The Gabriel-Popescu theorem tells us that every Grothendieck category with a generator is a left exact localization of a module category. I'm interested in a slightly different way of "representing" such categories:

Question: Let $\mathcal C$ be a Grothendieck category with a generator.

  1. Does there exist a Grothendieck topos $\mathcal X$ such that $\mathcal C \simeq Ab(\mathcal X)$ is equivalent to the category of abelian group objects in $\mathcal X$?

  2. Slightly less naively, does there exist a ringed Grothendieck topos $(\mathcal X, \mathcal O_{\mathcal X})$ such that $\mathcal C \simeq \mathcal O_{\mathcal X}\textrm{-}Mod$ is equivalent to the category of $\mathcal O_\mathcal X$-modules?

  3. Same as (2), but with $(\mathcal X, \mathcal O_{\mathcal X})$ being locally ringed?

  • $\begingroup$ You actually also get from Gabriel-Popescu that every Grothendieck category is a left exact accessible localisation of presheaves of abelian groups. This should solve 1. $\endgroup$ – user40276 Jun 13 at 21:26
  • $\begingroup$ @user40276 I don't think I follow. I can see how the Gabriel-Popescu theorem exhibits a Grothendieck category as a localization of a category of $Ab$-enriched presheaves on a certain $Ab$-enriched caetgory (viz. a certain ring) -- but I want to relate this to a topos -- i.e. a category of sheaves of sets on a certain $Set$-enriched category. $\endgroup$ – Tim Campion Jun 13 at 21:29
  • $\begingroup$ It seems that the smallest Grothendieck topology generated by the "additive Grothendieck topology" of presheaves of abelian groups on that pre-additive category should work. But, now, that I think again about it, I'm not so sure... $\endgroup$ – user40276 Jun 13 at 21:49
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    $\begingroup$ I'm pretty sure the answer is no, but I don't know a proof off the top of my head. In noncommutative algebraic geometry, they take Grothendieck categories of graded modules over a graded ring, modulo finite-dimensional modules. If there was a result like you have in mind, the theory would probably be very different. $\endgroup$ – arsmath Jun 14 at 7:28
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    $\begingroup$ In fact, Hovey shows (Thm 3.3) that if $R$ is a division ring which is not a field, then the category $RMod$ of left $R$-modules does not admit any additive closed symmetric monoidal structure. This is a counterexample to (1) and to (2,3) unless we allow $\mathcal O_{\mathcal X}$ to be noncommutative. If we assume our Grothendieck category is symmetric monoidal closed, I'd still be interested to see if maybe it must be of the form $\mathcal O_{\mathcal X}Mod$ with $\mathcal O_{\mathcal X}$ commutative. $\endgroup$ – Tim Campion Jun 15 at 17:11

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