# Is K(R-Mod) compactly generated when R is an artin algebra?

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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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. –  Greg Stevenson Apr 6 '10 at 21:17

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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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. –  Greg Stevenson Jul 31 '10 at 5:30