Let \(R\) be a PID and let \(0\neq x\in R\). Show that \(x\) is irreducible if and only if the ideal \(\gens{x}\) is maximal.
Solution
Let \(R\) be a PID and \(0\neq x\in R\).
1. If \(x\) is irreducible, then \((x)\) is a maximal ideal.
Since \(x\) is irreducible, it is a nonunit, so \((x)\) is proper. Let \(I\) be an ideal with \[ (x)\subseteq I\subsetneq R. \] Because \(R\) is a PID, \(I=(a)\) for some \(a\in R\). The inclusion \((x)\subseteq(a)\) gives \(x=ab\) for some \(b\in R\). Since \(I\) is proper, \(a\) is not a unit. Irreducibility of \(x\) therefore forces \(b\) to be a unit. Hence \[ (x)=(ab)=(a)=I. \] Thus there is no proper ideal strictly between \((x)\) and \(R\), so \((x)\) is maximal.
2. If \((x)\) is maximal, then \(x\) is irreducible.
Since \((x)\) is proper, \(x\) is not a unit. Suppose \[ x=ab. \] Then \[ (x)\subseteq(a)\subseteq R. \] Maximality of \((x)\) implies either \((a)=R\) or \((a)=(x)\). In the first case \(a\) is a unit. In the second case \(a=ux\) for some \(u\in R\), so \[ x=ab=uxb. \] Because \(R\) is a domain and \(x\neq0\), cancellation gives \(ub=1\), so \(b\) is a unit. Thus every factorization of \(x\) has a unit factor, and \(x\) is irreducible.
3. Therefore, for every nonzero element \(x\) of a PID,\[ x\text{ is irreducible}\quad\Longleftrightarrow\quad (x)\text{ is maximal}. \]
Combine 1 and 2.