Integral Closure is Integrally Closed

Theorem

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

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

Then $C$ is integrally closed.

Proof

Suppose $x \in B$ is integral over $C$.

Certainly $C$ is integral over $A$, so by Transitivity of Integrality, $C \sqbrk x$ is integral over $A$.

In particular, $x$ is integral over $A$, so $x \in C$.

$\blacksquare$