Lemma 15.11.3. Let $A = \mathop{\mathrm{lim}}\nolimits A_ n$ where $(A_ n)$ is an inverse system of rings whose transition maps are surjective and have locally nilpotent kernels. Then $(A, I_ n)$ is a henselian pair, where $I_ n = \mathop{\mathrm{Ker}}(A \to A_ n)$.

**Proof.**
Fix $n$. Let $a \in A$ be an element which maps to $1$ in $A_ n$. By Algebra, Lemma 10.32.4 we see that $a$ maps to a unit in $A_ m$ for all $m \geq n$. Hence $a$ is a unit in $A$. Thus by Algebra, Lemma 10.19.1 the ideal $I_ n$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_ nh_ n$ be a factorization of $\overline{f} = f \bmod I_ n$ with $g_ n, h_ n \in A_ n[T]$ monic generating the unit ideal in $A_ n[T]$. By Lemma 15.11.2 we can successively lift this factorization to $f \bmod I_ m = g_ m h_ m$ with $g_ m, h_ m$ monic in $A_ m[T]$ for all $m \geq n$. At each step we have to verify that our lifts $g_ m, h_ m$ generate the unit ideal in $A_ n[T]$; this follows from the corresponding fact for $g_ n, h_ n$ and the fact that $\mathop{\mathrm{Spec}}(A_ n[T]) = \mathop{\mathrm{Spec}}(A_ m[T])$ because the kernel of $A_ m \to A_ n$ is locally nilpotent. As $A = \mathop{\mathrm{lim}}\nolimits A_ m$ this finishes the proof.
$\square$

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