TITLE: Thick subcategories
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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.