Lemma 76.43.2. 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 morphism of algebraic spaces that is separated and 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 algebraic spaces with cartesian squares. Assume that
$Y_1 \to X_1$ is a finite morphism, and
$Y_1 \to S_1$ is proper.
Then there exists a finite morphism of algebraic spaces $Y \to X$ such that $Y_ n = Y \times _ S S_ n$ for all $n \geq 1$. Moreover, $Y$ is proper over $S$.
Proof.
Let's write $f_ n : Y_ n \to X_ n$ for the vertical morphisms. As the squares in the statement are cartesian we see that the base change of $f_ n$ to $X_1$ is $f_1$. Thus Lemma 76.10.2 shows that $f_ n$ is a finite morphism. Set $\mathcal{F}_ n = f_{n, *}\mathcal{O}_{Y_ n}$. 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 76.42.12, 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 76.42.11) 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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