Zero of a power series in a local field

Question

Let $f(z)=\sum_{n\ge1}a_nz^n$ be a power series of $\mathbb C_p[[z]]$ where the $a_n$ are such that $|a_n|=1$ for every positive integer $n$. Consider $z_0\in\mathbb C_p$ such that $|z_0|<1$. Can one assert that there exist $x_0\in \mathbb C_p$ with $|x_0|<1$ such that $f(x_0)=z_0$.

Answer 1

Sure, and all you need is $|a_1| = 1$ and $|a_n| \leq 1$ for $n \geq 2$.

Let $K$ be a field complete with respect to a nonarchimedean absolute value $|\cdot|$ and assume $a_1 \in \mathcal O_K^\times$ and $a_n \in \mathfrak m_K$ for $n > 1$. It doesn't have to be $\mathbf C_p$.

For $z_0 \in \mathfrak m_K$ we want to show the equation $f(z) = z_0$ has a solution in $\mathfrak m_K$. Let $g(z) = f(z) - z_0$, so $g(z)$ is a power series in $\mathcal O_K[[z]]$ with constant term $-z_0$. We have $g(0) = -z_0$ and $g'(0) = a_1$, so $|g(0)| = |-z_0| < 1$ and $|g'(0)| = |a_1| = 1$. If $g(z)$ were a polynomial then you'd know $g(z)$ has a zero in $\mathfrak m_K$ by Hensel's lemma. Well, Hensel's lemma works for power series in $\mathcal O_K[[z]]$ evaluated on $\mathfrak m_K$ in the same way as it does with polynomials in $\mathcal O_K[z]$ evaluated on $\mathcal O_K$: see Theorem 10.7 here. You just need to be a little more careful with convergence issues since you're evaluating a power series rather than a polynomial. By Hensel's lemma for power series, the conditions $|g(0)| < 1$ and $|g'(0)| = 1$ implies there is some $x_0 \in \mathfrak m_K$ such that $g(x_0) = 0$, so $f(x_0) = z_0$.