Timeline for When are infinite dimensional path algebras hereditary?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 6, 2013 at 22:34 | vote | accept | Oliver Straser | ||
| Sep 6, 2013 at 17:02 | answer | added | Dag Oskar Madsen | timeline score: 3 | |
| Sep 6, 2013 at 7:22 | comment | added | Dag Oskar Madsen | To reach a contradiction one could also argue $I/IJ \neq 0$. | |
| Sep 6, 2013 at 7:15 | comment | added | Dag Oskar Madsen | Proving that $J/IJ$ is projective might not be needed, only that it is a $kQ/I$-module. | |
| Sep 6, 2013 at 7:08 | comment | added | Dag Oskar Madsen | Here is a proof sketch, details need to be checked and filled in. There is an epimorphism of $kQ/I$-modules $J/IJ \rightarrow J/I \rightarrow 0$ with $J/IJ$ projective. If $J/I$ is projective, then this morphism splits and we get (I think) an isomorphism $J/IJ \simeq J/I$, also as $kQ$-modules. This leads to a contradiction since the finite dimensional spaces $J^m(J/IJ)/J^{m+1}(J/IJ)$ and $J^m(J/I)/J^{m+1}(J/I)$ have different dimensions. | |
| Sep 5, 2013 at 17:27 | comment | added | Dag Oskar Madsen | Then I believe $J/I$ is not projective, which contradicts $kQ/I$ being hereditary. (In a hereditary ring, submodules of projectives should be projective.) I have to think a bit to formalize the proof. | |
| Sep 5, 2013 at 14:00 | comment | added | Oliver Straser | Yes, you can assume that $I\subset J^2$ if that helps. | |
| Sep 5, 2013 at 12:54 | comment | added | Benjamin Steinberg | Note that the path algebra with no relations is always hereditary. | |
| Sep 5, 2013 at 10:22 | comment | added | Dag Oskar Madsen | Do you assume $I \subseteq J^2$ (where $I$ is the ideal generated by the relations and $J$ is the ideal generated by the arrows)? | |
| Sep 5, 2013 at 8:11 | history | asked | Oliver Straser | CC BY-SA 3.0 |