Module category equivalent to graded module category? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:36:02Z http://mathoverflow.net/feeds/question/85505 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85505/module-category-equivalent-to-graded-module-category Module category equivalent to graded module category? MTS 2012-01-12T17:02:09Z 2012-01-14T05:49:42Z <h2>Main Question</h2> <p>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 $\mathrm{gr}R-\mathrm{Mod}$?</p> <p>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?</p> <hr> <h2>Motivation</h2> <p>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.</p> <p>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.</p> <hr> <h3>Edit</h3> <p>I neglected to mention that I was hoping for a <em>unital</em> ring $S$ with <em>unital</em> 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.</p> <p>Theorem 1 of Chapter 4, Section 11 of the book <em>Categories and Functors</em>, 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.</p> <p>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 $n <p>It is the fact that all components of$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$cannot be a generator.</p> <p>I accepted Mariano's answer because I felt it was the most elegant, but I learned something from all of the answers posted. Thanks everyone!</p> http://mathoverflow.net/questions/85505/module-category-equivalent-to-graded-module-category/85508#85508 Answer by Fernando Muro for Module category equivalent to graded module category? Fernando Muro 2012-01-12T17:56:27Z 2012-01-12T17:56:27Z <p>The answer is yes. Let$T'$be the path$\mathbb{Z}$-algebra of the Quiver</p> <p>$$\cdots\rightarrow n-1\rightarrow n \rightarrow n+1 \rightarrow\cdots$$</p> <p>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</p> <p>$$a_n^2=a_n,\qquad b_na_n=b_n, \qquad a_{n+1}b_n=a_n,\qquad a_ma_n=0\;\text{ if }\;m\neq n.$$</p> <p>A right$R\otimes T'$-module$M$is the same as a diagram of$R$-modules</p> <p>$$\cdots\rightarrow M_{n-1}\rightarrow M_n \rightarrow M_{n+1} \rightarrow\cdots$$</p> <p>The correspondence is given by$M_n=a_nM$, and$M_n \rightarrow M_{n+1}$is left multiplication by$b_n$.</p> <p>Since you want complexes, let$T$be the quotient of$T'$by the additional relations</p> <p>$$b_{n+1}b_n=0.$$</p> <p>Right$R\otimes T$-modules are complexes of$R$-modules.</p> <p>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 <a href="http://ncatlab.org/nlab/show/unitalization" rel="nofollow">unitalization</a> of these rings ($\mathbb{Z}$-algebras) and unital modules over them.</p> http://mathoverflow.net/questions/85505/module-category-equivalent-to-graded-module-category/85515#85515 Answer by Mariano Suárez-Alvarez for Module category equivalent to graded module category? Mariano Suárez-Alvarez 2012-01-12T18:44:03Z 2012-01-13T20:41:26Z <p>$\newcommand\ZZ{\mathbb{Z}}$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.</p> http://mathoverflow.net/questions/85505/module-category-equivalent-to-graded-module-category/85521#85521 Answer by Neil Strickland for Module category equivalent to graded module category? Neil Strickland 2012-01-12T19:53:56Z 2012-01-12T19:53:56Z <p>This answer is just an elaboration of Moosbrugger's comment.</p> <p>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 <code>$F:\text{Mod}_S\to\text{GrMod}_R$</code>. 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\$.</p>