Timeline for Given a number field $K$, when is its Hilbert class field an abelian extension of $\mathbb{Q}$?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 21, 2010 at 19:10 | vote | accept | Scarlet | ||
| Feb 21, 2010 at 19:09 | vote | accept | Scarlet | ||
| Feb 21, 2010 at 19:09 | |||||
| Feb 19, 2010 at 15:55 | comment | added | Franz Lemmermeyer | For abelian extensions of the rationals: yes. In David's example, the field K generated by a square root of -5 has discriminant 20, and its genus class field is contained in the field of 20th roots of unity. More exactly, it's the maximal subfield in which 2 and 5 have the same ramification indices as in K. | |
| Feb 19, 2010 at 14:50 | comment | added | Scarlet | Sorry about the "unknown control sequence". I mean in the cyclotomic extension of $\mathbb{Q}$ one gets by adding the f(K)th roots of unity. | |
| Feb 19, 2010 at 14:25 | comment | added | Scarlet | With respect to the last comment, are you trying to say that if K has conductor $f(K)$, then the genus class field is contained in $\Q(\zeta_f(K))?$ | |
| Feb 19, 2010 at 7:28 | history | answered | Franz Lemmermeyer | CC BY-SA 2.5 |