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Let $X$ be an integral scheme with function field $K$. If $U\subset X$ is an open subscheme, we may consider the restriction functor $$\textsf{QCoh}(X) \to \textsf{QCoh}(U).$$ I don't know much about 2-categories (I'm probably thinking about (2,1)-categories, to be more precise), but I wonder if we may consider a 2-colimit of this and obtain an equivalence of categories $$\operatorname{2-colim}_U\textsf{QCoh}(U) \xrightarrow{\sim} \textsf{Vect}(K).$$ If this is true, I wonder moreover if the natural functor $$\textsf{QCoh}(X) \to \operatorname{2-colim}_U\textsf{QCoh}(U)\cong \textsf{Vect}(K)$$ is the one which sends a quasi-coherent module over $X$ to its stalk on the generic point.

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2 Answers 2

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Let $x$ be a point in a scheme $X$. There are two posets, namely the poset of affine opens containing $x$, $A(x)$, and the poset of opens containing $x$, $O(x)$.

The inclusion $A(x)^{op} \to O(x)^{op}$ is (homotopy) cofinal - by Quillen's theorem A, it suffices to show that for any open $U$, $A(x)_{/U}$ is weakly contractible. But it is a co-directed poset : for any $V,W \to U$, $V\cap W$ is an open and therefore contains some affine open $x\in O\subset V\cap W\subset U$; therefore it is weakly contractible.

It follows that, in the $(2,1)$-category of presentable $1$-categories, $colim_{U\in O(x)^{op}}QCoh(U) \simeq colim_{U\in A(x)^{op}} QCoh(U)$.

Now note that the latter diagram in fact lives in the category of $E_0$ presentable categories, that is, presentable categories with a chosen basepoint (namely $O_U$) that are equivalent to something of the form $(Mod_R,R)$ for some (commutative) ring $R$.

By a $(2,1)$-categorical analogue of theorem 4.8.5.11. in Lurie's Higher algebra (which says that $R\mapsto (Mod_R,R)$, as a functor from rings to $E_0$ presentable categories, is fully faithful and colimit preserving) (it's actually not an analogue, it follows strictly from this theorem by specializing to $Set$-modules in presentable $(\infty,1)$-categories), and using the fact that $A(x)$ is weakly contractible, so colimits over it in $E_0$-objects are computed underlying, it follows that $colim_{U\in A(x)^{op}} QCoh(U)\simeq colim_{U\in A(x)^{op}} Mod_{O_U(U)}$ is equivalent to modules over $colim_{U \in A(x)^{op}} O_U(U)= colim_{U\in A(x)^{op}} O_X(U)= O_{X,x}$.

Note that here the colimit is taken in ordinary associative rings, but it's a filtered colimit, so it can also be taken in commutative rings.

In conclusion, for any point $x$, $colim_{U\in O(x)^{op}} QCoh(U)$ is equivalent to modules over $O_{X,x}$.

If $x$ is a generic point, then $O(x)$ is the category of all opens, and so you get the corresponding statement for $colim_U QCoh(U)$.

Note that here I've used the following version of "$2-\mathrm{colim}$": homotopy colimits in the $(2,1)$-category (in the sense of "$(\infty,1)$-categories with $1$-truncated mapping spaces") of presentable $1$-categories. I'm assuming that this shouldn't matter too much

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    $\begingroup$ What do filtered colimits taken in the category of presentable categories look like? I am thinking about $\operatorname{colim}_n\operatorname{Mod}_{\mathbb Z[T_1,\ldots,T_n]}$ along natural base changes, and it seems that if I take the colimit in the category of categories, the result seems quite different from $\operatorname{Mod}_{\mathbb Z[T_1,\ldots,T_n,\ldots]}$ (for example, the former does not seem to contain $\bigoplus_n\mathbb Z[T_1,\ldots,T_n]$). $\endgroup$
    – Z. M
    Commented Jun 4, 2022 at 18:34
  • $\begingroup$ Oh, Lemma 7.3.5.11 tells us that if every category is compactly generated and transition maps preserve compact objects, then the (coconed) filtered diagram of subcategories spanned by compact objects form a colimit diagram in the category of categories, and applying Ind, we recover the colimit taken in compactly generated categories. $\endgroup$
    – Z. M
    Commented Jun 4, 2022 at 18:45
  • $\begingroup$ @Z.M. the colimit in presentable categories is very different from the one in categories. Your second comment is a possible description, another description is to pass to right adjoints for every morphism in the diagram and take an ordinary limit $\endgroup$ Commented Jun 4, 2022 at 19:21
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    $\begingroup$ @მამუკაჯიბლაძე : 4.8.5.11. is a statement about colimits in presentable categories (which happen to be computed in the way I described with right adjoints). That being said, maybe the question wanted a colimit in categories (and not presentable one). I have no idea what that looks like $\endgroup$ Commented Jun 4, 2022 at 20:35
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    $\begingroup$ The colimit in categories does not seem to lead to the equivalence that the OP wants, even in the simple case that $X=\operatorname{Spec}(\mathbb Z)$, if I am not mistaken: if it were, then the image of $M:=\bigoplus_n\mathbb Z/n\in\operatorname{Ab}$ in $\operatorname{Vect}_{\mathbb Q}$ should be zero, but $\operatorname{colim}_n\operatorname{End}(M\otimes\mathbb Z[n^{-1}])$ is nonzero. $\endgroup$
    – Z. M
    Commented Jun 4, 2022 at 21:17
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I don't understand what kind of higher category magic went on in the other answer. For me, the answer should be "no", as taking direct limits of rings is only compatible in passing to categories of finitely presented modules.

