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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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up vote 8 down vote accepted

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? – geometrystudent Jun 15 '12 at 12:03
Not all $\mathscr{O}_X$-modules are vector bundles. Not even if quasicoherent. – Zhen Lin Jun 15 '12 at 13:03
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. – geometrystudent Jun 15 '12 at 14:15
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.) – Zhen Lin Jun 15 '12 at 14:18
Dear Zhen, I'm impressed by your answer! Have you read much of that book by Demazure-Gabriel? – Georges Elencwajg Jun 15 '12 at 15:25

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