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Replies: 6   Last Post: Feb 4, 2013 7:45 PM

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 quasi Posts: 12,067 Registered: 7/15/05
Posted: Feb 4, 2013 2:18 PM

David Bernier wrote:
>
>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 ?

The elementary symmetric functions of the roots are
left invariant by _any_ permutation of the roots.

quasi

Date Subject Author
2/3/13 David Bernier
2/3/13 quasi
2/4/13 David Bernier
2/4/13 quasi
2/4/13 David Bernier
2/4/13 quasi
2/4/13 Leon Aigret