# When are the sheaf cohomology groups $H^i(X, \mathcal{F}) \otimes \mathbb{Z}/p^n \simeq H^i(X, \mathcal{F} \otimes \mathbb{Z}/p^n)$?

As the title suggests, I would like to understand when tensoring with a module, in particular $\mathbb{Z}/p^n$ commutes with taking sheaf cohomology - when we interpret this as tensoring with the constant sheaf $\mathbb{Z}/p^n$. It seems that people use statements like this a lot, but i'm having trouble justifying why it's true; any help would be greatly appreciated!

edit: Here's one such condition: If we can cover $X$ with connected affine sets $U_i$ (connected means $\Gamma(U_i, \mathcal{F} \otimes \mathbb{Z}/p^n) \simeq \Gamma(U_i, \mathcal{F}) \otimes \mathbb{Z}/p^n$) where $\Gamma(U_i, \mathcal{F})$ is $\mathbb{Z}/p^n$ flat, then we can compute cohomology on the Cech complex:

$$0 \to \prod \:\Gamma(U_i, \mathcal{F}) \otimes \mathbb{Z}/p^n \to \prod \Gamma(U_{ij}, \mathcal{F})\otimes \mathbb{Z}/p^n \to \ldots$$

and by flatness the tensor commutes with cohomology.

• Instead of editing answers into the question you should add answers! – Mariano Suárez-Álvarez Jan 10 '18 at 0:30

There is a short exact sequence of sheaves $$0\to\mathcal{F}\stackrel{p^n}\to\mathcal{F}\to\mathcal{F}\otimes\mathbb{Z}/p^n\to 0$$ which gives a long exact sequence of cohomology $$\dots\to H^i(X,\mathcal{F})\stackrel{p^n}\to H^i(X,\mathcal{F})\to H^i(X,\mathcal{F}\otimes\mathbb{Z}/p^n)\to H^{i+1}(X,\mathcal{F})\stackrel{p^n}\to H^{i+1}(X,\mathcal{F})\to\cdots.$$
The cokernel of $H^i(X,\mathcal{F})\stackrel{p^n}\to H^i(X,\mathcal{F})$ is $H^i(X,\mathcal{F})\otimes\mathbb{Z}/p^n$, and so the natural map $$H^i(X,\mathcal{F})\otimes\mathbb{Z}/p^n\to H^i(X,\mathcal{F}\otimes\mathbb{Z}/p^n)$$ is an isomorphism iff the map $H^{i+1}(X,\mathcal{F})\stackrel{p^n}\to H^{i+1}(X,\mathcal{F})$ is injective. That is, iff $H^{i+1}(X,\mathcal{F})$ has no $p$-torsion.