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The result of LMFDB claims (https://www.lmfdb.org/EllipticCurve/Q/1640/c/1 ) that (2-part of) Tate-Shafarevich group $\mathrm{Sha}(E/\Bbb{Q})$ of elliptic curve $y^2=x^3-8747x-314874$ has order $16$. In particular, the order of $2$-Selmer group is larger than $16$.

But Magma calculates as following,

A:=EllipticCurve([0,0,0,-8747,-314874]); 
Sel2:=TwoSelmerGroup(A); Sel2;

The output of this is,

Abelian Group isomorphic to Z/2 + Z/2 + Z/2 Defined on 3 generators Relations:
    2*Sel2.1 = 0
    2*Sel2.2 = 0
    2*Sel2.3 = 0

It seems contradiction because $2$-Selmer group is larger than $2$-part of Tate-Shafarevich group.

Where did I go wrong and how can I correctly calculate order of $2$-Selmer group using Magma ?

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    $\begingroup$ I suspect Magma's TwoSelmerGroup function is only computing the 2-torsion part of the Selmer group $Sel_2$, not the 2-primary part $Sel_{2^\infty}$. If the Tate-Shafarevich group has some elements of order 4 then that would explain the apparent discrepancy. $\endgroup$
    – David Loeffler
    Commented Jun 10, 2023 at 8:01
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    $\begingroup$ I don't understand the vote to close here. Although the question was based on a misconception, it's a pretty subtle and plausible mistake, and I think the discussion might still be informative for future readers. $\endgroup$
    – David Loeffler
    Commented Jun 10, 2023 at 8:51

1 Answer 1

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I don't see any contradiction: the Selmer group also has a contribution of rational points. Indeed, the group of 2-torsion rational points on this elliptic curve is isomorphic to $\mathbb Z/2\mathbb Z$ and generated by $(-54,0)$, as Magma will readily confirm (probably, I used GP/Pari).

UPDATE: As David suspected, the Tate-Shafarevic group of this elliptic curve is isomorphic to $(\mathbb Z/4\mathbb Z)^2$ so 2-descent misses the elements of order 4 in Sha.

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    $\begingroup$ It's the other way around: the problem is that the computations are (allegedly) saying that Sha is bigger than Selmer. $\endgroup$
    – David Loeffler
    Commented Jun 10, 2023 at 7:59
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    $\begingroup$ It is as you suspected in your comment: Pari confirms by 2-descent that Sha is $(\mathbb Z/4\mathbb Z)^2$, so it has elements of order 4, whereas the function used in the question only computes the 2-torsion part of the Selmer group. $\endgroup$
    – Olivier
    Commented Jun 10, 2023 at 8:30

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