The Stacks project

Proof. Lemma 98.13.8 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

is smooth¹. To finish the proof it suffices to show this map is surjective, see Criteria for Representability, Lemma 97.19.1 (this is where we use axiom [0]). By Criteria for Representability, Lemma 97.5.6 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. $\square$