One-one correspondence between maximal ideals and maximal elements. Let $S$ be a multiplicatively closed set, i.e. if $a\in S$ and $b\in S$, we must have $ab\in S$. Then $R_{S}=\left\{\dfrac{r}{s}|r\in R, s\in S\right\}$, where $R$ is a commutative ring containing $S$. $R_{S}$ can be thought of as a subring of quotient field of $R$. Now we want to show that there is a one-one correspondence between totality of maximal ideals of $R_{S}$ and maximal elements $A=\{\alpha|\alpha$ is an ideal of R and $\alpha\cap S=\emptyset\}$ with respect to set inclusion.

First of all, we can establish a homomorphism $\phi : R\rightarrow R_{S}$ where $\phi(r)=\dfrac{rs}{s}$. Now I for every proper ideal $I$ in $R_{S}$, we have $1\notin I$. So $I$ can't contain any element like $\dfrac{rs}{s}$ where $r\in S$. $\phi^{-1}(I)$ is an element in $A$. Now we have to prove that if $I$ is maximal ideal, then $\phi^{-1}(I)$ is a maximal element. Suppose $\phi^{-1}(I)$ is not a maximal element, we have $\phi^{-1}(I)\subset L$ where $L$ is another element in $A$. Now $&lt;\phi(L)&gt;$ is an ideal in $R_{S}$ generated by $\phi(L)$. I don't know how to show that this is a proper ideal. 

Conversely, for every maximal element $L$ in $R$, we have to show $&lt;\phi(L)&gt;$ is a maximal ideal in $R_{S}$. Suppose $&lt;\phi(L)&gt;$ is not maximal ideal, we have $&lt;\phi(L)&gt;\subsetneqq M$ where $M$ is maximal ideal in $R_{S}$. Then if the statement in the first paragraph is true, then we have $\phi^{-1}(M)=L$. How can we accurately prove that this is a contradiction. 

Appreciate it in advance.
We know the prime ideals of $R_s$ are 1:1 correspondence to the prime ideal of the prime ideals of $R$ which has no elements of $S$.So what you said before may be trivial.
We know the prime ideals of $R_s$ are 1:1 correspondence to the prime ideal of the prime ideals of $R$ which has no elements of $S$.So what you said before may be trivial.