In a PID a nonzero element is irreducible iff it generates a maximal ideal

Statement

Let $R$ be a principal ideal domain and let $p \in R$ with $p \neq 0$. Then $p$ is irreducible if and only if the principal ideal $(p)$ is a maximal ideal.

Remarks

The proof: an ideal $I$ with $(p) \subseteq I$ is $(a)$ with $a \mid p$. If $p$ is irreducible, then $a$ is a unit or an associate of $p$, so $I = R$ or $I = (p)$. A consequence is that $R/(p)$ is a field for each irreducible $p$. For example, $\mathbb{Q}[x]/(x^2 - 2)$ is a field. In a principal ideal domain, every nonzero prime ideal is maximal.

Explanations

L1 human

In a principal ideal domain, ideals that contain $(p)$ come from the divisors of $p$. An irreducible $p$ has only trivial divisors. So no ideal fits strictly between $(p)$ and the whole ring.