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Let $k$ be an algebraically closed field, $\ell$ is a prime different with characateristic of $k$, and consider the $\ell$-adic etale cohomology. We know the number of connected components of a scheme finite type over $k$ by looking at $H^0$, but how about the number of irreducible components?

Looking at the example $\{xy=0\}$ in $\mathbb P^2$ which has $1$-dim $H^0$ and $2$-dim $H^2$, it seems that the number has something to do with top dimensional etale cohomology.

So the question is: let $X$ be a connected equidimensional finite type scheme over $k$ (of dimension $n$), when do we know $\dim_k (H^{2n}_{c}(X))$ is equal to the number of irreducible components of $X$ ?

The complex case is partially discussed in https://math.stackexchange.com/questions/2393326/top-cohomology-and-irreducible-components, but the proof does not work in positive characterisitc case.

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The answer is yes (in fact the result also holds over separably closed fields). You can find this statement in Corollary 7.5.21 of:

Poonen - Rational points on varieties.

Poonen gives a sketch of a proof with references to details in SGA4.

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    $\begingroup$ Thank you. If I have a map between proper varieties with same dimension n, then what is the induced map on the set of irreducible components under identification with $H^2n$? Is it just taking the preimage? $\endgroup$
    – Zhiyu
    Commented Sep 19, 2019 at 19:38
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    $\begingroup$ I would guess so yes. $\endgroup$
    – Daniel Loughran
    Commented Sep 20, 2019 at 8:42
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    $\begingroup$ @sawdada You for $f: X\to Y$, you would send an irreducible component of $Y$ to the sum of the irreducible components in the preimage weighed by the degree of the map restricted to that irreducible component. This should be clear from following through the proof. In particular the weight can be zero if the map from one component to another is not dominant. $\endgroup$
    – Will Sawin
    Commented Sep 22, 2019 at 17:20

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