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!
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 $n$.