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Let $g: A \to B$ be a local ring morphism between local Noetherian (commutative) rings $A,B$ (so $g(m_A) \subset m_B$ for the unique maximal ideals of the corresponding rings). Assume that the induced map of completions of the two local rings $\widehat{g}: \widehat{A} \to \widehat{B} $ is an isomorphism.

What do I then know about the previous morphism $g: A \to B$ and the correspondence of the maximal ideals $m_A$ and $m_B$? So essentially which information can we derive/ "pull back" about $g,A$ and $B$ from knowing that $\widehat{g}$ is an isomorphism?

I know that we can deduce following informations instantly:

  1. the canonical maps $A/m_A^n \cong \widehat{A}/\widehat{m_A}^n \to \widehat{B}/\widehat{m_B}^n \cong B/m_B^n $ are surjective for all $n \in \mathbb{N}_{\ge 1}$

  2. $\widehat{m_B}=\widehat{m_A}\widehat{B}$

  3. for every $k$ there exist a $d_k$ with $ g(m_A)^{d_k} \subset m_B ^k$ and vice versa (so same topology)

The question is what do we know about $g:A \to B$ (injective, surjective? ... by a bunch of counterexamples we know that $g$ is almost never an isomorphism, but what "can nevertheless be saved"?) and how are related the maximal ideals $m_A$ and $m_B$?

When we can expect $g(m_A)B=m_B$?

What do we still know about $g$ if we neglect the Noetherian condition?

Remark: This question arises from following former MathSE question of mine and intends to generalize how the "tool box" of completions can be progressively applied to deduce some useful informations/relations about the initial morphism of local rings.

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    $\begingroup$ Since we know that the map $A \to \hat{A}$ is injective, it is easy to conclude that $A\to B$ is injective as well. Moreover, one can show that $A\to B$ is a flat morphism using the local flatness criterion. As $A\to B$ is local, it implies that it is faithfully flat. Also, we know that $\mathfrak m_A \hat{A}=\mathfrak m_{\hat{A}}$ and $\mathfrak m_{\hat{A}}\hat{B}=\mathfrak m_{\hat{B}}$. This implies that $(\mathfrak m_A B)\hat{B}=\mathfrak m_{\hat{B}}$. Since $B\to \hat{B}$ is f.flat we see that $(\mathfrak m_A B)=((\mathfrak m_A B)\hat{B})\cap B=\mathfrak m_{\hat{B}}\cap B=\mathfrak m_B$. $\endgroup$
    – gdb
    Commented Aug 6, 2019 at 23:23
  • $\begingroup$ @gdb: And $A \to \hat{A}$ injective follows by Artin-Rees lemma (so Noetherian is here essential)? $\endgroup$
    – user267839
    Commented Aug 7, 2019 at 11:51
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    $\begingroup$ The kernel of the map $A\to \hat{A}$ is equal to the intersection $\cap_{n=1}^{\infty}\mathfrak m_A^n$. So the map is injective iff $A$ is separated in the $\mathfrak m_A$-adic topology. This is automatic in noetherian setting but false in general. For example, consider $A=\mathcal O_{\mathbf C_p}$ with its maximal ideal $\mathfrak m$, it is easy to see that $\mathfrak m^2=\mathfrak m$. It implies that $\cap_{n=1}^{\infty}\mathfrak m^n=\mathfrak m\neq (0)$. (Note that this shows that $\mathcal O_{\mathbf C_p}$ is complete in the $p$-adic topology, but not in the $\mathfrak m$-adic topology) $\endgroup$
    – gdb
    Commented Aug 8, 2019 at 0:58

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