[summary: An integral domain is a ring in which the only way to make $0$ as a product is to multiply $0$ by something. For instance, in an integral domain like [48l $\mathbb{Z}$], $2 \times 3$ is not equal to $0$ because neither $2$ nor $3$ is.]

[summary(Technical): An integral domain is a ring in which $ab=0$ implies $a=0$ or $b=0$. (We exclude the ring with one element: that is conventionally not considered an integral domain.)]

In keeping with ring theory as the attempt to isolate each individual property of [48l $\mathbb{Z}$] 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

• $\mathbb{Z}$ is an integral domain.
• Any field is an integral domain. (The proof is an exercise.)

%%hidden(Show solution): Suppose $ab = 0$, but $a \not = 0$. 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 $\mathbb{Z}_p$ of integers mod $p$ is an integral domain.
• When $n$ is a [composite_number composite] integer, the ring $\mathbb{Z}_n$ is not an integral domain. Indeed, if $n = r \times s$ with $r, s$ positive integers, then $r s = n = 0$ in $\mathbb{Z}_n$.

Properties

The reason we care about integral domains is because they are precisely the rings in which we may cancel products: if $a \not = 0$ and $ab = ac$ then $b=c$. %%hidden(Proof): Indeed, if $ab = ac$ then $ab-ac = 0$ 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 $r, s \not = 0$. Then $rs = r \times 0$, but $s \not = 0$, 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. %%hidden(Show solution): Given $r \in R$, we wish to find a multiplicative inverse.

Since there are only finitely many elements of the ring, consider $S = \{ ar : a \in R\}$. This set is a subset of $R$, because the multiplication of $R$ is closed. Moreover, every element is distinct, because if $ar = br$ 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 $1 = ar$ for some $a$. %%