Re: deciding whether two padic extensions are isomorphic

On Thu, Jan 16, 2025 at 05:56:05PM +0000, John Cremona wrote:
> On Thu, 16 Jan 2025 at 15:09, Fernando Gouvea <[email protected]> wrote:
> >
> > Aha! Since it's C2xC2 even over Q, it should be something simple, and it is. Playing with my "new" tool,
> >
> > gp> pol=x^4 + 2*x^3 + 11*x^2 + 10*x + 4
> > %17 = x^4 + 2*x^3 + 11*x^2 + 10*x + 4
> > gp > polcompositum(pol,x^2+x+1)
> > %18 = [x^4 + 5*x^2 + 1, x^4 + 17*x^2 + 25]
> > gp > polcompositum(pol,x^2+7)
> > %24 = [x^4 - 2*x^3 + 11*x^2 - 10*x + 4, x^4 - 2*x^3 + 39*x^2 - 38*x + 172]
> >
> > So the field defined by f is Q(sqrt(-7),omega), where omega^3=1. Since -7 is congruent to 2 mod 3, over the 3-adics adjoining sqrt(-7) is the same as adjoining sqrt(2).
> 
> Excellent.  And it appears that we can test whether an integer a is a
> square in the p-adic field defined by a polynomial f by computing
> polcompositum(f,x^2-a) and seeing whether the result has the same
> degree as f (or double).
> 
> Does that look right to you, Bill?

You need to call factorpadic on the result of polcompositum.

Cheers,
Bill.