How to show that there is a bijective correspondence between two sets of prime ideals I'm trying to solve this Algebra Problem, and I'm not quite sure, if I'm on the right way.

Let $R$ be a commutative ring and $S \subset R$ a multiplicative subset. Show that $p \to pS^{-1}R$ defines a bijective map from the set of prime ideals of $R$ disjoint from $S$ to the set of prime ideals of the localization $S^{-1}R$.

This is what I did until now:
let $f$ be the map from the set of prime ideals over $R - S$ (let's call it P) to the set of prime ideals over $S^{-1}R$ (let's call it Q)
So $f(p)=pS^{-1}R$
to show bijectivity of $f$, we have to show that $f$ is injective and surjectiv.
for the injectivity, assume there are 2 prime ideals $A,B \in Q$ with $A=B$ ,therefor we could write $A=aS^{-1}R$ and $B=bS^{-1}R$ (this should follow from the surjectivity) and this implies $aS^{-1}R=bS^{-1}R$ and hence $a=b$, so $f$ is injectiv.
But now I don't know how to show the surjectivity. I mean why can I write an prime ideal in $S^{-1}R$ as a product $AS^{-1}R$ with $A$ a prime ideal in $R-S$ ???
Perhaps I should also mention how we defined the localization. Let $S \subset R$ be an multiplicative subset and R a commutative ring. then $S^{-1}R := S\times R / \sim$ where 
$(r,s)\sim (r',s') : \Leftrightarrow \exists p \in S$ s.t. $p(rs'-r's)=0$
Could somebody give me a hint how to show surjectivity?
 A: I'll assume $1\in S$, which is known to be not restrictive. The map can be more properly written
$$
P \mapsto S^{-1}P
$$
where
$$
S^{-1}P=\left\{\frac{x}{s}:x\in P\right\}
$$
It is easy to show that if $P$ is any ideal in $R$, then $S^{-1}P$ is an ideal in $S^{-1}R$.
Now, let's show that if $P$ is prime in $R$, disjoint from $S$, then $S^{-1}P$ is prime in $S^{-1}R$. Suppose
$$
\frac{x}{s}\frac{y}{t}\in S^{-1}P,\qquad(x,y\in R, s,t\in S)
$$
Then, by definition, there exist $z\in P$ and $u\in S$ with
$$
\frac{x}{s}\frac{y}{t}=\frac{z}{u}
$$
so there is $v\in S$ with $v(uxy-stz)=0$ or
$$
vuxy=stz
$$
By assumption $z\in P$, so also $stz\in P$. Since $P$ is a prime ideal, we must have $v\in P$ or $u\in P$ or $x\in P$ or $y\in P$. Since $P$ is disjoint from $S$, we are left with $x\in P$ or $y\in P$, so we conclude that either $x/s\in S^{-1}P$ or $y/t\in S^{-1}P$. We have to prove also that $1/1\notin S^{-1}P$, though. But $1/1=x/s$ with $x\in P$ implies $t(s-x)=0$ for some $t\in S$, hence that $tx=ts\in P\cap S$, which is a contradiction.
Now our map is surely well defined. In order to see it's bijective, we just need to find an inverse.
If $I$ is an ideal of $S^{-1}P$, consider $I^c=\{x\in R: x/1\in I\}$ (the contraction of $I$). We want to see that if $P$ is a prime ideal of $R$ disjoint from $S$, then $P=(S^{-1}P)^c$ and, if $I$ is a prime ideal of $S^{-1}R$, then $S^{-1}(I^c)=I$.
Note that, if $\lambda_S\colon R\to S^{-1}R$ is the canonical morphism, 
$I^c=\lambda_S^{-1}(I)$ is a prime ideal in $R$ whenever $I$ is a prime ideal in $S^{-1}R$ and also that $I^c$ is disjoint from $S$.
The inclusion $P\subseteq (S^{-1}P)^c$ is obvious, because for $x\in P$, $x/1\in S^{-1}P$ by definition. So, let $x\in (S^{-1}P)^c$; then there are $y\in P$ and $t\in S$ such that
$$
\frac{x}{1}=\frac{y}{t}
$$
that is, there is $u\in S$ with $u(tx-y)=0$. Thus $uy=utx\in P$ so $x\in P$ because $P$ is prime and $ut\notin P$.
Let's do the converse; suppose $I$ is a prime ideal in $S^{-1}P$. If $x/s\in I$, then
$$
\frac{x}{1}=s\frac{x}{s}\in I
$$
so $x\in I^c$ and, therefore, $I\subseteq S^{-1}(I^c)$. The converse inclusion is easy.
A: Hint: 
Observe any prime ideal $P$ different from $S$ gives a prime ideal $S^{-1}P$ in $S^{-1}R$.
Now take $p\in P$ then you will get the surjectivity.
