Re: Transforming general cubic to standard form

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

On Thu, 4 Nov 2021 at 13:54, Grechuk, Bogdan (Dr.) <bg83@leicester.ac.uk> wrote:
>
> Dear Bill, John, and everyone,
>
> Thank you very much for the answers and comments!
>
> May I have one follow up question? If I have a general genus 1 cubic and a rational point exists (so this is elliptic curve), but the transformation to Weierstrass form is rational but non-linear, it of course does not present the internality of the points. I can use Magma or SageMath to find integral points on the Weierstrass model. Is there any implementation for finding integer points on the original (general cubic) model?

(I assume "internality" is a typo for "integrality").

Integrality is a property of rational points in the affine plane, it
does not make sense for rational points in the projective plane, so as
it stands the question is not well-defined.  You need to specify a
line at infinity.  So I assume that you have an affince plane cubic,
not in Weierstrass form, and want to find its integral points.  That
is a perfectly good question, but I do not know an answer to it.
There are only finitely many (by Siegel's Theorem
https://en.wikipedia.org/wiki/Siegel%27s_theorem_on_integral_points).

John

>
> Sincerely,
> Bogdan
>
> ________________________________
> From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>
> Sent: 01 November 2021 18:36
> To: pari-dev@pari.math.u-bordeaux.fr <pari-dev@pari.math.u-bordeaux.fr>
> Subject: Re: Transforming general cubic to standard form
>
> On Mon, Nov 01, 2021 at 02:47:42PM +0000, John Cremona wrote:
> > These formulas can all be found in Tom Fisher's papers on genus one
> > models.  The binary quartic case is also in my book.   Note that the
> > question could mean two different things, given a genus 1 curve C
> > (e.g. given by one of the types of model you mention):   there is
> > always an elliptic curve J(C), the Jacobian, whether or not C has any
> > rational points;   but when C is an n-cover of an elliptic curve E
> > (with n=3,2,4 respectively in your cases), there is a degree n map
> > from C to E, and also *if* C has a rational point then C and E are
> > isomorphic.    To get J(C) you only need the invariants of C (e.g. I
> > and J of a binary quartic).  The degree n map from C to E, or the
> > isomorphism from C to E given a rational point on C, are more
> > complicated.
>
> There is a formula of F. Villegas-Rodriguez that gives the map.
> (which is the largest mathematical formula I know).
>
> Since Fernando web site does not work anymore, I have made a copy
> at <https://eur03.safelinks.protection.outlook.com/?url=http%3A%2F%2Fpari.math.u-bordeaux.fr%2F~bill%2Ffvr%2F&amp;data=04%7C01%7Cbg83%40leicester.ac.uk%7C99eee3799e544fc345b808d99de9b3a4%7Caebecd6a31d44b0195ce8274afe853d9%7C0%7C1%7C637714449330157408%7CUnknown%7CTWFpbGZsb3d8eyJWIjoiMC4wLjAwMDAiLCJQIjoiV2luMzIiLCJBTiI6Ik1haWwiLCJXVCI6Mn0%3D%7C3000&amp;sdata=2iBhzXKCtzRUyR3nzENyqGo7gdeoBPU3j0W9CbDD6HY%3D&amp;reserved=0>.
> (download both files).
>
> This is an example:
>
> ? W=getall(y^3+(x^3+z^3))
> %1 =
> [[0,0,9,0,-27],[(-729*y^3-729*z^3)*x^3-729*z^3*y^3,(13122*y^3-6561*z^3)*x^6+(-6561*y^6-78732*z^3*y^3+13122*z^6)*x^3+(13122*z^3*y^6-6561*z^6*y^3),27*z*y*x],x^3+(y^3+z^3)]
> ? [X,Y,Z]=W[2]
> %2 =
> [(-729*y^3-729*z^3)*x^3-729*z^3*y^3,(13122*y^3-6561*z^3)*x^6+(-6561*y^6-78732*z^3*y^3+13122*z^6)*x^3+(13122*z^3*y^6-6561*z^6*y^3),27*z*y*x]
> ? Y^2+9*Y*Z^3-(X^3-27*Z^6)*Mod(1,W[3])
> %3 = Mod(0,x^3+(y^3+z^3))
>
> Cheers,
> Bill.
>
>