# Rings with 'non-harmless' zero-divisors

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 $$_6,_6$$ of $$2$$ and $$4$$ are associates according to our definition, and $$_6 = _6\cdot_6$$, yet $$_6$$ is not a unit. However, $$_6 = _6\cdot _6$$ and $$_6$$ 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!)

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

## 1 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:

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. ).)