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Suppose you have a separated Deligne Mumford quotient stack $[V/G]$ over a field of characteristic $0$, where $V$ is a quasiprojective variety and $G$ is an algebraic group that does not necessarily act linearly on $V$.

What can we say about the coarse moduli space? What are necessary and sufficient conditions for the coarse moduli space to be a quasiprojective scheme? Same question for a proper Deligne Mumford stack of the same form.

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Isn't it true that $V/G$ is a quasi-projective scheme if and only if $V$ has a $G$ invariant ample line bundle? Is that what you are looking for? –  Jim Bryan Jul 23 '13 at 0:22
    
@Jim Bryan: Why is that statement true? By $V/G$ you mean either the uniform geometric quotient or the uniform categorical quotient in the sense of Keel-Mori? –  stacksgg Jul 23 '13 at 6:45
    
For $G$ finite, you can just take the set-theoretic quotient (and put an appropriate sheaf on this). See for example Section 3 of Conrad's notes on the Keel-Mori theorem: math.stanford.edu/~conrad/papers/coarsespace.pdf –  Lennart Meier Jul 23 '13 at 7:12
    
A source which deals more generally with quotients which are Artin stacks is Alper's Good Moduli Spaces: maths-people.anu.edu.au/~alperj/papers/good_moduli_spaces.pdf But be aware that his notion of a good moduli space is more general than that of a coarse moduli space and applies, for example, also to $[\mathbb{A}^1/\mathbb{G}_m]$. –  Lennart Meier Jul 23 '13 at 7:15
    
The OP is correct: Jim Bryan's question has a negative answer. For the action of $\mathbf{G}_m$ on $\mathbb{A}^2\setminus\{(0,0)\}$ via $t*(x,y) = (tx,t^{-1}y)$, the structure sheaf is ample and admits a $\mathbb{G}_m$-linearization. However, the coarse moduli space is the (non-separated) line with doubled origin. So I upvoted this answer. –  Jason Starr Jul 23 '13 at 13:25
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