# First Galois cohomology of Weil restriction of $\mathbb{G}_m$

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.

• I agree with René that the result is a simple application of Shapiro's lemma. What else would you like? Feb 3 '16 at 0:15
• Here's the summary (see also @nfdc23's answer): Weil restriction is pushforward. Finite morphisms have no higher pushforwards in the étale site. The result follows from the Leray spectral sequence and Hilbert 90. Mar 25 '17 at 19:01

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.

• I guess $R$ is a twist of $\mathbb G_m^{[L:K]}$ as opposed to $\mathbb G_m^2$. (I think I can tell which field extension you were thinking of...) Feb 3 '16 at 5:39
• @René: For the $K$-group $R$, the $L/K$-twists of $R$ are classified by ${\rm{H}}^1(G, {\rm{Aut}}(R_L))$, not by ${\rm{H}}^1(G, R(L))$; i.e., the coefficient group is the automorphism scheme of $R$ rather than $R$ itself. Also, twisted forms of $\mathbf{G}_m^n$ are $n$-dimensional tori over $K$ split by $L$, of which there are many non-split ones (no link to GL$_n(L)$). Rather, $R$ is the Aut-scheme of the underlying $K$-vector space of an $L$-line, so $H^1(G,R(L))=1$ since any such $L$-line has a basis. Feb 3 '16 at 6:42
• You're totally right. I edited the answer.
– RP_
Feb 3 '16 at 12:52
• Thank you for your answer! I think that it is basically what I outlined, and I just reproved Shapiro's lemma, or something similar. Feb 3 '16 at 18:16

One can do much better: it is not necessary to assume $L/K$ is Galois (merely separable is sufficient). And in fact one can formulate the result in a manner which works beyond that of fields, working over more general rings. Namely, let $f:S' \rightarrow S$ be a finite etale map of schemes, and let $G'$ be a commutative finitely presented and relatively affine $S'$-group scheme. Then the finitely presented and relatively affine $S$-group scheme $G = {\rm{R}}_{S'/S}(G')$ defines a sheaf for the etale topology on $S$, and one can contemplate the etale cohomology group ${\rm{H}}^i(S, G)$; when $S$ and $S'$ are spectra of fields and $G' = {\rm{GL}}_1$ and $i=1$ then this recovers the setup of the question with the Galois hypothesis relaxed to separability of the field extension.

So what? Well, I claim that naturally ${\rm{H}}^i(S, G) = {\rm{H}}^i(S', G')$. In fact, this is a special case of something more general, and which in greater generality is clearer to prove: if $\mathscr{F}'$ is any abelian etale sheaf on $S'$ then we get the pushforward $f_{\ast}(\mathscr{F}')$ as an abelian etale sheaf on $S$, and when $\mathscr{F}'$ is the functor of points on $G'$ on the category of etale $S'$-schemes then $f_{\ast}(\mathscr{F}')$ is the functor of points of $G$ on the category of etale $S$-schemes (since by definition Weil restriction is a "functorial pushforward"). Consequently, this is all a special case of the general claim that ${\rm{H}}^{\bullet}(S, f_{\ast}(\cdot)) \simeq {\rm{H}}^{\bullet}(S', \cdot)$ on the category of abelian etale sheaves on $S'$. But $f_{\ast}$ is exact between categories of abelian etale sheaves since $f$ is finite (i.e., higher direct images of $f$ for the etale topology vanish), so this is just the degenerate Leray spectral sequence.

[Of course, in the special case of a finite separable extension of fields this recovers the Shapiro Lemma identification by relating everything to edge maps in spectral sequences.]

It's overkill for the specific question posed, but since you asked for a way to do it "without calculation" (which I'll interpret to mean "by more conceptual means") it seemed worth mentioning. One virtue of this more general approach is that it illuminates the importance of considering the etale topology rather than the fppf topology (e.g., over fields perhaps one might want to consider coefficients in a non-smooth group schemes, and then fppf cohomology is more appropriate to consider than etale cohomology, though the two coincide in all degrees with coefficients in a smooth commutative relatively affine group scheme by section 11 of Grothendieck's Brauer III paper). The point is that the preceding argument breaks down for the fppf topology because pushforward of finite flat maps are not exact for abelian sheaves relative to the fppf topology, even for something as concrete as a non-separable finite extension of fields (and correspondingly degree-1 higher direct image sheaves on $\alpha_p$ and $\mu_p$ are generally nonzero when considering imperfect fields of characteristic $p$). So there is no version of Shapiro's Lemma for the fppf topology.

Addendum: Although there is no version of Shapiro's Lemma for the fppf topology, when the coefficients upstairs are a smooth relatively affine commutative group scheme then (by Brauer III, as noted above) fppf cohomology coincides with etale cohomology and so by exactness of pushforward for the etale topology relative to any finite morphism we do have a Shapiro Lemma with $f$ above merely finite locally free rather than just finite etale. In the special case of a finite (possibly non-separable) extension of fields and coefficients in a smooth commutative group scheme of finite type this recovers (by a different viewpoint of proof) the (unique) Corollary in section 2.3 of Chapter IV of Oesterle's beautiful paper "Nombres de Tamagawa et groupes unipotents en caracteristique $p$" in Inventiones Math 78 pp. 13--88 (1984) (which in sections 2.3 and 2.4 gives related useful results in the setting of local and global fields).

• It's important to note that Sasha asked for $\operatorname{Gal}(L/K)$-cohomology and not $\operatorname{Gal}(K)$ cohomology. In particular, this is the source of the assumption that $L/K$ be Galois! To check that your answer applies, we apply a spectral sequence to equation the $\operatorname{Gal}(K)$-cohomology of a group scheme to the $\operatorname{Gal}(L/K)$ cohomology of its $\operatorname{Gal}(L)$-cohomology, then observe that its $\operatorname{Gal}(L)$-cohomology is equal to its $L$-points in degree $0$ (obvious) and vanishes in higher degree (Hilbert 90). Feb 3 '16 at 3:32
• @WillSawin: Thanks for noting my oversight when reading the original question (and for noting why it ultimately addresses the question posed as a special case). Feb 3 '16 at 3:43