5' is a unit. In fact, 0' is real since

/la , and Ab are real. Therefore

2 /la)

(Pa Par +

Pb POP = 6/1- 2"1-17(P) P,

where 5 is a real unit. We now need the following.

9

172

The Second Case of Fermat's Last Theorem

Assumption II. rialrh, is a pth power of a unit of C1 ( p ) + .

Assuming this, we let , 2Ip 'l—a W 1 = ( ) Pafia, qb

el

and

= — Pbi5b3

i = Pô. Then col + OÇ = 6A 2 m - PÇ. Note that 6 is a real unit and 2m

—

AP

I)

—

1

)

—

1) =(I)

—

2

)P > P P

—

2 1.

Since the numbers co

+ Ca()

l — ca = rlaPfi,

1 < a < p — 1,

and (with /14 p o)

o.) + 0 = n o Am -(P - 1 " pg

are pairwise relatively prime, it follows that 0 1 , B i , i , A are pairwise relatively prime. We are now in the situation in which we started. Suppose now that had the smallest possible number of distinct prime ideal factors (not counted with multiplicity). We know from above that () = Bo B i . . . B p _ i and that these factors are relatively prime. But ( I) = (Pô) = B. Therefore B 1 = . • • = Bp_ 1

= (1),

SO

W + CV . is a unit, 1

—

1 < a < p — 1.

Ca

Let a = + 1. We find that a

=

(

a)

+ c o)(c o + cc -1- 1 -1

1— c

1

i

—

173

§9.2 The Theorems

is a unit with a5 = 1. By Lemma 1.6, a is a root of unity: a = + t;` for some c. But a -= 1 mod (1 — 'OP, hence mod(1 — ) 2 , by a previous calculation (formula preceding Lemma 9.1), so a = 1. Consequently U) +

w + C -1 0 1— 1•

1— A short calculation yields

(0 + oi) =

'(0 + w).

Since 0 + w 0 0 (otherwise = 0; this is where the trivial solutions are excluded), we have C2 = 1, which is false. This contradiction completes the argument.

§9.2

The Theorems

There are various methods of satisfying Assumptions I and II. We give three ways in this section.

Theorem 9.3. If p is regular then the second case of Fermat's Last Theorem has no solutions. If p is regular then p h + (Q(Cp)), so Assumption I is satisfied. From formulas in the previous section,

PROOF.

w + Ca0 _ r= 1 ca Pa P ( 0) _j_ -1- L'

(1) ± 0) 1—

Ca

WP a-P - mod(1 —

_

P. P

C)2172 P

.

Since a similar equation holds with b, we obtain

na = lb

(12-b-) P mod(1 — C) 2 m - P. P.

But 2m — p > p — 1 and (Pb/Pa)P is congruent to a rational integer mod p by Lemma 1.8. Therefore

— 11a

rational integer (mod p).

By Theorem 5.36, riaM b is a pth power. This proves Assumption II and completes the proof of Theorem 9.3.

9 The Second Case of Fermat's Last Theorem

174

Theorem 9.4. Suppose p 3 B pi for all even i, 2 i < p — 3, and assume p (CP( p)). Then the second case of Fermat's Last Theorem has no solutions. Remark. Since B 1/pi B i/i mod p, we have p 3 I B pi only if pl B i (but not conversely).

PROOF. Assumption I holds by hypothesis, so it remains to check Assumption II. Since p = 3 is covered by Theorem 9.3 (or the Exercises), we assume p > 3. We know that W Pa

1 — Ca

_1 — n—a

r-(i) ±

— 'la

C —a 0

1 — ça

O = no

1)/2

(P

=

n

1

—

and

Ca

p

Po•

Therefore P P—a —

P

,

cano e— (p— 1 )/ 2 ,P r--09 1 1 ± Ca

hence P- 1

II (Pa —

— a) =

i=o

(:)2m— Pg

where ij is a (nonreal) unit. Since pa and p _ a are relatively prime, it follows as at the beginning of the previous section that the numbers Pa — iP—a

(1

i

p — 1), and

Pa — P—a

1—

are relatively prime algebraic integers. Since 2m p > (p — 2)p > p (since p > 3), at least one (hence exactly one) of these numbers is divisible by 1 — C. It follows that (1 _

divides pa —

p _ a for some i,

0 < i < p — 1.

Consequently, —i12 pa

19—,

mod(1 —

)

2m-2p+ 1 .

In the previous section, pa and p _ a were determined up to roots of unity, subject to the restriction that fia = a . Therefore, we may replace pa by C - "pa and assume that Pa p _ a mod(1 — 02,2p+ 1.

175

§9.2 The Theorems

As before, there exist ideals Ci , 0 < i < p

CiP -a) = q

(Pa 1

-- Ci

P -a)

(Pa

—

1, of[] such that

i < < p— 1, and

,

( 1 — -)21n "- 2P ÷ 1 Cg •

Since (p a Cp _ a)1(1 — C) is real, it follows as in the previous section, since that C 1 is principal: Pir h + , —

Pa — 1—

—

a

p = 71a Pa,

where Pia is a real unit and pa e CP(C) ± . From the above congruence (pa p _ a), 0 2.- 2p, Pa aR mod(1 — so W

cal!)

1 — Ca = 1a 14 ilaPPa 2 mod( 1

02m-2p. —

A similar formula holds with b in place of a. Therefore

Ca°9 1

_

?la 'lb

(Pb) P2

Ca a) + b t9

1 ) P

— )2m - 2p in°d( 1

2

mod(1 — c) 2m - 2p,

Pb

(—

Pa

since

1 1

w + cic) 1 cb (co i ya 9 + (D)( ca co i + cb 0 = c a CO ± ç 1 — C b i—

-r-

—

1 mod(1 — C )2 m - P. Since 2m — 2p > p(p 1) 2p = p(p 3) > 2(p a congruence mod p2 . From Lemma 1.8, —

—

("Lill P

—

—

1), the above becomes

rational integer (mod p),

/la

from which it follows that 11

P

ri--' ) (

Pa

2 7=7 rational integer (mod p2 ).

Therefore a — is a pth power, .

rib 174

by Corollary 8.23, hence riaM b is a pth power. This verifies Assumption II and completes the proof of Theorem 9.4. 1=1

176

9 The Second Case of Fermat's Last Theorem

The disadvantage of Theorem 9.4 is that it requires us to show that p ,1' 17 + ( U(p)) The best way to do this is via Corollary 8.19. However, if the hypotheses of Corollary 8.17 are satisfied for a sufficiently small 1, we are very fortunate, since not only does but also the second case of Fermat's Last Theorem has no solutions, as the following result shows. ■

Theorem 9.5. Let the notation be as in Proposition 8.18 and Corollary 8.19. If there exists a prime 1 1 mod p with 1 < p2 — p such that 1 mod 1 for all i E {i 1 ,

,

then the second case of Fermat's Last Theorem has no solutions. PROOF. By Corollary 8.19, p /I' h+ (CP( p)), so Assumption I is satisfied. Suppose that XP +

VP

= ZP,

p)(xy,plz,z 0 0,

where x, y, z E Z are relatively prime. Let 1 be as in the statement of the theorem. Lemma 9.6. 1,1' xy. PROOF. Suppose / I y, hence

xz. Since P -1 a =o

= X P,

(Y

the standard argument shows that the numbers 0

a

p — 1,

are relatively prime in Z[(p], so there exist ideals A a , 0 < a < p that i:az) = A.

(y

Let a

1, such

0 mod p, and let

= (y =1

ci z)(Y

raz) - 1

—

y — C - az

1 mod(1

C)P,

since pl z.

Since (a) is the pth power of an ideal, a(a l /P, Ç)/(Ç) is unramified except possibly at (1 — C). Lemma 9.1 implies that this extension is also unramified at (1 —). Since p,1' h+ , a is a pth power by Lemma 9.2. As in the previous section, the ideal (y — 'C"z)(y —

=

177

§9.2 The Theorems

is the pth power of a principal ideal in (11(Cp) ÷ (since p,{/h), and by the argument used there, Y

—

Caz = Vaea,

where 'l'a is a real unit, aa E Z[Cp]. Since we are assuming li y,

Caz

y a o-Pa mod /.

Taking complex conjugates and noting that ya = _ a , we find that _ -a

ya o-P_ a mod 1.

Since 1,1/ z, we may divide and obtain C 2a =

cra ) Pmod 1. -

a

Let -/ be a prime of CC p) lying above 1. Since 2a 0 0 mod p, the equation p= — ci) implies that 2a # 1 mod 7. Therefore a -a/a- , has order 1) 2 mod 1. Since 1 1 mod p, 1[C]mod 1 has 1 elements; hence p 2 Il 1. Since 1 < p2 p, this is impossible, so 1,1/ y. Similarly, /)(x. This proves Lemma 9.6. —

—

Lemma 9.7. Il z (this is where 1 < p2

—

p is used most strongly).

Write 1 = 1 + kp with k

PROOF.

(Y

—

Caz)aa-P = Ya

Y

—a

—

= (Y

1. By the equations obtained in

a Z)(71g)

1

a

p

—

1.

Let 7 be a prime of 1[C p] above 1. Since Z[C]mod -/ has 1 elements, cr.kp

= -1 = 1 mod

(/,ex, so 1,1/o-a ). Therefore

(y

Multiply each side by ay k

k zy k— 1 +

—

Caz)k (y

—

C - az) k

mod I.

V a and expand to obtain ca(k— 1) z k

c— ay k

2a ,k— 1 +

—a(k+ 1)?.

Since k < p 1, only the term — kzyk -1 contains a trivial power of (i.e., C0) • Note that the above congruence also holds for a = O. Therefore we may sum for 0 a p 1. The powers of sum to 0, so we obtain —

—

—

pkzy'

0 mod I.

But 1 = 1 + kp, so LI' pk. Lemma 9.6 implies that 1,1/y. Therefore il z, hence liz. This completes the proof of Lemma 9.7. 0

1 78

9 The Second Case of Fermat's Last Theorem

Now we may work with the "basic argument" of the previous section. From the above, we find that we may start with the equation wP + OP yfrr%P with the added condition that /I Assuming that we can show that rialq b is a pth power, we obtain g = 2rn- pp (D where = . We want to show that / I p o , hence / Then we may assume that has the minimum number of distinct prime factors subject to the condition that /1 The last part of the "basic argument" then yields the result. Recall that + 0 = r1 o r -(P -1 "pg. If we can prove that / I (co + (9), then every prime divisor of / divides po . Since / is unramified in L(4), Po. Lemma 9.8. li co + O. PROOF.

Let '1 be a prime divisor of / in 11(4)k Since / I p-1 (0) fl =o

1 0)

0 mod

I.

Therefore co + 0 mod 1 for some j. Suppose j (co + CaO)/(1 — Ca) were pairwise relatively prime, for a

1,rco + CaO

O. Since the numbers

hence ?A'p a for a

j,

j.

Since ria was real, so ria = W =

Recall that 1 = 1 + kp, hence pk„P ±jmod p, (w (w + calk

(--ao 1 _ c- •

1 mod I for a

j. Therefore, if

a

ca

—

a

—

(

VW _ 1°

) k

mod l.

Since k is even, we obtain (co + c'0)k (Vw + 0)k mod I. But co

— CiB mod I, by the choice of j. Therefore (tca Ci#

Since co, 0, prime, so

(0(1

j))k mod 1

are relatively prime and /I we must have c,j)k

(I — c,a+1 mod 1.

/

and 0 relatively

180

9 The Second Case of Fermat's Last Theorem

Lemma 9.9. r1a/lb is a pth power. PROOF.

Let I be a prime of Q(ç) lying above 1. Then w + 11a

=

1— Wp a-19

Ca

P

mod

I, by Lemma 9.8.

1-c-i1-

( e-b- P MOd Pa

a (13 +ca 0)P P

= (1)

Therefore

qb

I.

Consequently 17,,M b is a pth power modulo every prime above 1. Since p,l/h + (0(Cp)), Corollary 8.15 implies that E + mod(E +)P is generated by the units Ei , i = 2, 4, ... , p — 3. Therefore

for some integers di . We want to show p Idi for all j. As in the proof of Theorem 9.3, rational integer (mod p). 1 1b

By Exercises 8.11 and 8.10, we have pi di if p,j/ B i , so we only need to consider the " irregular" indices . is for which p B i . In Proposition 8.18, the integer t determines a prime ideal t by t mod I. Henceforth, let I denote this prime ideal. Fix a generator yi for the multiplicative group (Z[ p] mod I) >< . If 1)f• x, define indi x by y itnclix

x mod

I,

so inch x is defined mod(/ — 1), hence mod p. Let a, e Gal(Q(C p)/0). Let y„..(i) = o-„(7.1) be the multiplicative generator mod a-„(1). Then ind,.(1) x = india; '(x). Since P-

Ei ---

FI

(c(b- bg)I 2

1

_

1—

, 1

cbg bP -1- '

CI' )

• (pth power),

it follows easily that a; '(E1)

ETP

(pth power).

Therefore a; '(E1)

cx -i indi Ei mod p.

181

§9.2 The Theorems

From above, we obtain

ind,y(i) E

cC i indi Ei mod p.

Since rid% was shown to be a pth power modulo each prir _ above 1,

0 _=

E di ind,..(1) Ei mod p,

hence

0

_= z di cc -i indi Ei mod p,

for all a # 0

mod p.

Since

det(a -

11

= 2, 4, . - 3

2 (p - 3)/ 2 = ((P —

3 ) !)

(ci _2 _ 13_ 2)

1.5_/3< a (p- 3)/2

# 0 mod p (essentially a Vandermonde determinant), we must have

di indi Ei _= 0 mod p. As mentioned above, di _= 0 if i {1 1 , , is). But ■2:` # 1 mod 1 implies, by Proposition 8.18, that indi Ei # 0. Therefore, for all i, d. _= 0 mod p. It follows that ria/rib is a pth power. This completes Lemma 9.9. 1=1 The proof of Theorem 9.5 is now complete. The verification of Fermat's Last Theorem is carried out on a computer as follows. First, the irregular indices are determined via congruences such as (3p- 2k ± 4p- 2k

6p-2k

1 )B 2k/4k

S p/6 <s
2k - 1 (mod p)

(There are several such congruences; see (Wagstaff [1]) for details). This is the longest part of the computations. Then Theorem 9.5 is used to verify the second case of Fermat's Last Theorem. This has been done for p < 125000. For the first case it is possible to use Theorem 6.23 since i(p) < 5 for all p < 125000. However, it is faster to use the Wieferich criterion: if 2P -1 #1 mod p 2 , then the first case of Fermat's Last Theorem has no solutions. In fact, if aP- # 1 mod p for some a < 31 then there are no solutions. For a = 2, the only values of p < 6 X 109 with 2P -1 # 1 mod p 2 are p = 1093 and 3511, and for both of these 3P -1 # 1 mod p 2. Therefore the first case holds up to 6 x 109 . The fact that th.t.re are only two "bad" primes should not be very surprising: the probability that 2'' mod p 2 should be 1/p. Therefore, the number of such p less than x should be approximately

E

1/p — log log x + 0.26.

p<x

Note that log log(6 x 109) 3.1; perhaps another example should be expected soon.

182

9 The Second Case of Fermat's Last Theorem

NOTES

The criteria in this chapter were developed by Kummer and Vandiver. Theorem 9.5 has been extended to allow p < (p2 p) by Inkeri. For more on Fermat's Last Theorem, see Vandiver [1] and Ribenboim [1].

EXERCISES 9.1. Let K be a number field, aEKx, and ne /. (a) Suppose K(a l I")is unramified. Show that (a) = r for some ideal I of K. (b) Suppose (a) r for some ideal / of K. Show that KO i')/K is unramified except possibly at the primes dividing n. (Hint: work locally. In a completion I is is principal, so the local extension can be obtained by adjoining u" with u a local unit.) 9.2. (This exercise proves part of Exercise 9.3(b)). Let p be prime and let K be a number field containing Ç. Let it = I and let fi be a prime ideal of K dividing Ir. Let Aa be the exact power of fi dividing n. Let aeKx with /' a. Let c be maximal such that X P a mod hc has a solution. Assume that c < pa, so xP a mod pit, hence mod p".

(a) Suppose b
with 0 < d < a and

0 < r < p.

(b) Suppose p is inert in the extension K(a u )/K. Let x a l lP mod fio, with g maximal. Show (i) g > 0; (ii) if g > a then xP a mod AP", which is impossible; (iii) g = d, with d as in (a). (c) Let notations and assumptions be as in (b). Let z E K be such that (z)fid is an integral ideal prime to fi. Show that z(x — a ljP) is an integer in K(a l lP) which is prime to fi but whose norm to K is divisible by fir, with r as in (a). (d) Show that (c) contradicts the assumption that fi remains prime in K(a 1/P). Conclude that p must ramify or split completely (in fact, by Exercise 9.3(a), p must ramify). 9.3. Let p be prime and let K be a number field containing Ç. Let IL = Ç — 1 and let A Let a E K x with a ct (Kx y, and assume a is relatively prime to p. The number a is called primary if xP a mod pit has a solution in Kx ;hyperprimary if xP a mod prrY has a solution; and singular primary if a is primary and (a) for some ideal I of K. (a) Show that a is hyperprimary if and only if all primes above p split completely in the extension K(a l IP)/K. (b) Show that a is primary if and only if K(Œ L i P)/ K is unramified at all primes above p. (c) Show that a is singular primary if and only if K(a lIP)/ K is unramified at all primes of K (one exception : if p = 2 then K could be real and there might be ramification at the infinite primes). (Hints: Exercises 9.1 and 9.2; also look at the proof of Lemma 9.1 and the second proof of Theorem 5.36.)

183

Exercises

9.4. (a) Let f (X) = (((Ç I, — 1)X + 1)P — 1)/(C f, — 1)". Show that f (X)

X mod((,) — 1)

XP

and thatf (1) = O. Conclude that p/((p — 1 ).P — 1 mod((p — 1). (b) Look at the terms with lowest p-adic valuation in the expansion of 0 = logp(1 +

— 1))

to obtain the result of (a). This also works for Cp„ • (c) (The easy way.) Find the minimal polynomial g(X) for CI, — I, compute g(Cp — 1) mod(Cp — 1)P, and obtain (a). This also works for

9.5. (The second case of Fermat's Last Theorem for p = 3). Recall that we needed p> 3 for part of the "basic argument." This exercise treats p = 3, Let = C3 It is well known that Q(C 3 ) = Q(/-3) has class number 1. Suppose we have x, y, z E 2, with 3,0cyz, and m > 1 such that x 3 + y 3 = (3mz) 3 . We of course may assume that x, y, and z are pairwise relatively prime. (a) Show that

x + y = q 0 3 3 m - 1 p8, + y = n1(1 — Op?, .7( +

y = 112(1 — C 2 )p3,

with p i , p 2 p 3 2 [ ] and pairwise relatively prime, and with q i , q 2 , q 3 units of 2 [C]. (b) Show that q 1 is congruent to a rational integer mod 3, hence q i = ± I = (± I). Therefore we may assume q i = 1. Similarly, we may assume q 2 = 1. (c) Show that qdrio is a cube. Using the fact that g o has the form + conclude that rio = +1; hence we may assume go = 1. (d) Show that we may assume that p o EL (Hint: a straightforward calculation shows that (s + t ) 3 E 1 s = t or st = O.) (e) Write p i — a + K. Show that (a, b) = 1, hence a, b, a — b are pairwise relatively prime. (f) Show that ab 0 and a b (in particular, a = b = I is excluded). (g) Show that x + y = 9ab(a — b). (h) Use the equation x + y = 3 3 m - 'p8 to show that there are nonzero rational integers a l , b 1 , c 1 such that some permutation of (a, b, a — b) equals

b?, 3 3 ' 3 cD, (i) Since al ± bl = ( ±3m- 1 c 1 ) 3 for some choice of signs, and since we know the first case has no solutions (by congruences mod 9), we are done by induction.

Chapter 10

Galois Groups Acting on Ideal Class Groups

Relatively recently, it has been observed, in particular by Iwasawa and Leopoldt, that the action of Galois groups on ideal class groups can be used to great advantage to reinterpret old results and to obtain new information on the structure of class groups. In this chapter we first give some results which are useful when working with class groups and class numbers. We then present the basic machinery, essentially Leopoldt's Spiegelungsatz, which underlies the rest of the chapter. As applications, Kummer's result "pill + pill' is made more precise and a classical result of Scholz on class groups of quadratic fields is proved. Finally, we show that Vandiver's conjecture implies that the ideal class group of CD(Ct„,) is isomorphic to the minus part of the group ring modulo the Stickelberger ideal.

§10.1 Some Theorems on Class Groups Since it is useful, we repeat the following result. Theorem 10.1. Suppose the extension of number fields L/K contains no unramified abelian subextensions F/K with F K. Then hic divides hL . In fact, the norm map from the class group of L to the class group of K is surjective. PROOF. The first statement is Proposition 4.11. The second is proved in the appendix on class field theory. 0 Theorem 10.2. Let n * 2 mod 4 be arbitrary and let kr = h(Q(C,r)). If 2Ih: then 2117 .

184

185

§10.1 Some Theorems on Class Groups

PROOF. U(C„)/CP(C„) + is totally ramified at infinity, so Theorem 10.1 implies that the norm map on the ideal class groups N: C -3 C -1-

is surjective, so h,T = h,,/h, = I ker NI. Suppose 21h: . Then there exists a nontrivial ideal class a E C * such that a 2 = 1. Lift a to C. Since Nix = a 2 = 1, e ker N. Since the map CI' C is injective by Theorem 4.14, a 0 1 in C, but a 2 = 1. Therefore ker N has even order, so 21h. 0 Remark. Note that this proof works for any CM-field K such that the map C+ C is injective. This theorem is useful when one looks for cyclotomic fi elds whose real subfields have even class numbers. Such fields arise in topology (see, for example, Giffen [1]). Theorem 10.3. Let K be a CM-field. The kernel of the map C + C from the ideal class group of IC to that of K has order 1 or 2. PROOF. Let I be an ideal of K+ and suppose ! = (a) in K. Then (1) = /// = (/ix), so à7c/a is a unit, hence a root of unity by Lemma 1.6. This root of unity does not depend on the class of I in K +. Let W be the roots of unity in K. We obtain a homomorphism 0: ker(C+

C)

W W/W 2 .

0(/) = 1, then [fc/a = C 2 ; SO Ca = Ca. This means Ca E K +. Since ! = (Ca) in K, unique factorization into primes implies I = (Ca) in K +. Therefore 0 is injective. Since W/W 2 has order 2, the proof is complete. Note that it is possible for the kernel to have order 2. See the example following Theorem 4.14. Theorem 10.4. (a) Suppose L/K is a Galois extension and Gal(L/K) is a p-group (p = any prime). Assume there is at most one prime (finite or infinite) which ramifies in L/ K. If p I h L then plhK. (b) If L/U is Galois, Gal(L/U) is a p-group, and at most one finite prime ramifies, then p hL•

PROOF. Assume plk. Let H be the Hilbert p-class field of L, so H is the maximal unramified abelian p-extension of L and Gal(H/L) is isomorphic to the p-Sylow subgroup of the ideal class group of L. Since L/K is Galois, the maximality of H implies that H/K is Galois. Let G = Gal(H/K). Let A be the prime (if it exists) of K which ramifies, let be a prime of H above A, and let I c G be the inertia group for Since H/L is unramified, III

deg(L/K) < IG I.

186

10

Galois Groups Acting on Ideal Class Groups

By a well-known result in the theory of p-groups, there exists a normal subgroup G 1 of G, of index p, with I gG 1 c G (proof: mod out by the subgroup generated by an element of order p in the center, then use induction on I G ). The inertia subgroups of the other primes of H above A are conjugates of I, hence lie in G I . Since A is the only ramified prime, no prime ramifies from K to the fixed field of G 1 . But the fixed field of G 1 is Galois of degree p over K, so K has an unramified abelian extension of degree p. Class field theory implies that prhK . This proves (a). The case K = Q is treated similarly, except that we ignore ramification at infinity. Therefore, if pIh r, we obtain an abelian extension of degree p which is unramified at all finite primes. But the Minkowski bound (see Lemma 14.3) implies that the discriminant of any nontrivial extension of Q is greater than 1, so at least one finite prime ramifies, contradiction (alternatively, if there is ramification at infinity then p must be even, and every quadratic extension of Q has ramification of at least one finite prime.) This completes the proof of Theorem 10.4. 1. Then plh(CP(Cp))<:-> P h(Q(Cp0).

Corollary 10.5. Let n PROOF.

Theorems 10.1 and 10.4.

Corollary 10.6. If Vandiver's conjecture holds for p then p)(h + (QG))for all n > 1. Corollary 10.7. Let p > 2 and let t„ be the unique subfield of 11: ( ,,,1) of degree p" over Q. Then p, /170 Ç) (note that IL co/It o is the cyclotomic. extension of of Q. For p = 2, the corresponding result is contained in Corollary 111 10.6, since 1:„ = Q(( 2 ,2) + ).

If A is a finite abelian p-group, then A

OZ/pul

for some integers a,. Let

= number of i with ai = a, ra = number of i with ai a.

Then

= p-rank A = dim(A /AP) and, more generally, ra = din-izpz(AP'

187

§10.2 Reflection Theorems

Theorem 10.8. Let LIK be cyclic of degree n. Let p be prime, p ,rn, and assume all fields E with K E L satisfy p ,rhE . Let A be the p-Sylow subgroup of the ideal class group of L, and let f be the order ofp mod n. Then

ro(A) n o(A)

0 mod f

for all a, where ro and no are as above. In particular, ifp IN, then the p-rank of A

is at least f and pf Ih L .

Let V = AP" - AP", so V has pr- elements. Let a generate Gal(L/K). Then a acts on V_ Let v E V, I) 0, and suppose the orbit of v under the action of Gal(L/K) has less than n elements. Then aiv = v for some i
fore (1 + a + o-2i + - • + igi(n")

1)1)

= Norm(v),

where the norm is induced by the norm from L to the subfield of degree i over K. Since p does not divide the class number of this subfield, by assumption, we have (nli)v = 0. But p n, so v = 0, contradiction. It follows that the orbit of every v 0 has n elements, so pi." 1 mod n. Therefore f I ro . Since = ro — r we obtain f in o . This completes the proof. LII Remark. It is easiest to apply this result when n is prime, so there are no nontrivial intermediate fields. In that case, we only need p)(n and p,rhE .

As an example for the theorem, consider (n29). It can be shown (Exercises for Chapter 11) that its class number is 8. Therefore the class group in Z/87, f/47 x 7L /2Z, or (7Z/27L)3. Which is it? ag 29) is of degree 28 over a hence has a subfield K of degree 4 over G. By Theorem 10.4, 2 Since 2)(n = 7 and there are no nontrivial intermediate fields between K and O ( 29 ), Theorem 10.8 applies. The order f of 2 mod 7 is 3, so the rank of the class group is at least 3. Therefore the class group is (7/27) 3.

§10.2 Reflection Theorems Let p be an odd prime and let LIK be a Galois extension with Gal(L/K) = G. We assume that E L. Let L' be the maximal unramified elementary (i.e., isomorphic to FlpZ x x 71p7) abelian p-extension of L. Then H = Gal(L'/L) A/AP, where A is the p-Sylow subgroup of the ideal class group of L. Note that L' /K is Galois and H is a normal subgroup of Gal(L'/K), so G can act on H by conjugation (let h E H, h E G. Extend g to "d E Gal(L'/K). Then hg = Oh6- , which is independent of the choice of 7 since H is abelian). H becomes a Z [G]-module. Z[G] also acts on AI AP, and in fact

ci,

H

AIAP

as I [G]-modules

(see the appendix on class field theory).

188

10 Galois Groups Acting on Ideal Class Groups

Since 4 E L, L'/L is a Kummer extension, so there is a subgroup B g_LX/(L such that L' = L( B), in the obvious notation. There is a pairing HxB— ■ Wp = pth roots of unity = h(bb 11P11 ) . It is easy to see that this pairing is nondegenerate ( = 1 -4=> h and = 1 <=> b =1) and bilinear. By Lemma 3.1, B

11

H

1,

A /AP,

though the second isomorphism is noncanonical and not G-linear. An easy calculation shows that <119, b9 > = 9, g e G. Let b E B (or more accurately b mod(L )P e B). Since L(b l /P)/L is unramified, (b) = P for some ideal I of L (Exercise 9.1). Changing b by an element of (L )P leaves the ideal class of I unchanged. We therefore have a map 0: B — ■ A p = {x E AIXP = 1). Clearly 0(1)9) = 0(b)9 for g E G. Suppose 0(b) = 1. Then (b) = (a)P, so b = ea" for some E e E =-- units of L and a E L. Therefore ker E(Lx )P/(L )< )P E/EP where the last isomorphism is G-linear. To summarize, we have B

A/AP,

0: B ker

Ap,

non-G-linearly, G-linearly, and

subgroup of E/EP,

G-linearly.

It is precisely the non-G-linearity in the first isomorphism which will make things work (see Exercise 10.8). The basic machinery is now complete; we are ready for the applications. Theorem 10.9. Let A be the p-Sylow subgroup of the ideal class group of

0(4)

and let P-

A i= o

LA

be the direct sum decomposition corresponding to the idempotents of the group ring Zp[Gal(0(4)/0)] (see Proposition 6.16). Let i be even and j odd with i + j 1 mod(p — 1). Then

p-rank

Ei A

< p-rank vi A < 1 + p-rank

(this strengthens the result "pi h +

plh - ").

Ei A.

189

§1O.2 Reflection Theorems

PROOF.

Let G = Gal(Q(4)/0) in the above. We have

H

ei H

as G-modules, so

AI AP

Ei(A I AP)

for all i.

Let h e EH. Then ah = fe lla) for all a e

pZ) X . Let b e Ek B. Then

E Wp) = = = = 1. Since the pairing between B = k B and H=OEH is nondegenerate, it follows easily that the induced pairing

Ei H x ei B W p ,

i +1 1 mod(p — 1)

is nondegenerate. By Lemma 3.1, Ei B

Now, 0: B

EiH

as abelian groups.

ei(AI AP),

A p is G-linear, so

Ei B

ei /1 p .

We also have

(ker 0) n Ei B

subgroup ot

From Propositions 8.10 and 8.13, Ei(E/EP)

{ZIpZ, j even, j 0 mod(p — 1); or ] 0, otherwise.

1 mod(p — 1);

Let dim denote dimension over 71g. Note that

p-rank E 1 4 = dim Ei (A/AP)

and

p-rank E J A = dim O p .

From the above,

dim(E i(A/AP)) = dim( 3 B) < dim(Ej(E/EP)) + dim(e y4p). If j is even and ] 0 mod(p — 1), we obtain

p-rank(E i A)

1 + p-rank(Ei A).

Ifj is odd and ] 0 1, then p-rank(E i A)

p-rank(Ei A).

If] 0 1, then we find

p-rank(e 0 A) < 1 + p-rank(E 1 ,4). However, we already know that e0 A = 1 A = 0. by Proposition 6.16. Putting everything together, we obtain the theorem. LII

190

10 Galois Groups Acting on Ideal Class Groups

The next result is classical, due to Scholz. We proved a weak form of it in Chapter 5, using p-adic L-functions.

Theorem 10.10. Let d > 1 be square-free. Let r be the 3-rank of the ideal class group of CP(N/cl) and s the 3-rank of the ideal class group of CP (\/ —3d) (=Q(N/-d13) if 3Id). Then r < s < r + 1. Let L =,N/ — 3d) and G Gal(L/0). There are three quadratic subfields: CD(j1), 0(. \/ — 3d), and (U(.\/ — 3). Let PROOF.

{1, = Gal(L/0(j1)), {1, a) = Gal(L/O( N/ — 3d)), {1, at} = Gal(L/Q( \/ — 3)). In Z 3 [G] we may decompose the identity as a sum of idempotents:

1 = Ei + 6 2 + 6 3 + 6 4 (1 + t)(1 + 0- ) 2 2

(1 + 1(1 — 2 2

0-)

(1

— 2

2)(1. ± a)

2

(1 — 1(1 — a Let A be the 3-Sylow subgroup of the ideal class group of L. Then A = (:) EiA. Since E i = (Norm L/0)/4, we have E 1 A = 0. Also, 64

so

64 A

± 07) =

(1 —

t)(Norm L/CD( N/ —

= 0, since h(CP(N./ — 3)) = 1. We now have A = 6 2 A C)

63A.

But 6 2 =. 4-( 1

a)(Norm L/C(/d)),

SO

62 A g Acpwil) ,

where the last group is the 3-Sylow subgroup of the class group of Q(j/). Let a e A e() . Then ta = a. Since 1 + a = Norm(0(j)/0), (1 + o-)a = 0, hence o-a = —a. It follows that E 2 a = a, so

AGGid) g 82 A.

191

§10.2 Reflection Theorems

Therefore A = Acpwo •

E2

Similarly We now use the machinery at the beginning of this section. We have B = C) .B. A calculation similar to that in Theorem 10.9 shows that

(ei H, ei B> = 1 unless i = 2, j = 3 or i = 3, ] = 2. For example, if h e ah = h' and ab = b 1 , so

= (h- 1 ,1) -1 > =

E2

H and b e 64 B then

ba> = '.

. Therefore = 1. But a 0 Gal(L/G(.\/ — 3)), so a(c 3) Since H x B W3 is nondegenerate, we must have X e3

B—*

W3,

63 H X 6 2

B—*

W3

E2

H

and

nondegenerate. Also, 0: 62 B

E2 A3,

63B

63A3,

(I):

(ker

n 62 B

(ker 0)

subgroup of E2(E/E 3), and

n E3 B

subgroup of e3(EIE 3).

Since e2 = 1(1 — a)(Norm LICD(/ d)), E 2 (EIE 3) is contained in the units of 0(j/) mod 3rd powers. Therefore

E2 (E/E 3 )

0 or Z/3Z.

Similarly,

t)(Norm L/41(.\/ — 3d)).

E3 = t.(1 —

Since d 1, Q(.\/ —3d) 0 ©(\/ — 3), so the units of 12(\/— 3d) are either {±1}or {-FL + 1}. Consequently

e3(EIE 3 ) = 0. Putting everything together, we obtain,

r = 3-rank = 3-rank

Au (vid ) = 62

3-rank

H = 3-rank

E3

62

A

B

3-rank E3(E/E 3) + 3-rank E3 A

= 0 + 3-rank Auw _ 3d) = s,

192

10 Galois Groups Acting on Ideal Class Groups

and similarly, s < 1 + r.

This completes the proof.

0

Remark. The cases r = s and r + 1 = s both occur. For d = 79, r = s = 1,

while for d = 69,r = 0, s = 1. Theorem 10.11. Let p be an odd prime. Let L be a CM-field with Ç e L, and let A be the p-Sylow subgroup of the ideal class group of L. Then

p-rank A + < 1 + p-rank A - . Let W be the roots of unity in L. If L(W i lP)IL is (totally) ramified, then

p-rank A + < p-rank A - . (As usual, A ± = {x e AI .R = x ±1 } and A + '2 A(L + )). PROOF. In the notation at the beginning of the section, let K = L + , the maximal real subfield of L. Then G = Gal(L/K) = {1, J}, where J = complex

conjugation, and A+

=1+J A 2 '

A- =

1—J A. 2

As in the above theorems = = 1 (since p 0 2), so H + x B - -- Wp is nondegenerate. Also, 4): B -

and (ker

0

)

n 13 - '-' subgroup of (E/E.

Since [E: WE] = 1 or 2 (Theorem 4.12), (E/E

= (W/WP) - -, ZlpZ.

Therefore p-rank A+ = p-rank H + = p-rank 13 1 + p-rank A - . If L(W l /P)/L is ramified then W n B = 1, since L(B l iP)/L is unramified. Therefore (Ker 0) n B - = 0 and the "1" disappears from the above inequality. This completes the proof. CI

193

§10.2 Reflection Theorems

If p = 2, the above result holds if we modify A + . Note that A+n A-

g fx E Alx 2

which does not allow us to conclude that the intersection is trivial for p = 2. In fact, for O ( ,./ 5), A + = A - = A -- Z/ZE. Also, observe that if x is an ideal class of L + with x 2 = 1 then xcA'n A - . However, we actually want to be able to transfer information from h - to h+ , so instead of A + we should be looking at the 2-Sylow subgroup of the class group of L +, whose order is the 2-part of h + . Instead of the decomposition A = A + 8 A - obtained for odd p, we have an exact sequence —

I —o A L- —0

A L —o

AL

+ —0

1,

which is induced by the norm (i.e., 1 + J) from L to L + (cf. Theorem 10.1). Proposition 10.12. Let L be a CM-field and let A L and A L + be the 2-Slow subgroups of the ideal class groups of L and L+, respectively. Then

2-rank A L + < 1 + 2-rank A L- . PROOF. Let i(A L +) denote the image in A L and let (A L + ) 2 and (A L- )2 denote the elements of order 2 in the respective groups. As noted above, i((AL-0 2 )

g

(AE )2.

Therefore 2-rank i((A L + ) 2 ) < 2-rank AT . But RAL,)21

= 1 or 2 I i((AL , )2)1 by Theorem 10.3. Since the elements of order 2 determine the 2-rank, — 1 + 2-rank(A L +) < 2-rank i((A L 02 ) < 2-rank A L- . This completes the proof.

0

Finally, we use the Kummer pairing to give another characterization of irregular primes. D Proposition 10.13. Let p be odd. Then p divides h(0(4)) if and only if there is an extension K 10(4) + , with K # 0(42) + and Gal(K/C(4) + ) '--' ZIpl, which is unramified at all primes not above p.

First assume that such a K exists. Let F be the maximal elementary abelian p-extension of CP(C p) + which is unramified outside p. Then F/0 is Galois and also F(4)/0 is Galois. As before, G = Gal(()/) acts on H = Gal(F(4)/0(4)). Since F is real, complex conjugation acts trivially, so PROOF.

194

10 Galois Groups Acting on Ideal Class Groups

= 1 for all odd i (where E i is the usual idempotent for G). Suppose that H = Eo H. Then hg = h for all g e G. Recall the definition hg = 074- 1 , where 'd is an extension of g to F(Cp). We find that jh = hg for all h e H, g e G. Since G is cyclic, we may choose the J's so they commute with each other. It follows that Gal(F(C p)/e) is abelian. By the Kronecker-Weber theorem (14.1), and since F(4)/0 is unramified outside p, F(Cp) g 0(4.) for some n. Therefore F g Q(Cp4 + . By the choice of K, this is impossible, so H 0 E0H, hence ;H 0 1 for some even i O. Since F()/0() is a Kummer extension, there is a subgroup B g 11(Cp) x /(Q(C p) x )P such that F(Cp) = C(Cp)(B l iP). There is a nondegenerate bilinear pairing HxB->Wp such that = g,

g c G.

As before, the fact that i H 0 1 for some even i 0 implies that Ei B 0 1 for some odd j 1 (i + j 1 mod p - 1). Choose b e Ei B, b 1. Then 0(4, b 11P)/(1(4) is unramified outside p, hence (b) = IP(Cp - 1)d for some ideal I and some integer d. Since b e Ei B, b(7. = ba'cP, with c c 0(4); for all a e (Z/pZ)x . Also, (Cp = (Cp - 1). Therefore way(cp o d = (b a (by (op = a y (c p od a (cy )° It follows that d - da ai0 mod p, hence d 0 mod p. We therefore have (b) = P for some ideal I of 0( C). Suppose now that I is principal, so I = (a). Then ilaP = b for some unit i. We may change b by a pth power, hence assume b = 17. Write b = Cprti l with real. Since Cpl. is in the E l component and n i is real, it is impossible for Crpti i to lie in the Ei component with odd j 1. This contradiction shows that I is nonprincipal. Since IP is principal, we must have plh(O(Cp)). Conversely, suppose plh(O(Cp)). Then plh - (Q(Cp)) so the Ei component of the class group is nontrivial of some odd j. By Proposition 6.16,1 0 1. Let I (nonprincipal) be an ideal representing a class of order p in the Ei component: IP = (b) and Pi I mod principal ideals. Let -

fi

mod(O(Cp)')P.

Since (fi) is the pth power of an ideal, Exercise 9.1 implies that 0(4, fi 1 /P)/0(Cp) is unramified outside p. If B 1 denotes the subgroup of CP(Cp) x /(LI( p) x )P corresponding to the maximal elementary abelian p-extension of (IQ unramified outside p (call it F 1 ), then fi E Ei B i . We claim fi is nontrivial. Suppose fi = ce. Then, modulo pth powers of principal ideals, 1

(a)P = (13)

(b ) i = JPJIP,

so IP is the pth power of a principal ideal; hence I is principal, which is a contradiction. This proves the claim that fi is nontrivial in Ei B. Therefore

195

§10.3 Consequences of Vandiver's Conjecture

ci B i # 1. Let H i = Gal(F i /Q(Cp)). Via the Kummer pairing H i x B 1 —> Wp , we find that v.H 1 is nontrivial for some even i 0. Let K 1 be the corresponding extension of 0(4); that is, Gal(K /Q(C p)) = Ei I- 11. It is clear that K 1 is Galois over Q. Since i is even, complex conjugation J commutes with e i H l . Therefore the group generated by J and ei H 1 has order 21E i H 1 1, so the fixed field must be Q(4)+. Since Kt is the fixed field of J, we find that Gal(K /Q(C p)+)

vH 1 # 1.

Since Gal(Q(C p2)+/Q(C p)+) is in the E 0 component and since i 0, we have Kt n U(42) = CP(Cp) + • Clearly Kt /Q() is unramified outside p. If we take a subfield K c KP such that Gal(K/Q(C p) + ) ZIpZ, we obtain the desired field. This completes the proof. LII

§10.3 Consequences of Vandiver's Conjecture In this section we assume that Vandiver's conjecture holds, namely that p does not divide the class number of U(Cp) By Corollary 10.6, this implies that p also does not divide the class number of Q(C p,,) + for all n > 1. It is not clear whether or not Vandiver's conjecture should be true in general, but, as we mentioned in Chapter 9, it seems that it should hold for a large majority of primes (at present it is known to be true for all p < 125000). Hence the following results are possibly a good approximation to the truth in general. Theorem 10.14. Let p be odd and assume p,' h(0()). Let A,,, = A , be the p-Sylow subgroup of the ideal class group of EP(Cpri + 1), let R p , n = 1p[Ga1el:Kp„,_

and let 4 be the Stickelberger ideal (see Chapter 6). Then A , is cyclic as a module over R p,„ and

R p-,11;„ as modules over R. In other words, the Stickelberger ideal gives all the relations in

PROOF. The main part of the proof involves proving the cyclicity. The rest follows easily. For n = 0 the result is an immediate consequence of Theorem 10.9, but in general we have to work a little more. We use the ideas of the previous section, but we must look more closely at the units. Recall that in the notation of the previous section, ker c E/EP. Since the structure of E is fairly well understood, it is convenient to use all of E/EP, rather than just a subgroup.

1 96

10

Galois Groups Acting on Ideal Class Groups

To do so, we must enlarge B. Therefore, let L" be the maximal elementary abelian p-extension of L = U(Cpn i) which is unramified at all primes except possibly A = (1 + 1), which is the prime above p. Then L" = 4\ •') for some subgroup 11 g L 'AL )P. Let H' = Gal(L"/L). There is a nondegenerate bilinear pairing H' x —> Wp ,

satisfying = g for g e G" = Gal(L7Q). Let b

E

B'. Then, as in the previous section,

(b) = I PA d

for some ideal I and some integer d. The A must be singled out because of possible ramification at A. We obtain a map Of: B' b

A p = {x

E

AI xP = 1}

class of I.

