Suppose $L/K$ is a finite extension of fields. Suppose $v$ is a non-archimedean absolute value on $L$ such that the restriction of $v$ on $K$ is non-trivial and discrete. Can we say that $v$ is discrete on $L$?
By "discrete" on $K$, we mean that $v(K^{\times})$ is a discrete subset of $\mathbb{R}$.
Thank you!
The answer to your question is yes: let $\widehat{K}$ be the completion of $K$ with respect to the absolute value $v$. Then by the extension theory for absolute values we know that there exists a field compositum $L\widehat{K}$ within the algebraic closure of $\widehat{K}$, such that
$$v(x)=(v(N(x))^{1/n},$$
where $N$ is the norm function of the finite (!) extension $L\widehat{K}|\widehat{K}$, and $n$ is the degree of the minimal polynomial of $x$ over $\widehat{K}$. Since $n$ is bounded from above by $[L\widehat{K}:\widehat{K}]$, the set $v(L^\times)$ is discrete.