Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $L(1,E \times \chi^i)$?

share|improve this question
add comment

3 Answers 3

up vote 7 down vote accepted

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.)

share|improve this answer
    
Actually, the Tamagawa factors are probably not such a big nuisance. For instance, if E is semistable, then the Tamagawa number for E/Q comes from primes with split multiplicative reduction. The Tamagawa number over K should then just be Tam(E/Q) together the contribution of non-split multiplicative primes over Q that become split over K. One then just has to determine which power of $\chi$ corresponds to adjoining the local quadratic unramified extension at each non-split multiplicative prime. –  Robert Pollack Mar 30 '10 at 13:57
add comment

First of all, I am not sure I fully agree with the notion that Tamagawa numbers are harmless factors.

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: $$|H^2(\operatorname{Spec}(O_{K}[1/p]),T_{p}E)|=|H^1(\operatorname{Spec}(O_{K}[1/p]),T_{p}E)/z_{Kato}|$$

In order to get the precise form you want, you will need a specific computation of $\exp^{*}(z_{Kato})$ (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 $$\chi^{i}\ker\left(H^{1}\left(\operatorname{Spec}(\mathbb{Z}[1/p]),T\right)\rightarrow H^1(\mathbb{Q}_{p},T)/H^1_{f}(\mathbb{Q}_{p},T)\right)$$ 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 $\chi^{i}$ to them.

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 $${R\Gamma}_{et}(\operatorname{Spec}(O_{K}[1/p]),T_{p}E)$$ (the complex computing étale cohomology of $T_pE$)

share|improve this answer
    
Could you briefly give an idea of how the last object is related to sha? –  Franz Lemmermeyer Mar 30 '10 at 12:37
    
Its second cohomology group is very close to Sha, but there are differences linked to units in the ring of integers, Tamagawa numbers and (I think) the index of Z[G] in its normalisation. This is also discussed in C.Wuthrich and R.Pollack answers. –  Olivier Mar 30 '10 at 13:04
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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