I'm trying to understand the proof of the following statement:
Let $f\in\mathbb{Z}_p[X]$ be a polynomial with coefficients in $\mathbb{Z}_p$. Suppose there is a $p$-adic integer $\alpha_1$ such that $$|f(\alpha_1)|_p<|f'(\alpha_1)|_p^2.$$ Then there is a unique $p$-adic integer $\alpha$ such that $f(\alpha)=0$ and $\alpha\equiv\alpha_1\;\text{mod }p^{\nu_p(f(\alpha_1))-\nu_p(f'(\alpha_1))}.$
To prove it in the case $p\neq 2$, we write $f(\alpha_1)=p^j\xi$ and $f(\alpha_1)=p^k\gamma$, with $0\leq 2k< j$ and $\nu_p(\xi)=\nu_p(\gamma)=0$, and we look for $\alpha_2=\alpha_1+b_2 p^{j-k}$ such that $f(\alpha_2)\equiv 0\;\text{mod }p^{j+1}$.
Using the Taylor expansion we have $$f(\alpha_2)\equiv f(\alpha_1)+f'(\alpha_1)b_2p^{j-k}\text{ mod }p^{j+1}=p^j\xi+b_2\gamma p^j\text{ mod }p^{j+1}.$$ Imposing $f(\alpha_2)\equiv 0\text{ mod }p^{j+1}$, we get $$0\equiv \xi+b_2\gamma \text{ mod }p,$$ which can be solved in $b_2$. Then we construct the following sequence ($\alpha_n$) by induction.
My question is about the claim. To apply induction we need that $$|f(\alpha_2)|_p<|f'(\alpha_2)|_p^2.$$ How can this condition be checked?
Thanks in advance.
Write $$f(\alpha_n+x)=a_0+a_1x+a_2x^2+\cdots $$ where $a_0=f(\alpha_n)$, $a_1=f'(\alpha_n)$. If $\alpha_{n+1}=\alpha_m+h$, then $$ f(\alpha_{n+1}+x)=b_0+b_1x+\cdots$$ where $$b_0=a_0+a_1h+a_2h^2+\cdots$$ and $$b_1=a_1+2a_2h+\cdots$$
We choose our $h$ such that $a_0+a_1h$ becomes small, namely as $0<|a_0|_p<|a_1^2|_p\le 1$, our pick $h\approx -\frac{a_0}{a_1}$ will make $|h|_p=\frac{|a_0|_p}{|a_1|_p}<|a_1|_p$, hence for all $k\ge2$, we have $|a_kh^k|_p\le |h^2|_p<|a_1h|_p$ so that $$|b_0|_p=|a_0+a_1h|_p<|a_0|_p.$$ On the other hand, at least for $2\le k<p$, we have $|ka_kh^{k-1}|_p\le |h|_p^{k-1}\le |h|_p<|a_1|_p$; and for $k\ge p>2$, we also get $|ka_kh^{k-1}|_p\le |h|_p<|a_1|_p$ because certainly $h^{k-1}$ is more divisible by $p$ than $k$. Thus $$|b_1|_p=|a_1|_p.$$
In other words, when using Newton's method in $p$-adics ($p>2$), surprisingly only the function value gets smaller while the derivative keeps its size.