The fibers of a finite morphism of affine varieties are all finite

I am trying to find the proof of :

The fibers of a finite morphism $\phi: X \rightarrow Y$ ($X,Y$ affine) are all finite.

Here, a morphism is called finite if $K[X]$ is integral over the image of $K[Y]$ under the comorphism $\phi^*$ of $\phi$.

If I can suppose this morphism to be dominant of irreducible varieties, then for any closed subset $V \subseteq Y$, the dimension of $\phi^{-1}(V)$ is no less than the dimension of $V$. Do I have to prove in this case that for any closed irreducible subset $V$ in $Y$, the reverse image in $X$ has equal dimension with $V$? How can I use the fact of $K[x]/\phi^*K[Y]$ being integral?

Thanks very much.

Solution 1:

You are asking why a finite morphism is quasi-finite. In the commutative-algebra language : For an algebra $A$, a finite $A$-algebra $B$, and a prime ideal $p$ of $A$, there finitely many prime ideals $q$ of $B$ such that $A\cap q = p$.

As Zhen Lin said, this is a local property. Indeed, is $S$ is a multiplicative subset of an algebra $A$, and if $B$ is a finite algebra over $A$, then $S^{-1} B$ is finite over $S^{-1} A$. More over, if $S\cap p = \varnothing$, then the prime ideals of $B$ above $p$ correspond bijectively to the prime ideals of $S^{-1}B$ above $pS^{-1}A$.

So we can assume that $A$ is local, with maximal ideal $p$. The prime ideals of $B$ above $p$ are prime ideals of $B/pB$. But this algebra $B/pB$ is a finite-dimensional $A/p$-vector space, and thus it is an noetherian and artinian algebra, and thus it has only a finite number of prime ideal [Bourbaki AC chap. II, §2, n˚ 5, prop. 9].

Edit — Thanks to Zach N for having pointed out the ridiculous mistakes I made. I don't know if there exists a direct proof for the last point.