John Cremona on Mon, 01 Nov 2021 17:09:09 +0100 |
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Re: Transforming general cubic to standard form |
Well, there is no known algorithm to determine whether or not a genus 1 curve over Q has any rational points, so you cannot expect miracles! You can expect some help, perhaps: for binary quartics, the easy case is when the quartic has a rational root x0 since then you can use (x0,0) as the base point; these represent the trivial element of the 2-Selmer group. Similarly, for plane cubics if there is a rational flex (and again, these represent the trivial element of 3-Selmer), as in that case you can just make a linear change of variables taking that flex to [0:1:0] and the tangent there to Z=0, which gives you a Weierstrass equation (perhaps after scaling the variables). I have forgotten what the trivial case is for quadric intersections. John On Mon, 1 Nov 2021 at 15:06, 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.) > > Sincerely, > Bogdan > > ________________________________ > From: John Cremona <john.cremona@gmail.com> > Sent: 01 November 2021 14:47 > To: Grechuk, Bogdan (Dr.) <bg83@leicester.ac.uk> > Cc: Bill.Allombert@math.u-bordeaux.fr <Bill.Allombert@math.u-bordeaux.fr>; pari-dev@pari.math.u-bordeaux.fr <pari-dev@pari.math.u-bordeaux.fr> > Subject: Re: Transforming general cubic to standard form > > 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. > > John Cremona > > On Mon, 1 Nov 2021 at 14:19, Grechuk, Bogdan (Dr.) <bg83@leicester.ac.uk> wrote: > > > > Dear Bill Allombert, > > > > I have found online your function > > > > ellfromeqn > > > > that "allows to recover a Weierstrass model for an elliptic curve given by a general plane cubic or by a binary quartic or biquadratic model." > > > > Is it possible to also get the actual transformation of variables? > > > > Sincerely, > > Bogdan