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Let $X$ be a proper scheme defined over an algebraically closed field of characteristic $p > 0$. Let $F : X\rightarrow X$ be the absolute Frobenius morphism. What is the dimension of $H^0(X, F_*\mathcal{O}_X)$?

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1 Answer

F is a finite morphism, so affine, so $H^i(X, \mathcal{O}_X) = H^i(X, F_*\mathcal{O}_X)$ for all i.

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You don't need any properties of $F$ for this; by definition $F_*\mathcal O_X(X) = \mathcal O_X(F^{-1}(X)) = \mathcal O_X(X).$ –  Matt E Mar 30 '11 at 19:15
    
For the $H^0$, indeed, but anon gives a result working for high degree cohomology too. –  Henri May 29 '11 at 22:18
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