TITLE: Another generalization of Hensel's lemma
QUESTION [6 upvotes]: I know this is a "dangerous" topic to ask a question about, since a lot of questions regarding Hensel's lemma have already been answered, but I searched for it and couldn't find this version of the lemma (please tell me when it is already answered!)
I am trying to find a proof to a generalized form of Hensel's lemma:
Let $F(x)$ be a polynomial with coefficients in $\mathbb{Z}_p$. If $a_0\in\mathbb{Z}_p$ satisfies $F'(a_0)\equiv0 \pmod{p^M}$ but $F'(a_0) \text{ not equivalent to } 0 \pmod{p^{M+1}}$, and if $F(a_0)\equiv0 \pmod{p^{2M+1}}$, then there is a unique $a\in\mathbb{Z}_p$ such that $F(a)=0$ and $a\equiv a_0\pmod{p^{M+1}}$. 
I have no clue where to start with this proof. It is probably something with induction to $M$, but I do not really know how to start, since there's so much information in the lemma! Could anyone help me out?
Thanx in advance!

REPLY [2 votes]: Let $a_n\in\Bbb Z_p$ such that $v_p(F'(a_n))=M$ and $v_p(F(a_n))>2M$.
Then
$$a_{n+1}=a_n-\frac{F(a_n)}{F'(a_n)}$$
satisfy $a_{n+1}\in\Bbb Z_p$, $a_{n+1}\equiv a_n\pmod{p^{M+1}}$, $v_p(F'(a_{n+1}))=M$ and $v_p(F(a_{n+1}))>v_p(F(a_{n}))$.
For let $e=v_p(F(a_n))$, then by Taylor's formula, we have
$$F(a_n+xp^{e-M})\equiv F(a_n)+xp^{e-M}F'(a_n)\pmod{p^{2e-2M}}$$
Since
$$v_p\left(-\frac{F(a_n)}{p^{e-M}F'(a_n)}\right)=e-(e-M+M)=0$$
we have
\begin{align}
a_{n+1}
&=a_n-\frac{F(a_n)}{F'(a_n)}\\
&\equiv a_n\pmod{p^{e-M}}\\
&\equiv a_n\pmod{p^{M+1}}
\end{align}
and
\begin{align}
F(a_{n+1})
&\equiv 0\pmod{p^{2e-2M}}\\
&\equiv 0\pmod{p^{e+1}}
\end{align}
hence $v_p(F(a_{n+1}))>e=v_p(F(a_n))$.
Moreover,
\begin{align}
F'(a_{n+1})
&\equiv F'(a_n)\pmod{p^{e-M}}\\
&\equiv F'(a_n)\pmod{p^{M+1}}
\end{align}
hence $v_p(F'(a_{n+1}))=v_p(F'(a_n))=M$.
Thus the assertion follows by induction on $n$.