## 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 étale 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 étale (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 $J_ m$. $\square$

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, $X$ should be $X_{/T}$.

Here is something a bit more serious. To infer that $J_{n-1}^{\wedge} J_m^{\wedge} \supset J_n^{\wedge}$ via the containment $J_{n-1}J_m \supset J_n$, it seems one needs to know that $J_{n-1}^{\wedge}J_m^{\wedge}$ is closed (so it contains the closure of the image of $J_n$ that in turn is equal to $J_n^{\wedge}$). Of course, once this holds then it follows (from the arbitrariness of $m$ and of $n \ge m+1$) that a product of open ideals in $A^{\wedge}$ is open (as is asserted and used in Case I of the proof of (1)). But why $J_{n-1}^{\wedge}J_m^{\wedge}$ 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 $J_n$ in $A^{\wedge}$ is contained in the ideal $J_{n-1}^{\wedge}I + J_{n+1}^{\wedge}$ that is manifestly open (hence closed), hence contains the closure $J_n^{\wedge}$ of the image of $J_n$.

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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