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Fix a ground scheme $S$ (a field say). By atlas for an algebraic stack I mean a smooth and surjective morphism $Y \to X$ from a scheme (or algebraic space or affine scheme) $Y$. If the stack $X$ is smooth then all atlases are smooth schemes over $S$.

What about when $X$ is smooth and proper? Can I find a proper atlas?

If in general the answer is no, what about for Deligne-Mumford stacks?

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It's worth nothing that this is true in the differentiable category; the result is in the thesis of Dorette Pronk. –  David Carchedi Feb 3 '13 at 22:10
    
thanks David, it's good to know. –  Jacob Bell Feb 4 '13 at 13:30

1 Answer 1

up vote 9 down vote accepted

The answer is "no". Consider a smooth, proper $1$-dimensional Deligne-Mumford stack over $\mathbb{C}$ which has coarse moduli space $\mathbb{P}^1$ and which has a single "stacky" point (with any nontrivial stabilizer group at that point).

Edit. A more precise definition of the stack is as follows. Let $m$ be an integer, $m>1$. Let $A$ be $\mathbb{A}^2 \setminus \{(0,0)\}$ with coordinates $x$ and $y$. Let $\mathbb{G}_m$ act on $A$ by $t\ast (x,y) = (tx,t^m y)$. The stack $Q_m$ is the quotient stack $[A/\mathbb{G}_m]$. The quotient morphism $A\to Q_m$ is a smooth atlas, implying that $Q_m$ is smooth (since $A$ is smooth). In fact, this morphism is naturally a $\mathbb{G}_m$-torsor; denote the associated invertible sheaf on $Q_m$ by $\mathcal{L}$. It is not hard to see that this is an $m$-torsion invertible sheaf; the $\text{m}^{\text{th}}$ tensor power of $A$ as a $\mathbb{G}_m$-torsor over $Q_m$ is the quotient stack $[(\mathbb{G}_m\times A)/\mathbb{G}_m]$ where the $\mathbb{G}_m$-action is $t\cdot (u,(x,y)) = (t^mu, (tx,t^my))$. This admits a "trivializing" morphism to $\mathbb{G}_m$, $(u,(x,y)) \mapsto u/x^m = u/y$, each defined on the appropriate open $\mathbb{G}_m\times D(X)$ or $\mathbb{G}_m\times D(y)$. The only nontrivial stabilizer group is $\mathbf{\mu}_m$ acting on the $\mathbb{G}_m$-orbit $\{0\}\times \mathbb{G}_m$ inside $A$. So $Q_m$ is a Deligne-Mumford stack. Finally, $Q_m$ is finite over its coarse moduli space $\mathbb{P}^1$, where the $\mathbb{G}_m$-invariant morphism $A\to \mathbb{P}^1$ is just $(x,y) \mapsto [x^m,y]$.

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I see. Could I ask you how one would go about building such a stack? –  Jacob Bell Jul 31 '12 at 12:22
    
You use the so called "root construction", which is (roughly) a way of creating stacky structure in codimension one. See Charles Cadman, "Using stacks to impose tangency conditions on curves". –  Dan Petersen Jul 31 '12 at 12:37
    
thanks, I'll check it out. –  Jacob Bell Jul 31 '12 at 12:42
    
This example is old. It is trivial to see that there is no finite \'etale atlas which is an algebraic space. It takes a little more work to see there is no proper, smooth atlas which is an algebraic space. –  Jason Starr Jul 31 '12 at 13:25
    
thanks a lot, it will take me a little bit to digest what you wrote (I fear our definitions or trivial may differ). –  Jacob Bell Jul 31 '12 at 13:49

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