TITLE: Finite ring extension and number of maximal ideals
QUESTION [8 upvotes]: I want to understand why the following is true:

Let $S \subseteq R$ be commutative rings with $1$ and assume that $R$ is finitely generated as an $S$-module by at most $k$ elements. For every maximal ideal $M$ of $S$ there are at most $k$ maximal ideals of $R$ lying over $M$.

So let $T$ be a maximal ideal of $R$ lying over $M$ then $T \cap S = M$. Now it can be shown that $R/T$ is a finitely generated $S/M$-module. Hence $R/T$ is Artinian and thus it has finitely many maximal ideals. From here I have two questions:
1) Why from this follows that there can be only finitely many $T$ lying over $M$? I suspect they are using some correspondence which I don't see.
2) Why exactly at most $k$ ?
Can you please explain?

REPLY [3 votes]: You said $R/T$ is Artinian. This is true: it's already a field. What you want is that $R/MR$ is a finite-dimensional $S/M$-vector space, hence Artinian, so it has finitely many maximal ideals. This implies that finitely many maximal ideals of $R$ lie over $M$.
To see why, and also to understand the bound on the number of maximal ideals, use the following fact: if $R$ is any commutative ring with $1$, $R$ maps into the product of all $R/\mathfrak{p}$, where $\mathfrak{p}$ runs over prime ideals of $R$, and the kernel is the nilradical of $R$. This, together with the fact that prime is equivalent to maximal in an Artinian ring, should give you what you need.