Is it known that certain extensions of the integers are UFDs?

Question

For an integer $n$, let $\omega=\sqrt n$ unless $n=\equiv 1 \pmod 4$ and in that case let $\omega = (1+\sqrt n)/2$. Then the ring $Z[\omega]$ is well known to be integrally closed. For $p$ a prime, let $Q_p$ denote the subring of the rationals consisting of those whose denominators are not divisible by $p$. I and my collaborators have shown that for all prime $p$ and all integers $n$, the ring $Q_p(\omega)$ is a UFD. It occurs to me that this might well known to number theorists. The proof involves something like 8 cases. The final conclusion is that $Z(\omega)$ is an inverse limit of UFDs, even though it is frequently not a UFD itself.

Answer 1

In fact the following is true.

Proposition: Let $\mathcal{O}_K$ be the ring of integers of a number field $K$. For any $p_0$, it is possible to find finitely many primes $p_1, ..., p_n \ge p_0$ such that $\mathcal{O}_K[p_1^{-1}, ..., p_n^{-1}]$ is a PID.

Proof. Find ideals $I_1, ..., I_m$ representing every ideal class in the ideal class group of $\mathcal{O}_K$ whose norms are not divisible by primes less than $p_0$ and invert all primes dividing their norms. $\Box$

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Edit (12/4/13): Regarding the last line, every Dedekind domain is an inverse limit of UFDs. Indeed, every Dedekind domain is the inverse limit of its localizations at every maximal ideal, and these are all DVRs (in particular PIDs, in particular UFDs).