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I'm stuck in the middle of a proof on page 19/20 of Local Fields, by Cassels.

Background: Definition: A valuation is a real-valued function $|.|$ on a field $k$ such that (1)$|b|\geq 0$ with equality only for $b=0$. (2)$|bc|=|b||c|$ for all $b,c\in k$. (3)There is some $C \in \mathbb{R}$ such that $|b|\leq 1 \Rightarrow |1+b| \leq C$.

Theorem: Suppose that $|.|_1$ and $|.|_2$ are two valuations on a field $k$. Suppose that $|.|_1$ is non-trivial and that $ |a|_1<1 \Rightarrow |a|_2 <1. $ I want to prove that these valuations are equivalent.

WHERE I GET STUCK We know that for b,c be non-zero elements of $k$ and $m,n \in \mathbb{Z}$ $ m \log |b|_1+n\log |c|_1 >0 \Rightarrow m \log |b|_2+n\log |c|_2>0$

$m \log |b|_1+n\log |c|_1 =0 \Rightarrow m \log |b|_2+n\log |c|_2=0$

$ m \log |b|_1+n\log |c|_1 <0 \Rightarrow m \log |b|_2+n\log |c|_2<0$

Then apparently given $|c|_1 \not=1$, "it readily follows that

$\log |b|_1=\lambda \log |b|_2$ where

$\lambda=\frac{\log|c|_1}{\log |c|_2}$".

I am not finding this result forthcoming.

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Maybe I'm missing an easier argument, but I think the point is that since $|c|_1 \neq 1$ we can find a $\mu \geq 0$ such that $|b|_1 = |c|_1^\mu$. Let's assume that $|c|_1 > 1$. If $p/q > \mu$ is a rational number then $|b|_1 < |c|_1^{p/q}$ and hence $|b^q/c^p|_1 < 1$, so your third inequality implies that $|b^q/c^p|_2 < 1$ and hence $|b|_2 < |c|_2^{p/q}$. Similarly, you can look at rational numbers $< \mu$, and we end up with $|b|_2 = |c|_2^\mu$. Does that help?

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That helps a lot, thanks! (Although it drops out a little easier if you start with $\log|b|_1=\mu \log |c|_1$, since you don't have to separate cases $|c|_1>1$ and $|c|_1<1$). –  CruiskeenLawn Jan 18 '12 at 8:47
    
@CruiskeenLawn Good point! I neglected that completely. Will edit in a bit. –  Dylan Moreland Jan 18 '12 at 14:24

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