Proof for an integral domain involving subrings Theorem: if $R,S$ are integral domains, $R\subset S$, and where $s_{1},...,s_{n}$ in $R_{S}$. Then there is a $m\in \mathbb{N}$ and $t_{1},..., t_{m}$ in $S$ (not all 0) so that $s_{i}N\subset N $ ($i=1,...,n)$ where $N=t_{1}R+...t_{m}R$  
I was ill yesterday and could not attend the lecture (where it was shown). Does anybody know where I can find a proof of this theorem on the internet or a book? Or does anybody know how to show this and is willing to write it here?  Thanks.  
I will post now the proof that Milne gives for another Theorem but I fail to use for the theorem of my lecturer:  
 Proposition 5.1. Let A be a subring of a ring B. An element $\alpha$ of B is integral over A if and only if there exists a faithful (i.e. if $a M = 0 $ then this implies $a=0$) $A[\alpha]$ submodule M of B that is finitely generated as an A-module.  
 Proof  $\Rightarrow$ : Suppose $$\alpha ^{n} + a_{1}\alpha^{n-1}+\cdots+a_{n}=0 $$
Then the A-submodule M of B generated by $1,\alpha,\ldots,\alpha^{n-1}$ has the property that $\alpha M \subset M$, and it is faithful because it contains 1 .
 $\Leftarrow : $ Let M be an A-module in B with a finite set $\{e_{1},\ldots,e_{n}\}$ of generators such that $\alpha M \subset M $ is faithful as an $A[\alpha]$ module. Then, for each i, $$\alpha e_{i} = \sum a_{ij}e_{j}; \text{ for some }a_{ij} \in A $$  
We can rewrite this system of equations as : 
$$\begin{align*}
(\alpha-a_{11}e_{1}-a_{12}e_{2}-a_{13}e_{3}-\cdots &=0\\  
-a_{21}e_{1}+(\alpha - a_{22})e_{2}- a_{23}e_{3}- \cdots&= 0\\  
\cdots &=0
\end{align*}$$
Let $C$ be the matrix of coefficients on the left-hand side. Then Cramers formula tells us that $\det (C)e_{i}=0$ for all $i$. As $M$ is faithful and the $e_{i}$ generate $M$, this implies that $\det(C)=0$. On expanding out the determinant, we obtain an equation:
$$\alpha^{n} + c_{1}\alpha^{n-1} + c_{2}\alpha^{n-2} + \cdots+ c_{n} = 0 ,\qquad c_{i} \in A$$

From this proof of the Proposition, can anybody please tell me the proof for the theorem of the lecturer? 
 A: For proofs like this, I think it's good to consider a single $s_1 = s$ first. I assume that for you $s \in S$ being integral over $R$ means that (INT 1) $s$ is a root of a monic polynomial with coefficients in $R$. But it is equivalent to require (we need another adjective if $S$ is not a domain) the following.

(INT 2) There exists a non-zero finitely generated $R$-submodule $M$ of $S$ such that $sM \subset M$.

I hope it's clear that (INT 2) is your desired statement for $n = 1$. The equivalence of these is well-known and is, for example, Proposition 5.1 of Milne's commutative algebra notes. Proving that (INT 1) implies (INT 2), which is all you need, is the easier direction: if $s$ satisfies
\[
s^k + r_{k - 1}s^{k - 1} + \cdots + r_1s + r_0 = 0 
\]
with $k \geq 1$ and each $r_i \in R$, then $R[s]$ is generated as an $R$-module by $1, \ldots, s^{k - 1}$.
Here are some preliminaries for the general case.


*

*Show that if $M_1, M_2$ are non-zero finitely generated $R$-submodules of $S$, then so is
\[
M_1M_2 = \{\sum x_iy_i : x_i \in M_1, y_i \in M_2\}.
\]
Indeed, if $M_1$ is generated by $u_1, \ldots, u_r$ and $M_2$ is generated by $v_1, \ldots, v_s$, then the $rs$ elements $\{u_iv_j\}$ generate $M_1M_2$ as an $R$-module.

*Given $s \in S$, show that if $sM_1 \subset M_1$ then $sM_1M_2 \subset M_1M_2$.
So, if you use (INT 2) for each $s_i$ to get a module $M_i$, then $N = M_1 \cdots M_n$ will do everything you want.
I'll prove the second preliminary exercise, just to show that nothing scary is happening here. I think the proofs are cleaner if you notice that $M_1M_2$ is generated by products, but I will avoid any sort of finesse. Suppose $sM_1 \subset M_1$, and let $\sum_{i = 1}^t x_iy_i$ with $x_i \in M_1$ and $y_i \in M_2$ be an element of $M_1M_2$. If we multiply by $s$, then we get $\sum_{i = 1}^t (sx_i)y_i$. As $sx_i \in sM_1 \subset M_1$, this sum is an element of $M_1M_2$.
