Timeline for A semisimple group ring
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 3, 2021 at 22:41 | answer | added | Waldeck Schützer | timeline score: 3 | |
| Sep 19, 2014 at 11:35 | vote | accept | Pablo | ||
| Jun 17, 2014 at 10:43 | comment | added | Pablo | Benjamin, this is a very good point. | |
| Jun 9, 2014 at 1:16 | answer | added | Geoff Robinson | timeline score: 4 | |
| Jun 7, 2014 at 14:58 | comment | added | Benjamin Steinberg | One might say the correct generalization of Maschke's theorem is that if $R$ is a commutative ring with unit (not necessarily associative) then $RG$ is a separable extension of $R$ iff $|G|$ is invertible in $R$ (see en.wikipedia.org/wiki/…). I am not sure if $R$ commutative is really needed. | |
| Jun 7, 2014 at 13:17 | answer | added | user26223 | timeline score: 3 | |
| Jun 7, 2014 at 11:20 | history | edited | Pablo | CC BY-SA 3.0 |
added 206 characters in body
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| Jun 7, 2014 at 11:16 | comment | added | Dima Pasechnik | perhaps you might want to edit the question to make it more "watertight"... | |
| Jun 7, 2014 at 9:16 | comment | added | Ben Webster♦ | @abx I totally disagree. If $R$ is a division algebra, then $R[G]$ is quite interesting (and semi-simple for the same hypotheses as Maschke's theorem for a field with the same proof). If $R$ is some completely arbitrary non-commutative ring, $R[G]$ is less familiar, but still makes perfect sense. | |
| Jun 7, 2014 at 8:50 | comment | added | abx | If $R$ is a field of characteristic $p$ and $|G|$ is prime to $p$, $R[G]$ is semi-simple: this is the usual Maschke theorem. On the other hand if $R$ is not commutative I don't think it makes much sense to look at $R[G]$. | |
| Jun 7, 2014 at 8:35 | comment | added | Pablo | Do you know what happens if $R$ is a semisimple ring of characteristic $p$? Is there some generalization of Maschke's theorem of this kind? | |
| Jun 7, 2014 at 8:32 | comment | added | Pablo | Yes that's very simple and I have missed it, thanks. | |
| Jun 7, 2014 at 7:51 | comment | added | abx | Try $G=\{1\}$ first. | |
| Jun 7, 2014 at 7:43 | history | asked | Pablo | CC BY-SA 3.0 |