Decomposition of Tate-Shafarevich groups in field extensions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T13:53:11Z http://mathoverflow.net/feeds/question/19760 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19760/decomposition-of-tate-shafarevich-groups-in-field-extensions Decomposition of Tate-Shafarevich groups in field extensions David Hansen 2010-03-29T19:33:30Z 2010-03-30T12:27:06Z <p>Suppose $E/\mathbb{Q}$ is an elliptic curve with rank zero. According to the conjecture of Birch and Swinnerton-Dyer, the special value $L(1,E_{/\mathbb{Q}})$ should be equal (up to some harmless factors) to the order of the Tate-Shafarevich group Sha$(E/\mathbb{Q})$. Now, suppose $\chi$ is a Dirichlet character, and $K/\mathbb{Q}$ is the cyclic extension cut out by $\chi$. My question is:</p> <p>What is the precise (conjectural) relation between the <em>individual</em> special values $L(1,E\times\chi^i)$ and Sha$(E/K)$?</p> <p>More precisely, we have $L(s,E_{/K})=\prod_{i=0}^{\mathrm{ord}(\chi)} L(s,E\times \chi^i)$. If $E/K$ has rank zero, is there a decomposition of Sha$(E/K)$ with respect to an action of $\mathrm{Gal}(K/\mathbb{Q})$ such that the individual pieces of this decomposition have orders given by the individual values $L(1,E \times \chi^i)$?</p> http://mathoverflow.net/questions/19760/decomposition-of-tate-shafarevich-groups-in-field-extensions/19810#19810 Answer by Olivier for Decomposition of Tate-Shafarevich groups in field extensions Olivier 2010-03-30T09:55:22Z 2010-03-30T09:55:22Z <p>First of all, I am not sure I fully agree with the notion that Tamagawa numbers are harmless factors.</p> <p>What you wish for exists, and here is roughly why. The Birch and Swinnerton-Dyer conjecture is a special case of the Equivariant Tamagawa Number Conjecture for elliptic curves. As $E$ is defined over $\mathbb{Q}$, it is modular so we have a good candidate for the conjectural zeta elements appearing in the ETNC, namely Kato's Euler system. If you are ready to admit this, then for all abelian extensions $K$ of $\mathbb{Q}$, there is a variant of BSD with coefficients which typically appears as an equality: $$|H^2(\operatorname{Spec}(O_{K}[1/p]),T_{p}E)|=|H^1(\operatorname{Spec}(O_{K}[1/p]),T_{p}E)/z_{Kato}|$$</p> <p>In order to get the precise form you want, you will need a specific computation of <code>$\exp^{*}(z_{Kato})$</code> (the dual exponential map of Bloch and Kato) over $K$. This looks like an exercise (now the case of $\mathbb{Q}$ has been treated by Kato), but this is one I never did, so there might be a hidden difficulty. Granting this, you should get a relation between the $p$-order of $L(1,E\times\chi^i)$ and the cardinal of <code>$$\chi^{i}\ker\left(H^{1}\left(\operatorname{Spec}(\mathbb{Z}[1/p]),T\right)\rightarrow H^1(\mathbb{Q}_{p},T)/H^1_{f}(\mathbb{Q}_{p},T)\right)$$</code> In the above $$T=T_{p}E\otimes_{\mathbb{Z}}\mathbb{Z}[G]\otimes \mathbb{Q}/\mathbb{Z}$$ with $G=\operatorname{Gal}(K/\mathbb{Q})$ and $H^1_f$ denotes Bloch-Kato Selmer group. The objects appearing are $\mathbb{Z}[G]$-modules so it makes sense to apply <code>$\chi^{i}$</code> to them.</p> <p>Now, if you do the above systemically, you will notice that there are non-obvious steps which tend to involve Tamagawa numbers, hence my first word of warning. Also, linking the second displayed equation with Sha is not so easy. A good reason, in my opinion, to stop worrying about Sha per se and to study directly <code>$${R\Gamma}_{et}(\operatorname{Spec}(O_{K}[1/p]),T_{p}E)$$</code> (the complex computing étale cohomology of $T_pE$)</p> http://mathoverflow.net/questions/19760/decomposition-of-tate-shafarevich-groups-in-field-extensions/19820#19820 Answer by Chris Wuthrich for Decomposition of Tate-Shafarevich groups in field extensions Chris Wuthrich 