I am trying to understand the following lemma
Lemma:
Let $f\in\mathbb{Z}_p[X]$ and let $f'$ be its derivative. Let $x\in\mathbb{Z}_p, n,k\in\mathbb{Z}$ such that $0\leq 2k<n, f(x)\equiv 0 \mod p^n, v_p(f'(x))=k$. Then there exists $y\in\mathbb{Z}_p$ such that $f(y)\equiv 0\mod p^{n+1}, v_p(f'(y))=k$, and $y\equiv x\mod p^{n-k}$.
Proof:
Take $y$ of the form $x+p^{n-k}z$ with $z\in\mathbb{Z}_p$. By Taylor's formula we have $f(y)=f(x)+p^{n-k}zf'(x)+p^{2n-2k}a$ with $a\in\mathbb{Z}_p$.
By hypothesis $f(x)=p^nb$ and $f'(x)=p^kc$ with $b\in\mathbb{Z}_p$ and $c\in U$ (where U denotes the set of units of $\mathbb{Z}_p$). This allows us to choose $z$ in such a way that
$b+zc\equiv 0\mod p$
From this we get $f(y)=p^n(b+zc)+p^{2n-2k}a\equiv 0\mod p^{n+1}$ since $2n-2k>n$. Finally Taylor's formula applied to $f'$ shows that $f'(y)\equiv p^kc\mod p^{n-k}$; since $n-k>k$, we see that $v_p(f'(y))=k$.
I have some questions to this proof. First of all, why can you applie Taylor's formula easily to functions in $\mathbb{Z}_p[X]$. And how do you applie the formula in general?
We take $y=x+p^{n-k}z$. Now f(y)=f(x+p^{n-k}z). I recalled the formula, but there is neither an intervall given, nor a development center. I do not see how the formula is used.
Why can we choose $z$ such that $b+zc\equiv 0\mod p$? Is $z$ unique?
This proof strikes me a little bit off-guard. My biggest problem is: Why can we use Taylor's formula, and how is it done?
Thanks in advance.
This is Hensel's Lemma (or at least one of the many variants of Hensel's Lemma).
Why does this work? Let $f(z)=z^m$. Then $$f(x+p^rz)=(x+p^rz)^m=x^m+mp^rx^{m-1}z+p^{2r}a =f(x)+f'(x)p^rz+p^{2r}a.$$ By linearity $$f(x+p^rz) =f(x)+f'(x)p^rz+p^{2r}a$$ for any polynomial with $p$-adic integer coefficients.
Solving $b+zc\equiv0\pmod p$ for $z$ is just the Euclidean algorithm. Each of $b$ or $c$ is congruent to an integer modulo $p$, so for the purposes of the congruence we may assume that $b$, $c\in\Bbb Z$. Also $p\nmid c$, so $\gcd(c,p)=1$ and we may solve $cz\equiv-b\pmod p$ for $z\in\Bbb Z$.