# 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.

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 ProofOfTheoremOnEquivalentValuations 2013-03-22 14:55:40 2013-03-22 14:55:40 rspuzio (6075) rspuzio (6075) 12 rspuzio (6075) Proof msc 13A18