Radical of an ideal

Let \(R\) be a commutative ring and \(I\trianglelefteq R\). The radical of \(I\) is \[ \sqrt I =\{r\in R:r^n\in I\text{ for some }n\ge1\}. \] It is an ideal containing \(I\).

The nilradical is the radical of the zero ideal: \[ \operatorname{Nil}(R)=\sqrt{(0)} =\{r\in R:r^n=0\text{ for some }n\ge1\}. \] Thus its elements are exactly the nilpotent elements of \(R\).

Equivalently, \[ \operatorname{Nil}(R) =\bigcap_{\mathfrak p\in\operatorname{Spec}R}\mathfrak p. \] Hence \(R\) is reduced exactly when \[ \operatorname{Nil}(R)=0. \] More generally, \[ \sqrt I =\bigcap_{\mathfrak p\supseteq I}\mathfrak p, \] where the intersection ranges over prime ideals containing \(I\).