If $2\leq p\in\mathbb{N}$ and $f\in\mathbb{Q}_p[x]$, then ($f$ has infinitely many roots $\Leftrightarrow$ $f=0$) $\Leftrightarrow$ $p$ prime power.

Question

Is the statement below true?

If $2\leq p\in\mathbb{N}$ and $f\in\mathbb{Q}_p[x]$, then ($f$ has infinitely many roots $\Leftrightarrow$ $f=0$) $\Leftrightarrow$ $p$ prime power.

Answer 1

Well, if $s$ is not a prime power, then the ring of $s$-adic integers $\Bbb Z_s$ is isomorphic to a ring $\Bbb Z_\sigma\oplus\Bbb Z_\tau$, where $\sigma$ and $\tau$ are integers with $1<\sigma<s$ and $1<\tau<s$. Take the polynomial $f(X)=(1,0)X\in\Bbb Z_\sigma\oplus\Bbb Z_\tau$, then $f(0,a)=0$ for all the infinitely many $a\in\Bbb Z_\tau$.

And of course if $s=p^m$ for a prime $p$, then $\Bbb Z_s=\Bbb Z_p$, and the parenthesized equivalence is true.