# 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.)

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 \(b=c\).

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.

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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