Let $R$ be a ring and $S$ a multiplicatively closed set. Let $R_S$ be the localization of $R$ at $S$.
Why is $$\operatorname{nilrad} R_S=\left(\operatorname{nilrad} R\right)_S ?$$
Recall that $\operatorname{nilrad} R$ is the set of nilpotent elements of $R$.
Let $I = \textrm{nil}(R) = \{ x \in R : x^n = 0 \textrm{ for some } n \geq 1\}$. We want to show $\textrm{nil}(S^{-1}R) = S^{-1}I$, where the right hand side is the ideal in $S^{-1}R$ generated by the image of $I$ in $S^{-1}R$. It is equal to the set of fractions $\frac{a}{s}$, where $a \in I$ and $s \in S$.
Let $\frac{a}{s} \in S^{-1}I$ for some $a \in I$ and $s \in S$. Since $a$ is nilpotent, it is obvious that $\frac{a}{s}$ is nilpotent.
Conversely suppose that $\frac{r}{s} \in S^{-1}R$ is nilpotent, say $(\frac{r}{s})^n = \frac{r^n}{s^n}$ is zero in $S^{-1}R$ for some $n \geq 1$. This means that there exists some $s' \in S$ such that $s'r^n = 0$ in $R$. Multiplying both sides of this last equation by $(s')^{n-1}$, we see that $s'r$ is nilpotent. Thus $\frac{r}{s} = \frac{rs'}{ss'}$ with $ss' \in S$ and $rs' \in I$, so $\frac{r}{s} \in S^{-1}I$.