how can I prove that $D^{-1}R$ has a unique maximal ideal? Let $R$ be a commutative ring without zero divisors & let $P$ be a prime ideal.
I have shown that $D=R\setminus P$ is a non-empty multiplicative closed set without zero-divisors.
Now how can I prove that $D^{-1}R$ has a unique maximal ideal?
 A: First determine the candidate for being the maximal ideal. Then prove that everything outside that candidate is a unit (has an inverse).
The candidate seems to be $\tilde{P}:=\{p/q:\text{ with }p\in P\text{ and }q\in D\}$.
If $a/b\notin \tilde{P}$ then $a\notin P$. Hence $a/b$ has an inverse $b/a$.
Why there is no other maximal ideal $M$. Take $x\in M\setminus \tilde{P}$. Then $x$ isa unit and that forces $D^{-1}R=M$ and $M$ can't be maximal. 
A: Let $M = D^{-1}P = \bigg\{ \frac{x}{d} : x \in P$, $d \in D\bigg\}$. Observe that since $P$ is a prime ideal $P \neq R$ so it follows that $D^{-1}P \neq D^{-1}R$ and thus $D^{-1}P$ is a proper ideal of $D^{-1}R$. Now suppose $z \notin M$. Write $z = \frac{y}{d}$ where $y \in R$ and $d \in D$. Since $z \notin M$ it follows that $y \notin P$. Hence $y \in D$. Now we construct $w = \frac{d}{y} \in D^{-1}R$ since $d \in R$. Hence $zw = 1$ and thus $z$ is a unit. This shows that any element that is not in $M$ must be a unit and this this is a unique maximal ideal.
