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 $a_0,a_1\not\in\mathfrak p$.
Best Answer
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
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.