Proof that prime elements are irreducible in an integral domain

Proof

Let $R$ be an integral domain and let $p \in R$ be a prime element.

1. By the definition of a prime element, $p \neq 0$ and $p$ is not a unit. These are the first two conditions for $p$ to be irreducible.
2. Let $a, b \in R$ with $p = a b$. Then $p$ divides $a b$, because $a b = p \cdot 1$.
3. Since $p$ is prime, $p \mid a$ or $p \mid b$. The roles of $a$ and $b$ are symmetric, so we may assume $p \mid a$.
4. Write $a = p c$ with $c \in R$. Then $p \cdot 1 = p = a b = p (c b)$.
5. Since $p \neq 0$, cancellation gives $c b = 1$. Since $R$ is commutative, also $b c = 1$, so $b$ is a unit.
6. Thus in every factorization $p = a b$, one factor is a unit. With step 1, $p$ is irreducible.