Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Recently I have heard somewhere that any (edit: small) abelian category can be expressed as the colimit of categories of projective modules over some rings. The remark was that this is "basically just idempotent completion". I have been trying to figure out myself how this works but got stuck. Maybe somebody knows how to do this or where I can find a good reference.

My idea so far was to consider the rings $R=End(A,A)$ for any object $A$ in the abelian category $\mathcal A$. Then I can consider the functor $h_A:\mathcal A\to R\text{-}mod$ which maps $X$ to the $R$ module $Hom(A,X)$, which gives by Yoneda an embedding $\mathcal A\to (R\text{-}mod)^\mathcal A$. $A$ itself and its coproducts are then precisely the free $R$-modules.

If I take the idempotent completion of the category of free $R$-modules, I get the category of projective $R$-modules.

But this all doesn't really seem to go anywhere. Am I on the right way?

Also this statement seems so strong that it is odd that I can't find a reference. If I have any funtorial property for projective $R$-modules which commutes with colimits (like so many things in K-theory) I automatically have it for any abelian category. Maybe this means that the statement should be weaker?

share|improve this question
3  
If your abelian category $\mathcal{A}$ is small, then it is the category of projective $\mathcal{A}$-modules, where $\mathcal{A}$-modules are contravariant additive functors from $\mathcal{A}$ to abelian groups. –  Fernando Muro Jan 31 '13 at 12:33
    
@Fernando Muro: And how do modules over categories relate to modules over rings? What do you mean by projective? Do you have a reference? (Also: is your comment basically an answer with gaps to be filled, or rather just a remark?) –  Simon Markett Jan 31 '13 at 15:37
    
[I hope this comment makes sense...I'm really tired right now, so no promises.] Once you're in $R$-mod, but before taking idempotent completion, why can't you just use the fact that every $R$-modules is a colimit of a chain of projectives (it's projective resolution)? As for Fernando's comment, I read about this once in a paper of Hovey (though there are probably references which are more algebraic): hopf.math.purdue.edu/Hovey-Lockridge/ssrs.pdf –  David White Jan 31 '13 at 19:44
add comment

1 Answer 1

I'll denote the abelian category by $A$ and an object in it by $a$. Any $a \in A$ gives rise to a ring $\text{End}(a)$. The inclusion $a \to A$ induces a functor from the category of finitely generated projective $\text{End}(a)$-modules to $A$, since $A$ has biproducts and is idempotent complete. These module categories, over all $a \in A$, fit into a diagram of module categories of shape $A$ with $A$ as a cocone, and the claim is that $A$ is the colimit of this diagram. But it is straightforward to verify that $A$ satisfies the requisite universal property because any additive functor preserves biproducts and splitting of idempotents (these are the absolute colimits for $\text{Ab}$-enriched categories).

Edit: If the OP wants a filtered colimit as in the title then we should look at full subcategories of $A$ on finitely many objects. Here the basic fact is that if $B$ is such a category then the category of right $B$-modules is equivalent to the category of right modules over the "category algebra" $\mathbb{Z}[B]$. This is the direct sum of all the Hom spaces in $B$ where the product is composition if that is defined and zero otherwise; equivalently, it's the endomorphism ring of the direct sum of all of the objects in $B$. We need $B$ to have finitely many objects in order for $\mathbb{Z}[B]$ as in the first definition to have a unit.

The category of tiny objects in the category of right $B$-modules is the Cauchy completion of $B$ (its completion under direct sums and splitting idempotents) and the category of tiny objects in the category of right $\mathbb{Z}[B]$-modules is the category of finitely generated projective $\mathbb{Z}[B]$-modules, so in particular the Cauchy completion of $B$ (which, as above, naturally admits a functor into $A$) is the category of finitely generated projective modules over some ring. Now we write down the diagram of module categories whose objects are all finite collections of objects in $A$ (the module category being given by the Cauchy completion of the full subcategory on these objects) and whose morphisms are all inclusions, and $A$ is the colimit of this diagram as above.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.