Lemma 30.28.2. Let $A$ be a Noetherian ring complete with respect to an ideal $I$. Write $S = \mathop{\mathrm{Spec}}(A)$ and $S_ n = \mathop{\mathrm{Spec}}(A/I^ n)$. Let $X \to S$ be a separated morphism of finite type. For $n \geq 1$ we set $X_ n = X \times _ S S_ n$. Suppose given a commutative diagram

\[ \xymatrix{ Y_1 \ar[r] \ar[d] & Y_2 \ar[r] \ar[d] & Y_3 \ar[r] \ar[d] & \ldots \\ X_1 \ar[r]^{i_1} & X_2 \ar[r]^{i_2} & X_3 \ar[r] & \ldots } \]

of schemes with cartesian squares. Assume that

$Y_ n \to X_ n$ is a finite morphism, and

$Y_1 \to S_1$ is proper.

Then there exists a finite morphism of schemes $Y \to X$ such that $Y_ n = Y \times _ S S_ n$. Moreover, $Y$ is proper over $S$.

**Proof.**
Let's write $f_ n : Y_ n \to X_ n$ for the vertical morphisms. Set $\mathcal{F}_ n = f_{n, *}\mathcal{O}_{Y_ n}$. This is a coherent $\mathcal{O}_{X_ n}$-module as $f_ n$ is finite (Lemma 30.9.9). Using that the squares are cartesian we see that the pullback of $\mathcal{F}_{n + 1}$ to $X_ n$ is $\mathcal{F}_ n$. Hence Grothendieck's existence theorem, as reformulated in Remark 30.27.2, tells us there exists a coherent $\mathcal{O}_ X$-module $\mathcal{F}$ whose restriction to $X_ n$ recovers $\mathcal{F}_ n$. Moreover, the support of $\mathcal{F}$ is proper over $S$. As the completion functor is fully faithful (Theorem 30.27.1) we see that the multiplication maps $\mathcal{F}_ n \otimes _{\mathcal{O}_{X_ n}} \mathcal{F}_ n \to \mathcal{F}_ n$ fit together to give an algebra structure on $\mathcal{F}$. Setting $Y = \underline{\mathop{\mathrm{Spec}}}_ X(\mathcal{F})$ finishes the proof.
$\square$

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