# Can $\mathcal O_X$ be recognized abstract-nonsensically?

This question has been asked by Teimuraz Pirashvili many years ago. I forgot about it after a while and remembered only now by accident. He probably knows the answer by now, but I still don't.

In the category of modules over a ring $R$, the module $R$ is a projective generator. This property does not determine it uniquely, even up to isomorphism, but it does when $R$ is commutative: it can be reconstructed as the ring of endotransformations of the identity functor. (For noncommutative $R$, it is still true that the category is equivalent to the category of modules over the endomorphism ring of any of its projective generators that may exist.)

Now what can be said about the structure sheaf $\mathcal O_X$ of a scheme $X$? Can it be detected inside the category of $\mathcal O_X$-modules without using any additional structure? Of course it is the unit of the monoidal structure, but can it be also characterized as an object in a plain category, without invoking any additional structures? Can in fact the category of $\mathcal O_X$-modules possess other non-isomorphic monoidal structures? Can it be equivalent to the category of $\mathcal O_Y$-modules for some other scheme $Y$? Or, say, some analytic space, or whatever?

Note that $\mathcal O_X$ is not even a generator - the subcategory generated by it is the category of quasicoherent sheaves. (NB As @HeinrichD points out in the first comment, even that is wrong - this is only true locally, in the appropriate sense.) In any case, also in this category $\mathcal O_X$ is typically not projective. So the same question arises - is it some particular kind of generator?

There are of course several reconstruction theorems but I still cannot figure out which (if any) of them provide answers to these questions.

LATER - as pointed out by Will Sawin, I should be more clear. Rather than modifying the question (I don't quite understand how) let me try to formulate it once more:

Consider the object $\mathcal O_X$ of the abelian category of sheaves on a scheme $X$. What are its properties formulable without the use of tensor product or any other additional structures, just as an object of this abelian category?

For example, as Will Sawin explains in his answer, a commutative ring inside its category of modules cannot be distinguished among rank one projectives. But at least it is a projective generator with a commutative endomorphism ring (in fact it is isomorphic to that ring), which is a very restrictive property. Are there some similar properties of $\mathcal O_X$?

• " the subcategory generated by it is the category of quasicoherent sheaves" is not correct. – HeinrichD Dec 9 '16 at 9:29
• @HeinrichD Really? Sounds important. Could you please elaborate? – მამუკა ჯიბლაძე Dec 9 '16 at 9:30
• The categorical center of Mod(R), which by definition is endomorphisms of the identity functor, is the center of R. Thus, when R is commutative you have an intrinsic way of characterizing R, however not as an object inside Mod(R), which is odd. This of course does not work globally. – Yosemite Sam Dec 9 '16 at 17:59
• We can view the category of $O_X$-modules as enriched over the category of sheaves on $X$, and I believe it should be the case that the sheaf of Sh(X)-natural endotransformation of the identity functor on $O_X$-modules is isomorphic to $O_X$ analogously. Not sure how helpful that is though. – Georg Lehner Dec 9 '16 at 23:28
• Interesting question, interesting answers, but the whole field has become quite confusing now, with many partial answers to many diiferent interpretations of the question. I think re-asking one or several independent questions may help bring some clarity. For instance, one of the question I'd like to see solved is: if $X$ and $Y$ are two schemes, is true that the category of quasi-coherent $O_X$-modules determines $X$? As the OP notes, this is true when $X$ and $Y$ are affine, and as Ben's answer shows, it at least determines $X$ as a topological space when $X$ is locally of finite type. – Joël Dec 11 '16 at 18:39

As Will points out, the best you can hope for is to recognize being a line bundle. Also, I find you're a bit vague about what category of sheaves you want work in; I'm thinking about coherent sheaves, which is an appropriate analogue for finite dimensional $R$-modules. As we'll see below, I'll also want to assume my scheme is locally of finite type (over a field, or other Jacobson ring, such as $\mathbb{Z}$) EDIT: nfdc23 also points out I should assume reduced.

I have an answer for this is which is probably ultimately equivalent to Heinrich's but which I like better (perhaps the difference is he's thinking of quasi-coherent sheaves?). The only simple objects in the category of coherent sheaves are the skyscraper sheaves at the different closed points. Thus, we can actually reconstruct the closed points of the scheme. You can also define the support of a sheaf by which skyscrapers it has maps to, and reconstruct the Zariski topology by defining a closed set to one that appears as the support of a coherent sheaf. A coherent sheaf is a line bundle if and only if the Hom space to every skyscraper sheaf is 1-dimensional.

