The Stacks project

85.9 Completion along a closed subset

Our notion of a formal algebraic space is well adapted to taking the completion along a closed subset.

Lemma 85.9.1. Let $S$ be a scheme. Let $X$ be an affine scheme over $S$. Let $T \subset |X|$ be a closed subset. Then the functor

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

is a McQuillan affine formal algebraic space.

Proof. Say $X = \mathop{\mathrm{Spec}}(A)$ and $T$ corresponds to the radical ideal $I \subset A$. Let $U = \mathop{\mathrm{Spec}}(B)$ be an affine scheme over $S$ and let $f : U \to X$ be an element of $F(U)$. Then $f$ corresponds to a ring map $\varphi : A \to B$ such that every prime of $B$ contains $\varphi (I) B$. Thus every element of $\varphi (I)$ is nilpotent in $B$, see Algebra, Lemma 10.16.2. Setting $J = \mathop{\mathrm{Ker}}(\varphi )$ we conclude that $I/J$ is a locally nilpotent ideal in $A/J$. Equivalently, $V(J) = V(I) = T$. In other words, the functor of the lemma equals $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/J)$ where the colimit is over the collection of ideals $J$ with $V(J) = T$. Thus our functor is an affine formal algebraic space. It is McQuillan (Definition 85.5.7) because the maps $A \to A/J$ are surjective and hence $A^\wedge = \mathop{\mathrm{lim}}\nolimits A/J \to A/J$ is surjective, see Lemma 85.5.6. $\square$

Lemma 85.9.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Then the functor

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

is a formal algebraic space.

Proof. Denote $F$ the functor. Let $\{ U_ i \to U\} $ be an fppf covering. Then $\coprod |U_ i| \to |U|$ is surjective. Since $X$ is an fppf sheaf, it follows that $F$ is an fppf sheaf.

Let $\{ g_ i : X_ i \to X\} $ be an ├ętale covering such that $X_ i$ is affine for all $i$, see Properties of Spaces, Lemma 64.6.1. The morphisms $F \times _ X X_ i \to F$ are ├ętale (see Spaces, Lemma 63.5.5) and the map $\coprod F \times _ X X_ i \to F$ is a surjection of sheaves. Thus it suffices to prove that $F \times _ X X_ i$ is an affine formal algebraic space. A $U$-valued point of $F \times _ X X_ i$ is a morphism $U \to X_ i$ whose image is contained in the closed subset $g_ i^{-1}(T) \subset |X_ i|$. Thus this follows from Lemma 85.9.1. $\square$

Definition 85.9.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a closed subset. The formal algebraic space of Lemma 85.9.2 is called the completion of $X$ along $T$.

In [Chapter I, Section 10.8, EGA] the notation $X_{/T}$ is used to denote the completion and we will occasionally use this notation as well. Let $f : X \to X'$ be a morphism of algebraic spaces over a scheme $S$. Suppose that $T \subset |X|$ and $T' \subset |X'|$ are closed subsets such that $|f|(T) \subset T'$. Then it is clear that $f$ defines a morphism of formal algebraic spaces

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

between the completions.

Lemma 85.9.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $T \subset |Y|$ be a closed subset and let $T' = |f|^{-1}(T) \subset |X|$. Then $X_{/T'} \to Y_{/T}$ is representable by algebraic spaces.

Proof. Namely, suppose that $V \to Y$ is a morphism from a scheme into $Y$ such that $|V|$ maps into $T$. Then $V \times _ Y X \to X$ is a morphism of algebraic spaces such that $|V \times _ Y X|$ maps into $T'$. Hence the functor $V \times _{Y_{/T'}} X_{/T}$ is represented by $V \times _ Y X$ and we see that the lemma holds. $\square$

The following lemma is due to Ofer Gabber.


Lemma 85.9.5. Let $S$ be a scheme. Let $X = \mathop{\mathrm{Spec}}(A)$ be an affine scheme over $S$. Let $T \subset X$ be a closed subscheme.

  1. If the formal completion $X_{/T}$ is countably indexed and there exist countably many $f_1, f_2, f_3, \ldots \in A$ such that $T = V(f_1, f_2, f_3, \ldots )$, then $X_{/T}$ is adic*.

