| Fernando Gouvea on Tue, 14 Jan 2025 22:47:34 +0100 |
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| 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
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, <Bill.Allombert@math.u-bordeaux.fr> 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 ...she wears a protective crystal under her shirt, "to absorb the energy of her fans' demands." -- Robin Roberts, in "Anne McCaffrey: A Life With Dragons"