I am looking for an elliptic curve $E$ with Weierstraß coefficients in $\mathbb{Q} $ so that for some prime $p$ the following conditions are satisfied:

(1) ${\large Ш}_{p^{\infty}}(\mathbb{Q},E)$ contains an element of order $p^{2}$.

(2) $\operatorname{Gal} (\mathbb{Q}(E_{p})/ \mathbb {Q} ) $ is nonsolvable, so in particular $E_{p}(\mathbb{Q})=0$ and $E$ has no CM.

I need this to verify (or falsify) a calculation of mine, but all curves I've found so far for which the exact structure of $\large Ш$ is known violate at least one of these conditions.

If such a curve exists, a Weierstraß equation or a Cremona label would be nice.

  • 1
    $\begingroup$ lmfdb.org/EllipticCurve/Q/… gives you three possible candidates 287175n1, 321398d1, 366100y1: . The Galois group is $GL_2(\mathbb{F}_5)$ and there are $5^4$ elements in Sha. Now you need to use Kurihara's results on the Fitting ideals of Selmer groups to check if these are 5-torsion or if there is a 25-torsion element. This can be done using modular symbols. - In general I believe there are many examples, just it could be that the conductor is too large to find them. $\endgroup$ – Chris Wuthrich Sep 19 '17 at 12:05
  • $\begingroup$ If you are happy to drop the assumption that the base field is $\mathbb{Q}$ then Iwasawa theory can provide you examples over some number field. $\endgroup$ – Chris Wuthrich Sep 19 '17 at 12:09
  • $\begingroup$ These curves have a Shafarevich group of order 625, but how do I know this group this group is not isomorphic to $(\mathbb {Z}/5)^{4} $? Is there some theorem of which I am not aware that tells me so? $\endgroup$ – The Thin Whistler Sep 19 '17 at 12:15
  • 1
    $\begingroup$ As I mentioned, there is a theorem of Kurihara. It is theorem 1.1.1 in the paper currently listed as 8 on his webpage "The structure of Selmer groups for elliptic curves and modular symbols". If I had time, I could do the calculations foryou, but I am afraid I can't do that. $\endgroup$ – Chris Wuthrich Sep 19 '17 at 12:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.