equivalent discrete absolute values

Question

If an absolute value $|\cdot|$ on a field $K$ is discrete, then the value group $|K^*|$ is a discrete subgroup of $\mathbb{R}_{>0}$. Hence $|K^*|=\lambda^{\mathbb{Z}}=\{\lambda^n \mid n \in \mathbb{Z}\}$ for some $0<\lambda<1$.

Therefore every discrete $|\cdot|$ corresponds to a discrete valuation $v:K \to \mathbb{Z}$ defined by $v(x):=\log_{\lambda}|x|$ for $x \neq 0$ and $v(x)=\infty$ for $x=0$.

Question: Show that if two discrete absolute values $|\cdot|$ and $||\cdot||$ are equivalent (that is, $||\cdot||=|\cdot|^c$ for some $c>0$), then they correspond to the same $v$ but with different $\lambda$.

Say $||K^*||=\lambda^{\mathbb{Z}}$ and $|K^*|=\rho^{\lambda}$ and $||\cdot||=|\cdot|^c$. Then they correspond to the same $v$ if and only if $c=\log_{\rho}\lambda$. How can I show this?

I have tried many ways but always ended up with an identity. Can anyone give me a hint?

Answer 1

Maybe I misunderstand the question, but if you have that

$||\cdot||=|\cdot|^c$ for some $c>0$

and $||\cdot||=\lambda^{v(\cdot)}$ with $v:K \rightarrow \mathbb Z$ a discrete valuation, then obviously

$|\cdot| = ||\cdot||^{1/c} = \lambda^{\frac1c v(\cdot)} = (\lambda^{\frac1c})^{v(\cdot)}$

and thus $|\cdot|$ corresponds to the same $v$ but with $\lambda^{1/c}$ instead of $\lambda$. And conversely, if

$||\cdot|| = \lambda^{v(\cdot)}$

and

$|\cdot| = \rho^{v(\cdot)}$ (you write $\rho^\lambda$ which makes no sense, maybe that's the source of confusion?)

with the same valuation $v$, then $||\cdot||=|\cdot|^c$ with $c := \log_{\rho}(\lambda)$ because by definition, $\rho^c = \lambda$. That's all there is to show.