Let $(A,m)$ be a commutative noetherian local ring such that $m$ is principal, say $m=(t)$. Let $(\hat A,\hat m)$ be its $m$-adic completion. Let $A\subset B\subset\hat A$ be any intermediate subring such that $n=tB$ is a maximal ideal of $B$.

The question is: Is it true that the localisation $B_n$ is contained in $\hat A$?

Does this follows simply by considering the isomorphims $\hat A\simeq A[[X]]/(X-t)\simeq A[[t]]$? This would imply that an element $g\in B\setminus n$ is a unit in $\hat A$. Am I right?

Thanks in advance!

  • $\begingroup$ I don't think that $\hat A$ is isomorphic to $A[[t]]$, what if $A$ is $m$-complete (e.g. $t$ nilpotent)? $\endgroup$ – Tommaso Centeleghe Mar 5 '12 at 12:54
  • $\begingroup$ I admit that I wasn't too sure of my "proof" precisely because of that "isomorphism". Anyway, I think Ralph's answer is conclusive. $\endgroup$ – user20544 Mar 5 '12 at 14:15

I think the following more general is true, that answers your question affirmatively:

Let $A$ be a comm. local ring, $B \le A$ a subring and let $n \trianglelefteq B$ be a maximal ideal such that $An$ is the maximal ideal of $A$. Then each element in $B \setminus n$ is a unit in $A$. In particular, $B_n \le A$.

Proof: First note that $B \cap An \supseteq n$ is an ideal in $B$ and since $n$ is maximal, $n = B \cap An\;$ (this step doesn't use that $A$ is local).

Now let $x \in B \setminus n$ and assume $x$ is not a unit in $A$. Since $A$ is local, the unit group is $A \setminus An$. Thus, $ x \in An$, i.e. $x \in B \cap An = n$, contradicting $x \notin n$. q.e.d.

In your example we have $n = Bt$ and $\hat{m} = \hat{A}t = \hat{A}n$. So the above applies.


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