If (6'(b) = 1 then b = E(1 — + 1)d • aP for some unit e and some a E Conversely, if b is of this form then L(b l /P)/L is unramified outside A, so b e B' and clearly 0'(b) = 1. Therefore ker (II is generated by E and 1 — Since we are assuming p h + , we have p (r[E+ :C + ] by Theorem 8.2, where C+ denotes the real cyclotomic units. Since E = <4.,+1> x E+ and

C =

x C +,

we also have p,i'[E :C]. Therefore C generates E/EP. It follows from Lemma 8.1 that ker (6' is generated over R p , by 1 — (note that = —(1 — (1 — --1 )). Consequently, (ker (6') + is generated by (1 — C p,, )(1 — In fact, {((1 —

1)Yra 1 1

1)(1

a < pfl (a, p) = 1} ,

is a basis for (ker 0') + as a vector space over ZIpZ. Note that G = Gal(Q(C p. i)/CD) acts transitively on the elements of this basis. If T411 + then A p+ = 1. Therefore

(Br

= (ker

so (Br is cyclic over R,1 (to obtain cyclicity is the reason we enlarged B). As before, the pairing (Hy x (B') + is nondegenerate. We claim that (HY is cyclic over R p,„. Let {h i , , be the dual basis of (H')+ corresponding to the basis of (Br constructed above (call it b 1 ,. , br 1). Then

197

§10.3 Consequences of Vandiver's Conjecture

Let H 1 be the R p, n-submodule of (H') generated by h 1 . Suppose H 1 0 (HY . Then, by Proposition 3.3 or 3.4, there exists b = xi bi 0 0 in B (x i c 2) such that = 1. In particular,

1 = =

= 1,

for all g e G.

Letting g = 1 we find

1 =

=

,

hence x 1 0 mod p,

and since G acts transitively on {b 1 , , kJ, we may use other choices of g to obtain xi 0 mod p for all i. Therefore b = O. It follows that H 1 = so (H) - is cyclic over R p,„, as desired. Returning to the beginning of the proof, we observe that L' g L", so H = Gal(L'/L) is a quotient of H' = Gal(L"/L). Consequently, 1-1 (21„/A ) - is cyclic over Let xo c A„ generate (A n/Af,) - . Let x c A„. Then X = roxo + Ph, with ro c R p,„ and But

= r i xo + py2 , etc. Therefore x = (r + pr + • • •)xo ,

so xo generates A n- over R p,„. Therefore A n- is cyclic as an Rp, n-module. Let x o be a generator for An- over R hence over R p-,„. Then we have a surjective R

rx o . By Stickelberger's theorem,

is contained in the kernel. Since [R;„: I;„] = IA,1

by Theorem 6.21, the kernel is exactly /1-,, n . This completes the proof.

0

Corollary 10.15. Let A be the p-Sylow subgroup of the ideal class group of 0(Cp) and let p-2

A i=0

be the decomposition according to idempotents. If p h(e(C p) + ) then

for i = 3, 5,

, p — 2.

198

10 Galois Groups Acting on Ideal Class Groups

PROOF. By Theorem 10.14, each EA is a cyclic group. By Proposition 6.16, annihilates E i A, so B1 ,

I riA I

p-part of B1, o,

Since p-

2

TI

IgiA I

i= 3

=

IA - 1

=

p-part of h -

i odd

p-2

= p-part of 2p

II

—p3 1 , co

o

odd

= p-part of

II (B 1

1#19 -

)

2

(see the proof of Theorem 5.16), the inequalities must all be equalities. This completes the proof. I=1 Remark. It has recently been proved unconditionally by Mazur and Wiles that Ei AI = p-part of B 1 , - L. It is a consequence of the "Main Conjecture" (see Chapter 13).

For the next result, recall that if = Gal(OGp .0/0(4)) and if T o is a topological generator of ro , then Zp[R1] Zp[[7 ], with T o corresponding to 1 + T. See Theorem 7.1. Hence a 7L p[[11]-module may be regarded as a Zpr [ T module. Let A n be as above, so A a 7L pr[17FP"]] -module. The norm map N n from A n to A n _ i commutes with the action of the group ring. Take the inverse limit lim A, with respect to the norm mappings. If ]

]] -

and .) e lim

(. . . , y. - 1

[FF ' ]

then N

n an) — Y n N n(an)

= y_ lan - 1

(since yn restricts to yn _ 1 ). Therefore ("

lan- 11 Ynan, • • •)

lim A

so lim A is a Z [EFT-module. We may also decompose each A n according to the idempotents 1

P

p- 1

,E

a - 1 E Zp[Gal(0(4)/0)]

Each component lirn E A n is also a Zp[EIT-module.

199

§10.3 Consequences of Vandiver's Conjecture

Theorem 10.16. Assume p h(0(4) + ). Let P(T) = (1 + T) " — 1. Then, for i = 3, 5, , p — 2, ci A

7Z p [[7]]/(P„(T), f(T , co'))

and

lirn Ei A

Z p [[T]]/(f(T,

-1 ))

as modules over Z p [[T]], where f (T, co l -1 ) is the power series satisfying f((1 + p)s — 1, co l ') = Lp(s, co') (see Theorem 7.10). For i = 1, E1 A„ = 0 for all n.

PROOF. By Theorem 10.1, the norm map A, An _ 1 is surjective, so e i A„ 1 is also surjective. Since p,l'h+, each E i A n is cyclic as an R,,,-module. If b„ generates Ei A„ over R p,„, then the norm of b„ generates E i A„_ over Rp,„_ 1 . This allows us to obtain arbitrarily long sequences (b 0 , . . . , b„) such that each bi is a generator for c. A. Since cA 0 is finite, there is some c/ o such that there are arbitrarily long sequences starting with a 0 . Similarly, there is an a 1 c E i A l whose norm is a() and such that there are arbitrarily long sequences starting with (a0 , a 1 ). Continuing, we obtain a sequence (ao , al , ...) e lim gi A„

such that a„ is a generator for Ei A n for each n. Let A = Gal(Q(4)/Q). Let Opil

C

O p[A][1-71-P1

be the Stickelberger element. Then / p, n is generated by elements of the form (c — ac)Op.+1 (Lemma 6.9). Therefore Ei n is generated by elements of the form (recall y, = o- „) ; see Chapter 7) (c

.

If we take c = 1 + p and i = 1, then this corresponds to an invertible power series by Lemma 7.12. This yields E 1 A n = 0 for all n. If i 1 and c is a primitive root mod p, then c — col(c)y, corresponds to a power series with constant term c — co i(c) 0 mod p, hence to an invertible power series which may be ignored. By Iwasawa's construction of p-adic L-functions (Chapter 7), corresponds to a polynomial fn( T,

) e Z p[T]

such that f(T, co l - i)

J,,( T, co l - i) mod P,,(T),

10

200

Galois Groups Acting on Ideal Class Groups

wheref is as in the statement of the theorem. From Theorem 10.14, '), P„(T))

Zp[T]/( fn( T,

g(T)

g(T)an ,

where a n is the generator obtained above. We claim this gives us an isomorphism lim Z P [[T]]/(f, P„) lim e i A n . Clearly an element on the left yields an element on the right side via the map defined above. Conversely, suppose (y o , y i , ...) Elim EiA n . Then yi,= gn(T)a„ for some g„. Since gn(T)a, = gn(T) Norm(a„) = Norm(g,,(T)a n) = 1 (T)a,, -- 1 , we must have gn(T) — g,,_ 1 (T) It follows that (g 05 g1,...) E lirn

Zp[[T]i/(f, P,,),

so (y o , Yi, ...) is in the image of the map. Since we have an injection at each level, we have an isomorphism as claimed. It remains to evaluate the inverse limit. Clearly there is a map

0: Zp[M]

lim Z p [[T]]/(f,

If 0(g) = 0 then, for each n, g = Bn f + BP,,, with /3„, B'„ G Zp[M].

Since P„ 0 in 7Lp[[7]], urn B,, = B exists. Therefore f divides g, so ker = (f). Now suppose (go- gi,...)

lirn 7LpUTD(./-3 PO.

Then gn + = g,, + Cn f +

with C„, D„ e Z p[M].

Let = g„ — (Z i) i
(g'0 ,

.) dim 7Z[[T]]/(13„)= Zp [[T]]

(see the proof of Theorem 7.1). Therefore there is a power series g such that g

gn mod P„,

200

10 Galois Groups Acting on Ideal Class Groups

wherefis as in the statement of the theorem. From Theorem 10.14, P(T))

Z p [T]/(f„(T,

g(T) —o g(T)a„, where an is the generator obtained above. We claim this gives us an isomorphism fim lim Z ETD( f, P, 4_ P Clearly an element on the left yields an element on the right side via the map defined above. Conversely, suppose (y o , y i ,...) c Jim g y4„. Then y, = gn(T)an for some g„. Since g(T)a_ =g (T) N Or M(a n) = Norm(gn(T)an) = we must have g(T) — g_ 1 (T)

E

(f-

1, Pn-1).

It follows that (g 05 g1, ...)

Lim 7LpECT111(.1; P,1 ),

so (yo , y i ,...) is in the image of the map. Since we have an injection at each level, we have an isomorphism as claimed. It remains to evaluate the inverse limit. Clearly there is a map 0: Zp[[7 ] —0 lim Z pETE/(f, P„) ]

If 0(g) = 0 then, for each n, g = B„ f + gi P„, with B,, B'„ c

Since P„ 0 in Z p[M], lim Bn = B exists. Therefore f divides g, so ker = (f). Now suppose (go, g 1 ,...) e inn Alp[Un(f, Pn). Then gn + 1 = g,, + C„ f + D,P„, with C„, D„ e Z pUTA. Let grt = grt

(c j )t;
Then g;,+1 = gfrt

DnPri,

SO

(do, g,...) dim Zp[[7 ]1(Pn) = - Zp[[T]i (see the proof of Theorem 7.1). Therefore there is a power series g such that g

g'„ mod Pn,

201

§10.3 Consequences of Vandiver's Conjecture

hence g # g, mod( f, P,),

for all n.

This proves that 0 is surjective. Therefore lim Z p[M]/(f, P„) lim

This completes the proof of Theorem 10.16. Remark. This result is rather amazing since it enables us to define an analytic object, namely the p-adic L-function, in terms of algebraic objects, namely ideal class groups. A similar situation exists for function fields (see Chapter 13).

A slightly weaker form of this theorem has been proved by Mazur and Wiles, without the assumption p h + . See Section 13.6. Corollary 10.17. Suppose p,' h(()). Let such that 2 i < p — 3 and pIB,. If

is be the even indices i

B 1 , 1 - # 0 mod p2 and

.Bi+p i mod p2 for all i e {i 1 , • • , t+p-1 -

t then

A„

(Z/pn+ 1 1)s

for all n > 0.

Remark. The above Bernoulli numbers are always divisible by.. p, but the above incongruences hold mod p2 for all p < 125000. But there does not seem to be any reason to believe this in general. The above yields, for p as above, p = 0, /1. = v = i(p)

where p, y are the Iwasawa invariants (see Theorem 7.14 or Chapter 13) and i(p) = s is the index of irregularity. PROOF. Letf (T, col) = ac, + al T + • • with ai E

Zp

for all p. Then, for s E 7p ,

Lp(s, coi) = f((1 + p)5 — 1, 0) # a 0 + ai sp mod p2 .

Since B2 = B

we must have i > 4, so B.

i # (1 — 13' 1 ) = — L (1 — 1

—a 0 —a 1 (1 — i)p

and B, + „_ 1 1

=

ao — a1(2 — —

—= —ao — a1(2 — OP.

202

10 Galois Groups Acting on Ideal Class Groups

We obtain

a 1 (1 — Op # a1 (2 — Op mod p2 . Therefore p al . Since p B i , we must have plao , so t = 1 for the power series f (T, w i). This means that f (T, cut) = (T — a i)EI i(T) with c, E pZ p and Eli E Z p UTE X (see Theorem 7.3). It follows that Z p ETE/(P„(T), T— ai) 7Z/P„(a1)7Z.

But P(cc) = (1 + ai)P" — 1 E p _iA n

pnai mod 13%7, so we already have the result

Z p/ pn+ fi

17'

where J1 =

Since —B

- --= f (0, w ) = — ai U i(0)

we find that v(x) = v p(B „„ -1) = 1.

This completes the proof.

NOTES For results related to the first section, see the papers of Masley. Corollary 10.5 was first proved by Furtwângler. For the general statement of Leopoldt's Spiegelungsatz, see Leopoldt [4]. For generalizations, see Oriat [2], Oriat-Satgé [1], and Kuroda [1]. There is some interest in class groups of cyclotomic fields because of their relations with class groups of group rings. See Kervaire-Murthy [1], Ullom [1], and McCulloh [2]. For divisibility properties of h: see Metsânkylâ [1], [2], [3] and Lehmer [3]. For parity questions, see the notes on Chapter 8, plus Cohn [1] and Uchida [4]. Proposition 10.13 has an elliptic analogue (Coates-Wiles [2]).

EXERCISES 10.1. Suppose LIK is an extension of degree n Show: = 1, then in I hr. • (a) If in I hi, and (in (b) If (n, = 1 then the map C K CL of ideal class groups is injective (this does not use class field theory or Theorem 10.1).

203

Exercises

10.2. Suppose Ll K is an abelian extension of odd degree. Show that if hi; is odd and hi, is even, then 4 divides h i, (it follows easily that the result is true for solvable extensions of odd degree; by the Feit Thompson theorem, it is therefore true for all Galois extensions of odd degree). 10.3. (a) It is known (for example, Kummer's Collected Papers, vol. I, p. 944) that the cubic subfield of 0(1.; 163 ) - has class number 4. Show that (13(1: 163 ) 4 has even class number. (b) Let p 1 mod 4 be prime. Show that the quadratic subfield of CD(C p ) has odd class number; so the technique of (a) will not produce even class numbers via quadratic subfields. This makes the computations more difficult. 10.4. Suppose p and q are distinct primes with p q 1 mod 4 and (p/q) = — 1. Let Ki,+g be the maximal real subextension of Min) such that [1(,+q : 0] is a power of 2, and let K i; be the similarly defined subfield of O(C,p). (a) Show that IC; has odd class number and that there is only one prime above q (Hint: Ki,÷ XI is cyclic of prime power order. What is the decomposition group for q?). (b) Show that K ptq has odd class number. 10.5. Consider (11(1-, p). Show that if i is even and ri A

0 then p B i . Is the converse true?

10.6. Let d> 0 be square-free. Let r be the 2-rank of the ideal class group of 8(‘,...d) and let s be the 2-rank of the ideal class group of 43(‘... —d). Show that r<s
(this was known to Gauss; it does not require the techniques of this chapter). 10.7. Let K be a CM- fi eld and suppose [E: WE} = 2. Show that the map C+ C of ideal class groups is injective. 10.8. Let L, B, H be as in the section on reflection theorems. (a) Show that H -= HotnE(B, Wp) as G-modules, where the action of G on a homomorphism f is defined by (gf)(b) = g(f(q l b)). (b) Let y be a 1-dimensional character of G such that the idempotent r, E Show that ,

Hom/(cx B, 14/p)

r;„x - Homk,(B, Wp).

This explains the condition i + j 1 (mod p — 1) of Theorem 10.9. The 1-dimensional character y may be replaced by a higher dimensional character cr, which is irreducible over C p . The idempotent must be replaced by r., where 0* is the character of the contragredient representation, and 0*(0-) = (1)(6-- I).

Chapter 11

Cyclotomic Fields of Class Number One

In this chapter we determine those ni for which Q(Li) has class number one. In Chapter 4, the Brauer-Siegel theorem was used to show that there are only finitely many such fields, but the result was noneffective: there was no computable bound on in. So we need other techniques. Since h„ divides hm if n divides ni, it is reasonable to start with ni prime. In 1964 Siegel showed that hp = 1 implies p < C, where C is a computable constant, but the constant was presumably too large to make computations feasible. In 1971, Montgomery and Uchida independently obtained much better values of C, from which it followed that hp = 1 <=> p < 19. Masley was then able to use this information, plus a table of hm- for 0(m) < 256, to explicitly determine all ni with hm = 1. Montgomery's original argument was for h; *, but Masley pointed out to me that the proof could be extended to composite indices. In the following, we use an adaption of Montgomery's method, though some of the less important estimates have been weakened. We obtain a finite list of prime powers for which h - = 1, and the estimates are also good enough to handle some composite cases, in particular in = 17 x 19, for which hm- has not been calculated. This information suffices for finding a finite list of possibilities for hm = 1. But we still must calculate hm+ , which is generally rather difficult. The original argument of Masley used some calculations plus some algebraic techniques. However, Odlyzko subsequently obtained rather precise lower bounds for discriminants, which allowed Masley to simplify the argument for limt It turns out that h ± = 1 for all those m with hm- = 1. So we obtain all m with hm = 1. Theorem 11.1. Let ni the following:

2 mod 4. Then h(CP(,)) = 1 if and only if ni is one of

1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 24, 25, 27, 28, 32, 33, 35, 36, 40, 44, 45, 48, 60, 84. 204

205

§11.1 The Estimate For Even Characters

§11.1 The Estimate For Even Characters We need to estimate h - from below, which will involve estimating L-series. It will be convenient to use imprimitive characters; so let z be a Dirichlet character of conductor f, with f I ni, and define zni(n)

= f AO, tO,

if (n, m) = 1 if (n, m) > 1.

We have

L(s, xm) n=1

L(s, ))

ns

ri (1 P Im

vcc — L(1, LO, ym) ' z-, zni(n) n— n=1

A 11 ( 1 Plm

The advantage of using these characters is that

E xm(n) . 0

if n

y mod m

1 mod m

and Z MO = 0

if n # ± 1 mod m.

)(even

This is not true if we use z (see Exercise 3.6). By using imprimitive characters we can take a sum involving L(s, z.) for all z and cancel many terms, and consequently obtain a better estimate than if we worked with each character separately. We know that hm- can be expressed in terms of the product of L(1, z) for odd characters z. But it works better to obtain a lower bound for the product over all nontrivial characters, and an upper bound for the product over the nontrivial even characters, then divide. The latter is perhaps a more delicate estimate in our case and we do it first. Let IT and I, denote respectively the product and sum over the nontrivial even characters z mod m. By the arithmetic-geometric mean inequality,

H L(1, z„ ) 2

21(4)(m)— 2)

<

2

L 2 + i(m)— L( 1 , Z.m)1 2 .

—0

We shall estimate the right-hand side. Note that Nm

L(1, x m ) =

hm N— ■ co n=1

zni

( n)

/1

206

11 Cyclotomic Fields of Class Number One

oo. Let xo denote We shall make estimates with the finite sum, then let N the trivial imprimitive character mod m ( 0(n) = 1 if (m, n) = 1; 0 otherwise). We have

E Znx(n)

Nm

n=1

2

n

n

x n=1

all even

Nm

)(JO 2

Nm

ZO(n)

n =.'1

2

n

def

= T(N) — To (N).

First, we estimate T(N): Nm Nm

T(N) =

x m cozmo')

E E even n1

nn'

n' = 1

1

00n) 2

n (n, in) = 1

0(m ) N m 1

„E =

2

N - 1 (N - j)m

n2

+ 0(m) E = 1 nE =1 j

(n, m) = 1

(n, m)-= 1

+

0(M) 2

1 n(n + jm)

fin

N

j1 =1n=1 (n, m)= 1

N - 1 Nm 1 O( M) + n(jm — n) 2 j=1 n=im n(Nm + fin — (n, m) =

(the first sum is for n = n'; the second, n —n') n

• 0011) n

2

N - 1 (N - j)m

2 6

0( m) +

1-

fin

2

1/1

n = 1 jm (n, m) = 1 i=

+

n

\ jm — 11)

Nm + jm n=

o( n) ;E= 1 Pi

n Nrn + jrn - n) Nm 1)

Nm 1

N-1 1

+ «m) j=E1 N1

+

n

1

(n, m)= 1

< 0(m)n2 — 12

1 ) n + jm

Nm

0(M) j‘f -v 1 2

(1

n = 1 jm m) = 1

i= 1

n'; the third and fourth,

n', n

n =1 (n, m) = 1

jm 1

—E n „= + (n, m)= 1 N-1

+ «m) E n =1 " (n, m) = 1

.

1

Nm v--,

1

L —n Nm + jm n =im (n, m) = 1

(in the second expression, we added some positive terms corresponding to (N — j)m
0(m)n2 + 20(m) N 1 1 7 12

m .J=1 r

«m) n+ m E 1

E

n= 1 -

(n, m) = 1

N - 1 1

-

N 4_ i

Nm 1

j.- 1 - , ., n= jm + 1 -

207

§11.1 The Estimate For Even Characters

Lemma 11.2. Let y = 0.577 ... be the Euler-Mascheroni constant and let A be a positive integer. Then

PROOF.

1 1 - — + y + log A.

A

y + log A

,1 a A

Since

A + 1)

1

A

A

log(

1 1 2A 2 3A 3

has alternating signs and decreasing terms, we have

1 A

1 1 (A + 1 \ < log < 4' 2A 2 A

Therefore

1 A +1

log

1 A +1

(A + 1 \ A )

1 A

+

1 < 0. 2A 2

It follows that A

a=1

a

—

log A

decreases monotonically to y. This proves the first inequality. Since A

y —

-

a=1

a

(

± log A

1

log

a+1

a=A

1 /( 1 a-A a+1 aJ cc

>

_ 1

A

the second inequality follows.

Lemma 11.3. Let n(m) be the number of distinct prime divisors of m. Then in,

- = (7 + logUm) n=1

n

(n,m)=-- 1

where —1 < O < 1.

+L pi.

log p) P

(

pim

1)

2 (m)0

1- + . P

208

11 Cyclotomic Fields of Class Number One

PROOF. We shall use induction on Tc(m). By Lemma 11.2, the lemma is true for z(m) = 0. Assume it is true for Tc(m) and then replace m by mq, with q

(q, m) = 1 (the cases mq 2 , etc., are obtained by varying j). We have jmq

1

imq

1

1n

n=1 (n, mq)= 1

n= (n, m)= 1

=

1 im 1 q n =1 n (n, m)= 1

log p p ) pFit 1

+ log(jm q) +

1)

pl.P

z

log q q

-

27r ( m ) 0

P Jmq

1 log — — + log( jm) + L

= + log(jmq) +

+

1\

r( m ) 01

p)

jmq

log py i

pim

p—

p1)

1

2")(6 — 0') pm! Since —2 < 0 — 0' < 2, the result follows. Lemma 11.4.

EN log j j

<0.11

=1

+ i(bog N) 2 for N > 1,

and log j

i=i PROOF.

<1(log N)2 for N

21.

A calculation shows that 21

log)

. < i(log 21) 2.

i=1

Since (log x)/x is decreasing for

x > e,

log j f log x j

22

21

dx = Ylog N) 2 — log 21)

X

The second part of the lemma follows easily. The first part follows from a calculation of the cases N < 21 (the worst case is N = 3).

209

§11.1 The Estimate For Even Characters

We now return to the estimation of T(N). For N > 21,

T(N)

log p) n ( (1)(m)72 _ _1) + 2(I)(m) i ! (7 + log( jm) + 1 12 Pi m j= 1 j \ pImP — 1 1 1;1 1 ,n \I. 20(m) N 2 7(m) 0 71) N x---,1 — log(j/N) 1 (1) +o + E . + N m A. 11 1 + j/N N m • 1 rn.12 (

< 0(1;172 + °(m) (log N)2 11 ( - -1 ) -m PIm P

20 (m)(7 + log m

log p)(

-- -

1

y + log N +

pImP », , m , r i _ log x dx 00) + 1P ) + ‘ ` 3m m Jo 1 + x (use Lemma 11.2 for the third term; use 2'") < m and fourth term)

n

_ _p) 1

Pm

j -2 7r2/6 for the

log p ) < (1)(m) 7r2 + [(log N) 2 + 2(y + log m + L — 12 pl.P da m \ Tr2 1) 2 ddin\ X 11 ( 1 + o(1) + m 12 3m Plm

+ log N)]

(use 0(m) = 41(1 — 1/p). Also, the integral is easily seen to equal 1 — ± —•= 6— r2 (6)). This is our estimate for T(N). We now estimate To (N):

To (N) =

1 2

= ((,) + log N + =

log p) n f 1

log m + L

pImP 1 1 P11 , log p)2 n i 1 + log N + log m + L pl.

P

pim

1\ 4. 21") 0 \ 2 Nm p 1 2 ± 0 (log N\ N P)

Therefore

T(N) — To(N) < 0(m)E2 + (P(m)7E2 12 3m + (7 2 — log ( m±

50(m)E2 + o(1) 12m »5 (m\Tr2 57 2 j + — + o(1) < 12 12 +

log p ) 2)

1)2

11 ( 1

plm

210

11 Cyclotomic Fields of Class Number One

(it is rather amazing that the coefficients of both (log N) 2 and log N disappear. It would have been easy to get rid of just the (log N) 2 term, but that would not suffice. This is why we called this estimate "delicate" at the beginning of this section). We now obtain

E I L (1 , x.)1 2 = iim N —.

(T(N)

—

To(N))

cc

+

+ 57r2

— 12 _(

12

0(m) — 2) (1.7) if 0(m) 220. 2

By an inequality at the beginning of this section,

11L(1, z.)

< (1 . 7)(4) (m) — 2)/4 .

We record this for future reference. Lemma 11.5. If 0(m)

220

then

ri

L(1, z,n)

< ( 1 . 7)((m) — 2 )/ 4..

E

x even

x# 1

§11.2 The Estimate For All Characters We now need to estimate 11 L(s, zni) from below, where z runs through all nontrivial characters mod nt We continue to use imprimitive characters. Surprisingly, we first need an upper bound. Lemma 11.6. If 0(m)

20 and Is — 21 '4

fi

x#1

then

L(s, x m) <

PROOF. By the arithmetic-geometric mean inequality,

n x

1

1,(0.) _ 1) L(s, x,,) 2

<

1

oon)

E 'us, z .)12. 1

— 1 x#

211

§11.2 The Estimate For All Characters

Let S(u, ;Cm) — for z 0 1,

<<x(n) =

Em
if u > m and z 0 1). Then,

E Zm(n)

L(s,

n=1"

= n=1 E 1 Zni(n) + xm(m)(m - s — (m + 1) - s) ns 1 )1( ( 11

± (41(0 ± Zm(ni

± 2 ) -s)

1) —s

+•••

(this is just partial summation) m-

E Z.(11) + s f s(u, z,n)u -5-1 du.

n= 1

118

Since I S(u, xm)I 0(m)/2 (but see Lemma 11.8), the integral converges for = Re(s) > 0, so by analytic continuation the above holds for a> O. Also, m—1

I L s, x . ) 1

xm(n)

ns

(

n=1

f oe u -a -1 du

2 + isi çb(m)

m—1

isi ( m) m . o- 2

n=1

Since Is — 21 <1, we have a > 4 and I s I/a < 2. Therefore m- 1

I Us, ZJI

n=1

x m (n) n5

The triangle inequality says that for real

2

a1 , b,

((al + b1) 2 + (a2 + b 2 ) 2 + • • •) 1/2 < (c4 + ai + --)1/2

( bi + 11.

In the present case this yields

E

x#1

x.)1 2

)1/2

2 ) 1/2

m—1 Zin( 11 )11— s n= 1

(

z (Om) \ 2 111-2/3 x

2)1/2

m—1

+ (0(m) — 1) 1 / 2

Zin( 11 )11— s n=1

) 2 ) 1/2

° (m) m -2/3 . 2

The first term is the square root of m-1

m-1 m-1 allx n=1 n'=1

OE z m (u)x,,,(n') = 0 if n ters). Since a > 4, m—

Zm(n)Xm(0/ 2- 10 —s = 95(m)

n=1 (n, m)= 1

if mod m, since we are using imprimitive charac-

1

ao /1-26 <

n= 1 (n, m)= 1

n -2a

du = 4.

< ± 1

212

11 Cyclotomic Fields of Class Number One

Putting everything together, we obtain 1 1(0(0

x

-

1)

fl L(s, x„,) 2 1

1

1 (4000/ 2 ± (002)

0(m) —

< (( 0(44(m) 1 ) 1 /2 + 4012) < 0(m)

Pen) -2/3) 2

1) 1 / 2 (

2

(m) 2/3)

if 0(m) > 20. El

The lemma follows easily. The next result uses the upper bound to get a lower bound.

Lemma 11.7. Suppose (a) f (s) is regular and satisfies I f(s)1 M in the disc Is — 21 < t (b)f (s)(s) = Inc°, l api n' for Re(s) > 1, with a l 1 and a„ > 0for n > 2, (c) 26/27 a < 1, and (d) f() > 0. Then

1(1 —

f(1)

Let F(s) = f(s)((s). Then F(2)

PROOF.

al

1 and

a„(log n)jn -2

(— 1)F(2)

0,

n=1

SO CO

F(s) = E b/2 — s)i i=0

(for Is — 21 < 1)

with

bc,

1

and b

.

0

for ./ > 1.

Also,

f (1) = E (bj — f (1))(2 — s)J for Is —21 < 1. S—1 j=o Since the left-hand side is regular throughout the whole disc 1 s — 21 <1, the right-hand side must converge in this disc (and equal the left-hand side). A short calculation shows that for Re(s) > 1, F(s) —

1 s

1+1

= = ((s). m

s fi

u — [u]

us+

du

m ml s (m +1 I y ) (

([u] = greatest integer < u) (1 - s — 2 - ) + 2(2 - s — 3 - ) + • •

213

§11.2 The Estimate For All Characters

Since both sides are regular for Re(s) > 0, s s. On the circle Is - 21 =

is_li

1, the equality holds for these

±i±,si

fi uo-+

1

du

< 3 + 1 + < 6. °-

(hence for Is - 21

Consequently, for I s — 21 =

f (1) < s-1

F(s)

f( 1 )6M+ < 9M. Is - 11

Therefore

lb; - f

=

ds f (1) s - lj (s - 2)i+1

(2rci fis-2=43 1/

3)j < 9M( . 4 With a as in the statement of the lemma, and for A > 0, we obtain co

f (1) - 1

F(a)

A

=o

(b - f (1))(2 - ay -

9 M (1)-1(2 - ay j=A+1

A

A±

• Eo (b; — (1))(2 _ ocy _ 9A4 (1.(2- -(2ce))- c,c) • bo — E f (1)(2 - oc)i - 32MGY j= A

0

(since b

0)

>1 Since f (a) have

f (1) a -1

(1 - (2 - a) A ± 1 ) - 32M(4)A .

0 and “oc) < 0, F() < O. Therefore, after some rearranging, we

f (1)

(2 - a) A +1 > 1 - 32 m()A 1 -

Let A = (note that M

[(log 64M)1 +1 log(4)

f (2) = F(2)g(2) 6/n 2 , so A > 0). Then f (1)

(2 - a)A +1 > -12 • 1-CL

214

11 Cyclotomic Fields o f Class Number One

Since log(4) > i, A < 1 ± 4 log(64M). Therefore (2 _ a )A — 1 < (e l -i)A -1 < (64M) 4(1- "). Also,

2(2 — oc) 2(64)4(1- ") < 2(2-8 )6 4 2 4/ 27 27

4.

<

Putting everything together, we obtain the lemma. Lemma 11.8 (Polya-Vinogradov). Let x Let S(u, x) = 10 <,,,
0

1 be primitive with conductor f.

I S(u, ;01 < fli 2 logf. (If x 1 is imprimitive mod m, the result holds with 2m 112 log m on the right. See Ellison [1 ], p. 344.)

PROOF. Let ç =-- e 21riif and let TOO = D..: z(c). By Lemma 4.7,

— / Ac)' .

Z(n)T(Z)

We may assume u is a positive integer, so

x) = E x(c)0..çn
T(c)s(u,

cr.

C

1

. t4c

E AO

=

C

Note that iff/2 e Z then x(f/2) = O. Therefore f—1

I TOO II SO49 )01

Z c=1 c# f /2 [(f — 1)/ 2]

f -1 sin(nuc/f) E sin(7rc/f ) c=1

2 Z c= 1

c# f /2

But sin x > 2x/7r for 0 < x < 7r/2. Therefore [(f — 1)/ 2] f

I T(x) II S(U,

)01

2

E c=1

1

--, = .4

cC

-C

We claim that [(f—l),21 1

E

c =1

— c

<

log f,

for f

3.

1 sin(7rc/f )•

215

§11.2 The Estimate For All Characters

It clearly suffices to prove the claim for oddf. The inequality holds for f = 3. Suppose it is true for f = 2n - 1. To change to f = 2n + 1, we add 1/n to the left and log(2n + 1) - log(2n - 1) to the right. Since

log(2n + 1) - log(2n - 1) = log(1 +

1 ) log(1 2n

- 2( 1 + 1 ( 1 ) 3 + 1 ( 1 ) 5 + 2 n 3 2n 5 2n 1 > -, n the inequality still holds. This proves the claim. Therefore

I TOO II S(14, )01 < f log f. By Lemma 4.8, I TOO I =-- f 112 , since x is primitive. This proves the lemma.

0

Lemma 11.9. If x is a primitive nontrivial character of conductor f and 1 > a > 1 - 1/4f, then

I L' (o- , PROOF.

x) I

(1.3)(log f )2

.

As in the proof of Lemma 11.6, we have for a = Re(s) > 0, f-1

L(s, x) =

E x(n)n - + s n=1

ccc

J S(u, x)u - s - 1 du.

Differentiate: f-1

L'(s, x) = -

E Anxiog n)n -

n=1

+ I S(u, x)u - 1 (1 - s log u) du.

By Lemma 11.8, f—1

I Oa, )01 <

co

E (log n)n - a +

f 112 log f

n= 1

(a log u - 1

a log f - 1 >

I Oa, ;01

f1

J

14 - 1 (0- log u - 1) du

log 3 - 1 > 0). Therefore f1 log n n=1

n

_ + f 112 '(logf) 2

f "f(0.11 + Y1ogf) 2 + f

2 (log f

) 2 )

(by Lemma 11.4)

< (1.3)(log f ) 2 . This proves Lemma 11.9.

LI

216

11 Cyclotomic Fields of Class Number One

Lemma 11.10. If x is a primitive quadratic character of conductorf, then

1 1 — 74 .

L(a, x) 0 for a

By the analytic class number formula (see the discussion following Theorem 4.9), PROOF.

2h log

x even,

Nff

L(1, x)

27rh

Z odd,

where h and E are the class number and fundamental unit of the corresponding quadratic field; and w = 2 iff 3,4; w = 6 iff = 3; w = 4 iff = 4. If x is real, f > 5, so a + b f 1 + .\/3 • 2 2

E

Since h > 1, we obtain, for all f,

L(1, x)

(0.96)f

2.

If 1 > a > 1 — 1/4f;

L(a, x)

L(1,

— (1 — a) Max I

Z)i

1 (1.3)(logf) 2 > 0. (0.96)f ' — 47 This completes the proof of Lemma 11.10.

Lemma 11.11. If x is a quadratic charactermod m, then

L(a, x„,) PROOF.

0 for a > 1 — 114M.

L( 0", Z,n) = L(a,

H (1 — x(P)) > o p

P6

I ni

since 1 — 1/4m > 1 — 1/4f.

Lemma 11.12. If 0(m)

20, then

[I L(1, xn,)

z *1

1

1 6 m0( m) 11 2

We shall use Lemma 11.7 withf(s) = 11, 1 L(s,x, i) and a = 1 — 1/4m. M is given by Lemma 11.6. Clearly (a) and (c) are satisfied. Since f (s)C(s) is the Dedekind zeta function of CD((m), with the terms removed which have a PROOF.

217

§11.3 The Estimate for 11;,

factor in common with m, we find that (b) holds. If x is real-valued, hence quadratic, Lemma 11.11 implies that L(Œ, xm) > 0. If x is complex then

L(a, xm)L(a, x m) = I L(at, xm)1 2

0.

Therefore f(Œ) > 0, so Lemma 11.7 applies. We obtain 1(1 -7 )0(M) ")12m . 4 4m

fl L(1, xm) x* 1 The lemma follows easily.

El

§11.3 The Estimate for 17,; 220 then

Lemma 11.13. If 0(m)

n L(1, X)

1

>

(1 . 7)( 2 – (15(m))/ 4 .

— 1 6MO(M) 1/2 '

x odd

PROOF. Lemmas 11.5 and 11.12.

0

The following lets us return to primitive characters. However, the estimate is not good enough to be of use for our purposes.

Lemma 11.14

n n (i ,

x odd

_

p )- 1 Ap) > e - 70(m)/ 24

.

iilm

Write m = nip m'p , where mp is a power of p and p)(m'p . Then z(p) o 0 fx I m'p . There are at most 1-0(tn'p ) such odd characters. Therefore the product is at least I —) ( 1 1 2 1 4)(m; ) . PROOF.

11 ( 1 ±

pl. P Take the logarithm (the minus sign reverses all the inequalities):

1

— 7)

1 E owp) 2p, m p

1)

1 --

z oon) log (1 + p-

- pi.,

>

0(m)

t

I

I

+ Ei n,

(1) -11 )P)

p

p>2 >

> This completes the proof.

00n) (1 2

4

+ p>2 E (p

0(n) (1 1) 2 4 3

1 -

1)p

)

7 0(m) .

24 El

218

11 Cyclotomic Fields of Class Number One

Proposition 11.15. If 0(m)

220 then 1 log dm - (1.37)0(m).

log h;

where dm is the absolute value of the discriminant of 0(Cm). From the class number formula (in particular, see the discussion preceding Theorem 4.17), PROOF.

log h; = I log(dn.') + log

d:

+ log Q -

jj L(1, x) + log w

x odd

0(m) log(27r),

2 where d: is the discriminant of CP(Cm) + , w = in or 2m, and Q = 1 or 2. By Lemma 4.19, dm > dm (d:)2 , so d: Using Lemmas 11.13 and 11.14, we obtain log h: > I log dm - log(16md)(m) 1 / 2 ) +

2 - 0(in) log(1.7) 4

On) 2 1og(27r) - 2'4 0(m) + log in i log dm - (1.37)0(m),

220.

if 0(m)

1:

This proves the proposition.

Since log dm - (k(m) log in (Lemma 4.17), it is clear that we are almost done. Also, note that we find that log h: grows at least as fast as predicted by the Brauer-Siegel theorem (see Theorem 3.20). It remains to estimate the discriminant. If in is a prime power, this is easy, but for composite in the estimates are harder. By Proposition 2.7, log dm

On)

= log in-

L

log p

plm P

,.

-

1

We can obtain an easy estimate as follows: If 21m then the right-hand side is at least

1 m - log 2 - -,, 4

1,

1

(m)

log p > log m - log 2 - jrlog log -i,

P1( 71/ 2 )

> logO 2 2. If in is odd, we obtain 4 log tn. Therefore

log h,T,

(-g-1 log(11127-) - 1.37)0(m) > 0

if in> 116000.

219

§11.3 The Estimate for /1,7,

So, in principle (i.e., with unlimited computer time), we are done. However, with a little work we can improve the situation. Of course, we could make some progress by estimating dm more carefully. But let's look back at the proof. The major terms are log(d.), which cannot be changed; /0(m) log(1.7), which comes from Lemma 11.5; -27,1 0(m), from Lemma 11.14; and 10(m) log(2n), which cannot be changed. If m is a prime power, then the estimate of Lemma 11.14 is very bad, since the left side is 1. Therefore, we bypass Lemma 11.14 and replace Proposition 11.15 with the following. Proposition

11.16. Assume 0(m) > 220. log hm-

If m is a prime power then

— (1.08)0(m).

If m is arbitrary then

— (1.08)0(m) —

log hm- >

1

10(M) pni

op p 2) •

PROOF. In Lemma 11.14, all the factors are 1 if in is a prime power. If in is arbitrary, we have, as in the proof of the lemma, log

fl

Al -

fl (i —

P

xodd Pirn

0(m) ( 1 1

>

2

(/)(4) 2 if

2 Im

+

1 p ni I

\

(p — 1)p

p>2

(omit the term for 2 if in is odd) 1

= Y(M)

2 •

p I m Cf/ GP

)

Using these estimates in the proof of Proposition 11.15, we obtain the result. Corollary 11.17. If 0(p")

220, then h;„ > 1.

PROOF. We have log(d) = 45(Pa)

log(pa)

log p P

log(pa) — log 2

1

> log(220) — log(2)

4.7.

Therefore log(Ç) (1(4.7) — 1.08)0(p") > 0. This proves Corollary 11.17 (in fact, we obtain hp-c, > 109).

0

Corollary 11.18. h;„ = 1 if and only if pa is one of the following: an odd prime p < 19, or 4, 8, 9, 16, 25, 27, 32 (one could also include pa = 1). PROOF. We know that 0(p") < 220. The table in the appendix yields the answer. LI

220

11 Cyclotomic Fields of Class Number One

Our strategy is now as follows: We know that hi,„ = 1 for at most those values listed in Corollary 11.18. By Exercise 4.4, or by Lemma 6.15 plus Theorem 10.1, h„Ihm if n divides m. Therefore, if hm = 1, all the prime factors of m are less than or equal to 19. In fact, if p" divides m, then p° is on the above list. This gives us finitely many possibilities, each of which can be checked individually. We give some of the details: 19. The table in the appendix shows that h; > 1, hence hm > 1, for in = 4 x 19, 3 x 19, 5 x 19, ... , 13 x 19. Corollary 11.18 takes care of 19 2 . However 0(17 x 19) = 288 > 256, so is not listed in the table. But Proposition 11.16 applies. We have log dm (km)

= log(323) —

log 17 16

log 19 18

> 5.4. Also 1 7

1 0(2p2)

1 ( 2

) 1 1 < 0 004 + x 17 18 x 19

Therefore log h 3-2 3 > (1(5.4) — 1.08 — 0.004)0(323) > 0. Consequently, h 19„ > 1 for 2

n > 1, so we may henceforth ignore 19.

17. From the table, 17. 7 „ > 1 for 1 < n < 17 (n o 2). Corollary 11.17 implies 17 89 > 1. Therefore h 17„ > 1 for n > 1. 13. From the table, h 3 > 1 for 1 < n < 13, so h 13„ > 1 for n > 1. 11. We obtain 11 3 = 1 and hT1.4 = 1, but 17 1p > 1 for 3 < p < 11. So if m is a multiple of 11, m = 11 • 2° • 3 b. Since Iç9 > 1,h 32 > 1, and hi8 > 1, we must have in = 33 or 44. 7. We have h. -8 = 1, h 1 = 1, and hi.5 = 1, so we have only eliminated multiples of 49. Next, consider 56, 84, 140, 63, 105, and 175. Only 84 gives = 1. Since 2 x 84, 3 x 84, and 5 x 84 are multiples of numbers already eliminated, we may stop here. 2,3, 5: These are treated similarly. We have proved the following. Proposition 11.19. If hm = 1 then in is one of the numbers given in the statement of Theorem 11.1. All these values have h; = 1.

221

§11.4 Odlyzko's Bounds On Discriminants

Remark. We have not yet calculated h;„ so we have not proved the converse of Proposition 11.19.

Masley proved that nIm h, I h; . Hence he was able to work exclusively with hni- in the above and show that Theorem 11.1 lists exactly those at with h; = 1.

§11.4 Odlyzko's Bounds On Discriminants The results of this section will be used in the next section to compute h„+7 . However, as we shall see, they are also useful in other situations. Let K be a number field of (absolute value of) discriminant D and degree n ri + 2r2 . Let K(s)

=

11( 1 - 1"-s)-1

Z(s) = —

(Dedekind zeta function of K),

d CK(s) = — — log C K (s), us Cic(s) d

Z 1 (s) = — — Z(s), ds F'(s) tli(s) = (F = gamma function). F(s) For a> 1, log NY

Z(a) =

N Ya — 1

>0

and since NY' increases with a, Z i (o) > 0.

The estimates we need will arise from the following.

Theorem 11.20. Let a = a>

—

— 0.28108 .... Suppose if, a > 1 satisfy

5 + 12a 2 — 5 and 6

a>

+ cus.

