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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?

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    Perhaps an example is in order? This is silly but maybe instructive. Take $S = F$ to be a field and $R = F^k$ to be the product of $k$ copies of $F$, where we view $F \subset F^k$ as a subring via the diagonal homomorphism $a \mapsto (a,\cdots,a)$. Now $F^k$ is generated as an $F$-module (vector space) by $k$ elements, and $F$ has precisely one maximal ideal $M = (0)$. All the maximal ideals of $F^k$ lie over $(0)$, and there are $k$ of them. Thus the bound is sharp. Note that $MR = (0)$ is still the zero ideal, and in particular not maximal for k > 1.2011-06-02

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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.

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    @Justin Campbell: thanks, I will study this in detail, need to get familiar with this kind of argument. Thanks again!2011-06-02