My question points in a direction similar to Qiaochu's, but it's not the same (or so I think). Let me provide you with a little bit of background first.

Let E be an elliptic curve defined over some number field K. The Tate-Shafarevich group of E/K consists of certain curves of genus 1, isomorphic to E over some extension, with points everywhere locally. In the simplest case of an element of order 2, such a curve has the form C: y2=f4(x) for some quartic polynomial f4(x) in K[X]; here, C does not have a K-rational point, but has points in every completion of K.

If we look at E in some extension L/K, this curve C still has points everywhere locally, but if it has a global point (in L) we say that the corresponding element in the Tate-Shafarevich group of E/K capitulates. Heegner's Lemma says that elements of order 2 cannot capitulate in extensions of odd degree, which is the analogue of the similarly trivial observation that ideals generating a class of order 2 cannot capitulate (become principal) in an extension of odd degree.

I gave a few talks on the capitulation of Tate-Shafarevich groups more than 10 years ago. A little later the topic became almost fashionable under the name of "visualizing" elements of Sha. I discussed the following question with Farshid Hajir back then, but eventually nothing came out of it. Here it is. For capitulation of ideal classes, there is a "canonical" extension in which this happens: the Hilbert class field. So my question is:

  • may we still dream about the existence of a curve with all the right properties, or are there reasons why such a thing should not exist?

We also know that capitulation is not the correct notion for defining the Hilbert class field, which is the largest unramified abelian extension of a number field. These notions do not seem to make any sense for elliptic curves, but we can characterize the Hilbert class field also in the following way: among all finite extensions L/K for which the norm of the class group of L down to K is trivial, the Hilbert class field is the smallest.

Taking the norm of Sha of an elliptic curve defined over L down to K does make sense (just add the equivalence classes of the conjugate homogeneous spaces using the Baer-sum construction or in the appropriate cohomology group). So here's my second question:

  • Has this "norm map" been studied in the literature?

(I know that the norm map from E(L) to E(K) was investigated a lot, in particular in connection with Heegner points).

Let me add that I do not assume that such a "Hilbert class curve" can be found among the elliptic curves defined over some extension field; if there is a suitable object, it might be the Jacobian of a curve of higher genus or an abelian variety coming from I don't know where.

  • $\begingroup$ Are you looking for a curve or are you looking for a number field which splits all elements of Sha? If the latter, you'll have to assume finiteness of Sha to even get started. Sha is related to the Brauer group of the elliptic curve and people have looked at splitting of algebras. $\endgroup$ Feb 21, 2010 at 21:32
  • $\begingroup$ I don't know what I'm looking for; there are many number fields that principalize Sha(E) - I'd like to have a nice one, or some nice curve with nice maps down to E. Of course I assume that Sha is finite; who doesn't? $\endgroup$ Feb 21, 2010 at 21:36

1 Answer 1


EDIT: This is a completely new answer.

I will prove that your specific suggestion of defining a Hilbert class field of an elliptic curve $E$ over $K$ does not work. I am referring to your proposal to take the smallest field $L$ such that the corestriction (norm) map $\operatorname{Sha}(L) \to \operatorname{Sha}(K)$ is the zero map. (I have to assume the Birch and Swinnerton-Dyer conjecture (BSD), though, for a few particular elliptic curves over $\mathbf{Q}$.)

Theorem: Assume BSD. There exists a number field $K$ and an elliptic curve $E$ over $K$ such that there is no smallest field extension $L$ of $K$ such that $\operatorname{Cores} \colon \operatorname{Sha}(L,E) \to \operatorname{Sha}(K,E)$ is the zero map.

Proof: We will use BSD data (rank and order of Sha) from Cremona's tables. Let $K=\mathbf{Q}$, and let $E$ be the elliptic curve 571A1, with Weierstrass equation $$y^2 + y = x^3 - x^2 - 929 x - 10595.$$ Then $\operatorname{rk} E(\mathbf{Q})=0$ and $\#\operatorname{Sha}(\mathbf{Q},E)=4$. Let $L_1 = \mathbf{Q}(\sqrt{-1})$ and $L_2 =\mathbf{Q}(\sqrt{-11})$. It will suffice to show that the Tate-Shafarevich groups $\operatorname{Sha}(L_i,E)$ are trivial.

Let $E_i$ be the $L_i/\mathbf{Q}$-twist of $E$. MAGMA confirms that $E_1$ is curve 9136C1 and $E_2$ is curve 69091A1. According to Cremona's tables, $\operatorname{rk} E_i(\mathbf{Q})=2$ and $\operatorname{Sha}(\mathbf{Q},E_i)=0$, assuming BSD. Thus $\operatorname{rk} E(L_i) = 0+2=2$ and $\operatorname{Sha}(L_i,E)$ is a $2$-group. On the other hand, MAGMA shows that the $2$-Selmer group of $E_{L_i}$ is $(\mathbf{Z}/2\mathbf{Z})^2$. Thus $\operatorname{Sha}(L_i,E)[2]=0$, so $\operatorname{Sha}(L_i,E)=0$.

  • $\begingroup$ So the only chance of keeping the dream alive is restricting the varieties to certain "nice ones" (modular curves?). I guess in this case we cannot hope to find counterexamples since the calculations are prohibitive. $\endgroup$ Feb 23, 2010 at 14:46

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.