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I have been reading a bit about the "functor of points" theory for schemes. There seem to be two ways of going about defining schemes this way:

  1. Equip the category $\textbf {Psh}=\operatorname{Fun}(\textbf{Ring},\textbf{Set})$ with a Grothendieck topology and define a scheme to be such an object $X$ in $\textbf {Psh}$ which has an open coverings with ring spectra (functors $\operatorname{Spec}(R)=h_R)$ and for which $h_X\colon\textbf {Psh}^{op}\to\textbf{Set}$ is a sheaf in the Grothendieck topology.

  2. Place a Grothendieck topology on $\textbf{Ring}^{op}$ and call an object $X$ of $\textbf {Psh}$ a local functor if it, viewed as a contravariant functor on $\textbf{Ring}^{op},$ is a sheaf. Then one defines open coverings in $\textbf {Psh}$ and defines shemes to be local functors wich have an open covering by spectra.

The Demazure-Gabriel volume (just about the only rather complete account of the functor-of-points approach I could find), which takes the second approach, proves what I believe is just that those two approaches are equivalent, ti. that the same objects of $\textbf {Psh}$ are schemes regardless of which the definition you follow. However, it does that by appealing to what it calls the "geometric realization" of a functor, which is just the associated ringed space. I find that very unsatisfactory and wonder if it can be proved directly in the functorial language without introducing ringed spaces?

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    $\begingroup$ What is the difference between (1) and (2)? You seem to have said the same thing twice. $\endgroup$ – Zhen Lin Aug 6 '14 at 19:02
  • $\begingroup$ In (1), the functor which must satisfy compatibility conditions is $h_X\colon \textbf{Psh}^{op}\to\textbf{Set}$. In (2), the functor which must satisfy compatibility conditions is $X\colon\textbf{Ring}=(\textbf{Ring}^{op})^{op}\to\textbf{Set}$ itself. Those are two functors from different categories, if nothing else. Though the Zariski topology on $\textbf{Ring}^{op}$ is the same as pulling back the Zariski topology of $\textbf{Psh}$ by the Yoneda embedding, I do not see how that directly implies that $X$ being a sheaf is equivalent to $h_X$ being one. $\endgroup$ – A Rock and a Hard Place Aug 6 '14 at 21:03
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    $\begingroup$ Ah. Well, that is just a general fact from topos theory. See here. $\endgroup$ – Zhen Lin Aug 6 '14 at 21:06
  • $\begingroup$ Thank you, that looks winderful. There is a detail I do not understand though: if I understand the linked answer correctly, you work there with sheaves on the site of all sheaves. Where as in (1), one considers sheaves on the site of all functors, sheaves or otherwise. I imagine that is not really a problem though? $\endgroup$ – A Rock and a Hard Place Aug 6 '14 at 21:24
  • $\begingroup$ Yes, there are some details to be filled in, but I'm sure it can be done. (You can use the comparison theorem, for example.) One has to use the appropriate Grothendieck topology on $\mathbf{Psh}$, of course. $\endgroup$ – Zhen Lin Aug 6 '14 at 21:30

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