Integrally Closed is Local Property

Theorem

Let $A$ be an integral domain.

For a prime ideal $\mathfrak p$ of $A$, let $A_{\mathfrak p}$ denote the localization at $S = A \divides \mathfrak p$.

Then the following are equivalent:

$(1): \quad A$ is integrally closed

$(2): \quad A_{\mathfrak p}$ is integrally closed for all prime ideals $\mathfrak p$.

$(3): \quad A_{\mathfrak m}$ is integrally closed for all maximal ideals $\mathfrak m$.

Proof

$(1)$ implies $(2)$

Let $\map Q R$ denote the field of quotients of an integral domain $R$.

We have by Localization Preserves Integral Closure that:

$\map Q {A_{\mathfrak p}} = \map Q A$

Hence $A_{\mathfrak p}$ is integrally closed for all prime ideals $\mathfrak p$.

The validity of the material on this page is questionable. In particular: What is the point of considering field of quotients? The claim seems exactly Localization Preserves Integral Closure You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

$\Box$

$(2)$ implies $(3)$

This is true because a Maximal Ideal of Commutative and Unitary Ring is Prime Ideal.

$\Box$

$(3)$ implies $(1)$

This theorem requires a proof. In particular: Need some more results on localization first You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.