Let $A$ be a commutative ring. The set of all nilpotent elements of $A$ (i.e., the set of all $x$ for which $x^n=0$ for some natural $n$) is called the nilradical of $A$. It is sometimes denoted $\mathfrak{N}$.

The nilradical is an ideal, for if $x^n = 0$, then for any $a\in A$, $(ax)^n = 0$; and if $x^n = y^m = 0$, then $$(x+y)^{m+n-1} = \sum_{k=0}^{m+n-1} \binom{n}{k} x^{m+n-1-k}y^{k} = 0 .$$

Proposition. The nilradical $\mathfrak{N}$ of $A$ is the intersection of all prime ideals of $A$.

Proof. Let $\mathfrak{p}$ be a prime ideal of $A$, and let $x$ be a nilpotent element of $A$. We claim that $x\in \mathfrak{p}$. Indeed, let $n$ be the least positive integer for which $x^n \in \mathfrak{p}$. (Such an integer exists, since $x$ is nilpotent.) Suppose that $n>1$. Then $x \cdot x^{n-1} \in \mathfrak{p}$; since $\mathfrak{p}$ is a prime ideal, then $x \in \mathfrak{p}$ or $x^{n-1} \in \mathfrak{p}$, a contradiction. This proves that $$\mathfrak{N} \subset \bigcap_{\mathfrak{p} \text{ prime}} \mathfrak{p} .$$ To show the converse, it suffices to show that for any non-nilpotent element $a$, there is some prime ideal that does not contain $a$.

So suppose that $a$ is an element of $A$ that is not nilpotent. Let $S$ be the set of ideals of $A$ that do not contain any element of the form $a^n$. Since $(0) \in S$, $S$ is not empty; then by Zorn's Lemma, $S$ has a maximal element $\mathfrak{m}$.

It suffices to show that $\mathfrak{m}$ is a prime ideal. Indeed, suppose otherwise; then there exist elements $x,y \notin \mathfrak{m}$ for which $xy \in \mathfrak{m}$. Then the set of elements $z$ for which $xz \in \mathfrak{m}$ is evidently an ideal of $A$ that properly contains $\mathfrak{m}$; it therefore contains $a^n$, for some integer $n$. By similar reasoning, the set of elements $z$ for which $a^n z \in \mathfrak{m}$ is an ideal that properly contains $\mathfrak{m}$, so this set contains $a^m$, for some integer $m$. Then $a^{n+m} \in \mathfrak{m}$, a contradiction.

Therefore $\mathfrak{m}$ is a prime ideal that does not contain $a$. It follows from the generality of this argument that $$\bigcap_{\mathfrak{p} \text{ prime}} \mathfrak{p} \subset \mathfrak{N} ,$$ as desired. $\blacksquare$