X,Y are varieties, and $A(X),A(Y)$ are coordinate rimg ,respectively. If$f:X\rightarrow Y$ is a finite surjective map, that is $f^{-1}(p)$ is finite for all $p\in Y$, then $A(X)$ is finitely generated $A(Y)$- module??
Some question about map of varieties
1 Answers
No.
Take for $X$ the hyperbola $xy=1$ in $ \mathbb A_k^2$ ($k$ an algebraically closed field), for $Y$ the affine line $ \mathbb A_k^1$ and for $f:X \to Y$ the projection $(x,y)\mapsto x$. The fibers are finite: they consist of exactly one point, except for the origin whose fiber is empty.
However the corresponding ring morphism $f^\ast :\mathcal O(Y)\to \mathcal O(Y)$ is the inclusion
$f^\ast :k[X]\to k[X,Y]/(XY-1)=k[X,X^{-1}]$, and $k[X,X^{-1}]$ is not a finitely generated module over the ring $k[X]$.
This is why algebraic geometers do not call morphisms with finite fibers "finite", but call them "quasi-finite" (up to some technical nitpicking). A finite morphism $X\to Y$ is defined (in the case of affine varieties) by the stronger property of module-finiteness of $\mathcal O(X)$ over $\mathcal O(Y)$.
Edit: An answer to Sang's actual question! I had overlooked that Sang wanted to know about surjective morphisms with finite fibers. The answer is still "no".
I am going to exhibit a surjective morphism of varieties $f:X \to Y$ with finite fibers whose corresponding ring morphism $f^\ast :\mathcal O(Y)\to \mathcal O(Y)$ is not module-finite.
The variety $Y$ is still the affine line $ \mathbb A_k^1$ with coordinate $x$. In $ \mathbb A_k^2$ with coordinates $\xi,\eta$, consider the equation $\xi=\eta^2$; it
defines a parabola $X'$ whose ring of functions is $k[\xi, \eta]/(\xi-\eta^2)=k[x,y]=k[y]$ (since $x=y^2$).
Now take for $X$ the punctured parabola $X=X'\setminus \{P\}$ where $P$ is for example the point $(1,1)$. This is still an affine variety, whose ring of functions is the localized ring $k[y,\frac {1}{y-1}]$
We again take for $f$ the projection $f:X\to Y:(x,y) \mapsto x$ : it is surjective with finite fibers, as required by Sang.
However the corresponding ring map $f^\ast: \mathcal O(Y)\to \mathcal O(X)$ is
$k[x]\to k[y,\frac {1}{y-1}]: x\mapsto y^2$
and it is not module-finite.
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0Sang required that $f$ was surjective. However, I have a harder question: if $f \colon X \to Y$ is a finite (and surjective) morphism of algebraic varieties, then $\mathcal{O}_X(X)$ is finite over $\mathcal{O}_Y(Y)$? (Of course the answer is yes if $Y$ is affine.) – 2011-08-13
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0Sorry, I had overlooked that Sang wanted a surjective map. I have posted a second (similar) counterexample which *is* surjective. – 2011-08-13
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0Just to make a quick comment, for Noetherian schemes quasifinite+proper is the same as finite. – 2011-08-13
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0Andrea, where I can see proof of your question??? – 2011-08-16