This section is the analgue of Section 87.14 for completions with respect to a closed subspace.
Definition 87.38.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $Z \subset X$ be a closed subspace and denote $Z_ n \subset X$ the $n$th order infinitesimal neighbourhood. The formal algebraic space (see Lemma 87.36.2) is called the completion of $X$ along $Z$.
\[ X^\wedge _ Z = \mathop{\mathrm{colim}}\nolimits Z_ n \]
For example, if $X = \mathop{\mathrm{Spec}}(A)$ and $Z$ is cut out by the ideal $I \subset A$, then we have
where $B = \mathop{\mathrm{lim}}\nolimits A/I^ n$ with the limit topology. Note that $B$ is a weakly adic topological ring but in general not adic.
Returning to the general case, if $T = |Z|$ then there is a canonical morphism $X^\wedge _ Z \to X_{/T}$ comparing the completions along $Z$ and $T$ (Section 87.14) which need not be an isomorphism (see Lemma 87.38.6).
Let $f : X \to X'$ be a morphism of algebraic spaces over a scheme $S$. Suppose that $Z \subset X$ and $Z' \subset X'$ are closed subspaces such that $f|_ Z$ maps $Z$ into $Z'$ inducing a morphism $Z \to Z'$. Then it is clear that $f$ defines a morphism of formal algebraic spaces
between the completions.
Lemma 87.38.2. Let $S$ be a scheme. Let $f : X' \to X$ be a morphism of algebraic spaces over $S$. Let $Z \subset X$ be a closed subspace and let $Z' = f^{-1}(Z) = X' \times _ X Z$. Then is a cartesian diagram of sheaves. In particular, the morphism $(X')^\wedge _{Z'} \to X^\wedge _ Z$ is representable by algebraic spaces.
\[ \xymatrix{ (X')^\wedge _{Z'} \ar[r] \ar[d] & X' \ar[d]^ f \\ X^\wedge _ Z \ar[r] & X } \]
Proof. Namely, suppose that $Y \to X$ is a morphism from a scheme into $X$ such that $Y \to X$ factors through $Z$. Then $Y \times _ X X' \to X$ is a morphism of algebraic spaces such that $Y \times _ X X' \to X'$ factors through $Z'$. Since $Z'_ n = X' \times _ X Z_ n$ for all $n \geq 1$ the same is true for the infinitesimal neighbourhoods. Hence the cartesian square of functors follows from the formulas $X^\wedge _ Z = \mathop{\mathrm{colim}}\nolimits Z_ n$ and $(X')^\wedge _{Z'} = \mathop{\mathrm{colim}}\nolimits Z'_ n$. $\square$
Lemma 87.38.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $Z \subset X$ be a closed subspace. The reduction $(X^\wedge _ Z)_{red}$ of the completion $X^\wedge _ Z$ of $X$ along $Z$ is $Z_{red}$.
Proof. Omitted. $\square$
Lemma 87.38.4. Let $S$ be a scheme. Let $X = \mathop{\mathrm{Spec}}(A)$ be an affine scheme over $S$. Let $Z \subset X$ be a closed subscheme corresponding to the ideal $I \subset A$. Then
The affine formal algebraic space $X^\wedge _ Z$ is weakly adic.
If $I$ is finitely generated, then $X^\wedge _ Z = \text{Spf}(A^\wedge )$ where $A^\wedge $ is the $I$-adic completion of $A$.
If $Z \to X$ is of finite presentation, i.e., $I$ is finitely generated, then $X^\wedge _ Z$ is adic*.
If $X$ is Noetherian, then $X^\wedge _ Z$ is Noetherian.
Proof. Omitted. $\square$
Lemma 87.38.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $Z \subset X$ be a closed subspace. Let $X^\wedge _ Z$ be the formal completion of $X$ along $Z$.
The formal algebraic space $X^\wedge _ Z$ is locally weakly adic.
If $Z \to X$ is of finite presentation, then $X^\wedge _ Z$ is locally adic*.
If $X$ is locally Noetherian, then $X_ Z$ is locally Noetherian.
Proof. Omitted. $\square$
Lemma 87.38.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $Z \subset X$ be a closed subspace and set $T = |Z|$. The canonical morphism $c : X^\wedge _ Z \to X_{/T}$ is an isomorphism if $Z \to X$ is of finite presentation, but not in general.
Proof. Both constructions commute with étale localization, hence it suffices to prove this when $X$ is affine. Say $X = \mathop{\mathrm{Spec}}(A)$ and $Z$ corresponds to the ideal $I \subset A$. If $I$ is finitely generated then both $X^\wedge _ Z$ and $X_{/T}$ are equal to $\text{Spf}(A^\wedge )$ where $A^\wedge $ is the $I$-adic completion of $A$, see Lemmas 87.14.6 and 87.38.4. If $A = \mathbf{Z}[x_1, x_2, \ldots ]$ and $I = (x_1, x_2, \ldots )$ then $\mathop{\mathrm{Spec}}(A/(x_ n^ n, n \geq 1)) \to X$ factors through $X_{/T}$ but not through $X^\wedge _ Z$ and hence $c$ is not an isomorphism. $\square$
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