# Pulling back roots from the Completion

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 $y^k = x$?

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

## 1 Answer

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.

• 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. – Laurent Moret-Bailly Sep 1 '11 at 11:32
• @Qing: Oui! Nice. – Georges Elencwajg Sep 1 '11 at 12:03
• 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! – Jesko Hüttenhain Sep 1 '11 at 12:29