# Equivalences of categories of complexes of modules

Let $$k$$ be an algebraically closed field of characteristic $$0$$. Let $$R, S$$ be two commutative $$k$$-algebras.

Let $$\operatorname{Ch}(R), \operatorname{Ch}(S)$$ be the categories of complexes of $$R$$-modules and $$S$$-modules, respectively.

Question If $$\operatorname{Ch}(R), \operatorname{Ch}(S)$$ are equivalences as $$k$$-linear abelian categories, then $$R \simeq S$$ as $$k$$-algebras?

It is well-known that $$\operatorname{Mod}(R) \simeq \operatorname{Mod}(S)$$ implies $$R \simeq S$$, where $$\operatorname{Mod}(R), \operatorname{Mod}(S)$$ are the categories of $$R$$-modules and $$S$$-modules, respectively. This is a corollary of Morita equivalence. I consider a variation of this fact.

In the future, I want to consider a generalization in the case of commutative differential graded algebras.

Any comments and references are welcome. Thank you!

The same question is in MSE.

The answer is yes by the same type of Morita theory, namely $$Z(Ch(R))\cong R$$, where $$Z(A)$$ is $$End(id_A)$$, the ring of endomorphisms of an abelian category
EDIT : sorry, I hadn't seen that you were asking about an isomorphism as $$k$$-algebras. Then this is clearly wrong, simply because a given $$R$$ can have two different $$k$$-algebra structures, and the category $$Ch(R)$$ has no way of knowing this. In particular, I answer below the question of whether $$R\cong S$$ as rings (the answer is yes). It is also not hard to adapt my answer to the case where $$Ch(R)\simeq Ch(S)$$ is assumed to be a $$k$$-linear equivalence, rather than an equivalence of abelian categories - everything I said sort of goes through $$k$$-linearly too.
It is clear that we have a map $$R\to Z(Ch(R))$$, given by degreewise multiplication by elements of $$R$$ (this is where I use that $$R$$ is commutative), and a map $$Z(Ch(R))\to R$$ given by evaluation at $$R[0]$$, or at $$(R\to R)$$.
Let $$D^{n+1} = (R\xrightarrow{id_R} R)[n]$$. We have that $$\hom(D^n,C) \cong C_n$$ , naturally in $$C$$, so the identity functor $$Ch(R)\to Ch(R)$$ is co-represented by a cochain complex $$D^n\to D^{n+1}$$ in $$Ch(R)$$ (that is, an object of $$CoCh(Ch(R))$$).
By a slight variant of the Yoneda lemma, we find that $$\hom(id,id)\cong \hom(D^\bullet,D^\bullet)$$. But now each $$\hom(D^n,D^n)$$ is isomorphic to $$R$$, and the differential $$D^n\to D^{n+1}$$ forces any endomorphism of $$D^\bullet$$ to be given by the same element of $$r$$ on every $$D^n$$ (that's because $$D^n\to D^{n+1}$$ is an isomorphism in degree $$n$$), so $$\hom(D^\bullet,D^\bullet)\cong R$$. I hope it's clear that all these identifications are compatible with the map $$R\to Z(Ch(R))$$. So we find the desired claim.
• Thank you for your wonderful answer. I should have assumed the categories are $k$-linear equivalence. (I have added this to the question.) Sep 28 at 16:23