What's the Hilbert class field of an elliptic curve? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:43:15Z http://mathoverflow.net/feeds/question/15990 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15990/whats-the-hilbert-class-field-of-an-elliptic-curve What's the Hilbert class field of an elliptic curve? Franz Lemmermeyer 2010-02-21T21:21:14Z 2010-02-22T04:35:21Z <p>My question points in a direction similar to <a href="http://mathoverflow.net/questions/10947/whats-the-analogue-of-the-hilbert-class-field-in-the-following-analogy" rel="nofollow">Qiaochu's</a>, but it's not the same (or so I think). Let me provide you with a little bit of background first.</p> <p>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: y<sup>2</sup>=f<sub>4</sub>(x) for some quartic polynomial f<sub>4</sub>(x) in K[X]; here, C does not have a K-rational point, but has points in every completion of K. </p> <p>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. </p> <p>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: </p> <ul> <li>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?</li> </ul> <p>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. </p> <p>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:</p> <ul> <li>Has this "norm map" been studied in the literature?</li> </ul> <p>(I know that the norm map from E(L) to E(K) was investigated a lot, in particular in connection with Heegner points). </p> <p>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. </p> http://mathoverflow.net/questions/15990/whats-the-hilbert-class-field-of-an-elliptic-curve/16018#16018 Answer by Bjorn Poonen for What's the Hilbert class field of an elliptic curve? Bjorn Poonen 2010-02-22T02:46:16Z 2010-02-22T04:35:21Z <p><strong>EDIT</strong>: This is a completely new answer.</p> <p>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 $\mathbf{Q}$.)</p> <blockquote> <p><strong>Theorem:</strong> Assume BSD. There exists a number field $K$ and an elliptic curve $E$ over $K$ such that there is <em>no smallest</em> field extension $L$ of $K$ such that $\operatorname{Cores} \colon \operatorname{Sha}(L,E) \to \operatorname{Sha}(K,E)$ is the zero map.</p> </blockquote> <p><strong>Proof:</strong> We will use BSD data (rank and order of Sha) from <a href="http://www.warwick.ac.uk/staff/J.E.Cremona//ftp/data/" rel="nofollow">Cremona's tables</a>. Let $K=\mathbf{Q}$, and let $E$ be the elliptic curve 571A1, with Weierstrass equation <code>$$y^2 + y = x^3 - x^2 - 929 x - 10595.$$</code> Then $\operatorname{rk} E(\mathbf{Q})=0$ and <code>$\#\operatorname{Sha}(\mathbf{Q},E)=4$</code>. 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.</p> <p>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 $\operatorname{Sha}(L_i,E)=0$.</p>