Lemma 10.154.8. A filtered colimit of (strictly) henselian local rings along local homomorphisms is (strictly) henselian.

**Proof.**
Categories, Lemma 4.21.5 says that this is really just a question about a colimit of (strictly) 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). It is easy to see that $\mathop{\mathrm{colim}}\nolimits \kappa _ i$ is separably algebraically closed if each $\kappa _ i$ is so; thus it suffices to prove $R$ is henselian if each $R_ i$ is henselian. 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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