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Suppose $K$ is a number field and $E$ is an elliptic curve defined over $K$. My question is: how do you compute the local cohomology group $H^1(K_v, \, E[p^{\infty}])$?

As to why I'm asking this, it comes up in Iwasawa theory for elliptic curves by Greenberg in page 74. In these notes, Greenberg is looking at the elliptic curve $E = 11a3$ defined over $\mathbf{Q}$. He claims that $$H^1(\mathbf{Q}_{11}, E[5^{\infty}]) \cong \mathbf{Z}/5\mathbf{Z}. $$ I'd like to understand how one would prove this result. Greenberg's notes say that this follows from the local duality theorems, but I'm not exactly sure how it follows from them. In terms of what I tried to figure this out, I tried to use the fact that $E[5^{\infty}]$ sits in an exact sequence of $G_{\mathbf{Q}_{11}}$-modules: $$0 \to \mu_{5^{\infty}} \to E[5^{\infty}] \to \mathbf{Q}_5/\mathbf{Z}_5 \to 0.$$ I tried applying the long-exact sequence in cohomology to this sequence, but I still couldn't show how to derive the result above. Does anyone know how to derive the fact above that $H^1(\mathbf{Q}_{11}, E[5^{\infty}]) \cong \mathbf{Z}/5\mathbf{Z}$? Is there some other fact that I need to complete the calculation?

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    $\begingroup$ I would guess that you need to use the mentioned class that the Tate period has valuation 1, which tells you something about the extension class of that exact sequence and should help in computing the differentials. $\endgroup$
    – Will Sawin
    Commented Dec 11, 2021 at 2:30

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The corollary 3.4 of chapter 1 Arithmetic Duality theorems by Milne says that: If $K$ has char 0 local field, there is a canonical pairing $$H^r(K,A^t)\times H^{1-r}(K,A)\to \mathbb{Q}/{\mathbb{Z}}$$ so because eliptic curves are self-dual for computing $H^1(K,E)$ it is enough to compute $E(K)$, It also implies $H^2(K,E)=0$. Now just look at the exact sequence $$0\to E(5^n)\to E\to E\to 0$$.

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  • $\begingroup$ I think your aproach could also work using some class field theory, and if you look at the proof in the milne book it is basically using simmilar ideas $\endgroup$
    – ali
    Commented Dec 11, 2021 at 13:37
  • $\begingroup$ Hmm, I'm not sure I follow. How would knowing $E(K_v)$ tell us what $H^1(K, E)$ is? I'm thinking one can try finding the image of $E(K_v)$ under the Kummer map, but that would only give us part of the full group $H^1(K_v, E)$, not the whole thing, right? $\endgroup$
    – Adithya Chakravarthy
    Commented Dec 11, 2021 at 14:41
  • $\begingroup$ The prefect pairing says that $H^1(K,E)$ is dual to $H^0(K,E)$ which is by definition the galios invariants of $E(K^{sep})$, which is $E(K)$ $\endgroup$
    – ali
    Commented Dec 11, 2021 at 16:39
  • $\begingroup$ Ah, I see now. Apologies, I got the exact sequence you wrote mixed up with Tate duality, which relates $H^i$ with $H^{2-i}$, whereas your exact sequence relates $H^i$ with $H^{1-i}$. This sequence is very helpful, thanks! $\endgroup$
    – Adithya Chakravarthy
    Commented Dec 11, 2021 at 18:09
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    $\begingroup$ tate duality is for multiplicative groups, this duality is for abelian varieties $\endgroup$
    – ali
    Commented Dec 12, 2021 at 7:22

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