Every proper ideal extends to a maximal ideal.Suggest a slogan
Let \(R\) be a commutative ring. Every proper ideal \(I\subsetneq R\) is contained in a maximal ideal of \(R\). In particular, applying this to \(I=0\), every nonzero commutative ring has a maximal ideal.
Let \(P\) be the set of proper ideals of \(R\) containing \(I\), ordered by inclusion; it contains \(I\). The union of a nonempty chain in \(P\) is an ideal containing \(I\), and it is proper because it does not contain \(1\), so it is an upper bound in \(P\); the empty chain has the upper bound \(I\). By Zorn's lemma, \(P\) has a maximal element, which is a maximal ideal of \(R\) containing \(I\).
What is Krull's theorem?