In an Elliptic Curves course, my lecturer states Hensel's Lemma as the following:
Let $k$ is a field that is complete with respect to a non-archimedian norm $|.|$ and $$R=\{x \in k : |x| \leq 1\}$$ a subring, then for any $f(x)\in R[x]$ and any $t_0 \in R$ such that $|f(t_0)|<|f'(t_0)|^2$ there exists $t\in R$ such that $f(t)=0$ and $|t-t_0|<|f'(t_0)|$.
He then says that, intuitively speaking,
Hensel's Lemma says that if a polynomial $f \in \mathbb Z_p[x]$ ($p$-adic integers) has a root in its reduction modulo $p$ (+ some other conditions), then it has a root in $\mathbb Q_p$.
My question is why this intuition holds.
I can see that it holds for simple quadratics like $f(x)=x^2-d$, where $d \in \mathbb Z$: Here if $t_0$ is a root of this modulo $p$ then, treating $t_0$ as an integer, we have $t_0^2 \equiv d \pmod p$ and so $|f(t_0)|_p=|t_0^2-d|_p<1$ but on the other hand $|f'(t_0)|_p=|2t_0|_p$ so provided $p \neq 2$ and $p$ does not divide $t_0$, we have $|f'(t_0)|_p^2=1>|f(t_0)|_p$ so the Lemma holds and there exists a root $t$ of $f$ in $\mathbb Z_p$.
I cannot, however, see how to do this with more general polynomials, and any help with this would be much appreciated.
For simplicity consider only the case of a polynomial $f(x)\in\mathbb{Z_p}[x]$ with a simple root $\alpha$ modulo $p$, that is $f(\alpha)=0\pmod p$ but $f'(\alpha)\neq 0\pmod p$. Now the conditions of the lemma are verified and the conclusion is that there exists $a\in\mathbb{Z}_p$ such that $a\equiv \alpha\pmod p$ and $f(a)=0$.
For example take $f(x)=x^3+7x^2+13$. Since $f(1)\equiv 0\pmod 7$ and $f'(1)\equiv 3\pmod 7$ we immediately know that $f$ has a root in $\mathbb{Z}_7$ which is of the form $1+7a$, $a\in \mathbb{Z}_7$.