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> [!proposition] Proposition. ([[taking integral closures is idempotent]])
>
Let $A \subset B$ be ([[commutative ring|commutative]]) [[ring|rings]], with $\overline{A}$ denoting the [[integral closure]] of $A$ in $B$. Then $\overline{A}$ is [[integral closure|integrally closed]] in $B$, i.e., $\overline{ \overline{A}}=\overline{A}$.
>
Thus, taking [[integral closure|integral closures]] is an [[idempotent]] operation.
^proposition
> [!proof]- Proof. ([[taking integral closures is idempotent]])
> $\supset$. The integral closure always contains the original set.
>
$\subset.$ Take $x \in B$ [[integral element of an algebra|integral]] over $\overline{A}$ (i.e., $x \in \overline{ \overline{A}}$) We want to show $x \in \overline{A}$, i.e., $x$ is $A$-integral. There is a chain of inclusions $A \subset \overline{A} \subset \overline{A}[x],$
where the notation $\overline{A}[x]$ means we are [[ring adjunction|adjoining]] $x \in B$ to $A \subset B$.
>
Now, the extension $A \subset \overline{A}$ is [[integral algebra|integral]] by [[integral closure|definition]] of $\overline{A}$. The extension $\overline{A} \subset \overline{A}[x]$ is integral, too, [[integral algebra#^equivalence|because]] $\overline{A}[x]$ is [[subalgebra generated by a subset|generated over]] $\overline{A}$ by finitely many integral elements (namely, by $x$). By [[transitivity of finiteness and integrality and finite-typedness for algebras]], it follows that $\overline{A}[x]$ is [[integral algebra|integral]] as an [[algebra]] over $A$, and in particular that $x$ is integral over $A$.
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
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> GROUP BY Tag
> ```
> [!frontlink]
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> ```