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

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

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

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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)