Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors

Let $R$ be a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides). An element $a \in R$ is a zero divisor when

$a \ne 0 \quad\text{and}\quad ab = 0 \ \text{ or }\ ba = 0 \ \text{ for some } b \in R \text{ with } b \ne 0 .$

The ring $R$ has no zero divisors when no element of $R$ is a zero divisor; equivalently, when $ab = 0$ implies $a = 0$ or $b = 0$, for all $a, b \in R$.

An integral domain, or simply a domain, is a commutative ring (Commutative ring) $R$ such that

• (D1) $1 \ne 0$ in $R$;
• (D2) $R$ has no zero divisors.

In a commutative ring the two clauses $ab = 0$ and $ba = 0$ of the definition of a zero divisor say the same thing, so there the notion is one-sided without ambiguity.

Two conventions fixed here, and kept. First, $0$ is not a zero divisor, because the clause $a \ne 0$ is part of the definition. This matters: $0 \cdot b = 0$ holds for every $b$ in every ring (In any ring $0 \cdot a = a \cdot 0 = 0$, $(-a)b = a(-b) = -(ab)$, $(-a)(-b) = ab$, $(-1)a = -a$ and $a(b - c) = ab - ac$), so without that clause $0$ would be a zero divisor in every ring with more than one element and the phrase "no zero divisors" would name the empty class. Second, (D1) is a hypothesis and not a consequence of (D2). The one-element ring, in which $1 = 0$, has no zero divisors at all — vacuously, since it has no nonzero element — so (D2) alone would admit it. It is excluded by (D1), and by nothing else; the companion page records that ring explicitly.

• Where the definition is used. Cancellation is what a domain really buys: in a commutative ring with $1 \ne 0$, the absence of zero divisors is equivalent to the cancellation law $ab = ac$ and $a \ne 0$ imply $b = c$, which is Cancellation characterises domains: in a commutative ring with $1 \ne 0$, the implication $ab = ac$ and $a \ne 0$ imply $b = c$ holds if and only if the ring has no zero divisors.

• Commutativity is part of the definition of a domain, and is not part of the definition of a zero divisor. A non-commutative ring may perfectly well have no zero divisors — the quaternions ($\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1} = \bar q / N(q)$ for $q \ne 0$, while $ij = k$ and $ji = -k$) are such a ring, since every nonzero element there is invertible — and it is still not called an integral domain here.

• Every field has no zero divisors (A field has no zero divisors: $ab = 0 \Rightarrow a = 0$ or $b = 0$), so every field is an integral domain once it is known to be a commutative ring; that is part of Every field is a commutative ring with $1 \ne 0$; it is an integral domain, and it is a commutative division ring. The converse fails, and $\mathbb{Z}$ is the standard witness, recorded on the companion page.