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

  • 1
    $\begingroup$ 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 '18 at 15:01
  • $\begingroup$ Uhm, yes I'm interested in the DVR case. But why? $\endgroup$ – manifold Jun 2 '18 at 15:21
  • 1
    $\begingroup$ 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 '18 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 '18 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 '18 at 20:19

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.

  • $\begingroup$ 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 '18 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 '18 at 20:04
  • 1
    $\begingroup$ @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 '18 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 '18 at 7:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.