Every maximal ideal is a prime ideal. The converse is true in a principal ideal domain – PID, i.e. every nonzero prime ideal is maximal in a PID, but this is not true in general. Let’s produce a counterexample.

\(R= \mathbb Z[x]\) is a ring. \(R\) is not a PID as can be shown considering the ideal \(I\) generated by the set \(\{2,x\}\). \(I\) cannot be generated by a single element \(p\). If it was, \(p\) would divide \(2\), i.e. \(p=1\) or \(p=2\). We can’t have \(p=1\) as it means \(R = I\) but \(3 \notin I\). We can’t have either \(p=2\) as it implies the contradiction \(x \notin I\). The ideal \(J = (x)\) is a prime ideal as \(R/J \cong \mathbb Z\) is an integral domain. Since \(\mathbb Z\) is not a field, \(J\) is not a maximal ideal.