Show that every proper ideal in a commutative ring with identity is contained in a maximal ideal.
Solution
Let \(I\subsetneq R\) and consider \[ \mathcal F=\{J\trianglelefteq R: I\subseteq J\subsetneq R\}, \] ordered by inclusion. This set is nonempty since \(I\in\mathcal F\).
If \(\mathcal C\subseteq\mathcal F\) is a chain, then \[ J=\bigcup_{L\in\mathcal C}L \] is an ideal: any two of its elements lie in a common member of the chain. It contains \(I\). Moreover \(J\) is proper, since \(1\in J\) would imply \(1\in L\) for some \(L\in\mathcal C\), contradicting \(L\ne R\). Hence every chain has an upper bound in \(\mathcal F\).
By Zorn’s lemma, \(\mathcal F\) has a maximal element \(\mathfrak m\). By definition, \(\mathfrak m\) is a proper ideal containing \(I\) and no larger proper ideal contains it. Therefore \(\mathfrak m\) is maximal.