Theorem. Let $R$ be a commutative ring with non-zero unity 1 and $\mathfrak{p}$ an ideal of $R$ . The quotient ring $R/\mathfrak{p}$ is an integral domain if and only if $\mathfrak{p}$ is a prime ideal.

Proof. $1^{\underline{o}}$ . First, let $\mathfrak{p}$ be a prime ideal of $R$ . Then $R/\mathfrak{p}$ is of course a commutative ring and has the unity $1+\mathfrak{p}$ . If the product $(r+\mathfrak{p})(s+\mathfrak{p})$ of two residue classes vanishes, i.e. equals $\mathfrak{p}$ , then we have $rs+\mathfrak{p}= \mathfrak{p}$ , and therefore $rs$ must belong to $\mathfrak{p}$ . Since $\mathfrak{p}$ is prime, either $r$ or $s$ belongs to $\mathfrak{p}$ , i.e. $r+\mathfrak{p}= \mathfrak{p}$ or $s+\mathfrak{p}= \mathfrak{p}$ . Accordingly, $R/\mathfrak{p}$ has no zero divisors and is an integral domain.

$2^{\underline{o}}$ . Conversely, let $R/\mathfrak{p}$ be an integral domain and let the product $rs$ of two elements of $R$ belong to $\mathfrak{p}$ . It follows that $(r+\mathfrak{p})(s+\mathfrak{p}) = rs+\mathfrak{p} = \mathfrak{p}$ . Since $R/\mathfrak{p}$ has no zero divisors, $r+\mathfrak{p} = \mathfrak{p}$ or $s+\mathfrak{p} =\mathfrak{p} $ . Thus, $r$ or $s$ belongs to $\mathfrak{p}$ , i.e. $\mathfrak{p}$ is a prime ideal.