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given a field $k$ and a proper integral scheme $f:X\rightarrow \operatorname{Spec}(k)$, is it true that $f_{*}\mathcal{O}_{X}\cong \mathcal{O}_{\operatorname{Spec}(k)}$?

Consider the normalization $\nu:X_1\rightarrow X$,let $g:X_1\rightarrow \operatorname{Spec}(k)$ be the structure morphism and assume that there is a quasi isomorphism $\mathcal{O}_{\operatorname{Spec}(k)}\cong \mathbb{R}^{\cdot}g_{*}\mathcal{O}_{X_1}$, what can I say about $\mathbb{R}^{\cdot}f_{*}\mathcal{O}_{X}$?


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can I relax proper+integral with proper+connected+reduced? – kethri Jan 24 '13 at 15:01
up vote 1 down vote accepted

The isomorphism $f_{*}O_X=k$ holds if $k$ is algebraically closed. Otherwise, take a finite non-trivial extension $k'/k$ and $X=\mathrm{Spec}(k)$ you will get a counterexample.

A sufficient condition over an arbitrary field is $X$ is proper, geometrically connected (necessary) and geometrically reduced. This amounts the prove the above isomorphism over an algebraically field. The this case, by general results $f_*O_X=H^0(X, O_X)$ is a finite $k$-algebra, integral because $X$ is integral, hence equal to $k$ (there is a direct proof using the properness of any morphism from $X$ to any separated algebraic variety).

For your second question, consider the case of singular rational curves. I think there is no much things one can say about the $H^1$. It could be arbitrarily large.

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