Ask Your Question
1

How to find a CM point with the image in the elliptic curve under modular parametrization given

asked 2019-08-11 23:05:54 +0100

Lee gravatar image

updated 2019-08-29 22:19:49 +0100

FrédéricC gravatar image

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?

edit retag flag offensive close merge delete

1 Answer

Sort by » oldest newest most voted
0

answered 2020-05-28 18:09:27 +0100

John Cremona gravatar image

I think that the modular functions giving the modular parametrization (the two power series in q returned by phi.power_series()) will converge very slowly.

edit flag offensive delete link more

Comments

Thank you very much, John.

Lee gravatar imageLee ( 2022-03-23 08:35:44 +0100 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower

Stats

Asked: 2019-08-11 23:05:54 +0100

Seen: 665 times

Last updated: May 28 '20