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The Elkies curve $$ E:y^2+xy+y=x^3−x^2−20067762415575526585033208209338542750930230312178956502x+34481611795030556467032985690390720374855944359319180361266008296291939448732243429 $$ conductor of Elkies curve is $f=33455601108357547341532253864901605231198511505793733138900595189472144724781456635380154149870961231592352897621963802238155192936274322687070=2\times 3^2\times 5\times 7\times 11\times 13\times 17\times 19\times 48463\times 20650099\times 315574902691581877528345013999136728634663121\times 376018840263193489397987439236873583997122096511452343225772113000611087671413$

We assume the Generalized Riemann Hypothesis (GRH), then $\text{rank}_\mathbb{Z}E(\mathbb{Q})=28$. If we add the BSD conjecture, can we calculate the order of $Ш(E/\mathbb{Q})$?

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    $\begingroup$ Sagemath for instance has E.sha().an() for the "analytic order" of Sha, ie the order according to BSD. But I doubt that you can calculate the leading term of the $L$-function of $E$ with such a huge conductor to any precision to determine what it is for this curve. $\endgroup$
    – Chris Wuthrich
    Commented Jun 11, 2023 at 15:19
  • $\begingroup$ Why are interested in this? $\endgroup$
    – Chris Wuthrich
    Commented Jun 11, 2023 at 15:22
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    $\begingroup$ I read an paper where the author gave some estimates of elliptic curve sha group and claimed some order assuming that BSD is true. For example, let's take the elliptic curve $E:y^2 + xy + y = x^3−x^2−911138880x−10586098442003$,the author asserts $\#Ш(E/\mathbb{Q})=169$ and explains that this is calculated using magma plus BSD. So I was curious to see if I could calculate the sha group Elkies curve by adding BSD. $\endgroup$
    – GuoJi
    Commented Jun 11, 2023 at 15:35
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    $\begingroup$ Oh but that is very different. That curve has a tiny conductor, in fact it is lmfdb.org/EllipticCurve/Q/14450ba2. You can prove that Sha is of that size without having to assume any conjecture. The speed of calculations of the L-function is going to be something like $O(\sqrt{N})$. For conductors as huge as Elkies' curve that is huge. $\endgroup$
    – Chris Wuthrich
    Commented Jun 11, 2023 at 15:41

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