• Aucun résultat trouvé

The cubics which are differences of two conjugates of an algebraic integer

N/A
N/A
Protected

Partager "The cubics which are differences of two conjugates of an algebraic integer"

Copied!
5
0
0

Texte intégral

(1)

are

an

algebraic integer

par TOUFIK ZAIMI

RÉSUMÉ. On montre

entier

algébrique cubique

sur un corps de nombres K, de trace nulle est la différence de deux

conjugués

sur K d’un entier

algébrique.

On prouve aussi que si N est une ex-

tension

cubique

normale du corps des rationnels, alors tout entier

de N de trace zéro est la différence de deux

conjugués

d’un entier

de N si et seulement si la valuation

du discriminant de N est différente de 4.

ABSTRACT. We show that a cubic

algebraic

integer over a number

field K, with zero trace is a difference of two conjugates over K

of an

algebraic

integer. We also prove that if N is a normal cubic extension of the field of rational numbers, then every integer of N

with zero trace is a difference of two conjugates of an integer of

N if and

only

if the 3-adic valuation of the discriminant of N is not 4.

1. Introduction

Let K be a number

an

number with

=

over K and L =

K (01, {32, ..., (3d)

the normal closure of the extension

In

Dubickas and

have shown that

can be

written

= a -

a’,

where a and a’ are

over K of an

if and

only

if there is an element a of the Galois group

of the extension

L/K,

of order n such that

= 0. In this

case

= a -

=

l)ai((3)/n

is an element

of L and the trace

of 0

for the extension

+

+ ... +

is 0.

Furthermore,

the condition on the trace of

/3

to be 0

is also sufficient to express

/3

= a - a’ with some a and a’

over

K of an

algebraic number,

when the extension is normal

e.

when L =

K(o))

and its Galois group is

cyclic (in

this case we say that the extension is

cyclic)

or when d 3.

Manuscrit requ le 20 décembre 2003.

(2)

Let D be a

rational

and

the

For

any number

K and

any

D over

is a

two

over K

an

then

is a

two

over K

an

In

P(D)

is true for all values of D. It is clear

that if = 0 and

E

is the

of

of

K, then 13

= 0 = 0 - 0 and is true. For a

extension

Dubickas showed that if =

is a difference

of two

over K of an

of

2 over

is true. In

Dubickas

proved

that if the minimal

of the

over

say

is of the form

p(xm),

where P E and m is a rational

than

then

13

is a difference of two

over K of an

algebraic integer.

Consider now the assertion For any number

K and

any

algebraic integer 13 of degree

D over K such that the extension

is

=

then

is a

two

over

K

an

algebraic integer.

The first aim of this note is to prove :

Theorem 1. The assertions

and

are

and

is

true.

Let Q

be the field of rational numbers. In

[5],

the author showed that if the extension

is normal with

then every

integer

of

N with zero trace is a difference of two

of an

of N if

and

if

TrN/Q(ZN) == ZQ.

It easy to check that if N = is a

field

is a

rational

then =

if and

if m =

1~4~.

For the cubic fields we have :

Theorem 2. Let N be a normal cubic extension

every

integer of

N with zero trace is a

two

an

N

and

the discriminant

of

N is not 4.

2. Proof of Theorem 1

First we prove that the

and are

It

is clear that

since

by

Hilbert’s Theorem 90

[3]

the

condition = 0 is sufficient to express

~:3

= a - a’ with some

a and a’

over K of an

number.

let

be

an

of

degree

D over K and which is a difference of two

over K of an

number.

By

the above result of Dubickas and

Smyth,

and with the same

notation,

there is an element a E

G(L~K),

of order n such that

~Z(,3) =

0. Let Q &#x3E; be the

of

and L~‘~~ _

E =

the fixed field

of &#x3E; .

K C L«~ c L«~

c

the

of

over is

(3)

D and

Artin’s theorem

[3],

the Galois group of the normal extension is a &#x3E; .

the extensions

and

are

cyclic

since their Galois groups are

respectively cr

&#x3E; and a factor group of

a &#x3E;.

Furthermore,

the restrictions to the field

L~a~(~3)

of the elements of the group a &#x3E;

belong

to the Galois group of

LO’&#x3E;(~3)IL’&#x3E;

and each

element of is a restriction of

exactly

d elements of the

group Q &#x3E;, where d is the

of

It follows that

and

/3

is a difference of two

over of an

algebraic

number.

Assume now that is true.

Then, /3

is difference of two

over

L~~,

and a fortiori over

of an

and so

is

true.

To prove that

P(3)

is true, it is sufficient to show that if

a cubic

algebraic integer

over a number field K with = 0 and such that the extension is

then

(3

is a difference of two

of an

since

P(2)

is true and the assertions

and

are

Let

and let a be a

of

G(K(,~)/K). Then,

p = and the discriminant of the

satisfies

r is of

degree

3 over K and

,,, &#x3E; I . - ,

As the

polynomial -27t+3+3p-26

is irreducible in the

Hilbert’s

theorem

[4],

there is a rational

such that

the

+

+

27s)

is irreducible in

if

since

E

Set a =

is a

of

a over

a fortiori over

K)

and

From the relations

obtain that a is a root of the

(4)

and a is an

3 over

E

A short

computation

shows that from the relation

+

=

we have 3

-

+

27 and

23 p

3 is a root of the

+

p2 -- q

whose coefficients are

integers

of K. D

Remark 1. It follows from the

of Theorem

that if

is a cubic

algebraic integer

over a number field K with zero trace, then

is a difference

of two

over K of an

of

3 over

The

following example

shows that the constant 3 in the last sentence is the best

Set K

and

3x - 1.

