I'm reading a paper where the author appeals to Hensel's lemma, but it is not clear to me quite how it is meant to be applied (or, for that matter, which version!). My commutative algebra background is unfortunately not as good as I'd like, and I can't find references between plain old Hensel's lemma for $p$-adics and Henselian rings.
The setup is this. I have a split exact sequence of (unital, and otherwise nice) rings $$ A \stackrel{r}{\to} B \to 0 $$ where $N = ker(r)$ is a nil-ideal, and $s$ is a section of $r$. Let $\Sigma$ be a set of primes (in the end it will probably just be a single prime), $\mathbb{Z}_\Sigma = \{a/b\in\mathbb{Q} \mid p\not| b,\ \forall p\in\Sigma\}$ and $R_\Sigma = R\otimes_\mathbb{Z} \mathbb{Z}_\Sigma$ for $R$ a ring (sim. for an ideal). Then we still have $N_\Sigma$ a nil-ideal, $r_\Sigma$ is still onto. Let $U_\Sigma = r_\Sigma^{-1}(1) \subset A_\Sigma$. We also have that $a\in A_\Sigma$ is a unit if and only if $r_\Sigma(a)$ is a unit.
Now comes the bit I don't understand.
The author claims, using Hensel, that given a unit $n\in \mathbb{Z}_\Sigma^\times$ with $n\gt 0$ and some $a\in U_\Sigma$, there is a unique $b\in U_\Sigma$ such that $b^n = a$.
I can think of several relevant points, but I can't connect them up into a proof in my mind.
As $N$ is a nil-ideal, so $A$ should be complete in the $N$-adic topology.
Since $a\in U_\Sigma$, I'm lifting the solution $1$ to $x^n - 1=0$ in $B_\Sigma$ through the various projections $A_\Sigma/N_\Sigma^{k+1} \to A_\Sigma/N_\Sigma^k$.
The rings $A$ and $B$ are quite nice ($B$ is isomorphic to $\mathbb{Z}^m$, for instance, but $A$ can be more complicated), so I can probably assume they're Noetherian...
Is the proof there under my nose? Or is a bit more subtle?
After inverting primes outside $\Sigma$, we can assume $A=A_\Sigma$. Hensel's lemma is being used as the following assertion: if $C$ is a finite étale $A$-algebra, then any section $C \to B$ lifts uniquely to a section $C \to A$. Apply this with $C = A[x]/(x^n - a)$, for any $n$ prime to $\Sigma$ and the section $C\to B$ corresponding to the identity element.
EDIT: To see why $A[x]/(x^n - a)$ is étale, one uses the differential criterion. The derivative of $x^n$ is $nx^{n-1}$. If $x$ is a solution to $x^n - a$, then it is n particular a unit, and therefore so is $nx^{n-1}$, which is then non-invertible everywhere.
In any case, the proof of the assertion that I gave is the same as that of the regular old Hensel's lemma via Newton's method, so it's really no different. But I prefer this formulation.