## 59.5 Feats of the étale topology

For a natural number $n \in \mathbf{N} = \{ 1, 2, 3, 4, \ldots \}$ it is true that

$H_{\acute{e}tale}^2 (\mathbf{P}^1_\mathbf {C}, \mathbf{Z}/n\mathbf{Z}) = \mathbf{Z}/n\mathbf{Z}.$

More generally, if $X$ is a complex variety, then its étale Betti numbers with coefficients in a finite field agree with the usual Betti numbers of $X(\mathbf{C})$, i.e.,

$\dim _{\mathbf{F}_ q} H_{\acute{e}tale}^{2i} (X, \mathbf{F}_ q) = \dim _{\mathbf{F}_ q} H_{Betti}^{2i} (X(\mathbf{C}), \mathbf{F}_ q).$

This is extremely satisfactory. However, these equalities only hold for torsion coefficients, not in general. For integer coefficients, one has

$H_{\acute{e}tale}^2 (\mathbf{P}^1_\mathbf {C}, \mathbf{Z}) = 0.$

By contrast $H_{Betti}^2(\mathbf{P}^1(\mathbf{C}), \mathbf{Z}) = \mathbf{Z}$ as the topological space $\mathbf{P}^1(\mathbf{C})$ is homeomorphic to a $2$-sphere. There are ways to get back to nontorsion coefficients from torsion ones by a limit procedure which we will come to shortly.

Comment #1700 by Yogesh More on

Very minor remark: I think it would be helpful to add, after the sentence "For integer coefficients, one has $H^2_{etale}(P^1_C, \mathbb{Z})=0$", the following:

By contrast $H^2_{Betti}(P^1_C, \mathbb{Z})=\mathbb{Z}$ since $P^1_C$ is topologically equivalent to $S^2$.

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