Let $R$ be a commutative ring. For any ideal $I$ of $R$ , the radical of $I$ , written $\sqrt{I}$ or $\operatorname{Rad}(I)$ , is the set $$ \{a \in R \mid a^n \in I \text{ for some integer } n>0 \ $$

The radical of an ideal $I$ is always an ideal of $R$ .

If $I = \sqrt{I}$ , then $I$ is called a radical ideal.

Every prime ideal is a radical ideal. If $I$ is a radical ideal, the quotient ring $R/I$ is a ring with no nonzero nilpotent elements.

More generally, the radical of an ideal in can be defined over an arbitrary ring. Let $I$ be an ideal of a ring $R$ , the radical of $I$ is the set of $a\in R$ such that every m-system containing $a$ has a non-empty intersection with $I$ : $$\sqrt{I}:=\lbrace a\in R\mid \mbox{if }S\mbox{ is an $m$-system, and }a\in S,\mbox{ then }S\cap I\ne \varnothing\rbrace.$$

Under this definition, we see that $\sqrt{I}$ is again an ideal (two-sided) and it is a subset of $\lbrace a\in R\mid a^n\in I \mbox{ for some integer }n>0\rbrace$ . Furthermore, if $R$ is commutative, the two sets coincide. In other words, this definition of a radical of an ideal is indeed a ``generalization'' of the radical of an ideal in a commutative ring.