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. 



Thick subcategories
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.


