LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-07-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Cancellation characterises domains: in a commutative ring with $1 \neq 0$ , the implication $a b = a c$ and $a \neq 0$ imply $b = c$ holds if and only if the ring has no zero divisors

Statement

Let $R$ be a commutative ring (Commutative ring) with $1 \neq 0$ . Consider the two conditions

(C) for all $a, b, c \in R$ : if $a b = a c$ and $a \neq 0$ then $b = c$ ;
(Z) $R$ has no zero divisors (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors).

Then (C) holds if and only if (Z) holds; that is, (C) holds exactly when $R$ is an integral domain.

Facts & Assumptions

Given: A commutative ring $R$ with zero $0$ and identity $1$ , and $1 \neq 0$ (Commutative ring, Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L1]

$(R, +, 0)$ is an abelian group and multiplication is commutative and distributes over addition (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Commutative ring, Group and abelian group).

[L2]

$a (b - c) = a b - a c$ and $a \cdot 0 = 0 = 0 \cdot a$ for all $a, b, c \in R$ (In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ ).

[L3]

Cancellation in the additive group: $x + z = y + z$ implies $x = y$ (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution, Group and abelian group).

[L4]

$a$ is a zero divisor when $a \neq 0$ and $a b = 0$ for some $b \neq 0$ ; $R$ has no zero divisors exactly when $a b = 0$ implies $a = 0$ or $b = 0$ ; and an integral domain is exactly a commutative ring with $1 \neq 0$ and no zero divisors (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors).

Proof

technique · direct

1.1

Assume (Z), and let $a b = a c$ with $a \neq 0$ . Then $a (b - c) = a b - a c = 0$ .

L2given

1.2

Assume (C), and let $a b = 0$ with $a \neq 0$ . Then $a b = 0 = a \cdot 0$ .

L2given

2.1

From step 1.1, (Z) gives $a = 0$ or $b - c = 0$ ; since $a \neq 0$ , we get $b + (- c) = 0$ . As also $c + (- c) = 0$ , cancelling $- c$ gives $b = c$ . So (Z) implies (C).

step 1.1L1L3L4

2.2

From step 1.2, (C) applied with $a$ , $b$ and $0$ gives $b = 0$ . So whenever $a b = 0$ and $a \neq 0$ we have $b = 0$ , which says exactly that $a b = 0$ implies $a = 0$ or $b = 0$ ; hence (Z). So (C) implies (Z).

step 1.2L4

3.1

By steps 2.1 and 2.2 the two conditions are equivalent, and (Z) together with commutativity and $1 \neq 0$ is the definition of an integral domain.

step 2.1step 2.2L4∎

Remarks

The hypothesis $1 \neq 0$ is used nowhere in the equivalence itself. It is carried in the statement only so that "(C) holds exactly when $R$ is an integral domain" is literally true, since (D1) of Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors is part of being a domain. In the one-element ring both (C) and (Z) hold vacuously and the ring is still not a domain.
Cancellation is by $a$ , not by $0$ . The clause $a \neq 0$ cannot be dropped: $0 \cdot b = 0 \cdot c$ holds for all $b$ and $c$ in every ring (In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ ), so cancelling $0$ would collapse the ring.
Multiplicative cancellation does not make the nonzero elements a group. It makes $(R ∖ {0}, \cdot, 1)$ a cancellative commutative monoid, and $Z$ shows that is strictly weaker than being a group: $2$ cancels and is not invertible. The rings where the nonzero elements do form a group are the fields (Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree).

Depends on

Used by

Every prime element of an integral domain is irreducible Lemma

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Integral domain (Wikipedia) (standard reference, not scraped)
Cancellation property (Wikipedia) (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §16.4: Integral Domains and Fields (standard reference, not scraped)

Cancellation characterises domains: in a commutative ring with 1≠01=0, the implication ab=acab=ac and a≠0a=0 imply b=cb=c holds if and only if the ring has no zero divisors