Integral morphism between varieties has finite fiber 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!
 A: 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.
A: 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.
