Bill Allombert on Mon, 01 Nov 2021 19:36:54 +0100 |
[Date Prev] [Date Next] [Thread Prev] [Thread Next] [Date Index] [Thread Index]
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.