The Stacks project

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 15.89.16 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 69.20.3 and 69.11.2). 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

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

See Cohomology of Spaces, Theorem 69.22.5. From the diagram we conclude that $J'$ maps to zero in $(A')^\wedge $. Hence $J' \subset J'' A^\wedge $. Consider the arrows

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 88.21.5 and 88.20.8. Applying Algebraization of Formal Spaces, Lemmas 88.21.14 and 88.21.6 we conclude that $\text{Spf}(A^\wedge /J''A^\wedge ) \to W$ is rig-étale and rig-surjective. By Algebraization of Formal Spaces, Lemma 88.21.13 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. $\square$