## 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 $n It is the fact that all components of$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$cannot be a generator. I accepted Mariano's answer because I felt it was the most elegant, but I learned something from all of the answers posted. Thanks everyone! - You have a forgetful functor to abelian groups: let$A$be the endomorphisms of this functor considered as a topological ring via the projective limit topology. Your category will be equivalent to discrete modules over this ring since the forgetful functor is exact, faithful, and commutes with colimits. You can see topologies are necessary already in the case of graded vector spaces, where the ring is$\mathcal{O}_{\mathbb{G}_m}^{\vee}$equipped with the algebra structure coming from the coalgebra structure on$\mathbb{G}_m$. – Moosbrugger Jan 12 2012 at 17:28 Notice that for graded$k$-vector spaces you can simply use$k^{(\mathbb Z)}$with pointwise operations. – Mariano Suárez-Alvarez Jan 12 2012 at 18:45 ## 3 Answers$\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$does what you want. -  What is the algebra associated to a (monoidal) category? – Martin Brandenburg Jan 12 2012 at 19:30 @Martin: It is the algebra generated by the Hom(X,Y) where multiplication is composition of arrows if defined and zero otherwise. – Johannes Hahn Jan 12 2012 at 19:35 Your functor from$Q$has values in abelian groups, not vector spaces, right? – Martin Brandenburg Jan 12 2012 at 19:35 Indeed, Martin. I tend to think about algebras over a field exclusively :) – Mariano Suárez-Alvarez Jan 12 2012 at 20:03 Very nice, Mariano! I think this is just what I was looking for. – MTS Jan 13 2012 at 0:47 show 4 more comments This answer is just an elaboration of Moosbrugger's comment. For simplicity, assume that$R$is just a field$k$concentrated in degree zero. Put$S=\text{Map}(\mathbb{Z},k)$, and let$e_n\in S$be the obvious idempotent supported at$n$. For any$S$-module$M$we can put$(FM)_n=M/((1-e_n)M)$; this gives a functor $F:\text{Mod}_S\to\text{GrMod}_R$. Some comments above suggest that this should be an equivalence, but it is not. To see this, let$J\leq S$be the ideal of finitely-supported functions. It is then easy to see that$S/J\neq0$but$F(S/J)=0$. This is the point behind Moosbrugger's remark about topologies. Suppose we give$S$the product topology, and consider only modules$M$such that the action map$S\times M\to M$is continuous with respect to the discrete topology. I think this just means that for all$m\in M$there is a finite set$K$such that$m=\sum_{k\in K}e_km$. Using this we find that$M\simeq\bigoplus_{n\in\mathbb{Z}}(FM)_n$, and thus that$F$gives an equivalence from the continuous module category to$\text{GrMod}_R$. - The answer is yes. Let$T'$be the path$\mathbb{Z}$-algebra of the Quiver $$\cdots\rightarrow n-1\rightarrow n \rightarrow n+1 \rightarrow\cdots$$ This$\mathbb{Z}$-algebra is defined by generators$\{a_n,b_n\}_{n\in\mathbb{Z}}$, where$a_n$is morally the$n^{\text{th}}$vertex of the quiver, and$b_n$is the morphism$n\rightarrow n+1$. The derining relations are $$a_n^2=a_n,\qquad b_na_n=b_n, \qquad a_{n+1}b_n=a_n,\qquad a_ma_n=0\;\text{ if }\;m\neq n.$$ A right$R\otimes T'$-module$M$is the same as a diagram of$R$-modules $$\cdots\rightarrow M_{n-1}\rightarrow M_n \rightarrow M_{n+1} \rightarrow\cdots$$ The correspondence is given by$M_n=a_nM$, and$M_n \rightarrow M_{n+1}$is left multiplication by$b_n$. Since you want complexes, let$T$be the quotient of$T'$by the additional relations $$b_{n+1}b_n=0.$$ Right$R\otimes T$-modules are complexes of$R$-modules. You may dislike that$T'$,$R\otimes T'$,$T$and$R\otimes T$don't have a unit. But this is easy to solve. Just take the unitalization of these rings ($\mathbb{Z}$-algebras) and unital modules over them. - Notice the question is for graded modules, not complexes :) – Mariano Suárez-Alvarez Jan 12 2012 at 18:37 There is a little trouble with your answer: the algebra$T'$has modules where all generators$a_n,b_n$act by zero. – Victor Ostrik Jan 12 2012 at 18:38 @Victor, on usually looks that the modules over the algebra$T'$which are locally unitary ($T'$does have local units) – Mariano Suárez-Alvarez Jan 12 2012 at 19:01 Mariano, you are surely right. All I say is that locally unitary modules is not the same as all modules; in particular modules over unitalization of$T'$is not the same as locally unitary$T'-\$modules. – Victor Ostrik Jan 12 2012 at 19:08
@Mariano: As you probably know, your local unitality is the same as my topology above! One won't be able to get this category as (unital) modules over a (unital) ring, and every answer in this thread finds some way to circumvent it :) – Moosbrugger Jan 12 2012 at 21:49
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