Lemma 76.43.3. 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 algebraic spaces 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
$X \to S$ is proper, and
$Y \to S$ is separated of finite type.
Then there exists a unique morphism of algebraic spaces $g : X \to Y$ 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 (Morphisms of Spaces, Lemma 67.4.7). Thus by Lemma 76.43.1 there exists a closed subspace $Z \subset X \times _ S Y$ 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 of Spaces, Lemma 67.40.6) whose base change to $S_ n$ is an isomorphism for all $n$. In particular, $p : Z \to X$ is quasi-finite on an open subspace of $Z$ containing every point of $Z_0$ for example by Morphisms of Spaces, Lemma 67.34.7. As $Z$ is proper over $S$ this open neighbourhood is all of $Z$. We conclude that $p : Z \to X$ is finite by Zariski's main theorem (for example apply Lemma 76.34.3 and use properness of $Z$ over $X$ to see that the immersion is a closed immersion). Applying the equivalence of Theorem 76.42.11 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.
$\square$
Comments (0)