Any abelian category as filtered colimit of categories of projective modules 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? 
 A: 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. 
