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?


2 Answers 2


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.

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    $\begingroup$ But the diagonal is affine in your example? $\endgroup$ Nov 8, 2018 at 13:49
  • $\begingroup$ Thank you very much for the answer! Just in case, here is a link to the mentioned paper. arxiv.org/pdf/1503.02891.pdf (top of page 540 in the journal formatting corresponds to the middle of page 3 in this text) $\endgroup$
    – SashaP
    Nov 11, 2018 at 0:53
  • $\begingroup$ I do not believe this example. The Hensel condition means that the open sets containing the closed point can be taken to be Zariski, but what about an etale map only hitting the generic point? Surely $\check{H}^2(X;\mathbb Z/n)$ is correct for the the $n$-fold cover of the generic point, just as the second cohomology of $\mathbb P^1$ can be detected using $\mathbb P^1-0$, $\mathbb P^1-\infty$, and the $n$-fold cover of $\mathbb G_m$. $\endgroup$ Nov 12, 2018 at 22:59
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    $\begingroup$ @BenWieland: if in the cover one component only maps to the generic point, just throw it out. The other opens will also contain the generic point, and refinements do not need to hit each component. $\endgroup$ Nov 12, 2018 at 23:02
  • $\begingroup$ @R.vanDobbendeBruyn thanks! It might have been more useful to reject my claim about $\mathbb P^1$. $\endgroup$ Nov 13, 2018 at 19:17

Čech cohomology fails for the plane with the doubled origin, $\mathbb A^2_{00}=\mathbb A^2\cup_{\mathbb A^2-0}\mathbb A^2$, with cohomology in the structure sheaf: $\check{H}^2(\mathbb A^2_{00};\mathcal O)=0\ne H^2(\mathbb A^2_{00};\mathcal O)$. Čech cohomology fails for both the Zariski and étale cohomologies.

Mayer-Vietoris shows that the punctured plane $\mathbb A^2-0$ has nontrivial cohomology in degree 1. Similarly, if we take any open subset of $\mathbb A^2$ that contains the origin, the same group appears in its first cohomology. Mayer-Vietoris pushes that into degree 2: $H^2(\mathbb A^2_{00};\mathcal O)=H^1(\mathbb A^2-0;\mathcal O)$. (In other words, if we push forward from the punctured plane to the affine plane $j\colon \mathbb A^2-0\hookrightarrow A^2$, then $R^1j_*\mathcal O$ is nontrivial and concentrated on the origin.)

The Čech cohomology for a particular cover amounts to pretending that the open sets of that cover and all of their $n$-fold intersections are acyclic for the sheaf. For a manifold and the constant sheaf, there do exist such “good covers” (eg, convex balls on a Riemannian manifold) and they are cofinal. For a separated scheme, affines are acyclic, so if intersections of affines are affine, then a single cover by affines computes quasi-coherent cohomology. This is why separated often appears as a hypothesis in cohomology (although it could be reduced to “semi-separated,” ie, affine diagonal).

If $U\to X$ is a cover and $U^n$ is the $n$-fold fiber product over $X$, then the Čech complex is: $$F(U)\to F(U^2)\to F(U^3)\to\cdots$$ This defines $\check{H}^*(U\to X;F)$. For any cover, the derived version of this computes the full cohomology: $$RF(U)\to RF(U^2)\to RF(U^3)\to\cdots$$ This gives the Čech-to-cohomology spectral sequence: $E_1^{pq}=H^q(U^p;F)\Rightarrow H^{p+q}(X;F)$ and $E_2^{pq}=\check{H}^p(U\to X;R^qF)$). The edge map is the comparison from Čech cohomology to real cohomology and the groups off of the edge are obstructions to Čech cohomology working. Of course, we are not interested in a single cover, but in the limit over all covers.

For any cohomology class there exists a cover such that the restriction to the cover kills the class. Refining by that cover moves that class off of $E_1^{p,0}=H^p(U;F)$ to at least $E_1^{(p-1),1}$. If $p=1$ then it has moved into Čech cohomology. That is why first Čech is always correct in the limit over all covers. But there is no guarantee that a cycle can be moved all the way to the Čech edge $E^{0,p}$.

Indeed, this cannot be done in the example from the beginning: the second cohomology of the affine plane with the doubled origin with coefficients in the structure sheaf. Switching back to the language of ordinary spaces, our cover must have two open sets, each containing one of the two origins; their intersection is punctured in codimension 2 and thus has cohomology. Moreover, it is essentially the same cohomology, independent of the choice of cover. It cannot be killed by refining the cover, only by leaving the cover the same and introducing an auxiliary cover of the intersection, ie, Verdier’s theory of hypercovers, a small change to Čech cohomology that always works. I have identified a cycle in $E_1^{1,1}$ that I claim obstructs Čech cohomology from being correct in this example. I have shown that it survives the limit of all covers. I have not shown that it survives to $E_2$, let alone $E_\infty$. For proofs of such pathologies, see the literature on hypercovers.


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