Henselian field and Monic Polynomials with - Neukirch exercise 5 (Henselian Fields) i tried to solve the following exercise:
Let $K$ be a nonarchimedean valued field, $\mathcal o$ the valuation ring, and $\mathfrak p$ the maximal ideal. $K$ is henselian if and only if every polynomial $f(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1 x+a_0\in\mathcal o[x]$ such that $a_0\in\mathfrak p$ and $a_1\not\in\mathfrak p$ has a zero $a\in\mathfrak p$.
Hint: The Newton-Polygon.
The one direction ist easy. If $K$ is henselian then the assertion follows immediately from Hensel's lemma. The other direction however i have no idea how to solve it. I tried to use the fact that, if for every irreducible polynom $f(x)\in K[x]$ the Newton-Polygon of $f$ has only a single line segment, then $K$ is henselian. But i can only proof this for $f(x)\in\mathcal o[x]$ irreducible such that $a_0,a_1\not\in\mathfrak p$. 
 A: I think you should just look at the proof of Proposition 6.7 in Neukirch and follow the same line of thought there. He credited the proof to (I think) Enric Nart who has been working on Newton's polygon for some time already. So for the other direction the idea (I think) is to 


*

*Assume you have an irreducible $f\in K[x]$ that is, without loss of generality, monic and with Newton polygon that consists of more than one segment.

*Show that (by some change of variables) you can even assume that $f\in o[x]$

*Because the Newton polygon has more than one segment: $f$ cannot be linear and $f(0)\in\mathfrak p$ 

*If $f'(0)\not\in\mathfrak p$ then we use our hypothesis and come to a contradiction (we get an $a\in o\cap \mathfrak p$ such that $x-a$ divides $f$)

*The difficult part is if $f'(0)\in\mathfrak p$ and this is where you use the argument of Nart:


5.1 Let $r$ be the length of the segment connecting to $(\deg(f),0)$
5.2 Find the minimal polynomial $g\in K[x]$ of $f'(0)^{-1}\alpha^r \in K(\alpha)$ where $\alpha$ is a zero of $f$ in its splitting field having least value in an extension of the valuation of $K$ to the splitting field.
5.3 You will then show that the value of $g'(0)$ is $0$ and that $g(x)$ has a Newton polygon with more than one segment
5.4 I think here you will practically repeat arguments 1-4 and arrive to your contradiction.
I personally don't think that using Newton's polygon is a good way of characterizing (or proving results about) Henselian fields. But maybe an expert could correct me here.
A: I think i have finally the solution: Assume that $K$ is not henselian, then there exists an irreducible polynom whose newton polygon has more than one line segment. Let $L$ be its splitting field and $v$ a fixed valuation on $L$. Let $G\subset Gal(L\mid K)$ be the subgroup that fixes $v$ and $M$ be the fixed field of $G$. Let $\sigma_1=\mathrm{Id},\sigma_2,\ldots,\sigma_n$ be the distinct embeddings of $M$ into $L$ then every $v_i=v\circ\sigma_i,2\leq i\leq n$ is a valuation on $M$ which is inequivalent to $v$. By weak approximation we get $\alpha\in M$ such that $v(\alpha)>0$ and $v_i(\alpha)=0$ for $2\leq i\leq n$. Thus $\alpha\not\in K$. Now note that $\alpha_i=\sigma_i(\alpha)$ are all the conjugates of $\alpha$ and from Neukirch[Proposition 6.3, page 145] we get that the minimal polynomial $g(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1 x+a_0$ of
$\alpha$ has coefficients in $\mathcal o$ and that $a_0\in\mathfrak p$ and $a_1\not\in\mathfrak p$. Hence $g$ hat a root in $\mathfrak p$, that is a contradiction to the irreducibility of $g$.
