Let $K$ be a field complete with respect to a discrete valuation $v$. Let $K_s$ its separable closure, $w$ the unique valuation extending $v$, $(\mathcal{O}_{K_s}$, $\mathfrak{M})$ respectively the ring of integers and its maximal ideal of $K_s$.
Let $p \in \mathbb{Z}$ be a prime number such that $w(p)=1$, and take the formal sum:
$$[p](t)=\sum_{i \geq1} a_it^i=pt + (\text{higher order terms})$$
where $a_i \in \mathcal{O}_K$ for all $i $, $a_i \in \mathfrak{M}$ for $i<p^2$, and for $i=p^2$ holds $a_{p^2} \notin \mathfrak{M}$.
Assume there exists an $x\in \mathfrak{M}$ such that $[p](x)=0$, that is
$$px + a_2x^2 + \cdots+a_{p^2}x^{p^2}+ \cdots = 0 $$
My question is: how can I compare the valuation to know the valuation of $x$?
The reasoning I done is the following:
$$w(px + \cdots + a_{p^2-1}x^{p^2-1})= w(a_{p^2}x^{p^2}+\cdots)$$ so since $a_i\in O_K$ then the first term became $1+w(x)$ while the second $p^2 w(x)$ and we finally get $$w(x)=\frac{1}{p^2-1}$$ Is it correct? I assume that the valuation of the sum is always the minimum. However I know this is true for a discrete valuation while $w: K_s \longrightarrow \mathbb{Q}\cup \{\infty\} $