Hensel's Lemma question with discriminant Let
$K$
be any field with a non-Archimedean valuation
$| \: |$
, and let $R= \{x \in K : |x| \leq 1 \}$.
Let $f(x)$ has discriminant $D$, and let $a_0 \in R$ satisfy $|f(a_0)| \leq |D|^2$.  Show that $f(X)$ has a root $a \in R$.
Can see the similarity to Hensel's lemma, but not sure how to use it!
 A: I assume that $f(x)\in R[x]$, and  $|f(a_0)|<|D|^2$.
(For the first assumption, consider $f(x)=x^2-p^{-1}\in\mathbb{Q}_p[x]$ with $p>2$, as then $|D|_p^2=|-4p^{-1}|^2_p=p^2$ and $|f(0)|_p=p<p^2=|D|_p^2$. However, if $a$ were a roof of $f$, $p^{2k}=|a^2|_p=|p^{-1}|_p=p$ for some integer $k$ ans so $f$ has  no root in $R$. The second assumption is needed crucially in this proof, and I think is necessary but do not have an example to show this.)
As $D$ is the resultant of $f$ and $f^\prime$, there are polynomials $p,q\in  R[x]$ such that $f(x)p(x)+f^\prime(x)q(x) = D$. Now, $$ |D|\leq\max\{|p(a_0)f(a_0)|,|q(a_0)f^\prime(a_0)|\}\leq 1$$ with the last inequality holding as $f,f^\prime,p,q\in R[x]$ and $a_0\in R$.
Next, $|f(a_0)p(a_0)|=|f(a_0)||p(a_0)|\leq|f(a_0)|<|D|^2\leq|D|$, as $p(a_0)\in R$
Therefore, $|D|=|q(a_0)f^\prime(a_0)|$, and so $|D|=|q(a_0)||f^\prime(a_0)|\leq|f^\prime(a_0)|$. Squaring both sides gives $|D|^2\leq |f^\prime(a_0)|^2$.
Putting this together with the hypothesis, we get $$|f(a_0)|<|D|^2\leq |f^\prime(a_0)|^2$$ to which we can apply Hensel's lemma, giving the desired root.
This answer is a bit late, but hopefully it can help people in the future.
