Integral Closure is Subring
Theorem
Let $R \subseteq A$ be an extension of commutative rings with unity.
"An extension" or "extensions"? Are both extensions (in which case "extensions") or is only $A$ an extension (in which case "commutative rings with unity" needs to be singular? As it stands the grammar is confusing.
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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 $R$.
$\blacksquare$
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