Proposition 5. Let $A$ be a ring contained in a field $L$. Let $В$ be the set of elements of $L$ which are integral over $A$. Then $В$ is a ring, called the integral closure of $A$ in $L$.
Proof. Let $x, у$ lie in $B$, and let $M,N$ be two finitely generated $A$- modules such that $xM \subset M$ and $yN \subset N$. Then $MN$ is finitely generated, and is mapped into itself by multiplication with $x \pm у$ and $xy$.
I forgot what $MN$ is defined as...