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Completion base change preserves completed local rings on the closed fibre

Statement

Assume AC. Let $(A, m)$ be a Noetherian local ring, $\hat{A}$ its maximal-adic completion, and $X$ a scheme locally of finite type over $A$ . Put $Y = X \times_{Spec A} Spec \hat{A}$ . The closed fibres of $X$ and $Y$ are canonically isomorphic. If $y \in Y$ lies on this fibre and $x$ is its corresponding point, then the local homomorphism $O_{X, x} \to O_{Y, y}$ induces an isomorphism of maximal-adic completions.

Facts & Assumptions

Given: A Noetherian local ring $(A, m)$ , its completion $\hat{A}$ , a scheme $X$ locally of finite type over $A$ , the base change $Y = X \times_{Spec A} Spec \hat{A}$ , and corresponding points $x \in X$ , $y \in Y$ on the closed fibre.

[F1]

def-axiom-of-choice. The Axiom of Choice (AC) is the following statement. > Every family of nonempty sets has a choice function > (def-choice-function). Written out: for every set $F$ all of whose members are nonempty, there exists a function $g$ with domain $F$ satisfying $g (S) \in S$ for all $S \in F$ . (The Axiom of Choice)

[F2]

thm-completion-of-a-noetherian-local-ring. Assume the Axiom of Choice. Let $(R, m)$ be a Noetherian local ring, and let $\hat{R}$ be its $m$ -adic completion. 1. $\hat{R}$ is a Noetherian local ring with maximal ideal $m \hat{R}$ . 2. The residue field is unchanged: $\hat{R} / m \hat{R} ≅ R / m .$ 3. The completion map $R \to \hat{R}$ is faithfully flat. (Completion of a Noetherian local ring is local with the same residue field)

[F3]

cor-completion-commutes-with-finite-quotients-and-submodules. Assume the Axiom of Choice. Let $R$ be a Noetherian commutative ring, let $I \subseteq R$ be an ideal, and let $N \subseteq M$ be finitely generated $R$ -modules. 1. The natural map $\hat{M} / \hat{N} ⟶ \hat{M / N}$ is an isomorphism. 2. (Completion commutes with finite quotients and induced submodules)

[F4]

thm-affine-fibre-product-tensor-ring. Let $A \to B$ and $A \to C$ be maps of commutative unital rings, allowing the zero ring. In the category of all schemes, $Spec B \times_{Spec A} Spec C ≅ Spec (B \otimes_{A} C) .$ The projections correspond to $b \mapsto b \otimes 1$ and $c \mapsto 1 \otimes c$ . (Affine fibre products are spectra of tensor products)

Proof

1.1F4given

The assertion is local on $X$ , so write $X = Spec B$ ; then $Y = Spec C$ with $C = B \otimes_{A} \hat{A}$ , because the fibre product of affine schemes is the spectrum of the tensor product.

2.1F3step 1.1

The completion of quotients identifies $C / m^{n} C = B \otimes_{A} (\hat{A} / m^{n} \hat{A})$ with $B / m^{n} B$ for every $n \geq 1$ ; for $n = 1$ this is the canonical identification of the closed fibres of $X$ and $Y$ .

3.1F3F4step 2.1

Let $p \subset B$ and $q \subset C$ correspond under the closed-fibre identification, so that $m B \subseteq p$ , $m C \subseteq q$ and the two primes have the same image in the common quotient $B / m B = C / m C$ . Since $m^{n} B \subseteq p^{n}$ and $m^{n} C \subseteq q^{n}$ , the isomorphism of step 2.1 passes to the quotients and gives compatible isomorphisms $B / p^{n} ≅ C / q^{n}$ for every $n$ .

4.1F2step 3.1

Localizing the compatible isomorphisms of step 3.1 at the corresponding primes gives compatible isomorphisms $B_{p} / p^{n} B_{p} ≅ C_{q} / q^{n} C_{q}$ ; passing to inverse limits produces an isomorphism of the maximal-adic completions $\hat{B_{p}} ≅ \hat{C_{q}}$ , which are the completed local rings of $X$ at $x$ and of $Y$ at $y$ , and the map is induced by the local homomorphism $O_{X, x} \to O_{Y, y}$ .

5.1F1step 4.1∎

The Axiom of Choice is inherited from the completion suppliers; no further choice enters, and the identification is canonical once the points are matched.

Remarks

The proof is the local computation behind the statement that completion of a Noetherian local ring commutes with base change for schemes locally of finite type.
The corresponding-points hypothesis is exactly the identification of the closed fibres in step 1.2; no separability or finiteness over the residue field is used.

Depends on

Used by

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Resolution of Surfaces, Lemmas 54.11.1 and 54.11.2 (complete proof read) (standard reference, not scraped)