# Localization of quotient ring

abstract-algebraring-theory

I'm working on the following exercise:
Let $$A$$ be a ring and $$I \subset A$$ ideal y $$S \subset A$$ multiplicative subset. Denote $$\pi(S) = \tilde{S}$$ where $$\pi$$ is the projection $$A \rightarrow A/I$$.
Show that $$\tilde{S}^{-1}(A/I) \simeq S^{-1}A/ IS^{-1}A$$.

I wanted to use the isomorphism theorem to solve this. I considered the ring morphism:
$$S^{-1}A \rightarrow \tilde{S}(A/I)$$
$$\frac{a}{s} \rightarrow \frac{[a]}{[s]}$$

I need to show that this map is well-defined, is ring morphism, is subjective and find the kernel.
I could show that this is well-defined and is a subjective ring morphism. However, I could not find that the kernel is $$S^{-1}(IA) = IS^{-1}A$$. I tried to find $$\frac{[a]}{[s]} = \frac{[0]}{[1]}$$. Of course, if $$IA$$ is in the kernel so I have to see the other inclusion. The idea I have to solve this is to note that $$\frac{[0]}{[1]} = \frac{[a']}{[s]}$$ such that there exists $$s' \in S$$ and $$s'a'=0$$. This tells me that the value of $$s$$ is not important. But, I can't see why $$a \in IA$$.

You don't need to show that $$a\in IA$$.
If $$\frac{[a]}{[s]}=0\in \tilde{S}^{-1}(A/I)$$, there exists an element $$u\in S$$ such that $$[u][a]=0\in A/I$$, which implies that $$ua\in I$$. Hence the element $$\frac{a}{s}=\frac{ua}{us}\in IS^{-1}A$$.