The problem is the following.
Let $R,S$ be rings such that $R\neq 0 $, $R\subset S$ and $S$ is finitely generated as an $R$-module, with $t$ generators. Let $\mathfrak{m}$ be a maximal ideal in $R$. Show that there are at most $t$ maximal ideals in $S$ lying over $\mathfrak{m}$.
My attempt to solve
First of all, we have that $S$ is integral over $R$. My idea was to suppose that there are $\mathfrak{m}_1,\ldots,\mathfrak{m}_{t+1}$ distinct maximal ideals in $S$ lying over $\mathfrak{m}$, and get a contradiction from that. It's possible to show that, for each $i=1\ldots t+1$, there is $a_i\in\mathfrak{m}_i$ such that $a_i\notin\mathfrak{m}_j$ for all $j\neq i$. We have that $a_1\cdot\ldots\cdot a_{t+1}\in\mathfrak{m}$ and $a_1\cdot\ldots\cdot a_{t+1}\in\mathfrak{m}_1\cap\ldots\cap\mathfrak{m}_{t+1}$. Also, by the chinese remainder theorem, we have $$\frac{S}{\mathfrak{m}_1\cap\ldots\cap\mathfrak{m}_{t+1}}\cong \frac{S}{\mathfrak{m}_1}\oplus \ldots\oplus \frac{S}{\mathfrak{m}_{t+1}}.$$
With this observations I tried to get some contradiction, like $S$ is not integral over $R$, or the Going Up property fails, or $S$ has more than $t$ generators and so on. Nothing is working, so I don't know what to do. I really appreciate a help here.
Thanks.
The question reduces to prove that there are at most $t$ maximal ideals in $S$ containing $mS$. We know that $S/mS$ is an $R/m$-vector space of dimension at most $t$, and therefore $S/mS$ is an artinian ring. One knows that $S/mS$ is isomorphic to a finite direct product of artinian local rings (which are the localizations of $S/mS$ at its maximal ideals), and then the number of these has to be at most $t$.