Torsors of pushforward group schemes 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:


*

*Am I interpreting $H^1(R,\pi_\ast F)$ and $H^1(S,F)$ correctly?

*Is Waterhouse correct that these are equal, or are they in fact different?

 A: 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.