EDIT: Why is this so? A coherent sheaf is a line bundle if and only every stalk is free of rank 1 over the local ring at each point. A module over a local ring is free of rank 1 if and only if the rank of its residue at the unique closed point and the generic point is 1. So, it's enough to check that you have residue of rank 1 at every point (including non-closed ones).

So, if you have rank 1 at all closed points, by semi-continuity, you have rank 0 or 1 at every point, and the set where you have rank 1 is closed. Thus, you must have rank 1 everywhere if the closed points are dense (if not, the structure sheaf on their closure will be a counterexample). This occurs whenever you are locally of finite type (as discussed here: https://math.stackexchange.com/questions/615709/is-the-set-of-closed-points-of-a-k-scheme-of-finite-type-dense). Presumably outside finite type, one can modify the construction a bit to include the non-closed points.

• for the criterion for invertible objects in Coh(X) you give, you need some assumptions on X, like smooth and proper. right? – Yosemite Sam Dec 11 '16 at 3:42
• @YosemiteSam Smooth and proper sounds way too restrictive. I might need something like finite type (since I need the closed points to be Zariski dense). – Ben Webster Dec 11 '16 at 16:08
• @მამუკაჯიბლაძე You're thinking about dimension wrong; I don't mean over a fixed base field (that makes no sense for a scheme like $\mathrm{Spec}(\mathbb{Z})$ anyways). Each closed point has a residue field, and you have to look at the dimension at that point over that field. That will be 1 for the structure sheaf by definition (and line bundles are indistinguishable from the structure sheaf for local purposes). – Ben Webster Dec 11 '16 at 18:14
• @BenWebster It is not true that a coherent sheaf is invertible if it has residual rank 1 at all points: consider non-reduced $X$ and $O_{X_{\rm{red}}}$ as an $O_X$-module. Also, what does it mean that a scheme is "of finite type"? Over what? It seems you assume $X$ is a Jacobson scheme (the class for which "closed point" is well-behaved such as being of local nature; schemes of finite type over a dvr are generally not Jacobson). For Jacobson affine schemes one has a "Nullstellensatz" with maximal ideals, so for reduced Jacobson schemes the residual criterion with closed points holds. – nfdc23 Dec 11 '16 at 22:25
• This is essentially done for quasicoherent sheaves on quasiseparated schemes by Martin Brandenburg in the link HeinrichD posted in the comments of my answer arxiv.org/pdf/1310.5978v3.pdf - one can use Martin's construction to define the internal Hom as a bifunctor from the abstract category to a category of sheaves, and a sheaf is invertible if and only if internal homs from it are an equivalence. – Will Sawin Dec 12 '16 at 13:20

I think you are a little confused about what your characterization of $R$ does, and this causes problems as you generalize to sheaves.

There is no characterization of $R$ as an element of the category of $R$-modules via abstract nonsense - this is because tensoring by any rank $1$ locally free module is an autoequivalence of this category. Instead, what you have is a construction of the element from the category. But there is no way of telling which element of the category actually corresponds to this ring you have constructed.

If you want to characterize $\mathcal O_X$, rather than construct it, you face the same problem - there are autoequivalences which cause trouble, so you need extra structure.

If you want to construct $\mathcal O_X$, you face a different problem - $\mathcal O_X$ is a much more complicated object than just a ring. It might be hard to construct all of $\mathcal O_X$ without already knowing something about how the open sets of $X$ relate to the category. For instance one could know, for each open set, the subcategory of sheaves that are pushforwards from that open set. Then I believe one can recover $\mathcal O_X(U)$ as the endomorphisms of the identity functor on the subcategory of sheaves that are pushforwards from $U$.

One certainly can't recover it just from the category and the topology on $X$, because there can be automorphisms of the topological space $X$ that do not fix $\mathcal O_X$.

So the only hope to construct $\mathcal O_X$ from just the category of $\mathcal O_X$-modules is to first construct the topological space $X$ from the category and then construct the sections on it. One would then not be able to construct the isomorphism of this topological space with a concrete $X$ unless given other data on $X$.

An idea for how to do this is that, for each closed set $Z$, the category of sheaves supported on $Z$ is a Serre subcategory, and the endomorphisms of the identity functor on the Serre quotient by that category should be the sections of $\mathcal O_X$ on the complement of $Z$. So it would be sufficient to characterize these subcategories among all Serre subcategories.

