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If $R$ is reduced, its nilradical $N$ is zero. By b, every localization at a prime ideal has zero nilradical, so in particular $R_{\mathfrak{m}}$ is reduced for every maximal ideal $\mathfrak{m}$.

Conversely, suppose every such localization is reduced. Take a nilpotent element $x\in R$. Then $x/1$ is nilpotent in every $R_{\mathfrak{m}}$, hence zero there. To see directly that $x=0$, suppose otherwise and consider the proper annihilatorChoose a maximal ideal $\mathfrak{m}$ containing this annihilator. The equality $x/1=0$ in $R_{\mathfrak{m}}$ would imply $sx=0$ for some $s\notin \mathfrak{m}$, by the vanishing criterion in a module localization. But then $s\in {\mathrm{Ann}}_R(x)\subseteq \mathfrak{m}$, a contradiction. This is the principle that localization detects zero elements. ThereforeThe argument also includes the zero ring, which is reduced and has no maximal ideals.

Let $N$ be the nilradical of $R$. By b, the reduced locus of a ring isThe Noetherian ring hypothesis makes the ideal $N$ finitely generated; write $N=(n_1,\dots ,n_t)$. Fix $\mathfrak{p}\in \mathcal{R}$. For every $i$, the vanishing criterion in a module localization gives $s_i\notin \mathfrak{p}$ with $s_i{n}_i=0$. The product $f=s_1\cdots s_t$ lies outside the prime ideal $\mathfrak{p}$ and satisfies $fN=0$. When $N=0$, take $f=1$.

For every prime ideal $\mathfrak{q}$ in the distinguished open setthe element $f/1$ is a unit in $R_{\mathfrak{q}}$. The equality $fN=0$ therefore implies $N{R}_{\mathfrak{q}}=0$, so $R_{\mathfrak{q}}$ is reduced. Thus $\mathfrak{p}\in D(f)\subseteq \mathcal{R}$, proving that $\mathcal{R}$ is a Zariski-open subset of the spectrum of a ring.

More precisely, let $A={\mathrm{Ann}}_R(N)$, the annihilator of a module $N$. The argument shows that $N{R}_{\mathfrak{p}}=0$ if and only if some $f\in A$ lies outside $\mathfrak{p}$. HenceEquivalently, the complement is the support of a module $N$. This is vanishing of a finitely generated module near a prime applied to $N$: its finite generation lets the separately chosen denominators be combined into one element annihilating all of $N$.