Re: deciding whether two padic extensions are isomorphic

Well, 2 is a square in the degree 2 unramified extension, which is 
contained in both fields, so that won't differentiate them. Since 2 is 
congruent to -1 mod 3, Hensel's lemma says -1 is a square too. Deciding 
on +/-3 seems harder, however.

Fernando

On 1/14/2025 4:27 PM, John Cremona wrote:
> 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.
>
-- 
=============================================================
Fernando Q. Gouveahttp://www.colby.edu/~fqgouvea
Carter Professor of Mathematics
Dept. of Mathematics
Colby College
5836 Mayflower Hill
Waterville, ME 04901

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