Let D be a princple ideal domain and let $\displaystyle p \in D$. Prove that <p> is a maximal ideal in D iff p is irreducible.
Proof.
Assume p is irreducible, and let p = ab, then either
(Actually we should also specify that $\displaystyle p$ is not a unit also).
Here is half of the argument:
Say that $\displaystyle \left< p \right>$ is a maximal ideal and $\displaystyle p=ab$. Then $\displaystyle \left< p \right> \subseteq \left< a \right> $. Thus either $\displaystyle \left< p \right> \subset \left< a \right> $ or $\displaystyle \left< p \right> = \left< a \right>$. If the first case then $\displaystyle p$ and $\displaystyle a$ are associate elements so $\displaystyle b$ is a unit. If the second case then $\displaystyle \left< a \right> = D$ (because $\displaystyle \left< p \right>$ is maximal) and thus $\displaystyle a$ is a unit.