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I have a question on the ''functor of points''-approach to schemes and $\mathcal{O}_X$-modules. Please let me first write up a defintion.

  • Let $Psh$ denote the category of presheaves on the opposite category of rings $Rng^{op}$. So $Psh$ is the category of functors from the category of rings $Rng$ to the category $Set$ of sets.

    Fix an $X\in Psh$. Demazure and Gabriel define in their book ''Introduction to Algebraic Geometry and Algebraic Groups" (page 58, I.2.4.1) an $X$-module $M$ to be an object $M\in Psh$ and a morphism $f:M\to X$ in $Psh$ (a natural transformation $f:M\to X$, that's what they call an $X$-functor) such that for every ring $R$ and every map $p:*\to X(R)$ the set $M(R,p):=*\times_{X(R)}M(R)$ has an $R$-module structure with the property that for any ring map $\phi:R\to S$ the induced map $\psi:M(R,p)\to M(S,\phi(p))$ is additive and satisfies \begin{equation} \psi(\lambda m)= \phi(\lambda)\psi(m) \end{equation} for all $m\in M(R,p)$ and $\lambda\in R$.

    They call $M$ quasicoherent if for any ring map $\phi:R\to S$, the induced map \begin{equation} M(R,p)\otimes_R S\cong M(S,\phi(p)) \end{equation} is an isomorphism.

I want to understand an $X$-module $M$ as a morphism $f:M\to X$ in $Psh$ for which some conditions are required to hold ''locally'', like a bundle, but let me more precise in what I mean: The map $p$ in the definition above corresponds by the Yoneda lemma to a map $p:R\to X$ (Here, I use the same notion for $R$ and its associated presheaf $\hom(R,-)$). Let the object $M_p'$ of $Psh$ be defined by the cartesian diagram \begin{eqnarray} M_p'&\to & M\\ \downarrow && \downarrow f\\ R&\xrightarrow{p} & X \end{eqnarray} in $Psh$. I want to formulate conditions on $M_p'$ (and not on $M(R,p)=*\times_{X(R)}M(R)$ as above) such that $M$ is an $X$-module. The set $M(R,p)$ is contained in the set $M_p'(R)$ but there are not equal, unfortunately. My question is thus: What are the conditions on the $M_p'$ such that $M$ (together with $f$) defines an $X$-module? How is quasicoherence defined in this situation?

  • To be more precise, I would like the above definition of a quasicoherent $X$-module to be the same as something like this: An object $M\in Psh$ and a morphism $f:M\to X$ in $Psh$ such that for every ring $R$ and every map $p:R\to X$ the set $M_p'(R)=(R \times_X M)(R)$ has an $R$-module structure with the property that for any ring map $\phi:R\to S$ the induced map $\psi'(R):M_p'(R)\to M_{\phi(p)}'(R)$ is additive and satisfies $\psi'(R)(\lambda m)= \phi(\lambda)\psi'(R)(m)$ for all $m\in M_p'(R)$ and $\lambda\in R$ and $M_p'(R)\otimes_R S\cong M_{\phi(p)}'(S)$ is an isomorphism.

I hope that I was able to clarify my question. Thank you in advance for any hints.

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The first step is to understand what presheaves are "really" about.

Theorem (Kan). Let $\mathbb{C}$ be a small category, and let $\hat{\mathbb{C}}$ be the category of presheaves on $\mathbb{C}$. Let $H_\bullet : \mathbb{C} \to \hat{\mathbb{C}}$ be the Yoneda embedding. If $\mathcal{E}$ is a locally small and cocomplete category and $F : \mathbb{C} \to \mathcal{E}$ is a functor, then there is a unique (up to isomorphism) cocontinuous functor $\tilde{F} : \hat{\mathbb{C}} \to \mathcal{E}$ such that $\tilde{F} H_\bullet = F$; in other words, $\hat{\mathbb{C}}$ is the free cocompletion of $\mathbb{C}$.

