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


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


but then




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


Title proof of theorem on equivalent valuations
Canonical name ProofOfTheoremOnEquivalentValuations
Date of creation 2013-03-22 14:55:40
Last modified on 2013-03-22 14:55:40
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 12
Author rspuzio (6075)
Entry type Proof
Classification msc 13A18