Two absolute values satisfy $|x|_1=|x|_2^t$ iff they satisfy $c_1|x|_1\leq |x|_2\leq c_2|x|_1$. Let $k$ be a field and $|\cdot|_1$, $|\cdot|_2$ be two absolute values on $k$. Consider the following propositions:
(1) There exists $t>0$ such that $|x|_1=|x|_2^t$ for all $x\in k$;
(2) There exists $c_1,c_2>0$ such that $c_1|x|_1\leq |x|_2\leq c_2|x|_1$ for all $x\in k$.
I know how to prove that (1) holds if and only if both absolute values induce the same topology. But two norms induce the same topology iff (2) holds. In other words, these propositions are equivalent.
I really think there should exist a simpler proof of this fact (with just some algebraic manipulations) but I don't know how. I appreciate any ideas!
 A: We assume that the field is non-trivially valued with valuation $|\cdot|$ defined on $k$.
Here is a sketch of a proof that proposition $(2)$ hold iff $|\cdot|_1$ and $|\cdot|_2$ define same topology on the non-trivially valued field $k$. 
Proof: It's easy to see that If proposition $(2)$ holds then the absolute values define the same topology. For the converse, Let $|\cdot|_1$ and $|\cdot|_2$ define the same topology on $k$. Since $|\cdot|$ is a non-trivial valuation on $k$, $\exists$ $\alpha\in k$ such that $|\alpha|>1$ and hence $\lim_{n\rightarrow\infty}|\alpha|^n=\infty$. Then every positive real number lies in between two consecutive powers of $|\alpha|$. Since the two absolute values define same topology on $k$, $\exists$ $r>0,s>0$ such that $B_2(0,r)\subset B_1(0,1)$ and $B_1(0,s)\subset B_1(0,1)$ ($B_{\epsilon}(v,\theta)$ is the open ball of radius $\theta$ and center $v\in k$ with respect to the norm $|\cdot|_{\epsilon}$ for $\epsilon\in\{1,2\}$). For any $v\neq0$ in $k$, $\exists$ $n\in\mathbb{N}$ such that $|\alpha|^n\leq\frac{|v|_1}{s}<|\alpha|^{n+1}$. Then $\left|\frac{v}{\alpha^{n+1}}\right|_1=\frac{|\alpha|_1}{|\alpha|^{n+1}}<s$. The $\left|\frac{v}{\alpha^{n+1}}\right|_2<1$ by assumption. Thus $|\alpha|_2<|\alpha|^{n+1}=|\alpha||\alpha|^n\leq\frac{|\alpha|}{s}|v|_1$. Setting $c_1=\frac{s}{|v|_1}$ we get $c_1|v|_2\leq|v|_1$. By similar methods one can show that $|v|_1\leq c_2|v|_2$ for $c_2=\frac{|\alpha|}{r}$. Hence we are done!
A: There is no such an absolute value relation in Proposition (2).

My motivation comes from noting that the relation in Proposition (2) has essentially two cases:

*

*$\exists c > 0;\;|x|_i = c|x|_j,$ with $i\neq j.$ [Proposition (21)]

*$\exists c_1, c_2 > 0;\;c_1|x|_1 < |x|_2 < c_2|x|_1.$ [ Proposition (22)]

Suppose that Proposition (1) holds, namely $\exists t > 0;\;|x|_2 = |x|_1^t,\;\forall x\in\kappa,$ and we're trying to show that (1) $\Rightarrow$ (2).

*

*If (21) happens, one is led, by considering absolute values of products, i.e. $c^2|x|_2|y|_2 = |x|_1|y|_1 = |xy|_1 = c|xy|_2 = c|x|_2|y|_2,\;\forall x, y\in\kappa,$ even in case of trivial absolute values, to $c^2 = c\Rightarrow c = 1,$ i.e. the two absolute values are the same, i.e. trivial equivalence.

*If (22) holds, then, $c_1|x|_1 < |x|_1^t < c_2|x|_1,$ for all $x\in\kappa.$ But this doesn't hold, e.g. when $x = 0_{\kappa}.$

P.S.

*

*where do you get this Proposition (2)? I would be grateful if you tell me a reference.


*I saw this relation in [1] and [2] but they are for norm and metric. I observe that an absolute value is essentially a metric but it comes with the extra condition of $|xy| = |x||y|.$

Refs.
[1] [pp. 157, 158]{Analysis I, Herbert Amann}
[2] https://en.wikipedia.org/wiki/Equivalence_of_metrics
