In keeping with ring theory as the attempt to isolate each individual property of and work out how the properties interplay with each other, we define the notion of integral domain to capture the fact that if $a \times b = 0$ then $a = 0$ or $b = 0$ . That is, an integral domain is one which has no "zero divisors": $0$ cannot be nontrivially expressed as a product. (For uninteresting reasons, we also exclude the ring with one element, in which $0 = 1$ , from being an integral domain.)

Examples

$Z$ is an integral domain.

Any field is an integral domain. (The proof is an exercise.)

Show solution

Suppose $a b = 0$ , but $a \neq 0$ . We wish to show that $b = 0$ .

Since we are working in a field, $a$ has an inverse $a^{- 1}$ ; multiply both sides by $a^{- 1}$ to obtain $a^{- 1} a b = 0 \times a^{- 1}$ . Simplifying, we obtain $b = 0$ .

When $p$ is a prime integer, the ring $Z_{p}$ of integers mod $p$ is an integral domain.

When $n$ is a composite integer, the ring $Z_{n}$ is not an integral domain. Indeed, if $n = r \times s$ with $r, s$ positive integers, then $r s = n = 0$ in $Z_{n}$ .

Properties

The reason we care about integral domains is because they are precisely the rings in which we may cancel products: if $a \neq 0$ and $a b = a c$ then $b = c$ .

Proof

Indeed, if $a b = a c$ then $a b - a c = 0$ so $a (b - c) = 0$ , and hence (in an integral domain) $a = 0$ or $b = c$ .

Moreover, if we are not in an integral domain, say $r s = 0$ but $r, s \neq 0$ . Then $r s = r \times 0$ , but $s \neq 0$ , so we can't cancel the $r$ from both sides.

Finite integral domains

If a ring $R$ is both finite and an integral domain, then it is a field. The proof is an exercise.

Show solution

Given $r \in R$ , we wish to find a multiplicative inverse.

Since there are only finitely many elements of the ring, consider $S = {a r : a \in R}$ . This set is a subset of $R$ , because the multiplication of $R$ is closed. Moreover, every element is distinct, because if $a r = b r$ then we can cancel the $r$ (because we are in an integral domain), so $a = b$ .

Since there are $| R |$ -many elements of the subset $S$ (where $| \cdot |$ refers to the cardinality), and since $R$ is finite, $S$ must in fact be $R$ itself.

Therefore in particular $1 \in S$ , so $1 = a r$ for some $a$ .