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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!

TP.TPPR
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    I personally think that version with the $p^M$ and $p^{2M+1}$ is ugly and hard to remember. It's much nicer to express this in terms of absolute values: the hypothesis says $|F'(a_0)|_p = 1/p^M$, so the congruence of $F(a_0)$ says $\boxed{|F(a_0)|_p < |F'(a_0)|_p^2}$. And the conclusion says $F(a) = 0$ where $|a - a_0|_p < |F'(a_0)|_p$. It turns out that for all $a_0$ near a simple root $a$ of $F(x)$ in $\mathbf Z_p$ we have the inequality $|F(a_0)|_p < |F'(a_0)|_p^2$: see Section 7 in https://kconrad.math.uconn.edu/blurbs/gradnumthy/hensel.pdf. – KCd Oct 31 '22 at 18:14

1 Answers1

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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 $n$.

Fabio Lucchini
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