Equivalences of categories of complexes of modules

Question

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.

Answer 1

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.