
Re: about the KroneckerWeber theorem
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

