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C OMPOSITIO M ATHEMATICA

o

1 (1987), p. 53-61

<http://www.numdam.org/item?id=CM_1987__62_1_53_0>

© Foundation Compositio Mathematica, 1987, tous droits réservés.

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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques

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Conditions for algebraic independence of certain power series of algebraic numbers

KUMIKO NISHIOKA

Department of Mathematics, Nara Women’s University, Kita-Uoya Nishimachi, Nara 630, Japan Received 4 June 1986; accepted 8 September 1986

00

In what

K denotes an

algebraic

number field. Let

f(z) = 03A3akzek

k=0

be a power series with nonzero coefficients

ak E K,

the convergence radius R &#x3E; 0 and

ek

the condition

where

=

and

is

such that

are

and

the

number

f(03B1)

is transcendental for any

a with 0

R. More-

over

and

the numbers

are

for any

algebraic

numbers aI’...’ an with 0

···

|03B1n|

R. In this paper, we shall establish necessary and sufficient conditions for

of the values

at

alge-

braic numbers aI’...’ an .

We say the

algebraic

numbers aI’..., as

are

(ek }-dependent

if there exist a number Y, roots of

i

and

numbers

d1,..., d,

not all zero such that

for any

k.

Then we have the

following

theorem.

THEOREM 1. Let aI’...’ an be

numbers with 0

i

Then the

three

are

i n,

are

over the

where

f (1)(z)

denotes the lth derivative

of f(z).

@ Martinus Nijhoff Publishers, Dordrecht - Printed in the Netherlands

(3)

54

ii )

There is a non-empty

such that

are

are

over the

numbers.

00

1. Let

f(z) = L zk!

and 03B11,..., an be

numbers with 0

i

Then

i n,

are

inde-

if and

if

is not a root of

for i

This result was

to be true

Masser.

00

2. Let

f(z) = zk!+k

and al, ... , an be

numbers with 0

1

Then

i n,

are

inde-

if and

if

for i

Theorem 1.

the

the

and the

the

We prove the

the

the

property i)

is satisfied. We may assume

1

are

and

are

algebraically independent

for any subset of n - 1 numbers al.

Changing

the indices of the

we may suppose

are roots of

are not roots of

Define U and

where

=

=

1 +

Then lim

Um

= U and

there is a nonzero

such that

F( U )

= 0. We may assume F has the least

among such

als and

coefficients.

the

on the

of

n, for

i t +

there exists an

1

1

(4)

such that

~F/~ y(l)iq

* 0. Since the total

of

~F/~y(l)iq

is less than the total

of

0.

we have

where

with

and

( U - Um )’

are defined in the usual way. There is a

positive

number 0 1 such that

=

what

the constants

in the

0 and

positive

constants cl, c2, ...

on

aI’’’.’ an and

Then we have

the fundamental

for any

a =1=

-

+

we have

+

Therefore

()em+l =

O( 1 am Il all |em03B8em)

and

Let g be the total

of F and d be a

with

i

Then

and

is an

Hence

by (1), (2)

and the fundamental

we have

= 0 for

m.

By (2),

where A is a number with max

For each

i t,

1

let

Q(l)l

be a maximal subset

such that

E

are

over the

numbers. Then

we have

where Q SI 3

~

(0,... ,0). By assumption,

for any i there exists 1

with

Let

(5)

56

and take a

with

and q E

We assert that

= 0 for any

k. This

the

On the

contrary,

suppose that there is an

a

such that em =

for

many m

p=1 0. Define

for each

q with

1

q ~

and

Then B is not

Let D be a

with

and define

for each

B. Since lim

there is a

constant M such

that

0 for any

any m &#x3E; M.

and

if

em ~

a(mod N),

then

We may assume 03B11,..., an, 03B31,..., -y,,

D(l)iq

and the coefficients of F are in

K if necessary.

