Theorem. In a commutative ring with non-zero unity, any maximal ideal is a prime ideal.

Proof. Let $\mathfrak{m}$ be a maximal ideal of such a ring $R$ and let the ring product $rs$ belong to $\mathfrak{m}$ but e.g. $r \notin \mathfrak{m}$ . The maximality of $\mathfrak{m}$ implies that $\mathfrak{m}\!+\!(r) = R = (1)$ . Thus there exists an element $m \in \mathfrak{m}$ and an element $x \in R$ such that $m\!+\!xr = 1$ . Now $m$ and $rs$ belong to $\mathfrak{m}$ , whence $$s = 1s = (m\!+\!xr)s = sm\!+\!x(rs) \in \mathfrak{m}.$$ So we can say that along with $rs$ , at least one of its factors belongs to $\mathfrak{m}$ , and therefore $\mathfrak{m}$ is a prime ideal of $R$ .