Re: question on mapping points from an elliptic curve back to a quartic

[John Cremona <john.cremona@gmail.com>]

Following on from Bill's reply, the image of the rational points on the quartic curve (which is a 2-covering of E) is one coset of 2E(Q) in E(Q). That is if the quartic has any rational points at all. That is why two-descent works: the number of 2-covers (up to appropriate equivalence) with rational points is equal to the number of such costs, i.e. the order of E/2E.  That is exactly how mwrank works.

Randall, I thought you had read my book which explains this!

John

On Wed, 9 Oct 2024, 18:49 Bill Allombert, <Bill.Allombert@math.u-bordeaux.fr> wrote:

On Wed, Oct 09, 2024 at 07:31:35PM +0200, Bill Allombert wrote:
> On Wed, Oct 09, 2024 at 09:54:10AM -0700, American Citizen wrote:
> > Suppose we consider a quartic with rational points
> >
> >   Q(x,y) : -x^4 + 39/380*x^3 + 39/380*x + 1 = y^2
> >
> > Question:
> >
> >   Why are most of the points in the elliptic curve pool L unmappable back to
> > Q(x,y)? This is surprising to me, as I believed that all the rational points
> > on E were mappable back to Q(x,y)?
>
> Q is a 2-cover of E, so only the points in [2]E(\Q) are mappable back to Q.

And to answer your question quantitatively:

E ~ Z^3 x Z/2Z 
[2]E ~ (2Z)^3 x 0Z/2Z

E/[2]E ~ (Z/2Z)^4

[E:[2]E]=16 so a point on E has proba 1/16 to be mappable back to Q.

Cheers,
Bill