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Topic: about the Kronecker-Weber theorem
Replies: 6   Last Post: Feb 4, 2013 7:45 PM

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Leon Aigret

Posts: 31
Registered: 12/2/12
Re: about the Kronecker-Weber theorem
Posted: Feb 4, 2013 7:45 PM
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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]
>>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
>>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.


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