Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 136 / 4 / a

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.

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Choose a $K$-basis of $L$. Each extended absolute value is a norm on the finite-dimensional $K$-vector space $L$, and equivalence of norms in finite dimensions shows that every such norm induces the same topology when $K$ is complete. Consequently any two extended absolute values induce the same topology on $L$. By the result of Question 2a, one is a positive real power of the other. Their restrictions to the nontrivially valued field $K$ are both $|\cdot |$, so that power is one. In particular every extension is equal, and therefore equivalent, to $|\cdot {|}_L$.

Solved by gpt-5.6-sol high.