proof that a domain is Dedekind if its ideals are invertible
Let $R$ be an integral domain^{} with field of fractions^{} $k$. We show that the following are equivalent^{}.

1.
$R$ is Dedekind. That is, it is Noetherian^{} (http://planetmath.org/Noetherian), integrally closed^{}, and every prime ideal^{} is maximal (http://planetmath.org/MaximalIdeal).

2.
Every nonzero (integral) ideal is invertible^{}.

3.
Every fractional ideal is invertible.
As every fractional ideal is the product^{} of an element of $k$ and an integral ideal, statements (2) and (3) are equivalent. We start by proving that (3) implies $R$ is Dedekind.
Lemma.
If every fractional ideal is invertible, then $R$ is Dedekind.
Proof.
First, every invertible ideal is finitely generated, so $R$ is Noetherian.
Now, let $\U0001d52d$ be a prime ideal, and $\U0001d52a$ be a maximal ideal^{} containing $\U0001d52d$. As $\U0001d52a$ is invertible, there exists an ideal $\U0001d51e$ such that $\U0001d52d=\U0001d52a\U0001d51e$. That $\U0001d52d$ is a prime ideal implies $\U0001d51e\subseteq \U0001d52d$ or $\U0001d52a\subseteq \U0001d52d$. The first case gives $\U0001d52d\subseteq \U0001d52a\U0001d52d$ and, by cancelling the invertible ideal $\U0001d52d$ implies that $\U0001d52a=R$, a contradiction^{}. So, the second case must be true and, by maximality of $\U0001d52a$, $\U0001d52d=\U0001d52a$, showing that all prime ideals are maximal.
Now let $x$ be an element of the field of fractions $k$ and be integral over $R$. Then, we can write ${x}^{n}={c}_{0}+{c}_{1}x+\mathrm{\cdots}+{c}_{n1}{x}^{n1}$ for coefficients ${c}_{k}\in R$. Letting $\U0001d51e$ be the fractional ideal
$$\U0001d51e=(1,x,{x}^{2},\mathrm{\dots},{x}^{n1})$$ 
gives ${x}^{n}\in \U0001d51e$, so $x\U0001d51e\subseteq \U0001d51e$. As $\U0001d51e$ is invertible, it can be cancelled to give $x\in R$, showing that $R$ is integrally closed. ∎
It only remains to show the converse^{}, that is if $R$ is Dedekind then every nonzero ideal is invertible. We start with the following lemmas.
Lemma.
Every nonzero ideal $\mathrm{a}$ contains a product of prime ideals. That is, ${\mathrm{p}}_{\mathrm{1}}\mathit{}\mathrm{\cdots}\mathit{}{\mathrm{p}}_{n}\mathrm{\subseteq}\mathrm{a}$ for some nonzero prime ideals ${\mathrm{p}}_{k}$.
Proof.
We use proof by contradiction^{}, so suppose this is not the case. As $R$ is Noetherian, the set of nonzero ideals which do not contain a product of nonzero primes has a maximal element^{} (w.r.t. the partial order^{} of set inclusion) say, $\U0001d51e$.
In particular $\U0001d51e$ cannot be prime itself, so there exist $x,y\in R$ such that $xy\in \U0001d51e$ and $x,y\notin \U0001d51e$. Therefore $\U0001d51e$ is strictly contained in $\U0001d51e+(x)$ and $\U0001d51e+(y)$ and, by the choice of $\U0001d51e$, these ideals must contain a product of primes. So,
$$(\U0001d51e+(x))(\U0001d51e+(y))={\U0001d51e}^{2}+x\U0001d51e+y\U0001d51e+(xy)\subseteq \U0001d51e$$ 
contains a product of primes, which is the required contradiction. ∎
Lemma.
For any nonzero proper ideal^{} $\mathrm{a}$ there is an element $x\mathrm{\in}k\mathrm{\setminus}R$ such that $x\mathit{}\mathrm{a}\mathrm{\subseteq}R$.
Proof.
Let $\U0001d52d$ be a maximal ideal containing $\U0001d51e$ and $a$ be a nonzero element of $\U0001d51e$. By the previous lemma there are prime ideals ${\U0001d52d}_{1},\mathrm{\dots},{\U0001d52d}_{n}$ satisfying
$${\U0001d52d}_{1}\mathrm{\cdots}{\U0001d52d}_{n}\subseteq (a)\subseteq \U0001d51e\subseteq \U0001d52d.$$ 
We choose $n$ as small as possible. As $\U0001d52d$ is prime, this gives ${\U0001d52d}_{k}\subseteq \U0001d52d$ for some $k$ and, as every prime ideal is maximal, this is an equality. Without loss of generality we may take $\U0001d52d={\U0001d52d}_{n}$. As $n$ was assumed to be as small as possible, ${\U0001d52d}_{1}\mathrm{\cdots}{\U0001d52d}_{n1}$ is not a subset of $(a)$, so there exists $b\in {\U0001d52d}_{1}\mathrm{\cdots}{\U0001d52d}_{n1}\setminus (a)$. Then, $b\notin (a)$ gives $x\equiv {a}^{1}b\notin R$ and
$$x\U0001d51e\subseteq {a}^{1}b\U0001d52d\subseteq {a}^{1}{\U0001d52d}_{1}\mathrm{\cdots}{\U0001d52d}_{n1}\U0001d52d\subseteq {a}^{1}(a)=R$$ 
as required. ∎
We finally show that every nonzero ideal $\U0001d51e$ is invertible. If its inverse^{} exists then it should be the largest fractional ideal satisfying $\U0001d51f\U0001d51e\subseteq R$, so we set
$$\U0001d51f=\{x\in k:x\U0001d51e\subseteq R\}.$$ 
Choosing any nonzero $a\in \U0001d51e$ gives $a\U0001d51f\subseteq \U0001d51f\U0001d51e\subseteq R$ so $\U0001d51f$ is indeed a fractional ideal. It only remains to be shown that $\U0001d51f\U0001d51e=R$, for which we use proof by contradiction. If this were not the case then the previous lemma gives an $x\in k\setminus R$ such that $x\U0001d51f\U0001d51e\subseteq R$. By the definition of $\U0001d51f$, this gives $x\U0001d51f\subseteq \U0001d51f$ and therefore $\U0001d51f$ is an $R[x]$module. Furthermore, as $R$ is Noetherian, $\U0001d51f$ will be finitely generated^{} as an $R$module. This implies that $x$ is integral over the integrally closed ring $R$, so $x\in R$, giving the required contradiction.
Title  proof that a domain is Dedekind if its ideals are invertible 

Canonical name  ProofThatADomainIsDedekindIfItsIdealsAreInvertible 
Date of creation  20130322 18:34:54 
Last modified on  20130322 18:34:54 
Owner  gel (22282) 
Last modified by  gel (22282) 
Numerical id  5 
Author  gel (22282) 
Entry type  Proof 
Classification  msc 13A15 
Classification  msc 13F05 
Related topic  DedekindDomain 
Related topic  FractionalIdeal 