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.

2.
for every maximal ideal^{} $\U0001d52a$,
$${R}_{\U0001d52a}\equiv \{{s}^{1}x:s\in R\setminus \U0001d52a,x\in R\}$$ is a principal ideal domain^{}.
For a given maximal ideal $\U0001d52a$ and ideal $\U0001d51e$ of $R$, we shall write $\overline{\U0001d51e}$ for the ideal generated by $\U0001d51e$ in ${R}_{\U0001d52a}$, which consists of the elements of the form ${s}^{1}a$ for $a\in \U0001d51e$ and $s\in R\setminus \U0001d52a$. It is then easily seen that $\U0001d52d\mapsto \overline{\U0001d52d}$ gives a bijection between the prime ideals of $R$ contained in $\U0001d52a$ and the prime ideals of ${R}_{\U0001d52a}$, with inverse^{} $\U0001d52d\mapsto R\cap \U0001d52d$. In particular $\overline{\U0001d52a}$ is the unique maximal ideal of ${R}_{\U0001d52a}$, which is therefore a local ring^{}.
Now suppose that $R$ is Dedekind, then the localization^{} ${R}_{\U0001d52a}$ 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 ${R}_{\U0001d52a}$ is a principal ideal domain for every maximal ideal $\U0001d52a$. In particular, ${R}_{\U0001d52a}$ is integrally closed and every nonzero prime ideal is maximal, so it contains a unique nonzero prime ideal $\overline{\U0001d52a}$.
We start by showing that every nonzero prime ideal $\U0001d52d$ of $R$ is maximal. Choose a maximal ideal $\U0001d52a$ containing $\U0001d52d$. Then, $\overline{\U0001d52d}$ is a nonzero prime ideal, so $\overline{\U0001d52d}=\overline{\U0001d52a}$ and therefore $\U0001d52d=\U0001d52a$ 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 $\U0001d51e$ be the ideal
$$\U0001d51e=\{a\in R:ax\in R\}.$$ 
We use proof by contradiction^{} to show that $\U0001d51e$ is the whole of $R$. So, supposing that this is not the case, there exists a maximal ideal $\U0001d52a$ containing $\U0001d51e$. Then $x$ will be integral over the integrally closed ring ${R}_{\U0001d52a}$ and therefore $x\in {R}_{\U0001d52a}$. So, $x={s}^{1}y$ for some $s\in R\setminus \U0001d52a$ and $y\in R$. Then, $sx=y\in R$ so $s\in \U0001d51e\subseteq \U0001d52a$, which is the required contradiction^{}. Therefore, $\U0001d51e=R$ and, in particular, $1\in \U0001d51e$ 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  20130322 18:35:27 
Last modified on  20130322 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 