Lemma 98.12.7. Let $S$ be a locally Noetherian scheme. Let $p : \mathcal{X} \to (\mathit{Sch}/S)_{fppf}$ be a category fibred in groupoids. Let $\xi = (R, \xi _ n, f_ n)$ be a formal object of $\mathcal{X}$ with $\xi _1$ lying over $\mathop{\mathrm{Spec}}(k) \to S$ with image $s \in S$. Assume

1. $\xi$ is versal,

2. $\xi$ is effective,

3. $\mathcal{O}_{S, s}$ is a G-ring, and

4. $p : \mathcal{X} \to (\mathit{Sch}/S)_{fppf}$ is limit preserving on objects.

Then there exist a morphism of finite type $U \to S$, a finite type point $u_0 \in U$ with residue field $k$, and an object $x$ of $\mathcal{X}$ over $U$ such that $x$ is versal at $u_0$ and such that $x|_{\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u_0}/\mathfrak m_{u_0}^ n)} \cong \xi _ n$.

Proof. Choose an object $x_ R$ of $\mathcal{X}$ lying over $\mathop{\mathrm{Spec}}(R)$ whose associated formal object is $\xi$. Let $N = 2$ and apply Lemma 98.10.1. We obtain $A, \mathfrak m_ A, x_ A, \ldots$. Let $\eta = (A^\wedge , \eta _ n, g_ n)$ be the formal object associated to $x_ A|_{\mathop{\mathrm{Spec}}(A^\wedge )}$. We have a diagram

$\vcenter { \xymatrix{ & \eta \ar[d] \\ \xi \ar[r] \ar@{..>}[ru] & \xi _2 = \eta _2 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & A^\wedge \ar[d] \\ R \ar[r] \ar@{..>}[ru] & R/\mathfrak m_ R^2 = A/\mathfrak m_ A^2 } }$

The versality of $\xi$ means exactly that we can find the dotted arrows in the diagrams, because we can successively find morphisms $\xi \to \eta _3$, $\xi \to \eta _4$, and so on by Formal Deformation Theory, Remark 90.8.10. The corresponding ring map $R \to A^\wedge$ is surjective by Formal Deformation Theory, Lemma 90.4.2. On the other hand, we have $\dim _ k \mathfrak m_ R^ n/\mathfrak m_ R^{n + 1} = \dim _ k \mathfrak m_ A^ n/\mathfrak m_ A^{n + 1}$ for all $n$ by construction. Hence $R/\mathfrak m_ R^ n$ and $A/\mathfrak m_ A^ n$ have the same (finite) length as $\Lambda$-modules by additivity of length and Formal Deformation Theory, Lemma 90.3.4. It follows that $R/\mathfrak m_ R^ n \to A/\mathfrak m_ A^ n$ is an isomorphism for all $n$, hence $R \to A^\wedge$ is an isomorphism. Thus $\eta$ is isomorphic to a versal object, hence versal itself. By Lemma 98.12.3 we conclude that $x_ A$ is versal at the point $u_0$ of $U = \mathop{\mathrm{Spec}}(A)$ corresponding to $\mathfrak m_ A$. $\square$

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