Then

log D

r 1 (log — (0) + 2r2(log 2m — tP(a))

6.) + r2 11/'(a)} + 2z(c) 14- 0/ G + (20- — 1 ){-11 + (2u — 1)Z i (ii) —

2 c

2 a —1

2u — 1 2u — 1 02 • (if

222

11 Cyclotomic Fields of Class Number One

We postpone the proof in order to show how to use the theorem. The idea is to fix n, r 1 , and r2 , and find optimal choices for a and 5% For small n it is best to take a -

5 + \/12c 2 — 5 6

The best choice for a will satisfy a > 1 + ow- for these cases. Fortunately, Odlyzko has determined in many cases the best value of a. We shall give the results below. For any a, a we obtain an estimate of the form log(D 1 /n)

2r 2 11 -1 A + — B — —

n

n

with A, B, C > O. Therefore, estimates for D lin for K of a given degree n are also valid for fields of higher degree, provided the ratios ri /n and r2/n are held constant (e.g., K is totally real, or totally complex). We also have, for any a > 1 and for any admissible A +

log(D 1 /n) >

2r,

B + 0(1)

where A = log it

+ 2c — 1 (62) 2 4

0 (a)

B = log 27E — 1/i(o-) + (2c — i)i'(a). We may let a be arbitrarily close to 1 and let other inequality). We find (50.66)nl/n(19.96)2r2/n

a =1+ +0

of (this satisfies the

).

We give an application. Let 1/ 0 = U( \ d) be a real quadratic field. Let H 1 be the Hilbert class field of U(d) and inductively let Hi+ 1 be the Hilbert class field of Hi . Does this class field tower stop (i.e., Hi = Hi+ = • • • for some i)? Equivalently, can CP(./) be embedded in a field of class number 1? (Exercise 11.4). Golod and Shafarevich have shown that for d = 3 x 4 x 7 x 11 x 13 x 19 x 23, the tower does not stop. Suppose that d is the discriminant (not disc.) of CP( d) and d < 2500. Since H1/G(d) is unramified, Di =

Q ")1

(see Lemma 11.22). Therefore, if ni = [111 :G],

d 112 < 50.

223

§11.4 Odlyzko's Bounds On Discriminants

If ni -> co then lim inf

> 50.66,

contradiction. So the class field tower stops. It can be shown that 1 - log D

y + 1og(47) + 1 - 8.317302n -2 / 3 (K totally real),

1 - log D n

y + log(47) - 6.860404n -2 13 (K totally complex).

(See Poitou[I]; also see Poitou[2]). These yield

(K totally real),

/" > 60.83 - o(1)

Du" > 22.38 - o(1)

(K totally complex).

Even better estimates are available if one assumes the generalized Riemann Hypothesis. We now give the table we promised. Keep in mind that an estimate for a given n works for a larger n; so for n = 18, for example, use the estimate for n = 15. Lower bounds for

De for [K: 0] > n

K totally real

K totally complex

D""

10 15 20 30 40 60 100 120 180 240

1.84 1.57 1.44 1.32 1.26 1.19 1.14 1.12 1.095 1.08

10.00 13.58 16.40 20.57 23.55 27.61 32.25 33.75 36.76 38.62

2.26 1.85 1.66 1.46 1.37 1.27 1.19 1.165 1.13 1.11

5.53 7.06 8.11 9.68 10.77 12.23 13.86 14.38 15.40 16.03

This table is copied from Odlyzko [1]. For a more comprehensive table, see Odlyzko [4] and Diaz y Diaz [1]. To show how the various terms contribute to the estimates, we give the calculation of the lower bound for n = 10 and K totally real. It is hard to estimate Z(0-) and Z i (a); but recall that they are positive, hence may be ignored. We have a = 1.84 and 5- = 1.828 .... Writing the terms in the same

224

11 Cyclotomic Fields of Class Number One

order as in Theorem 11.20 (leaving out terms with r2 = 0, and leaving out Z and Z 1 ), we have log D > 10(1.14 — ( — 0.72)) + (2.68)( 4Q 1.88)

— 1.09 — 2.38 — 0.80 — 3.91 = 23.02. Therefore D1/10 > e 2.302

9.99

(we lost a little to rounding errors). As one can see, most of the terms make significant contributions to the final answer. PROOF OF THEOREM 11.20. Let D

s ri

r(-2 )

g( S) = ( 2t ri(2 702r2)

r(5)r24(5)s(1

— s).

Then g(s) is an entire function of order 1 (see Lang [I], p. 332) and satisfies

g(s) = g(1 — s) (see the discussion preceding Corollary 4.6). By the Hadamard Product Theorem (see, for example, Lang [7], p. 253), there exist constants A and B such that

(1

g (s) = eA+Bs n

l eslp,

where p runs through the zeros of g(s)(= zeros of (K(s) with 0 < Re(s) < 1) counted with multiplicity. If g(p) = 0 then g(1 — p) = 0 (if g (+ ) = 0 then it has even multiplicity), so we pair p and 1 — p to obtain

g(s) =

eA + B s

11

) esip(1- p)

1 — `• 1

1—p

P, 1— )9( = exp(A + Bs + s E

1 P( 1

P))

H P (1 —

ps)(

I

19,1—

1 —s p)

(since g(s) is of order 1, E 1/p(1 — p) converges; therefore the product converges also; hence the rearrangement is easily justified). Since, for any p and s,

(1 -

P

s

1—p

= (1 \

1

s )( 1 P f\

1— 1—p

we have

1=

g(s) g(1 — s)

exp(B(2s 1) + (2s 1)

E

1 \ PO — P)) .

225

§11.4 Odlyzko's Bounds On Discriminants

Therefore B = — Z 1/p(1 — p), hence

g(s) = eA H (1 —1(1

S

1 — p) •

Recalling the definition of g(s) and taking the logarithmic derivative, we obtain

r i (s 1 1 D — 1 r1 log n — r2 log 2m +— 4/ —) Ilog + r2 1(s) — Z(s) + + ss—1 2 2 1

= l p , lp(S

p

+

1

• s — 1 + p)

This may be rearranged to yield log D = r i (log

s

7—

+ 2Z (s)

2r 2 (log 27 — tLi(s))

22

s

s—

+2 Z

1

p, I

1

(.5 — p

+

1

\

s—1+

This is valid for all s (except s = p, 1 — p, 0, 1). Differentiate with respect to s and let s = to obtain

2

—f, ii/G) + 2r2V(a) + 2Z1(a)

—

2

6. . 2

(6..

2 _

02

—1 —1 \ = 0. p(ar' — P)2 + (a — 1 + 02 ) (

Multiply by (2o- — 1)/2 and add the result to the previous equation, with S = CT:

o-

log D = ri (log m — 10) + 2r2 (log 2m —

+ (2a — 1){r61 V(1) + r 2 1V (a)} 2 cr — 1

+ 2Z(c) + (2c — 1)Z1(&) 2a 2c — 1 6..2

1 2c — 1,.) 1 (a _ 1)2 + - p, p (a _ p + a — 1 + p) .

—1

— ( 2' — 1 ) Z / -

— P)

2+

—1 (G. — 1 + 02 . -

226

11 Cyclotomic Fields of Class Number One

It therefore remains to show that

1 a —p

+

1

1\ v (a 2)

a — 1 + p)

—1 —1 — p) 2 (er- — + p)2 ) .

Since g(p) = 0 g(p) = 0, we may pair the terms for p and 15, which amounts to taking the real part of each side of the above inequality. Let p = x + iy. It suffices to prove the following. 32

Lemma 11.21. Let a =

. Suppose a, a> 1 satisfy

1 + X) 2

+

y

2

Therefore, it suffices to show (o7 —

2 1 + X) 2

y2 +

for < x < 1 and y Case

y2

y2

x)2

(y 2 + (6 —

_ 1 ± x )2

(y 2 + (6-- — 1 + x)2)2

.X) 2 ) 2

(*)

O.

I.ya—x(a - -1+ x).

In this case the right-hand side of (*) is negative, so the inequality is trivial. Case1I.6—x
A = (a — 1

x )2

5(6 x)

2

,

B = 1{17 Cri — x)4 — 14(6 — x)2(cr — 1 + x)2 + (a — 1 + x)4 1 ot 2 x)2 V(0- — 1 + x) 4{ (a —— 1 + )02 1 A) 2 _ B C ( y2 = y4 A y2 ± _ x) 2 (0., _ 1 ± x )2

}{

— )0 2

(c — 1 + x)2 x)4.

1

1772 }'

227

§11.4 Odlyzko's Bounds On Discriminants

A calculation shows that (5. _ x) 2 (y 2 ± _ x) 2 ) 2 y2

2 (a — 1 + x) 2 + y2

(( o.

_

y 2)(y2

(a.

x) 2 y •

We must show C > O. If A < 0, the second expression for C yields C > O. Suppose now that A > 0, which means that

(a — x )2 < (i — 1 + x) 2 . Since 17 2 < 5, 1

)

17a

Since a > 1 + ow by assumption,

hence _ x) 2 > Œ 2 (0. _ 1 ± x )2 .

The second expression for B yields B < O. The first formula for C shows that C > 0, as desired. Case III. O — 1 + x < y.

The right-hand side of (*) is bounded above by

2y2 —

(a — x) 2 — (a — 1 + x )2 (y 2 +

_ x) 2 ) 2

A short calculation yields

2y2

2 (0.

(0 x) 2 _

1 ± x )2 ± y2

(,2

y 2(5(6- — x) 2 ((cy

(a —

1 ± x) 2

±

_ 1 ± x) 2 _ x) 2 ) 2

1 + x)2 — 2(u — 1 + x) 2 ) y 2 y 2 ± (6. _ x) 2) 2

We must show the numerator is nonnegative. Let f (x) = 5(0 — x )2 + (a — 1 + x )2 — 2(cs — 1 + x) 2 . Then f ' (x) = 8(x —

a)

+ (2 — 4u) < 0,

for x < 1. Therefore f (x) > f(1) _ 5(0 0 2 + 5 2 2 0.2 > 09

228

11 Cyclotomic Fields of Class Number One

since ô-' > (5 + \/120-2 — 5)/6. This completes the proof of Case III, hence of Lemma 11.21. 11 The proof of Theorem 11.20 is now complete.

El

§11.5 Calculation of h m+ The estimates given in the table of the previous section may be used to calculate h: for small m, in particular for those in listed in Theorem 11.1. The main tools we need are the following two lemmas. Lemma 11.22. If L/K is an extension of degree n in which no finite primes ramify, then

D L = Dr K' where DL and respectively. PROOF.

DK

are the absolute values of the discriminants of L and K,

A well-known formula (see Lang [1 ], pp. 60, 66) states that

DL = IYIK W Li K where N.L/K is the norm of the relative different. Since no primes ramify, El .gLiK = (1). The result follows. Lemma 11.23. Let B(n) be the lower bound for D i ln for totally real fields of degree > n (as given in the table of the previous section). Let d: and h: be the discriminant and class number of O(,n) .+- . If
(14(m)\ 2 )

then

h: < h.

Let H be the Hilbert class field of Q(C.) ± , so H/Q(C,,) + is an unramified extension of degree h: , and PROOF.

=

[H:

0] =

By Lemma 11.22, Deli =

(ci:,- ) 2 / 0(m) < B (1 4) (M)) .

Therefore = nil < The lemma follows.

0

229

Notes

Since 2111,;,= 21 h„-, by Theorem 10.2, h: must be odd whenever h,T, = 1. Consequently, we only need to show li: <3. This may be done via Lemma 11.23. The value of d: may be calculated by Lemma 4.18 and Proposition 2.7, or by the conductor-discriminant formula. We give a few examples: m = 3, 4. CP(Cm) + = Q, so h: = 1.

5, 8, 12. UG) + = 0( m), so these class numbers may be calculated directly (via the analytic class number formula). in =

in = 7, 9. [0(C„,) + : 0] = 3 and only one prime ramifies, so 30:, by Theorem 10.4. Since h: is odd, h: = 1 or h: > 5. The discriminants are 7 2 and 9 2 . Both satisfy

(41) 1 / 3 < 10 = B(10) = B((Y)3). By Lemma 11.23, h: < If, hence h: = 1. in = 15, 16, 20, 24. These all have degree 4. The discriminants are 3 2 • 5 3, 11 2 , 24 • 5 3, 2 8 • 3 2 , respectively. The largest of these is dif-4 = 2 8 • 3 2 . Therefore (cri) 1/4 5... (2 8 . 3 2) 1/4 _ 4 3< B(10) = B(I • 4),

so h: < 1. Therefore h: = 1. in = 35, 45, 84. These all have degree 12. The discriminants are 5 9 • 710, 318 . 59 , 212 . -.6 . 10 .5 7 , respectively. The largest of these is 2 12 • 3 6 • 7 10 . As before, (d:)1/2 < (212 . 36 . 710)1/12

3 • 7 5/ 6 < 18 < B(30) = B(-1 • 12).

Therefore h: < 5/2, so h: = 1. The other values of m are treated similarly. So all the values of m listed in Theorem 11.1 have h: = 1. Since all of these have h,; = 1, and since Proposition 11.19 says that these are the only possibilities for h. = 1, the proof of Theorem 11.1 is complete. CI Remark. It is not always true that if h - = 1 for a CM-field, then h+ = 1. See Exercise 11.6. Nulls The papers of Masley contain several discussions of the results in this chapter. The estimation of h,; follows the method used in Masley [1]. For other methods, see Masley-Montgomery [1], Uchida [1], and Hoffstein [1]. The last paper applies to many CM-fields and does not rely on the factorization of the zeta function into L-series.

11 Cyclotomic Fields of Class Number One

230

For analytic estimates of h; and h;„, see Ankeny-Chowla [1], Lepist6 [1], and Metsdnkyld [4]. For a simple but accurate upper bound, see Carlitz [2]. For the calculation of hm+ , see van der Linden [1] and the papers of Masley. For examples of hm+ > 1, see Ankeny-Chowla-Hasse [1], S.-D. Lang [1], Cornell-Washington [1], and Takeuchi [1]. For Euclidean cyclotomic fields, see Masley [3], Ojala [1], and several papers of Lenstra. For more on Odlyzko's results, see his papers, plus Martinet [1], [2], Diaz y Diaz [1], and Poitou [1], [2]. EXERCISES 11.1. Use the Minkowski bound (Exercise 2.5) to show that D 1,n

e 212r2/n (6,2)ri 'n(

4

(1 + o(1)).

This is much weaker than Theorem 11.20. 11.2. Show that none of the fields 1(./ — 1), CP( \/ — 2), CP(.\/— 3), 0( \/ —7), 0(.\/— 11), (\/— 19), l(/-43), 0( — 67), 0(1— 163) has any nontrivial unramified extension. These are precisely the imaginary quadratic fields with class number 1, so we know that there are no such abelian extensions. The problem is therefore the other (not necessarily Galois) extensions. 11.3. (a) Show that if 211/ 9 then 8111 -9 (see the example following Theorem 10.8). (b) Show that h 9 = 1 (hence the class group of O(ç 2 ) is (Z/2,7) 3 by the example of Chapter 10). 11.4. Let K = Ho and let H i+ I be the Hilbert class field of H i . Show that the class field tower stops (H i = H 1 = • • • for some i) - > K is contained in a field of class number 1. 11.5. Let n divide In, so (1:1(„) g CI(: m). Let p be odd and let A n and A m be the p-Sylow subgroups of the ideal class groups. Show that the norm maps A,T, onto Conclude that h,; divides h„-„ except for possibly a power of 2 (Masley has shown that h; divides h). 11.6. Let (a) (b) (c) (d)

K = 0( — 1, .\/10). Show that K G ( 8 , 5) OG4o). Show that C(r,40)/e(ç8 , 5) is totally ramified at the primes above 5. Use Theorem 11.1 to show that 14 8 , \/3) has class number 1. Show that K has class number at most 2. In fact, use Theorem 3.5 to show that

the extension (1-1g8 , .\/)11‹ is unramified, so the class number is 2 and CP(C, 8 , \/3) is the Hilbert class field. (e) Show K + = 0(.\/10), which has class number 2. (f) Conclude that h- = 1 but h+ = 2 for K.

Chapter 12

Measures and Distributions

The concept of a distribution, as given in this chapter, is one that occurs repeatedly in mathematics, especially in the theory of cyclotomic fields. As we shall see, many ideas from Chapters 4, 5, 7, and 8 fit into this general framework. The related concept of a measure yields a p-adic integration theory which allows us to interpret the p-adic L-f unction as a Mellin transform, as in the classical case. Many of the extensions of the cyclotomic theory have used measures and distributions; see for example the work of Kubert and Lang on modular curves. For an approach to cyclotomic fields that is much more measuretheoretic than the present exposition, the reader should consult Lang [4] and [5]. In this chapter, we first introduce distributions and give some examples. We then define measures and give a p-adic integration theory, including the 1-transform and Mellin transform. We also give the relations between the present theory and the power series of Chapter 7. Finally we determine the ranks of some universal distributions, and consequently obtain a proof of Bass' theorem on generators and relations for cyclotomic units (Theorem 8.9). The second and third sections of this chapter are independent and may be read in either order.

§12.1 Distributions Let I be a partially ordered set. For technical reasons we assume that for each i,jeI, there is akeI such that k > i,k> j. Such sets I are called "directed." Let fX i ji

231

232

12 Measures and Distributions

be a collection of finite sets. If i > j we assume there is a surjective map no : X i —* X i . such that nii o ic 1 = nik whenever i j k. Suppose that for each i we have a function (Pi on X i , with values in some fixed abelian group, such that if ,

= RO (Y )

x

0,(y).

The collection of maps {O i } is called a distribution. EXAMPLES. ( 1) Let / be the positive integers with the usual ordering and let X. = Z/pt 1, with nii the obvious map. Fix a e Let Zpe

{

a mod pi, I, if y 0, otherwise.

Then {O i } forms a distribution, called the delta distribution. (2) Let / be the positive integers ordered by divisibility: i > j if j Ii. Let

X i = Z/iZ and icy :

Z/iZ

Z/jZ

y mod i y mod j. Let

Ci(a, s) =

z

mod i

n >0

be the partial zeta function, as in Chapter 4. Then {O i }, where 01(a) = s) is a distribution (with values in the additive group of meromorphic functions on C). (3) Let / be the positive integers ordered as in Example 2. Let X • = — Z/Z and let ic be multiplication by i/j. For k > 0, let B k(X) be the kth Bernoulli polynomial, as defined in Chapter 4. Let

233

§12.1 Distributions

where { } denotes the fractional part. To get an odd distribution for odd k, it is convenient to let =0

if k = 1.

Then {O i } forms a distribution, called the kth Bernoulli distribution. This follows from properties of Bernoulli polynomials. Even better, we know that Ci(a, 1 — k) = Bk({})

i

k

so the distribution relation follows from that of Example 2. One easily sees that the sets X i and maps no of Examples 2 and 3 are essentially equivalent: We have a commutative diagram Z/Z

2/0

1

/2

Z/jZ.

(4) Let I and X i be as in Example 3. Let C i be a primitive ith root of 1; we assume = (for example, with terrible notation, ç = e 2niii). Let

Since (i/j) -1 H (ça+bi — 1) =

O

— 1 (ifi

the (multiplicative) distribution relations are satisfied. But there is a problem. The function (P i takes values in C X u {0}, which is not a multiplicative group. We could allow monoid-valued distributions, but this causes problems with Theorem 12.18. It is more convenient to define a punctured distribution by omitting the relations with x = 0 in the defining relations for a distribution. The value Oi(0) may then be ignored. We could also let Oi(a/i) = I — 11 or Oi(a/i) = log I — 11. We then obtain punctured distributions (one multiplicative, the other additive) which satisfy

In other words, O i is even. Since Bk(1 — X) = ( — 1)kBk(X),

234

12 Measures and Distributions

the kth Bernoulli distribution is even or odd, depending on k (technical point: Since {—O} # 1 — 101, we also need the fact that Bk(0) = 0 for odd

k> 1). There is a special type of distribution, which will be studied in the third section of this chapter. Assume (f) is a function defined on C/1 such that 1 0 ( a\ + bj)

=

b =0

whenever a e 1 and i ii. Equivalently, if m e 1, m > 0, and y e e/l, then ç (y) = nix

=- y

0(x)

(let m = i/j, y = a/j). We call (/) an ordinary distribution. The distribution of Example 4 and the first Bernoulli distribution fit into this category if we let

= The main point is that if a/i = b/j then Oi(a) = (f);(b), so (f) is well-defined as a function on 0/1. The delta distribution of Example 1 does not arise from an ordinary distribution, even on ei,/4: 01(a) = 1 but 0i , l (pa) = 0 (unless a 0 mod p i + 1 ). Also, if k 1, the kth Bernoulli distribution is not ordinary. There is a second, equivalent definition of distributions. Consider the situation at the beginning of this section and let

X = lim 4-

Xi

(see the appendix for inverse limits). Since each X i is finite, X is compact. Let (f) be a finitely additive function on the collection of compact-open subsets of X. We shall show that (/) gives rise to a distribution. For each i there is a surjective (since each ni; is surjective) map

ni : X

X.

If a E X i then ij 1 (a) is a compact-open subset of X. All compact-open sets are obtained as finite unions of such t- 1 (a), as i and a vary (these sets form a basis for the topology of X). Suppose b e X i . For i j, 7r

1 1 (b) =

U

1 (a),

cie X i nii(a)=

b

and this is a disjoint union. Therefore

0(7Ei'(b)) =

rr,i(a)= b

0(ni- '(0),

so b 0(n1 1 (b)) satisfies the distribution relation. Conversely, any distribution {O i } on {Xi } gives a finitely additive function on compact-open sets of X.

235

§12.1 Distributions

Finally, we give a third formulation of distributions. A function f on X is called locally constant (or a step function) if for each x E X, there is a neighborhood U of x such that f is constant on U. Since X is compact, this means that f is a finite linear combination of characteristic functions of disjoint compact-open sets. In fact, f is a finite linear combination of characteristic functions of sets of the form ni- '(a). Call these characteristic functions x i , „. Let Step (X) be the set of locally constant functions on X. If (f) is a finitely additive function on compact-opens with values in a group W, then we may extend (f) by linearity to obtain a map

(f): Step (X) —> W If {O i } is the associated distribution, then a) = OP). Conversely, a linear function on Step (X) may be restricted to characteristic functions to yield a finitely additive function on compact-opens. In summary, we have the following one-one correspondences: distributions

finitely additive functions on compact-opens linear functionals on Step (X).

We now reinterpret the delta distribution of Example 1. Let U OE Z, be compact and open, and let a e 7Z. Let if a E U a( U)= {1°: if a çt U. Since ir

'(y mod pi) = y + piZ p ,

we see that 6,, corresponds to the delta distribution. Iff c Step (X), we have 6 a(f)

=

which is exactly how the classical delta function acts. There is a natural function on compact-opens of Zp , namely (/)(U) = meas(U), where meas is the Haar measure normalized by meas(Z p) = 1. We have

0(.1) + PiZp) =

1

so the associated distribution satisfies Oi(y mod pi) =

1 13`

236

12 Measures and Distributions

More generally, consider the spaces X i = l/il and maps 7rii of Example 2. In this case,

x

z/iz

=

11 i p ,

all p

where the isomorphism is obtained via the Chinese Remainder Theorem (/i fl zwizp). Again, we have a compact group, so we can let « U) = meas(U). This yields the distribution defined by Oi(y mod i) = 1/i. The Haar distributions will not be very useful to us when we develop p-adic integration in the next section. Consider the case of Zp As the sets y + piZ p become smaller (i oo), their Haar measures become p-adically larger. Clearly this is not desirable since a small change in a function could produce a large change in its integral. The distributions that will be of use will be those with bounded denominators, which we shall call measures. These will be studied in the next section. ,,

§12.2 Measures Let the notations be as in the first section. Consider a distribution {Oi }. Let (p be the corresponding functional on Step (X). For f E Step (X), denote

45(f)

f f dO. x

Assume that (1) takes values in Cp ( = completion of the algebraic closure of Op). We say that 4) (or c/(P) is a measure if there exists a constant K such that I 0(a)1

K

for all i and all a c X i . Let C(X, Cp) be the Cp-Banach space of continuous C,,-valued functions on X, where

11f11 =

sup 1 f (x)1. xeX

Then Step (X) (with values in

C„) is dense in C(X,

Proposition 12.1. If 4) is a measure, then

x

f d4): Step (X)

extends uniquely to a continuous Cr-linear map

Lx

f d4): C(X,

Cp.

Cr).

237

§12.2 Measures

PROOF. Since the step functions are dense, the map must be unique if it exists. Observe that if K is the constant used above and Xi ,„ is the characteristic function of the previous section, dç

K.

i(a)I

=

Since the absolute value is non-archimedean,

KII f II,

JE Step

(x).

If g e C(X, Cp) and {fn } is a Cauchy sequence in Step (X) converging to g, then

fx.

< K - fm IH O

in cO - x f„

as m, n -0 oc. Therefore, let xg

dO = lim Jfdc. LII

This has the desired properties, so the proof is complete. EXAMPLES. (1) Let a E

Zp

and let S a be the delta distribution. Then

fi

c1(5„ = f(a)

for f e Step (X), hence for f e C(X, Cp). (2) Let 0 be the Haar distribution on ZP = lirn Z/ pnZ. Then 0 is not a measure. What happens if we try to integrate anyway? Let f (x) = x be defined on lp • Recall that z„, „ is the characteristic function of a + p"Zp . Hence p" — 1

f(x) —

ax„(x)

p" P -> O

a=0

as n -+ oc.

also pn

But p ._

p.

E a meas(a + p n Zp) =

= a0

a

a v L., n 0

P

=

p" - 1 2

while p fl

a -- = 1

a meas(a + pnZ p) -> + - . 2

_

1 _

2'

238

12 Measures and Distributions

Therefore f x 61 4 is not well defined. This is why we require (/) to be bounded. However, it is possible to weaken this condition slightly (see Koblitz [1], p. 41). (3) If g e C(X, Cp) and dO is a measure on X, we may define a new measure di P = g (10

by

fx f dilf = f fg d4). x Clearly this gives a finitely additive linear functional on Step (X). Since X is compact, g is bounded. It follows that du' is a measure. Often we shall take g to be the characteristic function of a subset X' g X. We then write Sr f dO for Sx fg dO.

(4) If h: X —> Y is continuous and 0 is a measure on X, then we obtain a measure duP on Y by defining

Lf

du ji = 1 f(h(x)) dO.

x

This will allow us to obtain measures on the logarithm mapping.

lp

from measures on 1 + gp , via

The Bernoulli distributions are not measures. However it is possible to modify them. We treat only the case k = 1; the cases k > 2 are similar. Let (d, p) = 1 and let X n = (1/dnp+1Z)x . Then

X = lim X n '-- (Z/dpZ)x x (1 + pZ p) (if p 0 2; the modifications for p = 2 are left to the reader). We could also work with 7L/dpn +1Z, but the present situation fits into the framework of later results. Let c e Z, (c, dp) = 1. For n > 0, let c - 1 denote an integer such that cc -1 1 mod dp 2(n + 1) (the 2 in the exponent is for technical reasons: it is used to obtain (**) below; probably it can be avoided). Alternatively, c e X in a natural way, so let c -1 be the inverse of c in X and then reduce mod dp2(n + 1) when needed. For xn e X, define

E(x) = B1({ dpn+ xn 1 = f x n 11 ldp n + f

cB,

c lx 1n dpn+

cf c'x n 1 ± c — 1 . ldp" +lf 2

239

§12.2 Measures

It is easily seen that E, is a distribution. Since

f c - l x„1 7 c ldpn +1 f 6

Ç x„ 1 dpn +1 f

E(x) E Z p , so E, is a measure. If x is a Dirichlet character of conductor dpm for some m > 0, we may write xco -1 = Bik, where 0 is of the first kind, and ik is of the second kind (see Chapter 7). Then B is a function on (Z/dp Z)x and tp is a function on 1 + pZ p . Therefore we may regard xco -1 = Otp as a function on X. Also, <x> may be regarded as a function on X; it is just the projection onto 1 + pZ p . Theorem 12.2. Let x have conductor Om with (d, p) = 1 and m > O. For S E Zp ,

J

Za)'(a)s dE, = —(1 — x(c)s +1)Lp(—s, x)

(7/ Zr dp x (1 + plp)

We shall show later (Corollary 12.5) that the left-hand side is analytic in s, so it suffices to let s = k — 1, with k a positive integer. We may estimate a by b E Z on {x EXix b mod dp"}. We obtain the sum

PROOF.

dp" - 1

1 ZW -k(b)b k - 1 b=0 pt6

By Lemma 7.11, the term with ignore it. By the same lemma,

1 dp "

E z w - k(obk b

( b

c

dill

(C —

{c1 b

dp"

}+

C

— 1\

2 ).

1)/2 tends to 0 as n —* oo, so we may

(1 — zw _k(p)pk_ i )Bk,xco

def k -

TT

U.

The remaining term is the hardest to evaluate. Let

c -1 13 = bl + dpnb2 ,

with

0 < b l < dpn.

Note that

zw(b) = xco -k (c)xco - k (b i )

(if n _?: m)

and

bk --- ck(b 1 + dp"b2 )k -=- ck(bil + k dpnb2 bir 1 ) (mod

P2 n) •

Therefore L b

)(CO

k (b)b a

zw -k (c)ck 1 zw -k(b i )bil b

+ k dpnxw -k (c)ck 1 xco -k (b i )b 2 bki - 1 . b

240

12 Measures and Distributions

But b 1 runs through the same values as b, in a different order. Consequently, we obtain

zw _k(ock E zw -k (b

)b 2 bki - 1 1 1 E zw - k(b)bk (mod - pn). k dp„

k(c)c

(1 —

The remaining term in the original sum involves cfc - l b/dpnl = cbildp". We have

-

E zw -k (b)bk b

dp n b

E zark(ock - i(m. +

dPn b

( k — 1) dpnb

2 bi - 1 )

(by (**) with k replaced by k — 1) c

k

zw-koom

dpn

— (k — 1)c kzw -k (c)

w - k(b i )b bir 1 (mod

pi')

(by (*)).

By Lemma 7.11, the first term yields —

ZW - k (C)C k (1 — ZW

k

(p)pk 1 )13 k

xo

-

ktV

By the above calculations, the second term is congruent mod (1/k)pn to

- (k — 1)(1 - zw-k(c)ck) k—

„E

k dp -

k (b)bk

b

1 (1 — ZO) -k (C)Ck)(1

ZW - k (P)Pk 1 )Bk,

def k=

W,

as n oo . Addition of the relevant terms shows that the original sum approximating the integral becomes, as n oo, U + V + W = (1 _ zw -k (ock )(1 _ zw -k (p)pk= (1 — ZW - k (C)C k )L p (1 —

i ) Bk,

k, z)

= — (1 — z(c)k )L p(1 — k, This completes the proof. The reader probably noticed that there is a great similarity between this proof and that of Theorem 7.10. This is not a coincidence, as we shall see later. First, however, we note some consequences. If x 1, choose c so that X(c) 1. Then x(c)s 1. Otherwise s" = 1 for some N > 0, which

241

§12.2 Measures

implies s = O. Since z(c) # 1, we have z(c)° # 1, so the claim holds for all s. Consequently, we may divide by (1 — z(c)s). Assuming that the integral is holomorphic, we find that L p(s, x) is holomorphic. If z = 1, then 1 —
12.3. If m n mod pb-1 (p — 1), and m 0 0 mod p — 1, 13„, (1 — pm -1 ) — -- (1 — m

PROOF.

B" (1110d pn - i)—

n

then

ph).

Let d = 1, x = co', and s = m — 1. Then

(1 — cm)(1 — pm -1 ) 13m = —(1 — cm)L p(1 — m, of") = fa m -1 dEc . m zp' Since E, is Zp -valued, and am -1 0 an -1 mod ph,

L,';

a' dE,

f

2;

a' dE, mod ph.

Also,! — cm -_- 1 — c". Choose c so that cm 0 1 mod p. The result now follows easily, since m and n are interchangeable. 0 Theorem 12.2 has an analogue for the complex L-series. In the proof of Theorem 4.2 we showed that for a certain function Fb(t),

r(s)((s, b) = f a° Fb(t)ts -2 dt, o SO

F(s)L(s, x) = f m G(t)ts - 1 dt,

0

for some function G(t). The Mellin transform of a function f (t) is defined to be r tsf(t) dt .

t We write dt/t since this is the Haar measure on the multiplicative group of positive real numbers. Let A be a finite group and let (for simplicity, p 0 2)

X = A x (1 + gp) = lim A x (1 + p1p)/(1 + Pn+ 1 4). For a E X, represents the projection onto 1 + gp . Let 0 be a measure on X. Define the gamma transform of 0 by

( Fp 0 )(s) =

x

242

12 Measures and Distributions

If A = (ZIdp Z)' with (d, p) = 1, then X _-_- (Z1dZ) x x Z; . For a E X, write a = ad ap , corresponding to this decomposition. We may define the Mellin transform of 0 by 1 (M )(s) = f s — dO. aP x Of course, the gamma and Mellin transforms are almost the same: 1 1 0)* M O) = F p(-2P

From Theorem 12.2, we have —(1 — z(c)(c>' 1 )4( —s, x) = Fp(zco -1 E,)(s) = M p(ZE,)(s + 1). The gamma transform receives its name by analogy with the classical equation dt —.

F(s) = f Jo

t

Of course, this just the Mellin transform of e'. We now investigate the relation between measures and power series. Suppose X, = A x F, where A is a finite group and F„ ,' Z1117. We assume X, —* X. corresponds to ZIpnZ —* ZIpmZ. Then X = A x F,

with F

Zp.

Let (9 be the ring of integers of a finite extension of Op. Then

(9[ [ A x F]] = lim (9[A x Fn ]

=

lim (9[A] [F,].

Choose a generator 7 0 of F. Since

lim (9[F] '' O[[T]] by Theorem 7.1, with 7 0 1

-

3

1 + T, we obtain

(9[[A x 1 ] ] '-' 0[A][[7]]. Let (... , X n , .. .) E liM

(9[A.

X

Fr]*

243

§12.2 Measures

Write x„ = L

&(g)g.

geAxr n

It follows easily from the fact that x n i—ox n, if n > m that {0„} defines an 6-valued distribution, hence an (9-valued measure. Conversely, if {On} defines an (9-valued distribution on A x F, we obtain a corresponding element of (9[EA x 1-1]. We therefore have the following one-one correspondences: (P[[A x 11] (9-valued distributions 4 1 1 6-valued measures on A x F C[A][[7]] Therefore, measures correspond to power series. We shall investigate this correspondence. Assume the order of A is prime to p. Let 0 E A and let 1 EO ==

i

A I ccEA

be the idempotent. Henceforth, we assume (9 contains the values of all such 0, so Eo E (9[A]. We have (9[A] L"---* 0 6E0 Z 0 6

o

e

Therefore C[A][[T]] =*

CD 6[[7 ]] 0

L fe(T)E91-0 (- • •

0

Jan, ..

Consequently, (9-valued measures on A x F4-100 E A (9[[7 ]] = I A I-tuples of power series. EXAMPLES. (1) Let A = 1 and let y E F. Then y E Mil]. The corresponding distribution is the delta distribution, the measure is 6 ), . If y = A with s E Zp , then the power series is (1 +

T )s = Lc° j=0

Zan"

( S.)Ti E J

(2) Let A = (Z/dp Z)X with (d, p) = 1, and rn = ( 1 + p711)/( 1 +

pn+1zp)

(assume p 0 2; otherwise the theory needs a slight modification). Then Ax

Fn

244

12 Measures and Distributions

which we identify with Gal(Q(ç dp„ i)/Q). Let g o = dp, so y o = al +0 generates F = lim Fn . Let c = 1 + g o . Consider the measure E, of Theorem 12.2. is The corresponding element in MA x dp" 4- 1 - 1

lim

a

E

1 dPn+

a=0 (a. dp)= 1

= lim( 1

c

car)

E

c — 1)

{c on+ 1

2

a

aa

1

( dpn+ 1

al

which is essentially the Stickelberger element. We map this to Let 0 be even of conductor d or dp, so 0* = cur l is odd. In the 0*th component we have (a, = a + g o = To) 1 lim(1 — (1 + go)yo) — qn

E a (2)0 - 1 (a)y(a)

in the notation of Chapter 7. This is just — '1(02 0 -1 ) in that notation, except that y(a) replaces y n(a) -1 . Observe that y o-1 corresponds to 1/(1 + T), so when we change to power series we obtain 0( 1

e

d f

1, (2) 2 0 -1

1 +T

g0(T)

with g as in Chapter 7 (before Theorem 7.10). Note that g0 .((1 + goy — 1) = —g((1 + g o ys — 1, co 2 0 -1 ) = —(1 — (1 + g0) 1 +s)Lp(—s, c1) 0*). So the modified Bernoulli distribution E, corresponds to the vector of power series which give the p-adic L-functions (one technicality: the above calculations assumed that the character 0 had conductor exactly d or pd. The characters with smaller conductors yield slightly modified p-adic L-functions). These last two examples are special cases of a general phenomenon. Fix a generator y o of F. Let Ko correspond to yo under the isomorphism 1 + pZ p F (again, assume p 0 2 for simplicity), so

X=AxF , A x(1 +p4) y) i—*

Ka

For instance, in Example 2 above, yo = o i + q o and Ko = 1 + go . Recall that a character t/i of the second kind is one of conductor pn, n > 2, such that ,/j(a) = li(<0). Such a character may be regarded as a character on 1 + pZp.

Theorem 12.4. Let d(t) be an 6-valued measure an A x F and let (.

, go (T),

.) e

0€

C[[T]]

245

§12.2 Measures

be the corresponding power series. Let 0 c second kind. Then

r -(O0 do(s) =

Li and let 0

be a character of the

0(a)0(a)s dO = go(0(K 0)4 — 1). fAxo+plp)

PROOF. First consider

(. . . , ao,(1 + T),

.)

E

By the above, this corresponds to ao, g 0, 0'

=

I LAI

E

c/o , O'(x - 1 )a-yno

E

C[[A x 11].

cteA 0'

This yields a sum of delta distributions: dO = On 1 + pZ i„ ö

1

LAI

E ao, 0' ( C 1 ) 5 OE,8 . OE

is replaced by

0(a)t,l/(a) dO = Al 1

1

0,

We obtain

ao, 0(a)0'(oc - 1 )0(K ort4is

= a9(i//(K 0)4)" = ao(1 + T)" at T = 0(104 — 1 (by orthogonality of characters, the sum over 7 vanishes for 0 0'). By linearity, the theorem is true for polynomials. Since the polynomials are dense in CUR we need a continuity statement. For any fixed s E Z p , the function f (a) = 0(a)t,I (a)sis continuous on X. Let :> 0, There exists a step function S(a) such that I f (a) — S(a)I < e

for all a E X.

Suppose we are given a vector E Recall that integration of step functions was accomplished by evaluation at sufficiently large finite levels. Let N be large and let

go (T) = P(T)q° (T) + 4(T), where PN(T) = (1 + T)PN — 1 and deg 4, < pi' (Proposition 7.2). If dO corresponds to g and dON corresponds to rN = (. , ...), then

Therefore

fxf

Lx

S(a) dO f S(a) dO N for large N.

(a) dO — f f (a) dON < Max

f

x( f — S) d(/)

fx(s — f)dON

<E,

246

12 Measures and Distributions

since I f — SI < e and 0 and 0 N are 0-valued. But dO N corresponds to a polynomial, for which the theorem is true. Also I go(00( 0)Kso — 1) — 4(//(( 0 )00 — 1)1 1 PN(

O K0 ) Kso — 1) 1 = 1 11/(Ko)PKr

's

— 1)1

<

L

for large N (note tp(K 0 ) is a p-power root of 1). Therefore

Ix f d0 — g o(O(K0 )00 —


Since e was arbitrary, the proof is complete.

0

Corollary 12.5. Let 0 be a measure. Then (F )(s) is an analytic function of s.

PROOF. Let 0 --= til --= 1. Clearly any function of the form g(Kso — 1) is analytic. Theorem 12.4 gives us something stronger than analyticity. Functions of the form f (s) = g(le — 1) with g(T) E CHTE and functions. They satisfy

K E

1 + p7, (I + 472 if p = 2) are called lwasawa

f (s) f (0) mod pC for all s CFI,. This was the basis for Exercises 7.5-7.7. Not all analytic functions have this property, for example f (s) = s. Corollary 12.6. If IA „ 0 +1,zp) 01// dO = 0 for all 0 E A and all ti, of the second kind, then q5 = O. (In other words, a measure is determined by its values on characters of finite order. Note that 0(a)0(a)s is a character of infinite order ifs 0 0). PROOF.

Let go be one of the corresponding power series. Then

go(tp(K o) — 1) = 0 for all 0, hence

goGp. — 1) = 0 for all n. By the p-adic Weierstrass Preparation Theorem (see Corollary 7.4), a nonzero power series in 6[ [T ]] has only finitely many zeros. Therefore g o = 0 for all 0, so 0 = 0, as desired. El For f (T) E 6[[T]], let

D f (T) = (1 + T) f ' (T). Observe that when f (T) = (1 + T)", Dk f (0) = n k, which is a continuous p-adic function of k, if p ,i/ n and if we restrict k to a fixed congruence class mod p — 1. The next results will show that this holds more generally.

247

§12.2 Measures

We assume A = 1. Then X = 1 + pZ y . Let

There is an isomorphism p:Xi=o"

logy x logy Ko

x Alternatively,

y

If dq5 is a measure on X, define the measure d(7) on 'X- by =

f(Y)

1

7C E

f(p(x)) X

(see Example 4 at the beginning of this section). Suppose g(T) = a„(1 + T)" is a polynomial. Then the corresponding measure c/g5 is a sum of delta measures

= a„(5 8 (on ! + p) and

deb = an (5„

(on

4)0

Proposition 12.7. Let g c a[T]] and let di./;;9 be the corresponding measure on

g=

Zpo

For k > 0,

(Dk g )(0 ) = f

yk 49.

yEg

PROOF. First let g = (1 + T)". As mentioned above, the left-hand side is The measure di/3 is the delta measure 6 so

Ilk .

yk d = nk. By linearity, the result holds for polynomials. Let g(T) e C[[T]] be arbitrary. Let E > 0 and choose N so that p' As in the proof of Theorem 12.4,

g(T) = PN (T)q N (T) + rN (T). Therefore

Dkg(T) = ao (T)PN(T) + • • • + ak(T)PV(T) + D krN(T), where ai(T) c e[M]. But

PN(C) = 0

and

n)(0) 0 mod p"•

E.

248

12 Measures and Distributions

Therefore

I Dkg(0) — D krN(0)I <

E.

As in the proof of Theorem 12.4, we may approximate p(x)k = (log x/log KO" on X by a step function S(x), say within a Then

flyk dO f2 Y k cl(1),N —

< Max{

=

fx p(x)k dO — fx p(x)k d0,. N

fx (p(x)k — S(x)) d(1) fx S(x ) dO — fx S(x) dOriv

,

fx(

S( x )

— P(x)k ) d(15 .1,,

1.

For large N, the second expression vanishes (this would not necessarily have happened if we worked on X with d(T) and d(r),.N since the latter is not necessarily a finite sum of delta distributions; see Exercise 12.2). The first and third expressions are less than a Since we know the theorem is true for the polynomial rN , we obtain

L

yk d(lj — D kg(0) <

E.

This completes the proof.

1:1

To match the set-up of Theorem 12.4, we need an integral over z; instead of To obtain this we do the following. Let g(T) e C [ [7 ] ]. Define Zpe

1 Ug(T) = g(T) —.

L

g((1 + T) — 1).

P cP=1

One easily sees that N

U

L a„(1 +

N

L a„(1 +

Tr =

n=0

T) .

n=0 pl. ?.

