1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.