Problem Books in Mathematics Edited by P. R. Halmos
Problem Books in Mathematics Series Editor: P.R. Halmos
Unsolved Problems in Intuitive Mathematics, Volume I: Unsolved Problems in Number Theory by Richard K. Guy 1981. xviii, 161 pages. 17 iIlus.
Theorems and Problems in Functional Analysis by A.A. Kirillov and A.D. Gvishiani 1982. ix, 347 pages. 6 ilIus.
Problems in Analysis by Bernard Gelbaum 1982. vii, 228 pages. 9 iIlus.
A Problem Seminar by Donald J. Newman 1982. viii, 113 pages.
Problem-Solving Through Problems by Loren C. Larson 1983. xi, 344 pages. 104 iIlus.
Demography Through Problems by N. Keyfitz and J.A. Beekman 1984. viii, 141 pages. 22 iIlus.
Problem Book for First Year Calculus by George W. B1uman 1984. xvi. 384 pages. 384 iIlus.
Exercises in Integration by Claude George 1984. x. 550 pages. 6 illus.
Exercises in Number Theory by D.P. Parent 1984. x. 541 pages.
Problems in Geometry by Marcel Berger et al. 1984. viii. 266 pages. 244 iIlus.
D.P. Parent A Pseudonym for
D. Barsky F. Bertrandias G. Christol A. Decomps H. Delange J.-M. Deshouillers K. Gerardin J. Lagrange J.-L. Nicolas M. Pathiaux G. Rauzy M. Waldschmidt
Exercises in Number Theory
Springer Seience+Business Media, LLC
[SI
Series Editor Paul R. Halmos Department of Mathematics Indiana University Bloomington, IN 47405 U.S .A.
AMS Classification: OOA07, 12-01, 10-01
Library of Congress Cataloging in Publication Data Parent, D. P. Exercises in number theory. (Problem books in mathematics) Translation of: Exercises de theorie des nombres. Bibliography: p. Includes indexes. 1. Numbers, Theory of-Problems, exercises, etc. I. Title. II . Series. 512'.7 84-16056 QA241.P2913 1984
Title of the original French edition: Exercices de tMorie des nombres, © BORDAS, Paris, 1978. © 1984 by Springer Science+Business Media New York Originally published by Springer-Verlag New York, Inc. in 1984 All rights reserved. No part of this book may be translated or reproduced in any form without written permission from Springer Science+Business Media, LLC.
987654321 ISBN 978-1-4419-3071-2 ISBN 978-1-4757-5194-9 (eBook) DOI 10.1007/978-1-4757-5194-9
Preface
After an eclipse of some 50 years, Number Theory, that is to say the study of the properties of the integers, has regained in France a vitality worthy of its distinguished past.
More 'and
more researchers have been attracted by problems which, though it is possible to express in simple statements, whose solutions require all their ingenuity and talent.
In so doing, their work
enriches the whole of mathematics with new and fertile methods. To be in a position to tackle these problems, it is necessary to be familiar with many specific aspects of number theory. These are very different from those encountered in analysis or geometry.
The necessary know-how can only be acquired by study-
ing and solving numerous problems.
Now it is very easy to form-
ulate problems whose solutions, while sometimes obvious, more often go beyond current methods.
Moreover, there is no doubt
that, even more than in other disciplines, in mathematics one must have exercises available whose solutions are accessible. This is the objective realised by this work.
It is the collab-
orative work of several successful young number theorists.
They
have drawn these exercises from their own work, from the work of their associated research groups as well as from published work. Without running through all the areas in number theory, which
v
vi
PREFACE
would have been excessive, the exercises given here deal with those directions that now appear most important. are rarely easy, but this book always them.
give~
The solutions
a method to solve
The appearance of this volume is gratefully welcomed.
To all those who have made the effort to delve into the problems proposed here, it will solidify their attachment to the theory of numbers.
Ch. Pisot
Introduction
This book is the work of a group of mathematicians, including a large number of participants from the Seminaire DelangePisot-Poitou.
It was written as follows: first many exercises
were collected, mostly from university examinations.
Then a
choice was made, and they were grouped so as to constitute 10 chapters arranged in a random order.
Certain important areas in
number theory, notably the area of Diophantine Equations, have been left out. Many books on number theory have appeared recently in France, among them those of Y. Amice, A. Blanchard, W.J. Ellison and M. Mendes-France, G. Rauzy, P. Samuel, J.P. Serre and M. Waldschmidt (see the bibliography).
Readers of these books will
find here supplementary and illustrative material.
Conversely,
a neophyte mathematician, who is curious about, say, p-adic anaylsis, or distributions modulo 1, after having been interested by certain attractive exercises, will have a strong motive for learning the theory.
A part of our problems are accessible
to a reader at the level of good second or third year students of a university.
The introduction of each chapter is there to
recall the definitions and principal theorems that one will need.
vii
viii
INTRODUCTION
On behalf of the authors, I would like to thank those who have aided us, either by suggesting problems or in making up the solutions: J.P. Borel, P. Cassou-Nogues, H. Cohen, P. Damey, F. Dress, A. Durand, J. Fresnel, E. Helsmoortel, J. Martinet, B. de Mathan, M. Mendes-France, M. Mignotte, J.J. Payan, J. Queyrut, G. Revuz, G. Rhin, M.F. Vigneras.
We would equally like to
thank the Mathematical Society of France which helped in the publication of the original version of this book.
J.L. Nicolas
While reading the proofs of this introduction, I learned of the death of Professor Charles Pisot.
All the contributors
of this book dedicate this English translation to his memory.
Contents
PREFACE BY Ch. PISOT
v
INTRODUCTION
vii
CHAPTER 1: PRIME NUMBERS: ARITHMETIC FUNCTIONS: SELBERG'S SIEVE INTRODUCTION PROBLEMS SOLUTIONS
1
18 46
CHAPTER 2: ADDITIVE THEORY INTRODUCTION PROBLEMS SOLUTIONS
127 129 132
CHAPTER 3: RATIONAL SERIES INTRODUCTION PROBLEMS SOLUTIONS
144 148 153
CHAPTER 4: ALGEBRAIC THEORY INTRODUCTION PROBLEMS SOLUTIONS
170 182 198
ix
TABLE OF CONTENTS
x
CHAPTER 5: DISTRIBUTION MODULO 1 INTRODUCTION PROBLEMS SOLUTIONS CHAPTER 6: TRANSCENDENTAL NUMBERS INTRODUCTION PROBLEMS SOLUTIONS
241 246 261
308 311
324
CHAPTER 7: CONGRUENCES modp: MODULAR FORMS INTRODUCTION PROBLEMS SOLUTIONS CHAPTER 8: QUADRATIC FORMS INTRODUCTION PROBLEMS SOLUTIONS
359 363
373
408 411
414
CHAPTER 9: CONTINUED FRACTIONS INTRODUCTION PROBLEMS SOLUTIONS
426 429 439
CHAPTER 10: p-ADIC ANALYSIS INTRODUCTION PROBLEMS SOLUTIONS
466 470 486
BIBLIOGRAPHY
533
INDEX OF TERMINOLOGY
536
INDEX OF SYMBOLS AND NOTATIONS
540
CHAPTER 1
Prime Numbers: Arithmetic Functions: Selberg's Sieve
INTRODUCTION 1.1 PRIME NUMBERS With x a real number and x OF PRIME NUMBERS NOT EXCEEDING x
L
-&(x)=
0, we denote by
~
~(x)
the NUMBER
and we set
logp.
p~x
The essential result in the theory of prime numbers is the following: THEOREM: (Prime Number Theorem): When x ~(x)
'V
~
+00
x
-1-' ogx
This relation is equivalent to: -&(x) (as
'V
X
will be seen in Exercise 1·3). In fact, results quite a bit more precise than this are known
(cf., for example, [Ell]). We will also give estimates for certain quantities related to 1
2
CHAPTER 1: PRIME NUMBERS:
the primes, notably the sum
I land the product IT p.,;,x p.,;,x p
(1 - IIpJ
,
for which we have the celebrated MERTENS' FORMULA:
IT
p~x
(1 - 1.)p
e-Y logx
(x -+ +"'),
'V--
where y is Euler's constant (cf., Exercises 1-5 and 1-6). A FREQUENTLY USEFUL LEMMA
1.2
The following Lemma is often useful, in particular for the evaluation of the sum
I
p~x
l mentioned above. p
LEMMA: Let E be a set of real numbers sueh that for every real
En (-"', tJ
Let g be a real or eomplex funetion defined on G(t)
t~
is empty or finite.
=
I
usE
E~
and let
g(u)
u~t
If F is a C1 funetion on the elosed interval [a,b]~ then
I
usE
F(u)g(u)
F(b)G(b) - F(a)G(a) - JbG(t)F'(t)dt. a
a
This formula is obtained immediately after noticing that for all usE such that a < u ~ b we have:
F(b )g(u) - F(ll. )g(u)
=
where
Y(x)
{
0
for x < 0,
1
for x ;::. 0,
t a
g(u)y(t - u)F'(t)dt,
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
3
and adding. In fact, this is the formula for integrating by parts the integral JbF(t)dG(t) , which is equal to the sum a F(u)g(u) (cf., [Wid]). It is the best way to remember it.
Stieltjes
I
usE
a
F(b)G(b) - F(a)G(a) - J G(t)F'(t)dt. a EXAMPLE: Let
E.=~*,
~
Jx1 1t
I
1
g(u)
= 1, [a,b] = [I,x] , F(t) =t1
d[t] = ~ - 1 + JX !!l dt x 1 t2
logx - x - [x]
x
whence 1 I -n = logx
We have:
+ 1 -
(00 t
n~x
-I:
t
- [t] dt, t2
- [t] dt - x - [x] + J+OO t - [t] dt x t2 x t2
logx + y + n(x), where
y
1.3
~
1
x-
.
IDEA OF AN ARITHMETIC FUNCTION: THE USUAL ARITHMETIC FUNCTIONS
Any real or complex function defined
on~*
is called an ARITH-
METIC FUNCTION.
Usually, we consider functions for which
fen)
is determined
CHAPTER 1: PRIIlE NUIlBERS:
4
from arithmetic properties of the integer n. Some classical examples are the following:
d(n)
NUMBER OF DIVISORS of n;
cr(n)
SUM OF THE DIVISORS of n;
q>(n)
number> of integer>s m such that 1
~
m ~ n and (m,n)
(EULER'S TOTIENT FUNCTION); v(n)
n(n)
= NUMBER
OF PRIME DIVISORS
of n;
TOTAL NUMBER OF FACTORS in the decomposition of n
into pr>ime factors; (n(l)
=0
and, if n
= q~lq~2 ••• qak,
where Ql,Q2, ... ,qk are prime
numbers and a 1 ,a 2 , ... ,a k are some integers greater than zero, n(n) = a 1 + a 2 + ••• + a k ). (LIOUVILLE'S FUNCTION)
Other examples are: (_l)v(n) if n is divisible by the square of no ~(n)
={
prime number,
o
in the opposite case;
the MOBIUS FUNCTION, and VON MANGOLDT'S FUNCTION A defined by:
A(n)
J logp 1 0
if n
= Pk
. h p a prlme . numb er an d k
w~t
1
~,
i f n is not of that form.
We shall denote by z the fUnation whiah is identiaaZZy one on the integer>s, and denote by i the identity mapping of~* onto~*.
1
5
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
1.4
ADDITIVE FUNCTIONS AND MULTIPLICATIVE FUNCTIONS An arithmetic function is called an ADDITIVE FUNCTION if:
fCmn) = fCm) + fCn)
whenever Cm,n) = 1.
(1)
This clearly implies that f(l) = O. Furthermore, ifn1 ,n 2 , ... ,nk are pairwise relatively prime, then:
The functions v and is as well.
~
are additive.
The function log clearly
A function f is called a MULTIPLICATIVE FUNCTION if f(1)
1
and
fCmn)
=
fCm)fCn)
(The condition
fCI)
=
whenever Cm.n)
(2)
= 1.
1, ignored by certain authors, serves only
to avoid considering as multiplicative the function that is identically zero). If f is multiplicative and if n 1 ,n 2 , ••• ,n k are pairwise relatively prime, then:
fCn 1 n2 ···n k)
=
fCn1 )fCn2 ) •• ·fCn k) .
The functions d~o~~~A~~ are multiplicative. that 2 and i are as well.
It is obvious
An additive or multiplicative function is completely determined by its values on the powers of the primes. PROPOSITION:
6
CHAPTER 1: PRIME NUMBERS:
prime numbers and Q1,Q2, ... ,Qk are integers greater than zero, then: if f is additive, Q1 . Q2 Qk f(n) = f(q1 )f(q2 )ooof(qk)
if f is multiplicative,
which can be written more concisely as:
f(n)
f(n) =
or
II
f(pr').
pr'IIn
exists exactly one additive function and one multiplicative fUnction taking given values on the po~e~s of the p~ime numbe~s. A function f is a COMPLETELY ADDITIVE FUNCTION if: We see, moreover, that
the~e
f(mn) = f(m) + f(n), for arbitrary m and n, and not only when (m,n)
1
Thus the function n is completely additive. A function f is a COMPLETELY MULTIPLICATIVE FUNCTION if: f(l)
=I
and
f(mn)
= f(m)f(n),
for all m and n. A completely additive or completely multiplicative function is determined by its values on the prime numbers, for, r'
f(p ) = ~f(p) respectively.
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
1.S
7
ALGEBRAIC STRUCTURE ON THE SET OF ARITHMETIC FUNCTIONS The set A of arithmetic functions is made into an algebra
over
by defining addition and multiplication by scalars in the
~
usual way, and by taking as the product of two elements the CONVOLUTION defined as follows:
f*g is the arithmetic function h given by the formula: hen) =
2
din
2
[or hen)
f(d)g(J]
dd'=n
f(d)g(d')].
This convolution is commutative, associative, and distributive
with respect to addition. e defined by: e(l)
=1
and
e(n)
It possesses an IDENTITY, the function
=0
for n
> 1.
Moreover, for all f,g e A and A e~,
A is therefore a unitary commutative algebra. PROPOSITION: The group
G of invertible elements is the set of
arithmetic functions f such that f(l) If
+ o.
fe G the inverse of f will be denoted by f-"'.
PROPOSITION: The M8bius function
~
is the inverse of the func-
tion z. The relation
~*z
e, which says that V
as:
2
din
~(d)
for n
1,
for n
> 1.
Z-'!',
translates
8
CHAPTER 1: PRIME NUMBERS:
PROPOSITION: The Von Mangoldt function A is characterised by: A*,~
= log.
PROPOSITION: The set
Mof multiplicative fUnctions is a sub-group
of G. 1.6 THE FIRST
M~BIUS
INVERSION FORMULA
With a e G, it is clear that if f,g e A,
In particular, by taking a
z we see that for f,g e A we have
the equivalence:
in other words (as
g(n)
L
din
z*f = f*z),
f(d)
<=>
for all n e:N*
f(n)
=
L lJ(d)g(~)
din
This is usually called the FIRST
]
~6BIUS
for all n e:N 1<.
INVERSION FORMULA.
1.7 ARITHMETIC FUNCTIONS AS OPERATORS ON A VECTOR SPACE Let X be the vector space of complex functions defined on the interval [1,+00). Given a e A and Fe X we shall denote by a,-F the fUnction defined
on [1,+00) by:
a,-F
== G(x)
=
nL a(n)F(~)
.
Thus with a fixed, the mapping F
~
a,-F is a linear operator on X.
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
9
It has the following properties: For all FaX:
e,-F
= F;
(a.) (a
For all a,b a A, for all
+
b),-F
Fe X: {
a-r{b,-F)
= a,-F + b,-F, and (a*b ),-F;
(13)
For all a a A, for all ). a a:, for all FaX: ().a),-F = ).(a,-F).
1.8
THE SECOND MOBIUS INVERSION FORMULA With a a A, if F,G a X, it is clear that by Formulae (a) and
(a) we have:
G = a,-F
<=>
F
=
a-*rG.
In particular, upon taking a
G=
z,-F <=>
F
z we see that for F,GaX
= ~rG;
in other words: G(x)
L F(~)
for all x
~
1
n~
for all x This is usually called the SECOND 1.9
~6BIUS
~
1.
INVERSION FORMULA
SUMMATION FUNCTION OF AN ARITHMETIC FUNCTION
The SUMMATION FUNCTION Of an arithmetic function is the function A defined on the interval [l,+m) by:
10
CHAPTER 1: PRIME NUMBERS:
We observe that A is nothing but the function ing to F(x)
= 1 for all
x ~ 1.
a~
correspond-
Property (B) then gives the fol-
lowing result: PROPOSITION: Let a,b e A and a = ai,b. funations of a,b,a.
C
Let A,B,C be the summation
We have:
= aTi1 = bp4..
(2)
EXAMPLE 1: Let a
= [x]
B(x)
= ~,
b
= z.
and
C(x)= 1.
Then a
e, and for all x
~
1:
Therefore, 1 for all x .... 1. The same result would be obtained by applying the Second MBbius Inversion Formula with F(x)
=1
for all x
EXAMPLE 2: It is customary to denote by of von Mangoldt' s function l\. a
= log,
~
1.
Wthe summation function = A, b = z, one has
By taking a
and relation (2) gives:
EULER PRODUCTS The following result about multiplicative functions is fundamental.
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
11
THEOREM: Let g be a multiplicative function.
If: (3)
where p runs over the set of prime numbers and
r
runs
over~*,
then the series:
I
g(n)
n=l
is absolutely convergent, and we have: (4)
where the infinite product, in which p runs over the sequence of prime numbers, is absolutely convergent. (It is clear that each of its factors is well defined, for the series
I
r=l
g(pr) is absolutely convergent).
I g(n) is absolutely convergent, n=l relation (3) obviously holds (so that the absolute convergence Notice that if the series
of this series is equivalent to (3)).
I
n=l
Hence if the series
g(n) is absolutely convergent, relation (4) holds.
1.11 GENERATING FUNCTION OF AN ARITHMETIC FUNCTION To each arithmetic function a we associate the DIRICHLET SERIES
12
CHAPTER 1: PRIME NUMBERS:
It is known that there existso verges if Res
~ 0c
c
e R such that the series con-
and diverges if Res
ries diverges for all s e
<1:.
But if
0
c
holomorphic function of s for Res > o. c
< 0c. +00
If
0c;
+00 the se-
the series defines a
The analytic function
thus defined is called the GENERATING FUNCTION of the arithmetic
function a. is, by definition the genepating function of the function z: for Res> 1 we have: Thus the RIEMANN
~-FUNCTION
~(s)
PROPOSITION:
If the arithmetic functions a and b have genepating
functions, then the function a*b has as genepating function the ppoduct of the one fop a and the one fop b. In fact, for every value of s for which the Dirichlet series associated with a and b are absolutely convergent, the series associated with a*b is also absolutely convergent and has as sum the product of the sums of the preceding series. Thus we see, for example, that for Res> 1 we have:
~ (s)
L
Il(n) =
n=l
n
which shows that
has
1,
s
~(s)
+0
for Res> 1, and that the function 11
as genepating function. It is also seen that von Mangoldt's function A has genepating function: for Res> 1 we have: l/~
L
n=l
A(n)
--= s n
-~'/~
as
~' (s)
- ""Z
(since A*z = log).
This formula is fundamental in the theory of
the distribution of primes. The results of Section 1.10 give the following:
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
13
PROPOSITION: Let f be a multiplicative arithmetic function.
for some real
L
<
p,r in the
where~
runs
overE*~
;:00
tJ!!:l
n=l
n
l
S
sum~
oo~
p runs over the set of prime numbers and r
then the Dirichlet series associated with
I f(:)
f~
~
is absolutely convergent for Res we have:
n=l n
If~
a~
=
IT p
(1 + I
r=l
~ a~
and for these values of s
f(~:)) ~
(5)
p
and the infinite produat~ in which p runs over the sequence of prime numbers~ is absolutely convergent. In
fact~
we have the relation
which the Dirichlet series
(5)
for every value of s for
L i1!:!l is absolutely convergent.
n=l n S
As an example, by taking f
= Z,
EULER'S FORMULA is obtained:
1.12 IKEHARA'S THEOREM (cf., [Ell]) +00 a THEOREM: (Ikehara): Let L ~ be a Dirichlet series for which n=l n S the coefficients a are all real ~ O. Assume that this series n is convergent for Res > 1, with F(s) as sum~ and that there exists A A > 0 such that~ for all real t~ F(s) - s _ 1 tends towards a
14
CHAPTER 1: PRIME NUMBERS:
finite limit as s
1
7
+ it while remaining in the half-plane
Res> l.
Then when x a n~x
n
'V
7
+00
we have:
Ax.
It should be noted that the assumptions hold, in particular, when F is ho1omorphic at every point on the line Res = 1, apart from the point 1, and this point is a simple pole with residue A.
1.13 SELBERG'S SIEVE Here we shall be content with g1v1ng an inequality, due to Selberg, a proof of which the reader will find in Chapter IV of the book by Ha1berstam and Roth [Ha]; in the Exercises one will find various examples giving applications of this inequality. Let A = {a1, ... ,a n } be a family of positive integers and P a finite family of prime numbers; (we shall denote by II( P) the pro-
duct of the elements of P).
For each prime number p to P one chooses
w(p) distinct classes (Rip), ... ,R~~~)) modulo p.
w is made into
a completely multiplicative function by taking:
w(p)=Oifp+P. Then let d be a divisor of II(P); we denote by Ad the set of elements of A that are in a class Q~p) for at least one p dividing d and one i, 1
~
\cardA d -
i
~
w(p).
nW~d)\ ~
1
We make the assumption:
w(d).
(H)
We denote by S(A,P) the number of elements of A which are not
in any of the classes R~P); then for every positive real number ]. z, we have:
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
S(A. P) :;;: n
( 1: w<;::) J-1 m:;;:z
+ z2
II
peP
15 [1 -
~J
l-2
p
In Exercise 1.27 a lower bound will be given for the sum in certain special cases. BIBLIOGRAPHY [Har] Chapters I. II.XVI.XVII.XVIII.XIX.XXII; [Bla]; [Ell]; [Hal].
I m~z
w(m) m
16
CHAPTER 1: PRIME NUMBERS:
APPENDIX
This Appendix contains a Table of the usual arithmetic functions, giving their definition, their fundamental properties, and the generating function whenever it is simple, and in this case the abscissa of convergence of the associated Dirichlet series.
e
is the upper bound of the real parts of the zeros of the
~-function.
It is known only that
thesis is that 8 =
~.
!
~
e
~ 1.
The Riemann hypo-
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
17
Abscissa of
Generating Function
Fundamental
Definition
Properties
Convergence
of the Dirichlet Series
d(n) =
number of divisors of n
d =
sum of divisors
multiplicative
~(n)
=
~(n)
number of mls such = that 1 ~ m , nand
v(n)
=
of n
,(s),(s -1)
a = i*z
multiplicative cP = 1J.*i
[1 -!] pin IT
~(n)
number of prime
additive
total number of in the
completely additive
divisors of n
= factors
~(n)
1 =
factorisation of n
(_l)v(n) if n has no square factor 0
otherwise
= n
~ = 2 -,' "fl*Z = e
completely
multiplicative
completely
i(n) = n for all n
multiplicative multiplicative;
1 for n = 1
={ o
--1
for n
Unit element of the algebra A
> 1
olOgp i f n = otherwise
r
1)
multiplicative
multiplicative
z(n) = 1 for all n
ds -
~
p
completely
A(n)
A(n)
Zi,2
(m,n) = 1
n(n)
ern)
mul tiplicative
neither additive nor mul tiplicative A"z = log
1 ,(s)
,(2s)
0s)
,(s)
,(s - 1)
8
8
18
CHAPTER 1: PRIME NUMBERS:
PROBLEMS
2n 2 + 1. Show that Fn divides Fm - 2 if n is less than rn, and from this deduce that F and F n :n ifrn are relatively prime f n. EXERCISE 1·1: Set Fn
From the latter statement deduce a proof of the existence of an infinite number of primes. EXERCISE 1·2: Let S be a positive real number less
than
one;show that if the integer N is large enough there exists at 1east one prime number between S Nand N. EXERCISE 1·3: Show the equivalence of the Prime Number Theorem
(Section 1.1) with each of the following assertions: (i): .e(x)
(ii): Pn
'V
'V
x;
nlogn (Pn being the n-th prime number).
EXERCISE 1·4: Show that the series
if and only if the real number a is (Pn is the n-th prime number).
\ L
-1
n= ;5
Pn (loglogPn)
-a
greater than 1.
converges
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
2
EXERCISE 1-5: SOME SUMS
p ..x
19
fCp)
Given a prime number p and a real number x
I: (1) :
~
p, de-
p of p in the decomposition into prime factors of the product n. n ..x termine the exponent
Ct
IT
Show that:
max(~P
- 1
~)
• 2p
Show that for all x > 1 we have:
1:(2):
2logx,
p ..x and, for 1
(Note that
<
y .. x:
2
n ..x
I: (3): logn
logn .. xlogx, as well as that for each p ~ x
By now noticing
that:
[X]
~
J0 logtdt
show that for all x
>
~
J:lOgtdt - logx,
1:
and deduce from this the existence of a constant C
~>P ,.. logx -
C.
~
0 such that:
20
CHAPTER 1: PRIME NUMBERS:
11:(1): :>;.
2
2m
Let ~(x)
I
=
10gp.
Starting from the fact that
p:>;.x
, show that for each m eJN'~ : 2m10g2.
~(m) <
~(2m) -
From this deduce that for every x
>
1:
> 4x10g2.
~(x)
II: (2):
By using one of the inequalities of 1(1), show that
for every x>
1:
10gp
< 10gx
P
Set
III:
+ 410g2.
10gp p:>;.x
=
(x) = 10gx
p
and II(2),n(x) is bounded for x
>
+ n(x), so that by,I(3) and
1.
Using the Lemma of Section 1.2 of the Introduction to this Chapter, show that for x
x
> (x ) f I ! =
p:>;.x p
2
>
2 we have:
(t) 2 dt.
t(logt)
Deduce from this that there exists a real constant b such that, when x ..... +00:
I
1
p:>;.x
p
10glogx + b + 0 [101gxJ .
EXERC1SE 1-6: MERTENS' FORMULA (1):
I
[lOg
Show that the series:
~]
1 1 1 - x P P
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
is convergent for all real x >
21
!, and that if the sum is g(x)
then g is continuous on the interval [!,+oo). For x
(2):
~
0, set: and
P(x)
1
N(x)
n
n:fiix
Show that for 10g~(1
8
real> 0 we have:
+ s) - g(l + s)
= s f0
+oo -st
e
t
pee )dt,
and 1
log -"";::""--s1 - e (3):
By Exercise 1·5 we know that there exists a constant
b such that. when x P(x)
f+oo -stN(t)dt.
s 0 e
~ +00:
= loglogx +
b
+
0(_1_) logx
With Y being Euler's constant, show that when s
~
0 through
real values greater than zero:
From this deduce that b (4): As x
=y
- gel).
From the preceding deduce a proof of Mertens' Formula: ~
+00 we have:
IT
P:fiix
(Notice that:
( P1) 1--
y
e'V--. logx
CHAPTER 1: PRIME NUMBERS:
22
EXERCISE 1·7: INTEGERS n SUCH TIIAT ALL THE PRIME DIVISORS OF n ARE LESS THAN OR EQUAL TO In Let A be the set of those integers n > 1 all of whose priIT.e divisc satisfy p Bp
<
In.
For every prime number p set:
= {p,2p, ... ,(p
(1):
Show that A
- UpL
=
CN,.,(YBp ).
tinct prime numbers p and q, B ("'\B p
Let QA(X) = cardin
(2):
[x] -
L
p~%;
~
Verify that for dis-
1i'J.
q
x,neAL
(p - 1) -
L
x~p>~
Show that:
[:£p]
By using the formula:
(3):
L ! = loglogx
p~x P
+ b + O(101gx]
find a quantity equivalent to QA (x)when x
+
co.
EXERCISE l·S: A THEOREM OF HARDY AND RAMANUJAN In this Exercise we will need the results of Exercises 1·5:1(2) and 1·5:111, in particular the formula:
L
1
p~x p
loglogx + b +
O(101gx) ,
established in Exercise 1 S:1II,aswellasthefact that (1): v(n)
By starting from the property
L
1,
pin
show that as x
+
+co
we have:
rr(x)
=
that for each n eN*:
O(l~gxJ
.
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
L
v(n)
= xloglogx
23
+ bx +
n~x
'\'l
4 set P(x)
_1 •
(2):
For x
(3):
From the preceding formula deduce that as x
(Set P(x)
C
~
p~x p
Show that for x
~
~
+00:
loglogx + b + n(x) so that there exists M > 0
M such that In(x)1 ~ -oga:: 1 for x ~ 2. Notice that for p < x
loglog
~p = loglogx
and that there exists K
o~
u ~
+ 109(1 - 110gp) ogx >
0 such that Ilog(l - u)1 ~ XU for
!). Starting from the fact that for each n e:N*:
L
v(n)2
1,
P1/ n P2/ n show that when x '\'
l
(5):
v(n) 2
~
+00
\Ore
have:
= x(loglogx) 2
Show that when x
+ (2b + 1)xloglogx + o(x).
~
+00:
4:
CHAPTER 1: PRIME NUMBERS:
24
(v(n) - 10glogx)
2
= x10g10gx + o(x).
n~x
(6): 10glogx
>
Given A > O. we denote by nA(x). for x
0). the number of n
~
>
e (so that
x such that:
Iv(n) - 10g10gxl ~ A/10g10gx. -
1
Show that lim x++oo x (7): where a
>
Let g be a
~
n
~
A
positive increasing function on [a.+ oo ).
e. and such that g(x)
Denote by N(x). for x a
1 "2.
n A(x) ~
>
+
+00 when x
+
+00.
a. the number of n's
satisfying
x. such that:
Iv(n) - 10glognl ~ g(n)/10g10gn. 2~
(a) : Show that i f x > a • then for every n satisfying IX < n and such that Iv(n) - 10glognl ~ g(n)/10g10gn. we have:
Iv(n) - 10g10gxl ~ g(IX)/10g10gx - 10g2 - 10g2. (b):
Show 'that for any A > O.
From this deduce that N(x) x
+
for x large enough:
0 when x + +00.
EXERCISE 1-9: DIFFERENCE BETWEEN TWO CONSECUTIVE PRIME NUMBERS Set dn = Pn+l - Pn (n prime number. (a):
Show:
= 1.2 •••• \'
L
2~n~x
). where Pn denotes the n-th d
n
-l-'V
ogn
x•
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
(b):
25
From this deduce the inequalities:
lim inf[ld n ) n-ogn
~1~
lim SUp[ld n ) • n-ogn
(In Exercises 1·10 and 1·31
it will be shown that these two
equalities are, in fact, strict). EXERCISE 1·10: Denote by p
the n-th prime number, and by IT n n product of the prime numbers less than or equal to p •
the
n
(a):
Let n be an integer strictly greater than 2; show that
there exists a strictly positive integer i
p P 1 such that all n nthe integers in the interval [iIT 2 - P 1 + 1,iIT 2 + p 1 - 1] nnnnare divisible by at least one of the prime numbers P1' •.• 'P n • (b):
~
From the preceding deduce that:
lim SUp[ldn ] n-ogn
~
2
(with the notations of Exercise 1·9).
EXERCISE l'll: DIFFERENCE BETWEEN TWO CONSECUTIVE SQUAREFREE INTEGERS Let Q be the set of integers without square factors.
The
elements of Q are arranged in an increasing sequence:
Let S be the complement of Q inN; the elements of S are arranged into an increasing sequence:
Recall that (cf., [Har Ch. XVIII) we have: Q(x)
card{n e Q,n
~ x}
6x
"2 + 11
r-
O(vx).
26
CHAPTER 1: PRIME NUMBERS:
(1):
Show that when i
+
00
we have:
2
=T i
+ OCR).
q i+1 - q i
OCR).
qi and
(2):
Let
n1 .n 2 ••••• nk
be pairwise relatively prime.
Show
that there exists a number N satisfying:
N+ j
=0
2
(mod n,) ]
for 1
j ~
~
k.
Show that:
Find qi and qi+1 such that qi+1 - qi (3):
Show that:
'J
-[(q 1'+1 - q,1 )1og1ogq 1 11'm 1ogqi (4):
>.
~
1
2 .
Show that:
1 im( s, 1 - s,)
--
5.
~
1+
EXERCISE 1 -12:
1 and lim(s, 1 -
1
Let
1+
f
1
4.
be a multiplicative function.
the function g. defined by:
g(n) • { :
s,)
if
fen)
if
fen)
is also multiplicative.
~
D. D.
Verify that
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
27
EXERCISE 1·13: BELL'S SERIES FOR AN ARITHMETIC FUNCTION +00
\L
1'=0
Verify that if h = f"'g, then for each p the power series
h(pr)zr.1S t h e f orma 1 product +00
I
f(pr )zr
0
f t h e ser1es: .
and
1'=0
+00
If f(l) f 0, what can be said about the power series
I r'" (pr )zr?
1'=0
APPLICATION 1: Determine the functions "A."z and
"A -;'.
APPLICATION 2: With f the multiplicative function defined by:
1.3.5 ••• (21' - 1) 2.4.6 ••• 21'
determine the function f'·'f. EXERCISE 1·14: With h a completely multiplicative arithmetic function, show that for any arithmetic functions f,g:
and that for any f such that f(l) f
°we have:
(fh )-,', = f-'·'h.
What does this latter relation yield for f
z?
EXERCISE 1·15: Let f be a multiplicative function. Show that for all square-free n:
L
din
fed)
IT
pin
(1 + f(p)),
28
CHAPTER 1: PRIME NUMBERS:
and:
L
d/n
]J(d)f
IT J p/n
Show that if
f(pr)
(f(p) - 1).
~ 0 when r
>
1 the first of the above in-
equalities holds for a11 n eJN"'. EXERCISE 1·16: PERIODIC COMPLETELY MULTIPLICATIVE FUNCTIONS Let f be a periodic completely multiplicative function. Show that f has a period k such that: fen)
=
0 i f (n,k) > 1
and
fen)
+0
i f (n,k)
=
1
(so that f is a Dirichlet character (or a modular character), cf., [Bor]) • (For this, with h a period of f, set:
IT
k
v
p p
(h)
q
IT
p/h
p/h
f(p)=O
f(p )+0
v (h)
p p
where v (h) is the exponent of p in the decomposition of h into p
prime factors.
One would then show that k is still a period of
f by considering the product f(q)f(n + k»).
EXERCISE 1·17:1: With f a completely additive arithmetic function, show that, for all arithmetic functions f and g:
II:
To each arithmetic function f e G, where G is the group
of arithmetic functions invertible with respect to convolution, we associate the arithmetic function:
29
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
(flog is the product of the functions f and log, that is to say the function n 11:(1):
~
f(n)logn).
Show that if h = f*g one has Ah = Af + Ag • (Apply This says that the mapping f-+
Question I above, with! = log).
Af is a homomorphism of the group G into the additive group of arithmetic functions. What are the kernel and image of this homomorphism? What is the function A ? z
11:(2):
Show that if f is multiplicative then:
Af(n) = 0 if n is not a power of a prime number. (Notice that if n
> 1
(*)
and if n is not a power of a prime num-
ber n can be written in the form n 1n 2 , with n 1 > 1, n 2 > 1 and (n 1 ,n 2 ) = 1, and that all the divisors of n are obtained, each one only once, by forming all the products d 1d 2, where d1/n and
d/n). Show that ,if fell
1 and if relation (*) holds, then f is
mUltiplicative. (Notice that if g is the multiplicative function determined by g Cpr) = ICpr), then A A , and, by usinE; of Question IICl) a6ove, deduce tha€ g =ff).·
EXERCISE 1·18: SQUARE-FREE NUMBERS THAT ARE MULTIPLES OF A GIVEN NUMBER Let E be a finite set of prime numbers that has k ele-
I: ments.
Set P =
2
IT p, and
peE
denote by E the set of integers of the
form mq , where m,P and (q,P) = 1.
E can be empty; in which case
k = 0 and P = 1.
1:(a):
Show that for every x
>
0 the number of elements of
30
CHAPTER 1: PRIME NUMBERS:
E< X
TT (1 + ~J
does not exceed Ix
L
~,,2xk + 2:.... II IX peE
neE n " n>x
and that:
peE
(1 + 1r- J vp
(Use the fact that, for all u
>
0,
an inequality that is obtained by majorising the first term of the sum by l/u 2 and each of the others by
fn
n-1
I:(b):
dt ) . 2"
t
Let X be the characteristic function of the set of
those n e N* which are square-free. and prime to P. to P. Determine the function h
X*~.
(Notice that X is multiplic-
ative). Show that the series determine its sum.
h(n) . IS
L ---n
00
absolutely convergent, and
n=1
For which values of n is hen)
+ O?
What then is the absolute
value of hen)? I:(c):
Deduce from the preceding that for all x
>
Owe
have:
(Use the property X = h*z). II:
With m an arbitrary integer
~
1, let Qm(x) be the number
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
of square-free n sJN'"
31
divisible by m and.; m.
From result ICc) above, deduce that when x
-%- y(m)x
Q (x) m
+
~ +00:
O( v'x),
11
where y is the mu1tipicative function determined by:
f !
1
p
=
1
0
for
l'
for
l'
1, >
1,
and 0 is uniform with respect to m. More precisely, there exists a constant C
> 0
independent of
m such that: iQm(X) - - ; Y(m)xi <
c/X for all
x > O.
11
(The case where m does have a square factor is trivial. is square-free, the cases where x
p
When m
m and where x < m need to be
distinguished, the second being treated by a direct majorisation of the first member, without using I(c) above).
EXERCISE 1'19:
(1) :
SQUARE-FREE NUMBERS IN ARITHMETIC PROGRESSIONS
Show that every n SJN'" can be written uniquely in the form:
. h ' q square- f ree. n = m2q WIt m,qsN",
(2) :
L 2
d /n
Verify:
.Cd) "
i
1
if n is square-free,
0 otherwise.
32
CHAPTER 1: PRIME NUMBERS:
(Note that "d 2In" is equivalent to "dim"). (3): the n
S]N'"
Given k
S]N'"
t
and
let Q(x ;k,n be the number of
SZl,
which are square-free
~
x, and congruent to t modulo k.
Assume that (k,t) is square-free. Q(x:k,t)
(Otherwise it is clear that
= 0).
Show that when x
7
+00
we have:
where Ak £ is a constant depending upon k and
• with
form
t
and the 0 is uni-
respect to k and t.
Show that:
1 6 [ (1 - 1)-1 --IT k 2 Ik 2 1t
P
P
p-ti
(Write: Q(x:k,£)
L
ns;;x n=£(modk)
( 2L
d In
]..I
(d)J '
and reverse the order of summa tiona. In order to obtain the definitive expression for Ak £ it is • useful to observe that the characteristic function of the set of the d
Slt-P',
such that (d 2 ,k)lt is multiplicative).
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
33
EXERCISE 1·20: SUMS INVOLVING THE EULER FUNCTION
~
AND
DISTRIBUTION OF IRREDUCIBLE FRACTIONS WITH DENOMINATOR (1):
Show that
11.
q> = ].l,',i
(where
n for all n eJN"'), and
i(n)
deduce from this that the Euler function
q>
is multiplicative, and
that:
(Regrouping the m's such that 1
m ~ n according to the values
~
of (m,n), we see that i = z,.,q». (2):
I
n",-x
By using the property q>(n)
I
2
show that when
x -+ 00:
+ O(xlogx).
11
Given m,n e:N"', what is the expression:
(3):
dim din
3
"2 x
].l",i,
q>
].l(d)
equal to? (4):
Given x,Y
>
1, we denote by ¢(x,y) the number of pairs
[m,n] e:N'" xJN'" such that:
m '" x, n '" y
and
(m,n)
1.
Show that:
¢(x,y)
=
I
ddnf(x,y)
].l(d)
[J] [~],
and deduce from this that when X,y
-+
00:
CHAPTER 1: PRIME NUMBERS:
34
1
6
(x,y) xy
2" + 1t
O[lOg inf(X,y)} infCx,y) .
Can the result in (2) above be obtained by starting from this? Given n
(5):
e:N 1,
and t
> 0,
we denote by N(n,t) the number
of irreducible fractions with denominator n belonging to the interval (0, t], in other words the number of the m e:N 1: such that (m,n) =
1 and m
(a):
~ tn.
Show that if n
N(n,t) - t
Set B(x) = x - [x] >
1, for all t
>
!.
0:
L ].l(d)B[~) .
din
From this deduce that:
(b):
From the preceding result deduce that for any 0
there exists K
£
>
<
£
< 1,
0 such that for all n > I and for all t > 0:
EXERCISE 1'21: SMALL VALUES OF
We set: -&(x) =
L
logp
(CHEBYCHEFF'S FUNCTION).
p~x
We denote by Pk the k-th prime number, and set:
(1):
With
the Euler function and v(n) the number of prime
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
35
factors of n, show that: v(n) < k
and
>
Nk
< Nk :
(2):
Show that the only numbers N such that:
m < N
=>
m
are the integers (3):
--N--
Nk ,
k
~
1.
By Using Mertens' Formula:
where y is Euler's constant, show that:
I d be the sum of divisors of n. din Show that a(n)
Let a(n) =
and from this deduce the value of:
r-[ a(n) ) 1m nloglogn I
EXERCISE 1·22:
•
].J(n)
o(x)
n~x
(1):
Let
a
be an arithmetic function and let
A(x)
=
I n~x
a(n).
CHAPTER 1: PRIME NUMBERS:
36
By using the Lemma in Section 1.2 of the Introduction, show that if the function a is bounded
on~*,
then for x tending to
infinity:
L
n:i;x (2):
a(n)1og(n)
A(x)1og(x) + O(x).
Show that A = ]J",log, and, by writing down the value
of the function ]J",log, deduce from it that ]Jlog = -]J,',A. (3):
Show that for all x
where g(x) =
~(x)
~
1:
- [x].
L ]J (n) • n:i;x Show that the Prime Number Theorem implies that we have: (4) :
Set M(x) =
M(x) = o(x)
(x -+ +00).
(Notice that the Prime Number Theorem is equivalent to the relation
~(x) 'V x) .
EXERCISE 1·23: EVALUATION OF THE SUM (1) : F(s)
2
]J (n) ~ n:;;;x
L
We consider the Dirichlet series:
L
/(n)
n=l
where ]J is the
M~bius
function and
the Euler function.
Deter-
mine the abscissa of convergence of this series and put it into the form of an Eulerian product. (2):
Develop as an Euler product the function G(s)
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
~(!(!)l)
37
and show that this product converges uniformly on every
compact set K contained in the region Res >
-~.
Calculate G(O) and show that: G' (0) -
-
L peplogp - 1)
p
•
Show that G(s) is the sum of a Dirichlet series:
(3):
= L !lS!:l. s
G(s)
n
Calculate the abscissa of absolute convergence as well.
(4):
Show that:
= ~L.
with L(x)
n1
n~x
By using L(x) and In(x)1 ~ ~ L.
n~x
1.. x
2
.. (n)
~
Cj)\n,
= logx
+ y + n(x), where y is Euler's constant
for x > 1, show that:
=
~ logp logx + y + L. pep _ 1) +
p
0
( )
1 •
(The evaluation of this sum is useful in sieve methods).
EXERCISE 1·24: INTEGERS n SUCH THAT n(n) - v(n) = q (q GIVEN) +00 n(n)-v(n) (1): Show that the series L Z s is absolutely convergent for Res> 1 and Izl ~(s)
~(2S)
IT
n=1
<
n
2, its sum being equal to:
38
CHAPTER 1: PRIME NUMBERS:
(2):
Show that the infinite product:
is absolutely convergent for Res
> ~
and Izl
function of sand z holomorphic for Res With q
(3):
of the n
~
> ~
0 an integer, we denote by N (x) the number
~
q
x such that n(n) - v(n) = q.
From the preceding deduce that when x N (x)
q
1:2, and defines a and Izl < 1:2. <
= d qx +
where the numbers d
L
0
+00:
(x),
are determined by the property, for
q
~TI
q=O
~
[
1 -
1 -
11
Verify that all the d
ph]
>
!:..
Izl
<
2:
•
p
L
0 and that
d
=
1.
q=O q (By a theorem from the theory of functions of several complex
variables t , for Izl
<
q
1:2 and
Res
>
~ we can write:
+00
IT
L
q=O
where the A
q
are holomorphic for Res
> ~.
t If F is a function of two complex variables sand z, holomorphic in sand z for s belonging to a domain D of ~ and Izl < R,
there exist functions A holomorphic on D such that, for seD and Iz I < R: q F( B ,z)
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
39
Notice that for Res> 1:
L
ls
r/(n)-v(n)=q n
and apply Ikehara's Theorem). EXERCISE 1·25: THE NUMBER OF n's SUCH THAT Let
be the Euler function.
~
m
(1) :
x
For m
~
1 set:
Show that for every integer m
~
1 the set E is finite.
mL
{nl~(n) =
E
~(n) ~
Determine E when m is odd.
m
m
Set a
(2) :
cardEm, and:
m
A(x)
a m~x
m
=
card{nl~(n) ~ x}.
Show that there exists a constant C, independent of n, such that:
~~C ~~n}
L 1:.
din d n
Show that
~(n)
= O(logn), that x
~
A(x), and that A(x)
O(x1ogx) • (3):
Show that the series: 1
L -- -
00
n=l (~(n»s
co
and
a
L.2::.S
m=l m
are absolutely convergent for Res> 1, and that for Res> 1 the equalities:
L
00
a
m
m=l mS
1;(S)[(S)
CHAPTER 1: PRIME NUMBERS:
40
hold, where f(s)
~(s)
= IT P
denotes the Riemann
~-function,
and:
{1 + 1l)s (p _
Show that for every prime p and Res > 0 we have
(4) :
the inequality:
I(P_1 l)S
- pis
I <>
(p
lsi l)Res+l
From this deduce that f can be extended to a function morphic for Res> O.
j ho1o-
Calculate !,O)as a function of ~(2),~(3),
~(6) •
(5): Show that there exists a constant a such that when x-+ we have: 00
A(x)
'V
ax.
EXERCISE 1·26: PARTITIONS Let A CN*.
Denote by PA(n) the number of partitions of n
the summands of which are in A, and to which we associate the power series:
FA(x) where
= L PA(n)x n=O
PA(O) = 1.
IT neA
(1 -
x)
-1
,
(Always assuming that AC Ni,.)
By P(A) we
denote the property: PROPERTY: P(A):
T~ere
exists N such that n
~
N => PA(N) > O.
The object of this Exercise is to characterise the sets A ofN* with Property P(A). (1):
Show that i f A' C A, P(A') => P(A).
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
(2):
Let a,b
>
41
0 be relatively prime integers.
ab can be written as n = xa + yb with integers x,y ~ O. Show that if A contains two relatively prime elements, then Property P(A) holds. Show that every n
(3):
~
Show that if A is finite and equal to {al, ••• ,ar } then,
where g(n) is a polynomial of degree r - 1, for which the term of highest degree will be given, and q is the largest number of elements which can be taken from A, and whose greatest common divisor (g.c.d.) is greater than one.
Use the decompos-
ition into simple elements of FA(x). (4):
For A C
:N'~
define the properties:
PROPERTY: PI (A)
<=>
the g.c.d. of the elements of A is 1;
PROPERTY: P 2 (A )
<=>
For all prime p' s there exists a e A such that p~a.
Show that for all A C:N*. PI(A)
P2 (A). Show that for all A C :N*. P(A) => PI (A). Show that for every fini te subset A C :N"'. P(A) (5):
<=>
<=> PI (A).
Let A be an infinite set in:N* satisfying Property
P 2 (A ).
Show that there exists a fini te subset A' C A satisfying Property P 2 (A' ).
Show that for every set A C:N* we have P(A)
<=>
PI(A).
EXERCISE 1·27: LOWER BOUND FOR A SUM APPEARING IN THE SELBERG SIEVE (2):
Let N
~
2 be an integer, Q a set of prime numbers less
CHAPTER 1: PRIME NUMBERS:
42
than or equal to N, and A(N,Q) the set of those
integer~
less than
or equal to N that contain only prime factors belonging to Q. Show that there exists a positive constant not depending on N such that:
~
L
B
msA(N,Q) m
IT
psQ
(1 __
p1)-1
J
(One can induct on the cardinality of Q, after doing the case where Q is formed by all the prime numbers less than or equal to N) •
Let z
(b) :
~
4 be a real number, P be the set of prime num-
bers less than or equal to z, PI a subset of P, and P2 the complement of PI in Pj Pi (i = 1,2) will denote the product of the ele~· ments of
P., and w will denote the completely multiplicative func~
tion determined by:
w(p)
:
{
ifp divides PI' ifp divides P2 ,
ifp > z.
Show that there exists an absolute constant C such that: 1
T
EXERCISE 1-28:
~ C(1ogz)
2
Let x be a positive real number> 8, and m a
non-zero even integer with absolute value less than x. By using Selberg's inequality prove that there exist two absolute constants
D,E such that we have: card{plp ~
x,
p and Ip +
ml
prime numbers} ~
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
~
E _x_
IT
log 2x p/,m
EXERCISE 1·29:
43
(1 + ~JP
Deduce from Exercise 1·28 that the series of
the reciprocals of the prime numbers p such that p + 2 is a prime number converges. EXERCISE 1·30: GOLDBACH'S PROBLEM Let r(N) be the number of representations of an integer N
>
1
as a sum of two prime numbers. (a): F
Using Exercise 1·28, prove that there exists
not depending on n, such that: r(N) .:; FN( logN)
(b):
-2
L
~
diN d
From this deduce that there exists a constant F' such
that:
L
N.:;x
(c):
r
2
(N) .:;
3
F'x (logx)
-4
for all x > 4.
Deduce from the Prime Number Theorem that there exists
a positive constant F" such that:
L
r(N)
~
2 -2 F"x (logx)
for all x
>
4.
N~x
(d):
Let M(x) be the number of integers less than or equal
to x which are the sum of two primes; prove that there exists
44
CHAPTER 1: PRIME NUMBERS:
a positive constant C such that:
M(x)
~
(e):
Cx
for all x
~
4.
Deduce from this that the sequence formed by 0,1, and
the prime numbers, is a basis of the integers. EXERCISE 1 31: SMALL DIFFERENCES BETWEEN CONSECUTIVE PRIME 0
NUMBERS We return to the notations of Exercise 1 9. 0
By a and A we
denote two positive real numbers. (a):
Prove the inequality:
lim sup(logxl.card{ilp. 1 x-+<» x J 1. +
~
x,d. 1.
~
Alogx}
~
A-i.
M(x) denotes the number of integers i such that p.1. lies between !x and x, and d. between alogx and Alogx; by using Exer1. cise 1 28 prove that there exists a constant F such that for all (b): 0
x
~
2 we have:
M(x)
(c):
~
F
--T L IT (1 +~) . log x alogx
Deduce from this the existence of a constant F' such
that: . sup (M(X)lOgX) 11.m x-+<»
(d): s(x)
x
~
F , (A -
a )•
Write:
d.1. d.1. +
L
alogx
d.1.
+
L
d .>Alogx 1.
d.1.
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
Show that the hypothesis: "The first swn of the right-hand side is empty for x large enough" implies: S(x) 1 1m ln f ---0
0
x-+oo
x
~
2lA - F'(A -
(e):
By choosing a
M -
F'(A - a)2 > l2
<
holds, prove that: lim inf[ld i oJ :;:; a < 1. i-+oo og'Z-
a )2 •
1 and A so that
45
46
CHAPTER 1: PRIME NUMBERS:
SOLUTIONS
SOLUTION I-I: Let us write m
u =2
2n
•
n + k, where k e:N1', and let us set
We have:
Fn+k - 2
u
F
n
2k
u +
- 1 1
u
2k_l
- u
2k_2
+ ••• - 1,
hence F divides F - 2. Let d be the g.c.d. of F and F . n m n m' dlF n , hence dfFm - 2 and dlFm, so d12. Since Fn and Fm are odd, d = 1, and therefore Fn and Fm are relatively prime.
The mapping of :N'" into the set of prime numbers which assigns
to each integer n the smallest prime factor of F is therefore n
injective, and so there are infinitely many prime numbers. SOLUTION 1-2:
Let us fix real S N
n(N) ~ 10gN
and
n(SN) ~
1; by the Prime Number Theorem:
<
SN logN + logS
If N is large enough we have n(N) a prime number between SN and N.
>
SN 10gN
~--
n(SN), and hence there exists
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
SOLUTION 1'3: (i): -&(x)
= L
47
We have the following two relations:
L
logp ~ 10gx.
For all 0 e (0,1):
-&(x) ~
~
1
= 1I(x)1ogx;
L
o x
(1)
10gp
o
01l(x)10gx - x 10gx.
(2)
Assuming the Prime Number Theorem we deduce from these relations that: -&(x)
lim sup -x-
and
~ 1
1 im inf
-& (x)
x
~ 0
(for all 0 e (0,1)). Hence we certainly have: . f -&(x) l ;m ~ l.n -x-
and therefore -&(x)
1
~
•
~ x.
Conversely, if we start from the relation -&(x)
~
x,
using (1)
we find: lim inf 1I(x)1ogx ;,. 1, x
whence x
o
= o(1I(x».
lim sup
and from (2):
11 (x)1ogx
x
1
'8
~
and thus the Prime Number Theorem. (ii) :
For each n
~
1 we have 1I(p ) = n. n
If the Prime Number Theorem is assumed, we deduce that when n
-+
co:
CHAPTER 1: PRIME NUMBERS:
48
Pn n " 'logp --- • n
This implies: logn '" logpn
and
pn '" nlogp n '" nlogn.
Let us now observe that for all x
If, for "infinite" n, we assume p
~
n
2:
'" nlogn, we deduce from
this, for "infinite" x, that the extreme terms are equivalent to and consequently:
~(x)log~(x),
This implies: logx '"
log~(x)
SOLUTION 1·4:
and
By the Prime Number Theorem and by Exercise 1·3,
we have p n '" nlogn. As the given series has positive terms, the general term can be replaced by an equivalent one, which comes down to studying the series with general term u = (nlogn(1oglogn)
we see that the given series converges if and only if
>
1.
SOLUTION 1·5: 1:(1): The exponent of p in the decomposition of an integer n into prime factors is clearly equal to:
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
49
(The sum being zero if it is empty, i.e., if p%n). The exponent of p in the decomposition into prime factors of
IT
the product a
n,.x
rL [nL ,] [r~' ,] pr,.x pr/n pr/n
~
p
n is therefore
n~x
This is equal at most to
+00 ~ ~
L r 2"=1 P
~
2"~1
pr,.x
= p___x-___1 and
[:] . p
at least to
We have: x 2p ,
also for, if u
I, we have u ,. [u) + 1 ,. 2[u), and consequently
~
[u) ~ ~u.
1:(2): for each n
It is clear that we have
~ logn,. xlogx, since, n,.x However, by what has already been
x, logn ,. logx. shown, we have:
I
~
logn
log
n~x
~
~
P,.x
IT
n,.x
n =
x
2p logp
Hence we have:
~x
I
p~
~,. xlogx, P
I
p~x
a logp p
= ~x ~
p:t;;x
~ p
.
CHAPTER 1: PRIME NUMBERS:
50
whence: 2logx. If 1
~
Y
~
x one can also write:
I
logn ~
~ logp
y
and from this it follows that:
~xlogy
I .!. ~
y
xlogx,
P
whence:
1:(3):
For each n
fnn-l logtdt ~
~
1 we have:
logn.
Consequently:
I
logn
~
f0[X] logtdt
~
f:lOgtdt - logx
fXolOgtdt -
n~x
fX
[x]
(x - l)logx - x.
For each p < x, a p < _x_ we have: p - 1 •
L
p~x
~
a logp p
x
L ~.
p~x p -
Hence:
(x - 1 ) logx - x
~
x
~ L P _ 1 '
\'
P'x
logtdt
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
51
whence:
L
p10 gP1 ~ 10gx - 1 _ l~X
p~x
1 p(p - 1) ,
Since.!... = _1_
P
P - 1
\ .!.£BE. = \ P
L
p~x
~
L
p~x
P _ 1
_ \
L
10gp
p~x
p(p - 1)
~ 10gx - 1 - 10gx _
x
~
L p(p10gp - 1)
10gx - C,
where:
C (as
=1
10gx
x
+
!e
+ \
10gp
L p(p -
1)
has a maximum when x
11:(1):
= e).
Every prime number p satisfying m
( 2m ) ! but not m!, and. so divides ( 2m) __ (2m)! • m (m!)2
10gp But:
and so: 10g(2m) m
<
2m10g2.
Hence we have: .s(2m) - .s(m)
<
2m10g2.
= .s(2m)
- .s(m).
~
p
~
2m divides
Consequently:
52
CHAPTER 1: PRIME NUMBERS:
From this, for all k
1, it follows that:
~
8(Zk) = 8(Zk) - 8(1) = k
L
<
ZjlogZ
k
l,
j=l
(8(Zj) - 8(Zj-l))
Zk+l logZ .
<
j=l Given x> 1, there exists h ~ 0 such that Zh ~ x ~ Zh+l , and we have
8(x)
~
8(Z
h+l
)
~
h+Z Z logZ
<
4x log Z .
Given x > 1, since for each
II: (Z):
p.~
x we have:
a
;:,!E.- l , P
p '"
we obtain:
I
n::;;x
'i
logn =
ex logp ~ p
p~x
l,
logp= x
p~x
'i
p<:x
~ - 8(x). p
Therefore: logn + 8(x)
~
xlogx + 4xlogZ,
whence: logx + 4logZ. III:
Using the Lemma of Section 1.Z of the Introduction, for
x > Z we see that:
'i
p~
1:. = 1:. + P
Z
=
I
Z
~ P
_1_ = 1:. +
(Z) _ fX.(t) d (_l_Jdt logp Z logx logZ ~ dt logt
(x) + IX logx
(t) Z dt Z t(logt)
1 + n(x) + IX ~ + IX net) dt= logx Z tlogt Z t(logt)Z
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
(Contd)
1
53
fX2
nCr) + - - + loglogx - loglog2 + logx
n(t) 2 dt. t(logt)
Since, by 1(3) and 11(2) above, n(x) is bounded for x integral
n(t) 2 dt f+2OO t Oogt)
1 L -P = loglogx
+
1 -
loglog2 + f:OO
_ roo 2
If In(x)1 ~ M for all x M
~ logx
an
1, the
is convergent, and we can write:
p~x
In(x)1 logx
>
n(t) tOogt) n(t) tOogt)
2
dt + n(x) logx
2 dt.
1, we see that:
>
+OO n(t)dt I ~ M tOogt)2 x
f
d
One obviously obtains the formula indicated, with b
=
1 -
loglog2 +
I
[lOg 1
-:x]
1 1
P
n(t) 2 dt. t(logt)
The series:
(1) :
EXERCISE 1·6:
f+OO 2
x
is convergent for real x >
1
2,
because when p
+
+oo:
111
log --....,..... - 1 _ 1 pX
For all for x
~ xO.
Xo
> ~,
pX
'" - 2p2x
there is normal, hence uniform, convergence
In fact, log 1
=u
- u is a positive, increasing
54
CHAPTER 1: PRIME NUMBERS:
function of u for 0 for x
~
u < 1.
Consequently, for any p, we have
~ xO:
1
o :;: log -1-_--::1- p
x
1 x:;:
1
log ---:-11
p
p
xo
From this it follows that if g(x) is the sum of the series, the function g is continuous on the interval ]!,+oo[. For real s
(2):
Z;(1 + s)
IT
>
0 we have:
1
1 -
1
p
1+8
and consequently: logZ;(l + s) =
L log
1 1 -
p
1 1+8
From this it follows that:
L
logZ;(l + s) - g(l + s)
p
11+ 8
By applying the Lemma of Section 1.2 of the Introduction, we see that, for all T '\L p~e
Pee )
fo se +OO
~
-8t
0:
1- = '\L -1+8 -p1 e -81ogp = O
P( e T) e -8T
+
fT se -8tp( e t )dt • 0
+00 the first sum tends to l l~S' and j since p T -8T Consequently, the integral logT, P(e)e -+ O.
When T T
>
-+
t that is to Pee )dt is convergent and equal to '\L _1_ 1+8'
say, to logZ;(l + s) - g(l + s). Furthermore, for real s
>
0 we have:
P
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE +00
1 -ns n e • n=l '\
1
log -----s 1 - e
L.
Moreover as before, we have for T '\ L.
1 -
n~T n
e
55
-ns
L !
O
so, by letting T
>
0:
= N(T)e- sT
e- ns
+ fTose-stN(t)dt,
we obtain in the limit:
-+ +00,
log _.;;..1_ = f+ooose -stN(t)dt. 1 _e- s (3):
When t
we have:
-+ +00
P(e t ) = logt + b +
o(i]
logt + y + 0(1).
N(t)
and
Consequently:
P(e t ) - N(t) + Given
~
b
-+
o.
0, there exists to > 0 such that:
E >
Moreover, for 0 o
y-
t
P(e )
~
~
t
P(e
~
to
to' we clearly have: )
and
0
~
N(t)
~
N(t o )'
0) +
N(t o) +
and consequently:
~ P(e Then for all real s
>
t
Iy - bl = M.
0: (Contd)
56
CHAPTER 1: PRIME NUMBERS:
(Contd)
and from this we have: limI8f+OOe-S\p(e t) - N(t) + Y - b)dtl :;; 8-+0 8>0
As this is true for any
f+ e -st (P(e)t
8 0
when
8
-+
E.
0
E >
0, we see that:
- N(t) + y - b)dt
°through real
-+ 0
positive values.
The formulas established in the preceding paragraph show that the expression under consideration is equal to: 1
log<;O +8) - gO + 8) - log - - - + y - b, 1 - e- s that is to say: log«l - e
-s
)C;;(1 + 8»
- gO +8) + y - b, 1, because (1 - e- s ) ~
which tends to y - b - gel) when
8 -+
g is continuous at the point 1.
Hence we have y - b - gel)
whence b = y - gel). (4):
Since, for x > 0, we have:
P(x) + log
when x
-+
IT
p:;;x
[1 - ~J
+00 we see that:
p
I p:;;x
1
[lOg - - 1 1 -
P
8
and 0,
ARITHEMTIC FUNCTIONS: SELBERG'S SIEVE
P(x) + log
IT
(1 -
p~X
~)
p
-+ -
57
g(l),
which can be written: log
IT
p",X
(1 - ~)
=-
+ 0(1)
P(X) - g(1)
IogIogx - b - g(l) + 0(1) - IogIogx - y + 0(1). Thus one obtains the well known Mertens' Formula: e- Y 'V - -
Iogx
SOLUTION 1·7: (1):
ber p, p
>
In,
1 '"
~
<
n~A
if and only if there exists a prime num-
which divides n.
Then we have:
p, that is to say, n = ap with 1 '" a '" p - 1,
hence neB. p
Let us assume p
x
= ap = bq
<
q, and that x e B () B.
with 1
p
~
q
We must have:
a '" p - 1 and 1 '" b
~
q - 1;
but q divides ap, hence it divides a, therefore q '" a contradicts p (2):
<
<
p, which
q.
Let us define QB (x)
card{n '" x,n e B}. p
P
if x
<
2
p , 2
ifx~p.
By part (1) above,
We have:
58
CHAPTER 1: PRIME NUMBERS:
L
[x] -
pprime
L
[x] -
(3) :
equal to x.
Let
p~rx
R(X)
QB (X),
p
(p - 1) -
L
[~]
IX
•
be the number of prime numbers less than or
We know that
R(X)
~ ----Ix . ogx
We have then that:
On the other hand:
[p~]
= =
L
~+
IX
x[ loglogx
O( R (x ) )
+ b - loglogIX - b +
o(lOlgx) ]
+ O(l:gX)
= xlog2 + O(l:gX) • Finally, we have:
and:
SOLUTION 1·8: (1): n e:N*: v(n)
L
pin
By the definition of the function v, for all
1.
Consequently, for x
(1)
~
1:
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
I
v(n)
n~x
59
= I n~x
= I
[~] = x I .!.. + O( 11 (x)). P'X P
P~X P
since. for each P , x.
[~]
differs from
~ by
at most one.
Assuming the formula cited in the Problem. and that
o(l:gx)
lI(X)
• we obtain:
I
n,x (2):
v(n)
= xloglogx +
bx
The condition Pi P 2
+ O(l:gx) •
'x
implies:
If one adds up the sum of the terms for which Pi ' of those for which P 2 '
(2)
!:C and the sum
!:C. then the terms for which:
have been counted twice. We have:
and the same relation with Pi and P 2 interchanged. hand:
Thus one obtains:
On the other
60
CHAPTER 1: PRIME NUMBERS:
(3)
(3):
By expressing the function P as is done in the state-
ment of the problem we have:
because. for p
<
x:
lOglOg~) = log( 109:!:
- logp)
log (lOgX (1 -
.
~~g])
This gives: (loglogx +
b)
I
.!. +
p~rx p
I
.!.(lOg 1 -
p~rx p \' L
+
-1
p~/x p
(loglogx + b )p( IX) + +
I !
p~rx p
= loglog(/;X)
+ b + O(I:gX)
loglogx - log2 + b + O(l:gX] we find that:
1
logx
n (x] p
~J logx
log (1
(x)
L -np",rx p p
As we have: P(/;X)
\'
~]
.
61
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
(loglogx + b)P(/:;)
= (loglogX)2
lui
On the other hand, for
T=k=l +'"
logO - u)
Consequently, for 0
~
~
u
L
p~rx
u
k-1
LT.
k=l
u ~ ~: +'"
L
1 = Ku. - 1k
k=l k2 -
Since, for p ~ /:;, 0 "~ logp logx ~
I \'
1 it is known that: +'"
k
L
-
Ilog(l-u)1
<
+ (2b - log2)loglogx + 0(1).
1 P log [ 1 - ~)I logx
~
1
2, K
we see that: \' Lr- ~ plogx P,.YX
« _K_
'" logx
\'
LrP"Yx
~ « K p " ,
since, by Exercise 1·5:1(2): logx. For p ~ loX we also have ~ ~ loX, and consequently: p
M
2M
,.--~--
1
x
P
og
logx·
Therefore:
L 2:.n[~)1 Ip~rx p p
«~p(loX) ~ logx
o[lOg;OgX)
•
Finally, we have:
L
1
p,.IX P
p[x)P
=
Moreover, we have:
Ooglogx) 2 + (2b - log2 )loglogx + O( 1).
(4)
62
CHAPTER 1: PRIME NUMBERS:
(10 g 10 gX + b - log2 + (loglogx)
2
0(10~X))2
+ 2(b - log2)loglogx + 0(1).
(5)
Equations (3),(4),(5) give: Ooglogx)
2
+ 2b1og1ogx + 0(1).
From Equation (1) we deduce that for each n eJN", we
(4):
have:
Consequently, for x
~
1:
If P1 ~ P2 , "P 1 /n and p/n" is equivalent to "P1P/n", and we
have: i f P1P2 > x,
0
{
L
n.;;x P1/ n P2/ n
[p;P 2 ]
i f P1P2 < x.
Thus:
We have already seen that:
L [:£.]
P~x
P
=x
L ~+
P~x P
O(1t(x))
= x1og1ogx + bx + O(l:gx) .
(6)
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
63
It is sufficient to write here that:
I [~] = xloglogx
p:!,x
P
+
O(x).
On the other hand, the number of pairs [P1 ,P 2J such that
+
P1P2 ~ x and P1 P 2 is less than 2x, for each integer of the form P1P2 with P1 P 2 is written in this way in lreciselY two ways. Since, for each pair, the difference between ~] and ~ P1P 2 P1P2 is at most equal to one, we have:
+
= x
O(x)
whence, on taking account of (6):
I
P1P;i1>.x
[p ~ ] = x(loglogx) 2 + 2bxloglogx +
o(x).
1 2
Pl 'f'P2
Thus we arrive at: ~
L
v(n) 2 = x(loglogx) 2 +
(2b
+ l)xloglogx + O(x).
n~x
(5):
For each n ~ x (where x > 1)
(v(n) - loglogx) Consequently:
we have:
2 2 2 = v(n) - 2v(n)loglogx + (loglogx) .
(7)
64
CHAPTER 1: PRIME NUMBERS:
nli:x
(v(n) - loglogx)
I
2
v(n)2 - 2(
n~x
I
nli:x
v(n»)lOglOgX + [x] (loglogx)2.
By using Equations (7) and (2). we deduce that when x
(v(n) - loglogx)
2
= xloglogx
-+ +co
+ o(x).
(8)
n~x
It is clear, for x
(6) :
(v(n) - loglogx)
2
~
e, that:
2
~"n" (x )loglogx,
whence: Ii:
2
I
1
" xloglogx nli:x
2
(v(n) - loglogx) .
Using Equation (8) we get: -
1
1
lim -n,,(x) ~2. x-++oo x "
(7):(a):
For all n satisfying /:X
loglogx - log2
<
loglogn
Ii:
<
n
Ii:
x, we have:
loglogx.
From this it follows that:
iv(n) - loglogxi ~ iv(n) - loglogni - iloglogn - loglogxi ~ iv(n) - loglogni - log2
and:
g(n)/loglogn
~
g(/:X)/loglogx - log2
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE . ( SInce x > a 2 > e 2 , loglogx > log2 ) .
65
This certainly gives the
result in the Problem. (7):(b):
We have:
g(/;C)/loglogx - log2 - log2
= IlOglOgX[g(/;C)Vl For any A
>
log2 ) IloglogxJ
log2 loglogx
°the expression between parentheses is
large enough.
Then for n such that /;C
<
~
for x
n ~ x, and for which
Iv(n) - loglognl ~ g(n)/loglogn, we have:
Iv(n) - loglogxl ~ A/loglogx . The number of such n's is therefore at most equal to nA(x), and consequently:
whence: N(x)
-x-~
From this it follows, by what was shown in question (6) above, that:
As this is true for any A
>
N(x)
0, we see that -x-
7
0 when x
7
+00.
CHAPTER 1: PRIME NUMBERS:
66
SOLUTION 1 9: (a):
We have:
0
P[x]+l - 3 10gx
By Exercise 1 3(ii) this expression is equivalent to: 0
[x + 1] 10g[x + 1] logx that is to say to.x.. when x .... "'; hence:
For every positive number
1 we have:
~ <
dn
-+ logn
d
t:; _1_ (p
10g2
n
[X~]+l
+
+
- 3)
_1_ (p
~logx
[x+1
]
- P
From this one deduces (as above) the relation:
Since this re1at ion is true for all
I 2~nt:;x
d _n_'Vx. logx
~
e (0,1), we have:
N
[x~+l]
)
•
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
(b):
67
Let us assume that we have the relation:
d lim inf _n_ = 1 + a > 1. x-- 10gn
We would be able to find an integer nO such that for every n > nO' d 10~ ~ 1 + la. Then we would have: dn --+ 10gn
From this one would deduce: lim inf!.
x--
I
x 2~n~x
d 10ngn
~ 1 +~,
which contradicts (a) preceding. In a similar way we can prove the relation: d
n ' sup -1-l 1m
x--
ogn
~
SOLUTION 1·10: (a):
1.
For every integer l, each of the numbers:
lrrn- 2 + pn- 1 - 1,
••• J
is divisible by at least one of the numbers P1, .•• ,Pn-2' chooses an integer l such that: and lrrn- 2
= -1
One then
(mod Pn ).
By the Chinese Remainder Theorem (cf., [Har] Theorem 121) it is known that there exists such a positive integer l
~
PnPn- 1 (it is
68
CHAPTER 1: PRIME NUMBERS:
Then the number tIT n _2 1 is divisible by Pn - 1 and tITn _2 + 1 is divisible by Pn . (When n 5, one has t = 4). then unique).
We have 10gITn =
(b) :
tion.
We know that
~(x) ~
~(p
n
), where
~
is the Chebycheff func-
x, so it follows, using Exercise 1 3, 0
that for n large enough IT n - 2 > 2P n , whence tIT n _2 - Pn - > Pn ' and the integers in the interval [tIT n _2 - P n - 1 + 1,tITn _2 + P n - 1 - 1] are all composite. There exists A = A(n) such that:
logp,
1\
~
10gIT
n
= ~(pn ).
Using Exercise 1 3, for any 0
£
>
0 and n large enough, we have:
Hence:
[ d)
. A(n) h~~up 10gA(n)
~
2,
and a fortiori:
1~
1 im sup (1 d NN N-+
og ]
SOLUTION loll: (1): 6qi
2 > 1.
The formula given in the Problem yields:
-2 + 11
O(~).
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
From this we deduce that i ~
6qi 11
gives:
2
69 o( Ii),
, whence o(~) l
which
2
;- i + o(li).
qi
Next, we have: 2 11
6
(i + 1 _ i) + O(Ii+T) + o(li)
o( Ii)
. ". smce v
~
26 qi' 11
(2):
If the numbers n 1 ,n 2 , ... ,nk are pairwise relatively
prime, the same holds for the number n~,n~, ... ,n~, and by the "Chinese Remainder Theorem" the congruences: 2 ]
N _ a. mod n., ]
j
1,2, ... ,k,
have an unique solution N satisfying 0 That is true, in particular, if a. ]
<
N
k
~
-j.
2
IT n., J=l ]
for any a j .
Let us assume N to be chosen so that N + j will be divisible by n~ for 1 " j '" k. Let q i be the largest element of Q such that q.l ~ N. Then q.l+ 1
q. l
>
k + 1.
>
N + k, and q.l+ 1 -
As k can be chosen as large as one likes, from the
latter statement it follows that: lim (q. 1 - q.) l+
l
=+
00.
In order to find q.l such that q.l+ 1 - q.l
~
5 we solve the con-
gruences:
N
=0
mod 4,
N _ -1 mod 9,
N _ -2 mod 25,
N _ -3 mod 49.
70
CHAPTER 1: PRIME NUMBERS:
N = 0 (mod 4) and N = -2 (mod 25) are equivalent to N 100, or N = 48 + lOOk. The equation:
= -2
1+8 + lOok - -3 mod 1+9, or 2k has k
= 48
as solution.
1+81+8 + 1+90ok'
= 48
modulo
mod 1+9,
It remains to solve:
= -1
mod 9, or I+k' - 2 mod 9,
k'
or
= 5.
The smallest solution of the congruences required is therefore
N = 29,348. The numbers N,N + l,N + 2,N + 3,N + 4 have square factors. Hence we have:
= 29,347,
q.
~
which is a prime number, and
and
qi+1
= 29,353 = l49x197.
REMARK: The smallest solution of q. 1 - q.
q.~+ 1
~+
= 246. (3):
N
= -j
~
~
5 is q;
241 and
~
The congruence: 2 J
mod p.
for j
= l,2, ... ,k,
where p. is the j-th prime number, has a J
solut~on
Let qi be the largest element of Q such that qi
qi+1 - qi ~ k + 1.
We now have:
logqi ~ logN ~ 2
k
L
i=l
logpi
where e is the Chebycheff function.
10ge(Pk)
~
~
logk + loglogk
~
But we have:
logk,
N
~
N; then
71
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
and:
because the function ----Ix is increasing. ogx
With the formula:
this proves: lim (
q. 1 - qi)lOglOgqi] 1 l+ > _ logq. / 2 l
+00, this implies:
lim(s.l+ 1 - s.) l
--
1.
Let us now show that lim(s. 1 - s.) l+
4 are in S, we always haves. 1 - s. for i
s.l
~
= 4n;
io one has s.l+ 1 - s.l
l+ ~
this would imply s.l+ 1
3. ~
l
=
4.
As the multiples of
4.
Let us now assume that
When s
is a multiple of 4 let
l
~
4n + 3, and amongst the four num-
bers 4n,4n + 1,4n + 2,4n + 3, at least two would be in S. The density of the numbers in S would then be greater or equal to one half.
Now this density, by the Formula of Question (1) 6
above, is 1 -:2
= 0.39, ...
11
SOLUTION 1·12: (1): (2) :
f(m)f(n).
We have gO) = 1, since fO)
Let m and n be such that (m ,n) If f(m)
+ °and
f(n)
+ 0,
=
1.
then f(mn)
=1
+ 0.
Ne have f(mn)
+ 0.
°
We have then
that g(m) = 1, g(n) = 1, and g(mn) = 1. If f(m) = or f(n) = 0, then f(mn) = 0. Then g(m) = or g(n) = 0, and g(mn) = 0. In all cases, g(mn) = g(m)g(n).
°
72
CHAPTER 1: PRIME NUMBERS:
SOLUTION 1-13: h(pr)
We know that:
I
=
f(d)g(zf)
.
d/pr
As the divisors of pr are the numbers pj, where 0 ~ j ~ r, we have: r
I
f(pr)g(pr- j ).
j=o
· sows h Th IS t h at t h e power series of the power series:
+""
I
r=O
f(pr)zr
If f(l)
+0
\ h(pr)zr.IS t h e product L r=O
and
it is seen that the formal product of the power
series:
+""
I
r=O
f(pr)zr
+""
I
and
f-"'(pr )zr
r=O r
r
e(p )z , which reduces to its constant term, r=O +"" i.e. one. Hence, the series I f-*(pr)zr is the formal quotient r=O +"" of one by the series I f(pr)zr. r=O is the series
APPLICATION 1:
As the functions A and z are multiplicative, the
function A*Z is also multiplicative.
In order to determine it
it is sufficient to determine its values for the numbers pro
+""
I
r=O
Since A(pr) = (_l)r and z(pr) = 1, the series z(pr)zr are absolutely convergent for Izl
<
+""
I A(pr)Zr and r=O 1, with 1 z and
!
ARITHMETIC FUNCTIONS: SELBERG's sieve
and
1
~
73
The product series is therefore the expansion
as sums.
1 of --"-"""'2" , that is to say, 1 + 1 - z
z
2
+ z
4
+ ••• + z
2k
+ ••••
Hence
we have:
if r is even, i f r is odd.
It follows from this that A*Z is the function equal to one for the squares and zero for the non-squares.
+00
The function A-* is multiplicative, and
L
A -,.,(pr )zr
r=O expansion, in a neighbourhood of the origin, of 1/[1 1 + z. Hence we have: for r
1,
for r
1.
is the
~ z)
Therefore: A-'~(n)
=
1o
1 if n
APPLICATION 2:
is squarefree
otherwise,
For the function under consideration, 1
+
L
r=l
1.3.5 ••• (2l" - 1) 2 .4.6 .• • 2l"
z
r
is the expansion in a neighbourhood of the origin of (the branch equal to one for z = 0). If g =
[~r
f"'f the series 1
~
= z.
L
r=O
g(pr)zr is the expansion of
, that is to say,
g(pr) = 1 for all r.
n, i.e., g
+00
+00
L
r=O
z
r
Therefore we have
As g is multiplicative, g(n)
Thus f*f
1
~
= z.
1 for all
74
CHAPTER 1: PRIME NUMBERS:
SOLUTION 1·14:
Let
~
= f*g,
(fh)*(gh).
~
By the definition
of convolution, we have:
Since, for each d, h(d)h(J) = hen), this gives:
~(n) If f(l)
= hen)
+ 0,
L
by taking g = f-* one obtains:
(since h(l) = 1 and e(n) = (fh )-,~.
h(n)Hn) .
f(d)g(J)
din
If f = z, one has
°for n f-'~ = \.I.
>
1).
This shows that f-*h
Thus we see that "p" =
=
\.Ih
(as zh = h). SOLUTION 1·15: (1):
We have
L
din
fed) = (z*f)(n).
However, z and
f are multiplicative, so z*f is multiplicative, and for n square-
free we have: (z*f)(n) =
IT
pin
(z*f)(p).
With the divisors of p being one and p, we have: (Z1'f)(p) = z(l)f(p) + z(p)f(1) = f(p) + 1,
whence the result indicated. For arbitrary n: (z,~f)(n) =
IT
(z'~f)(p).
pr//n
As the divisors of pr are the numbers pj, where have:
°~ j
~ r, we
75
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
(f*z)(pr)
= 0,
If f(pr)
when
l'
> 1
l'
= L
j=O
•
•
f(pJ)z(pr- J )
this reduces to 1 + f(p).
We have
therefore:
(z*f)(n) = (2) :
IT
pin
(1 + f(p»
for all ne:N*.
One has:
L ll(d)f[~.1 = J
din
(ll*f)(n).
f is multiplicative, and consequently, for n square-free
IT
(ll*f)(n)
=
(ll*f)(p)
= ll(l)f(p)
pin
(ll'~f)(p).
But:
SOLUTION 1·16:
kq
=
IT
pin
+ ll(p)f(l)
= f(p)
- 1.
We have: v (h)
p P
= h.
For any n e:N* we have:
f(q)f(n + k) But f(q)
+0
= f(qn
+ qk)
= f(qn
+ h)
= f(qn) = f(q)f(n).
since for every p dividing q, f(p)
+ O.
Therefore
this implies f(n + k) = f(n). If (n,k) > 1 there exists a p which divides nand k. p divides k, f(p) If (n,k)
=1
= O. Consequently f(n) = O.
there exists m e:N* such that mn
=1
Since
(mod k).
76
CHAPTER 1: PRIME NUMBERS:
We have: f(m)f(n)
= f(mn) = f(1).
Therefore one cannot have fen) SOLUTION 1'17: I:
o.
We have:
On replacing len) by l(d l ) + l(d 2 ) we obtain:
((fl);'g)(n) + (f"'(gl) )(n).
II:
One sees immediately that the definition given for Af
is equivalent to defining it by the condition Af*f 11:(1):
By taking l
= flog.
log in the result of Question I
above, we obtain: hlog
the convolution being commutative and associative
By the above remark, this shows that Ah
+ Ag .
= 0 if and only if flog = 0, in = 0 for all n e:N;'. Since logl = 0 and
It is clear that we have Af other words f(n)logn
= Af
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
logn
+0
for n
=
fen)
0
>
77
1, this is equivalent to:
for all n
> 1,
which comes down to saying that f is of the form Ae, where A is a non-zero complex constant (by hypothesis f(l) of the homomorphism f where A €
0:".
+
+ 0).
The kernel
Af is therefore the set of functions Ae,
One sees that the image is the set of arithmetic
functions which are zero for n
=
1.
Indeed, on the one hand we always have:
On the other hand, if g is an arbitrary arithmetic function satisfying gel)
=
0 one can determine a function f by arbitrarily choos-
ing a non-zero value for f(l) and determining fen) for n
> 1
by
the recurrence relation: n) _1_ L g(d)f[-d logn din d>l
fen)
Thus: f(n)logn
= L
din
in other words g"'f
g(d)f[J)
for every n
€:JN",
= flog, which shows that Af = g. = log. = zlog, and from this it follows that
It is known that von Mangoldt's function A satisfies A*z This can be written A*z
Az
A. 11:(2):
Let us assume f to be multiplicative. We already
Af <1) = O. If, now, n > 1 and is not a power of a prime, there exist n 1 ,n 2 > 1 such that (n 1 ,n 2 ) = 1 and n 1n 2 n. By the definition of Af we have: know that
Af(n)
= L
din
f(d)lOgd.r'''[J)
(Contd)
78
CHAPTER 1: PRIME NUMBERS:
(Contd)
I
f(dld2)lOg(dld2)f-*(dlCt2]
I
f(d l )f(d2 ) (logd
dl/n l d/n2 =
d l /n 2
l
+
logd 2 )f- I"
(;t.) f-
d/n2
2
I"
(~l] 2
0,
since e(n l ) e(n 2 ) = O. Let us now assume that f(l)
=1
n is not a power of a prime number.
and that Af(n)
=0
whenever
Let g be the multiplicative
,function determined by:
= f(i') for all prime numbers p and all
g(pr)
We see that Ag
l'
eN;'.
= Af .
In fact, first of all, if n is not a power of a prime number, Af(n)
=0
and Ag(n)
= O.
Next, given a prime number p it follows from the relations Af*f
= flog
and Ag*g
= glog
that for all
l'
3 1:
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE l'
.
.I
.
Af(pJ)f(pr-J)
79
= 1'f(pr)logp
0),
J=l
and similarly: l'
r
I
1'g( P )1ogp,
j=l
that is to say: r
1'f(p ) logp.
By induction on
r , this shows that:
As Af = Ag' the function f,·,g-'·' belongs to the kernel of the homomorphism considered in Question 11(1). with Aeo:'·'. Hence
Since
=
f~,g-'"
(f~.g-~')O)
= Ae, we have A = 1.
Hence, f*g-*
= fO)g-~'(l) = 1, = g. Since
e, which shows that f
g is multiplic-
ative, f is multiplicative. SOLUTION 1-18: I:
Let us first not ice that each neE can be
written uniquely in the form:
n
= mq 2
with m/P and (q,P)
for this equality implies (n,P) I: (a):
= 1, = m.
The number of elements of E,
~
x, corresponding to
a given m is at most equal to the number of qe N such that
~
, hence at most to
of E ~ x is at most The sum
I
~. IX
I
Consequently, the number of elements
~.
m/P rm
~ is the value at P of the convolution of the
m/P 1m
q
2
~
80
CHAPTER 1: PRIME NUMBERS:
function z and the function n
+
~ Iii
This convolution is a mul-
tiplicative function whose value at each prime number p is 1
1 + I;P'
As
P
is squarefree, we have:
L ~ = IT
m/P
1m
piP
[1 + J:J ;p)
=
IT
peE
[1 + J:J ;p)
The number of elements of E ~ x therefore is at most:
The neE such that n > x are of the form m/, with m/P and q > ~ , and we have:
As:
this gives:
L
neE
~~
1
L
x m/P
1+~
L
~
IX m/P 1m
n>x
The last sum has already been calculated.
On the other hand, the
number of divisors of P is the number of subsets of E, i.e.
Zk.
Thus we obtain:
L
!
neE n
~
2kx + ~ IT [1 + ~) IX peE ;p)
n>x
I:(b):
The function X is multiplicative, for if (n 1 ,n 2 )
1,
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
81
n 1n 2 is square-free and prime to P if and only if this is true for n 1 and n 2 . The function h is multiplicative, being the convolution of two multiplicative functions. fices to determine h(pr). r
L
h(pr)
In order to determine it, it suf-
We have:
~(pj)x(pr-j) = x(pr) _ x(pr-1),
j=O
which gives: h(pr)
for r
0
1
-1
h(p)
=
>
2,
i f peE, i f p4E,
0
h(p2)
and
j
0
i f peE,
-1
i f P 4 E.
We see that: +00
L
r=1
ih({)i pr
{
~2
i f peE, i f P 4 E,
P
so that:
L
p,r
ih(Z{)i
p
< +00.
r~1
Consequently the series
+00
L
n=1
h(n) . n
--lS
absolutely convergent, and
we have:
y
n=1
h<;:) =
IT p
[1 +
Y
r=1
h«()]
[ITp [1 - p12]) [IT
peE
[IT
peE
IT
[1 - !))[ (1 - 12]) p p4E p
(1 + !r1] p
6
"2 1l
IT
peE
1 (1 + p-1r
CHAPTER 1: PRIME NUMBERS:
82
On the other hand, h(n) is non-zero if and only if the decomposition of n into prime factors does not contain numbers of E with exponents greater than one, nor numbers not belonging to E with exponents
+ 2,
in other words if and only if it is of the
form mq2, with m/P and q square-free and prime to P. more, we see that when h(n)
+ 0,
Ih(n)1
= 1.
Notice that the set of n's such that h(n)
+ °is
Furthera subset of
E. As h
I:(c):
X"'lJ, we have X
h*z, and consequently, for
all x > 0:
L
h(n) [~1
x(n)
n~x
Replacing
n~x
h(n)[~]
absol~te value equal Ih(n)l. Hence, replacing L h(n)[~J byx L h~n), n.;,x n..x in absolute value by at most L Ih(n)l, hence at most by of elements of E ~ x, and con~:C;uent ly by IX IT (1 + ~J
at most to one is off the number
by h(n)
~
. the error has
peE
h( ) ~ by
L
Next, replacing x
+00
L
h(n)
--n--' we introduce a new n.;,x n=l h error with absolute value at most equal to x L ~) ,hence
I
I
n>x
tox to
L
n
n>x
2k
+ IX
IT
peE
(1 + ~)p
n~x
x(n) -
x
L
+00 h( )
n=l
n>x
.
Finally, observe that:
IL
L
1, hence neE n
, and consequently at most equal to
I 2k+ 2& IT (1 ) 1 + rv'J .
~ ~
peE
p
Ip
83
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
Using the value of
L
n=l
h(n) , this yields: n
IT
[1 + !) -11
L 1n,x
x(n) - - ;
II:
If m is square-free, Q (x) is clearly O.
x
11
peE
p
rn
As
0,
y(m)
we have: Qrn (x) - - ; y (m )x
0•
11
Let us assume, therefore, that m has no square factor.
Then,
the square-free numbers that are divisible by m are the numbers of the form mn, where n is square-free and prime to m.
The num-
ber of those which are at most equal to x is therefore equal to the number of square-free n's such that n ~ ~ and (n,m) = 1. m Taking the set of prime divisors of m for E, Question ICc) gives:
or, since m
If x to:
But:
~
m one has
~~
1, and the second member is equal at most
CHAPTER 1: PRIME NUMBERS:
84
2 v (m)
2
-1m- I- T p/m Ii ' 2 and this is at most equal to 212 I'r ' as rp 2
212
2
Therefore if x
and 3, and /'['/3 = /3 .
~
IQm(x) - - ; y(m)xl
<
1 for all p except 2
~
m:
216&.
1t
If x
<
= 0,
m one has Qm(x)
IQm(X) -
6
2"
62 Y(m)xl
1t
since y(m) x
m
rm~iii
6 x
2" iii
x
< m
1t
But:
-.
m
IX 2- ~
y(m)x ~
1t
1
=r;;
and consequently:
IX.
<
Finally, we see that for any m and any x
~
IQm(X) - - ; y(m)xl
>
0:
216IX.
1t
SOLUTION 1-19: (1):
If n
=1
it is clear that there is the single 2'
solution m
=
q = 1.
If n
=
IT
J=l
cx. Pj]' where Pl'P 2 " .. 'Pr are dis-
tinct prime numbers and cx 1 ,cx 2 , ... ,cx r ~ 1 are integers, then m and q, divisors of n, must have the form: m
r
IT
J=l
13.
p.] ]
and
r
q
y.
IT p.], j=l ]
wi th 13., y. ~ O. ]
]
With m and q given by these formulae, then n = m2q, with q squarefree if and only if, for each j, 213. + y. = cx and y = 0 or 1. ] ] Therefore, there is an unique solution, obtained by taking 13. and ] y. equal respectively to the quotient and to the remainder when ]
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
85
a. is divided by 2. ]
(2):
We see that "d 2 /n" is equivalent to "d/m".
if n = 1, the only possible d is d = 1, and m = 1. r
IT
a. o. p.], d must be of the form ITp·] , with
Indeed,
If
o. ~ O. But J=l ] J=l ] ] the square of this number divides n if and only if 20. ~ a., that
n =
is to say
o.]
~
S. (as indicated above). ]
~(d)= L ~(d) f 1 =
dim
(3):
]
if m = 1,
1
ifm> 1.
0
But it is clear that m
]
This says that:
1 if and only if n is square-free.
By the preceding, we have:
Q(x:k,f)
n:::;:x n=f(modk)
Since, for d 2 1n with n ~ x, it is necessary that d 2 ~ x, this only occurs for d's at most equal to
IX.
On interchanging the order of summation, we obtain:
Q(x:k,f)
Since "d 2 /n" is equivalent to "n = md 2" with m 6]N1" 1 = n~x
n=f(modk) d 2 /n
L
m:::;:x/d 2 md 2 =f(modk)
1.
If (d 2 ,k)%f, the congruence md 2 and the sum is zero.
then:
= f(modk)
has no solution in m,
Hence one has only to consider the d's for
CHAPTER 1: PRIME NUMBERS:
86 2
which (d ,k)/!.
For such a d the congruence has exactly one sol-
ution in each sequence of ly for any X
>
0: 1
L
k
(d 2 ,k)
2
=
consecutive integers.
(d k,k) X + p (X) ,
m~X
Consequent-
with Ip(X)1 ~ 1,
md 2 :=:!(modk) since, if:
the sum is
~
r and
2
(d ,k) X and k
~
r + 1, and consequently the two numbers:
L
m('X
1
md2 :=:!(modk) belong to the interval [r,r + 1], and so differ by at most one. Hence:
Q(x:k,!)
x
=7< 2
~ ~ th e ser1es . L Il(d)(d ,k) is absolS1'nce ill(d)(d2,k)i 2 '" 2 ' (d2,k)/! d2 d d utely convergent, and we can write:
with:
87
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
A
k t
=!
k
2
\'
L
].l(d)(d ,k)
(d2,k)/i
d
2
and:
~].l_(d_)_(d-=-2.....:;,_k-,-)
x
- k
+
d2
We have:
f+oo
1
1
~ k ( x + IX t 2 dt
)
and:
and consequently: IRk t (x)
,
I
~
2;;
+ 1.
The formula which defines Ak t can be written:
,
+00
j
2
A =! L h(d) ].l(d)(d ,k) k,t k d=l d2
wh,", hid) •
It is immediately verified that the function d function h are multiplicative.
Thus:
~
1
2 i f (d ,k)/i,
0
2 i f (d ,k Vi. 2
(d ,k) and the
88
CHAPTER 1: PRIME NUMBERS:
where the function g, defined by: 2 g(d) = h(d) j.l(d)(d ,k) d2
is mUltiplicative.
As the series is absolutely convergent, its
sum is equal to:
r
We see that g(p)
1
0 for r
=
and:
if (p2,k)lf.
0
g ( p)
> 1,
-'-'(p,-2-<,__ k.:..)
2
i f (p ,k)/f.
-
p
Consequently: Ak 9- =
,
I
IT
(p
2
,k)/f
If plk, (p2,k) = 1, and therefore (p2,k)/f.
(p2,k) 2
(p ,k)
= =
If p/k and p2/k,
p and therefore (p2,k)/f provided that p/f.
p
2
and (p
2
~
,k)~f,
since (k,f) is square-free.
If p2/k, There-
fore:
II
(p
2
,k)/f
(IT p
h - p JJ[IT p/k 121
1 -
II
12r1J[ p p/(k,f)
Hk
p A
[1 - ~J) P
=
(Contd)
ARITHMETIC FUNCTIONS: SELBER'S SIEVE
89
(Contd)
By separating, amongst the p which divide k, those which do not divide t and those which divide (k,t), we obtain:
IT
(1_(p2 2k »)
(p2,k)/t
p
r
= 1162 (pIA
1
(1 - p12
pIt
)(p/N,t) p2/k
SOLUTION 1,20:(1):
The values that (m,n) assumes for l
the divisors of n.
If d is a divisor of nand n = dn', the number
of m's such that 1
~
m
=d
the condition (m,n)
~
nand (m,n)
=d
is equal to
is equivalent to m
= dm'
Thus by grouping together the m's such that 1
~
~(n'),
for
= 1.
with (m',n') m
~
n according
to the values of (m,n), we see that: n =
,
L ~ (~I
J
din
which says that i = z*~. The First Mtlbius Inversion Formula then shows that Since the functions
~
and i are multiplicative, so
~
= ~*i. Then,
is~.
~ is also multiplicative, and as it takes the value 1 - ~ for 'Z-
p
r
n = p , we have:
~<:) (2):
L n~x
=
~~~~
As ~(n)
~
=
IT (1 - .!.I pJ
pin
~*i,
for x
~
1 we have: where I(X)
L n~x
n.
(1)
90
CHAPTER 1: PRIME NUMBERS:
But for X ~(X
1, I(X)
~
- l)X
I(X)
~
Therefore, for X
I(X)
=
~
![X]([X] + 1), and consequently: ~ ~X(X
1 we have:
~X2 + R(X),
=
+ 1).
with IR(X)I ~ ~X.
Thus Equation (1) gives: Cjl (n)
x = - 2
2
n~x
=
L -(n) -+ L 2
].l (n)R
L
L
].l
n
n~x
+00
2 2
X
n~x
].l(~)
n=l
n
-
2 2
X
n>x
x (- )
n
].l(~) + n
L
].l(n)R
( )
~
n~x
Since:
IL
].l(~) I ~ L ~ =
n>x
n
n>x n
and:
~ ~
L ~ = O(x1ogx), n~x
and since +00 \'
L
n=l
].l
2 (n) = ~
n
2
6'
we certainly have the result indicated in the problem. (3) :
As "dim and din" is equivalent to "dl (m ,n)", we
have:
L
dim din
(4):
].l(d)
L
d/(m,n)
i f (m,n) = 1, ].l(d)
i f (m,n) > 1.
This allows us to write for x,Y
~
1:
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
91
L
L
].l(d)
d~inf(x,y)
1
m~x n~y
dim din ].l(d) [~]
L
d",inf(x,y) But if X,Y
~
[~]
.
1, we have:
XY - X - Y
+
1
= (X - l)(Y -
1) ~
[X] [Y]
~
XY,
and consequently:
= XY - R1 (X,Y)
[X] [Y]
with 0
~
R1 (X,Y)
Thus, by setting inf(x,y) = u, for x,y
-- xy
\,].l (d)
L d~u
d
-
2
\' L d~u
d>u
n
d
X + Y.
lone obtains:
].l(d)R 1 (~u.] d ' d
].l(~)
= -;. xy - xy L
~
<
-
L ].l(d)R1(~'~]
d~u
,
whence:
.1:.... 1xy
4>(x,y) - ...§...21 n
~
~ ~ <
1
u.
].l(d) 1 + .1:....1 L ].l(d)R [~ , d ) 1 d2 xy d~u 1 d
L
d>u
L
~ +xly
L
~+(~+~)Y
d>u d
d>u d
L ~t
L
d~u
[~+~) L
d~u
~
& L ~ = O[l:gu)
d>u d d~u The result of (2) can be deduced from this by noticing that for x ~ 1: ~(x,x)
2
L n~x
~(n) - 1.
92
CHAPTER 1: PRIME NUMBERS:
(In order to count the pairs [m,n] such that m
x, n ~ x and (m,n) = 1, we can take those for which m :>; n, then those for which n :>; m. Thus we have counted the pair [1,1] twice, which is the only one for which m = n). (5): (a):
N(n,t) =
For all n eJN1, and t
L (L
m:>;tn dim din
)J(d»)
L
din
>
~
0:
)J (d)
L
m:>;tn dim
1
L
din
)J(d)
[t;] .
In particular, for all n eJN",:
N(n,l) = ~(n)
L
din
)J(d)
J
Consequently, for all n eJN1, and all t > 0:
and i f n
>
1:
L )J(d) = O. As IB(x)1 :>; ~ for all din IN(n,t) - t~(n)1 ~ ~ L 1)J(d)l. din
since then
x,
this gives:
But the number of square-free divisors of n is equal to the number of subsets of the set of prime divisors of n, that is to say 2v (n). Hence, for all n > 1 and all t > 0 we have:
(5):(b):
By the preceding, for all n
>
1 and all t
>
0:
93
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
Icp(n) 1
N(n, t) - t
I~
2 v (n)-1 cp(n) •
In order to establish the desired result, it suffices to show 2v (n)n 1 - E that for all 0 < E < 1 the expression cp(n) is bounded for n > 1.
We see that g
Let g (n) E
2p.r(1-E) r r-1
p
E
is multiplicative and:
-rE
~ 1
- p
1 -
Consequently, for all n
~
p
> 1:
-E
IT~ pin 1 - P
-E
But when p
~l
-+ +00,
-+
O.
Therefore, there are only a finite
1 - -
P
-E
~l
number of p'S for which
p'S, then for all n g (n) E
~
IT
peE
E
1 _ _
If E is the set of these
1.
E
P
> 1:
2p
>
-E
--1 1 - -
P
SOLUTION 1·21: (1): position of n, with q1
~
q2
i
~
j, which implies:
2.3 . . . . . p.
~
n.
... , p.
1
N. ]
~
q. for 1 1
~
]
Since, by hypothesis, n
<
~
•••
~
qj'
Then, 2
~
q1' 3
~
q2'
Nk and the sequence Nk is strictly
CHAPTER 1: PRIME NUMBERS:
94
increasing, we deduce from this j
~
k - 1.
As v(n)
= j,
that
yields v(n) < k. Next:
rr (1 - ~) j
q>(n) = n '1-=1
qi
and since we have: q>(Nk - 1 )
Nk _1
1
-P1
1 -
q>(Nk ) q>(Nk ) --->
Nk
k
--w;;-
we obtain: q> (n) > n
(2):
q>(Nk )
--w;;-
Let N be a number different from all Nk's.
There
exists k such that:
Since N
<
Nk , Question (1) above shows that:
The number m
=
Nk _1 gives a counter-example to the implica-
tion: m < N => q>(m) > q>(N)
m
N
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
(3):
95
We have: 'V
-y
_e__ logp k
By the Prime Number Theorem:
That gives:
That is to say:
lim
k-+«>
----~~----
Nk
e
-y
.
Now let n be an arbitrary integer, and let k be such that:
By part (1) above:
and:
+ +00,
k + +00 also, and by taking the lower limit of the
two we obtain: lim
-
n
96
CHAPTER 1: PRIME NUMBERS:
(4):
IT
The function o(n);(n) is multiplicative, and for n n Pa with a = a(p,n) ~ 1:
pin
IT
o(n)
pin
cp(n)
p a+1 - 1 p - 1
IT
n
pin
nIT
pin
1 -
~
P
[1 - ~) p
from which we have:
IT [1
o(n)cp(n)
n
2
pin
a+1 J .
- _1
p
We now have: 1
~
1
IT
1;;(2)
pin
whence, for all n
~
6
= 2" ' 11
1, we deduce:
6 o(n)cp(n) 2"~ 2 ~1.
n
11
Next, for all o(n) .
€
>
<
nloglogn "
0: 1 ~ eY + n loglogn
which shows that:
Let us now calculate:
n
€
for n large enough,
=
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
97
~1 (1 - Pir1+1J = k . ~loglog(~) [IT (1 - ~J) 10glog(N~) ·~=1 p) ,..r)
o ( lVk
~
When k + +~ the numerator tends to nominator, by Mertens' Formula:
(PITk (1 -
~)) loglog(~)
~(r
= e-Y(1
+
+ 1)
As for the de-
loglog~
lO;l-OYgNk
'V
1
109!) loglogNk
Therefore we have:
o(~)
lim-----
dr
k~ ~loglog(~)
+ 1)
This shows:
~ o(n) eY 1m -n=-lo'-'g:';"l":"'o-gn- ~ ~ (r + 1) As this is valid for all r, and as
1, we have:
lim~(r) ~
o(n) 1m nloglogn
~
SOLUTION 1·22: (1):
I n~x
a(n)1ogn
e Y.
For all x
I
>
1 we have:
a(n)1og(n)
1
= A(x)1ogx - A(1)log1 - fXA(t) dt 1
t
= A(x)logx
- fXA(t) dt. 1 t
CHAPTER 1: PRIME NUMBERS:
98 If la(n)
I
~ M for all n SlN~:, we have for all t ~ 1:
IA(t)1 ~
M[t] ~ Mt,
and from this it follows for all x
I IJXA(t) t - dt 1-
~
JXtdt = M(x
- 1)
>
1 that:
= O(x).
(2): We know that z*A = log. From this it follows that II ,,:1 og . In other words, for all n SN~: :
A
= I
A(n)
din
=
ll(d)log ~
[I
= I
ll(d))lOgn -
din
din
I
din
ll(d)logn ll(d)logd
I
din
ll(d)logd
- I
din
ll(d)1ogd,
I ll(d) = 0 for n > 1, and log 1 = O. This says that din -z*lllog, and from this it follows that lllog = -ll*A.
since A
=
(3):
I
Since lllog
= -ll*A,
we have for all x
~
ll(n)1ogn
n~x
I
because
ll(n)
n~x
(4): g(x)
Given
E >
[;J
1.
The Prime Number Theorem implies that:
= o(x)
(x -+ +co).
0, there exists an X
>
1 such that:
1:
99
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
Ig (X) I
~
for x
EX
X.
>
On the other hand, there exists K
I
Ig(x)
for I ~ x ~
~ K
X.
= sup(~(X),[X]).
One can, for example, take K For x
>
X we can write:
~
EX
I
1
-+ K x n x
n~X
£
and
From that it follows that: lim
x-++oo
I
since As
+1 I
X
ogx
n~x
1
)J(n)g(~) ~ n
~ ~ log,x when x
+
I
1
~
Jtn~x
Ig(~J 1~ ~ for n ~ I '
because
0 such that:
>
Ig(~JI~~
EX
1 I n + Kx,
n~x
K for
x
X<
£,
+00.
n~x
£
is arbitrarily small, we have:
lim
+1 I
x-++oo x og,x
n~x
)J(n)g(~) n
1 o.
In other words: o(xlogx)
when x
+
+00.
The relation established in part (3) above then gives:
I
n::;x
)J(n)logn
o(xlogx).
n
~
x.
CHAPTER 1: PRIME NUMBERS:
100
However, by the result of Question (1) above, we have:
L
]J(n)logn
= M(x)1ogx +
= M(x)1ogx +
O(x)
o(xlogx).
n~x
Hence we have: M(x )1ogx
0
(xlogx),
whence: M(x)
o(x).
SOLUTION 1 .23 : (1) :
2
]J (n) . . Th e f unct10n ~ 1S mul tiplicative.
In
order to study the convergence look at the double sum:
L
]J
2( r)
P
p,r cp ( pr) p ro for r ~ 2, this sum reduces to:
=0
As ]J2(pr)
and it is convergent for all positive real o. The abscissa of convergence is therefore less than or equal to zero. For s
=
L (p = 1)
00
0 the series
diverges.
L
2
]J (~)) diverges, since the sub-series n=l cp
The abscissa of convergence of F(s) is there-
p
fore exactly zero. For Res> 0 we have the expansion of F(s) into an Euler pro-
duct:
F(s)
]J 2 (n) n=1 cp(n)n s
L
IT p
[1 +
(p
= 11 p
[1
11 )psJ
+
2 ]J (p) + ••• + J.l 2( Pr) + ••• J cp(pr)prs ql(p)ps
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
(2) : 1; (s)
101
Since:
IT _1-'1~' P
1 -
8
P
it follows from this that for Res
>
0 we have:
F(s)
1
+ -----,(p _ l) p 8+1
Let us show that this product converges uniformly on K.
This
is the same as showing that the series:
~ [(p - 1 8+1 - (p l)p
p
converges normally on K.
1 28+1)
- l)p
This series is extracted from the ser-
ies:
Since Res is minorised by
0
0 >
-~
on K, we have the inequali-
ties:
and: / (n _
~)n2S+1/ '" 1(n - 1200+11, 1)
and this latter series is normally convergent on K. have:
Now, we
102
CHAPTER 1: PRIME NUMBERS:
G( 0)
= IT P
»)
(1 + (p _1 1 )p - ( P -1 1 P
= 1.
Since the infinite product: G( s)
= IT
[1 +
1 - ___ 1--;:;---:-:;-) (p _ l)p8+1 (p _ l)p28+1
converges normally on every compact set K contained in the region Res >
-~,
its logarithmic derivative is given, for Res >
G'(s) _
[...2.!E - l)p28+1 28+1
~
'"G("8') - L
+ p8 ] 8 - 2 logp +p - 1
P (p - l)p
[
= L~
2 -
P (p - l)p
In particular, for s
88 ) logp.
P
28+1
+p
- 1
= 0:
2
2
logp G(O) loge p p(p - 1) - p p(p - 1)
G'(O)
For Res> 0 the Dirichlet series:
(3) :
F(s)
~,
=
2
co
n=l
2
~ (n) ~
~ n8
and
z;;(s
1
+
1)
I
= n= 1
~(n)
-8-
nn
are absolutely convergent, and their product has the value: G(s) =
~L as.!:l 8'
n=l
n
2
withg=L*~. rp 'l-
We have: _
g(p) -
and:
1
p(p - 1 ) '
2
g(p)
1
p(p - 1) ,
by:
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE k
o
g(p ) For real s
3.
~
-! we have:
>
1Q:(n) I 8
L
n
n~x
for k
103
IT
~
~ IT P
1 + (p - 1)p8+1 (p _ 1\p28+1)
[1 +
p~x
[1 +
1 ) < +00 1 8+1 + ' (p _ 1)p28+1 (p - l)p
which shows that the abscissa of absolute convergence of the series
L~ n
8
is less than or equal to -!.
The series
L Q:(~:)
p
, for s
p
= - L has the value
L (-~)
p
p
and
diverges; therefore the abscissa of convergence of the series
L
p
Q:(n)
n
8
(4):
isexactly-!.
Since g
2
Lcp ..... J:!.. i ' we have:
which yields:
Next we have:
L
n~x
g(n)(logx +
(logx + y)
L
n=l
y) -
g(n) -
L
n~x
L
n=l
g(n)log(n) +
L
g(n)
~
n (Contd) g(n)log(n) + Rl + R2 + R3 = n~x
104
CHAPTER 1: PRIME NUMBERS:
(logx + y)G(O) + G'(O) + Rl + R2 + R3 ,
(Contd) with:
Rl
=-
R2
= I
+ logx)
(y
n>x
I
n>x
g(n),
g(n)log(n),
Let c be a number such that
I n>x
-!
<
c
<
O.
Then:
Ig(n) I c c ~ Ig(n) I n 'x L c n>x n n>x n c
= I
Ig(n)1
Similarly:
IR21
~ I
n>x
Ig(n) Ilogn =
I Ig(~) I (logn)n
n>x
C
n
finally: n
for the series
I
Ig(n)1 is convergent. c n
2
I ~ In» = logx n~x n IjJ
+ y -
I P(~o~l)
p
Therefore we have: + 0(1).
l+c
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
SOLUTION 1-24: (1):
105
With z being fixed, the arithmetic function
f defined by:
is multiplicative, since n and v are additive, and for p prime and r
1 integral we have:
~
f(pr) If
z
r-l
Iz 1
< 2 the double series:
I
[({)
, that is to say,
ro p
p,r
r~l
r~l
is convergent for +00
I
r=l
1 Ir - 1
I
n=l
° > 1,
since, for each p, the series 1 1
, a geometric series with ratio ~ a
z ro p
with sum +00
Izlr-l ro p,r p
I
1
pO _ Iz
, and the series 1
I
1
P
~ pO - Izl
< 1,
is convergent
is convergent.
Consequently, for Res> 1 and IZI < 2, the series n(n)-v(n) z s is absolutely convergent and has as sum the value n
of the absolutely convergent infinite product: +00
r-l)
[1 + I - z 11 rs p r=l p
that is to say,
But we have:
1
1
+-~-
p
s
- z
p
s
+ p
s
1 -
- z
z
V[
1 + ps 1
J.
CHAPTER 1: PRIME NUMBERS:
106
It is easily seen that the infinite products
IT
[1 -+) , p
are absolutely convergent.
(pS ZZ+
1 -pS
The first two are, moreover, equal
~(~s) and ~(s).
respectively to
1- - l)J
IT
and
Hence for Res> 1 and Izi
<
2
we have: +00
L
z
lHn)-v(n)
n=l
n
~(s)
~ (2s)
S
IT
z
1 -
S
(E. + 1 -
1
(1)
z
-S
J
P
(2):
We see that the infinite product
is uniformly convergent on every compact set of
IT [1-1 (E._ ~
2
the set determined by pairs (s ,z) such that Res > ~ and R <
12 such that for
ZZ+s
l)J
p
contained in
IZ I
12.
<
In fact, if K is such a compact set there exists GO > satisfying 0 <
S
~
and
(s,z) eK, Res ~ GO' and Izl "'-R.
Then for (s,z) e K: 1 -
z S
(E. + 1 -
z
-S
1)
z
I(ps + l)(ps
- 1
p
(p
GO
Go
Go
G
_ l)(p 0 - R)
L ______R_ _ _ _
p
I
R
"'-
But the series
- z)
is convergent.
(p - l)(p - R) If F(s,z) is the value of the infinite product, the function
F is ho1omorphic in sand z for Res > ~ and factor is ho1omorphic in this domain.
Izl <
12, because each
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
(3):
For Res>
F(s,z)
+00
= L
q=O
!
and Iz I
<
107
12 we can write:
A (s)zq
(2)
q
where the functions A q are holomorphic in the half-plane Res >
=1
For s
+00
L
= IT
Moreover the equality in fact holds for Izl
<
2, for it is easily
seen that the infinite product converges uniformly on every compact set contained in the disc Izl < 2, and therefore represents a function holomorphic in this disc. Notice that the numbers A
q
(1)
are all real and posi-
In fact, for each p, for Izl
tive.
From this it follows, for Izl
IT
[
.
this gives:
A (l)zq q
q=O
21
<
2:
2, that:
<
J
_+z 1 - p 1 1 - ~
P
exp{
Y
q=l
a zq} , q
where:
because the double series convergent.
L zq(J:.... 1 ) is absolutely p,q q pq (p + l)q)
CHAPTER 1: PRIME NUMBERS:
108
Each a
is clearly real and greater than zero.
q
Now, the coef-
ficients of the expansion as a power series of exp{
L
q=1 are polynomials, with positive coefficients, in a 1 ,a 2 ,
azq} q
They are therefore real and greater than zero. +'" r:l(n)-v(n) '\ ---------z Whenever the series L is absolutely convergent,
... ,a q , •.••
n=l
nS
we can calculate its sum by grouping together the terms for which
r:l(n) - v(n)
has the same value, which gives:
Thus equalities (1) and (2) show that for Res
> ~:
The function on the right is holomorphic in the half-plane Res
except at the point 1, which is a simple pole with A (1) 6 residue = ~ Aq(U. > ~
n-rr-
Ikehara's Theorem of Section 1.12 of the Introduction shows that for x
+"':
-+
N (x)'V q
6 2" A 11
6
If we set -2 A 11
q
q
(l)x.
q
>
=d
(1)
d x +
N (x)
Clearly d
q
q
0
q
, this may be written N (x) q
'V
d x, or: q
(x ) .
O.
On the other hand, Equation (3) shows that for
Izl
<
2:
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
= 1 we obtain:
By setting z
d
109
~ IT [1
q
1l
-
(p
11 )]
1.
1 - -
P
Let D be the set of divisors of m. m a finite set whose cardinality is d(m).
SOLUTION 1-25:
Now let n
(1) :
IT pa
=
be an element of
E
m
It is
We have:
Therefore (p - 1) e D. There are therefore only a fini te number m of possible prime factors for n. For each of these factors a-I one must have p ~ m, so:
a ~ 1 + logm ,:; 1 + logm logp ... 10g2 ' and there are only a finite number of possible exponents.
Finally,
we have: cardEm
~
10gm]d() [1 + 10g2 m.
When m is odd, p - 1 must be a divisor of m, and therefore is odd. If n e Em' one must therefore have n = 2a . For a ~ 2, qJ( 2a ) is even.
Hence we have: and (2): n
qJ(n)
E m
¢
for odd m
>
3.
We have:
=
1
IT
pin
(1 - ~)p
IT pin
1 +
l:.
p 1 _ 1
2"
p
~
(Contd)
110
CHAPTER 1: PRIME NUMBERS:
(Contd)
~
11 pin IT (1 + !) = r; (2) IT (1 +!) . pin
IT
pprime 1 - 2
p
p
p
and:
IT
pin
(1 + !)p
~ ~ ~ ~.
~ din
din
ll(d)+o
Whence:
n
.:.
WiT" with
e
~
L
din
1 (I.
2
1t e = r;(2) = (r
.
Next: e(l
+ logn).
that is to say: n WiT = O(Iogn).
Now.
A(x)
= card{n:~(n)
~
x}.
If n.,; x then ~(n).,; n" x. so we conclude
the relation ~rn) n logn
= O(logn)
A(x)
yields:
= O(x).
On taking logarithms this becomes:
logn - loglogn
= O(logx).
and since loglogn is small compared with logn:
~~].
If ~(n).,;x
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
111
= O(1ogx).
logn
We have then: n
= O(x )1og(n) = O(xlogx),
and from this it follows that
= O(xlogx).
A(x)
Let us show that the series
(3) :
convergent for Res> 1. 1
1
1 cp(n)8
There holds:
(C(l
1
= cp(n)Re8 ~
L ___1__ is absolutely n=l cp(n)8
+n logn )lRe8
J
This series is convergent for Res> 1. Since the series L ___ 1__ is absolutely convergent, all the cp(n)8 terms with the same denominator can be grouped. For each m there will be exactly a terms with denominator m8 = cp(n)8, and m
we have: co
a
L : m=l m
co
L
1
n=l cp(n)
8
This shows that the series
a
L : m
converges for Res> 1.
To decom-
pose it into an Euler product the Theorem in Section 1.10 of the Introduction is used. (4):
1(p
We have: 1 -
1) 8
-
J:...I 8 = IfP
_s P- 1 x 8+1
P
~ fP
1s1
p-1 x Re8 +1
dx
I
dx
~
_--,-I.....,sI,---=(p _ 1)Re8+1
CHAPTER 1: PRIME NUMBERS:
112
Let K be a compact set contained in the region Res show that
1
>
O.
To
is holomorphic we will show that the product defin-
ing I(s) is normally convergent on K, or, which amounts to the same thing, that the series:
I [
P
1
(p _ 1)8
is normally convergent on K.
We have:
(on K, lsi is majorised by M, and Res is minorised by 00
>
0).
We have:
~J
IT
1 --(1 + p - 1
=
IT
p + 1 p(p - 1) (p + 1)
=
IT
]-'(1) =
p
3
ITE P
2
-E+1
p(p - 1)
IT 2 p p(p
6
- 1 3 - 1) (p - 1)
P
Since:
1;(k)
p 1
1
1
P
k
=
IT P P
pk k _ 1
we have: 1;(3)1;(2) 1;(6)
1.94······· .
(5):
Ikehara's Theorem, of Section 1.12 of the Introduction, 00 a is applied to the Dirichlet series I ~ = 1;(s){(s). The funcm=l m8
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
tion
~(s)j(s)
pole of
~(s)
is meromorphic for Res> 0 with a single pole, the for s = 1, whose residue is j(l).
L
A(x)
a
m:rox
m
~ j(l)x
1-26: (1):
SOLUTION
113
= ~(3)~(2)
We then have:
x
~(6)'
If A' C A every partition whose summands are
in A' is a partition whose summands are in A.
Hence:
and: P(A') => P(A).
(2):
n
By Bezout' s Theoremt there exists
x' ,y' a71
such that:
= x'a + y'b.
Let us assume x'
<
O.
o " x' + Ab
<
b
There exists a unique AalP', such that:
Then: n
= (x'
~
O. n
= xa
+ yb,
= x'
+ Ab, y y' - Aa. By the choice of A, we have We also have y > 0, for if y were negative, we would have:
on setting x x
+ Ab)a + (y' - Aa)b
= xa
+ yb
<
xa < ab,
which is contrary to the hypothesis.
= x'a
+ y'b with x' ~ 0 and y' < 0, one carries out a similar argument by constructing y y' + ~a. For A = {a,b} the number of partitions of n whose summands are If Bezout's Theorem gives n
t (Editor's Note): Bezout's Theorem is the French name for the proposition that the g.c.d. of a and b is a linear combination of a and b.
CHAPTER 1: PRIME NUMBERS:
114
in A is the number of solutions (x,y) of n
= xa + yb
with x
~
0
ab one has PA(n) > 0 and P(A) holds. If A contains two relatively prime elements a and b, we apply Question (1) above with A' = {a,b}, and we see that P(A) is true. and y
~
For n
O.
(3):
~
Let us decompose into partial fractions: 1
where
E~(x)
For
=
~
is the sum of those corresponding to the pole
~.
1, which is a pole of order r, we have:
and Ar
with:
A + (n + l)A
g(n)
1
2
+ ••• + (n
+
r - 1) (n + (r - 1) I
1) A
r'
We see that g(n) is a polynomial whose term of highest degree is:
Now let
~
be a pole different from 1.
It is a root of unity,
and let us assume that it is a primitive n-th root of unity. will be a pole of (1 - x pole
~
a.
1)
if n divides a.
~
The order of the
will therefore be the number s of elements of A that are
multiples of n.
These s elements are not relatively prime. There-
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
fore we must have s
~
115
q.
We have: E (x) 1;
IY. s _ x)s
(1;
The expansion of
E1;(x)
IY.s_1
+
( 1; - x)
8-1
+
...
+
1Y.1 (1; - x)
as a power series will be:
E1;(X)
I
with: h (n) 1;
1[1Y.1 (n+1)a 2 (n+l)(n+2) (n+s-l) - -+ + ••• + a 1; n n+1 (1)n+8-1 8 1; 1; s -
As 11; I = 1,
holds.
Finally, we have: g(n)
(4):
+ D(n
q-1
).
We have the following equivalences:
Not P 2 (A)
there exists a prime number p such that for all a eA
<=> pia
<=>
there exists a prime number p such that pl(g.c.d. of the elements of A)
<=>
the g.c.d. of the elements of A is greater than (])tle
<=>
Not P1 (A).
This shows: PleA)
P2 (A). In order to prove peA) => PleA) let us show that: <=>
.
116
CHAPTER 1: PRIME NUMBERS:
Not P1 (A) =9 Not P(A). Now, we have: P1(A) => the g.c.d. of the elements of A is d > 1.
If n is not a multiple of d it is impossible to write n as the sum of elements of A, and PA(n) =
o.
Property P(A) is therefore
not satisfied. It is now necessary to show that for a finite set A P1(A) <=> P(A). PA(n)
=
We use (3) above:
g(n) + O(n
q-l
If P1(A) is true, then q
~
{a • "a } 1 r
).
r - 1, and:
n
r-l
which shows that PA(n) is positive for n large enough.
The Property
P(A) is therefore satisfied.
Let aoeA and let P1 ,P 2 , ... ,P k be prime factors of a O. Since P 2 (A) is true, for every prime number P there exists a e A (5):
In particular, there exists a 1 , . .. ,a k such that Pi does not divide a 1 . (It may be that certain of the numbers such that
p~a.
a i are the same).
We set A' = {aO,a 1 ,· .. ,a k }. Property P2 (A') is satisfied. In fact let P be a prime number: If P f Pi for 1 ~ i ~ k, there exists a P does not divide a;
aO e A such that
If P = Pi' there exists a = ai such that P does not divide a. Let A be an infinite set
of~*
satisfying P1(A).
It also
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
satisfies P2 (A).
117
We construct a finite subset A' C A satisfying
P2 (A'). As A' is finite, peA') is true and peA) is true by Question (1) above. Thus P1 (A) => peA) has been shown, and as it was already known that peA) => P1 (A), we conclude that peA) <=> P1 (A). When Q is the set of prime numbers less than
SOLUTION 1-27: (a):
or equal to N, denoted QN' let us define EN by the relation:
EN
IT
p~N
(1 _~)-1 P
By Mertens' Theorem (cf., the Introduction) and a classical asymptotic estimate, we have:
From this it follows that: e- Y
>
0,
and the sequence EN has a lower bound E
O.
>
We will show that
E answers our Question. From this point on, N is to be a fixed integer greater than two.
Relation (*):
I
meA(N,Q)
1
(*)
- :> m ...
of course holds when Q
= QN'
Let us assume that it holds for a
non-empty set Q of prime numbers less than N; let q be an element of Q; we will show that Relation (*) still holds for Q' From the relations:
I
1
meA(N,Q') m
I
1
meA(N,Q) m
I
1
meA(N,Q) m qlm
= Q -{q}.
CHAPTER 1: PRIME NUMBERS:
118
L
L
!.<
L
2.=!
meA(N,Q) m ' m'eA(N,Q) qm qlm
!
q meA(N,Q) m
we deduce:
L
L
!~
meA(N,Q) m
meA(N,Q)
~.
By now using the induction hypothesis (*) we obtain:
!~B(l-!) IT
L
meA(N,Q') m
q
peQ
(1_!)-1 =B peQ' IT (1 __p1)-1. p
REMARK: This result gives a good lower bound, to the extent that one has the trivial (or almost trivial) upper bound:
1
L
-~lim
meA(N,Q) m (b):
N+oo
1
L
-=
meA(N,Q) m
IT
peQ
(1)-1 1--
p
.
Let us start by showing the first inequality.
be an integer less than or equal to z.
If one writes m2 T
IT
t=l whence:
We also have:
T
IT
t=l
q
Bt t
(B t + 1)
We can write:
we have:
and
T
w(m)
IT
t=l
2
Bt
,
Let m
119
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE 1
R
I
m::;,z
I
~ m
1
l::;,;IZ
lim
: ;, I ; m::;'z
I m~z
I
1
lim
(1.'P 1 )=1 _w<_m_)
m
Let us now prove the second inequality.
Because IZ? 2 there
exists· an absolute constant C1 such that:
By Exercise l·27(a) we have:
I
l::;,;IZ
1
T ~
B
TT
p~P 2
[11-1 1 - -)
P
B
IT
peP1
[1 - ~1
pJ pJJz
[1 __p11-1
By Mertens' Theorem (cf., Introduction) there exists a constant C 2 such that:
The required lower bound is clearly obtained (with C
BC1 C2 ).
J
120
CHAPTER 1: PRIME NUMBERS:
Here it will be assumed that m is positive. With the
SOLUTION 1- 28:
notations of Section 1.13 of the Introduction, let us choose:
n
= [x],
A
= {1,2, ... ,n},
If plm, set w(p)
=1
z
and R(P) 1
= n1/3 ,
P
= {p
::> z}.
{iii::: O[pJ}; {i Ii::: 0 [p J} ;
If pJm, set w(p)
= {iii:::
-m[p]}.
In order to verify relation (H) of Section 1.13 of the Introduction, we will verify that:
Let us now show the majorisation: card{plp
x, p + m a prime number}
::>
::>
S(A,P) + z.
(*)
Apart from the prime numbers of the first set which satisfy p (there are at most z) all the other pIS satisfy z
<
p
~
~ z
n; for
every element p. e P we have: l
p
l O[p.] l
and
p + m
l O[p.], l
which certainly shows that p is counted in S(A,P). From Selberg's Theorem and the upper bound (*) we deduce the upper bound: card{plp
::>
x, p + m a prime number}
From Exercise 1-27 we have:
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
121
It is easy to see that:
IT
IT
2
2
IT
2
The second product is majorised by the convergent product:
and the product
IT
~)
[1 -4- is maj orised (because of Hertens' 2
which is the first sought. The second is obtained by noticing that the product:
is majorised by the convergent infinite product
IT
1
1 P 1 - 2" p
SOLUTION 1·29:
By Exercise 1·28 above there exists an absolute
constant C such that: card I
x
~
x
C --2- ,
log x
CHAPTER 1: PRIME NUMBERS:
122
setting: I
{pip ~ x, p + 2 a prime number}.
x
By the formula of Section 1.2 of the Introduction:
L
pel
x
1 C -~ --2-+ log x
IX
p
c2
2 t10g t
dt = 0(1).
(This formulation is due to Viggo Brun (1919) who deduced it from the more coarse upper bound: cardI
x
~
c x(1og1ogx) 2 2
log x
which he had obtained by the first version of his sieve). SOLUTION 1·30: (a): r(N)
~
If N is odd or i f N<
7 it is clear that
2, and the upper bound stated is certainly satisfied.
If N is even and N > 7 we use Exercise 1·28 with x -N, from which we deduce: r(N) .. E
(b):
-+ II (1 logNplN
L
o
~J
pJ
< E _N_
L !
loiN diN d
From the preceding Question (a) we deduce: ~
Thus:
+
=
2
F
x
2
log4 x O
Nand m
123
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
From this follows:
L
( Y.
1)
2
O
whence the upper bound sought. (c):
L
We have: r(N)
=
N~
L
~
L Pl~
(d):
1
N~x
1I(~x) ~
x/2
2 11
2
(~x) ~ F" ~ i f x log x
~
4.
By the Cauchy-Schwarz Inequality we have:
With the help of the upper and lower bounds given in Questions (b) and (c) above, we have: M(x)
ex
if x
>
4.
124
CHAPTER 1: PRIME NUMBERS:
(e):
Let S be the sequence formed of 0,1, and the prime By Question (d) above the sequence S has a positive
numbers.
Snirelman density (greater than or equal to C because it contains 0,1,2,3, and 4); therefore the sequence S is an additive basis, and it follows that S is also a basis (cf., Chapter 2). REMARK: It was to obtain this result that Snirelman introduced the notion of density in 1930.
SOLUTION 1-31:
Let us set E
(a):
= {ilpi+1
let io be the largest element of E; we have:
L
d. ~
ieE
J.
d. 3 Alogx J.
L
ieE
~ x,d i 3 Alogx} and
= AlogxcardE.
1
Therefore we have: x cardE ~ A10gx By the definition of M(x):
(b):
L
M(x)
card{il~x < p. ~ x and J.
a10gx
d.J.
m.
If d. = m, the numbers p. and p. + m are prime numbers, and by J.
J.
J.
Exercise 1-28: card {i
I~x
<
pi
~
x and d. J.
= m} ~
IT
E _x_ 10g2x plm
(1 +!)p ,
and so the desired upper bound is deduced. (c):
IT
plm where
~
We have:
[1 + !)p
=
denotes the
I L I~(d) d
dTm
M~bius
function.
From this we deduce:
125
ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
I
[1 + ~)
II
alogx
p
=
I
I
alogx
I
I
d~Alogx
1
alogx
m=O(modd)
I
11-I~) I
(A - a)logx
+
o[ I
d~logx
(A -
~)
d~Alogx
1-1 (d) )
. \ a) [ L d=l
d 2
(1 + o(l)logx.
From this latter estimate and from the result of Question (b) above, the desired upper bound certainly follows: (One can take
=F
F'
I
d=l
(d): i
~
n(x)
M411). 2 d
Let us assume for fixed a that d. l
(i.e., p.
~
l
x)
> alogx
for all sufficiently large x.
whenever For
large enough, then:
I
sex)
alogx
d.l
I
+
d.>Alogx
d.l
for
n(~x) <
i
~ n(x).
l
The number of terms in the first sum is M(x), the number in the second is therefore: n (x) - n (~x) - M(x).
Hence: sex) ~ (n(x) - n(~x»Alogx - M(x)(A - a)logx.
By the Prime Number Theorem and the preceding Question (c) we obtain: · . f s(x) 1 lm In -;x;-+oo
x
~
126
CHAPTER 1: PRIME NUMBERS: ARITHMETIC FUNCTIONS: SELBERG'S SIEVE
(e): Let us choose A = 1 + t, a = 1 - t, with t = l6~> . We obtain:
M-
F' (A - a)2
; + ;t - F'4. t .
16~' = ; + ~t.
By the Prime Number Theorem:
L
• ( ) 'Z-:;:1I x
d.=p() l
11
x +1
-2'Ux,
from which one deduces Sex) lim inf Sex) ::: x--roo x
M-
'U
lx, which contradicts:
F'(A - a)2
The hypothesis of Question (d) above is therefore false.
There-
fore, there exist arbitrarily large x' s for which:
d. ::; alogx l
for at least one i satisfying lx d.
_ _l _ ~
logp i
<
Pi
~
d.l . . . ; a 1 ogx log~x lodx'
which implies:
d.
lim inf ~ ~ a < 1. i-TOO ogp i (By this method one can obtain a
15
16 ).
x.
For this i we have:
CHAPTER 2
Additive Theory
INTRODUCTION 2.1
DEFINITIONS
DEFINITIONS:
Let A be a strictly increasing sequence of non-
negative integers; the strictly positive elements of A arranged in increasing order will be denoted a 1 ,a 2 , ... (if 0 is an element of A it will be written a O = 0). For every positive real number X, we denote by A(X) the NUMBER OF POSITIVE ELEMENTS OF A which are LESS THAN OR EQUAL TO X, and one defines the SNIRELMAN DENSITY OF THE SEQUENCE A by the relation:
oA
= inf
Ne:N'~
(A(N») . N
If A and B are two sequences of integers, we denote by A + B the set of all the integers which are the sum of an element of A and an element of B. duction we define hA
If B = A we write 2A
= (h
= A + A, and by in-
- l)A + A.
We say that the sequence A is a BASIS if there exists an integer h such that hA
=~.
We call the ORDER OF A BASIS A the
smallest integer h such that hA
=:N.
necessarily contains 0 and 1). 127
(Notice that every basis
128
2.2
CHAPTER 2
RESULTS
THEOREM:
(~nirelman-Mann):
o(A + B)
~
If 0 belongs to both A and B then:
min(l,oA + oB).
By noticing that the only sequence containing 0 with
~nirel
man density one is the set of integers, it follows from the preceding Theorem that every sequence with strictly positive Snirel-
man density containing 0 is a basis.
2.3
BIBLIOGRAPHY The basic work in this field, rich in results and references,
and pleasant to read, is [Hal].
ADDITIVE THEORY
129
PROBLEMS
EXERCISE 2-1: (1): Show that if A is a basis there exist two pos-
itive real numbers Sand y such that for all n:
(2): Show that if a sequence A is such that:
A(n) = O(n E ) for any real number
E,
then A is not a basis. EXERCISE 2-2: Let A and
B be two increasing sequences of integers;
we denote by A(B) the SET OF ELEMENTS OF A which are INDEXED BY
B, that is to say, the elements of the form a with n n an element of B. ELEMENTS OF
Prove the inequalities: min(o(A),o(B»
~
o(A(B»
~
oA.oB.
Show by examples that these inequalities are best possible. Let f be a mapping ofN* into [0,1] such that
EXERCISE 2-3:
fen)
~
°as n
~
00
CHAPTER 2
130
Show that there exists a sequence A possessing the following properties: (i): A contains 0 and 1; (ii): For every integer N, A(N) > f(N).N; (iii): The sequence
A is
not a basis.
(We will be able to construct a sequence A satisfying (i), (ii) and the following Condition: For every integer j
I there
>
exists an integer n. such that A contains no element in the interJ
val [n.,j. n . )) . J
J
EXERCISE 2·4: Let A be a basis and
V a bounded sequence of inte-
gers ~ 0 such that do = d 1 = o. Show that the sequence B = {a. + 1.
d.11. i
e:N} is a basis.
EXERCISE 2·5: Let A be a sequence of integers containing 0 and 1,
and let a. be a positive real number; let B = {l} U {[a.a] Show that B is a basis if and only if A is.
la e A}.
Begin by estab-
lishing the relation:
EXERCISE 2·6: Show that for every pair of positive rational num-
bers a and b, a + b
<
1, one can construct two sequences A and B
such that: crA
= a,
crB
EXERCISE 2·7: (a):
b,
cr(A + B)
a + b.
Let A be a sequence of integers; assume that there exists an irrational number a. such that:
131
ADDITIVE THEORY
a.
lim{{aa })
n-+
n
Show that
(H)
A is not a basis.
(One should prove by induction on h that the density of the sequence hA is zero by using the equipartition of the
seque~ce
({an})nw' cf., [Rau]). (b):
The question as above, replacing (H) by (H'):
The set of fractional parts of the elements aa only a finite number of limit points.
has
n
(H' )
EXERCISE 2-8: Use Exercise 2-7(a) to give another proof of the result of Exercise 2-3. (An irrational number a should be chosen, and for every integer k
~
2 set: ~
2
k }.
By a diagonal process construct a sequence:
A = {a = aa
<
a1
=1
<
a2
< •••
}
satisfying Conditions 2-3(i),(ii), and such that {aa.} ~
~
0).
EXERCISE 2-9: Assume that every integer divisible by a prime number q congruent to 3 modulo 4, but not divisible by q2, is not the sum of two squares of integers ([Har], Theorem 355). Deduce from this that one can find arbitrarily long chains of consecutive integers which are not the sums of two squares; show that for no positive real number a is the sequence ([an 2 ]) a basis of order two.
CHAPTER 2
132
SOLUTIONS
SOLUTION 2 1:(1): Let us assume that 0
Every element x in the set E in the form:
= a. +
x
000
~1
+ a.
~h
A is a basis of order h.
= {O,l, ... ,a n -
with
0
~
i1
~
i2
I}
~
may be written
000
,
n - 1.
The number of such forms is the number of combinations with repetitions of the n objects a O,a 1 , ... ,a n_1 taken h by h, say
(n+~-l) . We must therefore have: cardE
= an
n+h-1) (n + h)h ~ ( h ~ h!
Now we have: a
n
h ' n
[1 + ~t hI
Since we have:
lim
n-+co
(1 + ~r 1 , hT hI
ADDITIVE THEORY
a
~ is bounded above; let B be an upper bound of this n h We have a ~ Bn .
the sequence sequence. (2):
133
n
From the above, if A is a basis there exist two posi-
tive real numbers Band y such that: a which can be written A(BnY)~n.
~ [~)
A(x)
Bn Y (n
<
n
By setting x
1,2, ... ),
= Bn Y
we obtain:
l/y ,
so -1. A(x)
1m
x
l/y
0,
>
which contradicts the hypothesis A(n)
SOLUTION 2-2: To say that an element a. say that i <
<=>
N is equivalent to
Hence:
A(ll).
N
~
~
1
h. ]
~
A(N)
<=> j
~
B(A(N))
Let us set C = A(B); hence C(N) = B(A(N)).
By the defin-
ition of the Snire1man density we have:
C(N)
=
B(A(N)) ;: A(B).aB ;: N.aA.aB.
From which we deduce:
a(A(B))
~
aA.aB.
On the other hand, A(N)
~
Nand B(M)
creasing function, we have that C(N) whence:
a(A(B))
~
min(aA,aB).
~
~
M.
Since B is an in-
B(N) and C(N)
~
A(N),
134
CHAPTER 2
EXAMPLE 1 : A
B A(B)
{1,4,S,6,7,8, ... },
oA = 3"
{1,2,4,S,6,7,8, ... },
oB
B A(B)
3
{1,2,4,S,6,7,8, ... },
oB
2 3
o(A(B) )
2 9
supf(n).
~(m) =
n;:,m
The function
min(oA,oB) .
1
oA
SOLUTION 2·3: Set bounds
1
{1,4,7,10,11,12,13, ... },
{1,4,10,11,12,13, ... },
of~*
2 3
o(A(B) )
{1,4,6, 7 ,8, ... },
EXAMPLE 2: A
mapping
1
3
~
oA,oB
is a decreasing
into [0,1], with limit zero at infinity, which
f.
By induction we define the sequence n i in the following way: nO = 0, n i (i ~ 1) is the smallest integer satisfying
ni
>
2(i - 1)n i _1 + 1 such that O:in.
J.
1
2i
One then sets:
n < n. l}. J.+
~
(i) is satisfied, as n 1
~(ni) <
1.
>
(ii) Let N be an integer; let us consider three cases: (a): 1
~
N
< n1,
A(N)
-N- = 1 ~ ~(N).
(b): There exists a positive integer i such that n.
J.
A(N)
A(n i )
-N- = -N-- >
1 A(n i )
i
-n-.J.
~
ni -
(i -
in. J.
1)n i _1
~
N
< ~n
i
ADDITIVE THEORY
135
(c): There exists a positive integer i such that in.
~
N - in.~ +
A(N)
-N-=
1 +
A(in.~ -
1)
A(in. -
1)
in. -
1 -
N
in. -
1 -
in.. II
~
> 1jJ( in.
in. -
1
A(in. -
N
<
1)
---~------~N~--~-----
A(in. -
~ 1 - --~~--~~--~~~~---
1 -
~
~
1)
1
- 1),
~
by using (a) or (b), whence: A(N)
-N--
~
1jJ(N).
(iii) Let h
The element (h + 1)nh+1 - 1 is not a sum of less than h elements of A; in fact the elements of ~
2 be an integer.
A which are less than it are bounded by n h +1 ; therefore A is not a basis. SOLUTION 2·4: Let h be the order of the basis A; for every inte-
ger N, at least one of the numbers N,N + 1, ... ,N + hd is an element of hB, where the maximum of the elements of V has been denoted by d.
a.
~1
In fact we can write N
+ d.
+ ••• + a.
~1
~h
+ d.
~h
lies between Nand N + hd.
o and
= a.
+ ••• + a. ; the element
~1
~h
is an element of hB, and it certainly Moreover, the sequence hB contains
1, and it therefore has positive density (greater than or 1
equal to hd + 1 ); it is therefore a basis, and the sequence B is also (of order h(nd + 1) at most). SOLUTION 2·5: First Step: By writing x.
o
~
{x.} ~
<
1, we have:
~
[x.] + {x.}, with ~
~
[x ] + ••• + [x ] ,::: x + x + ••• + x < [xl] + ••• + [x n] + R,. 1 R, ~ 1 2 R, ~
CHAPTER 2
136
Since [xl + ••• + xi] is the largest integer less than or equal to xl + x 2 + ••• + xi' and since [xl] + ••• + [xi] is an integer less than or equal to xl + x 2 + ••• + xi' we have:
and since both members of this relation are integers, we have:
Second Step: Let us first assume that A is a basis of order h. To every integer N we can associate an (unique) integer n such that:
[an]
~
N
<
[a(n + 1)].
One can then write:
o
~
N - [an]
~ 1
+ [a],
and: an
where the a
(i)
are elements of A.
By the inequalities proved in
the First Step, we have: ••• +
N can therefore be written as the sum of h elements [aa(i)] and
ADDITIVE THEORY
137
at most h + [a] elements equal to one.
By hypothesis, 1 e B,
hence B is a basis of order at most 2h + [a].
Third Step: Now assume that B is a basis of order h. Let N > ~ a be an integer. The integer [aN] - h is positve or zero, hence it is an element of hB; hence can be written as the sum of at most h elements [OO2(i)], i.e., there exists a real number ~ (with
o~
~ ~
1) and an integer l (with 0
[aN] - h
~
l
~
h) such that:
= [aa(l)] + ••• + [OO2(h-R,)] +
~R,
(where
a(i)
e A).
From this one deduces: 002(1) + ••• + OO2(h-R,) - h <
~
[aN] - h aa
(1)
+ ••• + aa
(h-R,)
+
h,
or again:
aa(l) + ••• + aa(h-R,)
<
aN
~
aa
(1)
+ ••• + aa
(h-R,)
+ 2h + 1,
hence one has:
N is therefore written as the sum of at most h elements of A and [ 2h ~+ ~
11]
1]
+ones; now 1 e A, hence N is the sum of at most h + [-2h- a
elements of A.
It is easily verified that the same holds when N ~ ~. A is therefore a basis of order equal at most a
138
CHAPTER 2
(The same method also leads to the upper bound: bound: sup
o~i~h
(h - i
+
[h + ! + 11) . ]
or.
h + 1 + [h :
and
SOLUTION 2·6: Set a
= p~
to saying that uq + vp
<
and b pq.
v
1] if a
to say that a + b
q Set:
A
{O.1.2 •...• uq.pq + 1.pq + 2 •... }.
aA
B
{ 0.1.2 •...• vp .pq + 1.pq + 2.... }.
aB
< 1).
< 1
reduces
u
p
and
v
q
Then:
A+ B
{O.1.2 ••..• uq + vp.pq + 1.pq + 2, ••• }
and a(A
+
B)
=~
uq + vp pq
p
+ ~ q
= aA
+ aBo
SOLUTION 2·7:(a): For every positive real number A'E =
{a.
~
e AI{aa ~.}
>
d
and
AE
E
we shall set:
CA A'. E
Notice that A'E is a finite set; let A'E be an upper bound for its cardinality.
For every integer N we have:
ADDITIVE THEORY
A(N)
139
= AE'(N) ~
+ A (N) E
~
A' + A (N) E E
A'E + card{n ~ NI{an} ~ E}
It is known that the sequence {an} is equidistributed mod 1 (cf., Exercise 5-8), and we have:
. sup --N-A(N) 11m
1 lim sup N card{n
~
N->-
~
N->-
NI{an}
~
E}
~
E.
This proves that the sequence A has a zero Snirelman density. A useful convention is to call the quantity lim sup A(NN) the N-'><1O
UPPER ASYMPTOTIC DENSITY of A.
Now let h
~
1 be an integer.
upper asymptotic density zero. (h + 1)A
= (h
+ l)(AUA') E E
Assume that the sequence hA has We can write: (A' + hA)U(h + l)A . E E
(In fact every element of (h + l)A is either the sum of (h + 1) elements of A , or the sum of an element of A' and h elements of E
A).
E
The sequence A' + hA, which is the union of a finite number E
of translates of the sequence hA, has upper asymptotic density zero. Now, every integer m which is an element of the sequence (h + l)A
have:
E
satisfies the relation
{am} ~
(h + l)E, hence we
1 card{n ~ Nln e (h + 1)A } lim sup -N N->-
E
~ lim sup ~ card{n ~ NI{aN} ~ (h + l)E} N-+oo
~
(h + l)E.
CHAPTER 2
140
The sequence (h + l)A
£
therefore has upper asymptotic density
zero, and the same holds for the sequence (h + l)A, since this is included in the union of two sequences with zero upper asymptotic density. (b): Let Xl' ... ,xk be the limit points of the sequence A. Set: A' = {aeAI'o'je [l,k]'llaa £
if 1
j
~
A(j) £
~
x.11 ]
>
d
k, and:
=
{aeAlllaa -
x.11 ~ d, ]
A
£
( .)
The sequence A' is finite, and each of the sequences A ] £
£
has
upper asymptotic desnity less than 2£; from this we deduce that the sequence A has upper asymptotic density zero. Let h be an integer such that the sequence hA has upper asymptotic density zero; as before, we have the (h + l)A
equali~y:
(A' + hA)U(h + l)A • £
£
Denote by E the set of points x.
1
+ ••• + x. 1 (h 1)
E has only a finite number (at most k +
1
taken modulo 1. h+1
) of elements, and
is included in the set of integers m for which there £ exists a yeE such that Ilam - YII < (h + 1)£. From this one de-
(h + l)A
duces the upper asymptotic density of the sequence (h + l)A (and £ therefore that of (h + l)A) is less than 2k(h+1)(h + 1)£, the sequence (h + l)A therefore has zero density.
ADDITIVE THEORY
141
SOLUTION 2·8: Let ex be a fixed irrational number, and let k be an integer.
Since
fen)
~
~
2
0 there exists an integer Mk such
that:
n
~
fen)
Mk =>
<
1
k .
Set Ak = {nelN!{an} :.: 2/k}; since the sequence (exn)nE!N is equidistributed modulo I (cf., [Rau] or Exercise 5·8), we can find an integer
M~
such that:
The sequence
and the sequence
(N')'~l l
l"",.
is defined by the relations:
A in the following way:
(i): The sequence A contains 0 and I by construction. (ii): Let N be an integer. - If N :.: N2 , A(N) = N ~ Nf(N). - If Nk :.: N < Nk +l we observe that the sequences Ak are included in each other, and we have: Akn [l,N] C An [l,N],
therefore
A(N)
~
N k> Nf(N).
CHAPTER 2
142
(iii): The sequence A is not a basis, for the sequence {aa.} 1
+
0 by construction; one then uses Exercise 2 7(a). 0
SOLUTION 2 9:(i): Let us show that there is an infinite number of 0
prime numbers congruent to 3 modulo 4. Let ql, ... ,qk be the first k such prime numbers; the number 4(Ql oooQ k) - 1 is congruent to 3 modulo 4, hence at least one of its prime factors is congruent to 3 modulo 4, and it is necessarily distinct from Ql, ... ,Qk' (ii): Let us consider the system of congruences:
where Q. is the i-th prime number congruent to 3 modulo 4; then 1
numbers Q~ and Q~ being relatively prime if i is distinct from j, J
J
this system has a solution ([Har] Theorem 121).
Each of the num-
bers n + l, ... ,n + k is divisible by a prime number congruent to 3 modulo 4, without being divisible by the square of this number; hence it is not the sum of two squares. (iii): Let a be a fixed positive real number. 2
We are going
to assume that the sequence ([an]) is a basis of order two; from this we are going to deduce that in every interval of the 3 form [Y,Y + -] there is a sum of two squares, which contradicts
Part (ii). Set X
a
=Y + ~
there exist two integers n 1 and n 2 such that
[a~
We then have: aX
2 2 2 2 ani + an 2 - ({ani} + {an 2 } - {aX}), 2
2 ani + an 2 - 2
<
2
2
aX < ani + an 2 + 1,
143
ADDITIVE THEORY 2
2 n1
+ n2
y
X-
2 ().
!
().
<
2
< n1
X<
2 n1
2 1 + n 2 +().
2 + n2 < X + ~ ().
3 = y +-
a
CHAPTER 3
Rational Series
INTRODUCTION 3.0
INTRODUCTION Let K be a field of characteristic zero.
ure of K will be denoted
R, K[X]
The algebraia alos-
will denote the polynomial alge-
bra of K, and K[[X]] the algebra of formal power series with aoeffiaients in K. If P and Q are polynomials with coefficients in K, we can exexpand the rational function P/Q as a power series if Q(D)
+ D.
Conversely, we will say that a power series is a RATIONAL SERIES, or RATIONAL, if it is the development of a rational function.
3.1
FIRST CHARACTERISATION
= L u Xn
Let f(X)
be a rational series with coefficients in n n K; on decomposing it into partial fractions one shows that the coefficients un can be written, starting from a certain index no' as: u
L
n
P.(n)CJ.~,
l~i~r
1 h were , 1 CJ..
1.
~
1.
~. v
~
1.
r, is a pole of f, and P.1. is a polynomial with
144
RATIONAL SERIES
145
coefficients in the algebraic closure
1....
the order of the multiplicity of
=
- Conversely: If f(X)
degree equal to
u Xn and if starting from a cern
n~O
tain index no' for all n
ai
R of K of
less 1.
no we have:
~
P.(n)a~,
un
].
where the ai' 1
~
].
i
~
r, are algebraic on K and distinct, and
the Pi' 1 ~ i ~ r, are polynomials with coefficients in
f =
I
n~O
X,
then
u xn represents in its disc of convergence a rational n
1
1
series which has r poles -, ... , --; the order of each of these al ar poles 1.... is equal to the degree of the polynomial Pi plus 1. a. ].
1
P'eoof: By expanding
we see that the series
(x - ~r I
, where s e:N, as a power series
P(n)anxn,
where
P
is a polynomial of
n~O
degree d, is a rational fraction with pole l with multiplicity a (d + 1). If, for n
un
~
nO' we have:
I
P. (n)a~,
].
l~i~r].
it follows that: f(X) =
I
n~O
is a rational series - with coefficients in K. When the u e K it n u remains to be shown that has its coefficients in K.
v
When K =~,
u I V
n~O
whence:
X=
we have:
~,
I
n~O
u
v'
146
CHAPTER 3
UV
UV
The fraction
UV.
U~ is equal to ~V and its coefficients are real.
VV
In the general case let K' be a Galois extension of K containing the coefficients of U and V.
The fraction ~ is invariant un-
der the elements a of the Galois group G. U
The fraction:
a(V)
11
IT
aeG"{id}
aeG
a(V)
is a quotient of polynomials with coefficients in K and is equal U
to V . 3.2
I
SECOND CHARACTERISATION [Sal]
A necessary and sufficient condition in order that reX) n u X , where u e K, be rational is that the coefficients u satn
n
n
isfy a recurrence relation starting from a certain index no: For all n
~
nO the un satisfy: 0,
where a. e 1
K.
The proof uses the formal series: If
then: 3.3
I
reX)
n
u Xn
r(X)Q(X)
n
e
and
Q(X)
K[X] •
THIRD CHARACTERISATION: HANKEL DETERMINANTS (cf., [Sal] and Exercise 3·1)
I u n xn , we call the HANKEL DETERMINANT D(s) of n n the series f(X) the following determinant: Let
reX)
=
RATIONAL SERIES
147 u
n+s
u n + s +1
In order that f(X) be a rational fraction it is necessary and sufficient that there exist two integers s and no such that foraHn ~noD~s) 3.4
=
o.
FOURTH CHARACTERISATION: KRONECKER DETERMINANTS We call the KRONECKER DETERMINANT of the series
determinant: U
/',
D(t)
t
L u n Xn
n
o u1
u1 u 2
0
ut
u t +1
In order that the series
L u n Xn
n
be rational, it is necessary
and sufficient that there exists to such that for all t we have
the
/',t
= o.
BI BL IOGRAPHY [Sal], [Ami], [Pis].
~
to
148
CHAPTER 3
PROBLEMS
EXERCISE 3-1: HANKEL AND KRONECKER DETERMINANTS (1): Let f(X) =
L
n~O
U n
Xn be a rational series with coeffic-
ients in K; show that there exist two integers for n
8
and
no
such that
~ nO : U
n u n +l
(s)
u n +l
U
u n+2
u n +s +l
n+s O.
D
n U
(2) :
n+s
Let D
in K, and let:
=
U
U
n+s+l
[a . . J
•
l,] l::;:l::;:n
n+2s
be a determinant with coefficients
l::;:j::;:n
d
a n- l ,n-2
Cd
is obtained by starting from D and deleting in D the first and
the last line and the first and the
149
RATIONAL SERIES
last column). Dendte the cofactor of a. . in D by A. " l.,] l.,]
Prove
SYLVESTER'S
IDENTITY
Dd
= AlIA , D,n
(3) :
- A
A
n,l 1,n'
Assume that for fixed B
D~S)
=
~
0 for n
nO'
Assume
that there exists n 1 9 no such that D~S-l) = 0 for n = n 1 . that Dn(s-l) = 0 for n >.r n l' (4):
Show that if one can associate with the series
two integers thi~
B
and nO such that for n
nO we have
D~s)
L
U XU
n~O
n
0, then
series represents a rationalfuuctiou_ (5):
Show that if the series
exists to such that for all t if ~t
~
Show
=0
~
to'
L n~O
u Xn is rational, there n
D~t)
~t
for t ~ to' then the series
L n~O
=
O.
Conversely,
u Xn is rational. n
EXERCISE 3-2: FATOU'S LEMMA (1):
A formal series in z::: [[X]] is called a
PRIMITIVE SERIES
if the greatest common divisor of its coefficients is one.
Show
that the product of two primitive series of z>;[ [X]] is primitive. (2) :
u Xn ez>; [ [X]] be a formal series which is a
Let n~O
n
. 1 f ·uuctl.OU. . p(X) Wl.t . h P ,Q ratl.ona Q(XT e P and Q inz:::[X] with Q(O)
=
m[X] . Show that one can choose
1.
EXERCISE 3-3: ANALYTIC CHARACTERISATION OF POLYNOMIALS AND RATIONAL FUNCTIONS Let f(z) n~O
z::;
u n zn be a power series with coefficients in
and let R > 0 be its radius of convergence.
150
CHAPTER 3
Assume R > 1.
(1) :
u zn reduces to a poly-
Show that
n
nomial. Assume R
(2):
function in a disc function.
Izl
mthe
Give
(3):
ber.
and that f is extended to a meromorphic
< 1
Show that fez) is a rational
p > 1.
< p,
p-adic valuations, where p is a prime num-
For this valuation the completion of
and the algebraic closure of
mwill
be denoted ID , 'p
mp by np .
Assume that f is extended into a meromorphic function in cr in the disc disc
1
z
1
p
Iz 1 <
p
and into a meromorphic function in n
Show that if pp
< p •
p
p
p
>
in the
1, then fez) is a rational
function. EXERCISE 3·4: HADAMARD'S PRODUCT
Let fez)
L u n zn
=
L v nz n be two rational series n Then the series h(z) = L (u x v n )zn is n n
and g(z) =
n
with coefficients in K. rational.
EXERCISE 3'5: HADAMARD'S QUOTIENT
L u n zn be a rational function, and g = L v n zn a n n rational function having a simple pole of absolute value strictly Let f(z) =
less than the others, with coefficients in Let a
n
m.
be the coefficient of a series defined in the following
way: -1
If:
vn
+0
set a n
u nv n ,
if:
vn
0
set a n
O.
L
a z n
If the series
n~O
n
has coefficients in :;z, then i t is rational.
RATIONAL SERIES
151
EXERCISE 3·6: FORMAL SERIES AND MATRICES Given a series f(X)
=
L
n~O
U
n
Xn e K[[X]].
We say that the
square matrix A = MN(K) is associated with the series f if for all n '" 1:
where [An]l N denotes the coefficient in the first row and the last column • of the matrix An . Show that if there exists a matrix A associated with the series L u Xn , then this series is rational. n n;;:O (1):
(2):
We want to prove the converse of Question (1) above.
Several steps are needed: (a): Show that to the polynomial:
we can associate the matrix:
o 1 o o o 1
o o
0
0
0
1
u N_ 2
0
0
0
0
u N- 1
0
0
0
0
0
...........
A
(b): Assume that the N-dimensional matrix A is associated with the series ies
L
n;;:O
u
n
xn , and the M-dimensiona1 matrix
L v Xn. Assume, furthermore, that n;;:O n Show that the product series:
U
o = o.
B with the ser-
152
CHAPTER
3
is associated the block matrix:
c
[:
~1
where A is the N x M ma,trix all the columns of which are zero except the first, which is equal to the last column of A.
I
(c): Let f(X)
n~O
unxn be a rational series and U o
A be a matrix, of order N
>
1, associated with f.
= O.
Let
Show that to
the rational series 1 _ ~(X) is associated the matrix A = A + where A is the N x N matrix defined, in Question (b), starting from A. (By induction on n prove the relation: For all j, 1 -n
~
j
~ N:
(A )1 .
,J
(d): Show that every rational function ~ with Q(O)
+ 0 can
be
put into the form:
where a e K and P1 and Q1 are polynomials with no constant terms. From this deduce that there exists a matrix associated to ~ .
Show that i f the fraction ~ e ~[[X]] there exists a matrix, with coefficients in~, associated to ~ .
A,
RATIONAL SERIES
153
SOLUTIONS
If f(X) is a rational series, for n
SOLUTION 3·1: (1):
there exists a recurrence relation of order s on the u Second Characterisation.
(2) :
nO
by the
This gives a relation on the rows of
D~s), which is therefore zero for n
the determinant
n
~
¢
no.
We have:
a n-l,l a n,l
a n-l,2
a n,2
Ai , 1 Ai , 2
A
l,n-l l,n
a n,n-l A
0
0
1
A
0
a n-l,n a n,n
n,l n,2
1
x A
a n-l,n-l
(Contd) 1
A
n,n-l
0
A
n,n
154
CHAPTER 3
D
al,n-l
o
0 0
D2d ,
(Contd)
o o
0
a n,n-l
D
whence:
D2d
D(A l , lA n,n - An, lAl ,n ),
therefore: If D
+ 0: Dd
we have:
= Al , lA n,n - An, lAl ,n
If D = 0: the determinants Al , 1,A D,D ,A n, . 1,A l
polynomials with coefficients (a. ')1' l,]
,D
"'l",n
' d and Dare .
1.;j..:n
We have:
D[Dd - Al ,n,n 1A - An, 1A1] ,n
0,
D being a polynomial in (a.l , ]')1' one considers the polynomial ",]",n 1
Dd - A1,1An,n + An,1A1 ,n' then we have: PCa.
. )Q(a.
l,]
.)
l,]
O.
RATIONAL SERIES
155
At all the points of Kn
2
where P f 0, we have Q
= O.
Now, the ring K[X l ,X 2 , ... ,X 2] is an integral domain, whence Q _ O. n
Hence we have:
= Ai ,n,n lA
Dd
- AlAi n, ,n
in all cases. Let us apply Sylve ter's relation to the Hankel deter-
(3):
minant.
We obtain:
which lets us prove by induction that, if D(s) = 0 for n n and if there exists n l ~ nO with D(s-l) = 0, then D(s-l) nl n all n ~ n l , in fact: 0,
as soon as D(s-l) n
=0
= O.
(this by induction) and D(s)
(4) : Let 8 be the smallest with D(s) = 0 for all n ~ nO' nO n surne that D(s-l) f 0 for all n ~ n Consider the following system
n
integer such that there exists The preceding allows us to asno'
of equations for n
- u n+s'
a sU n
1
asu n+s +l + as-lu n+ s +
~ no:
+ a u n+2s - 2
- u
n+2s-1'
where the unknowns are (a l ,a 2 , ... ,a s ) and the coefficients are (u , ... ,u 2 2); the determinants of the coefficients is not n n+ szero, this system possesses one and only one solution (al, ... ,a s ) in
KS
•
156
CHAPTER 3
Now,
D~S) = 0, hence this solution is compatible with taking
s + 1 equations.
Therefore there exists a recurrence relation
between the (u ), and by the Second Characterisation of rational n
series the series (5):
L u n Xn
n
is rational.
By the Second Characterisation for n
~
nO the un are
connected by the recurrence relation:
o. From this one deduces for t
~
nO + s a linear dependence relation
between the last s + 1 columns of the Kronecker determinant which is zero. Conversely, assume
°for
~t
t
~
to'
The Sylvester's
relation gives:
and one has
Di t ) = ° for
all t
~ to'
Using Sylvester's rela-
tion:
by induction on p one shows that for all p ~ 0, for all t ~ to' D(t) = 0. Then using the characterisation of rational series by p
the vanishing of the Hankel determinants gives the rationality of the series
L
n~O
SOLUTION 3-2: (1):
Let
L
n~O
a Xn and n
L
n~O
b Xn be two primitive n
series, and let us assume that the product:
157
RATIONAL SERIES
is not. There then exists a prime number p that divides all the a. This number does not divide all the a. Let h be the smalln n est index such that p does not divide a h . Similarly, let k be the smallest index such that p does not divide bk . We then have:
The terms in parentheses are divisible by p, but ahb k , and therefore a h+k , is not divisible by p, which contradicts the hypothesis. (2):
After possibly multiplying by a common denominator,
we can assume that P and Q are
in~[X].
Let us assume P and Q are relatively prime, there then exist A,B e~[X] and a e~, a
f 0, such that:
a,
PA + QB
whence: a
PA
Q=Q +
B;
P
as Q ,A,B e?Z [[X]], from this is deduced:
~ e ~[[X]]; by expanding ~ in ~[[X]] we obtain:
L
where
ane~
for all n ::: 0.
n~O
Let q (resp. a) be the g. c. d. of the coefficients of Q (resp. C). Hence:
158
CHAPTER 3
Q
qQ'"
where Q", is primitive, Q1' eZ'::[X) ,
C
eC",
where C,', eZ':: [[X)) and C1, is primitive.
From the relation ~Q = C we conclude ~ = C*Q*. qe above C1'Q'" is primitive, thus:
By Question (1)
~ = ± 1.
qe
If we write:
a
then eoqO
-=
P(X)
Q(X)
qe
±l, which implies qo
L
'\
L
n~O
This now gives:
u Xn n
n~O
qQ ;'
±1.
un
Xn
'
whence P(X) eZ'::[X) and q divides the coefficients of P, that is to say, that P(X) eZ'::[X). We then have: q
P(X) ~ with Q*(O)
1.
Which ends the proof. REMARK: In this Exercise one can replaceZ':: and ID by an arbitrary Dedeking ring A and its quotient field K.
In particular, K can
be a number field and A its ring of integers. SOLUTION 3'3: (1): en by the formula:
The radius of convergence of a series is giv-
RATIONAL SERIES
159
-R1 = -nVTiD lim Iu n '
~
now if R > 1, 1 (now, u
n
the series
(2):
1, whence lim n~ < 1, that is to say lunl <
<
= 0) starting from a certain rank nO; n n u 2 reduces to a polynomial.
e:
I
n~O
n
Let 1
< p'
< p.
In the disc
121
~ p' the function
1
has a finite number of poles and there exists a polynomial:
••• +
= 1(2)q(2) is hOlomorphic for 121
such that g(2)
< p'
We will
show that the Hankel determinants of the series 1(2) tend to zero; since these determinants are integers it follows that they are zero starting from a certain point.
We have: V 2 n
n
where:
whence for s
~
k:
un D(s) n
u
u n+l
u
u n+s
u
un
u n+l
u n+s
n+l
u n+k-l
n+2
u
n+k
n+s+l
u n+s+k-l
u n +2s
v n+k
v n+s
V
vn+s+l
u
n+k-l u n+k
.... u n+k-l+s
n+k+l
u n+s
u n+s+l
...... vn+k+s
vn+2s
x
( a )s-k+l .
o
160
CHAPTER 3
After writing out the
(s
+ I)! terms of this determinant we obtain:
Let A be a real number such that 0
and let B satisfy
Set:
1 < B < p' M1 =
< A < R
sup Ig(x) I, Ixl::::B
suplf(x)l,
Ixl::::A
and:
Apply the Cauchy inequalities: M
lu n I ...< - n A
The following upper bound for D(s) is obtained: n
where C is a constant depending upon M,s,k and a o (for example, one can take: C
As B
...:..CM_(.:..;S,--,;+~l):....;),-S_+_l > 1,
k s+l-k
s can be fixed large enough so that A B
whence: limIDCS)1
n-+<»
Ca 0 ) S - k+ 1) •
A 2ks
n
o.
> 1,
RATIONAL SERIES
161
With D(s) being an integer, therefore D(s) n
n
o
starting from a
certain index nO'
A < R 0 < B p p P' vergence of the series f in ised exactly as in Question isfying 0
<
ID(s)1 n
for Ap,Bp sat< p (where R is the radius of conp P n The determinants D(s) are majorn p (2) preceding:
The Cauchy inequalities are valid in n
(3) :
-nk
p
~ CA
p p
P
-n(s+l-k)
PB P
P
We always have:
Set: -k
P
C'
Since
-(s+l-k) P
A PB P P
~ (8)
C.C. P
pp
P
>
1 there exists an integer
ID ( s ) II D ( s ) I n
n
p
~ c' ~ n ,
One can therefore choose
As the coefficients un
elZ,
where ~ 8
8
such that:
M8) < 1.
such that:
the determinant D~s)
By using the relation that for all x e 91"':
elZ
is as well.
162
CHAPTER 3
Ixl
IT
p prime
we see that
Ixl p
Ixl Ixl p
1,
~ 1 if x is a non-zero integer.
It follows that D(s) = 0 for s fixed and n large enough, and the series
n
L un zn
is rational.
n
By the First Characterisation, as
SOLUTION 3-4:
fez)
is rational,
there exist k algebraic numbers a 1 , ... ,a k , and k polynomials P1 , ... , P k , such that: P.(n)a~
un
1
1
As g(z) is rational, there exist 1 numbers a 1 , ... ,a£, and 1 polynomials Q1, ... ,Q£, such that: V
n
whence for n u xv n
we have the relation:
~ sup(n 1 ,n 2 )
P.(n)a~x
n
1
L
l::;i::;k
Q.(n)s~ ]
1
]
p. (n)Q . (n)( a . S . ) n • 1
]
1
]
l::;j::;l
By the First Characterisation of rational series the series hex) is rational.
L an zn is meromorphn ic on arbitrary large discs, in particular on discs containing
SOLUTION 3-5:
We will show that the series
the unit disc.
Exercise 3-3 gives us that:
RATIONAL SERIES
163
- Starting from a certain index no the coefficients of g(z) can be written:
where the Pi (2
~
i
~
p) are polynomials and the a i are algebraic
numbers satisfying the condition: I:~I
1.
<
Starting from a certain index n 1 , the term smaller in absolute value than A < 1.
vn
+0
~
.In is
P. (n) [-2:.
Whence for n
~
a1 n 1 one has
and:
-1 u nv n
a
una -1 1-n[ 1
a
= una -1 1-n[ 1 +
2
Pi(n)
2
P.(n)
2~i~p
2~i.li:p
this expansion is valid for all p For all p ~ 1 the series t
2
2~i~p].
[a·rr a: 1
-2:. ]. (a·r a1 +
~
000
o. ai n p P. (n)(-) ] ] ]. al
n
is rational because it is a linear combination of Hadamard products of rational series (see 4). We first have to see the series
lJ
n
a i n]p+1
]. a
P. (n)
1
w zn
with:
1
(a.)n
n
----"'-----
1 -
2
P.(n)..2:. a1
2~i.li:p].
(Contd)
CHAPTER 3
164
(Contd)
p can be chosen large enough so that the radius of convergence of this series is greater than one. The series
L
n;m1
a zn is the sum of a rational function
n
and of the series
hl n~nl
than one.
As a
n
e~
L
t zn
nm 1 n
n
zn whose radius of convergence is greater
Exercise 3·3(1) says that this is a rational
function REMARK: If the rational series
L
n~O
v zn has a pole of modulus n
strictly greater than the others, if the series L v zn is ran n u tiona1 and if the series a = L v n zn has integer coefficients (or n n algebraic integers) then the series a is rational (cf., D. G. Cantor: 'On Arithmetic Properties of the Taylor Series of Rational Functions.
II', Pac if. J. Math., 41, No.2, (1972), pp. 329-334). SOLUTION 3·6: (1):
Assume that for n
~
1 we have:
The matrix A satisfies its minimal polynomial, the coefficients
u
n
of the series
L u n Xn
satisfy a recurrence relation, and this
series is therefore rational. Let {e 1 ,e 2 , ... ,e N} be the canonical basis of~. In this basis the matrix A is the matrix of the linear mapping, (2): (a):
denoted g, defined by:
RATIONAL SERIES
165
We have:
whence we obtain: i
(A )l,N Finally AN
= ui
for 1
~
i
~
N - 1.
0, and for i ~ N, (Ai)l,N
(2): (b):
O.
First of all we have: A n-l-kAB k
t L O~k~n-l
I
Bn The matrix A
n-1-k-
A has all its columns zero except the first,
and: [A
t L
l~i~N
Next,
[A
n-1-k
n-l-k
] 1 ,1.. [A].1., 1
]1 .[A]. N ,1.
1.,
[A
n-k
] 1 , N"
CHAPTER 3
166
Finally we have:
\' U V -w o~~~n n-k k - n'
as
U
o = O. (2): (c):
Let us prove the relation by induction.
For n
=1
this reduces to:
where I is the identity matrix. For j
+ 1:
For j = 1:
[A]l ,J. = [A]l ,J. certainly holds; we have:
Assume the formula to be true for n, and let us prove it for
n + 1.
We have:
[An ]l,k is given by the induction hypothesis, therefore:
I
O~ilii:n l~k~N
i
-n-i
[A ]1 N[A
,
]1 k[A]k ••
"J
Let us calculate the first sum, which has the value: (Contd)
RATIONAL SERIES
=
(Contd) For j
167
[An+l] 1 . + \ [An] [A-] L 1 k k·· ,] l:r.k:r.N ' ,]
+ 1:
For j = 1:
In two cases we have:
The second sum has the value:
s2 =
L
°"'i:r.n Finally we obtain:
n
For j = N, and setting bn = [A ]l,N' and b o we have just proved gives: b
n
= un
+
which proves that the series g(X)
= L
n~O
tion: g(x)
0, the relation
= f(x)
that is to say:
+ y(x)f(x),
b Xn satisfies the equan
168
CHAPTER 3
f(X) f
g(X)
1
1 - f(X) - 1.
1 -
1
The series 1 _ f(X) differs from g(x) only by the constant term; the matrix A associated with g(x) is also associated with it.
(2): (d):
Let Po
= P(O) and qo
Q(O).
We have:
with: Q (X) 1
= 2:.... qo
(q
0
- Q(X))
A matrix associated with
and
P1 Q1 will also be
1 -
ass~;iated
with
P
Q' Associate with the polynomial Q1 the matrix A of order N defined in Question (2)(a) above.
To the fraction 1
!Q
by Question (2)(c) above, be associated the matrix A + Question duct If
(P1)
~2)(b)
1
A.
~
2
will, And
above gives the matrix associated with the pro
II ! QJ .
~ e:
sentation
~ of ~ , with
P*,Q2'
= 1.
All the pre-
ceding calculations show that the associated matrix that we have just constructed has integer coefficients. REMARK: This relation between rational series and matrices allows us to generalise the notion of rational series. If X is a finite set, X* the free monoid that it generates, a rational series on X* with coefficients in:
MN~)
or MN(IIl) for a certain N
~
1.
If
RATIONAL SERIES
X has a single element, then X*
169 =~,
and one shows that the usual
definition of rational series on~ or mis obtained (cf., M.P. Schutzenberger: 'Properties of the Families of the Automata', Inf. and Control, 6, (1963), pp. 246-264).
CHAPTER 4
Algebraic Theory
INTRODUCTION 4.1
FIELD EXTENSIONS
4.1'1
FINITE EXTENSIONS: ALGEBRAIC EXTENSIONS
Let L be a field; if X is a subfield of L we also say that
L is an EXTENSION FIELD OF X. If L, as a vector space over X, has finite dimension n, we say that the extension Llx is a FINITE EXTENSION OF DEGREE
n, and this is written n = [L:X]; otherwise
we say that Llx is an INFINITE EXTENSION.
If MIL and LIX are
two finite extensions, Mix is then finite, and [M:X]
= [M:LJ x
[L:X] . Let Xl,X2 be two extensions of X contained in the same field
L.
The COMPOSITE OF Xl 'X 2 ' which is denoted by X1X2 ' is the sub-
field of L generated by Xl and X2 . Let
(ai)l~i~k
field of
X.
be k elements of a field
L
that is an extension
We denote by X(a1, ... ,a k ) the subfield of L gener-
ated by X and by the a. Cl < i < k). l
Let Llx be an extension of fields and let a be an element of L.
One says that a IS ALGEBRAIC Crespo TRANSCENDENTAL) OVER X
if the extension X(a)IX is finite Crespo infinite). An extension Llx is called an ALGEBRAIC EXTENSION if every element of L is algebraic over X.
Every finite extension is alge-
170
ALGEBRAIC THEORY
braic.
By an ALGEBRAIC NUMBER FIELD we mean a finite extension
of the field 4.1·2
171
~
of rational numbers.
IRREDUCIBLE POLYNOMIAL OF AN ALGEBRAIC ELEMENT
Let L IK be an extension of fields and let a be an element of L algebraic over K.
The set of polynomials P e K[X] such that P(a)
=
0 is
an ideal of K[X] generated by an irreducible polynomial, assumed unitary, which is called the IRREDUCIBLE POLYNOMIAL
K (the notation used is Irr(a,K)). to the field K[X]/(Irr(a,K)).
of a over
The field K(a) is isomorphic
The DEGREE OF AN ELEMENT a is the
degree of the polynomial Irr(a,K); this is also the degree
[K(a) :K] . Conversely, if P is an irreducible polynomial of K[X] there exists an extension of
K
in which
P
admits a root, and this exten-
sion contains a field K-isomorphic to K[X]/(P). 4.1·3
SPLITTING FIELD: ALGEBRAIC CLOSURE
Let P be a polynomial of K[X].
There exists an extension L
of K such that: (1): P splits in L[X] into a product of first degree polynomials; (2): If (a1, ... ,a k ) is the set of roots of P in L, L : K(a1, ... ,a k ) .
Such an extension L is unique up to K-isomorphism and is called a SPLITTING FIELD OF THE POLYNOMIAL P OVER K
A field ial of
~
~[X]
is said to be ALGEBRAICALLY CLOSED if every polynomof degree greater than or equal to one has a root in
~.
Let K be a field. that:
There exists an extension
~
of K such
(a): ~IK is algebraic; (b):
~
is algebraically closed.
The field K is unique up to K-isomorphism; BRAIC CLOSURE of K.
~
is called an ALGE-
172
CHAPTER 4
4.1'4
CONJUGATE FIELDS
n
In an algebraic closure
of X given two extensions L,L' of
X; we say that Land L' are CONJUGATE FIELDS over X if there ex-
ists a X-isomorphism of L onto L'. elements of
n we
Analogously, if a,a' are two
say that they are CONJUGATE ELEMENTS over X if
there exists a X-isomorphism
~
of X(a) onto X(a') such that
~(a)
=
Two elements of n are conjugate on X if and only if they
a'.
have the same irreducible polynomial over X. If X is a field of characteristic zero or a finite field, an irreducible polynomial of X[X] has all its roots distinct in an algebraic closure
n of X;
if L is an extension of X of finite de-
gree n there exist n distinct X-isomorphisms of L into 4.1' 5
NORM: TRACE: CHARACTERISTIC POLYNOMIAL
Let LJX be a finite extension. associate the endomorphism by
~ (x)
a
n.
~
= ax for all x e L.
a
To every element a e L we
of the X-vector space L defined
The NORM (resp. TRACE, CHARACTERISTIC POLYNOMIAL) OF AN ELEMENT a IN THE EXTENSION LJX is the determinant (resp. trace, char-
acteristic polynomial) of the endomorphism
~.
a
The norm is denoted
NLJK and the trace is denoted TrLJK' If the field X is of characteristic zero or is finite, on writing
k
=
[X(a):X] and (a')l l
'~k
~l ....
in an algebraic closure
for the roots of the polynomial Irr(a,X)
n of
L we have:
The characteristic polynomial of a is the polynomial Irr(a,X)n/k. 4.2
GALOIS THEORY We assume the field X is either characteristic zero or a
finite field.
ALGEBRAIC THEORY
4.2·1
173
GALOIS EXTENSION
Let
Llx
be a finite extension.
We say that
Llx
is a GALOIS
EXTENSION if for all a e L the polynomial Irr(a,X) has all its
roots in L.
The set of X-automorphisms of L has a group struc-
ture that is called the GALOIS GROUP OF THE EXTENSION
Llx,
and
is denoted Gal(LIX); its order is equal to the degree [L:X]. 4.2·2
FUNDAMENTAL THEOREM OF GALOIS THEORY
Let
Llx
be a finite Falois extension, and let G
= Gal(LIX).
If H is a subgroup of G, the set of elements of L invariant by every a e H is a field intermediate between X and L that is denoted
LH.
If F is a field intermediate between X and L the extension
LIF is Galois and Gal(LIF) is a subgroup of G. There exists a bijection between the set of fields F between
X and L and the set of subgroups H of G; this bijection is given by:'
F ~ H = Gal(LIF), the inverse map being:
The extension Fix is Galois if and only if Gal(LIF) is a normal
subgroup of G; in this case we have an isomorphism
Gal(FIX) ~ G/Gal(LIF). CYCLIC) EXTENSION if
We say that
Llx
is an ABELIAN (resp.
LIX is Galois and if Gal(LIX) is abelian
(resp. cyclic).
4.2·3
EXTENSIONS OF FINITE FIELDS
Let X
~
F
q
be a finite field with q elements and let L
be a finite extension of X of degree n.
The extension
Llx
~
is
cyclic and its Galois group is generated by the automorphism x ~ xq.
F
qn
174
CHAPTER 4
4. 2 . 4
NORM AND TRACE
If LIK is a finite Galois extension, then for all 0. eL:
IT
0(0.),
oeGal(LIK)
L
0(0.).
oeGal(LIK)
4.3
INTEGERS
4.3'1 ELEMENTS INTEGRAL OVER A RING
Let B be a commutative ring and A a subring of B.
An element
x e B is called an INTEGRAL ELEMENT OVER A if there exists an inte-
ger n and elements a O,a 1 , ••• ,a n _1 eA such that:
x
n
+a
n-l
x
n-l
+ ••• + a x + a 1
0
= O.
The set of elements of B integral over A is a subring of B which contains A and is called the INTEGRAL CLOSURE OF A IN B.
If the
integral closure of A in B is B itself, we say that B is
INTEGRAL OVER A. Let C be a commutative ring, B a subring of C, and A a subring of B. If C is integral over B and if B is integral over A, then C is integral over A. If a ring A is an integral domain and K is quotient field, we say that A is INTEGRALLY CLOSED if A coincides with its integral closure in K. 4.3'2
A principal ideal dom. in is integrally closed.
ALGEBRAIC INTEGERS. EXAMPLE OF QUADRATIC FIELDS
Let K be an algebraic number field, that is to say, a finite extension
of~.
An element of K is called an ALGEBRAIC INTEGER
if it is integral over the ring:iZ of rational integers. integers of K is a free:iZ-module with
[K:~]
The set of
generators, which are
ALGEBRAIC THEORY
175
called an INTEGRAL BASIS of K. EXAMPLE: K is a quadratic field, that is to say, of degree two There exists a square-free integer d, such that K =
over~.
If d _ 2 or 3 modulo 4, {1,1ci} is an integral basis of
~(Ici).
K.
If d
=I
modulo 4, {l,(l + 1ci)/2} is an integral basis of
K. 4.3·3
DISCRIMINANT
Let A be an integrally closed domain, K the quotient field of
A, L a finite extension of K, and B the integral closure of A in L.
If x e B, the polynomial Irr(x,K) has its coefficients in A;
NL!K(x) and TL!K(x) belong to A. The DISCRIMINANT OF A BASIS (ei)l
n
l ]
~l,J~n
If K is a field of characteristic 0 or a finite field, the discriminant of a basis of Lover K is not zero; if the n distinct K-isomorphisms of L into an algebraic closure
~
of K are denoted
0l, ... ,on' then: det(cJ.(e·»12 . . . l ] ~l,J~n Example:
If
L '"' KCa), and i f
we write n = [K(a):K], P = Irr(a,K), we have: D(l,a, ... ,a
n-1
) = NL!K(P'(a».
If (e.)l . N is a basis of Lover K formed of elements of B, the l
~l~
discriminant D(e 1 , ... ,en) eA. The DISCRIMINANT IDEAL of B over A, denoted by liB! A' is the ideal of A generated by the discriminants of the bases of Lover K which are contained in
B.
If B is a free A-module, llB!A is the ideal generated by the
176
CHAPTER
discriminant of any basis of B on A. free ant.
~-modu1e,
4.4
and all
~
bases of
B
If A
=~,
K
4
= m, B is a
have the same discrimin-
IDEALS
4.4·1
DEDEKIND DOMAINS
Let A be a commutative ring; we say that A is a NOETHERIAN RING if every ideal of A is an A-module of finite type.
A DEDE-
KIND DOMAIN is a Noetherian integrally closed domain of which
every non-zero prime ideal is maximal. Let A be a Dedekind domain, K the quotient field of A, L a finite extension of K, B the integral closure of A in L. of characteristic zero.
Assume K is
Then B is a Dedekind domain.
A principal ideal domain is a Dedekind domain.
The ring of
integers of a number field is a Dedekind domain. 4.4·2
IDEALS OF A DEDEKIND DOMAIN
If A is an integral domain with quotient field K a FRACTIONAL IDEAL of A is a sub-A-modu1e I of K such that there exists a non-zero element d of A satisfying dIe A.
The product
of two fractional ideals I,J is the set of finite sums
L xiYi'
where x.1 eI, y.1 eJ; this is a fractional ideal denoted IJ. Let
of A.
A
be a Dedekind domain,
P
the set of non-zero prime ideals
The set of non-zero fractional ideals of A has a group
structure.
Every non-zero fractional ideal
A of
A can be writ-
ten uniquely in the form: A
where p
n. .1
1
e
P
and n. e~. 1
The principal fractional ideals form a subgroup of the group of ideals of A; the quotient group is called the CLASS GROUP OF A.
177
ALGEBRAIC THEORY
4.4·3
DECOMPOSITION OF PRIME IDEALS IN AN EXTENSION
Let A be a Dedekind ring, X its quotient field is assumed to be of characteristic zero, L a finite extension of X of degree n, B the integral closure of A in L. If P is a non-zero prime ideal of A one has the decomposition:
where the Pi are distinct prime ideals of B, and where the e i ~ I are integers. The ideals Pi are the prime ideals P of B over p, that is to say such that pnA
= p.
The ring BIP i is a field which
is a finite extension of the field Alp; we write f. l
= [BIP.:AIP] l
and ca11:
f i the RESIDUAL DEGREE of Pion A,
e i the RAMIFICATION INDEX of Pi on A. We have the equality:
I
l~i~q
e.f. l
= n.
l
We say that p is RAMIFIED IN (THE EXTENSION) L if one of the ramification indices e. is l
~
2.
The prime ideals p ramified in
L are finite in number; these are the prime divisors of the dis-
criminant ideal 4.4·4
~BIA.
DECOMPOSITION IN A GALOIS EXTENSION The notations and hypotheses are the same as in Section 4.3;
we assume, further, that For every
0 EO
Llx
G one has oB
is Galois, and we write G
= Gal(LIX).
0; if P is a prime ideal of B, oP
is a prime ideal of B which is called the CONJUGATE IDEAL OF PRIME IDEAL) P.
THE
178
CHAPTER 4
The decom;:lOsition of a non-zero prime ideal p of A is written:
(IT
PB
l:;;i:;;g
Pi)e,
where the ideals P. are all conjugate, have the same residual 1
degree f, and the same ramification index e.
We have the equal-
ity:
efg
= n.
The DECOMPOSITION GROUP D OF (A PRIME IDEAL) P of B is the set of
0
e G such that oP = P.
D is a subgroup of G of order
!2. g
if P is one of the ideals p .• The DECOMPOSITION FIELD OF (THE PRIME D 1 IDEAL) P is the field L of elements of L invariant under D. Let
0
be an element of D; by passage to the quotient field
o defines an
Alp automorphism a of BIP.
The mapping
0 ~
a
is a
group homomorphism the kernel of which is called the INERTIA GROUP I OF (THE PRIME IDEAL). P. FIELD OF (THE PRIME IDEAL) P.
[L:L] I
4.4'5
e,
EXAMPLE:
The field LI is called the INERTIA We have:
f,
g.
DECOMPOSITION IN A QUADRATIC FIELD
Let A =::z, K = Ill, L = III ( id) (where d is an integer of::z with no square factor), and B the integral closure of::Z in L. If P is a prime number, the splitting of
P in B has one of
the three following forms:
(1): pB = P1 P2 , where PI and P2 are two distinct prime ideals
of residual degree one; it we say that p is decomposed in L;
(2): pB = p, where P is a prime ideal of residual degree two; we say that p is INERT IN L;
ALGEBRAIC THEORY (3): pB
179
= P,2
where P is a prime ideal of degree one; we
say that p is RAMIFIED IN L. Given an odd prime number p and an integer d prime to p, one says that d is a QUADRATIC RESIDUE MODULO P if the class of d modulo p is a square.
The LEGENDRE SYMBOL
[~)
is defined by:
if d is a quadratic residue modulo p, otherwise. The decomposition of p in B is given by:
(i)
P is decomposed in L if:
(ii)
P is inertial in L if:
either: p is odd and
1
p
or:
d
either: p is odd and
1
(iii) P is ramified in Lift
= 2 and
p
or:
1
= 2 and
d
(~)
= +1,
= 1 mod
(~)
8;
= -1,
= 5 mod
8;
either: P is odd and divides d, or:
p
=2
and d
=2
or 3 mod 4.
In order to determine the value of the Legendre symbol [dp)J we observe that this symbol is multiplicative, that is to say:
J (where a,b are integers prime with p). [apb)J __ rap)J [pb1 One also uses the QUADRATIC RECIPROCITY LAW: If p and q
distinct odd prime numbers3 then
[~)(~) = (_l)~(p-l)(q-l). 4.5
IDEAL CLASSES AND UNITS OF ALGEBRAIC NUMBER FIELDS
K denotes a number field and n = [K:ml its degree.
a!'e
two
180
CHAPTER 4
4.5'1
THE INTEGERS r 1 AND r 2
There exist exactly n distinct m-isomorphisms 0i (1 of K into the complex number field
0:.
~
i
~
n)
We denote by r 1 the number
of indices i such that 0i(K) is contained in the real number field lR. If i is an index such that o. (K) is not contained in lR, and 1
if
T
is complex comjugation in
0:, TO.
1
is also a m-isomorphism of
The number of indices i such
K in 0: and it is distinct from 0.. 1
that o.(K) is not contained inlR is therefore even, a number 1
which will be denoted 2r 2 .
4.5'2
IDEAL CLASSES OF
Let
~
We have:
B
be the discriminant of B (that is to say, the discrim-
inant of an integer basis of B).
The MINKOWSKI CONSTANT OF A
FIELD K is the number:
One can show that every class of ideals of B contains an integral ideal
A such that:
N(A) :;; M,
of the integer ideal
A is
where the norm
N(A)
by definition the
cardinality of
BfA. As the set of integral ideals of B for which
the norm is a given integer is finite, it follows that: The aZass
group of a number fieZd is finite. 4.5'3
THE UNITS OF K
Dirichlet's Theorem gives the structure of the group E of units of the ring B of integers of K: Let r = r 1 + r 2 - 1, and let U be the group of roots of unity
ALGEBRAIC THEORY
contained in K. U and of 1"
1"
181
The group E is the direct product of the group
groups isomorphic to:;z.
In other words, there exist
units £1" "'£r' called FUNDAMENTAL UNITS, such that:
REMARK: As every finite subgroup of the multiplicative group K'" of elements of an arbitrary field K is cyclic, in particular therefore, the group U is a cyclic group. BIBLIOGRAPHY [Bor] , [Bou] , [CaF] , [Che] , [Har] ,
[Has], [Joly] , [Lang 1],
[Mac], [Mor] , [Sam], [Ser 2], [Wei]. The book giving the quickest and most complete access to algebraic number theory is that of P. Samuel, cited above, and from which this Introduction has largely been drawn.
182
CHAPTER 4
PROBLEMS
EXERCISE 4·1: EXTENSION OF A FINITE FIELD Notations and revision:
k
K
=F
denotes the finite field with q
q
= F
q
p
r
elements (p a
prime number); n
denotes the extension, of degree n, of k.
We denote by
xq
the k-automorphism of K defined by o(x)
0
and we recall that
is a generator of Gal(Klk).
0
We denote by T(x) and N(x), respectively, the trace and norm, relati ve to the extension KI k of an element x e K. I:(l):
Show that the mapping T:K
I: (2) :
Let
-&
e
K
x
1
-&
T(u) (-&o(u)
k is surjective.
be an element of the form
(where x e K); show that T( -&) Conversely, let
+
-& = x - o(x)
= O.
e K be an element satisfying T( -&) = 0; set: +
(-&
+
2
0(-&»0 (u)
+ ••• +
+
0
n-2
(-&
(-&»0
+
0(-&)
n-l
+
(u»,
ALGEBRAIC THEORY
183
+ 0).
(where u e K is an element such that T(u)
Verify that -&
= xl
x - o(x), and that every xl e K such that -&
- 0(x1 ) is of the
form xl = x + A, where A e k. Let -& e K be an element of the form -&
II: (1) :
= 1.
x e K is a nm-zero element); show that N(-&)
x
o(x) (where
Conversely, let -& e K be an element such that N(-&) 1; show x that there exists an element x e K such that -& = o(x). Verify xl
that every element xl e K such that -& = ~ 0~X1J
.
1S
of the form xl
AX, where A e k is a non-zero element.
Show that the mapping N:K
11:(2):
In this Part it is assumed that n divides q - 1.
III: denote by
~n
We
the subgroup of the n-th roots of unity of k*.
III:(1):
Let A
= {aeK~':anek'~};
group of K* containing k*. by ~(a)
k is surjective.
+
= ata)
Denote by
A is a multiplicative sub~:A + ~
n
the mapping defined
Verify that ~ is a group homomorphism and induces
an isomorphism: iji:A/k'~
+ ~n .
111:(2): ~(a)
in
~n.
Let a be an element of A and let d be the order of Show that [k(a):k)
= d.
Give the factorisation of
the polynomial Xn - an as a product of irreducible polynomials of
k[X). 111:(3):
Show that the mapping W:A
+
k* defined by w(a)
is a group homomorphism which induces an isomorphism:
Show that every element of k is an n-th power of an element of K.
a
n
184
CHAPTER 4
EXERCISE 4-2: EXTENSIONS OF DEGREE P OF A FIELD OF CHARACTERISTIC P Let k be a field of characteristic p prime subfield.
+ 0,
and let n be its
Let P(X) = xP - X - a, where a e k.
Let a be a root of P(X) in an algebraic closure of K.
(1):
Express all the roots of P(X) as a function of a. (2):
Assume that there exists an irreducible polynomial
Q(X) of k[X] of degree r Show that r = p.
~
2 that is a factor of P(X).
(Consider the term of degree r - I of Q(X)).
Give a necessary and sufficient condition for P(X) to be irreducible over k. (3):
Assume P(X) to be irreducible on k.
Let a be a root
of P(X); show that K = k(a) is a cyclic extension of k of degree
p. Let P(X) =XP - X - a
(4):
and Q(X) = xP - X - b be two
polynomials of k[X] which are assumed to be irreducible. Show that in an algebraic closure of k the splitting fields of these two polynomials coincide if and only if b is of the form
b = ta + c P - c (where c e k, ten, t (5):
+ 0).
Let K be a cyclic extension of k of degree p.
Show that there exists an element a e K such that K = k(a), a being a root of a polynomial P(X) = xP - X - a (where a e k) . EXERCISE 4-3: EMBEDDING OF A QUADRATIC FIELD IN A DIHEDRAL EXTENS ION OF III
Notations: m denotes a rational integer different from I and -3 that is not divisible by any square p
1m (resp. -3);
2
of a prime number p.
~) denotes one of the square roots in ~ of
m (resp.
I-3m denotes the product ~Im ; j denotes the complex num-
ber I - ~ 2
ALGEBRAIC THEORY
185
PRELIMINARY QUESTION: splitting field over
Let L be the subfield of
mof
the polynomial X3 - 2.
extension Lim is Galois and non-abelian.
~
defined as a Recall why the
Frum this deduce that
the quadratic field m(~) possesses an extension L of degree three that is Galois and non-abelian over
m.
We propose to show that every quadratic field possesses this property. Let k
= m(i-3m). Show that the extension kim is Galois of degree two; write f for the m-automorphism of k different from (1):
the identity, and for ]J e k set ~ = f( w. Show that there exist elements ]J e k such that ]J2p is not the cube of an element of
k. In what follows]J denotes an element of k such that ]J2p is not the cube of an element of k. Let K be the subfield of ~ defined as the splitting field on k of the polynomial X3 - ]J2p. The roots of this poly(2):
nomial in ~ will be denoted a,ja,j2a . (a):
Show that the extension Klk is Galois of degree
(b):
Show that there exist two k-automorphisms
six. 0"
of K
such that:
/iii, ,( /iii)
- /iii,
(c):
1=3, ,( 1=3)
= Gal(Klk).
ja,
- 1=3,
Show that the group
exactly the group G
o(a)
(0,,)
~enerated by
Is it commutative?
° and,
is
Find all the
intermediate fields between k and K. (3):
Find the polynomial Irr(a,m); what are the conjugates
of a over Ill?
186
CHAPTER 4
Show that the extension Show that there exists a a( 1=3)
rm,
K/~
is Galois of degree twelve. a of K such that:
~-automorphism
- 1=3,
Ct
a(ct}
2
]..I
2
Calculate a . Show that the group G =
(4):
of the subgroup
(a)
Ga1(K/~)
is the direct product
(generated by s) and of the subgroup G.
Let
N be the subfie1d of K defined as the fixed field of the subgroup
(a) . Verify that ~(rm) C N and that [N:~(rm)] extension
N/~
is Galois and non-abelian.
=
3.
Show that the
From this deduce that
every quadratic field possesses an extension of degree three that is Galois and non-abelian over
~.
EXERCISE 4·4: BIQUADRATIC EXTENSIONS OF Let ml' m2 be two
~
distinct square-free integers different
from zero and one; kl (resp. k 2 ) denotes the extension obtained by adjoining to the rationals a root ~l (reap. ~2) of the equa° 2 0 ( 2 ) tlon x - m1 = reap. x - m2 = o. K d enotes t h e fOleI d composof kl and k 2 • The ring of integers of a finite extension E of the rational numbers is denoted 0E and the discriminant of ite
Recall that if Ct i , (i = 1,2, ... ,n = [F:E] )are elements of Or whose conjugates relative to . . -- 1 , 2 , .. . ,n, t h en ~(r/E) dOlVl°des E are d :note d Ct i(j) ' ~,J (det(Ct ~ J)))2. an extension F of E is denoted
~(r/E).
1
I: (1):
Show that KI~ is Galois.
I: (2):
What is the degree
I: (3):
Prove that there exists a
isfying:
[K:~]
of K relative
to~?
~-automorphism
a of K sat-
ALGEBRAIC THEORY
187
and am-automorphism
T T(~
of K such that: 2
=-
)
~
.
2'
deduce from this the Galois group of Kim. 1:(4):
Show that K contains one and only one quadratic field
k3 distinct from kl and k 2 , what is the discriminant as a function of m1 and m2 ? 1:(5):
Can a prime ideal of m different from (2) be rami-
fied simultaneously in k 1 ,k 2 ,k 3? Give an example showing the same does not happen with the ideal (2). 1:(6):
Show that ~K/k. divides ~k./m if j J
1
+ i.
From this
deduce that a prime ideal of m which is ramified in K is ramified in at least two of the fields k 1 ,k 2 ,k 3 . II:
Assume that m1
11:(1):
= -1
and m2
=1
mod 4.
What is the number of ideal classes (class
number) of Ok . 1
11:(2):
Is the Ok -module Ok free? 1
By using Question I(b) above, calculate the dis-
criminant of Klk 3 .
ramify in K?
What prime ideals of Ok 3
III: What are the possible different decompositions of a rational prime ideal in K? EXERCISE 4-5 STUDY OF A CYCLIC CUBIC FIELD WITH PRIME CONDUCTOR Let m be the field of rational numbers and p a prime number congruent to 1 modulo 3. I:
Let j
+1
be a cube root of unity, k the field m(j), A
the ring of integers of k.
188
CHAPTER 4
Verify that A is a principal ideal domain and that
1:(1):
in A, p becomes a product of two distinct prime ideals, therefore that there exist rational integers aO'SO for which p
= Nk/~(aO + jSO)'
Find, as functions of ao'So' all the pairs (x,y) of
1:(2):
2 2 3x 4+y
l 'lntegers satls ' fylng ' . ratl0na p
Deduce from this
that there exists one and only one pair (a,b) of rational integers for which: a > 0,
(1)
b - 1 mod 3,
(2)
27a 2 + b 2 4
p II: of
~
Let w
+1
(3)
be a p-th root of unity, and
obtained by the adjunction of w.
~'
the extension
Denote by G the Galois
*
group ~'/~ and its elements by si (i mod p, i 0 mod p), S being defined by the condition s.w' = w,i for every p-th root of unity
1
w' •
11:(1):
Verify
that~'
of degree three over 9). ure of:;z in
possesses one and only one subfield
Denote this field by K, the integral clos-
by
B,
and the discriminant of
K is cyclic over
~,
and that D
K
K
by
D.
Verify that
= P2
II: (2): Let t be a generator of (:?z,/p:;z)'" and H be the subgroup (:?z,/p:;z),.,3 of (:?z,/ :;z) 1, • Let us set: i
ieH
w
ti
ieH
w
ieH
Verify that 11,11' ,11" are elements of K conjugate over 9) and that
K=
~(11)
= ~(11') = 9)(11").
189
ALGEBRAIC THEORY
11:(3):
Calculate.fr +.fr' +.fr" and .fr.fr' + .fr.fr" + .fr'.fr".
II: (4) :
To calculate the product .fr.fr'.fr", set
and
l'
assume that in the expression:
I
i,jeH there are k terms equal to One equal to
(therefore there are
kx1'
terms
one in .fr.fr' .fr") .
Verify that the minimal polynomial of .fr is then:
P(x) = x
3
+x
2
-
1'X
+
l'
Let 6 be its discriminant.
2
-
3
k
P 2 2
Show that we have 6
P a
with
a elN.
In calculating 6 one finds that k must satisfy an equation of the second degree the discriminant of which is the square of an integral number b'.
Then verify that we have p
= 27a
2 + b2 4 an d
that:
P(x) = x
3
+x
2
p - 1
- --3- x
_ p(3 + b) - 1 27
a,b being the integers defined in Question I. N.B. The discriminant of the polynomial x
111:(1):
3
+ a 2x
2
+ a1x + a o is:
Let q be a prime number distinct from p.
Verify that the following conditions are equivalent: (i): q lies in K and is a product of three distinct prime ideals;
190
CHAPTER 4
m'
(ii): The number of prime factors of q in
is divisible
by three; (iii): q(p-l)/3
=1
mod p;
(iv): There exists x eZl such that q 111:(2):
Verify that for q
+p
cannot have a triple root in Zl/qZl.
=x 3
mod p.
the equation P(x)
=0
mod q
From this deduce that q de-
composes in K if and only if the congruence P(x)
=0
mod q has a
solution. 111:(3):
Show directly that 2 decomposes in K if and only
if a,b are even.
Deduce from this that in order to be able to
write a prime number p in the form p it is necessary and sufficient that p
= 27x2
=1
+ y2, with x,y eZl,
mod 3 and 2(p-l)/3
=1
mod p. III:(4):
Show that
3~
+ 1 is a root of the polynomial x
3px - pb. From this deduce that if q ing b, then q(p-l)/3 = 1 mod p. 111:(5):
~
3
5 is a prime number divid-
Show that if 3 divides a, 3(p-l)/3 _ 1 mod p.
EXERCISE 4·6: STUDY OF A FIELD m( 3~) The goal of this Problem is the study of certain fields obtained by adjoining to
mthe
not the cube of an integer.
cube root of an integer which is In order to simplify matters we re-
strict ourselves to the case where d is square-free and d
Cd = ±2,±3, or 3 polynomial f(x) = x
mod 9
I:
±4 mod 9).
We write
~
- d and K for the field
* ±l
for a root of the m(~).
INTEGERS (1):
Calculate the discriminant ~(1,~,~2).
(2) :
Show that the prime factors of d are totally rami-
ALGEBRAIC THEORY
fied in X.
191
Show that the same is true for 3 (consider the poly-
nomials f(x) ,f(x + 1) ,f(x - 1)). From the preceding deduce that 1,~,~2 is a basis of
1:(3):
the ring OK of the integers of X, and calculate the discriminant of
D
X.
II: DECOMPOSITION IN X OF PRIME NUMBERS
Notations: If p/3d we shall write p
= p3. P
Let p be a prime number which does not divide 3d.
11:(1):
Show that in order for p to be inert in X it is necessary and sufficient that the congruence d 11:(2):
Show that if p
= -1
=x 3
mod p be impossible:
mod 3, p decomposes in X.
addition verify that -3 is not a square in that a prime number p
Pp P'p' where
p
p
= -1
p
Let q
mod 3 decomposes in X into the form p
=1
In
From this deduce
is a prime ideal of degree one and p
ideal of degree two. 11:(3):
~.
is a prime
mod 3 be a prime number not dividing 3d.
Show that q remains inert or becomes in X the product of three prime ideals of degree one (we can show that -3 is a square in II! ). p
III: IDEAL CLASSES Recall, with the usual notations, that the Minkowski constant is: M
=
(~r2 n~ /iDT.
III: (1):
n
Show that:
192
CHAPTER 4
111:(2): Show that if a prime number does not remain inert in K it has a principal prime factor if and only if the equation:
has a solution with a,b,csZl. APPLICATION: Verify that for d
III: (3):
= 2,
Ok is a prin-
cipal ideal domain IV: STUDY OF THE CASE d IV:(l):
= ±2
Let m
7
or ±3 mod 7 be an integer.
Show that the equation:
N(a +
b~
+
c~
2
)
=±m
has no solutions with a,b,c sZl.
Verify that P2 P3 is principal. From this deduce that the class number of K is equal to three. 2
Denote the class of P3 by h (thus P3 s h, P2 s h ). IV:(2):
= -1
Let p
mod 3 be a prime number.
By considering the equations:
N(a + show that p mod 7.
b~
p
+
c~
2
)
= p,3p,9p,
and p' are principal ideals if and only if p p
In general determine the class of
the two cases p
= ±2
and ±3 mod 7.
in the method used for p IV:(3):
Let 7
+q
= ±1
= ±1
Pp by distinguishing
(Inspiration will be found
mod 7).
_ +1 mod 3 be a prime number which decom-
poses in K. By using the method of the preceding Question show that the prime factors of q are in the same ideal class and determine
ALGEBRAIC THEORY
193
this class by distinguishing the three cases q _ ±1,±2,±3 mod 7. EXERCISE 4-7: CLASS FIELD OF m(/65) Let m be the rational number field, k (1): (2)
= m(/65) and
W
= /65.
Show that the ideal (2) splits in k into the form
= P1 P2 where Pl,P2 are prime ideals.
Show that Pl is gener1 ated by 2 and - 2 - and P2 by 2 and - 2 2 W + 1 Show that Pl is a principal ideal generated b y 4 + 2W
(2):
+ 1
W -
Sho"T that the ideal Pl is not principal (assume that
P2 is principal and generated by 2a + b(W ; 1) , where a,b are integers, write Nk/ m(2a ~ tion of the form 8
=x2
+b(W +2 1))
-= ±2, and reduce to an equa-
- 6sy2, which does not have a solution
in integers). (3):
Show that the group of ideal classes of k is of order two.
(4):
Show that in k there exists a unit u with norm over
m
equal to -1. (S):
Let
E
be a fundamental unit of k such that the comple-
tions of k(l£) for the Archimedean absolute value are isomorphic to the real number
fie1d~.
Show that there exists an ideal of k which is ramified in k( 1£) (it suffices to notice
(6):
that k( 1£)
k( /U)
or k( 1£)
= k( vCU») .
Let K be a quadratic extension of k whose completions
at the Archimedean absolute values are isomorphic to
~
and whose
discriminant over k is (1) (that is to say, there are no ideals of k ramified in K). Letting -E
E
always be a fundamental unit of k, show that
E
and
are not squares in K. (7):
Let us denote the Galois group of Kover K by (1,0).
CHAPTER 4
194
Show that for every unit a of K, a + a(a)
+ O.
(8): Show that there exists a non-zero integer be K such that b = -a(b). (9):
Show that the principal ideal (b) is the extension of
a non-principal ideal
A of k, that is to say, that if Ok and OK
denote the integral closures
of~
in k and K then:
Let Kl ,K2 be two fields having the properties in Question (6) above, show that Kl = K 2 • (10):
(11):
Show that
m(IS,if3) is the only field K that has the
properties of Question (6) above.
EXERCISE 4·8: STUDY OF THE NORM IN AN UNRAMIFIED EXTENSION OF A LOCAL FIELD Throughout the Problem k denotes a loaal field (that is to say, a complete non-Archimedean valued field with discrete valuation and finite residue field), and
K
denotes a finite extension
of k, furnished with the absolute value extending that of k, that is unramified (that is to say, e(Klk)
= 1).
Notations: (af. The introduction of Chapter 10 on p-Adic Analysis): [K:k] = n, n the uniformiser of k; 1·1 the absolute value in k and K; Ok = {x e k: Ixl ~ 1l; OK = {x e K: Ixl ~ 1l; k. (resp. K) the residue field of k (resp. K) (7< will be identified with a subgroup of K); x ~ x the canonical mapping of OK into K; cardk.
= q = pro
(1) Verify that the extensions Klk following satisfy the conditions above: (1)
(a)
k
=m p
(the p-adic number field); K
=m (a), p
a be-
ALGEBRAIC THEORY
195
ing a root of the polynomial X2 - d (d a rational integer not divisible by p and not congruent to a square modulo p). (1) (b)
k =m 5 (S), S being a root of the polynom5; K = m
ial X4 - 2. (2)
Prove the following results:
(2) (a)
[R:k]
(2) (b)
Every element of
n
Xq
-
=
n. n
is a splitting field over k of the polynomial Xq
X; K
(2) (c)
-
X.
Klk is Galois; its Galois group G(Klk) is cyclic
and generated by the element T(W)
K is a root of the polynomial
T
defined by:
= wq for all we K.
(Recall that the multiplicative group of non-zero elements of a finite field is cyclic). (3) n
ial Xq
Show that K is a splitting field over k of the polynom- X, and that there exists a root ~ of this polynomial
such that K
= k(~).
From this deduce that the extension Klk is
Galois. Let a e G (K Ik) be a k-automorphism of K. Show that the mapping 0, which to x e K associates o(x) = ax (for an ele(4) (a)
ment x e OK)' is well defined and that it is a k-automorphism of
K.
(4) (b) ~(a)
= ais (4) (c)
Show that the mapping ~:G(Klk) ~ G(Rlk) defined by an isomorphism. From this deduce that the group G(Klk) is cyclic
and generated by the element a* determined uniquely by <
1 for all x e OK.
Ia*(x) - x q I
196
CHAPTER 4
Describe
(4) (d)
0*
for the examples (I) (a) and (I) (b).
(5) (a) Let m ~ I be a rational integer and let ]J e I + m m m It OK' that is to say ]Jm = I + alt (a e OK) . Show that:
Show that the mapping w:OK ~ k, which to an element aeOK associates w(a) = TrKlk(a), is surjective. From this m deduce the following result: For any Am e I + It Ok there exists (5) (b)
m
].lm e I + It OK such that: (mod
m+l
It
),
which can be written: A
Nrn Klk]Jm ( ) = -Am , m+l
Show that:
(5) (c)
(Uk
(resp.
where Ale I + Itm+l0 K· m+
UK)
denotes the multiplicative group of invertible
elements of Ok (resp. OK)). (5) (d)
Let d be as in Question (I) (a).
Show that every
p-adic number x with absolute value I can be written: x
= a2
(6)
- db 2
with a,be!!! . p
Denote by k* (resp. K*) the multiplicative group of
non-zero elements of k (resp. K). Show that the group k*/NrnKlk(K*) is a finite group isomorphic to Gal(KI k).
ALGEBRAIC THEORY
197
EXERCISE 4·9: PURELY TRANSCENDENTAL EXTENSIONS 1
Let K be the field of rational functions in an
terminate X over the field I
mof
in~e
rational numbers.
In the ring K[Y] show that the polynomial y2 + X2 + 1
(1)
is irreducible. Let E be the extension of K generated by a root of
I (2)
this polynomial. Show that E is not a purely transcendental extension of K. Let i e
I (3)
0:
be a root of X2 + 1.
Show that E(i) is a
purely transcendental extension of m(i). In the ring K[Y] show that the polynomial
I (4)
is irreducible.
Let F be the extension generated by a root of
this polynomial. II
Let L
i - X2 + 1
Is F a purely transcendental extension of
= o:(X)
m?
be the field of rational functions in an
indeterminate X over the complex numbers. II (1)
Show that the polynomial y3 + X3 + 1 is irreducible
in the ring L[Y]. II (2)
Let M be a splitting field of this polynomial over
L. Show that M is not a purely transcendental extension of
0:.
198
CHAPTER 4
SOLUTIONS
SOLUTION 4-1: 1:(1):
Every extension of a finite field is separ-
able; it is known that the trace is then surjective ([Bou] Chapter
V §10, or [Jo1y] Chapter
1).
This result can be proved directly; obviously it suffices to show the existence of an element of K with non-zero trace.
If
the degree n of the extension is prime to p it suffices to notice that T(l) = n, and therefore T(l)
+ 0.
m In the general case let us set n nOp , no being prime to p; nO m let L be the fixed field of (J ; then [K:L] = p and [L:K] = no ([Lang 1] Chapter 8). We have just observed that the trace in
the extension L/K is surjective; as Tr K/ k = Tr L/ k Tr K/ L ([Bou] Chapier V) it suffices to show that the trace of the extension K/L is surjective.
For the same reason, by considering the ex-
tensions intermediate between K and L, we can reduce to an extension of degree p. Therefore let us assume that K is an extension of degree p of a finite field L, and let us show that there exists an element of K with non-zero trace over L.
Let
lic group K* (cf., [Sam] Chapter
1), that is to sayan element
CI.
be a generator of the cyc-
of K of order q'P - 1 (where q' denotes the cardinality of L).
ALGEBRAIC THEORY
199
Let: -i + a.XP +
P(X)
1
be the irreducible polynomial of a over L. a2 = = a p _1 = 0, otherwise (by writing 000
ism of
K
defined by
is absurd, because
T(X
(:a)
~ = =
+ ap
000
We cannot have a 1 = T
for the L-automorph-
xq' ), we would have (:a) p = 1 which aP(q -1) f
l, since p(q' - l)
< q'
P- 1.
Consequently, if i denotes the smallest index such that a. f 0, 1
we have:
I: (2) :
T('O)
If '0 e K is of the form '0
=x
- 0' (x ) then obviously
= T(x) - T(O'(x)) = 0.
Conversely, let
'0
be an element with vanishing trace, and u
an element with non-zero trace.
x
1 = T(u)
('OO'(u) +
000
Set:
L
+
O'i('O))O'j(u)
+
000
o~i~j-1
L
+ (
i
0' ('0))0'
n-1
(u)).
O~i~n-2
We have: O'(x)
1
= T(u)
(0'('0)0'
2
(u) +
000
+
i+1
j+1
L
0'
L
O' i +1 ('O))u).
('0))0'
(u)
O~i~j-1
+ (
O~i~n-2
Whence: x - O'(x)
= _1_ T(u)
('O(
L
1~j~n-1
O'j(u))-
L 1~j~n-1
O'j('O))u)
+
000
200
CHAPTER 4
1 = T(u)
(.e(T(u) - u) -
If xl e K is such that xl - oX 1 xl - x
is such that ;\ =
0(;\),
=x
(T(.e - .e)u)
- ox,
.e.
then the element ;\
and therefore ;\ e k.
REMARK: The above proof applies in fact to every cyclic extension of an arbitrary field k ([Bou]
Chapter V §ll) (additive form of
Hilbert's "Theorem 90"). 11:(1):
If.e is an element of the form.e
x
o(x) ,
where x e K
is non-zero, then: N(.e )
N(x) N(o(x) )
1.
Conversely, let .eeK have unit norm. N(.e)
= .e 1 +q+"'+ q
n-1
We have:
n .e(q -1)/(q-1)
= 1.
Let a be a generator of the cyclic group K* and let us set .e As a has order qn - 1 in K*, the equality:
implies that q - 1 divides i. we have:
and therefore if x
If x
1
a
-m
e K* is such that:
Let us set i
= m(q
- 1).
Then
i a .
ALGEBRAIC THEORY
then A
Xl
= -X
201
..
satIshes
a(A)
= A,
and therefore
Aek.
REMARK: For every cyclic extension K of a field k the elements of
K with unit norm are the elements of the form atx) (where X e K* and a is the generator of Gal(KI k» (Hilbert's "Theorem 90", cf., [Bou] Chap ter V §11, or [Mac] Chapter 2 §S). II (2)
The image under the norm of the multiplicative group
K* is a multiplicative subgroup N(K*) of k*.
Let us denote by
kerN the set of elements of K* of unit norm; it is a subgroup of . 1 Ity . cardK* b y t h e preced Ing . • h card Ina K..• WIt problem. Since there cardk* is an isomorphism N(K*) ~ K*/kerN we have: cardN(K*)
cardK* card kerN '
and therefore: cardN(K*)
= cardk*.
From this it follows that N(K*)
= k*
(cf., [Joly] Chapter
1).
A is evidently a multiplicative subgroup of K* containing k*. It is clear that cp Isahomomorphismofgroups; if a' Aa , with Ae k*, then cp (a' ) = cp (a), and therefore by passage to the quotient cp defines a homomorphism ~:A/k* + ~. The homomorphn ism ~ is surjective, for a(a) = a implies a e h''<; ~ is surjective, because an element l; e ~n has norm one, hence there exists (by III (1)
Question II(1» III (2)
an aeK such that
l;
a = a(a)
and then a is in A.
