TITLE: Decomposition of Tate-Shafarevich groups in field extensions
QUESTION [12 upvotes]: 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:
What is the precise (conjectural) relation between the individual special values $L(1,E\times\chi^i)$ and Sha$(E/K)$?
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 $L(1,E \times \chi^i)$?

REPLY [8 votes]: 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 $\chi$. 
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 $\chi^i$.  
This $\chi^i$-component is then a reasonable candidate to compare to the $p$-adic valuation of the algebraic part of $L(E,\chi^i,1)$.  
Some further comments:
-Note that extending scalars to ${\mathcal O}$ increases the size of the modules so this has to be taken into account.
-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).
-To make a precise relationship between the $L$-value and Sha, you need to take into account the other terms in BSD.  Namely:
*The torsion-term should work out exactly as above (decomposing into $\chi$-components).  
*The periods have to be considered (which was ignored above in my vague phrase "the algebraic part of").  
*The Tamagawa numbers give me pause -- possibly there is an analogous $\chi$-decomposition, but I don't see it now.  
*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.)

REPLY [6 votes]: 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. 
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.
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.
The presence of Tamagawa numbers and torsion points in $E(K)$ will make this less precise and maybe there is no easy description.