Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 1 / 2 / iii / Solution

Let $S=R\setminus P$. If $Q\subseteq P$ is a prime ideal, then $Q\cap S=\varnothing$. Its extension isIt is proper: $1=q/s$ would give $u(s-q)=0$ for some $u\in S$, and hence $us=uq\in Q$, impossible because $u,s\notin Q$. If a product $(a/s)(b/t)$ lies in $Q{R}_P$, clearing a denominator gives $uab\in Q$ for some $u\in S$. Primeness and $u\notin Q$ imply $a\in Q$ or $b\in Q$. Thus $Q{R}_P$ is prime. Its contraction is $Q$, since $r/1\in Q{R}_P$ implies $ur\in Q$ for some $u\in S$, forcing $r\in Q$.

Conversely, contract a prime ideal $\mathfrak{q}$ of $R_P$ to $Q=\{r:r/1\in \mathfrak{q}\}$. This is a prime ideal of $R$. It contains no element of $S$, since such an element becomes a unit and cannot belong to the proper ideal $\mathfrak{q}$. Hence $Q\subseteq P$. Also $r/s\in \mathfrak{q}$ exactly when $r/1\in \mathfrak{q}$, because $s/1$ is a unit, so $\mathfrak{q}=Q{R}_P$. These extension and contraction operations are inverse, giving the prime ideal correspondence for localization: