Let $\mathbf{Z}_{(p)}$ denote the ring of all rational numbers whose denominators in lowest terms are not divisible by the integer prime $p$. (This is normally described as the localization at $p$.) Let $n$ be a square-free integer (positive or negative). Is the integral closure of $\mathbf{Z}_{(p)}[\sqrt n]$ a UFD? This integral closure will be $\mathbf{Z}_{(p)}[\sqrt n]$ itself unless $n\equiv 1 \pmod 4$, and in the latter case it is $\mathbf{Z}_{(p)}[(1+\sqrt n)/2]$.
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2$\begingroup$ Any discrete valuation ring is a PID, hence a UFD. $\endgroup$– Daniel LoughranCommented Jun 18, 2014 at 20:40
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$\begingroup$ Just to be clear, do you use the symbol $Z_{p}$ to denote the ring of $p$-adic integers, or the set of rational numbers $\{ a/b : \gcd(b,p) = 1 \}$? $\endgroup$– Jeremy RouseCommented Jun 18, 2014 at 20:40
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1$\begingroup$ @DanielLoughran: the integral closure has finitely many maximal ideals, not just one, so it need not be a discrete valuation ring (but I agree it is a PID). $\endgroup$– KConradCommented Jun 18, 2014 at 20:43
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$\begingroup$ @KConrad: I think Daniel understood $Z_p$ as the ring of $p$-adic integers, in which case the field of fractions of $R=Z_p[\sqrt{n}]$ is the local field $K=Q_p(\sqrt{n})$, and the integral closure of $R$ in $K$ is a discrete valuation ring, namely the maximal compact subring of $K$. See also the comment of Jeremy Rouse. $\endgroup$– GH from MOCommented Jun 18, 2014 at 23:19
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$\begingroup$ @GHfromMO: Since the original question involves square roots of plain integers ("positive or negative") rather than $p$-adic integers, it seemed to me that in the question localization really did mean the algebraic process rather than completion. $\endgroup$– KConradCommented Jun 18, 2014 at 23:27
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