Let \(R\) be a commutative ring and let \(\operatorname{nil}(R) = \sqrt{(0)}\) be its nilradical (the ideal of all nilpotent elements in \(R\)). Prove that the quotient ring \(R / \operatorname{nil}(R)\) is reduced (i.e. has no non-zero nilpotent elements).
Solution
Let \(N=\operatorname{nil}(R)\). Suppose \(x+N\in R/N\) is nilpotent. Then for some \(m\ge1\), \[ (x+N)^m=N, \] so \(x^m\in N\). Hence \((x^m)^k=0\) for some \(k\ge1\), and therefore \(x^{mk}=0\). Thus \(x\in N\), so \(x+N=0\) in \(R/N\).
Hence \(R/N\) has no nonzero nilpotent elements and is reduced.