 Solution

Use the convention that a multiplicative subset contains $1$ and is closed under multiplication. Usually $0$ is excluded; allowing it gives the harmless zero-ring localization. A prime ideal disjoint from $S$ is mapped toin the localization of a ring. This ideal is proper, andis an integral domain, since the denominators are nonzero in the integral domain $R/P$. Thus $S^{-1}P$ is a prime ideal.

Conversely, for a prime ideal $Q$ in $S^{-1}R$, its contractionis a prime ideal disjoint from $S$: each $s/1$ is a unit and cannot belong to a proper ideal. Membership in $Q$ is equivalent to numerator membership in $P$, since $a/s=(a/1)(s/1{)}^{-1}$. Hence $Q=S^{-1}P$.

To check the other composite, $a/1\in S^{-1}P$ implies $sa\in P$ for some $s\in S$, by the equality criterion for fractions in a localization of a ring. Since $s\notin P$, primality gives $a\in P$. Therefore extension and contraction are mutually inverse:If $0\in S$, both sets are empty. This is the prime ideal correspondence for localization.