Bill Allombert on Wed, 30 Mar 2022 21:56:57 +0200 |
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Re: Question related to an unramified cubic extension polynomial |
On Wed, Mar 30, 2022 at 09:14:11PM +0200, Eleni Agathocleous wrote: > Hello > > I have a question regarding the computation of the unramified cubic > extensions of the real quadratic field K_D with discriminant > D=4*(m)^3-27*(n)^2= 48035713, for m=229, n=3. I wrote the discriminant > like this since in cases where the class number is divisible by 3 the > discriminants can be written in this form, for finitely many pairs > (m,n), and the polynomials f=x^3-m*x+n define unramified cubic > extensions (when gcd(2m,3n)=1). > > The ideal class group CL(K_D) of K_D has 3-rank 2 and is of the form > (C6)x(C6). I computed the four polynomials giving me the 4=(3^2-1)/2 > unramified cubic extensions but only one of the polynomials has > poldisc equal to the discriminant of K_D, namely the polynomial f. The > following is a classical result of Hasse: > "For K a quadratic number field of discriminant D, and r the 3-rank of > the ideal class group of K, the number of non-conjugate cubic fields > of discriminant D is m = (3r − 1)/2." > > My question is two-fold: In this case for this D= 48035713, is it > because of computational limitations that PARI gives me only one > polynomial with discriminant D (namely the f=x^3-mx+n)? Hello Eleni, Could you clarify which command you issued that gives you a single polynomial ? You can proceed as follow ? bnf=bnfinit(quadpoly(48035713,'a),1); ? bnrclassfield(bnf,3) %2 = [x^3-5853*x-156580,x^3+(3891134060370*a-13486242851888223)*x+(245953599375323953710*a-852448134659276677806065)] But this gives you polynomials defined over K_D. If you want all absolute polynomials, the simplest is ? R = nfsubfields(bnrclassfield(bnf,3,2),3,1); Now you should reduce them: ? Rmin = Set(apply(polredabs,R)) %49 = [x^3-x^2-1662*x-5633,x^3-229*x-3,x^3-x^2-1858*x+3719,x^3-x^2-650*x+6016] So you get 4 disctincts polynomials as expected. ? disc=apply(poldisc,Rmin) %50 = [17340892393,48035713,25410892177,192142852] Only the second one has exactly the right discriminant. ? apply(nfdisc,Rmin) %51 = [48035713,48035713,48035713,48035713] but they all have the right field discriminant. Given a field, it is difficult and often impossible to find a polynomial with the same discriminant as the field. In this instance I do not think it is possible due to ... > Finding another polynomial of discriminant D would give a specific > elliptic curve, namely the curve E:y^2=x^3-27*16*D, a second point > (the first one is the point P=[12m,108n]) which would be independent > form P and this would result to the curve having rank>=2 instead of > rank=1 which is what I get now. ... alas, the rank of this curve is 1: ? ellrank(E) %47 = [1,1,0,[[2748,324]]] so you will not find new independent points. If you want a D for which this method gives more points, try D=-3321607 Thanks for your question! Bill