Bill Allombert on Tue, 24 Jun 2014 16:58:17 +0200

 Re: ellheegner

On Tue, Jun 24, 2014 at 09:44:01AM -0300, Ariel Martin Pacetti wrote:
>
> Dear Bill,
>
> >Hello Ariel,
> >Could you provide an example ?
>
> Take the elliptic curve 37a1 (which has rank 1, and prime
> conductor). In this case, take any d such that kronecker(-d,37)=1,
> and construct a Heegner point attached to d (for general N, you need
> each prime dividing N to split in Q[\sqrt{-d}] if you take the whole
> ring of integers):
>
> d= -3 --> P=[-1,0]
> d=-7 --> P=[0,0]
> ...
> d=-27 --> P=[2,-3]
> d=-4*33 --> P=[-1,0] (here the class number is not one, so you need
> to take the trace I was talking about)

Indeed, that it what PARI do, but it does not attempt to reconstruct
the point as an element of E(Q[\sqrt{-d}]), but only the trace as an
of E(Q). (Except that PARI applies the Atkin-Lehner involution to
the Heegner points)

> >This can be done with minimal change to the PARI source code.
> >
> >Hwoever, PARI definition of the Heegner point is
> >"A  non-torsion  rational  point  on  the curve,  whose canonical height is equal to the
> >product of the elliptic regulator by the analytic Sha"
>
> If I am not mistaken, then this implies that the point is a
> generator of the free part.

No, it is a the analytic Sha time a generator.

> element. I guess that you are computing some Heegner point, and use
> it to search for a generator (you have few choices), so it is a
> little tricky to call this routine "heegnerpoint".

It is called ellheegner, not heegnerpoint.

Cheers,
Bill.