3
$\begingroup$

$\newcommand{\Sha}{\operatorname{Sha}}$Let $E/\mathbb{Q}$ be an elliptic curve, and let $\Sha(E/\mathbb{Q})$ denote the Tate–Shafarevich group of $E/\mathbb{Q}$. It is known that the 2-primary component $\Sha(E/\mathbb{Q})[2]$ can become arbitrarily large as we vary $E$.

For example, Kramer found an elliptic curve over $\mathbb{Q}$ with discriminant $m(16m+1)$ and $\operatorname{\#Sha}(E/\mathbb{Q})[2] \geq 2^{2k-2}$, where $k$ is the number of prime factors of $16m+1$ ("A family of semistable elliptic curves with large Tate–Shafarevich groups," Proceedings of the AMS, 89 (1983)).

This elliptic curve has a discriminant $\Delta_E$ with a very large number of prime factors.

However, is it possible to enlarge the 2-part of the Tate–Shafarevich group while keeping the number of prime factors of the discriminant small? Specifically, can we increase $\dim_2\Sha(E/\mathbb{Q})[2] - \omega(\Delta_E)$ arbitrarily?

The largest value of $\dim_2\Sha(E/\mathbb{Q})[2] - \omega(\Delta_E)$ that I am aware of is 4. If anyone knows of an example with a larger $\dim_2\Sha(E/\mathbb{Q})[2] - \omega(\Delta_E)$, I would greatly appreciate the information.

Let $\omega(\Delta_E)$ denote the number of distinct prime factors of $\Delta_E$. The following two elliptic curves have a small number of prime factors of their discriminants, $\omega(\Delta_E) = 2$, and $\dim_2\Sha(E/\mathbb{Q})[2] - \omega(\Delta_E) = 4$:

  1. $E_d: y^2 = x^3 + dx$ when we let $d$ be $p = 2^{2^k} + 1$, then $\Sha(E_{65537}) \cong (\mathbb{Z}/8\mathbb{Z})^2$. For more information, refer to the discussion on MathOverflow here.

  2. Let $E_{p,n}: y^2 = x^3 + p^nx$ be an elliptic curve. According to LMFDB, for $(p,n) = (73,3)$, $\operatorname{\#Sha}(E_{p,n}) = 64$. This was mentioned in a MathOverflow discussion available here.

Edited

In Table 2 of the document available at Sza.pdf, the case of $(n,p) = (16,48)$ is cited as an example where $dim_2\Sha(E/\mathbb{Q})- \omega(\Delta_E) = 18 - 6 = 12$. I'm also interested in the strong problem:$ord_2\#Sha(E/\mathbb{Q})- \omega(\Delta_E)$ can be arbitrary large ?

This should probably be considered a separate issue, so after giving it some thought on my own, I might ask you about it another time.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Partial answer: the answer is no if $E$ has a rational 2-torsion point, which is basically just an application of genus theory. In this case the 2-Selmer rank is bounded explicitly by $2^{\omega(\Delta_E)}$ (up to some fudge factor). $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 2, 2023 at 16:19
  • 2
    $\begingroup$ For $E$ without rational $2$-torsion, I would look for examples among those curves for which the $2$-torsion of the class number of $\mathbb{Q}(E[2])$ is large. $\endgroup$
    – Chris Wuthrich
    Commented Nov 2, 2023 at 17:15
  • 1
    $\begingroup$ I think you are confusing $\dim_{\mathbb{F}_{2}} {\rm Sha}(E/\mathbb{Q})[2]$ and the $2$-adic valuation of the order of ${\rm Sha}$. For the $n=16$, $p=48$ curve mentioned in your post, a $2$-descent says that the $2$-Selmer group is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^{4}$. $\endgroup$
    – Jeremy Rouse
    Commented Nov 3, 2023 at 1:34

0

Reset to default

You must log in to answer this question.