Thick subcategories

I hope this question is not too trivial for mathoverfolw. Let $R$ be a commutative ring (with $0\neq 1$) and $D_{Perf}(R)$ the triangulated category of perfect complexes. Let $C$ be a thick subcategory of $D_{Perf}(R)$ is it true that there exists a (commutative ?) ring $A$ such that $C$ is equivalent to $D_{perf}(A)$ as triangulated category? If yes is there a precise description of $A$ ? Thank you.

• Not in general. If $R$ is a Dedekind domain (so every complex is a sum of shifts of modules) then no non-trivial thick subcategory of $\mathsf{D}^\mathrm{perf}(R)$ can have a tilting object. One can reduce to looking at what happens at a point and just see that all the finitely generated modules supported at that point have self-extensions so there is no object with the correct derived endomorphism ring. – Greg Stevenson Jun 8 '17 at 15:56
• @GregStevenson Is the answer still negative if we allow $A$ to be a differential graded algebra ? – M.O. Jun 8 '17 at 15:58
• No, you can construct a DGA easily, e.g. use the Koszul complex. – Leo Alonso Jun 8 '17 at 16:01
• @LeoAlonso I don't understand what do you mean by "No"? :) could you be more precise please. – M.O. Jun 8 '17 at 16:04
• As Leo says, if you allow DGAs then the answer will be yes, it is perfect complexes over a DGA, provided the thick subcategory has a generator. This translates to its support (the thing that classifies the thick subcategories of $\mathsf{D}^\mathrm{perf}(R)$) being a closed subset with quasi-compact complement (rather than a possibly infinite union of such). – Greg Stevenson Jun 8 '17 at 16:08

The answer is no, in principle. By Thomason's classification of thick subcategories, these correspond to certain stable for specialization subsets, i.e. arbitrary unions of closed subsets (with quasi-compact complement).

Even for a closed subset $Z \subset \mathrm{Spec}(R)$ with quasi-compact complement the corresponding thick subcategory has objects the perfect complexes supported at $Z$, something than one may denote $D_{Z}(A)_\mathrm{perf}$. If $I$ is the ideal of $Z$ this is bigger than $D(A/I)_\mathrm{perf}$. In algebraic terms, a complex $M^\cdot \in D_{Z}(A)$ if and only if its homologies are killed by any power of $I$.

If you allow DGA then it is possible, see

Dwyer, W. G.; Greenlees, J. P. C. Complete modules and torsion modules. Amer. J. Math. 124 (2002), no. 1, 199–220.

See, concretely, Theorem 2.1: $\mathbf{A}_\mathrm{tors}$ denotes $D_{Z}(A)_\mathrm{perf}$ and  $\mathcal{E}$ is the DGA you asked for.

• I agree, but this doesn't rule out the existence of some tilting object in $\mathsf{D}_Z^\mathrm{perf}(R)$ as far as I can see. I am fairly sure the answer is still no, but I don't currently see a proof. – Greg Stevenson Jun 8 '17 at 15:54
• Geometrically $\mathsf{D}_Z^\mathrm{perf}(R)$ corresponds to the category of torsion sheaves on the formal scheme $\mathrm{Spf}(\widehat{R})$. I can't give a precise argument but, in my experience, looks very different from a derived category of quasi-coherent sheaves. – Leo Alonso Jun 8 '17 at 15:58
• @LeoAlonso could you be more precise where should I look in your linked reference, please – M.O. Jun 8 '17 at 16:09
• @LeoAlonso Thanks I have accepted your answer but I don't have enough reputation to upvote :) – M.O. Jun 8 '17 at 16:16