| Fernando Gouvea on Thu, 16 Jan 2025 16:09:57 +0100 |
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| Re: deciding whether two padic extensions are isomorphic |
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).
Fernando
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 <fqgouvea@colby.edu> 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 ...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"
-- ============================================================= Fernando Q. Gouvea http://www.colby.edu/~fqgouvea Carter Professor of Mathematics Dept. of Mathematics Colby College 5836 Mayflower Hill Waterville, ME 04901 Families would not have the significance they do for us if they did not, in fact, give us a claim upon each other. At least in this sphere of life we do not come together as autonomous individuals freely contracting with each other. We simply find ourselves thrown together and asked to share the burdens of life while learning to care for each other. We may often resent such claims on our time and energies. We did not, after all, consent to them. -- Gilbert Meilaender