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 10.148.9. Let $(R, \mathfrak m, \kappa )$ be a complete local ring, see Definition 10.154.1. Then $R$ is henselian.

Proof. Let $f \in R[T]$ be monic. Denote $f_ n \in R/\mathfrak m^{n + 1}[T]$ the image. Denote $f'_ n$ the derivative of $f_ n$ with respect to $T$. Let $a_0 \in \kappa $ be a simple root of $f_0$. We lift this to a solution of $f$ over $R$ inductively as follows: Suppose given $a_ n \in R/\mathfrak m^{n + 1}$ such that $a_ n \bmod \mathfrak m = a_0$ and $f_ n(a_ n) = 0$. Pick any element $b \in R/\mathfrak m^{n + 2}$ such that $a_ n = b \bmod \mathfrak m^{n + 1}$. Then $f_{n + 1}(b) \in \mathfrak m^{n + 1}/\mathfrak m^{n + 2}$. Set

\[ a_{n + 1} = b - f_{n + 1}(b)/f'_{n + 1}(b) \]

(Newton's method). This makes sense as $f'_{n + 1}(b) \in R/\mathfrak m^{n + 1}$ is invertible by the condition on $a_0$. Then we compute $f_{n + 1}(a_{n + 1}) = f_{n + 1}(b) - f_{n + 1}(b) = 0$ in $R/\mathfrak m^{n + 2}$. Since the system of elements $a_ n \in R/\mathfrak m^{n + 1}$ so constructed is compatible we get an element $a \in \mathop{\mathrm{lim}}\nolimits R/\mathfrak m^ n = R$ (here we use that $R$ is complete). Moreover, $f(a) = 0$ since it maps to zero in each $R/\mathfrak m^ n$. Finally $\overline{a} = a_0$ and we win. $\square$


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