Help needed in understanding this proof of localization

This question arose while studying notes of a senior in the course commutative algebra.

Consider the proof of properties of localization => $S^{-1} \sqrt(I)= \sqrt {S^{-1} I}$: Let $x/s\in \ S^{-1}\sqrt I$ with $x\in \sqrt I$ , so $x^n \in I$=> $\frac{x^n} {1} \in S^{-1} I$=> $x/1 \in \sqrt{S^{-1} I}$. (Till this point proof is clear)

Next line is assuming the previous line we got, $x / s \in \sqrt{S^{-1} I}$ .

I am not able to deduce this line and need help.

Solution 1:

Assume $A$ is the ring and $I$ is its ideal. Notice that $\sqrt {S^{-1} I}$ is an "ideal" of $S^{-1}A$, by definition if $J$ is an ideal of ring $A$, then $ra\in I$ for $r\in J$ and $a\in A$.

Now you showed that $x/1\in \sqrt {S^{-1}I}$, and obviously $1/s \in S^{-1}A$, hence $x/s = (x/1)\cdot (1/s) \in \sqrt{S^{-1}I}$, by definition of ideal.