A single-segment Newton polygon implies henselian? I have a question about Newton polygons and henselian fields.
In p149 of Neukirch’s book(algebraic number theory:the beginning of Proposition 6.7), he says that
“We have just seen that the property of $K$ to be henselian follows from the condition that the Newton polygon of every irreducible polynomial $f(x)\in K[x]$ is a single segment.”
Question: In which place, this fact is shown?  Is it in the proof of 6.6? This fact is correct?
In Proposition 6.4, he showed that the uniqueness of extended valuations implies the condition of Newton polygons.
In Proposition 6.6, he showed that the equivalence between the uniqueness of extended valuations and henselianness.
However, I think that the proof of 6.6 uses not only the condition of Newton polygons but also the uniqueness(which I cannot remove).
On the other hand, there exists a monic irreducible polynomial such that its Newton polygon is a single segment but the splitting field of it has two extended valuations. (By an example $f(x)=x^2+1\in Q[x]$  and $p=5=(2+i)(2-i)$ given by Hagen at stackexchange/862893.)
 A: Where do you see a problem?
Theorem 6.6 says that a field is henselian (that is Hensel's Lemma holds) if and only if the valuation extends uniquely to every algebraic extension.
The proof of the implication $\Rightarrow$ is already given in Theorem 6.2.
For the proof of the implication $\Leftarrow$ one assumes uniqueness, in which case the Newton polygon of an irreducible polynomial is a line. The latter is a consequence of Satz 6.4.
OK, I see now that the proof of Satz 6.7 really starts a bit in a confusing way. The point here is, that he shows that EVERY irreducible polynomial has a Newton polygon with only one segment, which by the remark three pages before is equivalent to the fact, that all roots possess the same value taking any extension of the valuation to the splitting field.
Now if for some algebraic extension of $K$ there are two different extensions of $v$, then one can find an irreducible polynomial having two roots with different values with respect to one of the two extensions of $v$.
That's it.
A: This is typical Neukirch at work here, skipping steps as he pleases. The idea is the following:
You want to verify the validity of Hensel's Lemma 4.6 making the assumptions of Proposition 6.7. Now the first several lines of the proof of Proposition 6.7 say that IF in addition you know that the Newton polygon of an irreducible polynomial is a single line segment, THEN it is "trivial" to verify Hensel's Lemma. Consequently the proof is "reduced" to showing the statement regarding the segment.
I admit that the actual proof is never carried out, and that "trivial" is in the eyes of the beholder, but it is doable. What one ends up doing is:


*

*Split a primitive polynomial $f$ (the one that you want to verify the lemma for) into irreducible primitive factors in $\mathcal{o}[x]$ and reduce mod $p$. Thus $f = P\prod f_i$, where $P$ is the product of all the primitive factors with unit leading coefficient and $f_i$ are the factors whose leading coefficient is in $\mathcal{p}$

*Using the single segment information one has $f_i = c_i \mbox{ mod } p$. So one has $\overline{f} = c\overline{P} \mbox{ mod } p$. Where $c = \prod_i c_i$.   

*Then the factorization $\overline{f} = \overline{g}\overline{h}$, which is given in Hensel's lemma, translates into $c \overline{P} = \overline{g}\overline{h}$. 

*Since the leading coefficient of $P$ is a unit you can write $P = sP_1$ with $P_1$ monic and $s \in o^{\times}$. Consequently from 3. you have a factorization $\overline{P_1} = \overline{g_1} \overline{h_1}$ where $\overline{g_1}$ and $\overline{h_1}$ are both monic and differ from $\overline{g}$ and $\overline{h}$ by constant factors in $\kappa$.

*Now you apply the statement of Proposition 6.7 which tells you that you can lift and write $P_1 = g_1 h_1$. If you set $g = s_1g_1$ and $h = s_2h_1\prod f_i$, where $s_1 $ is a lift of $\overline{g}/\overline{g_1}$ to $o^{\times}$ and $s_2 = s_1^{-1}$ you can see that $g,h$ are a working pair for Hensel's lemma. 
The above is not particularly trivial but it is the shortest argument I can muster. Hope this helps.
