Bill Allombert on Thu, 27 Feb 2003 20:10:38 +0100

 Re: [Fwd: Re: nfgaloisconj]

```On Thu, Feb 27, 2003 at 07:25:48PM +0100, Markus Endres wrote:
> of course, here it is:
>
> ? K=bnfinit(y^2+7);
>
> %7 = x^4 + Mod(-y, y^2 + 7)*x^3 - 3*x^2 + Mod(y, y^2 + 7)*x + 1
>
> ? rnfequation(K,%)
> %8 = x^8 + x^6 - 3*x^4 + x^2 + 1
>
> ? L=bnfinit(%);
>
> ? aut=nfgaloisconj(L)
> %10 = [x, 1/2*x^7 - 1/2*x^6 + x^5 - 1/2*x^4 - 1/2*x^3 + x^2 - 1/2, x^7 + x^5 - 3*x^3 + x,
>  1/2*x^7 + 1/2*x^6 + x^5 + 1/2*x^4 - 1/2*x^3 - x^2 + 1/2, -1/2*x^7 - 1/2*x^6 - x^5 - 1/2*x^4 + 1/2*x^3 + x^2 - 1/2,
> -x^7 - x^5 + 3*x^3 - x, -1/2*x^7 + 1/2*x^6 - x^5 + 1/2*x^4 + 1/2*x^3 - x^2 + 1/2, -x]~
>
>
> now I need all the automorphisms in aut which leaves the elements of K fix, i.e. gal(L|K).
> but these automorphisms are defined absolute over Q, and I need them relative over K

Do not use nfgaloisconj, use galoisinit, it is far more powerful!
(by the way you do not need to use bnfinit).
do:
G=galoisinit(x^8 + x^6 - 3*x^4 + x^2 + 1);

This compute the abstract galois group.
The way you have build your field, K is the fixed field by G.gen[1],

So let do
? F=galoisfixedfield(G,G.gen[1],2)
%8 = [x^2 + 7, Mod(2*x^7 + 3*x^5 - 4*x^3 + 2*x, x^8 + x^6 - 3*x^4 + x^2 + 1), [x^4 - y*x^3 - 3*x^2 + y*x + 1, x^4 + y*x^3 - 3*x^2 - y*x + 1]]

F[1] is the defining polynomial for the subfield. You are lucky, this the
right polynomial, so you do not need to change.

x -> F[2] is the inclusion morphism from K to L.  We do not need it here.

F[3] is the factorisation of x^8 + x^6 - 3*x^4 + x^2 + 1 over K.

So let R=F[3][1]*Mod(1,y^2+7) a polynomial defining the extension L/K.