proof of theorem on equivalent valuations

It is easy to see that $|\cdot|$ and $|\cdot|^{c}$ are equivalent valuations for any constant $c>0$ — it follows from the fact that $0\leq x^{c}<1$ if and only if $0<x\leq 1$ .

Assume that the valuations $|\cdot|_{1}$ and $|\cdot|_{2}$ are equivalent. 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 $c_{1}$ and $c_{2}$ such that $|b|_{1}^{c_{1}}={1\over 2}$ and $|b|_{2}^{c_{2}}={1\over 2}$ .

We will show that show that $|x|_{1}^{c_{1}}=|x|_{2}^{c_{2}}$ for all $a\in K$ by contradiction.

Let $a$ be any element of $k$ such that $0<|a|_{1}<1$ . Assume that $|a|_{1}^{c_{1}}\neq|a|_{2}^{c_{2}}$ . Then either $|a|_{1}^{c_{1}}<|a|_{2}^{c_{2}}$ or $|a|_{1}^{c_{1}}>|a|_{2}^{c_{2}}$ . We may assume that $|a|_{1}^{c_{1}}<|a|_{2}^{c_{2}}$ without loss of generality.

Since $|a|_{2}^{c_{2}}/|a|_{1}^{c_{1}}>1$ , there exists an integer $m>0$ such that $(|a|_{2}^{c_{2}}/|a|_{1}^{c_{1}})^{m}>2$ . Let $n$ be the least integer such that $2^{n}|a|_{2}^{mc_{2}}>1$ . Then we have

2^{n}|a|_{1}^{mc_{1}}<2^{n-1}|a|_{2}^{mc_{2}}<1<2^{n}|a|_{2}^{mc_{2}}.

Since $2=|b^{-1}|_{1}^{c_{1}}=|b^{-1}|_{2}^{c_{2}}$ , this implies that

\left|{a^{m}\over b^{n}}\right|_{1}^{c_{1}}<1<\left|{a^{m}\over b^{n}}\right|_% {2}^{c_{2}},

but then

\left|{a^{m}\over b^{n}}\right|_{1}<1

and

\left|{a^{m}\over b^{n}}\right|_{2}>1,

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

Q.E.D

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