zeros of p-adic power series Suppose I have $f(x) \neq 0 \in \mathbb{Z}_p[[x]].$ If it can help it is of the form $g(x^p)+ph(x)$.


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*Under which conditions can I say that $f(x)$ has finitely many zeros in $\bar{\mathbb{Q}}_p$?  (Under which conditions are the solutions integral over $\mathbb{Z}_p$?)

*Is there a method like Newton polygons to read off information of the zeros by looking at the coefficients of $f(x)$?
 A: If $f(x)\in\Bbb Z_p[[x]]$, then you always have the result that there are only finitely many elements $\rho$ in the algebraic closure satisfying the two conditions $f(\rho)=0$ and $v_p(\rho)>0$, where I’m using the (additive) valuation $v_p$. The condition on absolute values would be $|\rho|_p<1$. The way to do this is to see that there’s a $\Bbb Z_p$ constant $c$ such that $f=c\bar f$ where $\bar f$ has a unit coefficient. Then you apply Weierstrass Preparation to $\bar f$, and then Newton Polygon to get information about those roots, all of them integral over $\Bbb Z_p$.
However, the series may have a wider domain of convergence than the maximal ideal in the ring of integers of an algebraic closure of $\Bbb Q_p$. Here’s an example:
$$
f(x)=\sum_{n=0}^\infty p^{n^2}x^n\,.
$$
You see that the Newton Polygon has vertices at $(n,n^2)$ for every $n$, and the segments all have width $1$, and the slopes are all the odd positive numbers. You also see at a glance that this series converges for every $\alpha\in\overline{\Bbb Q_p}\,$, indeed throughout $\Bbb C_p$.
A suitable generalization of Weierstrass Preparation tells you that each segment of the Polygon corresponds to a root, every one of them not only algebraic over $\Bbb Q_p$, but in fact elements of $\Bbb Q_p$. And of course none of them in $\Bbb Z_p$.
