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)\smallsetminus\{0\}$, then $ ab = cp$
with $ c \in D$. We write $ a,\,b,\,c$ as products of irreducibles:
$$\displaystyle a \;=\; p_1\cdots p_l, \quad b \;=\; q_1\cdots q_m,
\quad c \;=\; r_1\cdots r_n.$$ Here, one of those first two products may
be empty, i.e., it may be a unit. We have $$\displaystyle p_1\cdots
p_l\,q_1\cdots q_m \;=\; r_1\cdots r_n\,p\tag{1}$$
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.
It may be too simple, but why $ a \in (p)$ instead of $p_1 \in (p)$?
Is it because $p$ has to be an associate of one of the $ p_i$'s or $ q_j$'s? Let's say $p_2$ is an associate of $p$. So, $p_2=pw$, $w\in R$. Since $a=p_1p_2\cdots p_l$ then $a=p_1pwp_3\cdots p_l$ and $a=p(p_1p_3\cdots p_lw)$, $p_1p_3\cdots p_lw \in R$ so $a$ is divisible by $p$ hence $a\in (p)$?
Best Answer
It's trivial to show that primes are irreducible. So, assume that $a$ is an irreducible in a UFD (Unique Factorization Domain) $R$ and that $a \mid bc$ in $R$. We must show that $a \mid b$ or $a \mid c$. Since $a\mid bc$, there is an element $d$ in $R$ such that $bc=ad$. Now replace $b,c$ and $d$ by their factorizations as a product of irreducibles and use uniqueness.