Let $A$ be a ring, $S\subset A$ a multiplicative set, and $I\subset A$ an ideal not intersecting $S$. For any ideal $J$, let $r(J)$ denote the radical of $J$. Is $S^{-1}r(I) = r(S^{-1}I)$?
Certainly $S^{-1}r(I)$ is generated by elements of the form $\frac{x}{s}$, where $x^n\in I$. This implies that $\left(\frac{x}{s}\right)^n = \frac{x^n}{s^n}\in S^{-1}I$, so $\frac{x}{s}\in r(S^{-1}I)$. This shows that $S^{-1}r(I)\subseteq r(S^{-1}I).$
The other direction seems less clear. Certainly $r(S^{-1}I)$ is generated by elements of the form $\frac{x}{s}$ where $\frac{x^n}{s^n}\in S^{-1}I$. It isn't clear to me that this implies that $\frac{x}{s}\in S^{-1}r(I)$.
Let $\displaystyle\frac{x}{s}\in r(S^{-1}I)$, so $\displaystyle\frac{x^n}{s^n}\in S^{-1}I$ for some $n$ and therefore $\displaystyle\frac{x^n}{s^n}=\frac{i}{t}$ for some $i\in I, t\in S$
Then $ux^{n}t=us^{n}i\in I$ for some $u\in S$, so $(uxt)^n\in I\implies uxt\in r(I)$ and therefore
$\hspace {.25 in}\displaystyle \frac{x}{s}=\frac{uxt}{uts}\in S^{-1}(r(I))$