TITLE: Is $M_g$ finitely covered by a scheme over the integers?
QUESTION [27 upvotes]: This question was prompted by my almost-answer to the question Does smooth and proper over $\mathbb Z$ imply rational? , but I never got around to asking this until now.
It is well known that $M_g$, the moduli space of curves, is a smooth stack. In fact it is globally a quotient of a smooth variety by the action of a finite group: if we take curves with level $n \geq 3$ structure, then we get a fine moduli space, and the quotient by $\mathrm{Sp}(2g,\mathbf{Z}/n)$ recovers $M_g$. However, we know that $M_g$ is in fact smooth over $\mathrm{Spec}(\mathbf Z)$, but this construction does not work over the integers: we need $n$ not divisible by the characteristic. This leads to the first question:
Question (a): Could it be true that $M_g = [X/G]$ where $X$ is a scheme, smooth over the integers, and $G$ is a finite group?
It is very possible that this question is too naive. Maybe there is a standard argument for why this can not be true, involving structure of group schemes and wild inertia and whatnot. Nevertheless one can ask for something weaker:
Question (b): Is there a scheme $X$, smooth over the integers, with a finite map $X \to M_g$? What if we only ask it to be proper and generically finite?
A (presumably harder) question is what happens over the boundary, i.e. if one replaces $M_g$ with $\overline M_g$ or $\overline M_{g,n}$. As I mentioned in my answer to the question linked above, it is known that over certain $\mathrm{Spec}(\mathbf Z[\frac 1 d])$ one can write $\overline M_g$ and $\overline{M}_{g,n}$ as global quotients by actions of finite groups, now using non-abelian level structures.

REPLY [3 votes]: Here's a suggestion in the genus $g=2$ case. If one considers a genus 2 Riemann surface, then it is well-known that it is hyperelliptic, and the fixed points of the hyperelliptic involution are 6 Weierstrauss points. If one quotients by the hyperelliptic involution, the quotient is a Riemann sphere with 6 distinguished points which are the images of the Weierstrauss points. So moduli space is isomorphic to the space of 6 points on a sphere, up to conformal equivalence. There is a finite cover of this, which is the space of 6 marked points, up to conformal equivalence. Thus, moduli space is 
$$ S_6 \backslash ((\mathbb{CP}^1)^6- \Delta)\ /\ PSL(2,\mathbb{C}),$$
where $\Delta$ denotes the large diagonal, and is cut out by $\Delta = \{ (z_1,\ldots, z_6) | z_i \neq z_j\ for\ i\neq j\}$, and $S_6$ acts by permuting coordinates and $PSL(2,\mathbb{C})$ acts coordinatewise . Since $PSL(2,\mathbb{C})$ acts faithfully on ordered triples, we may normalize the last three coordinates to be $0, 1, \infty$, so that $$ ((\mathbb{CP}^1)^6- \Delta) / PSL(2,\mathbb{C}) \cong  \{ (z_1,z_2,z_3)\in \mathbb{C}-\{0,1\} | z_i\neq z_j, i\neq j\}.$$
This is isomorphic to an affine variety defined over $\mathbb{Z}$ by the usual trick of introducing three new coordinates and equations $(z_i-z_j)t_{ij}=1$, $i\neq j \in \{1,2,3\}$ and has a quotient by $S_6$ which is isomorphic to $M_2$. The group elements of $S_6$ act as integral fractional transformations. 
So the suggestion is to take the affine scheme defined by the spectrum of the coordinate ring of this affine variety defined over $\mathbb{Z}$. But I don't know enough about schemes or stacks to know if this works for what you want to do.