11

A "nice" category $\mathcal{C}$ should be (for the purposes of this question) locally presentable at a minimum, and maybe a bit more. One might require $\mathcal{C}$ to be (in roughly order of increasing restrictiveness)

  • ABn for some $n$.

  • Grothendieck

  • locally finitely presentable

  • the category $\mathsf{QCoh}(X)$ of quasicoherent sheaves on a scheme $X$ (possibly with further adjectives)

  • etc.

In particular: if $\mathsf{QCoh}(X)$ has enough projectives, then is $X$ a disjoint union of affine varieties?

Clarification: Just to be clear, I'm well aware of the Freyd-Mitchell embedding theorem. This is not a question about how close abelian categories are to module categories -- it's a question about how restrictive it is for an abelian category to have enough projectives. The local presentability hypothesis rules out the duals of categories of sheaves for instance.

Motivation / Evidence: I'm thinking of things like this result: the category of sheaves on a locally connected topological space has enough projectives iff that space is an Alexandroff space -- a very restrictive condition. I suspect that the category of sheaves on an Alexandroff space is a module category additive presheaf category. Although on reflection, the category of sheaves on an Alexandroff space need not be an additive presheaf category -- in particular, it need not be locally finitely presentable. So perhaps one should assume that $\mathcal C$ is locally finitely presentable for the purposes of this question.

For another example in this direction, consider the fact that the category of quasicoherent sheaves on a smooth projective variety of dimension >0 over a field never has enough projectives. (See my CW answer below for a more general result).

Alternative formulation: If we assume that $\mathcal C$, in addition to being a "nice" abelian category with enough projectives, has a compact generator, then is $\mathcal C$ a module category? This would follow from the formulation above, since an additive presheaf category with a compact generator is a module category; but it could conceivably be easier to show. It would also settle a form of the question about categories $\mathsf{QCoh}(X)$.

Tim Campion
  • 60,951
  • 3
    Are you aware of Mitchell's embedding theorem? – abx Jan 07 '18 at 15:36
  • 2
    Yes. And I'm aware that any cocomplete abelian category with a compact projective generator is a module category. Am I missing something? – Tim Campion Jan 07 '18 at 15:49
  • You need enough indecomposable projectives. I believe there are examples with every projective nontrivially decomposable into a coproduct. – მამუკა ჯიბლაძე Jan 08 '18 at 17:09
  • @მამუკაჯიბლაძე What does "enough indecomposable projectives" mean? For instance, even $\mathsf{Ab}$ doesn't have the property that every object admits an epimorphism from an indecomposable projective. Does it mean that there is a set $I$ of indecomposable projectives such that every object admits an epimorphism from a coproduct of objects from $I$? If so, I suppose Qiaochu's example of $\mathsf{Ab}^\mathbb{N}$ is an example with enough projectives but not enough indecomposable projectives. – Tim Campion Jan 08 '18 at 17:16
  • Er -- no, under my guessed definition for "enough indecomposable projectieves", $\mathsf{Ab}^\mathbb{N}$ does have enough of them. So I must be wrong. – Tim Campion Jan 08 '18 at 17:28
  • 1
    I would say this is an obstruction "in a different direction". In any presheaf category, the family of indecomposable projectives generates, in particular, every projective is a coproduct of indecomposables. In $\mathsf{Ab}^{\mathbb N}$, these are $(0,...,0,0,\mathbb Z,0,0,...)$. An example of enough projectives without any indecomposables is given, I believe, by sheaves over an atomless complete Boolean algebra. – მამუკა ჯიბლაძე Jan 08 '18 at 17:29
  • 1
    @მამუკაჯიბლაძე Why do the indecomposable projectives generate in an additive presheaf category? For example, suppose I start with a representable $X$ (which is projective), and decompose it into successively smaller retracts. If this process terminates after finitely many steps, then I can reconstruct $X$ via a direct sum. But if it takes infinitely many steps, I don't see how to get $X$ as a sum of indecomposables. – Tim Campion Jan 08 '18 at 18:24
  • 2
    @მამუკა ჯიბლაძე: I'm also confused by your claim. Take the category of modules over $\mathbb{F}_2^{\mathbb{N}}$. I believe every projective is decomposable, but this is still a module category. – Qiaochu Yuan Jan 09 '18 at 06:41
  • @QiaochuYuan Sorry I should be more clear. What I said applies to presheaves with values in abelian groups (more generally, in modules over rings with atomic Pierce spectra). However I must confess I am confused myself by the last comment of the OP. – მამუკა ჯიბლაძე Jan 09 '18 at 07:54
  • მამუკა ჯიბლაძე: Yes, that's what I'm talking about too; the category of modules over $\mathbb{F}_2^{\mathbb{N}}$ is the category of (linear) presheaves, valued in abelian groups, over the one-object linear category with endomorphism ring $\mathbb{F}_2^{\mathbb{N}}$... – Qiaochu Yuan Jan 09 '18 at 08:30
  • @QiaochuYuan I had in mind "set-theoretic" presheaves - just plain functors from a plain category to the category of abelian groups – მამუკა ჯიბლაძე Jan 09 '18 at 09:09
  • I'm not sure if you are aware about it http://win.ua.ac.be/~wlowen/pdf%20papers/JPAA-GabrielPopescu.pdf . – user40276 Feb 01 '18 at 03:44

2 Answers2

14

The category $[C^{op}, \text{Ab}]$ of $\text{Ab}$-valued presheaves on any (small, for simplicity) $\text{Ab}$-enriched category is about as nice as it gets - locally finitely presentable, Grothendieck, etc. - and all coproducts of representables are projective (these are the "free" objects), but it won't be a module category in general if $C$ has infinitely many isomorphism classes of objects.

