Let $K^{\bullet}$ be a bounded complex of abelian étale sheaves on a quasi-compact and quasi-separated scheme $X$.

  • For any étale cover $\mathcal{U} :=\{ U_i\to X\}_{i\in I}$, can we find a refinement $\mathcal{U}'$ of $\mathcal{U}$ with $\mathcal{U}'$ finite (the answer to this should be easily yes, but I just want to double check)

  • We have a Cech-to-cohomology spectral sequence

$$E_2^{p,q} := H^p(\text{Tot}(\check{C}(\mathcal{U},\underline{H}^q(K^{\bullet})))\Rightarrow H^{p+q}(X_{\rm ét},K^{\bullet})$$ for any $\mathcal{U}$.

I must be misunderstanding the meaning of convergence of this spectral sequence for any $\mathcal{U}$. Zariski covers are, in particular, étale covers. But if I choose $\mathcal{U}$ a Zariski cover, the spectral sequence should converge to Zariski hypercohomology of $K^{\bullet}$, which won't be étale hypercohomology.

What is the meaning of for any $\mathcal{U}$, then?

Thanks a lot.


1 Answer 1

  1. Any étale morphism is flat and locally of finite presentation [Tag 02GR, Tag 02GS], hence open [Tag 01UA]. Quasi-compactness now immediately shows that étale covers have finite subcovers.
  2. The presheaves $\underline{H}^q(K^i)$ are given by $U \mapsto H^q_{\text{ét}}(U,K^i|_U)$, which is not the same as the Zariski version $U \mapsto H^q_{\text{Zar}}(U,K^i|_U)$. Thus, there is no reason this should converge to the Zariski cohomology (e.g. think of the one-object covering $\{X \to X\}$).

You must log in to answer this question.