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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