Before

we must here

Evertse’s

which will

an

important

role to prove our theorem.

a

on K we mean an

equivalence

class of non-trivial valuations on K. We denote

the set of

all

on K

by S~

the set of all infinite

on K. For every

on K

above a

p on

we choose a

such that

Then we have the

formula:

For

(6)

the

this

height

is well-defined. Put

Then we have the

fundamental

inequality:

where S is any set of

primes

on K. Let S be a finite set of

on

K, enclosing S~,

and c, d be constants with

A

is called

if its

homogeneous

coodinates

xo, xl, ... , xn can be chosen such that

all xk

1 if v 5É

and

vES k=0

The

following

theorem is due to

[Evertse, 1984]:

let c, d be constants with

and let n be a

Then there are

many

but

for each proper,

subset

of

(0, 1,..., n}.

Let S be a finite set of

primes

on K which includes

and all

prime

divisors of 03B11,..., 03B1n. Then

are nonzero

S-integers

and

PROPOSITION 1. Let

E

=

=1=

i2

and m1 &#x3E; m2 &#x3E; M.

m1 is

then

the

proposition

is not true. Then

(7)

58 hence

for

many m1. Since

is not a root of

and so

&#x3E;

equality (1).

PROPOSITION 2. Let

any

subset

B.

m is

then

We prove the

proposition by

induction on the cardinal

of

then the

is true.

and suppose that

for

infinitely

many m. First we consider the case where

is the

such that

(io, 1, q ) E Bo

for some q and 1. Define

Then

for

infinitely

many m. When we divide both sides

by Plo(em )

and let m tend to

we have

Since the total

of

~F/~y(l0)i0q(q ~ Q(l0)i0)

are less than the total

of

(8)

are

over the

algebraic

numbers. Next we suppose there exist two

with

Let E be any

number 1.

By Proposition

1

and the induction

hypothesis on |B0|, applying

Evertse’s theorem to the

as c = 1 and d =

we have

for

many m.

the facts that

Yi

Il

v = 1 and there exists a

UES v on K such

&#x3E;

we have

Now we can

the

proof

of Theorem 1. For

many m with em =

we

where

and

are

S-integers.

First we consider the case where

is

the

such that

(io, 1, q ) E B

for some 1 and q. Define

Then

When we divide the both sides

by Plo(em)Ylôm

and make m tend to

we

have

since A

|03B1111 = |03B3i0 |.

This is a contradiction. Next we suppose that there

are two

B

=

Let E be any

positive

number 1. We may assume K is not a real field

for some

infinite

vo on K.

1 and

Proposition 2, applying

the

(9)

60

Evertse’s theorem to the

we have

for

with em ---

and

(8),

the left hand side

of the

is not

than

On the other

a

we have

Since

Then we have

(10)

for

infinitely

many m. As m tends to

we have

Since e is any

number

this

This

This

the

of Theorem 1.

In

[Nishioka,

to appear in J. Number

the

of

1 is

Here we have the

be a

number

and denote

the

of

with

to the valuation

of

in

be

For an

number a with 0

we denote

the value of

at z = a in

Cp.

THEOREM 2. Let aI’...’ an be

numbers with 0

i

Then the

three

are

i ) f ( aI) p’ ... , f(03B1n)p are algebraically dependent

over the rationals.

ii)

There is a non-empty subset

such that

are

over the

numbers.

Theorem 2 is

proved

in the same way as Theorem

but

the

derivatives of

L =

0).

References

Bundschuh, P. and Wylegala, F.-J.: Über algebraische Unabhängigkeit bei gewissen nichtfort-

setzbaren Potenzreihen. Arch. Math. 34 (1980) 32-36.

Cijsouw, P.L. and Tijdeman, R.: On the transcendence of certain power series of algebraic

numbers. Acta Arith. 23 (1973) 301-305.

Evertse, J.-H.: On sums of S-units and linear recurrences. Compositio Math. 53 (1984) 225-244.

Nishioka, K.: Algebraic independence of certain power series of algebraic numbers. J. Number Theory, to appear.

Nishioka, K.: Algebraic independence of three Liouville numbers. Arch. Math., to appear.

Nishioka, K.: Proof of Masser’s conjecture on the algebraic independence of values of Liouville series. Proc. Japan Acad. Ser. A 62 (1986) 219-222.

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