Timeline for Strict Class Numbers of Totally Real Fields
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 12, 2010 at 7:06 | comment | added | Chandan Singh Dalawat | The "product formula" was called the Schachtelungssatz in the old German literature. | |
| Jan 12, 2010 at 6:15 | comment | added | Kevin Buzzard | Aah so I have slipped up! Apologies. I still don't see my mistake---can someone explain? If I have K/L/Q with L/Q fixed and K of huge degree over L and unramified then...OK, I see my mistake :-) In the "product formula", or whatever it's called, for discriminants, I pick up a power of disc(L/Q)^{[K:L]}, right? I think I'll leave my original erroneous assertion up though because it will confuse the comments too much if I remove it. | |
| Jan 12, 2010 at 1:30 | answer | added | JSE | timeline score: 9 | |
| Jan 12, 2010 at 1:15 | comment | added | JSE | @buzzard: in a Golod-Shafarevic tower, the discriminant is exponential in the degree: it's the so-called root discriminant that's constant. (Describing the set of number fields with bounded root discriminant is an extremely interesting and mysterious problem!) The set of number fields with discriminant < X, on the other hand, is indeed finite. In an appendix to a paper of Belolipetsky, Venkatesh and I show that the log-size of this set is at most (log X)^{1+eps} (the finiteness is much older, as Ben Linowitz points out.) | |
| Jan 12, 2010 at 0:38 | comment | added | user1073 | @Buzzard - I'm afraid that I don't quite understand your argument. That only finitely many number fields have a given discriminant follows from Minkowski's Theorem; see Theorem 5 on page 121 of Lang's Algebraic Number Theory. | |
| Jan 12, 2010 at 0:25 | comment | added | Anweshi | How will you go from the wikipedia version of Golod-Shafarevich to your claim? | |
| Jan 11, 2010 at 23:14 | comment | added | Kevin Buzzard | No :-) In fact it's a result of Golod and Schaferevich that there aren't :-) You need to bound the degree too. Unless I've slipped up. | |
| Jan 11, 2010 at 22:15 | answer | added | Dror Speiser | timeline score: 3 | |
| Jan 11, 2010 at 21:53 | comment | added | Dror Speiser | Isn't it a classical result of Hermite that there are finitely many fields of bounded discriminant? | |
| Jan 11, 2010 at 21:33 | vote | accept | CommunityBot | moved from User.Id=1073 by developer User.Id=69903 | |
| Jan 11, 2010 at 21:15 | answer | added | Emerton | timeline score: 6 | |
| Jan 11, 2010 at 20:17 | comment | added | user1073 | @Buzzard: Yes, the degree of the field is bounded. The paper works with a totally real field F of degree n and it is assumed throughout that F has strict class number 1. I believe that this remark is meant to justify this restriction. – Ben Linowitz 0 secs ago | |
| Jan 11, 2010 at 20:06 | comment | added | Kevin Buzzard | Are you bounding the degree of the field? If you don't then it's hard to make sense of this question, because I don't see why there will only be finitely many fields of a given discriminant. | |
| Jan 11, 2010 at 19:54 | answer | added | Jonah Sinick | timeline score: 1 | |
| Jan 11, 2010 at 19:27 | history | asked | user1073 | CC BY-SA 2.5 |