Proposition 99.28.9. Let $\mathcal{X}$ be an algebraic stack. The following are equivalent

$\mathcal{X}$ is a gerbe, and

$\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is flat and locally of finite presentation.

Proposition 99.28.9. Let $\mathcal{X}$ be an algebraic stack. The following are equivalent

$\mathcal{X}$ is a gerbe, and

$\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is flat and locally of finite presentation.

**Proof.**
Assume (1). Choose a morphism $\mathcal{X} \to X$ into an algebraic space $X$ which turns $\mathcal{X}$ into a gerbe over $X$. Let $X' \to X$ be a surjective, flat, locally finitely presented morphism and set $\mathcal{X}' = X' \times _ X \mathcal{X}$. Note that $\mathcal{X}'$ is a gerbe over $X'$ by Lemma 99.28.3. Then both squares in

\[ \xymatrix{ \mathcal{I}_{\mathcal{X}'} \ar[r] \ar[d] & \mathcal{X}' \ar[r] \ar[d] & X' \ar[d] \\ \mathcal{I}_\mathcal {X} \ar[r] & \mathcal{X} \ar[r] & X } \]

are fibre product squares, see Lemma 99.5.5. Hence to prove $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is flat and locally of finite presentation it suffices to do so after such a base change by Lemmas 99.25.4 and 99.27.11. Thus we can apply Lemma 99.28.7 to assume that $\mathcal{X} = [U/G]$. By Lemma 99.28.6 we see $G$ is flat and locally of finite presentation over $U$ and that $x : U \to [U/G]$ is surjective, flat, and locally of finite presentation. Moreover, the pullback of $\mathcal{I}_\mathcal {X}$ by $x$ is $G$ and we conclude that (2) holds by descent again, i.e., by Lemmas 99.25.4 and 99.27.11.

Conversely, assume (2). Choose a smooth presentation $\mathcal{X} = [U/R]$, see Algebraic Stacks, Section 92.16. Denote $G \to U$ the stabilizer group algebraic space of the groupoid $(U, R, s, t, c, e, i)$, see Groupoids in Spaces, Definition 76.15.2. By Lemma 99.5.7 we see that $G \to U$ is flat and locally of finite presentation as a base change of $\mathcal{I}_\mathcal {X} \to \mathcal{X}$, see Lemmas 99.25.3 and 99.27.3. Consider the following action

\[ a : G \times _{U, t} R \to R, \quad (g, r) \mapsto c(g, r) \]

of $G$ on $R$. This action is free on $T$-valued points for any scheme $T$ as $R$ is a groupoid. Hence $R' = R/G$ is an algebraic space and the quotient morphism $\pi : R \to R'$ is surjective, flat, and locally of finite presentation by Bootstrap, Lemma 78.11.7. The projections $s, t : R \to U$ are $G$-invariant, hence we obtain morphisms $s' , t' : R' \to U$ such that $s = s' \circ \pi $ and $t = t' \circ \pi $. Since $s, t : R \to U$ are flat and locally of finite presentation we conclude that $s', t'$ are flat and locally of finite presentation, see Morphisms of Spaces, Lemmas 65.31.5 and Descent on Spaces, Lemma 72.15.1. Consider the morphism

\[ j' = (t', s') : R' \longrightarrow U \times U. \]

We claim this is a monomorphism. Namely, suppose that $T$ is a scheme and that $a, b : T \to R'$ are morphisms which have the same image in $U \times U$. By definition of the quotient $R' = R/G$ there exists an fppf covering $\{ h_ j : T_ j \to T\} $ such that $a \circ h_ j = \pi \circ a_ j$ and $b \circ h_ j = \pi \circ b_ j$ for some morphisms $a_ j, b_ j : T_ j \to R$. Since $a_ j, b_ j$ have the same image in $U \times U$ we see that $g_ j = c(a_ j, i(b_ j))$ is a $T_ j$-valued point of $G$ such that $c(g_ j, b_ j) = a_ j$. In other words, $a_ j$ and $b_ j$ have the same image in $R'$ and the claim is proved. Since $j : R \to U \times U$ is a pre-equivalence relation (see Groupoids in Spaces, Lemma 76.11.2) and $R \to R'$ is surjective (as a map of sheaves) we see that $j' : R' \to U \times U$ is an equivalence relation. Hence Bootstrap, Theorem 78.10.1 shows that $X = U/R'$ is an algebraic space. Finally, we claim that the morphism

\[ \mathcal{X} = [U/R] \longrightarrow X = U/R' \]

turns $\mathcal{X}$ into a gerbe over $X$. This follows from Groupoids in Spaces, Lemma 76.26.1 as $R \to R'$ is surjective, flat, and locally of finite presentation (if needed use Bootstrap, Lemma 78.4.6 to see this implies the required hypothesis). $\square$

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