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 $y^k = x$?



Pulling back roots from the Completion
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.


