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 $a \in IA$.

## Best Answer

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 $\frac{a}{s}=\frac{ua}{us}\in IS^{-1}A$.