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The Stacks project

87.37 Recompletion

In this section we define the completion of a formal algebraic space along a closed subset of its reduction. It is the natural generalization of Section 87.14.

Lemma 87.37.1. Let S be a scheme. Let X be an affine formal algebraic space over S. Let T \subset |X_{red}| be a closed subset. Then the functor

X_{/T} : (\mathit{Sch}/S)_{fppf} \longrightarrow \textit{Sets},\quad U \longmapsto \{ f : U \to X : f(|U|) \subset T\}

is an affine formal algebraic space.

Proof. Write X = \mathop{\mathrm{colim}}\nolimits X_\lambda as in Definition 87.9.1. Then X_{\lambda , red} = X_{red} and we may and do view T as a closed subset of |X_\lambda | = |X_{\lambda , red}|. By Lemma 87.14.1 for each \lambda the completion (X_\lambda )_{/T} is an affine formal algebraic space. The transition morphisms (X_\lambda )_{/T} \to (X_\mu )_{/T} are closed immersions as base changes of the transition morphisms X_\lambda \to X_\mu , see Lemma 87.14.4. Also the morphisms ((X_\lambda )_{/T})_{red} \to ((X_\mu )_/T)_{red} are isomorphisms by Lemma 87.14.5. Since X_{/T} = \mathop{\mathrm{colim}}\nolimits (X_\lambda )_{/T} we conclude by Lemma 87.36.1. \square

Lemma 87.37.2. Let S be a scheme. Let X be a formal algebraic space over S. Let T \subset |X_{red}| be a closed subset. Then the functor

X_{/T} : (\mathit{Sch}/S)_{fppf} \longrightarrow \textit{Sets},\quad U \longmapsto \{ f : U \to X \mid f(|U|) \subset T\}

is a formal algebraic space.

Proof. The functor X_{/T} is an fppf sheaf since if \{ U_ i \to U\} is an fppf covering, then \coprod |U_ i| \to |U| is surjective.

Choose a covering \{ g_ i : X_ i \to X\} _{i \in I} as in Definition 87.11.1. The morphisms X_ i \times _ X X_{/T} \to X_{/T} are étale (see Spaces, Lemma 65.5.5) and the map \coprod X_ i \times _ X X_{/T} \to X_{/T} is a surjection of sheaves. Thus it suffices to prove that X_{/T} \times _ X X_ i is an affine formal algebraic space. A U-valued point of X_ i \times _ X X_{/T} is a morphism U \to X_ i whose image is contained in the closed subset |g_{i, red}|^{-1}(T) \subset |X_{i, red}|. Thus this follows from Lemma 87.37.1. \square

Definition 87.37.3. Let S be a scheme. Let X be a formal algebraic space over S. Let T \subset |X_{red}| be a closed subset. The formal algebraic space X_{/T} of Lemma 87.14.2 is called the completion of X along T.

Let f : X \to X' be a morphism of formal algebraic spaces over a scheme S. Suppose that T \subset |X_{red}| and T' \subset |X'_{red}| are closed subsets such that |f_{red}|(T) \subset T'. Then it is clear that f defines a morphism of formal algebraic spaces

X_{/T} \longrightarrow X'_{/T'}

between the completions.

Lemma 87.37.4. Let S be a scheme. Let f : X' \to X be a morphism of formal algebraic spaces over S. Let T \subset |X_{red}| be a closed subset and let T' = |f_{red}|^{-1}(T) \subset |X'_{red}|. Then

\xymatrix{ X'_{/T'} \ar[r] \ar[d] & X' \ar[d]^ f \\ X_{/T} \ar[r] & X }

is a cartesian diagram of formal algebraic spaces over S.

Proof. Namely, observe that the horizontal arrows are monomorphisms by construction. Thus it suffices to show that a morphism g : U \to X' from a scheme U defines a point of X'_{/T} if and only if f \circ g defines a point of X_{/T}. In other words, we have to show that g(U) is contained in T' \subset |X'_{red}| if and only if (f \circ g)(U) is contained in T \subset |X_{red}|. This follows immediately from our choice of T' as the inverse image of T. \square

Lemma 87.37.5. Let S be a scheme. Let X be a formal algebraic space over S. Let T \subset |X_{red}| be a closed subset. The reduction (X_{/T})_{red} of the completion X_{/T} of X along T is the reduced induced closed subspace Z of X_{red} corresponding to T.

