Exercises in Number Theory

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)

=

~

, 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: