On Thursday, August 2, 2018 at 8:58:04 AM UTC-7, jianrong wrote:
>
> Dear All,
>
> Let $C$ be the following curve in $\mathbb{C}^2$.
> \begin{align}
> & 11664 {c_1}^3 {c_2}^2 + 536544 {c_1}^3 c_2 + 6170256 {c_1}^3 + 67068
> {c_1}^2 {c_2}^2 + 1542564 {c_1}^2 c_2 \\
> & + 3085128 c_1 {c_2}^2 - 32393844 c_1 c_2 + 3085128 c_1 + 17739486
> {c_2}^2 + 6941538 c_2 = 0.
> \end{align}
> I checked that this curve has genus $1$ using Sage. Therefore it is an
> elliptic curve. How to change coordinates such that the equation of this
> curve is of the form $y^2 = f(x)$, where $f$ is some polynomial. Thank you
> very much.
>
> A genus 1 curve is only an elliptic curve once you choose a rational point
on it. Your curve does have rational points (the singularities don't count
-- we need a point on the desingularization of the curve, but others aren't
hard to find either), but finding an isomorphism to a Weierstrass model
requires the choice of one.
Getting the model involves finding functions x,y on C that have pole orders
2 and 3 at the chosen point, and are regular everywhere else. You can then
solve for a linear relation between y^2, xy, y, x^3,x^2,x, 1 (which you
could do with interpolation at that point). Unfortunately, this hasn't been
implemented as a fully general automatic procedure in sage yet.
What you CAN do (although this is a little insane), is recognize the
j-invariant of the Jacobian:
S=C.riemann_surface(prec=400)
tau=S.riemann_matrix()[0,0]
CC=parent(tau)
q=exp(2*CC.0*CC.pi()*tau)
J=j_invariant_qexp(100).truncate(100)
sage: J(q).algdep(3)
6586060771822965534*x - 3927422869602298031
so the j-invariant is quite probably
j=3927422869602298031/6586060771822965534
which means that the elliptic curve you are looking for is almost certainly
a quadratic twist of:
sage: EllipticCurve_from_j(j)
Elliptic Curve defined by y^2 + x*y = x^3 + 120280*x - 864149826 over
Rational Field
Again, the purely algebraic operation of computing a Weierstrass model of C
with a point directly is a much better way to go about things.
In fact, with Magma (even the online calculator) it's easy to do. You'll
find the elliptic curve you're looking for is the twist by 23 of the curve
above.
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