Timeline for Ext in symmetric algebras and group algebras
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 1, 2018 at 21:37 | vote | accept | Mare | ||
| Jan 1, 2018 at 21:37 | comment | added | Mare | Thanks, that is clever and makes it easy to calculate $Ext$. | |
| Jan 1, 2018 at 21:30 | comment | added | Jeremy Rickard | @Mare A more general example would be to take two algebras (over an algebraically closed field, say, to avoid non-separability issues): $B$ a non-semisimple local algebra, and $C$ an algebra such that $\text{Ext}^1(S,S)=0$ for every simple $C$-module. Then take $A=B\otimes_kC$ and $M=B\otimes_kS$ for $S$ a simple $C$-module. | |
| Jan 1, 2018 at 21:25 | comment | added | Jeremy Rickard | @Mare A projective resolution for $M$ can be obtained by tensoring $M$, considered as a $kC_3$-module, with a projective $kS_3$-resolution of the trivial module, and then the calculation follows easily. | |
| Jan 1, 2018 at 19:41 | comment | added | Mare | thanks, how to see that $Ext^i(M,M)=0$ for $i=1,2$ quickly? The quiver of $KG$ should have loops at both points so this shows that $Ext^1(S,S)$ should be nonzero for the simples. | |
| Jan 1, 2018 at 19:31 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |