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Write $N=\sqrt{(0)}$. Every element of the extension of an ideal $N{R}_{\mathfrak{p}}$ 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 $(a/s{)}^n=0$, soConversely, let $a/s$ be a nilpotent element of the localization at a prime ideal $R_{\mathfrak{p}}$. For some $n\ge 1$, $a^n/s^n=0$. The vanishing criterion in a module localization supplies $t\notin \mathfrak{p}$ with $t{a}^n=0$. It follows thatso $ta\in N$. Since $ts\notin \mathfrak{p}$,This proves that the nilradical commutes with localization:The multiplication by $t$ is necessary: the numerator $a$ of a nilpotent element in a localization need not itself be nilpotent in $R$.

The subset of the spectrum of a ring where the localization at a prime ideal is reduced. Since the nilradical commutes with localization, its complement is the support of a module $N=\mathrm{Nil}(R)$. When $N$ is a finitely generated module, vanishing of a finitely generated module near a prime givesso the locus is a Zariski-open subset. In particular this holds for every Noetherian ring.