Timeline for For which pairs of distinct positive primes $p$ and $q$, the integral closure of $\mathbb{Z}$ in $\mathbb{Q}[\sqrt{pq}]$ is a UFD?
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| Mar 14, 2015 at 9:58 | comment | added | Censi LI | Well, it seems that I didn't realize how complicated the question is. Thanks a lot. | |
| Mar 13, 2015 at 21:02 | comment | added | KConrad | 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 $5$. | |
| Mar 13, 2015 at 20:21 | comment | added | Greg Martin | 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. | |
| Mar 13, 2015 at 18:58 | comment | added | Censi LI | 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? | |
| Mar 13, 2015 at 13:09 | comment | added | KConrad | 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. | |
| Mar 13, 2015 at 12:49 | comment | added | Censi LI | And the only possible case I knew is that $p=2$ and $q=3$. | |
| Mar 13, 2015 at 12:49 | review | First posts | |||
| Mar 13, 2015 at 12:56 | |||||
| Mar 13, 2015 at 12:47 | history | asked | Censi LI | CC BY-SA 3.0 |