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Let $ f:A\to B$ be an injective, local homorphism between two Noetherian local rings. Consider the completions $\hat A$ and $\hat B$ with respect the maximal ideals. We have an induced homomorphism $\hat f: \hat A \to\hat B$. What assumptions do we need in order to ensure that also $\hat f$ is injective?

My main interest is geometric, so for example the local map induced by a surjective morphism between Noetherian schemes.

Many thanks in advance

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    $\begingroup$ Trivia: if $f$ is an homeomorphism onto its image (e.g., $A$ and $B$ are DVR), so is $\hat{f}$. $\endgroup$
    – Luc Guyot
    Jun 2, 2018 at 15:01
  • $\begingroup$ Uhm, yes I'm interested in the DVR case. But why? $\endgroup$
    – manifold
    Jun 2, 2018 at 15:21
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    $\begingroup$ Assume that $A$ and $B$ are DVRs with uniformizing elements $\pi_A \in A$ and $\pi_B \in B$. Write $f(\pi_A) = u \pi_B^{\nu}$ with $u$ a unit of $B$ and $\nu$ a positive integer. Then $f(A) \cap \pi_B^{n \nu} B \subseteq f(\pi_A^{n}A)$ for every $n \ge 0$, hence $f$ is open. $\endgroup$
    – Luc Guyot
    Jun 2, 2018 at 18:46
  • $\begingroup$ The natural condition is that the topology on $A$ induced by powers of its maximal ideal and the topology induced by the powers of the maximal ideal of $B$ intersected with $A$ are the same. $\endgroup$
    – Mohan
    Jun 3, 2018 at 20:41
  • $\begingroup$ Trivia continued: If $\hat{f}$ is injective and if $\hat{A}$ is compact (i.e, the residue field of $A$ is finite), then $f$ is an homeomorphism onto its image. So the "natural condition" is necessary under the compactness assumption. $\endgroup$
    – Luc Guyot
    Jun 4, 2018 at 20:19

1 Answer 1

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This may be useful:

Proposition (Zariski) (see EGA I, (3.9.8) in Springer edition)

Let $f: (A,\mathfrak{m})\to (B,\mathfrak{n})$ be a local homomorphism of noetherian local rings. Assume that:

  • $f$ is injective.
  • $\hat{A}$ is a domain.
  • $f$ is essentially of finite type.

Then the $\mathfrak{m}$-adic topology on $A$ is induced by the $\mathfrak{n}$-adic topology on $B$.

Of course this implies that $\hat{f}$ is injective. If we don't assume that $\hat{A}$ is a domain (but $A$ is) it is easy to construct counterexamples.

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  • $\begingroup$ Wait, but what about the following example: $A=\mathbb Z[t]_{(p,t)}$ and $B=\mathbb Z[t]_{(t)}$. Clearly $A\subset B$, but $\hat A= \mathbb Z_p[[t]]$ and $\hat B=\mathbb Q[[t]]$ $\endgroup$
    – manifold
    Jul 18, 2018 at 19:54
  • $\begingroup$ Moreover I'm looking at EGA I (chapter 0 of Springer edition), and I cannot find any 3.9.8. $\endgroup$
    – manifold
    Jul 18, 2018 at 20:04
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    $\begingroup$ @manifold: your example is not a local homomorphism (the map $A_{\mathfrak p} \to A_{\mathfrak q}$ for $\mathfrak p \supsetneq \mathfrak q$ never is). $\endgroup$
    – R. van Dobben de Bruyn
    Jul 18, 2018 at 21:41
  • $\begingroup$ @manifold: it's 3.9.8 of chapter 1 (not chapter 0), on page 255. $\endgroup$
    – Laurent Moret-Bailly
    Jul 19, 2018 at 7:25

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