## Main Question

Let $R$ be a graded ring, graded by the nonnegative integers. Denote by $\mathrm{gr}R-\mathrm{Mod}$ the category of $\mathbb{Z}$-graded left $R$-modules with morphisms that preserve the grading. Is there a ring $S$ such that the category $S-\mathrm{Mod}$ of left $S$-modules is equivalent to $\mathrm{gr}R-\mathrm{Mod}$?

As the category of graded $R$-modules is abelian, the Freyd-Mitchell theorem guarantees an exact embedding into a module category, but this is not necessarily an equivalence of categories, right?

## Motivation

My motivation for the question is an offhand remark made to me indicating that for a given ring $A$, there is a ring $S$ such that the category of complexes of $A$-modules is equivalent to the category of $S$-modules. If you define the graded ring $R = A[t]/(t^2)$, graded by powers of $t$, then (I think) complexes of $A$-modules are equivalent to graded $R$-modules, so the question is reduced to the main question stated above.

Of course, it could be the case that the answer to the original question I asked is negative, and yet the offhand remark is still true, in which case I would be interested in hearing about why that is.

### Edit

I neglected to mention that I was hoping for a *unital* ring $S$ with *unital* modules. As several people have pointed out in the comments, this is not possible. I thought I would put up an argument to show why this is in case people come looking at this post in the future.

Theorem 1 of Chapter 4, Section 11 of the book *Categories and Functors*, by Bodo Pareigis, gives a complete characterization of when an abelian category $\mathcal{C}$ is a module category. The criteria are that $\mathcal{C}$ must contain a progenerator (i.e. a finite, projective generator; I had to look that up) and it must contain arbitrary coproducts of that generator.

Now let's see that $\mathrm{gr}R-\mathrm{Mod}$ cannot contain a finite progenerator. Take any finite (hence finitely generated) projective module $P = \bigoplus_{n \in \mathbb{Z}} P_n$. Since $P$ is finitely generated, there is some index $k_0$ such that $P_n = 0$ for $n<k_0$ (this uses the fact that $R$ is graded by the nonnegative integers).

It is the fact that all components of $P$ below $k_0$ vanish that prevents $P$ from being a generator. For example, take the graded module $M$ such that $M_{k_0 -1} = R_0$ and all other $M_n = 0$. Since the only map from $P$ to $M$ is the zero map, morphisms from $P$ to $M$ cannot distinguish morphisms originating from $M$. Hence $P$ cannot be a generator.

I accepted Mariano's answer because I felt it was the most elegant, but I learned something from all of the answers posted. Thanks everyone!