# Understanding etale cohomology versus ordinary sheaves

I am a physicist trying to understand etale cohomology from Shafaverich, and I would like to check a misunderstanding, undoubtedly.

When defining etale cohomology, it seems it is sheaf cohomology in the sense of right-derived functors, but with the etale site, as opposed to just concerning open subsets.

For concreteness, we fix an etale sheaf $$\mathcal F : U \mapsto \mathcal O_U(U)$$ where $$U$$ is a scheme which comes equipped with an etale morphism $$f:U\to X$$. We could then take an injective resolution, i.e.

$$0\to \mathcal F \to \mathcal I^0 \to \mathcal I^1 \to \mathcal I^2 \to \cdots$$

We can then take sections, i.e. apply $$\Gamma(X,-)$$:

$$\Gamma(X,\mathcal I^0) \to \Gamma(X,\mathcal I^1) \to \Gamma(X,\mathcal I^2) \to \cdots$$

Taking the cohomology then yields $$H^q(X,\mathcal F)$$. However, I do not see how this makes use of the "new" version of a sheaf, namely the etale sheaf.

We are applying the etale sheaves to $$X$$, which belongs to the site used in ordinary sheaf cohomology, so it seems like ordinary sheaf cohomology and etale cohomology should always agree? I don't see from the definition of etale cohomology, how we end up using anything extra, thanks to enlarging the site to etale maps.

• Being injective is not an intrinsic property of a module (say). A given module M, can be injective in a category A but not in a category B. So the whole "take an injective resolution" doesn't make sense unless a category is specified and once it is, one that will work for the Zariski site usually won't work for the etale site. Commented Apr 21, 2020 at 0:35
• @Ayoub Could you give me an example of a sheaf with a resolution, when the site is enlarged to the etale site, fails to be a resolution?
– JPhy
Commented Apr 21, 2020 at 11:49
• Because the open sets in the Zariski topology (and hence the Zariski site) are so huge, constant sheafs are flasque and therefor acyclic. So, any (finite) constant sheaf is an acyclic resolution of itself ! Therefore : $H^i(X,$ constant sheaf$)=0$ for $i>0$ for the Zariski topology. Of course, in the etale site, this won't be the case -- that's the whole point of moving to the etale site. Commented Apr 21, 2020 at 12:47
• @Ayoub can this be shown using the Cech complex, and if so what's the step wherein the results diverge for etale and ordinary constant sheaves?
– JPhy
Commented Apr 21, 2020 at 18:37

As mentioned in the comments, acyclic sheaves for the Zariski site are not the same as those which are acyclic sheaves for the étale site. (see the edit for the reason.)

Consider $$\mathbf{P}^1_{\mathbf{C}}$$ in the Zariski site and consider the constant sheaf $$\mathbf{Z}/n\mathbf{Z}$$. Since $$\mathbf{P}^1_{\mathbf{C}}$$ is irreducible this sheaf is flasque and hence acyclic for Zariski cohomology.

However the étale case is far more interesting. It is well known for cohomology associated to any site that for $$G$$ a sheaf valued in groups, $$H^1(X,G)$$ is the isomorphism classes of $$G$$-torsors (i.e. principal homogeneous spaces for $$G$$). This is true even if $$G$$ is non-abelian. It follows from analysis of cocycle conditions.

Now, if $$G$$ is finite, in the étale case this is equivalent to isomorphism classes of finite étale covers of $$Y\to X$$ with automorphism group $$G$$, in other words, étale $$G$$-torsors. This is not true in Zariski because the Zariski topology is far too coarse to have any non-trivial fibrations.

Thus $$H^1_{ét}(X,\mathbf{Z}/n\mathbf{Z})$$ classifies all finite étale covers of $$X$$ with automorphisms $$\mathbf{Z}/n\mathbf{Z}.$$

Using this one can compute $$H^i_{ét}(\mathbf{P}^1_{\mathbf{C}},\mathbf{Z}/n\mathbf{Z}).$$

