# Roots of Units in Complete $\mathbb{C}$-Algebras

$\newcommand{\cc}{\mathbb C}$ Let $R$ be a finitely generated $\cc$-algebra and ${\frak m}\subset R$ a maximal ideal. Denote by $\hat R$ the completion of $R$ with respect to $\frak m$. Assume that $x\in R^\times$ is a unit and let $\hat x$ denote the image of $x$ under $R\to\hat R$. I am wondering if $\hat x$ has an $n$-th root in $\hat R$, or in other words:

Is there an element $\xi\in\hat{R}$ such that $\xi^n = \hat x$?

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## 1 Answer

My commutative algebra is a little rusty outside of number fields, but I think (hope) this works.

First, $\hat R/\hat{\mathfrak m} = R/\mathfrak m$ is a finitely generated field extension of $\mathbf C$. By Zariski's lemma (the first corollary on pg 49 of Hochster's notes) it is even equal to $\mathbf C$. So there exists a unit $y \in \hat R \setminus \hat{\mathfrak m}$ which becomes a zero of $f(X) = X^n - \hat x$ in the residue field. Since $f'(y) = ny^{n - 1}$ is a unit in $\hat{R}$, we can apply Hensel's lemma.

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Exactly what I was hoping for. Thanks heaps! –  Jesko Hüttenhain Sep 3 '11 at 17:49
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