Is $\mathscr{M}_{1,1,\mathbb{Z}}$ isomorphic to a quotient stack by a finite group?

Let $\mathscr{M}_{1,1,\mathbb{Z}}$ denote the moduli stack of elliptic curves.

Does there exist a scheme $X$ and a finite group $G$ acting on $X$ such that $\mathscr{M}_{1,1,\mathbb{Z}}$ is isomorphic to the quotient stack $[X/G]$?

Remarks/thoughts: For any scheme $S$, set $\mathscr{M}_{1,1,S} := \mathscr{M}_{1,1,\mathbb{Z}} \times_{\operatorname{Spec}\mathbb{Z}} S$.

1. We have $\mathscr{M}_{1,1,\mathbb{Z}} \simeq [W/H]$ where $W = \operatorname{Spec} \mathbb{Z}[a_{1},a_{2},a_{3},a_{4},a_{6},\Delta^{-1}]$ where $\Delta \in \mathbb{Z}[a_{1},a_{2},a_{3},a_{4},a_{6}]$ is the discriminant and $H$ is a subgroup scheme of $\mathrm{GL}_{3,\mathbb{Z}}$ of relative dimension 4 over $\mathbb{Z}$, see [1, Tag 072S] and [6, Section 3].
2. By [3, 4.7.2], for $N \ge 3$, the restrictions $\mathscr{M}_{1,1,\mathbb{Z}[\frac{1}{N}]}$ are isomorphic to $[Y(N)/\mathrm{GL}_{2}(\mathbb{Z}/N)]$ where $Y(N) \to \mathbb{Z}[\frac{1}{N}]$ is a smooth affine morphism of relative dimension 1. (Thus there is an affine open covering of $\mathbb{Z}$ on which the restriction of $\mathscr{M}_{1,1,\mathbb{Z}}$ is a quotient stack by a finite group.) For $N=2$, see Remark 2.8 and the following paragraph of  and also Section 4 of .
3. Such scheme $X$ would have to be affine and smooth over $\mathbb{Z}$ of relative dimension 1. (Reason why $X$ is affine: Fix $N \ge 3$ and set $T_{N}' := Y(N) \times_{\mathscr{M}_{1,1,\mathbb{Z}[\frac{1}{N}]}} X[\frac{1}{N}]$; then $T_{N}' \to Y(N)$ is a $G$-torsor, hence $T_{N}'$ is affine; moreover $T_{N}' \to X[\frac{1}{N}]$ is a $\mathrm{GL}_{2}(\mathbb{Z}/N)$-torsor so $X[\frac{1}{N}]$ is affine; thus $X \to \mathbb{Z}$ is Zariski-locally on the target an affine morphism; thus $X$ is affine.)
4. Such scheme $X$ cannot have a $\mathbb{Z}$-point (otherwise $\mathscr{M}_{1,1,\mathbb{Z}}$ itself has a $\mathbb{Z}$-point, but there are no elliptic curves over $\mathbb{Z}$).

References:

 Stacks Project

 Olsson, "Algebraic Spaces and Stacks", Colloquium Publications 62, AMS (2016)

 Katz, Mazur, "Arithmetic Moduli of Elliptic Curves", volume 108 of Annals of Mathematics Studies, Princeton University Press, Princeton, NJ, 1985

 Conrad, "Isogenies and level structures", Notes for Stanford Math 248B, link

 Antieau, Meier, "The Brauer group of the moduli stack of elliptic curves", arxiv

 Fulton, Olsson, "The Picard group of $\mathscr{M}_{1,1}$", Algebra & Number Theory, vol. 4, no. 1 (2010) link

• The map $\mathscr{M}_{1,1,\mathbb{Z}} \to \mathbb{Z}$ is smooth, and $X \to \mathscr{M}_{1,1,\mathbb{Z}}$ is a $G$-torsor (hence smooth). Aug 31 '17 at 21:31
• Actually now I'm confused. Firstly, I presume you're using the etale topology? If $X\rightarrow\mathcal{M}_{1,1,\mathbb{Z}}$ is a $G$-torsor for the etale topology, then it is in fact finite etale, but $\mathcal{M}_{1,1,\mathbb{Z}}$ has no nontrivial finite etale covers. Have I made a mistake somewhere? Or perhaps you don't want to use the etale topology? Aug 31 '17 at 21:47
• actually even in the fpqc topology, being finite-etale is local on the base, so any fpqc G-torsor is finite etale, right? Aug 31 '17 at 21:59
• @oxeimon Thank you, I think your comment resolves my question. I assume you're referring to user22479's answer in mathoverflow.net/questions/105047/…? (I am indeed using the etale topology, but I guess (as you say above) it shouldn't matter since being finite-etale is fpqc local on the base.) Aug 31 '17 at 22:02
• Yes the link you gave is where I first learned of that fact. Aug 31 '17 at 22:13

The definition of the quotient stack makes $p : X\rightarrow [X/G]$ into a $G$-torsor (in whatever topology $\mathcal{T}$ one chooses). Since here we're working with a finite abstract group, $X\rightarrow[X/G] = \mathcal{M}_{1,1,\mathbb{Z}}$ is $\mathcal{T}$-locally isomorphic to a disjoint union of $|G|$ copies of $\mathcal{M}_{1,1,\mathbb{Z}}$, and hence since being finite etale is local on the target for pretty much any choice of topology $\mathcal{T}$, we find that $p$ is finite etale.
However, since the fundamental group of $\mathcal{M}_{1,1,\mathbb{Z}}$ is trivial, $X$ is itself a disjoint union of copies of $\mathcal{M}_{1,1,\mathbb{Z}}$, and hence cannot be a scheme.
EDIT: Actually, it seems all this also follows from Chapter 6 of LMB's book (again using the triviality of $\pi_1(\mathcal{M}_{1,1,\mathbb{Z}})$).