Are there rings where factorization is unique, but does not necessarily exist?

Question

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.

Answer 1

Let $k = {\Bbb F}_2$ and let $M$ be the multiplicative monoid defined on $\{ x \in \Bbb R \mid x \geqslant 1\}$. I claim that the monoid ring $k[M]$ contains no irreducible elements. The elements of $k[M]$ can be written as formal sums
$$ \sum_{m \in M} c_m m \quad \text{where $c_m \in k$ and $c_m = 0$ for almost all $m \in M$} $$ First, $1$ is the unique unit of $R$. Next, since we are in characteristic $2$ and each $c_m$ is idempotent, one has $$ \sum_{m \in M} c_m m = \Bigl(\sum_{m \in M} c_m\sqrt{m} \Bigr)^2 $$ and thus no $\sum_{m \in M} c_m m$ is irreducible.

Since there is no irreducible element, $1$ is the unique element to have a factorization, namely the trivial, but also unique, factorisation $ 1 = \prod_{i \in \emptyset} x_i $. Thus $k[M]$ answers your question.

Answer 2

I'm not familiar with the content, but this paper suggests yes:

Coykendall, Jim, and Muhammad Zafrullah. "AP-domains and unique factorization." Journal of Pure and Applied Algebra 189.1-3 (2004): 27-35.

It appears to be the main subject of the article. From the introduction:

The standard definitions of “factorization domains” (e.g., UFDs, HFDs, BFDs, etc.) always include the assumption that the domain is atomic (i.e., every nonzero, nonunit of the domain can be written as a product of irreducible elements or atoms). It is natural (and perhaps imperative) to consider the implications to the theory when this assumption is dropped. For example, one might declare more generally that a domain, R, is an unrestricted unique factorization domain (U-UFD) if every element that can be factored uniquely into irreducible elements has unique factorization.

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:

Def 2.1: Let R be an integral domain,

(1) We say that R is an U-UFD if every element that can be factored into irreducibles has a unique (up to units) factorization into irreducibles.