 Solution

Let $P$ be a prime ideal of $R$ disjoint from the multiplicatively closed set $Y$. In the localization $S=Y^{-1}R$, its extension of an ideal isIndeed, finite sums of such fractions have a common denominator in $Y$ and numerator in $P$, and every generator $p/1$ has this form.

We show that its contraction of an ideal recovers $P$:One inclusion is immediate. Conversely, if $r/1=p/y$ with $p\in P$ and $y\in Y$, equality of fractions in the field of fractions of the integral domain $R$ gives $yr=p\in P$. Since $y\notin P$ and $P$ is a prime ideal, $r\in P$.

If $S{P}_1=S{P}_2$, contracting would therefore give $P_1=P_2$, contrary to the hypothesis. HenceWe will also need the fact that $SP$ is a prime ideal, part of the prime ideal correspondence for localization. It is proper by the contraction identity, since otherwise $1\in P$. If $(a/u)(b/v)\in SP$, multiplication by the denominator gives $ab/1\in SP$, and contraction gives $ab\in P$. Thus $a\in P$ or $b\in P$, so one of the two fractions belongs to $SP$.