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 [4] and also Section 4 of [5].
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:

[1] Stacks Project

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

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

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

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

[6] 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). Commented Aug 31, 2017 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? Commented Aug 31, 2017 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? Commented Aug 31, 2017 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.) Commented Aug 31, 2017 at 22:02
• Yes the link you gave is where I first learned of that fact. Commented Aug 31, 2017 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 $\pi_1(\mathcal{M}_{1,1,\mathbb{Z}})$).