• Thank you for the very informative answer. I still have to digest it, but here are some quick questions/comments. I agree I should be more accurate, but I don't understand already the part about $R$. (1) is not it correct that if an abelian category is equivalent to the category of modules over a ring, then this ring is isomorphic to the endomorphism ring of a projective generator? (2) if further one knows that the ring in question is commutative, is not it true that it is necessarily isomorphic to the ring of endotransformations of the identity functor? – მამუკა ჯიბლაძე Dec 9 '16 at 9:17
• @მამუკაჯიბლაძე I completely agree with (2). I'm just saying that is not a way of detecting $R$ inside the category of $R$-modules. It's a way of constructing $R$, outside the category of $R$-modules, using the category of $R$-modules. – Will Sawin Dec 9 '16 at 9:20
• I see, thanks. I will think about modifying the question accordingly; although I still do not understand well enough the difference you pointed out, I acknowledge that there is a difference. – მამუკა ჯიბლაძე Dec 9 '16 at 9:22
• Yes, the group of auto-equivalences of $\mathsf{Mod}(R)$ is $\mathsf{Pic}(R) \rtimes \mathrm{Aut}(R)$. Something similar holds for quasi-separated schemes. See arxiv.org/abs/1310.5978 – HeinrichD Dec 9 '16 at 9:31
• What Will Sawin has sketched in the last paragraph has been worked out for the category of quasi-coherent sheaves by Gabriel, Rosenberg and others. See again arxiv.org/abs/1310.5978 . For the whole category of modules there is a similar construction using idempotents (unpublished work). – HeinrichD Dec 9 '16 at 9:38

There won't be any property which really distinguishes $\mathcal{O}_X$ inside $\mathsf{Mod}(X)$, since any invertible $\mathcal{O}_X$-module $\mathcal{L}$ induces an auto-equivalence of categories $\mathcal{L} \otimes -$. Instead, we may hope for properties of invertible $\mathcal{O}_X$-modules (which only have to be proven for the special case $\mathcal{O}_X$ as soon as they are categorical). I hope that you don't mind that I switch to $\mathsf{Qcoh}(X)$ when appropriate, because $\mathsf{Mod}(X)$ is too large and does not really incorporate the condition that $X$ is a scheme.

1) Let $\mathcal{L}$ be an invertible $\mathcal{O}_X$-module. Then every epimorphism $\mathcal{L} \to \mathcal{L}$ is an isomorphism.

2) The endomorphism ring $\mathrm{End}(\mathcal{L})$ is commutative. In particular, the group $\mathrm{Aut}(\mathcal{L})$ is commutative.

3) If $X$ is quasi-compact and quasi-separated, then $\mathcal{L}$ is a finitely presentable object of $\mathsf{Qcoh}(X)$.

4) If $X$ is separated, then every quasi-coherent $\mathcal{O}_X$-module is a subquotient of a direct sum of copies of $\mathcal{L}$. I don't know a classical reference for this (anyone?), but it is proven in Proposition 3.18 here (let $I=0$ there).

5) We know that there is an open covering $\{X_i \to X\}$ such that $\mathcal{L}|_{X_i}$ is a projective generator of $\mathsf{Qcoh}(X_i)$ with a commutative endomorphism ring (and this characterizes invertible modules $\mathcal{L}$). This can be formulated categorically as follows (see here) when $X$ is quasi-separated: There is a family of thick subcategories $\mathcal{T}_i \subseteq \mathsf{Qcoh}(X)$ with $\bigcap_i \mathcal{T}_i=0$ such that the images of $\mathcal{L}$ in the quotient abelian categories $\mathsf{Qcoh}(X)/\mathcal{T}_i$ are projective generators with a commutative endomorphism ring.

• (1) is true for every finitely generated module over a commutative ring – Denis Nardin Dec 9 '16 at 18:29
• Yes, hence it is also true for quasi-coherent modules of finite type. – HeinrichD Dec 9 '16 at 20:00
• @მამუკაჯიბლაძე: Yes, it is equivalent (if $X$ is quasi-separated and we also include 2), as mentioned), and I also wondered about a first-order condition, but couldn't find one so far. Actually I am not sure if there is such a condition. The problem is really that we are not allowed to use the tensor product, and hence neither internal homs. – HeinrichD Dec 9 '16 at 21:59
• @მამუკაჯიბლაძე I think the problem is that a sheaf may have commutative endomorphism ring but noncommutative endomorphism rings on an open cover. – Will Sawin Dec 12 '16 at 13:09
• @მამუკაჯიბლაძე A stable rank $2$ vector bundle on a projective curve of positive genus should do the trick. On any affine subset its endomorphisms ring is large, but on the curve there are no nonscalar homomorphisms. – Will Sawin Dec 12 '16 at 17:05