# Integral Closure is Subring

## Theorem

Let $A$ be an extension of a commutative ring with unity $\left({R, +, \circ}\right)$.

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

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

## Proof

For each $r \in R$, we have $r$ is a zero of $a - r = 0$.

Hence $R \subseteq C$, and in particular, $C$ is non-empty.

By Subring Test it is sufficient to prove that if $x,y \in C$ then $x + \left({-y}\right), x \circ y \in C$.

Let $x,y \in C$.

By Equivalent Definitions of Integral Dependence: $(1) \implies (2)$, $R \left[{x}\right]$ and $R \left[{y}\right]$ are finitely generated over $R$.

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

Therefore, we have finitely generated extensions:

- $\displaystyle R \hookrightarrow R \left[{x}\right] \hookrightarrow R \left[{x}\right] \left[{y}\right] = R \left[{x, y}\right]$

Hence, by Transitivity of Finite Generation, $R \left[{x, y}\right]$ is finitely generated over $R$.

So by Equivalent Definition of Integral Dependence: $(2) \implies (1)$, every element of $R \left[{x, y}\right]$ is integral over $R$.

In particular, $x + \left({-y}\right)$ and $x \circ y$ are integral over $R$.

$\blacksquare$