I am trying to learn about the functor of points approach to algebraic geometry. Given the category of locally ringed spaces $GSp$ (geometric spaces) we have a functor $$\mathcal{G}: GSp \to Set^{Rng} \\ X \mapsto Hom(\operatorname{Spec}-,X)$$

It is clear to me that if $X \cong \operatorname{Spec}A$ then $\mathcal{G}(\operatorname{Spec}A)\cong Hom(A,-)$. I am having problems to understand what happens when $X$ is a general scheme. First of all, a scheme has an affine cover by $\{ \operatorname{Spec}A_i\}$ which is mapped by $\mathcal{G}$ to a family of open subfunctors of $\mathcal{G}X$ in the sense of this Trying to understand open (closed) subfunctors.

Also since $\{ \operatorname{Spec}A_i\}$ is a covering it is clear that the induced morphism $\bigsqcup \operatorname{Spec}A_i \to X$ is an epimorphism. If the notion of covering by open subfunctors is analogous to the topological one then we should have an epimorphism

$$\bigsqcup Hom(A_i,-) \to \mathcal{G}X.$$

This condition is translated in the book of Demazure by saying that for every field $K$, $\bigsqcup Hom(A_i,K) \to \mathcal{G}X(K)$ is an epimorphism.

I think this condition is essentially the same as the one I stated before but since we are working with sheaves on the Zariski site instead of general presheaves the condition of being a sheaf epimorphism can be checked locally on fields. Am I right? In this case how can I proof the fact that both ideas are equivalent? A general statement for this would be:

Given two sheaves $\mathcal{F},\mathcal{G}$ in the Zariski site is it true that $\mathcal{F} \to \mathcal{G}$ is a sheaf epimorphism $\iff$ $\mathcal{F}(K) \to \mathcal{G}(K)$ is surjective for every field?

  • $\begingroup$ It's true for schemes but not for general sheaves. $\endgroup$ – Zhen Lin Jun 19 '16 at 7:39

This is not true.

Intuitively: When thinking of sheaves of functions on spaces, the surjectivity on fields is a statement about the value types of the functions only, while being a sheaf epimorphism would incorporate the local nature of the functions. Hence, if we seek to formalize something like the inclusion of continuous/regular/well-behaved functions into all set-theoretic functions, we should find a counter-example.

Formally: Consider on the one hand the sheaf ${\mathscr F} := {\mathbb A}^1$, i.e. ${\mathbb A}^1(R) := R$. You can view this as a sheaf of 'regular' functions on $\text{Spec}(-)$ by associating to $f\in {\mathbb A}^1(R)=R$ the function on $\text{Spec}(R)$ mapping the point ${\mathfrak p}$ to the image of $f$ in $k({\mathfrak p})=A_{\mathfrak p}/{\mathfrak p}A_{\mathfrak p}$, i.e. a section of the family ${\mathfrak p}\mapsto k({\mathfrak p})$. On the other hand, considering all such set-theoretic sections also gives rise to a sheaf ${\mathscr G}$, and the just explained construction constitutes a morphism of sheaves ${\mathscr F}\to {\mathscr G}$. On fields, this morphism is an isomorphism, but it is not an epimorphism of sheaves: If that was the case, we would have an epimorphism on all $R$-valued points for local $R$, but this is not true: For example, if $R$ is a DVR with residue field $k$ and quotient field $K$, then ${\mathscr G}(R) = k\times K$, while ${\mathscr F}(R)$ would only map to pairs of the form $(\overline{r},r/1)$.

  • $\begingroup$ Thanks @Hanno for this really interesting answer. Then how can i understand the covering condition with the fields without using that the map $\bigsqcup Hom(A_i,-) \to \mathcal{G}X$ is an epimorphism? $\endgroup$ – Abellan Jun 18 '16 at 13:27
  • $\begingroup$ @Abelian I'm sorry, could you restate precisely what you are interested in? $\endgroup$ – Hanno Jun 18 '16 at 16:43
  • $\begingroup$ @Abellan The difference is that the $\mathrm{Hom}(A_i, -)$ are open subfunctors. For a family $(F_i \to F)_i$ of open subfunctors (all sheaves for the Zariski topology), the following statements are equivalent: (1) For any field $K$, the family $(F_i(K) \to F(K))_i$ is jointly surjective. (2) For any local ring $R$, the family $(F_i(R) \to F(R))_i$ is jointly surjective. (3) The family $(F_i \to F)_i$ is jointly epimorphic in the category of Zariski sheaves. $\endgroup$ – Ingo Blechschmidt May 9 '17 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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