The following excerpt is from pp. 246–247 of Paolo Aluffi's Algebra: Chapter 0:
1.2. Prime and irreducible elements. Let $R$ be a (commutative) ring [with $1$], and let $a,b\in R$. We say that $a$ divides $b$, or that $a$ is a divisor of $b$, or that $b$ is a multiple of $a$, if $b\in(a)$, that is
$$
(\exists c\in R), \quad b = ac.
$$
We use the notation $a \mid b$.Two elements $a,b$ are associates if $(a) = (b)$, that is, if $a\mid b$ and $b\mid a$.
Lemma 1.5. Let $a,b$ be nonzero elements of an integral domain $R$. Then $a$ and $b$ are associates if and only if $a = ub$, for $u$ a unit in $R$.
[Proof omitted.]
Incidentally, here the reader sees why it is convenient to restrict our attention to integral domains. This argument really shows that if $(a) = (b) \ne (0)$ in an integral domain, and $b = ca$, then $c$ is necessarily a unit. Away from the comfortable environment of integral domains, even such harmless-looking statements may fail: in $\Bbb Z/6\Bbb Z$, the classes $[2]_6,[4]_6$ of $2$ and $4$ are associates according to our definition, and $[4]_6 = [2]_6\cdot[2]_6$, yet $[2]_6$ is not a unit. However, $[4]_6 = [5]_6\cdot [2]_6$ and $[5]_6$ is a unit, so this is not a counterexample to Lemma 1.5. In fact, Lemma 1.5 may fail over rings with 'non-harmless' zero-divisors (yes, there is such a notion) [emphasis added].
Since at this point, Aluffi does not say what such rings are called, I was hoping someone might know what type of rings Aluffi is referring to. (And hopefully provide a little context as to why they are interesting!)
Best Answer
Just to post an answer, citing from the papers in Bill Dubuque's helpful comment.
From Ornella Greco's paper "Equivalence of three different definitions of irreducible element," and I am paraphrasing a bit here:
From Christopher Frie's and Sophie Frisch's paper "Non-unique factorization of polynomials over residue class rings of the integers," we have the same definition, stated slightly differently, and again, only paraphrasing slightly: