I'm trying to understand the solution to this problem. It came up when doing some revision. It is essentially to show that the conclusion of Hensel's Lemma holds if we have take a valuation ring $R$ attached to a field $K$ complete with respect to a non-Archimedian valuation $|.|$, an $a_0 \in R$ and $f(t) \in R[t]$ such that
- $|f(a_0)|=|f'(a_0)^2|$
- $0<|f'(a_0)|<1$
- $|\dfrac{1}{2} f''(b)|<1$ for all $ b \in R$
Has anyone got an suggestions for this? I'm trying to modify the usual proof for hensel's lemma, but it isn't working out so well for me.
Update:
I think I might have a solution to this, but it may be completely wrong.
Taking a Taylor series expansion of $f$ about $a_0$, we write:
$$ f(a_0+h)= f(a_0)+hf'(a_0)+\frac{h^2}{2!}f''(a_0)+... $$
Setting $h = -\frac{f(a_0)}{f'(a_0)}$, then note that $|h|=|f'(a_0)|<1$ so that $a_1:=a_0+h \in R$ and $h \ne 0$ or else we have a solution. Then
$$ f(a_1)=\frac{h^2}{2!}f''(a_0)+\frac{h^3}{3!}f'''(a_0)+...$$
Under condition 2. then $$|\frac{h^3}{3!}f'''(a_0)+...| \leq \max(|f(a_1)|,|\frac{h^2}{2!}f''(a_0)|)<\max(|f(a_1)|,|h^2|)$$
So then $$|\frac{h}{3!}f'''(a_0)+...| <\max(\frac{|f(a_1)|}{|h^2|},1)$$
Therefore as,
$$ \frac{|f(a_1)|}{|h^2|} \leq \max( |\frac{1}{2}f''(a_0)|,|\frac{h}{3!}f'''(a_0)+...|)< \max(1,\frac{|f(a_1)|}{|h^2|}) $$
so that $$|f(a_1)|<|h^2|=|f(a_0)|$$.
Also, then we have that
$$|f'(a_1)-f'(a_0)| = |hf''(a_0)+\frac{h^2}{2}f'''(a_0)+...|=|f'(a_0)||f''(a_0)+...|$$
So then $$ |f'(a_1)-f'(a_0)|< |f'(a_0)||2|\leq |f'(a_0)|$$ and $|f'(a_1)|=|f'(a_0)|$.
So we have found an $a_1 \in R$ such that $|f(a_1)|< |f(a_0)|=|f'(a_0)|^2=|f'(a_1)|^2$ and we can use Hensel's Lemma with $a_1$. So we find a unique $a \in R$ such that $f(a)=0$ and $|a-a_1| \leq \dfrac{|f(a_1)|}{|f'(a_1)|}$. Then $|a-a_0|=|a-a_1+a_1-a_0|\leq \max( \dfrac{|f(a_1)|}{|f'(a_1)|}, \dfrac{|f(a_0)|}{|f'(a_0)|})= \dfrac{|f(a_0)|}{|f'(a_0)|}$.
Am I on the right track with this or way off course?