Let $R \subset S$ be commutative rings. Let $s \in T$ where $T$ is a finitely generated $R$-module in $S$ with $R \subset T \subset S$. I want to show that $R[s]$ is finitely generated as an $R$-module.
Let $(t_i)_{i = 1\dots n}$ be a minimal spanning set for $T$ over $R$. Then $s^k = \sum_i s^k_{i} t_i$. And so we have
$$ 1 = \sum_i s_i^0 t_i \\ s = \sum_i s_i^1 t_i \\ \dots \\ s^n = \sum_i s_i^n t_i $$
Don't know where to go from here.
I assume that $T$ is a ring and $t_1,\dots,t_n$ is a set of generators of $T$ as an $R$-module. Then $st_i=\sum_{j=1}^n a_{ij}t_j$, $a_{ij}\in R$, $i=1,\dots,n$. Set $A=sI_n-(a_{ij})$. The previous relations can be written as follows: $A(t_1,\dots,t_n)^T=0$. (Here $^T$ denotes the transpose.) Then $(\hbox{adj}A)A(t_1,\dots,t_n)^T=0$ and since $(\hbox{adj}A)A=(\det A) I_n$ we get $(\det A)t_i=0$ for all $i$. Since $1\in T$ we can write $1$ as a linear combination of the $t_i$'s with coefficients in $R$, so $\det A=0$. This gives a monic polynomial with coefficients in $R$ that has $s$ as a root, that is, $s$ is integral over $R$. Now it's easy to prove that $R[s]$ is a finitely generated $R$-module.