For which pairs of distinct positive primes $p$ and $q$, the integral closure of $\mathbb{Z}$ in $\mathbb{Q}[\sqrt{pq}]$ is a UFD? I've proved that neither $p$ nor $q$ can be congruent to $1$ modulo $4$.
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$\begingroup$ And the only possible case I knew is that $p=2$ and $q=3$. $\endgroup$– Censi LICommented Mar 13, 2015 at 12:49
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5$\begingroup$ You expect an explicit description? That is hopeless. There are more, probably infinitely many, but it's not realistic at present to expect a list. Other examples include $pq$ equal to 14, 21, 22, 33, 46, 57, 62, 69, and 77. $\endgroup$– KConradCommented Mar 13, 2015 at 13:09
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$\begingroup$ Thanks, I'm a bit surprised. Do you know if the converse holds? That is when $p$, $q$ are disticnt positive primes, $q\equiv 3\pmod 4$, $p=2$ or $p\equiv 3\pmod 4$, then $\mathbb{Q}[\sqrt{pq}]$ must be UFD? $\endgroup$– Censi LICommented Mar 13, 2015 at 18:58
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4$\begingroup$ The converse surely does not hold, I believe. Being a UFD in this context is equivalent to having class number 1, and that is quite a difficult problem to address. $\endgroup$– Greg MartinCommented Mar 13, 2015 at 20:21
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3$\begingroup$ The converse doesn't hold, and as Greg briefly put it you really shouldn't expect any simple rule predicting an infinite list of real quadratic fields with class number $1$. Examples where the converse does not hold include $pq$ equal to $142$, $254$, $321$, $326$, $469$, $473$, $817$, $1957$, and $2021$. In these cases the class number is $3$ except for $817$, where the class number is $5$. $\endgroup$– KConradCommented Mar 13, 2015 at 21:02
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