If an integral domain $R$ has a factorization basis, is it a UFD? By a factorization basis for an integral domain $R$, let us mean a subset $\xi$ of the commutative monoid $R^\times = R \setminus \{0\}$ such that firstly, no two elements of $\xi$ are associates, and secondly, letting $\cong$ denote the relation of being associates on $R^\times$ and writing $q$ for the quotient mapping $R^\times \rightarrow R^\times/\!\cong,$ we have that the image of $\xi$ under $q$ freely generates $R^\times/\!\cong$.
Suppose an integral domain has a factorization basis. Is it necessarily UFD?
 A: I think freely generating $R^\times/\cong$ is what you're looking for.
The requirement that the image of $\xi$ freely generates the monoid $R^\times/\cong$ implies that the elements of $\xi$ are irreducible. (If an element of $\xi$ were reducible, factorization of its factors would lead to a nontrivial relation among the generators.)
Another observation is that all irreducibles should already lie in $\xi$. I don't think units would be permitted since the identity is already uniquely generated as the empty product of generators, so adding any other unit would be adding an associate.
Then the fact that the image of $\xi$ freely generates $R^\times/\cong$ means that every element of $R^\times$ has a factorization, an the freely part guarantees it is unique.

So it seems to me that the answer to the question is yes, since this just seems to rephrase the definition of a UFD. But I'm prepared to be told I overlooked a facet of the problem.
If there is indeed no problem with my reasoning, it seems like the form of $\xi$ is pretty predictable. It would always consist of a full set of representatives of the equivalence classes.
