Let \(R\) be a ring with \(1\ne0\). Show that any element \(x\) having no left inverse is contained in a maximal left ideal. In a commutative ring, this applies to every nonunit.
Solution
If \(x\) has no left inverse, then the principal left ideal \[ Rx \] is proper: otherwise \(1\in Rx\), so \(rx=1\) for some \(r\in R\).
Consider the set of proper left ideals containing \(Rx\), ordered by inclusion. It is nonempty. The union of a chain is again a left ideal containing \(Rx\), and it is proper because if it contained \(1\), then some member of the chain would contain \(1\) and hence equal \(R\). Thus every chain has an upper bound.
By Zorn’s lemma there is a maximal element \(M\) in this poset. Any proper left ideal strictly containing \(M\) would also contain \(Rx\), contradicting maximality. Hence \(M\) is a maximal left ideal and \[ x\in Rx\subseteq M. \]
In a commutative ring, an element is left-invertible exactly when it is a unit, so every nonunit is contained in a maximal ideal.