= Solution
Write $N=\sqrt{(0)}$. Every element of the <extension of an ideal> $NR_{\mathfrak p}$ is a fraction $a/s$ with $a\in N$ and $s\notin\mathfrak p$: a finite sum of such fractions has a common denominator and numerator still in the <ideal> $N$. If $a^n=0$, then $(a/s)^n=0$, so
$$
NR_{\mathfrak p}\subseteq\sqrt{(0)_{R_{\mathfrak p}}}.
$$
Conversely, let $a/s$ be a <nilpotent element> of the <localization at a prime ideal> $R_{\mathfrak p}$. For some $n\geq1$, $a^n/s^n=0$. The <vanishing criterion in a module localization> supplies $t\notin\mathfrak p$ with $ta^n=0$. It follows that
$$
(ta)^n=t^{n-1}(ta^n)=0,
$$
so $ta\in N$. Since $ts\notin\mathfrak p$,
$$
\frac as=\frac{ta}{ts}\in NR_{\mathfrak p}.
$$
This proves that the <nilradical commutes with localization>:
$$
\boxed{\operatorname{Nil}(R_{\mathfrak p})=NR_{\mathfrak p}.}
$$
The multiplication by $t$ is necessary: the numerator $a$ of a <nilpotent element> in a <localization> need not itself be <nilpotent> in $R$.
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