Re: Transforming general cubic to standard form

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

On Mon, 1 Nov 2021 at 16:33, Georgi Guninski <gguninski@gmail.com> wrote:
>
> On Mon, Nov 1, 2021 at 5:06 PM Grechuk, Bogdan (Dr.)
> <bg83@leicester.ac.uk> wrote:
> >
> > Dear John Cremona,
> >
> > Thank you for the answer.
> >
> > What I am looking for is the implementation that, given genus 1 equation as an input, produces its  Weierstrass model (preferably with integer coefficients) together with actual isomorphism (change of variables). If there are no rational point, the function may return the degree n map you described, but in fact I am ok if it returns just error message that the input has no rational point and is therefore not an elliptic curve. There is a similar function in Magma but the inconvenience is that it requires explicit rational point as an input, so I need first to search for it. I then found function ellfromeqn in PARI that does not require a rational point as an input, and wanted to know if it can output the actual isomorphism (change of variables), but from your answer I understood that not (Because J(C) can be found just from invariants, while isomorphisms are more complicated.)
>
> For at least cubic curve you can do this with sagemath, https://sagemath.org
>
> IIRC point on the curve is not needed.
>
> Probably you can find the formulas from the source.

I wrote that code in Sage!   The reason why you can *sometimes*  not
give a point is that in the code I find rational flexes, if any.  For
a cubic with no rational flexes, if you do not give a rational point
then it will raise an error.

John

>