Skip to main content
4 events
when toggle format what by license comment
Aug 29 at 14:39 comment added Najib Idrissi I cannot tell where, in the proof of Lemma 5.12, they would use that $F(\mathrm{Ass}, A)$ with some differential is a resolution of $A$. Intuitively this doesn't look right - in any case, it's concentrated in degree 0 if $A$ is, so it cannot even have a nontrivial differential...
Aug 13 at 12:52 comment added JD1874 Hm, perhaps it's okay to simply choose my almost-free resolution to be the free algebra with the differential which forces it to be acyclic. The underlying thing is certainly free and it's weakly equivalent to the thing we started with.
Aug 12 at 13:43 history edited Connor Malin CC BY-SA 4.0
deleted 35 characters in body
Aug 12 at 12:55 history asked JD1874 CC BY-SA 4.0