Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Recently, I was doing exercise 2.3 (b) in chapter 2 of Hartshorne's book celebrated book on algebraic geometry. The exercise is as follows:

Let $(X,\mathcal{O}_X)$ be a scheme. Define a presheaf by $U\mapsto (\mathcal{O}_X(U))_{red}$, where the reduction of ring $A$ is defined to be $A$ quotient by its nilradical, ie $A_{red} = A/\sqrt{0}$. Define $(\mathcal{O}_X)_{red}$ to be the sheaf associated to the previous presheaf. Show that $(X,(\mathcal{O}_X)_{red})$ is a scheme (there is more to the exercise, but my question does not concern it).

I managed to do this exercise in this exercise by reducing to the affine case and showing that $(\text{Spec }A,(\mathcal{O}_{\text{Spec }A})_{red})\cong (\text{Spec }A/\sqrt{0}, \mathcal{0}_{\text{Spec }A/\sqrt{0}})$ by actually defining an isomorphism on basic open sets. To me, this is the most straight forward way to do this exercise. However, there are quite a few things to check.

I asked my lecturer about it and he suggested that instead of defining the isomorphisms on basic open sets, I could define the map by looking at the following diagram of sheaves on $\text{Spec }A$:

$$ \begin{array}{ccc} 0 & & \\ \downarrow & & \\ \sqrt{0} & & \\ \downarrow & \searrow & \\ \mathcal{O}_{\text{Spec }A} &\rightarrow & \mathcal{O}_{\text{Spec }A/\sqrt{0}} \\ \downarrow & \nearrow & \\ (\mathcal{O}_{\text{Spec }A})_{red} & & \\ \downarrow & & \\ 0 & & \\ \end{array} $$

(sorry about the terrible diagram, I don't know of a better way to draw them on math.SE)

Here the map $\mathcal{O}_{\text{Spec }A} \rightarrow \mathcal{O}_{\text{Spec } A/\sqrt{0}}$ is the map coming from the ring map $A\rightarrow A/\sqrt{0}$. And $\sqrt{0}$ is the sheaf of ideals $U\mapsto \text{nilrad}({\mathcal{O}_X(U)})$.

Now, my lecturer said that to show that the $\begin{array}{c}\nearrow \\ \end{array}$ map exists it is enough to show that the $\begin{array}{c}\searrow \\ \end{array}$ map is $0$ (then to show it is an isomorphism we can just look at the maps on stalks). My question is: why does the $\begin{array}{c}\searrow \\ \end{array}$ map being 0 imply that the $\begin{array}{c}\nearrow \\ \end{array}$ arrow exists. It looks identical to the case of the universal property of the quotient of abelian groups, rings (etc). Does anyone know where this is written down? I have looked in Hartshorne and Qing Liu's book but can't find anything about a universal property of quotients of sheaves. Thanks for any help.

share|improve this question
    
The universal property is the same, because it is a cokernel! –  Zhen Lin Feb 12 at 8:43
    
@ZhenLin is this because the category of sheaves over a space is an abelian category? (though I am not sure exactly what an abelian category is!) And do all abelian categories have such quotient universal properties? –  Sebastian Feb 12 at 9:17
    
The point is that the explicit construction of sheaf quotients has the universal property of a cokernel; if it didn't, then it would be the construction that is wrong. –  Zhen Lin Feb 12 at 10:02

1 Answer 1

up vote 1 down vote accepted

Let $X = \operatorname{Spec} A$ be an affine scheme. There is a covariant functor $^\widetilde{\hspace{2mm}} : \textbf{A-mod} \to \operatorname{QCoh}(X)$ that sends a module $M$ to the quasi-coherent $\mathcal{O}_X$-module $\widetilde{M}$ defined as follows. On basic open sets, $\widetilde{M}(D(f)) = M_f$. Check this defines a presheaf of $\mathcal{O}_X$-modules and furthermore (a sheaf!) of $\mathcal{O}_X$-modules. That it is a sheaf basically follows from the same proof that setting $\mathcal{O}_X(D(f)) = A_f$ is a sheaf of rings on $X$.

The miraculous thing is that $^\widetilde{\hspace{2mm}}$ is an exact functor! This basically boils down to the fact that localization is exact. The observation that $\mathcal{O}_{\operatorname{Spec} A} = \widetilde{A}$ now shows given

$$0 \to \sqrt{0} \to A \to A/\sqrt{0} \to 0$$

we may apply $^{\widetilde{\hspace{2mm}}}$ to get the exact sequence

$$0 \to \widetilde{\sqrt{0}} \to \mathcal{O}_{X} \to \mathcal{O}_{\operatorname{Spec}(A/\sqrt{0})} \to 0.$$

Then $\mathcal{O}_{\operatorname{Spec}(A/\sqrt{0})} \cong \mathcal{O}_{X}/\widetilde{\sqrt{0}} \cong (\mathcal{O}_X)_{Red}$ as desired. Note $\widetilde{\sqrt{0}}$ is the same as your sheaf of ideals above. IMHO the way Hartshorne sets up the exercise is very bad, he rigs the structure sheaf to be the sheafification of something which to me is not natural. The best way I believe to do this is to first define the associated reduced subscheme for affines (As $\operatorname{Spec} A, \mathcal{O}_{\operatorname{Spec}(A/\sqrt{0})}$), then cover $X$ with affines and use the exercises on gluing sheaves and schemes.

To answer your question directly: Let us now prove the following. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a morphism of sheaves. Given a subsheaf $\mathcal{H} \subseteq \ker \varphi$ (by this we mean for every $U$, there is an inclusion of subgroups) there is a morphism of sheaves $$\mathcal{F}/\mathcal{H}\to \mathcal{G}$$ making everything commute.

Proof: For every $U$, the map $\mathcal{F}(U) \to \mathcal{G}(U)$ factors through the quotient group $\mathcal{F}(U)/\mathcal{H}(U)$ by the universal property of ordinary quotients. Check this is actually a morphism of presheaves (show compatibility with restriction by drawing a 3-D diagram like I drew in your house the other day). Now use the universal property of the sheafification to furnish the existence of a morphism of sheaves $\mathcal{F}/\mathcal{H} \to \mathcal{G}$.

share|improve this answer
    
You were in my house? –  Sebastian Feb 14 at 0:48
    
Joking, thanks Ben. The managed to do the exercise using the same reasoning as the second part of your answer. Modulo understanding what $QCoh(X)$ is, the first part makes sense too. –  Sebastian Feb 14 at 0:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.