Karim Belabas on Wed, 30 Mar 2022 22:17:22 +0200 |
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Re: Question related to an unramified cubic extension polynomial |
Hi Eleni, in complement to Bill's nice answer, you can also use 'nflist' as a shortcut to list number fields with given discriminants (and specified Galois closure; in this case it must be S_3: (22:09) gp > F = nflist("S3", 48035713) %1 = [x^3 - x^2 - 1858*x + 3719, x^3 - x^2 - 1662*x - 5633, x^3 - x^2 - 650*x + 6016, x^3 - 229*x - 3] (22:10) gp > apply(nfdisc, F) %2 = [48035713, 48035713, 48035713, 48035713] (22:10) gp > apply(poldisc, F) %3 = [25410892177, 17340892393, 192142852, 48035713] Cheers, K.B. P.S. There exist 4 cubic (integral) cubic polynomials defining these 4 non-isomorphic fields *and* with polynomial discriminant D ... but non-monic ones (Delone-Fadeev / Davenport-Heilbronn). * Bill Allombert [2022-03-30 21:57]: > 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 -- Karim Belabas, IMB (UMR 5251) Tel: (+33) (0)5 40 00 26 17 Universite de Bordeaux Fax: (+33) (0)5 40 00 21 23 351, cours de la Liberation http://www.math.u-bordeaux.fr/~kbelabas/ F-33405 Talence (France) http://pari.math.u-bordeaux.fr/ [PARI/GP] `