Re: deciding whether two padic extensions are isomorphic

In your example, could you not also check whether -3 and 2 are both squares
in each of the fields?

John

(Who remembers Fernando telling him how to compile gp from source on a DOS
laptop with 2M of RAM, in about 1992.)

On Tue, 14 Jan 2025, 20:32 Bill Allombert, <
[email protected]> wrote:

> On Tue, Jan 14, 2025 at 02:42:49PM -0500, Fernando Gouvea wrote:
> > In my book on the p-adic numbers, I mention the GP command padicfields,
> > which lists out the (finitely many) extensions of a given Q_p of a given
> > degree. With the flag 1, it lists the polynomial that generates the
> > extension, followed by the ramification index e, the residue degree f,
> the
> > (power of 3 in) the discriminant, and the number of different embeddings
> in
> > an algebraic closure.
> >
> > gp > padicfields(3,4,1)
> > %14 = [[x^4 + 13*x^3 + 64*x^2 + 61*x + 40, 1, 4, 0, 1],
> >        [x^4 + 2*x^3 + 11*x^2 + 10*x + 4, 2, 2, 2, 1],
> >        [x^4 + 2*x^3 + 8*x^2 + 13*x + 7, 2, 2, 2, 1],
> >        [x^4 + 3, 4, 1, 3, 2],
> >        [x^4 + 6, 4, 1, 3, 2]]
> >
> > Earlier in the book I had introduced a field F obtained from Q_3 by
> > adjoining a cube root of 1 and a square root of 2. That is an extension
> of
> > degree 4 with e=f=2, so it is either the second or the third in this
> list.
> > How might one decide which? In other words, given two polynomials of
> degree
> > 4, is there a way to use GP to decide whether they define the same
> > extension?
>
> Yes, but I do not know the best way to do it.
> One way which is simple but not very efficient:
>
> ? P=polcompositum(x^2+x+1,x^2-2)[1]
> %32 = x^4-2*x^3-x^2+2*x+7
> ? L=padicfields(3,4,1)
> %33 =
> [[x^4+13*x^3+64*x^2+61*x+40,1,4,0,1],[x^4+2*x^3+11*x^2+10*x+4,2,2,2,1],[x^4+2*x^3+2*x^2+7*x+16,2,2,2,1],[x^4+3,4,1,3,2],[x^4+6,4,1,3,2]]
>
> ?
> foreach(L,l,print(l[1],":",[poldegree(f)|p<-polcompositum(l[1],P);f<-factorpadic(p,3,10)[,1]]))
> x^4+13*x^3+64*x^2+61*x+40:[8,8]
> x^4+2*x^3+11*x^2+10*x+4:[4,4,4,4]
> x^4+2*x^3+2*x^2+7*x+16:[8,8]
> x^4+3:[8,8]
> x^4+6:[8,8]
>
> So we see the right polynomial is the second one (we find a compositum of
> degree 4).
>
> (this relies on the fact that irreducibility over Qp implies the
> irreducibility over Q).
>
> Cheers,
> Bill.
>
>