In a UFD irreducible elements are prime

Statement

Let $R$ be a unique factorisation domain and let $p \in R$. Then $p$ is irreducible if and only if $p$ is prime.

Remarks

Prime elements are irreducible in every integral domain. The other direction needs unique factorisation. The proof: suppose $p$ is irreducible and $p c = a b$. Factor $a$, $b$ and $c$ into irreducibles. By uniqueness, $p$ is an associate of an irreducible factor of $a$ or of $b$. Conversely, an integral domain in which every nonzero nonunit is a product of irreducibles, and every irreducible is prime, is a unique factorisation domain.

Explanations

L1 human

In the integers, Euclid's lemma says that a prime that divides a product divides a factor. In a unique factorisation domain, the same thing is true for every irreducible element. In $\mathbb{Z}[\sqrt{-5}]$ this fails, and so factorisation is not unique there.