# Commutative rings with same derived category

Let $R,S$ be two commutative rings, assume there is an equivalence $F:D(R-mod) \cong D(S-mod)$ as triangulated categories, is there a simple way to show that $R,S$ are isomorphic?

(I do not assume $R,S$ are finite dimensional algebras over a field, as this paper shows derived morita equivalence is the same as morita equivalence in such case.)

As $Z(D(A-mod))$ may not equal to $Z(A)$ for general ring, it's not good to just consider the center of each category as in the usual Morita equivalence. Also, assume $T=F(R)$ then $T$ must be compact object as well hence is a perfect complex (i.e $T$ is a bounded complex of finitely generated projective modules), and $R \cong End_{D(B-mod)}(T)$ is commutative, will this help?

For a ring $R$, the unbounded derived category $D(R)$ doesn't have enough information to recover the module category $R$-mod. In order to do this we need a $t$-structure.
Suppose $R$ and $S$ are derived morita equivalent and there is a $t$-structure on one of their derived categories. Assuming triangulated equivalences preserve $t$-structures we can recover equivalent module categories $R\mathrm{-mod}\simeq S\mathrm{-mod}$ which as $R$ and $S$ are commutative means morita equivalnce implies isomorphism.
Here is a reference I read in order to get these ideas. I don't know too much about $t$-structures, however it seems like a sufficient condition for proving $R$ and $S$ are isomorphic. Is it necessery? I do not know.