$\DeclareMathOperator{\h}{H}$Apologies in advance if this is overly stupid. Let $k$ be a field and $X$ a variety over $k$. Let $n$ be an integer which is invertible in $k$. One often looks at the groups $\h^i(X_{\bar k},\mathbb Z/n)$, i.e. the cohomology of the constant sheaf $\mathbb Z/n$ on the etale site of $X_{\bar k} = X\times_{\operatorname{Spec} k}\operatorname{Spec}{\bar k}$. We give this a $G_k=\operatorname{Gal}(\bar k/k)$-action via the action of $G_k$ on $\bar k$. I'm wondering if there is a simpler way to do this. The category of continuous $G_k$-modules is equivalent to the category of etale sheaves on $\operatorname{Spec} k$. If we let $f:X\to \operatorname{Spec} k$ be the structure morphism, then for any sheaf $\mathscr F$ on $X$, we have $G_k$-modules $\mathsf R^i f_\ast \mathscr F$. My question is:

Is $\mathsf R^i f_\ast \mathscr F$ naturally isomorphic (as a $G_k$-module) to $\h^i(X_{\bar k}, \mathscr F)$?


2 Answers 2


This is true if $X/k$ is proper (SGA 4.5, Les points de départ, IV. I, Théorème 1.1).

  • $\begingroup$ This is exactly what I was looking for. Thanks! $\endgroup$ Dec 3, 2013 at 2:16
  • 1
    $\begingroup$ @DanielMiller My pleasure! It is not a trivial theorem. :) $\endgroup$ Dec 3, 2013 at 2:17

Of course, this isomorphism follows from the proper base change theorem. However, the proper condition is superfluous in the situation, since you are just extending the scalar from $k$ to its separable closure. Comments on Roland's answer in the following link might be helpful: How does Galois group acts on etale cohomology?

Another reference is B. Conrad's Stanford note, page 44.


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