# Integral domain

In keeping with ring theory as the attempt to isolate each individual property of $$\mathbb{Z}$$ 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.)

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 $$b = 0$$. <div><div>

• When $$p$$ is a prime integer, the ring $$\mathbb{Z}_p$$ of integers mod $$p$$ is an integral domain.

• When $$n$$ is a 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 $$b=c$$.

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. <div><div>

## Finite integral domains

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

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 $$a$$. <div><div>

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