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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.

We first establish the nilradical as the intersection of prime ideals:Every nilpotent element belongs to every prime ideal. In the other direction, if $a$ is not nilpotent, the multiplicative subset $T=\{1,a,a^2,\dots \}$ avoids zero. By the Zorn lemma, there is an ideal $J$ maximal among those disjoint from $T$. It is a prime ideal: if $uv\in J$ and neither $u$ nor $v$ belongs to $J$, then $J+(u)$ and $J+(v)$ each meet $T$. Multiplying elements in these two intersections gives an element of $T$ lying in $J$, a contradiction. Since $a\notin J$, this proves the identity. The identity also holds for the zero ring, with the intersection over an empty family interpreted as the whole ring.

A Zariski-closed set $V_R(I)$ contains the image of $f$ exactly whenThe last equality follows because $\phi (r)$ is a nilpotent element exactly when $r^n\in \ker \phi$ for some $n\ge 1$. Consequently the smallest Zariski-closed set containing the image, namely its closure, isThis is all of $\mathrm{Spec}R$ precisely when the kernel of a ring homomorphism is contained in every prime ideal of $R$. Hence