Proof. It follows from Lemma 87.12.1, Properties of Spaces, Definition 66.12.5 (which uses Properties of Spaces, Lemma 66.12.3 to construct Z), and the definition of X_{/T} that Z and (X_{/T})_{red} are reduced algebraic spaces characterized the same mapping property: a morphism g : Y \to X whose source is a reduced algebraic space factors through them if and only if |Y| maps into T \subset |X|. \square

Lemma 87.37.6. Let S be a scheme. Let X be an affine formal algebraic space over S. Let T \subset X_{red} be a closed subset and let X_{/T} be the formal completion of X along T. Then

  1. X_{/T} is an affine formal algebraic space,

  2. if X is McQuillan, then X_{/T} is McQuillan,

  3. if |X_{red}| \setminus T is quasi-compact and X is countably indexed, then X_{/T} is countably indexed,

  4. if |X_{red}| \setminus T is quasi-compact and X is adic*, then X_{/T} is adic*,

  5. if X is Noetherian, then X_{/T} is Noetherian.

Proof. Part (1) is Lemma 87.37.1. If X is McQuillan, then X = \text{Spf}(A) for some weakly admissible topological ring A. Then X_{/T} \to X \to \mathop{\mathrm{Spec}}(A) satisfies property (2) of Lemma 87.9.6 and hence X_{/T} is McQuillan, see Definition 87.9.7.

Assume X and T are as in (3). Then X = \text{Spf}(A) where A has a fundamental system A \supset I_1 \supset I_2 \supset I_3 \supset \ldots of weak ideals of definition, see Lemma 87.10.4. By Algebra, Lemma 10.29.1 we can find a finitely generated ideal \overline{J} = (\overline{f}_1, \ldots , \overline{f}_ r) \subset A/I_1 such that T is cut out by \overline{J} inside \mathop{\mathrm{Spec}}(A/I_1) = |X_{red}|. Choose f_ i \in A lifting \overline{f}_ i. If Z = \mathop{\mathrm{Spec}}(B) is an affine scheme and g : Z \to X is a morphism with g(Z) \subset T (set theoretically), then g^\sharp : A \to B factors through A/I_ n for some n and g^\sharp (f_ i) is nilpotent in B for each i. Thus J_{m, n} = (f_1, \ldots , f_ r)^ m + I_ n maps to zero in B for some n, m \geq 1. It follows that X_{/T} is the formal spectrum of \mathop{\mathrm{lim}}\nolimits _{n, m} A/J_{m, n} and hence countably indexed. This proves (3).

Proof of (4). Here the argument is the same as in (3). However, here we may choose I_ n = I^ n for some finitely generated ideal I \subset A. Then it is clear that X_{/T} is the formal spectrum of \mathop{\mathrm{lim}}\nolimits A/J^ n where J = (f_1, \ldots , f_ r) + I. Some details omitted.

Proof of (5). In this case X_{red} is the spectrum of a Noetherian ring and hence the assumption that |X_{red}| \setminus T is quasi-compact is satisfied. Thus as in the proof of (4) we see that X_{/T} is the spectrum of \mathop{\mathrm{lim}}\nolimits A/J^ n which is a Noetherian adic topological ring, see Algebra, Lemma 10.97.6. \square

Lemma 87.37.7. Let S be a scheme. Let X be a formal algebraic space over S. Let T \subset X_{red} be a closed subset and let X_{/T} be the formal completion of X along T. Then

  1. if X_{red} \setminus T \to X_{red} is quasi-compact and X is locally countably indexed, then X_{/T} is locally countably indexed,

  2. if X_{red} \setminus T \to X_{red} is quasi-compact and X is locally adic*, then X_{/T} is locally adic*, and

  3. if X is locally Noetherian, then X_{/T} is locally Noetherian.

Proof. Choose a covering \{ X_ i \to X\} as in Definition 87.11.1. Let T_ i \subset X_{i, red} be the inverse image of T. We have X_ i \times _ X X_{/T} = (X_ i)_{/T_ i} (Lemma 87.37.4). Hence \{ (X_ i)_{/T_ i} \to X_{/T}\} is a covering as in Definition 87.11.1. Moreover, if X_{red} \setminus T \to X_{red} is quasi-compact, so is X_{i, red} \setminus T_ i \to X_{i, red} and if X is locally countably indexed, or locally adic*, pr locally Noetherian, the is X_ i is countably index, or adic*, or Noetherian. Thus the lemma follows from the affine case which is Lemma 87.37.6. \square


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