Prime elements are irreducible in an integral domain

Statement

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

Remarks

The proof: let $p = a b$. Then $p \mid a b$, so $p \mid a$, say. Write $a = p c$. Then $p = p c b$, and cancellation gives $c b = 1$, so $b$ is a unit. The converse is false in general: in $\mathbb{Z}[\sqrt{-5}]$, $2$ is irreducible but not prime. In a unique factorization domain, irreducible elements are prime.

Explanations

L1 human

Suppose that a prime $p$ factors as $a b$. Then $p$ divides one factor, so that factor is at least as large as $p$ for divisibility. The other factor must then be a unit.