Bill Allombert on Tue, 01 Oct 2013 18:43:30 +0200

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 Re: Galois subextensions question

```Ariel,

You are still posting to the old list.cr.yp.to address
See http://pari.math.u-bordeaux.fr/lists-index.html
for the new address.

> I might not have stated the problem correctly. Say I have a degree n
> polynomial and want to compute the fixed field by the A_n subgroup?

This one is easy: let denote by P your polynomial and by K its splitting field.
The field you are looking for is given by x^2-poldisc(P).
However A_n is a very special case: this is the only non-trivial
normal subgroup of S_n (except for n=4).

All other subgroups have lots of conjugates, and unless you somehow
specify which one you want (e.g. by giving the action on the roots in
a Galois closure), your problem defined only to Q isomorphisms not
K isomorphisms.

> Or an S_m subgroup (m<n), etc. (you can give the subgroup in any way you
> want).

For m=n-1, then K is one of the solution. Of course all other solutions
are Q-isomorphic to K, but not K-isomorphic to K.

> I didn't mean bnfinit, but you need to run first galoisinit(), which
> needs nfinit... I do not want to compute the Galois closure group,
> but a fixed subextension.

galoisinit does not require nfinit either, but it is true it will not be usable
for your application. On the other hand, computing the Galois closure can help
you to compute the solution.

I would do as follow:

I would start by computing a set S of polynomials R(X1,...,Xn) over Z that
generate the ring  of polynomials invariant by your subgroup when permuting the
X1...Xn.

S only depends on the subgroup G.

Then I would compute the roots a1..an of P in some Galois or algebraic closure of K.
R(a1,...,an) will belong to the field you search.

You can get a minimal polynomial for R(a1,...,an) by computing
F = Prod_{c in S_n/G} (X-R(a_c(1),...,a_c(n))) where c is a coset representative.

(For A_n, R is prod_{i<j}(Xi-Xj), for S_{n-1}, R is X_n)
For small degree, you can find some such polynomial in the GTM 138, section 6.3.

Cheers,
Bill.

```