Timeline for Is the tensorproduct of a triangulated category with a ring again triangulated?
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Jul 8, 2010 at 9:13 | vote | accept | Manuel Koehler | ||
| Jun 2, 2010 at 23:37 | comment | added | Greg Stevenson | I should have made it clear that by $D^b(\mathbb{Z})$ I meant the bounded derived category of finitely generated abelian groups. So in fact one can split $C_x$ as a finite sum of shifts of $\mathbb{Z}$ and $\mathbb{Z}/p^n\mathbb{Z}$ for various primes. This follows from the structure theorem for finitely generated abelian groups together with the fact that $\mathbb{Z}$ has global dimension 1 which can be used to show that complexes are quasi-isomorphic to the direct sum of their cohomology groups. | |
| Jun 2, 2010 at 10:50 | vote | accept | Manuel Koehler | ||
| Jun 2, 2010 at 10:56 | |||||
| Jun 2, 2010 at 10:47 | comment | added | Manuel Koehler | This is a neat counterexample, thanks. The only thing I do not understand is why $C_x$ is decomposable in a free and a torsion part. But one could also argue that the assumption that T is triangulated leads to the contradiction that Z decomposes into a nontrivial direct sum of abelian groups. The situation you mention in the end handles Z[1/n]. I think this even works without the assumption that T is tensor triangulated - the morphisms which become isomorphisms after multiplication with some power of n form a system which arises from a cohomological functor (Weibel, 10.4) | |
| Jun 2, 2010 at 2:03 | history | answered | Greg Stevenson | CC BY-SA 2.5 |