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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.

calearner
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  • So far I can say the following: if D is a Dedekind domain with countable many non-zero prime ideals and infinite class group, then it has an overring which is not a ring of fractions, but all its overrings are the intersection of two rings of fractions. (However I still don't have a concrete example of such overring. In fact, it's not easy to give examples of Dedekind domains with infinite class group.) – user26857 Mar 30 '16 at 20:42
  • @user26857 Why would every overring of such a Dedekind domain have to be the intersection of two rings of fractions? Is it related to the fact that every ideal is at most 2-generated? – calearner Mar 31 '16 at 01:38
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    This is a non-trivial result of Gilmer: "Finite intersections of quotient rings of a Dedekind domain". – user26857 Mar 31 '16 at 06:44
  • That's great! I guess you mean "non-torsion class group", not "infinite class group". Thank you! – calearner Mar 31 '16 at 16:01
  • Actually I've meant "cyclic infinite class group", that is, isomorphic to $\mathbb Z$. But, of course, one can extend this to non-torsion class group. – user26857 Mar 31 '16 at 20:15
  • @user26857 I guess you could use Claborn's theorem, Leedham-Green's version (Trans AMS (1972), Theorem 1.4): http://math.uga.edu/~pete/leedhamgreen.pdf . Let $K=\mathbb Q$ and $G = \mathbb Z$. – neilme Mar 31 '16 at 21:03

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I don't know the full answer to the question, but here's a partial answer:

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

  2. 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 $x, y \in D \setminus \{0\}$?

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neilme
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