I haven't learned that much about primary decomposition, but from I understand about Dedekind domains, we have that all fractional ideals are invertible and all (plain old) ideals factor uniquely into a product of prime ideals, so that Dedekind domains should satisfy this condition. Are these the only such rings, or is there a weaker condition we can give?
[Math] When does the group of invertible ideal quotients = the free abelian group on the prime ideals
ac.commutative-algebra
Best Answer
First some responses to comments above:
One can (and should) define fractional ideals for any integral domain $R$. A fractional $R$-ideal is a nonzero $R$-submodule $I$ of the fraction field $K$ such that there exists $x \in K \setminus \{0\}$ such that $xI \subset R$. The product of two fractional ideals is again a fractional ideal, and the multiplication is associative and has $R$ itself as an identity: in other words, the set $\operatorname{Frac}(R)$ of fractional ideals of $R$ forms a commutative monoid.
We have the notion of a principal fractional ideal: this is a submodule of the form $xR$ for $x \in K^{\times}$. The set of principal fractional ideals forms a subgroup $\operatorname{Prin}(R)$ of $R$. One could take the quotient $\operatorname{Frac(R)}/\operatorname{Prin(R)}$, but this is a bit of a false step, as it is not a group in general.
In the case of Dedekind domains there is nothing to worry about:
Theorem: For an integral domain $R$, the following are equivalent:
(i) $R$ is Noetherian, integrally closed of dimension at most one (a Dedekind domain).
(ii) Every nonzero integral ideal of $R$ factors into a product of prime ideals.
(ii)' Every nonzero integral ideal of $R$ factors uniquely into a product of prime ideals.
(iii) The fractional ideals of $R$ form a group.
So for a Dedekind domain, certainly $\operatorname{Frac}(R)/\operatorname{Prin(R)}$ is a group, called the ideal class group of $R$.
In general, there is an easy way to remedy the problem that $\operatorname{Frac}(R)/\operatorname{Prin(R)}$ need not be a group. Namely, instead of taking the full monoid of fractional ideals, we restrict to the unit group $I(R)$, the invertible fractional ideals. For any domain $R$, we may define the Picard group
$\operatorname{Pic}(R) = I(R)/\operatorname{Prin}(R)$.
Because of the theorem above, for a non-Dedekind domain the Picard group isn't capturing any information about the prime ideals in particular. There is however a different construction -- coinciding with $\operatorname{Pic}(R)$ when $R$ is a Dedekind domain -- which does just this. For (a tiny bit of) motivation: even in the case of a Dedekind domain we don't take the free abelian group on all the prime ideals: we omit $(0)$.
Now let $R$ be any Noetherian domain. One can define the divisor class group $\operatorname{Cl}(R)$ as follows: let $\operatorname{Div}(R)$ be the free abelian group generated by the height one prime ideals $\mathfrak{p}$ (these are ideals so that there is no prime ideal $\mathfrak{q}$ properly in between $(0)$ and $\mathfrak{p}$). One can also define, for each $f \in K^{\times}$, a principal divisor $\operatorname{div}(f)$. (I don't want to give the exact recipe in the general case: it involves lengths of modules. If $R$ happens to be integrally closed, then the localization $R_{\mathfrak{p}}$ at a height one prime $\mathfrak{p}$ is a DVR, say with valuation $v_{\mathfrak{p}}$, and then one takes $\operatorname{div}(f) = \sum_{\mathfrak{p}} v_{\mathfrak{p}}(f) [\mathfrak{p}]$.) Again the principal divisors $\operatorname{Prin}(R)$ form a subgroup of $\operatorname{Div}(R)$ and the quotient $\operatorname{Div}(R)/\operatorname{Prin}(R)$ is the divisor class group.
There is a canonical homomorphism $\operatorname{Pic}(R) \rightarrow \operatorname{Cl}(R)$, which is in general neither injective nor surjective. However, the map is an isomorphism in the case that $R$ is a regular ring.
These constructions are the affine versions of more familiar constructions in classical algebraic geometry: they are, respectively, Cartier divisors and Weil divisors, which agree on a nonsingular variety but not in general.
Finally, one can also define analogues of these groups for certain non-Noetherian domains (the Noetherianity is used to ensure that $v_{\mathfrak{p}}(f) = 0$ except for finitely many primes $\mathfrak{p}$), e.g. Krull domains and Prufer domains. The latter is a domain in which each finitely generated nonzero ideal is invertible. Both are natural and interesting classes of rings.
For more details on this material, see e.g. the (rather rough and incomplete) notes
http://alpha.math.uga.edu/~pete/classgroup.pdf
[Addendum: also see Section 11 of factorization2010.pdf.]
For much more detail see the references cited therein, especially Larsen and McCarthy's Multiplicative Ideal Theory.