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Let $S$ be a noetherian scheme and let $X$ be a "nice" algebraic stack over $S$. For instance, let's say $X$ is a finitely presented algebraic stack over $S$, or that $X$ is a finite type separated DM-stack over $S$.

Let $P$ be an absolute or relative property. More precisely, let $P$ be "flat over $S$", "smooth over $S$", "etale over $S$", "Deligne-Mumford over $S$", "regular algebraic stack", "normal algebraic stack", "reduced algebraic stack", "separated over $S$", "proper over $S$", ...

Assume $X$ has $P$. Does $I_X$ have $P$?

I know that these are a lot of questions in one. I am only looking for hints on some of the difficult ones to verify, so that I (hopefully) get a feeling how to verify others myself.

Example. If $X$ is DM over $S$, then $I_X\to X$ is DM.

Example. If $X$ is proper over $S$, then $I_X\to X$ is proper and thus $I_X$ is proper over $S$.

Example. If $X$ is separated over $S$, then $I_X\to X$ is proper and thus $I_X$ is separated over $S$.

If $X$ is smooth (flat, etale, unramified) over $S$, is $I_X$ smooth (flat, etale, unramified) over $S$?

Edit: What if $S$ is the spectrum of a field?

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    $\begingroup$ It is helpful to recall that the "inertia stack" is defined to equal the $2$-fibered product of the diagonal with itself, $I_X=X\times_{\Delta,X\times_S X,\Delta} X$. Thus, for instance, if $X$ is a quotient stack $[Y/G]$ of a scheme $Y$ by the action of a finite, etale group $G$, then the pullback of $I_X\to X$ to $Y$ is the closed subgroup $Y$-scheme of $G\times Y$ that is the stabilizer of the action. Thus, even if $Y$ is smooth, flat, or etale over $S$, this subgroup $Y$-scheme need be none of these. $\endgroup$
    – Jason Starr
    Commented Mar 8, 2018 at 11:30
  • $\begingroup$ @JasonStarr I see. You are right that flatness will fail over positive dimensional base schemes. But, what if $S$ is the spectrum of a field? I have a feeling that smoothness of $X$ then implies smoothness of $I_X$. Is that right? $\endgroup$
    – Pan Da
    Commented Mar 8, 2018 at 13:50
  • $\begingroup$ "Is that right?" No, that is not right. Let the finite cyclic group $G=\langle g \rangle$ of order $p$ act on $Y=\mathbb{A}^2_{\mathbb{F}_p}=\text{Spec}\ \mathbb{F}_p[x,y]$ by $g\cdot(x,y) = (x,y + x^2)$. The stabilizer group scheme over $\mathbb{A}^2$ is a disjoint union of the identity section $\{e\}\times Y$ and $(G\setminus\{e\})\times \text{Zero}(x^2)$. This is nonreduced. $\endgroup$
    – Jason Starr
    Commented Mar 8, 2018 at 13:55
  • $\begingroup$ @JasonStarr Thank you for this example. I think I agree with you. What's confusing me is that Corollary 3.1.4 (and its proof) in "Gromov-Witten Theory of stacks" seems to claim otherwise; see arxiv.org/pdf/math/0603151.pdf . $\endgroup$
    – Pan Da
    Commented Mar 8, 2018 at 13:59
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    $\begingroup$ There are no nonconstant morphisms of group schemes from $\mu_r$ to the cyclic group scheme $G$ of order $p$ over any field of characteristic $p$. So there is no contradiction between the example above and Corollary $\pi.$ Abramovich, Graber, and Vistoli are well-aware of the issue in my example, i.e., the issue of "non-tame stacks". I believe there was a follow-up article about Gromov-Witten theory of tame stacks that explicitly addresses this issue. $\endgroup$
    – Jason Starr
    Commented Mar 8, 2018 at 14:05

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