Algebraic closures are henselian?

Question

Let $(K,v)$ be a nonarchimedean valued field and $(\widehat{K},\widehat{v})$ be its completion.

Let $o$ and $\widehat{o}$ be the valuation rings of $K$ and $\widehat{K}$.

Let $K_v$ be the separable closure of $K$ in $\widehat{K}$. Let $o_v:=K_v\cap \widehat{o}$.

Let $f(x)=a_nx^n+\cdots+a_0\in o_v[x]$ be a primitive polynomial, i.e. $\max\{|a_n|,\cdots,|a_0|\}=1$.

Let $g,h\in \widehat{K}[x]$ be polynomials with $f=gh$ and assume that the highest coefficient of $g$ is a unit in $o_v$.

If $K_v$ is algebraically closed in $\widehat{K}$, then all coefficients of $g$ and $h$ are in $o_v$?

Answer 1

I found this question while searching for clarification on that same section of the book. Have you found a satisfying explanation?

By the way from my understanding, the argument is that the coefficients are polynomial functions of the roots of $f$, which are obviously algebraic over $\mathcal{O}_v$ by definition of $f$, and so the coefficients are algebraic as well (because algebraic numbers form a field).

For $g$, after multiplying by a suitable constant (and dividing $h$ by the same constant) such that the highest degree coefficient of $g$ is in $\mathcal{O}_v^*$, we can then use Vieta's formulas to argue that the coefficients equal up to units the elementary symmetric polynomials in the roots of $g$ (which are roots of $f$). Hence they are also algebraic. Then $h = f/g$ so its coefficients will also be algebraic by the division algorithm.

So $f$ factors over $K_v$ (because it was assumed to be algebraically closed, i.e. $K_v = \overline K$), and then by Gauss' lemma $f$ also factors over the ring of integers $\mathcal{O}_v$.

What I am wondering is, wouldn't this argument imply that not only the separable closure of $K$, but also the algebraic closure of $K$, should be henselian?

Answer 2

Yes, they are. The reason is that, as Neukirch says, we choose the leading coefficient of $g$ from $\mathcal{O}_v$.

More precisely, if we look back at the proof of Hensel's lemma $(4.6)$, we see that $$g=g_0+p_1\pi+p_2\pi^2+\dots,$$ where $g_0\in \hat{\mathcal{O}}[x]$ is any lift of $\overline{g}\in(\hat{\mathcal{O}}/\hat{\mathfrak{p}})[x]$ and $\deg(p_i)<\deg g_0$. So, we are able to change the leading coefficient of $g$ (which is just the leading coefficient of $g_0$) in a way that does not affect the reduction of $g$: since $\deg g = \deg \overline{g}$, the leading coefficient of $\overline{g}$ is represented by a unit in $\hat{\mathcal{O}}$, which by the isomorphism between residue fields $\mathcal{O}_v/\mathfrak{p}_v=\hat{\mathcal{O}}/\hat{\mathfrak{p}}$, can be chosen to be a unit in $\mathcal{O}_v$.

The remaining steps are easy: since $K_v$ is algebraically closed, all roots of $f$ are in $K_v$. In particular, all roots of $g$ are in $K_v$ and $g\in K_v[x]\cap \hat{\mathcal{O}}[x] = \mathcal{O}_v[x]$ because, by Vieta's formulas, the coefficients of $g$ are polynomials in the roots of $g$ and the leading coefficient of $g$. Since $f,g\in \mathcal{O}_v[x]$ and the leading coefficient of $g$ is invertible in $\mathcal{O}_v$, it follows from Euclidean algorithm that $h\in \mathcal{O}_v[x]$. So, $f$ factors over $\mathcal{O}_v$.