When is the sheaf of isomorphism classes of a reasonable moduli problem an algebraic space? a scheme?

Question

Let $\mathcal{M}$ be a reasonable moduli problem (ie, at least a separated Deligne-Mumford stack, flat over some base scheme $S$).

Let $M$ be the sheafification of the presheaf: $$(U\rightarrow S)\mapsto \{\text{isomorphism classes of objects in }\mathcal{M}(U)\}$$

I'd like to understand when $M$ is an algebraic space, or a scheme, and if not, what sorts of things can go wrong.

Here, I'd like to stick to the conventions used in the stacks project, except that I'd like to stick to the etale topology, if it turns out to be relevant. Ie, a DM stack is a stack in groupoids with unramified diagonal representable by algebraic spaces, and admitting an etale surjective cover by a scheme, and an algebraic space is a sheaf with diagonal representable by schemes, also admitting an etale surjective cover by a scheme.

In the particular case I'm currently looking at, $\mathcal{M}$ is in addition proper and quasi-finite over $S$.

Answer 1

As Piotr Achinger observes (but I shall stick to your notation), $\mathcal{M}$ has a coarse moduli space, say $M_0$, and we have morphisms $\mathcal{M}\to M:=\pi_0(\mathcal{M})^\mathrm{sh}\to M_0$. If $M$ is an algebraic space, then $M\to M_0$ has a section, hence $\mathcal{M}\to M_0$ must be an epimorphism for the étale topology (I assume here that $^\mathrm{sh}$ means "associated étale sheaf", but see below).

Here is an example where this fails (but at the moment I cannot think of one which is flat and quasifinite over a scheme). Let $k$ be a field of characteristic $\neq2$, and let $\mathcal{M}=[\mathbb{A}^2_k/G]$ where $G=\{\pm1\}$ acts in the obvious way. Then $M_0$ is the usual quotient scheme, and $\mathcal{M}\to M_0$ is an epimorphism iff the quotient map $f:\mathbb{A}^2_k\to M_0$ is. This is clearly not the case: this map has no section locally for the étale topology.

In fact, let me prove the stronger result that $\mathbb{A}^2_k\to M_0$ is not an epimorphism of fppf (or even fpqc) sheaves. In other words, we still have a counterexample if $^\mathrm{sh}$ means "associated fppf sheaf". Put $X=\mathbb{A}^2_k$, let $U$ (resp. $V$) be the complement of the origin in $X$ (resp. $M_0$), and $j:V\to M_0$ the inclusion. Then $\mathcal{A}:=f_*(\mathcal{O}_X)$ is a finite $\mathcal{O}_{M_0}$-algebra, étale of rank 2 on $V$ but not locally free on $M_0$ since its rank at the origin is 3. Moreover we have $$j_*(\mathcal{A}_V)=\mathcal{A}\quad\text{ and }\quad j_*(\mathcal{O}_V)=\mathcal{O}_{M_0}$$ because both $X$ and $M_0$ are $\mathrm{S}_2$. Note that these properties are preserved by flat base change.
Now assume $p:Z\to M_0$ is faithfully flat and lifts to $p':Z\to X$. Put $W:=p^{-1}(V)\subset Z$, and let $w:W\to Z$ be the inclusion. The étale double cover of $W$ induced by $U\to V$ has a section, so it is trivial and in particular $\mathcal{A'}:=p^*\mathcal{A}$ is free on $W$. Hence, using the above-mentioned base change property,
$$\mathcal{A'}\cong w_*\mathcal{A'}_W\cong w_*\mathcal{O}_W^2\cong \mathcal{O}_Z^2$$ which implies by descent that $\mathcal{A}$ is locally free over $M_0$, a contradiction.