The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.31.8. Let $R$ be a ring. Let $I \subset R$ be a locally nilpotent ideal. Let $n \geq 1$ be an integer which is invertible in $R/I$. Then

  1. the $n$th power map $1 + I \to 1 + I$, $1 + x \mapsto (1 + x)^ n$ is a bijection,

  2. a unit of $R$ is a $n$th power if and only if its image in $R/I$ is an $n$th power.

Proof. Let $a \in R$ be a unit whose image in $R/I$ is the same as the image of $b^ n$ with $b \in R$. Then $b$ is a unit (Lemma 10.31.4) and $ab^{-n} = 1 + x$ for some $x \in I$. Hence $ab^{-n} = c^ n$ by part (1). Thus (2) follows from (1).

Proof of (1). This is true because there is an inverse to the map $1 + x \mapsto (1 + x)^ n$. Namely, we can consider the map which sends $1 + x$ to

\begin{align*} (1 + x)^{1/n} & = 1 + {1/n \choose 1}x + {1/n \choose 2}x^2 + {1/n \choose 3}x^3 + \ldots \\ & = 1 + \frac{1}{n} x + \frac{1 - n}{2n^2}x^2 + \frac{(1 - n)(1 - 2n)}{6n^3}x^3 + \ldots \end{align*}

as in elementary calculus. This makes sense because the series is finite as $x^ k = 0$ for all $k \gg 0$ and each coefficient ${1/n \choose k} \in \mathbf{Z}[1/n]$ (details omitted; observe that $n$ is invertible in $R$ by Lemma 10.31.4). $\square$


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