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I'm reading William Waterhouse's "Discriminants of etale algebras and related structures", and he makes a basic claim I'm struggling to justify.

Suppose $S/R$ is etale of rank $n$... and let $\pi$ denote the map $\mathrm{Spec}(S)\to\mathrm{Spec}(R)$... Now on the other hand let $F$ be any affine group scheme over $S$; since $S$ is finite over $R$, we also have a direct image group scheme (or Weil restriction) $\pi_\ast F$ over $R$. As a functor, it is defined by $(\pi_\ast F)(U) = F(U\otimes_R S)$. Almost automatically we have then $H^1(S,F)=H^1(R,\pi_\ast F)$.

Here $H^1(S,F)$ is the set of isomorphism classes of $F$-torsors on $S$; i.e. sheaves of $F$-sets locally (over $S$) isomorphic to $F$ acting on itself. I see how to produce a map $H^1(R,\pi_\ast F)\to H^1(S,F)$, but it seems to me that the only $F$-torsors on $S$ you can obtain this way are those $F$-torsors that have a trivialization over some base change to $S$ of a covering of $R$.

My questions:

  1. Am I interpreting $H^1(R,\pi_\ast F)$ and $H^1(S,F)$ correctly?
  2. Is Waterhouse correct that these are equal, or are they in fact different?
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  • $\begingroup$ These matters are discussed in SGA3, Exposé XXIV, section 8. Proposition 8.4 there is the statement you want and is specific to the étale topology, so you may want to assume that $F$ is smooth. $\endgroup$ Commented Dec 4, 2017 at 20:38

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You don't say how you're interpreting $H^1(R,\pi_* F)$, but these are indeed naturally isomorphic. The short answer: if $\pi$ is a finite map, then $\pi_*$ is exact and so $R^1 \pi_* F$ vanishes; then the Leray spectral sequence shows that your map is indeed an isomorphism. The version of this in most books only works if $F$ is Abelian, but I think you can find what you want for the non-Abelian case in Giraud's Cohomologie non abelienne. (I am at home so don't have the reference right now.)

This can be seen as a generalisation of Shapiro's lemma in group cohomology. For example, take $S/R$ to be a quadratic extension of fields, and take $F=\mathbb{Z}/2\mathbb{Z}$. The sheaf $\pi_* F$ is the sheaf corresponding to the induced module $\mathrm{Ind}_{S/R} (\mathbb{Z}/2\mathbb{Z})$, which is two copies of $\mathbb{Z}/2\mathbb{Z}$ interchanged by the action of the Galois group of $S/R$. Shapiro's lemma says precisely that $H^1(S,F)$ and $H^1(R,\pi_*F)$ are isomorphic; they both classify quadratic extensions of $S$.

In many cases, such as the example above, $F$ is a sheaf which itself is the restriction of a sheaf on $R$. In that case you also have the map $H^1(R,F) \to H^1(R,\pi_* F)$ coming from the natural morphism $F \to \pi_* \pi^* F$. That map has no reason to be an isomorphism, which is maybe why you are confused.

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    $\begingroup$ > if $\pi$ is a finite map, then $\pi_*$ is exact You need to work in etale site to have this result. $\endgroup$
    – gdb
    Commented Dec 5, 2017 at 7:55

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