5

For a ring $R$, which is a finite-dimensional algebra over a field, the category of finite-dimensional, projective, right $R$-modules, $\mathcal{P}_R$ is generated by the indecomposable projective modules, in the sense that every object is isomorphic to a direct sum of them. Is there a similar statement for the bounded-above derived category $\mathcal{D}^-_R=\mathcal{K}^-\left(\mathcal{P}_R\right)$? I want to say that $\mathcal{D}^-_R$ is generated by projective resolutions of the irreducible modules. Is that right?

flag
No, it's wrong. If that were so then we would expect to be able to write the indecomposable projectiles as a direct sum of the projective resolutions of the simple modules so that taking 0th homology we'd get that they were decomposable. – Eitan Chatav Feb 29 2012 at 2:07

1 Answer

2

In a triangulated or dg-category, this is not the usual notion of "generators." One definition is that there is no smaller triangulated subcategory containing the objects. In this sense, the indecomposable projectives do generate the homotopy category over all projectives. This is also true for the projective resolutions of the simples; since the algebra is finite dimensional, there's a finite iterated cone of the resolutions of the simples which is quasi-isomorphic to any indecomposable projective, even if the algebra doesn't have finite global dimension.

link|flag
1 
«Indecomposable projectiles»... I have many times wished to throw them far, far away! – Mariano Suárez-Alvarez Feb 29 2012 at 5:09
Haha, stupid spell checker changed "projectives" to "projectiles" :-) – Eitan Chatav Feb 29 2012 at 12:46
Yeah, I hadn't even realized that was what happened until it did it again (actually while I was writing you an email). – Ben Webster Mar 1 2012 at 3:54

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.