It can be proven that if in a quasicompact scheme $X$ any finite subset is contained in an affine open subset then for any sheaf $\mathcal{F}$ on $X$ its Cech cohomology $\hat{H\_{et}^{\bullet}}(X,\mathcal{F})$ is naturally isomorphic to etale cohomology $H^{\bullet}\_{et}(X,\mathcal{F})$.


Is there a counterexaple for an arbitrary scheme? The assumptions of the theorem I cited fail, for instance for Hironaka's example of a threefold with two points not contained in any affine open subset but does it give a counterexample?



Etale cohomology can not be computed by Cech
Let $k$ be an algebraically closed field. Glue two copies of $\text{Spec}(k[[x]])$ along $\text{Spec}(k((x)))$. This gives a scheme $X = U \cup V$ such that any etale covering of $X$ can be refined by the Zariski covering $X = U \cup V$. (Hint: use that $k[[t]]$ is strictly henselian.) Thus the etale Cech cohomology is the usual Cech cohomology. For a constant sheaf with value an abelian group $A$ we get $\check{H}^i\_{et}(X, A) = 0$ for $i > 0$. On the other hand, if $A$ is finite of order invertible in $A$ then we have $H^2\_{et}(X, A) \cong A$ as you can see by using the Cech to cohomology spectral sequence and using the fact that $H^1\_{et}(U \cap V, A) \cong A$ as abelian groups (noncanonically).


Currently I am blanking on a separated example. If I remember, I will post another answer.


**Edit:** In the paper by Stefan Schoer, "Geometry on totally separably
closed schemes", the author conjectures that for separated and quasi-compact schemes etale Cech cohomology should agree with etale cohomology. (See remarks at the top of page 540.) He even conjectures it should suffice if the diagonal is affine. I have no idea how reasonable this conjecture is.