Let a e K* be such that an e k 1., and let d be the
order of cp(a) in ~n' By Question 111(1) the integer d is also the smal1est integer greater than zero such that ad e k 1•• The conjugates ai(a) = l;ia (0 ~ i ~ n - 1) (where l; = cp(a» satisfy ai(a)
= aj(a)
if and only if i - j is a multiple of d.
Since a
has d distinct conjugates, the extension k(a)la has degree d.
202
If
CHAPTER 4
is an n-th primitive root of unity, we have:
~
and the polynomials Xd - ~iad are irreducible polynomials of
k[X] . 111:(3):
The mapping
Wis
= Anw(a) (where
such that w(Aa)
evidently a group homomorphism aeA, Aek'~).
By passage to the
quotient one therefore obtains a homomorphism:
The homomorphism
¢
is injective, for ~(a)
= an is in k*n if and
only if a e k",; on the other hand k-J'/k,·,n is a cyclic group isomorphic to
~n'
and therefore the finite groups ¢(A/k*) and k*/k* ,
having the same cardinality (Question 111(1)), coincide.
Conse-
quently every element of k is the n-th power of an element of k. REMARK: The study above (Question III) is a particular case of Theory, that is to say, K/k is abelian of degree n, and
KUmmer
k is a field, whose characteristic does not divide n, which con-
tains the n-th roots of unity (cf., [Mac] Chapter 2 §7). SOLUTION 4·2: (l):
Let t e 1£; we have t P
= t.
Therefore:
P(a + t) = (a + t)p - (a + t) - a = a P - a - a = P(a) = O. If a is a root of P(X), (a + t)t (2):
e1£
is the set of roots of P(X).
Let us assume that P(X) possesses an irreducible di-
visor Q(X) of degree r
~
2.
Q(X) is equal to a product of r
polynomials X - (a + t), where t e 1£.
The terms of degree r - 1
of Q(X) therefore has a coefficient of the form -ra + u, where U
e 1£.
But then, if r
<
p, a e k and Q(X) is not irreducible.
ALGEBRAIC THEORY
= p.
Hence r
203
Consequently, P(X) is irreducible in k[X] if and
only if P(X) has no root in k; in fact by the preceding either P(X) is irreducible or P(X) has all its roots in k. If p(X) is irreducible on k, with a denoting a root of
(3)
P(X), one sees (Question (1)) that all the roots of P(X) belong to k(a); k(a) is therefore a normal extension of k ([Lang 1] VII As a is separable over k the extension k(a)/k is separable
§3).
([Lang 1] VII §4).
Consequently k(a)/k is a Galois extension of
prime degree p, therefore cyclic. Let K = k(a) and let
(4)
such that o(a) = a + 1.
0
be the generator of Gal(Klk)
Let us look for the elements 6 e K for
which the irreducible polynomial on k is of the form Irr(6,X) We have
X2 - X-b.
o(~] = %+
t6 -
a a =t
(X ;
0(6)
= 6 + t, where t e It, t '" 0, whence
1, and therefore where a e k.
0(% - a] = ~ - a.
Consequently
6 is therefore a root of the polynomial:
r _ a) - a.
a
(X ;
Whence: (X - a)p - (X - a) - at
Irr( 6,X)
xP
- X - (at + a P - a),
and hence: b
= at + a P
- a.
Conversely, if 6 is a root of the polynomial: (X - a)p - (X - a) - at we have 6 = ta + a, and therefore 6 e K and generates (5)
Let
TrKlk(-l) = O.
K
K
on
be a cyclic extension of k of degree p.
k.
We have
Consequently (by the additive form of Hilbert's
204
CHAPTER 4
Theorem 90, cf., Problem One) there exists a e K such that
=1
(0 being a generator of Gal(Klk)).
00. -
a
Whence:
(O~i~p-l).
As a has p distinct conjugates we have K = k(a). Furthermore, a is a root of a polynomial xP - X - a, where a e k; in fact:
and therefore:
SOLUTION 4·3: PRELIMINARY QUESTION:
The
splitting
polynomial X3 - 2 is the field L
= ~(3/:2,~).
field over
(j}
of the
Let us denote by
~ the ~( 3/:2)-automorphism of L such that ~(~) = -13, by ~ the ~(~)-automorphism of L such that ~(3/:2) = j 3/:2 ([Mac] Chapter 1). One verifies that ~~( 3/:2) = j 3/:2 and ~~( 3/2) = / 3/2. Consequently the group Gal(LI~) is non-abelian.
This is a group of order six
whose distinct elements are:
In terms of generators and relations Gal(LI~) is defined by generators ~
~,~
2
and relations: 3
= ~ = e,
Up to isomorphism such a group is the only non-abelian group of order six (the dihedral group of order 2m is by definition the group defined by the generators ~~
=~
-1
~,~
~).
and relations:
ALGEBRAIC THEORY
205
The extension ~(!-3m)l~ is evidently Galois of de-
(1)
gree two since m is different from -3 and square-free as well. Let A denote the ring of integers of k and let p be a prime number split in k ([Sam] Chapter
5), that is to say that pA
=
pp', where p,p' are prime ideals distinct from A. The image under the
~-automorphism
f of k of the prime ideal
p is evidently a prime ideal of A dividing p, hence we have
f(p) = p or f(p) = p'.
If we had f(p)
= p,
then f(p')
Let, then, x be an element of p' not in p.
NKI~(x)
= p'.
= xxepA,
and therefore xx e p without either x or x belonging to p, which is absurd (cf., [Mac] Chapter 5 §4).
Consequently f(p)
p', and
= p.
thus f(p')
Let ~ be an element of p that is neither in p2 nor in pp' , (p2 U pp' is distinct from p since p Up' is distinct from A); this
*
We have ~A = pq, where the choice of ~ implies that ~ p' . ideal q is divisible neither by p nor p'. From this we deduce
pA = p'q', where q' is divisible by neither p nor p'. Therefore 2-A . not d'1V1S1 . . bl e b y p 3 As a resu 1t ~ 2_. ~ = p2, p q 2q , 1S ~ 1S not the cube of an element of k. ~
Let a b e a cub e root 0 f
(a)
~
2_. ~
1n
~.
The extension
k(a)jk is of degree three and not Galois since k does not contain cube roots of one. ial
X3 -
The splitting field K of the polynom-
~2p on k is the field K = k(a,j) = k(a,~) of degree
six over k. (b) that o(a)
There exists a k(~)-automorphism a of
= ja
-~ ([Mac] Chapter I §2).
we have o(rm) have 'T (G )= (c)
K
such
and a k(a)-automorphism T of K such that T(~)
= G. -G.
Since o(~)
As T(r-3)
= -r-3
=~
and aU-3m)
and TU-3m)
= I-3m,
=I-3m we
It is clear that a is or order three and T of or-
der two; furthermore, aT
= TO
(for oT(a)
= ja
and To(a)
j2a ).
The group (O,T) is therefore isomorphic to the dihedral group of
206
CHAPTER 4
order six; this is a subgroup of Ga1(Klk) and so they must be the The subgroups of G distinct from G and {e} are the subgroups (0),(T),(TO),\T0 2). There are same since they have the same order.
thus four intermediate fields between k and K: k(~), the fixed field of (0); k(a), the fixed field of (T); k(ja), the fixed field of \ TO); k(j2 a ), the fixed field of \ T02) ([Lang 1] Chapter 8). The extension k(a)l~ has degree [k(a):k] [k:~] = 6;
(3)
therefore a has degree over
~
a divisor of six.
The irreducible polynomial of a over k, Irr(a,k) = X3 - ].I2iJ P(X) does not have its coefficients in ].I
+ iJ.
The polynomial
Irr(a,~)
~,
for by the choice of ].I,
is therefore a multiple of P(X)
and of strictly greater degree; so, necessarily: .
d
o
Irr(a,~)
=
[~(a):~]
= 6.
Let P(X) = X3 - ].I2iJ be the image under f of the polynomial P(X).
The polynomial P(X)P(X) has coefficients in
a multiple of the polynomial
Irr(a,~).
~,
hence it is
As these two polynomials
have the same degree we have:
2
2
We see that the roots of the polynomial P(X) are ~ , j a].l , .2 a 2 . . .2 a2 J - - ; the conJugates of a on ~ are therefore a,Ja,J a, -- , ].I ].I 2 2 . a .2 a Th 1" f' 1d f h 1 . 1 J -- , J e sp lttlng le on ~ 0 t e po ynomla ].I ].I Irr(a,~) is thus the field ~(a,j,].I) = k(a,j) = K (cf., Question (2)(a)).
Consequently ([Lang 1] Chapter VII) the extension
KI~ is Galois; its degree is [K:~]
=
[K:k] [k:~]
= 12.
The field ~(Irn) is contained in K, and as [~(Irn):~] = 2, [K:~(Irn)] = 6.
We have K = ~(Irn,~,a).
The degree of a over
~(Irn,~) is therefore equal to three and Irr(a,~(Irn,~»
X3 - ].I2iJ.
Irr(a,~(Irn»
=
This polynomial is not in ~(Irn)(X), consequently is a proper multiple of X3 - ].I2iJ.
The degree of a
ALGEBRAIC THEORY
207
over ~(fIn) is therefore greater than or equal to four and divides six, whence:
and:
From this we deduce the existence of a ~(fIn)-automorphism s of 2
K such that sea)
=
~
= P and
~p e~, s(~)
Moreover, s2(a)
~
,
whence s(~2p)
sU-3m)
-I-3m.
= s(a~2) = ~ = a. 2_ ~
(4):
=
s(p2~).
Therefore since
We have s( r-'3) = -1=3. As K
= ~(a,~) we can con-
~
The group G = Ga1(KI~) has order twelve; it has sub-
groups G = (O,T), of order six, and (s) which has order two. The automorphism s is not one of the three elements of order two of G, T,TO,T0 2 , because these leave l-3m invariant. Therefore s.G, and so G = G U sG . To show that G is the direct product of to show that os
= so
and
TS
= ST,
(s)
and G it suffices
which is easily established
by looking at the action of these automorphisms on fin, r-3 and a. G is therefore the direct product of G and
(s).
Let N be the fixed field of s. six, and it is Galois since tain ~(fIn).
(s)
The extension NI~ has degree is a normal subgroup of G; N con-
Furthermore, Gal(NI~)
and therefore Ga1(NI~) ~ G.
G/(s) ([Lang 1] Chapter
VIII)
Thus we have obtained an extension
N of ~(fIn) that is Galois over ~, and such that Ga1(NI~) is isomorphic to the dihedral group of order six. (Cf., J. Martinet and J.J. Payan: 'Sur 1es extensions cubiques non-ga1oisiennes des ratione1s et leur cloture ga1oisienne', J. reine angew. Math., 228, (1967), 15-37).
CHAPTER 4
208
SOLUTION 4-4
(1)
As the composition of two Galois extensions
is Galois ([Lang 1] Chapter VIII), and as k 1 ,k 2 are quadratic and thus Galois, K = klk2 is therefore Galois. I (2) I (3)
From K
= k2(~1)
it follows that every element of K
can be written uniquely as A + I.l~ l' A, I.l e k2' and that the mapping is a k 2 -automorphism of K (and therefore a m-automorphism of K). Similarly one sees that T is the non-trivO:A +
I.l~l ~
A-
I.l~l
ial k1-automorphism of K.
It is clear that Gal(Ki m) is made up of
the m-automorphisms idK,o,T,OT.
Ga1(Klm) is therefore isomorphic
to the Klein four group. I (4)
Let k3 be the subfie1d of K formed by the elements
of K invariant under OT
= TO;
k3 has degree two on m.
This is
the only quadratic subfie1d of K outside of kl and k2' because
Ga1(Klm) only possesses three subgroups of index two. m( Im 1m2 ) is contained in K, is quadratic over m, and is distinct
= m(lm1m2 ). Let us denote by m3 the square-free part of m1m2 . It is clear that k3 = m(~) and that ~k3/m = m3 (resp. 4m 3 ) if m3 = 1 mod 4 (resp. m3 1 mod 4).
from kl and k2' hence we have k3
*
The calculation of ~k 1m shows that ~k Im'~k 1m' 312 1m can have no other common prime divisor than two. A prime I (5)
~k
3
ideal (p) of m, with p
+2
k2,k3 simultaneously. The example m1 = 2, m2 same does not hold for 2. I (6)
cannot, therefore, be ramified in k1 , -1, and thus m3
Let Cl,a i ) be a Zl-basis of Ok.. ].
-2, shows that the
We have:
ALGEBRAIC THEORY
209
It is known that ~K/k, (if j J
+i)
divides ~k,/m (by the review ~
at the beginning of the section), if j
+i,
~K/k,
ified in Kim.
J
From this it follows that
Let (p) be a prime ideal of
divides ~k,/m' ~
Let us assume it is not ramified in k,/m. J
mram-
At
least one of its prime divisors in k, is therefore ramified in
Klk j and consequently divides for i with
+ j, and i + j.
~K/k,'
J
As a result p divides
~k./m
J
~
therefore p is ramified in the two extensions k./m. ~
The class number of Ok ' the ring of Gaussian in-
II (1)
1
tegers, is equal to one,
The Ok -module OK' which is of finite 1
type and torsionfree over a principal ideal domain, is therefore free ([Sam] Chapter II (2)
1),
We have k3
The ideals of Ok
= m(!-m 2 )
and ~k3lm
= -4m 2 ,
which are possibly ramified in the extension 3
Klk3 are the prime divisors of (2) in Ok ; indeed, as es
~k
1
1m = -4.
sion Klk 3 .
As
3
~K/k
3
divides
No prime ideal of Ok III
msplits
3
divid-
only a divisor of two can be ramified in the exten~k
the prime divisors of two in Ok
Klk 3 .
~K/k
2
1m = m2 ,
which is prime to two,
are not ramified in the extension
3
3
is ramified in the extension Klk3'
As the extension Kim is Galois a prime ideal (p) of in OK into the form:
with the degree of R.~ equal to f and efg
=4
([Sam] Chapter 6),
As the group Ga1(Klm) is not cyclic p cannot be inert in the extension of Kim (that is to say, e
= g = 1)
([Sam] Chapter 6).
210
CHAPTER 4
+2
On the other hand, a prime number p the extensions k./!J1 (i 1
=
ly the extension K/k. (j Therefore if p
J
+2
is not ramified in one of
1,2,3) (Question 1(5) above).
+ i)
one has e
Consequent-
is not ramified over p ([Wei] 3.4.6).
= 1 or e = 2.
It is then easily shown that, depending on the values of p, one of the following must hold:
pOK
ppop TpOT
(e
1,[
= l,g
4),
pOK
pp'
(e
1,[
=
2,g
2),
p2p.2
(e
2,[
= l,g
2) ,
2,[
=
pOK
2
pO K = P
(e
=
2,g
=
1).
In addition, 2 may have the decomposition: (e
SOLUTION 4'5
I (1)
4,[
= l,g
1) .
The discriminant of the field is -3, and
the Minkowski constant is M = ~
} 13. The ring A is therefore a
principal ideal domain (moreover, it is known to be Euclidean, cf., [Har] Chapter XII). Since p
=1
mod 3 we have l~J
Reciprocity Law, [Sam] Chapter
= 1.
From this (by the Quadratic
5) we have:
As a result p is split in the extension k/!J1.
p-.
Let us write pA
=
As A is a principal ideal domain the ideal p is generated by
an element of A that can be written a O + jSO ' aO'SO (l,j) is a basis of the ~ -module A. We have:
Z , since
ALGEBRAIC THEORY
211
and from this it follows that:
J
(2)
Let (a,S) be a pair of rational integers such that
Nk/'fJ(a + jS)
p; the ideal (a + jS)A is equal to p or to p; as
the only units
£
of the ring A are
1,±j,±j2 there are onlyafin-
ite number of pairs (a,S), given by:
A pair of rational integers
(x,y) such that p
a pair such that Nk/'fJ (y +2 xvC'S) 2
=p
(where
is a
vC'S = 2j + 1). There-
2
3x + y . «y - x) + 2jx fore p 4 is eqUlvalent to Nk/'fJ 2 = p. Hence there are a finite number of pairs (x,y). Starting from
the above pairs (a,S) they are given by:
= 2a,
y - x
x
= S.
One looks for pairs (x,y) such that x=:O mod 3 and y =: 1 mod 3, and this reduces to looking for pairs (a,S) such that
S=:O mod 3 and a =: -1 mod 3.
Twelve pairs (a,S) arise, by
multiplication by -1 and by symmetry (a,S)
~
(S,a), from the
three pairs (ao,So)'(-So,a o - So),(So - ao,-a o )' Oneknowsthat 2
that a o and So satisfy the congruence a o
2 + So - aoS o =: 1 mod 3.
From this one deduces, by enumerating all possible cases, that there are exactly two pairs (a,S) such that S=:O mod 3 and a -1 mod 3.
Hence, by imposing the additional condition S
>
0,
we have the existence and uniqueness of a pair (a,S).
II (1) Chapters
The extension 'fJ'/'fJ is cyclic of degree p - 1 ([Sam]
2 and 6, or [Wei]
Chapt~r
7, or [Has]
27); as p =:
CHAPTER 4
212
1 mod 3 its Galois group has an unique subgroup of index three, hence there is an unique subfie1d K of m' of degree three over that is cyclic. fied in
m'lm,
m
On the other hand p is totally and tamely rami-
and similarly in
K,
whence
ID I
=
P
3-1
=
P
2
(cf.,
[Wei] Chapitre 3 and the references above: p is the only ramified prime in Kim).
Furthermore, K is totally real, for it is contain-
Y
ed in the maximal real subfield m~ of gl', which has degree over m. Thus the discriminant D is positive, and therefore D =
2 P .
II (2)
Let us denote by i
~
s. the isomorphism of 1
onto the Galois group of m'/m defined by s.(w) 1
=
. w1 •
~/p~*
The group
Ga1(m'IK) is the subgroup H' of Gal(m'/m), the image under this isomorphism of the group H = Since .e
= I
3
. s.(w), .e is invariant under every element of 1 ~/p~*
sieR' The same holds for.e' and .e", therefore .e,.e' ,.e" e K.
H' .
are conjugate, for St(.e)
= .e'
and S 2(.e) = .e".
10wsthatmC.e)=m(.e') = m(.e")c K.
They
From this it fol-
t
It remains to see that .e+m.
If this were the case we would have .e = .e' = .e", whence 3.e =
wi
~
L
-1 (for w is a root of
l~i~p-l
L
X
i
= 0), and .e would
O~i~p-l
not be integral. II (3)
~
We have just seen that .e + .e' + .e"
-1.
On the
other hand we have:
I
i,jeH We have a sum of powers of w containing each term wh the same number of times, in fact this sum can be written I niw i , with l~i~p-l
n. an integer, and it is invariant under every element S of 1 Ga1Cm'/m). From this one deduces (for example by considering the linear system in the corresponding n.'s (1 ~ i ~ p - 1), 1 that n 1 = n 2 = ••• = n p _1 ' This sum consists of 3«p - 1)/3)2
ALGEBRAIC THEORY
213
terms; thus we have:
+
~~.
.. + ~..~ = __1_3
~.~
p - 1
n
-
1
2
i w
~
L.:.J:.
3
i w
3
L.:.J:. 3
We have:
II (4)
L h,i,jeH
. 2. wh +tJ.+t ]
and we consider the terms in this sum equal to 1,w, ... ,w
p-l
The number of terms equal to one in the expression: h
w
L
(h e H)
i,jeH
is independent of the choice of heH, as is seen by applying the automorphism s , where u e H. u
there are therefore k - r This
pro~uct
- kr = r
With the notations of the problem terms equal to one in
comprises (p ; 1]3 terms; hence there remain
(p; 1)3
- kr other terms, and we see that each of these occur
the same number of times. kr -
Thus we have:
1 (r 3 - kr) p - 1
kp - r
2
3
The minimal polynomial of
P(x)
~~ '~".
~
is thus the polynomial:
3 2 2 k x+x-rx+ r - p 3
214
CHAPTER 4
It is known that the discriminant of P is equal to the product 2
of p , the discriminant of the field, with the square of an integer a; let us calculate it directly. ~
= r2
_ 18r r
We have:
2 2 2 - kp _ 4 r - kp + 4r 3 _ 27 r - kp 3 3 3
222 4 4 312 - 3k P + 6r + 6r + "3 kp - (3r + 2r +"3 r ).
One finds the second degree equation:
o. As kp is a rational number the discriminant of this equation is the square of a number b' defined up to sign, which after expanding leads to the relation:
Replacing r by 16
3
"9 P
= b'
L..:.....!:. 3 2
then yields:
2 2 + 12p a ,
or again:
The number 3;; is an integer b, and we have the relation 4p b 2 + 27a 2 . On the other hand, the calculation of kp gives: kp
l.6
(6r 2 + 6r + 4 + b')
"3
2
P - 2p + 9
1
+
~ 3
2
+'9+
2
pJ:.. _ P + P + pb 9
-
9
ALGEBRAIC THEORY
215
Whence: (p - 1)
2
-
P2 - P -
- 3p - pb + 1 9
pb
9
which gives the result. (i) <=> (ii): q is a product of three factors in
III (1)
K if and only if the decomposition field of a in Q' contains K ([Mac]) Chapter 5, or !Che] Chapter 2). that is to say. if the index of the decomposition group of q in G is divisible by three (the index is the number of prime factors of q in Q'). (ii) <=> (iii): Let f be the smallest positive integer satisfying the congruence q f
=1
mod p, and let g
of cyclotomic fields shows that q in ideals of degree f ([Wei] Chapter q
(p-1)/3 - 1
=
m'
7).
= P-.:..1.. f
The theory
is the product of g prime We have:
mo d P.
if and only if f divides ~ • which is equivalent to 31g. (iii) <=> (iv): If q - x
by Fermat's Little Theorem. class modulo p generates
= Pm mod
has q
3
mod p we have:
Conversely, let
ez/p~*.
p
be an integer whose
For a suitable integer m we have
P. whence:
p-1
-3- m P
-
1 mod p;
P - 1 then divides take x = pm/3. III (2)
n - 1
~m.
therefore 3 divides m. and we can
Let us assume that P has a triple root a modulo q.
216
CHAPTER 4
There we have: p(3
+ b) -
1
27
- (X - a)
3
mod q,
which implies the congruences 3a - -1 mod q and 3a mod q.
From this we deduce p
As the extension
K/~
2
_
p - 1 -3-
= q.
is Galois a prime number q decomposes in
K into a product of prime ideals in either of the following ways:
(q)
(q)
or
the q.'s being ideals of inertial degree i. l
The number p de-
composes following the first form, and a prime number q
~
p, not
being ramified, decomposes in one of the remaining ways. Let q
+p.
If the congruence P(x)
=0
mod q has a solution at
least one of the roots is simple, and is recovered as a simple root in the q-adic number field.
q then has a prime factor of
degree one in K, and q splits completely.
If, on the contrary,
the congruence does not have a solution modulo q, P is irreducible in the q-adic number field, and therefore q is inert in K ([CaF] Chapter 2, or [Wei] Chapter
2).
REMARK: Modulo p, P has a triple root equal to III (3) mod 2.
P(l)
The integer ~ is even, and p(3 +2~) - 1 _ b
Therefore P(X)
=b
mod 2.
b (and thus a)
31
= X3 + X2 + b mod 2, and hence P(O) _
P therefore has a root modulo 2 if and only if is even.
The rest of the Question immediately
follows from Question 111(1). REMARK:
If P has a root mod 2, 0 is a double root and 1 is a
simple root. Let us set P (X) = 27P ~ The roots of PI 1 3 are 3-& + 1, 3-&' + 1, 3-&" + 1. It is immediate that: III (4)
ALGEBRAIC THEORY
For q
+ 3,
217
P and PI have the same number of roots modulo q.
this i t immediately follows that for q
+3
From
the condition "q divid-
es b" is sufficient for q to split, and therefore for: q(P-l)/3 _ 1 mod p.
Let us set b
III (5)
1 + 3b'.
Without difficulty one
shows the congruences:
y
==
-
(b'
+
1)
mod e:
and:
p(3 + b) - 1 _ b,3 _ b,2 + b' + a 2 mod 3. 27
Hence we have:
P(X)
==
X3 + X2 + (b' + l)X - (a 2 + b,3 - b,2 + b') mod 3.
If 3 divides a we see that P(O) and then that P(l) is zero for b'
== ==
°mod 3 if b' 1 mod 3.
==
0,-1 mod 3,
Consequently P has
a root modulo 3. From this one deduces that 3 splits in-X, and therefore that 3(p-l)/3 == 1 mod p. In addition we see that 3 does not divide a, andP has no root modulo 3.
Consequently for
= 2 and q = 3 the condition "q divides ab" is necessary and " f or t h e congruence q (p-l)/3 == 1 mod b ff su lClent p to e " satls f"led .
q
REMARK 1: Question 111(1) can be solved using the Cubic Reciprocity Law (cf., [Mor] Chapter 15). REMARK 2:
For arbitrary q, the condition "q divides ab" is suf-
218
CHAPTER 4
ficient for the congruence q
(p-1)/3
=1 3
mod p to hold.
this consider the polynomial Q(X) = X - pX - pa.
To see
Its discrimin-
ant is p 2b 2 ; the splitting field K' of the polynomial Q on therefore an abelian field of dgree 3 over
m.
mis
The prime number
P is totally ramified in K' (because q is an Eisenstein polynom2 ial for p); 2 is not ramified since Q(X) = X(X - 1) mod 2 when 2 divides b.
If
q is a prime factor of b other than 2 we have
q =t 3. 3
X - 3X - 2 mod q, a congruence which has the simple root X
= 2,
hence q is split.
Consequently the field K' has p2 as discriminant; as it is an abelian field over
mit
rem, cf., [Wei] Chapter
is contained in
m'
(Kronecker-Weber Theo-
7); from this it follows that K' = K.
One concludes by noticing that modulo the prime factors q of a, Q(X) = X(X 2 - p), and thus q is decomposed in K. REMARK 3:
The equivalence "q divides ab" i f and only i f /P-1)/3
- 1 mod p" is false in general, but is valid for small values of q (notably q
= 5). 3
The discriminant of X + pX + q is -4p 27q2, whence ~(l,~,~2) = _27d 2 . SOLUTION 4·6
I (1)
3
If P divides d the polynomial X3 - d is an Eisen-
I (2)
stein polynomial for p ([Wei] Chapt-er that p is totally ramified and that
3). l,~,~
If 3 does not divide d then:
f(X +
1)
f(X -
1)
= X3
+ 3X2 + 3X + 1 - d,
and:
X3 - 3X2 + 3X -
(1 +
d).
2
From this one deduces is a local basis at p.
ALGEBRAIC THEORY
219
Therefore one of these polynomials for which the constant term is divisible by 3 is an Eisenstein polynomial at 3, which completes the proof. The lattice with basis 1,a,a 2 coincides locally with
I: (3):
the ring of integers 0
for every prime number (for 3 notice that K 2 2Z[a + 1] = 2Z[a - 1] = 2Z[a]). Therefore OK = 2Z[a] and D = l\{l,a,a )
=
-27d 2 .
REMARK 1: In the general case d can be assumed to have no cubic
factors, which allows us to write d
= fg2, with
f,g relatively
2
~ is then a root of the polynomial, g
prime and square-free.
which is an Eisenstein polynomial at p for every prime divisor of g.
Thus when d
= ±l mod 9
a2
l,a, -
g
and hence: l\{l,a,
D
a2 g
g
-2
2
l\{l,a,a)
.
1S
an integral basis of K,
2 - 27(fg) .
a2 we can assume that 3 does not
Moreover, by replacing a with divide g.
g
Moreover we can further assume g
REMARK 2: Let us assume d
= ±l
mod 9.
=1
mod 3.
a f7 2
It is known that l,a,
is a basis locally for every prime number p f 3.
As the exponents 2
of 3 in the discriminant D of K and the discriminant 6(1,a, ~ ) g
is not even, we have: D
=-
27(fg)2
or
D
=-
3(fg)2.
Thus in all cases 3 is ramified in K. (3)
3
P
Hence:
or
(p,p' ,p" being prime ideals of degree one).
In the first case
the extension is not tamely ramified, it then follows that D -27(fg)2 ([Wei] Chapter
7).
=
In the second case the extension
CHAPTER 4
220
is tamely ramified, we then have D
= -3(fg)2.
Let us show that the first case is impossible. ing d to -d, we may also assume d
=1
therefore ~3 (~ -
1)
3
+
=1
mod 9.
Begin by chang-
We have
~
3
= d,
mod p6, which may be written: 3~(~
- 1)
6
=0
mod p ,
whence:
That implies ~ - 1
=0
mod p, and therefore ~ - 1
(otherwise (~ - 1)3 + ~(~ - 1)
* 0 mod
= 3~ is divisible by p3 but not p4 .
p4).
=0
mod p2
Consequently (~_ 1)3
Hence we have (~- 1)
=0 mod p3,
therefore ~ ; 1 is integral in K, which is impossible, because: ~
1
- 1 3
2
- 2 l'l(1,~,~ ) 27
is not an integer. We are therefore in the second case. ~3 _ d
_ (
~
- 1) +
(~ _ 1)2
proves the congruence
~
3~)
mod
The identity:
32
_ 1 mod p'p".
But there exists an integer
~2
1
of the form 3 (a + b~ + c GI ) with a,b,c not all divisible by 3. As we assume that g _ 1 mod 3,
a +
b~
~2
+ c -
g
=a
+ b + c mod p' p" ,
and necessarily therefore a + b + Since
~
C
=0
none of the a,b,c is divisible by three. 1
that 3 (1 +
mod 3.
* ±l mod 3 by the resoning used ~
+
~2
GI
in the first case. From this it follows
is integral, and consequently that
l,~,
ALGEBRAIC THEORY
1
- (1 + .e + 3
.e 2
221
) is an integral basis.
g
If the congruence x 3 - d
II (1)
=0
mod p has a root in ~
this root is simple (since 3x 2 $ 0 mod p), and thus lifts into
~ p as a simple root of x 3
-
d.
This implies the existence of
a prime factor of degree one of p ([Wei] Chapter
2, or [CaF]
Conversely, if p splits in K it has a prime factor
Chapter 2).
of degree one; the polynomial X3 - d then has a root in ~ , whence a root
p
Therefore p is inert in K if and only if
modulo p.
in~
the congruence x II (2)
3
- d mod d does not have a solution in~.
Let p
= -1
mod 3.
The mapping x
t+
x 3 of (~/p~)'"
into itself is an inj ecti ve homomorphism since card ~/p~)'" is not divisible by three.
=d
to say, there exists x~
=
p - 1
It is therefore surjective, that is Hence p splits in K by Quest-
mod p.
tion II(l). Since 3 does not divide p - 1, -3 is not a square in if P f 2, -3
_
2
=A
mod p implies:
[ - 1 2 + AJ 3 =_ 1
P
; for
mod p,
and so an element of order three in in
~
~/p~)1';
*
-3 is not a square
~2' for -3 1 mod 8. The discriminant D _27d 2 is therefore not a square in ~2'
consequently X3 - d has only one root in ~ , which gives the rep
suIt. If q
II (3) three; let
p
=1
mod 3,
be this element.
(p - 1) ( p 2 + p +
1) = 0
~/q~)*
has an element of order
Then: and
( 2 p+ 1) 2 = 4 ( P 2 +
P +
1) = - 3 .
The polynomial X2 + 3 is therefore the product of two polynomials of degree one Chapter
3) in
in~/~[X] ~
q
[X].
and so (by Hensel's Lemma, cf., [Wei]
From this it follows that -3 is a square
CHAPTER 4
222
in
, so D is a square in ~ . q There are, therefore, two possible cases: either the polynomial ~
p
X3 - d does not have a root in ~ and q is inert, or the polynomq ial X3 - d is the product of three polynomials of degree one of and q in K is the product of three prime ideals of degree
~q[X],
one. Let f = a + b~ + algebraic closure of K are: III (1)
(i
1, j
2
a~.
The conjugates of f in an
+ 1).
We immediately have:
whence:
REMARK: One also has TrK/~(f'~')
= 3(a 2
- bad), which permits the
verfication of the result of Remark 2. III (2)
If P has a principal prime factor p in K it has a
principal prime factor of degree one, for if p is of degree two pp-l is principal and of degree one. Therefore let p = (n) be a principal prime factor of degree one.
a
3
+ db
323 + d a - 3daba
has therefore a solution.
The norm of p is the prime
Hence we have N(n)
ideal generated by NK/~(n). N(±n) p. The equation:
=p
Conversely, if:
= ±p,
and so
ALGEBRAIC THEORY
223
has a solution in ~, let us set: K
=a
+
+
b~
K
4 6 = "it 27
= 2, M <
For d
; As
K
is an integer, the principal ideal
is a prime ideal of degree one dividing p. We have:
II I (3) M
2
= p.
we have NK/m(K) generated by
c~
~
3dY3 < 1.48d.
3.
(~)3.
But (2)
As every class contains an
ideal with norm at most M, we deduce that OK is principal. shows similarly that OK is principal for d 5 one uses the equalities N(2 NO
+
~)
~)
= 3,
=3
N(l +
or d ~)
= 5;
(One
if d
= 6).
=
= 6).
IV (1)
For a e
~,a
f 0 mod 7, we have a 3 - ±l mod 7, hence
the impossibility of the congruence:
= ±2 or ±3 mod 7. In particular, = 3 we see (Question 111(2) above)
whenever m sult to m
3
is not principal (we have (3) = P3 ). By Question 11(2) above we have (2)
=0
b
1,
6.
Therefore
M < 11.
C
by applying this rethat the ideal P3
= P2P;.
in Question 111(2) above it is seen that (~
- 1)
= P2P3
holds.
= -1,
By taking a N(~
- 1)
On the other hand, for d 2
=
=7
Let us denote the class of P3 by h; we have P2 e h ,
P; e h. As NK/ (2 + ~) = 15 we have (2 + -8) = P3PS' whence and P; are the only ideals of norm 4; P s e h 2 , P; e h. Moreover,
m
P;
P2P3 is the only ideal of norm 6; P7 is the only ideal of norm 7,
since (7) = (-8)3;
P; and
(2) are the only ideals of norm 8;
P; is
CHAPTER 4
224
the only ideal of norm 9; lastly P2 P5 is the only ideal of norm 10. One can therefore conclude that the class number is three, and the class group is cyclic, generated by the class of P3' IV:(2):
By the preceding, one and only one of the three
ideals Pp ,p3 pp ,P 32 pp is principal.
NK/~(f) = 3p
and
NK/~(f) = 9p
=±
If p
(where f
=a
mod 7 the equations
+
b~
+
c~2)
are imposs-
Therefore P3 Pp and P~Pp are not principal. From this it follows that Pp is principal, and therefore P'p as well, since Pp P'p = (p). Conversely, i f P p is principal the equation NK/~(f) = P has ible.
a solution, which implies p that P and P~P p
t
±l mod 7.
If P
= ±2
mod 7 we see
are not principal, and hence P3 Pp is, from whjch
p 2
we conclude P e hand P' e h. p
If P
p
2
= ±3
mod 7 we see that P and P
2
P3 Pp are not principal, and hence P3Pp is, and so Pp e h, p;eh . IV:(3):
Let us write qOK ,
"
r'
r"
d
2r
2r',
be the 2r""
The pro ucts P3 q,P 3 q 'P 3 q From this it follows that the equations:
respective classes of q,q ,q. are principal.
r
= qq q and let h ,h ,h
(resp. 3 2r ' q, 3 2r" q)
have solutions. 3 2r q
= 2rq
Thus Ne have: _ ± 1 mod 7 (analogous congruences in r' and r"),
whence: 4r _ ± q mod 7,
4r '
=±
q mo d 7,
4r "
=±
q mod 7.
has order three in ~/7'7L.)~' i t is seen that r = r' = Since 4 r" mod 3, hence the fact that q,q' ,q" are in the same class. We can then apply the method of the preceding question.
In partic-
ular we find that q,q' ,q" are principal if and only i f q
= ±l
mod 7. The resul ts of these two questions can be expressed in the following
ALGEBRAIC THEORY
way: Let p (p
+3
225
and p
+7
be a prime number split in K, and let p
be a prime factor of p in K of degree one. 2
Let a
p
be a represent-
1 , Pp then belongs to the class h
ative modulo 7 of p ;
-a
p
EXAMPLES: Looking modulo 13 at the polynomial X 3 - 7 we see that thirteen is inert. On the other hand, modulo 19 we have 4 3 - 7 = 57
= 0,
consequently nineteen is the product of three factors of
degree one, the class of which is h 2 • COMMENTARY: (Using Class-Field Theory): First of all let us recall the following result: Let L be an extension
In order
of~.
that there be a positive integer f such that all prime numbers prime to f and congruent to a modulo f have the same decomposition in L,
L/~
must be abelian.
Furthermore, only the prime divisors
of f can be ramified in L (cf., [Che] Introduction, [CaF] Chapter
XI). As a result, the extension
K/~
of the Problem not being abel-
ian, the fact that a prime number p congruent to one modulo three is inert or split in K cannot be decided with the aid of congruences modulo an integer f. Furthermore, Class-Field Theory (cf., references above) enables us, for all number fields, to construct a bijection between the subgroups of the class group of K and the unrarnified abelian extensions of K (at every prime ideal and at infinity), the prime ideals of the subgroups considered being precisely those which are completely split in the corresponding unramified extension, the relative Galois group being isomorphic to the subgroup. The results of Part IV of this Exercise show that the principal prime ideals of which the norm to
~
K
(other than
P7)
are the prime ideals for
is congruent to ±l modulo 7.
of the ideal classes of
K
As the group
is of order three, class field theory
says that the field K has an abelian extension C of degree three (the "class field" of
K)
in which the prime ideals of
K
completely are exactly the principal prime ideals of K.
that split
226
CHAPTER 4
Let us see whether there is a field L, abelian and of degree three over
~,
such that cr coincides with K.L, the field composed
of K and of L. Let L be an abelian field and let it be of degree three
Let p be a prime ideal of Kover
over~.
p of degree one.
The prime ideal p is completely
split in K.L if and only if p is completely split or ramified in L (in fact if P is a prime divisor of p in K.L p is completely split if and only if
P is of degree one in the extension
K.L/~).
The field L there-
fore satisfies the following conditions: 7 must be the only prime number ramified in L, and the prime numbers p split in L must be the prime numbers p congruent to ±l modulo 7.
Exercise 4·5 shows
that there is only one field L possible, the cubic subfield of the field of seventh roots of unity. Let us show that K.L coincides with C; it comes down to showing that the extension K.L/K is unramified. be a prime ideal of Kover 8. L/~,
p is unramified in the extension
and therefore ([Wei] Chapter
tension K.L/K.
Let p f 7 and let p
3) p is unramified in the ex-
With the fields K and K.L being real it remains
to verify that the prime ideal P7 is unramified in the extension
K.L/K. Let k = ~(~), N = K(~). The Galois group G of NL/k is the product of two groups of order three. Let p be an ideal of k over 7. Since the quotient of the inertia group GO of 7 by its higher ramification subgroup G1 is cyclic, and as G1 = {l} because 7 is prime to 9 (cf., [Ser 2] Chapter IV), NL/k is not totally ramified at p. Therefore there exists an unramified extension L1 of degree three of k at p; as N.L = N.L 1 , NL/N is not ramified at the ideals over p. From this one easily shows that K.L/K is unramified at P7' which shows K.L
= C.
The results of Part IV concerning the determination of the class of an arbitrary prime ideal of K are connected with a reciprocity law.
ALGEBRAIC THEORY
227
SOLUTION 4·7 (1)
It is known that the integral closure of
k is~[a], where a
= !(w - 1). The irreducible polynomial of a
over ~ is X2 + X - 16. has two roots in
~
in
The image of this polynomial in~/2~[X]
~/2~,
0 and 1.
It follows then ([CaP] Chapter
III, Appendix) that the ideal (2) splits into two prime ideals,
PI generated by 2 and a - 1, and P2 generated by 2 and a. letting:
Hence we have:
now,
~(w
+ 1)2
16 +
~(w
+ 1)
3.4 + 4 + ~(w + 1),
therefore:
Moreover: 4
=
(4 + ~(w + 1))(4 - ~(w - 1)),
and therefore: 4 e (4 + ~(w + 1)),
whence: ~(w +
1) e (4 +
~(w +
1)).
Consequently:
P~C
(4
+
Hw
+ 1)),
and therefore we have the equality;
So
228
CHAPTER 4
(4
(2):
+
+ 1).
~(w
If Pi
automorphism of
P2 = (2a +
=
(2a
+ b!{w + 1), by applying the non-trivial
Ga1(k/~)
b~ ( - w
we would have:
+ 1»,
whence: 1 b)2 _ 645 b 2 ). ( (2a+"2
So there exists a unitE elZ such that:
and thus:
. Now, t h e equatlon x 2 - 65 y 2
cause x 2 P
65y2
. b e= ±8 h as no·lntegra1 solutlons
is congruent to 0,1 or 4 modulo 5.
Consequently
is not principal. (3):
Given a number field k it is known that in each ideal
class there exists an integral ideal B such that:
(where n is the degree of the field k, d is the discriminant of k, and 2r 2 is the number of complex conjugate fields of k, cf., [Sam] Chapter IV §4). Here r 2 = 0, n = 2, d = 65, thus:
It therefore remains to determine the ideals with norm 2,3,4.
ALGEBRAIC THEORY
229
P2 • There is no kim. The
The ideals with norm 2 are the ideals PI and
ideal of norm 3 because 3 is inert in the extension 2
2
ideals with norm 4 are the ideals 2,P I ,P 2 . We know (Questions (1) and (2) above) that Cl(PI ) f 1 and 2
Cl(PI )
= 1.
PI P2 = (2), we have:
Furthermore, since
1.
The ideal class group of k is therefore of order two; CI(PI ) is a generator of this group. Let u be a unit of k.
(4): u
=a +
+
b~ (w
Let us set:
(a, b e 7Z) ;
I)
-1 if and only if:
· . S lnce t h e equatlon x 2 - 6sy2
-1 has x
8, Y
1 as a solution,
tion, we can take:
u = 8 + w. If
(S):
€
-1, Nk/m (€)
± u
=
€
is a fundamental unit, then u
= fEn.
-1 and n is odd, n
Hence we have:
2k+1
Consequently, k(iE) is real, k(iE) ~
=
v'+_u
and =
=
2k + 1. €
k Y , €.
k(IU), where k(iE)
k(~) necessarily.
in the extension
Since Nk/m(u)
k(~)lk
= k(~).
Since k(iE)
The discriminant of
is equal to 4(8 + w).
A
ram-
ified prime ideal in the extension k(~)lk must be a divisor of 2. Let us show that PI is ramified in this extension.
Let a
230
CHAPTER 4
The irreducible polynomial of a over k is the poly-
~ - 1.
nomial:
J?
+
2X - 7 -
w
= J?
+
2X - 6 - 2 ~
(w + l).
It is clear that P 1 -adic valuation of - 6 - 2· ~ ·(w + 1) is equal to unity (in fact P~ divides2'!'(w + 1) and 6 is divisible by Pl
but not by P~).
The irreducible polynomial of a is therefore an
Eisenstein polynomial for Pl' consequently ([Wei] Chapter is ramified in k(~). The P2 -adic valuation of - 7 - w = - 8 - 2
The irreducible polynomial over k of
8).
Ia = I (~
is the polynomial X2 + X - 2 - 1
- 1)
therefore has
P2 -adic valuation zero. This polynomial remains
irreducible in kp [X] (where kp 2
for kp :;c
(w - 1) is equal
P; and not by p~, and
to 2 (in fact 2 I (w - 1) is divisible by
P~ divides
I
3) Pl
e kp
2
is the P2 -adic completion of k)
is isomorphic to the 2-adic number field 2 2 :;c
+:;c
2
=0
mod P2'
ramified in k(~).
m2 ,
and thus if
Consequently P2 is either inert or
Now, the discriminant of ~ in this exten-
sion is equal to 8 + w, which is an unit.
Consequently P2 is not
ramified. There is therefore a single ramified ideal in the extension k(~); it is the ideal (6) :
If
or K = k(/:3).
E
or
-E
Pl'
were squares in
K
we would have
K = k( 13)
Thus the ideal Pl of k would be ramified in K,
which is contrary to the definition of K. If a is an unit of K such that a + a(a) = 0, a is therefore a root of X2 + aa(a). aa(a) is an unit of k, and we (7):
have:
aa(a) = ± E2k
or
aa(a) = ±
E
2k+l
One cannot have aa(a) = E2k because k(a) would not be real.
If
ALGEBRAIC THEORY
231
= -£ 2k ,then a = ±£ k ,
we had aa(a)
and so a contradiction.
b
Let
C
= ±£2k+1, for this
e K be an integer that does not belong to k.
Set
°and ab = -b.
Then b t
ac - c.
One cannot have aa(a)
= £2k
= ±£, which is impossible by Question (6) above.
implies (a£-k)2 (8)
and consequently aa(a)
Let us consider the decomposition of the ideal bOK into a
(9)
product of prime ideals of OK'
A prime ideal of OK is either an
ideal R above an inert ideal of Ok or an ideal q over a split ideal We have aP
of OK'
=P
and aq t q.
Since the ideal bO K is invari-
ant under a, and ideal q and its conjugate aq appear with the same exponent in the decomposition of bOK' Now P (resp. q.a(q)) is the extension of an inert (resp. split) prime ideal of Ok'
Consequently bOK is the extension of an ideal
= AOK. If A were a principal ideal, that is to say = xOk (with x e Ok) one would have bO K = xO K. There would thus exist an unit a of K such that x = abo But since ax = x, ab = -b, and b t 0, this implies a + a(a) = 0, which is impossible, by
A of Ok' bO K A
Question (7) above. (10)
Let us assume that there exist two distinct fields
K1 and K2 possessing the properties of Question (6) aobve. us write
K3
Let
for the third field of degree two over k contained in
the composition K1 .K2 ; let 0 1 (resp. O 2) be the K1 - (resp. K2-) automorphism of K1 .K2 . Let b 1 (resp. b 2 ) be a non-zero element of OK
1
(resp. OK) such that a2 (b 1 ) = -b 1 (resp. a1 (b 2 ) = -b 2 ). 2
There exists a non-principal ideal Al (resp. A2) of OK such that:
Since
° has
such that Al
only two ideal classes there exists an element ye k
= yA 2 • Consequently A10K K
yb 20 K K' 1 2
1 2
yA 2 0 K K ' whence 1 2
Therefore there exists an unit u of KIK2
232
CHAPTER 4
= b 2 y. We have 0lo2(b i ) = -bi' 0lo2(b 2 ) = -b 2 , = y. Consequently 0lo2(u) = u, and therefore ueK 3 . Now, 0i(b i ) = b i , 01(b 2 ) = -b 2 , 0l(Y) y, which implies that 0i(u) = such that biu
0lo2(Y) u.
It is clear that the field K3 possesses the properties of Question (6) above (in fact, the extension KiK2/k is unramified for every prime ideal of k).
By Question (7) above no unit of K3 has
a vanishing trace over k, which contradicts the preceding result.
(11)
The field ~(IS,II3) contains the field k; ~(IS,II3) Let us verify that no prime ideal of k is ramified in
is real. ~(IS,II3) .
Let Q be a prime ideal of k and let q be the prime number of such that
Q
n?Z = q~.
~
If Q is ramified in ~(IS, 113), then q is rami-
fied in the extension ~(IS,m)/~.
Consequently q is ramified
either in the extension ~(IS)/~ or in the extension ~(If:3)/~. Now, 5 is the only ramified prime in ~(IS); 13 is the only ramified prime in ~(If:3); 5 and 13 are the only ramified primes in ~(165) =
k.
Consequently the ramification index of 5 in the ex-
tension ~(IS,If3) coincides with its ramification index in the extension
k/~.
The prime ideal over 5 in k is therefore unrami-
fied in the extension
~(I5,If3)/k.
The same argument shows that
the prime ideal over 13 in k is unramified in ~(I5,/f3).
Thus
the field ~(I5,If3) satisfies the conditions of Question (6) above, and by Question (10) above it is unique.
(One calls
~(IS,II3) the "class field" of k, cf., [Che] Introduction).
SOLUTION 4-8
(1) (a)
The polynomial X2 - d is irreducible in
k[X] because its image in k[X] =F p [X] is irreducible.
The ex-
tension of k by the polynomial K2 - d is therefore of degree two, and it is an unramified extension because the degree of the residual extension is two (cf., [Wei] Chapter
3).
ALGEBRAIC THEORY (1 )
(b)
233
Since 2 is not a square modulo 5, X4 - 2 is ir-
reducible inF 5[X]. Consequently, the extension of k ID by X4 - 2 is of degree four, and is unramified because the degree of the residual extension is four. (2)
(a)
This follows immediately from the definition.
(2)
(b)
This is because K is isomorphic to the finite
field F n q
(2) (c)
This is a property of extensions of finite fields.
(Cf., [Joly] Chapitre 1). (3)
In K[X] the polynomial Xq
n
- X factorises into a pro-
duct of factors of the first degree, all relatively prime. Lemma ([Wei] Chapitre 2) shows that Xq
n
-
Hensel's
X also factorises in
K[X] into a product of factors of the first degree.
Let ~ be a qn_l root in K of the polynomial X - I that is primitive (that is to say, of order qn - 1).
The extension k(~) coincides with K,
for these two extensions have the same residue field, and the extension K/k is unramified. n
of Xq
- X on k; the extension K/k is Galois.
(4) (a) <
K is therefore a splitting field
If x,y e OK are such that x - y, we have
Ix - yl
1, whence la(x - y)1 = Ix - y I = 1, and ax = ay.
o:x
~
The mapping ax is therefore well defined and is a k-automorphism of k.
(4) (b)
We have:
and therefore ~ is a homomorphism of G(K/k) into G(K/k). id R, for all x e OK we have la(~)
- ~I
<
Ia(x)
- xI
1, which implies a(~)
<
= ~,
a
If
1, and in particular and hence a
= id K.
The
234
CHAPTER 4
homomorphism ¢ is therefore injective, and consequently surjective, because G(K/k) and G(X/k) have the same order.
G(Klk) being isomorphic to G(Xlk) is cyclic and
(4) (c)
generated by the element 0*, whose image under T
of G(Xlk) (Question (2)(c) above).
0*
(4) Cd)
is the generator
is therefore the k-auto-
= xq (for all
morphism of K such that a'-'(x) say, 101,(x) -xql < 1 (forallxeOK).
¢
x e X), that is to
In example (1) (a) we clearly have
01,(/d)
-/d,
which suffices to define 0*. In example (1) (b) we have k = ID 5, K = ID5 (4/2). 01' is defined i f 0*( 4/2) is known. It is c1 ear that 0*(4/2) = i4/2, where i is a square root of -1.
Of the
two square roots of -1 in k one chooses the one determined by:
that is to say:
Ii -
11
<
1,
(one has i 2 - 22 = (i - 2)(i + 2) = -5, which shows that the two square roots of -1 in IDs are congruent, modulo the maximal ideal, one to 2, the other to -2). (5)
Ca)
If
].Jm = 1 + aTtm (with a e OK) we have:
IT
oeG(Klk) (5) (c)
Tr K/ k (a)
m
(1 + o(a)Tt )
=1
+ TrK/k(a)Tt
m
mod Tt
We have:
L
oeG(K/k)
o(a)
L
oeG(K/k)
o(a)
=
L-
oeG(K,k)
m+1
ALGEBRAIC THEORY
235
Since the mapping TrR/k:K ~ k is surjective (cf., Exercise 4 1) 0
the mapping
is surjective.
~
m
Consequently, for any Am e 1 + n OK'
that is to say Am = 1 + bnm with be Ok' there exists a e OK such that TrK/k(a) = b, and therefore if ~m = 1 + anm we have: MnKlk(~m)
-
m+i 1 + bnm mod n
and therefore: m MnKlk(~m) - Am mod n (5) (c)
We evidently have Mn Klk(1 + nO K ) C 1 + nOk • Let A = Ai belong to 1 + nOk ; by Question (5)(b) preceding there exAi 2 ists ~i e1 + nO K such that MnKlk(~l) = ~where A2 e1 + n OK· 2
By iterating this process one shows the existence of a sequence (Am)m~i
of Ok and of a sequence (~m)m~i of OK such that: and
~m e 1
m
+ n OK'
and:
From this one deduces:
The sequence with general term urn an element
~
~1~2ooo~m
e 1 + nO K , because urn - u m+i
converges towards m
mo~ n.
1
We have
MnK/k(~)
A, since the sequence with general term -A--- converges towards m+i Ai = A, and because the norm is a continuous mapping, whence the
=
equality:
236
CHAPTER 4
n
Let VK (pesp. Vk ) be the multiplicative group of (q - l)-th roots of unity in K (pesp. of (q - l)-th roots of unity in k). We have: UK Uk
= VK x = Vk x
(1
(1
+ 1I0 K) + 1I0k )
(direct product), (direct product)
}
VK is isomorphic to the multiplicative group is isomorphic to k*; since the norm in an extension of
([Has] II §lS). and Vk
(1)
R*
finite fields is surjective (cf., Exercise 4·1), from this it follows that: (2)
Relations (1) and (2) imply:
(S):(d): a +
This is obvious because a 2 - db 2 is the norm of
bid and because every unit
x e
mp (Id).
(6):
We have the equalities: 11
k*
11
liZ
x UK
(direct product),
liZ
x Uk
(direct product).
From this we have:
whence:
mp is the norm of a unit of
ALGEBRAIC THEORY
k": / Nm
237
KIk (K1:) '" 'Zl/n'Zl.
Hence we have just shown that if Klk is an unramified extension of local fields, the group of norms of K is a multiplicative subgroup, of finite index, of the group k*, the quotient being isomorphic to the Galois group of the extension.
This is a partic-
ular case of local class field theory (cf., [Ser2] Chapter
V,
SII, XIV).
SOLUTION 4'9: (cf., [Bou] Chapter 1:(1):
V §S):
If the polynomial y2 + X2 + 1 is not irreducible in
K[Y], as it is of degree two it possesses a root in K = m(X); furthermore, this root is then a polynomial P(X) of m[X], for with the ring m[X] being principal it is integrally closed ([Sam] Chapitre 2). Now the equality P(X)2 + X2 + 1
=0
is impossible because the
polynomial X2 + 1 is irreducible in m[X]. II: (2): of E on
m([Bou]
mis
The algebraic dimension (or transcendence degree) If E is a purely transcendental extension of
one.
Chapitre 5 §S) there then exists Z e E such that E =
Let ~ e E be a root of the polynomial y2 + X2 + 1 e K[Y].
m(Z).
Let us write that
P1 ,Ql,P2 ,Q2 e
x
X,~
~[Z]
e m(Z).
There exist non-zero polynomials
such that:
and
P1,Ql being relatively prime, and P2,Q2 as well. ~2 + X2 + 1 = 0 can be written:
The relation
We see that Q2 divides Q1 (since Q2,P2 are relatively prime and
CHAPTER 4
238
Q2 divides P 2Ql) and that Q1 divides Q2 (since Ql,P1 are relatively prime and Q1 divides P1 Q2)' Hence there exists a polynomial Q e m[Z] such that:
Examining the terms of higher degree, and the impossibility of solving the equation: a2
+ b 2 + (/
=0
3 , (a,b,c) f (0,0,0), with (a,b,c) e m
gives the desired contradiction. The possibility of parametrising the ellipse X2 + a by:
I (3)
y2 + 1
and suggests using the identity: i(l - Z2))2 + [ [
1 + Z2
2iz
1 + Z2
Let us look for Z (in
)2 + 1 O. such that:
m(i,X,~))
2iZ 1
We easily obtain Z
=i
!X
+ Z2
which shows that E(i) is a purely
transcendental extension of m(i):
I (4)
If the polynomial
y2
+ X2 + 1 were irreducible in
K[Y], since it is of degree two it would have a root in K = m[X] ,
ALGEBRAIC THEORY
239
and this root would then be a polynomial P(X) e \Il[X] (cf., Question Now the equality P(X)2 - X2 + 1
(1)).
=0
is impossible because
X2 - 1 = (X - l)(X + 1) is not a square in \Il[X]. Finally, F is a purely transcendental extension of Q, because of the identity: (
2Z
1 - Z2
]2 _
(1 + Z2] 2 + 1= 0, Z2 1 _
which shows, if we denote a root of y2 - X2 + 1 in F by n, that we have:
II
Let l,j,j2 be the cube roots of unity in ~.
II (1)
If the polynomial y3 + X3 + 1 were not irreducible
. P e L such tat: h in L[Y] there would eXIst Q
with P,Q beingrelativelyprimein~[X].
Hence X + l,X + j,X + /
divide P, and if:
P(X) = (X + l)(X + j)(X + j2)P1 (X) we have:
(Q(X»3 + (X 3 + 1)2(P1 (X»3
0,
3
and consequently X + 1 divides Q, a contradiction. II (2)
If M were a purely transcendental extension of
~,
there would exist three polynomials u,v ,w e ~ [Xl, not all constant, pairwise relatively prime, and such that u 3 + v 3 + w3 = 0, that is to say, w3
=-
(u + v)(u + jv)(u + j2v ).
Let n be the maximum
of the degrees of U,V,W, and let us choose these polynomials so
240
CHAPTER 4
that n is a minimum.
We may assume that w is of degree n).
If pea: [X] is an irreducible polynomial which divides
u + v,
then it does not divide u + jv (becauseu,varerelativelyprime), nor u + j 2v, and p divides w3 , therefore p divides w; from this we deduce that p3 divides u + v.
Thus u + v is a cube in a: [X] :
there exists rea: [X] such that u + v
=r3
s,tea:[X] such that u + jv = s3, u + j 2v =
Similarly there exist
t 3 , whence:
- w.
r.s.t
As u,v,w are not all constant, at least two of the polynomials r,s,t are not constants, hence the maximum of the degrees r,s,t is strictly less than n.
r
3
. 3
+ JS
.2 3 + J t
The relation:
o
shows that there exist in a:[X] three polynomials u l WI
=
3.(2
Ij~t,
= r, vI = 3/.7s,
not all constant, and of degree strictly less than n
satisfying:
contradicting the minimality of n.
CHAPTER 5
Distribution Modulo 1
INTRODUCTION 5.1
EQUIDISTRIBUTION DISCREPANCY: (Exercises 5 '1-5 ,10)
Let (u n ) n<:&, ~~ be a sequence of elements in the interval [0,1], and let [a,S] be an interval within [0,1]. For every integer
N e:N'" we denote by sN(a, S) the number of integers k such that
°
~ k < N, uk
e [a,S].
The DISCREPANCY is the sequence
(D N )
de-
fined by:
DN =
sup O~a~S~l
A sequence (u n )n .... ~~ of elements of the interval [0,1] is said to , be EQUIDISTRIBUTED in [0,1] if: For all a,S such that [a,S] C [0,1]
sN(a,S)
1 im --'N:;--N->«>
S - a.
A sequence (u n ) n<:&, ~~ of real numbers is said to be EQUIDISTRIBUTED MODULO 1 if the sequence ({u }) ~~ of fractional parts is equidistributed in [0,1]. and {x}
=x
-
n n""" (If x elR we write [x]
[x]).
241
= sup{k e~,
k
~
x}
CHAPTER 5:
242
Let S be a finite set, and let (un)neN be a sequence of elements of S. For all a e S and every integer N e:N~' let us denote by sN(a) the number of integers k such that 0
~
k
<
N and Uk
= a.
We say that the sequence (u) is EQUIDISTRIBUTED IN A SET S if: nnSN
sN(a) For all a e S lim - N N
1
= -c-ar';;;d'"'(-=S"<"")
A sequence (un)nSN of elements UTED MODULO q (q
is said to be EQUIDISTRIB-
of~
> 1 an integer) if the sequence (u n ) n..... ~T is equi-
distributed in the set
~/q~
(where
x denotes
the class of x modulo
q). 5.2
CRITERIA FOR EQUIDISTRIBUTION MODULO 1: (Exercises 5·11-5·20) The sequence (u n )n .... ~~ of elements of [0,1] is equidistributed ,
on [0,1] i f and only i f for every Riemann integrable function f on [0,1] we have: 1 N
2
O.r;n
f(u) ~ n
Ii
0
as N ~
f
co.
WEYL'S CRITERION: The sequence (u )
of elements
~~
n n ....,
of~
is equi-
distributed modulo 1 if and only if: For a 11 p
where e (x) p
e~"~
_1
l.\
N O:(.n
e (u ) p
n
~
.c
0 as
N
~ co,
= e2i~px.
These results lead to: DEFINITION: Let S be a compact space,
~
a positive measure of
total weight one on S, (u n )n .... ~'T a sequence of elements of S. The , sequence (un)nSN is said to be DISTRIBUTED IN ACCORDANCE WITH A MEASURE ~ if for every continuous function f of S into ~ we have:
DISTRIBUTION MODULO 1
~
5.3
243
L f(u n) + (~,f)
O.;;n
as N
+
00.
PARTICULAR SEQUENCES: (Exercises 5·21-5·28)
FEJER'S CRITERION: Let g be a monotonic differentiable continuous
function with monotonic derivative.
g'(t)
+
00,
tg'(t)
+
If, as t
+
g(t)
00,
+
00,
then (g(n»neN is equidistributed modulo 1.
00,
VAN DER CORPUT'S CRITERION: A sufficient condition for the se-
quence (u ) h
~~
n n=,
~
5.4
to be equidistributed is that for any integer
1 the sequence (u
n+
h - u) is equidistributed. n neN
SEQUENCES WITH MORE THAN POLYNOMIAL GROWTH: (Exercises 5·29-5·39) For almost all
onR) the sequence
~ >
(~
n
1 (in the sense of the Lebesgue measure
) is equidistributed modulo 1 (a result of
Koksma) . If a sequence (u n ) nCo'~~ satisfies the property: There exists "
U
such that 0.;; U < 1 and limnu(u n+ 1 - u n ) > 0, then for almost all 1 the sequence (Au ) ~~ is equidistributed modulo 1 (a result of n n"""
H. Weyl).
A real number A is said to be NORMAL IN BASE g (g teger) if the sequence (Ag n )
neN Therefore almost all numbers are normal. NORMAL SET B«u » n
~
2 an in-
is equidistributed modulo 1. More generally, the
associated with the sequence (u ) is by definn
ition the set of real numbers A such that (Au) is equidistribnneN uted modulo 1. It is known that given a set E a necessary and sufficient condition for there to exist a sequence (un)neJN such that B«u n » is that: (i) 0 • E, (ii) 'ZZ,;'E C E, (iii) E is a FGO (that is to say it is the countable intersection of a countable union of closed sets).
=E
244
5.5
CHAPTER 5:
CONSTRUCTION OF SEQUENCES STARTING FROM GIVEN SEQUENCES: (Exercises 5·40-5·46) Let (u )
~~
n neL'
be a given sequence,
0
a mapping
of~ into~.
We propose to deduce distribution properties for the sequence
(u o ( n )) neL. ~~ from distribution properties of the sequence (u) . n ne1N Let 0 be an increasing mapping, and let X be the associated GENERALISED CHARACTERISTIC FUNCTION of
0,
given by x(n) =
A particularly important situation is where X is
card{o-l(n)}.
almost periodic in the sense of Besicovitch, that is to say it belongs to the closure of the trigonometric polynomials P of the form:
I
pen) =
Aell.
aAe(An)
is the finite part of lR)
(II.
with the semi-norm:
Ilfll
= lim
N-+oo
~ I
k
If(k) I·
The following result is known: If for every integer
function which associates e(ru ) with n n
e~
.e.
~
I the
is pseudo-random3 and
if X is almost periodic with non-zero mean, then the sequence (u ()) is equidistributed modulo 1. (A PSEUDO-RANDOM FUNCTION onneN is a function f ofN into ~ admitting a correlation function y defined by: 1 n lim - - I f(k)f(k + p) n-+oo n + 1 k=O
yep)
satisfying, for example, the condition: __
1
lim - - 1
n-+oo n +
n
I k=o
Iy(k) I
2 =
0).
A method widely used for the construction of sequences is the method of "blocks", an example of which is the object of Exercise 5·45.
DISTRIBUTION MODULO 1
5.6
245
ERGODIC THEOREM: (Exercises 5-47-5-49) Let
be a probability space, and let
(S,ll,~)
~
be a measure-
preserving mapping of S into S, that is to say, such that: For all A eft, Then i f fe Ll(S,Jf1,~) for almost all xeS, the quantity: 1 N
\'
L
k
f(~
k
x)
~
g(x)
as
N
~
"".
The function g(x) e Ll(S,.,~) (furthermore, the convergence also holds in the norm) and is invariant under
~,
and:
is ergodic (that is to say, if the only sets
In particular, if
~
invariant under
are of measure zero or measure one) then for
~
almost all xeS:
Let S be a compact topological space, total mass one on S,
~
~
a Radon measure of
a continuous mapping of S into S such
that the unique Radon measure invariant under mass one is
~,
S we have:
lim N~
supl~ L f(~nx)
xeS
~
and of total
then if f is a continuous function from S into
k
-
~(f) I o.
BIBLIOGRAPHY [Rau], [Kui], [Cas], [Sal].
246
CHAPTER 5:
PROBLEMS
EXERCISE 5'0:
Tony lives very near a bus station from whence two
services leave with the same frequency: Route 41, which takes him to his blonde girlfriend Bridget, and Route 63, which takes him to his other girlfriend, Claudia, a brunette.
On Saturday morn-
ings Tony gets up at a time that varies with his mood or with the time it takes to go round to one or the other of his girlfriends.
But feeling at a loss as to how to decide between the
blonde and the brunette, he prefers to trust to luck; so when he arrives at the bus station he takes the first bus that leaves. Doing this he arrives at Claudia's more often. Can you explain why? EXERCISE 5'1:
Let (u n ) nco., ~~ be a sequence of elements in [0,1].
Prove that (u ) ~~T is equidistributed on [0,1] if and only if n nc.o.' DN -+ as N -+ '"
°
EXERCISE 5'2: N e:N"'. D'" N
Set: sup
0.,
a., 1
Let (u )
~~T
n nc.o.'
be a sequence of elements in [0,1],
DISTRIBUTION MODULO 1
247
(1) :
Show that D'" ~ DN ~ 2Dn· N
(2) :
Show that i f
D--',
sup
N
o.. k
1
Uo ~
~I
suP(lU k -
2N + sup O
IU k
u1
-
2k + 2N
~
...
IU k
-
< u N- 1 we have:
~I)
11
From this deduce that for every sequence (u n ) n.,.., ~~l and for every integer N ~ 1 we have: (3) :
EXERCISE 5'3:
Define a sequence (u) by setting: nner< . m and 1f 2 .. n
0,
<
m+l 2 (m e:N)
(1):
Show that if 2m
(2) :
From this deduce that for any n
nD;', n
~
1+ ~[10gn) 2 log2
<
U
1
Un_.JIl + --IL 2m+
n
2m we have:
n ~ 2m+l, on setting N
~
1
we have
.
Let (u) be a sequence of real numbers. nnW Show the following are equivalent:
EXERCISE 5,4:
(i): (u) nna;)' is equidistributed modulo 1; (ii): For every integer q
~
1 the sequence ([qu ]) _, is
equidistributed modulo q;
n
nco..
248
CHAPTER 5:
(iii): The set of integers q
~
1 such that the sequence
([qun])neN is equidistributed modulo q is infinite. For every pair of integers (N,k) such that
EXERCISE 5-5: (1):
°k
N, denote the interval [k/(N + l),(k + l)/(N + 1)] by I(N,k) . Let (u n ) ~~ be a sequence of elements in [0,1] such that ~
~
n~,
if:
NeN +
1) ~ n <
eN + l)eN +
2
2)
2
we have: un s I(N ,n - N(N 2+ Is the sequence (u n )
l)J . equidistributed on [O,l]?
~~
n~,
The same question, but this time calling I(N,k) the interval [k/2 N,(k + 1)/2 N] (0 ~ k < 2N) and assuming that if (2) :
N
2
- 1
~
N+1
n < 2
EXERCISE 5-6:
- 1, then u sI(N,n - 2N + 1). n
Let p be an integer, as [0,1].
of integers n such that p ~
in ~
Evaluate the number
p + a, and deduce from this that
the sequence (in)neN is equidistributed modulo 1. EXERCISE 5-7:
Is the sequence (log(n + 1)neN equidistributed
modulo I? EXERCISE S-S: (1): integers with q
~
Let as [0,1], and let p,q be relatively prime 1.
Evaluate the number of integers n such that:
°"' n ~ q
and
0
<
n
~-
[n
~]
., a.
DISTRIBUTION MODULO 1
249
From this deduce bounds for the number of integers such that:
n~ Nand
o~ (2):
0
<
n~ - [n~]
~
a
(N
~
1).
By using a result in Diophantine approximation theory
deduce from the preceding (1) that the sequence distributed modulo 1 when
~
(n~)
is irrational.
n e ..ThY, is equi-
Show that the Fibonacci sequence (F )
EXERCISE 5·9:
Fo = 0, FI = 1, and for n uted modulo 5. EXERCISE 5·10:
~
0 Fn+2
= Fn+I
~T'
n n"""
where
+ Fn' is equidistrib-
Let A be the measure on {O,ljN, that is the pro-
duct of the measure
~
on {O,l} defined by
~({O}) = ~({l}) =
!.
On the other hand, let ~k be the function from {O,l}~ into {-l,+l} that to
U
=
(un)n~
-1 if uk = 1.
(1) :
associates
= +1
~k(u)
if uk
=0
and
~k(u) =
Set:
Evaluate the quantity PN =
JlfN(U) 12 dA(U)
and show
that the series with the general term (p 2) is convergent. N
(2):
Show that for almost all
ure A) f 2(u)
~
0 as N
~
U
(with respect to the meas-
00
N
(3):
From this deduce that every sequence (with respect to
the measure A) with terms in {O,l} is equidistributed in {O,l}. EXERCISE 5·11: (1): Let (u) and (v ) ~~ be two sequences of n na:! n n""" real numbers such that (u - v ) ~ 0 as n ~ 00. n
n
250
CHAPTER 5:
Show that (un)neN is equidistributed modulo 1 if and only if (Vn)n~ is.
(2) :
Do the same question, this time assuming u
log(n + 1).
un - v n
What happens if we have:
= (log(n
+ 1))8
- V
n
n
with 8 > I?
Let (u ) ~'T be a sequence of real numbers, and n nco.. let (Nk ) be a sequence of strictly increasing integers such that Nk+ INk ~ 1 as k ~ 00. Assume that: EXERCISE 5-12:
Show that the sequence (u )
~'T
n n .....
EXERCISE 5-13:
is equidistributed modulo 1.
Let a,b,E be real numbers such that 0
b - a < b - a + 2E < 1.
< E,
0
<
Denote by F the function with period
one, which is also affine on the intervals [a - E,a], [a,b] ,
[b,b + E], [b + E,a - E + 1], and which has the value zero on [b + E,a - E + 1] and the value one on [a,b]. Show that if F(x)
= I
kS72:
Fourier series, then for k
akek(x) is the expansion of F as a
+ 0 we
have the upper bound:
Let (u ) AaT be a sequence of real numbers. n nCL' every integer q set: EXERCISE 5-14:
a (N) q
L
O
e (u ), q n
and for every [a,S] C [0,1] denote by BN(a,S) the number of
For
DISTRIBUTION MODULO I
251
integers n such that:
o~ n
{u } e [a.,B]. n
< N,
Using the result of the preceding Exercise 5-13, show that for
Q ? 1 we have the upper bound:
L
IsN(a.,B) - N(B - a.)1 < 2
1~q~Q
4N
10 q (N) 1 + - 1t
Thus recover Weyl's Criterion and the result of Exercise 5-1. EXERCISE 5-15:
Using the results of Exercise 5-2 show that if
(u) is a sequence of elements of [0,1] and f is a continuous nnet'l function from [0,1] into~, we have the inequality:
I
I -N
L
f(u n ) - flofl
O~n
~
w(D*)
N '
where w denotes the modulus of continuity of f, that is to say the function defined for r w(r)
=
sUJil
It-sl
EXERCISE 5-16: Let u
in {O,l}.
