# Zariski's main theorem for non-representable morphisms?

Let $$f:\mathcal{X}\to \mathcal{Y}$$ be a separated quasi-finite map of qcqs Deligne-Mumford stacks. Is there a version of Zariski's main theorem that makes sense in this context? Rydh proved a version of this in the case where the map $$f$$ is also assumed to be representable, in which case we recover a stacky version of the classical factorization of $$f$$ as the inclusion of an open substack into a finite stack over $$\mathcal{Y}$$.

Obviously we can't hope for exactly such a factorization, since this would make $$f$$ automatically quasi-affine (and therefore representable). But I was wondering if maybe there is a factorization of $$f$$ into something like a (locally constant?) gerbe over an open substack of a finite stack over $$\mathcal{Y}$$.

## 1 Answer

You can take the relative coarse map to get a factorization of $$f$$ into $$\mathcal{X} \to X \to \mathcal{Y}$$ where $$g : X \to \mathcal{Y}$$ is representable and $$\pi : \mathcal{X} \to X$$ is proper + quasi-finite with $$\mathcal{O}_X \to \pi_*\mathcal{O}_{\mathcal{X}}$$ an isomorphism. Then you can apply the representable case of ZMT to $$g$$ to obtain a factorization $$X \hookrightarrow \overline{X} \to \mathcal{Y}$$ where $$X \hookrightarrow \overline{X}$$ is an open immersion and $$\overline{g} : \overline{X} \to \mathcal{Y}$$ is finite.

Putting this together, we get that any such $$f$$ factors into $$\mathcal{X} \xrightarrow{\rho} \overline{X} \xrightarrow{\overline{g}} \mathcal{Y}$$ where

1. $$\overline{g}$$ is finite and in particular representable,
2. $$\mathcal{X}$$ is proper + quasi-finite over an open substack $$i : X \subset \overline{X}$$, and
3. $$\rho_*\mathcal{O}_\mathcal{X} = i_*\mathcal{O}_X$$.

I think conditions $$2 + 3$$ can be replaced by something like $$\mathcal{O}_\overline{X} \to \rho_*\mathcal{O}_\mathcal{X}$$ is injective and integrally closed.

The existence of the relative coarse space is guaranteed under your assumptions by Theorem 3.1 here. Indeed the relative inertia stack is proper over $$\mathcal{X}$$ by the separated assumption and quasi-finite by the DM assumption.

I think by universality of the relative coarse map this is essentially the best you can do. In general the kernel of the map on inertia can jump so I don't think you can expect the first map to be a gerbe over an open substack, e.g., if $$f$$ itself the coarse space of a separated DM stack that is not a gerbe.

• This is great! If I had assumed that the map was finite presentation étale rather than quasi-finite, would I get a factorization such that the relative coarse map $\mathcal{X}\to X$ is proper and étale? – Harry Gindi Oct 26 '20 at 19:01
• I think both questions are local on $\mathcal{Y}$ because taking coarse space commutes with flat base change and "separated" and "étale" are both étale (and in fact fpqc) local on the target, so we can assume that $\mathcal{Y} = \mathrm{Spec} A$ is affine in which case relative coarse space = usual coarse space. Now separatedness of the coarse space is one of the conclusions of the Keel-Mori theorem. – Dori Bejleri Oct 28 '20 at 3:40
• The étale question seems more subtle. I think the coarse map $\mathcal{X} \to X$ is étale if and only if $\mathcal{X}$ is a gerbe. So the question is essentially if you have a DM stack which is étale over a scheme, is it a gerbe? This feels true but I'm not totally sure. – Dori Bejleri Oct 28 '20 at 3:56
• The composition of an etale gerbe and a finite etale cover of a scheme is not necessarily a gerbe, right? – Ariyan Javanpeykar Oct 28 '20 at 12:54
• There's a comment in one of Rydh's papers that a proper étale map $\mathcal{X}\to \mathcal{Y}$ factors canonically as a proper étale gerbe $\mathcal{X}\to X$ over a finite étale map $X\to \mathcal{Y}$. The thing I wanted to know is if a separated finite presentation étale map factors as a proper étale gerbe over a separated representable finite presentation étale map (and if this corresponds to the Keel-Mori factorization). – Harry Gindi Oct 28 '20 at 13:53