Suppose $T$ is a free finite rank $\mathbb{Z}_p$-module with a continuous action of $\operatorname{Gal}(\overline{K} / K)$, where $K$ is a number field. There is a definition of local Tamagawa numbers $\operatorname{Tam}(T / K_v)$ for each prime $v$ of $K$, going back to Fontaine and Perrin-Riou (or to Bloch and Kato for $K = \mathbb{Q}$). For $v \nmid p$ this is the order of the torsion subgroup of $H^1(I_v, T)^{D_v}$, where $D_v$ and $I_v$ are the decomposition group and intertia group at $v$; for $v \mid p$ it is something more complicated using the Bloch-Kato exponential map. (I'm led to believe that if $T$ is the Tate module of an abelian variety, this recovers the usual description in terms of Neron models.)

If $K_{v, n} = K_v(\mu_{p^n})$, is it true that the factors $\operatorname{Tam}(T / K_{v, n})$ are eventually constant for large enough $n$?

(EDIT: In the light of Rob's comment, maybe I should add the assumption that my Galois representation is crystalline at $p$. I'm chiefly interested in the case of the p-adic representation of a modular forms of level prime to $p$ and non-ordinary at $p$.)

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    $\begingroup$ I think the p-part of Tamagawa numbers at p can blow up as you go up the cyclotomic Z_p-extension (all the same p here). Consider an elliptic curve E/Q with split multiplicative reduction at p and with Tate uniformizer q_E satifying ord_p(q_E)=p. This should be the same as saying that the p-part of Tam_p(E/Q) equals p. However, to compute the p-part of Tam_p(E/Q(mu_{p^n})) one needs to be computing ord_pi(q_E) where pi is some uniformizer of Q_p(mu_{p^n}). As n grows, so does ord_pi(q_E). $\endgroup$ – sibilant May 19 '11 at 19:34
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    $\begingroup$ After your edit, I think you are on better ground, but I am still unsure. By Sen and the crystalline property, you have eliminated one of the potential source of Tamagawa number explosion (i.e Frobenius of weight 0), but I worry about the possibility that your representation might have weights -2 (in which case there might be Tamagawa explosion in the dual, and this should cause Tamagawa explosion in the original representation). Anyways, I would tend to think it much easier to prove the result in the specific context you have in mind than trying to formulate a general theorem. $\endgroup$ – Olivier May 19 '11 at 20:19

As stated, the answer to your question is certainly no.

For instance, an elliptic curve $E/\mathbb Q$ with split multiplicative ordinary reduction at $p$ will have unbounded Tamagawa number at $p$ in the cyclotomic extension of $\mathbb Q$. To see this, you can check that the Tamagawa number is the order of $H^{2}({\mathbb Q_{\infty,p}},(T_{p}E)_{p}^{+})$. This is what B.Mazur calls the phenomenon of anomalous primes in his Inventiones 18 paper, which is the algebraic counterpart of the existence of exceptional zeroes for $p$-adic $L$-functions.

However, perhaps you meant for the Tamagawa factors outside $p$ to be eventually constant? In that case, I think the answer is yes.

First, $\operatorname {Gal}(K(\mu_{p^{\infty}})/K)$ has finite prime-to-$p$ part so there exists a large enough $n$ so that $\operatorname {Gal}(K(\mu_{p^{\infty}})/K(\mu_{p^n}))$ is unramified outside $p$ (EDIT: Of course this first step is unnecessary). Let $v\nmid p$ be a place of $K(\mu_{p^{n}})$ and let $w|v$ be a place of $K(\mu_{p^{n'}})$ with $n'≥n$. Then $H^{1}(I_{v},T)\simeq H^{1}(I_{w},T)$ so the Tamagawa number is constant.

  • $\begingroup$ Surely $K(\mu_{p^\infty})/K$ is unramified outside $p$ in all cases? $\endgroup$ – David Loeffler May 19 '11 at 19:58
  • $\begingroup$ Sure. For obscure reasons, I wrote the answer for a general Zp[G]-extension, where G is a finite group. Sorry about that. $\endgroup$ – Olivier May 19 '11 at 20:20
  • $\begingroup$ I wondered about this myself but I couldn't see why the torsion subgroup of $H^1(I_v, T)$ need be finite. It's since been pointed out to me (off-line) that this is in Appendix B of Rubin's Euler systems book. $\endgroup$ – David Loeffler May 19 '11 at 20:22
  • $\begingroup$ @Olivier: I thought anomalous primes were primes p for which a_p was congruent to 1 modulo p. I don't see the connection here... $\endgroup$ – sibilant May 20 '11 at 5:13
  • $\begingroup$ Dear Rob, If a_p is congruent to 1, then one of the eigenvalues is actually equal to 1, isn't it? So E is Steinberg at p and there is Tamagawa number explosion. $\endgroup$ – Olivier May 20 '11 at 6:08

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