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There is an analogy of Chow's lemma for a DM stack $X$ written in the Laumon's book 'Champ algebrique'. There exists a generically finite, proper surjective morphism $Y \to X$ from a quasi-projective scheme $Y$. Furthermore, assume that there be a connected linear algebraic group $G$ acts on $X$. Then, is there a such quasi-projective scheme $Y$ with $G$-action, and an equivariant morphism $Y \to X$?

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    $\begingroup$ If $G$ is a torus and you are ok with working etale locally, you can find an answer to your question in Alper-Hall-Rydh's Section 2.3 of sites.math.washington.edu/~jarod/papers/luna.pdf $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 27, 2017 at 12:24
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    $\begingroup$ Let $G=\mathbb{G}_a$ and consider the action of $G$ on the weighted projective line $\mathcal{P}(1,n)$ with $n\geq 2$ over $\mathbb{C}$. Note that $\mathcal{P}(1,n)$ is a smooth proper DM-curve. Let $Y\to X$ be a generically finite, proper surjective morphism with $Y$ a normal integral scheme. Then $Y$ is a smooth projective connected curve. If the $\mathbb{G}_a$-action on $X:=\mathcal{P}(1,n)$ lifts to $Y$, then $g(Y) =0$, so that $Y =\mathbb{P}^1$. Now, the question becomes whether there is a $\mathbb{G}_a$-equivariant morphism $f:\mathbb{P}^1\to \mathbb{P}^1$ which ramifies at infinity. $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 27, 2017 at 12:27
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    $\begingroup$ More precisely, write $Y:=\mathbb{P}^1$ and $X = \mathbb{P}^1$ and consider $X$ with the natural action of $\mathbb{G}_a$ (via the inclusion $\mathbb{A}^1\subset \mathbb{P}^1$). My previous comment shows that the question now becomes whether there is a $\mathbb{G}_a$-action on $Y$ and a $\mathbb{G}_a$-equivariant morphism $Y\to X$ which ramifies at infinity. If this data does not exist, then the answer to your question is negative. $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 27, 2017 at 17:00
  • $\begingroup$ Thank you for your kind explanation. Your comments help me very much. Also results in the reference you mentioned are exactly what I was finding these days. $\endgroup$
    – keaton
    Commented Nov 28, 2017 at 13:17
  • $\begingroup$ You're welcome. I will come back and write down an answer once I'm sure that what I say is really correct. Or maybe somebody else will find another (easier?) example. $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 28, 2017 at 13:23

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