# equivalent discrete absolute values

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?

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.