Vanishing of higher direct image of finite morphisms relative to the fppf topology

Question

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?

Answer 1

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.

Answer 2

Here is another coutnerexample, this time in characteristic $0$.

Let us first note that it is enough to exhibit sheaves $F$, $G$ on $X_\mathrm{fppf}$ and a surjective morphism $\psi:F\to G$ such that the push-forward $f_*\psi: f_*F\to f_*G$ in $\mathbf{Sh}(Y_\mathrm{fppf})$ is not an epimorphism --- in this case, $R^1f_*(\ker \psi)\neq 0$.

In fact, it is enough to exhibit an epimorphism of set-valued sheaves $\psi:F\to G$ on $X_\mathrm{fppf}$ whose push-forward to $\mathbf{Sh}(Y_\mathrm{fppf})$ is not an epimorphism. Indeed, given such a morphism, let $F'$ denote the sheafification of the presheaf $ U\mapsto \mathbb{Z}[F(U)]$, where the right hand side denotes the free abelian group spanned by $F(U)$. Defining $G'$ similarly, there is an induced morphism of abelian sheaves $\psi':F'\to G'$. One readily checks that $\psi'$ is an epimorphism, but $f_*\psi'$ is not.

To construct $\psi:F\to G$ as in the previous paragraph, fix a field $k$, let $X=\mathrm{Spec}\, k[\epsilon\,|\,\epsilon^2=0]$, $ Y=\mathrm{Spec}\,k$, $F=G=\mathcal{O}_X$ (viewed as a set valued sheaf) and define $\psi:\mathcal{O}_X\to \mathcal{O}_X$ by $$ \psi (x)=x^2$$ on sections. It is easy to check that $\psi$ is an epimorphism. However, $f_*\psi :f_*\mathcal{O}_X\to f_*\mathcal{O}_X$ is not an epimorphism.

To see this, it is enough to show that $\epsilon\in k[\epsilon\,|\,\epsilon^2=0]=\Gamma(Y,f_*\mathcal O_X)$ is not a square in $(f_*\mathcal O_X)(Y')$ for any $\mathrm{fppf}$-covering $Y'\to Y$. Writing $Y'=\mathrm{Spec}\,A$, this amounts to showing that $\epsilon$ is not a square in $A[\epsilon\,|\,\epsilon^2=0]$ when $A$ is nonzero, which is routine to check.