What is $M_g$ over a finite field, really?

Let $M_{g} \to \mathbb{Z}$ be the coarse moduli scheme of non-singular genus g curves over $\mathbb{Z}$. That is, suppose that $M_{g}$ co-represents the functor $M^{\sharp}_{g} : \text{Sch} \to \text{Sets}$ whose Yoneda points are (flat) families of non-singular, genus $g$ curves.

What is $M_{g} \times \mathbb{Z}/p$?

One would expect that $M_{g} \times \mathbb{Z}/p$ co-represents $M^{\sharp}_{g} \times \mathbb{Z}/p$, but in general, the formation of GIT quotients can fail to commute with passing to fibers. Does that failure occur here?

In other words: does $M_{g} \times \mathbb{Z}/p$ co-represent $M^{\sharp}_{g} \times \mathbb{Z}/p$?

Some references for the construction of $M_g$ over $\mathbb{Z}$ can be found here.

Added: Torsten Ekedahl had concerns about the use of the term co-represents." I intended it to mean that there is a natural transformation $M^{\sharp}_{g} \to M_g$ that is universal with respect to transformations into a scheme. Of course, this should be the same as asking that $M_{g}$ is the coarse space of the moduli stack $\mathcal{M}_{g}$, for a suitable notion of coarse space.

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I think that this might not be known; I was interested in this question a few years ago but was not able to find anything in the literature. (For a fixed $g$ it is true for large $p$ but I don't know if this can be made effective.) –  ulrich Aug 15 '11 at 8:18
I think you are misusing the notion of corepresentable (which as I understand it is just the dual of "representable" and hence doesn't even make sense for contravariant functors). So the question should be if $M_g\times\mathbb Z/p$ is the coarse moduli space of the algebraic stack $\mathcal M_g\times\mathbb Z/p$. –  Torsten Ekedahl Aug 15 '11 at 8:32
@ulrich: I think you get an effective bound by using the effective bounds on the order of automorphism groups of curves (by Stichtenoth). As the statement is true if $p$ doesn't divide the order of any automorphism group of a curve (of genus $g$). (On the other hand I think this is if and only if so in order to have it true for large $p$ on needs to know that the automorphisms groups of curves in char. $p$ is not divisible by $p$ for large $p$.) –  Torsten Ekedahl Aug 15 '11 at 8:45
It seems hard to prove since it seems to require knowing something about the action of the automorphism group of all curves on their versal deformation spaces. On the other hand, it might not be hard to show that it is false by finding an explicit curve over (say) $\mathbb{Z}/p$ giving rise to a non-normal point on $M_g \times \mathbb{Z}/p$. –  ulrich Aug 15 '11 at 9:29
@Torsten: Thanks for the reference to Stichtenoth. My argument for large $p$ was that $M_g \times \mathbb{Z}/p$ is normal if $g$ is fixed and $p$ is large... –  ulrich Aug 15 '11 at 9:34