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.