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Let $k$ be a number field. Let $M$ be a (continuous) $\text{Gal}(\overline{k}/k)$-module.

One can define two subgroups of the Galois cohomology group $H^i(k,M)$:

  • the group of elements of $H^i(k,M)$ mapping to zero in $H^i(\langle g \rangle, M)$ for all $g \in \text{Gal}(\overline{k}/k)$,

  • the group of elements of $H^i(k,M)$ mapping to zero in $H^i(k_v,M)$ for almost all places $v$ of $k$.

Is it true that these groups coincide? (The answer should be yes.) And how does one prove this?

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A bit of context perhaps? For instance, I'm guessing you wouldn't mind assuming that $M$ is a continuous $\operatorname{Gal}(\bar{k}/k)$-module but you don't really say. – Olivier Oct 18 '11 at 5:30
Yes, of course I assume continuity, thanks. – Wanderer Oct 18 '11 at 10:41
The context: various texts about arithmetic duality theorems... Also, this might have something to do with Chebotarev's theorem. – Wanderer Oct 18 '11 at 10:43

Some ideas: I it holds when $M$ is a trivial Galois module (and $i=1$, $M$ finite) . For, if $\phi \in H^1(G_k,M) = Hom(G_k,M)$ restricts in almost all decomposition groups to $0$, it is $0$ by Chebotarev (it even suffices for the densitiy of the set of primes to be greater than $1/p$, $p$ the smallest prime divisor of $|M|$. On the other hand, if it maps all generators (all Frobenii suffice) to $0$, it must be $0$. So the groups coincide in this case and are equal to $0$.

Now for a counterexample: Take $M = \mu_n$, $i=2$. Then $H^2(k,M) = Br(k)[n] \hookrightarrow \oplus_v Br(k_v)[n] = \oplus_v \frac{1}{n}\mathbf{Z}/\mathbf{Z}$. Take $x = (\frac{1}{n}, \frac{1}{n}, 0, 0, 0, \ldots)$. We have $0 \neq x_v = \frac{1}{n} \in Br(k_v)[n] = \frac{1}{n} \mathbf{Z}/\mathbf{Z}$ $= Br(k_v^{nr}/k_v)[n] = Hom(<Frob_v>, \frac{1}{n} \mathbf{Z}/\mathbf{Z})$. Can someone finish from here (perhaps might help.

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