Show that if \(R\) is a nonzero ring where every element is either a unit or nilpotent, then \(R\) is local.
Solution
Let \(N(R)\) denote the nilradical, i.e. the set of nilpotent elements of \(R\).
1. No nilpotent element of the nonzero ring \(R\) is a unit.
If \(a^m=0\) and \(a\) were a unit, then multiplying by \(a^{-m}\) would give \(1=0\), contradicting that \(R\) is nonzero.
2. The set of nonunits of \(R\) is exactly \(N(R)\).
By hypothesis every element of \(R\) is either a unit or nilpotent. By 1 these possibilities are disjoint. Hence an element is a nonunit exactly when it is nilpotent.
3. The set of nonunits is a proper ideal.
Because \(R\) is commutative, its nilpotent elements form the nilradical \(N(R)=\sqrt{(0)}\), which is an ideal. It is proper because \(1\) is not nilpotent in a nonzero ring. By 2, this ideal is exactly the set of nonunits.
4. The ring \(R\) has a unique maximal ideal.
2. Thus \(N(R)=\mathfrak m\).
If \(\mathfrak m'\) is any maximal ideal, then every element of \(\mathfrak m'\) is a nonunit, hence lies in \(N(R)=\mathfrak m\). Maximality gives \(\mathfrak m'=\mathfrak m\). Thus the maximal ideal is unique.
5. Therefore \(R\) is local.
A nonzero commutative ring is local precisely when it has a unique maximal ideal. By 4, that ideal is \(N(R)\).