Re: deciding whether two padic extensions are isomorphic

Bill's solution using polcompositum is certainly the best, being a
general method.   And certainly my suggestioin required being able to
test for being a square in an extension of Q3.

 But I cannot resist pursuing this:

f = x^4 + 2*x^3 + 11*x^2 + 10*x + 4 has invariants I=109, J=-646 and
hence cubic resolvent x^3-327*x-646, which factors even over Q (so
certainly over Q3) as (x-19)*(x+2)*(x+17), so the Galois group of f is
c2xc2, that field is biquadratic, so must be the right one as Q3 only
has one biquadratic extension.

For the other candidate, the cubic resolvent factors over Q3 as
linear*quadratic, which means that the Galois group of f has order 8.

John

On Tue, 14 Jan 2025 at 22:02, Fernando Gouvea <[email protected]> wrote:
>
> Thank you! I had not really internalized polcompositum; clearly very useful here.
>
> Fernando
>
> On 1/14/2025 3:32 PM, Bill Allombert 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.
>
> --
> =============================================================
> Fernando Q. Gouvea         http://www.colby.edu/~fqgouvea
> Carter Professor of Mathematics
> Dept. of Mathematics
> Colby College
> 5836 Mayflower Hill
> Waterville, ME 04901
>
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>