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everyone! Let $E:y^2+y=x^3-61$ be the minimal model of the elliptic curve 243b. How can I find the CM point $\tau$ in $X_0(243)$ such that $\tau$ maps to the point $(3\sqrt[3]{3},4)$ under the modular parametrization? Can anyone tell me the answer or how to use sagemath to find it?

I use the sagemath code

EllipticCurve([0,0,1,0,-61])
phi = EllipticCurve([0,0,1,0,-61]).modular_parametrization()
f=phi.power_series(prec = 10000)[1]
f.truncate(20000)

to get the parametrization of y coordinate, then I use

q=var('q')
f(q)=
df=diff(f,q)              
NewtonIt(q)=q-(f/df)(q)
xn=e^(2*pi*I*a/20.031)                      
for i in range(1000):
    xn=N(NewtonIt(xn),digits=2000)
print xn

to get the numerical $e^{2\pi i \tau}$. After taking log and dividing by $2 \pi i$, I get the numerical $\tau$. But if I use 

z=
p=z.algebraic_dependency(100)

I get the wrong polynomial. Why?

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  • $\begingroup$ any method is welcome！ $\endgroup$ Commented Aug 10, 2019 at 7:41
  • 1
    $\begingroup$ In principle you can compute a fundamental domain of $X_0(N)$ which reduces to the problem to that of finding a root of an analytic function on some compact domain. The standard way to compute the image $\varphi(\tau)$ under the modular parametrisation is to integrate the differential form $\omega_f =2\pi if(\tau) d\tau$ from $\infty$ to $\tau$, which gives you a complex number and thus a point of $E$ via the Weierstrass functions. You may also be interested by the PhD theses of Christophe Delaunay and Hao Chen, see arxiv.org/abs/1412.2827v2 $\endgroup$ Commented Aug 10, 2019 at 10:10
  • $\begingroup$ @Brunault Thanks a lot, Francois! I find the numerical approximation using the Newton method, but I can not find the exact CM point. It should be an algebraic number, but I can not recover it from the numerical approximation. $\endgroup$ Commented Aug 10, 2019 at 11:39
  • $\begingroup$ If you have a numerical approximation you can try using PARI/GP algdep(), Mathematica RootApproximant[], or equivalent function of some CAS for find the irreducible polynomial for the number. Maybe your approximation is wrong? $\endgroup$ Commented Aug 10, 2019 at 12:52
  • $\begingroup$ Not all preimages are CM points, so one needs a priori to locate all the preimages (possible in theory using the residue theorem applied to finer and finer subdivisions of the fundamental domain). As you see, this problem is not easy. $\endgroup$ Commented Aug 10, 2019 at 16:11

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