# Moduli problem of stable nodal curves over the integers

Over an algebraically closed field of characteristic zero, e.g. $\overline{\mathbb{Q}}$, the Deligne-Mumford stack $\overline{\mathcal{M}}_{g,n}$ represents the functor $$\overline{\mathcal{M}}_{g,n}(S)$$ assigning to a scheme $S$ the groupoid of proper stable nodal curves $C\to S$ over $S$ with genus $g$ and $n$ marked sections, taken up to isomorphism. If we are careful about Galois actions, it becomes clear that in fact $\overline{\mathcal{M}_{g,n}}$ is defined over $\mathbb{Q}$, and classifies families of curves together with certain descent data to $\mathbb{Q}$. Using GIT, Mumford defined certain natural models of $\mathcal{\overline{M}}_{g,n}$ over $\mathbb{Z}$. I would like to know whether these models represent any meaningful moduli functor, or if not, whether there is still a notion of stable nodal curve with marked points over an integral base that has some of the nice properties of the Deligne-Mumford functor.

• Could you please add a reference to the "natural models" that Mumford defined? There is a Deligne-Mumford stack of families of genus-$g$, $n$-pointed stable curves over an arbitrary base. There is a coarse moduli space of this stack. That coarse moduli space is a projective scheme over $\text{Spec}\ \mathbb{Z}$. Since you speak of "natural models", I suspect that you are talking about different schemes than the coarse moduli space of this Deligne-Mumford stack. May 6 '18 at 15:23
• I was looking at Mumford's "Stability of Projective Varieties", where one of his examples is pluricanonically embedded curves. May 6 '18 at 15:57
• Part of the point of that paper is that by varying the "GIT realization" of the moduli space, you get new ample divisor classes on the moduli space, even if the moduli space itself is independent of the GIT realization. At the end of the paper, Mumford reviews what was known at that time about the intersection of the cone of ample divisors and the subgroup generated by $\lambda$ and $\delta$. May 6 '18 at 16:14

Already Deligne and Mumford's original paper constructs $\overline{\mathcal M}_g$ over $\operatorname{Spec} \mathbb Z$ [DM, §5]. They do not do the pointed version, so they restrict themselves to $g \geq 2$ (otherwise it will only exist as an Artin stack). Note also that they write $\mathcal M_g$ for what is now commonly known as $\overline{\mathcal M}_g$. Another reference is of course the Stacks project; in particular [Stacks, Tag 0E99].

The same should true for the pointed version $\overline{\mathcal M}_{g,n}$, and I think it should not be so hard to modify the proof from the unpointed to the pointed case (especially if one also studies the versions for $g \in \{0,1\}$). I suspect that this has been carried out in the literature somewhere, but I do not know a reference off hand.

References.

[DM] Deligne, Pierre; Mumford, D., The irreducibility of the space of curves of a given genus. Publ. Math. Inst. Hautes Étud. Sci. 36 (1969), p. 75-109. ZBL0181.48803.

[Stacks] A.J de Jong et al, The stacks project.

• According to SGA, Grothendieck had already constructed $\overline{M}_{0,n}$ earlier. It is straightforward to embed $\overline{\mathcal{M}}_{g,n}$ in $\overline{\mathcal{M}}_{g+n}$ by attaching a curve of genus $1$ (of geometric genus $0$, just for definiteness) at each of the $n$ marked points. May 6 '18 at 15:49
• Thanks! I did not realize how much that paper does. May 6 '18 at 15:55
• Maybe Theorem 3.1 in this paper arxiv.org/pdf/0710.3374.pdf could be of some help. May 6 '18 at 18:19
• @JasonStarr doesn't $\overline{\mathcal{M}}_{g,n}$ parametrize curves with ordered marked points? It seems like your "embedding" would have image which is the quotient $\overline{\mathcal{M}}_{g,n}/\mathfrak{S}_n$. This doesn't seem like too big of a problem, but I don't necessarily see an easy fix. May 7 '18 at 1:15
• @Vaiprond "I don't necessarily see an easy fix." There is an easy fix. At the $r^{\text{th}}$ marked point, attach a chain of $m_r$ genus $1$ curves where $m_r$ is some sufficiently "unique" integer, say $m_r = r(g+1)$. May 7 '18 at 1:19