Given integral domain $A$ and multiplicatively closed subset $S$ of $A$ (which we may assume to have $1_A$), let $A_S = \{\frac{a}{s}: a \in A, s \in S \}$. Now, I want to show that the maps:
$$Q' \mapsto Q = Q' \cap A$$ $$Q \mapsto Q' = QA_S$$
yields a bijection between prime ideals $Q'$ of $A_S$ and prime ideals $Q$ of $A$, which are disjoint from $S$.
I tried to do the following. Choose a prime ideal $Q$ of $A$, and consider $QA_S$. To show this is prime ideal of $A_S$, suppose $\frac{aa'}{ss'} \in QA_S$ - we want to show that either of these are in $QA_S$. So, suppose $\frac{a'}{s'} \notin QA_S$. Then, to show $\frac{a}{s} \in QA_S$ is equivalent to showing that $\frac{a}{s} = q\frac{a''}{s''}$ for $q \in Q, a'' \in A, s'' \in S$.
Now, we know that $\frac{aa'}{ss'} = q\frac{a''}{s''}$. Since $s'$ is a unit of $S$, $\frac{aa'}{s} = q\frac{sa''}{s''} $. But where do I go from here, since I don't know how to cancel out the $a'$ on the LHS?
P.S. this is a theorem in appendix B of Fourier Analysis on Number Fields by Ramakrishnan and Valenza.
P.S. This question is different from the tagged duplicate because the map defined is different. The other question involves a (notation substituted to make sense in this question) map $A \rightarrow \text{prime ideals of }A_S$, and I am not sure how to connect that to what I have here. Moreover, I am unable to get feedback on my own method that I have outlined here.