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!)

  • $\begingroup$ maybe there is a misprint. it should read "if $(a)=(b) \ne 0$ and $a = bc$ then $c$ is a unit. $\endgroup$ – David Holden Jan 7 '19 at 2:09
  • $\begingroup$ @DavidHolden: Thank you for the catch. $\endgroup$ – Alex Ortiz Jan 7 '19 at 2:10
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    $\begingroup$ You may find helpful the papers I cite here which discuss generalizations of "associate" and related notions to non-domains. $\endgroup$ – Bill Dubuque Jan 7 '19 at 2:12
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    $\begingroup$ You can find definitions of "harmless" zero-divisors here and here $\endgroup$ – Bill Dubuque Jan 7 '19 at 2:15
  • $\begingroup$ A zero-divisor $z$ is a harmless zero-divisor if $1-z$ is a unit. $\endgroup$ – lhf Jan 7 '19 at 10:34

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:

Let $R$ be a commutative ring with unity and let us denote by $Z(R)$ the set of all zero divisors in $R$.

We say that $r\in R$ is a harmless zero divisor if $r\in Z(R)$ and there exists a unit $u$ such that $r = 1-u$. A ring is said to be a ring with only harmless zero divisors if every zero divisor in $R$ is harmless.

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:

Let $R$ be a commutative ring with unity, and let $Z(R)$ denote the set of zero-divisors of $R$. We say that $R$ is a ring with harmless zero-divisors, if $Z(R)\subset 1-U(R) = \{1-u \mid u\ \text{a unit in $R$}\}$. (These rings are called présimplifiable by Bouvier (cf. [2]).)


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