irreducible of a UFD is prime

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

Proof.  Let $p$ be an arbitrary irreducible element of $D$.  Thus $p$ is a non-unit.  If $ab\in (p)\setminus \{0\}$,  then  $ab=cp$  with $c\in D$.  We write $a,b,c$ as products of irreducibles:

$$a=p_1\mathrm{\cdots }{p}_l,b=q_1\mathrm{\cdots }{q}_m,c=r_1\mathrm{\cdots }{r}_n$$

Here, one of those first two products may me empty, i.e. it may be a unit.  We have

$p_1\mathrm{\cdots }{p}_l{q}_1\mathrm{\cdots }{q}_m=r_1\mathrm{\cdots }{r}_{n}p.$(1)

Due to the uniqueness of prime factorization, every factor $r_k$ is an associate of certain of the $l+m$ irreducibles on the left hand side of (1).  Accordingly, $p$ 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.