The nilradical is the intersection of all prime ideals, i.e. \[ \nilrad{R} = \Intersect_{\mathfrak{p} \in \spec(R)} \mathfrak{p} \]
Solution
\(\nilrad{R} \subseteq \intersect \mathfrak{p}\):
\(x \in \nilrad{R} \implies x^n = 0 \in \mathfrak p\), and induction on \(n\) using that \(\mfp\) is prime gives \(x\in \mathfrak{p}\).
\(R\sm \nilrad{R} \subseteq \union_{\mfp} (R\sm \mathfrak{p})\):
Fix \(a\) non-nilpotent and define \(S = \theset{I\normal R \suchthat a^n\not\in I \text{ for any } n\geq 1}\).
Then \(\gens{0}\in S\), and the union of a chain in \(S\) is again in \(S\), so Zorn’s lemma gives an ideal \(\mfp\) maximal within \(S\).
Such a \(\mfp\) is prime: if \(a', b'\not\in \mfp\) then \(\mfp + \gens{a'}\) and \(\mfp + \gens{b'}\) strictly contain \(\mfp\), hence leave \(S\), so some \(a^m \in \mfp + \gens{a'}\) and \(a^n \in \mfp + \gens{b'}\); then \(a^{m+n} \in \mfp + \gens{a'b'}\), which forces \(a'b'\not\in\mfp\).
Note \(\mfp\) need not be a maximal ideal, only maximal among the ideals avoiding all powers of \(a\). Since \(\mfp\in S\) we have \(a\not\in \mfp\), so \(a\) escapes the intersection.