Proposition 12.8. Let g E C [[T]] and let dO, and dcb, be the corresponding measures on X = 1 + pZi, and k- = Zp Let z )z. (y) be the characteristic i; c X. Then function of X x = r A

dCf. u9 =

'Liz.

d(j),,

SO

(Dk Ug)(0) = fyk dCk g . Y ; E2

249

§12.2 Measures

PROOF. By Proposition 12.7, it suffices to prove the first equality. As usual, let N 1 be large and write g(T) = P(T)q(T) + rN (T).

Then Ug = U(PN qN) + UrN .

But PN(41 + T) — 1) = PN(T),

N> 1,

So Ug — PNUqN + UrN .

Let rN(T) =Ian@ +

Tr,

hence

E a„(1 + T).

UrN(T) =

P -{' n

On )7, rN corresponds to the measure

E an 6„ n

and UrN corresponds to

E a,1 6„ = Zg. En an 6,.

pn

The same argument as was used at the end of the proof of Proposition 12.7, with 321` replaced byfzi. for any continuous function f on )7, shows that

J2

f di-ljur,

= f2 fzg. cl(T),.,,,-- fg fx5,-- d(T) g

and since Ug -- UrN mod

PN,

as N -- oo ,

we similarly have

fi f cb

r —>

f

g.

f d ug .

Therefore clibug = zgx ckb g .

This completes the proof.

El

Corollary 12.9. Let g E (9[[T]]. Fix a congruence class a mod p — 1. Then there exists h(T) E C[[T]] such that (Dk Ug)(0) = h(4 —1) for k a a mod p — 1,k

O.

250 PROOF.

12 Measures and Distributions

Decompose = (ZIpi) X x (1 +

Then (Dk Ug)(0) =

--„

w(y)"dcbg

= Fp(w" d(b g)(k) = h(4 — 1)

1:

for some h E 6[[7 ] ], by Theorem 12.4. We give an application. Let c E Z, (c, p) = 1, and let g(T) --=

1 (1 + T) —

C

(1 + T )C — l •

Since 1

C

we have g(T) E Z p[[T]]. Using the relation i YP —

t

1 >

we easily find that Ug(T) = g(T) — g((1 + T)P — 1). Observe that Dg((1 +T) — 1) = p(1 + T)Pg'((1 + T)P — 1), which is just p times Dg, with 1 + T replaced by (1 + T)P. It follows by induction that (Dk Ug)(0) = (1

pk )(Dkg)(0).

To calculate Dkg(0) we change variables. Let T = ez — 1 = Z + 1Z 2 + kZ 3 + - - - e 0 pf[Z]i. Let f(Z) = g(e z — 1) c

251

§12.3 Universal Distributions

Then d f(Z) = ezg'(ez — 1) = (1 + T)g'(T) = Dg(T). dZ

Therefore, \

k

12 d )f"

(

=

D

"g".

But 1 c e — 1 ec z — 1

g(ez — 1) = z

= — E (1 — Z

n=0

cn)Bn — n!

co

=E (1 _ c n+i ) n=0

Bn + 1 Zn

n + 1 n!

Therefore Dkg(0) = ( dk ) kf (0) = (1 — c k+ 1 )

kBk

;

11

,

SO

(Dk Ug)(0) = (1 — c k +1 )(1 — p k ) _

k+ Bk÷ 11

(1 _ co k-h 1 (wo k+ 1 )Lp( _ k, co k + 1 ).

By Corollary 12.9, we find that for k ce (mod p — I), this extends to an analytic function, in fact to an Iwasawa function. The reader might find it interesting to start with Theorem 12.2 and deduce that g(T) is the power series we should use to obtain the above. The use of differentiation to obtain values of L-functions was also implicitly used in the proof of Theorem 5.18 (evaluation of L (1, wk)). This technique was probably first used by Euler, later by Kummer, and more recently by Coates and Wiles.

§12.3 Universal Distributions The main purpose of this section is to prove Bass' theorem on generators and relations for cyclotomic units. But to do so, we consider the general question of universal ordinary distributions (sometimes punctured, even, or odd) on

252

12 Measures and Distributions

e/Z. We restrict to the subset (1/n)Z/Z. Let An be the abelian group with generators

and relations

a r

=g

((n1r)a\ n

Lg

(a + rk)

, for rin.

Then An is called the universal ordinary distribution on (1/n)Z/Z. The map

a

g ( a\ n)

defines an ordinary distribution on (1/n)Z/Z. If 0 is another ordinary distribution on (1/n)Z/Z, then there is a map An -+ group generated by {0 ()} (a\

(a\

g n) 1-4) 17)' so A n is universal in the sense of category theory. We shall show that An is a free abelian group of rank 0(n). We start with an upper bound.

Proposition 12.10. There is a set of 0(n) elements which generates An

.

PROOF. Let n = fl Ka. We may write, for any a E Z,

a

ai

mod Z,

with 0 < ai < Ka. We first show that for each i, either ai = 0 or (ai , pi) = 1} generates A n . This is not yet a minimal set of generators, but it gets things started. Note that if ai = 0 then pi does not appear in the denominator of aln, while if (ai , pi) = 1 then the full power Ka is in the denominator. Consider an arbitrary aln. If ai = 0 for some i then by induction we may conclude that g(aln) is in the group generated by Bnipe, c Bn (This induction starts with the case n = 1, which is trivial). Therefore assume ai 0 0 for all i. Write a = ct with t n and (c, n) = 1. This is possible since ai 0 0 implies a, so t divides K which divides n. If t = 1 then aln = cln has

n

253

§12.3 Universal Distributions

denominator exactly n, so (ai , p i) = 1 for all i. Hence g(aln) c B„ and we are done. Therefore assume t > 1. As mentioned above, t divides fi , so pi I(nIt) for each i. By the distribution relation,

g (t) = tzi g (c + (nIt)k)

g (a)

k=0

Since (e, n)

1 and since pi I(nIt) for each i, we must have + (1 ) k, n) = 1,

for all k. Therefore all fractions involved in the last sum have denominator exactly n, so g(aln) is in the group generated by B. Therefore B A„ . But there are relations among the elements of B. Let C,, {g( an) for each i, ai

1 and either ai = 0 or (ai , p) = 1 . }

We claim that C,, generates A. By induction, we may assume C„11,7, generates Anip , for each i. Note that Cnipn C„. Let g(aln) E B. Suppose b i = 1. Let ai Y= i t 1 P7' Then

k\

—1

+ rie = g(Pel l rl i

k=0

and 1 -11

pji

p i k\ n e = g (Pel l

g( Y

k=0

11

1 -1))*

Since pelly and 14 1-1 y do not have p in their denominators, g(ply) and g(per y) lie in = the group generated by C nipyi . Subtraction yields pi-

1 /

k=0 p k

gy

ne -)

6

rl i

1

hence g(3)

1 e\ =

E

g(y + e )mod.

k#1

Notethat a 1 = 1 is changed to a sum with a l = k 1 and (a l , p 1 ) = 1, but y is left unchanged (this is important). Now consider 1k

g(Y

e) Pi

e

Pi

ai \

E i* Ki

254

12 Measures and Distributions

If another ai 1 then we may perform the above operations again. Note that ai for j i is left unchanged (in particular no such ai is changed to 1). Continuing, we eventually get ai # 1 for all i, and also (ai , p) = 1 or ai = 0. Therefore all g(aln) in 13„ are expressible in terms of C„, so C„ generates Since C„ contains 0(120 = On) elements, the proof of Proposition 12.10 is complete. Proposition 12.11. The universal punctured ordinary distribution ,4° on (11n)212 requires at most 0(n) + n(n) - 1 generators, where n(n) equals the number of distinct prime factors of n.

PROOF. A n° is generated by

al

E 212, ic nn n with relations g (a\

= ("I')-1 g (a + rk) L

whenever r In and

a - # 0. r

So we have taken the distribution A n and removed g(0) and also eliminated the relations g(0) =

E g(x).

g(x) = g(0) (fr)xao(2)

(njr)x 0 x#0

We see that whenever g(0) appears in a relation for a nonpunctured distribution, it appears equally on both sides. So we really have the relations

E

g(x) = 0,

rIn.

(nilr)x a 0 x#0

We claim that such relations follow from those with nIr prime. Let m = n/r and let pInt Then

E g (x) = E E g(x) + E mx a 0 x 0

py 0 (m1p)x ay y 0

E PYC)

y 0

g(X)

(m1p)x a 0 x 0

g(x).

g(y) + (pm p)x 0 x 0

Choose a prime dividing mlp and continue. Eventually mxa 0, #0g(x) is expressed as a sum of expressions of the form 0 ,y 0 g(y) with p prime. This proves the claim.

255

§l2.3 Universal Distributions

We now see that to obtain A„ from A n° it suffices to add a generator g(0) and add the relations g(y) = 0 py _=

y*()

for each pin. Let p— 1

a) E g(— p divides n, p prime } . a= 1 P

R

We have just shown that we have a natural isomorphism (24,? 0 4j(0)) mod

An .

We already have the set of generators C„ for A,,. Let = C u R — {g(0)}. ,,

Clearly Dn generates A. Since D„ has 0(n) + 7r(n) — 1 elements, Proposition 12.11 is proved. To show that the sets of generators in Propositions 12.10 and 12.11 are minimal, we shall produce concrete examples of the desired ranks. By the rank of a distribution 4) we mean the 2-rank ( = number of summands isomorphic to 2, in the usual decomposition) of the abelian group generated by {0(a/n)I0 a < n} . (Omit 0(0) if is punctured). From now on, we assume n > 2 (n = 1 and n = 2 are trivial, of course). Assume first that n 2 mod 4. Let C„ be a primitive nth root of unity. Consider the punctured even distribution defined by a h(n

) = (.

log I C:r — 1 I, ...) EC 4° )

where r runs through the integers with (r, n) = 1, 1 < r < n (we could have used half of these r's). Various combinations, call them y i , of the vectors h(a/n) give the logarithms of the 14)(n) — 1 independent units of Theorem 8.3 (in the first component; the other components are the Galois conjugates). We may also take a/n = 1/p for p dividing n and obtain a generator for the ideal of E:D(C p) lying above p. We claim that the group generated by the h(a/n) has rank at least 10(n) + n(n) — 1. In fact the vectors h(1 /p) for pin and the vi are independent over 2: Suppose

E aph(-1 ) + E ai v i = 0. Add the components of the vectors. For each ri , we get the logarithm of the norm of a unit, hence 0. For h(1 /p) we get the logarithm of a nontrivial power of p. But the logarithms of primes are linearly independent over 2, so ap = 0 for all p. Since the ./),'s are independent over 2, ai = 0 for all i. This proves the claim.

256

1 2 Measures and Distributions

If n -_-- 2 mod 4, then we may use the same distribution h(a/n) defined above. We know that the group generated by { h(2a/n)} has rank at least

1(1)(

11) ± 41 — 1 = 10(n) + 70) — 2.2

But h(1) = (log 2, ... , log 2),

which is independent of the vectors used to get this estimate on the rank. Therefore the rank is at least 10(n) + n(n) — 1. Therefore, for all n(> 1), we have a punctured even distribution of rank at least 40(n) + n(n) — 1. We now produce an odd distribution of rank 10(n). As in the case just completed, the construction will depend on the fact that L(1, x) 0 0; but this time it will be in the form B i , x o 0 for odd x. We shall be using the first Bernoulli distribution, but our preliminary calculations will be valid more generally. Let h be an ordinary distribution on (1/n)Z/Z, and let x 0 1 be a Dirichlet character of conductorfx . The proofs of the following lemmas are essentially the same as the arguments given for Lemmas 8.4-8.7. Simply replace log I 1 — „a I by h(a/n). The fact that ',; = fl when n = mm' corresponds to the fact that h is ordinary. We also use the fact that h is periodic mod 1. (Of course, assuming Bass' theorem, essentially any relation satisfied by log I 1 — a I is also satisfied by h(a/n), with the possible exception of evenness). ,

Lemma 12.12. Suppose min. Wei' (n/m) then n

0

E z (a)hr-) =_- o n

a=1 (a,

n)= 1

Lemma 12.13. Let n = mm' with (m, m') = 1, and suppose him. Then

i

z(a)h( am) ,_ 0(fril) n —

a= 1 (a,

n)= 1

i

z(b)h(-b-).

El= 1

:

M

(b, m)= 1

Lemma 12.14. Suppose F, g, t are positive integers with fx l F, g I F, and Ftln. Then F a —) = 1 z(b)h(—b ) 1 Z(a)h( Ft F b= 1 a= 1 Ft

(a, g)--= 1

(b, 9)= 1

(we require Ft In since h is not defined for larger denominators).

f=1

257

§12.3 Universal Distributions

Lemma 12.15. Assume Lim and min. Then

E b= 1 (b,m)= 1

x(b)h (- ) = m

z(b (Fit (1 — Am))h h= 1

Lemma 12.16. Let n = mm' with him. Then

E

x(c)h(

am'

a=1

fy

= O(n')(1-1(1 — AP)))oh E Ac

— ).

a= 1

Pm

(a, n)= 1

PROOF.

See the calculations following the proof of Lemma 8.7.

FP= 11

Proposition 12.17. Let n = iP7`. Let I run through all subsets of {1, , s}, and let n1 = i€1K i Then except {1,

, s},

•

Ix

Aci)h( ani) = (11 (0(10 ) + 1 — I a= 1 p, fx (a, n)= 1 PROOF.

) AciA(1. a= 1

See the end of the proof of Theorem 8.3. This is where we need , s} and the result holds). 0

x o 1 (if x = 1, include I = {1,

As in the case of even distributions, it is this last formula which will prove useful. Consider G = (ZIn.Z ) X = Gal(Q()/Q) (we allow n -= 2 mod 4). Let aa be the automorphism corresponding to a mod n. Let b(aln) be a complex-valued ordinary distribution on (11n)ZIZ. Define C)

H( n

= v b faC\ aa_ i c (E[G]. aL =1

n

(a, n)= 1

We claim that H is an ordinary distribution on (11n)ZIZ. It suffices to prove this for b(ac In) for each a. Let ri n. Since (a, n) = 1, { rj

n } { ark — mod 1 0 < j < — = — mod 1 0 n

k < —n} .

Therefore,

b k=0

This proves the claim.

(a(c + rk))

=

(n/r) — 1 ( ac +

E

b n =o ac = b(--), as desired.

258 Let

12 Measures and Distributions

E

be a Dirichlet character mod n of conductorfx. . Let E— X

1 0(n)

E z (a)a-a-i (

a, n)=

be the corresponding idempotent. Since Ex a; 1 = 14(1 451 Ex il(C ) =

1 E z(a)b (ac) E x . d) (n) a = 1 (a,n)= 1

Of course, Hz is a distribution. Let H be the abelian group generated by {H(c/n)10 < c < n) and let Hc be the C-subspace of C[G] spanned by H. Then rank H > dim Hc (we have an inequality since elements which are independent over Z could become dependent over C (e.g., 1, 2)). Since aa H = H

,

Hc is stable under G, so

=

xH

hence dim Hc =

E dim Ex ile .

Observe that Hx(c/n)e Ex ile for each c. We now choose the distribution b. Let B i (X) = X — Bernoulli polynomial, so

c) = B1 ({cn}) B(

{ cn }

n

c

be the first

0; B(0) = 0,

is the corresponding distribution. Let n = 11 K and I be as in Proposition 12.17. Then we let

b(

= B( n i ).

Clearly b(c/n) is odd, hence so is H(c/n). By Proposition 12.17,

1 E z(a) O e (1)(n) \n n 1 0 (pill, (0(pie.i) + 1 1

H

)

(n)

a=I

fx z

259

§12.3 Universal Distributions

If x is even, the sum vanishes. If x is odd,

o o. Since the product over p i does not vanish (each factor has positive real part),

1 0 # Hx ( )e gx 1-1c , n so ex lic is non-trivial. Since there are 10(n) odd characters, rank H

dim Hc > 10(n).

We now have an odd distribution of rank at least 10(n). Note that H(0) = 0, which does not affect the rank. If we ignore H(0) and some of the relations (e.g., E,= 0 H(y) = 0), then we may consider H as a punctured odd distribution, which still has the same rank. Therefore we have a punctured even distribution of rank at least 10(n) + E(n) - 1 and an odd one of rank at least 10(n). We want to put them together. Suppose h + and h - are any two punctured distributions, with h + even and h- odd. Let H ± be the groups generated by h± and define

ci)

a\, h _

e

e H_. n

h(

h. Then h(-an) + h(- -a ) = (2h+(a), 0) e H n n

Let H ç H + ir be the group generated by

and

+a)n - h(- -an ) = (0, 2h - (-1) e H. Consequently

2H+ (i) 21/ - ç H

H+ e

W,

SO

rank h = rank h + + rank h - . Using the distributions obtained above, we obtain a punctured distribution of rank at least On) + n(n) - 1. Since the universal punctured ordinary distribution 4, is generated by 4)(n) + g(n) - 1 elements (Proposition 12.11), and maps surjectively onto the group generated by the values of this distribution A,? is generated by (P(n) + n(n) - 1 elements (Proposition If we had an even punctured distribution of rank greater than 10(n) + n(n) - 1, or an odd one of rank greater than /(n), we could obtain a punctured distribution of too large a rank. Therefore the universal punctured even

260

12 Measures and Distributions

distribution (A) + has rank 1-0(n) z(n) — 1, and the odd distribution (4) - has rank -1-0(n). However, we cannot conclude that (4)± are free abelian. We know from the above that 2(A,D+ 2(4) so 2(4) 1 have no torsion. But there is the possibility of 2-torsion. In fact, (A F(:)+(12)000+ n(n) — 1 0 (z/21)2

I—r

and (An - —

Z(1 / 2} '" G (1/21)2' '-

where r = n(n) if n A 2 mod 4, r = n(n/2) if n -= 2 (mod 4). (See C.-G. Schmidt [4], K. Yamamoto [2]). We now consider nonpunctured distributions. A n is a quotient of A,°, 0 ig(0) by a subgroup of rank at most n(n), hence An

with e

Ze C) torsion

O(n). By Proposition 12.10, V (n).

Since (A) + C) ig(0) modulo yields an even distribution, rank A:-

-10(n).

Also, we already have an odd distribution (constructed via B 1 0) of rank at least -10(n). As in the case of punctured distributions, we must have (z/2z ) for some integers C±. In this case it is easy to see that the 2-torsion actually occurs (Exercise 12.4). We summarize what we have proved: Theorem 12.18. Let n > 2. For some integers a, b,c, d, we have Universal punctured

1,

=A

Universal even punctured Universal odd punctured Universal

= (A) + = (Ay

Z (112)(1) (")+" - 1 C) zoi2)000 0

(Z/2Z) b,

=An 7(11 2)0(n)

Universal even = Universal odd = A

(1/2)0(n) ,

(1/2z )C , (zp

v.

The proof of Bass' theorem is now immediate. Since (An +

group generated by {log I — 1 I, 0 < a < n}

261

Exercises

is surjective, and the latter is free abelian of rank (at least, hence exactly) 10(n) + n(n) 1, we must have —

—i

(4) + /( 1/ 2Z) =
I

».

This is Bass' theorem (8.9).

NOTES For more on measures, see Koblitz [1], Mazur—Swinnerton-Dyer [1], and Lang [4], [5]. For the concept of pseudo-measures, which can handle denominators, see Serre [3]. For other versions of the F-transform, see Leopoldt [10], Iwasawa [23], and Lichtenbaum [4]. For more on Iwasawa functions, see Serre [2]. The theory of universal distributions was developed by Kubert-Lang. Theorem 12.18 was proved, in more generality, by Kubert. The fact that 2-torsion must be considered in Bass' theorem was first recognized by Ennola [1], [2].

EXERCISES 12.1. Give another proof of Corollary 12.6 by showing that the characters 0 of the second kind and the O c A span Step (X). 12.2. In the second section, we started with a power series g E Π[[T]], obtained a measure dO, on 1 + pZ, (assume A = 1), then a measure 4, on Z p , which restricted to c* 9 on (i/pi)' x (1 + pi p). Therefore d(i), corresponds to a vector of power series (... , 0( (T),

Show that if g(T) =1„N- 0,0 +

<

<

p — 2.

Tr is a polynomial, then

an w 2 (n)(1 + T) fo gpn, lo gpKo) 12.3. Let be as in the chapter. (a) Let u c 1 + pZ „ (or 1 + 42 2 ). Show that there exists h(T) E i„[[T]] such that u = h(Ks0 — 1). (b) Suppose 11,(00 — 1) is a Cauchy sequence (in the sup norm on continuous functions on Zp) of Iwasawa functions, with hp(T) c ([M]. Show that there exists h(T) E C[[T]] such that lim h n(reo — 1) = h(F%-.4 — 1). (H int : let s be close to O. Show successively that each coefficient converges mod pn for all n). (c) Show that the Iwasawa functions are the closure of the span of the functions of the form f (s) = us, with u c 1 + pip.

262

12 Measures and Distributions

12.4. (a) Let p be an odd prime and let Ap+ be the universal even ordinary distribution on (1/p)/. Show that (p-

1)/2

(a )

g — P

a=1

0

but 2x = O.

Therefore A p+ has 2-torsion. (This idea may be extended to arbitrary n > 2). (b) Let A, be the universal odd ordinary distribution on (1/n)Z/Z. Show that g(0) 0 but 2g(0) = 0, hence A,. has 2-torsion.

12.5. ([Ennola 2]) Let n = 105. Let a x ------ g(

x ) e (AT 0 5 )± . 105

(a) Show that all relations among the ax are generated by the relations

ax = a, a3x = ax + a + 35 + 0x+70 a5 x

=

ax

± a x + 2 i + • • • + a x + 84

a7, = ax + a„15 + • • • + a x + 90 • (b) Show that in all such relations, the number of x with x $ 0 mod 3 is even (count both sides of the equation). (c) Show that r = 0, where

r = a l + a 2 + a 17 + a43 + a44 + a46 — a 3 + a 9 + a 36 + a 25

± a40 ± a28 ,

is not a relation in (AT 05 ) + . (d) Show that 2r = 0 is a relation. This shows that 2-torsion must be considered in Bass' theorem.

Chapter 13

Iwasawa's Theory of 4-extensions

The theory of 4-extensions has turned out to be one of the most fruitful areas of research in number theory in recent years. The subject receives its motivation from the theory of curves over finite fields, which is known to have a strong analogy with the theory of number fields. In the case of curves, it is convenient to extend the field of constants to its algebraic closure, which amounts to adding on roots of unity. There is a natural generator of the Galois group, namely the Frobenius, and its action on various modules yields zeta functions and L-functions. In the number field case, it turns out to be too unwieldy, at least at present, to use all roots of unity. Instead, it is possible to obtain a satisfactory theory by jusl adjoining the p-power roots of unity for a fixed prime p. This yields a Zp-extension. The action of a generator of the Galois group on a certain module yields, at least conjecturally, the p-adic L-functions. In the present chapter, we firs1 prove some preliminary results on 7/,,extensions. We then determine the structure of modules over the ring A = As a result, we obtain the beautiful theorem of Iwasawa which describes the behavior of the p-part of the class number in a Z,,-extension. We then discuss the Main Conjecture, relating certain Galois actions to p-adic L-functions. Finally, we use logarithmic derivatives to prove a result of Iwasawa, which could be considered as a local version of the Main Conjecture, which describes local units modulo cyclotomic units in terms of p-adic L-functions. Extensions of this theorem to elliptic curves have proved very useful in the work of Coates and Wiles on the conjecture of Birch and Swinnerton-Dyer. In this chapter we use more class field theory than in previous chapters. A summary of the necessary facts is given in an appendix.

Zan].

263

264

13 Iwasawa's Theory of 2p-extensions

§13.1 Basic Facts A 1p-extension of a number field K is an extension K co / K with Gal(1( 00/K) Zp , the additive group of p-adic integers. As Proposition 13.1 below shows, it is also possible to regard a Zr-extension as a sequence of fields c= K. = Kn

K = Ko K i with Gal(Kn/K)

In Chapter 7 we showed that every number field has at least one Zp- extension, namely the cyclotomic Zp-extension. It is obtained by letting K. be an appropriate subfield of K(4.°). Proposition 13.1. Let K ./K be a 7L-extension. Then, for each n > 0, there is a unique field K n of degree p" over K, and these K n , plus K., are the only fields between K and K.

The intermediate fields correspond to the closed subgroups of Zp Let S 0 be a closed subgroup and let x e S be such that v(x) is minimal. Then x1, hence x1 p , is in S. By the choice of x, we must have S = xZ p = poZp for some n. The result follows. PROOF.

Proposition 13.2. Let K./K be a 7L-extension and let -I be a prime (possibly archimedean) of K which does not lie above p. Then K./K is unramtfied at 1. In other words, 7L-extensions are "unramtfied outside p."

Let I Gal(K oo /K) Zp be the inertia group for 1. Since / is closed, I = 0 or / = p"Z p for some n. If I = 0 we are done, so assume I = p"Zp . In particular, / is infinite. Since I must have order 1 or 2 for infinite primes, we may assume 1 is non-archimedean. For each n, choose inductively a place 1,, of K n lying above 1„_ 1 , with 1 0 = 1. Let K„ be the completion, and let K G° = K. Then PROOF.

/ Gal(K oo/K).

Let U be the units of K. Local class field theory says that there is a continuous surjective homomorphism pnz p .

But U

(finite group) x

a G Z,

where / is the rational prime divisible by 1 (proof: log / : U l -N C for some N; the kernel is finite and (9( = local integers) is a finitely generated free 11-module). Since /Y2 1, has no torsion, we must have a surjective and continuous map z

pn z p pn zphon +

1

z

p.

265

§13.1 Basic Facts

However, 7;' has no closed subgroups of index p, so we have a contradiction. This completes the proof. The proposition may also be proved without class fi eld theory. See Long [1 ], p. 94. Lemma 13.3. Let K co/K be a 4-extension. At least one prime ramifies in this extension, and there exists n 0 such that every prime which ramifies in IC.311<,, is totally ramified.

Since the class number of K is finite, the maximal abelian unramified extension of K is finite, so some prime must ramify in K ojK. We know that only finitely many primes of K ramify in K oo/K by Proposition 13.2. Call them A i , , A„ and let I, , I N be the corresponding inertia groups. Then PROOF.

n I. = [P4 for some n. The fixed field of pn?4 is K ,1 and Gal(K/Kn) is contained in each I. Therefore all primes above each A i are totally ramified in K.,IK„. This completes the proof. D However, it is possible to have K n/K unramified for some n (see Exercises 13.3 and 13.4). We already know that every number field K has at least one 4-extension, namely the cyclotomic 4-extension defined in Chapter 7. However, there could be more. Let E l be those units of K which are congruent to 1 modulo every prime A of K lying above p. Let U 1 , A denote the local units congruent to 1 mod A. There is an embedding E1

-k• U1 -

U

sAIP 0-

The closure E l is a Zp- module. Leopoldt's conjecture predicts that the 7p-rank is r 1 + r2 1, where r 1 , r2 have the usual meanings. We know this is true for abelian number fields (Corollary 5.32). r 2 — 1 — (5, with 6 > O. Theorem 13.4. Suppose the 4-rank of E l is r Then there are r2 + 1 + 6 independent 4-extensions of K. In other words, if IZ is the compositum of all 4-extensions of K, then Gal(k/K) .42+ t +6. Let IZ be as above and F the maximal abelian extension of K which is unramified outside p. Then k' F. Let J denote the idèles of K. By class field theory, there is a closed subgroup H with PROOF.

Kx =1-1..qJ

such that JIH

Gal(F/K).

266

13 lwasawa's Theory of 74,-extensions

Let Ul- denote the local unit group at a finite prime 1 of K, and U1 = Kr if 1 is archimedean. Let

U' = n LT, fil P

U" =

n Ut,

U = U' x U".

1 -t" P

All of these may be regarded as subgroups of J by putting a 1 in all the remaining components for U' and U". U is an open subgroup. Since F/K is unramified outside p, U" g H. Since F is maximal, we must have

H = K x U" (technical point: we need J/K x U" to be totally disconnected; but this will follow from the fact that this is true for U 1 ). Let

J' = J/H, and

J" = IC U/H = U'H/H'-.' U'/U' n H. Let U1 = 11, 11,U", be as in the discussion preceding the statement of the theorem. Then

U' = U1 x (finite group). Therefore

J"-'' U'/U' n H = (finite) x U i (U' n

H )/( C!' n H)

(finite) x U 1 /U 1 n H. We have a map as above, but note that tii(e) has component 1 at all 1,1/p. So this is not the same as K x c-* J. Lemma 13.5. U 1 nH = Ui nK x U" = PROOF. Let

c e E l . Then OW e U 1 . Also (e) = (O(

c

)) e K x U"

since t/j(e)/e has component 1 at all A I p . Taking closures, we obtain one inclusion. The reverse inclusion is more difficult. Since U has a "nice" topology, we may obtain the closure of an arbitrary subset S by taking the intersection of (a cofinal subset of) the closed neighborhoods of S. If Un, i, denotes those units congruent to 1 mod An and Un = I p Un , A (put 1 in all components for A ri, p ) then

fi,

Kx U" =

nn Kx U" U,.

267

§13.1 Basic Facts

Also, we have

= n Ip(Eoun.

tp(E1)

It suffices to show that

U1 n

U"Un

Let x E Kx , u" E U",ue Un . Suppose

xu"u E U 1 . Then xu" e U 1 . Since u" has component 1 at all Alp, x must be a principal unit at these primes. Since U 1 has component 1 at 1,1/p, and u" is a unit at these places, x must also be a unit at these places. Therefore x is a unit everywhere, so x is a global unit, in fact XE E l . To summarize, at Alp we have xu" = x E E l . At 1,1/p, xu" = 1. This is exactly what it means for xu" to be in 114E 1 ). Consequently

xu"u E tii(E 1 )Un .

LI

This completes the proof of Lemma 13.5.

A n CA for large enough n, by Proposition The logarithm maps U 5.7. But CA Zep,4 44 where eA ,fA denote the ramification and residue class degrees. Also, [K : = eA f. We obtain ,, A

U1

(finite) x z[K P , Q] •

Therefore

U 1 /U 1 n H = U1/1/01)

(finite) x Zrp2+ 1 +6

A similar statement holds for J". We want information about J'. But

J'/J" J/K x U

ideal class group of K

(see the appendix on class field theory; better: prove it yourself). Consequently IZ rp2+ 1+6

finite group.

This is approximately what we want. However, we need the quotient of J' by a finite group to be Zrp2+1+6, since then the fixed field of the finite group is L. Let N be the order of the finite group in the last equation above. Then N Z r; + 1 + 6 C NJ'

+ 1 +69

SO

NJ' — Z'2+ 1 +6 ,

as a 4-modu 1 e

268

13 lwasawa's Theory of , p-extensions

(we are writing J' additively). Let J'N = {x E J'INx = 0 }. Then JN is closed and Fp2 +1 '. NJ' J'/JIN It is easy to see that J'N is finite: if it had order larger than N, then two elements of JN would have the same representative in the finite group above. Hence their difference, which is killed by N, would be a nontrivial element of finite order in 7rp2 + 1 '. This is impossible, so J'N is finite. The fixed field of J;N, c J' = Gal(F/K) must be g, so the proof is complete.

El Corollary 13.6. Let H be the Hilbert class field of K and let F be the maximal abelian extension of K unramified outside p. Then Gal(F/H) (11 U 0)/E, IP where E is the closure of E, embedded in

El

U 0 diagonally.

J', and the closed subgroup J" corresponds to H. Hence Gal(F/K) Gal(F/H) J" U'/U1 n H. The same proof as for Lemma 13.5 shows that U' n H = 114E). The result follows. PROOF.

§1 12 The Structure of A-modules Let A = Tp[Ul Recall that a nonconstant polynomial P(T) distinguished if ].

P(T) = Tn + a, _ Tn - 1 + • • • + ao,

ai ,

0

By the p-adic Weierstrass Preparation Theorem (7.3), if f (T) then we may uniquely write

i E

E

A is called n — 1.

A is nonzero,

f (T) = p"P(T)U (T)

with i > 0, P(T) distinguished, and U(T) E A • By Lemma 7.5, if f is a polynomial so is U. Also, there is a division algorithm (Proposition 7.2) for distinguished polynomials: if f(T) E A and P(T) is distinguished then (uniquely) f (T) = q(T)P(T) + r(T)

with r(T) E Zp[1 , deg r(T) < deg P(T) (let deg 0 = — oc, for convenience). It follows from the above that A is a unique factorization domain, whose irreducible elements are p and the irreducible distinguished polynomials. The units are the power series with constant term in 7;. ]

Lemma 13.7. Suppose fgEA are relatively prime. Then the ideal (f g) is of finite index in A.

269

§ 13.2 The Structure of A-modules

Let h e (f, g) be of minimal degree. Then h = ell with H = 1 or H distinguished. Suppose H 0 1. Since f and g are relatively prime, we may assume H does not divide f. But PROOF.

f = Hg + r,

deg r < deg H = deg h,

SO

psf = lui + Since deg(psr) < deg h and psr G (f; g), we have a contradiction. Therefore H = 1 and h = ps. Without loss of generality, we may assumefis not divisible by p and is distinguished. Otherwise, use g or divide by a unit. We have

(f, g)

2 (ps,

By the division algorithm, any element of A is congruent mod f to a polynomial of degree less than deg f Since there are only finitely many such polynomials mod ps, the ideal (ps, f) has finite index. This completes the proof. Lemma 13.8. Suppose f, g E A are relatively prime. Then (1) the natural map

Mfg) —> AV)

o Mg)

is an injection with ‘finite cokernel; (2) there is an injection A/(f )

A/(g) —> A/(fg)

with finite cokernel. PROOF. (1) Since A is a unique factorization domain, the map is an injection. Consider (a modf, b mod g). Ifa — b e (f, g), then a—b= fA+ gB, for some A, B. Let c=a— fA=b+gB. Then c===..- a modf,

c

b mod g,

so (a, b) is in the image. Now let r 1 , „ . rn G A be representatives for A/(f g). It follows that {(0 mod fr .; mod g)Il < j < n is a set of representatives for the cokernel of the above map. This proves (1). (2) From (1), A/(fy)

M g A/( f)

o A/(g) d=‘"-1 N

270

13 lwasawa's Theory of Zp -extensions

with M of finite index in N. Let P be any distinguished polynomial in A which is relatively prime to fg. If (x, y) E N, then (P)(x, y) _= (Pi)(x, y) mod M for some i < j. Since 1 — Pi' G A x ,

we have Pi(x, y) E M.

It follows that P&N M for some k. (Alternatively, this follows from the fact that Pk —) 0 in M Suppose Pk(x, y) = 0 in N. so f I ex, giPky. Since gcd(P, fg) = 1, f ix and g I y; so (x, y) = 0 in AL Therefore N 11') M '3' AAfg)

is injective. The image contains the ideal (Pk, fg), which is of finite index by Lemma 13.7. This completes the proof. 0 Proposition 13.9. The prime ideals of A are 0, (p, T), (p), and the ideals (P(T)) where P(T) is irreducible and distinguished. The ideal (p, T) is the unique maximal ideal. PROOF. All the above are easily seen to be prime ideals. Let

A

0 be prime. Let h E A be of minimal degree. Then h = ell with H = 1 or H distinguished. Since A is prime, p E A or H E A. If 10 HEA then H must be irreducible by the minimality of the degree of h. Therefore, in both cases, (f ) L A where f = p or f is irreducible and distinguished. If (f ) = A, then A is on the above list so we are done. Therefore assume (f) 0 A, so there is a geA with f ,f/ g. Since f is irreducible, f and g are relatively prime. Lemma 13.7 implies that A is of finite index in A. Since A/A is a finite 4-m odule, ply E A for large N, hence p E A since A is prime. Also, T i a Ti mod A for some i < j. But 1 — Ti -i E A ' , so T i E A. Therefore T c A, so (p, T) L A. But A/(p, T)._ 7/p7, so (p, T) is maximal and A = (p, T). Since all the prime ideals are contained in (p, T), this is the only maximal ideal. This completes the proof. 0 Lemma 13.10. Let f E A then A/(f ) is infinite. PROOF. We may assume f 0 O. It suffices to consider f = p and f =

distinguished. If f = p, A/(f) '--' 7/p7[[1]]. If f is distinguished, use the division algorithm. 0 Lemma 13.11. A is a Noetherian ring. PROOF.It is known (Lang's Algebra) that if A is Noetherian then so is A[[7]]. One could also use the Hilbert basis theorem (A Noetherian A[T] Noetherian) since the generators of an ideal may always be assumed to be polynomials. 0

271

§13.2 The Structure of A-modules

Definition. Two A-modules M and M' are said to be pseudo-isomorphic, written M-M' , if there is a homomorphism M --0 M' with finite kernel and co-kernel. In other words, there is an exact sequence of A-modules

with A and B finite A-modules. Warning. M M' does not imply M' M. For example, (p, T) A, obviously. But suppose A --0 (p, T). Let f (T) be the image of 1 e A. Then the image of A is (f) (p, T). But A/(f ) is infinite, so (p, T)/(f) is infinite. Hence, the cokernel is infinite. However, it can be shown that for finitely generated A-torsion A-modules, M M' .4=> M' — M. Lemma 13.8 says that if (f, g) = 1 then

A/(fg) A/( f)

A/(g)

and

A/(f ) A/(g)

A/( fg).

We shall need to know the structure of finitely generated A-modules. The following theorem was first proved by Iwasawa in terms of the group ring lp[Ern Serre observed that the group ring is isomorphic to A and deduced the structure theorem from some general results in commutative algebra. Paul Cohen showed that one could give a proof via row and column operations, just as is done for modules over principal ideal domains. In the following, we follow Lang's treatment [4] of Cohen's proof. Theorem 13.12. Let M be a finitely generated A-module. Then M A GAjpni)) G (CD A/(fi(Tr)), = i= where r, s, t, ni , m e Z, and fi is distinguished and irreducible.

that the result is the same as for modules over principal ideal domains, except that there is only a pseudo-isomorphism. The proof will use an extension of the techniques employed in that theorem (the reader who has not seen the p.i.d. theorem proved via row and column operations should immediately consult an algebra text). Suppose M has generators u l , , un , with various relations ). 1 u 1 + • • + A n ur, = 0, 21 E A. PROOF. Note

Since the relations R are a submodule of An, and A is Noetherian, the relations are finitely generated. So we can represent M by a matrix whose rows are of the form (A 1 , , An), where > A. ui = 0 is a relation. By abuse of notation, we call this matrix R. We first review the basic row and column operations, which correspond to changing the generators of R and M.

272

13 Iwasawa's Theory of 2p-extensions

Operation A. We may permute the rows or permute the columns. Operation B. We may add a multiple of a row (or column) to another row (column). Special case: if A' = gA + r then

(

Operation C. We may multiply a row or column by an element of A*. The above operations are used for principal ideal domains. However, we have three additional operations, which are where the pseudo-isomorphisms enter. Operation!. If R contains a row (A i , 0.2 , ... ,p)) with p,1/.1. 1 , then we may change R to the matrix R' whose first row is (\Al, A 21 An) and the remaining rows are the rows of R with the first elements multiplied by p. In pictures: • • • 9

PA2

Al

A2

•••

12

•••

--°• P 1 1

12

••••

fl2

• • •

mgi

fl2

•••

As a special case, if A2 = • • • = A„ = 0 then we may multiply a l , fi t , ... by an arbitrary power of p. PROOF.

In R we have the relation A 1 u 1 + p(A2 u 2 + • • • + Au) = O.

Let M' = M C) vA, with a new generator v, modulo the additional relations (

—

u 1 , pv) = 0,

(A 2 u 2 + • • • + )L n un , A i v) = O.

There is a natural map M -- M'. Suppose m 1— O. Then in lies in the module of relations, so (m, 0) = a( with a, b

E

—

u i , pv) + b(A2u2 + • • • + A,, u,, A 1 y)

A. Therefore ap = —bA l .

Since p)(/1. 1 by assumption, pib. Also, A 1 1a. In the M-component, al

=

=

a a — — (A i u i ) — — p(A 2 u 2 + • - • + Au) A1 Ai a

Al

273

§13.2 The Structure of A-modules

Since the images of pv and A l v in M' are in the image of M, the ideal (p, A 1 ) annihilates M'/M. Since A/(p, A 1 ) is finite and M' is finitely generated, M'/M is finite. Therefore MM'. The new module M' has generators v, u 2 , . , un . Any relation a l u l + • • + a„u„ = 0 becomes pa l v + • • • + a n un = 0, so the first column is multiplied by p, as claimed. We also have the relation A l + • • • + .1.„u„. So the new matrix R' has the form stated above (we removed the redundant row (1)/1-1, • • •

Operation 2. If all elements in the first column of R are divisible by pk and if there is a row (p' A 1 , , Pk)) with p A i , then we may change to the matrix R' which is the same as R except that (p' A 1 , . . . , p k A n) is replaced by (A 1 , . . , A n). In pictures:

(pk k P PROOF.

pk A2 ...\ (1 2

• •)

( A1 2 k P

'2

22

Let M' = M 0 Av modulo the relations

(pk u i , — p k v) = 0,

(A 2 u 2 + • • + /ton , A i v) = O.

As before, the fact that p,f'A i allows us to conclude that M embeds in M'. Also, the ideal (p(, A 1 ) annihilates M'/M, so the quotient is finite. Consequently M M'. Using the fact that pk(u i — v) = 0 and the fact that plc divides the first coefficient of all relations involving u 1 , we find that M' = M" 0 (u 1 — v)A, where M" is generated by v, 14 2 , . , u„ and has relations generated by (A 1 , , An) and R. Therefore M" has R' for its relations. Note that (u 1 — v)A which is already of the desired form. So it suffices to work with M" and R'

LI

, pk A n), and for some A with p A, Operation 3. If R contains a row (p" A 1 , , AA„) is also a relation (not necessarily explicitly contained in R), then we may change R to R', where R' is the same as R except that (pk A l , , , kA) is replaced by (A 1 , , A n). PROOF. Consider the surjection

u,,)A.

M/(Aqui M The kernel is annihilated by the ideal (A, pk ). Since M, hence the kernel, is finitely generated, and since A/(A, p) is finite, the kernel must be finite; so M M'. Clearly M' has R' as its relation matrix. LII + • • • +

274

13 Iwasawa's Theory of Zp-extensions

This completes our list of operations. We call A, B, C, 1, 2, 3 admissible operations. Note that all of them preserve the size of the matrix. We are now ready to begin. If 0 f e A, then

f (T) = p#13 (T)U(T), with P distinguished and U e

. Let

deg, f = {

oo,

>0

deg P(T), R = 0;

this is called the Weierstrass degree off Given a matrix R, define dee(R) = min deg(a)

for j,] k,

where (du) ranges over all relation matrices obtained from R via admissible operations which leave the first (k — 1) rows unchanged (we allow aij for i > k and all j to change; we also allow operations such as B which use, but do not change, the first (k — 1) rows). If the matrix R has the form

O *

/11- - 1 , r - 1

•

(D r _ i 0\ 13) 2,1

0

*

with Akk distinguished and deg Akk = d eg A

1 < k < r — 1, =for deg(k) (R),

then we say that T is in (r — 1)-normal form.

Claim. If the submatrix B 0 then R may

be

transformed, via admissible operations, into R' which is in r-normal form and has the same first (r — 1) diagonal elements.

The "special case" of Operation 1 allows us to assume, when necessary, that a large power of p divides each .1,i with i > r and j < r — 1. That is, pN I A, with N large (large enough that p" B). Using Operation 2, we may assume that pl' B. We may also assume that B contains an entry Aii such that PROOF.

deg„, Aij = deg(r) (R) < co.

If Aii = P(T)U(T), then multiply the jth column by u-1. Therefore we may assume Aii is distinguished. (Since the first r — 1 rows have 0 in the jth column, they do not change). Operation A lets us assume A ii A„ (again, the O's help us).