The simplest example is to take $C = \mathbb{N}$: $\text{Ab}^{\mathbb{N}}$ has the property that any generator must be supported at every element of $\mathbb{N}$, and no such presheaf can be compact. In fact the compact projectives are precisely the finitely supported sequences of finitely generated free abelian groups. Note that this is also the category of sheaves on an Alexandrov space, namely $\mathbb{N}$ with the discrete topology.

Qiaochu Yuan
  • 114,941
  • 3
    This is a module category if you are willing to loosen your definition of rings from unital rings to rings with local units. A ring R has local units if it is a direct limit of unital rings via homomorphisms that are not necessarily unit preserving. The category of unitary R-modules consists of those modules M with RM=M. If C is a category the category algebra ZC has local units ands its unitary modules are just the functors on C to Ab (up to equivalence of categories). – Benjamin Steinberg Jan 07 '18 at 22:07
  • 5
    I would rather just think about $\text{Ab}$-enriched categories directly as a generalization of rings (they are "rings with many objects," or "ringoids" if you really want). – Qiaochu Yuan Jan 07 '18 at 23:57
  • 2
    Darn, you're right. I suppose I asked the wrong question after all. – Tim Campion Jan 08 '18 at 00:49
0

Let $\mathcal C_0$ be a small additive category with countable coproducts, and let $\mathcal C$ be the category of additive functors $\mathcal C^{op}_0 \to \mathsf{Ab}$ preserving countable products. Then by Lemma 1.3 and Prop 2.5 here, $\mathcal C$ is a cocomplete abelian category which is easily seen to be locally $\aleph_1$-presentable. Moreover, cokernels are computed "levelwise", so that the representables are a generating set of projective objects; in particular, $\mathcal C$ has enough projectives. But $\mathcal C$ is pretty far from being an additive presheaf category.

Along with მამუკა ჯიბლაძე's example in the comments of sheaves on a complete boolean algebra, this makes it pretty clear that there are plenty of locally presentable abelian categories with enough projectives which are not additive presheaf categories. I'm still not sure about categories of quasicoherent sheaves, though.

Tim Campion
  • 60,951
  • 1
    I'm not sure I understand your proof. $\mathcal{C}$ having a compact projective generator is not saying anything about $X$ being affine. It is only saying that $\mathcal{C}$ is the module category for a (possibly noncommutative) ring. – Denis Nardin Jan 09 '18 at 20:18
  • @DenisNardin Shoot -- you're right. I don't know much about noncommutative geometry -- what's an example of a noncommutative ring $R$ (not Morita equivalent to a commutative one) such that $Mod(R) = QCoh(X)$ for a scheme $X$? – Tim Campion Jan 09 '18 at 20:22
  • Unfortunately, I don't know any. In general, there aren't that many schemes with enough projectives, so it might very well be that the only such schemes are affines. – Denis Nardin Jan 10 '18 at 08:19
  • How do you go from compact projective to locally free of finite rank? As far as I know the equivalence projective=locally free is not usually true for non-affine schemes – Denis Nardin Jan 18 '18 at 08:45
  • @DenisNardin I only need the implication (compact) projective $\Rightarrow$ locally free. My thinking was that this follows from the fact that projective implies free over a local ring. But now I'm suddenly doubting the implicit assumption that localization preserves the property of being projective... – Tim Campion Jan 18 '18 at 17:37
  • To close the gap: Let $i: U \to X$ be the inclusion of an open affine. Then $i$ is a quasicompact, quasiseparated morphism of schemes. Hence the direct image $i_\ast$ takes quasicoherent sheaves to quasicoherent sheaves. It's easy to check that the inverse image $i^\ast$ (which is just restriction) also preserves quasicoherence. Moreover, because $U$ is open, $i_\ast$ is exact. So $i^\ast: \mathrm{QCoh}(X) \to \mathrm{QCoh}(Y)$ has an exact right adjoint, and hence preserves projectives. Every projective over the affine $U$ is locally free, and so every projective over $X$ is locally free. – Tim Campion Jan 23 '18 at 17:35
  • 1
    Neeman's example in his famous paper is very interesting, but if I remember correctly, it is not a Grothendieck category (in Roos's article correcting the one for which Neeman found this counter-example, I think that it is proven that less than Grothendieck category is enough to make sure that all works). It would be interesting to have strange phenomena like the previous one, but for Grothendieck categories (or theorems saying that it is impossible...). For example, can one find a Grothendieck category with enough projectives, locally finitely generated, but beeing not a presheaf category? – Aurélien Djament Jun 04 '19 at 19:13