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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 $k > 1$. –  Justin Campbell Jun 2 '11 at 2:24
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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, in your notation, what is $MB$? If you mean the ideal $M$, isn't $B/M$ a field as well? page $68$ , corollary $5.8$ of Atiyah's and MacDonald's book says that if $T$ is maximal then so is $M$. Sorry, getting confused here. –  user6495 Jun 2 '11 at 1:31
    
No worries. You might want to look at what Atiyah and MacDonald have to say about extending ideals. From what I remember, they write $M^e$ for what I am denoting by $MB$: in either case, this is the ideal of $B$ generated by $M$ (note $M$ is an ideal in $A$, not $B$!), which concretely looks like all finite $B$-linear combinations of elements of $M$. In general there is no reason to expect $MB$ to be maximal or even prime, but the maximal ideals of $B$ lying over $M$ are precisely those which contain $MB$. Let me know if I can explain anything else. –  Justin Campbell Jun 2 '11 at 1:37
    
P.S. I could understand why this seems mysterious: my best shot at an explanation is that we need to show something is a finite-dimensional $A/M$-vector space and hence Artinian to make the argument run, and the correct ring is $B/MB$. Note $B/MB$ is the "largest" quotient of $B$ such that the map $A \to B$ factors through $A/M$ into $B/MB$, and the maximal ideals of $B/MB$ correspond to the maximal ideals of $B$ lying over $M$. –  Justin Campbell Jun 2 '11 at 1:42
    
@Justin Campbell: thanks, I will study this in detail, need to get familiar with this kind of argument. Thanks again! –  user6495 Jun 2 '11 at 2:18
    
No problem. Not to be rude, but a great way to say thanks is to accept the answer. If it helps I can flesh out the proof, but I thought a hint would be more helpful. –  Justin Campbell Jun 2 '11 at 2:28
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