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The Stacks project

Lemma 30.28.1. 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 separated morphism 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 schemes with cartesian squares. Assume that

  1. Z_1 \to X_1 is a closed immersion, and

  2. Z_1 \to S_1 is proper.

Then there exists a closed immersion of schemes Z \to X such that Z_ n = Z \times _ S S_ n. 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 Morphisms, Lemma 29.45.7 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 30.27.2, 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 30.23.4) we see that the cokernel \mathcal{Q} of \mathcal{O}_ X \to \mathcal{F} has vanishing completion. Since \mathcal{F} has support proper over S and so does \mathcal{Q} this implies that \mathcal{Q} = 0 for example because the functor (30.27.0.1) is an equivalence by Grothendieck's existence theorem. 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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