Proposition 17

Let $A$ be a ring and $\mathfrak{N}$ its nilradical. For every multiplicative subset $S$ of $A$, $S^{-1}\mathfrak{N}$ is the nilradical of $S^{-1}A$. In particular, if $A$ is reduced, then $S^{-1}A$ is reduced.

If $x \in A$, $s \in S$ satisfy $(x/s)^n = x^n/s^n = 0$, there exists $s' \in S$ such that $s'x^n = 0$ (no. 1, Remark 3) and a fortiori $(s'x)^n = 0$, hence $s'x \in \mathfrak{N}$ and $x/s = s'x/s's \in S^{-1}\mathfrak{N}$; the converse is immediate.