Let $A$ be a ring. If $e$ is an idempotent, then there is an abelian recollement involving the categories $A\text{-}\mathrm{Mod}$ and $eAe\text{-}\mathrm{Mod}$. This is Example 2.7 in Homological theory of recollements of abelian categories by Chrysostomos Psaroudakis.

There is a "restriction" funtor $e(-):A\text{-}\mathrm{Mod}\to eAe\text{-}\mathrm{Mod}$ and its right adjoint "direct image" functor $\mathrm{Hom}_{eAe}(eA,-),$ so if we think of $A\text{-}\mathrm{Mod}$ as sheaves on something noncommutative, then $eAe\text{-}\mathrm{Mod}$ behaves at least somewhat like sheaves on an open chart.

Now let's say we have a collection of orthogonal idempotents $e_1,\ldots,e_m\in A$ with $1=e_1+\ldots + e_m.$ Also for $I\subset\{1,\ldots,m\}$ denote $e(I)=\sum_{i\in I} e_i.$ I have in front of me such a situation in which it seems there is a sort of descent for modules over $A$: the category $A\text{-}\mathrm{Mod}$ seems to be equivalent to the category of descent data, the objects of which are collections of modules $M_j$ over $e(I_j)Ae(I_j)$ for some collection of indexing sets $I_j$ such that $\bigcup_j I_j=\{1,\ldots,m\}$ with isomorphisms $e(I_i \cap I_j)M_i\simeq e(I_i \cap I_j)M_j$ satisfying the usual cocycle conditions.

For now, in my partial proof of this equivalence I do use the properties of my algebra $A$, but maybe that's not really necessary? Has this sort of descent for module categories been worked out before? I have a gut feeling that this is something known and I just don't know what algebraists would call it.


1 Answer 1


If $m=2$, your data gives no information about the relationship between $e_1M$ as an $e_1Ae_1$-module and $e_2M$ as an $e_2Ae_2$-module. So this is false. For example, take a quiver with two vertices, and one edge in each direction, with relations saying that every composite of three of these edges is zero. Then you can't distinguish between an indecomposable of length two and the sum of its two composition factors. It's unlikely to be true for $m>2$ either, for similar reasons.

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    $\begingroup$ I see, so one does really need something additional from the algebra. Thank you! $\endgroup$ Sep 30 at 10:47

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