Are rings of fractions of integral domains closed under finite intersection? Let $D$ be an integral domain with fraction field $K$.  Let $V$, $W$ be multiplicatively closed subsets of $D$.  Consider the rings of fractions $V^{-1}D$ and $W^{-1}D$ as subrings of $K$.  Is $(V^{-1}D) \cap (W^{-1}D)$ also a ring of fractions of $D$?  That is, does there exist a multiplicative subset $S$ of $D$ such that $(V^{-1}D) \cap (W^{-1}D) = S^{-1}D$?
It seems unlikely, but I am unable to come up with a counterexample.
 A: I don't know the full answer to the question, but here's a partial answer:


*

*We always have the inclusion $D_{V \cap W} \subseteq D_V \cap D_W$.


*The reverse inclusion holds, provided that both $V$ and $W$ (or their saturations) are complements of finite unions of primes.

I claim (1) is obvious.  Here's my proof of (2):

Say $V=D \setminus \bigcup_{i=1}^n {\frak p}_i$ and $W = D \setminus \bigcup_{j=1}^m {\frak q}_j$.  Let $\alpha \in D_V \cap D_W$.  Let $I := \{d\in D : d\alpha \in D\}$.  Clearly $I$ is an ideal of $D$.
Suppose $I \subseteq (\cup_i {\frak p}_i) \bigcup (\cup_j {\frak q}_j)$.  Then by prime avoidance, either $I \subseteq {\frak p}_i$ for some $1\leq i \leq n$ or $I \subseteq {\frak q}_j$ for some $1\leq j \leq m$.  But the first statement contradicts $\alpha \in D_V$ (since there is some $v\in V$, hence not in any ${\frak p}_i$, with $v\in I$) and the second contradicts $\alpha \in D_W$ for analogous reasons.
Thus, $I \nsubseteq (\cup_i {\frak p}_i) \bigcup (\cup_j {\frak q}_j) = D \setminus (V \cap W)$, which means that $I \cap (V \cap W) \neq \emptyset$.  Let $d\in V \cap W$.  Then $d\alpha \in D$, so that $\alpha \in D_dfd \subseteq D_{V \cap W}$.

However, I don't know whether your question has a positive answer when $V$, $W$ correspond to infinite unions of primes.  An interesting special case would be when $V$, $W$ correspond to (the saturations of) the powers of single elements.  That is, is $D_x \cap D_y$ always a ring of fractions of $D$, where $x, y \in D \setminus \{0\}$?
