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单词 ProofOfTheoremOnEquivalentValuations
释义

proof of theorem on equivalent valuations


It is easy to see that || and ||c are equivalent valuations for any constant c>0 — it follows from the fact that 0xc<1 if and only if 0<x1.

Assume that the valuationsMathworldPlanetmathPlanetmath ||1 and ||2 are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath.Let b be an element of K such that 0<|b|1<1. Because the valuations are assumed to be equivalent, it is also the case that 0<|b|2<1. Hence, there must exist positive constants c1 and c2 such that |b|1c1=12 and |b|2c2=12.

We will show that show that |x|1c1=|x|2c2 for all aK by contradictionMathworldPlanetmathPlanetmath.

Let a be any element of k such that 0<|a|1<1. Assume that |a|1c1|a|2c2. Then either |a|1c1<|a|2c2 or |a|1c1>|a|2c2. We may assume that |a|1c1<|a|2c2 without loss of generality.

Since |a|2c2/|a|1c1>1, there exists an integer m>0 such that (|a|2c2/|a|1c1)m>2. Let n be the least integer such that 2n|a|2mc2>1. Then we have

2n|a|1mc1<2n-1|a|2mc2<1<2n|a|2mc2.

Since 2=|b-1|1c1=|b-1|2c2, this implies that

|ambn|1c1<1<|ambn|2c2,

but then

|ambn|1<1

and

|ambn|2>1,

which is impossible because the two valuations are assumed to be equivalent.

Q.E.D

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更新时间:2025/5/5 1:33:09