If I have a Deligne-Mumford stack $\Pi : X \to (\mathrm{Sch}/k)$ for some field $k$, can it be reconstructed from $\Pi^{-1}(C) \subset X$ for some "small" subcategory $C \subset (\mathrm{Sch}/k)$? For example, let's say we assume the fppf topology (to be concrete), then does the restriction to the fully faithful subcategory $T$ with objects

$\mathrm{Ob}(T) := $ { $I_n := \mathrm{Spec~} k[\epsilon]/(\epsilon^n) \mid n \in \mathbf{N}$ }

completely determine the stack? I think I've read that the restriction to $(\mathrm{Aff}/k)$ is enough, but since the fibers over the $I_n$ determine the $n^{th}$-order formal neighborhoods ($\mathrm{HOM}(I_n, X) \cong X(I_n)$), I wonder if the restriction to $T$ is enough?

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    $\begingroup$ What does $Xh$ mean? $\endgroup$ – Martin Brandenburg Oct 7 '10 at 16:52
  • $\begingroup$ If I'm not mistaken, Deligne-Mumford stacks are, by definition, stacks w.r.t. the étale topology, not fppf. $\endgroup$ – Qfwfq Oct 7 '10 at 17:22
  • $\begingroup$ Is it easy to show it's not true even for a scheme? @unknown: In general, no, I don't think you can assume the topology, though I could be mistaken. My cue is from Behrend-Manin defining the stack of stable maps when they call it DM and use the fppf topology. $\endgroup$ – expz Oct 7 '10 at 18:40

Restricting to Aff is certainly enough, but Aff isn't small (there are e.g., polynomial algebras on arbitrary sets). If your DM stack is finitely presented over $k$ (which is probably good to include in the definition, to avoid these issues), then it is determined by it's restriction to finitely-presented affines (which is essentially small).

Without some finiteness hypothesis, no set of finitely-presented algebras can suffice (even for affine schemes, nevermind DM stacks). (And I suppose no small category of test objects can suffice: Take Spec of a field generated by a set of cardinality larger than that of global sections of any of your test schemes.)

The set you give is insufficient even for smooth varieties over an alg. closed field: you will have a morphism whenever you have an (arbitrary) map on $k$-points.

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    $\begingroup$ And if $k$ is not alg. closed, you have lots of different $k$-schemes of finite type with no $k$-points, hence no $I_n$-points at all, for any $n>0$. The question makes more sense if you "vary $k$" in the definition of $T$ (taking for example finitely generated extensions of $k$, if your stacks are of finite type). $\endgroup$ – Laurent Moret-Bailly Oct 8 '10 at 9:33

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