Proof.
Lemma 98.13.8 applies to F. Using this we choose, for every finite type field k over S and x_0 \in F(\mathop{\mathrm{Spec}}(k)), an affine scheme U_{k, x_0} of finite type over S and a smooth morphism U_{k, x_0} \to F 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
U = \coprod \nolimits _{k, x_0} U_{k, x_0} \longrightarrow F
is smooth1. To finish the proof it suffices to show this map is surjective, see Bootstrap, Lemma 80.12.3 (this is where we use axiom [0]). By Criteria for Representability, Lemma 97.5.6 it suffices to show that U \times _ F V \to V is surjective for those V \to F where V is an affine scheme locally of finite presentation over S. Since U \times _ F V \to V is smooth the image is open. Hence it suffices to show that the image of U \times _ F V \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 the composition \mathop{\mathrm{Spec}}(k) \xrightarrow {v_0} V \to F. Then (u_{k, x_0}, v_0) : \mathop{\mathrm{Spec}}(k) \to U \times _ F V is a point mapping to v_0 and we win.
\square
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