Let $R$ be a commutative ring. An element $x \in R$ is said to be nilpotent if $x^n = 0$ for some positive integer $n$ The set of all nilpotent elements of $R$ is an ideal of $R$ called the nilradical of $R$ and denoted $\operatorname{Nil}(R)$ The nilradical is so named because it is the radical of the zero ideal.

The nilradical of $R$ equals the prime radical of $R$ although proving that the two are equivalent requires the axiom of choice.