Show that \(R\) is a local ring iff for every \(x\in R\), either \(x\) or \(1-x\) is a unit.
Solution
Assume \(R\) is a commutative ring with identity.
Suppose first that \(R\) is local with unique maximal ideal \(\mathfrak m\). The nonunits of \(R\) are exactly the elements of \(\mathfrak m\). If both \(x\) and \(1-x\) were nonunits, then both would lie in \(\mathfrak m\), hence \[ 1=x+(1-x)\in\mathfrak m, \] a contradiction. Thus for every \(x\), at least one of \(x\) or \(1-x\) is a unit.
Conversely, assume that for every \(x\in R\), either \(x\) or \(1-x\) is a unit. Let \(\mathfrak m,\mathfrak n\) be maximal ideals. If they were distinct, choose \(a\in\mathfrak m\setminus\mathfrak n\). Since \(\mathfrak n\) is maximal and \(a\notin\mathfrak n\), we have \(\mathfrak n+(a)=R\), so \[ 1=b+ra \] for some \(b\in\mathfrak n\), \(r\in R\). Put \(x=ra\). Then \(x\in\mathfrak m\), so \(x\) is not a unit, while \(1-x=b\in\mathfrak n\), so \(1-x\) is not a unit. This contradicts the hypothesis.
Hence \(R\) has a unique maximal ideal, so \(R\) is local.