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. $\square$

Comment #3651 by Brian Conrad on

On line 3 of the proof, replace "radical" with "Jacobson radical". Near the end of the proof, it should be explained via the "locally nilpotent" hypothesis on the kernels of the surjective maps that the condition on $g_m$ and $h_m$ generating the unit ideal for $m=0$ is inherited for $m=1$ and so on for invoking Lemma 0ALI to "successively" lift as indicated.

Comment #3748 by on

OK, I replaced "radical" by "Jacobson radical" in all places where appropriate and whenever we used the notation $\text{rad}(R)$ I have added text saying this denotes the Jacobson radical. See the corresponding changes here.

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