Bill Allombert on Mon, 01 Nov 2021 19:36:54 +0100

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 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 <http://pari.math.u-bordeaux.fr/~bill/fvr/>.
(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.

```