# Is the inertia stack of a Deligne-Mumford stack always finite?

Let X be a DM stack over a field k. We follow the definition in Laumon and Moret-Bailly's book, so that its diagonal is quasi-compact (and hence diagonal is of finite type). Then is the diagonal necessarily finite?

Edit: I meant to ask if the inertia $I\to X$ is finite. Recall that the inertia stack $I$ is defined to be the 2-fiber product of $X$ with $X$ over $X\times X,$ where the two maps $X\to X\times X$ are both the diagonal map. This question is equivalent (I think) to the following. Let $G\to S$ be an etale $S$-group scheme of finite type, where $S$ is a $k$-scheme of finite type. Then $G$ is finite over $S.$

I'm not sure about the suggested equivalence in the last two sentences of your question, but at least the statement about etale group schemes has a negative answer.

That is, it is possible to have an etale group scheme $G \rightarrow S$, with $G$ and $S$ both finite type over a field $k$, but $G$ not finite over $S$.

For example, let $H$ be the constant group scheme ${\mathbb Z}/2{\mathbb Z}$ over $S$, let $s$ be some fixed closed point of $S$, and let $G := H \setminus 1_s,$ where $1_s$ is the non-zero element of the fibre $({\mathbb Z}/2{\mathbb Z})_s$.

Then $G$ is open in $H$, hence etale over $S$. Assuming that $S$ is positive dimensional, it is certainly not finite (we deleted one point of one fibre), and it is a subgroup scheme of $H$. (If $T$ is an $S$-scheme, then $G(T)$ is the subgroup of $H(T)$ consisting of points whose values at points of $T$ lying over $s$ are trivial.)

• At any rate, it you now take the classifying stack of the scheme G over S, it would be a DM stack with diagonal that is not finite – t3suji Jan 21 '10 at 3:08
• You beat me to it. Oh well. – Anton Geraschenko Jan 21 '10 at 3:12

No, the inertia stack of a DM stack need not be finite over the stack. The stack I constructed at the end of this answer is an example. I'll explain the example in this context.

Let $G$ be the affine line with a doubled origin, regarded as a group scheme over $S=\mathbb{A}^1$. Then $G\to S$ is an etale $S$-group scheme of finite type which is not affine, and therefore not finite. Let $X=[S/G] = B_SG$. Then all the squares in the following diagram are cartesian:

G ---> S
|      |
v      v
I ---> X
|      |
v      v
X --> X×X


Since finite morphisms are stable under base change, $I\to X$ cannot be finite since $G\to S$ isn't.

Related note: Theorem 8.1 of Laumon and Moret-Bailly's book states that an algebraic stack is DM if and only if its diagonal is unramified. I'd be surprised if it were possible to say anything stronger in general.

The inertia stack of a separated Deligne Mumford stack is finite. In general algebraic stacks with finite inertia have coarse moduli space. Some references:

• Gerd Faltings and Ching-Li Chai, Degeneration of abelian varieties, Ergebnisse der Mathematik und ihrer Grenzgebiete (3) [Results in Mathematics and Related Areas (3)], vol. 22, Springer-Verlag, Berlin, 1990.
• Sean Keel and Shigefumi Mori, Quotients by groupoids, Ann. of Math. (2) 145 (1997), no. 1, 193-213.
• Dan Abramovich, Martin Olsson, and Angelo Vistoli, Tame stacks in positive characteristic, Ann. Inst. Fourier (Grenoble) 58 (2008), no. 4, 1057-1091.