Let \(R\) be a commutative ring and let \(\operatorname{Nil}(R)\) denote its nilradical. Show that if \(a + \operatorname{Nil}(R)\) is nilpotent in the quotient ring \(R/\operatorname{Nil}(R)\), then \(a \in \operatorname{Nil}(R)\).
Solution
1. There is \(n\ge1\) with \(a^n\in\operatorname{Nil}(R)\).
Multiplication of cosets gives \((a+\operatorname{Nil}(R))^n=a^n+\operatorname{Nil}(R)\) for every \(n\ge1\). Since \(a+\operatorname{Nil}(R)\) is nilpotent, \(a^n+\operatorname{Nil}(R)=\operatorname{Nil}(R)\) for some \(n\ge1\), that is, \(a^n\in\operatorname{Nil}(R)\).
2. \(a^{nm}=0\) for some \(m\ge1\).
By step 1, \(a^n\) is nilpotent, so \((a^n)^m=0\) for some \(m\ge1\), and \(a^{nm}=(a^n)^m\).
3. Q.E.D.
By step 2, \(a\) is nilpotent, so \(a\in\operatorname{Nil}(R)\). Equivalently, \(R/\operatorname{Nil}(R)\) has no nonzero nilpotent elements.