Integral Closure is Subring

Theorem
Let $R \subseteq A$ be an extension of commutative rings with unity.

Let $C$ be the integral closure of $R$ in $A$.

Then $C$ is a subring of $A$.

Proof
Clearly $R \subseteq C$, so $C \neq \emptyset$.

By Subring Test it is sufficient to prove that if $x,y \in C$ then $x \pm y, xy \in C$.

Let $x,y \in C$. By 1. $\Rightarrow$ 2. of Equivalent Definitions of Integral Dependence, $R[x]$ and $R[y]$ are finitely generated over $R$.

Because $R \subseteq R[x]$, $R[y]$ is also finitely generated over $R[x]$.

Therefore we have finitely generated extensions


 * $\displaystyle R \hookrightarrow R[x] \hookrightarrow R[x][y] = R[x,y]$

Therefore, by Transitivity of Finite Generation, $R[x,y]$ is finitely generated over $R$.

So by 2. $\Rightarrow$ 1. of Equivalent Definitions of Integral Dependence, every element of $R[x,y]$ is integral over $R$.

In particular, $x \pm y$ and $xy$ are integral over $R$.