Let $K$ be a finite extension of $\mathbb{Q}_p$, and let $C=\widehat{\overline{K}}$ be the completion of the algebraic closure of $K$. Let $\mathscr{O}_C$ be the ring of integers in $C$, and let $G_K$ be the absolute Galois group of $K$. I would like to know the following:

What is the annihilator of $H^1_{\mathscr{O}_{C}-mod}(G_K, \mathscr{O}_C(i))$, for $i\in \mathbb{Z}, i\not=0$?

Here $\mathscr{O}_C$-mod is the category of continuous $\mathscr{O}_C$-semilinear representations of $G_K$.

The (proofs of) Tate-Sen theory tell us that this is a torsion module annihilated by some element $\mathscr{O}_K$. I've convinced myself that in principle one could extract such an element from the proofs, though not that the existing proofs would give the best possible result. That said, I'm hoping that someone has already worked this out. In particular, I would like to know:

Is the answer (for fixed $i$) independent of $K?$


I think it's not too hard to reduce this to computing $$H^1_{\mathscr{O}_{K_\infty}-mod}(\Gamma, \mathscr{O}_{\widehat{K_\infty}}(i))$$ where $K_\infty$ is a cyclotomic $\mathbb{Z}_p$-extension of $K$ with $\mathbb{Z}_p\simeq \Gamma:=\text{Gal}(K_\infty/K)$. In other words, if $\sigma$ is a topological generator of $\Gamma$, this boils down to computing the annihilator of the cokernel of $$\sigma-\text{Id}: \mathscr{O}_{\widehat{K_\infty}}(i)\to \mathscr{O}_{\widehat{K_\infty}}(i),$$ if I'm not mistaken.

The annihilator of this cokernel is not quite the same as the annihilator of $H^1_{\mathscr{O}_{C}-mod}(G_K, \mathscr{O}_C(i))$, but if I've worked things out correctly, they differ by an amount which is independent of $K$.


I think the answer is "no", and that the minimal power of $p$ annihilating this module can be unbounded for fixed $i$ and varying $K$.

More precisely, fix a choice of a Galois-equivariant continuous $\mathcal{O}_K$-linear surjection $\widehat{\mathcal{O}_{K_\infty}}\to \mathcal{O}_K$ (e.g. by playing with Tate's normalized traces). This induces a surjection of $\mathcal{O}_K$-modules $H^1(\Gamma,\widehat{\mathcal{O}_{K_\infty}}(i)) \to H^1(\Gamma,\mathcal{O}_K(i))$. Now the point is that $H^1(\Gamma,\mathcal{O}_K(i))$ is easy to compute in most cases (probably all cases but I am too lazy right now). In particular, let $n=n_K$ be the largest integer such that $K$ contains a primitive $p^n$th root of unity, and assume that $n>0$. Then $H^1(\Gamma,\mathcal{O}_K(i))$ is isomorphic to $\mathcal{O}_K/((1+p^n)^i-1)$; this is an easy computation, using the fact that the cyclotomic character sends any generator of $\Gamma$ to something of the form $1+p^nu$ for some $p$-adic unit $u$. Since the $p$-adic valuation of $(1+p^n)^i-1$ is clearly unbounded for any fixed $i$ and varying $n$, this proves my claim.

  • 2
    $\begingroup$ BTW, $H^1(\Gamma, \widehat{\mathcal{O}_{K_{\infty}}}(i))$ and $H^1(G_K, \widehat{\mathcal{O}_{C}}(i))$ are uniformly comparable: there is an injective map from the first guy to the second, and the cokernel is killed by any element in the maximal ideal of $\mathcal{O}_K$, so in particular by $p$. $\endgroup$ Nov 16 '18 at 22:21
  • $\begingroup$ OK, great, I knew I was missing something simple. This seems to work--thanks! $\endgroup$ Nov 17 '18 at 2:18
  • $\begingroup$ Hmm; I think the claim in your comment is not quite right if K is very ramified...but the cokernel is indeed killed by p. $\endgroup$ Nov 19 '18 at 18:58
  • 1
    $\begingroup$ By inflation-restriction, the cokernel injects into $H^1(G_{K_{\infty}},\mathcal{O}_C)$, and the latter is almost zero as an $\mathcal{O}_{K_{\infty}}$-module. $\endgroup$ Nov 30 '18 at 17:35
  • $\begingroup$ Are you sure? I was only able to prove it was killed by some explicit element... $\endgroup$ Dec 1 '18 at 1:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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