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Consider an elliptic curve $E$ defined over $\mathbb Q$. Assume that the rank of $E(\mathbb Q)$ is $\geq2$. (Assume the Birch-Swinnerton-Dyer conjecture if needed, so that analytic rank $=$ algebraic rank.) How do you construct a point of infinite order on $E(\mathbb Q)$?

(If the rank were $1$, then the Gross-Zagier construction would do the job. If the rank were $0$, then, of course, there would be no such point.)

Implicit in a paper of Mazur and Swinnerton-Dyer ("Arithmetic of Weil curves", Invent. Math., 25, 1-61 (1974); see especially section 2.4) there is a construction that seems to work a positive proportion of the time, though not always. Here is what the construction would be according to my understanding: take a modular parametrisation $\phi:X_0(N)\to E(\mathbb C)$, consider its points of ramification on the imaginary axis (there is at least one), take the image $\phi(z)$ of one such point $z$; due to standard magic, $X_0(N)$ has an algebraic model that makes phi into an algebraic map; the trace of $\phi(z)$ is a point of $E(\mathbb Q)$ that might be non-torsion, and sometimes is).

Has any further work been done on this? (In particular, has it been proven that this works infinitely often?) Are there any other constructions for which similar statements have been conjectured or proven?


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I don't know of any further work on this; my impression is that this construction has been to some extent neglected. – Emerton Feb 14 '11 at 17:22
This article… by Christophe Delaunay talks about this construction. – Chris Wuthrich Feb 14 '11 at 17:53
@Helfgott: Section 5.2 of [Mazur-Swinnerton-Dyer] discusses another related construction. This turns out to be related to recent research of Zhang and his students: see… and preprints/triple.pdf Their triple product L-function result explains why this construction must give torsion points when the analytic rank is 2 or higher. – William Stein Feb 14 '11 at 20:09
@Helfgott: I'm certainly interested in helping with this computational project... – William Stein Feb 16 '11 at 16:16
@Stein: Cool. Let's take this over to email. (Kevin B., are you in?) – H A Helfgott Feb 16 '11 at 20:35

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