Lemma 76.43.5. Let $(A, \mathfrak m, \kappa )$ be a complete local Noetherian ring. Set $S = \mathop{\mathrm{Spec}}(A)$ and $S_ n = \mathop{\mathrm{Spec}}(A/\mathfrak m^ n)$. Consider a commutative diagram
\[ \xymatrix{ X_1 \ar[r]_{i_1} \ar[d] & X_2 \ar[r]_{i_2} \ar[d] & X_3 \ar[r] \ar[d] & \ldots \\ S_1 \ar[r] & S_2 \ar[r] & S_3 \ar[r] & \ldots } \]
of algebraic spaces with cartesian squares. If $\dim (X_1) \leq 1$, then there exists a projective morphism of schemes $X \to S$ and isomorphisms $X_ n \cong X \times _ S S_ n$ compatible with $i_ n$.
Proof.
By Spaces over Fields, Lemma 72.9.3 the algebraic space $X_1$ is a scheme. Hence $X_1$ is a proper scheme of dimension $\leq 1$ over $\kappa $. By Varieties, Lemma 33.43.4 we see that $X_1$ is H-projective over $\kappa $. Let $\mathcal{L}_1$ be an ample invertible sheaf on $X_1$.
We are going to show that $\mathcal{L}_1$ lifts to a compatible system $\{ \mathcal{L}_ n\} $ of invertible sheaves on $\{ X_ n\} $. Observe that $X_ n$ is a scheme too by Lemma 76.9.5. Recall that $X_1 \to X_ n$ induces homeomorphisms of underlying topological spaces. In the rest of the proof we do not distinguish between sheaves on $X_ n$ and sheaves on $X_1$. Suppose, given a lift $\mathcal{L}_ n$ to $X_ n$. We consider the exact sequence
\[ 1 \to (1 + \mathfrak m^ n\mathcal{O}_{X_{n + 1}})^* \to \mathcal{O}_{X_{n + 1}}^* \to \mathcal{O}_{X_ n}^* \to 1 \]
of sheaves on $X_{n + 1}$. The class of $\mathcal{L}_ n$ in $H^1(X_ n, \mathcal{O}_{X_ n}^*)$ (see Cohomology, Lemma 20.6.1) can be lifted to an element of $H^1(X_{n + 1}, \mathcal{O}_{X_{n + 1}}^*)$ if and only if the obstruction in $H^2(X_{n + 1}, (1 + \mathfrak m^ n\mathcal{O}_{X_{n + 1}})^*)$ is zero. As $X_1$ is a Noetherian scheme of dimension $\leq 1$ this cohomology group vanishes (Cohomology, Proposition 20.20.7).
By Grothendieck's algebraization theorem (Cohomology of Schemes, Theorem 30.28.4) we find a projective morphism of schemes $X \to S = \mathop{\mathrm{Spec}}(A)$ and a compatible system of isomorphisms $X_ n = S_ n \times _ S X$.
$\square$
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