1 f(
Let X
>
°by:
t) - f( s )
1.
= {O,lfN be the set of sequences with terms
= (u n ) n.,., ~~ be an element of X and let
0
be the
function fromID intoN defined by: 0,
0(0) A
mapping
1t(u)
1t
o(n + 1)
= o(n)
+ I + u .
n
of X to itself is defined by associating the sequence to the sequence u = (u ) _, such that for all
= (v n )n.,., _,
n eN:
1,
n n.,.,
and if o(n + 1) - o(n)
2 then
v a (n)+1
o.
252
CHAPTER 5:
Show that the map
(1):
a point w satisfying
~(w) =
~
has a unique fixed point w (i.e.,
w) that is w
= (1,0,1,1,0,1,0,1, ...
).
Show that there exists a measure A on X such that the
(2):
sequence (TnW)neN is distributed in accordance with the measure
A, where T is the mapping of X to itself which to the sequence u
= (u) n nE!IiJ
associates the sequence Tu
= (u n+ 1) n",,-, ~~.
Let R be the set {0,1, ... ,9} and let u = (u ) ~~ nne>" be a sequence with terms in R JN such that for every integer n, if . t h e express10n . b ase . a k x1Ok + a k _1 x 10k-l + --- + a O 1S 0f 2n 1n EXERCISE 5-17:
ten, then un as a sequence of elements of R begins as (ak,a k +1 ,
... ,a O' · "
).
Does the sequence u admit a distribution in R JN ? Show that:
EXERCISE 5-18:
lim N+oo
jl
L
eitlog(n+l)I = IJleitlOgXdxl. 0
O~n
What can be concluded from this for the sequence (log(n + l))nEN modulo I? Denoting the n-th prime number by Pn , what can be said about the equidistribution modulo 1 of the sequence (logp) . nneN (The formula: EXERCISE 5-19:
lim [ -P- -J
n->-oo n10gn
1
may be assumed. EXERCISE 5-20:
Let U3 be the multiplicative group of 3-adic units.
When given the 3-adic topology it is a compact group admitting a Haar measure m that can be assumed normalized (i.e., Show that the sequence (Sn)
neN
m(U 3 ) =
1).
is distributen in accordance
253
DISTRIBUTION MODULO 1
with the measure m in U3 • for n;;. 1, 5
k
:: 1 mod 3
EXERCISE 5'21:
n
(One could begin
i f and only i f 2x3
by establishing that
n-l
divides k).
Show that if 0 < Y < 1 the sequence CnY)nSN is
equidistributed modulo 1 in a form analogous to Fejer's Criterion. EXERCISE 5·22:
Using Exercise 5·18 as inspiration, give a cri-
terion, analogous to Fejer's Criterion, for not being equidistributed modulo 1. EXERCISE 5'23: sequence
CPCn»
Give a necessary and sufficient condition for the neN
P is a
to be equidistributed modulo 1, where
generalized polynomial: + ••• + a x s
EXERCISE 5·24:
Ys
Let a,S be real numbers.
Denote by t
inality of the set of integers k such that 0
~
k
~
n
n, 0
the card~
{ak}
~
{sk} .
What conditions must a,S satisfy in order that: tn
1
'C-n-+-::-l"') -+
EXERCISE 5'25: {z e~,
Iz 1 <
fCz)
2" as n
-+ co?
Show that the function f from D into
~,
where D
I}, defined by:
L
n=O
{n~ }zn
admits the circle
Izl
1 as a natural boundary if and only if
~
is irrational. EXERCISE 5'26.:
Using the result of Exercise 5·13 show that the
254
CHAPTER 5:
discrepancy of the sequence
(n1J
)neN satisfies
D~
e
D( 11m),
(i): for almost all 1J; (ii): for all algebraic 1J. (One could use the results of Khinchin and Roth on the approximation of real numbers by rational numbers). EXERCISE
5 - 27:
Let
(jl
be a periodic function from lR into a: with
period one that is Riemann integrable. Let a,S be two real numbers such that S does not belong to the vector space generated on III by 1 and a. Show that as n
~
00
the quantity
has a limit that can be calculated. nO
<
n1
< ---
From this deduce that if
denotes the sequence of integers n such that {na}
belongs to a given interval I of [0,1], then the sequence (Snk)keN is equidistributed modulo 1 (I is of length greater than zero). EXERCISE
5-28:
Let a,S be two real numbers such that 1,a,aS are
linearly independent on Ill. Show that the sequence (S[an])neN is equidistributed modulo 1. EXERCISE
5-29:
If (u)
n nelN
satisfies limn(u
n+
1 - u ) n
>
0, does it
follow from this that for almost all A the sequence (Au) equidistributed modulo I? EXERCISE
5-30:
nnelN
is
Let 1J be a normal number in base g_
Show that g~l is also normal in base g. EXERCISE
5-31:
Let P be a polynomial with real coefficients and
DISTRIBUTION MODULO 1
g ? 2 an integer.
255
Compare the following two assertions:
n
(i): (xg )neN is equidistributed modulo 1; (ii): (xg (x
f
n
+ P(n»
neN
°is real).
EXERCISE 5'32:
Let g
is equidistributed modulo 1;
1 be an integer and A, 0
>
~
A ~ 1, a real
number with expansion in base g:
a k e {O, ... , g - I}.
Show the following are equivalent: (i): The sequence (Agn)neN is equidistributed modulo 1; (ii): For every integer
B ~
0 the sequence «an, ... ,an+s»neN
is equidistributed in the finite set Gs +1 , where G = {O, ... ,g -
EXERCISE 5'33:
n.
Let (u n )n .... ~~ be a sequence of strictly positive ,
real numbers, and let I tervals of
JR
such that:
n
=
[a n ,S n ] be a sequence of compact in-
For all n eN
Show that there exists A such that: For all n eN
AU
n
e I
n
EXERCISE 5'34: From Exercise 5'33 preceding deduce that if the sequence (u ) satisfies: n
256
CHAPTER 5:
Un +1 -u-- ;::. 1',
For all n e:N
where
l' >
2,
n
then there exists A such that: For all n
{AU } e [0, 1/ (1' - 1)], n
eJN
and that by also assuming n
eN
l'
> 3,
the set of A'S such that for any
we have: {Au } e [0,2/(1' - 1)] n
has the power of the continuum. EXERCISE 5 35:
By using the preceding result, Exercise 5 34, for
0
0
a sequence (V), where v = U band a,b are suitable integers, n n an+ prove that if a sequence (u ) SN of real numbers greater than zero n n
U
satisfies lim
n+1 u
>
1, then the complement of B«u )) is everyn
n
where dense and has the power of the continuum. Let A,0 be real numbers such that A t 0 and 0
EXERCISE 5 36: 0
>
1.
Assume that there exists a sequence of integers (un)neN and a sequence of real numbers (En)neN such that: limE n-- n
o.
Show that starting from a certain index we necessarily have:
~ u l n+2 u~+ll u
<
-
n
2 .
From this deduce that the set of such pairs (A,0) is countable. Now show that the set of pairs (A,0) such that the sequence
DISTRIBUTION MODULO 1
({A~n})
neN
257
has a finite number of limit points is countable.
Show that the pair (1, 1 +2
~
belongs to this set.
(For a gen-
eralization to Pisot's numbers, cf., [Sal]). Show that to any sequence (u) of elements of nneN [0,1) one can associate a real number a such that:
EXERCISE 5-37:
lim({n!a} - u ) n
n-+=
= a.
In particular, show that lim{n!e} n-+=
Give examples of sequences (u) of real numbers nneN
EXERCISE 5 - 38 : such that:
B«u )) n
=
O.
JR - {a},:?Z - {a},lR - g),¢.
EXERCISE 5-39:
Give an example of a strictly increasing sequence
of integers (u) such that B«u )) =JR - E, where E is the vecn neN n tor space over generated by 1 and a (a irrational). You may
m
need to use the results of Exercise 5-27. EXERCISE 5-40:
Show that from every sequence (u)
nneJN
of elements
of [0,1] that is dense in [0,1] one can extract a sequence (uo(n))neN (0 strictly increasing) equidistributed modulo 1. EXERCISE 5 -41:
Prove that the function X from IN to:N is almost
periodic in each of the following cases: (i): x(n)
= card{k e:N: [ak + S] = n}, where a, S eJR, a
> 0;
(ii): X is the characteristic function of the set E of integers n such that: There exist a e A and a In,
258
CHAPTER 5:
A a set of integers such that
L aeA a
1
-<00
(Example:
The set of square-free integers). EXERCISE 5·42:
Let g , 2 be an integer, A a normal number in
Show that for any integer l r 1 the function which associates e(lAg n ) with n is pseudo-random. From this deduce, in
base
g.
particular, that A is normal in base gS for any integer s ~ 1. Comparing this result with that of Exercise 5'30, deduce from it that if A is g-normal, then the same is true of every number of the form 1'A, where EXERCISE 5'43:
l'
e
gJ'~.
For integral g ? 2 let B(g) be the set of normal
numbers with base g. Under what condition on the integers g,h 7 2 does the equality
B(g) = B(h) hold? EXERCISE 5,44:
Prove that if (u )
~~T
n n=,
is a sequence of real num-
bers such that for every integer p ? 1 (u
- u ) ~~T is equidisn+p n n=, tributed modulo 1, then the function which associates e(lu ) to
n (where l e:N"') is pseudo-random.
n
In this way recover Van der
Corput's Criterion. EXERCISE 5·45: that Nk
~
00
as k
of real numbers.
Let (Nk)keJN be a sequence of integers Nk 7 1 such Let (u ) ~~ and (v ) ~~ be two sequences n n""" n n""" Construct a sequence (Wn)neJN in the following
~
00.
manner:
More generally, (wO,w 1 " " where:
is a sequence of "blocks" (CO,C 1 " "
)
DISTRIBUTION MODULO 1
c
2n
=
(U 0' ... ,uN
259
C
),
2n
(VO"",V N
2n+1
2n+1
).
Prove that if f is a function fromF into F we have the in equality: __
1
lim - - 1
n
n +
n
L
k=o
~ EXERCISE 5·46:
f(W k )
~1 k=o I
sUP[lim n n ~
f(u k ),
~1 k=o I
lim n n +
f(V k
») .
By using the result of Exercise 5·45 and that of
Exercise 5·5(1), show that, given a function f from [0,1] intoF, a necessary and sufficient condition for it to be Riemann integrable is that for every sequence (u) of elements of [0,1] equidistribnneN uted on [0,1] the quantity 1
n+1
n
L
k=o
f(u k )
has a finite limit as n
+
(Begin by showing that this limit
00.
is independent of the sequence chosen). EXERCISE 5·47:
Let a e F -
F defined by sending x
~
~
~
be the transformation on
x + a.
Show that the only measure sure.
and let
on~
invariant under
~
is Haar mea-
Thus recover the result on the equidistribution modulo 1
of the sequence (na) EXERCISE 5·48: of ~ onto
~
neN
.
Apply the Ergodic Theorem to the transformation
defined by x
+ gx,
where g
e~"'.
260
CHAPTER 5:
EXERCISE 5,49:
Let m be a positive measure
on~
that has total
mass one, and let II be the product measure on S = 'll':IiI. Show that the transformation
~:S +
S defined by
) = nneN From this de~«u)
(u n+ 1) n ...... ~~ preserves the measure II and is ergodic. duce that with respect to II almost all the sequences
m-distributed.
on~
are
DISTRIBUTION MODULO 1
261
SOLUTIONS
SOLUTION 5,0:
The buses leave, for example, every thirty minutes,
the 63 according to the schedule 9.00,9.30, 10.00, etc., ... , and the 41 with the schedule 9.05, 9.35, 10.05, etc.,... age, five out of six times Tony will visit Claudia.
On aver(See also,
Landau: GrundZagen der Mathematik, ... ). SOLUTION 5·1:
It is evident that if DN
~
°then (un)
is equidis-
tributed on [0,1]. Conversely, let us assume (u ) is equidistributed on [0,1] and let
£
>
O.
n
Let P be an integer.
By hypothesis, for every pair
of integers (h,k) such that 0
~
h
['2 ~J - ['P2 - ~)P) I ~ I!N s NP'P)
0
as N
~
k
~
~
P, '"
As the set of such pairs is finite, it follows from this that there exists an integer No such that for all N ? NO and for every pair (h,k) we have:
.!.IN s N('P2 , ~JP
- ('2 - ~J I P
P
~
< 2'
If [a,S] is an arbitrary interval there exists a pair (h,k) with
262
CHAPTER 5:
o~ h
k
~
P such that:
~
[Ph , pk]
::> [a,S]::>
[h- p + 1 k - 1] -, -p-
From this, for N ? No' we deduce:
It suffices to choose P such that P
~ in order to deduce the I':
>
result, Clearly
SOLUTION 5'2:(1):
D~ ~ DN ,
Furthermore, for any
I':
>
0
we have:
Making
I':
-+
(2):
0 one obtains DN .;;; 2DB' The function from [0,1] intoR which with
0.
associates
meN) -_ ISN(oN,a) - NI1"S Y ~ u affine on every interval of the form
(U k 'U k +1 ).
Its maximum is therefore of the form
Whence the first result, mains true if Uo
~
u1
~
By passing to the limit this result re•••
~
UN-I'
The second result is obtained by noticing that if Uk is on the "d k+ i ( , ) relative to t h e centre 2k+l Sl e as -N-'l- = 0 or 1 2N
same
of the interval
IU k (3) :
- k ;
[~
il .; ;
, k ; 1] , we have:
IU k
- k +
~
1
il
= IU k - 2k 2;
11
+
2~
The inequality DB ? 2N clearly follows from the pre-
DISTRIBUTION MODULO 1
ceding inequal ity.
263
It remains to observe that i f uk
equality is obtained. By induction on m and by setting 2m
SOLUTION 5-3:
2k + 1 2N
the
N, we see
that:
(and in particular that for all n, uke [O,l[). (1):
From this it follows that:
From the equalities:
and: card{k:N
~
k < n,uk e [O,a.]}
= card{k:N we deduce the desired equality. sN(O,~)
= 0 if
(2) :
-Iii
k < n,u k _N e [O,a. - 1/2N]}
(By convention one assumes that
~ < 0).
By induction let us show that if 2m < n < 2m+1 , for
all a. such that - ~ ~ a. < 1 we then have: 2m+
ISn(O,a.)
- na. I
<
3m
1 + :2
The formula is evidently true for m = O. Let us assume that it is true for all n -Iii 2m and let n be such that 2m < n ~ 2m+l. The
264
CHAPTER 5:
formula is obviously true if a
<
O.
show that it is true for 0
~
1 also.
a
~
Therefore it suffices to Now,
and \(
I
a - _ 1 (n _ 2m) - a(n - 2m) \ 2m+1j
~
21 .
By using the formula of (1) above, the result is obtained, whence the result of (2) is obtained immediately. REMARK: In fact much better upper bounds can be obtained. SOLUTION 5-4:
that q t
~
1, 0
q •r •s
(n)
Let q,r,s be integers such
Clearly (ii) => (iii). ~
r =
s
<
~
q, and let:
card{k:O
~
k
(~q , ~q
-
oJ
n,r
<
~
[qu k ]
<
s}.
Clearly, t q.r.s (n) = s
,
and [qu ] is equidistributed modulo q if and only if: n
n~ s n (~ q , ~ q -
0)
~~ q
as n
~
00.
From this it immediately follows that (i) => (ii) and that an argument analogous to the one carried out in Exercise 5-1 enables us to show that (iii) => (i). SOLUTION 5-5: (1):
that:
Let tN(a) be the number of integers n such
DISTRIBUTION MODULO 1 N(N + 1) 2
"n<
265
(N + 1)(N + 2) 2
Evidently i f a e I(N,k), k
tN(a)
~
and ~ k
u n e [a,a].
+ 2 and k
~ (N
+ 1) "
k + 1, whence:
If now: N(N + 1)
2
~
n
<
(N + 1)(N + 2)
2
and
M = N(N + 1) 2
then:
whence:
whence: Isn (a,a) - nal " 2(n - M) + 2N" 4N + 2 " 41:2n + 2, so: D* 4& + 2 N " n -+
a as
n -+
00.
The sequence (u n ) is therefore equidistributed on [0,1]. (2):
The preceding argument used the fact that the interval
l),!(N + l)(N + 2» was of approximate length !N, and therefore "negligeable" in comparison with the interval [0, !N(N + 1» [~N(N +
(which would allow sn(O,a) to be approximated by sM(O,a»). The same is no longer true in the case considered here.
For
266
CHAPTER 5:
N
example, let us assume n
2
- 1 + 2
N-l
N
and let M = 2
- 1. Then:
By the preceding argument one easily shows that when N ~ 00:
From this it follows that:
The sequence (u ) is therefore not equidistributed on [0,1]. n
Neither does it admit a distribution with an arbitrary measure. If t
SOLUTION 5'6:
card{n:p'" In ~ p + cd, since p ... In ...
p
P + a is equivalent to p
t
P
2
~
n ... (p +
a)
2
one evidently has:
2
+ [2pa + a ],
1
so:
ItP -
2pa I
...
2.
Now let n ? 1 and let p = [In] and P
Furthermore, p-l
L
q=O whence:
t , q
2
P ,
DISTRIBUTION MODULO 1
Isn (O,a)
- nal
267
7p + 2
<
7m + 2.
<
From this one certainly deduces that D* n
proof. SOLUTION 5·7:
~
0, which completes the
If one tries to apply the preceding argument to
the sequence log(n + 1) the same difficulty is met as in passing from the first to the second part of Exercise 5·5. s (O,!) for n of the form [e n
k+l
By evaluating
2] (k an integer) we will show that
this quantity does not tend towards
!,
and therefore as before
the sequence (log(n + l))neN is not equidistributed modulo 1, nor does it admit a distribution measure. With p,q being relatively prime, multiplica-
SOLUTION 5·8:(1):
tion by p is a bij ection of 'ZZ./ q'ZZ. onto itself. card{n:O
~ n <
{~q}
q,
card{h:O
e [O,a]},
h h < q, - e [O,a]}
~
q
If:
TN =
card{n:O
~
n
< N,
{~}
then we have: T
q
= 1
Setting r T
rq
[qa].
+ =
[N/q], we deduce:
~ TN ~ T
rq + q,
whence finally we have:
Hence we have:
e [O,a]},
1
+ [aq].
268
CHAPTER 5:
ITN
-
(2):
No.1 " ~ + Let
q.
0 be given.
E >
By a classic theorem on Diophan-
tine approximation we know that there exist two integers p,q such that: E
1"q~3N,
In addition, with
E
sufficiently small p
+0
certainly holds, and
(p,q) = I can be assumed. If SN(a) and TN(a) denote respectively the numbers of such integers n such that:
°~ n
{n~}
and
< N
e [0,0.]
or
{n~}
e [0,0.],
we clearly have:
whence, on using the result of Exercise 5·7:
SN(a)
!- N -
~
- a
!
~
1
N
q + ~ + q[NE/3]
being irrational, when N +
ever N is large enough:
whence:
< E.
00
so does q.
In particular, when-
DISTRIBUTION MODULO 1
269
SOLUTION S-9: Modulo S we have FO = O. Fl = 1 •••.• F20 = O. F21 = 1. And therefore for n = 0 and n = 1 Fn+20 = Fn' By induction one deduces from this that the sequence is periodic with
period twenty.
It only remains to establish by a further direct
calculation that whenever n e {O •...• 19} F
n
value modulo S four times.
takes exactly every
REMARK: More generally. F is periodic modulo Sk (k ~ 1 an inten
ger) with period 4xSk. and in each period it takes each value modulo Sk four times. hence is equidistributed modulo Sk
One
can interpret this result in terms of a distribution
(S-adic
integers).
in~5
In addition. if F is equidistributed modulo q (q n
>
1
an integer) q is necessarily of the form Sk. SOLUTION S-lO: (1):
Clearly: if k
= h.
otherwise. 1
From this one immediately deduces PN = N • whence the result. (2):
The series:
being convergent. from the Fatou-Lebesgue Theorem it immediately follows that the series
I
N=l
where. and in particular that (3) :
Ifn (u)
f
2(u) 2 is N f 2(u) -+ 0 N
By noticing that. for N2 -
f
2 (u)
N
I"
n - N2
4;
4;
convergent almost everyas N
-+
co.
2 n < (N + 1) :
1 + 2m.
we deduce that for almost all u. with respect to the Lebesgue
270
CHAPTER 5:
measure >..: 1 f (u) n n
~
as n
0
~
00
Now, the space of functions from {O,l} to
~
is generated by
the constant function which is equal to one and by the function ~
such that
~(O)
= 1,
~(l)
= -1.
Thus for almost all u, we have
with respect to >..:
~
L ~(u
N k
k
) ~ 0
whence the result: Let p be an integer, and let
SOLUTION 5-11: (1):
exists 8
£
>
O.
There
0 such that:
>
Ix-yl
<8=> le(x)-e(y)1 <2£ p p
Hence there exists an integer NO such that for n
Ie p (u n )
- e (v )1 p n
whence, if N
I ~ n
~
~
No:
£
<'2'
4N o/£:
e (un) p
~ L ep(vn)1 < %+ 2~O n
<
£.
The result follows by applying Weyl's Criterion. (2):
Let us assume p
equidistributed modulo 1.
~ n
e (v ) p
n
~
o.
~
1 is an integer and that (v ) is n
Weyl's Criterion implies that:
DISTRIBUTION MODULO 1
271
In particular, for any 0 For all
0 there exists Co such that:
~
N > I
or: For all M,N, N > M ~ 0,
L 1Mtr.n
e (v )1 < 20N + 2C O' p n
Furthermore, there exists Cl > 0 such that:
Ie p (x)
Let then
o> a
I
- e (y) p
£
>
a
< Cl (x - y).
and M be an integer such that
2C l
~ <
£
2'
let
be such that: e
11M
£
40 e l I M _ 1 < -2 '
and let (Nk ) be the sequence defined by:
Let us assume that the integer N satisfies the inequali ty Nk < N ~ N
k+l
Then:
1
L
n
[e (u ) - e (v)] 1 <0 p n p n
K~l k=O +
If n is such that Nk
IIOg(n
+ 1) -
~I
n
<0
<
<
Nk +l :
k'
1
N
L
~n
k'"
1
L
Nk
k+l
e (u ) - e (v ) 1 p n p n
[e (u ) - e (v)] 1 . P n P n
•
272
CHAPTER 5:
whence: [e (u ) - e (v)] P n P n
- e (v ) [e [75:.) P n p M
- 1]
whence:
so:
IN
I
~n
k"
k+l
[e (u ) - e (v )] p n p n
I~
2;
Nk +l
- Nk )
Similarly:
But: K-l
I k=o
Nk + l ~
K-l
I k=o
(k+l)/M
e
~
e(K+1)/M 11M
e
- 1
~
e
N + 211M 11M e
- 1
and finally:
In
~
00,
[e (u ) - e (v)] p n p n
(N + 2)IN
~
I
CIN
11M
~ 2 ~ + 48(N + 2) ~e~ __ e l/M _ 1
1, (K + l)IN
~
0, therefore as soon as N
DISTRIBUTION MODULO 1
273
is large enough:
2:. I
1N n
[e (u ) - e (v)] 1 < p n p n
which implies that ~
I
e (u ) p
n
n
7
E,
0, and therefore, by Weyl's
Criterion, that (u ) is equidistributed modulo 1. n
Conversely,
it is evident that if (u ) is equidistributed modulo 1 we show n
in the same way that the same holds for
(v ). n
The essential fact used in this proof is that we have divided the interval [0,1] into intervals [Nk,Nk +1 ) such that for n e [Nk,Nk+1 ), (un - v n) is almost constant, the sum I Nk is boundNk.;,N ed by CN where N
7
approximation of u
(C being a quantity depending only upon the
00
- v n sought).
1 it would therefore (k/n 1/o be necessary to take a sequence of the type Nk ~ e , which n
When 0
>
is not increasing fast enough to obtain the result sought.
In
fact, a criterion close to that of Fejer shows that the sequence u
n
(log(n + 1»0 is equidistributed modulo 1 for all 0 > 1;
then taking v when 0
>
1.
n
=
°we
see that the preceding result does not hold
Let us assume that given p
SOLUTION 5 12: 0
e~*
exists K such that: For all k
~
K
First of all, by induction on k we have: For all k ;;. K
And consequently if k
~
K, and Nk
< n ~
Nk +1 ,
and
E >
°there
274
CHAPTER 5:
So finally, if k
~
K, Nk < n , Nk+1 :
so again:
NK L e p (u n ) I <-+-+ I~ O'h
K+1- 1 ) (N -Nk
As soon as n, and therefore k, is large enough: Nk +1
£
-- - 1 < -
NM<
4 '
and thus:
The sequence (u ) thus satisfies Wey1's Criterion and is theren
fore equidistributed modulo 1. SOLUTION 5·13:
ok
f1+a
We have: -£
a-£ 1
4a 2 k 2 £
e(-kx)F(x)dx
[e(-ka) - e(-ka + kc) + e(-kb) - e(-kb - kc)],
whence the upper bound:
DISTRIBUTION MODULO 1
275
The upper bound Ickl ~ 1 evidently results from the fact that: For all x
elR
1.
Taking a = a, b = S,
SOLUTION 5·14: F
Ie( -kx)F(x) I ...
>
£
0, we define a function
If S - a + 2 £ < 1 we then
as in the preceding Exercise 5 ·14.
have:
= L
sN(a,S)
n
L
F(u n )
qe~
C 0
q q
(N),
whence the upper bound:
sN(a,S) ~ coN + 2
L q~l
and taking into account
Ic q 110 q (N)I,
Co = S
- a
+
£:
L Ie q 110 q (N) I .
sN(a,S) - N(S - a) ... £N + 2
q~l
If S - a + 2£
~
1 we obviously have:
and thus:
And therefore in all cases:
sN(a,S) - N(S - a) ... 2£N + 2
L q~l
Ic q 110 q (N)I·
Replacing the function F by the function G, obtained this time by taking a
=a
+ £, b
=
S -
£,
ing in the opposite direction:
one would obtain an inequality go-
276
CHAPTER 5:
sN(a,B) -
a) ? -
N(B -
L Ie q 110 q (N) I.
2
2£N -
q?1
So finally, for the discrepancy we have: For all
>
£
DN " 2£
0, for all N ? 1,
L
+ 2
)1 (N)I 0
1
min (1, 2""2 ~
q?1
q
11
£
(by using the upper bound of the preceding Exercise 5·12). One can further write:
2£ + 2
DN "
L
0
(N)I
~ +
1
1~q"Q
2
L
q>Q
1
2""2 11
q
£
(N)I --7.
10
Taking:
£
1
= -11
L
10 q (N)/NI q
q
2
one finds: 0 (N) 1 1--..9..-
N
4
+_ 11
L
q>Q
10 q (N)/NI q
2
which is the required result. Let us notice that
10 q (N)/NI
.. 1, and that:
whence: DN 4t. 2
L
14tq.. Q
1
0qN(N)
I + 7T1Q • 4
DISTRIBUTION MODULO 1
277
In particular, if a (N)/N any Q we will have:
Therefore DN
~
q
0 as N
equidistributed.
~
~
1 for a sequence for all q
1, for
~
00, and, in particular, the sequence is
Conversely, if a sequence is equidistributed
one knows that for all q
~
1 0q(N)/N
~
00, therefore
DN ~ 0, which
is the result of Exercise 5'1. Let us reorder the first N terms of the sequence
SOLUTION 5·15:
(un)' say: Vo
vI
~
~
••• < v N- I ' then we have: and
But as f is continuous there exists
I
(k+I)/N kiN
I I
f =
o
N-I I(k+I)/N
L
k=o
sk e [kiN, (k
kiN
+ 1 )/N] such that:
_!.
f - N f(sk)'
whence:
But:
by Exercise 5'2.
whence the result.
f·
From this it follows that for all k:
278
CHAPTER 5:
Let us introduce the metric on X defined,
SOLUTION 5·16: (1): if u
+V,
by:
d(u,v)
where v(u,v)
= 2- V(U,v), = inf{n,un +v n }.
It is easy to see that restricting p to the set
Y of sequences
u such that U o = 1 gives an operation on Y. More precisely, if u begins with (1, ... ) then n(u) begins with (1,0, ... ) and: For all u,v e Y
d(1t(u),1t(v»
Y is a contraction it admits an unique fixed
As 1t restricted to point.
~ ~d(u,v).
Furthermore, i f
W
is a fixed point of 1t one has 1t(w) e Y
and as 1t(w) = W from this we deduce that we Y - which gives the result we want. In fact it is easily seen that the mapping 1t consists of making correspond to the sequence u = (u O.u 1 •... the sequence w = 1t(u) formed by juxtaposing the blocks V = (B o,B1 •... ), where
and the block (1) if uk = 0. Thus one obtains the beginning of w by these successive substitutions.
Bk is the block (1.0) if uk
w
= (1, ...
),
whence:
w
= (1.0 •...
),
whence:
w
= (1,0,1, ...
),
whence:
w = (1,0,1,1,0, ... ), .................. whence:
=1
DISTRIBUTION MODULO 1
279
whence:
w = (1,0,1,1,0,1,0,1, ... ), ......, ......,......, 1
0
1
0
etc., ... (2): (Tnw)
neN
To say a distribution measure exists for the sequence is equivalent (with the functions depending only upon a
finite number of coordinates forming a dense subset of the set of continuous functions on X) to showing that being sequence f = (f 0' ... ,fm) the set of n
has a density.
el'l
given a finite
such that:
This is a classic result of permutation theory,
the associated matrix here being the matrix
[~
i] , the square
of which has strictly positive coefficients. In this particular case this result can be recovered directly. In fact, let a =
3-15 . 2 be the root of the equatl0n
lying between 0 and 1.
x
2
- 3x+ 1
=0
Let XO,X l be the intervals in [0,1] de-
fined by: [1 - a,l),
Xl
=
[0,1 - a),
and let T be the transformation of [0,1) into itself defined by: Tx
{x + a}.
To every point xe [0,1) one can associate the sequence (v(Tnx» with terms in {O,l}, where one sets vex) if x e Xl' Given a sequence f that:
(fo""
= 0
if xeX o ' vex)
E!IN
= 1
,fm ) a priori, the set of n el'l such
280
CHAPTER 5:
... , that is to say, furthermore, that: na e (X f
o
- x) n(Xf
- x - a)
1
n
n(Xf
-
m
x - m) modulo 1
evidently admits a density, because of the equidistribution of the sequence (na).
Therefore it suffices to show that there ex-
ists an element a e [0,1) such that w = (w(Tna) )n~' On noticing that the transformation S induced by T on the interval Xl' that is to say the transformation defined by:
is again a rotation: Sx
=x
- a(l - a) modulo (1 - a)
such that if ~(O)
we have of
~
=1
~oT
~ (say, a
denotes the affine mapping defined by: - a,
H1)
= So~.
0,
It is easily seen that the unique fixed point
=~ )
= ;
=
answers the question.
(Notice that, quite
n
generally, for all x e Xl the sequence (v(T x)neN) is obtained from the sequence
(v(~
SOLUTION 5-17:
-1 n
S x))neN by the mapping n).
In order for u to admit a distribution in~ it
is necessary and sufficient that for all sequence (cO"" ,c ) aklO
k
S
cO'···, a k _ s
8 ~ 0
and every finite
the set of integers n, where 2
+ --- + a o' such that a k
Now, a k
so:
eJR
s+l
= cO'···,
= C s means:
a k _s
= Cs
s
=
has a density.
281
MODULO 1
DISTRIBUTIO,'~
so, moreover:
n~-ke [~ ~) 10g10 10g10 '10g10 on setting A
= Co
+ ••• +
C
8
'
10- 8 and B
The n's
therefore satisfy:
{n ~} 10g10
e
[~ ~I 10g10' 10glC)
,
and such an n satisfies the question.
As /00l120 is irrational the
100"2}.1S equ1' d " b ute d on [0 , 1) and t h e se t 0 f n I s sequence {n ~ 1str1 therefore has a density equal to: 10gB - 10gA
10g10
SOLUTION 5'18:
Let J = J:eXP(itlOgX)dx.
A classic result in
Riemann integration theory shows that: J
= lim ~ NIl N--
k=o
exp[itlog[k ; 1)) )
From this we deduce:
~ n
e
itlog(n+1)
JeitlogN + 0(1).
In particular:
liml~ L
N--
n
e itlog (n+l)I
IJI.
In order to show that for no t is the sequence (tlog(n + 1)) equidistributed modulo 1 it suffices to show that J is never zero.
282
CHAPTER 5:
e -u shows that:
The change of variable x
1
1 +
SOLUTION 5·19:
it f O.
Let us assume that the sequence (logp ) admits a n
Denote by Nk and Mk the integers defined
distribution modulo 1.
by:
Now let X be the periodic function with period one which is equal to one on [O,!) and to zero on [!,1). x(1ogp n ) =
I
Clearly:
x(1ogp), n
n
and the hypothesis that (logp ) admits a distribution implies n
that the two sequences: X(logp )
and
n
have the same limit
t as
k
+
x(1ogp ) n
00.
If this limit is not zero, this
implies, in particular:
Now, if
~(x)
denotes the number of prime numbers less than or
equal to x we know that: ~(x)
X
'V - -
logx
Therefore, as k
+
as x 00
+
00.
we have:
DISTRIBUTION MODULO 1
283
which is a contradiction.
It remains to show that if the limit
were to exist it would necessarily be different from zero.
L
x(logp) ~ card{p:k - 1 ... logp
n
n
<
Now:
k - ;},
and:
k-; [
k
: _ ; 1 - k
= !1 e -21]
+ 0
k _~ ,
(k-; )
and so: x(logp ) n
SOLUTION S-20:
~
1 - e
_l 2
>
O.
It suffices to show that for every integer n
~
1
the sequence (Sk) is periodic modulo 3n , and that in the fundamental interval the number of times it has a residue which is invertible modulo 3n is independent of this residue (it is clear that it is always invertible).
As the number of invertible resi-
dues modulo 3n is equal to 2x 3n - 1 , it suffices to establish that the fundamental period of the sequence Sk is equal to 2X3 n - 1 Therefore let T(n) be the smallest positive integer such that:
Clearly, i f Sk
Sh mod 3n
k - h is a multiple of T(n).
In par-
284
CHAPTER 5:
ticular, T(n) divides T(n + 1). 1 + 8x3.
Now, we have T(l) = 2 and ST(l)=
From this one then easily deduces by induction that:
where 3 does not divide A(n). SOLUTION S-2l:
function t
+
It suffices to apply Fejer's Criterion to the
t Y.
SOLUTION S-22:
Let
there exists a ~(k
satisfying:
> 0
+ 1) -
be a strictly increasing function such that
~
~(k
+
~) ~ a~(k
whenever k is large enough. the sequence
(~(n»neN
+
~)
~
is the inverse function of
If
cannot be equidistributed modulo 1, for
denoting by Sex) the number of integers n such that n ~(n)
-
[~(n)]
S(~(k
~,
~
x and
e [O,!), for any large enough k we would have:
+ 1»
=
S(~(k
>
~
+
~»,
whence: S(~(k
+ 1»
~(k + 1)
1
+ 1» th ere f ore S(~(k ~(k + 1)
an
S(~(k
+
~»
~(k + ;)
d S(~(k + !» ~(k
+
!)
cannot
b
h ot· converge to
1
2·
To obtain a criterion analogous to Fejer's, one takes, for example,
~
if M =
supt~'(t)
ue
J
u
to be differentiable and
t~'(t)
for any u we have:
1/(2M) ~'(t)dt ~
1
2'
bounded
(~'
~
0).
Then
DISTRIBUTION MODULO 1
285
or again: ( 1/(2M» cp ue
Ifu
~(k +
-
cp(u) .,..
3"1
!):
cp(e 1 /(2M)cp(k
+~»""k+
1
2:
+
l2
+ 1,
k
whence:
e 1 /(2M)
which is the relation sought, with c
_ 1.
SOLUTION 5·23:
By assuming Y1 > Y2 > ••• > Ys' and by applying Van der Corput's Criterion a sufficient number of times we are led to studying sequences of the type: v
= a P ••• p kY1 (y 1
nil
- l)···(y
y -k
1
where wn tends to a limit, and where
- k + l)n 1
°
<
Yl - k.,.. 1, and Pi
+°
are integers. If Yl is not integral, or if a 1 is irrational, (V n ) is equidistributed modulo 1, hence un is also. If Y1 is integral and Y
a 1 is rational, the sequence (a 1n 1) is periodic modulo 1. If T is its period it then suffices to show that each sequence of the type: (te{O, ... ,T - 1})
is equidistributed modulo 1.
By induction one thus shows that (u ) is equidistributed modulo 1 if one of the y's is not inten gral or if one of the a's is irrational. In the contrary case
286
CHAPTER 5:
(u ) is periodic modulo 1, therefore is not equidistributed.
We
n
have thus obtained a necessary and sufficient condition. If (I,a,B) are
SOLUTION 5·24:
independent the equidistribution of the sequence (an,Bn)neN inR 2 mod~2 implies that ~-linearly
the limit oft /(n + 1) exists and is equal to the measure of the n
set of (x,y)'s such that 0
x
~
~
a,B are rational numbers let a
y
~
1, which is equal
a/q, B
= b/q
with (a,b,q)
then if we denote by P n ,3 n the integers such that 0
o~
3
= na (mod
Pn
~ Pn <
1,
q,
q, then:
<
n
If
to~.
and
q)
B
n
nb (mod q).
We have: P
n q
{na}
3
n q
{nB}
therefore:
{an}
{Bn}
~
<=> P
3
~
n
n
= P n and 3 n+q = 3 n , and if we respectively den+q note by t' ,t" the number of integers k such that 0 ~ k .. nand Furthermore, n
{an}
n
{Bn} or {an}
<
t
P
n
= {Bn} t'
n
t ' + t"
n
we see that:
-,---::-,<""
n'
(n
+ 1)
-+
t"
t'
n
-q-' -,---::-,<"" (n + 1)
t"
-+ -
q
(t' + t") q t' ,t" respectively denoting the number of integers n such that
o~
n
< q
and
Pn < 3n
or
Pn
= 3n·
But
P q- n
=q
- P n and 3 q _n
q - 3 n , therefore if P < 3 we have P > 3 ,from which we n n q-n q-n
deduce:
DISTRIBUTION MODULO 1
287
= q,
t" + 2t'
so: 1
t
"2
t" + 2q
If d is the g.c.d. of b - a and q (with the convention (b - a,q)
=d
q i f q divides b - a) clearly t" never tends to
~
=
1 holds, therefore t I(n + 1) n
!.
If a is rational and S is irrational, then {Sn} is equidistributed modulo 1 and {an} periodically takes the values {O,
q - 1
--q--- }, each sequence (S(qn + r»
neN
l , ... , q
is equidistributed modulo 1
when r runs through the set of residues modulo q, and for each of these sequences the frequency when ~~ {S(nq + r)} is therefore h q Finally: 1 q
t
n
- - - - - -+
n + 1
l
q-l
I
1 _ h
q
q h=O
The same is true if a is irrational and S is rational. Lastly, there remains the case where a,S are both irrational, but where a,S,l are not independent
over~.
Arguing as in Exer-
cise 5·8(1) we show that this time t ICn + 1) Specifically, t I(n + 1) n
tiona1. SOLUTION 5·25:
Let a
=p
+
-+
!
q~,
n
f(re(a) ) If we show that:
I
n=O
q l 0, and p,q be integers.
Izl = 1,
is not bounded whenever r {n~}rne(na).
!.
if and only if a,S are both irra-
the set of points e(a) is dense on the circle to show that f(re(a»
-+
-+
1.
Since
it suffices Now,
288
CHAPTER 5:
n
1
-- t
n +
1
k~O
{ne}e{nOl.} -+ A
+ 0,
we will be done, for by virtue of a classical lemma, as
~ -+
1 we
would have: (1 -
~)f(re(a)) -+
A.
Now, with {ne} being equidistributed on [0,1]: 1
__
n t
n + 1 k~O
1 {ne}e(na) - _ _ - n + 1
-+
SOLUTION 5·26:
5·13.
n t
k~O
{ne}e(nq~)
J1oXe(X)dX = ~ nq
Let us take Q
=
+ O.
°in the upper bound in Exercise
We then have:
D ,,~ n 1t
it then suffices to show that:
In the case u
n
Ia q (n) I
= ne: if
II x II = min Ix ke:/!Z
- k I.
Therefore it is sufficient to show that the series is convergent.
t L
1
q>l q2l1q~11
DISTRIBUTION MODULO 1
289
II q.e II
.e,
Now, by Khinchin' s Theorem, for almost all
>
q -4/3 for
q sufficiently large, and by a theorum of Roth the same upper
bound is valid for all algebraic
.e.
Let us then consider the sequence qo such that II q.e II < q -3/4 As the series it suffices to shown that the series
I
< q
I
< •••
1
.
of integers
5/4 1S convergent, q 1 -::---- is convergent.
q~l
q~lIqn.e1l
n~l
therefore starting from a certain index
1
2q~3/4
>
(qn+1 -
qn)-4/~
so, furthermore: c Since 9h6 ists y
II qn.e II
>
>
>
V2 ,
>
o.
it follows from this by induction that there ex-
0 such that whenever n is large enough qn
q~4/3, therefore q~ II qn.e II
>
q~/3
>
yn
2
But
> //3n 4/3, and the
series is certainly convergent. SOLUTION 5·27:
Let us assume a to be irrational.
The given hy-
pothesis then implies that the sequence (na,nB) is equidistributed modulo 72: 2 in]R2, from which it follows that: 1
n
lim - - 1 I rp(ka)e(kB) n-+oo n + k=o In particular, let us take rp to be the characteristic function of
I and let us replace B by qB (q
and clearly:
= 0).
We have:
290
CHAPTER 5:
lim _1_ + 1
n-+<» n
L
1
.t(I)
nk~n
From the second formula it follows that the sequence (n k ) has a density equal to .t(I), that is to say (taking n = n k ) K + \ nk + .t(I) ~ O. Putting this back into the first formula, for q ~ 0 gives:
which implies that the sequence (nkS) is equidistributed modulo 1. These results clearly hold when a is irrational. SOLUTION 5·28:
Letusassume, to simplify matters, thata> 1. We
then immediately verify that since we are given two integers n, k,
~
e
(k - ~ ,k)
<=>
n
=
[ka].
Therefore the sequence (n k ) of [ka] 's is the sequence of integers n such that: 1
a
,1) .
We can apply the preceding Exercise 5·27 if 1, ~-linearly
S are
independent, that is to say, if a,l,aS are.
SOLUTION 5·29:
example. SOLUTION 5·30:
that:
1 a'
The sequence u
n
log(n + 1) furnishes a counter-
By Van der Corput's Criterion it suffices to show
DISTRIBUTION MODULO 1
°
For all k >
291
( n+k _ (_.e_ g - 1 g
gn)) nE!N
is equidistributed modulo 1.
Now, this sequence is of the form (p.eg n ) with p = 1 + g + ••• + k-1 . naN g e~", and Wey1' s Criterion shows that if .e is normal in base g, then so is
p.e.
SOLUTION 5·31:
By induction on the degree of P and by applying
Van der Corput's Criterion, it is easily seen that (i) => (ii). As opposed to this, by taking, for example, x
=1
and P as a
polynomial with an irrational coefficient (not the constant term) it is seen that (ii) can be true without (i) being true.
SOLUTION 5·32: Let E be the set of functions f of~ into ~ s,c that are periodic with period one, which for every pair (s,c), s+l se:N, c = (co, ... ,cs)eG have as restriction to [0,1) the
characteristic function of:
The functions in E are Riemann integrable, and it is easy to see that for every periodic function f with period one that is Riemann integrable, and for every of the vector space
g :::
over~
f ::: h,
On the other hand:
flofCs,c)
1 =
gs+l '
E >
0, there exist elements g,h
generated by E such that:
292
CHAPTER 5:
and if Ae [0,1) has as its expansion:
The quantity
n
L
f
k=o
S,c
(Agk) is equal to the number of integers k
such that:
o
~
k
~
n
and
If property (i) holds,
1
~ n-+:r .~-
n
L
k fs C(Ag )
k=o'
=
f1
fs c 0'
1
= g s+l'
and consequently we certainly have (ii).
Conversely, if (ii)
holds for every function f of E,
lim
n--
1
n-+:r nL k=o
f(Ag)
=
f1 0
f·
This is still true for every f of. the generated vector space, therefore by a classical argument (see Section 5.2 of the Introduction) it is true for every Riemann integrable function f, which show that (i) holds.
= [an ju n ,a n ju]. The hypothesis implies n n J n+ 1 C J. The decreasing closed sets (J n )n~. ~'T have a non-empty n intersection. Every Ae J works. SOLUTION 5·33:
Let J
n
ne:N
SOLUTION 5·34:
Po
+
I
z;--::-T
e z: and an
n
Let Po be an integer, let us set a o = Po' a o We will define [a ,a ] by induction so that a
= an
I
+ z;--::-T .
n
n
n
Let us assume that [an ,an ] has been
DISTRIBUTION MODULO 1
293
-u:-
un+1 u n+1 ] The interval [an un ,8 n has length:
defined.
Un +1
(8 n - a ) - - > r( 8n - a ) n u'" n n
n +1 ,8 ---. . Un+1~ , therefore there eXIsts an Integer p n+ 1 e [aU ---n un u n
and:
n
--_1-1-
If we set a n+ 1 = Pn+ l' 8n+ 1 = Pn+ 1 + r the relations from the preceding Exercise 5·33 certainly hold. But then there exists A such that AU e (p ,p n
n
n
+
~l)
r-
for any n eli!.
(1)
(In addition we see
that there exists such a A in every interval [p/uo'p + ~ /u o ] for any integer p. Since p e7l and --_1--1 < 1, we certainly have {Au }
~
n If r
_1-1 . r>
n
r
3 let us give a sequence
and let us define the intervals I lowing way:
n
E
-
=
(E)
Assume: I n (E)
is defined. q
n
=
[p n (E),p n (E) + ~1] r Then let q be the integer
un+l un+1 ) e [p n (E) u ,Pn(E) -u- + 1 n n
(E ) elN' where E e {O,l} n n n by induction in the fol-
294
CHAPTER 5:
Take
and then we certainly have I
n+
l(E) C I (E). for: n
u n+1 () + u n+1 _2_] [UUn+1 Pn (E) '-u--Pn E u 1'-1 n n n 2
has length equal to
2
.z;-::-y = 21' + .z;-::-y
.[p
= [pn (E)/U n n (E) + 1'~l)/U n ] n it is easily seen that the condition l' > 3 implies that if E + n
On the other hand. if we set J (E)
there exists n such that J n (E)nJn (n) = ¢. The points A associated in this way with two distinct sequences E and n will be distinct. The set in question therefore has the cardinality of the continuum. SOLUTION 5·35:
for all n pa
= l'
~
> 3.
By hypothesis there exists
N Un+ l/u n
~ p.
Then let b
~
p >
1 and N such that
N and let a be such that
By the preceding Exercise 5·34 the set of A's such that {Av n }e In addition there
[0,2/(1' - I)] has the cardinality of the continuum.
exists such a A on every interval of the form [P/Vo.(p + l' : l]/V O]· The sequence (an + b) has a density l/a in the set~, from which it follows that the sequence (Au n ) cannot be equidistributed mod-
ulo 1 i f ~l l' -
<
.!.
a
(because for this sequence:
The condition ~l < .!. holds as soon as a is chosen large enough l' a for pa > 2a + 1, which is always possible. Therefore the set of A's such that (Au) is not equidistributed n modulo 1 has the cardinality of the continuum. In addition there
DISTRIBUTION MODULO 1
295
~ l)/U p]' where Since b, and therefore u b ' can be chosen arbitrarily large,
exists such a A in every interval [p/u b , (p + r P eZl.
the set of such A's is dense SOLUTION 5-36:
As n
-+
in~.
We have:
00,
Hence:
_ U~+1J-+ U
o.
n
'
in particular, whenever n is large enough:
J
u
n+2
-
U:+ 1
J
n
With every pair
1 < 2
(A,~)
let us associate one of the triples
(n,u,u 1) such that: n n+ For all k
~ n
The set of triples (n,a,b) elN 2 being countable, it suffices to prove that the correspondence so defined is injective. (n,a,b)
is given, if there exists (A,a) and
(~,S)
Now, as
such that:
296
CHAPTER 5:
if:
For all k
~
Vk2 +1 I IVk +2 - -v;-
n
1
<
2 '
<
!,
and if: U
V
n
b.
a,
n
As there exists an unique integer c satisfying we see that we must have
U
by induction we show that:
n+
2
= vn+2 = c.
Ie - ba2
1
Proceeding by induc-
For all k ;: n But then:
and:
The correspondence is certainly injective and the set of pairs (A,~)
is countable.
REMARK:
Starting from a pair (a,b) , where b
the sequence U
(un)
defined by
U
o = a,
u1
n+l n
!
(Un elN) ,
a, we show that
= b, and for n ;: o by
2
1 -2 ::; Un +2 - - - < U
>
is such that
U n+l - - - + Ct. U
n
DISTRIBUTION MODULO 1
297
The set of such a's is countable.
Nevertheless, it is not known
how to characterize it other than by its definition. 2 If-&=l +15 2 the other root of the equation x - x - 1
IS
1 -
0
-&n + -&,n we have U o 2, n u 1 = 1, and (un) satisfies the recurrence relation (of the Fibonacci sequence) u n+2 = u n+1 + un· In particular u is integral. for -& is -&'
2
If we set u
n
= un -
As -&n
-&,n and as 1-&'1
ing way: set Ao a
u
n
0, and, with n
An having been defined, set: (n + 1)(A
'
n
+ a ). n
He then have: For all n
la n 1 :;:
eN
L
If we set a
n=O
a
n7
1,
A
n
+ a
n
[A ] + u .
whence: -
I
k=o
~: I ~
ak
o.
It is easily seen that: n
L k=o
n!
a k IT = A
whence the result.
n
n
a classical majorisation shows:
1 n(n!) ,
In!a
=
Let us specify a sequence (A n ,an ) in the follow-
- {A }
n
1, it follows from this that -&n
o.
un + En' with En ~ SOLUTION S· 37:
<
+ a
n
In particular we see that:
n
298
CHAPTER 5:
o
~
n!e - [n!e]
1
< -
n
•
SOLUTION 5·38: By Exercise 5-6, for example, if u By Exercise 5-8(2), if u On the other hand, if u
n
n
n, B((u
= n,
n
=
B( (u
0,
B((u
n n n
»
= lR
- {a}.
»
= lR
- Zi':.
»
=
¢.
Finally, if (u ) is a sequence of elements of [0,1] equidisn
tributed on [0,1], let A elR''', q eZi':''', and we have:
lim ___1___
I
n-+= n + 1 k=o
eq(AU k ) =
f10e (A QX)dx
e(AQ) - 1 Aq
This quantity vanishes for all q eZi':'" if and only if A eZi':"', therefore B( (u » n
= Zi':"'.
If (n k ) is the sequence of integers defined in Exercise 5-27 we already know that if A e lR - E, (An k ) is equi-
SOLUTION 5-39:
Then let A e E, there exist integers a ,b ,q
distributed modulo 1. 9 1 such that Aq
= aa
+ b.
Keeping the same notation we then
have:
nk + 1
1
k+1n k +1
n=l
e(aan)q>(na)
(for a.lJ2 => (na) is equidistributed modulo 1).
I = [O,F,;], where e(aF,;) - 1
If
and
21tia
if a
+0,
F,;
.lJ2, the integral has the value
F,;
if a
=
in any case it is a number different from zero,
°
DISTRIBUTION MODULO 1
299
and the sequence (An k ) cannot be equidistributed. certainly have B«n k » = ~ - E. SOLUTION 5'40:
Therefore we
Using the result of Exercise 5'5(1) this is ob-
One may also be given a sequence (v ) _. of elements of
vious.
n nco..
[0,1] equidistributed on [0,1]; one then takes:
SOLUTION 5·41: (i):
q-1
L
Since x(n)
1'=0
xr(n), where
X (n) = card{k:[aqk + a + 1'a] = n}, r
We can always assume that a > 1 (taking q large enough). let
~
be the periodic function
of~ into~
Then
with period one, the
restriction of which to the interval (a/(a - l),a/a] is the characteristic function of the interval (a/(a - l),a/a], almost periodic and x(n) = (ii):
~(n/a),
~
is clearly
therefore X is as well.
Since X being almost periodic implies that 1 - X is
almost periodic, we can limit ourselves to proving it for the characteristic function X of EC Let 1 ~ a O < a 1 < < ak <
...
~k(X)
1 _
~
a -1 kL
a k 1'=0
e(1'x) a
...
be the set A, and let us set: k
xk(X)
IT
h=O
~h
(x);
by construction Xk is a trigonometric polynomial. Clearly we have Xk ~ X, and the set J of neE such that Ixk(n) - x(n)1 = 1 is therefore eontained in the union U J h where J h is the set of multiplies of a h .
h>k J h has density l/a h , and therefore:
300
CHAPTER 5:
In particular, as the series
I
~ is convergent, we can take
a
a~l
the set of square-free integers for E. If ~(n)
SOLUTION 5·42:
p
= l(gk
~(n + k)~
= e(pAgn)
with
- 1), its mean is therefore zero by an earlier remark
(Exercise 5·30) if k that i f h
= e(lAgn ),
+ 0: 1
lim - n-+oo n + 1
n
I
If a(n)
=
0,
is pseudo-random.
~
= sen)
From this it follows (Exercise 5·30)
~(n + k)cp(n)
k=o
which shows that
+ 0.
the associated characteristic function is peri-
odic and therefore almost periodic, and by virtue of the result
(Aga(n))n~ is equidistrib-
given in the Introduction the sequence
uted modulo 1, therefore A is gS-normal.
Conversely, it is easy
to see that every gS-normal number is g-normal. Using the preceding result and Exercise 5·30, it follows that if A is g-normal so is every number of the form where s,t,p are integers satisfying s
~
1, t
~
A
t
g (g 0,
P
sP
-
+ 0.
1)
, As
every rational number is of this form, the result is proved.
SOLUTION 5·43:
First of all let us assume that the ratio logg/logh
is rational, say, for example: ~b
rogn = a
One then has ga
with a
= hb •
~
1, b ~ 1, (a,b)
1.
If P is a prime, let a(p),S(p) be the ex-
ponents defined by: a(p) = sup{a:p
a
divides g},
S(p)
sup{S:pS divides h}.
DISTRIBUTION MODULO 1
Then we have aa(p)
301
= b8(p),
and consequently a divides 8(p), b
= B~)
divides a(p), and if in addition y(p) If ~ denotes the number
II pY(p)
,
then y(p)
we then have g
p
= ~b,
=~ = ~a,
h
and by the result of Exercise 5'42 above: B(g)
= B(~) = B( h) •
Conversely, it can be shown that if the ratio logg/logh is irrational there exist g-normal numbers which are not h-normal. The proofs known to date are quite delicate. In the same way as above, if cp (n)
SOLUTION 5·44:
for p
~
°we have: 1
n
lim - - L q>(k + p)~ n n + 1 k=o
therefore
q>
is pseudo-random for
e (fun)'
.e. eN*,
= 0, .e. eN*,
and in particular the se-
quence (u ) ~T is equidistributed modulo 1. n nCL. Let:
SOLUTION 5·45: A
Then for any B
>
A there exists C such that:
For all n elN
and
n
L
k=o
n
f(u k ) ~ B(n + 1) + C
L f
ho
B(n + 1) + C.
If we set Mo = 0, M1 = No' M2 = No + then if Mk < n ~ Mk +1 we have:
N1 ,
... , Mk
No +'"
+ Nk - 1 ,
302
CHAPTER 5:
whence:
2
h
f(W h )
<
Bn + (k + l)C
But the hypothesis Nk
~
00
implies that
k~ 1 ~ 0, whence:
Since this can be done for all B > A we certainly have:
SOLUTION 5·46:
2
Taking lim!
n h
f(w h ) and -lim! 2 (-f)(w h ) we n h
see that i f (wh ) is the sequence constructed by starting from (un) and (v ), as in the preceding Exercise 5 ·45, we have for every real n
function (and therefore for every function having values in and
lim ~
~
Then:
lim!
2
~ n h
2
h
~)
f:
f(v h ) exist and are the same,
f(w h ) also exists and is equal to their common values.
In particular, if (u ) and (v ) are equidistributed modulo 1, it n
n
follows from Weyl's Theorem that (w ) is also equidistributed modulo 1. On
the other hand, if:
n
DISTRIBUTION MODULO 1
for all
E >
303
0 there exists C such that:
Ih
For all n e:N
h ) - Ani < En + C,
Ih
~nl
<
En + C.
We then have:
< £M 2k
~(N 2
+ N + ••• + N )1 4 2k
+ 2KC,
and:
<
£M2k +1 +
(2k + l)C.
If the sequence (N k ) is chosen so that:
Nl +
N2
+ ••• + Nk - 1
Nk (for example Nk = k!), it follows from this that:
in particular, if A + ~,~ In this Exercise, if ~
L
h
L
a
f(w h ) does not have a limit as f(w h ) has a limit for every equi-
304
CHAPTER 5:
distributed sequence, it follows that this limit is independent Let l be this limit.
of the sequence (W h ) chosen.
Let us now
construct a sequence (u ) as in Exercise 5·5(1) by taking u
I(N,n - N) such that: f(u n ) >
I f for all
n
sup f(x) xeI(N,n-N)
n
in
£.
1
N
N we denote by SN the quantity N1 I sup f(x), + k=O I(N,k)
by considering the limit of N(N 2+ that we have the inequality:
. 1xSo + ••• + NSN_1 11m ~N(N + 1) -
1)
I
-.-
E
f(u) we deduce
n<~N(N+1)
n
1xS
~ l ~ 11m
0
+ ••• + NS
~N(N + 1)
N-1
and so:
1XSO + ••• + NSN_1
~N(N +
1)
but it is known that SN ogous argument we deduce
+
~
l;
ff. ff ~
Hence we have l; consequently
ff ~ l; f
by an anal-
is Riemann inte-
grable, and its integral is equal to l. SOLUTION 5·47:
As Haar measure is (up to a proportionality fac-
tor) the only measure invariant under translation, it suffices to show if
~
is a measure invariant under
~,
and B a real number,
then:
Now, as a is irrational it follows that for all
E >
0 there
DISTRIBUTION MODULO 1
305
= minlx
exists n such that Iina - all < e: (whence IIxll
f is uniformly continuous, therefore for all
ke71
- kl).
e: > 0 there exists
n > 0 such that Ix - yl < n => If(x) - f(y)1 < e:. From this it follows that for all n > 0 there exists n such that: sup If(x - a) - f( ex) I
<
n.
xE!R
Since:
we certainly have the inequality sought. SOLUTION 5-48: ~
defined by
We must show that the transformation
~(x)
e
of~
into
= gx preserves the Haar measure on 'Il'.
If [a,a] C [0,1), the inverse image under of the g intervals [a/g + h/g,a/g + h/g] , h
~
of [a,a] is made
= O, ... ,g
- 1, and the measure of the union is the same as that of [a,a]. This is also true for every finite union of intervals, therefore also, by passing to the limit, for every Borel set. On the other hand, we must show that the transformation e is ergodic.
Let then A be a set invariant
under~.
For any e:
>
0
there exists a set B made up of a finite union of intervals of the type:
such that m(A~B) < e: (denoting the Haar measure by m and the symmetric difference by ~). We then have for all integral N ~ 1:
306
CHAPTER 5:
therefore:
If N is chosen larger than the largest of the n's entering into the definition of the intervals that form B, we have: m(Bri.&
-N
B)
= m(B)m('& -NB) = m(B) 2 •
Since:
we have:
and similarly: (MiB)ll(AriA) =
An (AlIE)
CAllE,
hence: Im(AnB) - m(A)1 <
£.
So finally: Im(A) - m(Bn.& -NB)
On the other hand,
I
<
2£.
Im(A) - m(B)1 < £,
therefore
Im(A)2 - m(B)21 <
2£, or again: Im(A) - meA)
2
I
< 4£.
As this holds for any or 1.
£
>
0 we have meA)
2 meA) ,
so
meA)
o
DISTRIBUTION MODULO 1
SOLUTION 5·49:
307
It suffices to verify that
ure on finite unions of the "cylinders" A this is clearly true.
1
preserves the meas-
~
x··· x An
X'l['
x'II' x •••
'
The ergodicity is proved exactly as in the
preceding Exercise 5·48, by approximating every invariant set by a finite union of cylinders. continuous f of
=1
'II'
into
~
From this it follows that for any
there exists a set A C S such that
~(A)
and for any (u) e A: nneN
Let us take a countable set of functions, say (f ), dense with respect n
to uniform convergence (or even such that the vector space they generate is dense) in the set of continuous functions. To each function f we associate a set A , and if A = (lA n
we have
n
(u n ) n.,.., ~~ e A, then for any continuous f and any
an integer p such that
I(m,f} - (m,f p ) I
< £
~ (A) £
>
= 1.
0 there exists
and:
For all n eJN
From this it follows that for all
I
liml---l--+ 1 k=o
n-+<» n
f(u k )
-
£
(m,f) I <
>
n
But i f
0:
2£,
therefore that:
and consequently every element of A is m-distributed.
CHAPTER 6
Transcendental Numbers
INTRODUCTION NOTATIONS ~
denotes the field of rational numbers,
~
the complex num-
bers, and il} the algebraic numbers (il} is the algebraic closure of
min
We say that a complex number is a TRANSCENDENTAL NUMBER
~).
if it does not belong to 00. If P € ~ [X] is a non-zero polynomial in one variable with com-
plex coefficients, we denote by d(p) the DEGREE of P, we denote by H(P) the HEIGHT of P (the maximum of the absolute values of the coefficients of P); lastly, s(P) the SIZE of P.
= d(P) + logH(P) is called
The notation [x] denotes the INTEGRAL PART of a real number x. The first result about transcendence (Liouville, 1844) was based on an approximation theorem: an algebraic number a cannot be approximated too well by rational numbers.
One can, using
this, construct transcendental numbers (Exercises 6'1 and 6'2), but these numbers are "artificial"; it is much more important to show the transcendence of a naturally occurring number, like e (the base of Napierian logarithms) or the
number~.
Hermite
(1873) shows that e is transcendental and Lindemann (1882) showed that
~
is. 308
TRANSCENDENTAL NUMBERS
309
THEOREM (1): (Hermite-Lindemann): Let ex. e
number that is algebraic over
m.
be a non-zero complex Then the number exp(ex.) is trans0:
cendental. Another, classical, statement, that had already been mentioned by Euler in 1748, and again by Hilbert in 1900 (Hilbert's Seventh Problem) was proved in 1934: THEOREM (2): (Gel' fond-Schneider): Let R.
+ 0 and b. mbe
two com-
plex numbers. At least one of the three numbers a
= e R...
b..
a
b
= e bR...
is transcendental over
m.
This Theorem was generalized by A. Baker in 1966, and his work has had some important consequences in diverse areas of number theory (cf., [Bak]). THEOREM (3): (Baker): Let R.l....... R. n be m-linearly independent comR.l
R.
plex numbers. If the numbers e ....... e n are all algebraic .. then l .. R.l....... R. n are linearly independent over the algebraic closure
iii of m in
0:.
The proofs of these three Theorems may be found in the books [Wall and [Bak].
The first two are also proved in [Sch] and
[Lang 2]. Many of the problems in transcendence theory are as yet unsolved. A very general conjecture about the transcendence of the values of the exponential function has been given by S. Schanel ([Lang 2]). CONJECTURE (S): Let xl....... xn be m-linearly independent complex
numbers.
The transcendence degree over
m(x l , ... ,xn ,e
is at least n.
Xl
, •.• ,e
xn
)
mof the
field
310
CHAPTER 6:
Let K be an extension of the field k.
We say that the n ele-
ments al, ... ,a n of K are ALGEBRAICALLY INDEPENDENT oVer kif, for every polynomial P e k[X l , ... ,Xn ], P(a l , ... ,an) = 0 => P = O. The TRANSCENDENCE DEGREE of Kover k is the maximum number of algebraically independent elements over k that can be taken from K. Modern proofs of transcendence have in common a preliminary construction resting upon the "Pigeonhole Principle": If objects
are put into
pigeonholes~
than the number of
and if the number of objects is greater
pigeonholes~
tains at least two objects.
then one of the pigeonholes con-
In other words a mapping of a set of
n elements into a set of m elements is not injective if m < n. This result will be used in the following form: PIGEONHOLE PRINCIPLE: Let Fl~ ••• ~Fm be disjoint sets~ E a set
having at least m + F =
U
l",j",m
under
~
F.. ]
1
elements and
~:E ~
There exist two elements x
are in the same subset F. of F. ]
BIBLIOGRAPHY (Chronologically Ordered) [Sch], [Lang2], [Wal], [Bak].
F a mapping of E into ~
x' of E whose images
TRANSCENDENTAL NUMBERS
311
PROBLEMS
EXERCISE 6-1:
Let a be an
algeb~aia numbe~_
The DEGREE d(a)
of a, the HEIGHT (a) of a, and the SIZE s(a) of a are the degree, height,
and size of its minimal polynomial
over~.
Recall that for every non-zero algebraic number a there exists a smallest positive integer q
~
integer; we call q the DENOMINATOR of a. the NORM of an
algeb~aia numbe~
q.a is an algebraic Lastly, N(a) denotes
1 such that
a.
(1):
Let a be a non-zero algebraic number of degree
(a):
By using that:
B.
if S is a non-zero algebraic integer show that there exists a constant a
a (a) > 0 such that for every polynomial P
n and height H P(a)
(b):
o
~
e~
[Xl of degree
1 we have:
or
From this deduce that there exists a constant a
~
0
312
CHAPTER 6:
depending only upon a such that for every non-zero polynomial P e2Z [X] we have: pea)
=
(2) :
or
0
e
Let .e ea:.
-as(P)
Assume that there exists a sequence (.e) n
of pairwise distinct algebraic numbers such that -k s(.e )
n
I.e - .en I wi th 1im k
n
for all n
~
n~
1
1,
= +00.
n-++oo n
Show that .e is transcendental. Let x be a real number; assume that there ex-
EXERCISE 6-2: (1):
ists a sequence (p /q ) of rational numbers and a sequence (A ) n
n
of real numbers, with 1 im A n~oo
-~
A
n
n
=
+00,
n
such that:
.
Show that for every integer a ~ 2 the number aX is transcendental.
(Use part (1) (a) of the preceeding Exercise 6-1).
(2):
From this deduce the existence of a constant C
>
0
such that for every rational number p/q we have:
(3):
Let
APPLICATION:
~
e Ill.
>
0 such that:
Let a1, ... ,an , ... be a sequence of integers such
that the sequence stationary).
Show that there exists C( 0
lan I
be increasing (in the large , but not
TRANSCENDENTAL NUMBERS
313
Show that the number: n
L II
1
n=l i=l a i is irrational. EXAMPLE: e
l/q
For q an integer, the numbers: ,
1 q
cos
.
sm
1
q'
1
cosh - , q
sinh ~ q
are irrational. EXERCISE 6·3:
APPLICATIONS OF DIRICHLET'S PIGEONHOLE PRINCIPLE
(1): A LEMMA OF SIEGEL: (a):
Let u . . (1 1,J
~
i
~
v,l
~
j
~ ~)
be real numbers, and let V1 , ••. ,V II be positive rational integers such that: v
.. 1 ~ L lu1,J
i=l
.e. ••• .e. 1
JJ
~
v
IT i=l
(1 ~ j
V.
]
~ ll).
(l+X.). 1
Show that there exist elements such that:
lei 1
~ X.1
I~
u. .
(b)
Let a . . (1
/;1"'" /;v e~,
not all zero,
and:
i=l
el
1,] 1
1,]
~
i
~
n,l
~
j ~ m) be complex algebraic
314
CHAPTER 6:
integers.
For 1 , j
~
m let us denote by K. the subfield of
obtained by adjoining to ID the n numbers:
J
~
a 1 ,j'··· ,an,j'
and let:
o.
[K.:ID]
J
J
be the degree of the number field Kj •
Let A1 , ... ,Am be positive
integers satisfying: (1 , j
, m),
where: (j)
(j)
°1
,···,°0. J
are the different embeddings of K. in J
Assume
n
~
(1 , j 'm).
> ]l
Show that there exist rational integers x 1 , ... ,xn , not all zero, satisfying: n
L
i=l
a . .x. ~,
J
~
o
(1 , j
, m),
and:
(2): CONSEQUENCES: (a):
Let uo, ... ,um be real numbers and
TRANSCENDENTAL NUMBERS
315
H a positive integer. Show that there exist rational numbers so, ... ,sm' not all zero, such that: max 1s.1 ~ H
O",j~m
]
and:
(b) :
Let u o ,··· ,urn be complex numbers and H a positive in-
teger. Show that there exist rational integers so' ... , sm' not all zero, such that: max 1s·1
O",j",m
]
~
H
and: luos o + (c):
... + umsml
<
I2clu o l +
...
2 + 1u m1 )H-lCm-l)
Let Nl, ... ,Nq,H be positive integers, and xl, ... ,x q
complex numbers. Show that there exists a non-zero polynomial P eZ': [Xl' • .. ,Xq] of degree at most Nh with respect to Xh (1 less than or equal to H, such that:
where:
and:
~
h
~
q) and of height
CHAPTER 6:
316
I
(d):
Show that a complex number x is transcendental if and
only if, for every real number n
>
°such that o
<
la
o
~
>
0, there exists an integer
the inequality:
+ a x + ••• + a xnl 1
n
< (
max la.1 )-~
O~i~n
l
has an infinite number of solutions:
EXERCISE 6·4: (1):
CONSEQUENCES OF THE GEL'FOND-SCHNEIDER THEOREM
Show that there does not exist a real number u + +R defined by f (x) = e Ux sends
such that the mapping f U ~
°
U
every real algebraic number into an algebraic number. REMARK: One can prove that the additive group of the field
m(]R
of real algebraic numbers is isomorphic to the multiplicative group
9i (]R~.
The above is used to prove that there is no such
isomorphism which is locally increasing, cf., J. Dieudonne: Alg~bpe lin~aipe
et
g~om~tpie ~l~mentaipe,
Enseignement des Sci-
ences, (Hermann, Paris), (1964), p. 164. (2) :
Let P cZl [X ,Y] be an irreducible polynomial such that:
' -lT 0 , PX
Let
U
P~ ,. 0,
P(O,O) ,.
0,
PO,l) ,. O.
be an irrational algebraic number.
Show that the equation in z: P(z,ZU)
=
0
TRANSCENDENTAL NUMBERS
317
does not have roots in OJ (that is to say, i f JI. Jl.o. P(e,e )
.e e a:
satisfies:
= 0,
then eJl. is transcendental). (3): in a:, and
Let Me M (a:) be an n n
0. 1 ,0. 2
x n
square matrix with coefficients
two algebraic numbers such that the matrices:
and belong to the general linear group GL (iii) of invertible n x n matrices with algebraic coefficients.
n
Show that if M is not nilpotent, then
0. 1 ,0. 2
are
~-linear1y
dependent.
