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

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 136 4 a
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i
One direction follows by restriction. Conversely, suppose ∣⋅∣ is a Non-Archimedean absolute value. Then ∣m∣L​=∣m∣≤1 for every integer m. For x,y∈L, the binomial theorem and the ordinary triangle inequality give
∣x+y∣Ln​≤∑j=0n​(jn​)∣x∣Lj​∣y∣Ln−j​≤(n+1)max(∣x∣L​,∣y∣L​)n.
(1)
Taking nth roots and letting n→∞ yields the ultrametric inequality for ∣⋅∣L​. Thus an extension of an absolute value is non-Archimedean exactly when its restriction is.

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