  2. The conclusion of (1) is wrong if we omit the assumption that $T$ can be cut out by countably many functions in $X$.

Proof. The assumption that $X_{/T}$ is countably indexed means that there exists a sequence of ideals

\[ A \supset J_1 \supset J_2 \supset J_3 \supset \ldots \]

with $V(J_ n) = T$ such that every ideal $J \subset A$ with $V(J) = T$ there exists an $n$ such that $J \supset J_ n$.

To construct an example for (2) let $\omega _1$ be the first uncountable ordinal. Let $k$ be a field and let $A$ be the $k$-algebra generated by $x_\alpha $, $\alpha \in \omega _1$ and $y_{\alpha \beta }$ with $\alpha \in \beta \in \omega _1$ subject to the relations $x_\alpha = y_{\alpha \beta } x_\beta $. Let $T = V(x_\alpha )$. Let $J_ n = (x_\alpha ^ n)$. If $J \subset A$ is an ideal such that $V(J) = T$, then $x_\alpha ^{n_\alpha } \in J$ for some $n_\alpha \geq 1$. One of the sets $\{ \alpha \mid n_\alpha = n\} $ must be unbounded in $\omega _1$. Then the relations imply that $J_ n \subset J$.

To see that (2) holds it now suffices to show that $A^\wedge = \mathop{\mathrm{lim}}\nolimits A/J_ n$ is not a ring complete with respect to a finitely generated ideal. For $\gamma \in \omega _1$ let $A_\gamma $ be the quotient of $A$ by the ideal generated by $x_\alpha $, $\alpha \in \gamma $ and $y_{\alpha \beta }$, $\alpha \in \gamma $. As $A/J_1$ is reduced, every topologically nilpotent element $f$ of $\mathop{\mathrm{lim}}\nolimits A/J_ n$ is in $J_1^\wedge = \mathop{\mathrm{lim}}\nolimits J_1/J_ n$. This means $f$ is an infinite series involving only a countable number of generators. Hence $f$ dies in $A_\gamma ^\wedge = \mathop{\mathrm{lim}}\nolimits A_\gamma /J_ nA_\gamma $ for some $\gamma $. Note that $A^\wedge \to A_\gamma ^\wedge $ is continuous and open by Lemma 85.4.5. If the topology on $A^\wedge $ was $I$-adic for some finitely generated ideal $I \subset A^\wedge $, then $I$ would go to zero in some $A_\gamma ^\wedge $. This would mean that $A_\gamma ^\wedge $ is discrete, which is not the case as there is a surjective continuous and open (by Lemma 85.4.5) map $A_\gamma ^\wedge \to k[[t]]$ given by $x_\alpha \mapsto t$, $y_{\alpha \beta } \mapsto 1$ for $\gamma = \alpha $ or $\gamma \in \alpha $.

Before we prove (1) we first prove the following: If $I \subset A^\wedge $ is a finitely generated ideal whose closure $\bar I$ is open, then $I = \bar I$. Since $V(J_ n^2) = T$ there exists an $m$ such that $J_ n^2 \supset J_ m$. Thus, we may assume that $J_ n^2 \supset J_{n + 1}$ for all $n$ by passing to a subsequence. Set $J_ n^\wedge = \mathop{\mathrm{lim}}\nolimits _{k \geq n} J_ n/J_ k \subset A^\wedge $. Since the closure $\bar I = \bigcap (I + J_ n^\wedge )$ (Lemma 85.4.2) is open we see that there exists an $m$ such that $I + J_ n^\wedge \supset J_ m^\wedge $ for all $n \geq m$. Fix such an $m$. We have

\[ J_{n - 1}^\wedge I + J_{n + 1}^\wedge \supset J_{n - 1}^\wedge (I + J_{n + 1}^\wedge ) \supset J_{n - 1}^\wedge J_ m^\wedge \]

for all $n \geq m + 1$. Namely, the first inclusion is trivial and the second was shown above. Because $J_{n - 1}J_ m \supset J_{n - 1}^2 \supset J_ n$ these inclusions show that the image of $J_ n$ in $A^\wedge $ is contained in the ideal $J_{n - 1}^\wedge I + J_{n + 1}^\wedge $. Because this ideal is open we conclude that

\[ J_{n - 1}^\wedge I + J_{n + 1}^\wedge \supset J_ n^\wedge . \]