Note that any vector space over the field of rational functions $K$ can be regarded as a quasi-coherent sheaf on $X$ (by taking pushforward along $\operatorname{Spec}(K)\to X$ - see Hartshorne AG Prop II 5.8). Denote by $\mathcal{K}$ the quasi-coherent sheaf on $X$ corresponding to $K$. Since the stalk of $\mathcal{K}$ at the generic point is $K$, which is the same as the stalk of $\mathcal{O}_X$, we have an isomorphism of stalks at the generic point $\mathcal{K}_\eta\to (\mathcal{O}_X)_\eta$. However, usually this isomorphism can not be spread out to any Zariski neighborhood of the generic point (except in trivial situations, for example if the point $\eta$ is already open). So the direct limit of categories does not see this map; in other words, $\mathcal{K}$ and $\mathcal{O}_X$ are not isomorphic in the direct limit of categories, but are isomorphic at the generic point.

As I hinted above, this issue is fixed if we work with coherent sheaves (assuming the schemes are Noetherian). Indeed, more generally if $R=\varinjlim R_i$ is a direct limit of rings (indexed by some filtering category), then the category $\operatorname{Mod}_{\rm fp}(R)$ of finitely presented modules is the direct limit of the $\operatorname{Mod}_{\rm fp}(R_i)$. To see the essential surjectivity of the map, take a finitely presented $R$-module $M$ and write is at the cokernel of a finite matrix $[r_{\alpha\beta}]$. We can find $i\gg 0$ such that all $r_{\alpha\beta}$ "come" from $R_i$, and take the cokernel there, obtaining a finitely presented $R_i$-module $M_i$ which is isomorphic to $M$ after we tensor with $R$ (because tensor product is right-exact). Full faithfulness can be proved analogously by presenting morphisms by lifting them to presentations.

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    $\begingroup$ Dear Piotr, perhaps what you say is related to @Z. M's comment on the other answer. If $R$ is the filtered colimit of $R_i$, then perhaps $\textsf{Mod}(R)$ is not the 2-colimit of $\textsf{Mod}(R_i)$ in the (2,1)-category of categories, but it is in the (2,1)-category of presentable categories. $\endgroup$
    – Gabriel
    Commented Jun 5, 2022 at 9:36
  • $\begingroup$ This property is usually called continuity. This link jep.centre-mersenne.org/item/10.5802/jep.153.pdf Appendix A may be helpful (to both of you) to understand the relation between the "small" and "big" (presentable) category versions. Higher category magic is indeed not relevant, so no need to be scared. Do note that the point about reduction to affine opens addressed in the other answer (glossed over here because you only talk about the affine case) is important though. And of course, no noetherianness is necessary for either small or big version. $\endgroup$ Commented Jun 5, 2022 at 10:18
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    $\begingroup$ Look at Definition A.2 (colimit of presentables), A.3 (limit of presentables, but taken over right adjoints), and A.10 (colimit of small categories [of compact objects]). We have A.2 = A.3 and in the compactly generated case, A.2 = A.10; follow the references to Lurie's HTT for the proofs. This has nothing to do with ∞-categories, so you can ignore the "∞"s; or if you prefer, you may track down the corresponding statements in some books of Adamek and Rosicky on locally presentable 1-categories (note that "presentable" ∞-categories in [HTT] correspond to "locally presentable" 1-categories) $\endgroup$ Commented Jun 5, 2022 at 10:57
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    $\begingroup$ @crystalline : I agree with your comments, although there is a bit of higher categories involved, namely $(2,1)$-categories : I'm not considering a strict colimit of a strict functor :) $\endgroup$ Commented Jun 5, 2022 at 13:06
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    $\begingroup$ @Piotr : as crystalline pointed out (and as I maybe didn't make clear enough in my answer) I was taking a colimit in the category of presentable categories. Passing to compact objects recovers your claim about ordinary categories of finitely presented modules (and conversely) $\endgroup$ Commented Jun 5, 2022 at 13:09

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