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 $S.$ 
 A: 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.)
A: 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.
A: 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.

*Sean 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.

