Denis Simon on Tue, 09 Jun 2015 17:29:45 +0200

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Re: qfisom over Q

Dear Jeroen,

You can use qfsolve() in the following way:

Let q1(x,y)=a1*x^2+b1*x*y+c1*y^2 and q2(x,y)=a2*x^2+b2*x*y+c2*y^2 be the quadratic forms.
Solve q1(x1,y1) = a2 over Q using qfsolve([a1,b1/2,0;b1/2,c1,0;0,0,-a2])
Then complete the 2x2 matrix T=[x1,...;y1,...] to have det=1.
Then T(q1) and q2 have the same 1st coeff and same det.
It should then be easy to finish the job.

Denis SIMON.

----- Mail original -----
> De: "Jeroen Demeyer" <>
> À: "pari-users" <>
> Envoyé: Mardi 9 Juin 2015 16:02:42
> Objet: qfisom over Q
> Dear pari-users,
> In qfisom, I am guessing that the isomorphism is supposed to be defined
> over Z, i.e. the transformation matrix is unimodular. This is not so
> clear from reading ??qfisom.
> Is there any way to do this computation over Q, i.e. allowing the
> transformation matrix to be any non-singular rational matrix?
> Cheers,
> Jeroen.