The second Tate-Shafarevich group of a permutation module is trivial

Question

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) ?

Answer 1

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 $G$.

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.