EXERCISE 6-5: CONSEQUENCES OF THE HERMITE-LINDEMANN AND GEL' FONDSCHNEIDER THEOREMS Let Me M (iii) be an n x n matrix with algebraic coefficients. n
Assume that M is not nilpotent. (1):
Show that for all a. e!ii, a.
+ 0,
the matrix:
expMo. does not belong to GL (m). n
(2):
Assume that there exists u e a: such that:
expMu e GL
n
Show that the
(iii).
~-vector
subspace of a: generated by the eigen-
values of M has dimension equal to one, and that M is diagona1izable.
318
CHAPTER 6:
EXERCISE 6·6: CONSEQUENCES OF BAKER'S THEOREM (1):
Let 8 . . (1 ~,J
~
i , h,l
j
~
~
k) be algebraic numbers,
and let y l' ... 'Y k be non- zero algebraic numbers. For 1 $!, j.!;. k let log\ be a non-zero determination of the logarithm of Yk . Assume: k
L
o
8 . . logy. ~
j=l
,J
]
(1 ~ i
~
h).
Show that there exist rational integers c 1 , ... ,c k , not all zero, such that: k
L
8 . . c.
j=l
~,J
(2):
J
o
(1 ~ i ~ h).
Show that Baker's Theorem (3) is equivalent to the
following:
Let a 1 ••••• a n be non-zero aLgebraic numbers; choose a determination of the logarithm at each of the points a 1•..•• a n • and assume that the numbers 10ga1••••• loga n are m-LinearLy independent. If 8 0••.•• 8n - 1 are aLgebraic numbers. then the equaLity
8.
(where a.~ = exp(8.laga.). 1 ~ i ~ n - 1) is possibLe onLy if ~
~
~
81••••• 8n - 1 are aLL rationaL numbers. and 8 0
(3):
= O.
Show that the following four statements are equivalent,
and that they follow from Baker's Theorem (3): (i): If i 1 , ... ,in are m-linearly independent logarithms of algebraic numbers, then i 1 , ... ,in are m-linearly independent; (ii):
Let 10ga1 , ... ,10ga n be non-zero logarithms of algebraic numbers, and 81 , ... ,8 n be algebraic numbers. If:
TRANSCENDENTAL NUMBERS
319
are W-linearly independent, then the number:
is transcendental; (iii):
Let logal, ... ,logan be W-linearly independent logar-
ithms of algebraic numbers, and Sl, ... ,Sn algebraic numbers.
As-
sume that at least one of the numbers Sl, ... ,Sn is irrational. Then the number:
is transcendental. (iv):
Let L
gebraic numbers.
{leo::eR- em } be the set of logarithms of al-
The injection of L into 0: can be extended to a
OJ-linear injective mapping of 91@L into 0:. W
(4):
Show that the following two statements are equivalent,
and are consequences of Baker's Theorem (3): (v):
m.L
Every non-zero element of the set: R,.
= {Sl.e.l + ••• + S.e. :13. eiji,e leiji,n ~ O} n n l
is transcendental; (vi):
Let logal, ... ,logan be logarithms of algebraic num-
bers, and J3 o , •.. ,J3 n algebraic numbers with So f O. ber: exp(J3 0 + is transcendental.
n
2
i=l
J3.loga. ) l
l
Then the num-
320
CHAPTER 6:
(5): (v).
Show that Baker's Theorem (3) is equivalent to (i) and
(Other pairs suffice). (6):
Let MeM (a:) be a matrix that is not nilpotent. m
Let
tl, ... ,tn be- m-linearly independent complex numbers such that: exp(Mt.) e GL (m) J m
(1 ~ j ~.n).
Show that the numbers tl, ... ,t n are m-linearly independent. Show that if MeMm(!ID, then the numbers l,t l , ... ,tn are ii}-linearly independent. Let ~:a:n
(7):(a):
+
GL (a:) be a non-rational analytic homom
morphism; assume ~~n) C GLm(m), and let semn, S = (Sl, ... ,Sn)+O, be such that
~(S)eGLm(ii}).
Show that 1,Sl, ..• ,Sn are m-linearly dependent. Let ~:a:n for all neiiin , n (b) :
+
GLm(a:) be an analytic homomorphism such that
+ 0,
the homomorphism
z ~
q>(nz) of a: into GL (a:) m
is non-rational. If al, ... ,a£ are m-linearly independent elements of mn such that ~(aj) e GLm(iii), then a l , ... ,a£ are a:-linearly independent. Show that the number:
flo 1 dx+ x 3 is transcendental. EXERCISE 6-7: LOWER BOUNDS OF LINEAR FORMS IN LOGARITHMS Let al, ... ,an be positive rational integers and bl, ... ,bn rational integers. Assume that the number:
321
TRANSCENDENTAL NUMBERS
is non-zero.
Denote:
Then:
IAI
> A
-nB
•
REMARK: This inequality is very precise as a function of A, but not as a function of B. Using his transcendence method, Baker was able to prove a more precise inequality in terms of B:
IA I
>
exp{ -
(8n)
400n OogA) n+l 10gB}.
This kind of inequality is what makes this method applicable to many problems in number theory (see, A. Baker: 'The Theory of Linear Forms in Logarithms', Transcendence Theory, Advances and
AppZications, (Academic Press). (1977)). EXERCISE 6-8: CONSEQUENCES OF SCHANUEL'S CONJECTURE (S) Show that the following conjectures are consequences of Schanue1's conjecture. (1):
Let 10gal •...• 10gan be m-1inear1y independent logarithms of algebraic numbers. Let Bl •...• Bn be m-1inear1y independent algebraic numbers. Then the numbers:
are .a1gebraica11y independent (over m. or over
m.
which amounts
to the same thing). (2) :
Let log a be a non-zero logarithm of an algebraic num-
ber a. and let 1.B l •... ,B n be m-linear1y independent algebraic numbers. Then the numbers:
322
CHAPTER 6:
are algebraically independent. Let ul, ... ,un be m-1inear1y independent complex num-
(3):
bers, and v a transcendental complex number. ence degree over m(e
is
~
Ul
mof
un
, ... ,e
,e
Then the transcend-
the field:
vU l
, ..• ,e
n - 1.
(4):
The sixteen numbers:
nee nne iii 10g3 e,n,e ,logn,e ,n ,n ,log2,2 ,2 ,2 ,e ,n ,log3,(log2) ,
212, are algebraically independent over (5):
m.
If x is a complex number different from 0 and 1, one
of the two numbers:
x,x
e
is transcendental. (6):
If x is an irrational complex number, one of the two
numbers ~,~
2
is transcendental.
EXERCISE 6·9: Assume the following result: There exists an absolute constant Co > 0 such that for every non-zero polynomial P e 1l; [X] of degree", n and size", 8, we have: logIP(n)1 > - C n8(l + logn).
a
TRANSCENDENTAL NUMBERS
(1):
323
Show that i f I; is a complex number, algebraic over m( 1t),
there exists a constant C1
= C1 (1;) such that for every non-zero
polynomial P e?2: [X] of degree less than or equal to n and of size less than or equal to s, we have:
(2):
Let (a) 1 be a sequence of rational numbers, an inm m;: Consider the function f de-
finite number of which are not zero. fined by: a
f(z)
Let
I;
Z m
m
1.
be a non-zero complex number such that the sequence
is convergent. (a):
L
m;:O
a I;m m
Show that there exists an infinite number of m's such
that the number:
is non-zero. (b):
Denote by Dm the lowest common denominator of the num-
ber a 1 , ... ,am, and assume: logla 11 lim m+ m+oo m(m + logDm)(logm) Show that if f(l;) is algebraic on are algebraically independent on
m.
m(l;)
(Hint: Consider the norm of Cb(I;) over (1) above).
the two numbers I; and 1t
m(l;)
and use Question
324
CHAPTER 6:
SOLUTIONS
Let us denote the denominator of a by q.
SOLUTION 6'1: (l):(a): Let P e71 [Xl,
an
~
0,
be a non-zero polynomial of degree n and height H.
If P(a)
0
~
n
the number q P(a) is then a non-zero algebraic integer, therefore: (1)
Let a
a
(1)
,a
(2)
, .•. ,a
(8)
be the conjugates of a.
This
yields:
IT
qnp(a (i»
i=l
qn8P (a)
IT
p(a(i».
(2)
i=2
Now, n
L
k=o
,a
k
"
a (i),k
~
H(n + l)max{l, , a (i),n }
~
(Contd)
TRANSCENDENTAL NUMBERS
325
(Contd) with: 1 +
sup
la
(
.)
1.
I.
1::;i~8
And so: (3)
From relations (1),(2),(3) we deduce: a
n
s-l '
H
where a
= (2~) l-sq -s
(b):
> 0
is a constant depending only upon a.
By the preceding, i f
Pe~[X]
is a non-zero polynomial
of degree n and height H, we have: P(a)
=0
or
n
-[n10g1/c + (s-l)logH]. e ~ -E-1 s- =
IP(a) I
H
The result required is obtained by setting: a
max{log
(2):
braic.
a1 ,8
-
1}
~
o.
Argue by reduatio ad absurdum and assume
~
to be alge-
By Question (1) above there exists a constant a
that for every non-zero polynomial P(~)
=0
Pe~[X]
~
0 such
we have:
or
Let Pn be the minimal polynomial of
~
n
; let us denote its degree
326
CHAPTER 6:
by d n and its height by hn , so that: s(~
n
) =d
+ 10gh
n
n
= s(Pn ).
For large enough n, as the
~n
are distinct, the numbers
are not conjugate, hence we have P
n
-as(~
n
+ 0,
(~)
~
and
~n
and consequently:
)
(4)
If:
d
n
L k=o
P (X) n
k
a kX , n
this yields:
Ipn (~) - Pn (~ n )1
d
~
L la II~k _ ~kl
k=l
nk
lim k
=
n
By hypothesis we have: I~
-
~n
I
-k
n
s(~
n
)
with
For large enough n this yields I~ for 1
~
k
~
dn :
Hence, by relation (5):
n+t co n
-
~
n
I
~ 1, and consequently,
TRANSCENDENTAL NUMBERS
327 (6)
e
~
e
210gd +logh +d log(l+l~
n
-(k
n
n
n
I)-kn s(~ n )
) n ,
-d)s(~
with:
=2
d
+ log(l + I~I).
From relations (4) and (6), for n large enough we conclude: (k
- d)s(~ )
n
n
~ as(~ ),
n
or, again, k
n
~
a + d,
which contradicts lim k
=
n++"" n
Consequently
~
+"". is transcendental.
REMARK: We can show that this condition, which is sufficient for a number
~
to be transcendental, is also necessary, and similarly
that we can choose k
n
= n/4
(n
~
1) and
d(~
n
)
~
n.
Let us assume a~2 is an integer, with aX=~ an algebraic number of degree d and height h. Let P(X) = Xq - a P.
SOLUTION 6·2: (1):
By Question (l)(a) of the preceding Exercise 6·1 there exists B > 0 (independent of p and q) such that for ~ satisfying: q
~em
q
,."
328
CHAPTER 6:
we have: (*)
For P
=
Pn and q
=
qn' by using the inequality:
we obtain:
~ a ('X'+3)~lx whenever
I:: -xl ~
_:nl n
1, and so:
<
-~
B
for sufficiently large n, which contradicts (*). REMARK: This proof extends to the case a e Ill, a > 0, a
(2):
+1
It suffices to remark that i~!~ is irrational, and
that 31og2/1og3 = 2 is not transcendental. REMARK: This result extends to the case where a is a positive
and algebraic number, S loga. . 1 logS 1rrat10na . (3):
+1
a positive rational number, with
Let ~ = ~ , with a,b integers, b
have laq - bpI ~ 1, therefore: 1
Ii
>
O.
If
r..q + ~
we
TRANSCENDENTAL NUMBERS
APPLICATION:
Let
~
n
1
L IT - .
00
~
n=l i=l a i '
denote:
Whenever a N+2
329
for N a positive integer, we
2 we have:
2 ~ 0 when N ~ +00, we deduce from this that ~ is irralaN+11 tional. Using the classical expansions of the functions expz,cosz,sinz, coshz,sinhz as power series finishes off the numbers mentioned. For example, for exp(l/q) we will take a = nq.
Since
n
SOLUTION 6·3: (l):(a):
Let us consider the mapping
:N(X1, ••• ,X)
= {(~l, ... ,~)e:;z\l:o ~ ~i ~
which to
... '~\I)
(~l'
Xi,l:lO i:lO associates (nl, ... ,n~) with:
~ \I}
of the set intolRlJ
\I
L
i=l
u . . ~. ~,J
~
(1 ~ j
:10 lJ).
For 1 ~ j ~ lJ denote by -v. (resp. w.) the sum of the negative J J (resp. positive) elements of the set: {U 1
., ... ,u .}. ,J \I,J
Hence we will have:
v.
J
+
w.
J
~
u. J
(1 ~ j
~ lJ).
330
CHAPTER 6:
We notice that i f (1;1, ••. ,1;) e:N(X1 , ••• ,X) then the image (n 1 ,·.·.nJ = ~(l;l •.•. ,l;v) belongs to the set:
=
with X
max X ••
For 1 ::; j
~
1::; ili: v
~
the interval [-v .X,w.X] is divid-
\.I
]
]
ed into P. equal intervals (of length less that or equal to U//R. j ), ~ which means that E is divided into R.1 o oR.\.I subsets Ek (1 ::; k ::; 0
R. 1 o oR.
v
0
\.I
<
II
i=l
(1 + X.) ~
= card:N(X1 •. .. ,X
v
)
we can use the Pigeonhole Principle to show there exist two distinct elements 1;* and 1;** of:N(X1 , ••• ,Xv ) whose images under ~ belong to the same subset Ek of E. Let us write I; as the difference (componentwise) of 1;* and 1;**; we will have
= (1;1, ... ,1; v ) f
I;
and the image n
1n·1 J
~
t.u. j
(n 1 , ...• n
X
(1
J = ~(I;)
~ j ~
with 11;.1 ~ ~
(0, ... ,0) of
I;
under
~
(j)
01
For 1 , ... ,0
~
j
~
\.I),
m the different embeddings:
(j)
OJ
of K. into a: are numbered in such a way that we have: J
for 1
~
h
~ X'., ]
~
(l~i~v).
satisfies:
hence the result. (b):
X.
TRANSCENDENTAL NUMBERS
331
a(j) = ~ r .+8 .+k r .+k J
where r. and J
O.
J
(complex conjugate of a(j )k) for 1, k ~ 8., r.+ J
J
J
J
are two integers satifying:
8.
= r. +
J
28 ••
J
(j) Define the mappings Th , 1 ' h , OJ' of Kj a(
.) T J (
h
J
h
j)
for 1 , h
for r. + 1
Im(a~j))
for
h
J
r. +
J
h
~
~
+ 1
8.
J
by:
r.,
~
Re(a(j))
into~
J
r. + J
~
8
j,
h ~ o .. J
Define a positive integer X by: X
where
01 + ••• + om.
~
therefore, because A. J
(1
+ X)n >
~
Thus we have:
1 for 1
~
j
m,
~
mo.
IT
(1
j=l
+ I2A .X) J J
°,
The preceding exercise (with v = n, ~ = 01 + ••• + and t. = m J [1 + I2A.X], each t. being repeated o. times, 1 ~ j ~ m) shows J
J
J
that there exist rational integers x 1 , ... ,xn , not all zero, satisfying:
332
CHAPTER 6:
max
l~i~n
Ix. I
~ X,
1
and: (1 ( j ~ m).
From this we deduce:
A.X
~
1
(1 ( j ( m),
J
+ [fiA.X] J
and:
max
~ r.+ 1~ h~u.
J
J
I.~1 ah(a 1,J . . )x.1 ~ ( J X ) 1 1 + [ 2A.X]
And so:
I K/lIl ( nL
a . .x ) i=l 1,J i
N
(1 ~ j ( m).
J
~=
I ( 2s.( 1 + A.X [12A .X] J
J
OJ )
(1 ~ j ( m);
J
in this last inequality the left hand side is a rational integer n (since L a . .x. are integers of K.), and the right hand side i=l
is bounded
1, J
1
, because s . ..;; J
I2A.X
(1 + [k l
])
!
°., by:
J
J
OJ < 1.
Hence: n
L
i=l
with:
a . .x.
1,J 1
0
(1 ~ j
( m),
TRANSCENDENTAL NUMBERS
333
Hence the result. Let us notice that if I is an integer, 0
~
I
~
m, such that
the fields K1 , ... ,K£ are totally real (that is to say C :R for 1 ::; h ::; 0., 1 ~ j ~ I), then the coefficient
; 2-wn
a~j)(Kj)
]
-]J
can be replaced by (0
12
H1
+···+0 )/(n-jJ) m
which is one if I = m). (2): (a):
The result required is an immediate consequence of .JD+1 jJ 1, v = m + 1, Xi = H, and I1 = it ).
Question (1) (a) (with (2):(b):
Question (1) (a) is used again, separating the real
and imaginary parts.
v.]
Reu.
jJ
= 2,
]J
max
I ~.I ::; ]
j
= m + 1,
exist rational integers O::;j::;m
~
~
m let:
w.]
and
]
1et us choose
For 0
Xi
~O""'~m'
= H,
I1
= I2 = H(m+1)/2 ; there
not all zero, such that:
H
and:
m
I ,: I Iv i I m m ~ i=l
+ ••• + V ~
H
H(m+1)/2 '
m
i~l IWil
H(m:i)/2
334
CHAPTER 6:
As:
lu os0
+ ••• + u
s 12
mm
Iv 0"0 <"
+ ••• + +
Iwos0
V
<"
m"m
12
+ ••• +
w
s 12
mm '
and as: m
I Iv·1~
i=o
m
m
I lu.l, i=O
~
I Iw·1 i=o ~
~
~
m
I lu·l,
i=o
~
the result is obtained. (2):(c):
We use the preceding Question with:
therefore: m + 1
=
it
h=l
(1.+ Nh )
and there exist integers
= M, so, ... ,sh' not all zero, such that:
now: m
I lu·1
i=o
~
we majorise Nh + I by e
Nh
for I
~
h
~
q, and the result follows:
TRANSCENDENTAL NUMBERS
(2):(d):
335
Let x be a complex number and w a positive real
number; an integer N
~
2w + 3 (for example N
= [2w
+ 4]) is chos-
en, and we define:
By the preceding Question, for all H
>
0 (therefore for all
H > Ho) there exists a non-zero polynomial P e:?l [X] of degree less
than or equal to N and height less than or equal to H, such that:
if H
> Ho
we will therefore have:
where H(P) denotes the height of the polynomial P. Let us assume that x is transcendental, it is then easy to construct by induction on k a sequence
(Pk)k~O
of polynomials
of :?l [X], of degree less than or equal to N and height H(Pk)' satisfying: and for all k
~
0 (if po •... ,Pk are constructed, choose an integer H > H- w; this is possible since x is transcend-
such that IPk(x)1
ental; then apply the preceding construction). Therefore if x is transcendental, for every real w > 0 there exists an integer N 0<
Ip(x)1
>
0 such that the inequality:
< (H(P))-w
has an infinite number of solutions Pe:?l[X], with the degree of P..::: N.
336
CHAPTER 6
Conversely, let us assume x to be algebraic of degree cover 1Il; let d be a denominator of x.
Let us select w = C, and let n be
an arbitrary positive integer. the (n + I)-tuples (a o'
Let us notice that the set of ... ,an) e zo n+l , with:
max la.1 ~ dcn(n + l)C-lmax(l,lo(x)l)n(c-l), O~i~n l Ofl (where {a} denotes the set of embeddings of K in n+l Let (a o' .•. ,an) eZO be such that:
~)
is finite.
max la.1 > dcn(n + l)C-lmax (1,lo(x)l)n(c-l) , O~i~n l Ofl and: + a xn n
t o.
We are going to show that: la
o
+ a x + ••• + a xnl > ( max la.I)-w. 1
n
O~i~n
l
In fact the number:
is a non-zero algebraic integer of lIl(x) overZO, hence its norm satisfies:
hence:
TRANSCENDENTAL NUMBERS
For
337
+ 1:
0
I Ii=o
a. (0 (x) ) i J.
I
is majorised by (n + 1)maxla.lmax(1,lo(x)1 )n, J.
0+1
hence:
\i~O
aix i \
~
d-on(n + 1)-0+1(maX(1,lo(x)1 »-n(0-1)
x max
O,.i..,n
-(0-1)
Ia. 1
,
J.
and the choice of a. gives the desired result. J.
SOLUTION 6·4:
(1):
for all algebraic x.
Let us assume that a
+0
and eax is algebraic
For an algebraic, but irrational, number x,
one of the three numbers: a
e,
x,
eax
is transcendental.
It can only be ea.
But for x
1, eax
e
a
would have to be algebraic. (2) :
Let.t e a: be such that: (1)
The field ~(e~,e~a) then has transcendence degree over ~ ~ 1 (by
Theorem (2) and the hypothesis P(l,l) + 0). Finally, as P is irreducible (1) shows that if e~ is algebraic, then e~a is algebraR.
ic, hence that in this case the transcendence degree of m(e,e
R.a
)
over II! is zero. (3): nilpotent).
Let A be a non-zero eigenvalue of M (since M is not Then for a11 tea: exp(At) is an eigenvalue of exp(Mt).
338
CHAPTER 6
If expMa e GL (Q), exp(Aa) is then algebraic. n Schneider Theorem. if:
a
b
the Gel' fond-
b
+
are algebraic numbers. a 0, Aa 1 = loga hence a 1 ,a 2 are m-linearly dependent. SOLUTION 6- 5: (1):
By
+ 0,
then b is rational,
Let a em be such that expMa e GL n (Q.i).
Let P e
GLn(~) be such that P- 1MP is a diagonal matrix (if M is diagonalizable) or reduced to the form of diagonal blocks (Equation (2)). Let A1 ••.• ,A n be the eigenvalues of M. with 1.1
+ O.
The diagonal
-1
of exp(P MPz). for z ea:, is:
Therefore expA 1a
e~,
whence a = 0 by Theorem (1).
Let us assume that M is not diagonalizable.
(2):
Then,
with the notations of Question (1), one of the blocks A1 , ... ,Ar of the matrix:
o A
belongs toM (m) with v v A. ].
=
[",
~
1
(2)
r
2; let A.]. be one of these blocks: 0
1
0
iJ i
1
,
and:
expAiz
exp"i' [
1
z
0
1
0
0
z
2
2T z
0
1
TRANSCENDENTAL NUMBERS
339
From this we deduce that u e part (1).
Ql,
which contradicts the preceding
Hence Mis diagonalizable:
= [
p-1MP
A1
0
0
A n
1
,
and:
expP
e A1z
0
0
AZ e n
-1
MPz
Since all the number e
A.U 1
(1 ~ i ~ n) are algebraic, Theorem
(2) shows that the dimension of the Ql-vector subspace of
~
gen-
erated by A1 , ... ,A n is equal to one. Let us denote by xl" ",xr a basis over Ql of the vector space generated by Ql by logY1"" ,logy k . Let bj,s e Ql (1 ~ j $ k,l ~ 8 ~ r) be defined by: SOLUTION 6-6: (1):
r
L
logy.
J
b.
(1 :::: j :::: k).
x J,s s
8=1
By Baker's Theorem x 1 , ... ,xr are linearly independent over 00; now: k
L
j=l
r
L
8 . . logy. 1, J
J
8=1
( kL 8 . .b. ) x j=l 1,J J,S S
o.
Hence: k
L
j=l
8 . . b. 1,J J,S
o
(1
< i : : h,l ::::
8 :::: r).
As the b. are not all zero, there exists J,S
8,
1 ::::
8 ~
r, such
340
CHAPTER 6
that one of the numbers b I , ... ,b k is not zero. ,8 ,8 by a common denominator gives the result. (2):
Multiplying
If Baker's Theorem were false, there would exist
~
linearly independent logarithms of algebraic numbers such that 1,logaI, ... ,loga n were
~-linearly
dependent: (S. e gj), 1
with one of the S. (0.::; i .::; n) being non-zero, for example S
n
1
Start by dividing by -S n , assume Sn
-1, and:
If at least one of the numbers S , ... ,S O
lation: a
1 is irrational, the re-
(3)
n
gives the desired result. So
n-
+0
+o.
If So, ... ,Sn_I are all rational, then
because of the independence of logaI, ... ,logan over
~,
and relation (3) again gives a contradiction to the property stated in part (2). Conversely, let us assume that Baker's Theorem is true and that: a
n
(4)
with logaI, ... ,loga n ~-linearly independent and So, ... ,Sn-I algebraic. By taking the logarithms of two members of (4) we establish that there exists a rational integer k such that: (5)
Therefore 1,2in,loga I , ... ,logan are ID-linearly dependent, there-
TRANSCENDENTAL NUMBERS
341
fore (by Baker's Theorem) 2in,loga1 , ... ,logan are m-linearly dependent; from this we deduce the existence of rational numbers
b1 , ... ,bn such that: + b loga . n
n
By using Equation (5) we obtain:
o
(8.]. + kb.]. Hoga.].
with
-1.
Since 1,loga1 , ... ,loga n are m-linearly independent (Baker's Theorem) we find: and
kb.]. e
8].•
m
Notice that the two numbers Ii of 2) and I2
= 2I1
(1 ~ i ~ n).
= log2
(the Napierian logarithm
+ 2in are m-linearly independent logarithms of
algebraic numbers, and yet:
which shows that the relation a out the 8.]. being all zero. (3):
n
81 8n - 1 80 a 1 ···an e can hold with-
As (i) is obviously a special case of Baker's Theorem,
it suffices that we show the equivalence of the four statements (i),(ii),(iii),(iv). ent. Let us show (i)
=>
It is clear that (i) and (iv) are equival(ii) and (i)
=>
(iii).
Let Il, ... ,ln be non-zero logarithms of algebraic numbers, and 81 , ... ,8 n algebraic numbers, such that the number:
342
CHAPTER 6
is algebraic, that is to say, such that:
is a logarithm of an algebraic number.
= -1;
Let So .. . ,l
n
L
we have
S.l. = 0, therefore the numbers lo'
i=o
1
1
are @-linearly d pendent (hence m-1inear1y dependent, by
n
(i)) .
To prove (iii) we notice that if 11, ... ,ln are m-1inearly independent, then lo belongs to the vector space m generated by
11'··· ,.in: n
L
i=l
k.l. ; 1
1
from this we deduce S.1 tiona1 numbers. To prove (ii) we denote by AI' ... ,Aq a basis of the m-vector space generated by lo, ... ,ln. l.
k . . A. 1, J
1
J
(0
Thus: i
~
n),
~
hence: n
L s.l.
i=o
1
1
f [Ii=O S.k . . ]A.
j=l
1 1,J
J
o·,
by (i) AI' ... ,A q are @-linearly independent: n
.L
"2-=0
S.k . . 1
1, J
o for 1
~
j
~
q.
If the k. . were all zero we would have l. 1,J 1
=0
for 0
~
i
~
n,
which is contrary to the hypothesis that 11' ... ,In are not zero.
TRANSCENDENTAL NUMBERS
343
Hence Bo = -1,B 1 , ... ,B n are
~-linearly
dependent.
Let us show that (ii) => (i).
t 1 , ... ,tn be
Let
numbers, and
independent logarithms of algebraic
~-linearly
~-linearly
dependent, then we have:
n
L B.L
i=1
~
0,
~
with B1 , ... ,Bn algebraic numbers not all zero. We can reduce to the case where the B. are ~-linearly independent, in fact, ~
if:
n-1
L
i=1
k. e ~,
k. B., ~
~
~
set:
and
t'.~
t.~ + k.t ~ n
t~,
... ,t~_1 are
(1
~
i
n - 1),
~
independent logarithms of algebraic
~-linearly
numbers, but are m-linearly dependent.
B1 , ... ,Bn can be assumed to be are therefore non-zero; for 1
~
i
~
~-linearly
n - 1 let
have:
tn
n-1
L
i=1
n-1
L
i=1
b. = -B./Bn , and we ~
~
b.t. ~
~
with l,b 1 , ... ,bn _1 now, exp
independent; they
~-linearly
independent and
t 1 , ... ,tn - 1 non-zero;
b.L ~ ~
which contradicts (ii). that (iii) => (i).
Hence (ii) => (i).
It remains to verify
344
CHAPTER 6
Let il, ... ,in be W-linearly independent logarithms of algebraic numbers; if they are iii-linearly dependent we have: n
L
(3.i.
i=l
~
~
0,
One of the (3.'s is non-zero, for example (3n i 0; let b.~ ~ (1 , i , n - 1); thus:
-(3./(3 ~ n
as il, ... ,in are W-linearly independent, at least one of the numbers (31, .•. ,(3n_1 is irrational, which contradicts (iii). (4):
Let us show (vi) => (v).
If (v) is false there exists
a non-zero element:
since the numbers (31, .•. ,(3n are not all zero, we can assume (3n i 0, and then:
[
exp - (30 +
e~n il
~-1] + ••• + -(3in-1 n
contradicting (vi). Let us show that (v) => (vi).
Assume (v) to be true, and let
(30,(31, ... ,B n egj, il, ... ,ineL be such that:
Let us show that 13 0 is zero. then ln+ i e Land: (3
o
=-
13 i
1 1
-
(3
i
n n
Let i n+ 1
- i n+ 1 e
13
+···+(3i· o +(3i l i n n'
(lij.L)(')lij,
hence (30 = O. Finally, (v) is obviously a consequence of Baker's Theorem.
TRANSCENDENTAL NUMBERS
(5):
345
It remains to show that (i) together with (v) imply
Baker's Theorem.
Let .e1 , ... ,.en e L be m-linearly independent; let 60 ,6 1 , ... ,6 n e ID be such that 60 + 61.e 1 + o. We + 6n.e n then have 60 = 0 (by (v)) , and then 61 6n = 0, by (i) . (6):
...
The argument is exactly the same as for parts (1) and
(2) of Exercise 6'5. (7): ism 1/I:a:
-+
First of all let us remark that an analytic homomorphGLm(a:) is of the form 1/I(t) = exp(Mt), with MeMm(rr.), in
fact, let M = 1/1' (0), then 1/I(t)exp(-1/I'(0)t) is constant, and equal to I, the identity matrix, at O. Therefore a homomorphism
~:rr.
n
-+
GLm(rr.) is of the form:
since, as (e 1 , ... ,en ) is the canonical basis of rr. n , we have:
~(t)=~[.~
J=l
and t.
J
~ ~(t.e.)
J J
t.e
J
.J J
n
IT
j=l
~(t.e.),
J J
is a homomorphism of a: into GL
(a:).
m
Furthermore,
the matrices M1 , ... ,Mn commute with each other, because:
and:
and identifying the terms in t 2 gives:
346
CHAPTER 6
M2.+2MM J j k
M2
+ k'
As the matrices M1 , ... ,Mn commute with each other
(7):(a):
and are not all nilpotent (otherwise
+ (0, ... ,0).
Since
expM. e GL (m)
J
n
(1 ~ j ~
n),
A.
therefore e J, which is an eigenvalue of expM., is algebraic. J
I
A.S.] , which is the characteristic value of j=l J J
n
I
logy o
j=l
I
S.logy., J
J
and part (1) above shows that there exist rational integers c O,c 1 , ... ,c n , not all zero, such that:
n
I
j=l
S.c .. J J
Therefore 1,Sl" ",Sn are (7): (b):
~-linearly
The result is proved by induction on n.
we have:
dependent.
exp(Mt) ,
MeM (a:), m
For n
= 1
TRANSCENDENTAL NUMBERS
and M is not nilpotent.
347
We now use Question (3) of Exercise 6'4.
Let us assume the result to have been proved for a homomorphism of ~v into GLm(~) for v ~ n - 1.
If a1""'~ are ~-linearly in-
dependent and: h
1:
j=l
S.a. , J J
~ ~(a1t1 + ••• + aht h ) of ~h into GL (~) takes algebraic values at the points of~h and at n
then the homomorphism (t1, ... ,t h )
the point (Sl"",Sh)' As a1, ... ,a h are ~-linear1y independent, --h for (n 1 , ... , nh ) sOl, n f 0, ~«a1n1 + ••• + ahnh)z) is an irrational function of z S ~. By the induction hypothesis, since
+Olh ,
n; moreover, by (a) the numbers we have h 1,Sl,,,,,Sn are Ol-linearly dependent:
(Sl' ... , Sh )
bn+1
n
1: b.J S., J
j=l
with b1, ... ,b n +1 SOl, and let us say that b n f O. a'.
b a.
b.a
J n
J
n J
(1
~
j
~
Set:
n + 1),
so that: a
But
n-1
,
n+1
a~,
...
1:
j=l ,a~_l
S.a~.
J J
are
~-linear1y
independent, and the preceding
argument applied to the homomorphism (t1, ... ,t n _1 ) ~ ~(a~t1 + ••• + a' t ) leads to a contradiction of the induction hypon-1 n-1 thesis. (8):
As
J:
1
:Xx 3
t[lOg2 +
Ii) ,
and as in is a logarithm
348
CHAPTER 6
of -I, by Baker's Theorem the integral has a transcendental number as its value. REMARK: More generally, we can show the following result: Let P,Q be two relatively prime polynomials of m[X). Assume that the degree of P is strictly less than the degree of Q and that the zeros of Q are all distinct. Let r be a path in the complex plane; assume that r is closed or that r has algebraic or infinite limits. If the integral: I
=J
P(z) dz
r
Q(z)
exists, then I vanishes or is transcendental. (ct., A.J. Van der Poorten: 'On the Arithmetic Nature of Definite Integrals of Rational Functions', Proa. Ame~. Math. Soa., 29, (1971),451-456).
SOLUTION 6-7: (a):
In order to simplify matters, let us first
show: ~
A
-2nB
By the inequality:
valid for all
satisfies:
Now,
IX:
elR; the number:
TRANSCENDENTAL NUMBERS
IAI
349
:> nBlogA.
On the other hand, y
a
b1 1
- - -a
b n
n - 1
is rational, and has a denominator d
~
I satisfying:
Therefore dy is an integer, and it is non-zero because A is a non-zero real number, therefore:
We obtain:
and:
(b):
The proof that:
is essentially the same. n
First of all A
~
0 implies A
~
2.
For
= I we have:
IAI
Ihloga I
~
log2
>
21
~
1 1 A ~ :B
A
therefore the result is true in this case. Let us assume that n ~ 2.
We can also assume that
IAI
<
I
(otherwise the required inequality is true), and that at least
350
CHAPTER 6
of the b.ls are positive or zero (otherwise change A to -A).
[~n]
]
We can then choose:
and since: Gn]B1ogA > 0,
we have: IAI
[~nJB1ogA.
<
Therefore:
IAI
~ exp{-2[~n]B1ogA} > A
-nB
.
REMARK: Under the same hypothesis we have:
On the other hand, an analogous proof shows that if a1, ... ,a n are
non-zero algebraic numbers, if loga. is the value of the logarithm of a. (1 ]
~
j
~
n), if
]
b1, ... ,bn are rational integers such that
the number:
is not zero, then:
A
= max{e, max expI1oga.l, max l~j~n
]
H(a.)}.
l~j~n
]
TRANSCENDENTAL NUMBERS
SOLUTION 6·S: (1):
351
We observe that Schanue1's Conjecture is
In fact, if loga l , ••. ,loga n are m-1inear1y independent logarithms of algebraic numbers, then the
stronger than Baker's Theorem. transcendence degree over
m of
the field:
is equal to n, therefore loga l , ... ,logan are algebraically independent over Oi, and in particular 1, loga l , ..• , logan are iP - L - I. Let us define xl" .. ,xm+n by: (1 ~ i ~
X.
1
x
n),
. n+]
(1 ~ j
~ m).
By Baker's Theorem the numbers xl""'xm+n are m-1inear1y independent, therefore, by using Schanue1's Conjecture (S), the transcendence degree over 00 of the field: m(xl, ... ,xm+n ,e
xl
, ... ,e
xm+n
)
is equal to m + n. (2):
Let us choose:
X.
1
(1 ~ i ~
n),
and: xn+l = loga. By hypothesis the numbers xl"",x n+l are m - L - I, therefore the transcendence degree over mof the field:
352
CHAPTER 6
is equal to n + 1, that is to say that the numbers loga,a Sn a are algebraically independent on
Sl
, ... ,
m.
REMARK:
1-/22/2 = 2 shows that it is not enough to
The number 2
assume Bl, ... ,Bn are irrational and m-linearly independent. (3): subspace of
Let n + ~
t (0
t
~
~ n)
be the dimension of the m-vector
generated by the numbers;
Ul, ... ,U n are ordered in such a way that:
is a basis of this m-vector space. Let us write vut+l, ... ,vun in this basis: n
L i=l
t
a.
L J=l
.u . +
k,~
1
bk . VU .
,J
J
(1 ~ k ~ n - t).
These relations may also be written:
where 5h,k+t is the Kronecker 5 (which has the value 1 if h k + t and 0 if h k + t). Let us consider these relations as a
+
system of equations in ut+l' ... ,u n ; the determinant of this system is non-zero, because of the transcendence of v (it would be enough, if v is not algebraic of degree for
t + 1 ..
h
~
n we have:
~
n - t). As a consequence,
TRANSCENDENTAL NUMBERS
353
and the transcendence degree over m of the field:
is
~
i + 1.
Now, by Schanuel's Conjecture (S) the transcendence
degree over m of the field:
is at least n + i. (4):
Hence the result.
Let K be the field obtained by adjoining the sixteen
given numbers to m, and let L = K(loglog2).
Let us consider the
following seventeen numbers x l ' ... ,xl7 : 1, in, n, logn, e, elogn, nlogn, log2, nlog2, elog2, ilog2, ilogn, log3, loglog2, log3loglog2, I:2log2. Xl
x l7
Since the field m(x l , ... ,xl7 ,e , ... ,e ) has the same transcendence degree over m as L, by Schanuel's Conjecture (S) it is sufficient for us to show that xl'" .,x l7 are m-linearly independent. Now a non-trivial linear form in x l , ... ,x l7 is also a nontrivial polynomial in: n, logn, e,log2, log3, loglog2.
Thus we are led to showing that these six numbers are algebraically independent.
The same argument starting from the numbers:
1, in, logn, log2, log3, loglog2
leads the problem to the algebraic independence of the five numbers: n, logn, log2, log3, loglog2,
354
CHAPTER 6
therefore to the linear independence of the numbers: iR, logR, log2, log3, loglog2. On considering the exponentials of the real parts of these numbers,
we have to show that R,2,3,log2 are multiplicatively independent. Now, by Question (1) above, Schanuel's Conjecture (S) implies the algebraic independent of Rand log2.
Hence the result.
If logx and e are algebraically independent, then 1,
(5):
logx,elogx are m-linearly independent, therefore one of the two numbers: x, x
e
is transcendental over m(e,logx), hence over m. If logx and e are algebraically independent, the transcendence degree over m of the field: e mOogx,elogx,x, x )
being ~ 2 (by Schanuel's Conjecture (S)), one of the two numbers x, x e is transcendental over !I?(e,logx). (6):
If x is algebraically irrational, x x is transcendental
(for example, by Question (2) above, or more simply by the Gel'fond-Schneider Theorem (2)). If x is transcendental, the numbers: 2
logX, xlogx, x logx are w-linearly independent, hence at least one of the two numbers x x2 x,x is transcendental over W(x,logx).
SOLUTION 6·9: (1): nomial:
If
~
is algebraic on m(R) there exists a poly-
355
TRANSCENDENTAL NUMBERS
uox
a
+ ••• + u o'
wi th coefficients u j in IZ [11:], such that:
By induction, for every integer r
~
1 there exist elements VO,r'
""Vo-l,r oflZ[1I:] such that: r~o+r-l
u~s u
= v O,r + v l,r ~ + .•• +
furthermore, the degrees of the v.
],r
orised by: max
d(v.
],r
V~-l,r~
0-1
u
;
(with respect to 11:) are maj-
) ::: r max d (u . ) ,
o:::i:::o
1
and their height by: max
O~:::o-l
H(v.
],r
Let P(X) = a n and size
B
o
)::: 2r(l
+ max d(u.){( max H(u.){. o~i:::o
o:::i:::o
1
+ a X + ••• + a Xn be a polynomial of degree n
1
with coefficients inlZ. 0-1 \' L
j=O
b.
.
- ] ~] U
n-o o
We have:
'
where bo, ... ,b o_l are elements oflZ[1I:] defined by
b.
+
]
The norm
R(1I:)
1
n-o n-o-r
L
r=l
u 0:
ar+ a- lV'] ,r .
of u~(~) on ~(11:) can be written:
356
CHAPTER 6
R(lt)
02
= Uo
IT
(0-1
L
b.~ ]
j=O
{a}
jJ
,
{a} being the set of m(lt)-isomorphisms of m(lt,~) into~.
Conse-
quently R(lt) is a polynomial in
It, of degree ~ kin, and of size k2B (where k 1 ,k2 are independent of the polynomial P), with coefficients in~. Furthermore, we have:
~
IT
afl
'J
( 0-1 L b.~] j=O ]
.
Therefore:
where k3 depends neither upon
B
nor upon n.
deduce the existence of a constant C1
(2):(a): numbers Qm
such that:
If mO is an integer such that a
mo
-l(~),Qm (~)
o
(2):(b):
From this we easily
= Cl(~)
0
Since
is not zero (since
f(~)
an irreducible polynomial
is algebraic over ~,
~
f 0, one of the
f 0), because:
m(~),
there exists
in one variable, with coefficients
in2Z[~]:
U. ]
such that
~(f(~))
0.
6 ~[~],
Let n1 , ... ,n o be the roots of
o ~
~
j
~
0,
(that is
TRANSCENDENTAL NUMBERS
357
to say, the conjugates of
f(~)
over
m(~)).
The norm R
o
II (- ni
R (0 m
i=l
(~)
m
is:
of
+ 1 + a 1 ~ + •.• + am~m)
-1 I0 Uo
j=o
'J
u. ( m I ai~l j . J
i=o
Therefore:
o
D u~R (0 mum
is a polynomial in
with coefficients
~,
k'D o
in~,
of degree
~ k~m,
and of height '" 2 m' where k~,k; do not depend upon m. Furthermore, if mR, is a sequence of integers for which Q (~) f 0, we mR,
have P
mR,
f 0; let us show that the hypothesis on the am's im-
(~)
plies: -
(this will show that Let C
>
~
is not algebraic over
0; there exists M = M(C)
>
00
m(~)).
0 such that, for m
have:
ia m+k +1 i
<
e -2C" m+k,
with: "m+k
=
(m + k)(m + k + logDm+k)log(m + k).
Clearly we have:
" k~ (1 + !::...)" m m+
m;
,
~
M, we
358
CHAPTER 6
as we have Am
>
m2 , we obtain:
IhOL am+k+l sm+k+ll [e whenever m is sufficiently large.
-2CA
1m
m
(lsi
+ l)]k ~ e
-CA
m
Consequently:
Finally, loglp (s)1 is at most equal (up to a multiplicative conm stant) to logl~(s)l.
CHAPTER 7
Congruences Modulo p: Modular Forms
INTRODUCTION 7. 1
HODULAR FORMS
L
Let f(q)
AsqS be a Laurent series that is convergent
8;'8 0
in the disc 0
<
Iql
< l.
Let H be the Poincare (upper) half-plane,
that is to say, the set z e a: with Irnz for z e H set F(z) OF WEIGHT
=
f(q).
>
Set q = e
O.
2illZ
,and
We say that F is a MODULAR FUNCTION
k, for k an even integer, if:
Since we have F(z + 1) = F(z), and because the group of linear " az ++ d b ,wlt "h coe ff"" " ~ . 1 trans f f ractlona ormatlons z ~ ez lClents ln satisfying ad - be = 1, is generated by z
~
z
+ 1 and by z
~
-liz,
it follows that we have:
F[az + b) ez + d}
= (ez
+ d) k F(z)
for a,b,e,d e:
359
f is holomorphic
360
CHAPTER 7
at the origin (A
s
=0
for s
and that F is a CUSP FORM if,
< 0),
in addition, f vanishes at the origin (A = 0 for s ~ 0). s The set of modular forms of weight k is a vector space whose dimension, when k is even, is equal to:
[l~]
ifk
[ lk2] + 1
if k
0
if k
-
2 mod 12, k
>
0,
2 mod 12, k
~
0,
* o.
<
The set of cusp forms of weight k
+
12 is a vector space iso-
morphic to the vector space of modular forms of weight k. EISENSTEIN SERIES:
For even k
L
(m ,n )ez:;xz:;
(m,n)+(O,O)
~
1
4, the series:
(mz + n)
k
is convergent for z e H, and is a modular form of weight k. .
expansl0n at q = e
2illz
Its
is given by:
2(2ill)k 21;(k) + (k - 1) where: 1;(k)
and
L
din
d
k-l
.
The vector space of modular forms of even weight k, has as a basis the monomials of the form G~G: with a,13 ~ 0 integers satisfying 4a + 613 If we set: L'l(z)
= k.
CONGRUENCES MODULO p: 110DULAR FORMS
361
the function 6(Z) is a cusp form of weight 12. 7.2
SUBGROUPS OF THE MODULAR GROUP If A is an unitary ring, we write the group of 2
with coefficients in A and determinant one as SL 2 (A). has the trivial subgroup {±I}.
The group r
on the Poincare upper half plane
-+
x
2 matrices This group
= SL2~)/{±I}
operates
H by:
az + b cz + d
On the other hand r is generated by T and J with: T
T and J satisfy the relations: J2 deduce that every point of
= I,
(TJ)3
= I.
From this we
H is transformed by an element of r
to a point of the fundamental domain of r: D = {z = x + iy:lxl "~,Izl ~ 1},
the transformations T and J, which generate
r, permit one to iden-
tify the boundaries of this domain. A subgroup of r, of index n, will have a fundamental domain formed by n copies of D, and the boundaries of this domain will be identified by elements of this subgroup. The genus of a Riemann surface can be found by taking a triangulation, and we have the formula: 2 - 2g
Number of Triangles - Number of Sides
+ Number of Vertices. If we want to apply this to the quotient of H by a subgroup of
CHAPTER 7
362
r, we first have to compactify this quotient, that is to say, to look at what happens at the cusps: to do this we take a local uniformization-for example, for the point at infinity (denoted i) the uniformization is q e 2i1tx/h , where h is the smallest integer such that T
h
belongs to the group.
If a Rieman surface has genus zero it is known that its field of functions is isomorphic to the field of rational functions over
H by a group gives a surface of genus zero there exists a meromorphic function f on H (includ-
~;
in other words, if the quotient of
ing the cusps) invariant under the group such that every function having the same properties as
f
is a rational function of
f.
This
is what happens with the group r, the function f then being the "modular invariant" function
~(z),
which we shall denote
avoid confusion with the number j such that j3 BIBLIOGRAPHY [Serl] , [Gun].
=
1.
~(z)
to
363
CONGRUENCES MODUW p: MODULAR FORMS
PROBLEMS
EXERCISE 7-1: IRREDUCIBLE POLYNOMIALS WITH COEFFICIENTS INF (1):
P
Let P be an irreducible polynomial of degree n with
coefficients in F
p
= 'Zllp'Zl, where p is a fixed prime number.
Show, by studying the field generated by one of its roots, that P divides the polynomial TI (2) :
n
n
= xP - X.
Show that an irreducible polynomial ofF [X] divides P
TIn if and only if its degree divides n. (3):
Deduce from this that the number
~
P
(n) of monic irre-
ducible polynomials ofF [X], of degree n, is given by the formP
ula: ~ (n)
P
1
L
n din
n] pd]1 [-d
where ]1 denote the M5bius function. EXERCISE 7-2: SOLUTION OF THE EQUATION
xP - X
a IN A FINITE
FIELD (OF CHARACTERISTIC p) (1) :
Set P(x)
= xP -
x, Q (x) r
=x
r-l
+ x P + --- + x P
364
CHAPTER 7
Show that the mapping ofF r to itself defined by: p
u(a)
P(a)
v(a)
and
= Qr (a)
are F -linear, and satisfy: p
Ker(v) = Im(u). Establish the relation:
(2):
II
aeF
- x. p
Given an element a that is algebraic overF , determ-
(3):
p
ine which field the roots of the polynomial x P - x - a lie. EXERCISE 7·3: THE RING'll./ph'll. Let a
+ 1 be
an integer not divisible by p.
Denote by w(h)
the order of the element a in the multiplicative group (1)
Show that w is not constant and moreover satifies:
w(h)lw(h + 1) (2): w(h + 1)
h
~/p ~*.
Assume p
+w(h
w(h + l)lpw(h)
and
+ 2).
+ 2.
Show that w(h)
+ w(h
+ 1) implies
From this deduce that there exists ho such
that: For a11 h Define h o' (3):
~ h
o'
w(h)
Study the case p
(1)
= 2.
p is again odd; show that w(h) is equal to the order
CONGRUENCES MODULO p: MODULAR FORMS
of
h
365
(~/ph~* for any h if and only if this holds for h
1 and
= 2. Show that there exists a satisfying the conditions of Ques-
(4):
(~/ph~),., is cyclic. Verify that
tion (3), and deduce from this that
the numbers a satisfying the conditions of Question (3) belong to distinct classes modph, the number of these being equal to the number of generators of
(~/ph~),.,.
Determine the generators of
(;z/2h~),., is not cyclic.
Show that for h ? 3
(5):
(~/5h~),~.
EXERCISE 7·4: BERNOULLI NUMBERS
L
We say that a series
a
n=O
n
L
00
n.
Two H-entire series
a ~ and n
n=O
CONGRUENT mod m if a
n
_ b
n
n
~ is H-ENTIRE if a e ~ for all n n
n
L
n=O
b
n
~ are said to be
n n
(mod m).
Show that if fez) and g(z) are H-entire series, then
(1):
the same is true of:
fez )g(z),
f'(z) ,
m!
if f(O)
O.
Show that for any non-prime m > 4:
(2):
(e Z
fez )m
_
1)m-1
-
=0 2
(mod m).
2k+1 \ z (mod 4). k~l (2k + i)!
For p prime, by using the periodicity mod p of the coefficients show that:
L k=l
z
k(p-1)
[k(p - I)]!
(mod p).
366
CHAPTER 7
Set:
(3) : z
e
z
2n (_1)n-1 B z n (2n) ! n=l
z
-'2+ L
1
- 1
(the Bn are the Bernoulli numbers). Show that if the Pi are the distinct prime numbers such that p. - 1 divides 2n, then (_l)n-1 B + L ~ is an integer. ]. n i Pi Let q
(4) :
Set:
n
be the denominator of B
n
=2
Calculate a (n) (seperate the cases P p
Show that for P
(5) :
If 3 ~ P ~
>
2 a (n)
a (n) p ~
P
12n,
p
~
(in lowest terms).
and P
+ 2).
0, and that;
(2n) 3/2 ,
and lastly that:
L
p>12n
(6):
~~
a (n) ~ P
2n
I2n -
1
Calculate: nQ1Q2°"qn (2n)!
EXERCISE 7·5: RAMANUJAN'S IDENTITY FOR THE NUMBER OF PARTITIONS Recall that the following identities between formal series (cf., [Har] Chapter
XIX):
CONGRUENCES MODULO p: MODULAR FORNS
cp(x) =
IT
367
L (_1)nx~(3n2+n),
(1 - xm)
1
n=-oo
00
L
3 cp (x)
n=O
(-1)n(2n + l)x,
(2)
where pen) is the number of number of partitions of n. (1):
For
0,1,2,3,4 set:
B
(_1)nx~(3n2+n)
L
G (x) s
~(3n2+n)=B(mod5)
Show that we have:
D=
By
G4
G3 G2 G Go 4 G3 G2 G1 Go G4 G3 G2 G1 Go G4 G3 G2 G1 Go G1
(2) :
G3
G1 G2 G3 G4 Go
G1
and
=-
using Equation (2) show that: GOG2 +
p
S
(x)
[cp(x 5 )]6 cp(x 25 )
Show that:
= G4 = 0
(3):
(1)
2
G1
O.
Set:
=
~ L p(5n + n=O
B
)x 5n+s .
xcp(x
25
).
(3)
368
CHAPTER 7
Show that:
Conclusion? EXERCISE 7·6: CALCULATION OF DETERMINANTS mod p: CHARACTERISATION OF LOGARITHMIC DIFFERENTIALS Throughout this problem K is a field of characteristic p. With P a polynomial of K[x] , of (at most) degree p - 1,
(1):
calculate (in
K):
p-l
I
P(n).
n=O
With a1, ... ,a p elements of K, calculate the determin-
(2): ant:
a
p
(Begin by showing that if a. = b. + c., the determinant formed J.
J.
J.
with the a.'s is the sum of that obtained by starting from the J.
bi's and that obtained by starting from the ci's). (3):
Calculate the determinant:
a1
a
p
a2
a1
-
1
....... .... .... a1 - i
....
a1
a a
-
p p-l p + 1
CONGRUENCES MODULO p: MODULAR FORMS
(4):
Define an operator
~
369
on the field K«t)) of formal se-
ries with coefficients in K by:
~
[I a n=-N
n
=
t n)
I
n=-N
ria
n
tn.
Show that there exists an element b of K«t)) such that: ~b
a t n =1) n
L
a
n=-N
i f and only i f a
(a )p for all n. n
np
EXERCISE 7·7: MULTIPLICATIVITY JF THE RAMANUJAN ,-FUNCTION (1): Let p be a prime number, let m be an integer satisfying l..;;m 0 we have: 1
-z+m
au+b au + d
p
z + m'
with u
p
(2): Let F be a modular function of weight k. sociate the function w defined for Imz > 0 by: p
w (z) P
= p k-1F(pz) + -1 p-1 L
P m=O
[z + m)
F--
P
Show that:
w
p
(z + 1)
(3) : F(z)
w (z) p
and
wp (
with q
e
-;)
Assume that:
L
n~nO
Anq
n
2iltz
z
k
w (z). p
To it as-
370
CHAPTER 7
Show that
p
(z)
L
may be written as
n~n
as a function of the coefficients A .
o
n
Show that if F is a modular form then is a modular form of p the same weight, and that if F is a cusp form then also is a p
cusp form. (4) :
Set:
F(z)
L
lI( z )
n=l
n -r(n)q •
We know that F is a cusp form of weight twelve. Show that p and II are proportional. have: -r(p
a+1
)
N = -r(p)-r(p")
- p
11
N-1
-rep"
Show by induction on a that for a
Show that for a
~
1 we
). ~
0 and p not dividing
8
we
have:
From this deduce that -r is a multiplicative function. (5):
Now choose F(z) = Gk(z)
an Eisenstein series of
weight k. Show that p and Gk are proportional. EXERCISE 7·8: STUDY OF A SUBGROUP OF THE MODULAR GROUP (1) :
Show that the set:
{i.[: :]. a.[::]
b
d
[: :] [: :]
,
a
=
e
=
[: :] [: :]}
CONGRUENCES MODULO p: MODULAR FORMS
forms a subgroup of the group
371
[SL2~/4~/±I].
(2) :
4 -1 Denote by r 4 the subgroup of r generated by T ,J,TJT
(a) :
Show that:
r
Find a fundamental domain for r 4 and the genus of the
(b) :
Riemann surface r 4' Hu {i co }. li1lZ
Set q = e 2
(3):(a):
and show that there exists an unique
H, such that:
function H(z), holomorphic on li1l
e4
--;:::: + 2f2q
H(z)
H(yz)
for q
O( q)
3
II
Calculate
T
+ h)]
[x - H(z
h=o
¢(x,z).
Let D be a positive integer congruent to 3 modulo
(4): (a): 16, and set
0,
for all y e r 4.
H(z)
(b):
-+
= !(l + ilV).
Show that there exists integers a,S,8 such that: a8
D+ 1 4
S :=
° (mod 4),
a > 0,
which satisfy: H(T +!)
(b):
H ( a (T +
Show that:
f)
u
+
sJ
if and only if! - 2 (mod 4).
372
CHAPTER 7
IT
a,a,o
[X - H(az
0-
a)]
ao=~(D+1)
0", a<40 a>0,a=Omod4 is a polynomial in x and H(z), we will denote it (c):
~(x,H(z)).
Show that H(T + 2) is the unique root common to the
two equations:
(x - 3)(x + 1)3 { q,(x,x) =
o.
0,
373
CONGRUENCES MODUW p: MODULAR FORMS
SOLUTIONS
SOLUTION 7-1:(1): Let a be a root of P; since P is irreducible, JF (a) is an n-dimensional vector space overJF p ' because it is a p field it can only be F n. By the rules for calculating in char2 n-l p are also roots acteristic p, it is clear that aP,a P , ... ,a P of P; now, these numbers are distinct, for
k
aP = a
is equivalent
to a e JF k' therefore P has all its roots inJF n' that is to say P
pi TI n .
Let Q be an irreducible polynomial of degree m that
(2) :
divides TI
P
By Question (1) above Q splits completely inF
; pm F m must therefore be included inJF n ,which implies that m diP
n
P
vides n.
Conversely, if m divides n it is clear that TI
TIn' hence that Q divides TIn. (3):
m
divides
TIn is divisible by all the irreducible polynomials of
degree m with m dividing n, and by them alone. It is not divisible by any power of these polynomials since it only has simple roots. that: p
By examining the degrees of these polynomials we see
n
I
min
m1jJ (m).
P
CHAPTER 7
374
The
M~bius
Inversion Formula then gives the desired result (cf.,
chapter 1). SOLUTION 7·2: (1):
As we are in characteristic p, the mapping
h
a ~ a P is F -linear, therefore u and V are linear also. P have: vou(a)
= aP
- a + aP
2
- a P + •••
We
- a
0,
that is to say, Im(u) C Ker(v). Ker(u) is made up of elements satisfying a P = a, that is to is l'-dimensional over IF and u has r P P r-l rank l' - 1, Im(u) contains therefore p elements. On the other r-l hand, as Ker( v) is made up of roots of a polynomial of degree p 0 lOt contaIns at most pr-l e 1ements, hence Ker(v) = Im(u). say Ker(u) = F.
Since IF
P
(2):
We see that u
F , therefore Im(v) P
V
= 0,
= Ker(u).
that is to say Im(v) C Ker(u) Therefore
II
aEiF
P
[Q (x) - a] is a r
polynomial of degree pr which vanishes at all the elements of r
IF r' therefore it is equal to k(x P - x); examining the coefficiP r ent of the term of degree p shows that k ~ 1. (3):
If b is a root of the polynomial x P - x - a, the other
roots of this are b +
Cl, o
lie in the same field.
with
Cl
eFP , they are a11 distinct and
From Question (1) it follows that the
two conditions: The roots of the polynomial x P - x - a lie inF r (a e Im(u»),
1
°
Qr (a) =
are equivalent. a elF h. P
If l'
P (a
e Ker(v»),
Let then h be the smallest integer such that
= sh +
k, with k < h, we have:
375
CONGRUENCES MODULO p: MODULAR FORMS
2h-1 h-1 sh-1 + ••• + a P + ••• + a P a + ••• + a P
Q (a) r
sh+k-1
+ ••• + a P
with the convention QO(a) =
o.
Now, if k
+ 0,
Qk(a) is not in
F , for it follows from Question (2) above that the only numbers P
a such that Qk(a) lies in
are those of F k' Since Qh(a) eF p ' P P Qr(a) can vanish only if sQh(a) = Qk(a) = O. That is to say, if k = 0, either Qh(a) 0 or p divides s. From this we conclude: F
0, the roots of x P - x - a lie in F h' P
SOLUTION 7·3: (1):
w(h) = k would imp1ya k
=1
(mod ph) for all
1, which is impossible if a + 1. If w(h+1) h+1. . a = 1 (mod p ), 1t 1S clear that a w(h+1) = 1 (mod p h),
h, that is to say, a k
that is to say, w(h) divides w(h + 1). (mod ph), apw(h) pw(h).
=1
=1
Therefore, either w(h + 1) = w(h), or w(h + 1) = pw(h).
(2):
Assume w(h + 1)
+ w(h),
with a not divisible by p.
w(h+1)
=
a
pw(h)
=
that is to say, a
w(h)
=l+ap
By Question (1) above we know that
w(h + 1) = pw(h), therefore: a
Convers1y, if aw(h)
(mod ph+1) , that is to say, w(h + 1) divides
h p (1 + ap )
h
376
CHAPTER 7
As P is different from two, ph is larger than 2h + 1; on the other hand, as (~) is divisible by p for 2 ~ i ~ p - 1, it follows that:
which shows that w(h + 1) f w(h + 2). If we choose as ho the smallest number such that w(h o + 1) f
w(h o) (h o exists, by (1) above), we see that w(h + 1)
= pw(h)
for
h 3 ho' which gives the formula.
If p = 2, the same argument gives:
which shows that if h f 1, w(h + 1) f w(h + 2). result is obtained by taking ho (3):
~
Hence the same
2.
The condition is trivially necessary. It is immedi\ , .1S ph _ ph-l ph-l(p - 1). 0f ( Z2:/ P Z2:,"
ate 1y seen t h at t h e ord er
Now, by Question (2) above, w(l) = p - 1 and w(2) = p(p - 1) certainly implies that w(h) = ph-l(p - 1). (4):
We must choose a in such a way that w(l) = p - 1, that
is to say, a = a + AP with a a primitive (p - l)-st root of 1 mod p-l 2 p, and that w(2) = p(p - 1), that is to say that a 1 (mod p ).
*
Now: (a
+
Ap)
p-l
where we have set a P- 1
=1
+ pv (mod p2).
Since (p - 1) and a
are non-zero (mod p), for each a there exists a single value of
A,
say (-
(p -
v
-2)'
l)a P
mod p, such that
_ 1 (mod p 2 ).
377
CONGRUENCES MODULO p: MODULAR FORMS
For the (p - 1) other values the resulting a's give us our result. It is clear that a generates ez/ph~*, therefore this group is cyclic. (p - l)~(p - l)incongruent solutions mod p2 have been found, that is to say, ph-2(p - l)~(p - 1) = ~[~(p )] solutions
= cardez/ph~~ ~[~(ph)]
modulo ph; since ~(ph)
is the number of
its generators. EXAMPLE:
For p
5 the primitive roots are 2 and 3.
Now,
(2 + 5A)4 _ 1 + 5(2 + 4.8A) - 1 + 5(3 + 2A)
*
A l l mod 5
=1
+ 5(1 + 4.27A)
*
* 3 mod 5;
(3 + 5A)4
A
the generators are therefore the numbers a congruent (mod 25) to 2,12,17,22;3,8,12,23. (5): If a = 4n + 1 we find w(1) = w(2) = 1, hence w(h) ~ 2h - 2 for h ~ 2; if a 4n + 3 we find w(l) = 1, w(2) = w(3) = 2, and therefore w(h) ~ 2.2 h- 3 for h ~ 3. w(h) can therefore never for h ~ 3 be equal to 2h- 1 = the order of ez/2h~*. SOLUTION 7·4: (1):
n
00
fez)
= L
n=O
Let us set:
a
n ~
00
~ n n!
and
g(z)
\'
L n=O
b
n n!
we then find:
r(z)
L
z
n
a n+l -n! ' n=O
L
z
n
a n-l -n! ' n=l
378
CHAPTER 7
These series are therefore clearly H-entire. We now assume that [(0)
0 and that
fez )m-l
Gn _ 1)!
is H-entire;
fez )m-l , since [ and [' are H-entire, the same is true of (im _ 1)! [ (z), therefore also for: m-l fo let) m-l)! z
r(t)dt
.Jf!f_\
= LJEl m!
Which proves the last result by induction. (2):
By pa~t (1) we see that (e z - l)m-l = (m - l)!g(z),
where g(z) is H-entire, since for non-prime m > 4(m - 1)! = 0 Z m-l (mod m), we find (e - 1) = 0 (mod m). (To show that (m - 1)! = 0 (mod m) set m = pq. If P q, as p,q < (m - 1) the result is obvious. If P = q we have the case
+
+
m = p2 with p prime; if p 2 we then see that p and 2p are sma112 er than (p - 1), whence the result). We have:
in particular, ex>
L
n=l
ex>
L
n=2
n
[3 + 3n]~ n
(mod 4).
CONGRUENCES MODULO p: MODULAR FORMS
Now 3 2
379
(mod 4), hence 3 + 3 2p+l - 2 (mod 4) and 3 + 3 2p -= 0
=1
(mod 4), which yields: _ 2
L
z
2k+1
(mod 4).
k=l (2k + i)!
We now apply the formula with m
L
n=l
the formula hP- 1
a
p - 1; if we set:
zn
n
n!
(mod p) implies that a n+p- 1 = a n (mod p), and the coefficients are periodic; on the other hand, we know that (p - I)!
=1
= -1
(mod p), hence: zp-l + •••
Zp-l - (-1) (p _ 1)! + ...
(mod p),
which certainly gives:
L hi (3) :
k(p-l) z [k(p - i)]!
It is easy to verify that the function -z-::..z- + e
even, hence the given expansion.
le z
and for
z
Now set z
= loge 1
- 1
+
eZ
-
t is 1),
11 < 1 this yields:
L
( e Z _ l)m-l (_l)m+l...:...:..._........:::...:......._ m
m=l As f(z)
(mod p).
- 0 (mod m) is equivalent to ~ being H-entire, we find: m z
1
1 - 2
zk \' 7.7, - 3.. L\' k=l K! 4 k=l 00
L
2k+l z (2k + 1)!
"7"::"''---~~
(Contd)
CHAPTER 7
380
L
(Contd)
1
L
pprime p k=l p>2
k(p-l)
Z
[k(p - i)]! + f(z),
where f(z) is H-entire, which finally gives: 1
L
pprime p>2 (p-1)/2n
- + integer, p
which is the required result (2 - 1
=
1 always dividing 2n);
this is the Von Staudt Clausen Theorem. (4):
From Question (3) it follows that:
IT
p>2 (p-1)/2n
p,
2 divides qn with multiplicity 1 and [2n]! with multiplicity [2;] + [2;] + ••• , from this it follows that
- [~] - [~l p divides q that is to
,
with multiplicity 1 if and only if 2n
~ay
n
= k(~)
= k(p
- 1),
therefore p divides (q1, ... ,qn)
. h mu 1· Wlt tlP 1·· lClty [2n] p-:-T , an d consequent 1y:
a (n) = [p 2_n 1] P
(5) :
We find:
a (n)
~[p2~1]
p
_[~] _[2~] P
2n P
2n 2 p
- ----
[p
2~ 1]
2n 1 P 1 -
----~
!
p
381
CONGRUENCES MODULO p: MODULAR FORMS
)
[
1-
2n
p-:l
2n - - > - 1.
P - 1
As ap(n) is an integer, it certainly follows from this that a (n) ~ o. p r r+1 We define r by p ~ 2n < p that is to say r [log (2n)] p
_... _[2;] P
~ ~ _ (2n _ P - 1
P
1) _... _ (2nr _ 1) p
+ ~ _ 2n [ 1 p-1
p
-~] pr
1-~
P
and by using the fact that 2n a (n) " r
P
<
p
r+1
,for p
~
3 we find:
3 + ----EP - 1 " r + -2 •
As a (n) and r are integers, we deduce from this that: p
a (n)
E;:
P
r + 1.
Therefore, for 3 a (n)
p p If P
>
"p
~
P
~
fin we find:
log (2n)+1
p
" 2np" 2nf2n
fin then r " 1, therefore:
= (2n)3/2.
382
CHAPTER 7
Now, it is clear that
[k 2-:: 1] - [2;]
is positive or zero for all
k and is zero for k > 2n + 1, therefore:
~
[k 2~ 1]
)'
k?[I2n] +1
- [2;]
[[~]]
We have:
(6):
n,----Q1" 'Qn
IT
(2n) !
3... p~12n
a (n)/n pP
n
p>?2ri
a (n)/n pP
We are going to calculate the limit of each of the three factors. For all h ? l: n n - "2 - 4" ••• . ;;
a 2 (n) < -
n n n "2 + 1 - 4" + 1", - - + 1, 2n
or:
- n
~
a 2 (n)
< -
n +
n
~
2 +
+ h,
that is to say, for all h we have:
2
-1
.;
a 2 (n)/n 2(h/n)+(1/(2 h+1» .; 2- 1 2
that is to say,
~
"
lim2
a 2 (n)/n
_ a 2 (n)/n h +1) .; lim2 .. lX21/(2 2
CONGRUENCES MODULO p: MODULAR FORMS
383
Since this holds for all h we find that
lim2
a 2 (n)/n
=
~.
Also, by Question (5) above we find: 1 ~
a (n)/n ~ p P ~ [( 2n ) 3 / 2n] 2n
IT
3~p ... /2n
and because:
lim(2n)3/~
n-+ oo
= 1,
we have: a (n)/n
lim p P 3... p.. 12rl
1.
It is known that if p
1
~
1T
p>./2n
a (n)/n
p P
~
>
2n + 1, a (n) P
(2n +
1)
1T
n-+oo p>72n
a (n)/n
p P
1.
Taking all the results together we find:
lim
n,.--Q1·" Qn
(2n) !
l2 •
0, hence:
(p>~n ~)
and therefore: lim
=
n
384
CHAPTER 7
SOLUTION 7-5: (1):
D is a circular determinant ("circulant").
Hence we find:
where
~
runs over the fifth roots of unity.
D = IT cp(~) =
IT m=1
~
This can be written:
IT (1 - ~rnxrn). ~
Now it is clear that:
On the other hand, since for m
* 0 (mod 5), ~rn also runs over
the fifth roots of unity, we find: D
(1 - x Sm )
IT ml~odS)
IT
(1 - xm)S
m==O(modS)
IT (1 - xSm)s IT--(-1-_-x"""2'""s'-m-) m
m 2
3n 2+ n == 8 (mod 5). We multiply both sides by 24 (which is invertible mod 5), and we (2):
Let us try to solve the congruence
find: 36n 2 + 12n _ 248
- 8 (mod 5),
or: (6n + 1)2
==
1 -
8
(mod 5).
In order that this congruence have a solution it is necessary and
CONGRUENCES MODUW p: MODULAR FORMS
sufficient that 1 say that Gs 8
=3
or 4.
=0
8
if (1
Hence G3
385
be a quadratic residue mod 5.
~
8J
= -1,
= G4
O.
(mod 5), which implies 6n + 1
=0
which gives 1 For
8
=1
That is to
8 _ 2 or
3, so
we find (6n + 1)2
(mod 5), that is to say n
=0 = -1
(mod 5), therefore we find:
G1 (x)
- xcp(x
25
).
It follows from what has gone before that:
L (-1)n(2n
2
+ l)x(n +n)/2
o
2
Let us now try so solve n ; n _
8
(mod 5); by multiplying by
8 we find:
(2n + 1)2 = 38 + 1 (mod 5),
which is impossible if 38 + 1 = 2 or 3, that is to say, if 8 2 or 4. 3
Now, in CGo + G1 + G2 ) the terms giving an exponent of x con2 2 gruent to 2 (mod 5) arise from 3G OG1 + 3G OG2 · This term must therefore be zero, that is to say G2 - GOG2 · 1 (3):
We observe that:
and by decomposing the series according to the congruences modulo 5 of the exponents, we find:
386
CHAPTER 7
Dx
1
Po Pi P2 P3 P4
0 0 0 0
therefore if the fifth column of D is mUltiplied by P4 , and to it we add the first column multiplied by Po' the second by Pi' the third by P 2 , and the fourth by P 3 , we find:
G4 G3 G2 1 Go G4 G3 0 G2 Gi Go G4 0 G3 G2 Gi Go 0 G4 G3 G2 Gi 0 Go Gi
P 4D
and by making use of Question (2) we find:
Gi
P4 D
G2 0 G2 Gi Go 0 G2 Gi 0
0 0
GO G2 Gi
0
2 G4 - Gi (2G i G2GO) - G2 (G OGi 1
From this we deduce:
which shows, in particular, Ramanujan's Congruence:
p(5n + 4)
=0
(mod 5).
