proof that a domain is Dedekind if its ideals are products of primes
We show that for an integral domain^{} $R$, the following are equivalent^{}.

1.
$R$ is a Dedekind domain^{}.

2.
every nonzero proper ideal^{} is a product^{} of maximal ideals^{}.

3.
every nonzero proper ideal is a product of prime ideals^{}.
For the equivalence of 1 and 2 see proof that a domain is Dedekind if its ideals are products of maximals. Also, as every maximal ideal is prime, it is immediate that 2 implies 3. So, we just need to consider the case where 3 is satisfied and show that 2 follows, for which it enough to show that every nonzero prime ideal is maximal.
We first suppose that $\U0001d52d$ is an invertible^{} (http://planetmath.org/FractionalIdeal) prime ideal and show that it is maximal. To do this it is enough to show that any $a\in R\setminus \U0001d52d$ gives $\U0001d52d+(a)=R$. First, we have the following inclusions,
$$\U0001d52d\u228a\U0001d52d+({a}^{2})\subseteq \U0001d52d+(a).$$ 
Then, consider the prime factorizations^{}
$\U0001d52d+(a)={\U0001d52d}_{1}\mathrm{\cdots}{\U0001d52d}_{m},$  (1)  
$\U0001d52d+({a}^{2})={\U0001d52e}_{1}\mathrm{\cdots}{\U0001d52e}_{n}.$  (2) 
We write $\overline{a}$ for the image of $a$ under the natural homorphism (http://planetmath.org/NaturalHomomorphism) $R\to R/\U0001d52d$ and $\overline{\U0001d51e}$ for the image of any ideal $\U0001d51e$. Equations (1) and (2) give
$${\overline{\U0001d52e}}_{1}\mathrm{\cdots}{\overline{\U0001d52e}}_{n}={(\overline{a})}^{2}={({\overline{\U0001d52d}}_{1}\mathrm{\cdots}{\overline{\U0001d52d}}_{m})}^{2}.$$  (3) 
As $\U0001d52d$ is strictly contained in $\U0001d52d+(a)$ and $\U0001d52d+({a}^{2})$, it must also be strictly contained in ${\U0001d52d}_{k}$ and ${\U0001d52e}_{k}$. So, ${\overline{\U0001d52d}}_{k},{\overline{\U0001d52e}}_{k}$ are nonzero prime ideals, and by uniqueness of prime factorization (see, prime ideal factorization is unique) Equation (3) gives $n=2m$ and ${\overline{\U0001d52d}}_{k}={\overline{\U0001d52e}}_{k}={\overline{\U0001d52e}}_{k+m}$, after reordering of the factors. So ${\U0001d52d}_{k}={\U0001d52e}_{k}={\U0001d52e}_{k+m}$ and,
$$\U0001d52d+({a}^{2})={\U0001d52e}_{1}\mathrm{\cdots}{\U0001d52e}_{n}={\left({\U0001d52d}_{1}\mathrm{\cdots}{\U0001d52d}_{m}\right)}^{2}={\left(\U0001d52d+(a)\right)}^{2}={\U0001d52d}^{2}+a\U0001d52d+({a}^{2}).$$  (4) 
Then, $a\notin \U0001d52d$ gives $\U0001d52d\cap ({a}^{2})\subseteq a\U0001d52d$ and taking the intersection^{} of both sides of (4) with $\U0001d52d$,
$$\U0001d52d={\U0001d52d}^{2}+a\U0001d52d+\U0001d52d\cap ({a}^{2})\subseteq {\U0001d52d}^{2}+a\U0001d52d=\left(\U0001d52d+(a)\right)\U0001d52d.$$ 
But $\U0001d52d$ was assumed to be invertible, and can be cancelled giving $R\subseteq \U0001d52d+(a)$, showing that $\U0001d52d$ is maximal.
Now let $\U0001d52d$ be any prime ideal and $a\in \U0001d52d\setminus \{0\}$. Factoring into a product of primes
$${\U0001d52d}_{1}\mathrm{\cdots}{\U0001d52d}_{n}=(a)\subseteq \U0001d52d,$$  (5) 
each of the ${\U0001d52d}_{k}$ is invertible and, by the above argument, must be maximal. Finally, as $\U0001d52d$ is prime, (5) gives ${\U0001d52d}_{k}\subseteq \U0001d52d$ for some $k$, so $\U0001d52d={\U0001d52d}_{k}$ is maximal.
Title  proof that a domain is Dedekind if its ideals are products of primes 

Canonical name  ProofThatADomainIsDedekindIfItsIdealsAreProductsOfPrimes 
Date of creation  20130322 18:35:07 
Last modified on  20130322 18:35:07 
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  PrimeIdeal 