LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)

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Surface finite completion factors

Statement

Assume AC. For a finite map $R \to S$ of Noetherian rings and $p \in Spec R$ , $\hat{R_{p}} \otimes_{R} S ≅ \prod_{q \cap R = p} \hat{S_{q}}$ . The finitely many factors use their maximal-adic completions. Thus formal fibres for finite extensions are factors of residue-field base changes of the original formal fibres.

Facts & Assumptions

Given: A finite map $R \to S$ of Noetherian rings and a prime $p \in Spec R$ .

[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]

def-dependent-choice. Let $X$ be a set and let $R \subseteq X \times X$ be a binary relation on $X$ . Call $R$ entire on $X$ when $for every x \in X there is y \in X with x R y .$ The Axiom of Dependent Choice, written $D C$ , is the following statement. (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain)

[F3]

thm-completion-is-exact-on-finite-modules. Assume the Axiom of Choice. Let $R$ be a Noetherian commutative ring, let $I \subseteq R$ be an ideal, and let $0 \to M^{'} \to M \to M^{''} \to 0$ be a short exact sequence of finitely generated $R$ -modules. Then the induced sequence of $I$ -adic completions $0 \to \hat{M^{'}} \to \hat{M} \to \hat{M^{''}} \to 0$ is exact. (Adic completion is exact on finite modules over a Noetherian ring)

[F4]

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)

Proof

1.1F3given

For each $n \geq 1$ the ring $S_{p} / p^{n} S_{p}$ is a finite module over the Artinian local ring $R_{p} / p^{n} R_{p}$ , hence Artinian; its finitely many maximal ideals correspond to the primes $q$ of $S$ with $q \cap R = p$ , and the Chinese remainder decomposition of an Artinian ring gives $S_{p} / p^{n} S_{p} ≅ \prod_{q} S_{q} / p^{n} S_{q}$ .

2.1F4step 1.1

In each factor the radical of $p S_{q}$ is $q S_{q}$ , so the $p$ -adic and $q$ -adic topologies coincide and the inverse limit of the factors is $\prod_{q} \hat{S_{q}}$ , the product of the maximal-adic completions; the product is finite, so it commutes with the inverse limit.

3.1F3step 1.1step 2.1

The left-hand side of the inverse limit is the $p$ -adic completion of the finite $R_{p}$ -module $S_{p}$ , which by exactness of completion on finite modules is $\hat{R_{p}} \otimes_{R_{p}} S_{p} = \hat{R_{p}} \otimes_{R} S$ .

4.1F3step 2.1step 3.1

Combining the two computations of the inverse limit gives the asserted isomorphism $\hat{R_{p}} \otimes_{R} S ≅ \prod_{q \cap R = p} \hat{S_{q}}$ , with finitely many factors.

5.1F1F2step 4.1∎

Tensoring the isomorphism with the residue field identifies the formal fibre of the finite extension at $q$ as a factor of the base change of the formal fibre of $R$ at $p$ ; the Axiom of Choice and the Axiom of Dependent Choice are inherited from the completion suppliers.

Remarks

The decomposition is the Chinese remainder decomposition of an Artinian quotient, not a general formal-gluing statement.
Finiteness of $R \to S$ is used to make the quotients finite and the number of primes over $p$ finite.

Depends on

Used by

Dependency tree · two levels

18 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: full proof imports for normal-surface resolution, lemma-completion-finite-extension (standard reference, not scraped)