It feels like this should be a well-known question, but I can't find any related questions on this site by searching; apologies in advance if this is a duplicate.
I assume rings are commutative with multiplicative identity. Most examples of non-UFDs involve cases where factorization is not unique, e.g. $\mathbb Z[\sqrt{-5}]$ is not an UFD because $$ 2 \cdot 3 = 6 = (1+\sqrt{-5}) \cdot (1-\sqrt{-5}) $$
and none of the divisors are associate. Is there an integral domain $R$ where:
- Factorization is unique in the sense that if $p_1p_2\cdots p_n = q_1q_2\cdots q_m$ where $p_i$ and $q_i$ are irreducible elements in $R$, then $m=n$ and we can reorder such that $p_i$ is an associate of $q_i$ for each $i=1,2,\dots, n$.
- Factorization may not exist: there is some non-unit $x$ that cannot be written as the product of irreducible elements.
Best Answer
I'm not familiar with the content, but this paper suggests yes:
It appears to be the main subject of the article. From the introduction:
So, you might find examples there.
I am a little puzzled by "if every element that can be factored uniquely into irreducible elements has unique factorization." in which "uniquely" seems like it might be included erroneously.
The first definition seems to say what you are looking for, though: