Jtdylktuy

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:
$$displaystyle a ;=; p_1cdots p_l, quad b ;=; q_1cdots q_m,
quad c ;=; r_1cdots r_n.$$ Here, one of those first two products may
be empty, i.e., it may be a unit. We have $$displaystyle p_1cdots
p_l,q_1cdots q_m ;=; r_1cdots r_n,ptag{1}$$

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, $p_2=pw$, $win R$. Since $a=p_1p_2cdots p_l$ then $a=p_1pwp_3cdots p_l$ and $a=p(p_1p_3cdots p_lw)$, $p_1p_3cdots p_lw in R$ so $a$ is divisible by $p$ hence $ain (p)$?

The proof given above is probably the standard one to show that a factorial domain is an AP-domain. But there is another proof using the following application: for an irreducible $pin D$, let's define $e_pcolon Dsetminus {0}rightarrow Bbb{N}$ given by $amapsto e_p(a)=#$ of times that $p$ or its associates appear in the irreducible factorization of $a$.

We notice that because $D$ is factorial the application given above it's well defined. Moreover, if $ain D^{times}$, then $e_p(a)=0$ for every irreducible $p$, and if $ain Dsetminus{D^{times}_0}$, then $e_p(a)=0$ iff $pnotmid a$. Equivalently, $e_p(a)>0$ iff $pmid a$.

We have the following:

Lemma: Let $D$ be a factorial domain and $a,bin Dsetminus {0}$. Then $$e_p(ab)=e_p(a)+e_p(b)$$ for every irreducible $pin D$.

Proof: As $D$ is factorial, we can write $a=p^{e_p(a)}ldots $ and $b=p^{e_p(b)}ldots $, then $$ab=(p^{e_p(a)}ldots)(p^{e_p(b)}ldots)=p^{e_p(a)+e_p(b)}ldots $$
Hence, $e_p(ab)=e_p(a)+e_p(b)$.

Now we're going to prove that if $D$ is factorial, then $D$ is an AP-domain. Let $p$ be an irreducible element in $D$ and let $a,bin D$ such that $pmid ab$. If $ab=0$, then $a=0$ or $b=0$, so in this case clearly $pmid a$ or $pmid b$. If $abneq 0$, since $pmid ab$ we have $e_p(ab)>0$, so by the above lemma we find $$e_p(ab)=e_p(a)+e_p(b)>0.$$
Therefore we deduce that necessarily $e_p(a)>0$ or $e_p(b)>0$, i.e., $pmid a$ or $pmid b$. Hence, $p$ is prime.

As a remark, this kind of ideas applied to $Bbb{Z}$ can be found in the first pages of the book "A Classical Introduction to Modern Number Theory" by K. Ireland and M. Rosen.

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 $amid bc$, 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.