Every proper ideal is contained in a maximal ideal.Suggest a slogan
Let \(R\) be a ring and \(I\subsetneq R\) a proper ideal. Then \(I\) is contained in a maximal ideal of \(R\).
Let \(\mathcal S\) be the set of proper ideals of \(R\) containing \(I\), ordered by inclusion; it is nonempty since \(I\in\mathcal S\). If \(\mathcal C\subseteq\mathcal S\) is a nonempty chain, then \(J\coloneqq\bigcup_{J'\in\mathcal C}J'\) is an ideal containing \(I\), and \(1\notin J\) because \(1\) lies in no proper ideal, so \(J\in\mathcal S\) is an upper bound for \(\mathcal C\). By Zorn’s lemma \(\mathcal S\) has a maximal element \(\mathfrak m\). Any ideal strictly containing \(\mathfrak m\) contains \(I\) and is not in \(\mathcal S\), so it equals \(R\); thus \(\mathfrak m\) is a maximal ideal.