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.

$\begingroup$ Not in general. If $R$ is a Dedekind domain (so every complex is a sum of shifts of modules) then no nontrivial 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 selfextensions so there is no object with the correct derived endomorphism ring. $\endgroup$ – Greg Stevenson Jun 8 '17 at 15:56

$\begingroup$ @GregStevenson Is the answer still negative if we allow $A$ to be a differential graded algebra ? $\endgroup$ – M.O. Jun 8 '17 at 15:58

$\begingroup$ No, you can construct a DGA easily, e.g. use the Koszul complex. $\endgroup$ – Leo Alonso Jun 8 '17 at 16:01

$\begingroup$ @LeoAlonso I don't understand what do you mean by "No"? :) could you be more precise please. $\endgroup$ – M.O. Jun 8 '17 at 16:04

1$\begingroup$ 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 quasicompact complement (rather than a possibly infinite union of such). $\endgroup$ – 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 quasicompact complement).
Even for a closed subset $Z \subset \mathrm{Spec}(R)$ with quasicompact 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.

$\begingroup$ 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. $\endgroup$ – Greg Stevenson Jun 8 '17 at 15:54

$\begingroup$ 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 quasicoherent sheaves. $\endgroup$ – Leo Alonso Jun 8 '17 at 15:58

$\begingroup$ @LeoAlonso could you be more precise where should I look in your linked reference, please $\endgroup$ – M.O. Jun 8 '17 at 16:09

$\begingroup$ @LeoAlonso Thanks I have accepted your answer but I don't have enough reputation to upvote :) $\endgroup$ – M.O. Jun 8 '17 at 16:16