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 10.149.7. A filtered colimit of henselian local rings along local homomorphisms is henselian.

Proof. Categories, Lemma 4.21.5 says that this is really just a question about a colimit of henselian local rings over a directed set. Let $(R_ i, \varphi _{ii'})$ be such a system with each $\varphi _{ii'}$ local. Then $R = \mathop{\mathrm{colim}}\nolimits _ i R_ i$ is local, and its residue field $\kappa $ is $\mathop{\mathrm{colim}}\nolimits \kappa _ i$ (argument omitted). Suppose that $f \in R[T]$ is monic and that $a_0 \in \kappa $ is a simple root of $\overline{f}$. Then for some large enough $i$ there exists an $f_ i \in R_ i[T]$ mapping to $f$ and an $a_{0, i} \in \kappa _ i$ mapping to $a_0$. Since $\overline{f_ i}(a_{0, i}) \in \kappa _ i$, resp. $\overline{f_ i'}(a_{0, i}) \in \kappa _ i$ maps to $0 = \overline{f}(a_0) \in \kappa $, resp. $0 \not= \overline{f'}(a_0) \in \kappa $ we conclude that $a_{0, i}$ is a simple root of $\overline{f_ i}$. As $R_ i$ is henselian we can find $a_ i \in R_ i$ such that $f_ i(a_ i) = 0$ and $a_{0, i} = \overline{a_ i}$. Then the image $a \in R$ of $a_ i$ is the desired solution. Thus $R$ is henselian. $\square$


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