I have to prove that $\sqrt{S^{-1}I}=S^{-1}\sqrt{I}$.
The part $\sqrt{S^{-1}I}\supset S^{-1}\sqrt{I}$ is easier, but I am stuck at the other part.
Let $\frac{a}{s}\in S^{-1}A$ s.t. for some $n\in\mathbb{N}$ we have $(\frac{a}{s})^n=\frac{a^n}{s^n}\in S^{-1}I$.
I am confused about definitions, because I am beginner: this means that $a^n\in I$ and $s^n\in S^{-1}$? If yes, then is easy, but I do not know if I can state this. Only know that there is $\frac{i}{t}\in S^{-1}I$ s.t. $(a^nt-is^n)u=0$, for some $u\in S$.
Many thanks in advance.