Say $I = (g_1, \ldots , g_ t)$. Pick $f \in J_{m + 1}^\wedge $. Using the last displayed inclusion, valid for all $n \geq m + 1$, we can write by induction on $c \geq 0$

\[ f = \sum f_{i, c} g_ i \mod J_{m + 1+ c}^\wedge \]

with $f_{i, c} \in J_ m^\wedge $ and $f_{i, c} \equiv f_{i, c - 1} \bmod J_{m + c}^\wedge $. It follows that $IJ_ m^\wedge \supset J_{m + 1}^\wedge $. Combined with $I + J_{m + 1}^\wedge \supset J_ m^\wedge $ we conclude that $I$ is open.

Proof of (1). Assume $T = V(f_1, f_2, f_3, \ldots )$. Let $I_ m \subset A^\wedge $ be the ideal generated by $f_1, \ldots , f_ m$. We distinguish two cases.

Case I: For some $m$ the closure of $I_ m$ is open. Then $I_ m$ is open by the result of the previous paragraph. For any $n$ we have $(J_ n)^2 \supset J_{n+1}$ by design, so the closure of $(J_ n^\wedge )^2$ contains $J_{n+1}^\wedge $ and thus is open. Taking $n$ large, it follows that the closure of the product of any two open ideals in $A^\wedge $ is open. Let us prove $I_ m^ k$ is open for $k \ge 1$ by induction on $k$. The case $k = 1$ is our hypothesis on $m$ in Case I. For $k > 1$, suppose $I_ m^{k - 1}$ is open. Then $I_ m^ k = I_ m^{k - 1} \cdot I_ m$ is the product of two open ideals and hence has open closure. But then since $I_ m^ k$ is finitely generated it follows that $I_ m^ k$ is open by the previous paragraph (applied to $I = I_ m^ k$), so we can continue the induction on $k$. As each element of $I_ m$ is topologically nilpotent, we conclude that $I_ m$ is an ideal of definition which proves that $A^\wedge $ is adic with a finitely generated ideal of definition, i.e., $X_{/T}$ is adic*.

Case II. For all $m$ the closure $\bar I_ m$ of $I_ m$ is not open. Then the topology on $A^\wedge /\bar I_ m$ is not discrete. This means we can pick $\phi (m) \geq m$ such that

\[ \mathop{\mathrm{Im}}(J_{\phi (m)} \to A/(f_1, \ldots , f_ m)) \not= \mathop{\mathrm{Im}}(J_{\phi (m) + 1} \to A/(f_1, \ldots , f_ m)) \]

To see this we have used that $A^\wedge /(\bar I_ m + J_ n^\wedge ) = A/((f_1, \ldots , f_ m) + J_ n)$. Choose exponents $e_ i > 0$ such that $f_ i^{e_ i} \in J_{\phi (m) + 1}$ for $0 < m < i$. Let $J = (f_1^{e_1}, f_2^{e_2}, f_3^{e_3}, \ldots )$. Then $V(J) = T$. We claim that $J \not\supset J_ n$ for all $n$ which is a contradiction proving Case II does not occur. Namely, the image of $J$ in $A/(f_1, \ldots , f_ m)$ is contained in the image of $J_{\phi (m) + 1}$ which is properly contained in the image of $J_ m$. $\square$

Comments (4)

Comment #1909 by Antoine Chambert-Loir on

First sentence : "... along a closed subset."

Comment #1947 by Brian Conrad on

At the end of the proof of Case I, should be .

Here is something a bit more serious. To infer that via the containment , it seems one needs to know that is closed (so it contains the closure of the image of that in turn is equal to ). Of course, once this holds then it follows (from the arbitrariness of and of ) that a product of open ideals in is open (as is asserted and used in Case I of the proof of (1)). But why closed?

For the purpose of the preliminary discussion in the proof of (1) this isn't needed, since the argument given does show that the image of in is contained in the ideal that is manifestly open (hence closed), hence contains the closure of the image of .

So it seems something needs to be expressed either differently or a bit more clearly (unless I am overlooked something obvious).

Comment #1976 by on

Yes, this is a mistake in the proof. Looking back at the handwritten notes Gabber sent me, this mistake was entirely my fault. I have fixed it according to the discussion we had on email, see here. But really you should just wait till later today, when I will update the Stacks project and the new version of the proof will be here.

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