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I read Matsumura's Commutative Ring Theory. I cannot understand the proof of Theorem 8.14 (2) $\Longrightarrow$ (3) on page 62. Here is the statement.

Theorem 8.14. Let $A$ be a Noetherian ring and $I$ an ideal. If we consider $A$ with the $I$-adic topology, the following conditions are equivalent:
(1) $I \subset \mathrm{rad}(A)$;
(2) every ideal of $A$ is closed set;
(3) the $I$-adic completion $\widehat{A}$ of $A$ is faithfully flat over $A$.

And here is (2) $\Longrightarrow$ (3) part of its proof

(2) $\Longrightarrow$ (3) Since $\widehat{A}$ is flat over $A$, we need only prove that $\mathfrak{m}\widehat{A} \neq \widehat{A}$ for every maximal ideal $\mathfrak{m}$ of $A$. By assumption, $\left\{ 0 \right\}$ is closed in $A$, so that we can assume that $A \subset \widehat{A}$, and by Theorem 11, $\mathfrak{m}\widehat{A}$ is the closure of $\mathfrak{m}$ in $\widehat{A}$. However, $\mathfrak{m}$ is closed in $A$, so that $\mathfrak{m}\widehat{A} \cap A = \mathfrak{m}$, and so $\mathfrak{m}\widehat{A} \neq \widehat{A}.$

Why can we conclude $\mathfrak{m}\widehat{A} \cap A = \mathfrak{m}$ because $\mathfrak{m}$ is closed in $A$?

Here is what I know or guess to solve this question.
In this case, we can say $\psi \colon A \rightarrow \widehat{A}$, the natural hom, is injective because $\ker \psi = \overline{\left\{ 0 \right\}} = \left\{ 0 \right\}$. And if we can say $\psi$ is homeomorphism to its image ($\ast$), $\psi(\mathfrak{m})$ is closed in $\widehat{A}$, so that $\mathfrak{m}\widehat{A} = \overline{\psi(\mathfrak{m})} = \psi(\mathfrak{m})$. Then, since $\psi$ is injective, $\psi^{-1}(\mathfrak{m}\widehat{A}) = \mathfrak{m}$. It is we wanted.
To say ($\ast$), it is suffice to show $\psi$ is an open mapping. However, I think, in general, we cannot say $\psi$ is an open mapping. For example, $\psi \colon \mathbb{Q} \rightarrow \mathbb{R}$ is not an open mapping. But it is not an ideal-adic topology…

And independently of this proof, I know we can prove (1) $\Longleftrightarrow$ (2) and (1) $\Longleftrightarrow$ (3) like the Excercise 10.7 in Atiyah-Macdonald Introduction to Commutative Algebra. However I want to understand Matsumura's proof.

Additional note: At first, I mistakenly wrote $\psi$ as $\phi$. So I corrected them.
And I may be able to prove $\psi$ is an open mapping to its image, $\psi(A)$, under this condition. The following is my proof about the openness of $\psi$.

It is suffice to show the openness for arbitrary neighborhoods of $0$, $I^n = I^nA$. By Theorem 8.13, the topology of $\widehat{A}$ is the topology whose neighborhoods of $0$ are $\left\{ I^n\widehat{A} \cap \psi(A) \right\}$. And by Artin-Rees’s Lemma, this topology on $\psi(A)$ is the same as the topology whose neighborhoods are $\left\{ I^n\psi(A) \right\}$. Thus we can check $\psi(I^nA) = I^n\psi(A)$ is an open set in $\psi(A)$.

Riley
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  • Have you tried working directly with the definition of a closed subset of the given topologies? – Martin Brandenburg Nov 23 '24 at 23:44
  • @MartinBrandenburg Thank you for your comment. Do you mean if I tried proving $\mathfrak{m}\widehat{A} \cap A = \mathfrak{m}$ by using just general topology? – Riley Nov 24 '24 at 08:56
  • What I mean is if in general $J \widehat{A} \cap A = J$ holds for every $I$-adically closed ideal $J$, just by taking an element in the left side etc. But I am just guessing. – Martin Brandenburg Nov 24 '24 at 08:58
  • @MartinBrandenburg I do not know if it holds in general. I want $J\widehat{A} \cap A = \overline{J}$ to hold. However, I have never seen this statement. At least, I think Matsumura does not say if it holds or not. – Riley Nov 24 '24 at 09:31
  • It seems what you want is the following: If $Y$ is a topological space, $A\subseteq X\subseteq Y$ subsets, then $\overline A^Y\cap X = \overline A^X$ (here the superscript denotes which space we are taking the closure in). This follows from the definition of closure – leoli1 Nov 24 '24 at 22:37
  • @leoli1 Thank you for your comment. In the proof, Matsumura says we can assume $A \subset \widehat{A}$. If this is true, I understood your method would go well. However, to assume $A \subset \widehat{A}$, I think we need to prove $\phi$ is an open mapping. – Riley Nov 25 '24 at 10:09
  • $\phi$ is not an open mapping in general. $\phi$ is injective because $0$ is closed (so that $\cap I^n=0$), so the question is why under $\phi$ the subspace topology of $\hat A$ on $A$ coincides with the $I$-adic topology on $A$ and this follows e.g. from the fact that both topologies are subspace topologies from $\prod_{n=1}^\infty A/I^n$. – leoli1 Nov 25 '24 at 22:48
  • I edited the question because I may be able to prove $\psi$ is an open mapping to its image. – Riley Nov 27 '24 at 10:37

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