2
GOG2 )
CONGRUENCES MODULO p: MODULAR FORMS
387
The numbers nO, ... ,p - 1 are the solutions of the equation x(x - l)···(x - p + 1) = x P - x. As the sums of
SOLUTION 7·6: (1):
the powers of the roots of a polynomial can be expressed as symp-1 metric functions of these roots, we see that I n h = for h < n=O p - 1. On the other hand,
°
p-1
I
n
n=O
p-1
p-1
I
- 1.
1
n=l
From this therefore, we deduce:
p-1
I
- (coefficient of degree p - 1 of the polynomial P).
pen)
n=O (2) :
The usual method of calculating this determinant gives
nothing, because 1 does not have a p-th root in any extension of K.
By using the distributivity of determinants we find: b1 + c 1 b + c
b
b2 +
b1 +
P
P
+ c
f 1 (1)
P
I
P
C2
f
C1
f (p) P
f (1) P
f 2 (1)
f1 (p)
where f runs over the set of functions from {l, ... ,p} to the If f is considered as a function on
set of two elements {b,c}. '7.l/p'7.l, we see that for all
from the function f
1"'
l'
the determinant obtained by starting
defined by f (i) = f(i + 1"'
1')
is equal to that
obtained by starting from f (it suffices to make a cyclic permutation of the columns and the same permuation on the rows). Let us assume now that f 1"' (i)
k we have f(k1') = f(O); since
l' ~
f(i) for l' ~ 0, then for every is invertible in '7.l/p'7.l, 0,
388
CHAPTER 7
and kr runs over the whole of 'O./p'O. when k runs over 'O./p'O.. From this it follows that fCi) = constant. Therefore, i f fC i) is not a constant there are p distinct elements, e.g. f , which r give the same value to the determinant, that is to say, which give zero to the total in the summation. b
bi + c i
b
P
+
C
P
+
C
bi
P
Hence we find: b
P
P
ci
C
c2
ci
P
+
b2 + c 2
bi + c i
b2
bi
We notice then that the system Cai, ... ,a p ) is the sum of the systems (O, ... ,ai, ... ,O), and that:
o
a.1
0
o o
o • • • • • • •
a.
a.
o
1,
a.·····
o
1
a~,
1
1
0
which shows that the determinant has the value:
1 a~.
a=i
(3):
D
1
We use the same method, and we find: ai
, , , ,
a
ai - 1
P
a2
a
p
CONGRUENCES MODULO p: MODULAR FORMS
f
f1 (1)
p
(p)
II
2
f
f 2 (1)
389
(1 -
keK
f1 (p)
k),
where, this time, f(k) either has the value a or fh(k) Kronecker symbol), and where K
= {k:f(k) f a}.
= 0hk (the
This then leads
to:
a
p
D=
IT
+
1
o
o
1
(1 - k)
k=1
f \'
L
p~1 L
IT
class of f 1'=0 keK
p
(p)
(1-k+r)
f1 (p)
where we sum only over a set of elements f such that f( i) f fer + i) for all i.
But by part (1),
r0
p-l
2 IT
1'=0 keK
(1 - k +
1')
i
p-1
I
IT
1'=0 k=2
if cardK (1 - k + 1')
< p -
1,
-1 otherwise.
Hence we find:
D=
a .•• a 1 P a P a 2 ••• a 1 (Continued)
+
a1
o ••• 0
a
1 ••• 0
P
a2
1
0
IT
(-1) +
o •••
1
0
1
k=1
(1 - k)
390
CHAPTER 7
~
(Contd)
i=l
REMARK: When, for i f 1, a.
0, we recover the well known equality:
l
(4):
If a e k( (t)), the field of "meromorphic" [that is having
only finitely many terms of negat i ve order] formal series, we can wri te: p-l
L
a =
bo
a
(k)
with a(k) =
L
n:=kmodp
If b is another element of k«t)) we see that:
L
(ab)(k)
a(h)b(h'),
h+h':=k(modp)
which can be written: a a a
(0)
a
(1) (p-l)
a
(1)
(ab ) ( 0 )
(0)
On the other hand:
L
n:=kmodp
k
L
n:=k
If we look for a non-zero solution of the equation: ab - SJb
= 0,
in k«t)),
we are led to solving the equation:
391
CONGRUENCES MODULO p: MODULAR FORMS
(0)
a (1) a a
a
(p-l)
a
(1)
1
a(O) -
O.
(p-l)
a (0)
-
b(p-l)
(p - 1)
Since k«t)) is a field, this system has a non-zero solution if and only if its determinant vanishes, which by Question (3) gives . (0) _ (p-l) _ (p-2) _ (1) _ p . (wlth a - ai' a - a2 , a - a 3 , ••• , a - a ).
0
p-l
~
i=o
(a(i))p _ a (0)
aP - a
(0)
that is to say, we finally obtain: np ~ aPx n
aP
a
(0)
np ~ a npx .
REMARK: The coefficient a O satisfies a~ = a O and therefore is in F , that is to say it can be considered as an integer; it corresp
ponds to a factor xn in b, for we see that: !Jb
n + b
in particular, the solution is not unique. SOLUTION 7"7: (1):
Using the homomorphism between the product
of matrices and the composition of fractional transformations, the equality required:
- z1 + m P
can be written:
mz - 1 pz
au + b au + d
with u
z + m' p
392
CHAPTER 7
Solving for
we obtain
[: :
m
p
-mm~
- 1
1
-m'
In order that b be an integer it is necessary that m' be chosen in such a way that mm' + 1 is divisible by p.
m,
As 1
~
m
p - 1,
the class of m in 'lZ/p'lZ, is invertible, and there exists an
unique m', 1 ~ m' ~ p - 1, which represents the class ~1
(2):
As F if a modular function, F(z + 1)
¢
P
(z
+
1)
pk-1F(pz +
p)
+ ~ p-l L F (z + ~ + m) , p m=O
and on setting n = m + 1, ¢
p
(z
+ 1)
Pk-1 F(pz) +
~
1 F(~E.) .
P n=1
p
¢ (z + 1) p
F is
gives:
modular of weight
¢
k,
we have
F( -;-)
p
in
F(z), and we
obtain:
As
~
(z).
393
CONGRUENCES MODULO p: MODULAR FORMS
by using
part (1) above.
As au + d
z,
When m takes all the values between 1 and p - I, m' takes all the values between 1 and p - I, and we have:
k z ~ (z). p
If:
(3):
A e2illnz n
F(z)
we have: ~ (z) p
=p
k-l
L n:.mo
A e 2 i llnpz +! n p
L
n~n 0
by the definition of the sum of a series. we find:
p-l
L
m=O
e 2illn (z+m)/p
e2illnz/p
p-l m L
m=O
~
.
A
p-l
L
n m=O
e
Setting
2i lln( (z+m)/p)
~
e 2i1ln/p ,
CHAPTER 7
394
p-l
L ~m has the value zero if ~ + 1 (that is to say, if p m=O does not divide n) and has the value p if ~ = 1 (that is to say, if p divides n). We then obtain:
The sum
p
(z)
with:
If P does not divide n, the second term alone gives: a
n
= Apn
(1)
If P does divide n, we obtain: a
(2)
n
Using part (2), if F is a modular form we have nO
=0
also a modular form; if F is a cusp form we have nO
and
is also a cusp form. (4):
We have seen that
Since the vector space of cusp forms of weight twelve is onedimensional,
p
and
~
are proportional.
In Formulae (1) and (2), by setting A
n
a
1
= T(p),
and since T(l)
P
= 1, we deduce from this: and
a
n
T(ph(n),
T(n)
we obtain:
CONGRUENCES MODULO p: MODULAR FORMS
395
which gives us the relations: Ifp does not divide n:
T(pn) = T(p)T(n),
Ifp does divide n:
p 11 T[n)
a
p , a
For n
P
+ T(pn)
(3)
= T(p)T(n),
(4)
1, relation (4) gives:
~
a
T(p)T(p ) - p
11
T(p
0.-1
(5)
).
Let us now prove by induction on a the formula: a
T(p s)
= T(p a )T(S)
if P does not divide s;
(6)
= 1 this is given by relation (3), and for a = 0 it is ob-
for a vious.
Let us assume it to be true for a and a - 1 with a
and let us prove it for a + 1.
T
Formula (4) with n
= pax
~
1,
gives:
( P0.+1 S )
=
T(p
0.+1
)T(s),
when we use Formula (5). To prove the multiplicativity of T, let n
a.
= II p.l. . l
"
Let us
calculate T(n) by applying Formula (6) successively to all the prime factors of n, we find:
which proves that
T
is multiplicative.
396
CHAPTER 7
(5):
We know that:
with: and
If we set:
Formula (1) gives for the first coefficient of
To show that
~
p
~
p
:
and F are proportional we must show that:
°k_lCp)F,
~p
so for all n, set:
If n
If n
0, Formula (2) gives:
>
°and if pln,
Formula (1) gives:
taking into account the multiplicativity of the function
CONGRUENCES MODULO p: MODULAR FORMS
n
1+
397
ok_len).
If n ~ 0 and if pin, Formula (2) gives, with n an
k-1A
P
nip
+A
pas, p%s:
pn
By the definition of the function 0k_l we have:
°k-l(p
a-l
)
and substituting these in the square brackets we find: a
n
The functions Gk and
~p
are therefore proportional.
Recall that SL 2 w,/ 4'lZ) is the set of 2 x 2
SOLUTION 7·8: (1):
matrices with entries in'lZ/4'lZwith determinant 1 (mod 4). SL 2 W,/4'lZ)/±1 the two matrices
[a
[-a
In
b] and -b] are identical. e d -0 -d An elementary calculation allows us to establish the following
multiplication table: d
e
aid e b bel d e del a e 0 abe d e boa 1
e a b
1
abe
1
d
398
CHAPTER 7
for example:
ed • [
2
I
o
I
I
I
i.
Therefore our group is isomorphic to S3' The relations [JT]3 = I and J2 = I with I =
(2):(a): I
o
: ] are already known.
From this we immediately deduce:
and consequently:
From this it follows that for all n there exists
~(n)
such that:
Since every element of r (the modular group) can be written: n
n
n
y = T IJT 2 ••• JT r,
that is to say:
CONGRUENCES MODULO p: MODULAR FORMS
399
and therefore that: n -~(n
Ye r T r 4
r-l
-~(
•••
» '
T4 e r 4' 'f b e 1ongs to one 0 f t h e r 4Ti (."Z- = 0 '" 123) . . and , s 1nce Let us show that the sum is direct, that is to say that there nl
do not exist y l ,y 2 er 4 and n l I n 2 (mod 4) such that ylT n l -n 2
=
y~lY2 would belong to r 4 , it there-
In this case T
for is enough to show that T,T 2 ,T3 are not in r 4 . Now, if we reduce our matrices (mod 4) we see that J becomes the a of Question (1), T4 the i and TJT- l
= [1
1
-2] the b. -1
By reduction (mod
4) r 4 therefore becomes a subgroup of the group in part (1) (since it contains a and b it is the same group as in part (1)). Now, it is easy to establish that the latter contains neither the reduction of T, nor of T2, nor of T3.
[This would allows us to con-
struct the 48 matrices of 3 6l
i=o (2) : (b) :
(group of 1)
? 1).
x {±1}].
Let V be the "classical"
fundamental domain of r Izl
i
xT
(-!
~
Re(z)
~
!,
The fundamental domain of
r 4 (a subgroup of r) is made up of a certain number of "copies" of V.
In
particular, by the decomposition above, every point of H is congruent by r 4 to a point in V u TV U T2V U T3 V• We now make the triangulation indicated in the Figure above, where we show how the sides are identified by the action of [4 [the other points are not equivalent, for example the transformations of r which send j into j + 3 are T 3 , T3 (T- 1J) and
400
CHAPTER 7
T3 (T- 1J)2, which are, mod r 4 , equivalent to T3 , those which send
i into i + 1 are T and TJ which do not belong to r 4 ]. We then count 5 vertices: j ('" j + I, j + 2 and j + 4), j + 3, i, i + I, i
oo '
6 vertical sides and 3 horizontal sides (j,i)
["'(i,j + 1)], (j + l,i + 1) ["'(i + l),(j + 2)] and (j + 2,j + 3) ['" (j
+ 3),(j + 4)], and finally 6 triangles.
Therefore we have
a surface of genus g with: 2 - 2g - 5 - 9 + 6
2,
that is to say, of genus zero. (3):(a):
q is the local uniformizer at the point ioo of the
fundamental domain of r 4 . Since r 4"H is of genus zero, there exist functions having a simple pole at the point at infinity, and holomorphic elsewhere, moreover; they are of the form AH +
~
if H is anyone of them. The two conditions correspond to the particular functions having a residue ei~/4/(21:2) (which fixes A) and zero constant term (which fixes
~),
it is therefore uniquely
defined. (3):(b):
By Question (2)
i~
is clear that
r, the full
modular group, permutes the four functions H(z + h) h h' h' [H(yT (yz» = H(y'T z) = B(T z)]. The polynomial in x,
3
II
h=O
H(z + h», therefore has coef-
(x -
ficients which are functions of
= H(Thz)
z
invariant under r and having
poles only at infinity, they are therefore polynomials in j (z). On the other h an d we h ave e (i~/2)(z+h) = q (i~/2)h e = q (~)h ~ , that is to say:
~(x,z)
CONGRUENCES MODUW p: MODULAR FORMS
ilt 4
e / 1 -----.-n are
[by noticing that the
401
the roots of X4
2/2q
1
--].
64q4
Now, it is known that:
~(z)
~+
q
744 +
L a(n)q4
1 4" +
0(1),
q
~(z)
therefore a non-constant polynomial in
cannot be
o( ~)
and q a constant could not be 0(1) without being zero, from which it follows that:
~(x,z)
:4 j(z),
x4 + ax 2 + bx + a +
where a,b,a are constants. By the construction of the domain we see that:
= H(j
H(j)
+ 1)
that is to say that
= H(j ~(x,j)
+ 2),
has a triple root; now,
~(j)
0,
therefore: (x -
ex) (x
-
. x 3 must b e zero, ex an d as t h e term In
3
6) ,
= -36.
On the other hand we see that H(i + 2)
say, that
~(x,i)
has a double root; now,
hence (x + 36)(x - 6)3 + 27
°must
= H(i ~(i)
+ 3), that is to
64x27
= 1,728,
have one double root in com-
mon with its derivative [3x + 96 + x - 6](x - 6)2; this can not be 6 (27
+ 0),
therefore it is x
= -26,
hence it is necessary to have ( - 26 + 36)(26 - 6)3 + 27 = 0, that is to say 64 1; it remains to choose a suitable fourth root (the choice of S corresponds to the four functions ±H,±iH).
402
CHAPTER 7
We draw the graph of the function y
y
= (x + 3)(x - 1)3 (see the Figure
on the right). line j
On the vertical half-
the function
~ 00
~(z)
real values decreasing from
takes
°to
Imaginary Branches
the function (±H,±iH), a root of
i4 '
(H' + 3)(H' - 1)3 = will be such that (H'(z) + 3)(H'(z) - 1)3 will increase from
°to
+00.
On the graph we see that the branch arising from the simple root
= H'(z)
is the branch where x
varies from -3 to
_00;
by the form
of the fundamental domain this branch corresponds to the halfline (j + 3)
The root of the equation:
~ 00
(H' + 3)(H' -
j - 64
1)3
is therefore real and negative on this half-line; now, on the latter, q
=e
(in/2)«5/2)+ih)
therefore if q
H(z)
~
e in / 4
5in/4
with a €lR +,
0,
'\,---x
212
ae
e
-5in/4
1 a
x -
and this number certainly belongs tolR (x,z)
f
Hand
(x + 3)(x - 1)3 + j~:)
(4):(a): main of
therefore H'
H is an univalent function on the fundamental do-
4 , therefore the condition H(w)
= H(aw;
8)
is equival-
ent to: There exi,,, [:
that is to say:
:
1'
f
4 which transforms
aw + 8 . Into w, 0
CONGRUENCES MODULO p: MODULAR FORMS
403
which leads to: aaw
2
+ (as + do - aa)w - ob - as
O.
The discriminant of this equation is:
( - as - do + aa)
2
+ 4aaob + 4aa8a
(aa + as + do)2 - 4aaod + 4aoa (aa + as + do)
because ad - ba
= 1.
We can always assume
a~ 0
2
- 4ao,
[otherwise take
even a > 0, for otherwise w would be real. longing to
w
= aa
[-a -b] ] and
-a -d The solution w be-
H is then: - as - do + il4ao - (aa + as + do)2 2aa
in order that this be
T
1 +
+t
U 2 + ili5 lOt is therefore neces-
sary that we have:
that is to say, with the conditions imposed: D + 1 - (aa + as + do)
which implies:
2
= D(aa) 2 ,
404
CHAPTER 7
aa
1,
aa +
as
+
=±
do
that is to say a
1,
= a = 1,
since a,a
~
0, and moreover it is nec-
essary that: 1 +
2i = aa -
as - do
D+ 1 -4-
and
Hence we must find a matrix [: :] such that: a
= 1,
a+B+dx~=±1 4
ad - b
'
1,
~ = 1 + 2i, 4
a - S - d
which gives:
a
f + i, 1i,
f - i, 1-
1
1 -
i;
D + 1.1S pr1me . now, ~ 4 -= 1 (mod 4) , t h ere f ore --4--to 4 , t h ere f ore
the second relation has a solution in d and S with S a multiple
°< S
D + 1), therefore for every i the problem is solvable in r, it is enough to check that the solution is in r 4 ,
of 4 (and
<
therefore it is enough to check if, by reducing mod 4 the solution belongs to the group of 1); since S a = 1, which gives for:
f
a =
1 +
1i,
i,
d,
=0
(mod 4) we have:
1-: _ t
(mod 4)
b
ad - 1,
CONGRUENCES MODULO p: MODULAR FORMS
i -
o
(mod 4),
the matrices
i - 1 (mod 4),
the matrices
i - 2 (mod 4),
the matrices
i - 3 (mod 4),
the matrices
We see that only when i
405
: ~O' I [: I [: [: [: [: I [: ~: I [: [
~l
or
~31
-2
~31
or
-1
~71
-1
- d
~l
-3
-2
or
~71 -3
- e,
or
=2
do the solutions belong to r 4 , in this case the solutions certainly belong to r 4 , because by reducing (mod 4) we have a matrix equal to d or e. (4):(b):
We are going to prove that the coefficients of
the given polynomial in x are invariant under r 4 , as it is clear that these coefficients are ho10morphic in H, from which it follows that these are polynomials in H(z), which proves the result. Let us then consider the set E of 2 x 2 matrices with integral entries and determinant D ; 1 (which is congruent to 1 modulo 4), and which are, modulo 4, congruent to an element of the group of 1).
We shall say that two elements A,B e E are equivalent if there
exists y e r 4 such that A = yB; we are going to prove the each class contains an unique representative of the form:
so, in effect, A
[:
:] e E, then a
AY +
y'
with
Iy'l
<
Iyl.
406
CHAPTER 7
Now in r 4 we have at our disposal four types of matrices:
and:
0
-1
1
4p
[:
[:
2 + 4p
4p + 3
I
1 - Bp 4p + 1
i
[
-1
- 3 - 4p
1
4p + 3
I'
,
that is to say that for one of them we will have:
a - "Ay
which means every matrix of
E
is equivalent to a matrix having an
entry y whose absolute value is strictly less than the one you start with, and by iterating this procedure we will obtain an equivalent matrix having an entry y
= 0;
if necessary we can always assume that a otherwise
ao
D; 1 );
multiplying by [-1
0]
o
>
-1
0 (non-zero,because
but since:
we see that we could just as well choose 0
~
S
<
40; finally,
the matrix obtained must be congruent modulo 4 to one of the elements ±i,±a,±b,±c,±d, or ie, and it can only be ii, hence S (mod 4), Let us show the uniqueness.
Let us assume 0'
with yer 4 ; multiplying both sides by [ 0
-
[ ao :u 1
s' ]
a'
we find:
[ao'
=0 u :,'
1
407
CONGRUENCES MODUW p: MODULAR FORMS
y
o a
= a'o',
but since ao
that is to saYar
0 = 8' '
these two numbers,
the product of which is one, must have value ±l.
= a'
positive one finds a
therefore
S - S'
0'
As a,a' are
must be integral, and
by reduction (mod 4) one sees that it must be a multiple of 4. Since 0
S
~
<
40, and since (4,0)
It is clear that H(Bz).
II (x
A
= 1,
we see that S
is equivalent to
B
= S'.
if and only if
H(Az)
Therefore the polynomial can be written in the form - H(Az)),
where the product runs over the equivalence
classes; but if A hence Ay
= y'By],
~
B it is obvious that Ay
~
By [in fact A = y'B,
that means right multiplication by y (an invert-
ible operation) permutes the equivalence classes, which completes the proof. (4):(c):
The roots of the first equation are B(T), H(T + 1),
H(T + 2) and H(T + 3), by Question (3). Let us assume that x is a root of the equation Since
H(z)
is univalent, there exists
action of r 4 ) such that x
= 0,
= B(z),
z
~(x,x)
= o.
e H (unique up to the
therefore we have
~(B(z),H(z))
that is to say, there exist a,S,D, satisfying the conditions,
such that:
Now, we have seen in Question (4)(a) above that this is true for
+ i if and only if i = 2 (mod 4), from which we obviously have that B(T + 2) is the unique solution common to the two equations.
z
=T
CHAPTER 8
Quadratic Forms
INTRODUCTION
The goal of this Chapter is to determine all the rational numbers which are represented by a quadratic form f with rational coefficients. (A QUADRATIC FORM K REPRESENTS Y
eK
fCX 1 .X2 ••••• x n ) with coefficients in a field
IN K
if there exist a 1 .a 2 •...• a n e K. not all
zero. such that fCal.a2 •...• an) 8.1
= y).
NOTATIONS
We denote by V the union of the set P of prime numbers and the symbol 00. and the convention moo = ~ is adopted. Let ve
V
by:
and a. S e m*. v
1
Ca.S)v
= -1
The
HILBERT SYMBOL
Ca. S) v is defined
° 1Of ax 2 + Sy 2 represents 1 1n
my.
otherwise.
222 Let f = a1x 1 + a 2x 2 + ••• + anxn • where a i e my. be a non-singular diagonalised quadratic form in n variables. We denote by
dCf) the determinant of f and by by:
E v
408
HASSE SYMBOL.
defined
409
QUADRATIC FORMS £
IT
d(f)
8.2
i=1,2, ... ,n
a.
1
and
1 £
v v
n
(f)
-(
(f)
(a.,a.) 1
]
V
ifn
>
ifn
1
2, 1.
PROPERTIES
The following two properties of the Hilbert symbol: and allow, in all cases, the calculation of the values of the Hilbert symbol, since we know that:
if peP - {2},
(p,
if
d p
£
is an unit of ID p '
£0 eZl, £ " £0 (p ) ;
ifp
2,
1
if
£
and n are units of ID ; p
if
if P = 2,
and n
are units of £
v
£
(a,S)
=
11 -1
= £0(4),
ID 2 , £0' nO eZl, n = no (4);
if a or S > 0, if a
<
O,S
<
0.
THEOREM (I): Let f be a non-singular diagonalised quadratic form
in n variables~ with coefficients in ID ; in order that f represents p zero in ID it is necessary and sufficient that one of the following p
410
CHAPTER 8
aonditions hold: (1):
n =
(2):
n
(3)
n = 4 and
and'-d(f) is a square in
2
111 ; p
= 3 and (-l,-d(f»p = Ep(f); either: d(f) is not a square in IIlp3 or:
d(f) is a ffquare in IIlp and €p (f) = (~l,-l)p;
(4): n
~
5.
THEOREM (II): Let f be a non-singular diagonalised quadratia form
in n variables 3 with aoeffiaients in 111p ; in order that f repreit is neaessary and sUffiaient that one of the p p following aonditions hold:
sents a e 111* in 111
(1)
n =
(2)
n
= 2 and (a,-d(f»p = Ep(f);
(3)
n
=3
1
and ad(f) is a square in 111p; and - either: ad(f) is not a square in IIl p3 - or:
ad (f) is a square in 111 and p
e: (f) = C-l,....d(f)) ; p
(4): n
~
p
4.
THEOREM (III): (Hasse-Binkowski): A quadratia form with rational aoeffiaients represents a e 111 in 111 if and only if it represents a in 111v for all v e V. PRODUCT FORMULA:
If
ve V and:
IT
veV
a,b e 111"', we have (a,b) v
1 for almost a11
(a,b)v = 1.
The theory of quadratic forms may be found in the following works: [Bor], (Ser 1].
QUADRATIC FORMS
411
PROBLEMS
EXERCISE 8·1: Let f 2
= a1x 12
2
b 2x 2 + ••• + bkx k be two diagonal non-singular quadratic forms with p-adic coefficient. Write f .;. g for the f01'l7l in n + k vari-
ables defined by:
Show that: £
p
EXERCISE 8·2:
m7
£ (f)£ (g)(d(f},d(g» . p p p
(f .;. g)
Determine all the elements of
by the form 3x
EXERCISE 8'3:
2
+
7y
2
m7
represented in
In this problem a e?2: is a non-zero integer that
is written in the form:
where a, S by S.
~
0 are integers and b e?2: is divisible neither by 2 nor
412
CHAPTER 8
(1): Characterise by conditions on a,B,b those numbers a such that 30a is the square of a 2-adic number. Give the list of those numbers a whose absolute values are less than or equal to 50. (2):
The same questions, replacing the field
numbers by the field
m5
2-adic
of 5-adic numbers.
In the remainder of the problem f denotes the quadratic
(3):
form in three variables on
X~
f
m2 of
3X~
+
-
m:
lOX;,
and d denotes its discriminant. Calculate (4):
E
v
and the Hilbert symbol (-l,-d)
For which
V
v
for all
Ve V.
e V does f represent every element of
~*
in mv? (5):
Describe the elements
of~
which are not represented
by f in m, and give the list of thos whose absolute value is less than or equal to 50. EXERCISE 8-4:
This problem studies the rational integers repre-
sented in
the quadratic form:
mby
5X~ - 7X~.
f (1):
Does the form f represent 0 in m?
(2):
Show that the form f represents the non-zero rational
integer a in (5,-7)
p
mif
and only if the two Hilbert symbols (a,35)
are equal for all odd prime numbers p.
p
and
(3): Assuming the rational integer a to be non-zero and to be square-free, characterise by conditions on Legendre symbols
QUADRATIC FORMS
413
those a's that can be represented by f in
m,
distinguishing the
following four cases:
a is divisible neither by 5 nor by 7; a is divisible by 5 and not divisib1y by 7·, a is divisible by 7 and not divisible by 5; a is divisible by both 5 and 7. (4):
What is the list of rational integers between -14 and
+14 which are represented by f in
m?
414
SOLUTIONS
SOLUTION 8·1:
We argue by induction on n.
If n
=1
there cer-
tainly holds:
(1)
so:
Let us assume, therefore, the property to be true for forms in n - 1 variables.
We then write f = ax~
ona1ized form in n - 1 variables.
by Equation (1). g
p
(f'
+ g) =
On the other hand:
+ f',
Then we have:
By the recurrence property g
p
(f')g (g)(d(f'),d(g» p
where f' is a diag-
p
.
QUADRATIC FORMS
415
So: e:p(f.j. g)
SOLUTION 82 .:
i)
e: (f')(a1,d(f')) e: (g)(a1d(f'),d(g)) p p p p
7. not square In . 1P7 · not represente db O IS , ecause - "31S
since V 7 [ co 1. By Theorem (II) y is represented if and only if: (1)
Or, in 1P7 y
y n
admits an unique expansion in the form: i
7 (a o + 7a 1 + ••• + 7 a i + ••• ),
where: n ez-::, a i e {O,1,2,3,4,S,6}, a O
+ o.
Let us set y = 7ny', where y' is a 7-aqic unit.
since y' is a 7-adic unit congruent to a O modulo 7. If n is odd, one obtains similarly:
But:
If n is even:
CHAPTER 8
416
(_1)~(7-1)
-1,
and condition (1) becomes:
(a70)
-1
if n is even,
a0 (71 J
1
i f n is odd,
or: aO
e {l,2,4}
i f n is odd,
{
a o e {3,S,6}
if n is even.
The elements of m7 represented by the form 3x 2 + 7y2 are therefore those whose expansion have the form: y = 7 2k (a o
+ 7a 1 + ••• + t na n + ••• ) ,
where: aoe
{3,S,6}, an e {O,1,2,3,4,S,6} for n
~
1,
or of the form: 7 2k +1 (a o + 7a 1 + ••• + 7na n + ••• ) ,
y
where: aoe
{l,2,4}, an e {O,1,2,3,4,S,6} for n
SOLUTION 8'3: (1):
m2 it
In order that 30a
~ 1.
= 2a+1S8+13b
is necessary and sufficient that we have:
be a square in
QUADRATIC FORMS
a odd and 5
417
a+1
3b square in
m2 .
This second condition is equivalent to:
a odd
and 3b square in
m2 ,
or:
a even
and 15b square in
m2 .
A necessary and sufficient condition for a unit x of a square in
m2 is
m2 to
be
that:
x :: 1 (mod 8).
As 3.3
=1
(mod 8) and 15(-1) - 1 (mod 8), we finally obtain
the following characterisation: A necessary and sufficient condition for 3Qa to be a square in
m2 is
that we have:
(i): a odd; and: (ii): either: or:
a odd
a even
and b :: 3 (mod 8), and b :: -1 (mod 8).
This allows us to obtain the (a,a,b) corresponding to such integers a, with if:
a
lal
=1
~ 50:
a
0,
b
a
1,
b
a
2,
b
-1,
-17,-9,-1,7,23,
=3,
a
=3
a
0,
b
-1,
a
=5
a
0,
b
-1.
CHAPTER 8
418
lal
The set of numbers a
+0
such that 30a is a square in
m2 ,
with
~ 50, is:
{-50,-34,-32,-18,-8,-2,14,30,46}, (2):
In order that 30a
= 2a +15 a+1 3b
be a square in m5 it is
necessary and sufficient that we have:
a odd
and 2a +1 3b square in
m5 ,
This condition is equivalent to: a+l 2 3b square modulo 5,
or
a 2 b square modulo 5,
Now the only squares ofF 5 , the field with five elements, are +1 and -1; on the other hand, if a is even: 2a
= ±l
modulo 5,
and this finally leads to the following characterisation: A necessary and sufficient condition for 30a to be a square in m5 is that we have: (i):
a odd;
and: (ii): either: a odd and b or:
a even and b
= ±2 = ±l
mod 5, mod 5,
This permits us to obtain the (a,a,b) corresponding to such integers a, with
a=1
lal
a a a
~ 50:
= 0, =1 = 2,
b
= -9,-1,1,9,
b
-3,+3,
b
-1,+1,
419
QUADRATIC FORMS
and the set of numbers a such that 30a is a square in
lal
Ws with
~ SO is:
{±5,±20,±30,±45}. (3) :
E
V
(1,3) (1,-10) (3,-10) v v v (3,-10)
v
as 1 is square in Wv for all v; on the other hand, inE: E""
= (3,-10)..,
1,
as 3 is positive. For p an odd prime, if u and v are two p-adic units (u,v) the value 1, and from this it follows that: E
P
If P
p
1 for p ; 2,3,5. 3:
(-~o)
(Legendre symbol)
-1,
as -10 is not a square modulo 3. If P
E
S
= 5: = (3 ' -10) S = (~) = -1 ' S
as 3 is not a square modulo 5. By
t he Product Formula we obtain
E2
1, and so we deduce
that: E
v
1 for all v different from 3 or 5,
has
CHAPTER 8
420
and:
The discriminant d of f is equal to -30:
= (-1,30) v ,
(-1 , -d) v
(-1,30)=
in~,
If P
f:
=1
as 30 is positive. (-1,30)
2,3 or 5
P
= I,
for -1 and 30 then are p-adic units. If P
=3
( -1,30) 3 = (-31)
-I,
for -1 is a square modulo S. If P
=5
(-1,30)5
I,
for -1 is a square modulo S. We then obtain (-1,30)2 = -1 by the Product Formula, which allows us to conclude that: (-1,30)
=1
v
for all v different from 2 Or 3,
and: (-1,30)2
(-1,30)3
-1.
= =: The coefficients of f do not all have the same sign, therefore f represents in~ every element of~. (4):
Fop v
Fop peP, E
p
p
+2,
p
+ 5:
= <-l,d
,
We have:
421
QUADRATIC FORMS
hence, by Theorems (I) and (II) f represents in of mp ' therefore every element of ~ . For p = 2 and p = 5: We have:
=-
e: (f) p
(-i,-d(f»
p
mp
every element
,
hence by Theorems (I) and (II) f does not represent zero in nor the elements a of~'" such that 30a be a square in m •
mp
p
f does not represent zero in
m,
for f does not represent zero in m 2 and m 5; and by Theorem (III) f represents the integer a non-zero in mif and only if 30a is not square in m 2 and m5 . The integers lying between -50 and 50 not represented by f are therefore: (5):
-50,-45,-34,-32,-30,-20,-18,-8,-5,-2,0,5,14,20,30,45,46. SOLUTION 8·4: (1):
square in
m.
f does not represent zero in
msince
~ is not
By Theorem (II) f represents a in m if and only if p (a,35) = (5,-7). Now, (a,35)"" = (5,-7)"" = 1 for all a em. By p P the Product Formula: (2):
IT
peV
(a,b)
P
=1
i f (a,35)
also we will have (a,35)2 (a,35)
p
= (5, -7) p
p
= (5,-7)
P
for all p e V - {2},
= (5,-7)2' The condition:
for all peP - {2}
(1)
is therefore necessary and sufficient for f to represent a in all the mv for v e V, and therefore, by Theorem (III), for f to represent a in m. (3):
Let us spell out condition (1); we have:
422
CHAPTER 8
(S,-7)
p
(S,-7)5
=1
+5
if P
=
and 7,
and
(5,-7)7
= (5,7)7 = <~) = -1.
Therefore the condition is equivalent to: (a,3S)p = 1
for all peP - {2} - {5} - {7},
{
(1)
(a, 3 S)
Now i f
P.
P
= -1
for p e {5, 7} .
{2, 5, 7} and i f p does not divide a, then (a,35)
Therefore (1) is further equivalent to: (a,35)
=1
(a,3S)P
= -1
But i f pia and (a,3S)
p
P.
(1)
i f pe{5,7}.
{2,5, 7}, then, since a has no square factors,
= (p,3S)
p
= (3S\ ,
p
J
and finally we obtain the condition: (a) :
(a,35)
{
(b) :
=1.
i f P divides a and p+{2,5,7},
{
P
P
(a,3S)
p
P
=1 = -1
for all p dividing
a
and
P.
{2,5,7}, (1)
for p e {5, 7}.
Let us make condition (lb) explicit in the different cases of the problem:
a i8 divi8ible by neither 5 nor 7:
and the conditions are therefore:
423
QUADRATIC FORMS
(a): { (b) :
(~5) = 1 for (~)
= -1
and
all p dividing a and p+ {2,S,7}, (2a)
(~) = -1.
a is divisible by S and not divisible by and 7ia' ~ Sia': then:
7~
i.e.~
a
= Sa'
(-1)(1)(~) = - (~), and:
(a,35)7
= (5a'
,7)7
=
(f)[c;)
the conditions are therefore:
{
(a): (b) :
(~5) = 1
(~) = 1
for all p dividing a and p+{2,S,n, and
(a;)
(2b)
= 1.
a is divisible by 7 and not divisible by 7ia'~ Sia': then:
S~
i.e. a
( -1)
(c;) (a,35)5 = (7a'
,5)5 = (~)(a;)
the conditions are therefore:
_ (a:) ,
= 7a'
(a; )(
-1 )
and
CHAPTER 8
424 (a) :
(
{
~5) = 1
for all p dividing a and p 4 {2 ,5, 7},
(~) = 1
(b) :
and
(2c)
(a~) = -1.
a is divisible by 5 and 7~ i.e.~ a ~ 35a' and 51a'~ 7tb': then:
= _
[a;) ,
and the conditions are therefore:
(~)
(a): {
[a;)
(b):
(4):
= 1 = 1
for all p dividing a and p4 {2,5,7},
(a;)
and
(2d)
= -1.
Let us look for the prime number p different from
2,5,7 and < 13 which satisfy the condition
(~) = 1.
Therefore if: p = 3 p = 11 p = 13
(t)(t] (i~) (t](*] (i;) (t] (~1) (~5)
(-1) = (-1)(-1)(-1)
(-1) = -1,
= (-1)(-1)
1.
-1,
We have:
QUADRATIC FORMS
425
Therefore the only prime number p
~
13 different from 2,5,7 satis-
fying (3;) = 1 is P = 13. Hence amongst the integers with absolute value less than or equal to 14 not represented by f we find {±3,±6,±11}, besides these there are numbers multiplied by a square
from~,
that is
to say {±3,±6,±11,±12}. After examining the other numbers we see that for them the conditions of (a) are realised. the conditions of (b).
It remains, therefore to verify
One then sees that ±l is not represented,
therefore ±4,±9 are not represented, 2,-5,7,8,±10,-13,±14 are not represented, whilst -2,5,-7,-8,13 are represented. In conclusion, the list of integers lying between -14 and +14 representable by f in {-8,-7,-2,5,13}.
mis:
CHAPTER 9
Continued Fractions
INTRODUCTION The problem is to determine the best approximations of a real number a by rational numbers.
DEFINITION: The fraction E., 'With (p,q) e q
APPROXIMATING FRACTION,
or a
CONVERGENT,
2lX:W"
is called a
BEST
of the number a if q >
1~
and if, furthermore: For all (p' ,q') e
2l
x]N"',
q'
<
q =>
Iqa - pi
<
Iq'a - p'l·
It is easy to see that the definition allows us to classify Pn Pn the convergents of a as a sequence -- which tends to a, with -qn qn is in lowest terms.
2.1
CALCULATION OF CONVERGENTS We define the sequence (a ) of partial quotients of a by the
formulae:
n
426
CONTINUED FRACTIONS
et
n
an
427
1
with an > l, a OeZl and an e:N if n inates at index n if [et ] = et n
~
l, and the sequence an tenn-
n
The convergents are then calculated by the following recurrence relations:
if n ~
2.2
°with
(P-2,q-2)
(0,1),
(P-l,q-l)
<1,0) .
1
FORMULAE From the recurrence relations we deduce the formulae: (_l)n+l,
et
Pnet n +1 + Pn - 1 qnetn+l + qn-l
and: 1
THEOREM (I): The sequence (a ) is finite (resp. periodic starting n
from a certain index) if and only if et is rational (resp. quadratic) . THEOREM (ll): Let et
a + bid, with a,b e \11 and de \11>", be a quad-
428
CHAPTER 9
ratic number, and let o(a)
=
a - blci be its conjugate.
The se-
quence (a ) is periodic starting from index 0 if and only if a and -1
n
<
o(a)
<
0, that is to say if a is reduced.
THEOREM (III): (Lagrange's Criterion): If the pair
and a satisfy
2.3
> 1
1a - £1q . . ~, 2q
BIBLIOGRAPHY [Har] , [Niv].
then
£q is a
(p,q)e~x:N*
convergent of a.
CONTINUED FRACTIONS
429
PROBLEMS
EXERCISE 9·1:
Let a be a quadratic irrational whose expansion as
a continued fraction has period of a and
P n
Pn
Let -- be the convergents
B.
qn
the vector with components (p ,q ) in~2 n
n
Show that there exists a linear transformation T and
nO e:N
such that: P {
n+s
detT
EXERCISE 9·2:
Let a
= [aO,a1, ... ,ak_i]
which is periodic starting from index n fine two fractions in lowest terms ~ and
v
u - =
and
v
and: q
o
ifk
1.
be a continued fraction
= 0 with E by: q
period k.
if k
De-
~
2,
430
CHAPTER 9
Pn Let -- be the convergents of a, and P the vector in the plane qn n
B2 with components (Pn,qn)' (1):
Find a recurrence relation between P(n+2)k' P(n+l)k
(2):
Show that there exist real numbers
A,~,p,C,C'
such
that the following inequalities are satisfied for all m SN*:
EXERCISE 9-3:
Let d
>
1 be a rational number that is not a per-
fect square. (1):
Show that the development of I.d has the form:
with: and (2):(a):
a.1
a B-1. for i
1,2, ... ,s - 1.
Show that for n > 1 we have: (1)
(2):(b):
Show that for 1
~
m ~ ns we have: (2)
CONTINUED FRACTIONS
431
(3)
where the components of Pn inF 2 are (Pn,qn) and those of Qn are (dq ,p ), n
n
(2):(c):
From this deduce that:
(4)
2 2 P ns- 1 + dq ns- I'
(3):
Apply Newton's method to the calculation of I.d; that
is to say, set:
Ps - 1
Calculate xn when we set x 0 = - - and give an upper bound for qs-l the error made by replacing I.d with x , n
EXERCISE 9-4: DEVELOPMENT OF e AS A CONTINUED FRACTION (1): u
v
With a,n,k being positive integers, a 2 n (n + k)! k! (2n + 2k) !
n+k
n
I
k=O
Show that lJ
n
(2n
u n,k' lJ
n
+ 0,
set:
1
2i<" , a
av lJ
n
n
Vn+l
satisfies the recurrence relation:
+ l)a +
1
lJn +1
From this deduce the expansion as a continued fraction of
432
9
CHAPTER
coth
i as well as that of : ~ i (2):
Taking b.a as strictly positive integers. g
n
as a se-
quence of strictly positive integers (with the exception of go which may be negative) we set:
[go(ba +
a' p
1) +
b + a.g1 (ba +
1) +
b +
0 ••••
].
p'
Let -..E. (pesp. -..E. ) be the n-th convergent of a (pesp. a'). qn qn Show that:
q3n+2
= (ba
+ l)q~.
Calculate a' as a function of a.
From the expansion of : ~
i
as a continued fraction deduce the one for e.
EXERCISE 9·5: THE POINT-OF-INTERROGATION FUNCTION The function we are going to study was introduced by
~1inkow
ski under the name of "point of interrogation". here we shall denote it by
is a. numeric function defIned on the real numbers in
[0.1] and satisfying the conditions: cp(o)
cP
= o.
cp(l)
l',
(1)
p')
+ (p q + q'
for any adjacent fractions
(2)
£q
n' q
'7 (Le., p'q - q'p
= ±1);
CONTINUED FRACTIONS
433
q> is continuous.
(3)
Show that every fraction in lowest terms ~ in (0,1) + ' , can be written uniquely in the form PP, where E , P, are two q+q' q q adjacent fractions in [0,1]. Deduce from this that conditions (1):
(1) and (2) completely determine
q>(~)
Pn
Let -- be the n-th convergent qn
(2) :
of
~ and
set q>(::J
un·
Show that:
Un
.
[1 - ! )Un2 n
1 +
+ un_ 2 n
2·
Deduce from this an eXp'ression for quotients an and write (3) :
ql(f)
ql
(f)
as a function of the
in binary.
Prove the existence and uniqueness of
ql
defined by the
condi tions (1), (2), and (3). Calculate q>(x) as a function of the partial quotients a
n
of
the expansion of x as a continued fraction, and write q>(x) in binary.
Determine the set of x's such that q>(x) is rational. Prove that for x e [0,1], q> satisfies the identities:
(4):
q>(x) + q>(1 - x) = 1, Aq>(x) + B, a
c
where b ' d are two adjacent fractions in [0,1], and A,B are independent of x.
CHAPTER 9
434
EXERCISE 9-6: SYMMETRIC DEVELOPMENT With a O,a 1 , ••• ,an' ... indeterminates, define two sequen-
(1):
ces of polynomials with coefficients
in~,
PnCao,al, ... ,a n ) and
QnCao,al, ... ,a ), by the relations: (Pn Ca o'· .. ,an)
= a nPn- lCaO,···,a n- 1)
{
+ Pn- 2 Ca O,···,a n- 2) (n
PoCa o )
=
2,3, ... ),
aO' an Qn- lCaO,···,a n- 1) + Qn- 2 Ca O,···,a n- 2) (n
2,3, ... ),
Denote by Mi the matrix:
a.l.
1
1
0
1
Prove the following identity, called the FUNDAMENTAL IDENTITY, valid for m
~
2, n
~
1:
+ Pm- 2 Ca O,···,am- 2)Pn- lCam+l,···,am+n ).
CONTINUED FRACTIONS
(2):
Let
A B >
435
1 be a fraction in lowest terms, [ao, .•• ,aNl one of
its expansions as a continued fraction.
Pn Denote by -- the n-th qn
convergent of this expansion. Show that:
Show that if the expansion of ~ is symmetric (i.e., an n
= a N_n ,
= 0,1, ... ,
) then A divides B2 + (_l)N-l. Conversely, show that if A divides BZ ± 1, one of the expan-
sions of ~ as a continued fraction is symmetric.
In the case
where A divides B2 + 1 show, with the help of the fundamental identity, that A is a sum of two square. valid i f
~
Is the result still
< I?
EXERCISE 9'7: CONVERGENTS OF THE SECOND KIND Let a be an irrational number. Set a = [aO,a 1 ,··· ,an"" ]'. We call the SECONDARY CONVERGENTS of a assoaiated with the aonve~gent
Pn
form -- the fractions: qn if n
and Define A', A" by the equalities:
Ia
(l):(a):
P"
-E..I
- q" n
P' P" Show that q~' ~ are irreducible. n
qn
~
2.
CHAPTER 9
436
Calculate A' ,An as functions of ct ct S S n n n' n+l' n' n+l' qn [a ,a 1, ... ,a1J = n nqn-l
( l):(b): with S n
=
Deduce from (I) (b) that every secondary convergent
(1):(c):
~I < ~ q
satisfies Ict
By calculating A' + An show that at least one of
(1): (d):
n
n
the secondary convergents associated with the same convergent satisfy: < -
1
q
2
Let ~ be a rational number whose expansion as a q continued fraction is chosen in such a way that: (2):(a):
with b n
~
2.
Show that if ct lies between the rational numbers: and then ct admits ~ as a convergent or secondary convergent. q (2):(b):
From this deduce that if
~
satisfies Ict -
then ~ is a convergent or a secondary convergent of ct.
~I ~ ~ q
EXERCISE g·S: GOLDEN NUMBERS Let ct o
=
IS 2
I
Pn .
'Q 1 ts
fixed once and for all. C
m
15 + 2
1
P 2m - 1
n
+--Q2m-l
Set:
n-th convergent, m
~
I an integer
,
CONTINUED FRACTIONS
437
With a a given irrational number, we want to study the number of solutions of the inequality: \ a _ E.\ q
"
(1)
_1 C 2
mq
(p,q relatively prime integers, q > 0); for linguistic convenience a solution of (1) will be called the fraction ~ . (1) :
Let [a O,a 1 , •.. ,an , ... ] be the expansion of a as a p
continued fraction, ~ its n-th convergent, and define A by the qn
n
equality:
Show that:
(2):(a):
Verify that: 8 3
< -
(2):(b): of a.
Show that every solution of (1) is a convergent
From this deduce that (1) is equivalent to: A
n
~
C .
(2)
m
(3): Take a = aO' Show that (2) has exactly m solutions and that the inequality: 1
" -2- ,
Cq
with C > Cm'
CHAPTER 9
438
has fewer than m solutions. (4):
Take a
(4):(a):
+ ao'
Show that if an
~
3 for an infinite number of n's
then (2) has an infinite number of solutions. (4):(b):
Show that the same holds if we have a
n
~
2 from a
certain index on but a n = 2 for an infinite number of n's (we 8 will show that if an+1 = 2, an ~ 2, a n+2 ~ 2, we have An > '3 ) . (4):(c):
Assume that a
n
=1
be the largest integer such that a
from a certain index on, let r r
>
1.
Show that:
... , are m solutions of (2). (5): (1) .
State a theorem on the minimum number of solutions of
CONTINUED FRACTIONS
439
SOLUT IONS
SOLUTION 9'1: Let TeM 2 (lR)be the matrix of the mapping T with respect to the canonical basis. Let us assume that the development of a is periodic starting from the index nO' ine T
Let us determ-
= [ ac db] by the relations: aPn -2 + bqn -2'
o
aPn
-1 +
o
0
bqn -1' 0
These two systems have a solution (a,b,c,d) e~4 since the determinant of these two systems is equal to: n -1 (-1) 0
Let T be the transformation determined in this way. by induction that Pn + s
T(Pn ) if n and
~
no - 2.
Let us show
We have:
440
CHAPTER 9
and if we assume that: P
n+s-2
= T(pn-2 )
and
Pn+s- 1
= T(Pn- i)'
we then have:
and so Pn+s = T(pn ), since an+s = an if n > nO. On the other hand, from the equality: =
a
b
c
d
we can show that:
SOLUTION 9-2: (1):
Since the development is purely periodic,
there exists a linear transformation T such that Pn+ k = T(Pn ) for n ~ -2; we determine T with the help of the equalities Pk - 2 = T(P_ 2 ) and Pk- 1 = T(P_ 1 ). If T is the matrix of T with respect to the canonical basis (cf., Exercise 9-1), we then obtain: T
=[
Pk-1 Pk - 2 ]
qk-1 qk-2
hence:
from the relations:
and
T- 1 =
(_l)k[ qk-2 -qk-1
-Pk - 2 ]
Pk-1
441
CONTINUED FRACTIONS
and from the preceding inequality we deduce that:
o. (2): x If k
~
2
(1)
Suppose we have the equation: ( u + q)x + (-1 )
k
o
2 we have:
and, if k
= 1:
Therefore this equation has a real root p such that p real root
p'
such that
!p'!
<
>
1 and a
1.
From Equation (1) we deduce that: Pmk
, m = AP
+ A " P ,m ,
)1Pm +)1 "m P
with: >. + >.'
= a 0'
)1 + )1' = 1
and meN.
>'P + >.'p' = P k , >',>",)1,)1' can be calculated with the aid of the two systems, because the determinant has the value p - p' We then have:
+0
442
CHAPTER 9
Hence we obtain: lA' I IA'I
Ipmk - Apml and similarly: Iqmk - llpml
I Ill'p,ml =Ill' -m p
=
If we set a In fact:
SOLUTION 9·3: (1):
a 1 is a convergent.
Ill'I
c' = --, qmk
I.d, with the usual notations,
I.d=a=a +..l..
o
a1 '
therefore: a'
=-
.fd
= ao
1
+ -,- , a1
and therefore: 2.fd> 2,
and since ~ a1
< 1,
lr >
we deduce from this that - a 1
1, and a 1 is
certainly the convergent. The expansion of a is (cf., Theorem (II)) therefore of the form:
and we have a s +1 = a 1 . We can obtain an equation of the second degree satisfied by a by using the formula given in the introduction:
CONTINUED FRACTIONS
443
Ps (a - a ) +
p s a s +1 + Ps - 1
a
o
Ps - 1
q s a s +1 + qs-l
which gives: 2
a q s- 1 - a(aoq s- 1 - q s + Ps- 1) + aOp s- 1 - Ps
= O.
This polynomial must be proportional to the polynomial a 2 - d. Hence:
so:
or, again: [a s ,a s- l' ... ,all
so:
and by the uniqueness of the expansion as a continued fraction we obtain:
(2) : (a) :
Since a ns +1 = a 1 for all n elN we can also write:
a o) a = qns + q l(a - a ) , ns0
Pns + Pns- l(a
-
CHAPTER 9
444
so:
a2q
ns-l
- a(a q
0 ns-l
- q
ns
+P
ns-l
) + a_v ns _1 - P ~ ns
O.
Using that the polynomial is proportional to a 2 - d we obtain:
{
aOqns-l - qns +
Pns - 1
= 0, (1)
Pns + aOP ns - 1
= dqns-l'
Multiplying the first equation by Pns - 1 and the second by qns-l' and then adding them, we obtain:
2
Pns - 1 -
(2): (b):
d 2
qns-l
Let us prove the following relation by induction on m.
= a ns-m if 1 ~ m ~ ma. m If m = I, since qo = I, q-l = D, and (2) is certainly satisfied. Therefore it suffices to prove that:
using the equality a
q
m-lPns-m +qm-2 Pns-m-l =qP m ns-m-l +qm-lPns-m-2'
or again:
-which obviously holds.
)P m m-l + q m-2 ns-m-l + qm-lPns-m-2'
(a q
CONTINUED FRACTIONS
445
Similarly, let us prove (3) by induction; if m
1 we must
therefore show that:
Now the formulae (1) are equivalent to:
Now: PDS
= 2a oPDS- 1
+ Pns- 2'
since a DS 2a o' and (3) is certainly satisfied if m Next, it suffices to verify the equality:
1.
which can be shown in the same way as formula (2). (2):(c):
From (2) we deduce (by replacing n by 2n and m by
ns) that:
(5)
but: 2q
DS
+ (q
Ds-2
by formula (1), and therefore: q
2ns-l
= 2pDs-l qns-l
since
- q
DS
)
CHAPTER 9
446
similarly:
by
ormulae (1), so:
2dqns-l -p - ns +pns-2' Therefore we have:
so, with Question (2)(a) above: P2 ns- 1
2 + dq ns- l'
By induction let us show that:
(3):
x
2
= Pns- 1
P n 2 s-l n
If n
q n
2 s-l
=0
P n
the formula holds; and if x n _l
from this that:
= ![P:n - 1S_l
x
n so:
2
+
dq~n-1S_l]
q n-l P n-l 2 s-l 2 s-l
'
2 s-l -'-'--"-we deduce q n 2 s-l
CONTINUED FRACTIONS
x
447
P n 2 8-1 q n 2 8-1
n
by formulae (4). Therefore we deduce from this that: limx
n
= Id,
and that: 1
q nan 2
8-1
2
8
1 "'n--2q n
2 8-1
But by formulae (5):
so:
But q28-1 ~ 2, for if 8 ~ 2 then q28-1 > q3 > 2, and if 8 = 1 then a 1 = 2a o and q28-1 = q1 = a 1 ~ 2. Therefore the error in replacing Id by x is less that _..:;.1__ n 2n 2 + 1 9·4: (1): The convergence of the series giving vn is shown using d'A1embert's Criterion:
SOLUTION
Un ,k+1 -"--- =
1
(k
1
+ 1)(2n + 2k + 1) - 2 ' 2a
448
CHAPTER 9
whence: lim k-+-+co
U
n.k+1
O.
un • k
For positive nand k we have: 1
Un • k = (2n + 1)Un+1 •k + a 2 u n+2 •k - 1 On summing over V
(with u n+2 .- 1 ·
0).
k we obtain:
n = (2n + 1)V n+1 +
1
'2 v n+1 ' a
and by multiplying by ~ , n+1 =(2n+1)a+
1
Wn+1
We can apply the continued fractions algorithm, for since W n is positive we have: W n
7
(2n
+
l)a ~
1
and
o
1
< - - < 1.
Wn+1
From this algorithm we deduce:
Wo
[a,3a,5a, .•. ].
The expansions of the hyperbolic functions as power series give: cosh whence:
!a
. h 1, aSln
a
CONTINUED FRACTIONS l.U O
= coth
by taking a
e + e -
449
1
- . a '
=2
we obtain:
1
- - 1 = [2,6,10,14, ... ].
(2):
For the indices 3n - 2,3n - l,3n,3n + l,3n + 2 write
the recurrence relation satisfied by p P3n-2
bP 3n - 3 + P3n - 4 ,
1,
P3n - 1
eP 3n - 2 + P 3n - 3 ,
- b,
P 3n
gnP3n-l + P3n - 2 ,
P3n+1
bP 3n + P 3n - 1 ,
P3n+ 2
= eP 3n+1
+ P3n '
be + 1, e, 1.
Multiplying by the integers indicated above, and adding, we obtain:
The sequence P3n+2 satisfies the recurrence relation:
We know that the set of solutions of this equation in u
n
forms a
two-dimensional vector space, a basis of which is P' and q'. n n Hence there exist constants A,~ such that:
CHAPTER 9
450
A and
are calculated by taking n
~
-1 and n
0; we obtain:
P3n+2 = Pn' - b qn'.
An analogous calculation gives:
= (be
q3n+2
+ l)q~.
By division we obtain the relation:
(be +
1)
P~
P
..1E.!3.. = qr- b. q3n+2
n
and by passing to the limit we obtain:
= (be
a'
taking b
+ l)a + b;
= e = 1.
e + 1 --1• whence: e a
= a'
- 1 2
n
IOn
= 2n •
1 -----1
e -
by Question (1) above we have a'
= [0.1.1.2.1.1.4.1.1 •.•. 1.
and:
1
because ---e - 2 = a 2. SOLUTION 9,5: (1):
We must find two fractions
that: p'q - pq'
= 1,
p + p'
= P,
Eliminating p' and q' we obtain:
q + q' = Q.
E and ~ such q q
CONTINUED FRACTIONS
= 1,
Pq - Qp
451
with 1" q ... Q - 1.
Using Bezout's Theorem we can verify the existence and uniqueness of p and q. Next we have the existence and uniquess of p' and q'. An induction on Q then shows the existence and uniqueness of
~ (~)
1
='2' (2):
=
3
4" '
For fixed n, set: for h eE.
The fractions: and
(h - 1)Pn_1 + Pn-2
(h - l)q n- 1 + qn-2
are adjacent, as is shown by the relation:
Hence we have: =
From this we deduce:
which certainly gives:
~
Pn-1) (- ). qn-1
452
CHAPTER 9
u
n
=
(1 - 2J + _1_ (u - u ) a J n-1 a n- 2 n-2 u
2 n
2 n
Writing this relation in the form:
we have: (_1)n-1
= a + •• '+a -1 ' 2 1
n
hence: 1
1
(_1)N-1
-a:T - -a-1-+a---2 1°- + ••• + -a-+-a"--+.....-.-.-+-a--"="l 2 1
2
2 1
2
N
p
REMARK: Q has two expansions as a continued fraction, of which only one ends in a 1; we could verify that this would require
that ~(~) not depend upon the expansion chosen. Taking N even and grouping the terms in pairs, we have in binary: O. 00'" 0 11"'1 00'" 0 ... 11'" 1 ~~~
a1
a2
a3
digits digits digits
~
aN digits
(3): It follows from the preceding that the restriction of ~ to the rational numbers in [0,1] is continuous and increasing, this restriction can therefore be extended in an unique way to a map on the real numbers in [0,1]. If x = [0,a 1 ,a 2 , ... ,an , ... ] we have:
CONTINUED FRACTIONS
ql(X) =
co
I n=1
453 (-1 )n~l
a +a + •• ·+a ...1 2 1 2 n
and the binary development of ql(x) is written as before. ql(x) is rational if and only if its binary development is finite or periodic. therefore if and only if the development of x as a continued fraction is finite or periodic. The set sought is formed by the union of the rationals and quadratic irrationals in [0.1]. (4): In order to prove the identity ql(x) + ql(l - x) can assume that 0 < x < 1. and we have: with a 1
>
1 we
1.
and we can verify that: 1 -
x
[O.1.a 1 - 1.a 2 •••• ].
The identity clearly follows from this. Let us set y = ax + ~ ; first of all we show that x e [0.1] ax + implies ye [0.1]. In order to prove that cp(y) = Aql(x) + B we can
£
use the following result (cf .• [Har] Theorem 172]: if ~ and are two adjacent fractions. then they are two consecutive convergents in the expansion of y. and x is a complete quotient of them. More precisely. if y = [0.bo.b 1•... ] then there exists an index k such that:
-aa = and:
454
CHAPTER 9
We then have:
I n=l
q>(y)
I n=l
(_l)n-l
b +b +---+b -1 ' 2 1 2 n (_l)n-l
b
2 k+1
+b
k+2
+---+b
k+n
-1'
where the summations may be finite_ By
writing: k
I n=l
q>(y)
(-1) n-l
b +---+b -1 2 1 n (_l)k
+
00
I
(_l)n-l
b +---+b b +b +---+b -1' 2 1 k n=l 2 k+1 k+2 k+n
we find:
A
B=
k
I n=l
(_l)n-l
b +- - -+b -1 2 1 n
SOLUTION 9-6: (1):
We show by induction that: p n- l(a O,···,a n- 1)
Qn- l(a O,··_,a n- 1) whence:
CONTINUED FRACTIONS
455
P (a , ... ,a o) MM
n n-l
oM
00
n
n
0 = [
P
l(an , ... ,a o)
Q
l(a, ... ,a O) n
n-
n-
and by transposition:
M Moo oM
o
1
n
=
[
p(a, ... ,a o )
Q (a , ••• ,a o)
Pn - l (an'··· ,a l )
Q l(a , ... ,a l ) nn
n
n
n
n
Comparing the two expressions for the product MoMloo·Mn we obtain the two identities required. The fundamental identity expresses the associativity of the matrix product:
Taking the entry in the first row and first column we obtain: m+n (a o'
P
... ,am+n )
P
m- l(aO, ... ,am- l)Pn (a m, ... ,am+n )
+ Pm- 2(a O, ... ,am- 2)Qn (a m, ... ,am+n ). Since: Q (a , ... ,a
n
m
m+n ) =
P
n- l(am+ l, ... ,am+n ),
we obtain the fundamental identity. (2):
A B> 1 implies a O ? 1, therefore the continued fraction
[aN, ..• ,a o] exists.
By the results of Question (1) above we have: PN(a W .•• ,a o ) QN(a N , ••. ,a o )
456
CHAPTER 9
if the development of ~ is symmetric we have:
hence:
The identity:
can be written: A
so
A
2 qN-l - B
= (_l)N-l,
certainly divides
B2
+ (_l)N-l.
Conversely, if A divides
B2 ±
1 we choose the parity of N
(choosing aN = 1 if need be) in such a way that A divides B2 + (_l)N-l and we set:
(B' an integer). On the other hand we know that:
Hence by addition:
and since (A,B) = I, A divides B - PN- 1 . However, A > B and A > PN- 1; whence A > we deduce B - PN- 1 = 0, therefore:
IB -
PN-11.
From this
CONTINUED FRACTIONS
457
and the uniqueness of the expansion as a continued fraction shows that this expansion is symmetric. EXAMPLE: ;0 = [1.2.3]. the expansion is not symmetric; however. 101<72 + 1). We can write = [1.2.2.1]. and the development is symmetric. We now assume that A divides B2 + 1 and we choose N odd. say N 2M + 1. Apply the fundamental identity with m = M + 1. n = M:
;D
and since an
= a 2M+1 - n
thus S is a sum of two The result is still exist two integers Q.R divides R2 + 1. A is a
SOLUTION 9·7: (l):(a): qn
>
for n
= D•...• 2M
+ 1 we have:
squares. valid if ~ < 1. Dividing B by A. there such that B = AQ + R with ~ > 1. Since A sum of two squares. p'
p"
~ and ~ are well defined because
qn qn qn-l' and they are irreducible because:
and: 1.
458
CHAPTER 9
We have:
(1) : (b) :
la -
tl
IPn an+1 qn a n+1
=
P n + Pn - 1 1 qn + qn-1
+ Pn-1 + qn-1
so:
A'
n
and so: 1
1 + _~1_7 a - 1
n+1
.
qn
SInce - qn-1
=
S .
similarly,
n'
P"
la - .22.1 q" n
P n a n+1 + P n - 1 qn a n+1 + qn-1
I
so:
A" n
(a n +1 + 1)(qn - qn- 1)
and so:
A"
1
1
1 + Sn - 1
(l):(c):
a n +1 + 1 .
We have: 1
A' > 1 > 1 n Sn + 1
since a n +1
> 1
and Sn
> 1,
~
l
2,
and similarly:
>(
qn
an+1 - 1 an +1qn + qn-1 )
,
CONTINUED FRACTIONS 1~ > 1 -
n
459
1
a n +1 +
i,
1 >
and therefore every secondary convergent ~ satisfies: q
(l):(d):
We have:
l' + 1"
2 +
n
n
22
en -
1
+ -a'=2-2__-1- > 2; n+1
therefore l'n or 1" is strictly greater than one; and therefore n we have:
la
Either:
-
P~I
< _1
q'n
q'
2 '
or
n
1 q,,2 n
If a lies between the rational numbers [b o' ... '
(2):(a):
bn- 1,bn + 2] and [bo, ... ,bn- 1,bn - 1], then a can be written as:
with: b
n
- 1
~
a
n
= a
but a n e:N* and a n+ 1 b
n
a
n
+
£
1
n
+ - - , bn + 2, a n +1
>
1, therefore:
with
E
e {-l,O,l}.
p. But if ~ denotes the convergents of a which are also the conqi
460
CHAPTER 9
vergents of ~ for i ~ n - 1, we have: q a nPn-1 + Pn - 2 + gP n - 1 a nqn-1 + qn-2 + gqn-1
so: P
q
Pn + gPn - 1 = qn + gqn-1
with g e {-I,D,!}, that is to say, that ~ is either a convergent q or a secondary convergent of o. (2):(b):
By (2)(a) above it will be enough to prove that if
~ = [b o,b 1 , ... ,bn ], with bn ? 2, then B1 and B2 , where:
0
lies between the numbers
and Because ~ lies between B1 and B2 it suffices to show that:
1
£1 .
inf B. i=1,2 ]. q Now,
(bn + 2)P n _1 + Pn - 2 (bn
+ 2)qn_1 + qn-2
(b n - 1)P n _1 + Pn - 2 (b n
p. where ~ denotes the convergents of ~ , or again, qi q P + 2Pn _1 q + 2qn_1
1)qn-1 + qn-2
461
CONTINUED FRACTIONS
We obtain:
because 2qn-i similarly:
~
qn
1
q(qn Therefore, if
Ia
-
1
qn-l) ~ q2
la - ~I ~ ~
r..1 ~ q
q
inf lB. i=1,2 1
we have:
- r..1 ' q
and therefore r.. is consequently either a convergent of a secondq ary convergent of a. SOLUTION 9·8: (1):
From the equality: 1
q (q a + q ) n n n+i n-i we deduce that:
A-
n
so: A- n
[a n+ i,a n+ 2' ... ] + [O,a,a n n- i,···,a i ]·
(2):(a):
The convergents of odd order of a number form a
decreasing seqeunce tending towards the number.
Therefore:
462
CHAPTER 9
15= 15+ 2
1
+ 152
1
because the development of Now
315
15 +
7, therefore:
<
em < 15+ 2
<
1
Pi 15+ + -= Q 2
3
1
15- 1 is [0,1,1, ... 2
]
.
8 3
3
< -
2
(2):(b):
By Lagrange's Criterion, because
Pn
em
>
2, every
solution of (1) is a convergent of a, and - will be a solution qn of (1) if and only if A ? e . n
m
If a = a O let us evaluate An' where:
(3):
[1,1, ... ] + [O,!,l,: .. ,J], n times
An so:
Inequality (2) can therefore be written:
which is equivalent to n e {I, 2, ... ,2k - 1, ... , 2m - I}.
Therefore
in this case Inequality (2) has exactly m solutions, and there fore the inequality:
la
-
e.1 q
< _1
e/
with
e
>
em
CONTINUED FRACTIONS
463
has fewer than m solutions. From Question (1) above we deduce that i f a 1 ~ 3, n+ then A- n > 3, and that i f a n ~ 3 for an infinite number of n's, then A- ~ 3 ~ Cm for an infinite number of n's, and therefore n that the inequality: (4):(a):
la - lllq .. cl
_1
has an infinite number of solutions if C .. 3. From Question (1) above, we deduce that if n
(4):(b): A-
n
>
a
n+l
+
1
a n+2 + 1
n
> 2 + 2 3
1
+ ----:;an + 1 '
and if a n+ 2" 2, a n .. 2, a n+l A-
~
= 2,
then:
8 3
Since these conditions are satisfied for an infinite number of n's, Inequality (1) has an infinite number of solutions for arbitrary m, and the inequality:
has an infinite number of solutions for arbitrary C .. (4): (c): A-
We then have:
r- 1 > [a r ,1,1, ... ].
38
1:
464
CHAPTER 9
so: A + 1 r-l > a r [1,1, ...
Pr - 1
Therefore ---- is certainly a solution of (1). qr-l
Pn
- If n ? r, in order that -- be a solution of (1) it is necesqn sary and sufficient to show that: A
n
=
[1,1, ... ] + [O,l, ... ,l,a , ... ,a 1 ] '---v---'
(n - r)
r
times
~
IS +
1
--:2::--- + [0,1,1, ... ,1] , '----v------'
(2m - 1)
times or again, [1, ... ,l,a , ... ,a 1 ] " ~
(n - r)
r
[1,1, ... ,1]. '----v------'
(2m - 1)
times
times
- If n - r is odd and 2m - 3 ? n - r, this last inequality is equivalent to: [a r , ... ,a1 ]? [1,1, ... ,1], (2m-n'+r-l)
times which is realised because a [a , .•. ,a 1 ] r
> ar
?
r
? 2 and:
2 > [1, ... ,1].
CONTINUED FRACTIONS
Therefore for j e {O, 1, ... ,m - 1} the Inequality (1).
465
Pr-1+2j qr-l+2j
are solutions of
(5): The results of Questions (3) and (4) therefore allow us to deduce that for an arbitrary irrational number Q Inequality (1) has at least m solutions.
CHAPTER 10
p-Adic Analysis
INTRODUCTION Denote a prime number by p;
:N,
!
JR, JR
have their usual
meanings. We know that for every prime number p there exists on III an non-Archimedean (ultrametric) absolute value (called the p-ADIC
1·1 (or 1·1 if following wa~: let x e Ill,
ABSOLUTE VALUE) denoted
no confusion can arise)
defined in the
x
= ~n
ph be an expression
for x in lowest terms, with h e!
p
= p -h.
We
also define the p-ADIC VALUATION of x, denoted v (x) (or v(x) if no confusion can arise), by v (x) p
= h.
pletion of III (resp. !
IT
pprime
PROPOSITION 2 :
of integers
(0
p
(resp.!
Let x e Ql, and let us denote the usual absolute
value on Ql by Ixl • ""
Ixl""
p
~
Ixl
Then we have the product formula: P
=
1.
Let x e III , there exists an unique sequence (b n ) n?n P x n ~ b n < p, n e!
p-ADIC ANALYSIS
467
This sequence is called the HENSEL
converges p-adically to x. EXPANSION of x.
We note that if K is a complete non-Archimedean field, whose valuation is denoted v(·), associated with it is an absolute
value, denoted:
Ix I
=
p
-v(x)
where
p
ell\ and 0 < p < 1,
its valuation ring A, that is to say the set of elements with absolute value less than or equal to one, and its maximal ideal
M in the valuation ring A, that is to say the set of elements with absolute value less than one. The residue field of K will be denoted K = AIM, and i f yeA the image of y in AIM under the canonical surjection will be denoted y. Let L be an extension, of finite degree n, of a complete non-Archimedean field K. The relation: PROPOSITION 3:
defines the unique valuation w of L extending the valuation v of K.
(NL/K(x) ~
p
denotes the norm of x
e
Lover K.
is defined to be the completion of the algebraic clos-1
is normalised by Ipl = p . P If P(X) e K[X] the degree of P is denoted degP, and the image of
ure of ~ , the absolute value on ~ p
P in K[X] by P. LEMMA: (Hensel): Let K be a complete non-Archimedean field, and PeA [X] a non-zero polynomial of degree d. Let us assume that there exist two monic polynomials g,h in A[X] such that P = gh, g and h are relatively prime in K[X], degg + degh ~ d. Then there exist G,H in A[X] such that: G
= g,
li = h,
degG
=
degg,
and
P = GH.
CHAPTER 10
468
EISENSTEIN'S CRITERION: Let K be a non-Archimedean field with
valuation ring A and maximal ideal M, and let P be a monic polynomial in A[X] :
P(X)
= Xn
If a. eM for J.
n-l
+ a1X
i
+ ••• + an.
