Timeline for Indecomposable objects in bounded derived category of $\mathbb C[x]/x^2$-mod
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 10, 2019 at 0:13 | comment | added | Mare | By arxiv.org/pdf/1910.01494.pdf , it seems that K[x]/(x^n) is not derived tame for n>=3 (but it is for n=2). | |
| Dec 9, 2019 at 20:09 | comment | added | Zhiyu | I think this may be accessible, as we can compute all ext between finitely generated modules. | |
| Dec 9, 2019 at 11:02 | comment | added | Jeremy Rickard | @sawdada I don't think so. I'm not sure, but I suspect that for most $n$ this is a wild classification problem. | |
| Dec 9, 2019 at 5:50 | vote | accept | Zhiyu | ||
| Dec 8, 2019 at 23:52 | comment | added | Zhiyu | OK, I see. Can we classify things over $\mathbb C[x]/x^n$ using your method? | |
| Dec 8, 2019 at 21:23 | comment | added | Jeremy Rickard | @sawdada That's why I truncated the projective resolution to the left of its homology. If, in degrees $n$ and below, a projective resolution of an object looks like $P_n\stackrel{d_n}{\to} P_{n-1}\to P_{n-2}\to\dots$, and if the object has zero homology in degrees $n$ and above, then the object is quasiisomorphic to $\dots\to0\to\ker{d_n}\to P_n\to P_{n-1}\to P_{n-2}\to\dots$. | |
| Dec 8, 2019 at 18:31 | comment | added | Zhiyu | Thank you, I have a question: $\mathbb C$ is indecomposable but not perfect in the derived category, so we can't find a projective resolution in the bounded derived category. How to rule out such possibility? | |
| Dec 8, 2019 at 13:48 | history | answered | Jeremy Rickard | CC BY-SA 4.0 |