Lemma 97.17.1. Let $S$ be a locally Noetherian scheme. Let $p : \mathcal{X} \to (\mathit{Sch}/S)_{fppf}$ be a category fibred in groupoids. Assume that

1. $\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable by algebraic spaces,

2. $\mathcal{X}$ satisfies axioms [-1], , , ,  (see Section 97.14),

3. every formal object of $\mathcal{X}$ is effective,

4. $\mathcal{X}$ satisfies openness of versality, and

5. $\mathcal{O}_{S, s}$ is a G-ring for all finite type points $s$ of $S$.

Then $\mathcal{X}$ is an algebraic stack.

Proof. Lemma 97.13.8 applies to $\mathcal{X}$. Using this we choose, for every finite type field $k$ over $S$ and every isomorphism class of object $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_{\mathop{\mathrm{Spec}}(k)})$, an affine scheme $U_{k, x_0}$ of finite type over $S$ and a smooth morphism $(\mathit{Sch}/U_{k, x_0})_{fppf} \to \mathcal{X}$ such that there exists a finite type point $u_{k, x_0} \in U_{k, x_0}$ with residue field $k$ such that $x_0$ is the image of $u_{k, x_0}$. Then

$(\mathit{Sch}/U)_{fppf} \to \mathcal{X}, \quad \text{with}\quad U = \coprod \nolimits _{k, x_0} U_{k, x_0}$

is smooth1. To finish the proof it suffices to show this map is surjective, see Criteria for Representability, Lemma 96.19.1 (this is where we use axiom ). By Criteria for Representability, Lemma 96.5.6 it suffices to show that $(\mathit{Sch}/U)_{fppf} \times _\mathcal {X} (\mathit{Sch}/V)_{fppf} \to (\mathit{Sch}/V)_{fppf}$ is surjective for those $y : (\mathit{Sch}/V)_{fppf} \to \mathcal{X}$ where $V$ is an affine scheme locally of finite presentation over $S$. By assumption (1) the fibre product $(\mathit{Sch}/U)_{fppf} \times _\mathcal {X} (\mathit{Sch}/V)_{fppf}$ is representable by an algebraic space $W$. Then $W \to V$ is smooth, hence the image is open. Hence it suffices to show that the image of $W \to V$ contains all finite type points of $V$, see Morphisms, Lemma 29.16.7. Let $v_0 \in V$ be a finite type point. Then $k = \kappa (v_0)$ is a finite type field over $S$. Denote $x_0 = y|_{\mathop{\mathrm{Spec}}(k)}$ the pullback of $y$ by $v_0$. Then $(u_{k, x_0}, v_0)$ will give a morphism $\mathop{\mathrm{Spec}}(k) \to W$ whose composition with $W \to V$ is $v_0$ and we win. $\square$

 Set theoretical remark: This coproduct is (isomorphic to) an object of $(\mathit{Sch}/S)_{fppf}$ as we have a bound on the index set by axiom [-1], see Sets, Lemma 3.9.9.

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