proof that a Noetherian domain is Dedekind if it is locally a PID

We show that for a Noetherian (http://planetmath.org/Noetherian) domain $R$ with field of fractions $k$ , the following are equivalent.

1.

$R$ is Dedekind. That is, it is integrally closed and every nonzero prime ideal is maximal.

for every maximal ideal $\mathfrak{m}$ ,

$R_{\mathfrak{m}}\equiv\left\{s^{-1}x:s\in R\setminus\mathfrak{m},x\in R\right\}$

is a principal ideal domain.

For a given maximal ideal $\mathfrak{m}$ and ideal $\mathfrak{a}$ of $R$ , we shall write $\mathfrak{\bar{a}}$ for the ideal generated by $\mathfrak{a}$ in $R_{\mathfrak{m}}$ , which consists of the elements of the form $s^{-1}a$ for $a\in\mathfrak{a}$ and $s\in R\setminus\mathfrak{m}$ . It is then easily seen that $\mathfrak{p}\mapsto\mathfrak{\bar{p}}$ gives a bijection between the prime ideals of $R$ contained in $\mathfrak{m}$ and the prime ideals of $R_{\mathfrak{m}}$ , with inverse $\mathfrak{p}\mapsto R\cap\mathfrak{p}$ . In particular $\mathfrak{\bar{m}}$ is the unique maximal ideal of $R_{\mathfrak{m}}$ , which is therefore a local ring.

Now suppose that $R$ is Dedekind, then the localization $R_{\mathfrak{m}}$ will be a Dedekind domain (localizations of Dedekind domains are Dedekind) with a unique maximal ideal, so it is a principal ideal domain (Dedekind domains with finitely many primes are PIDs).

Only the converse remains to be shown, so suppose that $R$ is a Noetherian domain such that $R_{\mathfrak{m}}$ is a principal ideal domain for every maximal ideal $\mathfrak{m}$ . In particular, $R_{\mathfrak{m}}$ is integrally closed and every nonzero prime ideal is maximal, so it contains a unique nonzero prime ideal $\mathfrak{\bar{m}}$ .

We start by showing that every nonzero prime ideal $\mathfrak{p}$ of $R$ is maximal. Choose a maximal ideal $\mathfrak{m}$ containing $\mathfrak{p}$ . Then, $\mathfrak{\bar{p}}$ is a nonzero prime ideal, so $\mathfrak{\bar{p}}=\mathfrak{\bar{m}}$ and therefore $\mathfrak{p}=\mathfrak{m}$ is maximal.

We finally show that $R$ is integrally closed. So, choose any $x$ integral over $R$ and lying in its field of fractions. Let $\mathfrak{a}$ be the ideal

$\mathfrak{a}=\left\{a\in R:ax\in R\right\}.$

We use proof by contradiction to show that $\mathfrak{a}$ is the whole of $R$ . So, supposing that this is not the case, there exists a maximal ideal $\mathfrak{m}$ containing $\mathfrak{a}$ . Then $x$ will be integral over the integrally closed ring $R_{\mathfrak{m}}$ and therefore $x\in R_{\mathfrak{m}}$ . So, $x=s^{-1}y$ for some $s\in R\setminus\mathfrak{m}$ and $y\in R$ . Then, $sx=y\in R$ so $s\in\mathfrak{a}\subseteq\mathfrak{m}$ , which is the required contradiction. Therefore, $\mathfrak{a}=R$ and, in particular, $1\in\mathfrak{a}$ and $x=1x\in R$ , showing that $R$ is integrally closed.

Title	proof that a Noetherian domain is Dedekind if it is locally a PID
Canonical name	ProofThatANoetherianDomainIsDedekindIfItIsLocallyAPID
Date of creation	2013-03-22 18:35:27
Last modified on	2013-03-22 18:35:27
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	4
Author	gel (22282)
Entry type	Proof
Classification	msc 13F05
Classification	msc 13A15