Date: Feb 4, 2013 2:05 PM
Author: David Bernier
Subject: Re: about the Kronecker-Weber theorem
On 02/04/2013 04:04 AM, quasi wrote:

> David Bernier wrote:

>>

>> I have a further question about conjugate roots ...

>>

>> The non-trivial third roots of unity

>> -1/2 +i*srqrt(3)/2 and -1/2 -i*srqrt(3)/2

>> are complex conjugates.

>>

>> I don't know of a definition where, for example, in the

>> setting above,

>> 2^(1/3) is said to be conjugate to

>> 2^(1/3) * (-1/2 +i*srqrt(3)/2).

>

> Let H be an algebraically closed field, and let K be a

> subfield of H. Let K[x] denote the ring of all polynomials

> in the indeterminate x with coefficients in K. If f in K[x]

> is irreducible in K[x], the roots of f in H are said to be

> conjugates of each other over K.

>

> Thus, the 3 cube roots of 2 are conjugate over Q since

> they are roots of the polynomial x^3 - 2 which is irreducible

> over Q.

Thank you, quasi.

Let's suppose the base field is Q, and P(x) is an irreducible

polynomial of degree n over Q. Let alpha_1, ... alpha_n

be the n conjugate roots in the splitting field L (subfield of

C, the complex numbers) of P(x) over Q.

If sigma: {alpha_1, ... alpha_n} -> {alpha_1, .. alpha_n}

is a permutation of the n conjugate roots,

then according to me if a field automorphism of phi of L exists

which acts on {alpha_1, ... alpha_n} the same way

the permutation sigma does,

all the elementary symmetric polynomials in n indeterminates

must be invariant under the application of such

elementary symmetric polynomials:

[wikipedia, with def. of elementary symmetric polynomials]

http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial

In the other direction, if we have a sigma, permutation as above,

and all the elementary symmetric polynomials are left

invariant, does it follow that for the splitting field L,

there is a field automorphism phi of L such that

phi(alpha_j) = sigma(alpha_j), 1<=j<=n ?

In other words, phi acts on the alpha_j the same way sigma does.

If the elementary symmetric polynomials are left invariant by

sigma, does it follow that some automorphism phi of L

acts on {alpha_1, ... alpha_n} the same way sigma acts ?

David Bernier

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