If $R$ is commutative ring, $P_1, P_2, \dots, P_n$ prime ideals of $R$ with the property $P_i \not\subseteq \bigcup _{j \not = i} P_j$, $\forall 1\le i\le n$, and $S:=R\setminus(P_1 \cup \cdots \cup P_n)$, then show: $$S^{-1}R \text{ has exactly } n \text{ maximal ideals}.$$
Definition. $S^{-1}R=${${r \over s} : r \in R , s \in S $}.