Hensel's Lemma in $\mathbb{C} (\!( s )\!)$

Question

Reading a proof about regular $S_n$-extensions I got stuck here:

Let $g(s,X)=X^n+(s-\frac{n^n}{(1-n)^{n-1}})X+(s-\frac{n^n}{(1-n)^{n-1}}) \in \mathbb{C} (\!( s )\!)$

Specializing at $0$ we get $$g(0,X)=X^n+-\frac{n^n}{(1-n)^{n-1}}X-\frac{n^n}{(1-n)^{n-1}}$$

and after scaling we get $$ h(X)=X^n-nX+(n-1) \in \mathbb{C}[X] $$ $h(X)$ has $1$ as a double root, and $n - 2$ simple roots. Thus, by Hensel’s Lemma $g(s, X)$ has $n - $2 simple roots in $\mathbb{C} (\!( s )\!)$.

I have one question:

How can I apply Hensel's Lemma in $\mathbb{C} (\!( s )\!)$? I know only the version for $p$-adic fields..

Answer 1

Hensel's Lemma works for any complete valued field. $\mathbb{C}((s))$ is such a field. Moreover, it works exactly the same. You write your valuation ring as an inverse limit $R \cong \lim\limits_{\gets} R/(\pi^n)$ where $\pi$ is your uniformizer and then you do Newton's method mod $\pi$ then $\pi^2$ then $\pi^3$ etc.

For example, let's lift the roots of $x^2 + sx - \frac{1}{1-s} \equiv (x+1)(x-1) \pmod s$. We start with $x_1 = 1$ and then set

$$ x_{n + 1} = x_n - \frac{x_n^2 + sx_n - \frac{1}{1-s}}{2x_n + s} \pmod{s^{n+1}}.$$

So for example,

$$ x_2 = 1 - \frac{1 + s - (1 + s)}{2 + s} = 1 \pmod{s^2}$$

and

$$ x_3 = 1 - \frac{1 + s - (1 + s + s^2)}{2 + s} = 1 + \frac{s^2}{2} \pmod{s^3}$$

and so on and so forth.