4
$\begingroup$

Consider the following diagram of regular local rings

$\begin{matrix} \hat{A} & \xrightarrow{\quad\hat\varphi\quad} & \hat{B} \\ \ \uparrow\scriptstyle\alpha & \circlearrowleft & \ \uparrow\scriptstyle\beta \\ A & \xrightarrow{\quad\varphi\quad} & B \end{matrix}$

where $\widehat{\\,\dot\\,}$ denotes the completion functor. Let $m\subset A$ and $n\subset B$ be the respective maximal ideals. Assume that $\varphi$ is injective and makes $B$ integral over $A$, in particular all morphisms are inclusions.

Given $\hat y\in\hat n\setminus \hat n^2$ such that $\hat y^k=x \in m\setminus m^2$, can I find $y\in B$ such that $y^k = x$?

$\endgroup$
1
  • 1
    $\begingroup$ The question is very pleasantly presented (mathematically interesting, clear and precise, elegant notations, nice and sober diagram, correct spelling,...) +1 $\endgroup$ – Georges Elencwajg Sep 1 '11 at 11:57
6
$\begingroup$

Non. Let $A$ be the localization of $\mathbb C[x]$ at the maximal ideal $x\mathbb C[x]$ and let $B=A[z]/(z^2-x(1+x))$. Then $1+x$ is the square of a unit in $\hat{A}$. Hence $\hat{B}=\hat{A}[y]/(y^2-x)$. But $x$ is not a square in $B$ because otherwise $1+x$ would be a square in $B$, but one can check directly that this is not the case.

$\endgroup$
3
  • 1
    $\begingroup$ The answer is more likely to be yes if $A$ and $B$ (maybe just $B$) are henselian and excellent, by some version of Artin/Popescu/Spivakovsky approximaion. $\endgroup$ – Laurent Moret-Bailly Sep 1 '11 at 11:32
  • $\begingroup$ @Qing: Oui! Nice. $\endgroup$ – Georges Elencwajg Sep 1 '11 at 12:03
  • 1
    $\begingroup$ Nice example! Thanks a bunch. @Laurent: I don't think I can get $A$ or $B$ to be Henselian, and I suppose that rules out this way. Thanks anyway! $\endgroup$ – Jesko Hüttenhain Sep 1 '11 at 12:29

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.