# Is the residual gerbe really independent of the choice of a representative?

My question is about the passage (11.1) in the book of Laumon and Moret-Bailly on algebraic stacks. There we have a scheme $S$, an algebraic stack $\mathscr{X}$ over $S$, and a point $\xi$ of $\mathscr{X}$. The residual gerbe $\mathscr{G}_\xi$ at $\xi$ is defined as follows: choose an $S$-field $K$ and an $S$-morphism $x\colon \mathrm{Spec}(K) \rightarrow \mathscr{X}$ representing $\xi$, and let $\mathscr{G}_\xi \subset \mathscr{X}$ be the smallest substack (for the fppf topology) through which $x$ factors. The claim is that this $\mathscr{G}_\xi$ does not depend on the choices of $K$ and $x$. How does one verify this?

I am especially having trouble in the case when $K'$ is a big overfield of $K$, when I cannot figure out how to make contact with the fppf topology that is used in defining $\mathscr{G}_\xi$. It seems to me that $\mathscr{G}_\xi$ could shrink after replacing $K$ by this $K'$ in the defining procedure.

• The definition should require $\xi$ admits a representative $(x,K)$ with $K$ essentially of finite type over the residue field at a point in a smooth scheme cover $X \rightarrow \mathscr{X}$ ("ess. finite type" defined via Rem. 5.3), and only use such $K$ in 11.1. Then for such $\xi$ the same gerbe is obtained using any representative $(x,K)$ without finiteness conditions on $K$ (because ${\rm{Spec}}(K)\times_{\mathscr{X}} X$ is a non-empty smooth algebraic space, so admits a $K'$-point for $K'/K$ finite separable)! For the appeal to 3.7 near the start of 11.1, they should also point out 10.7. – grghxy Sep 10 '15 at 0:49
• @grghxy: Thank you. Could you clarify your first sentence? Any $\xi$ admits a representative $(x, K)$ with $K$ even of finite type over the residue field at a point of a smooth scheme cover (because already the classes of these residue fields exhaust $\xi$). Do you mean that one should use only such fields $K$ in 11.1 (for arbitrary $\xi$)? – O-Ren Ishii Sep 10 '15 at 2:43
• Sorry yes, I meant that in 11.1 one should only consider such $K$. It is more robust to impose "essentially finite type" (i.e., finitely generated as field extension) rather than "finite type" (= "finite" by Nullstellensatz) because a typical point of the algebraic space $x \times_{\mathscr{X}} X$ has residue field ess. finite type over $K$. – grghxy Sep 10 '15 at 4:53
• @grghxy: Thanks for clarifying. I am still puzzled: consider the case $\mathscr{X} = \mathrm{Spec}(F)$ for some field $F$. Then choosing $x = \mathrm{id}_{F}$ I get that the residual gerbe is $\mathscr{X}$ itself. However, I could choose a smooth cover $X \rightarrow \mathrm{Spec}(F)$, choose a generic point $\mathrm{Spec}(K)$ of $X$, and look at $x' \colon \mathrm{Spec}(K) \rightarrow \mathrm{Spec}(F)$. Now I should be getting that $x'$ is surjective for the fppf topology, but it is not: there is no fppf cover of $\mathrm{Spec}(F)$ that would factor through $\mathrm{Spec}(K)$. – O-Ren Ishii Sep 10 '15 at 15:50
• Sorry, you are absolutely correct. So focusing on finite extensions of residue fields at points of a smooth scheme cover is the right thing to do. Looking back at my copy of L-MB, in the margin I had made the incorrect note to myself that $K'/K$ is an fppf cover for a finitely generated extension field, which as you say is false; that was where my error crept in. I have now fixed my margin comment to say "finite extension" (in 11.1, after which one bootstraps to the general case). – grghxy Sep 11 '15 at 19:29

To see that it is algebraic (following Rydh and in the quasi-separated case, i.e., in the case of algebraic stacks as discussed in Laumon and Moret-Bailly), you can look Section Tag 06UH. Essentially the idea is to reduce to the case where $x$ is the generic point and $\mathcal{X}$ is an integral algebraic stack, then reduce to the case where $\mathcal{X}$ is a gerbe (flattening stratification for inertia), then pull back from the algebraic space that the algebraic stack is a gerbe over. Cheers!