1, ... ,n and a • M2, the polynomial P is irreducn
ible in K[X].
If K is complete, P determines a totally ramified
extension of K. Let L be a finite extension, of degree n, of the complete nonArchimedean field K. Let w be the unique extension of the valuation v of K.
By the proposition on extending valuations, if the
set of invertible elements of K is denoted K*, then: w(L*) c-n -1 v(K>',). If v(K*) is a discrete subgroup
of~,
then:
V(K*) ~ w(L*) C ! v(K'~) -n implies that v(K*) is a subgroup of w(L*) of finite index e which divides n and which is called the RAMIFICATION INDEX of Lover K. Let f be the dimension of
L,
since
the RESIDUAL DEGREE of Lover K.
R is
a vector space f is then
We have the following result:
Let K be a complete non-Archimedean field. r
e~+,
Let a e K and let
then: {xeK:
Ix - al
< d.
469
p-ADIC ANALYSIS
THEOREM: Let F(X) =
L
n~O
a (x - a)n be a Taylor series of K[[X]] n
convergent in B(a,r+)~ then: sup IF(x) I xeB(a,r+)
If K is also algebraically sup IF(X)I xeB(a,r+)
closed~
= supla n~O
n
then:
Irn.
Notice that if Be K and if F is a function from B into K, then:
supIF(X) I. XeB If Band C are two topological spaces we denote by C(B,C) the space of continuous functions from B to C with the norm given by uniform convergence on B. If m,n e72:, (m,n) denotes the greatest common divisor of m and n. U denotes the group of units of m, U = {x em: p p
Ix I
= 1}.
p
p
PROPOSITION 4: In order that a complete valued field K be loc-
ally compact it is necessary and sufficient that the subgroup v(K) of~ be discrete and that the residue field~ K be finite. PROPOS ITION 5: 72:p is a compact set of mp.
BIBLIOGRAPHY [Ami], [Bac], [Bor]. [Sam]. [Serl].
470
CHAPTER 10
PROBLEMS
EXERCISE 10-1: VALUATIONS ON A FINITE FIELD Let k be a finite field and let L be a finite (i.e., algebraic) extension of k. Show that every absolute value on L is trivial. EXERCISE 10-2: VALUATIONS ON THE RATIONAL FUNCTIONS Let K be a field, L
= K(X)
the field of rational functions
in an indeterminate over K, P(X) an irreducible polynomial of
K[X], and V the valuation on L associated with P. p Determine the valuation ring, the valuation ideal and the residue field of vp' Describe the completion of L for the valuation Vp when P = X. From this deduce a description of the completion in general. EXERCISE 10-3:
The assumptions are the same as in Exercise 10-2.
When is the compltion of K(X) locally compact for a P(X)-adic valuation?
471
p-ADIC ANALYSIS
EXERCISE 10·4: VALUATIONS OVER K(X,Y) Let K be a field, and a,b two positve real numbers. Let K(X,Y) be the field of rational functions in two variables OVer K. Let P(X,Y) = L a jffiyn be a polynomial. m,n m,n If P f 0, set v(P) = inf(am + bn) where m and n run through 2 the set of pairs (m,n) eN such that a + 0; i f P = 0, set m,n v(O) = +00. If R(X,Y)
= P(X,Y)/Q(X,Y)
where P and Q are polynomials; set
v(R) = v(P) - v(Q). (a):
Show that v is a valuation on K(X,Y).
(b):
Determine the valuation ring, the valuation ideal, as
well as the residue field of K(X,Y). (c):
Determine the completion K(X,Y) of K(X,Y) for this
valuation.
Under what conditions on parts (a) and (b) and K is K(X,Y) locally compact?
EXERCISE 10·5: STUDY OF ~*/(~*)p p
p
Let G be a multiplicative group; the subgroup of G formed by x P , where x e G, is denoted GP • (a):
If G is the direct product of the subgroups H1 and H2 ,
Hi x H~.
GP is the direct product
(b):
Let a
(1 + pX)p has a root in (c) :
We propose to study ~1./(~,·.)p. P P
2
1 + P b,
be~
P
, show that the equation:
=a ~p'
From this, deduce, for p f 2, that ~,•• /(~,.• )p P P
(~/p~)
2
•
472
CHAPTER 10
EXERCISE 10-6: NORMAL BASIS FOR A FINITE EXTENSION OF A COMPLETE NON-ARCHIMEDEAN FIELD Let K be a complete non-Archimedean field with discrete valuation, and let L be a finite extension of K with the absolute value extending that of K. Under what condition does L have a normal basis over K? Let B be the valuation ring of L; what are the normal bases of L relative to the A-module B? EXERCISE 10-7: STUDY OF THE MAPPING n ~ an Let a e A (a Banach algebra over II! ). p
(a):
Show that:
The mapping n ~ an of~ into A can be extended to a
continuous mapping f a of ~ into A when la -
11
<
1.
The mapping cp:a ~ f a of L = {a e A : Ia-II < 1} to Cezp ,A) is a continuous injective homomorphism of L into a multiplicative subgroup of Cez ,A). (b) :
p
EXERCISE 10-S: CAUCHY'S INEQUALITIES Let K be a complete non-Archimedean field, and let f(X) = L a be a Taylor series of K[ [Xl] convergent for 0 ~ I xl, < 1"0
n~O
n
xn
with 1"0
+ O.
For 1"
<
1"0 let us set: M(f,l") = supla 11". n~D
n
If the residue field k of K is finite and if v(K*) is a discrete subgroup of~, show that we do not always have M(f,l") = sup If(X) I. Ixl~l"
EXERCISE 10-9: AN ENTIRE FUNCTION BOUNDED ON II!
p
We want to construct an entire function bounded on II! • p
p-ADIC ANALYSIS (a) :
Let
= 1 - XP-1 ; determine:
P(X)
=
M(P.K)
473
inf
Xe lllp
v(P(X».
ke:?2:.
v(x)=k
(b):
Determine a sequence of integers h(k) such that if:
we have: sup
xe lllp v(x) ..-k (c):
1.
\Pk(x)i
From this deduce an example of a bounded entire
function on III . p
EXERCISE 10·10: A CONGRUENCE IN :?2: 2
h
is divisible by 2
(h eN) for n large enough.
EXERCISE 10'11: HENSEL'S EXPANSION FOR RATIONAL NUMBERS We say that we have a PERIODIC HENSEL EXPANSION for a number
aelll
i f there exist two integers nO e:?2: and
l'
>
0 such that a
an for n > nO' where: + •••
(m
e:?2:).
n+r
=
474
CHAPTER 10
Show that the Hensel expansion of only i f
e III
e Ill.
p
is periodic i f and
EXERCISE 10-12: p-th ROOTS OF UNITY IN III
P
Show that for p
+2
that 1 is the only p-th root of unity
in III . p
EXERCISE 10-13: A CALCULATION IN 1113 In 1113 determine the first five terms of the Hensel expansion of: 10g4
log(l + 3)
EXERCISE 10-14: UPPER BOUND FOR THE NORM OF THE POLYNOMIALS (~) IN
l?Z
P
Show that in II1p Pk(x) = (~) , with Po(x)
EXERCISE 10-15: 111*2 IS OPEN IN III p
Show that 111,·,2 =
(p
+ 2).
p
{x
e III
p
1, satisfies:
p
{O}:x
= Y 2 ,Y e II1p}
is open in II1p
EXERCISE 10-16: A FORM OF HENSEL'S LEMMA Let F be a complete valuation field for a discrete valuation
v of the valuation ring O.
Let f(x) be a monic polynomial in
O[x] such that there exist
<
1 and If'(
475
p-ADIC ANALYSIS
f«(J,l) (J,2
... ,
(J,l - r«(J,l)
f.
converges in 0 to a root of
EXERCISE 10'17: Find a rational number in Q that approximates 5 1 .32 a solution to P(x) = x •. x •. 2x ~ 8 within ~. 5
EXERCISE 10'18: STUDY OF THE LOCUS OF THE ph_th ROOTS OF UNITY IN 91p Show that if ~h is a ph_th primitive root of unity in ~p' wi th heN>", then we have:
11 -
~hl
(
1) lip
h-l
= pJ
(p-l)
.
EXERCISE 10'19: SOLUTION OF CONGRUENCES mod pk Let F(x l ,x 2 ,··. ,xn ) e~[xl"" ,xn ]. Show that the congruences: F(x l , ••. ,xn )
can be solved
in~
= 0 mod
p
k
(for a11 k eN*)
if and only if the equation F(xl, ... ,x n )
0
has a solution in ~ • p
EXERCISE 10'20: REDUCIBILITY OF POLYNOMIALS IN 91 2 Show that the polynomial X2 + X + a, where a e~, has a solution in 91 2 if and only if a is even.
CHAPTER 10
476
EXERCISE 10·21: LIOUVILLE NUMBERS IN Let
(1) :
ct
e:iZ
rational integer.
mp
be algebraic over
p
mof
degree n, but not a
Show that there exists a constant C depending only on
ct,
such
that:
Ict
- A
IAI
where
Ip
for all A e:iZ'" ,
is the ordinary absolute value on:iZ.
(2):
Let a. e:iZ.
Show that if, for an infinite number of
p
values of N, we can find a rational number A (A
o
<
Ict
-
A
I <~ PIAl
We say that
,
then the number
ct
+ +1)
such that
is transcendental over
m.
is a LIOUVILLE NUMBER if it satisfies the preceding
ct
property. Let
(3) :
ct
e:iZ
P
and 1 et
ct =
L
k=l
a~
bk
be its Hensel expan-
sion. Assume that 1 ~ a k ~ p - 1, and that the sequence is strictly increasing, and, moreover that: -I" 1m
k-++'"
b k+l -b-- = + '"
Show that Example:
k
ct
is a Liouville number.
L
ct =
k=l
P
k!
EXERCISE 10·22: m-th ROOTS IN
mP
Let p be a prime number, and
mP the
Let m be an integer prime to p, and let that v
P
(bk)k~O
(£ -
1)
>
O.
field of p-adic numbers. £
be a p-adic unit such
477
p-ADIC ANALYSIS
Show that
is an m-th power in
£
mp .
EXERCISE 10-23: m-th ROOTS OF ELEMENTS OF (1):
mp
Recall that the series: n
00
1 +
2 n=l
converge for 11;1
~,
10g(1 + 1;)
and
.
2
n~l
~!, that: p
p
Ilog(1 + I;) 1
p
and that: e 1og(1+1;)
Let m
~
=1
~
+ ".
2 be an integer; set v (m) P
=f
(f) 0).
Show that i f y e m satisfies v (y - 1) ~ f + 1, then there P m P m 11m • exists p e m such that p = y. We shall set p = ;.y = y P
(2):
From this show that there exists a finite algebraic
extension K of m containing the m-th roots of all the elements P P of P Give an upper bound for the number of m-th roots of elements
m.
of
~
P
to be adjoined to
~
P
in order to obtain K . P
EXERCISE 10-24: m-th ROOTS OF ELEMENTS OF
~
P
We use the same notation as in Exercise 10-23. (1):
Let S e
b + PS1' where 1 Show that:
~
~
and satisfying v un = O. P b < p - 1 and V(Sl) ? O.
p
Let us set S
478
CHAPTER 10
wi th
V
p
(a') > o.
Note that, for k > 1:
= !!!.(m - 1) (m) k k k-1 (2) :
Let
a
pf+1af +1 , where
•
e!D be such that v p
V p (a f +1 )
~
p
(a)
0, and set a
a +
o.
Show that a necessary and sufficient condition for a 1 / m em p is that there exists an integer b satisfying 1 ~ b < p - 1 such m f+1 that the congruence a = b mod p holds. EXERCISE 10·25: p-th ROOTS OF ELEMENTS OF !D
p
(1): Let Show that v
(2) : v(a')
y
P
e!D
satisfy v(y) > O. > 1, and prove that, for k > 1,
p (yP - y)
Let p,a,a' be elements of m satisfying v(p) = p
= o.
v(a)
=
Show that if there exist two distinct integers k,k' such that:
then: v(p P-1 - 1) > 2. (3):
Show that if there exist integers
~
such that:
p-ADIC ANALYSIS
then for p p
=2
~
479
3 we can choose
we have p
= 3.
satisfying 1 <
p
p
< p - I, and for
Establish that the numbers pkbP, for 0 < k
~ P - I, 1 ~ b < p - I, form, modulo p2, a system of residues. From this deduce that the extension p-th root of every element of
-&
eU •
mp [pl/p]
p
reducible in; [X].
P
contains a
P
mp (p + 2)
EXERCISE 10·26: SQUARE ROOTS OF ELEMENTS IN Let y be a unit of;
of;
such that the polynomial X2 - y is ir-
P
Show that a11 the polynomials
x2 - cS,
where cS e
ible in ; [/Y].
U ,
P
are reduc-
P
(2):
Show that the field; ([/Y])[/;p] is a splitting field p
for every second degree polynomial of; [X]. is identical with; [/y + P
IF]
p
Show that this field
(Use Exercises 10·23, 10·24, ~0.25).
EXERCISE 10·27: SQUARE ROOTS OF ELEMpNTS OF ;2 (1):
o or 1)
Show that the equation X2
=a
has solutions in ;2 if and only if
(2):
lal 2 2- i , la - 11 < 12 123 .
(where
i
Show that there are exactly seven different quadratic
m
extensions of 2 generated respectively by adjoining a root of the following polynomials:
(3): Show, by a simple change of variables, that the extensions defined by X2 - 3 and X2 - 7 are ramified. (4): fied.
Show that the extension defined by X2 - 5 is not rami-
480
CHAPTER 10
mp OF
EXERCISE 10·28: EXTENSIONS OF
DEGREE TWO
Consider the polynomial X2 - X-I and denote its two roots in an algebraic closure of m by ~ and ~. p
Show that if p
(1) :
= 11
or 19, then 8,
e belong
to
mp .
(2): Show that ~ and ~ are not in m if p = 7. Find the p residual degree and the ramification index of the extension m7[~'~]·
(3):
Calculate the S-adic valuation of:
N(~ -
a)
= (~ -
a)(e - a)
for a
= 0,1,2,3,4.
Find the residual degree and the ramification index of the extension of
m5
generated by
EXERCISE 10·29: EXTENSIONS OF Let
~
~
and
~.
mP BY
A SEVENTH ROOT OF UNITY
be a primitive seventh root of unity.
(1): Show that the extension m7(~) of m7 is completely ramified. Find a uniformizer for m7[~]. Show that the equation X7 - 1 = 0 splits completely into factors of the first degree in m if and only if p = +1 p mod 7. (2):
(3) :
mp
if p
=
Show that the extension mp [~] is of degree two over -1 mod 7.
Show that the extension mp[~] is of degree three over mp if p = 2 or p = 4 mod 7. (4) :
Show that the extension mp [~] is of degree six over mp = 3 or p = S mod 7. (S) :
if P
481
p-ADIC ANALYSIS
EXERCISE 10-30: CONTINUOUS FUNCTIONS AND GENERATING FUNCTIONS (1):
= 1, ... ,
Set Po(x)
Show that if x e:7l;
p
= (~) , ....
Pk(x)
we have:
Let a k be a sequence of numbers of
Show that
(2) :
I
k=o
Let
~p
with:
akPk (x) defines a continuous function for x e:7l; . P
f e
C(:7l; ,~ ).
F(X)
P P
Show that the analytic function:
defined on B( 0,1 -)
is in fact an analytic element on ~ that F(X) is the uniform limit on ~
P
P
rational functions without poles in ~ Deduce from this that every function the form:
c
,
- B(l,l-) of a sequence of
P
- B(l,l-), cf., [Ami]).
f e
C(:7l; ,~ ) can be put into p P
Show that: lim ..!...(pn] n h p
n-+o:>
e:N'~.
P
- B(l,l-) (that is to say
EXERCISE 10-31: THE p-ADIC LOGARITHM
exists for all h
~
Calculate this limit.
482
CHAPTER 10
(2) :
Let A (x) = 1-[(1 + x)p n
p
n
n
- 1].
Find the p-adic absol-
ute values of the zeros of A (x) in ~. Is A (x) irreducible n P n over lI! ? Find the value of IA (x) I for x e ~ with Ix I = r, n P P P P where 0 < r < 1 and r
m.
+p
-m. J
for every integer j, with:
--:-.--=1=--_
J
pJ(p _ 1)
Deduce from this that lim A (x) = !(x) exists in ~ for all x n-- n P satisfying Ixl < 1. Show that for all x,y e ~ with Ix - 11 < 1 P P P and Iy - 11 < 1 we have: P
!(xy - 1)
= !(x
- 1) + fey - 1).
Give the expansion of lex) as a Taylor series, and give its radius of convergence. (3):
For n > 1 set:
A (x) B (x) _ ..,.......;n::....,.~ n -An _ 1 (x)
Show that B (x) is an irreducible polynomial in x on lI!. Show n P that the zeros of B (x - 1) are the primitive pA_th roots of unity In ~
n
P
what are the zeros of lex) inside Ixl
P
<
I?
EXERCISE 10-32: DIFFERENTIAL EQUATIONS Let F be a continuous mapping of ~ x~ into lI!. We propose P P P to show that there exists a differentiable function ! defined on ~ to lI! such that: P P rex)
= F(x.!(x»
with
teo)
o.
483
p-ADIC ANALYSIS
If x eZ';
p
then x can be written:
+ •••
with 0
~
a. < p - 1. 1.
Set:
Set: M
=
sup IF(x,y) I , xeZ'; P yeZ';
and let k
P P
e:N
M -(k+1) p
be such that:
" 1.
Set:
and for n > k define f
n
by the recurrence relation:
f n (x) = f n- l(x) + (x - x )F(x ,f l(x ». n n n nn (1) :
Show that f (x) is defined for x eZ'; .
(2) :
Find an upper bound for:
P
n
If n- l(x)
- f n (x)1 p ,
and deduce from it that limf (x) exists for all x eZ': . n+oo n
p
CHAPTER 10
484
Set: I(x)
= nlimfn (x).
Find an upper bound for: II(x) -
f n (x)
p
and verify that 1(0)
O.
(3) :
Show that:
(a) :
For all e: > 0 there exists N e:N such that n '" Nand:
Ix - x'i
< p-(n+l)
implies:
If
In
(x) -
f
x - x
~
If m
(b):
(x')
>
- F(x,l(x))
I -<
e:.
N '" N, then the inequality:
implies:
If m
(x) -
f
x - x
~
(x')
_ F(x,l(x))
Deduce from this that
I -<
e:.
1 is differentiable
on~
p
and that:
f' (x) = F(x ,I(x) ). EXERCISE 10·33: CONGRUENCES FOR BELL NUMBERS Let P be the n-th BELL NUMBER, that is to say the number of n
485
p-ADIC ANALYSIS
partitions of a set of n elements (Po
1).
We know (cf., [Com])
(1 -
(ph - l)X)
that:
Show that:
(1):
L
n>,.O
(1 -
X)"'(l -
nX) •
From this deduce that: h
P -1
L
n~O
P
n
L
r
Xn (l -
(n
+ l)X)'"
~n~=~O ________________________~ ____ modph~[X].
(1 -
X)"'(l - (p
Using this deduce that the (P) n
h
- l)X) -
h
xP
h
n~
0 form, modulo p , a linear
recurrent sequence, that is to say, that there exist integers Ai h,···,A k h such that: , h' + ••• +
(2):
Show that if:
k(p) _ pP -
1
- P - 1 '
then: P
n+k(p)
- Pn mod p.
- 0 mod p
h
486
CHAPTER 10
SOLUTIONS
Let q = q 1 cardK. let x e k*. We know that x - = 1. therefore i f the absolute value on k is denoted 1·1 we have Ixl q- 1 = 111 = 1. hence Ixl = 1 for all x e k*. If L is an algebraic extension of k then L is itself a finite field. and the preceding argument can be applied. SOLUTION 10·1:
On
k every absolute value is trivial.
The valuation ring A of K(X) for the P-adic valuation is the set of rational functions ~ such that (R.S) = 1 and (S.P) = 1. The valuation ideal I is the set of rational functions R S such that (S,R) = 1. (S.P) = 1 and P/R. The residue field is A/I. and by Bezout's Identity: SOLUTION 10·2:
_ ARK[X~ _ KIX] A/I - I K[X - P(~)K[X] . If P(X) = X. the completion of L is the set of Laurent series with coefficients in K having only finitely many negative indices. If P is an irreducible polynomial the completion of L is the set of Laurent series in P(X) with coefficients in a complete system of repre~entatives of K[X]/P(X)K[X] (hence in a finite extension
487
p-ADIC ANALYSIS
of K) and having only finitely many negative indices. For this to be so, it is necessary and sufficient
SOLUTION 10-3:
(cf., [Ami]) that the group of values of v
p
be discrete (which is
the case here since it is~) and that the residue field be finite. Now, the residue field is a finite extension of K. Therefore it is necessary and sufficient that K be a finite field.
SOLUTION 10-4: (a):
The valuation of P(X,Y) is, in fact, the
T-adic valuation of the polynomial P(~,~). of v is the additive subgroup
= a~ +
hence v(K(X,Y» (b):
of~
The value group
generated by a and b
on~,
b~.
The valuation ring is the set of rational functions
P(X,Y) Q(X Y) suc h·· th a t ,
P(~,~) -h Q(~,:r)
has T-ad'lC va l ' uatl0n
possibly after simplifying, such that Q(O,O)
~
+ 0.
°, hence,
The valuation
ideal is the set of rational functions P(X,Y) such that P(~,~) Q(X,Y) Q(Ta,~) has T-adic valuation greater than zero, hence, possibly after simplifying, such that Q(O,O)
+ °and
P(O,O)
= 0.
The residue
field is K. (c): The completion K(X,Y) is isomorphic to the set of Laurent series:
I
a
(m,n)e~
2
tnr
m,D
such that: lnf
a
+0 m,D
(am
+
bn) > -
00.
488
CHAPTER 10
is locally compact if K is finite and if the group gen-
~(X,Y)
erated by a and b on z::: is discrete, hence if a and bare Z:::-linearly dependent. SOLUTION 10'5: (a):
2
If a
(b):
(1 +
This is immediate.
1 + P b, bez::: p ' then:
p)p - a - 0 mod p 2 ,
and:
p(l + p)p-1 _ 0 mod p, and:
Hensel's Lemma (cf., [Ami]) shows that there exists a solution of the equation:
+ pX)p - a
(1
in
mp '
and, furthermore,
x(c): m~'
p
hence:
now,
1 mod p. We have: Z:::xTx
(1
+ pz::: p ) with T '" (z:::/pz:::)"',
489
p-ADIC ANALYSIS
and
+
(1
p'll )
P
1 +P
2
'llp'
therefore:
2
'" rp,/P'lZJ . L has a normal basis over K if and only if L is
SOLUTION 10·6:
not ramified (cf., [Sam]) over K (by Condition N, [Ami]). If the valuation on L is normalised in such a way that its value group as well as that of K is'll, then L has normal bases over K formed of elements of L integral over K. SOLUTION 10·7: (a):
f a (x)
I
n~O
We have: e C('ll ,a) p
In fact this formula is true for all x e
if
~
a e L.
and therefore, by
continuity, for all x £ 7 (since la - 11 < 1). P (b): The mapping ~ is injective, since f (1) a
continuous, since:
a; i t is
We have: for all x eN, and therefore, by continuity, for all x e'llp; in fact, if a,b e L, then so does abo
490
CHAPTER 10
SOLUTION 10'S:
q.
Let us set cardk
Ixq - xl
< 1
M(f,l)
1
if
Ixl.;
We have:
1,
but:
SOLUTION 10'9: (a): if k
<
0, M(P,k)
(b):
f(x)
if
= xq
-
x.
If k > 0, M(P,k) > 0; i f k
= k(p
= 0, M(P,k)
- 1).
Let us assume that h(O)
1,h(1), •.. ,h(k - 1) are
determined; then: 1
for xe(ll
p
and
v(x) ~
-k.
Therefore: k(p -
1)
= h(k),
+ (k - l)(p - l)h(l) + •••
which determines h(k) uniquely if h(l), ... ,h(k - 1) are known.
(c):
Let us show that the sequence of polynomials Pk(x)
converges uniformly on every disc D of Pk + 1 (X) - Pk(X)
We have:
= Pk(X)(
(Il.
p
In fact,
- 1 + (1 _ pk+1 X )p-1)h(k+1».
p-ADIC ANALYSIS
491
For k large enough we have: and The sequence k
+ Pk
sup Ip k+1x I xeD
..
E.
is therefore a Cauchy sequence in the topol-
ogy of uniform convergence on
every bounded subset of
~
p
, and
therefore has a limit f which is an entire function bounded by one on
~
p
. Let us consider the identity:
SOLUTION 10'10:
o
for all k >
o.
We have: 1 - (1- 2)2
k
2k
_ (_1)n 2 n 2:...[2 ) 2k n n=l
2k
Now, ifn - k 2n 2k
-~
~
h, then:
2h ,
therefore, for n
~
k + h, is divisible by 2h
(_l)n 2n[2k) 2k n
(since
k
L
[~) e:
On the other hand:
(_1)n-1 n
2 + ,M(n), n.
where M(n)e:
492
CHAPTER 10
Now, the largest power of 2 which divides n! is:
And therefore if n - k k -
I [~]
i,l
2l
< h
we have:
> k - n > h,
hence: n 2k +n M(n) (_l)n- 1 _2 + __
n
n!
Lastly, let us notice that:
-
Cl
2
h k mod 2
i f k ~ h,
and that:
where: :A(n,m) =
Inf(n -
[~] log 2
'm -
[~]) log2'
hence the result. m
+... , and
Let Cl e ID p ' Cl = aJ? expansion of a is periodic. that the Hensel SOLUTION 10,11:
ist
nO
and r such that a n+r
Cl
n for
n ~ nO'
let us assume
Therefore there exTherefore:
493
p-ADIC ANALYSIS
+ ••• ) + •••
+p
n o+2r-l
+ ••• )
no nO+I'-l a p + ••• + an +r-1P + __n~o____________~o~_________ 1 _ pI' therefore (l e !D. Conversely, if a e!D, then a = ~ with m,n e~. There exists n' n such that if n > 0, then n'n (pI' - l)ps (this is an application cp(n) of Lagrange's Theorem which says that p - 1 is divisible by n if (n,p) = I, where cp(n) is Euler's TOTIENT function). So let us assume that m > 0 and n < 0 (in order to fix our notation). Theremn' By possibly fore there exists n' > 0 such that a = ---------pS(1 _ pr) modifying n' we may assume:
Therefore: a
=M +
R
1 _ pS
+ mn' pS
WI. th
0
R < PS -
L'
~
1 , M e .... .... . T
Hence, if we set:
mn ,
with k
< 8,
for l + 1 a multiple of
8
= mo" + m1p
and l
~
, j , + ••• + m.p J
sup(i,j), we have:
494
CHAPTER 10
+ ••• +
+ •••
+ ••• • The Hensel expansion of a is therefore periodic with period s for n~.e.+1.
SOLUTION 10·12: y - 1
= u.
Let y
+ 1 be
a p-th root of unity in
Hence u is a root of the equation:
Now,
therefore we must have:
by the u1trametric inequality, and therefore:
which is impossible i f u SOLUTION 10·13:
la We have:
10g41 ....-
em. p
We want to find ae:N such that:
(_13) 5
m. p
Set
495
p-ADIC ANALYSIS
since:
15 J
n
13n l < (_13
whenever n > 5.
Now,
3 -
32 33 34 "2 + "3 - '"4 = 12
-
99
'"4 = 12 +
243 - 99 mod 35 4
48 mod 35
SOLUTION 10·14:
We know that
(~)
e:N for all n e:N for all n e:N.
Therefore:
Now: Z?:
P
= {x e IIIP : Ix I <
1}
and:N is dense inZ?:p' and finally Pk(x) is a continuous function of Z?:
p
into III , therefore: p
i f x eZ?: . p
SOLUTION 10'15:
It suffices to show that if:
up = {x eZ?:p : Ix I = 1} ,
496
u2
P
CHAPTER 10
is open in~
P
and to end using a homothety. 2
fices to show that if the equation X
=0
. X3 - (x + h) t h en t h e equat10n
- x
=0
Therefore it sufhas a root (x e U ). P
has a root for every suffici-
ently small h. . X2 - x ybe a root 0f h t e equat10n
L et
the sequence:
a
... ,
o-
2
a
n
a
a n - 1 - (x - h)
n-l -
2a n _1
Hence: h y + -
2 h2 a 1 - (x + h) = ---
and
2y
4/
Let us assume that: Y + h + A h 2 + ••• + A 2y
n-l
2
and: 2 a n - 1 - (x + h)
2a n _1
therefore: a
n
and: a 2 - (x + h)
with
- An
hn - 1
o.
Let us consider
497
p-ADIC ANALYSIS
+ h)
(x
with I~n+ll < 1. Ihl
Therefore the sequence (an)n~ is convergent if
1 and its limit satisfies:
<
a 2 - (x + h) a - --"""'"'2"""a---'-
or, again,
o.
a 2 - (x + h)
SOLUTION 10'16:
We have:
By Taylor's formula, with n being the degree of f, we have: n-2 (a 2 - a
Now, (k)
f k! (X) e O[X] , therefore:
Hence:
1
)
2[r(a1 )
2!
(a 2 - a 1 )
+ ... + -=----,,..=--- f(n)(Nu 1 ) ] · n!
498
CHAPTER 10
and:
Now, la 2 - all
<
I, therefore:
Then by induction we show that: Ir(a.)1 ~
= 1,
and that: la.~ - a.~- 11
<
If(a1 )l i - 1
Therefore:
o.
limla. - a. 11
i-+<»
~
~-
The sequence (a.).~, is therefore a Cauchy sequence in F, so it ~ ~C&, has a limit a. We have: a
f(a) = a -f'(a) ,
therefore f(a)
=0
because If'(a)1
= 1.
Use Hensel's Lemma in the form in Exercise 10'16 (Newton's Method). Thus we are looking for a 1 e1l 5 such that:
SOLUTION 10'17:
It suffices to look for a 1 in the set {O,l,2,3,4} which forms a complete system of representatives for 1l /51l. We show that
p-ADIC ANALYSIS
a1
= 1,
499
= -10
because P(l)
and P'(l)
there exists a e2Z 5 such that ition (Exercise 10-16), if:
a2
a
1
P(a 1 )
-
pi (a 1 )
-
P' (a 2 )
P(a)
=0
= -1. and
Then (Exercise 10-16) Ia -
a 1 15
~
t.
In add-
- 9,
and if:
a3
a
2
P(a 2 )
1,531
- 2'59
then: la 3 - al 5 ~
1
?"
,
and therefore:
Furthermore, we can replace a 3 by any number 83 e2Z 5 satisfying:
and by the u1trametric inequality we would also have:
and so by any number of the form: -
1,531
~
+ 125
\
i L a.5 ,
i~o
~
with 0 < a. < 4 (cf., Proposition 2 of the Introduction). ~
CHAPTER 10
500
In particular, we can always choose 63 e!iZ and 0 .. 63 .. 124. = 31 + 1,500, and 259 = 9 + 250, therefore:
Now, 1,531
1,531
25'9
= -
31 + 1,500 9 + 250
1 and
31
- """9
1915
31 = 1 - 10
31 - - + 125k',
where k' e!iZ 5
9
= 1);
on the other hand:
31(1 + 10 + 10 2 ) + k".125,
where k e!iZ5. H
What is left is to find a representation of 3,441 modulo 125 lying between 0 and 124.
The remainder from dividing by 125 gives
Therefore a solution in m5 to (1/5)
63 = 66.
3
P(x) = x 3 - x 2 - 2x - 8
is 63
= 66.
SOLUTION 10·18:
xP
Let us set y
~h
~h
- 1.
- 1
X - 1
then
h
'
is a root of:
We show by induction on h that: yP
and that:
h-1
(p-1) mod p!iZ[X],
Let:
near to
501
p-ADIC ANALYSIS
therefore we necessarily have: 1
= -p
1 lip
and
h-1
(p-1)
p
(k)
(k)
10·19: Let (xl ••••• x h ) be a solution of the congruk n ence F(x 1 ••.. •x n ) :: 0 mod p. As (:ll:p) is compact we can extract from the sequence k ~ (xik) •...• x(k» a subsequence converging n n towards (xl •..•• X n ) e:tZ p ' and we have F(x 1 •...• x n ) = O. Conversely. if there exists (xl ..... X n ) e:tZ~ such that F(x 1 ..... x n ) = O.
SOLUTION
then for all k e:N one can find F( xl(k) •••• •x (k» n
(xik ) •••. •x~k»
e:tZ n such that
- 0 mo d p. k f or:tZ n.IS d ense In:tZ . n an d F IS . a = p
continuous mapping of:tZ n into:tZ • whose restriction to:tZ n has p
p
image contained in:tZ (cf .• [Bor] p. 44). 2
10·20: If X + X + a has a root in W2 this root must belong to:tZ 2 • for the polynomial is monic and has coefficients in :tZ.
SOLUTION
Let a be a root. a e:tZ 2 . Therefore Ia I '" 1. If Ia I < 1. then the equality a = - a - a 2 implies lal < 1. therefore a is even. If
lal
1. then:
and therefore:
Conversely. if a is even. the image of the polynomial X2 + X + a in (:tZ/2?4[X] = F 2 [X] is the polynomial X2 + X. which splits in
F 2 [X] into a product of two monic. relatively prime factors. Therefore by Hensel's Lemma the polynomial X2 + X + a is reducible in W2 [X].
502
CHAPTER 10
SOLUTION 10·21: (1):
P(x)
= Xn
Let:
+ a 1~-1 + ••• + an
be the minimal polynomial of a over
m.
Set a O
For A ez*
1.
we have:
P(A) =
P(a)
+
=
(a -
A)F(a,A),
(a -
A)p'(a) + ••• +
and:
F(a,A) since P(A)
+ 0, + 0,
for P is irreducible over
m.
Therefore:
Now, by the Product Formula,
and:
therefore: where c
i 1 = max IF(a,A>1 na;T' A eZ
and c
+ 0,
since:
P
~
p-ADIC ANALYSIS
max
A e:iZ
503
IF(a,A) I < + co, P
because:iZ C:iZ , which is compact. P
(2):
The result is now proved.
Let us assume that for an infinite number of values of
+ ±l)
N e:N there exists A e:iZ (A
such that:
O
+±l) o
<
IAI
in fact
o
such that:
<
> 1, and therefore if m
< N we
have:
la - AI P
Lastly, this implies that for every constant e e R+ -
{a}
and for
every n e:N there exists A e:iZ* such that:
o
<
Ia
- A
I < _e_ PlAin
,
so there exists m(n) such that:
The latter is the opposite of the result proved in Question (1) above, therefore a is transcendental. (3):
We have:
504
CHAPTER 10 1
b N+1
P
Now,
and therefore by hypothesis there exists N large enough such that: bN n 1 - n b N+1 - b N+1
0,
>
which implies that:
-b-~-+-l < p
-)-n
-(-N;;--.:;:l-"'-bk
L
a~
k=l
Hence we have found an infinite number of integers n, and for each n an A eZl such that: O<\a-A\
EXAMPLE:
(k
<_1_
PlAin
Clearly:
k! +
1
1)!
SOLUTION 10·22:
P(X)
k+T+
O.
This is a matter of showing that the polynomial:
~ - e:
p-ADIC ANALYSIS
505
To the polynomial P we associate its image P
has a root in W. in
F
p
p
We have
[X].
P(X) =
xn - 1.
The identity:
x.lD - 1 = (X - l)(x.,rn-1 + x.,rn-2 + ••• + 1) shows that 1 is a root of the second polynomial only if
m
* 0 mod p, the polynomials: X - 1
.lD-1 + x ···
and
+ 1
are relatively prime inF [X].
xn -
clude that
p
Hensel's Lemma allows us to con-
is divisible by a polynomial of the first de-
€
gree in W [X]. P
The inequality:
SOLUTION 10'23: (1):
v
P
(y - 1) ~
f +
1
implies that: y
1 + p f+1 Y1
with
IY1 Ip
~
1.
Therefore: logy
= log(l
+ P
f+1
Y1 )
is meaningful, and hence: 1 / /~
p
f+1/ ~ /~ =p1
Consequently e
logy/m
makes sense, and we clearly have:
506
CHAPTER 10
Therefore e logy/m l'S an m- th roo t
0f
y.
(2):
We are going to separate the proof into two steps: (a): If an extension Kp of IIlp contains p1/m and a 1/m for all a e U , then K contains 811m for all 8 e III . p
p
In fact, i f 8 e IIl p ' 8
h
P
P a where a e U. p
Therefore, since:
him 11m a ,
p
and since: P h 1m
=(
p
1/m)h e K
p'
as is a 1/m , we deduce that 811m e K • P
(b): Let us show that there exists a finite extension of IIIp .. contalnlng a 11m f or a 11 a e U . p
Let A be the following set: A 1 ... i ..
If an extension K
of III
p
p
contains a 1/m for all a e A, then for
all a e U we can find a e A such that: p
1) v p (~a
~
therefore: e III , p
P
f+1
,
f.
507
p-ADIC ANALYSIS
by Question (1) above, and therefore ex Now A contains pf(p - 1) elements.
11m
eK •
There~ore
by adjoining to m the m-th roots of the p (p - I)-elements of A and p 11m ,we f
p
have an extension of m containing the m-th root of every element f P of m. Therefore mp (p - 1) + 1 is an upper bound for the number P of m-th roots of elements of m to be adjoined to m so as to obP P tain an extension K of m containing all the m-th roots of the P P elements of m . P
Because of the equality:
SOLUTION 10a24: (1):
we have: am __ bm + ~
[ml]palbm-l ~
+
aaa
+
(m]k Pk 61kbm-k
+
aaa
+ p mam ~1'
Now the equality:
= (m - 1]~ (m] k k -1 k implies:
and therefore for k ? 1:
Now, if k
v (k) P
= pha, =h
"
with (a,p)
= 1,
we have:
log k '" k - 1, P
because the curve x
~
log x - x + 1 has a maximum onE* at x P
1.
508
CHAPTER 10
Therefore:
Hence we have proved that:
a
(2) : 11m ;
Let us asswne that there exists Sell! the equality
v(a)
=0
implies v(S)
p = 0,
S can be written:
S
=b
and V(Sl)
+ PS1 ~
with 1
~
b < p - 1,
0; the preceding question shows that:
with v(S') whence the congruence: m
b
:: a mod p
f+1
.
Conversely, if the congruence: bm :: a mo d pf+1
holds then a can be written: with V (a') p
and since V (b) p
~
0,
~
O.
= 0 we have:
a - = 1 + p f+1 a "
with
Then by Exercise 10·23(1):
V
P
(a")
~
0,
such that and therefore
509
p-ADIC ANALYSIS
e Q1
P
and a 1 / m therefore belongs to
~
shows
e + PY1 with 0 (e
~
P - 1
0, we obtain:
yP = e P + p2Y2
(since
P
By writing y
SOLUTION 10-25: (1):
and Vp (Y1)
Q1 •
with Vp (Y2) ~ 0
Vp«(~]) ~ 1 for 1 that a P = a mod p,
v (yP - y) P
( k (P - 1).
Fermat's little Theorem
therefore:
1.
?
Let us set:
It follows that:
The term inside the square brackets contains (yp-1 - 1) as a factor, therefore:
H
k
= y (y
p-1
- 1)
2
hence v (H) ? 2 when k P
(2):
H
1 ?
2.
(For k = 1, H = 0).
The preceding inequality shows that if, for an integer
satisfying 1 ( k
~
p - 1, the inequality:
510
CHAPTER 10
is satisfied, then:
v
P
(pp - p) ~ 2.
Now,
implies the inequality:
v
P
(ph - oP) > 2
if k > k', and we set k - k' = hand 0 = 8'/8. By writing 0, which is a unit of m, in the form: P
o = d o ... po 1 we deduce from this:
From the inequality:
v
P
~ 2
(ph - oP)
we deduce: p
h
= d Po +
with vp(n 1 )
~
p
2
n1 h
0 and p
= do
+ pn', with v(n')
~
O.
Hence:
511
p-ADIC ANALYSIS
and therefore:
v
P
(php _ ph) > 2
(3):
with 1
For p = 2, that r
consider (p - 1)
(p - 1)
p-1
h
~
~
p - 1.
3 is obvious.
If P
+2
let us
:
P-1
all the terms [ p j- 1) p p-1-j are divisible by p 2 for 1
~
j ~ p - 3,
and for j = p - 2 the corresponding term is divisible by p and not by p2 Therefore v «p - 1)p-1 - 1) 1. If there were p
k,b,k',b' satisfying the hypotheses of the Problem and such that k k' 2 r bP = r b'P mod p , we would then have: v (r P - 1 - 1) ;. 2 P
by Question (2) above). Then letting ~ be a unit of ~ , the numbers rkb P forming, 2
P
modulo p , a complete system of residues, there exist band k with 1
~
b
~
p - 1 and 0
or again:
By Exercise 10·24:
~
k
~
p - 1 such that:
512
CHAPTER 10
-& [k
r
Jlip em, P
therefore
-&
e m (p G) . P
SOLUTION 10'26: (1):
Let us set:
As X2 - y is irreducible over Wp ' Co is not a quadratic residue mod p (cf., Exercise 10·24(2)). Let 0 e U be such that: p
and
First: do may be a quadratic residue mod p, and then (cf., Exercise 10,24(2)) the polynomial X2 - 0 is reducible on
mp ;
otherwi·se dOis not a quadratic residue mod p, but then cod o is a quadratic residue mod p (this is a property of the Legendre symbol), and therefore X2 and therefore
ro e mp [/Y].
yo
is reducible in W [X], hence p
The polynomial X2 - 0 is therefore reducible over (2):
P(X)
mp [/Y].
Let:
= X2
+ 2XX + ~
be a polynomial of W [X]; put it into the canonical form: p
P(X)
= (X
+ X)2 + ~ _ X2.
Let us set: ~
2
- X
= Pk £
where k
e~
and
£
eU • P
/Y6emp ,
513
p-ADIC ANALYSIS
= 0,
If k
then
reducible over
+ 0,
If k ~
~
mp [IY].
= 2h,
and k
2
2
2h
- A =P
e U , and therefore the polynomial P is
- A
P
where h e:iZ, then:
£,
therefore:
and
P(X)
If k
and
is still reducible on
+ 0,
and k = 2h + 1, then:
is reducible over
P(X)
mp [IY
mp [IY].
+ IP]
c mp [IY]
mp [IY] [vp]
[IP] .
But conversely, since:
2/YP =
[IY + vp]
2
- y - p,
we see that:
/YP
e
mp [IY
+
IPL
and that:
/yp(1Y +
vp)
e
mp [IY +
vp],
and from this we deduce that:
IY
e Qlp [IY +
IPL
the same holds for Ip .
.
It is obvious that:
514
CHAPTER 10
The splitting field of all the second degree polynomials of
[X] is a fourth degree extension of polynomial of ;.y + Ijp, which is: ~
p
SOLUTION 10·27: (1):
~
p
defined by the minimal
We can either use Exercise 10·23(1), or we
may use a slightly different method. By hypothesis we have a = 1 + 23b where x
=1
Y
+ 2y. 2
Ibl2
< 1.
Let us set
The equation becomes:
+ Y - 2b =
o.
Passing to the residue field the equation becomes:
y(y +
1) =
o.
Hensel's Lemma then implies that the polynomial y2 + Y - 2b factors in
~2'
and therefore that a is a perfect square.
ly, if a is a perfect square, then: and
a
=
(1
+
2c)
2
Therefore: a-I
Now in
F
a(1
= 4c(1
+ c).
2 we have: + c) = 0
therefore:
for
a 0 or I,
where
1c 12 <
1.
Converse-
p-ADIC ANALYSIS
515
and therefore:
(2):
Let us assume that
lal 2 = Ibl2 = 1.
The extensions
determined by X2 - a and X2 - b are identical if and only if: or or again: 3
b - a mod 2 ::Z2'
Now ::Z3/23::Z2 has four elements and one can choose 1,3,5,7 as 2 2 representatives in::z 2 . The extensions of ID2 by X - 1, X - 3, X2 - 5, X2 - 7 are therefore distinct. Let us note that the ex. tens10n 0 f ID2 by X2 - l '1S ID 2 •
Assume that la!2 = Ib l 2 = !. We then obtain four distinct ex2 2 tensions defined by X - 2, X - 6, X2 - 10, X2 - 14, and each of these extensions is ramified, for the polynomial which determines it is an Eisenstein polynomial.
In fact ID2 [ra] = ID2 [IF] i f and
only if ID 2 [ra/2] = ID 2 [1F/2] , and we finish using the result proved in the first part of this Exercise. Therefore altogether we have seven disinct quadratic extensions of ID 2 • (3):
In X2 - 3 and X2 - 7 let us make the change of vari-
able X 1 + Y, which gives the Eisenstein polynomials y2 + 2Y - 2 and y2 + 2Y - 6. (4) : . glves y2
Making the change of variable X = 1 + 2Y in X2 - 5
Y - 1. InF 2 this polynomial becomes y2 + Y + 1, which does not have roots inF 2 ; it therefore defines an extension of degree 2 ofF 2 .
The inertial degree f is therefore two,
516
CHAPTER 10
and the ramification index e is one, because ef is equal to the degree of the extension of
m2 by
X2 - 5.
In conclusion, we note that there are six ramified extensions and one non-ramified extension of
IF.
p
If P
degree two.
Let us examine the polynomial X2 - X-I
SOLUTION 10·28: (1):
in
m2 of
= 11 we want to show that:
b)
X2 - X - I = (X - a)(X -
with ab = -1 and a + b
1.
Hence we must find two integers a,b such that: ab =: 10 mod 11
a +.b _ 1 mod 11.
and
It suffices to take a
=
4 and b
=5
and b (2) :
in
By Hensel's Lemma the poly-
and -\1,.e: e m 11 . 15, and the same result holds.
nomial X2 - X-I factors in a
8.
=
If p
'Zl./7'Zl. = IF 7.
m11 ,
For
mIg
we take
7 let us show that XL - X-I has no roots In fact it would be necessary to have two integers
a,b such that 1 ~ a mod 7, therefore:
~
6, 1
b _ 8 - a mod 7
~
b < 6, a + b = 1 mod 7, and ab =: 6
a(l - a) =: 6 mod 7,
and
which is impossible for 1 ~ a ~ 6.
The polynomial X2 - X-I
therefore determines an extension of degree 2 oflF 7 , therefore the residual degree of index is one. (3):
m7[-\1]
over
m7
is two, and its ramification
Therefore the extension is not ramified.
We have:
N(-\1 - e)
Then: 1
ife
3 and 0 otherwise.
p-ADIC ANALYSIS
517
Making the change of variable X
=3
+ Y, gives y2 + 5Y + 5.
This polynomial is an Eisenstein polynomial. is therefore totally ramified over
The extension
~5[~]
~5'
Let us make the change of variable X = 1 + Y.
SOLUTION 10· 29: (1): . X7 For t h e equatlon
1 satisfied by
this gives us:
~,
The polynomial:
is an Eisenstein polynomial, it is therefore the irreducible polynomial for degree six, and
- 1).
(~ ~
Hence
~7[~]
is totally ramified and of
- 1 is a uniformizer for
since:
~7[~]'
1 6
(2) :
In order that the equation X7 - 1 be solvable in
it is necessary that it be solvable inlF contain the seventh roots of
Therefore IF
p
I, and so OF
P
)* contains a subgroup
of order seven, and hence that 7 divides p - 1.
Conversely, if
7 divides p - 1 the polynomial X7 - I factors inlF of seven factors of the first degree.
p
~p
must
p
as a product
In fact, the elements of
OF p )* satisfy the equation:
which therefore splits completely inlF [X], and X7 - I divides p-l p 7 the polynomial X - I. By Hensel's Lemma X - I has seven distinct roots in
~
p
.
518
CHAPTER 10
(3):
If
is of degree two on
~p[~]
mp '
the residue field of
m [~],
IF is of degree one or two over IF (hence s ps p
In
we have:
p
JF
p
S
~7
If s
=
I
=
1 or 2).
I,
and
1, by the Question (2) preceding we have
contradicts the hypotheses [OJ
7 divides p2 - 1, so by
[~]:m
] = 2.
Gauss~s Le~a,
~ e
OJ , which p
= 2,
Therefore s
and
7 divides p - 1, or 7 di-
vides p + 1; i f 7 divides p - 1 then by part (2) above
~
em, p
hence 7 divides p + 1, or, again: p =: -1 mod 7.
Conversely, if P =: -1 mod 7, then 7 divides p + 1, therefore 7 divides p2 - 1, hence X7 - I has seven distinct roots inlE' 2' and p
therefore X7 - 1 determines a non-ramified extension of degree two of
mp .
(4):
Similarly to the preceding Question (3), in the residue
p
mp [~]
~7 = I
and
fie1dlE' 2 of
we must have:
1,
therefore 7 divides pS - 1. then s = 1, 2 or 3.
If the extension is of degree three,
Therefore 7 divides p3 - 1, and so;
p3 =: 1 mod 7, which implies: p =: 2 or 4
Conversely, if
mod 7.
71 p3
- 1, the polynomial X7 - 1, which determines
519
p-ADIC ANALYSIS
the extension of degree three.
mp ,
Therefore:
[m [.e]:111 ] = P
defines an extension of the residue field of
P
3,
since X7 - 1 splits completely in
mp [.e],
by Hensel's Lemma.
6, then necessarily: p
=3
or 5 mod 7,
for otherwise we are in the situation of parts (1),(2),(3) or (4). Conversely, if p
=3
or 5 mod 7, the smallest s such that 7 divides pS - 1 is s = 6. Therefore the polynomial X7 - 1 determines an extension of degree six OfF, since we always have [Ill [.e]:m ] ~
6.
And so we have equality.
SOLUTION 10-30: (1):
p
p
We notice that i f x e:N, then
is a rational integer, therefore: i f x e:N C :?Z
(cf., Exercise 10-14).
and it suffices to set x
P
Now:N is dense in:?Z , therefore: p
= k.
If (ak)k>O is a sequence of elements of
then the series:
~p
such that:
p
520
CHAPTER 10
converges uniformly 1y on
~
p
towards f e
(2) :
Now
f e
C(~
that:
on~
p
, and consequently it converges uniform-
C(~ ,0: )
p
p
since, for all
neN:
We have:
p
,0: ), therefore for all p
In-n'l
E
> 0 there exists h eN such
~p-h
P
implies:
If(n)
- f(n')
I
P
< E.
Therefore if we set: h
P -1 ~
Fh(X)
n=O 1 -
f(n)r h
xP
we have:
Notice that the sequence h -+- F h converges uniformly on O:p - B( 1,1-) . In fact:
521
p-ADIC ANALYSIS
h
p-l P -1
I
I
k=o
(f(n + kph) - f(n)
~
P
- B(l,l h+l
11 - xP
h
n=O
1 - xP
and on
xn +kp
h+l
we have:
Ip
1
Ip
Ixl PP
if
Ixl P
~ 1,
or:
11 -
xP
h+l
h+l
(cf., Exercise 10 18).
~
h.
p
- B(l,l-) towards an analytic element that
again is denoted by F.
Ix -
1
>
The sequence (F h )h~O of rational functions there-
for converges on ~ for
Ixl P
Therefore:
0
whenever h'
if
So there exist coefficients c. such that J
11 ~ 1 we have:
F(X)
c.
\
J
'+1 j~O (1 - X)J L
. h
Wlt
limlc.1
j-+oo
J
=
O.
Expanding the second member as a Taylor series in a neighbourhood of zero, and identifying it with: F(X)
I
f(n)X n ,
n .. O
we deduce that: f(n) =
I c.(n+~-lJ
j .. o
J
J
for all n e:N,
522
CHAPTER 10
and therefore, as f e C(~
f(X) =
L
j~O
p
,0: ),
p
c.P.(x + j ] ]
1).
From the identity:
we deduce:
f(X)
= L
ho
akPk(x)
with:
_ .L~
ak -
J~k
(j - 1)
O.
c .. - k ] J
SOLUTION 10-31: (1):
We have:
Let us set: .t(n)
[~J logp
.
Then:
as is easily seen using the identity:
p-ADIC ANALYSIS
523
Therefore: (_l)h-l
lim ..!...(pnJ n-- p n h
h
The p-adic absolute values of the zeros of A are, on
(2) :
n
the one hand,
_pi] 1/ (p-l)p (
h
with 0 " h
n,
<
A (x) is not ir-
and 0 on the other hand (cf., Exercise 10·18).
n
reducible on W , because: p
= ..!...
A (x) n
p
n
[(1
+ x)p
n-l
(p-l) + (1 + x)p
n-2
x [(1
+ x)p
x [(1
+ x) p-l + ••• + l]x.
-ffi.
+P
J
with
m.
(p-2) + ••• + 1]
(p-l) + ••• + 1] •••
If:
l'
n-l
1
J
then:
=P
·+1 -J·-l pJ l'
524
CHAPTER 10
where j is defined by: -m.
] >
p
Ixl
> p
m.+1
]
this is seen either by considering the Newton polygon of An (cf .• [Ami]). or by writing: A (x) n
= ...!.. IT n
yer
p
(x
+
1 - y),
where r h is the group of p-th roots of unity, and using Exercise 10·18. nlimA __ n (x) exists for all x
ea:p
such that
Ix IP
1, as the Tay-
<
lor series of the An (x)'s converge coefficientwise towards the series: (_1)h-1
h
x
h
and the A (x)'s are bounded for fixed x independently of n,
Ixl p
<
1.
n
We have:
f(x - 1)
= limA
(x - 1),
f(y - 1)
= limA
(y - 1).
n+oo n
n-- n
f(xy - 1) = limA (xy - 1), n-- n Now, A (x - 1) + A (y - 1) - A (xy - 1)
n
since:
n
n
(xp
n
-
l)(yP pn
n
- 1)
525
p-ADIC ANALYSIS
where M does not depend upon nand: n
1imlyP n-+
-
11 P
0,
and we have:
1im{A (x - 1) + A (y - 1) - A (xy - i)} n-+ n n n
0,
The radius of convergence of the series (_1)h-1
h
x
h
is one, this is seen either because the sequence A (x) converges for all x
e a: P
such that
xl P
1
< 1,
or because:
n
(3) : (1
is irreducible on
~
p
+ x)p
n-1
(p-1) + ••• + 1
, since it is an Eisenstein polynomial.
The
roots of B (x - 1) are the primitive pn_th roots of unity, for n
these are the zeros of An (x - 1) which are not zeros of A n _1 (x - 1). If x - 1 is a pn -th root of unity, then for m :;. n we have A (x) = 0, m
and therefore: f(x)
=
limA (x)
n-+ n
0.
526
CHAPTER 10
h
The p -th roots of unity (h f(x - 1) =
Conversely, let
~
0) are therefore roots of:
o.
Uh
x ,
r h , then IA
n
(x)
I is constant for
n
large
enough, and it is different from zero, therefore: limA (x) i n4-+OO n
o. h
The zeros of f(x - 1) are therefore exactly the p -th roots of unity for h " O.
SOLUTION 10·32: (1):
for all x that (x
e:
p
It suffices to prove that if f
then so is f
1,f (x
n
is defined
1·' and therefore it suffices to verify n+ 1)) belong to the domain of definition of F, that
n+ n n+ is to say that f (x 1) e:
f n - 1 (x n ) + (x
Now, we have: n+
1 - x
)F(x ,f
n
n
n-
1(x )),
n
therefore: f n (x n +1 ) p ~ max( Ifn- 1(xn ) Ip , Ix n+1 - x n Ip IF(x,f n n- 1(xn )) Ip ) ~
and as n
~
max(1,Mp
-( n+1)
),
k we have:
Ifn (x n+ 1)1 p ~1. (2) :
we obtain:
By replacing f
n+
1(x) and f (x) by their expressions,
n
x [F(x,f
n
- F(x n '1,fn (x n+1))]' n- 1(x)) n T
p-ADIC ANALYSIS
527
whence the upper bound:
If n+l () x
- f
n
(x)1 p
Mp-(n+2).
<'
...
The series:
L
(fn + 1 (x) - f
n~k
n
(X))
is therefore uniformly convergent for x e?l;. p
exists and is a continuous function on?l;. inequality we deduce that if m If (x) - f
n (x)
m
I
>
p
n
~
~
k,
Therefore limf (x) n-+oo n
From the ultrametric
k we have:
M -(n+2) p ,
~
and therefore that: If(x) - f
n
(x)
I
.. Mp -(n+2) ,
Furthermore, as: f
n
f n _ 1 (0)
(0)
for all n
we have: f(O)
o.
(3): (a):
Ix
- x 'I
implies xn = f
n
p
The inequality: .. p -(n+l)
x~,
(x) - f
n
therefore:
(x')
=
f
n-
l(x ) - f
n
n-
l(x') + (x - x
n
- (x' - x' )F(x' ,f
n
n
n-
n
)F(x , f
n
l(x'))
n
n-
l(x ))
n
= (Contd)
528
CHAPTER 10
(Contd)
n
On z::
x
p
z:: p , which is compact,
Hence for all
Ix
n- l(x n ».
(x - x' )F(x , I
- x'
Ip
F(x ,y)
0 there exists n
E >
~ n
0 such that:
>
Iy -y'l p .;
and
is uniformly continuous.
n
implies: IF(x',y') - F(x,y)1
P
~
E.
Let N eN be such that and
p
-(N+l)
~
n,
then:
II(x)
-
1n- l(x n ) Ip .; ~
Therefore, for n
Ix -
x n Ip ~
~
n
max ( II(x) max (Mp
1n (x) Ip ,IIn (x)
-(N+2) M -(N+l» , p
~
-
n.
N we have:
II(x)
and
-
1n- 1 (x n ) Ip
~ n,
hence: IF(x,/(x»
n- l(x»1 n p ~
- F(x , I
n
(b):
The equality:
f m(x)
-
1m(x') = 1m(x)
-
1n (x)
E-
+
1n (x)
- 1n (x') + 1n (x') implies:
- Irn(x')
1n- 1 (x n ) Ip )
529
p-ADIC ANALYSIS
If(X)-f(X') m x _ x~ f
- F(x,f(x))
I p
(x) - f (x'')
~ max [ I n x _ x~
- F(x ,f(x))
I p
,
I
Ifm(X) - fn, (x) x - x
p
'
Now, the equality:
x [F(x +l,f (x +1)) - F(x ,f
n
n
n+ 11 p
n
implies: If+ 1 (x)-f(x)1
n
n
and therefore for m
p
> n:
If(x)-f(x)I"Elx-x'l
m
n
p
p
similarly: Ifm(x')-f(x')I .. Elx-x'l n p p
Hence for m
Ix implies:
> n
- x '1
p
the inequality .. p -(n+l)
n
n- l(x))] n
530
CHAPTER 10
By making m
1x
7
00
we thus obtain:
- x 'I
~
p -(n+1)
which implies:
I[(X)x
n
-
- x
x ') - F(x'[(X))1
~
e:,
hence: 1·
1m
= F( x, [( x )) .
[(x) - [(x')
x-x'
x '-+x
Therefore [ is differentiable [,(x)
on~p'
and:
= F(x,[(x)).
SOLUTION 10'33: (1):
A formal calculation shows that:
and therefore that:
L
n~O
(1 -
r
X)···(l -
(Decompose into partial fractions). h
Modulo P 72:[[X]] we have:
L (1 n .. O
-
r
X)"'(l - nX)
nX)
p-ADIC ANALYSIS
531
n=O
(1 - X)···(l -
I
Therefore and so mod p
p
n ..O n h
(p
h
-
h
xP
l)X) -
Xn is a rational function modulo ph~[[X]],
the (P)
n n ..
0 form a linear recurrent sequence (cf.,
[Ami]) .
(2):
Modulo p we have:
p-1
I
n=O
X (1 - (n + l)X)···(l (1 - X)···(l -
p-1
I
n=O
satisfy: ~l+p+ ••• +p
p-1
xp- 1
F. p
-
xP
l)X)
is irreducible over
m, P
IF
p
for it its roots
= 1.
.,rk(p) = 1 , ~
(p -
In the algebraic closure of
(It suffices to calculate the norm of
and therefore in
l)X)
l)X) - xP
Xn (l - (n + l)X)···(l -
The polynomial 1 is irreducible over
(p -
(p -
P
its roots satisfy:
~
to
IF). p
Therefore:
532
CHAPTER 10
In addition, i f I',;i and I',;j are two distinct roots of 1 - Xp- 1 XP , then:
I1',1;.
- 1',;·1 ]
= 1,
and the Mittag-Leffler Theorem (cf., [Ami]) then gives the result:
F1 (X) =
where
IA.I 1
p-l n L X (1 - (n + l)X)···(l n=O 1 - Xp-l - xP
~ 1.
Consequently:
(p -
l)X)
Bibliography
[Ami]
Y. AMICE: Les nombres p-adiques, Collection Sup. P.U.F., (Paris), (1975).
[Bac]
G. BACHMANN: Introduction to p-adic Numbers and Valua-
tion Theory, Academic Press (NY, London), 1964. [Bak]
A. BAKER: Transcendental Number Theory, (Cambridge University Press), (1975).
Cambridge, NY.
A. BLANCHARD: Initiation
a
[Bla]
la theorie analytique des nombres premiers, (Dunod, Paris), (1969).
[Bou]
N. BOURBAKI:
[Bor]
Z.I. BOREVITCH and I.R. CHAFAREVITCH: Theorie des Nombres,
[Cas]
(French Translation), (Gauthier-Villars, Paris)i (1967). tng1ish trans., ~umber Theory, Academic Press, 956. J.W.S. CASSELS: An Introduction to Diophantine Approx-
[CaF] [Che]
chapitres IV et V, (2nd edn.), (Hermann, Paris), (1959). Alg~bre,
imation, (Cambridge Tract No. 45), (Cambridge Univ. Press) (1957). Canbridge and HY J.W.S. CASSELS and A. FROHLICH: Algebraic Number Theory (Academic Press), (1967). New York and London C. CHEVALLEY: "Sur la theorie du corps de classes',
J. Fac. Sci. Tokyo, II, (1933), 365-476.
533
BIBLIOGRAPHY
534
[Com]
L. COMTET: Analyse combinatoire, Vol.s 1 and 2, Collection Sup. P.U.F., (Paris), (1970).
[El1]
W.J. ELLISON and M. MENDES-FRANCE: Les nombres premiers, (Hermann, Paris), (1975).
[Gun]
R.C. GUNNING: Lectures on Modular Forms, Annals of Mathematics Studies No. 48, (Princeton University Press), (1962).
[Hal]
Princeton.
H. HALBERSTAM and K.F. ROTH:
S~quences,
Vol. I, (Oxford
University Press), (1966). Reprint Springer 1983 [Har]
G.H. HARDY and E.M. WRIGHT: An Introduction to the Theory
[Has]
of Numbers, (4th edn.), (Oxford University Press), (1960). ()"xford, New York H. HASSE: Zahlentheorie, (Akademie Verlag), (1963). English trans., Number Theory, Springer-Verlag 1981. J.R. JOLY: 'Equations et varietes algebriques sur un corps
[Joly]
fini', Enseignement [Kui]
Math~matique,
XIX, No.s 1-2, (1973).
L. KUIPERS and H. NIEDERREITER: Uniform Distribution of
Sequences, (J. Wiley
&Sons),
(1974).
[Langl]
S. LANG: Algebra, (Addison Wesley), (1965). Reading, Mass.
[Lang2]
S. LANG: Introduction to TPanscendental Numbers, (Addison Wesley), (1966). Reading, Mass.
[Lev]
W.J. LEVEQUE: Topics in Number Theory, (Addison Wesley), (1965).
[Mac]
Reading, Mass.
P.J. MacCARTHY: Algebraic Extensions of Fields, (Blaisdell), (1966).
[Mor]
Reprint Chelsea 1981 (NY).
L.J. MORDELL: Diophantine Equations, (Academic Press, London and New York), (1969).
[Niv]
I. NIVEN and H. ZUCKERMAN: An Introduction to the Theory of Numbers, (3rd edn.), (J. Wiley and Sons), (1972).
[Pis]
C. PISOT: 'FamilIes compactes de fractions rationelles et ensembles dermes de nombres algebriques', Ann. Sci.
Ecole Normale Sup., Ser. 3, 81, (1964), 165-188.
BIBLIOGRAPHY
[Rau]
G. RAUZY:
535 Propri~t~s
statistiques de suites
arithm~tiques,
Collection Sup. P.U.F., (Paris), (1976).
des nombres, Collect
[Sam]
P. SAMUEL:
[Sal]
Methodes, (Hermann, Paris), (1967). Trans. Algebraic Theory of Numbers, Hermon 1969 R. SALEM: ALgebraic NUmbers and Fourier AnaLysis, (Heath
Th~orie aLg~brique
Mathematical Monographs, Boston), (1973). [Sch]
T. SCHNEIDER: Introduction aux nombres transcendants, (Gauthier-Villars, Paris), (1957).
[Ser1]
J.P. SERRE: Cours
[Ser2]
(Paris), (1970). Trans. A Course of Arithmetic, Springer N.Y. and Heidelberg. J.P. SERRE: Corps Locaux, (Hermann, Paris), (1962). Trans. Local Fields, Springer-Verlag M. WALDSCHMIDT: Nombres ~anscendants, Lecture Notes
[Wa1]
d'arithm~tique,
Collection Sup. P.U.F.,
Series No. 42, (Springer-Verlag), (1974). [Wei] [Wid]
E. WEISS: ALgebraic NUmber Theory, (McGraw-Hill), (1963). Reprint, Chelsea, 1980. D.V. WIDDER: The Laplace Transform, (Princeton University Press), (1946). Princeton.
Index of Terminology
additive function 5
class field 193, 225, 232, 237
algebraic 171
Clausen 380
algebraic closure 171
conjugates 172
algebraically independent 309
continued fraction 426
almost periodic 244
convolution 7
arithmetic function 3
cusp form 360, 370
arithmetic progression 31
cyclic cubic field 187
asymtotic superior density 139
cyclic extension 200
Baker 309, 318, 321
decomposition 177
basis 127
Dedekind domain 176
Bell 27, 484
determinants 146, 147, 148, 368, 384, 408
Bernoulli 365
Dieudonne 316
Besicovitch 244
difference between primes 24, 44
biquadratic extension 186
difference between squarefree numbers 25
Brun 122
differential equation 482
Cantor 164
dihedral 184
Cauchy (inequalities of) 472
Dirichlet 11, 28, 180, 190
character 28
discrepancy 241
characteristic polynomial 172 Chinese remainder theorem 67, 69 536
discriminant 175
537
INDEX
e 431
Hi! bert 200, 408
Eisenstein 360, 370, 468
Hilbert symbol 408
equally distributed 241 ergodic 245, 259, 260
ideal classes 176
Euler 13, 306
Ikehara 13
Euler product 10
incomplete quotient 426
extension 168, 182, 466
inertia 178 integer 174
Fatou 149
integer basis 175
Fejer 243, 253, 273, 284
integral closure 177
Fibonacci 249, 297
integrally closed 174
finite field 173, 182, 470
irreducible fraction 33
formal series 369
irreducible polynomial 171, 363
fractional ideal 176 fundamental domain 361, 371
Khintchin 254 Koksma 243
Galois 173
Kronecker 147, 148
Ge1'fand 308, 316, 317
Kummer 202
generating function 11, 481 genus 362, 372
Lagrange 428
Goldbach 43
Landau 261
golden number 436
Lengendre symbol 179
group 370
Lindemann 308, 317 Liouville 4, 306
H-entire series 365
local field 194
Haar (measure) 252, 259
locally compact 469
Hadamard 150
logarithmic differential 368
Hankel 146, 148 Hardy 22
Mann 128
Hasse 410
Martinet 207
Hasse symbol 408
Mertens (formula) 20
height 308
Minkowski 180, 410, 432
Hensel 467, 473
Mittag-Leffler 532
Hermite 308, 317
Mtlbius 4, 7, 9, 363, 374
INDEX
538
modular 359, 369
rational fraction 145, 470
modular invariant 362
rational series 144
multiplicative 5, 369
reducibility 475 residual degree 176, 468
Newton polygon 524
residue class field 467
norm 172, 474
Riemann 12, 16, 362, 371
normal 243, 254, 258, 472
Roth 254
normal basis 472 number field 170
Schanue1 309, 321 Schneider 308, 316, 318
order of a basis 127
SchUtzenberger 169 Selberg 14, 41
p-adic 466
Siegel 313
p-adic logarithm 481
sieve 14, 41
partition 40, 366
size 308
Payan 207
Snire1man density 127
periodic (completely multiplicative function) 28
Snire1man 124, 128
'IT 323
pigeonhole principle 310 Pisot (numbers of) 257 prime nubmer theorem 1, 18 primitive 149 primitive roots 376, 480 product formula 410, 466
splitting field 171 square free numbers 25, 27, 29 Stie1tjes 3 sum L:f(p) 19 sum function 9 Sylvester 149 symmetric development 434
pseudo-random 244, 258
trace 172
p-th root 474, 475, 476
transcendental 170, 308, 476
quadratic (form) 408 quadratic field 174, 184 quadratic reciprocity 179
transcendental degree 176 transcendental extension 197 ultrametric 467
quadratic residue 179
unit 180
Ramanujan 22, 366, 369
valuation 466
ramification index 177, 468
INDEX
valuation ring 467 Van der Corput 243, 258, 285, 291 Van der Poorten 348 Von Mangoldt 4, 8 Von Staudt 380 Wey1 242, 251, 270, 291
539
Index of Symbols and Notations
N, 'll, IQ, lR, ([: The entire, natural, rational, real, complex numbers. N *, 'll *,
~*,
IF P = 'll /p'll; 'll 7n'll .
lR *, m*, The same sets, without zero. ('ll /n'll ) *
= The set of invertible elements of
(W) Binomial coefficient i, j, k, i, m, d Elements of N p, q Prime numbers, except for Ch. 9, where p/z denotes a convergent. (m,n), [m,n) g.c.d. and l.c.m. of m and n din; dIn; prll n; exactly divides n [x)
=
d divides n; d does not divide n, pr
entire part of x; {x}
=
x - [x); II xII
=
min Ix-al.
K : Algebraic closure of the field K, or residue class field.
IT (x);
e(x)
Ch. 1; 1
den), a(n), ~(n), v(n), Q(n), A(n), Ch. 1; 4, 7, 17
540
~(n),
A(n), zen), i(n)
INDEX
f
541
* G,
e(n), G, f-*
Ch. 1; 7
Ch. 1; 8
aT" F
s(A,p) Ch. 1; 14
°A
A(X),
Ch. 2; 127
K[X], K[[X]] Ch. 3, 144
LIK [L:K], K(a1, ... ,au), Irr(a,K) NLIK , Tr LIK , Gal(LIK)
Ch. 4; 170, 171
Ch. 4; 174
r 1 , r 2 Ch. 4, 180
ON' sN(a,S), sN(a) ep(x) , e(x) B((un ))
Ch. 5; 242
Ch. 5; 243
d(P), H(P), s(P)
W Ch. N(a)
Ch. 5; 241
Ch. 6; 308
6; 309 Ch. 6; 311
GLn(Q)
Ch. 6; 317
H Ch. 7, 359 SL 2 (A), r, 0k(n),T (a,S)v' d(f),Ev(f) [ao,al' ... ,an ]
Ixl p ' vp(x),
Ch. 8; 408
Ch. 9; 429
~p' ~p'
B(a,r-), B(a,r+) e,f
Ch. 7, 361
K, A, M, X,
Ch. 10; 468
Ch. 4; 178, Ch. 10; 468
Y
Ch. 10; 466, 467