Any irreducible element of a factorial ring $D$ is a prime element of $D$ .

Proof. Let $p$ be an arbitrary irreducible element of $D$ and let $ab \in (p)$ . So $ab = cp$ with $c \in D$ . We can write $a,\,b,\,c$ as products of irreducibles: $$a = p_1\cdots p_l, \quad b = q_1\cdots q_m, \quad c = r_1\cdots r_n$$ Accordingly, $$p_1\cdots p_l\,q_1\cdots q_m \,=\, r_1\cdots r_n\,p,$$ and due to the uniqueness of prime factorization, $p$ has to be an associate of one of the $p_i$ 's or $q_j$ 's. It means that either $a \in (p)$ or $b \in (p)$ . Thus, $(p)$ is a prime ideal of $D$ , and its generator must be a prime element.