Nilradical is the intersection of the prime ideals

Remarks

One inclusion is direct: if $r^n = 0$, then $r^n$ lies in every prime ideal $P$, so $r \in P$. For the other inclusion, let $f$ be an element that is not nilpotent. Then the set $S = \set{1, f, f^2, \ldots}$ does not contain $0$, and Zorn's lemma gives an ideal $P$ that is maximal among the ideals disjoint from $S$. Such an ideal is prime, and $f \notin P$. Example: in $\Z/12\Z$ the prime ideals are $2\Z/12\Z$ and $3\Z/12\Z$, and their intersection $6\Z/12\Z = \set{0, 6}$ is the set of nilpotent elements. If $R$ is the zero ring, it has no prime ideals, the empty intersection is $R$, and every element of $R$ is nilpotent, so the statement still holds. Applied to $R/I$, the statement shows that the radical of an ideal $I$ is the intersection of the prime ideals that contain $I$.

Going deeper

For a student

The theorem is the starting point of the link between ideals and closed subsets of the spectrum. It gives $\sqrt{I} = \bigcap_{P \supseteq I} P$, so two ideals have the same zero locus in $\operatorname{Spec} R$ exactly when they have the same radical. The proof is a standard use of Zorn's lemma, in the same style as the existence of maximal ideals. Over a polynomial ring over an algebraically closed field, Hilbert's Nullstellensatz sharpens it: maximal ideals alone already suffice. Compare the Jacobson radical, which is the intersection of the maximal ideals only and contains the nilradical.