# Tamagawa numbers of abelian varieties and torsion.

Let $A$ be an abelian variety defined over a number field $K$. Fix a prime $v \subset \mathcal{O}_K$, with underlying rational prime $p$. What relationship, known or conjectural (if any), should there be between the local Tamagawa number $c_v(A) = [A(K_v): A_0(K_v)]$ and the cardinality of the $p$-primary subgroup $A(K_v)(p)$ of $A(K_v)$?

I am aware of the recent work of Lorenzini, which considers possible cancellations of the ratio \begin{align*} \frac{\prod_{v \subset \mathcal{O}_K} c_v(A)}{\vert A(K)_{\operatorname{tors}}\vert}, \end{align*} and shows for instance this ratio for $A$ an elliptic curve defined over $K ={\bf{Q}}$ is always greater than or equal to $\frac{1}{5}$.

The question comes up naturally in a certain Euler characteristic computations (e.g. for the $p^{\infty}$-Selmer group of $A$ defined over some profinite Galois extension $K_{\infty}$ of $K$, where certain primes of $K$ are known to split completely, and hence where the torsion subgroup $A(K_{\infty})_{\operatorname{tors}}$ is known to be finite). In particular, granted the refined conjecture of Birch and Swinnerton-Dyer, it is apparent from these computations that $c_v(A)$ for a prime of bad (multiplicative) reduction $v$ is given essentially by the quotient \begin{align*}\frac{\vert H^1(G_w, A(K_{\infty, w}))(p)\vert }{\vert H^2(G_w, A(K_{\infty, w}))(p)\vert }. \end{align*} Here, $K_{\infty, w}$ is the union of all completions at primes above $v$ in $K_{\infty}$, and $G_w$ denotes the Galois group $\operatorname{Gal}(K_{\infty, w}/K_v)$. If $A$ has good ordinary reduction at all primes above $p$ in $K$, then it is possible to use the Coates-Greenberg theory of deeply ramified extensions to characterize the local factor at $v$ in the Euler characteristic formula coming from this quotient as $\vert \widetilde{A}(\kappa_v)(p)\vert^2$, where $\widetilde{A}$ is the reduction of $A$ mod $v$, and $\kappa_v$ the residue field at $v$ (which is consistent with B+S-D). But, this characterization seems to break down when $A$ does not have good ordinary reduction at $v$, and I have not found any calculations in the literature for this case of bad reduction. Hence why I ask about any possible known relation to torsion. Sorry if the question is silly!

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There is no general relation between the local $p$-primary torsion and the Tamagawa numbers. I believe one can have $p$-torsion points that map to non-trivial or to the trivial element in the group of components of the Neron model.
This should indicate you that, in your Euler characteristic formula, you can not hope to replace the square of the local $p$-primary part with just the Tamagawa numbers when the reduction is not good ordinary.
If I understand correctly what you want to do is to compute the cokernel of the universal norms on the local points at a place above $p$. I would start by looking at papers about the case of a $\mathbb{Z}_p$-extension. If the reduction is multiplicative, then the computations were carried out by John W. Jones in Compositio 73 (1990). He has also a paper about the additive case. If your reduction is good but not ordinary, then it is a completely different story since the cokernel is not finite. I am sure there are many other references
Then you should look at papers of Kabayashi and Iovita-Pollack. Also Perrin-Riou has computed the Euler characteristic of the dual of Selmer group over a cyclotomic $\mathbb{Z}_p$-extension in her Asterisque book. But you will find "Arithmétiques des courbes elliptiques à réduction supersingulière en p" a better place to start. ps: Are you jvo ? –  Chris Wuthrich Mar 14 '11 at 9:17
Hi again, and thanks for these too! I have already done and written the full calculation assuming $A$ has good ordinary reduction at all primes above $p$ in $K$ for this particular $K_{\infty}$ (which does not contain the cyclotomic ${\bf{Z}}_p$-extension of $K$). I will definitely have a look at the papers of Kobayashi (?), Iovita-Pollack and Perrin-Riou now. I usually avoid the supersingular setting for simplicity ... and yes, I am jvo :) –  jvo Mar 14 '11 at 9:26