# Compare cohomology of tensor product and exterior product

Let $X$ be a smooth projective variety over a field $K$ of positive characteristic. Let $\mathcal{F}$ be a sheaf of $K-$algebras. Is there any criterion when the natural map from $H^i(\otimes^n\mathcal{F})$ to $H^i(\wedge^n \mathcal{F})$ for some $n>0$ is an isomorphism for all $i>0$?

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It seems strange to me to just ask for the cohomology groups to be isomorphic. Do you want the canonical map between them to be an isomorphism? –  Andreas Blass Jun 14 at 13:23
What does $\mathcal F \wedge \mathcal G$ even mean? The only exterior product I know is $\wedge^2 \mathcal F$. –  Dan Petersen Jun 14 at 13:25
@Blass: Yes. I am sorry for my imprecision. –  Jana Jun 14 at 13:50
Aside from $n = 1$? If $\mathcal F$ is the sheaf of sections of a line bundle then $\wedge^n \mathcal F = 0$ for $n > 1$, while $\otimes^n \mathcal F$ can have both lots of sections and nontrivial cohomology. –  Gunnar Magnusson Jun 14 at 14:34
The kernel $K$ of $\otimes^n\mathcal{F}\to \wedge^n\mathcal{F}$ is the $n$th component of the ideal generated by $f\otimes f$. It's enough to assume that $H^i(K)=H^{i-1}(K)=0$. Without more information, this is the best answer one can give. –  Donu Arapura Jun 14 at 15:38