# Are there any nontrivial abelian categories with only finitely many objects?

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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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. –  Zhen Lin Nov 10 '12 at 8:26
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. –  François Brunault Nov 10 '12 at 14:10

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? –  François Brunault Nov 11 '12 at 19:04
@Simone : But $R$ is not commutative. –  Jeremy Rickard Nov 22 '12 at 19:59