I guess, that this is a very basic question, but still I cannot find a source to answer it nor am I able to give a solution by myself.
If $ f \in \widehat{\mathbb{Z}_p}[x]$ is a non-constant polynomial with coefficient in the $p$-adic integers for some prime number $p$, does $f$ always have a zero in $\widehat{\mathbb{Z}_p}$?
I already read that the question above should be equivalent to asking whether $f$ has a root in $\mathbb{Z}/p^k\mathbb{Z}$ for all $k \in \mathbb{N}$. Also it should have to do something with the Newton polygon, which I am not quite familiar with.
I would be very thankful for any help!
Best Answer
$x^2-p$ has no root. The first thing to do is to show that when $f \mod p$ has a root and $f'(a)\ne 0\bmod p$ then the gradient descent converges to the unique root $A\in \Bbb{Z}_p$ such that $A=a\bmod p$.
It is true that $f$ has a root iff it has a root in every $\Bbb{Z/p^kZ}$, this is because if $f$ has no root then in a finite extension of $\Bbb{Q}_p$ $$f(x) = c\prod_{j=1}^d (x-\alpha_j)$$ so $$|f(n)|_p= |c|_p\prod_{j=1}^d |n-\alpha_j|_p \ge |c|_p A^p$$ where $A=\inf_{n\in \Bbb{Z},j} |n-\alpha_j|_p$ which is $> 0$ since otherwise some sequence of integers would converge to some $\alpha_j$ ie. $\alpha_j$ would be in $\Bbb{Z}_p$.