MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
What does $Xh$ mean? – Martin Brandenburg Oct 7 '10 at 16:52
If I'm not mistaken, Deligne-Mumford stacks are, by definition, stacks w.r.t. the étale topology, not fppf. – Qfwfq Oct 7 '10 at 17:22
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. – expz Oct 7 '10 at 18:40
up vote 7 down vote accepted

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.

share|cite|improve this answer
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). – Laurent Moret-Bailly Oct 8 '10 at 9:33

Your Answer


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.