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Added another example without the condition $\mathbb{Q}\subseteq A$
Johannes Hahn
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For some abelian categories it is also very easy to describe such a ring quite explicitly if the category you start with is similar enough to a module category.

Let's say you consider $\mathsf{Ch}(A\mathsf{-mod})$ and that $A$ is a $\mathbb{Q}$-algebra. Then $\mathsf{Ch}(A\mathsf{-mod})$ embeds as a full subcategory into $B-\mathsf{mod}$ where $$B:=A \otimes \mathbb{Z}\langle z,d\rangle / ([z,d]=d, d^2=0)$$ by mapping a chain complex $(X_\ast,\partial)\in\mathsf{Ch}(A\mathsf{-mod})$ to $\bigoplus_{n\in\mathbb{Z}} X_n$ and defining the $B$-action by $$\forall x\in X_n: z\cdot x := nx, d\cdot x := \partial(x).$$

In particular $z$ acts diagonalisable on this $B$-module, has spectrum $\mathbb{Z}$ and $X_n$ is exactly the eigenspace of $z$ for the eigenvalue $n$. It is easy to verify that every chain-map defines a $B$-linear map and vice versa so that this construction gives an equivalence between $\mathsf{Ch}(A\mathsf{-mod})$ and the full subcategory of $B\mathsf{-mod}$ consisting of those $B$-modules on which $z$ acts diagonalisable with spectrum $\mathbb{Z}$.


One can remove the condition $\mathbb{Q}\subseteq A$ by using slightly more complicated algebras, like $$C:=A \otimes \mathbb{Z}\langle e_n, d | n\in\mathbb{Z}\rangle / (e_i e_j = \delta_{ij} e_i, d e_n = e_{n-1} d, d^2=0)$$ This also algebra also acts on $\bigoplus_n X_n$: $e_n$ is the projection onto the $n$-component and $d x_n =\partial(x_n)$ for all $x_n\in X_n$.

In this way $\mathsf{Ch}(A\mathsf{-mod})$ gets identified with the full subcategory of $C\mathsf{-mod}$ of all those modules that satisfy $M=\sum_{n\in\mathbb{Z}} e_n M$.

This second construction can be generalised: $C$ is a special case of a category algebra of an $\mathsf{Ab}$-enriched category $\mathsf{Ch}$: It has object set $\mathbb{Z}$ and $Hom(n,m)=\mathbb{Z}\partial$. The category of chain complexes of $A$-modules is nothing else than the functor category (of additive functors) $Fun_\mathbb{Z}(\mathsf{Ch},A\mathsf{mod})$.

Generally: Every category that happens to be the functor category (of $k$-linear functors) $Fun_k(\mathsf{C},A\mathsf{-mod})$ from some $k$-linear category $\mathsf{C}$ into a module category can be embedded into $(A\otimes_k k[\mathsf{C}])-\mathsf{mod}$ in a similar spirit. And even more generally: If $\mathsf{A}$ is an abelian category which you already know how to embed into a module category, then you can adapt this construction to get an embedding of $Fun_k(\mathsf{C},\mathsf{A})$ into a module category as well.

Johannes Hahn
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