Thus, we may think of a presheaf on $\mathbb{C}$ as a formal specification for gluing together objects of $\mathbb{C}$. Making this idea completely precise is the essence of the proof of this theorem. Indeed, if $P$ is a presheaf on $\mathbb{C}$, then there is a category $\int^{\mathbb{C}} P$ whose objects are pairs $(c, x)$, where $c \in \operatorname{ob} \mathbb{C}$ and $x \in P(c)$, and arrows $f : (c, x) \to (c', x')$ are those arrows $f : c \to c'$ of $\mathbb{C}$ such that $P(f)(x') = x$; equivalently, $\int^{\mathbb{C}} P$ is the comma category $(H_\bullet \downarrow P)$ (by the Yoneda lemma). Let $X : \int^{\mathbb{C}} P \to \hat{\mathbb{C}}$ be the functor $c \mapsto H_c$. It is straightforward to show that $P \cong \varinjlim X$, and from here one deduces that $\tilde{F} : \hat{\mathbb{C}} \to \mathcal{E}$ must be defined by $P \mapsto \varinjlim A$, where $A : \int^{\mathbb{C}} P \to \mathcal{E}$ is the functor $c \mapsto F c$. (Verifying that this works is slightly tricky, but not very interesting.)


Now, the setup of Demazure and Gabriel calls for a universe axiom, but this isn't really necessary if one is willing to restrict the category of schemes under investigation. Let $R$ be any ring, and let $\mathcal{R}$ be any essentially small full subcategory of the category of $R$-algebras which is closed under principal localisations (i.e. localisations of the form $A [1/f]$) and finite colimits (i.e. the initial algebra $R$, tensor products, and coequalisers). For example, we could take $\mathcal{R}$ to be the category of finitely-presented $R$-algebras.

Let $\mathcal{A} = \mathcal{R}^\textrm{op}$. We think of this as being a full subcategory of the category of schemes over $S = \operatorname{Spec} R$. Then, the category of presheaves on $\mathcal{A}$ includes as a full subcategory the category of all $S$-schemes locally modelled on $\mathcal{A}$, i.e. all those $S$-schemes that have an open cover by schemes isomorphic to objects in $\mathcal{A}$.

Given a presheaf $P$ that represents a $S$-scheme $X$, how do we know what $X$ "looks" like? Of course, we use the canonical diagram given in the proof of the above theorem, but the geometric setting means we have a very concrete interpretation of what is happening. Elements of $P$ should be thought of as $S$-scheme morphisms to $X$: to be precise, if $x \in P(A)$, then $x$ corresponds to a unique $S$-scheme morphism $\operatorname{Spec} A \to X$; after all, that is what it means to be represented by $X$: there is a presheaf isomorphism $P \cong \textbf{Sch}_S (\operatorname{Spec} (-), X)$. But we know every scheme admits a cover by open affine subschemes, so this is more than enough information to reconstruct $X$ – as claimed.

What about presheaves on $X$? Well, there's only one thing they could be in this language: a presheaf $E$ together with a presheaf morphism $p : E \to P$; in other words, it is an object of the slice category $(\hat{\mathcal{A}} \downarrow P)$. (Actually, this isn't literally true: presheaves in this sense are strictly more general that presheaves on $X$; but the important thing is the idea.) As before, given an element $x$ of $P(A)$, we think of the set $E(A, x) = \{ e \in E(A) : p(e) = x \}$ as being the set of "sections" of $E$ over the "neighbourhood" $x$. Now suppose $E$ represents an $\mathscr{O}_X$-module. Then, $E(A, x)$ would have to be an $A$-module, in such a way that the various "restriction" maps become generalised module homomorphisms – exactly as in your definition. So, you see, it is already a local condition, just perhaps not in the way you expect!


If, however, you are dead-set on seeing this in terms of Yoneda, then you can use the following fact:

Proposition. Let $\mathbb{C}$ be a small category, and let $\hat{\mathbb{C}}$ be the category of presheaves on $\mathbb{C}$. If $P$ is a presheaf on $\mathbb{C}$, then the slice category $(\hat{\mathbb{C}} \downarrow P)$ is isomorphic to the category of presheaves on $\int^{\mathbb{C}} P$.

