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A scheme is equivalent to a functor $\mathcal{F}:\textbf{AffSchemes} \rightarrow \textbf{Set}$ such that it admits a cover of affine schemes and is a sheaf of rings on the Zariski site.

Suppose $\mathcal{C}$ is a subcategory of $\textbf{AffSchemes}$ and $\mathcal{T}:\mathcal{C}\rightarrow \textbf{Set}$ is a functor

Question, can we find a condition on $\mathcal{T}$ and $\mathcal{C}$ which tells us if wether $\mathcal{T}$ extends to a functor $\mathcal{F}:\textbf{AffSchemes} \rightarrow \textbf {Set}$ and is represented by an scheme.

Example for example we can consider the category of Noetherian complete local rings (which is in fact the one that interests me. $\underline{\text{EDIT}}$: it is a bit more, I'm interested in Noetherian complete local rings with a fixed residue field).

Question Same question for algebraic spaces and stacks.

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    $\begingroup$ Sorry, your definition of schemes does not seem to coincide the one that I knew. The one I knew depends on the concept of open subfunctors, which does not seem to be equivalent to be Zariski-covered by an affine. $\endgroup$
    – Z. M
    Commented Oct 26, 2022 at 12:15
  • $\begingroup$ @Z.M we are talking about the same thing. This is just a quick recall to put the question in context. The characterisation of a scheme by its functor of points is known and is the same for everybody, the point of my question is what follows. $\endgroup$ Commented Oct 26, 2022 at 12:24
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    $\begingroup$ For local rings, see math.stackexchange.com/questions/4339719/… $\endgroup$
    – Watson
    Commented Jan 20, 2023 at 13:40
  • $\begingroup$ Thank's @Watson :) $\endgroup$ Commented Jan 20, 2023 at 22:46

1 Answer 1

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For Noetherian complete local rings with a fixed residue field, any map to a scheme factors through the spectrum of the completed local ring at some point defined over that field. So the functor from the category of Noetherian complete local rings with a fixed residue field only sees the formal neighborhood of each point.

It is impossible in general to reconstruct a scheme from the formal neighborhoods of a bunch of points, so I don't think it will be possible to get a nice criterion for the existence of a scheme extending a functor, and you will basically never have uniqueness.

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  • $\begingroup$ thank you for your answer. Sorry to keep asking, I trust you but I need to know at least the possible cases. If we consider an affine scheme whose underlying ring is a complete Noetherian ring, then can we recover the scheme only with its functor restricted on the complete local Noetherian rings? $\endgroup$ Commented Oct 26, 2022 at 15:24
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    $\begingroup$ But there might be criteria for (pro-)representability by formal schemes, I guess? This setups look very similar to deformation theory, where the "affines" are complete guys. $\endgroup$
    – Z. M
    Commented Oct 26, 2022 at 15:27
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    $\begingroup$ @MarsaultChabat I think not, unless you restrict the ideal with respect to which the ring underlying the affine scheme is complete. For residue field $k$ infinite, every smooth affine curve with infinitely many $k$-rational points gives the same functor. $\endgroup$
    – Will Sawin
    Commented Oct 26, 2022 at 17:33
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    $\begingroup$ @Z.M Yes, I think if you look in foundational works on deformation theory you will find such a theorem in there somewhere! $\endgroup$
    – Will Sawin
    Commented Oct 26, 2022 at 17:34

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