The nilradical is the intersection of all prime ideals

This is [@AM18, Proposition 1.8]. Let \(R\) be a commutative ring and let \(P\) be the intersection of all prime ideals of \(R\).

\(\nilrad{R} \subseteq P\): Suppose \(r\in \nilrad{R}\) so \(r^n = 0\) and let \(\mfp \in \spec R\). Then \(r^n = 0 \in \mfp\), and induction on \(n\) using that \(\mfp\) is prime gives \(r\in \mfp\).

\(\nilrad{R}^c \subseteq P^c\): Fix \(f\) non-nilpotent; we want to produce one prime ideal that does not contain \(f\). Set \(S\) to be the collection of ideals \(I\) such that \(f^n \not\in I\) for every \(n\geq 1\). Apply Zorn’s lemma: \(S\neq \emptyset\) since \(\gens{0}\in S\), because \(f\) is not nilpotent. Ordering \(S\) by inclusion, a union of a chain in \(S\) is again in \(S\), so \(S\) contains a maximal element \(\mfp\), which we claim is prime. If \(a,b \in \mfp^c\) then \(\mfp + \gens{ a }\) and \(\mfp + \gens{b} \supset \mfp\) strictly, and by maximality they aren’t in \(S\). So there exist \(m,n\) such that \(f^m\in \mfp + \gens{ a }\) and \(f^n \in \mfp + \gens{b}\). Then \(f^{m+n} \in \mfp + \gens{ab}\), so \(\mfp + \gens{ab}\) is not in \(S\), which forces \(\mfp + \gens{ab} \supsetneq \mfp\). Thus \(ab\not \in \mfp\), and \(\mfp\) is prime. Since \(\mfp\in S\) we have \(f\not\in \mfp\), so \(\mfp\) is a prime ideal missing \(f\) and therefore \(f\not \in P\).