Let $Q$ be a Dynkin quiver. Let $\mathbb CQ$ be its complex path algebra. It is defined in a way such that modules over $\mathbb CQ$ are the same as representations of the quiver $Q$. Let's write $\mathrm{mod}(\mathbb CQ)$ for the finite-dimensional $\mathbb CQ$-modules. There is a beautiful description of the bounded derived category $D^b(\mathrm{mod}(\mathbb CQ))$ in terms of translation quivers and mesh relations, see Happel: On the derived category of a finite-dimensional algebra.

I would be interested in a similar description of the bounded derived category $D^b(\mathrm{mod}(\mathbb ZQ))$ of finitely generated modules over the path ring $\mathbb ZQ$. More precisely, I am looking for a result along the lines "There is a nicely characterized full subcategory in $D^b(\mathrm{mod}(\mathbb ZQ))$ which, after tensoring with $\mathbb C$, becomes equivalent to $D^b(\mathrm{mod}(\mathbb CQ))$." Since Happel's paper is from 1987, I believe someone must have tried to generalise his results to integer representations since then.

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    $\begingroup$ @Rasmus: great general question. If I understand your more preciese question, I think that the entire category $D^b(mod(\mathbb{Z}Q))$ has the property you want. $\endgroup$ – Benjamin Antieau Mar 23 '13 at 0:50
  • $\begingroup$ @Benjamin: I think you are right. I was probably aiming for something like a full subcategory $\mathcal C\subset D^b(\mathrm{mod}(\mathbb ZQ))$ such that the obvious functor from $\mathcal C\otimes\mathbb C$ to $D^b(\mathrm{mod}(\mathbb CQ))$ is not only an equivalence but also induces a bijection of isomorphism classes. For $Q=\bullet$, I suppose we could choose the subclass of complexes with at most one non-vanishing entry which is free. $\endgroup$ – Rasmus Mar 23 '13 at 7:55

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