# Reference request: number of irreducible components and top dimension etale cohomology

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.

• 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? Sep 19 '19 at 19:38
• @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. Sep 22 '19 at 17:20