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?


1 Answer 1


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.

Now the idea that a tilting complex gives back the original ring is interesting. Given how rare commutative endomorphism rings are, yet I don't think that allows us to say much given how little we understand about what it means for an endomorphism ring to be commutative. Even for the case where we have endomorphism rings for abelian groups we very little idea what it means.

However I may be wrong, in that it could help but from what I can gather and what we know (or what I think we know) I don't think so.


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