The Stacks project

Lemma 97.27.8. In Situation 97.27.1 assume given a closed subset $Z \subset S$ such that

  1. the inverse image of $Z$ in $X'$ is $T'$,

  2. $U' \to S \setminus Z$ is a closed immersion,

  3. $W \to S_{/Z}$ is a closed immersion.

Then there exists a solution $(f : X' \to X, T, a)$ and moreover $X \to S$ is a closed immersion.

Proof. Suppose we have a closed subscheme $X \subset S$ such that $X \cap (S \setminus Z) = U'$ and $X_{/Z} = W$. Then $X$ represents the functor $F$ (some details omitted) and hence is a solution. To find $X$ is clearly a local question on $S$. In this way we reduce to the case discussed in the next paragraph.

Assume $S = \mathop{\mathrm{Spec}}(A)$ is affine. Let $I \subset A$ be the radical ideal cutting out $Z$. Write $I = (f_1, \ldots , f_ r)$. By assumption we are given

  1. the closed immersion $U' \to S \setminus Z$ determines ideals $J_ i \subset A[1/f_ i]$ such that $J_ i$ and $J_ j$ generate the same ideal in $A[1/f_ if_ j]$,

  2. the closed immersion $W \to S_{/Z}$ is the map $\text{Spf}(A^\wedge /J') \to \text{Spf}(A^\wedge )$ for some ideal $J' \subset A^\wedge $ in the $I$-adic completion $A^\wedge $ of $A$.

To finish the proof we need to find an ideal $J \subset A$ such that $J_ i = J[1/f_ i]$ and $J' = JA^\wedge $. By More on Algebra, Proposition 15.89.15 it suffices to show that $J_ i$ and $J'$ generate the same ideal in $A^\wedge [1/f_ i]$ for all $i$.

Recall that $A' = H^0(X', \mathcal{O})$ is a finite $A$-algebra whose formation commutes with flat base change (Cohomology of Spaces, Lemmas 68.20.3 and 68.11.2). Denote $J'' = \mathop{\mathrm{Ker}}(A \to A')$1. We have $J_ i = J''A[1/f_ i]$ as follows from base change to the spectrum of $A[1/f_ i]$. Observe that we have a commutative diagram

\[ \xymatrix{ X' \ar[d] & X'_{/T'} \times _{S_{/Z}} \text{Spf}(A^\wedge ) \ar[l] \ar[d] & X'_{/T'} \times _ W \text{Spf}(A^\wedge /J') \ar@{=}[l] \ar[d] \\ \mathop{\mathrm{Spec}}(A) & \text{Spf}(A^\wedge ) \ar[l] & \text{Spf}(A^\wedge /J') \ar[l] } \]

The middle vertical arrow is the completion of the left vertical arrow along the obvious closed subsets. By the theorem on formal functions we have

\[ (A')^\wedge = \Gamma (X' \times _ S \mathop{\mathrm{Spec}}(A^\wedge ), \mathcal{O}) = \mathop{\mathrm{lim}}\nolimits H^0(X' \times _ S \mathop{\mathrm{Spec}}(A/I^ n), \mathcal{O}) \]

See Cohomology of Spaces, Theorem 68.22.5. From the diagram we conclude that $J'$ maps to zero in $(A')^\wedge $. Hence $J' \subset J'' A^\wedge $. Consider the arrows

\[ X'_{/T'} \to \text{Spf}(A^\wedge /J''A^\wedge ) \to \text{Spf}(A^\wedge /J') = W \]

We know the composition $g$ is a formal modification (in particular rig-étale and rig-surjective) and the second arrow is a closed immersion (in particular an adic monomorphism). Hence $X'_{/T'} \to \text{Spf}(A^\wedge /J''A^\wedge )$ is rig-surjective and rig-étale, see Algebraization of Formal Spaces, Lemmas 87.21.5 and 87.20.8. Applying Algebraization of Formal Spaces, Lemmas 87.21.14 and 87.21.6 we conclude that $\text{Spf}(A^\wedge /J''A^\wedge ) \to W$ is rig-étale and rig-surjective. By Algebraization of Formal Spaces, Lemma 87.21.13 we conclude that $I^ n J'' A^\wedge \subset J'$ for some $n > 0$. It follows that $J'' A^\wedge [1/f_ i] = J' A^\wedge [1/f_ i]$ and we deduce $J_ i A^\wedge [1/f_ i] = J' A^\wedge [1/f_ i]$ for all $i$ as desired. $\square$

[1] Contrary to what the reader may expect, the ideals $J$ and $J''$ won't agreee in general.

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