Show that if \(R\neq 0\) is a ring in which every non-unit is nilpotent then \(R\) is local.
Solution
Assume \(R\) is a nonzero ring with identity and every nonunit is nilpotent.
Let \(x\in R\). If \(x\) is a unit there is nothing to prove. If \(x\) is not a unit, then \(x^m=0\) for some \(m\ge1\), so \[ (1-x)(1+x+\cdots+x^{m-1})=1=(1+x+\cdots+x^{m-1})(1-x). \] Hence \(1-x\) is a unit. Thus for every \(x\in R\), either \(x\) or \(1-x\) is a unit.
Maximal left ideals exist by Zorn’s lemma. Suppose \(M\) and \(N\) are two distinct maximal left ideals. Choose \(a\in M\setminus N\). Since \(N\) is maximal and \(a\notin N\), \[ N+Ra=R, \] so \[ 1=b+ra \] for some \(b\in N\) and \(r\in R\). Put \(y=ra\). Then \(y\in M\) and \(1-y=b\in N\). Neither \(y\) nor \(1-y\) can be a unit, because a proper left ideal contains no units. This contradicts the preceding paragraph.
Therefore \(R\) has a unique maximal left ideal, so \(R\) is local.