Suppose I have a global field $K$ and a finite Galois extension $L/K$ of Galois group $G$. It is often written without proofs (it seems that this is a very common statement) that for every $G$-module $M$ of permutation (that is a free $\mathbf{Z}$-module that admits a basis permuted by the action of $G$), the Tate-Shafarevich group of degree $2$
$$ \mathrm{Sha}^2(G,M)=\mathrm{Ker}\left(H^2(G,M)\longrightarrow\prod\_{v\in\Omega\_K}H^2(G\_v,M)\right) $$
where $\Omega\_K$ is the set of places of $K$ and $G\_v=\mathrm{Gal}(L\_w/K\_v)$ for $w$ a place of $L$ that extends $v$, is trivial. Do you have any reference for this ? or a proof (if it's not too long) ?



The second Tate-Shafarevich group of a permutation module is trivial
We write $G\_w={\rm Gal}(L\_w/K\_v)$.



> 
> **Definition.** For $n\ge 1$, we denote
> $$Ш\_\omega^n(G,M)=\ker\Big(H^n(G,M)\to\prod\_C H^n(C,M)\Big)$$
> where $C$ runs over the cyclic subgroups of $G$.
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> 
> 



> 
> **Remark.** $Ш^2(L/K,M)\subseteq Ш\_\omega^2(G, M)$. Indeed,
> by the Chebotarev density theorem, for any cyclic subgroup $C\subseteq G$
> there exist $v$ and $w$ such that $G\_w=C$.
> 
> 
> 



> 
> **Definition.** A permutation $G$-module is a torsion free $G$-module admitting a $G$-invariant (as a subset) ${\mathbb Z}$-basis $B$.
> 
> 
> 



> 
> **Theorem.** If $M$ is a permutation $G$-module, then $Ш\_\omega^2(G,M)=0$, and therefore $Ш^2(L/K,M)=0$.
> 
> 
> 


*Sketch of proof.*


*Step 1.*
We have
$$ Ш\_\omega^2(G,M\_1\oplus M\_2)=Ш\_\omega^2(G,M\_1) \oplusШ\_\omega^2(G, M\_2),$$
which reduces the theorem to the case when $G$ acts on the basis $B$ transitively.
Then using Shapiro's lemma we reduce the theorem to the case $M={\mathbb Z}$ (with trivial $G$-action).


*Step 2.* We have a short exact sequence of $G$-modules (with trivial $G$-action)
$$0\to {\mathbb Z}\to{\mathbb Q}\to {\mathbb Q}/{\mathbb Z}\to 0,$$
which gives rise to cohomology exact sequences
\begin{align\*}
&0= H^1(G,{\mathbb Q})\to H^1(G,{\mathbb Q}/{\mathbb Z}) \to H^2(G,{\mathbb Z})\to H^2(G,{\mathbb Q})=0\\
&0= H^1(C,{\mathbb Q})\to H^1(C,{\mathbb Q}/{\mathbb Z}) \to H^2(C,{\mathbb Z})\to H^2(C,{\mathbb Q})=0.
\end{align\*}
Therefore, we may identify
$$Ш\_\omega^2(G,{\mathbb Z})=Ш\_\omega^1(G,{\mathbb Q}/{\mathbb Z}),$$
and it suffices to show that $Ш\_\omega^1(G,{\mathbb Q}/{\mathbb Z})=0$.
By definition,
$H^1(G,{\mathbb Q}/{\mathbb Z})={\rm Hom}(G,{\mathbb Q}/{\mathbb Z})$ and $H^1(C,{\mathbb Q}/{\mathbb Z})={\rm Hom}(C,{\mathbb Q}/{\mathbb Z})$.
It remain to show that
$$\ker\Big({\rm Hom}(G,{\mathbb Q}/{\mathbb Z})\to \prod\_C{\rm Hom}(C,{\mathbb Q}/{\mathbb Z})\Big)=0,$$
but this is obvious because $G$ is the union of its cyclic subgroups $C$.


*Reference:* Lemma 1.9 of J.-J. Sansuc, Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres, J. Reine Angew. Math. 327 (1981), 12–80.


