Timeline for Is K(R-Mod) compactly generated when R is an artin algebra?
Current License: CC BY-SA 2.5
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| Sep 14, 2014 at 21:57 | comment | added | Todd Trimble | (continuing guest's comment April 7, '10 0:29) Therefore, by the above argument R-mod would be generated by all shifts of finitely generated R-modules (which is a compact set). Or am I wrong? For a general artin algebra R my hope would be that if a surjection d:A \to B (read M^i \to Im d^i) is non-split then there is a finitely generated submodule B' of B such that d^{-1}(B') \to B' doesnt split, because then the funtor Hom(B', ) would detect the failure to split. But I dont know if this hope is true. | |
| Jul 31, 2010 at 5:30 | comment | added | Greg Stevenson | Some/all credit should go to Daniel Murfet. I mentioned this question to him and he told me that Šťovíček had answered it which reminded of the paper. | |
| Jul 31, 2010 at 5:26 | history | answered | Greg Stevenson | CC BY-SA 2.5 | |
| Apr 7, 2010 at 0:47 | comment | added | Greg Stevenson | It was early when I wrote that and thinking about it again I do see a problem (I wasn't embedding the modules as stalk complexes). I'll think about this some more and see what I can come up with... | |
| Apr 7, 2010 at 0:29 | comment | added | user5189 | I don't understand Gregs argument. In general it is true that K(R) is generated by the class of all R-modules and shifts of them. Because if Hom_K(x[i],M) = 0 for shifts of all R-modules x then taking x=R we see that the complex M = (M,d) is exact. On the other hand, if M is exact we can use that Hom_K(Im d^i[-i-1], M) = 0 to conclude that the projection M^i \to Im d^i splits for all i. This shows that M is homotopic to 0. Finally, if R has finite representation type (which btw doesn't imply that K(Inj R) = K(R-mod)) it is known that any R-module is a direct sum of finitely generated ones. |