Intersection of height one localisations of a normal Noetherian domain

Question

Exercise 8.3 in Kemper's A Course in Commutative Algebra:

Let $R$ be a commutative normal Noetherian domain. Prove that $$R=\bigcap_{\substack{P\in\operatorname{Spec}(R)\\\operatorname{ht}(P)=1}}R_P$$ Hint: For $a/b\in\operatorname{Quot}(R)\setminus R$ consider an ideal $P$ that is maximal among all colon ideals $(b):(a')$ with $a'\in(a)\setminus(b)$.

Well because $R$ is Noetherian its easy to see that there is such an ideal $P$ that is maximal among colon ideals. Otherwise given all $a'\in(a)/(b)$ there would exist $a''\in(a)/(b)$ such that $(b):(a')\subsetneq (b):(a'')$ then we would have an infinite strictly ascending chain of ideals. So pick $(a')$ such that $P:=(b):(a')$ is maximal. Then $P$ is prime because if $xy\in P$ with $y\notin P$ then we have $x\in (b):(ya')$, $ya'\in(a)\setminus(b)$, and $P\subseteq(b):(ya')$, so $x\in P$.

So now let $Q$ be a minimal prime satisfying $(b)\subseteq Q\subseteq P$ then I believe I want to show that $a/b\notin R_Q$, and to do that it's sufficient to show that $a/b\notin R_P$.

But I have no idea how to go about doing this.

Answer 1

A priori it does not suffice to show $\frac{a}{b} \notin R_P$.

But note the following: At some point, you have to use the normal assumption. The idea is that one easily sees that $P=(b):(a')$ does not contain a regular sequence of length $2$. By the normal assumption on $R$ (use Serre's criterion for normality), we get that $P$ has height $1$. So it suffices to show $\frac{a}{b} \notin R_P$ and this is easily done:

If $\frac{a}{b} \in R_P$, we get some $s \notin P$ with $sa \in (b)$. In particular $sa' \in (sa') \subset (sa) \subset (b)$, i.e. $s \in (b):(a')=P$, contradiction!