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May 16, 2021 at 20:58 answer added Joe Silverman timeline score: 10
May 16, 2021 at 18:27 comment added user166831 Rank 0 means the points are torsion, and there are only finitely many by Mordell's theorem. If $E$ is in Weierstrass form with integer coefficients and $(x:y:1)$ is a torsion point, then either $y=0$ or it divides the discriminant (Lutz-Nagell).
May 16, 2021 at 17:01 comment added folenn @NoamD.Elkies I see. What if we just require the rank to be zero?
May 16, 2021 at 17:01 history edited folenn CC BY-SA 4.0
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May 16, 2021 at 16:43 comment added Noam D. Elkies Can't happen if you put $E$ in Weierstrass form $y^2 = P(x)$ because the 2-torsion points are just $(x,0)$ where $x$ is one of the zeros of $P$.
May 16, 2021 at 14:11 review First posts
May 16, 2021 at 14:26
May 16, 2021 at 14:07 history asked folenn CC BY-SA 4.0