Non-archimedean absolute value of a locally compact field is discrete?

Question

I want to prove that for a locally compact field $F$ equipped with a non-archimedean absolute value, that absolute value must be discrete.

Let $O$ be the valuation ring, $M$ its maximal ideal, then $O/M$ is finite, with this being finite I feel it could prove that the absolute value is discrete, but I somehow can't figure it out, help please.

Answer 1

$q = \inf_{ x \in F^*, |x| < 1 } |x|^{-1}$. If $q > 1$ then $\forall x \in F^*, |x| = q^n$ for some $n \in \mathbb{Z}$.

If $q = 1$ then $M = \{ x \in F, |x| < 1\}$ is open and non locally compact as well as $O^\times = O/M( 1+ M) = \{ x \in F, |x|= 1\} $.