Integral extensions and number of generators

here's a doubt which arised from a previous question:

Suppose $R$ is a ring and $S \subseteq R$ is a subring. Moreover suppose $R$ is integral over $S$ and $R$ is finitely generated as an $S$-module. In a previous post Matt E showed that we always have Specmax(S) is bounded above by Specmax(R). My question is: is it true that for any maximal ideal of S there are at most t maximal ideals of R?

Thanks again

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Do you mean, the number of elements in the maximum spectrum of $S$ is bounded above by $t$ times the number of elements in the maximum spectrum of $R$? Because, as stated, it's wrong with $R=S$ provided $R$ has more than one maximal ideal (as then $t=1$). – Arturo Magidin May 16 '11 at 3:14
@Arturo Magidin: Sorry I meant that if (is it true) that for any maximal ideal of $S$ there are at most $t$ maximal ideals of $R$ ? – user6495 May 16 '11 at 3:18

Let $\mathfrak{m}$ be a maximal ideal of $R$; we want to show that the fiber of the map $\mathrm{Spec} S \to \mathrm{Spec} R$ has cardinality at most $t$, where $S$ is generated by $t$ elements as an $R$-module. By making the base-change $R \to R/\mathfrak{m}$, $S \to S/\mathfrak{m}S$, we are reduced to the case of $R$ a field. In this case, $S$ is a $t$-dimensional vector space. We need to bound the size of $\mathrm{Spec} S$ by $t$.
Now $S$ is an artinian ring, and it factors as a product of local artinian rings (uniquely). Each of these local factors is an $R$-vector space of dimension at least one, so there are at most $t$ factors. By definition, each factor has one prime ideal. Since the spectrum of a direct product is the disjoint union, this shows that $\mathrm{Spec} S$ has at most $t$ elements.