Karim Belabas on Wed, 30 Mar 2022 22:17:22 +0200

 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

* 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
>
> ? 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
>