275

§13.2 The Structure of A-modules

By the division algorithm (special case of B), we may assume that A n.; is a polynomial with deg A n < deg An-,

j

deg An.; < deg

j < r.

r,

and

Since A„ has minimal Weierstrass degree in B, we must have p I A ri for j > r. By 1, we may assume pi' I j < r, for some large N. Suppose A ri 0 0 for some j > r. Operation 1 lets us remove the power of p from some nonzero Ani with j > r (the O's above are left unchanged). Then deg, Ari = deg Ari < deg A„ = deg„ which is impossible. Consequently, Ari = 0 for j > r. If some Ari 0 0 for j < r, use Operation 1 to obtain p,1' A ri for some j. But then deg, A ni deg A n < deg A. = deg, All . Since deg„ Aii = deg(i)(R), this contradicts the definition of deg (R). Therefore A ri = 0 for all j This proves the claim.

r.

LI

If we start with a matrix R and r = 1, we may successively change R until we obtain a matrix 0)

(A11

A rr A

0

with each A. distinguished and deg Aii = dee(R) for j r. By the division algorithm, we may assume that A i; is a polynomial and deg A u < deg

for i j.

Suppose A i; 0 0 for some i j. Since deg„ Aii is minimal, p I A ii ; so we have a nonzero relation (A 1 ,..., Ain , 0, , 0) which is divisible by p. Let A = A rr . Then p)(A, since the Ails are distinguished; and Aii (il l

1,•••,

1 -

, 0, • . . , 0)

is also a relation, since Au = 0. By Operation 3 we may assume p does not divide A ii for some j, so deg„ A ii < deg A i; < deg Au = deg (R).

276

13 Iwasawa's Theory of

2p -extensions

This is impossible. Therefore = 0 for all i and j with i j. This means A = O. In terms of A-modules, we have

MA1 i) 0 • • 0 AAA„) 0 A. Putting back in the factors A/(p k) which were discarded in Operation 2, we obtain the desired result, except that the Â,ii are not necessarily irreducible. Lemma 13.8 takes care of this problem. This completes the proof of Theorem 13.12.

§13.3 Iwasawa's Theorem The purpose of this section is to prove the following result. Theorem 13.13. Let K ao/K be a 7Z-extension. Let pen be the exact power of p dividing the class number of K. Then there exist integers A > 0, p > 0, and y, all independent of n, and an integer no such that en = An + pp" + y for all n

no .

Zp and let y o be a topological generator of PROOF. Let F = Gal(K 00/K) F, as in Chapter 7. Let L n be the maximal unramified abelian p-extension of K n , so X, = Gal(L n/K n) A n = p-Sylow of the ideal class group of K. Let L = Un > L,, and X = Gal(L/K,0). Each L n is Galois over K since Ln is maximal, so L/K is also Galois. Let G = Gal(L/K). We have the following diagram.

G/X =1—

The idea will be to make X into a F-module, hence a A-module. It will be shown to be finitely generated and A-torsion, hence pseudo-isomorphic to a direct sum of modules of the form A/(e) and A/(P(T) k ). It is easy to calculate what happens at the nth level for these modules. We then transfer the result back to X to obtain the theorem. We start with the following special case.

277

§13.3 lwasawa's Theorem

Assumption. All primes which are ramified in K1 K are totally ramified.

By Lemma 13.3, this may be accomplished by replacing K by K m for some m. By our assumption, K,1 , 1

Kn,

=

SO

Gal(Ln/K)

Gal(L„K,, i lK„, i ),

which is a quotient of X,, 1 . We have a map X n+1

n'

X

This corresponds to the norm map A, 1+ A n on ideal class groups (see the appendix on class fi eld theory). Observe that Gal(L,K 0,1K),

X SO

lim Gal((j L„ /(,)/K on ) = Gal(L/K „) = X. Let y x by

E

F„ = F/FP n. Extend y to

E

Gal(L„/K). Let x

E

X,,. Then y acts on

x' = Since Gal(L n/K,,) is abelian, x' is well-defined. (This action corresponds to the action on A„). Therefore X,, becomes a Zp [F] -module. Representing an element of X lim X,, as a vector (x o , x l , ..), and letting l p [F,J act on the nth component, we easily find that X becomes a module over A lim Zp [F,J. (The only thing to be checked is that x' e X, and this is easy to do). The polynomial 1 + T e A acts as y e F. We have

53x57)',

for y

E

F,

XE

X,

where .)) is an extension of y to G. Let A i , As be the primes which ramify in K 03 1K, and fix a prime of L lying above A i . Let I i g G be the inertia group. Since LIK„, is unramified,

/ i n X = 1. Since K,,,IK is totally ramified at

I

GIX = F

is surjective, hence bijective. So G=I

= XI i ,

1,

, s.

Let ai E I. map to y 3 . Then ai must be a topological generator of / i . Since Ii

XI 1,

278

13 Iwasawa's Theory of 2p - extensions

we have

ai = aiai for some a, E X. Note that al = 1. Lemma 13.14 (Assuming the above "Assumption"). Let G' be the closure of the commutator subgroup of G. Then G' = X" -1 = TX. PROOF. Since r r i G maps onto F = G/X, we may lift y E r to the corresponding element in / 1 in order to define the action of F on X. For simplicity, we identify F and I so x = yx[ 1 . Let a = cex, b = fly,

with a, 13

F, x, y

G

E

X,

be arbitrary elements of G = FX. Then

aba- l b -

l y -1 /3 axflyx = xaut3yx - l a - y - y-1 = xa(yx -i )2ti (t p)u - y - y-1 = x'(yx -1 )4i(y -1 )11 (since F is abelian) = (x1) 1 -13(YIT -1 .

Let p = 1, a = y o . We find that y"° -1 e G', so

X"0-1 For fl arbitrary, there exists c

E Zp

G'.

with /3 = yco , so

1 — = 1 — yco = 1 —(1 + Ty

1—

" (c

E

rr= o n

TA.

Since y o — 1 = T,(x") 1- I' G X'° - 1 . Similarly, (1) 1- ' E X'° -1 . Since ro = X'') -1 . This proves TX is closed (it is the image of the compact set X), G' the lemma. Lemma 13.15. (Assuming the "Assumption"). Let Yo be the la-submodule of X generated by fa i I 2 < i < sl and by X'°' TX. Let Y, = v„Y o , where

y gn — 1 = 1 + TY "' — 1 (

V,,= 1 + Then X,, -= X /Y„

for n

O.

First, consider n = O. We have K L o ç L. Since Lo is the maximal abelian unramified p-extension of K, and since L/K is a p-extension, L o/K is the maximal unramified abelian subextension of L/K. Therefore Gal(L/L o) must be the closed subgroup of G generated by G' and all the inertia groups PROOF.

279

§13.3 Iwasawa's Theorem

I i , 1 < i < s. Therefore Gal(L/L 0) is the closure of the group generated by / 1 , and a 2 , , a„ so

X 0 = Gal(Lo/K) = G/Gal(L/L o) = Xli/Gal(L/Lo)

X /<XY° - 1 , 0 2 , . Now, suppose n o-P". Observe that ar-

a,> = X /Yo .

1. Replace K by K n and T o by y

= oio. ok+ =

i o.f aio.

r. Then 0-i becomes

2

k o./i+ 1

= ai1 +al -1- ••• + o- k+1 1 . Therefore

= (vn ai)o- r,, so ai is replaced by vn ai . Finally, X) -1 is replaced by ( y gn — = v„ x70 — 1. Therefore Yo becomes v n Yo , which yields the desired result. This completes the proof of Lemma 13.15. The above result is a very crucial step since it allows us to retrieve information about X n from information about X.

Lemma 13.16 (Nakayama's Lemma). Let X be a compact A-module. Then

X is finitely generated over A <=> X/(p, T)X is finite. If xl , , xn generate X /(p, T)X over Z, then they also generate X as a Amodule. A special case:

X /(p, T)X = OX = O. U of 0 in X. Since (p, T)` —> 0 in A, that (p, T)U, U for large n. Since X is compact, finitely many U, cover X. Therefore (p, T)X g U for large n, so ((p, 1) 1X) = 0 for any compact A-module X. Now assume x l , , xn generate X /(p, T)X. Let Y = Ax + • • • + Axn g X. Then Y is compact (image of M), hence closed, so X/Y is a compact A-module. By assumption, Y + (p, T)X = X. Therefore PROOF. Consider a small neighborhood each z E X has a neighborhood U, such

n

(p, T)(X /Y) = (Y + (p, T)X)/Y = X /Y , hence

(p, TAX / Y) = X /Y

for all n > O.

It follows from the above that X/ Y = 0, so X = Y and {xi} generates X (this could also be proved more explicitly by successively considering x e X mod(p, T), then mod(p, T) 2 , etc.). The other parts of the lemma follow easily.

280

13 lwasawa's Theory of I4- extensions

Lemma 13.17 (With the "Assumption," but see Lemma 13.18). X = Gal(L/K 0o) is a finitely generated A-module. PROOF. Clearly y1 E (p, T), so Y 0/(p, T)Y 0 is a quotient of Y0/v 1 Yo = Yo/Yi g X/Y 1 = X 1 , which is finite. Therefore Yo is finitely generated. Since X/Yo = X 0 is finite, X must also be finitely generated. This proves the lemma. 111

Arbitrary K. We now remove the Assumption. Let KIK be a lp- extension and choose e > 0 such that in K oDIK, all ramified primes are totally ramified. Then Lemmas 13.15 and 13.17 apply to K colK e . In particular, X, which is the same for K e and K, is a finitely generated A-module. For n e, 1

e + yg +

A Pe

+ • • +))

15t

= 11 tf V " Ve

e

This replaces y, for K G,3 1K e , since yfr generates Gal(K 03/K e). Let Y, be " Yo for K e ," as defined in Lemma 13.15. Then

Y„ = y„, e Ye , and

X/}, for n > e.

X

We have proved the following.

Lemma 13.18. Let KIK be a 4-extension. Then X is a finitely generated A-module, and there exists e > 0 such that

X„

X lv„

Ye ,

for ail n > e.

LI

We can now apply Theorem 13.12 to X. We can also apply it to Y, with the same answer, since X/Ye is finite. So we have X

A

fi( T)mi)).

(C) A/(p)) (

We shall calculate Viy„, e V for each of the summands V on the right side. (1) V = A. By Lemma 13.10, A/(v e) is infinite. Since Ye/v,,, Ye is finite,

it follows easily that A does not occur as a summand.

(2) V = A/(p k). In this case, e

V

A/(p', V e ).

It is easy to show that if the quotient of two distinguished polynomials is a polynomial, then it is distinguished (or constant). Therefore

=

y,

((I + T)P" — I)/T ± T)Pe — I)/T =

e

is distinguished. By the division algorithm, every element of A/(pk, Vue) is represented uniquely by a polynomial mod p' of degree less than deg v„, = — pe. Therefore „kp" + c VIv r,, „V = p k(p"I

for some constant c.

P

281

§13.3 Iwasawa's Theorem

(3) V = A/(f (T )m) . Let g(T) = f (T)m. Then g is also distinguished, say of degree d. Hence T d pQ(T) mod g for some polynomial Q(T), so

T ic (p)(polynomial) mod g

for k > d.

If pn > d then

(1 + T) Pn = 1 + (p)(poly.) + T Pn 1 + (p)(poly.) mod g. Therefore

(1 + T)""- '--_ 1 + p2 (poly.) mod g. It follows that P + 2 (T) = (1 + T)Pn+2 — 1 = ((1 + T) (13-1)Pn+1 + • • • + (1 + T) Pn+1 + 1)((1 + T)

= ( 1 + • • • + 1 + (P 2 )(NlY •))(P n ± 1( T )) -=- p(1 + (p)(poly.))Pn+ 1 (T) mod g. Since 1 + (p)(poly.)

E

Ax,

n +2 acts as (p)(unit) on V = A/(g), Pn+1 d. Assume no > e, p" > d, and n no . Then P

for pn

Vn+2,e

Vn+2

Pn+2

Vn+1,e

Vn+1

Pn+l'

and T1

Vn+2,e r —

Pn+2 I

Pn+ 1

lVn+1,e

V)

— PVn-i-i,e

V

Therefore I Vivn+2,e VI = I V/PVIIPV/Pvn+1,e VI for n

no . Since (g, p) = 1, multiplication by p is injective, so IPV/Pvn+i,e VI = I V/vn+i,e VI.

Since

V/ pV ,.' A/(p, g) = A/(p, T d), we have

IV/pVI --,-- pd. By induction, i=

p d(n— no — 1 ) 1 V/ 1 Vno+i,e VI I V/Vn , e V for n > no + 1. If V/v n, e V is finite for all n, then I p d n +c, n > no + 1 I ViVn,e V

' — 1)

282

13 Iwasawa's Theory of 2p-extensions

for some constant c. If V/vn, e V is infinite then V cannot occur in our case. This happens only when (v,, f) 1, by Lemma 13.7. Putting everything together, we obtain the following.

Proposition 13.19. Suppose E = A' ()

A/(p k ))

(i t A/(g i(T))),

i=

E

where each g i(T) is distinguished (not necessarily irreducible). Let m = k i and 1 = deg gi • If E/v„,,E is finite for all n, then r = 0 and there exist n o and c such that E/v„,eEi = pmp.+In+c, for all n > n0.

E

We interrupt the proof of Theorem 13.13 to give the following, which will be used in the next section.

Lemma 13.20. Assume E is as in Proposition 13.19, with r = O. Then m = 0 p-rank (E/v, e E) is bounded as n

co.

PROOF. Recall that the p-rank of a finite abelian group A is the number of direct summands of p-power order when A is decomposed into cyclic groups of prime power order. It is also equal to dim2/p2 (A/pA). Recall that v„, is distinguished of degree pn — pe, so if deg V

El(p, v„, e)E =

( =I

Al(p, vn,e)) C)

= 0 A/(p, TP" - P')) (t= 1

e A/(19,

.=

max deg gi ,

vn,e))

(0 A/(p, reggi))

(Z/Pl) 5(Pn- p ' )+1 .

Therefore the rank is bounded <=>s = O. This proves Lemma 13.20. We now return to the proof of Theorem 13.13. We have an exact sequence 0

0

where A and B are finite and E is as in Proposition 13.19. We know the order Of E /v E for all n > n0 , It remains to obtain similar information about 1Ç. At the moment, all we can conclude is that en = mpn + In + c„, where c„ is bounded. The following lemma solves our problem.

Lemma 13.21. Suppose Y and E are A-modules with Y — E such that Y /v„,,Y is finite for all n > e. Then, for some constant c and some n o , I Y /V ne 'I

=

pc

I E/v„, e El

for all n

no .

283

§13.3 Iwasawa's Theorem

PROOF.

We have the following commutative diagram 0

> Y/v , e Y

Y

O'n

Orf;

E/vE

vE

0

0

0

There are the following inequalities. (i) (ii) (iii) (iv)

I Ker 01 I I Ker I Coker 0 I I Coker 44, I I Coker 44 I I Coker 0 I 'Coker 0 I. I Ker I Ker

Inequality (i) is obvious. (iii) holds because representatives of Coker 0 give representatives for Coker O n". For (ii), multiply the representatives of Coker 0 by v,. By the Snake Lemma (see Clayburgh [1 ], or any book on homological algebra), there is a long exact sequence

0 Ker 0,, Ker —> Ker On" —> Coker 0, —> Coker 0 —> Coker O n" —> O. Everything is straightforward except the map Ker O n" —> Coker 0' . Let x e Ker On". There exists y e Y which maps to x. Since 0 (i2) maps to 0 in E/v n,,E by the commutativity of the diagram, we must have 0(y) E v„,,E. One checks that 0 (i2) mod On'(v, Y) depends only on x. The map xi-4 0(y) is the desired one. It remains to check exactness. This is left to the reader. It follows that

Ker 011Coker

I Ker

Ker

I I Coker I,

by (ii). This proves (iv). Now suppose m

n > O. We have the following inequalities.

(a) I Ker 0,, I I Ker 0.1 I Coker 0.1 I (b) 'Coker I Coker 0;'„ I. (c) ICoker (p;; I

e/v„, ,)v n, . Therefore V m, e Y Ç v n,, Y, so For (a), observe that v. e = ker 4 Ker 0' . For (b), let v m,e y e V m E. Let z E v„,,E be a representative for v n, e y in Coker 0' . Then ,

v n, e y

—

e

z = 0(v„,,x)

for some x e Y.

Multiply by v., e /v,,,, to obtain

vin, e y

(

Vm e

: , z = 0(v,n x)= cy,„(v,n,,x). ,

vn

e

284

13 lwasawa's Theory of . p -extensions

So (v„,, /' n, e ) times representatives for Coker 0„' gives representatives for Coker This proves (b). Since v,E v„ . ,E, inequality (c) follows easily. By (i), (ii), (iii), (a), (b), (c), the orders of Ker Coker 0'„, and Coker 0',: are constant for n > n o , for some n o . It remains to treat Ker 0'„'. By the Snake Lemma, I Ker 0;,I I Ker 0,; I I Coker 0I = I Ker I I Coker 0;, I I Coker

I.

(In any exact sequence, the alternating product of the orders is 1; proof: replace 0 —0 A —o 3 —0 • • by 0 --■ B/ A —o • • and use induction on the length of the sequence). It follows that I Ker 0',: I must be constant for n no . Lemma 13.21 follows easily. 111 We therefore have E as in Proposition 13.19, integers A. y, and an integer n o such that

0, > 0, and

pe- = IX„I = IX1Y„ = (const.)IE/v„, e El =

n

pp"

for all n > no .

This completes the proof of Theorem 13.13.

Li

§ 1 3.4 Consequences Proposition 13.22. Suppose K,,/K is a 7 p - extension in which exactly one prime is ramified, and assume it is totally ramified. Then A n X,

X/((1

T)P n — 1)X

and p ,}/ h o <-> p 1 n for all n > 0. PROOF. Since K,e /K satisfies the "assumption" in the proof of Theorem 13.13, we may use Lemma 13.15. We have s = 1, so Yo = TX and Yn + - ox. This proves the first part. If p,i'ho , then X/TX = 0, so X/(p, T)X = 0. By Lemma 13.16, X = 0. This completes the proof.

Tr

Of course, the last statement of the proposition also follows from Theorems 10.1 and 10.4, and, in a special case, from Exercise 7.4. Proposition 13.23. u = 0 <=> p-rank(A„) is bounded as n —0 oc. PROOF. We have ye E with E as in Lemma 13.20. By the lemma, U = 0<=> p-rank(E/v„ ,E) is bounded. From the proof of Lemma 13.21, we have an exact sequence C„ --> Ye/ V,,, e Y —>

—o B„ —o 0

285

§13.4 Consequences

with I C„ I and IB„ I bounded independent of n. It follows that it = 0 <=> p-rank( Ye/v„, e Ye) is bounded.

But

A„ '' X„ = Xlv„,, Ye and X/ }', is finite. The result follows easily.

El

Suppose each K„ is a CM-field. Then K o3+ IK + is a 4-extension (cyclotomic if Leopoldt's conjecture is true, by Theorem 13.4). If p is odd we may decompose the p-Sylow subgroup A„ of the class group of K„ as A n = A: 0 A n- .

Also, X n = X,-,1- (-.) X n- ,

hence

X = X+ (D X - . We obtain, as in the proof of Theorem 13.13,

A,ç± '-.' X,;

X I lv,,,j ± .

If pe is the exact power of p dividing h., - , then en = et-,1- + e.

We obtain

n6-+ ,

+ = /1 ±n + 1.2--+ pn + v + ror n - f ew

with V=V

+

-1-

-

V.

The analogue of Proposition 13.23 applies, so it + = 0 <->p-rank(A„± ) is bounded.

If p = 2, we cannot decompose A. However, if

= {alJa =

—

a}

(J = complex conjugation)

then everything in the proof of Theorem 13.13 works for A,; , X,, etc. We may obtain en+ by looking at the class group A n(K„±) of K„+ , rather than A„± (cf. Proposition 10.12). We again obtain en± = A ± n + 12± 2" ± v ± . From the exact sequence

286

13 Iwasawa's Theory of Z r-extensions

we have p = p

+ p. - , etc. We also have, as above, p + = 0 <=> 2-rank A(K) is bounded, = 0 <=> 2-rank A is bounded.

Proposition 13.24. Let p be prime. Suppose K is a CM :field with Cp el( and let K x / K be the cyclotomic 7Z-extension. Then =0<=>

=0

is trivial. For " we know that p: = 0 p-rank A n- is bounded. By Theorem 10.11 and Proposition 10.12, p-rank A: (or 2-rank A(K)) is bounded, which implies p + = 0. This completes the proof.

PROOF. "

This result also completes the proof that p = 0 for abelian number fields (Theorem 7.15), since in Chapter 7 we showed that = 0 for all such fields. Proposition 13.25. Suppose K cc / K is a 4-extension and assume p. = O. Then X

C) (finite p-group)

lim A n

as 4-modules. PROOF.

We have X E

A/(gi(T))

where each gi is distinguished and

E deg gi = A_

A/(g j(T))

By the division algorithm,

Zdp eg

Therefore E Since X is a Zp- module, which is finitely generated since E is finitely generated, the result follows from the structure theorem for modules over principal ideal domains. Proposition 13.26. Let p be odd. Suppose K is a CM-field and K co/K is the cyclotomic 7Z-ex tension of K. Then the map An —4 An+

1

is injective.

Remarks. The map A n+ — ■ A n++ is not necessarily injective (Exercise 13.4). If p = 2, A,-; +, is not necessarily injective (Exercise 13.3). Since the map of ideal class groups C„ C„.+. followed by the norm is the pth power map, the kernel is always in the p-Sylow subgroup. PROOF. Suppose / is an ideal in A„ which becomes principal in K n+ 1, so

I = (a)

287

§13.4 Consequences

Let u be a generator for Gal(K„ 1 /Kr). Then

= ( 1 ).

(a6-1 ) = Consequently e

En + = units of

E

Let N be the norm for K n+ 1/K u . Then 1 = 1.

Ne = (NŒ)

For those who know cohomology of groups: we easily obtain an injection

Ker(A„ —0 A

1

—4

H i (Gal(Kn+ i /K„), En+ 1 ).

Now suppose I represents a class in A,-,- . Let J denote complex conjugation. Then

/ 1+J = (#),

with fi

E

K n (so fi = fl),

hence a l +J

Pi

=

with ij E

Let

and 8

8 1 = 0(71

2

1 =

eE n +1 .

rI

Then 1+J

(Œ

Y) 1 1+Jr-1

02y—

=

(

cr— 1)1 —J

+1.

But

En-+ 1 =

+ = roots of 1 in Kn+

by Lemma 1.6. Therefore some power of e i is killed by 1 + J, hence is a root of 1, again by Lemma 1.6. Consequently

Ei E Wn+1. (Alternatively, since p 2 we may assume fi E Kn+ (see Theorem 10.3) and consequently j = +j/fi is real. Therefore q' = 1, so Lemma 1.6 applies directly to c i ). Also, observe that

Ne i = (Na i ) ' = 1. Lemma 13.27. If giE W1 -1-1 and Ne i = 1 then e l = e'2-1 with e 2 E 1/ 1 (Gal(K„ i /Kn), W„ + 1 ) = 0).

Wn +i

(so

288

13 Iwasawa's Theory of

Zp -extensions

Hilbert's Theorem 90 tells us that ei = y6-1 with y e K n+1, but we already know this with y = ai . We want v e Wn + 1. Consider the following two sequences: PROOF.

J

1

14in

1

Wn+1 n Ker N

1

14in +

Kin + 1 W„ 1.

The first is obviously exact. The second is slightly more difficult. I f 4 It K0 then 4 Et K m for all m. Otherwise, a nontrivial subgroup of (/p1))< would be in Gal(K co/K0), which is impossible. Since N: Wn is the pth power map, it is surjective in this case, hence N: Wn -4 Wn is surjective (in fact, = 4M for some 111 Wn+ = W). If 4 e Ko then K n+1 = Kn( ), where n 1. Also, Wn+ = x at > for some t with (p, t) = 1, and Wn = aP > X A trivial calculation shows that Nç = and Nç = U, hence = ar >. Therefore N is surjective in this case, so the second sequence is exact. We obtain

at>.

,

'WI =

Wn

=1141,,+1 n

Ker NI.

Since Wncr ill OE Wn+ln Ker N,

we have equality. This proves Lemma 13.27. We can now complete the proof of Proposition 13.26. We have cx cri =__ with E2 e W,1. + 1.

Therefore (al" = al 62

82

SO Oli —E

K

.

E2

But ( al

= (cci) = (a 2 ) =

r2

in

E2

By unique factorization of ideals, we must have (4 1)

/2

in K.

e2

Since p is odd and I has p-power order in A we must have I principal in K. This completes the proof of Proposition 13.26. 0 ,,

289

§13.4 Consequences

Proposition 13.28. Let p be odd, let K be a CM-field, and let K adK be the cy clotomic Zp -extension. Then X = urn A„ contains no finite A-submodules. Therefore there is an injection, with finite cokernel, X - =-*

Al(p k )

e Al(g i(T)).

PROOF. Suppose F g X - is a finite A-module. Let y o be a generator of Gal(K 0,11C). Since F is finite, yg n acts trivially on F for all sufficiently large n, n0 . Suppose say n

ThenX rn +l x„., under the appropriate norm map, and x. 0 0 for all sufficiently large in, say rn> mo . Let m be larger than m o and n o . By Proposition 13.26, x. o 0 when lifted to A m- Apply the map + 7E - + y + • • • + T (cr - "Pto x. Since in no , it acts as p on x. Also, it is the norm from K., 1 to so it maps x„ to x.. Therefore 1

px, = x. 0 0,

in

1,

SO

px O. It follows that multiplication by p is injective on the finite p-group F, so F = O. This completes the proof. LII Corollary 13.29. Let p be odd. Let K be a CM-field and let K colK be the cyclotomic 7Z-extension. If p - = 0 then

XPROOF.

Vp •

Proposition 13.28 plus the analogue of Proposition 13.25 for X - .

Usually the finite kernel and cokernel in the pseudo-isomorphism make it difficult to obtain much information at finite levels of the l u-extension. Proposition 13.28 is useful since it eliminates half the problem. For a situation where there is also a trivial cokernel, see Theorem 10.16. In the above we have used the decomposition X = X + X - for odd primes. More generally, suppose we have the following situation

KY

y

290

13 lwasawa's

Theory of 4-extensions

where A is a finite abelian group and Gal(K ap/K) = A x F. For example, K = U ( r ), K co = (L1Gp .), k = U, and A = (Z/pE)X . Then A acts on X, since A x F acts on X by conjugation in the same way as the action of F on X was defined. If p does not divide the order of A and if the values of the characters X e A are in Zp (rather than an extension), then we may decompose X according to the idempotents Ex of

E3 E x X. z For example, in the above we used A = Gal(K/K +) and e + = (1 + J)/2. In X

the present case we obtain Ex X-

A/(p5 0

A/(g 1) (T))

for some integers kf and distinguished polynomials gl(T). We also have = /ix , etc. Returning to the general case, we consider the Cr-vector space

V=X If

X then it is easy to see that

e A/(p)

e A/(g i(T))

V -'1CD C p[T]Ag ja», which is a finite-dimensional vector space. The group F acts on generator 70 acts as 1 + T. So g(T) = fi g J(T)

V; the

is the characteristic polynomial for yo — 1. If A = Gal(K/k) is as above, with no assumption on IA I or the values of E A, then we may decompose V = Then

g(T) =

g(T),

where g(T) is the characteristic polynomial of yo — 1 on , V. We shall discuss the significance of these polynomials when we treat the main conjecture.

§13.5 The Maximal Abelian p-extension Unramified Outside p Often it is more convenient to work with an extension larger than the p-class field and allow ramification above p. This is what we did in the proof of Theorem 10.13. In many respects, the theory is more natural in this context,

§13.5 The Maximal Abelian p-extension Unramified Outside p

291

especially from the point of view of Kummer theory. In this section we sketch the basic set-up, leaving the details to the reader. The proofs are very similar to those in Chapter 10. We start with a totally real field F. Let p be odd, let K o = F(Cp), and let K oo/K 0 be the cyclotomic 4-extension. Let M oo be the maximal abelian p-extension of K oo which is unramified outside p, and let X„o = Gal(M cojK oo ). Then X is a A-module in the natural way (just as for X = Gal(L 13 /K 03 )). Let M „ be the maximal abelian p-extension of K is unramified outside p. Clearly M n K. We have Gal(Mn/K) where con = yg" — 1 = (1 + T)P" — 1. The proof is essentially the same as for Lemma 13.15, namely computing commutator subgroups, but in the present case we do not have to consider inertia groups. From Corollary 13.6 we know that Gal(Mn/K 0 )

Fp2Pn+ 1+6"

x (finite group),

where r 2 = r2 (K0) and 6„ is the defect in Leopoldt's Conjecture (see Theorem 13.4). Therefore Ep2P"'n x (finite group). 1. 00 /w„X By Lemma 13.16, X is a finitely generated A-module, so (A-torsion) for some a > O.

Lemma 13.30. 6„ is bounded, independent of n. Suppose 6„ > 0 for some n. Let E i , , Er be a basis for E l = E i (K n). We may assume £6. 1_ 1, Er are independent and generate E l over 4. Then PROOF.

El d, with aii e Zp , > 6n for 1 < i < 6„. Let a'i; be the nth partial sum of the p-adic expansion of aii . Let

Ei =

rh = ei

e E1 ,

1 < i < bn .

Then i is a path power in F 1 Flo p U 1 , A, and /I I , , lb," generate a subgroup (Z/paZ) 6- of K nx /(K nx)P". Since Cp E Ko by assumption p „ E K n . Therefore the extension Kn(triPP"})/K. has Galois group (Z/p"Z ) n. Clearly this extension is unramified outside p. Since each q i is a path power in U 1 o for each hip, these primes split completely hence do not ramify. Therefore the Galois group X n of the Hilbert p-class field of K n has a quotient isomorphic to (Z/paZ) 6". In the decomposition of

292

13 lwasawa's Theory of Zr-extensions

X, the terms of the form A/(e) cannot account for this for large n. The term of the form ei A/(g i(T)) can only yield (/p"), where ). = E deg Therefore 5 , < i This completes the proof. III If Cp K o , the lemma is still true. Simply adjoin Cp and use the easily proved fact that if K g L then (5(K) < 6(L). The above result perhaps could have been conjectured from Theorem 7.10 (although we already know 6, = 0 in that situation). Intuitively, the number (5„ should be approximately the number of occurrences of L(1, = 0 for K„+ . Since each series f (7; 0) has only finitely many zeros,

Lp(1, Ot)i) = f((1 + q0) — 1, 0)

0

when 1,i/ has large enough conductor. So the number of z with L p(1, bounded. By the lemma,

0 is

7p-rank X„,/to„X„ = r 2 pn + 0(1). By the structure theorem for X„ , we see that the A-torsion contributes only bounded 7p-rank (at most /1) and Aa/con Aa yields apn. Therefore we have proved the following. Theorem 13.31.

A r2 CI (A-torsion).

One advantage of using &1 than X is that it is easier to describe how L c,, is generated. Since all p-power roots of unity are in K oo , Mcolk, is a Kummer extension. There is a subgroup

V g K:, V = {a

UplZ p

p' I various n 0 and a e K1 }

(it is not hard to see that all elements of such that

Upl7p are of the form a

p')

M, = K ao (fa i n). There is a Kummer pairing

x V W p. = p-power roots of unity, just as in Chapter 10. In particular,

(ax, o-v) = (x, v ) ,

a E Gal(K

Let I. be the group of fractional ideals of K. and let / = U Since a p' gives an extension unramified outside p, and since a e K. for some in, it follows that

(a) =

Bir •

B2 in some /.,

§13.5 The Maximal Abelian p-extension Unramified Outside p

293

where B 1 e /,„ and B2 is a product of primes above p. Since all primes above p are infinitely ramified in a cyclotomic 4-extension, B2 is a path power in I. Hence we may assume (a) = Br.

We obtain a map

V -4 A = lim> A „ a 0 r" I-0 class of B,.

It is not hard to see that this map is well-defined, i.e., independent of m and the representation a 0 p. It is also surjective, since A e A AP" = 1 for some n (see Exercise 9.1). As in Chapter 10, the kernel is contained in

E.

02

Q p/Zp,

where E. = U E(K„). Since we are allowing ramification above p,

E. 0/ Q,/7Z 1,

V,

so it follows that this gives the kernel (cf. Theorem 10.13, where the situation is essentially the same). We now have an exact sequence

1 -4 E co

02

Qp/Z p -4 V-4/1.-41.

Let A = Gal(K o/F), which is a subgroup of (Z/pZ) x = Gal(()/). For E a is a character of A (i j mod I A I <=> w1 = wi on A). Let

ei =

1

E co - 'OA

II SGA

Everything decomposes via these idempotents. Wp . is in the e l component. If i is odd and i 1 mod I A I, then ei(E Up/Zp) = 0, since [E : W = 1 or 2 for each K. We obtain £iV

£iA x ,

i

odd, i

1 mod I A I.

Note that by Proposition 13.26, Ei A oc, = U ei A n . As in Chapter 10,

is nondegenerate, hence

Ei X ao X Ei t4 Œ,

Wp . ,

+1 1 mod odd, i # 1 mod I I,

is nondegenerate. Therefore

Hom2,(e i A co , Wp .), where Gal(K co/F) acts via (o- f )(a) = o- ( f(a 'a) (cf. Exercise 10.8). This last equation is often written in another form. Let

294

13 lwasawa's Theory of Zr -extensions

where the inverse limit is taken with respect to the pth power map (which is the same as the norm map from Q(CpPI + 1) to Q(Cp„)), Then T

lp , as abelian groups,

but the Galois group acts via ria (t) = at for ae A x (1 + pl p ) where we are writing T additively. Let T(-1) = Hom 2,(T, Zp) with the Galois action on Hom as above. Then T" ) Z p , as abelian groups. If f c T" ) and t

E

T then, since ria acts trivially on Zp ,

(a a f)(t)= o- a(f (a

= f (a- t) =

SO

cra f = a 1 f It follows that T 0 2F 11-1-) Define the "twist"

lp with trivial Galois action.

co(-1) by r

This is the same as changed:

Do as a

Zp- module

but the Galois action has been

cr a(x 0 f) = a(x) 0 a 1 f = coi(a)a'(x 0 f). Proposition 13.32. Hom2 ,(e 1 A co , Up/Zp) as A-modules, where +j 1 mod II, is odd, and i # 1 mod II.

PROOF. We shall show more generally that Hom2p(B, Up/Ip) ="' HOM2 p B, Wp ) (

for any A-module B. There is an isomorphism of abelian groups Op/Zp

a Pn Choose a generator t o for T" ) as a Zp- module. If we ignore the Galois action, we obtain an isomorphism by mapping

h

(Oh) 0 to,

295

§13.6 The Main Conjecture

for h E H0111 210, U p/Z p ). Let a = aa E F. Then (crh)(b) = a(h(a- I))) = and O()h 0 to) = GrOh6 - 1 0 ato

= a oha.- 1 0 a - 1 to = Oha - 0 t 0 . Therefore ah 1-+ a(g5h 0 to) under the above isomorphism, so the Galois actions are compatible. This completes the proof. The proposition says that the discrete group ei A co and the compact group e. co (— 1) are dual in the sense of Pontryagin.

03.6 The Main Conjecture For simplicity, we assume p A 2 in this section. Consider the Zp -extension 1)(C,0 )/0(4). In Theorem 10.16 we showed that if Vandiver's Conjecture holds for p then E,X for i = 3, 5,

Ai( f(T,w''))

, p — 2, where p) .S — 1, co'

f((1

= Lp(s, co l -').

Factor f (T, co l ) = pgi(T)U ,(T) with gi distinguished and U1 e Ax. We know that = 0 by Theorem 7.15. Therefore ei X

A/(g,(T)),

which is in the form of Theorem 13.12. So in this case the distinguished polynomial in the decomposition of 8,X is essentially the p-adic L-function. This is conjectured to happen more generally. Let F be totally real and let K o = Fg), K Fgp .,). Let A = Gal(K o/F) c (Z/pZr . Let x e

A be odd (i.e.,

x(J) = —1). Then A/(p 4 ) IOC) A/01(T ))

ex X

with finite cokernel. Let

=

E kf

and let

gx (T) = lY tx

H gl(T).

296

13 lwasawa's Theory of 2p-extensions

It has been shown (see Barsky [4], Cassou-Nogués [4], Deligne-Ribet [1]) that there exists a p-adic L-function L p(s, cox') for the even character wx -1 . If F = Q, this is the usual p-adic L-function. For larger F, the existence is more difficult to establish. Let 70 be the generator of Gal(K co /K 0 ) corresponding to 1 + T Define K o E 1 + pZ p by ), C0 Cpri = C7,2 for all n > 1. It has been shown that there is a power series fx E A such that L p(s, tox -1 ) = f;(4 — 1),

x 0 co.

The Main Conjecture (First Form). h(T) = gx(T)U x(T) with U x(T) E A x . We may also state a slightly different form. For simplicity, assume F = Q, though any totally real field F could be used. Let x 0 co be an odd Dirichlet character of the first kind (see Chapter 7, let K x be the associated field (see Chapter 3), and let K o = K, = (Cp 4 Let 0 ? Zp contain the values of x and let fx E O[[T]] satisfy fx (Kso — 1) = Lp(s, cox -1 ), as in Theorem 7.10. Then

MT) = p''' fx (T)U x(T) with U x E CHTEx ,f-X distinguished, and !ix > 0 (in the present case, /ix = 0 by Theorem 7.15). Consider the Cr -vector space V = X 02 p Cp as at the end of Section 13.4, and let g7(T) be the characteristic polynomial for To 1 acting on Vx = E x V —

The Main Conjecture (Second Form). fx (T) = gx (T). The advantage of this form is that we may consider a larger class of characters x. The disadvantage is that we are no longer requiring the tt obtained from Ex X (when this module is defined) to equal the /ix obtained from f. For abelian extensions of Q this makes no difference since both are 0. The motivation for the main conjecture comes from the theory of curves over finite fields (or, function fields over finite fields). Let C be a curve (complete, nonsingular) of genus g over a field k of characteristic I 0 p and let J be its Jacobian variety (if we were working over C, J would be Cg modulo a lattice). Let J,. be the points on J of p-power order defined over the algebraic closure k of k. This is essentially the analogue of A lim A n (or of A - = H ■..i A„-- ) for cyclotomic 4-extensions. Then Jp =---- (Q p/Zr )", as abelian groups.

297

§13.6 The Main Conjecture

Therefore (compare Corollary 13.29)

HomE,(Jp,

Op/Zp) 429

and

OE,

F101112 p (fp , Q 1,/7L)

Op29.

The Frobenius automorphism of k over k acts on this last space, and a classical theorem of Weil states that the characteristic polynomial is the numerator of the zeta function of C (see Weil [5]). Therefore, the Main Conjecture is an attempt to extend the analogy between number fields and function fields to this important situation. The Main Conjecture (first form) has been proved by Mazur and Wiles for F = (11, Ko = 0(4). In fact, they proved a slightly stronger statement, which we now briefly describe. Let R be a commutative ring and let M be a finitely generated R-module. For some r E Z and B ç IV', there is an exact sequence

0

B (2> Rr

11

O.

Consider the r x r matrices of the form

(01 41) = 00r) where (b 1 , . , b,.) runs through all r-tuples of elements of B. The Fitting ideal FR (M) (see Fitting [1], Mazur-Wiles [1] or Northcott [1]), is defined to be the ideal in R generated by the elements det(0) for all such 4:1). It may be shown that FR (M) is independent of the choices of r and tP, hence depends only on M. It is not hard to show that if

Ann(M) = {a e RiaM = then

(Ann(M)y ç FR (M)

g

Ann(M).

EXAMPLES. (1) R = Z, M = a finite abelian group. Then FR (M) = IMIL

(2) M = R/I, where I is an ideal of R. Then FR (M) = I. (3) R = A and M satisfies

0 -> M

Al(g i(T))-> (finite) -> O.

Then it can be shown that

F A(M) = (119)A. (4) If M -> N is a surjective map of R-modules then FR (M) ç FR(N).

298

13 lwasawa's Theory of 4-extensions

(5) If I is an ideal of R then

FR AM/IM) = FR (M) mod I. (6) If R = Z p [G] with G Z/p"Z, then

FR(M) = F R(HOME E,(M ) where Horn is an R-module via (of )(m) = cr(f(a 1 m)) = f(a 1 m)

for a

E

G.

Let i # 1 mod p — 1 be odd. Then there exists f,,,,(T) e A such that f((1

p)s — 1) = L(—s,

Let ei be the idempotent and recall that

= lim Ei =

On .

Define

X (c; = Hom z, (e i A co , (lap/Zp), It can be shown that if Ei X

A/(g i(T)) then X (coi )

A/(/T)), where

gi(T) = gi((1 + T) -1 — 1) (i.e., y o is replaced by y,; 1 ). See Iwasawa [25, p. 250], where a slightly different Galois action is used.

Theorem (Mazur-Wiles). Let j # 1 mod p — 1 be odd. Then (i)

FA XLJ = (fa(T)),

(ii) F mp„(1.)(ciAn) = F ni (p„(T))(XVP,r(T)X (1 ) ) = (i(T) mod P(T)), where P(T) = (1 + T)P" — 1.

By Example 3, part (i) yields the main conjecture. Note that (f(T) mod 13,(T)) is essentially part of the Stickelberger ideal (see Chapters 6 and 7; the difference is that 7 0 is replaced by y o-1 ). So this result identifies the Stickelberger ideal. If we assume Vandiver's conjecture, then we are in the situation of Example 2 above. But there are often many noncyclic modules which give the same fitting ideal as a cyclic module gives, as shown by Example 1. So the theorem does not imply Vandiver's conjecture or the cyclicity of the class group as a module over the group ring. The proof uses delicate techniques from algebraic geometry and the theory of modular curves to construct unramified extensions of (11(4,i) for each n. This makes X big enough that FA (Xn (f,,,(T)) for each f. If ,

G

no (T))

299

§13.7 Logarithmic Derivatives

with finite cokernel, then def

FA (X) = (H MT)) = (goi(T))• Hence

J(T) I jwi(T). Therefore deg„fo, < deg„ "4„„ where deg„ denotes the Weierstrass degree, which is the degree of the distinguished polynomial in the Weierstrass decomposition. But = deg, fi by Theorem 7.14. Also

Ei

E deg, as in the proof of Theorem 13.13. Therefore

deg„ f = deg„ SO

= ()(unit). This yields (i). Of course, the main part of the proof is the construction of sufficiently many unramified extensions. The techniques are an extension of those used to prove Ribet's converse to Herbrand's theorem (Theorem 6.18). For further details we must refer the reader to the paper of Mazur-Wiles [1] or to Coates' Bourbaki talk [8]. One application is the following result (compare Proposition 6.16):

kiAoi = p-part of B 1 , „

# 1 mod p — 1, i odd).

This follows from (ii). Since A/(P o(T)) = order, as in Example 1. But

the Fitting ideal gives the

f(T) —= f(0) = L(0, w 1- `) = —B 1 ,

mod Po(T),

which yields the result.

§13.7 Logarithmic Derivatives This section provides the machinery needed for the next section. We first give a classical homomorphism, due to Kummer, and then present its generalization, due to Coates and Wiles. Let p be odd and let U 1 be the local units of 0)7(4) which are congruent to 1 mod( 4 — 1). If u e U 1 (or U), we may write

u = f

(

-

1),

with f (T) e A'.

Let

D = (1 + T)

d dT

300

3 Iwasawa's Theory of lp -extensions

as in Chapter 12, and define

(u)= Dk 1 (1 + T)4

J T=0

mod p, for ! k p — 2.

If g E A is another such power series, then (f/g) — 1 as a zero. Write

E

A and has

4

1

f - - 1 = pPP(T)A(T)

with P(T) distinguished and A

E

A' . Then P

1

(

— 1) = 0, so

p

+ T)P — 1)

divides P(T). Therefore

f(T) = g(T)(1 + h(T)B(T)) for some B

E

A. We obtain

—=—+ Y

f

Since h

(1 + hB) .