Consider the étale cover $$\mathbf{A}^1_{\mathbf{C}}\coprod \mathbf{A}^1_{\mathbf{C}} \to \mathbf{P}^1_{\mathbf{C}}$$ and there is a Mayer-Vietoris sequence (it follows from generalities about spectral sequences, you can look it up in Milne's textbook)

$$\ldots\to H^{i}_{ét}(\mathbf{P}^1_{\mathbf{C}},\mathbf{Z}/n\mathbf{Z})\to H^{i}_{ét}(\mathbf{A}^1_{\mathbf{C}},\mathbf{Z}/n\mathbf{Z})\oplus H^{i}_{ét}(\mathbf{A}^1_{\mathbf{C}},\mathbf{Z}/n\mathbf{Z})\to H^{i}_{ét}(\mathbf{G}_{m,\mathbf{C}},\mathbf{Z}/n\mathbf{Z})\to H^{i+1}_{ét}(\mathbf{P}^1_{\mathbf{C}},\mathbf{Z}/n\mathbf{Z}) \to \ldots$$

Here $$\mathbf{G}_{m,\mathbf{C}}$$ is $$\operatorname{Spec} \mathbf{C}[t,t^{-1}].$$

By Riemann's existence theorem $$\mathbf{A}^1_{\mathbf{C}}$$ is simply connected. Note that this step exploits the connection between étale and analytic topologies, for this too refer to Milne's textbook.

Now by Riemann-Hurwitz $$\mathbf{P}^1_{\mathbf{C}}$$ has no non-trivial finite étale covers. This is an exercise in Hartshorne and in fact is something you should expect given that $$S^2$$ is itself simply-connected in the euclidean topology.

Thus the first cohomology groups vanish. The second cohomology groups of all affine schemes vanish as a general result.

We are left with computing $$H^{1}_{ét}(\mathbf{G}_{m,\mathbf{C}},\mathbf{Z}/n\mathbf{Z})$$. But this is the same as classifying finite étale covers of $$\mathbf{G}_{m,\mathbf{C}}$$. These correspond to degree $$n$$ maps $$\mathbf{G}_{m,\mathbf{C}}\to \mathbf{G}_{m,\mathbf{C}}$$ sending $$z\mapsto z^n.$$ It is an instructive exercise to see that there are exactly $$\mu_n(\mathbf{C})\cong \mathbf{Z}/n\mathbf{Z}$$ (the covers needn't connected!) isomorphism classes. Here $$\mu_n(\mathbf{C})$$ is the $$n$$-th roots of unity.

Thus we see that $$H^2_{ét}(\mathbf{P}^1_{\mathbf{C}},\mathbf{Z}/n\mathbf{Z})=\mathbf{Z}/n\mathbf{Z}.$$

(I learned this computation from de Jong's lectures on étale cohomology. )

As you can check this is the same as the singular cohomology of $$\mathbf{P}^1_{\mathbf{C}}$$ in the analytic topology. Note that $$H^0_{ét}(\mathbf{P}^1_{\mathbf{C}},\mathbf{Z}/n\mathbf{Z})=\mathbf{Z}/n\mathbf{Z}$$ since $$\mathbf{P}^1_{\mathbf{C}}$$ is connected.

If you move to a field in char $$p>0$$ then it is no longer true that the affine line is simply connected (see here). So this method will no longer work, but it hints at being the correct generalisation.

Edit: (A more geometric viewpoint)

I think the confusion of the OP lies in viewing the definition abstractly.

In general one can look at the category of sheaves on any site. This category is called a topos. One may define the cohomology associated to that topos by looking at abelian group objects quite generally.

However, the geometry lies in what the geometric points of the topos look like. If you buy into the philosophy of sheaf cohomology measuring obstructions to extensions of local to global sections, then the difference between the Zariski cohomology and étale cohomology is that the geometric points carry distinct information. More precisely, in the category of abelian Zariski sheaves a sequence of sheaves is exact if and only if it is exact on stalks. On taking global sections, one loses some information about local sections. For a constant sheaf however, there are 'enough' sections to patch up to global information.

In the étale case the stalk local condition is still true. However étale stalks are distinct from Zariski stalks. For a complex variety $$X$$, the local rings for points of the étale topos are strict Henselization of the local rings for the Zariski topos. In particular, if $$\mathcal{F}$$ is a coherent Zariski sheaf, and $$\mathcal{F}^{ét}$$ the associated étale sheafification, then at a geometric point $$\bar{x}\to X$$ an étale (i.e. geometric) point of $$X$$ one has (we denote the image of the $$\bar{x}\to X$$ by $$x$$) $$\mathcal{F}^{ét}_{\bar{x}}=\mathcal{F}_x\otimes_{\mathcal{O}_x}\mathcal{O}^{sh}_x.$$ where $$\mathcal{O}^{sh}_x$$ is the strict Henselization of $$\mathcal{O}_x$$, the local ring at $$x$$. Thus the stalk local condition is explicitly different!