Timeline for What is the Hilbert class field of a cyclotomic field?
Current License: CC BY-SA 4.0
18 events
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| S Sep 22, 2020 at 12:28 | history | suggested | user11333 | CC BY-SA 4.0 |
some changes with tex
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| Sep 22, 2020 at 11:24 | review | Suggested edits | |||
| S Sep 22, 2020 at 12:28 | |||||
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Oct 22, 2009 at 22:35 | comment | added | Ilya Nikokoshev | It's some computation I think. Perhaps Hilbert 90? | |
| Oct 18, 2009 at 20:49 | answer | added | Jonah Sinick | timeline score: 1 | |
| Oct 18, 2009 at 14:19 | vote | accept | Ben Webster♦ | ||
| Oct 18, 2009 at 7:20 | answer | added | Joel Dodge | timeline score: 3 | |
| Oct 18, 2009 at 4:32 | answer | added | user631 | timeline score: 14 | |
| Oct 18, 2009 at 4:00 | comment | added | Akhil Mathew | Oops, I realize I wasn't clear earlier: Each character of G is an integral combination of characters induced from 1-dimensional subgroups. One reduces by the previous comment to the case of 1-dimensional representations, which is clear. | |
| Oct 18, 2009 at 3:31 | comment | added | Akhil Mathew | Ben- if a character of a representation V is an integral linear combination of characters of representations definable over a field k, then V is definable over k (Prop. 33 on p.91, Linear Representations of Finite Groups by Serre). This seems to be how Serre proves the result in question (p.94, Corollary). | |
| Oct 18, 2009 at 3:02 | comment | added | Ben Webster♦ | Akhil- This isn't something you can prove just studying the characters. You have to think about the representations themselves. | |
| Oct 18, 2009 at 3:00 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
added 241 characters in body; added 1 characters in body
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| Oct 18, 2009 at 2:14 | comment | added | Akhil Mathew | I believe this is a consequence of Brauer's theorem: this is true for 1-dimensional representations, and every representation's character is an integral linear combination of characters of 1-dimensional representations (by Brauer's theorem and the theorem on representations of supersolvable groups). | |
| Oct 18, 2009 at 1:51 | comment | added | Ben Webster♦ | My recollection is it's not super easy. You can see the characters are in a cyclotomic extension by noting that all the eigenvalues of a matrix of finite order are roots of unity, but your representation doesn't have to be defined over the field gotten by adjoining character values. | |
| Oct 18, 2009 at 1:49 | answer | added | Andreas Holmstrom | timeline score: 1 | |
| Oct 18, 2009 at 1:29 | answer | added | David Zureick-Brown | timeline score: 2 | |
| Oct 18, 2009 at 1:24 | comment | added | Qiaochu Yuan | It's still not obvious to me (although I believe it) why all representations of finite groups are defined over cyclotomic fields. Could you sketch that argument? | |
| Oct 18, 2009 at 1:04 | history | asked | Ben Webster♦ | CC BY-SA 2.5 |