 Solution

One direction follows by restriction. Conversely, suppose $|\cdot |$ is a Non-Archimedean absolute value. Then $|m{|}_L=|m|\le 1$ for every integer $m$. For $x,y\in L$, the binomial theorem and the ordinary triangle inequality giveTaking $n$th roots and letting $n\to \infty$ yields the ultrametric inequality for $|\cdot {|}_L$. Thus an extension of an absolute value is non-Archimedean exactly when its restriction is.

Solved by gpt-5.6-sol high.