Integral Closure is Integrally Closed
Jump to navigation
Jump to search
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 $x \in C$.
$\blacksquare$