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$C/ \Bbb{Q}: 3X^3 + 4Y^3 + 5Z^3 = 0$ is known to be a nontrivial element of the Tate–Shafarevich group of the elliptic curve $E/\Bbb{Q}:X^3 + Y^3 + 60Z^3 = 0$. It is also an example of an abelian variety for which finiteness of Sha is known. In fact, $|\mathrm{III}(E/\Bbb{Q})| = 3^2$.

But I have never seen the proof of $|\mathrm{III}(E/\Bbb{Q})| = 3^2$.I don't either know the proof of $[C]$ has order $3$ in $\mathrm{III}(E/\Bbb{Q})$. Maybe this is first proved by Rubin, but I cannot find article which discusses this example. Could you give me an reference for the proof?

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$\DeclareMathOperator{\sha}{Ш}$ I am not sure that the proof that Sha has order 9 is anywhere spelled out in full. Here the ideas how to do it.

First, that the order of $C$ is three in the $\sha$ is just saying that it has a point over a field of degree 3 (index=period), which is obvious, and none of degree 1, which was first proven by Selmer. See Cassel's lectures, chapter 18.

A $2$-descent, proves that $E(\mathbb{Q})$ has rank $0$ and that $\sha[2]$ is trivial. Calculating the torsion subgroup of $E$ (trivial), the Tamagawa numbers (all trivial and the value of $L(E,1)$, either using cm-theory or modular symbols, reveals that the Birch and Swinnerton-Dyer conjecture is equivalent to $\sha$ having $9$ elements.

Rubin's work on the main conjecture of elliptic curves with CM, as in "Tate-Shafarevich groups and $L$-functions of elliptic curves with complex multiplication" for instance, shows that the order of the $p$-primary part of $\sha$ is correctly predicted by BSD for all primes not dividing the order of the units in $\mathbb{Q}(\sqrt{-3})$. For our curve this means $\sha[p]=0$ for all primes $p>3$. But this does not cover the very bad prime $3$; though maybe has done this since, I don't know.

Selmer's original work proved that $\sha[3]$ contains $9$ elements, by doing a second descent $E\to E'$ via an isogeny of degree $3$. This exact example appears in magma's documentation of its ThreeDescent function: https://magma.maths.usyd.edu.au/magma/handbook/text/1515#17622 . More interestingly this calculation also shows that the $3$-Selmer group of the curves $E'$ at the other end of the $3$-isogeny is trivial. Hence the $3$-part of BSD is true for that curve and since this is invariant under isogeny, it is also true for ours. Therefore $\sha$ has order $9$ with the structure being $\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}$.

This last step should also be possible using Heegner points with $D=-17$. Sage uses this in "prove_BSD" for the curve $E'$.

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  • $\begingroup$ Thank you for your really helpful answer. I have two questions. Firstly, $ E(\Bbb{Q})/2 E(\Bbb{Q})$ is trivial in this case, so $sha(E/ \Bbb{Q})=Sel^{(m)}(E/ \Bbb{Q}))$, and $m-$Selmer group is computable. Why do you need more complex argument ? $\endgroup$
    – Duality
    Commented Feb 10, 2023 at 10:38
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    $\begingroup$ 1. The $m$-Selmer group is computable for $m=2$ and $m=3$. That (2-descent) is what I use to prove that $E(\mathbb{Q})/2$ is trivial. It does not help for any $m$ with a prime divisor larger than 3. The 3-descent only tells you about $Ш[3]$, but not $Ш[9]$ without further work. $\endgroup$
    – Chris Wuthrich
    Commented Feb 10, 2023 at 11:18
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    $\begingroup$ 2. For an element in $Ш$ it is equivalent to ask if it non-trivial or that the curve has a rational point. So all methods to show that a given element is non-trivial prove that $C(\mathbb{Q})$ is empty. Though sometimes, like when you use the Cassels-Tate pairing it is hidden that you are really just solving a diophantine equation. $\endgroup$
    – Chris Wuthrich
    Commented Feb 10, 2023 at 11:18
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    $\begingroup$ Read a standard text on 2-descent, like chapter 3 in Cremona's book johncremona.github.io/book/fulltext/index.html. Or use software like I did. $\endgroup$
    – Chris Wuthrich
    Commented Feb 10, 2023 at 20:37
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    $\begingroup$ Lastly. Selmer's proof is reproduced in Cassels' lectures as mentioned. More general study n-descent. Kolyvagin's result that helps you proving triviality of $p$-primary part of $Ш$ is explained in places like Gross' survey. $\endgroup$
    – Chris Wuthrich
    Commented Feb 11, 2023 at 0:10

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