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

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 Leon Aigret Posts: 31 Registered: 12/2/12
Posted: Feb 4, 2013 7:45 PM

On Mon, 04 Feb 2013 14:18:44 -0500, quasi <quasi@null.set> wrote:

>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.

But the splitting field can be of dimension less than n! over Q, and
then not all permutations are linked to an automorphism.

For P(x) = x^4 +1 for example, the four roots are powers of each
other, so there are only 4 automorphisms, complely defined by the
image of a single root.

Leon

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