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Let $R$ be an associative ring. Let $K(R)$ be the category of chain complexes of $R$-modules and chain homotopy classes of maps between them, and let $D(R)$ be its localization with respect to acyclic complexes.

Is there a ring $S$ such that $K(R) = D(S)$? Is the question more likely to have a positive answer if I allow $S$ to be a differentially graded ring, or don't require that $S$ has an identity?

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7  
I expect this is usually false. For instance, $D(S)$ always has a compact generator, and it seems unlikely that $K(R)$ does (except maybe if $R$ is Artinian or some condition of that sort). Indeed, it's not clear to me that $K(\mathbb{Z})$ even has a set of generators, though I don't have very strong intuition for this. –  Eric Wofsey Jul 6 at 3:07

2 Answers 2

up vote 9 down vote accepted

Eric is right, $K(\mathbb{Z})$ has no generating set. This is Lemma E.3.2 in Neeman's Triangulated Categories. Presumably, the proof will apply to many other rings, but I can't comment on that.

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As Eric and Karol have noted it is usually not the case that there exists such an $S$ (which I'll take to be a dga; I don't know what happens off the top of my head if one asks for $S$ to be an honest ring).

Indeed for $K(R) \cong D(S)$ one needs $K(R)$ to be compactly generated. But by a result of Stovicek (see Theorem 2.5 here) in order for $K(R)$ to even be well generated it is necessary that $\mathrm{Mod}\;R$ be pure semisimple i.e., every $R$-module is a direct sum of finitely presented modules. In fact this is also sufficient: $K(R)$ is well generated if and only if $R$ is right pure semisimple.

This allows one to write it as a localisation of the derived category of a small dg-category (so a non-unital dga if one prefers). I don't know exactly when $K(R)$ is moreover compactly generated though.

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If $R$ is an artinian ring of finite representation type (and I think it's still an open question whether this is equivalent to being pure semisimple?) then I think $K(R)$ is equivalent to $D(S)$ for $S$ the Auslander algebra of $R$ (i.e., the endomorphism ring of the direct sum of all indecomposable modules, or its opposite ring, depending on your choice of conventions). I'm confident of this if $R$ is an artin algebra (i.e., finitely generated as a module over an artinian centre), but I might be missing a problem for general artinian rings. –  Jeremy Rickard Jul 6 at 19:42

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