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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    $\begingroup$ This article archive.numdam.org/ARCHIVE/JTNB/JTNB_2005__17_1/… by Christophe Delaunay talks about this construction. $\endgroup$ Feb 14, 2011 at 17:53
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    $\begingroup$ @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 terrytao.wordpress.com/2007/05/04/… and math.columbia.edu/~yxy 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. $\endgroup$ Feb 14, 2011 at 20:09
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    $\begingroup$ @Helfgott: I'm certainly interested in helping with this computational project... $\endgroup$ Feb 16, 2011 at 16:16
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    $\begingroup$ @Stein: Cool. Let's take this over to email. (Kevin B., are you in?) $\endgroup$ Feb 16, 2011 at 20:35
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    $\begingroup$ Has there been any recent progress on this very interesting set of questions? $\endgroup$ Feb 6, 2016 at 18:52


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