Main Question
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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?




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Motivation
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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.




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### 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!



Module category equivalent to graded module category?
$\newcommand\ZZ{\mathbb{Z}}$Let $R=\bigoplus\_{n\in\mathbb N\_0}R\_n$ be your graded ring. Construct a category $Q$ with objects $\ZZ$ and where $\hom\_Q(n,m)=R\_{m-n}$ with composition coming from the multiplication in $R$. A graded left $R$-module is the same thing as a functor from $Q$ to abelian groups. The (non-unital) ring associated to $Q$ does what you want.


