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The question is stated in the title. I think BCnrd states in a comment here

Is every (Artin/DM) algebraic stack fibered in sets an algebraic space?

that while the answer is not found in Laumon & Moret-Bailly, but that it is nevertheless true. Does anyone know of a reference? (Also: I only care about stacks over $\mathbb{C}$, if that makes a difference.)

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BCnrd uses geometric points, not closed points. –  S. Carnahan Feb 23 '12 at 11:33
Over $\mathbb{C}$, are closed points different from geometric points? –  David Steinberg Feb 23 '12 at 23:27
If you take $C[x]$ and the morphism $C[x] \to C(x)$, that is the inclusion of the generic point, you can continue by composing $C[x] \to C(x) \to k$, where $k$ is an algebraic closure of $C(x)$. This is a geometric point of $A^1$ which is not closed. On the other hand closed points correspond to $C$ points when working in the locally of finite type case ($C$ points are closed, residue fields at closed points are finite extensions of $C$, which is again $C$ as it is algebraically closed). But for some things checking on closed points is enough (i.e. the ultrascheme determines the scheme) –  Yosemite Sam Feb 24 '12 at 8:46
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2 Answers

up vote 3 down vote accepted

In case you're still looking for a reference I Lemma 2.3.9 of Abramovich and Hassett stable varieties with a twist should do it (consider the map to a point, which has no automorphisms, which by assumptions is injective on automorphism groups)


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It appears that the lemma you are referencing requires injectivity of automorphism group schemes for all geometric points, not just those mapping to closed points. –  Andrew Niles Feb 21 '12 at 17:55
yes, that is indeed true. I assumed (out of habit) the OP was working with a locally of finite type stack over an algebraically closed field. (that is indeed enough no?) –  Yosemite Sam Feb 21 '12 at 18:03
I am unable to see where in the proof they pass from the assumption that geometric points have trivial stabilizers to the statement that the fibers over all objects have trivial automorphism groups. The step that they skip seems to be precisely the statement that is the subject of the question. –  S. Carnahan Feb 23 '12 at 11:43
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Here's a counterexample to your title question, that is not a counterexample to BCnrd's claim:

Let $Y =\operatorname{Spec} \mathbb{C}[[t]]$. This scheme has a closed point and an open generic point. Let $Z$ be the scheme formed by gluing two copies of $Y$ by the identity map on generic points. This is a non-separated scheme with an action of a group of order two that fixes the generic point and switches the two closed points. Let $X$ be the stack quotient of $Z$ by this action.

$X$ is not an algebraic space, since the image of the generic point of $Z$ has a nontrivial stabilizer. In particular any geometric point (meaning a map from the spectrum of an algebraically closed field) that factors through the generic point of $Z$ will have nontrivial stabilizer. The topological space $|X|$ (defined in Champs Algebriques 5.5) only has one closed point, and the automophism group of any element in its equivalence class is trivial.

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@scott: I thought that for finite type things checking this on closed points was enough. Am I wrong in assuming that? Does this play any role in your counterexample? Or is it more about the non-separatedness? –  Yosemite Sam Feb 24 '12 at 8:55
Based on my hazy intuition, I think closed points suffice in the lemma you cite, for locally finite type morphisms, but I do not know a proof. I have tried to look in a lot of places for a general statement about the triviality of a group algebraic space over a scheme whose geometric fibers are trivial, but without success. There is a statement in EGA4.3 section 9 about local constructibility of the isomorphism locus of a map, but I couldn't push it through. –  S. Carnahan Feb 24 '12 at 11:01
this is so annoying. I had a similar problem with geometric bijections (morphisms which induce bijections on geometric points) and bijections on C-points. They should be the same for locally of finite type schemes but never could seem to prove it. I wish I were less ignorant. I've never had a look at jacobson stuff and ultraschemes, perhaps the time has come to do so. –  Yosemite Sam Feb 24 '12 at 11:13
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