Exercise 6.4

Suppose \(m \geq 2\) and \(\mathbb{Z}_m\) has a zero-divisor. That is, there exist integers \(a, b\) such that \(\overline{a} \neq \overline{0}, \overline{b} \neq \overline{0},\) and \(\overline{ab} = \overline{0},\) or equivalently, neither \(a\) nor \(b\) is a multiple of \(m\), but \(ab\) is. The only way this can happen is if \(m\) is composite i.e. not prime, as in this case there must exist integers \(1 < k, l < m\) with \(kl = m\) such that \(k\) divides \(a\) and \(l\) divides \(b\).

Conversely, suppose \(m \geq 2\) and \(\mathbb{Z}_m\) is an integral domain, i.e. it has no zero-divisors. That is, for any integers \(a, b\) with \(1 < a, b < m,\) we have that \(ab\) is not a multiple of \(m\). The only way this can happen is if \(m\) is prime.

Therefore, \(\mathbb{Z}_m\) is an integral domain if and only if \(m\) is prime.