Let $f:X \to Y$ be a finite morphism of schemes.
Let $\mathcal{F}$ be a sheaf of abelian groups on the the etale site of $X$ then we know that $R^{i}f\_{\*} \mathcal{F} = 0$. Is this statement also true when $\mathcal{F}$ is sheaf of abelian groups on the fppf site?



Vanishing of higher direct image of finite morphisms relative to the fppf topology
No, it is not true. Let $k$ be an algebraically closed field of characteristic $p > 0$ and set $k' := k[x]/(x^2)$. Let $f \colon \mathrm{Spec}(k') \rightarrow \mathrm{Spec}(k)$ be the corresponding map. Then the sequence $0 \rightarrow \mu\_p \rightarrow \mathbb{G}\_m \rightarrow \mathbb{G}\_m \rightarrow 0$ will show you that $(R^1f\_\*(\mu\_p))(k) \cong k'^{\times}/k'^{\times p} \neq 0$. 


As far as I know, it is an open question whether such vanishing is true when $f$ is a closed immersion.


