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In this MO question it is claimed that a catchphrase for "algebraic spaces" could be that they are "the result of looking at the orbit space of the action of a finite group on a scheme".

Hence my question:

Given an algebraic space $X$, is it true that there exists a scheme $S$ and an action of a finite group $G$ on $S$ such that $X=S/G$ ?

If I remember correctly, every algebraic space is the quotient of an affine scheme by an étale equivalence relation, so I tend to think that there could exist such equivalence relations that are not "implemented" by a group action...

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Do you mean a group scheme which is of finite type over $S$? –  Martin Brandenburg Jun 29 '11 at 20:10
    
I don't know :) Probably, I mean that everything is -say- over $\mathbb{C}$ and $G$, as a $\mathbb{C}$-scheme, is a finite discrete collection of closed points. Anyway, I mean any situation that resembles the setting in the question quoted above. –  Qfwfq Jun 29 '11 at 22:50
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If X is a geometric quotient for G acting on S, then won't it be the coarse moduli space for the stack quotient of G acting on S? Not all DM stacks are global quotients (even if you allow groups which aren't finite - there is an example in the paper by Edidin, Hassett, Kresch, and Vistoli - Brauer Groups and Quotient Stacks) so this would give a negative answer to your question. –  mdeland Jun 30 '11 at 18:03
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@mdeland: The coarse moduli space of the example you mention (I assume you mean Example 2.2.1) is actually a scheme. –  ulrich Jul 1 '11 at 10:43
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@ulrich - you are right, I hadn't looked at the paper again.... there must be other examples, but none that I can think of off the top of my head. –  mdeland Jul 1 '11 at 14:46

3 Answers 3

As I just remembered, the answer is yes if $X$ is quasiseparated, noetherian and normal: see Laumon and Moret-Bailly, Champs algébriques, (16.6.2). The quesiseparated assumption is needed (see Scott Carnahan's answer). Without the normality condition, I don't have a counterexample but my guess is that there is one.

EDIT after Chris' and Jason's comments: in fact the proof in the case of normal algebraic spaces can be made substantially simpler than in the book (which proves a more general result about noetherian Deligne-Mumford stacks). It goes like this:

Assume $X$ noetherian, integral, normal and, to simplify, separated (I am not sure how much this helps). Cover $X$ by étale maps $X_i\to X$ with each $X_i$ integral and affine. There is a dense open subspace $U$ of $X$ which is a scheme and such that each induced map $U_i:=X_i\times_X U\to U$ is finite. Let $V\to U$ be an étale Galois cover, with Galois group $G$, dominating all the $U_i$'s. Now let $\overline{V}\to X$ (resp. $\overline{X_i}\to X$) be the normalization of $X$ in $V$ (resp. in $X_i$); we have dense open immersions $V\subset \overline{V}$ and $X_i\subset \overline{X_i}$. By functoriality, $G$ acts on $\overline{V}$, with quotient $X$.

I claim that $\overline{V}$ is a scheme. Indeed, for each $i$, $\overline{X_i}$ is also the normalization of $X$ in $U_i$. In particular there is an $X$-morphism $f_i:\overline{V}\to\overline{X_i}$ (deduced from $V\to U_i$) which must be finite surjective (everyone is integral, finite and surjective over $X$). Put $V_i:=f_i^{-1}(X_i)$: this is an open subspace of $\overline{V}$ which is finite over $X_i$, hence an affine scheme. So, the union $W$ of the $V_i$'s is an open subspace of $\overline{V}$ which is a scheme and maps surjectively to $X$ (since $V_i\to X_i$ is surjective), hence $\overline{V}$ is covered by $\{gW\}_{g\in G}$.

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@Chris: Are you asking Laurent to repeat the arguments from his book? –  Jason Starr Feb 7 '12 at 11:15

One class of counterexamples arises from quotients by actions of infinite discrete groups acting freely, but it only works under some definitions of algebraic space. For example, if you specify an action of $\mathbb{Z}$ on $\mathbb{G}_m$ generated by multiplication by an infinite order element (say, over a characteristic zero field), the sheaf quotient $\mathbb{G}_m/\mathbb{Z}$ is an algebraic space that is not a quotient of a scheme by a finite group. Since it is non-quasi-separated, it does not satisfy Knutson's definition of algebraic space, but it does satisfy Demazure-Gabriel's. This appeared a couple years ago in Chris Schommer-Pries's question.

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I am not too familiar with algebraic spaces, but maybe this is an example of an algebraic space that cannot be a quotient of a scheme by a finite group. The property of such quotients that I use is that if an irreducible algebraic space $X$ of dimension at least two is the quotient of a scheme $S$ by a finite group $G$, then there is no point on $X$ with the property that every closed curve in $X$ contains that point. The reason is that given any point in $S$, it is certainly possible to find a curve in $S$ missing the $G$-orbit of that point, and thus we obtain a curve in $X$ missing any given point.

On the other hand, there are algebraic spaces with a point contained in every closed curve. A classical example is obtained as follows. Let $X'$ denote the blow up of $\mathbb{P}^2$ at 10 points lying on a smooth plane cubic. Suppose that the points on the cubic are chosen so that the 10 divisor classes determined by them and the class of a line in the Picard group of the cubic are independent. It is a fact (proved by Artin, I believe) that a curve with negative definite self-intersection on a smooth surface can be contracted to an algebraic space. In particular, denoting by $C$ the strict transform of the cubic in $X'$, the curve $C$ can be contracted on $X'$ to an algebraic space $X$. It is also a consequence of the construction that the curve $C$ intersects every curve in $X'$, and therefore every curve in $X$ contains the point to which $C$ is contracted.

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"given any point in $S$, it is certainly possible to find a curve in $S$ missing the $G$-orbit of that point": why is that? –  Laurent Moret-Bailly Jan 27 '12 at 10:46
    
@Laurent: I was thinking $S$ of finite type over a field and a finite set of closed points. If $S$ is a surface, resolve it and reduce to the quasi-projective case (which should be clear). If $S$ is not a surface, restrict to a divisor and proceed by induction. Does this not work? –  M P Jan 27 '12 at 15:37
    
@MP: "If S is a surface, resolve it": but then what? If $p:S'\to S$ is a resolution, and $C$ is a curve in $S'$ avoiding some finite set $F$, $p(C)$ need not avoid $p(F)$. –  Laurent Moret-Bailly Jan 27 '12 at 20:14
    
@Laurent: although I cannot find a counterexample, I cannot also find a proof of the fact that on a surface you can always find a closed curve missing any finite subset. Thus, for the moment, the above is not an argument. –  M P Jan 30 '12 at 23:33

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