Which primes can divide orders of Tate-Shafarevich groups?

Question

Heuristic arguments due to (I believe) Delauney predict that every prime divides the order of the Tate-Shafarevich group of infinitely many elliptic curves over $\mathbf{Q}$. However, is it even known that for every prime $p$, there's at least one elliptic curve over $\mathbf{Q}$ whose Tate-Shafarevich group is finite and has order divisible by $p$?

Answer 1

Just to record an answer: yes, it is believed to be true that for each prime $p$ there is an elliptic curve $E_{/\mathbb{Q}}$ whose Shafarevich-Tate group contains an element of order $p$. But this is wide open.

One can however prove the following:

Theorem (R. Kloosterman): If you fix a prime $p$, then there is a number field $K_p$ -- i.e., depending on $p$ -- and an elliptic curve $E/K_p$ whose Shafarevich-Tate group contains an element of order $p$.

As Kloosterman remarked (in his 2005 paper on the subject, and also just now in the above comments) via a simple "Weil restriction" argument, this implies that there exists an abelian variety $A_{/\mathbb{Q}}$ whose Shafarevich-Tate group contains an element of order $p$. But now the dimension of $A$ goes to infinity with $p$.

Others have since worked out various improvements of Kloosterman's Theorem. Here are three different directions:

1) What is the minimal degree $d_p = [K_p:\mathbb{Q}]$ of a number field $K_p$ in above theorem? Kloosterman's argument gave $d_p = O(p^4)$. In a 2005 paper I showed that one can take $d_p = 2p^3$. In a 2010 paper with Shahed Sharif, we showed that one can take $d_p = p$. (Note that whether $d_p = p$ is in fact best possible is an open question. At one point I thought I had an argument to show that it was, but this was wrong. I now suspect that this is only best possible by a method proceeding along the lines of our construction. For instance, I believe it is conceivable that $d_p = 2$ for all $p$, and this has something to do with how Sha behaves in quadratic twists...)

2) What kind of elliptic curves $E$ can be used to produce these elements in Sha? Sharif and I showed that in fact one can start with any elliptic curve $E_0$ over $\mathbb{Q}$ and take its base change to $K_p$ a degree $p$ number field. (Similarly, one can start with any elliptic curve $E$ over any number field and get order $p$ elements of Sha in an extension field of degree $p$.) And in fact $p$ does not need to be a prime number here: it holds for every integer $n$. And in fact you can get as many elements of order $n > 1$ as you want.

3) What kind of control can one get over the field extension $K_p/\mathbb{Q}$? Can one for instance choose it to be Galois, abelian, or cyclic? Matsuno proves that for any cyclic degree $p$ extension $K_p/\mathbb{Q}$ one can find elliptic curves $E_{\mathbb{Q}}$ such that the base change to $K_p$ has as many order $p$ elements in its Shafarevich-Tate group as one wants. There are similar results along these lines (with dihedral extensions and elements of the Selmer group) by Alex Bartel.

P.S.: As Tim Dokchitser points out, replacing $\mathbb{Q}$ by any fixed number field does not help to answer the OP's question, although the results of Sharif and myself hold equally well in this relative setting. (I was going to ask "In what way could an arbitrary but fixed number field be any better than $\mathbb{Q}$?" but then I remembered an amazing result about $2$-Selmer parity that only holds over certain imaginary number fields...due to Dokchitser and Dokchitser.)