# 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.

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This is a question in commutative algebra, really: all you are asking is, if $B$ is an integral extension of $A$, why are there only finitely many primes of $B$ lying above a given prime of $A$? The answer has very much to do with integrality and is local on $A$, so it suffices to consider the case where $A$ is a local ring... – Zhen Lin Feb 23 '12 at 7:54
@Zhen Lin: Thanks for reminding me of the essence of the question. Also, thank you for editing the tag~ – ShinyaSakai Feb 27 '12 at 16:04

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.

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Thanks a lot for the answer~ Would you please tell me why is $B/pB$ finite-dimensional over $A/p$? Also, as you said, "a finite $A$-algebra, $B$", is $K[X]$ necessarily finite over $\phi^*K[Y]$? I was just assuming $K[X]/\phi^*K[Y]$ to be integral. Thanks again~ – ShinyaSakai Feb 27 '12 at 16:02
@ShinyaSakai — If you consider finitely generated $k$-algebra, then yes, integral is equivalent to finite. So the $A$-module $B$ is finite. Of course, a generating family of $B$ over $A$ is, after reduction, a generating family of $B/pB$ over $A$, hence over $A/p$. The point is that $A/p$ is a field, so you have a vector-space. Am I clear ? – Lierre Feb 28 '12 at 9:28
@Lierre: I don't follow the last part of your argument. I think you want to argue that B/pB is artinian since B/pB is finite over A/p, so B/pB has only finitely many maximal (= prime, for artinian rings) ideals. Surely you don't mean to say that a finite-dimensional vector space can have only finitely many subspaces? Am I missing something? – Zach N Jun 5 '12 at 18:03
I should point out that you can prove directly (perhaps this is what you were trying to get at) that a ring finite over a field has only finitely many prime ideals, but I think the argument is still nontrivial. – Zach N Jun 5 '12 at 18:14
@ZachN — Oh... What the heck did I say ?? Thanks for pointing out that. – Lierre Jun 5 '12 at 19:00