# Cohomology of constant sheaves

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 $$f:X_{fppf} \to X_{et}$$.

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