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Let $X$, $Y = \mathrm{Spec}(A)$ be Noetherian schemes and $f: X \to Y$ be proper with geometrically integral fibres. I want to show this implies $\mathcal{O}_Y = f_*\mathcal{O}_X$.

My idea was to reduce to $A$ local, use the theorem on formal functions and that the completion is faithfully flat and that for a connected reduced scheme $X$ over an algebraically closed field $k$, one has $H^0(X,\mathcal{O}_X) = k$.

I can show that $H^0(X,\mathcal{O}_X)$ is a finite local $A$-algebra. It probably suffices to show that $H^0(X \times_A A/\mathfrak{m}^n) = A/\mathfrak{m}^n$ (then apply the formal function theorem and the faithful flatness of the completion).

user5262
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1 Answers1

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I have to redraw the proof, because the result in Hartshorne I thought to requires $f$ to be flat.

There is in fact a counterexample without further assumption on $f$:

Suppose $A$ has non-trivial nilradical $I$. Let $X=\mathrm{Spec}(A/I)$ and let $f : X\to Y$ be the closed immersion. Then for all $y\in Y$, $X_y=\mathrm{Spec}(k(y))$ is geometrically integral. But $f_*O_X=A/I$ is not equal to $A=O_Y$.

Suppose we are in the situation where $Y$ is local with closed points $y$. As $X_y$ is geometrically integral, we have $H^0(X_y, \mathcal O_{X_y})=k(y)$. So the canonical map $$ H^0(X, \mathcal O_X)\otimes_A k(y)\to H^0(X_y, \mathcal O_{X_y})=k(y)$$ is surjective. This implies that the canonical map $$ H^0(X, \mathcal O_X)\otimes_A M\to H^0(X, \mathcal O_{X}\otimes_A M)$$ is an isomorphism for any finitely generated $A$-module $M$ (see Hartshorne, III, ยง12). Now apply this to $M=A/\mathfrak m^n$ and you can conclude as in your post.

Cantlog
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