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If $X$ is a complete nonsingular curve over $k$, $Y$ is any curve over $k$, $f: X \to Y$ is a morphism not map to a point (so $f(X)=Y$), then $f$ is a finite morphism.

This is the assertion prove in Hartshorne Chapter2, Prop6.8. But the proof is a little sketchy at the point of the inverse image of an affine set is also affine. I quote it here:

...Let $V=\rm{Spec}B$ be any open affine subset of $Y$. Let $A$ be the integral closure of $B$ in $K(X)$. Then $A$ is a finite $B$-module, and Spec$A$ is isomorphic to an open subset $U$ of $X$. Clearly $U=f^{-1}(V)$...

Can anyone explain why "Spec$A$ is isomorphic to an open subset $U$ of $X$. Clearly $U=f^{-1}(V)$"?

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Presumably R=B ...? –  Matt Apr 30 '12 at 16:16

1 Answer 1

up vote 7 down vote accepted

Since $X$ is complete, the morphism $f$ is proper.
For any $y\in Y$, the fibre $F=f^{-1}(y) $ is closed and strictly included in $X$ , because $f$ is not constant.
Hence $F$ is finite, i.e. $f$ is quasi-finite.
But a proper and quasi-finite morphism is finite, and so $f$ is indeed a finite morphism.

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Thank you so much! However, I have two questions: 1) why $F$ is closed? 2) why proper+quasi-finite imply finite? I have looked at Wikipedia, it referred this a result as a consequence of Stein factorization. Suppose $Y'$ is the intermediate scheme in Stein factorization, do we necessary have $X=Y'$? Or, maybe you can point out the reference where I can find the proof. –  Li Zhan May 1 '12 at 21:47
Dear Li, a) $F$ is closed because it is the inverse image of the closed set$\lbrace y \rbrace$ under the continuous morphism $f$. b) The result quoted in my last line can be found in Vakil's wonderful on-line notes, 11.3.8. page 267 –  Georges Elencwajg May 2 '12 at 6:32

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