A sufficient condition for a domain to be Dedekind? We know that in a Dedekind domain, every nonzero ideal admits a unique factorization into a product of prime ideals. 
I was wondering if this condition is sufficient for a domain to be Dedekind, i.e., if in a domain $D$ every non-zero ideal admits a unique factorization into a product of prime ideals, then is the domain Dedekind?
It seems that "Noetherianness" of $D$ follows (via the ascending chain condition) if we can show that every non-zero prime ideal of $D$ is maximal. But, I don't know how to prove the latter assertion.
Perhaps, one needs Noetherianness of $D$ to prove that $D$ is a dimension $1$ domain.
Any help or hint would be appreciated.
 A: Jacobson's Basic Algebra, volume II (second edition) has:
Definition. A domain $D$ is called a Dedekind domain if every $D$-fractional ideal of $F$ (the field of fractions of $D$) is invertible.
Theorem. (Theorem 10.6, page 630) Let $D$ be a domain. The following are equivalent:

*

*$D$ is Dedekind.

*$D$ is integrally closed, noetherian, and every nonzero prime ideal of $D$ is maximal.

*Every proper integral ideal of $D$ can be written in one and only one way as a product of prime ideals.

*$D$ is noetherian and $D_P$ is a discrete valuation ring for every maximal ideal $P$ in $D$.

The proof that 3 implies 1 is credited to Zariski and Samuel:
First, every invertible prime $P$ of $D$ is maximal in $D$; let $a\in D-P$, and assume that $aD+P\neq D$; then $aD+P$ can be written as a product of primes,
$$aD+P = P_1\cdots P_m$$
and likewise, $a^2D+P\neq D$ so
$$a^2D+P = Q_1\cdots Q_n$$
Since $P$ is properly contained in both $aD+P$ and in $a^2D+P$, then $P$ is properly contained in each of the $P_i$ and each of the $Q_j$. Now look at $D/P$, which is also a domain, and in which $\overline{aD} = \overline{P_1}\cdots\overline{P_m}$ and $\overline{a^2D}=\overline{Q_1}\cdots \overline{Q_n}$. The principal ideals $\overline{aD}$ and $\overline{a^2D}$ are invertible, and $\overline{P_i}$ and $\overline{Q_j}$ are prime ideals. Since $\overline{a^2D} = \overline{a}^2\overline{D}=(\overline{aD})^2 = \overline{P_1}^2\cdots\overline{P_n}^2$, and  since factorizations for invertible integral ideals are unique, $\{\overline{P_1},\overline{P_1},\ldots,\overline{P_n},\overline{P_n}\}$ is the same sequence as $\{\overline{Q_1},\ldots,\overline{Q_m}\}$, except perhaps for the order. Hence, the corresponding sequences of ideals of $D$ that contain $P$, $\{P_1,P_1,\ldots,P_n,P_n\}$ and $\{Q_1,\ldots,Q_m\}$ are the same up to order. Hence $(aD+P)^2 = a^2D+P$, so
$$P\subseteq (aD+P)^2 = a^2D+aP+P^2 \subseteq aD+P^2.$$
So if $p\in P$, then $p=ax+y$ with $x\in D$ and $y\in P^2$; hence $ax\in P$, and since $a\notin D$, then $x\in P$.  So $P\subseteq aP+P^2\subseteq P$, so $P=aP+P^2$. Since we are assuming $P$ is invertible, then $D = P^{-1}P = P^{-1}(aP+P^2) = aD+P$, a contradiction.
So $aD+P = D$ for every $a\notin P$, hence $P$ is maximal.
Now let $P$ be any nonzero prime ideal, and let $b\neq 0$ be an element of $P$. So $bD\subseteq P$ and $bD=P_1P_2\cdots P_m$ where the $P_i$ are prime and are invertible. So the $P_i$ are maximal. Since $P$ is prime, $P_i\subseteq P$ for some $i$; hence $P=P_i$, so $P$ is invertible.
Since every proper ideal is a product of primes, and every prime is invertible, it follows that every integral ideal is invertible.
Given a fractional ideal $I'=aI$, with $I$ integral and $a\neq 0$, we have $I'(I')^{-1}=II^{-1}=D$, so every fractional ideal is invertible. $\Box$