=

34

and the extension is

since

(32 - 2

is also a root of

Theorem 3 of

[5], /3

is not a difference of two

of an

of

is

and if

= a -

where a is

an

of

2 over

and a’ is a

of a over

Q((3),

then there exists an element T of the group

such

that

=

= a and

=

= a’ - a =

-/3.

Remark 2. With the notation of the

of Theorem 1

second

we have: Let

be a cubic

algebraic integer

over K with zero trace and such that the extension is

is a

two

an

and

there exists a E

ZK

such that the two nurrzbers and

°3+2pa+b

are

K.

suppose that

= a - where a E

= a -

then

/3 = a+Q(a) -Q(a+Q(a)~~~ Then, a - a(a) = 3 -Q(3), a- 3 = Q(c~- 3), a - 3

E K and there exists an

a of K such that

= a.

3 - E K((3)’ 3 -

+ 3

a3+2pa+b

27 ~

Z [X]

K

and so the numbers

2013

are

integers

of K. The converse is

since

/3 = 3 - ~ ( 3 ) - ~ 3 ° - ~ ( ~ 3 ° ) for all integers

a of K. It follows

in

when

diS;6(,6)

E

that

/3

is a difference of two

of

an

of

=

Note

finally

that for the case where K

a

more

explicit

condition was obtained in

(5~.

3. Proof of Theorem 2

With the notation of the

of Theorem 1

second

and

K =

Q,

let N be a cubic normal extension

of Q

with discriminant A and let v be the 3-adic valuation.

Suppose

that every non-zero

integer (3

of N

with zero trace is a difference of two

of an

of N.

N = and

Theorem 3 of

4. Assume also

= 4.

Then, v( disc((3))

&#x3E; 4 and hence

&#x3E;

since E

and

disc((3)

is a square of a rational

It follows

is an

(5)

integer,

since its minimal

is

E

and

be written

= a - where a

is an

of N with zero

trace.

Thus, v(disc(a)) &#x3E;

6 and there is an

integer il

of N with zero trace, such that a =

It follows that

= q -

r~ -

+

is also an

integer

of N with zero trace.

E

for all

rational

n.

suppose

v(0) ~

4. Assume also

that there exists an

integer j3

of N with zero trace which is not a difference

of two

of an

of N.

N =

and

Theorem 1

of

we have =

3Z,

since = 3 and is an

ideal of Z. If

is an

basis of

1V,

then from the relation A =

we obtain

3 and hence

v(0) &#x3E; 6,

since A is a

square of a rational

The last

inequality

in this case we have 6 and

E D

This work is

partially supported by

the research center

(N°

References

[1] A. DUBICKAS, On numbers which are differences of two conjugates of an algebraic integer.

Bull. Austral. Math. Soc. 65 (2002), 439-447.

[2] A. DUBICKAS, C. J. SMYTH, Variations on the theme of Hilbert’s Theorem 90. Glasg. Math.

J. 44 (2002), 435-441.

[4] A. SCHINZEL, Selected Topics on polynomials. University of Michigan, Ann Arbor, 1982.

[5] T. ZAIMI, On numbers which are differences of two conjugates over Q of an algebraic integer.

Bull. Austral. Math. Soc. 68 (2003), 233-242.

Toufik ZAIMI

King Saud University

Dept. of Mathematics P. O. Box 2455

Riyadh 11451, Saudi Arabia E-mail : zaimit ouOksu. edu . sa

Références

Documents relatifs

Capes externe de math´ ematiques, session 1997, 1` ere composition.. Danc ce probl` eme, α d´ esigne un nombre r´ eel

Calculer la probabilité qu'il retrouve sa montre dans le cas où il cherche sur la plage ; puis dans le cas où il cherche sur le chemin.. Il a cherché la montre sur le chemin mais ne

La difficulté majeure consiste à obtenir les comparaisons permettant de conclure, en particulier celle de la question

Cela est dû au fait que, pour obtenir s(k), plus petit entier qui a exactement k diviseurs positifs y compris 1 et lui-même, on doit, comme nous l'avons relevé maintes fois,

Cette construction progressive peut se faire à partir d’une suite d’un seul terme

Les nombres supérieurs à N 0min deviennent immédiatement positifs car constitués d’une collection de a i =0, (distance positive car la soustraction des chiffres devient nulle

Remarque 1 : on écarte les nombres divisibles par 10 (appliquer 2 fois l’opération miroir à un nombre divisible par 10 ne redonne pas le nombre de départ).. Donc aucun des deux

C'est là une condition seulement nécessaire car deux voisins immédiats n'ont évidemment pas toujours le même nombre de chiffres.. Mais, pour k≥5, une autre