# Induced morphism of completions of local rings

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.

• 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 $(\mathfrak m_A B)=((\mathfrak m_A B)\hat{B})\cap B=\mathfrak m_{\hat{B}}\cap B=\mathfrak m_B$.
– gdb
Aug 6 '19 at 23:23
• @gdb: And $A \to \hat{A}$ injective follows by Artin-Rees lemma (so Noetherian is here essential)? Aug 7 '19 at 11:51
• 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)
– gdb
Aug 8 '19 at 0:58