Lemma 106.2.10. In Situation 106.2.1 let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism, where $k$ is a finite type field over $S$. Let $A$ be a versal ring to $\mathcal{X}$ at $x_0$. Then the morphism $\mathop{\mathrm{Spec}}(A) \to \mathcal{X}$ of Remark 106.2.9 is flat.

Proof. If the local ring of $S$ at the image point is a G-ring, then this follows immediately from Lemma 106.2.8 and the fact that the map from a Noetherian local ring to its completion is flat. In general we prove it as follows.

Step I. If $A$ and $A'$ are two versal rings to $\mathcal{X}$ at $x_0$, then the result is true for $A$ if and only if it is true for $A'$. Namely, after possible swapping $A$ and $A'$, we may assume there is a formally smooth map $\varphi : A \to A'$ such that the composition

$\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A) \to \mathcal{X}$

is the morphism $\mathop{\mathrm{Spec}}(A') \to \mathcal{X}$, see Lemma 106.2.4 and Remark 106.2.9. Since $A \to A'$ is faithfully flat we obtain the equivalence from Morphisms of Stacks, Lemmas 100.25.2 and 100.25.5.

Step II. Let $l/k$ be a finite extension of fields. Let $x_{l, 0} : \mathop{\mathrm{Spec}}(l) \to \mathcal{X}$ be the induced morphism. Let $A$ be a versal ring to $\mathcal{X}$ at $x_0$ and let $A \to A'$ be as in Lemma 106.2.5. Then again the composition

$\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A) \to \mathcal{X}$

is the morphism $\mathop{\mathrm{Spec}}(A') \to \mathcal{X}$, see Remark 106.2.9. Arguing as before and using step I to see choice of versal rings is irrelevant, we see that the lemma holds for $x_0$ if and only if it holds for $x_{l, 0}$.

Step III. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then we can choose a finite type point $z_0$ on $Z = U \times _\mathcal {X} x_0$ (this is a nonempty algebraic space). Let $u_0 \in U$ be the image of $z_0$ in $U$. Choose a scheme and a surjective étale map $W \to Z$ such that $z_0$ is the image of a closed point $w_0 \in W$ (see Morphisms of Spaces, Section 66.25). Since $W \to \mathop{\mathrm{Spec}}(k)$ and $W \to U$ are of finite type, we see that $\kappa (w_0)/k$ and $\kappa (w_0)/\kappa (u_0)$ are finite extensions of fields (see Morphisms, Section 29.16). Applying Step II twice we may replace $x_0$ by $u_0 \to U \to \mathcal{X}$. Then we see our morphism is the composition

$\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u_0}^\wedge ) \to U \to \mathcal{X}$

The first arrow is flat because completion of Noetherian local rings are flat (Algebra, Lemma 10.97.2) and the second arrow is flat as a smooth morphism is flat. The composition is flat as composition preserves flatness. $\square$

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