In a PID a nonzero element is irreducible iff it generates 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.