Proof. Given a presheaf morphism $p : E \to P$, one constructs a presheaf on $\int^{\mathbb{C}} P$ in the obvious way: set $E(c, x) = \{ e \in E(c) : p(e) = x \}$. Conversely, given such a presheaf on $\int^{\mathbb{C}} P$, we may recover $E$ by setting $E(c) = \coprod_{x \in P(c)} E(c, x)$ and defining $p : E \to P$ to be the obvious projection. Checking that these constructions are functorial and mutually inverse is straightforward. $\qquad \blacksquare$

Thus, one sees that $E(A)$ is in general not an $A$-module, but rather a disjoint union of $A$-modules. (And I really do mean disjoint union, not coproduct!) So we must work in the category of presheaves on $\int^{\mathbb{C}} P$ instead, but the only thing we gain from doing this is notational convenience.

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    I can follows the arguments but don't understand the upshot. Is it that one should define $P$-modules in the category $Psh/P$ ($=\hat{\mathbb{C}}\downarrow P$) and then apply $\alpha_\star:Psh/P\to Psh$ for $\alpha:P\to\star$ rather than defining them in $Psh$? Is then $M(R,p)=(R\times_PM)(?)$ in $Psh/P$? The reason why I want to understand it like is: Isn't a *vector bundle* over $P$ (representing $X$) a morphism $M\to P$ in $Psh$ such that there exists an open cover $U_i$ of $X$, $U_i\times_P M=\mathbb{A}^n\times U_i$ in $Psh$ and all the induced $\mathbb{A}^n\to \mathbb{A}^n$ are linear?2012-06-15
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    Not all $\mathscr{O}_X$-modules are vector bundles. Not even if quasicoherent.2012-06-15
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    Yes, but all vector bundles are $\mathcal{O}_X$-modules and hence I was trying to understand the definition of an $\mathcal{O}_X$-module as a generalization of this definition of a vector bundle.2012-06-15
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    The correspondence is a little bit more sophisticated than what you seem to be suggesting. A vector bundle is, first and foremost, a scheme. There is a functorial construction that builds an $\mathscr{O}_X$-module out of a vector bundle and vice-versa, but they are not literally the same kinds of objects. (An $\mathscr{O}_X$-module is not a scheme.)2012-06-15
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    Dear Zhen, I'm impressed by your answer! Have you read much of that book by Demazure-Gabriel?2012-06-15
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    @Georges: I _tried_ to read it some months ago, but I gave up quite quickly. I got as far as learning their definition of scheme and getting some vague idea of how they were equivalent to the definition via locally ringed spaces.2012-06-15
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    Dear @Zhen: I was asking because I don't like that book at all, partly because of its hilariously non-standard and absurd notations! (Which doesn't prevent me from admiring its authors, who are great mathematicians) Fortunately nobody has to read it nowadays, when we have Vakil's and De Jong et al.'s wonderful online notes as well as the books by Hartshorne, Eisenbud-Harris, Qing Liu, Görtz-Wedhorn, ... Anyway, you seem to have a real gift for absorbing quite abstract concepts. Bravo!2012-06-15
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    @Zhen: Thanks to your answer, I am beginning to understand better. Please let me ask one more thing for clarification: If $P\cong Hom_{Rng}(r,-)$ is represented by a ring $r$, is then $E(c,x)$ for $x\in P(c)$ the set of $r$-module homomorphisms from $E(r,id_r)$ to $c$ where $c$ is a $r$-module via $x$?2012-07-06
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    What's the connection between $E$ and $c$? $E(c, x)$ is a $c$-module, and the "restriction" map induced by $x : r \to c$ restricts to a map $E(r, \textrm{id}_r) \to E(c, x)$ that is a $r$-module homomorphism when $E(c, x)$ is considered as an $r$-module via $x$. But I don't see why $E(c, x)$ should itself be a set of module homomorphisms.2012-07-06