 Solution

A nilpotent element belongs to every prime ideal: if $a^r=0\in P$, repeated primality forces $a\in P$. Therefore $N\subseteq {\bigcap }_{P}P$.

Conversely, if $a\notin N$, none of its positive powers is zero. The multiplicative subset $S=\{1,a,a^2,\dots \}$ is disjoint from $(0)$. Part (b) supplies a prime ideal disjoint from $S$, hence not containing $a$. Thus $a$ is outside the intersection, proving the nilradical as the intersection of prime ideals identity:In the zero ring there are no prime ideals; the empty intersection is interpreted as the whole ring, which is also its nilradical. For nonzero rings this argument proves existence of a prime ideal whenever a nonnilpotent element is available.