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
Comments (0)