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Maybe for some straightforward, but not for me:

Let $(R,\mathfrak{m})$ be a commutative Noetherian local ring with maximal ideal $\mathfrak{m}$. Why is the completion $\widehat{R}$ of $R$ with respect to the maximal ideal $\mathfrak{m}$ again a Noetherian ring?

Thanks.

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  • $\begingroup$ $\def\fra{\mathfrak{a}}$More generally, if $A$ is any Noetherian ring and $\fra\subset A$ is some ideal, then the $\fra$-adic completion of a finitely-generated $A$-module $M$ is a Noetherian $\hat{A}$-module. Note that it suffices to show it for $M=A$, since $\fra$-adic completion is exact on finitely generated $A$-modules (thus, an exact sequence $A^{\oplus n}\to M\to 0$ maps to an exact sequence $\hat{A}^{\oplus n}\to \hat{M}\to 0$), see Atiyah, Macdonald, Introduction to Commutative Algebra, Proposition 10.12 (the category of finitely-generated modules over a Noetherian ring is abelian). $\endgroup$
    – Elías Guisado Villalgordo
    Oct 5 at 8:39

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I assume you are talking about the completion of a local Noetherian ring $A$ with respect to the topology induced by its maximal ideal $m$. Then $\hat{A}$ is again Noetherian local ring with maximal ideal $m \hat{A}$. Reference: Matsumura's Commutative Ring Theory p. 63.

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    $\begingroup$ Thanks. I found it in Theorem 8.12. $\endgroup$
    – Paul Kuhn
    May 28, 2013 at 5:06
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Yes; if $A$ is a Noetherian ring, and $\mathfrak a$ is an ideal of $A$, then the $\mathfrak a$-adic completion of $A$ is Noetherian. This is Theorem 10.26 in Atiyah-MacDonald. (I would write the proof here for you, but it is rather involved; hopefully, this standard reference will suffice!)

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  • $\begingroup$ $\def\fra{\mathfrak{a}}$It's interesting to note that all the preparation AM do to prove Theorem 10.26 actually shows that the $\fra$-adic completion of a finitely-generated $A$-module is a Noetherian $\hat{A}$-module. I wonder why do they reaped it only for the very simple case of $M=A$. $\endgroup$
    – Elías Guisado Villalgordo
    Oct 5 at 8:33

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