Show that \(I\normal R\) is maximal iff \(R/I\) is a field.
Solution
Assume \(R\) is a commutative ring with identity and \(I\subsetneq R\) is an ideal. By the correspondence theorem, ideals of \(R/I\) are in inclusion-preserving bijection with ideals \(J\) of \(R\) satisfying \[ I\subseteq J\subseteq R, \] via \(J\mapsto J/I\).
If \(I\) is maximal, the only such ideals are \(I\) and \(R\). Hence the only ideals of \(R/I\) are \(0\) and \(R/I\). A nonzero commutative ring with identity having no nontrivial ideals is a field, so \(R/I\) is a field.
Conversely, if \(R/I\) is a field, its only ideals are \(0\) and \(R/I\). By correspondence, the only ideals of \(R\) containing \(I\) are \(I\) and \(R\). Thus \(I\) is maximal.
Therefore \[ \boxed{I\text{ maximal}\iff R/I\text{ is a field}.} \]