Let \(R\) be a commutative ring with \(1\). We say \(R\) is a local ring if \(R\) has exactly one maximal ideal, \(M\). Prove that, in a local ring \(R\), any \(r \in R\) is either a unit or an element of the maximal ideal \(M\).
Solution
Let \(r\in R\) with \(r\notin M\).
1. The ideal \((r)\) equals \(R\).
Every proper ideal of a commutative ring with \(1\) is contained in a maximal ideal, by Zorn’s lemma [@DF04], and \(M\) is the only maximal ideal of \(R\). Since \(r\in(r)\) and \(r\notin M\), the ideal \((r)\) is not contained in \(M\), so it is not proper.
2. Q.E.D.
By step 1, \(1=sr\) for some \(s\in R\), so \(r\) is a unit. Hence every element of \(R\) is a unit or lies in \(M\).