Can the image of the absolute value of a locally compact ultrametric field contain an interval in $\mathbb{R}$? Let $k$ be a field equipped with a nonarchimedean absolute value
$$|\cdot| : k\rightarrow\mathbb{R}$$
w.r.t which it is locally compact.
Is it possible for the image of $|\cdot|$ to contain a nonempty interval $(a,b)\subset\mathbb{R}$?
In all the cases I can imagine, this is impossible - for example, even for $\overline{\mathbb{Q}_p}$, the image is $\mathbb{Q}\subset\mathbb{R}$.
 A: No, the range of a non-Archimedean absolute value on a locally compact field
can only be $\{0\} \cup \{ q^n \mid n \in \mathbb{Z} \}$ for some $q\in \mathbb{R}^+$.
The restriction $|\cdot| : k^* \to \mathbb{R}^+$ is a group homomorphism,
so its range must be a subgroup of $\mathbb{R^+}$. Since $\mathbb{R}^+$ is
an Archimedean ordered group, any subgroup is either trivial or cyclic 
or everywhere dense. To rule out the latter case it 
is sufficient to show that there is an open interval that contains no value of $|\cdot|$.
Assuming $k$ is locally compact w.r.t. $|\cdot|$, there is a compact neighbourhood $V$ of $0$ and an $\varepsilon > 0$ such that $B_\varepsilon(0) \subset V$. Since open balls in ultrametric spaces
are closed sets, $B_\varepsilon(0)$ is also compact. The absolute value,
being a continuous real-valued function, therefore attains a maximum on it,
say $\delta$, which must be less than $\varepsilon$. It follows that there
is no $x \in k$ with $|x| \in (\delta, \varepsilon)$, QED.
