Suppose that \(R\) is a commutative ring. Show that an element \(r\in R\) is not invertible iff it is contained in a maximal ideal.
Let \(R\) be a commutative ring with \(1\). Show that \(r\in R\) is noninvertible if and only if \(r\) lies in a maximal ideal.
Solution
Suppose first that \(r\) is noninvertible. Then the principal ideal \((r)\) is proper: if \((r)=R\), then \(1=sr\) for some \(s\in R\), making \(r\) a unit. By Zorn’s lemma, every proper ideal is contained in a maximal ideal, so there exists a maximal ideal \(\mathfrak m\) with \[ r\in(r)\subseteq\mathfrak m. \]
Conversely, suppose \(r\in\mathfrak m\) for some maximal ideal \(\mathfrak m\). If \(r\) were invertible, then \[ 1=r^{-1}r\in\mathfrak m, \] which would imply \(\mathfrak m=R\), contradicting maximality.
Thus \(r\) is noninvertible exactly when it belongs to some maximal ideal.