2010-03-30T12:25:00Z 2010-03-30T12:25:00Z <p>There is the Stickelberg element $\Theta$ considered by Mazur and Tate which gives more information in this direction. It is conjectured to be in the Fitting ideal and hence in the annihilator of the Selmer group. </p> <p>To simplify the notation suppose $K/\mathbb{Q}$ is of odd prime degree $d$. Then evaluating a non-trivial character $\chi$ on $\Theta$ gives $$\overline{\chi}(\Theta) = \frac{G(\overline{\chi})\cdot L(E,\chi,1)}{\Omega} \qquad\text{ in }\qquad\mathbb{Q}[\chi],$$ I believe. Here $G(\chi)$ is the Gauss sum and $\Omega$ is the Néron period. The element $\overline{\chi}(\Theta) \in \mathbb{Q}[\chi]=\mathbb{Q}[\zeta_d]$ is of norm $$\frac{\sqrt{\Delta_K}\cdot L(E/K,1)}{\Omega^d}\cdot \frac{\Omega}{L(E/\mathbb{Q},1)},$$ which has a conjectural BSD-expression.</p> <p>Suppose that the Tamagawa numbers of $E/K$ are trivial and that $E(K)$ is the trivial group. Then $\Theta\in\mathbb{Q}\bigl[\textrm{Gal}(K/\mathbb{Q}]\bigr]$ annihilates the Tate-Shafarevich group of $E/K$. In other words one should look at the prime decomposition of the fractional ideal generated by $\overline{\chi}(\Theta)$ of $\mathbb{Z}[\zeta_d]$ to extract information about the size of the $\chi$-parts of Sha. So that will link the number $L(E,\chi,1)$ to Sha, but only up to units in $\mathbb{Z}[\zeta_d]$. It seems difficult to pin down the correct unit.</p> <p>The presence of Tamagawa numbers and torsion points in $E(K)$ will make this less precise and maybe there is no easy description.</p> http://mathoverflow.net/questions/19760/decomposition-of-tate-shafarevich-groups-in-field-extensions/19821#19821 Answer by Robert Pollack for Decomposition of Tate-Shafarevich groups in field extensions Robert Pollack 2010-03-30T12:27:06Z 2010-03-30T12:27:06Z <p>Fix a prime p which doesn't divide the degree of K over ${\mathbb Q}$, and let ${\mathcal O}$ denote the ring of integers of ${\mathbb Q}_p(\chi)$ i.e. an extension of ${\mathbb Q}_p$ containing the values of $\chi$. Then the group algebra ${\mathcal O}[G]$ decomposes into a direct sum of 1-dimensional pieces over ${\mathcal O}$, one for each power of $\chi$. </p> <p>Then $Sha(E/K)[p^\infty] \otimes {\mathcal O}$ being an ${\mathcal O}[G]$-module inherits such a decomposition. Concretely, the $\chi^i$-component of $Sha(E/K)[p^\infty] \otimes {\mathcal O}$ is the subset where $G$ acts by $\chi^i$. </p> <p>This $\chi^i$-component is then a reasonable candidate to compare to the $p$-adic valuation of the algebraic part of $L(E,\chi^i,1)$. </p> <p>Some further comments:</p> <p>-Note that extending scalars to ${\mathcal O}$ increases the size of the modules so this has to be taken into account.</p> <p>-The component corresponding to the trivial character is the invariants under $G$, and when $G$ has size prime-to-p this is simply $Sha(E/{\mathbb Q})[p^\infty] \otimes {\mathcal O}$ (which is good).</p> <p>-To make a precise relationship between the $L$-value and Sha, you need to take into account the other terms in BSD. Namely:</p> <p>*The torsion-term should work out exactly as above (decomposing into $\chi$-components). </p> <p>*The periods have to be considered (which was ignored above in my vague phrase "the algebraic part of"). </p> <p>*The Tamagawa numbers give me pause -- possibly there is an analogous $\chi$-decomposition, but I don't see it now. </p> <p>*Lastly, if K is ramified over ${\mathbb Q}$ then the discriminant of K appears in the BSD quotient (in the denominator which increases the size of Sha). To handle this, I imagine what should be done is that rather then considering the L-value alone, consider the L-value times the Gauss sum of the character. (By the conductor-discriminant formula this should give exactly the extra powers of p needed.)</p>