Lemma 76.43.1. 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{ Z_1 \ar[r] \ar[d] & Z_2 \ar[r] \ar[d] & Z_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
Z_1 \to X_1 is a closed immersion, and
Z_1 \to S_1 is proper.
Then there exists a closed immersion of algebraic spaces Z \to X such that Z_ n = Z \times _ S S_ n for all n \geq 1. Moreover, Z is proper over S.
Proof.
Let's write j_ n : Z_ n \to X_ n for the vertical morphisms. As the squares in the statement are cartesian we see that the base change of j_ n to X_1 is j_1. Thus Limits of Spaces, Lemma 70.15.5 shows that j_ n is a closed immersion. Set \mathcal{F}_ n = j_{n, *}\mathcal{O}_{Z_ n}, so that j_ n^\sharp is a surjection \mathcal{O}_{X_ n} \to \mathcal{F}_ n. Again 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 map \mathcal{O}_ X \to \mathcal{F} of coherent \mathcal{O}_ X-modules whose restriction to X_ n recovers \mathcal{O}_{X_ n} \to \mathcal{F}_ n. Moreover, the support of \mathcal{F} is proper over S. As the completion functor is exact (Lemma 76.42.3) we see that \mathcal{O}_ X \to \mathcal{F} is surjective. Thus \mathcal{F} = \mathcal{O}_ X/\mathcal{J} for some quasi-coherent sheaf of ideals \mathcal{J}. Setting Z = V(\mathcal{J}) finishes the proof.
\square
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