Lemma 30.28.3. 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 $X$, $Y$ be schemes over $S$. For $n \geq 1$ we set $X_ n = X \times _ S S_ n$ and $Y_ n = Y \times _ S S_ n$. Suppose given a compatible system of commutative diagrams

$\xymatrix{ & & X_{n + 1} \ar[rd] \ar[rr]_{g_{n + 1}} & & Y_{n + 1} \ar[ld] \\ X_ n \ar[rru] \ar[rd] \ar[rr]_{g_ n} & & Y_ n \ar[rru] \ar[ld] & S_{n + 1} \\ & S_ n \ar[rru] }$

Assume that

1. $X \to S$ is proper, and

2. $Y \to S$ is separated of finite type.

Then there exists a unique morphism of schemes $g : X \to Y$ over $S$ such that $g_ n$ is the base change of $g$ to $S_ n$.

Proof. The morphisms $(1, g_ n) : X_ n \to X_ n \times _ S Y_ n$ are closed immersions because $Y_ n \to S_ n$ is separated (Schemes, Lemma 26.21.11). Thus by Lemma 30.28.1 there exists a closed subscheme $Z \subset X \times _ S Y$ proper over $S$ whose base change to $S_ n$ recovers $X_ n \subset X_ n \times _ S Y_ n$. The first projection $p : Z \to X$ is a proper morphism (as $Z$ is proper over $S$, see Morphisms, Lemma 29.41.7) whose base change to $S_ n$ is an isomorphism for all $n$. In particular, $p : Z \to X$ is finite over an open neighbourhood of $X_0$ by Lemma 30.21.2. As $X$ is proper over $S$ this open neighbourhood is all of $X$ and we conclude $p : Z \to X$ is finite. Applying the equivalence of Proposition 30.25.4 we see that $p_*\mathcal{O}_ Z = \mathcal{O}_ X$ as this is true modulo $I^ n$ for all $n$. Hence $p$ is an isomorphism and we obtain the morphism $g$ as the composition $X \cong Z \to Y$. We omit the proof of uniqueness. $\square$

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