One to one correspondence between the set of prime ideals of $S^{-1}R$ and the set of prime ideals of $R$ which do not intersect $S$ I am trying to prove that there is a one to one correspondence between the set of prime ideals of $S^{-1}R$ and the set of prime ideals of $ R$ which do not intersect $S$. 
But I don't know what the map would be for this map to be bijective. Actually I don't see why there can be a bijective map at all.  
 A: Suppose $P$ is a prime in $R$ with $P \cap S = \emptyset$. Then let $\tilde{P} = \left\lbrace\frac{a}{b} \mid a \in P \text{ and } b \in S\right\rbrace$. By the definition operation in the localization, it is quick to show that $\tilde{P}$ is a prime ideal in $S^{-1} R$. The requirement that $P \cap S = \emptyset$ is simply that we don't want $\tilde{P}$ to contain a unit, as if $s \in P \cap S$, then $\frac{s}{s} = 1 \in \tilde{P}$, but then $\tilde{P} = S^{-1}R$.  
Now suppose $Q$ is prime in $S^{-1}R$. Then $Q = \left\lbrace \frac{c}{d} \mid c \in T \text{ and } d \in S\right\rbrace$, where $T \subset R$. Again using the definition of the operation in the localization, it is quick to show that $T$ is a prime ideal of $R$, and modulo a concern about $T \cap S$, we are finished. If $T\cap S \neq \emptyset$, we have some $s \in T \cap S$, and so $\frac{s}{s} = 1 \in Q$, so since $Q$ is an ideal, $Q = S^{-1}R$. Since prime ideals are by definition not the entire ring, we must have that $T \cap S = \emptyset$.
This establishes the correspondence. 
