Let be a commutative ring. The set of all nilpotent elements of (i.e., the set of all for which for some natural ) is called the nilradical of . It is sometimes denoted .
The nilradical is an ideal, for if , then for any , ; and if , then
Proposition. The nilradical of is the intersection of all prime ideals of .
Proof. Let be a prime ideal of , and let be a nilpotent element of . We claim that . Indeed, let be the least positive integer for which . (Such an integer exists, since is nilpotent.) Suppose that . Then ; since is a prime ideal, then or , a contradiction. This proves that To show the converse, it suffices to show that for any non-nilpotent element , there is some prime ideal that does not contain .
So suppose that is an element of that is not nilpotent. Let be the set of ideals of that do not contain any element of the form . Since , is not empty; then by Zorn's Lemma, has a maximal element .
It suffices to show that is a prime ideal. Indeed, suppose otherwise; then there exist elements for which . Then the set of elements for which is evidently an ideal of that properly contains ; it therefore contains , for some integer . By similar reasoning, the set of elements for which is an ideal that properly contains , so this set contains , for some integer . Then , a contradiction.
Therefore is a prime ideal that does not contain . It follows from the generality of this argument that as desired.