Timeline for discriminant of subfield of $\mathbb{Q}(\zeta_p)$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 21, 2019 at 12:50 | history | edited | Aurel | CC BY-SA 4.0 |
better wording
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| Oct 20, 2019 at 0:44 | comment | added | user145307 | @GHfromMO The extension $\mathbf{Q}_p(p^{1/e})$ with $e > 1$ prime to $p$ (for example) is certainly totally tamely ramified and is never generated by a root of unity. Perhaps you are thinking about the fact that an abelian extension of $\mathbf{Q}_p$ is cyclotomic (local Kronecker-Weber)? | |
| Oct 20, 2019 at 0:33 | comment | added | user145307 | @dessind'efantterrible To split hairs, I would say that this is a computation of the conductor in this case rather than a definition. The distinction is that this if you actually want to prove the Führerdiskriminantenproduktformel (in the abelian case) for a local extension $L/K$, you really need to grapple with $N_{L/K}(L^*)$ in some way (say via basic local class field theory), whereas one could compute (with a full proof) the different of a totally tamely ramified extension $L/K$ in a single MO comment. | |
| Oct 19, 2019 at 22:52 | comment | added | dessin d'enfant terrible | @ElectricPenguin theres a very elementary definition of conductor in this special case: the smallest n s.t. $\chi$ factors through (Z/pZ)*-->(Z/nZ)* | |
| Oct 18, 2019 at 21:29 | comment | added | user145307 | This is (IMO) a better answer. First, the different (and discriminant) of a totally tamely ramified extension is certainly a simpler notion than understanding the conductor of an abelian extension. Second, it is more general. Of course, both answers are correct. | |
| Oct 18, 2019 at 16:01 | history | answered | Aurel | CC BY-SA 4.0 |