TITLE: When does the group of invertible ideal quotients = the free abelian group on the prime ideals?
QUESTION [6 upvotes]: 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?

REPLY [4 votes]: When one considers the ring of algebraic integers R of a number field, the Picard group Pic(R)=Inv(R)/Prin(R) of invertible fractional ideals modulo principal fractional ideals is a measure of how far the ring is from being a UFD (the finiteness of the group is a classical theorem of algebraic number theory). In particular, Pic(R)=0 precisely when R is a UFD. If the cardinality of Pic(R) is h>1, then the hth power of every invertible fractional ideal factors uniquely into the product of prime ideals. 
Perhaps unfortunately, there exist rings R which are not UFDs but for which the Picard group is trivial. An example of such a ring is $\mathbb{Q}+x\mathbb{R}[x]$. This is not to say that the Picard group does not convey important information when one leaves the world of Dedekind Domains. 
A ring is a Bezout domain if it is an integral domain in which every finitely generated ideal is principal. For example, if the ring is Noetherian, then being a Bezout domain is equivalent to being a PID.
Now define a ring R to be a Prufer domain if it is an integral domain in which every non-zero finitely generated ideal is invertible. It is a theorem that for a Prufer ring R, Pic(R) is trivial if and only if R is a Bezout domain.
Let T be a non-empty set of indeterminates and R be an integral domain. Mathematicians have long been interested in exploring the relationshipa between (1) Pic(R) and Pic(R[T]) and (2) Cl(R) and Cl(R[T]). It was proven by Gabelli that Cl(R)=C(R[T]) if and only if R is integrally closed. Call an arbitrary commutative ring seminormal if $b^2=c^3$ implies that there exists an element $a$ such that $a^3=b$ and $a^2=c$. Then  Pic(R)=Pic(R[T]) precisely when R is seminormal.
Finally, it makes sense - and can in fact be incredibly fruitful - to define the Picard group of a noncommutative ring. The most common set up in which this group is defined is as follows. Let R be an integral domain with quotient field K, let A be a finite-dimensional semisimple algebra over K and O be an order of A. Then one can define a group $Pic_R(O)$, which is a certain quotient of the group of invertible fractional ideals, and a homomorphism $\varphi: Aut_R(O)\rightarrow Pic_R(O)$. One can adapt techniques  from K-theory in order to study $Pic_R(O)$, and hence the automorphism group of the order O as well. 
An excellent reference for this is the follwoing paper of Frolich.
A. Frolich The Picard Group of Noncommutative Rings, in Particular of Orders, Transactions of the AMS, Vol 180 (Jun 1973), pp. 1-45