Assume $\mathcal{A}$ is a small cocomplete abelian $\otimes$-category (see here for a definition). Is there a cocontinuous, full, faithful, exact $\otimes$-functor $\mathcal{A} \to \text{Mod}(R)$ for some ring $R$?

See here for a related question. The reason why I'm asking is that I want to prove that a certain (perfectly natural!) construction works in such a category $\mathcal{A}$ and it seems to me that the axioms of $\mathcal{A}$ are not enough to check that a certain sequence is exact. In module categories there are no problems.

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    $\begingroup$ What tensor structure do you put on $\operatorname{Mod}(R)$? When you say ring you mean commutative ring? $\endgroup$ Nov 25 '10 at 17:24
  • $\begingroup$ "Martin, could you edit the question with the detail I asked in my previous comment? Otherwise your question is not quite answerable! $\endgroup$ Nov 25 '10 at 20:53
  • $\begingroup$ $R$ commutative, $\otimes_R$ the usual tensor product. I also suppose that $\otimes$ is symmetric on $\mathcal{A}$. $\endgroup$ Nov 25 '10 at 22:17

Let ${\bf 1}$ denote the unit object of ${\mathcal A}$. If $F: {\mathcal A} \rightarrow Mod(R)$ is a tensor functor, then $F( {\bf 1} ) \simeq R$. If $F$ is fully faithful, then you can recover $R$ as $End({\bf 1})$, and the functor $F$ is given by $A \mapsto Hom( {\bf 1},A)$. This is rarely a fully faithful embedding. For example, if ${\mathcal A}$ is the category of complex representations of a finite group, then $F$ annihilates every nontrivial irreducible representation.

  • $\begingroup$ Of course, thank you. Another counterexample is then $Qcoh(X)$ for some non-affine scheme. I should have seen this. $\endgroup$ Nov 25 '10 at 18:25
  • $\begingroup$ Wait, why is $F$ isomorphic to $Hom(1,-)$? $\endgroup$ Nov 25 '10 at 22:14
  • $\begingroup$ F(A) = Hom(R, F(A)) = Hom( F(1), F(A) ) = Hom(1, A) if F is fully faithful. $\endgroup$ Nov 26 '10 at 12:00
  • $\begingroup$ Thanks. Luckily I solved my original problems without using an embedding. $\endgroup$ Nov 26 '10 at 21:32

With one modification, the answer to your question is yes. Namely, don't try to land in $\operatorname{Mod}_R$ for $R$ a commutative ring, but in $_R\operatorname{Mod}_R$, the categoy of $R$,$R$ bimodules, for $R$ an arbitrary ring. This you can do:

Phùng Hô Hài, An Embedding theorem for abelian monoidal categories, http://arxiv.org/abs/math/0004160

Well, maybe with another modification. A small cocomplete category is necessarily a poset, and so necessarily not abelian. I assume you want your category to be small, abelian, and monoidal?

  • $\begingroup$ I think the condition is meant to be small-cocomplete (ie with colimits for small diagrams) not (small and cocomplete). $\endgroup$ Nov 25 '10 at 20:14
  • $\begingroup$ @Neil Strickland: Whenever I hear "cocomplete" I assume it means "contains all colimits of small diagrams". The point is that any category that contains all colimits for diagrams that are about the same size as the category itself is necessarily (equivalent to) a poset. As a trivial example to see that a small (nontrivial) abelian category cannot be cocomplete in this sense: pick some cardinality that is much larger than the total number of objects in your category, and take the coproduct of some non-zero object with itself that many times. $\endgroup$ Nov 25 '10 at 22:29
  • $\begingroup$ Alternately, the Hai paper works if your category is cocomplete and has ... I don't remember, a projective cogenerator or an injective generator or something? $\endgroup$ Nov 25 '10 at 22:32
  • $\begingroup$ The proof is wrong: journals.cambridge.org/action/… $\endgroup$ Dec 30 '10 at 11:31
  • $\begingroup$ @Martin: Good call. I was not aware of the correction. $\endgroup$ Jan 3 '11 at 3:12

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