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I wonder if the triangulated category K(R-Mod) is compactly generated when R is an artin algebra? R-Mod denotes all left R-modules. I understand this would be true if R has finite representation type since R-modules then are direct sums of finitely generated ones, but I am interested in the general case. Could it be that a generating set are finitely generated R-modules and shifts of them. (This would not be true for general rings, e.g. Neeman showed that K(Z-mod) is not compactly generated.)

Thanks.

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  • $\begingroup$ I don't believe that the shifts of finitely generated modules could form a generating set in general. The bounded derived category of finitely generated modules generates (up to equivalence) the homotopy category of complexes of injective R-modules, but the inclusion of K(R-Inj) into K(R-Mod) preserves coproducts, so it won't generate K(R-Mod) unless the inclusion is an equivalence. In general I don't believe that this is the case. $\endgroup$
    – Greg Stevenson
    Commented Apr 6, 2010 at 21:17

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The answer is in general no - $K(R\text{-}\mathrm{Mod})$ can fail to be well generated even when $R$ is artinian. As you mention $K(R\text{-}\mathrm{Mod})$ is compactly generated if $R$ is of finite representation type. It turns out that the converse holds. This is a result of Jan Šťovíček which occurs as Proposition 2.6 in this paper. The precise result is:

Proposition Let $R$ be a ring. The following are equivalent:

(i) $K(R\text{-}\mathrm{Mod})$ is well generated;

(ii) $K(R\text{-}\mathrm{Mod})$ is compactly generated;

(iii) $R$ is left pure semisimple.

In particular, when $R$ is artinian this occurs precisely when $R$ has finite representation type.

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  • $\begingroup$ 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. $\endgroup$
    – user5189
    Commented Apr 7, 2010 at 0:29
  • $\begingroup$ 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... $\endgroup$
    – Greg Stevenson
    Commented Apr 7, 2010 at 0:47
  • $\begingroup$ 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. $\endgroup$
    – Greg Stevenson
    Commented Jul 31, 2010 at 5:30
  • $\begingroup$ (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. $\endgroup$
    – Todd Trimble
    Commented Sep 14, 2014 at 21:57

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