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I'm looking for a proof/counterexample of the following fact:

Theorem Let $X \subseteq k^n$ and $Y \subseteq k^n$ be algebraic varieties over a field $k$ and let $\phi$ be a morphism from $X$ to $Y$. If the induced morphism on the Coordinate Rings is integral, then the fibers of $\phi$ are finite.

Please, help me to prove or disprove this fact or give me some references.

Thank you!

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up vote 3 down vote accepted

Let $A$ be the coordinate ring of $X$ and let $B$ be the coordinate ring of $Y$. Since $\phi : X \to Y$ is a morphism, we get a homomorphism $\phi^* : B \to A$. To show that the fibres of $\phi : X \to Y$ are finite, it is enough to show that $A$ is a finite $B$-module (via $\phi^* : B \to A$), because finiteness is preserved under base change.

So suppose $A$ is integral over $B$. We know $A$ is finitely generated as a $k$-algebra, so it is finitely generated as a $B$-algebra a fortiori, say by $a_1, \ldots, a_r$. But each $a_i$ is integral over $B$, so there exist a natural number $N$ such that $a_1, \ldots, a_1^N, \ldots, a_r, \ldots a_r^N$ generate $A$ as a $B$-module. Thus $A$ is indeed a finite $B$-module.

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It seems the OP is asking why a finite map is quasi-finite. Perhaps you could explain a little more why finiteness being preserved under base change implies quasi-finiteness (just to be perfectly clear) – zcn Feb 25 '14 at 4:42
@user115654 Not quite. He's asking why an integral map of varieties is quasifinite. Of course, in general, integral maps need not be quasifinite, but if they are also finite type then they do. Finite type comes for free since they are $k$-varieties and the maps are $k$-morphisms. – Alex Youcis Feb 25 '14 at 7:43

A morphism with finite fibers is called quasi-finite. In general, integral + finite type = finite (which was what Zhen Lin showed in his answer), so your question is why a finite map of affine varieties is quasi-finite. There are direct proofs (see e.g. here) for finite maps, but quasi-finiteness also holds under much weaker conditions:

Proposition: Let $\varphi : R \to S$ be a ring homomorphism which makes $S$ integral over $R$. If $S$ is Noetherian, then the induced map $\varphi : \text{Spec}(S) \to \text{Spec}(R)$ is quasifinite (hence so is the map of varieties $\text{mSpec}(S) \to \text{mSpec}(R)$). $\newcommand{\Spec}{\operatorname{Spec}}$

Let $p \in \Spec R$ be a prime ideal; it remains to see that there are only finitely many primes $q \in \Spec S$ such that $\varphi^{-1}(q) = p$. Such a prime $q$ must contain $\varphi(p)$, so we need only consider primes in $S/\varphi(p)S$. By incomparability of primes in an integral extension, $q$ must be a minimal prime of $S/\varphi(p)S$. But $S/\varphi(p)S$ is Noetherian, hence has only finitely many minimal primes.

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Alternatively, the fiber $\varphi^{-1}(p)$ is Noetherian by assumption, and is dimension $0$ since it's integral over $\text{spec}(k(p))$. Thus, it's Artinian, and so has finitely many primes. – Alex Youcis Feb 25 '14 at 7:50
@AlexYoucis: It's perhaps oversimplifying to say that $\varphi^{-1}(p)$ is Noetherian by assumption. Indeed, the tensor product of two fields can fail to be Noetherian: see here. It does work in this case though, because $S \otimes_R k(p)$ is a localization of a quotient of $S$ – zcn Feb 25 '14 at 8:36
That is, of course, what I meant :) Also, it is much easier to see that $\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}}$ is not Noetherian--else it would have finitely many points (since it's dimension $0$). Sorry for any mixup! – Alex Youcis Feb 25 '14 at 8:44

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