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 $F^1(K, M)$?
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2 Answers
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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.
answered May 20, 2016 at 9:25
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1$\begingroup$ This is clearly explained in the first paragraph of the proof of Lemma 1.1 of Sansuc's paper. Read it again! Good luck! $\endgroup$ Commented Jun 11 at 19:02
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1$\begingroup$ Your formula $$ Ш^1(K/k,B)=\ker\Big[{\rm Hom}_{\mathfrak g}\big({\mathfrak g},B(K)\big)\to \prod_{v\notin\Sigma} {\rm Hom}_{{\mathfrak g}_v}\big({\mathfrak g}_v,B(K_v)\big)\Big]$$ is erroneous. The right formula is $$ Ш^1(K/k,B)=\ker\Big[{\rm Hom}({\mathfrak g},B(K))\to \prod_{v\notin\Sigma} {\rm Hom}({\mathfrak g}_v,B(K_v))\Big].$$ $\endgroup$ Commented Jun 12 at 10:40
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1$\begingroup$ We write $B$ for $B(K)=B(K_v)$. We are interested in $$ \ker\Big[{\rm Hom}({\mathfrak g},B)\to \prod_{v\notin\Sigma} {\rm Hom}({\mathfrak g}_v,B)\Big].$$ $\endgroup$ Commented Jun 12 at 10:40
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1$\begingroup$ Let a homomorphism $\varphi\colon {\mathfrak g}\to B$ be an element of this kernel. Then the restriction of $\varphi$ to any decomposition group ${\mathfrak g}_v$ for $v\notin\Sigma$ is zero. By the Chebotarev density theorem, any cyclic subgroup $C\subseteq {\mathfrak g}$ appears as ${\mathfrak g}_v$ for some $v\notin\Sigma$. $\endgroup$ Commented Jun 12 at 10:41
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1$\begingroup$ It follows that the restriction of $\varphi$ to any cyclic subgroup $C\subseteq {\mathfrak g}$ is zero. It follows that $\varphi(g)=0$ for all $g\in{\mathfrak g}$. Thus $\varphi=0$. $\endgroup$ Commented Jun 12 at 10:42
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For a refined statement, see also https://www.mathi.uni-heidelberg.de/~schmidt/NSW2e/index-de.html (9.1.9) (i).