Proof that in a UFD irreducible elements are prime

Proof

Let $R$ be a unique factorisation domain and let $p \in R$. In particular $R$ is an integral domain.

1. If $p$ is prime, then $p$ is irreducible, because prime elements are irreducible in an integral domain.
2. For the converse, let $p$ be irreducible. Then $p \neq 0$ and $p$ is not a unit. It remains to show the divisibility condition.
3. Let $a, b \in R$ with $p$ dividing $a b$, say $a b = p c$ with $c \in R$.
4. If $a = 0$, then $a = p \cdot 0$ and $p \mid a$. In the same way, $b = 0$ gives $p \mid b$. So we may assume $a \neq 0$ and $b \neq 0$.
5. If $a$ is a unit, then $b = a^{-1} a b = p (a^{-1} c)$, so $p \mid b$. In the same way, if $b$ is a unit, then $p \mid a$. So we may assume that $a$ and $b$ are not units.
6. By unique factorisation, $a = a_1 \cdots a_k$ and $b = b_1 \cdots b_l$ with all $a_i, b_j$ irreducible and $k, l \ge 1$.
7. Since $R$ is an integral domain, $a b \neq 0$. So $p c \neq 0$ and $c \neq 0$.
8. Suppose that $c$ is a unit. The element $p c$ is not zero, and it is not a unit, since otherwise $p = (p c) c^{-1}$ would be a unit.
9. If $p c = x y$ with $x, y \in R$, then $p = x (y c^{-1})$. So $x$ is a unit, or $y c^{-1}$ is a unit and then $y$ is a unit. Hence $p c$ is irreducible.
10. Then $p c = a_1 \cdots a_k b_1 \cdots b_l$ are two factorisations into irreducibles, of lengths $1$ and $k + l \ge 2$. This contradicts uniqueness, so $c$ is not a unit.
11. By unique factorisation, $c = c_1 \cdots c_r$ with all $c_s$ irreducible and $r \ge 1$. Then $p c_1 \cdots c_r = a_1 \cdots a_k b_1 \cdots b_l$.
12. Both sides are products of irreducibles. By uniqueness, the factor $p$ on the left is an associate of some factor on the right.
13. Suppose that $p$ is an associate of $a_i$, so $a_i = u p$ with $u$ a unit. Then $a = p \cdot u \prod_{s \neq i} a_s$, so $p \mid a$.
14. In the same way, if $p$ is an associate of some $b_j$, then $p \mid b$. So $p \mid a$ or $p \mid b$, and $p$ is prime.