Timeline for Subalgebra of a group algebra
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jun 3, 2019 at 19:52 | history | suggested | Marco Farinati | CC BY-SA 4.0 |
improved formating
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| Jun 3, 2019 at 18:56 | comment | added | Alex M. | @AndrásBátkai: The weird thing is that both accounts are registered and actively used. I join you in asking Marco Farinati to merge them. | |
| Jun 3, 2019 at 18:18 | comment | added | András Bátkai | @MarcoFarinati: register and merge your accounts. Then you can edit your own posts. | |
| Jun 3, 2019 at 17:37 | review | Suggested edits | |||
| S Jun 3, 2019 at 19:52 | |||||
| Jun 3, 2019 at 17:35 | comment | added | Marco Farinati | in the grading/comodule case, the subcoalgebra $A$ verifies $(A)_g\subseteq (k[G])_g$ (because it is a graded subobject). But $(k[G])_g=kg$, so, $A_g=0$ or $A_g=kg$, because a sub-vector space of a 1-dimensional vector space is zero or everything. Again, being algebraically closed is not important, the point is that $k$ is a field. | |
| Jun 2, 2019 at 3:53 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
added 48 characters in body
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| Jun 1, 2019 at 1:40 | comment | added | Konstantinos Kanakoglou | Ok. I was mainly refering to the grading/comodule part but in any case both arguments seem nice and clear. +1 ! | |
| May 31, 2019 at 19:04 | comment | added | Marco Farinati | No, you don't. The álgebra structure on $k^G$ is just $t\times k\times \cdots\times k$, the only ideales are puting zeros in some coordinates | |
| May 31, 2019 at 18:54 | comment | added | Konstantinos Kanakoglou | Don't you need algebraic closure of the field for either argument to work? Or am i missing something ? | |
| May 31, 2019 at 17:55 | review | Late answers | |||
| May 31, 2019 at 18:01 | |||||
| May 31, 2019 at 17:36 | history | answered | Marco Farinati | CC BY-SA 4.0 |