TP -1 mod pA, f'

g' - mod(TP -2 , p).

f

g It follows that O k(u) is well-defined for 1 < k < p — 2, so we obtain a homomorphism 0 k: U1

1/pl.

These maps were first defined by Kummer. However, he worked with polynomials, let T = ez — 1, and considered f(ez — 1). Then D = (1 + T)(d/dT) corresponds to didz, so

dy

log f(ez — 1 )1. = 0 mod p. OA) = (— dz For an application due to Kummer, see Exercise 13.9. Lemma 13.33. Let u E U 1 and let 1 < nu a 1 modg, — '.

< p

2. Then Ok (u) = 0for 1 •k

PROOF. If u 1 modg, — 1)''' we may take f (T) = 1 + Tn ±1g(T) with g E A. Since f '(T) E T"A, O k (u) = 0 for all k n. Conversely, suppose O k (u) = 0 for all k < u. Let u = f 1). Write

f' (1 + T) —

ao + a1 (1 + T) + • • +

1 (1 + T'

Tn A.

301

§13.7 Logarithmic Derivatives

Then 0 1 (u)

ac, + a l + • • • + a 1 and O k (u) al + 2k -1 a2

(n _ 1)k- l an _ i,

for 2 < k < n. The determinant det(

0

j < n — 1, 1 < k < n,

with 0 0 = 1, is a Vandermonde determinant and is nonzero mod p. Therefore ai 0 for 0 < i n — 1. It follows that f'(T) E (T", p). Since f e A (so TPlp does not occur), integration yields f (T) a mod(T 1 , p) with a e Zp • Since a u 1 mod(C p — 1), we have a 1 mod p. Therefore

u = f(Ç —

1)

a mod(C p — 1)"

1 mod( p —

1)11+1

.

This completes the proof. Lemma 13.34. Let aa e Gal(()/). Then 41)&7 au) = akOk (u).

PROOF. Let u = f ( — 1). Define

g(T) = f((1 + T)a — 1). Then

(T a u

= g(4 — 1), and (1 + T)?- = a(1 + T)a

((1 + T)a — 1).

By induction, D k- 1

(1 + T)

=

ak Dk -1 (1 + X) Li X = (1 +

-1

The lemma follows easily. Lemma 13.35. Let ei , 0 < i < p — 2, be the idempotents of Z p[Gal(O(C p)I0)]. If u G U 1 then ifk {0 O k (e i u) = Oi(u), i f k =

j.

PROOF. By Lemma 13.34,

O k(e i u) =

1 /)

P-1 I a=1

(1) - i(a)ak

10 mod p, Ok(14) mod p,

if k if k = j.

Lemma 13.36. If i 1, mod p — 1 then e i U is cyclic as a Zp - module. For i = 1, el U 1 ap> x (cyclic Zp -module). If 2 i p — 2 and u EU 1 then u generates U, (u) 0.

302 PROOF.

13 Iwasawa's Theory of 7I-extensions

First, let i > 1 be arbitrary. Since gap — 1)/ ( p — 1)

a mod( p —

1),

p— 1 =

i dg ( 1 —

—

(cap

—

i(a"(P

1)

a= 1 1

1—

p-

1

P — 1 a= 1

co - L(a)a'( p — 1) i

)i MOd(Cp

FE1 —

1 )i+ 1.

From the p-adic expansions, we easily see that , rip generate U 1 mod( p — 1)P', which is easily seen to be U 1 /E/ 1;. By Nakayama's Lemma (see Lemma 13.16), they generate U 1 over Since th e ei U 1 for each i, th must generate ei U, for i o 1,p, and 17 1 Cp and rip together generate 61

Ut.

Now suppose 2 2 and let U E 6U 1 . Then u = for some b e Z p , and u is a generator .4=> (p, b) = 1. By Lemmas 13.33 and 13.35, NO 0 0. Therefore O i(u) = bOi(rii) 0 0 <=>(p, b) = 1 <=> u is a generator.

This completes the proof. Corollary 13.37. Let 2 < i p — 2. There exists 1 = 1 0 1 with such that ei

AP

=

(0)2(A— —CPO)

generates 6U1 . (We divide by co(1 — 1) to get an element of U 1 ).

By Lemma 13.36, it suffices to find 2 such that Ogi) o 0, and by Lemma 13.35, we can work with (2 — j,)1co(1 — 1). Let PROOF.

f(T) =

—1—T co(/' — 1) •

Then -

P f (p- 1) = 0)(2 —C1)

and f' = 1 + (1 + T)-7

+ T—

(1 + T\ )

[1 +

(1 + Ty -1 1+T—

303

§13.7 Logarithmic Derivatives

It follows that Di-1 (1 + T)-r_

Er.

P-1

i

(1 ±

(1 ± 7)P Di-1(

A

;=-_1

A)

A mod pA. 1 + T — A)

For 2 < i p — 2, we obtain, since A P-1 = 1,

(1 _1 A

(

PV I

—

'

=

p(A) ,

1)

where P(X) = X2

21— 1 x-p— 3 +

(p

1 )1

1.

Since P(X) has degree p — 2, and since p-1

P(1)

j=1

co(j)i- 1

0 (mod p)

(in fact, 1 is a multiple root), at least one A 0 1 satisfies P(A) # 0 mod p. Then Oi((.1. — p)/0)(/1 1 )) 0, SO i generates ei U 1 . This completes the proof. We now consider the generalization to higher levels. Let Of) be the local units of O pG,.+1) which are congruent to 1 mod( p., — 1). The norm Apn 4- 1) to p(Cpn) maps U(in +1) into U(f), so we define Nn,n-1 from

U = lim U(f ). Then U is A-module in the usual way and is also a Zp [Gal(Q(C p)/Q p)]module. Theorem 13.38. Let u = (u n) e U. Then there exists a unique fu e A such that fig pn+ — 1) = un for all n > 0. The map

U

—>

AX

u fu gives a bicontinuous isomorphism between U and the subgroup off E A satisfying f(0) 1 mod p f((1 + T)P — 1) = fl f((1 + T) — 1). v=1

304

13 lwasawa's Theory of 4-extensions

PROOF. Corollary 7.4 implies the uniqueness of fn .

For simplicity, let vn = 1. Assume for the moment that fu exists. Since fu(0) fu(Cp — 1) = uo 1 mod ( Cp 1) (;

n

and since fu(0) E Zp , we have L( 0) 1 mod p. Observe that the conjugates of p fl+1 under Gal(aGpn , i)/0(40) are {apn, I‘', " = 1) . Therefore

H1 fu (1 + v„) - 1) = (

N„,„_ fu (v„) =

Also

f 1 + ((

v„)P

— I) = fu(v„ -1) = un- 1.

Therefore, again by Corollary 7.4, we obtain (*). Observe that if f satisfies (*) then

N„,„ _1 f (vn) = SO

(f(v, i)) e Jim We therefore have a homomorphism X= { f eAX satisfying (*)} 4 U. Clearly X is closed in A X, hence compact. The topology on U is induced from the product topology on FL, uv, so U is compact. Any neighborhood of 1 in U contains a neighborhood of the form Vndc = {(14) l ti n

1(MOd Pk ), n

N).

If

f 1 mod(p,

T

)

(

PN "

then

f (V n)

1 MOd(p, v) k' 1 MOd pk

1)

for n < N.

So the map g is continuous at 1, hence everywhere, since it is a homomorphism. Now assume the existence part of the theorem, so g is bijective. Any closed set of X is compact, hence its image under g is compact, hence closed. So g sends closed sets to closed sets. Therefore g -1 is continuous, so g is bicontinuous (this argument shows that any continuous bijection between compact Hausdorff spaces is bicontinuous). It remains to prove the existence. We need several lemmas. Lemma 13.39. There exists a unique map N: A A such that

(Nf)((1

T)P — 1) = fl f(C(1 yP = 1

T) —

1).

305

§13.7 Logarithmic Derivatives

Let g(T) be the power series on the right, defined by the product. Observe that for CP = 1, g(C(1 + T) — 1) = g(T). PROOF.

Suppose we have a 0 ,

E Zp and q(T) e A such that

, 1

g(T) =

L ai((1 +

T)P

1)! + ((1 + T)P — l)g„(T).

i= o

For n = 0 this is trivial, which gets the induction started. Since + T) — 1) = g„(T)

(everything else satisfies this, hence so does g), for CP = 1.

grg — 1) = g„(0),

Using the Weierstrass Preparation Theorem, we see that T and the minimal polynomial of C — 1 both divide g(T) — g(0), so g „(T)

g„(0) = ((1 + T)P

1)g„ 1 (T)

for some g„ 1 e A. Letting an = g„(0), we obtain the above for n + 1. Continuing, we get 00

g(T) —

L a1((1 + T)P

1)i E n (p, TY = O.

i= o

rr0

Therefore we may let tO

(N f)(T) =

ai T i • 1=0

Uniqueness follows from Corollary 7.4. This proves Lemma 13.39. Of course, the condition (*) of the theorem says that f(0) N f = f, for f corresponding to an element of U.

1 and

Lemma 13.40. Let f E A. Then (N f)(v,- -1) = N PROOF.

1( f (1 )).

(N f )(v„__ 1 ) = (N f)((1 + vn )P — 1)

f (1 + v n) — 1) = N,,,_ as in a previous calculation.

=H

(

LI

Lemma 13.41. Let f E A and assume f((1 + T)" — 1) =- 1 mod p" A. Then f(T) =- 1 mod /M. PROOF. Let tO

f (T) = 1 + 1)1'

L ai

i= o

306

13 Iwasawa's Theory of 4-extensions

for some it > 0, with it maximal. Let a„ be the first coefficient such that p )(' an . Then

E ai(( 1 +

T)P

1= o

ai TP1 # 0 mod pA.

— 1)1 a„TP" + i>n

Since CO

p"

ai((1 +

— 1) 1 a- 0 mod pkA,

i=o

we must have ti > k. This completes the proof. Corollary 13.42. N: A

—> A> is continuous.

PROOF. Since N is a homomorphism it suffices to check continuity at 1.

Lemma 13.41 and the definition of N yield the result. Lemma 13.43. Suppose f e

. Then N kf

for all k

= 1 mod pA

O.

PROOF. Since

N kf N(N k- if) N 1f

N(f) f f

it suffices to consider k = 1. We have f((1 + T) — 1) _ f (T)P _ (Nf)((1 + T)P — 1) 1 mod( p — 1), f (TP) — f((1 + T)P — 1) — f((1 + T)P — 1) therefore mod p. The result now follows from Lemma 13.41. Lemma 13.44. Let k > 1. Then

f PROOF.

1 mod pkA

Nf

1 mod

p k+ 1 A.

Write f(T) = 1 + pkfi (T). Then f((1 + T) — 1) —= 1 + pkfi (T) mod(C p —

Therefore (N f)((1 +

— 1) # (1 + p kfi (T))P # 1 mod(C p —

hence mod pk + 1 . Lemma 13.41 completes the proof. Corollary 13.45. Let m k > 0 and let f e Ax . Then

N'nf N kf mod

pk+1.

307

§13.7 Logarithmic Derivatives

PROOF.

Lemma 13.43 implies Nm - / f

1 mod p A. Lemma 13.44 yields the

result.

Corollary 13.46. Let f e Ax. Then Ai' f = lim NV' exists. PROOF.

Corollary 13.45, plus the completeness of A.

We can now prove Theorem 13.38. Let u = (u„) e U. For each n, choose f(T) e A such that

fn(vn) = un.

Let g(T) = (Nmf2„,)(T). By Lemma 13.40

(Nkfn)(v,,-k) = N„,„ - k f,,(v„) = tm nk ,

for 0 < k < n. Therefore (N m- "g)(v„) = (N2m- "f2.)(vn) = un,

for all m > n. By Corollary 13.45, = N 2?" - "f2m N mf2m = gm mod pm + 1 A.

Letting T = vn , we obtain

u„g(v„) mod pm + 1 ,

for all m > h e A, and

n. Since A X is

g, —> h Since

n

N>

compact, the sequence

gm has

a cluster point

as mi —> oo through a subsequence.

0(P, v„)" = 0 in Cp , it follows that g„(v„)

h(v)

for each n. Therefore u„ = h(v„) for all n. This completes the proof of Theorem 13.38.

The above proof of the existence is an adaptation of Coleman's proof for formal groups. I would like to thank John Coates for supplying the details. We can now define the generalization of the Kummer homomorphism. Let u e U and let f„ be the associated power series. Recall that D

d = (1 + T) dT .

For k > 1 define the Coates-Wiles homomorphism (5k: U ff(T) k(u) = D k- 1 (1 + T) "

fu(T)T=0.

Zp

by

308

13 lwasawa's Theory of

Zp- extensions

Since fu 1 mod(p, T), log fu (T) e p [M] is defined, so 61, ( 4) also equals Dk log fu (T)I T . 0 . If u = (u 0 , .) E U then

2

Ok(uo) = k (u) mod p,

k p — 2.

Lemma 13.47. 6k(u) is a continuous function of u (it is "almost" continuous in k; see Proposition 13.51). PROOF. This follows immediately from the continuity of u F-0 fu . Identify Z; following.

Gal(Q(Cp .)/Q), a

ca . An easy calculation yields the

Lemma 13.48. Let u e U be associated tofu , and let X = Ebu o-u e Zp[GalellGp .)/0)]. Then

f(T)

= 1J f (1 + (

T)" — 1) b-.

It is trivial to check that everything above is defined; for example, (1 + T)" = E(an )Tn e A. Lemma 13.49. Let a e Z; and o-a e Gal(O(Cp.)/0). Then, for k

1,

6 k(crau) = ak6k (u). PROOF. See the proof of Lemma 13.34.

Lemma 13.50. If u e U then 6k( ")

SO,

if k i mod p — 1

1.5k (u),

if k i mod p — 1.

PROOF. See the proof of Lemma 13.35. Note that Ei =

1

P-1

E co - i (a)o-.0)

P — 1 a=1

is the idempotent since Gal(( p)/ID) corresponds to {co(a)} in 7L,.

Let 2 < i < p — 2 and let A = A 1 be as in Corollary 13.37. Let ) =

— coo,

1)).

+1

309

§13.7 Logarithmic Derivatives

Then N„,„- 1(V ))

/A - ap"1) w(A - 1 )

g1

AP —

Pn

= (n — 1)

0)(A — 1) P i( SO

= (cI")) e g i U. Let y o be a topological generator of Gal(O(Ç0)/0(C p)), and define K0 e 1 + plp by Cpa = CpK g for all n > 1. The following result will be crucial in the next section. Proposition 13.51. Let 2 s i < p - 1. There exists h i(T) e A X such that

(1 - pk-1 )6k( °) = hi(K ko - 1) for k PROOF.

i mod p - 1.

By Lemma 13.50 we may compute 6k

- (1 + T)\ w(A - 1) I

Let 1

+ T)\

d log(A ( f (T) = (1 + T) — w(A - 1) ) dT =1+

A (1 + T) -

Since A P = A, 1 A A (1 + T)P - A = Pi C(1 + T) - A (see the example at the end of Section 12.2). Let U be as in Proposition 12.8. Then 1 U f(T) = f (T) - P

f ((1 + T) - 1) =

A 1 + T - A (1 + T)P - A • We obtain (Dk- Uf)(0) = (1 - p k — 1)(Dk —if)(0)

(1

_

By Corollary 12.9, (Dc 1 U f)(0) =

1 - 1),

for k i mod p - 1,

for some e A. Letting hi (T) = k 1(K;27 1 (1 + T) - 1) e A,

310

13 lwasawa's Theory of 4-extensions

we have

44)

—

1) =

— 1) = (1 —

It remains to show that hi E AX. Let k =

i,

so 2

k < p — 2.

O mod

p,

Then

by the proof of Corollary 13.37. Since 1 — E Z px , — 1) is a unit. Since h i(K io — 1) a hi(0) mod p, h i(0) E . Therefore hi e A'. This proves Proposition 13.51. LI Lemma 13.52. Let h(T) e A and u E U. Then 6k(h(T)u) = h(i4 —

Since both sides are continuous in h, we may assume h is a polynomial; by linearity, we may assume h(T) = (1 + T). Then h(T)u = y u. By Lemma 13.49,

PROOF.

6 k(au) = ic'e6k(u) = h( Kt

1)5(u).

LI

This completes the proof.

§13.8 Local Units Modulo Cyclotomic Units In this section we prove a beautiful theorem, due to Iwasawa, that relates the p-adic L-funct ions to the Galois structure of the local units modulo the closure of the cyclotomic units. We continue to assume p > 3. Let N m, „ be the norm from p(C pV i) lo (1:P pg p, + and let N„ be the norm from p(C ) to Up . Recall that Ur denotes the local units of Up(Cp.+L) which are congruent to 1 mod(Cp..., — 1). Let 1)

= fu e UriNn u =

Lemma 13.53. Let un e UV_ Then u n E U;,<=>for all in > n, there exists

um e Ur' with N m,„(um) =

For simplicity, let K m = O1,(Cp..+1). An element a e 1 + pZ r, yields cra e Gal(K ni/Up), and 1 mod py" 1- . cr = la PROOF.

For un E

UV,

there is a corresponding element

cr(m) = cr(tin , K m / K,,)

E

Gal(K,n/Kn) g Gal(K„,/Qp),

and, by a property of the Artin symbol, ofu n , K m/ Kn) = a(Nn un , K m/Qp) = aNnun.

311

§13.8 Local Units Modulo Cyclotomic Units

We also have o-(m) = 1 -=- u n E N., n (K,nx ).=- un E N„,, „( Ur) (see the appendix on class field theory). The last equivalence follows since the norm of a nonunit is a nonunit, and the norm of a (p — 1)st root of 1 is itself, hence not congruent to 1 mod(4'pn+ i — 1). Therefore, putting everything together, we obtain N„u„ = 1 .- N„u„

1 mod 'n +1 for all m > n

= 1

for all m > n

U e N„,,„(Ur)

for all m > n.

This completes the proof.

0

Note that "s=" also follows from setting T = 0, then dividing by f(0), in (*) of Theorem 13.38. Also, for i 0 0, et U'n = c, U" since Nn = Nn, o No = N n, o((p — 1)e0 ) and eo Ei =

O.

Theorem 1334. Let 2 < i < p — 2 and let ,?'' be as above. Then A

and A/((1 +

Tr - 1) = ei Ll (in)

We start with the second assertion. As above, let K n = Up(Çpn +1). By Corollary 13.37, c i = d°) generates v i U'i = ci U(P ) over Zp = A/(T). Now let n > O. Let un e ei LPin). Then PROOF.

Nn,o(un) e WI' = Ad°) = A(AT 7., o(r)) = Nfl, o (Ar). Therefore N„ , 0 ( g tidnn ) ) = 1 for some g e A. By Hilbert's Theorem 90, clef ;inn)

t,

1 CX = -- = 137o —

for some fi E IC: . We want fi E ei Win). As abelian groups K nx = 7E2 x {AP -1 = 1} x Of),

312 where TE

13 Iwasawa's Theory of 4-extensions

= 4ri+

1. Therefore, for N > 0, K„x /(K„x )PN = elPN2 X U (in)1(UnPN

We may let ei act on this last space (it could not act on K). Since i 0, (7r) mod(K: )PN is represented by a unit; since el = ei , it is represented by an element of ei U. Therefore (K>< /(K,< )PN) = ei(Uin)/( U ))PN ).

Consequently, for some vN E ei /31,0 -

= ei cx

ei v ko -

mod(Knx )pN .

Since a and v N are units, this yields a congruence mod(U)0N. Since Vin) is compact, the sequence ei v N has a cluster point v E ei UV, and 0-1 G (T o — 1)E i Pin) = T; U.

Œ= Therefore

un g V) mod T 1 UT) .

By Nakayama's Lemma (13.16), 1" ) generates e i UV over A, since u„ was arbitrary. Since (1 + = y gn fixes 1"), A/((1 + Tr — l)— E is surjective. UP contains a subgroup of finite index which is mapped isomorphically via the logarithm to ir"Z p Rpr,+ il for some N (see Proposition 5.7). By the normal basis theorem, there is an element of K, which we may assume to be in n"Z p Vpn+ i], whose Gal(K n/Op)-conjugates are linearly independent over Up. Therefore 7r"Z p Vpn,,] contains a submodule of finite index isomorphic to the group ring. Consequently 7p -rank e,U(in)

7p -rank ei

= Zp-rank ei Zp[Gal(Kn/Qp)] = lp-rank Zp[Gal(Kn/Ko)]

= pn. By the division algorithm, A/((1 + T)" — 1) as Zp-modules. Therefore the above surjection must have trivial kernel, so

A/((1 + T)Pn — 1) Now let u = (u n) E ei U = lim ei U. Then u„ = fn(T)")

ei UT).

313

§13.8 Local Units Modulo Cyclotomic Units

for some In e A, uniquely determined mod(1 + mutative,

fn-1(T )

1) = un _ =

1(un) =

Tr —

1. Since A is com-

f(T)N n,n i(r) = n(T)C'

Consequently, fn - i(T) f n(T) mod(1 +

— 1,

so ( fn(T)) determines an element f (T) e A = lirn A/((1 + T)Pn — 1), and

for all n. If u = 1 then f (T) Therefore

= f (T) 0 mod(1 + T)P" — 1 for all n, hence f = O. A —> ei U

is an isomorphism. This completes the proof of Theorem 13.54.

111

Remark. Usually this theorem is proved by using local class field theory, plus the structure theorem for A-modules, to show that ei U A. It is then not hard to use Corollary 13.37 to show that is a generator (see Lang [4]). In some ways, this give a "better" proof, since it applies to extensions of the theory where ei U is not quite cyclic over A. We now consider cyclotomic units. Let On) denote the cyclotomic units of 0(,i), CÇ = On) n UÇ''), and Cr = closure in Win). (See Exercise 13.8 for a description of Cr). Then Cin) is a Zp [Gal(O pGp,,+ i)/0p)] module. If E e On) then e - 1 e Cin) and E

since 1/(p — 1) e Z p . Note that (0 -1 ) 1 /(P -1) = <=> C e C(in). In general, we get the analogue of <e>, where c = Fix a primitive root g mod p2 . Then g is a primitive root mod p" for all n > 1. By Lemma 8.1 and Proposition 8.11.

O

pn + 1

1 —1 -

and

—

4/1

+ I

generate C(") as a Zp [Gal(UpGp,,±1)/C4)] module. But they are not in OP). Let pin

_ ''(1p„ : g)/ 2 C

( -

*g n

p

„—1

pn

+i—1

P

– l

Then ti n and Cpn, generate (C"))P . It follows that they generate Cin) as a Zn [Gal(0,(4,,+1)/U p)] module'. Let C i = lim CT), with respect to the norm map. Then C i is a A-module and a Gal(O p(Cp)/O p)-module. Decompose C i according to the idempotents: 2

c1 = i=0

314

13 Iwasawa's Theory of 4-extensions

Then each C i is a A-module. It is easy to see that i odd, i

ei C i = 0,

1,

el C i = lim
Then u„ = fn(T)e i rin

for some fn(T) e A. An easy calculation shows that N n,n- i(eirIn) =

i•

Therefore e iq = (eirin) E

C1C1,

and un-1 = fn(T)eiqn- 1.

Consider the set Sn

f (T)lu k = f (T)ei qk for all k

n} .

Then S n is closed in A and nonempty since fn E S„. Clearly So

Si

•••.

Since A is compact, n S„ 0 (nested set property from topology). Let f e n S„. Then un = f(T)En,,

for all n, so we obtain the following. Lemma 13.55. Let i be even. Then eC 1 = Aei ri. (This actually holds for all i 1). We are now ready to prove the main result of this section. Theorem 13.56. Let i

0 mod p — 1 be even. Then

ei Ulei C i

A/(f i(T)),

where

fi(Kso — 1) = L p(1 — s, (K 0 is defined by yo Cp. = CpK,9, for all n

1).

315

§13.8 Local Units Modulo Cyclotomic Units

PROOF.

From Theorem 13.54, = g1(T)°

for some gi e A, hence by Theorem 13.54 and Lemma 13.55

= gi(T)e i U.

ei

It remains to evaluate gi(T). To do this, we use the Coates-Wiles homomorphism (5k with k i mod p — 1. By Lemma 13.52, 6 k(Eir/)

=

—

By Proposition 13.51, we know that (1 — pk-1 )(5k( .°) = hi(Kt — 1) with h i e A'. By Lemma 13.50, 5(g) = 6k (q), so

— 1) - 1 6k(n).

gi(Kt — 1) = (1 — P To identify g1 , it suffices to evaluate 5(n). Let

f(T) = ((1 + T)(1-

012 (1 + (1

— 1 p— • T) — 1

Then f(T) = f i(T) in the notation of Theorem 13.38. We have

f' (T) = (p — 1) ( 1 — g (1 + T)— + 2

1+T

g(1 + T)g

1).

Let T = ez — 1. Then D = (1 + T)(d/dT) = d/dZ. We have

geqz eg z

1 gZ Z +g 1 Z eg z — 1 e z — 1

ez ez _

= g

l

OD

Bn Zn n (n 1)!

(g n

n=

Therefore

f' ) SIN -= Dk- ((1 + T)— f

(o ) = (g ok—

= — (gk — 1)(1 — plc -1 ) -1 Lp(1 — k, co 1), since k i mod p — 1. Returning to the above, we obtain

gi(Kt — 1) = — (g k — 1)h 1 (Kt — 1) - 1 L(1 — k, Let

a

(logo g) (log p KO'

316

13 lwasawa's Theory of 4-extensions

and let

V(T) = — w(g) i(1 +

+ 1.

Then

V(0, — 1) = — and V(0) = 1 —

— 1)

0 mod p, so V

for k t mod p — 1, E

Ax . Let

fi(T) = gi(T)11,(T)V(T) - . Then fi and gi generate the same ideal in A and

fi(4 — 1) = L p(1 — k, w 1). Since both sides are analytic in k, we may replace k by s the proof.

E Zp •

This completes

NOTES The basic references for this chapter are Iwasawa [25], Coates [7], and Serre [1]. The other papers by Iwasawa and those by R. Greenberg should also be consulted. For arbitrary (non-cyclotomic) Zp - extensions, see Bloom [1], BloomGerth [1], Cuoco [1], Cuoco-Monsky [1], Monsky [1], [2], R. Greenberg [1], and Babaicev [2]. For relations with K -theory, see Candiotti [1], Coates [1], R. Greenberg [6], and Kramer-Candiotti [1]. For determining how to start a 4-extension, see Carroll [1], CarrollKisilevsky [1], and Bertrandias-Payan [1]. For capitulation (ideals becoming principal), see Ferrero [3] and Kuroda 1]. For a Hurwitz-type formula for the ,-invariant, see Kida [2], Iwasawa [29], and Kuimin [3]. It has been conjectured that A = 0 for all totally real fields. See Greenberg [ 5 ]. For the main conjecture, see Coates [7], [8], Greenberg [4], MazurWiles [1], Ribet [5], and Oesterlé [2]. Theorem 13.56 is from Iwasawa [13]. The above proof comes from the proof used in the elliptic case by Coates-Wiles [4] in their work on the conjecture of Birch and Swinnerton-Dyer. For an extension to abelian fields, see Gillard [6], part II. For more references, see the notes on Chapter 7. EXERCISES 13.1.

Suppose K„.IK is a Z r-extension such that each K„ is a CM-field. Show that if Leopoldt's conjecture holds for K then K w IK is the cyclotomic Zp -extension.

317

Exercises 13.2.

(a) Show that in a cyclotomic 4-extension no prime splits completely. (b) Suppose Gal(F/K) f p x f p . Let 1 p be a prime. Show that the decomposition group of a prime above 1 is either trivial or isomorphic to 4 (Hint: look at Frobenius). Conclude that there is a subextension F 1 F such that Gal(F 1/K) f p and such that some prime above 1 splits completely in F1/K.

13.3.

(a) Show that if x K then K( v 2)/K is the first step of the cyclotomic 1 2 extension of K. _ (b) Show that Q(,,/ II , 6,‘.:2)/G( ‘: —6) is unramified. Hence it is possible that K i /K t, is unramified in a 1-extension (see also Exercise 13.4). (c) Show that the ideal (2, ) is not principal in G( — 6) but is principal in _ - 6, .12). Show also that it represents a class in A - This shows that Proposition 13.26 does not hold for p = 2 (see also Ferrero [3]).

13.4.

Let 09 be a Dirichlet character of conductor 9 with ipZ = I. Let x, be of conductor 7 with x; = L Let K o be the field corresponding to 0 9 in the sense of Chapter 3. Show that K o is totally real. Show that the first step K 1 of the cyclotomic 13-extension of K o corresponds to the group generated by x, and 09 . Show that K 1 /K0 is unramified of degree 3. Hilbert's Theorem 94 (Hilbert [2]) states than in an unramified cyclic extension of odd prime degree p, at least one ideal class becomes principal. Conclude that the map A 0 —) A t is not injective. This shows that Proposition 13.26 does not hold for A n+ .

13.5.

Let M be a finitely generated A-module. Show that there is a unique (Hausdorff) topology which makes the action of A continuous, and show that M is compact with respect to this topology.

13.6.

Let a be algebraic and irrational. Let K c (-- algebraic closure) be a maximal extension of CD not containing a. Show that Gal(C/K) 1/21 or 4 for some p (cf. Lang [6], Ch. 8, Exercise 3).

13.7.

Show that for each n 1 there are infinitely many cyclic extensions of G of degree pit which are not contained in the 1p-extension of O. Show this is also true with G replaced by any number field K. This shows that not every such extension starts a f p- extension

13.8.

Show that Cr (Section 13.8) is the set of products of the form

H with p,rai ,In i 0, and 13.9.

n

c

- ir 1 mod p.

(a) Observe that the Kummer homomorphism can be defined for any unit u (i.e., we do not need u 1 mod(C r — 1)). Suppose E c §[c p] is a global unit which is congruent to a rational integer mod p . Show that Ok(E) 0, 2 k p — 3. (b) Let p be a regular prime. Show that for i, k = 2, 4, ... , p — 3,

Ok(Ek) Ok(Ei) = 0, where

Ek

is as in Chapter 8.

k,

318

13 Iwasawa's Theory of le-extensions (c) Let e be as in (a) and assume p is regular. Show that c is the pth power of a unit of Z[ p ].

13.10. Show that if p is regular then Theorem 13.56 is trivially true. 13.11. Let K/K 0 be a 4-extension. Suppose K. L, L/K o is Galois, and L/K. is unramified. Let G = Gal(L/K o), X = Gal(L/K), and F = G/X = Gal(K/K0). (a) Let A be a prime of L which is totally ramified in 1‹,./K 0 and let 1 E G be its inertia group. Show that for n > 0, P" is the inertia group in Gal(L/K,) for A, and that P' Fr" under the map G — > F. (b) Suppose F/K o is a finite extension with F L. Show that P" acts trivially on F for n sufficiently large . Conclude that, for n large, the (possibly trivial) extension FK„/K, is unramified at A. (c) Suppose X is abelian. Show that each finite subextension of L/K. is obtained by lifting an abelian extension F/K,, for some n, to K. Use (b) to show that we may assume the extension F/K„ is unramified. (d) Conclude that the field L = H L in the proof of Theorem 13.13 is the maximal unramified abelian p-extension of K. 13.12. Let M be a A-module. Define M r = frn c M m for all 'y E T1 and Mr= M/TM. Then M r and M r are the largest submodule and quotient, respectively, on which F acts trivially. Observe that M r is the kernel of multiplication by T. Suppose M r and M r are finite. Define Q(M)

Mn I

(this is a Herbrand quotient. See Lang [1], p. 179). TM). (a) Show that if M is finite then Q(M) = 1 (Hint: M/M` (b) Suppose 0 —> B —> C 0 is an exact sequence of A-modules. Show that Q(A)Q(C) = Q(B) in the sense that if two factors are defined, so is the third and equality holds (Hint: Apply the Snake Lemma to two copies of the sequence, with vertical maps multiplication by T). (c) Suppose M = A/(f ) withf(0) 0. Show that Q(M) = I f (0) P • (d) Extend (c) to M 0 A/(fi). (e) Suppose M is a A-module with M A/(fi). Let F = {1 f, and suppose F(0) 0. Show that Q(M) = F(0)1,(f) Show that the Main Conjecture for QD(Cp ) implies that e,A 0 I = 1B1,.-.1p 1 (compare p. 299. Hint: Prop. 13.22. The analytic class number formula changes the inequalities to equalities, so you do not need to know Mr).

Chapter 14

The Kronecker-Weber Theorem

The Kronecker-Weber theorem asserts that every abelian extension of the rationals is contained in a cyclotomic field. It was first stated by Kronecker in 1853, bul his proof was incomplete. In particular, there were difficulties with extensions of degree a power of 2_ Even in the proof we give below this case requires special consideration. The first proof was given by Weber in 1886 (there was still a gap; see Neumann 11]). Both Kronecker and Weber used the theory of Lagrange resolvents. In 1896, Hilbert gave another proof which relied more on an analysis of ramification groups. Now, the theorem is usually given as an easy consequence of class field theory. We do this in the Appendix. The main point is that in an abelian extension the splitting of primes is determined by congruence conditions, and we already know that p splits in U(„) if and only if p 1 mod n. The purpose of the present chapter is to give a proof of the KroneckerWeber theorem independent of class field theory. Our argument is a modification of one of Shafarevich (see Narkiewicz [1]), where the global result is deduced from the corresponding result for local fields Q. Except for a few minor references, this chapter is independent of the rest of the book. One significance of the Kronecker-Weber theorem is that it shows how to generate abelian extensions of U via an analytic function, namely e2' evaluated at rational x. For abelian extensions of imaginary quadratic fields, this is accomplished with elliptic modular functions in the theory of complex multiplication. In general, the situation is called Hilbert's Twelfth Problem. Theorem 14.1 (Kronecker Weber). If K/Q is a finite abelian extension, then —

K g Ugn) for some n.

319

320

14 The Kronecker-Weber Theorem

Theorem 14.2. If K/U p is a finite abelian extension, then K

for some n. PROOF.

We first show it suffices to prove 14.2.

14.2 (for all p)

14.1.

Assume K/0 is abelian. Let p be a prime which ramifies in this extension. Let Kp be the completion at a prime above p. Then K p/O p is abelian, so Kp

p(C„p)

for some np . Let peP be the exact power of p dividing np and let

np

n= p

ep .

ramifies

We claim K g Q(C). Let L = K(ç), so L/U is abelian and if p ramifies in L/U then p ramifies in K/U. Also, if L p denotes the completion at a suitable prime of L above p, L p = K(4)

g U p(Cp.p• n , )

with (n', p) = 1.

Let I, be the inertia group for p in L/0. Then /p may be computed locally, SO

/p Gal(U p(40/Up), which has order 0(pei9). Let I g Gal(L/0) be the group generated by all Ii,, with p ramified (p finite). Since Gal(L/0) is abelian,

1/1

fl 141 = 11 0(peP) =

0(0 = CO (4 n) : 01

Lemma 14.3. If F/0 is an extension in which no finite prime ramifies, then F = U. PROOF. A theorem of Minkowski (see Exercise 2.5) states that every ideal class of F contains an integral ideal of norm less than or equal to n! flY 2 N/ciF n7 .E)

where n = [F: 0], d F is the absolute value of the discriminant, and r2 < n/2 is the number of complex places. In particular, this quantity must be at least 1, so n n (n\r2 n n iz \n/2 d f N// dF

n! 4)

n! zi)

=-e- bn.

321

14 The Kronecker-Weber Theorem

Since b 2 > 1 and

b + = (1 b„

1) n 7T TT —> 2—>1

n

4—4

we must have, if n > 2,

dF > 1. Consequently there exists a prime p dividing dp, which means p ramifies. This proves Lemma 14.3. LI Returning to the above, we consider the fixed field F of I. Then F/O is unramified at all finite primes, so F = Q. Therefore

I = Gal(L/0), hence

[L :0] = II

[E(ç„) : 0].

Since

O ( n)

K ( „) = L,

we have equality, so

K This proves "14.2

14.1."

LI

We are now reduced to the local situation, where the structure of extensions is much simpler. We shall often use the following well-known result.

Lemma 14.4. Let K and L be finite extensions of 0 p such that K/L is unramified. Then

(a) K = L() for some n with p )(n, and (b) Gal(K/L) is cyclic. Also, for fixed L and for every integer m > 1, there exists a unique unramified extension K of L which is cyclic of degree m. We sketch the proof. First consider (a) and (b), and assume K/L is Galois. Let & K and A K be the integers and maximal ideal for K and define CL and A L similarly. Since K/L is unramified, there is a canonical isomorphism

Gal(K/L) Gal(C K mod

As( L mod A L)

(if there were ramification, we would have to mod out by the inertia group on the left). The right-hand side is an extension of finite fields, hence cyclic. If K/ L is not necessarily Galois, then the Galois closure yields a cyclic Galois group; so K IL is already Galois and cyclic. This proves (b). Since every

322

14 The Kronecker-Weber Theorem

nonzero element of a finite field is a root of unity of order prime to p, we may choose r,„ E ( • K mod AK with (n, p) = 1 which generates the extension of finite fields. Since X" — 1 = 0 has a solution mod AK, Hensel's lemma (note p,en) yields a solution in (•K, which generates the extension K/L because of the isomorphism of Galois groups. This proves (a). To prove the last statement, let ‘;, with p,i'n generate an extension of • L/A L of degree m. Then L(C,r)/L is unramified and, by the isomorphism of Galois groups, is cyclic of degree m. If there are two such extensions, then the composition is unramified, hence cyclic by (b). Therefore the two extensions must coincide. This proves the lemma. In the proof of Theorem 14.2, it will be convenient to know what the answer will be. Unramified extensions of Qp will be given by Lemma 14.4. In particular, we already have a good supply of unramified extensions. The ramified extensions of Q p will be subfields of QpGp„). Since our list of abelian extensions already includes such subfields, we can produce totally ramified extensions K/Q p with any group

Gal(K/Up)

<Ç /(p — 1).Z x Z/pn,Z, p # 2, z p z x z/27, p = 2,

for all n > O. The fact that we can get (Z/2Z) 2 for p = 2 will cause slight problems. We now start the proof of Theorem 14,2. Observe that it suffices to assume Gal(K/Q p) We consider three cases : p

Z/eZ,

g = prime, m

1.

g, p = g # 2, and p = g = 2.

Case Lemma 14.5. Let K and L be finite extensions of Ck, and let A L be the maximal ideal of the integers of L. Suppose K/L is totally ramified of degree e with e (i.e., K/L is tamely ramified). Then there exists ire L of order 1 at A L and a root ot of X e — IL = 0 such that K = L(a).

PROOF. Let Ix I be the absolute value on Cp = completion of the algebraic closure of Up). Let no e AL be of order 1. Choose fl e K to be a uniformizing parameter, so that

Then pe =

n o u with uG UK = units of K.

323

14 The Kronecker Weber Theorem

Since K/L is totally ramified, the extension of residue class fields is trivial. Consequently

u

u0 mod

A K with u 0 e U L .

Therefore

u = u o + x with x

AK.

Let n = no uo , so

fie = no uo go x = 7r + no X and

IPe — n1
, ae be the roots of

f (X) = Xe — Since the a's differ by roots of unity, I gi I = I of; I

for all i, j,

so

lad) But 1-1 I —

i# 1

= Igi

= Inu1)1 =

Consequently

= Since

1J Ifl—I=lf(13)l
< 1111Therefore

# 1.

113 — gil < Igi — By Krasner's lemma (Lemma 5.3), L(a i ) g L(f3)

K.

But f (X) is irreducible by the Eisenstein criterion, so [L(a l ) : L] = e = [K : L]. This completes the proof of Lemma 14.5.

0

324

14 The Kronecker-Weber Theorem

Lemma 14.6. Op(( — p) 11(P - 1) ) =

Up(Cd•

PROOF. Let (X + 1)P — 1

g(X) =

X

= XP -1 +

pX P- 2 + • • • + p.

1)

1)P

Then 0 = gGp

p mod(C — 1)P,

SO u=

OP-1

(Cp —

= 1 MOd(C p — 1).

P

It follows that (1 + P+— + P")

14 1 = liM

n-■ oo

exists in LP p(Cp) and satisfies U P-1 l =

u.

Therefore (— p) 1 I(P -1) G Up(C p)* Since XP -1 + p is irreducible over Up by the Eisenstein criterion, the lemma follows easily. I=1 Now assume K/LP p is abelian of degree gin. Let L/O p be the maximal unramified subextension ( = fixed field of the inertia group). Then L g 02D AO

for some n, by Lemma 14.4. Let e = [K : L]. Since e is a power of g, p ,{/ e, so K/L is totally and tamely ramified. By Lemma 14.5, K = LOr lie).

for some n of order 1 in

L.

Since L/U p is unramified, p has order 1 in L, so

7r= —

up

for some unit u E L. Since u is a unit and p, 0, the extension L(u l /e)/L is unramified; hence by Lemma 14.4 g L m) g (Up(CIVIn) (

for some M. In particular,

Q p(ti lle)

p(Cmn ), so

g_

Qp(u 1 /e)/Q p is abelian. Since K

p is abelian, U p(n l ie)/U p is abelian. It follows that

o (( p

'

To

p

325

14 The Kronecker Weber Theorem

is also abelian. But Xe + p is irreducible over U p . It yields an abelian, hence Galois, extension of Up, SO

(Upq _ 01/9 = (U p(Ce ( _ 01/e) for a primitive eth root of unity Çe. Therefore eG Since Upq P ) 1 /9/Q p is totally ramified, so is the subextension U p(Ce)/U p . But p /i/e, so the latter extension is trivial and e e U p s Therefore elp — 1 (if p = 2 we obtain e = 1 or 2, but since q 0 p,e = 2 is excluded). We may now put things back together. By Lemma 14.6,

Up(( — 0 11e) Therefore K = L(17 1 /e)

L(u l i e ,

( — 0 1/8 ) g

Up(CMnp)-

This finishes Case I. Case II. p = q 0 2 Lemma 14.7. Let F be a field of characteristic Op, let M = F(Cp), and let L = M(a l ) for some a e M. Define the character co: Gal(M/F) Zp by o-Cp --- C p . Then

a(0( ') mod(M ) )P

LIF is abelian = a(a) for all a E Gal(M/F).

Remarks. Zp acts on M x/(M x )P in the obvious way, so a 0 (1) mod(M x )P is defined. The converse of the lemma is also true but will not be needed. One

first shows that LIF is Galois, then reverses the proof below to obtain abelian. The lemma is sometimes stated as LIF is abelian <#. there exists where a,:

cE

M such that ag a = agcP,

14 Cgp generates Gal(M/F).

PROOF OF LEMMA 14.7. Let G = Gal(M/F) and H = Gal(L/M). Then G acts on H as follows. If a e G, extend it to an element of Gal(L/F). Then define If = aha- 1 . This is well-defined since H is abelian. In fact, since LIF is abelian, this action is trivial. Let A be the subgroup of M x l(M X ) 13 generated by a. We have the Kummer pairing H x A —> Wp = pth roots of unity, h(a l lP) all!,

.

326

14 The Kronecker-Weber Theorem

It is bilinear and nondegenerate. As in Chapter 10, we have =
ac

G.

Since G acts trivially on H and acts on Wp via w, = =

for all h. Since the pairing is nondegenerate, act") aa mod(M

This completes the proof of Lemma 14.7.

