Grechuk, Bogdan (Dr.) on Thu, 04 Nov 2021 14:54:23 +0100

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Re: Transforming general cubic to standard form

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?


From: Bill Allombert <>
Sent: 01 November 2021 18:36
To: <>
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 <"">>.
(download both files).

This is an example:

? W=getall(y^3+(x^3+z^3))
%1 =
? [X,Y,Z]=W[2]
%2 =
? Y^2+9*Y*Z^3-(X^3-27*Z^6)*Mod(1,W[3])
%3 = Mod(0,x^3+(y^3+z^3))