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 ( 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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