0

Let K/U p be cyclic of degree pm. We already have a totally ramified cyclic extension K r/U p of degree pm contained in O(+ 1), namely the fixed field of the subgroup of order p — 1 in the Galois group. There is also an unramified cyclic extension K u/O p of degree pm (Lemma 14.4) which equals Cl p(„) for some n. Since K r n K u Gal(K r K u /44) = (7//pni7/) 2 . Suppose K

Then Gal(K(4.+1, „)/O p) = (Z/pmZ) 2 x Z/puil

for some in' > 0. This group has (Z/pZ) 3 as a quotient, so there is a field N such that Gal(N/Op)

(Z/pZ) 3 .

The following lemma finishes the proof of Case II. Lemma 14.8. Assume p (Z/PZ) 3 . PROOF.

2. There areno extensions N/O p withGal(N/O p)

If we have such an N, then N(4)/U p is abelian and Gal(N(4)/U p(4)) (Z/pZ) 3

This is a Kummer extension so there is a corresponding subgroup B

p(4) x AU Xp) x )P

with B (Z/pZ) 3 and p(4)(B 1 P) = N(4). Let a c B and let L= UpGp a 11"). Then L/O p is abelian, so

ria

a'( ') mod(O p(4)')P,

a- e Gal(O p( p)/Op).

Let y be the valuation on (1p(4) with v(4 — 1) = 1. Then v(a) = v(ria)

w(a)v(a) mod p,

Since a4 0 4 if a 0 1, we have w(a) v(a)

for all a.

1 mod p for such a. Therefore

0 mod p.

327

14 The Kronecker-Weber Theorem

Since 0 p( C pY = (C p 1 ) x Wp _ , x U,

where U, = {u 1 mod 4 a e U 1 . So we may assume

-

1}, we may change a by a pth power to obtain B U 1 /U,

and Gal(O p(4)/14) acts via the character co. Let it = 4 1 and let u = 1 + InE + • • - e U, (with b e Z). Since p Ei b

we have u, = (; bu 1 mod nil+ 1 , hence uP

1 + bn mod 7 2 ,

1 mod n 2 . An easy calculation shows that /4 1 mod 7E1' 1 . Conversely, if u 2 1 mod 7L P + j, then 1 — logo u 2 0 mod n2 , P

hence 1

u = exp(- logy u 2 ) P

converges to an element of U, and uP = u 2 (see Section 5.1; alternatively, the binomial series for (1 + u 2 — 1)'/P converges). Therefore t.l ' = fu

Again, let u e U 1 . Write u =

1 mod nti +1 } .

4b u1 with u,

au u 6" ) mod UÇ,

1 mod n 2 . If u e B then

a E Gal(0y(4)/0,)

Since C hi, already satisfies this relation, so does u l . Write u, = 1 + cnd + • • • with e E Z, (c, p) = 1, and d

2. Since (an)In

w(a) mod 7E,

au, = 1 + cco(a)dnd + • . . . But uT(a) = 1 + cw(a)nd + • • • .

Since au, uel)(') mod Ul for all a, we must have either d > p + 1 or d 1 mod p — 1. The first means that u, is in UÇ, the second that d = p. Clearly 1 + nP generates modulo Ur; the subgroup of u, 1 mod it". Putting everything back together, we obtain B ç_

U,IU 1;.

where <x, y> denotes the subgroup generated by x and y. Since B --' (Z 1 pZ) 3 , 0 we have a contradiction. This proves Lemma 14.8.

328

14 The Kronecker-Weber Theorem

Case III. p = q = 2 We already have a totally ramified abelian extension Kr = at rr _2.72m+2) with Gal(K,./0 2) = 1/2Z x 1/2"Z.

We also have an unramified extension K u with Gal(K u/0 2 )

Since K r n K u =

02,

Gal(K,K u/0 2 ) = Z/2Z x ( 7/ /2"Z) 2 .

Let K/0 2 be cyclic of degree 2' and suppose K K r K u . Then Gal(K K r K u/0 2) has exponent r, requires at most 4 generators, one of which has order 2, and has Gal(K, K u/0 2 ) as a quotient by a nontrivial subgroup. Therefore {Z/22 x (Z/2"Z) 2 x Z/2"1, Gal(KK,K u/0 2 ) =

with m' > I,

or (Z/2"Z) 2 (D (Z/2"1),

with in > m' > 2.

Therefore there is a field N with {(Z/2Z)4 Gal(N/0 2 ) =

or (Z/4Z) 3 •

We shall show this is impossible. The first corresponds to four independent quadratic extensions of 02. We have Z/2Z x { --F 1 } x U i /Ui

where U 1 = {u 1 mod 4 } . As in Case II, it is easy to see that i_I = {u 1 mod 8 } , hence

U 1 /U

/2Z.

Therefore (Z/2 l) 3 .

By Kummer theory, the first possibility is now eliminated. Suppose now that Gal(N/0 2 ) = (Z/4Z) 3 . Then i = —1 E N, otherwise we could add it to N and obtain a subfield whose Galois group would be (Z/22)4, which we just excluded. It follows easily that there is a field L with 0 2 (i) c LOEN

329

Notes

and

Gal(L/0 2) Z/4Z (proof: every subgroup of (2/4Z) 3 of order 32 contains a subgroup of the form (Z/4Z) 2 • Let L be the fixed field). Let a be a generator of Gal(L/0 2 ). Then 0-2 generates Gal(L/0 2 (0) and a(i) = —i. We may write

L = 0 2 (i, oz) with CX 2 E 02(4 We also have L = 0 2 (i, aa) and (0.0)2 = a(a 2 ) 0 2(0/0 2 is Galois. Therefore

0-2 (a) = —a It follows that al/a is fixed by a2 , COL

a

and

6

2

E

0 2 (i), since

((roc) = — aa.

SO

= A + Bi 02(i)

and 2 0' 0C

o-a

= a(A + Bi) = A — Bi.

We obtain 2

—1 =

GC

a2 g 0-(X

= A 2 +B2.

Lemma 14.9. A 2 + B2 = —1 has no solutions in

02.

PROOF. We may transform this to

A? + Ai + Ai = 0 with A i E 2 1 < i < 3, and 2,1' Ai for some i. But there are no nontrivial solutions mod 8. This completes the proof of the lemma. 1=1 The lemma shows that we have a contradiction. Therefore K c K,.K u 2 (Cm ) for some M. This finishes Case III, so Theorem 14.2 is completely 0 proved. NOTES The Kronecker-Weber theorem was first stated by Kronecker [1] and was proved by Weber [1]. Later proofs were given by Hilbert [1] and Speiser [1]. See also M. Greenberg [1] and Ribenboim [2]. Our use of Lemma 14.5 in Case I, which allows us to avoid using higher ramification groups, is similar

330

14 The Kronecker Weber Theorem

to the proof of Abhyankar's lemma in Cornell [1]. A global proof of Theorem 14.1 which has the same flavor as the present proof may be found in Long [1]. A proof, similar to the above, of a more general local KroneckerWeber theorem has recently been given by Rosen [2]. For two more proofs of the global theorem, plus some interesting historical remarks, see Neumann W.

Appendix

In this appendix, we summarize, usually without proofs, some of the basic machinery that is needed in the book. The first section, on inverse limits, is used in Chapters 12 and 13. Infinite Galois theory and ramification theory are used primarily in Chapter 13. The main points of the section are that the usual Galois correspondence holds if we work with closed subgroups and that we may talk about ramification for infinite extensions, even though the rings involved are not necessarily Dedekind domains (much of this section comes from a course of Iwasawa in 1971). The last section summarizes those topics from class field theory that we use in the book. The reader willing to believe that the Galois group of the maximal unramified abelian extension is isomorphic to the ideal class group (and variants of this statement) will have enough background to read all but certain parts of Chapter 13.

§1 Inverse Limits Let / be a directed set. This means that there is a partial ordering on /, and for every i, j E / there exists k E / with i < k, < k. For each i E /, let A i be a set (or group, ring, etc.). We assume that whenever i j there is a map Op: A i —■ Ai such that q5 ii = id and CA.; = Oki whenever i j k. This situation is called an inverse system. Let A = fl A i and define the inverse limit by

lirn A i = {(. , ai , ..)

E

A Oki(ak) = ai whenever ]

For each i, there is a map (Pi : lim A i —■ A i induced by the projection A —■ A. Clearly (p ii = Assume now that each A i is a Hausdorff topological space. Then A is given the product topology and lim A i receives the topology it inherits from A.

331

332

Appendix

We assume the maps Oft are continuous. The maps (/), are always continuous: If Ui is open in A, then Oi-1 (U 1) is the intersection in A of an open set of A (definition of product topology) and Jim A i , hence open. The topology of lim A, is generated by unions and finite intersections of such sets (10. In fact, every open set contains Ok- 1 (Uk) for some k and some Uk (proof: it suffices to show thatJ 1 (U 1 ) n OT 1 (U 1) = Ok- l (Uk) for some k. Choose and let Uk = Ok-i l (Uf)). k> We claim that lim A i is closed in A. Suppose a = (. • , ai , ..) liM Ai. Then 4 » (a) ai for some i, j. Let U 1 and U2 be neighborhoods of Oii(ai) and ai , respectively, such that U i n U2 = 0. Let U3 = Ori i (Ui) and let U = U2 X U3 X

n

Ak

A.

k*i,j

Then a E U but U n I,jrn A i = 0. Since U is open, it follows that lim A. is closed. Suppose now that each A. finite, with the discrete topology. Then A is compact, hence lim A i is compact. Also, lim A i can be shown to be nonempty and totally disconnected (the only connected sets are points). An inverse limit of finite sets is called profinite. If each A i is a finite group and the maps Op are homomorphisms, then Af is a compact group in the natural manner. It can be shown that all compact totally disconnected groups are profinite. Also, if G is profinite then G = lim GIU , where U runs through the open normal subgroups (necessarily of finite index, by compactness) of G, ordered by inclusion. a mod pl EXAMPLES. (1) Let / be the positive integers, A i = ZIp1 2, a mod p1 . Then lim Z1p 1Z = 2p , the p-adic integers. The maps Oi are the natural maps zp zIp1.1. In essence, the ith component represents the ith partial sum of the p-adic expansion. (2) Let / be the positive integers ordered by m < n if in I n. If in I n, there is a natural map ZInZ ZlinZ. Let 2 = 2InZ. It can be shown, via the Chinese Remainder Theorem, that na,ip Zp s For more on inverse limits, see Shatz [1] or any book on homological algebra.

§2 Infinite Galois Theory and Ramification Theory Let Klk be an algebraic extension of fields and assume it is also Galois (normal, and generated by roots of separable polynomials). As usual, G = Gal(K/k) is the group of automorphisms of K which fix k pointwise. Suppose k F K with Flk finite. Then GF = Gal(K/F) is of finite index

333

§2 Infinite Galois Theory and Ramification Theory

in G. The topology on G is defined by letting such GF form a basis for the neighborhoods of the identity in G. Then G is profinite, and G

lim G/GF lim Gal(F/k),

where F runs through the normal finite subextensions Flk, or through any subsequence of such F such that U F = K. The ordering on the indices F is via inclusion (F 1 ç F 2 ) and the maps used to obtain the inverse limit are the natural maps Gal(F 2/k) Gal(F i/k). The fundamental theorem of Galois theory now reads as follows: There is a one-one correspondence between closed subgroups H of G and fields L with k..g Lg K:

H 4-4 fixed field of H,

Gal(K/L)

L.

Open subgroups correspond to finite extensions, normal subgroups correspond to normal extensions, etc.

(1) Consider U(Cp 4/0. An element a e Gal(144.)/0) is determined by its action on Cp. for all n > 1. For each n we have aCp. = 47t for some a, e (ElpriZ) x , and clearly a, an _ 1 mod p" - . So we obtain an element of EXAMPLES.

= lim(ZIp"Z ) = lim Gal(Q(C p”)/U).

Conversely, if a e 2; then a = Cam, defines an automorphism. The closed (and open) subgroup 1 + p7L p corresponds to its fixed field Q(Cip).

(2) Let F be a finite field and let F be its algebraic closure. For each n, there is a unique extension of IF of degree n, and the Galois group is cyclic, generated by the Frobenius. Therefore Gal(F/F) lim Z/n7 = Now suppose that k is an algebraic extension of 1), not necessarily of finite degree. Let Ck be the ring of all algebraic integers in k and let A be a nonzero prime ideal of ek • Then h n 2 is nonzero (if a e A, Norm ) , ( ) G A n Z) and prime, hence A n = pZ for some prime number p. Therefore 7/PZ

( 7 + A )/ A g-

It is easy to see that (9 k/A is a field and is an algebraic extension of ZIpZ (since C k is integral over 2). In fact, Gal((9 k/A)/(2/p2)) is abelian since any finite extension of a finite field is cyclic, and an inverse limit of abelian groups is clearly abelian. Let Klk be an algebraic extension, again not necessarily finite. Let be a nonzero prime ideal of O K and let A = n «k, which is a prime ideal of ek.

334

Appendix

Then OK/Y is an extension of Ok/h; in fact, it is an abelian extension since OK/Y is abelian over 1/p1 Conversely, suppose we are given a prime ideal A of O k . Then there exists Y in CK lying above A; that is, A = n Ck (see Lang [6], Chapter 9, Proposition 9; or Lang [1], Chapter 1, Proposition 9). Lemma. Suppose K/k is a Galois extension. Let Y and Y' be primes of K lying above A. Then there exists a E Gal(K/k) such that (TY = We know the lemma is true for finite extensions (see Lang [6], Chapter 9, Proposition 11, or Lang [1], Chapter 1, Proposition 11). Choose a sequence of fields PROOF.

k = Fo

•

K

F„

such that K =L F,, and such that each F„/k is a finite Galois extension. Such a sequence exists since the algebraic closure of is countable. Let

hr, = °

=

Since F,,/k is finite, there exists t„ E Gal(Fn/k) such that 1-„(An) =An1 . Let a„ E Gal(K/k) restrict to Tn . Since Gal(K/k) is compact, the sequence {a„} has a cluster point a. There is a subsequence {a-„,} which converges to a (a priori, we would have to use a subnet. But subsequences suffice since Gal(K/k) satisfies the first countability axiom. This follows from the fact that the set of finite subextensions of K/k is countable). For simplicity, assume lim an = a. Let m be arbitrary. Since Gal(K/F„,) is an open neighborhood of 1, a'a,,E Gal(K/F,n) for n > in sufficiently large. Hence, Q 'a/i = A., so aft,,, = A„, n CF ) = n CFpn = . Since = U A n, and Y' = A;„, we have (TY = Yfi. This completes the proof. E We now want to discuss ramification. However, Ck and OK are not necessarily Dedekind domains. For example, if k = OG p .) and A = 1, Cp2 - . then AP = it, since G p.+1 — 1)P = 1). This means that we cannot define ramification via factorization of primes. Instead we use inertia groups. Let K/k be a Galois extension, as above, and let Y lie above A. Define the decomposition group by Z = Z(Y/A) = {a E Gal(K/k)I aY = 3}. We claim Z is closed, hence there is a corresponding fixed field. Let the notations be as in the proof of the lemma and let Z„ = {a I a(h„) = A„}. Then Z Z„ for all n, and since PI) = U An we have Z = n Z. - Since Gal(K/F„) Z„, we have Z„ open, hence closed (it is the complement of its open cosets). Therefore Z is closed, as claimed. Now define the inertia group by T = T(7 /i) = {61

e Z, a(g) g mod Y for all a e CO.

§2 Infinite Galois Theory and Ramification Theory

335

It is easy to show that T is a closed subgroup. As with the case of finite extensions, we have an exact sequence

1—>T—Z—* Gal((( K/3))/(e k/A)) .-- 1. The surjectivity may be proved by using the fact that we have surjectivity for finite extensions (Lang [1] or [6], Proposition 14). Suppose now that K/k is an algebraic extension but not necessarily Galois. Let 0 be the algebraic closure of G. Then 0/K and 0/k are Galois extensions. Let Y be a prime of K lying over the prime A of k. Choose a prime ideal g of ea lying above Y. We have

T(g/A) g_ Gal(0/k), T(g/Y) g_ Gal(0/K) g_ Gal(0/k), T(g/Y) = T(g/h) n Gal(Q/k). Define the ramification

index by e(Y/h) = [T(/h):T(g/Y)],

which is possibly infinite. If g' is another prime lying above Y then g' = ag for some a e Gal(0/K), and

T(g'/A) = a T(g/h)a -1 , T(g'/Y) = Therefore the index e(Y/A) does not depend on the choice of g. If K/k is Galois then there is the natural restriction map

Gal(0/k) —> Gal(K/k) with kernel Gal(0/K). It is easy to see that the induced map T(g/A) —> T(Y/A) is surjective, with kernel equal to T(/?P). Therefore

T(g/h)/T(g/Y)' .- ' T(//) and

e(Y / A) = I T(Y /A)I I. So the ramification index equals the order of the inertia group, for Galois extensions. It follows that the definition agrees with the usual one for finite extensions. To consider archimedean primes, we proceed slightly differently. An archimedean place of k is either an embedding 0: k —> R or a pair of complexconjugate embeddings (p, ), with ifi 0 and 0: k —> C. Since C is algebraically closed, any embedding 0 or tp may be extended to an embedding O -> C (use Zorn's lemma). In particular, we can extend to K. If K/k is Galois and 0 1 and 02 are two extensions of 0, then 0j 1 0 1 E Gal(K/k). Hence 0 1 = 4 2 a for some a. If (0 1 , 'P i ) and (0 2 , fi2 ) extend 0, we have 0 1 = 02 a,

336

Appendix

hence (0 1 , IT/ 1 ) = (0 2 , IT/2 )o-, for some a. A similar result holds for extensions of complex places, so the Galois group acts transitively on the extensions of a given place. If K/k is Galois, w is an archimedean place of K, and y is the place of k below w, then we define T (w/v) = Z(w/v) = {a

G

Gal(K/k) I wa = w}.

It is easy to see that T is nontrivial only when vis real, w = (0, 0) is complex, and a 0 1 is the "complex conjugation" 0 - ( = -1 0), which permutes 0 and kW and has order 2. Therefore T(w/v)I = 1 or 2. We may now define the ramification indices for archimedean primes just as we did for finite primes. For more on the above, see Iwasawa [6], §6.

§3 Class Field Theory This section consists of three subsections. The first treats global class field theory from the classical viewpoint of ideal groups. The second discusses local class field theory. In the third, we return to the global case, this time using the language of idèles. We only consider some of the highlights of the theory and give no indications of the proofs. The interested reader can consult, for example, Lang [1], Neukirch [1], Hasse [2], or the articles by Serre and Tate in Cassels and Frehlich [1].

Global Class Field Theory (first form) Let k be a number field of finite degree over U. Let 00 = fl Ki denote an integral ideal of k and let M., denote a formal squarefree product (possibly empty) of real archimedean places of k. Then 9J1 = 9J10 9J1,, is called a divisor of k. For example, 9J? = 1, 9J1 = oo, 9J7 = 5 3 172. oo, and 9J1 = 3 - 37 •103 are divisors of U. If a ekx, then we write a 1 mod* 9J/ if (i) — 1) > ei for all primes A, (with ei > 0) in the factorization of 9N 0 , and (ii) a > Oat the real embeddings corresponding to the archimedean places in 901,,, . Let Pwrt denote the group of principal fractional ideals of k which have a generator a 1 mod* 9J1. Let 41 be the group of fractional ideals relatively prime to 9J1 (note that /911 = /90. The quotient /aR/P92 is a finite group, called the generalized ideal class group mod 9)1. For example, let k = U, let n be a positive integer, and let 9J1 = n. The group /„ consists of ideals generated by rational numbers relatively prime to

337

§3 Class Field Theory

n. Let (r) be such an ideal. Then (r) is generated by + r and by — r. If (r) E P„ then we must have ± r E__-- 1 mod n, hence r + 1 mod n. It follows that

I„/P„'-'. (Z/nZ)x /{± V. Now suppose 9)1 = noo. The group I, is the same as In , but if (r) E Pnoo then we must be able to take a positive generator congruent to 1 mod n, so we need (unless n = 2), so the — 1 mod n then WO P 1 mod n. If I ri I ri archimedean factor makes P931 smaller. It follows easily that

Inco/P„.0 ':_-_' (Z/nZ)x . The effect of the archimedean primes is apparent in the case of a real quadratic field k. Let gil o = 1 and let Moo = oo 1 oo 2 be the product of the two (real) archimedean places. Suppose the fundamental unit e has norm — 1, so E is positive at one place and negative at the other. Let (a) = ( — a) = (ea) = (— ea) be a principal ideal of k. One of the generators for (a) is positive at both co l and oo 2 , so every principal ideal has a totally positive generator, and P = P1 =„ 2 . Of course,

1 1 /P1 = ideal class group. By definition,

I coic,,,,/P.,, 2 = narrow ideal class group. So we find that the narrow and ordinary class groups are the same. It will follow from subsequent theorems that the narrow ideal class group corresponds to the maximal abelian extension of k which is unramified at all finite places. Now suppose E has norm + 1. Choose a E k such that a > 0 at oo 1 and a < 0 at oo 2 (for example, a = 1 + \/d). Then (a) has no totally positive generator, hence P 0, 1 ,02 P 1 (the index is easily seen to be 2). Therefore the narrow ideal class group is twice as large as the ordinary class group in this case. We return to the general situation, so k is a number field of finite degree over Q. Let ek denote the ring of integers of k. Consider a finite Galois extension Klk. Let h be a prime of ek and Y a prime of OK above A. Let N A = IGIAI = norm to 0 of A. The finite field C K/Y is a finite extension of (0k/A with Galois group generated by the Frobenius (xi—) xNfi). Let Z(Y/h ) be the decomposition group and T(Y/h) the inertia group. There is an exact sequence

1 -+ T (Y /A ) ---). Z@1/fl ) --+ G a lO C KIY KGI

»

1.

Suppose Y is unramified over A. Then T = 1, so Z is cyclic, generated by the (global) Frobenius ag , which is uniquely determined by the relation a[”, x

x NA mod Y for all x E eK.

338

Appendix

Suppose T is an automorphism of K such that T(k) = k. Then TY is ( 7.— l x )Nle mo d unramified over TA. Since 66" t_ ix y we have Tag.-c - i x x N A mod TY. Since NA = Nth, we obtain ,

= T1T0.-c - 1 . If K/k is abelian then a tg = a g for all T c Gal(K/k). Hence cr g. depends only on the prime A of k, so we let 0-h = aga.

We may extend by multiplicativity to obtain a map, called the Art in map, 'b -- Gal(K/k),

where b is the relative discriminant of K/k. What are the kernel and image? Theorem 1. Let K/k be a finite abelian extension. Then there exists a divisor f of k (the minimal such divisor is called the conductor of K/k) such that the following hold: (i) a prime h (finite or infinite) ramifies in K/k <=> A I f. (ii) If991 is a divisor with iI9J1 then there is a subgroup H with Poi g H g I 93-1 such that

/9R/H'-.- Gal(K/k), the isomorphism being induced by the Artin map. In fact, H = Pffli N Kik( 1 9N(KO, where Ian(K) is the group of ideals of K relatively prime to 9J1.

Theorem 2. Let 9J1 be a divisor for k and let H be a subgroup of I 9j1 with Pan g H g / 9m . Then there exists a unique abelian extension K/k, ramified only at primes dividing 9Y (however, some primes dividing 9JI could be unramified), such that H = Pgi N Kik (I yA (K)) and /931/H -.- Gal(K/k) under the Artin map.

Theorem 3. Let K i /k and K 2/k be abelian extensions of conductors i l and i2 , let 9j1 be a multiple of i 1 and f2, and let H 1 , H2 g 1.931 be the corresponding subgroups. Then H1 __ H2<>1(1 2

K2.

The above theorems summarize the most basic facts. We now derive some consequences. In Theorem 2, let 9N = 1 and let H = P9x = P. We obtain an abelian extension K/k with

Gal(K/k) -.- IIP -.' ideal class group of k.

339

§3 Class Field Theory

By Theorem 1(i), K/k is unramified, and any unramified abelian extension has f = 1 and corresponds to a subgroup containing P 1 = P. By Theorem 3, K is maximal, so we have proved the following important result.

Theorem 4. Let k be a number field and let K be the maximal unramified (including co) abelian extension of k. Then Gal(K/k)

ideal class group of k,

the isomorphism being induced by the Artin map. (The field K is called the Hilbert class field of k).

We note an interesting consequence. Let A be a prime ideal of k. Then A splits completely in the Hilbert class field <=> the decomposition group for A is trivial <:=> 6 = 1 <1=:,A E P A is principal. Similarly, for a prime number p, we may choose H P such that H/P = non-p-part of I/P. Then I/H p-Sylow subgroup of I/P. The field ( = Hilbert p-class field) corresponding to H is the maximal unramified abelian pextension of k. We now justify a statement made in Section 10.2. Let K be the Hilbert class field (or p-class field) of k, let F k, and suppose k/F is Galois. Then K/F is also Galois, by the maximality of K. As in Chapter 10, G = Gal(k/F) acts on Gal(K/k) (let I" E G; extend to iz E Gal(K/F); then at = to-t--1 ). Also, G acts on the ideal class group of k. Let A be a prime ideal of k. Then under the Artin map, and TA = = (c h)r, by a formula preceding Theorem 1. Therefore Gal(K/k)

ideal class group of k

as Gal(k/F)-modules, as was claimed in Chapter 10. We now need another property of the Artin map. Suppose we have fields F, k, M, and K, as in the diagram, with K/k and M/F abelian. Mk

A

K

F

(we do not assume M n k = F). Let A be a prime ideal of k, unramified in K/k, and let g' lie above A. Similarly, let A- and be the primes of F and M lying below A and Y, respectively. We also assume that is unramified in M/F. Letf = [Ck /A;C F/i] be the residue class degree. Then Normk/F A= g and NA = (N A.)1 . Since em _Ç (9 K , we have im x x N A mod for x E Cm. I

340

Appendix

We have used the notation a"' to mean "o- for the extension K/k, restricted to M." But "Norm AAJ

_ — x N Amod

— M/Ff ‘)X=X

:0.

Therefore ,MIF uM — "

Norm

•

We give an application. Suppose M is the Hilbert class field of F and K is the Hilbert class field of k. Furthermore, assume M n k = F. Then Gal(Mk/k) Gal(M/F), via restriction; hence Gal(K/k) Gal(M/F) surjectively via restriction. We have the following diagram (Ik/Pk = ideal class group of k, etc.): Gal(K/k)

tk/Pk

Norm

restr. Gal(M/F).

IF/PF

The horizontal maps are the Artin maps. The diagram commutes by what we just proved. Since our assumptions imply that the arrow on the right is surjective, Norm is also surjective. So we have proved the following.

Theorem 5 (= Theorem 10.1). Suppose the extension of number fields k/F contains no unramified abelian subextensions L/K with L 0 K. Then the norm map from the ideal class group ofk to the ideal class group ofF is surjective and the class number hp divides hk . We now relate the above theorems to abelian extensions of O. Let n be a positive integer and consider Q(). Let p/en. As we showed in Chapter 2, the Frobenius a-p is given by a p(ç„) = Cri . Thus we have a map

I,

Gal(O(C„)/0.).

If (a, n) = 1 and a > 0, then (a) a, so the map is surjective (in fact, by Dirichlet's theorem, it is surjective when restricted to prime ideals). We now Then, as determine the kernel. Let r E Q with (r)E In . Write Irl = H ideals, (r) = H SO

a(r) = H 0-bp = a lrl , where o-i r i (Cn) = C ni ri (In I mod n is a well-defined element of (Z/nZ)x ). Therefore

a ( r)

1 mod n nI <=> (r) E Pn„

= 1 <=> I

•

341

§3 Class Field Theory

Since I „ = I,we obtain

I n co/Pn co under the Artin map. This of course agrees with the fact that I n „/P,,„ (Z/nl) .

What happens if we leave off op and consider I„/P„? By Theorem 1(i), we cannot have ramification at po and it is not hard to show that the corresponding field is UG„) + . This agrees with our previous calculation that .1„/P„ (/fl )" /{±1} . Suppose now that K is a number field and K/U is abelian. By Theorem 1, there exists a divisor 911 and a subgroup H with Po H .1931 . We may assume 9N = noo, with n E Z. By Theorem 3, K is contained in the field corresponding to P„ namely U(C„). We obtain the following.

Theorem 6 (Kronecker-Weber). Let K be an abelian extension of U. Then K is contained in a cyclotomic field.

Let K/U be abelian and let H P Since in.1Pno,

the group 1-1//3 Since

the corresponding subgroup.

( 7Z/n 1) x

corresponds to a subgroup of congruence classes mod n. (p) splits completely <=>

=

1 <=>

(p) G H,

we find that the primes that split completely are determined by congruence conditions mod n. In fact, this property characterizes abelian extensions. Let p 1 (mod 4) and let q z p be an odd prime. Then q splits in U(.,5) <=> (p/q. = 1 <=>(by Quadratic Reciprocity) (q/p) = 1 <=> q is a square mod p, which is equivalent to q lying in certain congruence classes mod p. Let = Gal(U(fp)/U). Since q splits .<:=> aq = 1, we have shown that aq = 1 if q is a square mod p, a = t if not. Now let r E U with (r) E Ip (i.e., (r,p)= 1). Write r ri = jJ qb and am = fl c. It is easy to see that I r I is a square mod p

a (r) = 1

<-> r is a square mod p

(since p 1 mod 4). Let H denote the group of ideals in II, generated by squares mod p. We have shown (the main step was Quadratic Reciprocity) that H is the kernel of the Artin map. In particular, Pp g H.

Conversely, the fact that Pr, H implies Quadratic Reciprocity for p: Since H Ip has index 2, it must consist of the squares mod p, because I plPp

(.1/pZ)" /{+ 1}

342

Appendix

is cyclic. Therefore - = 1 ---=> q splits <=> a-g = 1 <=> q is a square mod p •4> (q-)

1.

In general, the fact that the kernel of the Artin map contains P 9x (Theorem 1(ii)) is one of the most important parts of the theory. For example, it was the major step in the above proof of the Kronecker-Weber theorem.

Local Class Field Theory Let k be a finite extension of Op. We may write

V< =ex U=e x W' x U 1 , where n = a uniformizing parameter for k, 7E 2 = {7E111 G Z} U = local units,

W' = the roots of unity in k of order prime to p, U 1 = {xEUlx -.E 1 mod ii}. Theorem 7. Let Klk be a finite abelian extension. There is a map (called the Artin map)

Gal(K/k) a 1—) (a, K I k) which induces an isomorphism k x KikK x

Gal(K/k),

where N Km denotes the norm mapping. Let T denote the inertia subgroup of Gal(K/k). Then U kIN K ikU K T.

If Klk is unramified then Gal(K/k) is cyclic, generated by the Frobenius F, and (a, K I k) =

Theorem 8. Let H g kx be an open subgroup q ffinite index. Then there exists a unique abelian extension Klk such that H = N KIkK x •

Theorem 9. Let K 1 and K2 be finite abelian extensions of k. Then K 1 NKI/kKlx 2 NK2/kK2N.

g K2 <=>

343

§3 Class Field Theory

The Artin map satisfies the expected properties. For example, if a is an automorphism of the algebraic closure of k then (aa,aKlak) = a(a, K/k) j. Also, if Klk and MIF are abelian, with F _g_ k and M g K (see the diagram in the previous subsection), then, for a c le, (a, Klk)I m = (NkiF a, MIF).

The above theorems may be modified to include infinite abelian extensions Klk. Let kx be the profinite completion of k x . This means

kx (1_11. lim kx IH where H runs through (a cofinal subsequence of) open subgroups of finite index. Write k x'-.' 7E2 x W' x U 1 , as above, and let H be of finite index. By taking a smaller H if necessary, we may assume k x 1H n-- (ZImZ) x W' x U i /Urr for some m and n. It is easy to see that U 1 = lim U i /Uir,

W' = lim 4--- W'.

But

=2

tim z/niz

n zp P

(see the section on inverse limits). Therefore, we may formally write 7 2 x W' x U 1 '.- ' 7c2 x U.

Theorem 10. Let k be a finite extension of Up and let kab denote the maximal abelian extension of k. There is a continuous isomorphism Gal(kab/k). This induces a one-one correspondence between abelian extensions Klk and closed subgroups H g kx . If H corresponds to K,

Gal(K/k). Let g Kik (U K ) = Klk. Then

n LN LIk(U L), where L runs through all finite subextensions of 11

Uk t1S-1 K Ik( UK)

T(K.1 109

the inertia subgroup of Gal(K/k).

We give an example. Let k = Up Then 0

zpX . Q; ^.-, pl x Wp _ 1 x (1 + p2 p) ,:-, p2 x

344

Appendix

Let (n, p) = 1 and let c > O. We have the following diagram: Up(Crtp.)

Let

a = pb14

E

.

Then (a, Up(C.)/0p) = (p h,

Vir

= Fb :

(F = Frobenius). The group U maps to the inertia subgroup, which is isomorphic to Gal(Q p(Cpc)/00. It can be shown that (u, U(C„pc)/O p) yields the map CpC. H+ ClipC 1 , where Cup: is defined in the usual manner. It is now easy to see that Wp _ corresponds to the (tamely ramified) extension ()/Q p and that 1 + pZ p corresponds to the (wildly ramified) extension

Op(CO/Op(Cp). Now consider the infinite extension Claph/O p . We have

Gal(Qapb/0 p) 0;p 7L x We know (Chapter 14) that Qapb

Ç4,...)

=

= Up(Cp4Up({Cni(P,

= 1 1).

We have z;.

Gal(O p(Cp .)/0 p)

Since Galois groups of unramified extensions are isomorphic to Galois groups of extensions of finite fields, it follows that

Gal(Up({C„ I (p. n) = 1 } )/U p)

Gal(Fp/Fp)

2

p.

Global Class Field Theory (second form) Let k be a number field and let A be a prime (finite or infinite) of k. Let k0 and U 0 denote the completion of k at A and the local units of k0 , respectively. If h is archimedean, let U A = k;. Define the idele group of k by k = {(.. , X o ,

G

fl k Ix A

UA fer 'arnost all AI

("almost all" means "for all but finitely many"). Topologize ik by giving

u =JJ u,

345

§3 Class Field Theory

the product topology and letting U be an open set of J. Then locally compact group. It is easy to see that there is an embedding kx

Jk

becomes a

k

(diagonally) and it can be shown that the image is discrete. The image is called the subgroup of principal idèles. Let klk x

Ck =

be the group of idèle classes. Let K/k be a finite extension. If is a prime of K above the prime A of k, then we have a norm map on the completions N K g k fi . Let x = J K. Define (... , x g , .) E NK/k(x) = (• • • ,

yi„ • • •) 6

41

where yie =JJ

N gpi/e Xg)•

It is not hard to show that if x = (... , x, ...) is principal, then (• IN Klk x, • • •), which is also principal. Therefore we have a map

N KikX =

••

N CK

C k'

Theorem 11. Let Kik be a finite abelian extension. There is an isomorphism Jk/k x N KikJK =

Gal(K/k).

CkIN KikCK

The prime A (finite or infinite) is unramified in K/k-U it , (UA embeds in Jk via tiA F-0 (1, , 1)).

kx NK00 1 K•

Theorem 12. If H is an open subgroup of Ci, of finite index then there is a unique abelian extension Kik such that N KACK= H. Equivalently, if H is open of finite index in Jk , and k x H, then there exists a unique abelian extension Kik such that kx N K ik1K = H. Theorem 13. Let K 1 and K1

K2

be finite abelian extensions of k. Then

K2 <=>

k x -N

k x N K2001 K2.

The above theorems may also be stated for infinite extensions. Let denote the connected component of the identity in Ck .

Dk

Theorem 14. (a) If K/k is abelian, then there is a closed subgroup H with Dk H Ck, such that Ck/H The prime

A

Gal(K/k).

is unramified <1=> kx UA /k x E H.

346

Appendix

(b) Given a closed subgroup H with Dk H g Ck (equivalently, Ck/H is totally disconnected), there is a unique abelian extension corresponding to H. in (a). as As a simple example, let K be the Hilbert class field of k. Since K/k is unramified everywhere, U = 11 Uo k x Nic ik fic. Since K is maximal, le U is the subgroup corresponding to K, hence Jk/k ' U -.- Gal(K/k).

There is a natural map Jk — ■

ideals of k

(• • • , xh , • • .) /-4 finite

h

The kernel is U. If we consider the induced map to the ideal class group, we obtain Jk/kx U ':-_- ideal class group of k.

Therefore Gal(K/k) is isomorphic to the ideal class group, as we showed previously.

Tables

§1 Bernoulli Numbers This table from H. Davis [1], pp. 230-231, gives the value of (-1 B2 for 1 n < 62. In this book we have numbered the Bernoulli numbers so that Bo == 1, 13 1 == --I, 13 2 == == ami 13 2n == 0 for n >1. Same authors use different numbering systems and a different choice of signs. For more Bernoulli numbers, see H. Davis [1] and Knuth-Buckholtz [1]. For prime factorizations, see Wagstaff [1]. )'

Denominator

Numerator 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

1 1 1

1 5 691 7 3617 43867 1 74611

2 861 770 257 26315 27155 2 92999 2 61082 71849

'

8 2363 85 37494 58412 93210 76878 30534 39138 64491

54513 64091 53103 61029 76005 41217 58367 77373 41559 22051

6 30 42 30 66 2730 6 510 798 330

1 2 3 4 5 6 7 8 9 10

138 2730 6 870 14322 510 6 1919190 6 13530

11 12 13 14 15 16 17 18 19 20

347

348

Tables

Numerator 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51

02691 35023 77961 27547 77525 91853 12691 63229 36021 41491

1806 690 282 46410 66 1590 798 870 354 56786730

21 22 23 24 25 26 27 28 29 30

00585 43408 68585 41953 03985 74033 86151 30147 86652 98863 85444 97914 26479 42017 00221 26335 65405 16194 28551 93234 22418

6 510 64722

31 32 33

29 2479 84483 121 52331 40483

560 49 80116 14996 39292 61334 75557

15 278 5964 94033 50572 57181 36348 93132 88800 20403

20097 33269 51111 68997 05241 35489 84862 26753 41862 04994

Denominator 64391 57930 59391 81768 07964 95734 42141 68541 04677 07982

80708 10242 21632 62491 82124 79249 81238 57396 59940 02460

10 1 99101 7877 1505 60235 58279 18442 34152 83087 246 78288 41 37537 4 57872

123 67838 47260 ... 31308 38134 ... 54961 60429 41728 ... 55088 65801 48463 24004 60378 75128

58718 72814 19091 49208 47460 62443 47001 73333 67003 80307 65673 77857 20851 14381

30 4686

34 35

66994 41104 38277 24464 10673 65282 48830 ... 92211 68014 33007 37314 72635 18668 83077

140100870

36

6

37

82593 ... 65575 83487 42994 89196

53727 07687 19604 05851 99904 36526

30

38

40082 82951 79035 54954 20734 92199 ... 79457 64693 55749 69019 04684 97942 56867 ...

3318

39

230010

40

1 36027 20 73911 660 86653 13114 55720 117 94966 129 48371 122 41520 2 39543 67 95728 945 15330

67701 57025 24576 98090 71461 38893 26488 28644 90572 96471 55859 94351 08138 26966 11600 89060 90826 50448 98037 66936

41491 34148 19593 87728 94176 17619 67401 29691 79021 16231 48207 43023 06579 21436 44959 23415 06729 75090 81912 13338

85145 81613 52903 10839 78653 96983 75079 98905 08279 54521 53752 31632 74446 21510 72665 06387 05495 73961 21252 56962

498

41

83682 ... 60231 32477 57384 ... 95511 74047 98841 57279 79894 68299 96073 52846 13097 33420 62405 24999 95227 04311

31545 09786 26990 02077 13116 01117 31009 98991 ... 78474 26261 49627 78306

3404310

42 6

43

61410

44

50837 75254

272118

45

67496 59341

1410

46

12039 58508

6

47

24233 67304 87259 ... 60560 73421 ... 21872 70284 47711 ...

