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The title says pretty much what I want. Of course, the abelian categories should contain at least one nonzero object.

In particular, is there an abelian category containing only one nonzero object? On the one hand, this is equivalent to construct a ring which is the endomorphism of the nonzero object. On the other hand, this is equivalent to construct a special module by Freyd–Mitchell theorem.

This seems silly for that's not what abelian category is invented for, but I really want to know the answer.

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    $\begingroup$ If there is an abelian category $\mathcal{C}$ with only one non-zero object $A$, then its endomorphism ring must fail to have the invariant basis number property: because then $\mathcal{C}(A, A) \cong \mathcal{C}(A, A \times A) \cong \mathcal{C}(A, A) \times \mathcal{C}(A, A)$ as right $\mathcal{C}(A, A)$-modules. $\endgroup$
    – Zhen Lin
    Commented Nov 10, 2012 at 8:26
  • $\begingroup$ Considering the multiplication by $p$ on the nonzero object with $p$ prime, it seems the base ring $R$ of the hypothetical special module can be taken to be either a $\mathbf{Z}/p\mathbf{Z}$ algebra or a $\mathbf{Q}$-algebra. $\endgroup$
    – François Brunault
    Commented Nov 10, 2012 at 14:10

1 Answer 1

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Take the category of (at most) countable-dimensional vector spaces over your favourite field. Then take the quotient by the Serre subcategory of finite-dimensional vector spaces. (And take a skeletal subcategory so that it strictly has only two objects.)

Then this is an abelian category with only one non-zero object, whose endomorphism ring is the endomorphism ring of a countable-dimensional vector space, localized at the set of endomorphisms with finite-dimensional kernel and cokernel.

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    $\begingroup$ Ah, this looks good! I hadn't hit on the idea of quotienting by the Serre subcategory. $\endgroup$
    – Todd Trimble
    Commented Nov 10, 2012 at 14:48
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    $\begingroup$ Just to clarify, while this category is the localization at a Serre subcategory, the endomorphism ring of a countably-infinite dimensional vector space is factored out by the set (ideal) of endomorphisms with finite rank. Correct? $\endgroup$
    – Manny Reyes
    Commented Nov 10, 2012 at 20:10
  • $\begingroup$ In this particular example, I think the localization at the Serre subcategory is the same as factoring out the finite rank maps, since any linear map defined on a subspace of finite codimension can be extended to the whole space, and the dual statement is alao true, so any map in the localization is induced by a genuine linear map. So yes, that gives a more elementary construction, although saying it in terms of localzation means that less explanation is needed for why you get an abelian category. $\endgroup$
    – Jeremy Rickard
    Commented Nov 11, 2012 at 11:48
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    $\begingroup$ Nice construction. I wonder if there is an example which is a full subcategory of $R-\mathrm{Mod}$ where the base ring $R$ is commutative. Is the ring you construct isomorphic to $\mathrm{End}(M)$ for some module $M$ over a commutative ring? $\endgroup$
    – François Brunault
    Commented Nov 11, 2012 at 19:04

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