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Let $X= spec(k)$ where $k$ is an algebraically closed field. Consider the constant sheaf $\mathbb{Z}$ on the fppf site of $X$. I'm interested in computing $H^1_{fppf}(X, \mathbb{Z})$. I know that $H^1_{et}(X, \mathbb{Z})$ is $0$. Is $H^1_{fppf}(X, \mathbb{Z})=0$ as well?

I think this is equivalent to showing $R^1f_{*} \mathbb{Z} = 0$ where $f$ is the canonical morphism of sites $f:X_{fppf} \to X_{et}$.

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    $\begingroup$ Since every $k$-scheme of finite type has a $k$-rational point, $Spec(k)$ has no nontrivial fppf covers, hence it has no cohomology. $\endgroup$
    – Marc Hoyois
    Commented Feb 6, 2019 at 4:37

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