4501770

48

6

49

33330

50

32040 19410 86090 70782 43020 78211 62417 75491 81719 71527 17450 67900 25010 86861 53083 66781 58791

4326

51

42401 ... 23351 22535 27828 46247 01679 49447 59772 05066 11175 29470 43306 39541

93118 43345 75027 24921 ... 53857 ... 41320 ... 81098 83499 46403 58239 94937 51972

349

§1 Bernoulli Numbers

Denominator

Numerator 52 53 54

55

56

57

58

59

60

61

62

31 60607 3637 64058 34 67857 65973 7645 13412 17515 26508 91925 35526 217 66790 86185 30 12806 19959 36 36108 77635 515 03532 93670

95336 82727 39031 16900 69342 60491 38057 99294 30790

31363 38327 72617 38930 24784 14687

83001 10347 41440 81637 78287 00058

79602 09896 00109 37832 90831 62649 95539 43298 57547 69631 06992 45170 07486 62675 77682

15509 45178 ... 31936 36080 16571 44245 52259 19969 04301 95799 53507 21309 67780

84297 50404 ... 71311 05944 . 91090 28691 32820

49 90311 13678 95876 13297 10268

63366 06013 09803 77533 33681 47268

60792 97703 79685 42471 95535 90946

62581 09311 64431 28750 12729 09001

12871 01628 81518 82818 89552 91371

03352 49568 20151 79873 08865 50126

79617 36554 59342 38620 93238 63197

42746 97212 71692 23465 52541 24897

71189 24053 31298 72901 39976 59230

n

1590

52

642

53

209191710

54

1518

55

97133 52597 21468 51620 14443 15149 84276 80966 75651 48755 15366 78120

1671270

56

91633 33310 76108 66529 91475 72115 61101 14933 60548 42345 93650 90418

42

57

69125 13458 03384 14168 69004 57210 08957 52457 19682 71388

1770

58

15349 47151 58558 50066 84606 06764 14485 04580 64618 89371

6

59

61843 99685 78499 83274 09517 67199 29747 49229 85358 81132 70131

2328255930

60

26374 75990 75743 87227 06832 14100 43132 90331

6

61

31075 42444 62057 88300 33544 35944 41363 19436

30

62

04847 42892 24813 42467 24347 50052 87524 66835 93870 75979 76062 69585 77997 79302

91253 79315 ... 77490 35859

350

Tables

§2 Irregular Primes This table lists the irregular primes p < 4001 along with the even indices 2a, 0 < 2 cr < p — 3, such that p I B 2a . It is essentially the table of LehmerLehmer-Vandiver-Selfridge-Nicol which is printed in Borevich-Shafarevich [1], but there are four additional entries (for p = 1381, 1597, 1663, 1877), which were originally missed because of machine error and which were later found by W. Johnson (see Johnson [1]; this paper gives a list of irregular primes for p < 8000). In order to obtain information about generalized Bernoulli numbers and about class groups, see Corollary 5.15 and Theorems 6.17 and 6.18. For a report on the irregular primes p < 125000, see Wagstaff [1]. P

2a

P

37 59 67 101 103 131 149 157 233 257 263 271 283 293 307 311 347 353 379 389 401 409 421 433 461 463 467 491 523 541 547 557

32 44 58 68 24 22 130 62, 110 84 164 100 84 20 156 88 292 280 186, 300 100, 174 200 382 126 240 366 196 130 94, 194 292, 336, 338 400 86 270, 486 222

577 587 593 607 613 617 619 631 647 653 659 673 677 683 691 727 751 757 761 773 797 809 811 821 827 839 877 881 887 929 953 971

2a

52 90,92 22 592 522 20, 174, 338 428 80, 226 236, 242, 554 48 224 408,502 628 32 12,200 378 290 514 260 732 220 330, 628 544 744 102 66 868 162 418 520,820 156 166

P

2a

1061 1091 1117 1129 1151 1153 1193 1201 1217 1229 1237 1279 1283 1291 1297 1301 1307 1319 1327 1367 1381 1409 1429 1439 1483 1499 1523 1559 1597 1609 1613 1619

474 888 794 348 534,784,968 802 262 676 784, 866, 1118 784 874 518 510 206, 824 202,220 176 382, 852 304 466 234 266 358 996 574 224 94 1310 862 842 1356 172 560

351

S2 irregular Primes

P

2a

P

2a

P

2a

1621 1637 1663 1669 1721 1733 1753 1759 1777 1787 1789 1811 1831 1847 1871 1877 1879 1889 1901 1933 1951 1979 1987 1993 1997 2003 2017 2039 2053 2087 2099 2111 2137 2143 2153 2213 2239 2267 2273 2293 2309

980 718 270,1508 388,1086 30 810,942 712 1520 1192 1606 848,1442 550, 698,1520 1274 954,1016,1558 1794 1026 1260 242 1722 1058,1320 1656 148 510 912 772, 1888 60, 600 1204 1300 1932 376, 1298 1230 1038 1624 1916 1832 154 1826 2234 876,2166 2040 1660, 1772

2357 2371 2377 2381 2383 2389 2411 2423 2441 2503 2543 2557 2579 2591 2621 2633 2647 2657 2663 2671 2689 2753 2767 2777 2789 2791 2833 2857 2861 2909 2927 2939 2957 2999 3011 3023 3049 3061 3083 3089 3119

2204 242,2274 1226 2060 842,2278 776 2126 290, 884 366, 1750 1044 2374 1464 1730 854,2574 1772 1416 1172 710 1244 404,2394 926 482 2528 1600 1984,2154 2554 1832 98 352 400, 950 242 332,1102,2748 138, 788 776 1496 2020 700 2522 1450 1706 1704

3181 3203 3221 3229 3257 3313 3323 3329 3391 3407 3433 3469 3491 3511 3517 3529 3533 3539 3559 3581 3583 3593 3607 3613 3617 3631 3637 3671 3677 3697 3779 3797 3821 3833 3851 3853 3881 3917 3967 3989 4001

3142 2368 98 1634 922 2222 3292 1378 2232,2534 2076,2558 1300 1174 2544 1416, 1724 1836,2586 3490 2314,3136 2082,2130 344, 1592 1466 1922 360, 642 1976 2082 16,2856 1104 2526,3202 1580 2238 1884 2362 1256 3296 1840, 1998,3286 216,404 748 1686,2138 1490 106 1936 534

352

Tables

§3 Class Numbers The following table gives the value and prime factorization of the relative class number h, of U(„) for 1 < 0(n) < 256, n # 2 (mod 4). It is extracted from Schrutka von Rechtenstamm [1], which also lists the contributions from the various odd characters in the analytic class number formula. Some of the large factors were only checked for primality by a pseudo-primality test, so there is a small chance that some of the "prime" factorizations include composites. For values of hp- for 257 < p < 521, see LehmerMasley [1]. A few of the factorizations below have been obtained from this paper. Since the size of I.'ç depends more on the size of 0(n) than of n, we have arranged the table according to degree. For h + there are the following results (see van der Linden [1]): (a) If n is a prime power with 0(n) < 66 then Ian+ = I. (b) If n is not a prime power and n < 200, 0(n) < 72, then 17,4- = 1, except for h i'36 = 2 and the possible exceptions n = 148 and n = 152. Also, we have 1,6'65 = 1. If we assume the generalized Riemann hypothesis, then the following hold: (c) If n is a prime power with ci)(n) < 162 then hn+ = 1. We have ht63 = 4. (d) If n is not a prime power and n s 200, then 11,:- = 1, with the following exceptions: ht36 = 2, ht45 = 2, 1483 = 4. It is possible to obtain examples of h; > 1 using quadratic subfields (Ankeny-Chowla-Hasse [1], S.-D. Lang [1]), or using cubic subfields (see the tables in M.-N. Gras [3] and Shanks [1]), or using both (CornellWashington [1]). See also Takeuchi [1]. Kummer determined the structure of the minus part of the class group of (11G,) for p < 100. By (a) above, this is the whole class group for p < 67; by (c), it is the whole class group for p < 100 if we assume the generalized Riemann hypothesis. All the groups have square-free order, hence are cyclic, with the following possible exceptions: 29, 31, 41, and 71. In these cases, 29 yields (2) x (2) x (2), 31 yields (9), 41 yields (11) x (11), and 71 yields (72 . 79241), Here (m) denotes the cyclic group Z/mZ. See Kummer [5, pp. 544, 907-918], Iwasawa [16], and Section 10.1. For more techniques, see Cornell-Rosen [1] and Gerth [5].

353

§3 Class Numbers

0(n) 1 3 4 5 8 12 7 9 15 16 20 24 11 13 21 28

1

2 2 4 4 4 6 6 8 8 8 8 10 12 12 12

n 140 144 156 168 180 53 81 87 116 59 61 77 93 99 124 85 128 136 160 192 204 240 67 71 73 91 95

0(n)

48 48 48 48 48 52 54 56 56 58 60 60 60 60 60 64 64 64 64 64 64 64 66 70 72 72 72 111 72 117 72

h-

h-

n 36 17 32 40 48 60 19 27 25 33 44 23 35 39 45 52

12 16 16 16 16 16 18 18 20 20 20 22 24 24 24 24

1 1 1 1 1 1 1 1 1 1 1 3 1 2 1 3

h-

q5(n) 56 72 84 29 31 51 64 68 80 96 120 37 57 63 76 108

24 24 24 28 30 32 32 32 32 32 32 36 36 36 36 36

2 3 1

8=2 9 32 5 17 8 = 23 5 9 32 4 = 22 37 9 _ 32 7 19 19

h-

O(n) 41 55 75 88 100 132 43 49 69 92 47 65 104 105 112

40 40 40 40 40 40 42 42 44 44 46 48 48 48 48

121 = 11 2 10 = 2.5 11

55 = 5 • 11 55 = 5 11 11 211 43 69 = 3. 23 201 = 3. 67 695 = 5 139 64 = 2 6 351 = 3 3 • 13 13 468 = 2 2 • 3 2 13

11 -

n

95(n)

h-

39 = 3 • 13 507 = 3 • 13 2 156 = 22 • 3 - 13 84 = 2 2 • 3 .7 75 = 3 • 5 2 4889 2593 1536 = 2 .3 10752 = 29 3 7 41421 = 3 • 59 233 76301 = 41 • 1861 1280 = 2 8 .5 6795 = 3 2 - 5 • 151 2883 = 3.312 45756 = 2 2 3 2 31 • 41 6205 = 5 • 17 73 359057 = 17 .21121 111744 = 2 7 3 2 97 31365 = 3 2 • 5 • 17 41 61353 = 3 2 • 17 - 401 15440 = 2 . 5 193 6400 = 2 8 - 5 2 853513 = 67 • 12739 3 882809 = 7 2 • 79241 11 957417 = 89. 134353 53872 = 2 4 • 7 . 13 . 37 107692 = 2 2 . 13 • 19 . 109 480852 = 2 2 3 2 192 . 37 132678 = 2 . . 7 13

135 148 152 216 228 252 79 123 164 165 176 200 220 264 300 83 129 147 172 196 89 115 184 276 141 188 97 119 153

72 72 72 72 72 72 78 80 80 80 80 80 80 80 80 82 84 84 84 84 88 88 88 88 92 92 96 96 96

75961 = 37 2053 4 827501 = 3 2 • 7 • 19 .37 109 1 666737 = 3' • 19' 1 714617 = 3 2 • 19 37 271 238203 = 3 2 7 19 199 71344 = 7 3 . 13 100 146415 = 5 53 .377911 8 425472 = 2 12 11 2 17 82 817240 = 2 3 .5 11 2 • 71 • 241 92620 = 2 2 . 5 11 421 29 371375 = 5 3 11 . 41 . 521 14 907805 = 5 . 11 2 . 41 601 856220 = 2 2 .5 31 • 1381 1 875500 = 22 .53 • 11 2 .31 1 307405 = 5 • 11 2 . 2161 838 216959 = 3 279405653 37 821539 = 7 . 29 . 211 883 5 874617 = 7 29 - 43 673 792 653572 = 2 2 .43 .211 21841 82 708823 = 43 71 27091 13379 363737 = 113 • 118401449 44 697909 = 3.331 .45013 1486 137318 = 2.3 .23 .672 .2399 131 209986 = 2.3.23 2 • 67 • 617 1257 700495 = 5 .47. 1392 277 24260 850805 = 5 47 139 742717 411322 842001 = 577 • 3457 • 206209 1238 459625 = 34.53 • 13.97 2 2416 282880 = 2 8 . 5 11 2 . 15601

354

n 195 208 224 260 280 288 312 336 360 420 101 125 103 159 212 107 109 133 171 189 324 121 113 145 232 348 177 236 143 155 175 183 225 231 244 248 308 372 396 127 255 256 272 320 340 384 408 480 131 161 201

Tables

(1)(n) 96 96 96 96 96 96 96 96 96 96 100 100 102 104 104 106 108 108 108 108 108 110 112 112 112 112 116 116 120 120 120 120 120 120 120 120 120 120 120 126 128 128 128 128 128 128 128 128 130 132 132

h 22 151168 = 2' .1 • 13 2 29904 190875 = 3 3 . 5 3 . 13 3 .37. 109 14989 501800 = 23 .32 • 5 2 • 72 • 13 • 17 • 769 531 628032 = 2 2 3 3 • 13 2 265 454280 = 2 3 • 3 3 5 7 • 13 . 37 . 73 3 5 . 13 2 457 . 1753 32899 636107 1621 069632 = 2 6 • 3 3 7 13 3 61 930 436416 = 3 3 . 7 13.61 • 97 523 952100 = 2 2 . 3 2 . 5 2 . 7 2 • 109 2 10 229232 24 • 3. 13 3 .97 3 547404 378125 5.101 601 .18701 57708 445601 = 2801 .20 602801 9 069094 643165 = 5 . 103 . 1021 . 17 247691 223233 182255 5 . 53 2 . 3251 . 4889 6 789574 466337 3 . 13 . 1093 . 4889 32579 63 434933 542623 = 3 • 743 • 9859 • 2 886593 161 784800 122409 = 17 • 1009 • 9431 866153 157577 452812 = 2 2 . 3 1° . 13 . 19 . 37 . 73 503009 425548 = 2 2 . 36 .7 19 . 73 109 163 105778 197511 = 7 37 109 127 163 . 181 5 770749 978919 = 19 2593 117 132157 12 188792 628211 = 67 353 20021 25741 1612 072001 362952 = 2 3 17. 11 853470 598257 1 467250 393088 2 14 .281 .421 757 248 372639 563776 = 2 18 • 3 .7 • 13 • 43 2 • 1877 .32 • 5 • 7 . 71317 5 889026 949120 = 81 730647 171051 = 3 . 59 . 233 .523 . 3 789257 4509 195165 737013 3 59 233 109337 677693 36 027143 124175 = 5 2 . 7 . 61 2 .661 • 83701 84 473643 916800 = 2 . 34 .52 • 631 . 129121 4 733255 370496 = 2 8 .61 .271 .601 . 1861 767 392851 521600 = 26 . 5 2 • 31 3 .41 211 .1861 1 5 175377 535571 = 11 .61.331 • 2791 24481 298807 787520 = 2 16 • 3 2 5 . 11 . 61 . 151 30953 273659 007535 3 3 . 5 . 11 . 41 . 61 .691 . 1861 .6481 12239 782830 975744 --= 2 8 . 3 2 • 112 .312 .41 .211 '5281 12 767325 061120 = 2 21 . 5 . 7 .312 . 181 307 999672 562880 = 26 .32 .5 • 31 .41 2 • 151 • 13591 44 485944 574929 = 3 11 . 13 .3l .181 .19231 2 604529 186263 992195 = 5 • 13 43 • 547 .883 • 3079 .626599 16 881405 898800 = 2 4 . 3 .52 • 17 2 . 73 .353 . 1889 10 449592 865393 414737 = 17 21121 .29 102880 226241 239445 927053 918208 = 2' 5 • 3 2 • 13 . 17 . 41 97 .577. 1601 39497 094130 144005 3 2 . 5 . 174 . 41 . 97 337 7841 1212 125245 952000 = 2 12 . 5 3 . 17 . 73 .593 .3217 107878 055185 500777 = 3 2 • 17 401 1697 21121 49057 4710 612981 841920 = 216 .32 . 5 . 41 • 97 • 193 • 2081 617 689081 497600 = 2 1 ' . 34 . 5 2 • 7 4 . 17 . 41 .89 28 496379 729272 136525 = 3 3 • 5 2 • 53 • 131 • 1301 - 4673 706701 17033 926767 658911 — 3 2 . 11 . 67 3 22111 25873 252655 290579 982532 = 2 2 11 .23 2 .67 2 • 12739 • 189817

355

§3 Class Numbers

n

0(0

207 268 137 139 213 284 185 219 273 285 292 296 304 315 364 380 432 444 456 468 504 540 149 151 157 169 237 316 187 205 328

132 132 136 138 140 140 144 144 144 144 144 144 144 144 144 144 144 144 144 144 144 144 148 150 156 156 156 156 160 160 160

352 400 440 492 528 600 660 163 243 249 332 167 203 215 245 261 344

160 160 160 160 160 160 160 162 162 164 164 166 168 168 168 168 168

392

168

57569 648362 893621 ---- 3 2 • 23 • 67 • 727 • 17491 • 326437 28 431682 983759 502069 723. 672 • 1607 • 12739 • I 921657 646 901570 175200 968153 = 17 2 •47737 - 46 890540 621121 1753 848916 484925 681747 = 3 2 472 277 2 967 1188 961909 20 748314 966568 340907 = 7 2 • 41 • 43 • 281 .421 • 25621 • 79241 1858 128446 456993 562103 ---- 72 2 . 71 113. 281 79241 7 319621 13 767756 481797 006325 ---- 5 2 • 72 13 37 2 53 2 9433 • 23833 219 406633 996698 095616 = 2 12 3 • 17 2 -37 .89 -46549. 134353 21198 594942 959616 ---- 2 20 • 3 2 7 32 19- 37 2 73 34397 734347 893592 ---- 2 3 • 34 13 • 9 • 372 • 73 1092 • 181 26883 466789 548427 261560 = 23 .3 2 5 •7-89• 109 • 181 2 • 433 -577. 134353 8269 489911 111632 618625 = 32.53 • 7 3 17 2 19 372 109 397 • 65881 1764 209801 444986 506285 = 3 5 5- 193 • 37 3 -73 109 • 525241 3990 441973 190400 = 28 • 34 52 73 132 372 97 - 7 • 13 5 . 37 2 153601 104578 560000 = 2' 4 37 3 118301 079203 997232 = 7 - 13 - 19 2 - 53 2 - 73 - 109 433 613 859 095743 251563 370449 ---- 3 2 • 13 2 • 19.372 • 109 - 271 • 541 • 1 358821 55 382724 129516 879312 = 2 4 34 • 7 . 19 3 • 372 • 1092 • 54721 2 2 • 3 • 72 • 19 4 . 199 • 487 • 3259 17 643537 152468 843364 2 1° 3 1° - 5 2 - 7 11 2 134 • 181 6 618931 810639 948800 2 077452 902069 895168 = 2' 6 • 3 13 76 13 2 1 892923 169092 229025 ---- 3 2 • 5 2 19 2 • 37 • 73 - 109 • 2053 • 38557 687887 859687 174720 123201 = 32 • 149 • 512 966338 320040 805461 2 333546 653547 742584 439257 = 7.112.281 . 25951 - 1 207501 312 885301 56 234327 700401 832767 069245 = 5- 13 2 • 157 2 • 1093 • 1873 . 418861 -3 148601 546489 564291 684778 075637 ---- 313 • 1873 4733 196 953296 289361 130445 289884 021402 281355 = 5 - 7 • 13.53. 157 3433 4421 6007 377911 22 036970 003952 429517 953845 -= 5- 13 2 • 53 79 -2393 • 377911 -6887474101 38816 037673 830728 480329 = 17 2 -41 -241 -4801 -299681 -9447601 78821 910689 378365 476000 = 25 . 3 2 • 5 3 . 11 2 • 41 • 101 2 661 4261 • 15361 82 221729 062003 473169 480000 = 26 5 4 11 2 - 17 31 -71 -101-241-521-35801081 5578700230786671358855375 = •1 I 41 2 • 113 -281 .521 • 1801 -2801 -28921 1 692044 042657 239185 550625 = 5 4 • 11 4 • 41 • 61 .101 601 26261-46381 3690 827552 653792 584000 = 2 6 • 3. 5 3 .11 31 2 61 2 181 1381 - 15641 331431 584848 686177 320960 -= 2 20 • 5.11 2 • 17 - 41 -71 -241 -1321 • 33161 20215 309155 022994 375000 = 2 3 5' 11 2 -31 -41 -61 101 -521 -65521 7166 325608 289022 528100 = 2 2 52 • 11 3 -41 - 101 • 131 -601 -2161 76421 20 090237 237998 576000 = 2 7 5 3 • 11 2 31 181 • 421 • 1381 • 3181 2708 534744 692077 051875 131636 = 2 2 .181 .23167 .365473 • 441 845817 162679 14 948557 667133 129512 662807 = 2593 • 5764 966319 758245 087799 (composite) 13 898958 132089 743179 099753 = 3 - 279 405653 16581 575906 876567 2233 138758 192814 382133 816279 = 3 • 80279 • 612377 -54 192407 • 279 405653 28121 189830 322933 178315 382891 = 11 - 499 - 5 123189 985484 229035 947419 4 413278 155436 385292 173312 = 2' 4 • 3 2 • 7 2 • 29 • 3907 • 26041 • 207 015901 8 562946 718506 556895 170449 ---- 7 2 -19 29 -37 -211 -757 -2017 -22709 I 171633 122845 138181 874350 560487 = 13 2 43 • 127 . 631 • 43793 -4816871221 18 379288 588511 605529 995776 ---- 2 9 32 61 -421 -883-10753-38011 430333 10789 946893 536931 852748 197440 = . 3 • 5 .7-29.43• 197 • 211 • 21841 -929419-1 525987 112 070797 379361 142494 415714 ---- 2 • 43 2 -71 617 -953 -27091 • 28393 -943741 -

•

•

•

356

n

Tables

(b(n)

516 168 38 888604 320171 861798 243568 = 516 - 32 7 - 29 - 43 2 .71 .211 883. 21841 .2 490307 588 168 482059 253351 850013 395157 = 7 -29 -43 -71 - 673 2017 -3571 - 5923 - 27091 173 172 1 702546 266654 155847 516780 034265 = 5 .20297 • 231169 - 72 571729 362851 870621 267 176 12963 312320 905811 283854 380235 = 5 23 - 113 • 1123 - 5237 -26687 53681 -118401449 -4159 34517608 519=3-•21 45013 2152 502881 356 176 4 707593 989354 615385 004311 705592 = -3- I I 23 - 113 - 463 - 15269 - 19207 426757 - 118 401449 368 176 243320 115114 433657 103908 135020 22 - 3 - 5 11 2 -23 .672 - 89 • 2069 2399 - 8537 - 162713 460 176 197 739166 909616 827795 207545 3 .5 - Ii - 67 - 331 .617 - 17029 - 45013 - 114 259861 552 176 767 354245 926929 350377 606384 24 3 235 • 67 2 -617 - 2399 - 10781 -34673 179 178 77 281577 212030 298592 756974 721745 = 5 - 1069 - 14458 667392 334948 286764 635121 181 180 211 421757 749987 541697 225501 539625 = 5 3 - 37 • 41 - 61 - 1321 - 2521 - 5 488435 782589 277701 209 180 4551 326160 887085 824176 768000 = 2 1° - 5 3 11 - 61 - 271 - 264 250891 - 739 979551 217 180 3724 911233 451940 358045 813517 = 35 - 7-11 - 37 -241 -541 -571 -691 -2161 -2791 17341 279 180 18164 714706 446857 534815 843195 = 36 -5 -7 13.151.211 1321 -2551 -4591 5011 -22171 297 180 1078 851803 253231 276755 717661 =3 2 -31 2 - 199 -8191 - 1674991 -45687081331

235184769 305471= 5. 139- 1657 .453377 • 156604 975201 463093 376 184 237 637802 564280 802840 123241 975060 = 22 - 5 .47 - 139 - 18493 • 742717 - 3 536987 .37437 658303 564 184 431950 475833 835326 053345 383630 = 2 - 5 - 47 3 139 3 -277-599-742717- I 257089 191 190 165008 365487 223656 458987 611326 929859 = 11 - 13 - 51263 - 612 771 091 -36733950669733713761 193 192 546617 105913 568165 545650 752630 767041 — 6529 - 15361 -29761 -91969 - 10 369729- 192026 280449 221 192 5 562629 629465 863945 291002 496000 = 2 1 0 - 36 5 3 - 17 31 2 61 - 73 - 113 - 193 - 1297 - 3529 - 8209 291 192 161 230789 161196 289366 922423 524464 = 24 7 - 13 2 - 17 2 - 577 1489 3457 5641 206209 - 8 531233 357 192 1504 490803 465665 772083 088125 = 3 4 54 • 74 132 37 97 3 - 1873 - 1 157953 388 192 145666 644086 003914 044409 030660 616112 = r • 32 • 7 2 - 13- 19 • 37 - 577 - 3457 - 5857 13441 -206209-69761089 416 192 1370 350108 087898 680332 276597 421875 — - 5 1 - 7 2 135 - 37 - 73 - 97 - 109 -241 -409 17401 448 192 327 965590 186830 575092 883770 837200 = 24 3 2 52 72 13 17 2 5 -2I _ 769 - 13697 - 299569 - 471073 476 192 I 099745 163233 204819 353212 762000 — 24. 3o.5 11 2 - 13 - 47 2 974 - 241 -1489-6833 42 34 520198 425 7 096= 4 • 7 3 133 • 17 • 37 2 - 73 .

.

.

357

§3 Class Numbers

I)

0(n)

560

192

576

192

612

192

624

192

672

192

720

192

780 840 197

192 192 196

199

198

275

200

303

200

375

200

404

20 0

500

200

h

54738 664378 286829 420235 392000 = 2" • 3 5 • 53 • 7 • 132 • 17 • 37 • 73 • 97 2 • 181 • 193 • 241 .409 1157 874338 412588 470629 857952 431771 = 35 • 13 2 • 17 • 401 • 457 • 1753 • 1873 • 1 751377 1573 836529 4 600831 021854 761317 711337 226240 = 22° • 3 5 11 2 61 73 • 97 • 193 • 241 • 15601 . 7 712737 2 180486 664807 803314 987752 000000 = 29 .37 .5 6 .7.13 5 .17 3 .37.61.97.109.409 438246 323791 968232 985203 468800 = 29 • 37 • 5 2 • 7 3 13 . 17 • 61 • 73 . 97 769 • 8761 70969 222312 165238 308958 816217 760000 = 28 34 54 • 72 • 1V 192 • 372 1092 • 277 • 3132 409 113496 073931 085358 039040 = 2 46 3 5 • 135 61 109 - 157 84 878288 737639 882168 320000 = 2' 4 . 34 . 5 4 . 7 2 . 134 19 • 37 2 - 73 • 97 397 5 532802 218713 600706 095993 713290 631720 = 2 3 • 5 - 1877 • 7841 • 9398 302684 870866 656225 611549 18 844055 286602 530802 019847 012721 555487 = 34 • 19 727 25 645093 207293 548177 3 168190 412839 18 124664 091430 165276 567871 093750 = 2 5 12 11 3 41 2 61 • 71 101 241 •461 •541 • 631 32442 006711 177310 012824 426376 953125 = LC 61 101 601 5701 • 6701 • 18701 • 1255 817401 22 533972 115769 639175 905217 196211 = 11 • 2801 • 12101 • 244301 20 602801 12007 682201 28 160409 852152 369458 876449 426375 546875 = 57 • 7 41 • 61 • 101 2 601 • 2351 • 18701 .40351 - 1892 989601 20244 072859 233305 618155 148176 257775 = 5 2 • 11 401 • 2801 20 602801 94 315301 33728 676001 360807 306655 167078 388646 788532 317360 = 24 • 5 • 17 • 103 2 • 239 • 1021 • 3299 • 233683 • 7 707223 • 17 247691 311 393365 861041 316591 357682 493761 574005 = 5 7 • 103 • 1021 • 2347 • 306511 • 17 247691 • 54 115489 • 125 998867 169406 792495 647432 946133 820476 066925 =5 2 53 1093 • 4889 • 12377 19813 11 452741 8519 216837 1435 850573 295225 659918 796765 068953 277637 = 34 • 13 79 677 • 1093 .4889 • 13469 • 32579 • 2 805713 • 3875 328913 1 127233 629616 849856 487768 072597 188295 = 3 • 5 • 13 3 • 532 • 1093 2 - 3251 • 4889 • 32579 • 19684 564069 49238 446584 179914 120276 706365 116286 443831 = 3 2 • 7 2 • 41 • 71 • 181 .2812 421 • 1051 • 12251 113 981701 • 4343 510221 41 597545 536058 643707 857919 997509 485501 = 3 • 743 9859 • 2 886593 • 10 109009 64868 018727 424243 70300 542035 941044 246482 693928 842589 712617 = 3 • 743 3181 9859 2 886593 • 348390 669416 638151 886259 13 453389 127871 713260 541632 243338 018775 = 3 9 5 2 72 132 192 73 2 rUV■-• 2 127 • 157 • 163 • 181 397 613 • 1009 15 168897 693915 178656 178325 215530 382842 = 2 • 32 0 • 76 132 17 2 193 - 37 73 3 • 271 • 14149 503 374795 561927 637884 794232 382274 404226 = 2 • 37 • 13 17 • 37 - 379- 1009 2377 -47629 - 34 465933 -9431 866153 •

•

.

•

•

309 204 412 204 265

208

424

208

636

208

211

210

321

212

428

212

247

216

259

216

327

216

•

358

Tables

333 216 84 239369 799126 310123 807613 556409 560000 = 26 . 5 4 . 72 1 3 195 • 372 .43 • 73 . 523 • 111637 • 561529 351 216 2 881839 794389 013705 029278 932481 257394 = 2. 3 12 .7. 13 . 19 6 37 2 • 73 631 , 2341 31393 - 136657 399 216 1178 892414 491021 808120 869355 574272 2' 13 . 3 2° . 7 . 13 • 19 2 . 37 - 61 • 73 2 • 577 . 829 . 1747 405 216 289942 114683 805443 433002 828021 894577 = 37 487 - 541 . 2053 . 2593 . 1 583767 3527 772707 308141 436 216 893749 713826 042123 652446 227238 954966 290576 = 24 .37 . 17 . 19 2 • 163 757 . 1009 3 016927 . 1174 772971 9431 866153 532 216 1 995278 293629 608216 703343 220411 633664 = 2 12 • 31 0 .73. 13 19 3 .31 372 .732 • 109. 1693 2377 .2719 648 216 4207 762445 242777 294033 981083 075596 417079 = 19 . 37 271 2 . 2593 . 117 132157 . 157 470427 . 63112 572037 684 216 9 549392 972039 711651 917872 649044 836352 = 2 14 36 .72 13. 192 37 2 . 73 109 . 127 163 . 199. 1693 .3637 . 12583 756 216 434848 520210 868494 245767 938408 147152 = 24 7 3 13. 19 3 . 37 3 . 109 127 2 .163 181 2 • 271 • 757 9109 253 220 256 271685 260834 247944 985594 908530 991952 = . 3 .11 4 • 1409 . 3301 . 26951 . 79861 . 13 962631 . 2608 886831 363 220 23 207253 826992 628179 863710 751562 290176 = 2 ° .67 . 89 353 . 20021 - 25741 . 20 891667 283264 099631 484 220 29678 406487 322012 695719 894464 039435 383271 = 67 353 • 1441!. 20021 . 25741 - 167971 . 1 005892 255694 569981 223 222 217 076412 323050 485246 172261 728619 107578 141363 7 , 43 • 17 909933 575379 - 11 757537 731851 3424 804483 726447 339 224 87309 027165 405351 637092 447907 404827 688960 = 35 17 - 71 113 - 127 • 281 . 2137 - 14449 . 99709 • 11 853470 598257 435 224 299190 086533 933244 039620 216234 180608 = 239 . 3 13 29 2 • 113 2 .281 - 421 . 757 • 1289 . 11257 452 224 229 865767 233324 575111 010848 122335 548084 846592 = 223 . . 7 . 13 2 . 17 29 . 281 . 24809 168617 374669 . 11 853470 598257 464 224 12 164820 242320 422627 042467 644729 294439 055360 = 23° .3 .5 7 - 13 • 17 2 • 295 .432 • 1877 4621 .226129 386093 580 224 776 785847 831995 632448 594543 440172 154880 = 29 . 281 . 421 .463 - 757 - 1 131397 - 1 413077 239 • 3 5 . 696 224 6438 349938 668172 599554 162206 096280 780800 = 238 • 3 3 . 52 .72 13 . 29 . 43 3 113 1093 . 1429 • 1877 • 71317 227 226 2888 747573 690533 630075 559971 022165 906726 932055 = 5 • 2939 3 • 1692 824021 974901 13 444015 915122 722869 229 228 10934 752550 628778 589695 733157 034481 831976 032377 13 - 17 , 457 - 7753 , 705053 , 47 824141 - 414153 903321 692666 991589 233 232 348185 729880 711782 527290 176798 948867 695747 163449 = 233 • 1433 . 1 042818 810684 723912 819200 922459 107271 266041 (composite) 295 232 670508 644900 926208 004253 553219 885108 451604 = 2 2 • 3 .59 .233 . 349 . 41413 . 9 342293 . 3483 942493 . 8 640296 021597 472 232 19371 983746 349662 149124 469187 254723 339443 284387 = 3 2 . 29 . 59' 233 .42283 - 135257 - 168143 . 4 237829 • 109337 677693 708 232 7 622833 744450 532364 757064 890176 317824 613409 = 3 59 , 233 - 523 , 2 069383 , 3 789257 109337 677693 , 412212 149161 .

§3 Class Numbers

359

239 238 19 252683 042543 984486 813299 844961 436592 191498 141760 = . 3 5 511123 . 14 136487 123373 184789 .22497 399987 891136 953079 241 240 74 361351 053524 744837 764467 869162 082791 741351 378657 = 472 • 13921 - 15601 .2 359873- 126 767281 .518123008737871423891201 287 240 75 414262 624860 852745 819151 571359 184834 222400 = 26 - 5 2 - 7 • 11 7 . 13 . 31 2 .61 521 • 1201 -1609 -2521 -8641 -20673617161 30524189 6037241 5680= 23 32 54 •1 -.2 37 • 41 4 • 61 3 1861 -2281 3061 -24061 -37501 63841 J 325 240 958286 131671 211592 542476 979144 265746 218304 = 61 3 • 101 • 1201 -2141 -7681 • 11701 • 194521 • 849721 • 17 098621 369 240 528 852535 797845 727358 844974 839889 196910 080000 = 212 • 54. 11 6 • 17. 19 • 31 • 271 421 4801 16921 • 1256 507775 765241 385 240 18 696191 070960 590983 421400 100896 768000 = 2 31 - 32 . 53 . 11 .192 31 . 157 - 1021 -9661 .16141 . 2 514961 429 240 1880 049931 342806 129486 552279 849583 657000 = 23 • 5 3 .7. 11 3 • 31 • 61 3 • 181 .571 .661 . 39521 -83701 • 126 901681 465 240 6056 875285 186558 003929 869566 624727 040000 = 2 19 3 6 • 54. 7 -31 -61 151 - 181 631 - 1481 • 1801 • 129121 - 322501 488 240 3 971856 968532 956975 396384 265567 521800 430781 628875 = 3 3 . 5 3 - 11 2 - 31 2 •41 2 .43.61.101 151.421.691.1861.4721-6481 34171.265892761 495 240 151 284295 307196 895954 238278 778191 913580 = 22 - 3 - 5 - 7 11 3 292 - 31 3 - 181 2 -229 .241 2 421 2131 3361 8221 496 240 686038 372620 782033 886901 075737 481803 287781 408768 = 2 15 - 32 • 11 4 - 31 4 -37-41 .612 -97 -211 .241 -601 -4621 -5281 -14281 -29501 525 240 29 585677 490787 726928 862791 955910 586368 = 212 • 34. 11 13- 31 2 • 61 4 • 271 . 331 .601 1861 -467 - 132041 572 240 5 290237 648692 385160 711880 570308 851548 534375 = 3.55 7 . 192 31 -41 61 2 .421 -661 . 27631 . 72271 . 83701 1015 122781 616 240 894031 197420 910862 005847 489304 819295 846400 = 24° .52 7 11 5 • 13- 31 4 181 211 -2161 . 4621 -6301 620 240 19 441064 004704 709948 640099 632484 806819 840000 = 2 26 • 34 • 54 . 11 . 31 . 41 - 61 . 421 - 631 - 5821 - 66931 - 129121 -502081 700 240 126016 649965 778239 405605 204267 365457 285120 -2 12 • 3 5 . 5 • II • 13 - 31 59 2 . 61 2 - 271 -601 - 1861 -9181 -44641 .3549901 732 240 1339 692320 604469 611903 838974 531410 116492 800000 = 2 12 3 3 . 5 5 • 11 • 13 • 19 2 • 31 3 .41 • 61 2 .211 . 691 • 1861 .6481 .25301 .371341 744 240 181 082733 783181 938577 850646 686177 657202 278400 = 2 17 3 5 - 5 2 - 11 5 • 31 2 - 41 2 -101 -131 -151 -211 -541 -5281 -13591-53401 7924056813 7 295134670= 22 • 32 5 3 • 11 2 • 132 19 • 31 5 . 61 2 - 181 • 1381 -5521 -5791 -19231 - 176161 900 240 744248 582096 150452 589487 856013 489542 134375 = 3.55 11 2 . 61 211 -331 -811 -2161 -2791 -24481 334261 -3847430341 92408165 493820571 = 23' . 3 5 . 52 7 . 11 2 • 31 4 . 61 . 101 . 151 - 18L 691 - 751 251 250 95469 181654 584518 651828 574432 658888 070113 445087 403827 = 7 • II - 348270001 - 9 631365 977251 - 369631 114567 755437 243663 626501 301 252 205430 142293 947345 943779 193986 871148 546394 604544 — . 7 - 19 . 43 2 . 211 . 631 -6301 -14827-16843.19531 -122599-511939 V°. 381 252 11 479286 278091 328075 258484 555696 616781 110509 888215 = 3 2 -5-72 -13.37.43 2 .547.631.673.883.3079.6007-626599 -2 185471 - 1126755757

360

n

Tables

0(n)

11 —

387 252 1 348400 009635 509434 335776 865706 103793 086610 214753 = 7 3 • 13 2 • 19 2 • 29. 43 .211 2 • 463 • 883 .967 .1933 • 3067 • 3319 • 4621 .125287 257713 441 252 2427 799098 355426 760759 007408 851329 652222 396831 = 7 4 • 29 • 43 5 • 127 • 337 •673 • 2731 • 11173 • 43051 •1 271383 -4930381 508 252 103042 170932 346966 742775 797541 839182 084871 642467 503360 = 2 8 .5.7 2 .13 3 .19.43 3 ,547.757.883 2 .2143.3079.626599.2664901.139 159441 257 256 5 452485 023419 230873 223822 625555 964461 476422 854662 168321 = 257 • 20738 946049 • 1 022997 744563 911961 561298 698183 419037 149697 512 256 6 262503 984490 932358 745721 482528 922841 978219 389975 605329 = 17 • 21121. 76 532353 • 29 102880 226241 • 7830 753969 553468 937988 617089 544 256 4584 742688 639592 322280 890443 396756 015190 545059 020800 = 2 30 3 8 5 2 7 4 13 17 6 ••,si•2 41 4 .97 • 353 • 433 • 577 • 929 • 1601 640 256 112 066740 284710 541318 559132 951039 771578 615246 011365 = 3 2 • 5 •17 4 • 41 • 97 2 • 337 - 7841 • 9473 • 21121 • 376801 69 470881 • 5584 997633 680 256 77483 560514 02244 288033 941979 251535 291351 040000 = 24 ' • 3' • 54 13 • 17 3 • 41 • 73 .97 • 593 • 977 • 3217 19489. 38273 768 256 1067 969144 915565 716868 049522 568978 331378 093561 484521 = 3 2 .17.401 .1697 , 13313 , 21121 -49057475361.198593.733697 , 29102880226241 816 256 793553 314770 547109 801192 086472 747224 274042 880000 = 2 38 • 3 8 • 5 4 • 13 • 17 4 • 41 2 • 97 • 113 • 193 • 577 • 1601 2081 94849 960 256 20130 907061 992729 156753 037152 064135 304760 934400 = 2' • 3 4 • 5 2 • 7 6 • 17 • 41 -89 • 97 . 337 . 401 .433593 . 7841 - 130513 1020 256 11 412817 953927 959213 205123 673154 912256 000000 = 242 • 3 3 • 5 6 • 17 3 • 73 • 193 • 353 • 593 • 1889 • 3217 • 69857

Bibliography

The following concentrates mainly on the period 1970-1981, since the period 1940-1970 is covered in Reviews in Number Theory (ed. by W. LeVeque; American Mathematical Society, 1974), especially Volume 5. For very early works, see the references in Hilbert [2]. The reader should also consult Kummer's Collected Papers for numerous papers, many of which are still valuable reading. The books of Narkiewicz and Ribenboim [1] also contain useful bibliographies. A note "MR 12: 345" refers to a review in Mathematical Reviews (similarly for " LeVeque" or " Zentralblatt"). These are given mostly for articles in less accessible journals, for untranted articles in Japanese or Russian, when the review lists errors or add .-nal information, or when the review gives a good summary of a difficult 1Adachi, N. 1. Generalization of Kummer's criterion for divisibility of class numbers. J. Number Theory, 5 (1973), 253-265, MR 48:11041. Adleman, L., Pomerance, C., and Rumely, R. 1. On distinguishing prime numbers from composite numbers (to appear). Amice, Y. L Interpolation p-adique. Bull. Soc. Math. France, 92 (1964), 117-180. 2. Les Nombres p.Adiques. Presse Universitaire de France, 1975. Amice, Y. and Fresnel, J. 1. Fonctions zétap-adiques des corps de nombres abéliens réels. Acta Arith., 20 (1972), 353-384. Amice, Y. and V !, 1. Distributions p-adiques associées aux séries de Hecke. Astérisque, 24-25 (1975), 119-131. Ankeny, N., Artin, E., and Chowla, S. 1. The class number of real quadratic number fields. Ann. of Math. (2), 56 (1952), 479-493. ,

361

362

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List of Symbols

nth root of unity, 9 conductor, 19 character group, 21 annihilator, 22

L(s, z) Lp(s, x)

TOO B„ x

B(X) 'C(s, b) K+ h+ hRK RK , p Cp

exp logo

q w(a)



(Xn

9(x) 386

L-series, 29 p-adic L-function, 57 Gauss sum, 29 Bernoulli number, 30 generalized Bernoulli number, 30 Bernoulli polynomial, 31 Hurwitz zeta function, 30 maximal real subfield, 38 class number of IC , 38 relative class number, 38 unit index, 39 regulator, 41 p-adic regulator, 70 completion of algebraic closure of Op, 48 p-adic exponential, 49 p-adic logarithm, 50 4 or p, 51 Teichmiiller character, 51

51 52 Gauss sum, 88

List of Symbols

J(x 1 , x2 ) Jacobi sum, 88 Stickelberger element, 93 0 fractional part, 93 {x} Ez , E i

Ai

AK A A —B v„

con

idempotents, 100 ith component of class group, 101 minus component, 101, 192 Iwasawa invariants, 127 7L-extension, 264 7L p[[7]], 268 pseudo-isomorphism, 271 276 278 280 291

387

Index

Adams, J. C., 86

Ankeny Artin-Chowla, 81, 85 Artin map, 338, 342 Baker- Brumer theorem, 74 Bass' theorem, 151, 260 Bernoulli distribution, 233, 238 numbers, 6, 30, 347 polynomials, 31 Brauer-Siegel theorem. 42 Capitulation of ideal classes, 40, 185, 286, 317 Car litz, L.. 86 Class field theory, 336ff. towers, 222 Class number formulas, 37ff., 71, 77ff., 151ff. CM- fi eld, 38ff., 185, 192, 193 Coates-Wiles homomorphism. 307 Conductor, 19, 338 Conductor-discriminant formula, 27, 34 Cyclotomic polynomial, 12, 18 units, 2, 1431I, 313 Zp- extension, 128, 286 Davenport- Hasse relation, 112 Dirichlet characters, 19ff. Dirichlers theorem, 13, 34 Discriminant, 9 388

Distinguished polynomial, 115 Distributions, 231ff., 251ff. Eichler, M., 107 Ennola, V., 262 Even character, 19 Exponential function, 49 Fermat curve, 90

Fermat's Last Theorem, 1, 107 , 167ff. First factor, 38 Fitting ideal, 297 Frobenius automorphism, 14, 337 Function fields, 128, 129, 296 Functional equation, 29, 34, 86 Gamma transform. 241

F-extension, 127 Gauss sum, 29, 35, 36, 87ff. Generalized Bernoulli numbers, 30 Herbrand's theorem, 102 Hurwitz zeta function, 30, 55 Idéles, 344 Imprimitive characters, 205 Index of Stickelberger ideal, 103 Infinite Galois theory, 332ff. Integration, 237ff.

Index

Inverse limits, 331 Irregular primes, 7, 62, 63, 165, 193, 350

Iwasawa algebra (= A), 268 function, 69, 246, 261 invariants, 127, 276 theorem, 103, 276 Jacobi sums, 88 Krasner's lemma, 48 Kronecker -Weber theorem, 319ff., 341 Ku bert's theorem ( = 1118), 260

Kummer congruences, 61, 141, 241 homomorphism, 300 lemma ( = 5.36), 79, 162 pairing, I88ff., 292 A, 127, 141, 201, 276 A-modules, 268ff. L-functions, 29ff., 57ff. Lenstra, 1-1. W., 18 Leopoldt's conjecture, 71ff., 265,291 Local units, 163, 310ff Logarithm, 50 Logarithmic derivative. 299 Mahler's theorem, 52 Main conjecture. 146. 198. 199. 295ff.

Masley, J., 204 Maximal real subfield, 38 Measures, 236ff. Mellin transform, 242 Minkowski bound, 17, 320 unit, 72 Montgomery, 11, 204 p, 127, 130, 276, 284, 286 Nakayama's lemma, 279 Normal numbers, 136, 142

389 p-adic regulator, 70ff., 77, 78, 85, 86 Parity of class numbers, 184, 193 Partial zeta function, 30, 95 Periods, 16 Polya-Vinogradov inequality, 214 Primitive character, 19, 28 Probability, 62, 86, 108, 112, 159 Pseudo-isomorphic, 271 Punctured distribution, 233 Quadratic

fields, 17, 45, 46, 81ff., 111, 190, 337 reciprocity, 18, 341 Ramachandra units. 147 Rank, 186 193 Reflection theorems, 187ff. Regular prime, 7, 62, 63 Regulator, 40, 70, 77, 78, 85, 86 Relative class number, 38 Residue formula, 37, 71, 165 Ribet's theorem. 102 Scholz's theorem, 83, 190 Second factor, 38 Sinnott, W., 103, 147 Spiegelungsatz (= reflection theorem), 187ff. Splitting laws, 14

Stickelberger element, 93, 119 ideal, 94, 195, 298 theorem, 94 Stirling's series, 58 Teichmilller character ( = Twist, 294

51, 57

Uchida, K., 204 Uniform distribution, 134ff. Universal distribution, 251ff. Vandiver's conjecture, 78, I57ff., 186, 195ff. Von Staudt-Clausen, 55. 141

Odd character, 19

Odlyzko, A., 221 Ordinary distribution, 234 p-adic class number formula, 71, 77ff. p-adic L-functions, 57ff., I17ff., 199, 239, 251, 295, 314

Wagstaff. S.. 181 Weierstrass preparation theorem, 115 Weyl criterion, 135 Zeta function for curves, 92, 128, 296 7L-extension, 127, 263ff.