Timeline for global dimension of invariant algebra
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21, 2014 at 9:57 | comment | added | Jeremy Rickard | @Aleksa: No, my second example shows that's not necessarily true when $\operatorname{gldim}(A)=2$. | |
| Jan 20, 2014 at 12:38 | comment | added | Jason Starr | @JeremyRickard: Great, thanks a lot! | |
| Jan 20, 2014 at 10:31 | comment | added | Jeremy Rickard | And there's a paper of George Bergman (Groups acting on hereditary rings. Proc. London Math. Soc. (3) 23 (1971), 70-82) that proves that if $A$ is hereditary and $|G|$ is invertible in $A$, then $A^G$ is also hereditary. | |
| Jan 20, 2014 at 10:16 | comment | added | Jeremy Rickard | @JasonStarr: Yes, that's true. A Maschke-like proof shows that for any (right) $A^G$-module $M$, the natural $A^G$-module map $M\to M\otimes_{A^G}A$ is a split embedding. If $A$ is semisimple, then $M\otimes_{A^G}A$ (and hence $M$) is projective. | |
| Jan 19, 2014 at 18:52 | comment | added | Jason Starr | @JeremyRickard: That is cool. I guess my intuition comes from thinking about semisimple rings. I still suspect that for a semisimple $k$-algebra $A$, if $|G|$ is prime to $\text{char}(k)$, then $A^G$ is semisimple. | |
| Jan 19, 2014 at 18:30 | comment | added | Dag Oskar Madsen | The second example also shows that $A$ can be basic. | |
| Jan 19, 2014 at 17:37 | comment | added | Jeremy Rickard | @abx: I've edited my answer to add an example where $\operatorname{char}(k)$ is prime to $|G|$. | |
| Jan 19, 2014 at 17:35 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
added 747 characters in body
|
| Jan 19, 2014 at 16:32 | comment | added | Jason Starr | @abx: I suspect that the answer is positive if $\text{char}(k)$ is prime to $|G|$. Obviously it would suffice to prove that $A$ is a projective $A^G$-module. | |
| Jan 19, 2014 at 15:35 | comment | added | abx | What if the characteristic of $k$ is prime to $|G|$? | |
| Jan 19, 2014 at 15:11 | comment | added | Jeremy Rickard | @JasonStarr : And I gave you an hour's head start, too! :-) | |
| Jan 19, 2014 at 15:07 | comment | added | Jason Starr | Shucks, you beat me to it. | |
| Jan 19, 2014 at 15:05 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |