Say we have a number field $K$. Let $G\_K = \text{Gal}(\overline{K}/K)$. Let $M$ be a discrete $G\_K$-module. We know that $H^1(K, M) := H^1(G\_K, M)$, i.e. profinite group cohomology. For each place $v$ of $K$, let $K\_v$ be the completion of $K$ at $v$; restriction to a decomposition group $G\_v$ at $v$ defines a homomorphism $H^1(K, M) \to H^1(K\_v, M)$. Define$$F^1(K, M) := \text{Ker}\left(H^1(K, M) \to \prod\_v H^1(K\_v, M)\right).$$If the $G\_K$-action on $M$ is trivial, what can we say about $F^1(K, M)$?



Group cohomology question, trivial Galois action on discrete Galois module means we can say what about kernel of map
If the $G\_K$-action on $M$ is trivial, then
$$H^1(K,M)=\mathrm{Hom}(G\_K,M),$$
and by Chebotarev's density theorem
$$ F^1(K,M)=0.$$


For details see Lemma 1.1(i) of Sansuc, J.-J. Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres. J. Reine Angew. Math. 327 (1981), 12–80.


