Lemma 10.97.3 (00MC)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.97: Completion for Noetherian rings
Lemma 10.97.3 (cite)

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Lemma 10.97.3. Let $I$ be an ideal of a Noetherian ring $R$ . Denote $R^\wedge$ the completion of $R$ with respect to $I$ . If $I$ is contained in the Jacobson radical of $R$ , then the ring map $R \to R^\wedge$ is faithfully flat. In particular, if $(R, \mathfrak m)$ is a Noetherian local ring, then the completion $\mathop{\mathrm{lim}}\nolimits _ n R/\mathfrak m^ n$ is faithfully flat.

Proof. By Lemma 10.97.2 it is flat. The composition $R \to R^\wedge \to R/I$ where the last map is the projection map $R^\wedge \to R/I$ shows that any maximal ideal of $R$ is in the image of $\mathop{\mathrm{Spec}}(R^\wedge ) \to \mathop{\mathrm{Spec}}(R)$ . Hence the map is faithfully flat by Lemma 10.39.15. $\square$

Comments (3)

Comment #8360 by Rankeya on May 01, 2023 at 16:02

Tag 00MC doesn't need $(R,\mathfrak m)$ to be local. The result holds as long as $I$ is contained in the Jacobson radical of $R$ . Perhaps it could be stated this way since the proof does not require any serious modifications?

Comment #8965 by Stacks project on April 13, 2024 at 13:38

Much better. Thanks! Changes are here.

Comment #9476 by Elías Guisado on June 26, 2024 at 03:06

Alternative suggested proof: By Lemma 10.97.2 it is flat. Thus, if $M$ is some non-zero $R$ -module, to see that $M\otimes_R\hat{R}\neq 0$ it suffices to see that $N\otimes_R\hat{R}\neq 0$ , for finitely generated submodules $N\subset M$ . By Lemma 10.97.1, $N\otimes_R\hat{R}\cong\hat{N}$ . Since $\hat{N}=\varprojlim_n N/I^nN$ , the map $N\to \hat{N}$ is injective by Krull's intersection theorem (Lemma 10.51.4), so $N\neq 0$ implies $\hat{N}\neq 0$ .

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