Let $L/K$ be a finite Galois extension, write $G:= Gal(L/K)$. Denote by $R = Res(\mathbb{G}_m)$ the Weil restriction of $\mathbb{G}_m$, from $L$ to $K$. I want to show that its first Galois cohomology vanishes: $H^1(G, R(L)) = 0$.
Question: Is there a simple way to do this, without calculation (something formal; Any torsor must be trivial because…)?
I think that I checked that this is true, but I just calculate: The Galois module in question is $(L \otimes_K L)^{\times}$, with action on the first coordinate only. Using Galois theory, I rewrite it as $\prod_{\sigma\in G}L^{\times}$ with some action. Writing $M:=L^{\times}$, this is just $Hom_{\mathbb{Z}}(\mathbb{Z}[G] , M)$ (inner $Hom$ in $G$-modules). Now I check that such modules are acyclic: As a functor of $M$, this has an exact left adjoint, and hence sends injectives to injectives. Thus, $$R^1Hom_{\mathbb{Z}[G]} (\mathbb{Z} , Hom_{\mathbb{Z}}(\mathbb{Z}[G],M)$$ equals to the first derived functor of $Hom_{\mathbb{Z}[G]} (\mathbb{Z} , Hom_{\mathbb{Z}}(\mathbb{Z}[G],M))$, which equals to the identity functor.
Best Answer
There is a general argument that is slightly more elementary than what you wrote. By standard properties of Weil restrictions, we have $R(L) = \prod_{\sigma} \mathbb{G}_m(L)$, where the product is taken over the different $K$-linear embeddings of $L$ into some algebraic closure $\overline{K}$ and where $G$ acts on the factors in a natural way. In other words, we have that $R(L) = \mathbb{Z}[G] \otimes \mathbb{G}_m(L)$, as $G$-modules. By an easy exercise, we have that $$\mathbb{Z}[G] \otimes \mathbb{G}_m(L) = \operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L)),$$ giving $R(L)=\operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L))$. (Indeed, take the isomorphism $$ f:\mathbb{Z}[G] \otimes \mathbb{G}_m(L) \rightarrow \mathbb{Z}[G] \otimes (\mathbb{G}_m(L))_0 \stackrel{\operatorname{def}}{=} \operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L)), $$ where $\mathbb{G}_m(L)_0$ denotes the underlying abelian group of $\mathbb{G}_m(L)$, satisfying $f(g \otimes x) = g \otimes g^{-1}x$.)
In conclusion, $R(L)=\operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L))$. Finally, by Shapiro's lemma, the $\operatorname{H}^1$ (and in fact, all higher cohomology) of this vanishes.