= Solution
One direction follows by restriction. Conversely, suppose $|\cdot|$ is a <Non-Archimedean absolute value>. Then $|m|_L=|m|\leq1$ for every integer $m$. For $x,y\in L$, the binomial theorem and the ordinary triangle inequality give
$$
|x+y|_L^n
\leq\sum_{j=0}^n\binom nj|x|_L^j|y|_L^{n-j}
\leq(n+1)\max(|x|_L,|y|_L)^n.
$$
Taking $n$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.
Back to article page
