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It seems that Lan Nguyen proved in a preprint on arxiv of 2013 that the Tate-Shafarevich group of a rational elliptic curve is finite. However, I couldn't find any published version thereof. So is it now known that the Tate-Shafarevich group of a rational elliptic curve is finite?

Many thanks in advance.

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  • $\begingroup$ Linking the arxiv preprint might be useful. $\endgroup$
    – Wojowu
    Commented Nov 24, 2016 at 14:02
  • $\begingroup$ Here it is: arxiv.org/abs/1309.7675 $\endgroup$
    – Sylvain JULIEN
    Commented Nov 24, 2016 at 14:05
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    $\begingroup$ the paper only deals with plane cubics locally isomorphic to $E$, so it's if anything a proof that the 3-torsion in Sha is finite (the error is p12, six lines up: there are all sorts of crazy homogeneous spaces that can contribute to Sha and it's certainly not true that they're all plane cubics). It also claims to prove that if Sha has an element of order 3 then $E$ has CM by a cube root of 1, which is surely not true. $\endgroup$
    – znt
    Commented Nov 24, 2016 at 19:58

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MO is not the place to discuss the validity of preprints, but I think it is safe to say that the finitiness of the Tate-Shafarevich group for elliptic curves over $\mathbb{Q}$ is considered an open problem for rank $\geq 2$.

As far as I can tell the paper was never published.

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    $\begingroup$ In fact, for every single elliptic curve over $\mathbb{Q}$ of rank at least $2$, the finiteness of its Tate-Shafarevich group is an open problem. $\endgroup$
    – Alex B.
    Commented Nov 24, 2016 at 20:31
  • $\begingroup$ Both of you are talking about analytic rank, right? We could currently have an elliptic curve with algebraic rank 0, analytic rank 2 and sha infinite, so the most general statement is perhaps that no ell curve with analytic rank 2 or more has provably finite Sha. $\endgroup$
    – znt
    Commented Nov 24, 2016 at 22:39

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