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Let $(R,m)$ be a local Noetherian ring. Denote by $\hat R$ the $m$-adic completion of $R$. Is the following statement correct?

If $\hat R$ is a finitely generated $R$-module, then $R=\hat R$.

Alex
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  • Yes, as $\hat{R}$ is $R$-flat module and $\hat{E} = \hat{R}\otimes_R E$ where $\hat{E}$ is $m$-adic completion of finitely generated module $E.$ Consider exact $R$-sequence $R\to\hat{R}\to \hat{R}/R.$ – dsh Dec 13 '23 at 01:41
  • @dsh What does that have to do with finite generation of $\hat{R}$? All the statements in your comment hold for any Noetherian ring $R$ – math54321 Dec 14 '23 at 19:29
  • @math5432,If $\hat{R}$ is not finitely generated, then $\hat{R} \otimes \hat{R} = \hat{R}$ does not follow. – dsh Dec 15 '23 at 01:56
  • $\hat{R}$ is faithfully $R$-flat module in this case. – dsh Dec 15 '23 at 02:10
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    @dsh I see your point - the isomorphism $\hat{E} \cong \hat{R} \otimes_R E$ only holds when $E$ is finitely generated. Now a finitely generated flat module over a Noetherian local ring is actually free, so $\hat{R} \otimes \hat{R} \cong \hat{\hat{R}} \cong \hat{R} \implies \hat{R} \cong R$ – math54321 Dec 15 '23 at 17:10
  • @math54321. Reasoning using finitely generated modules over local rings seems easier. From faithfully flat $\hat{R}$ one can derive $\hat{R}\otimes_R (\hat{R}/R) = 0$ and $\hat{R}/R = 0$ – dsh Dec 16 '23 at 01:23
  • @dsh perhaps, you can combine your comments into an answer. I'll gladly accept it. – Alex Dec 17 '23 at 15:18

1 Answers1

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Reference: Bourbaki, "Commutative algebra, ch II, III."

  1. $\hat{R}$ is faithfully flat $R$-module.
  2. $\hat{E} = \hat{R}\otimes_R E$ for finitely generated $R$-module $E.$
  3. From exact sequence of finitely generated $R$-modules $$ R\to\hat{R}\to\hat{R}/R $$ one derives exact sequence ($\hat{R}$ is flat) $$ \hat{R}\to\hat{\hat{R}}=\hat{R}\to\hat{R}/R \otimes_R \hat{R} $$
  4. $\hat{R}/R \otimes_R \hat{R} = 0,$ because first arrow maps 1 to 1.
  5. $\hat{R}$ is $R$-faithful, then $\hat{R}/R = 0.$
dsh
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