= Solution
{c}
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> is
$$
SP=\{p/y:p\in P,\ y\in Y\}.
$$
Indeed, 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 $P$:
$$
\boxed{\{r\in R:r/1\in SP\}=P.}
$$
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 $SP_1=SP_2$, contracting would therefore give $P_1=P_2$, contrary to the hypothesis. Hence
$$
\boxed{SP_1\ne SP_2.}
$$
We 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 $SP$.
Back to article page
