2
$\begingroup$

I'm having some trouble to understand the concept of valuations (and exponential valuations) and would be very grateful if one of you could tell me whether the following is right:

First of all, there's a difference between a valuation and an exponential valuation (which I also found in Neukirch).

An exponential valuation is defined as a map $v_p: K \rightarrow \mathbb{Z} \cup \{ \infty \}$ for a field $K$ with the properties that $v_p(x)= \infty \Leftrightarrow x=0$, $v_p(xy)=v_p(x)+v_p(y)$ and $v_p(x+y) \geq min \{ v_p(x), v_p(y)\}$ (1).

In contrast a valuation is given by a map $\vert . \vert: K \rightarrow \mathbb{R}_{\geq0}$ such that $|x|=0 \Leftrightarrow x=0$, $|xy|=|x||y|$ and $|x+y| \leq |x|+|y|$ or $\leq max \{ |x|,|y|\}$ if it's non-archimedean. (2)

Hence, if I have a map given by $$ |.|_{p,c}: K \rightarrow \mathbb{R}_{\geq0}, |x|_{p,c}= c^{v_p(x)} \text{if} ~x \neq 0 \text{ and } 0 \text{ if } x=0$$ for some $c \in (0,1)$, $p$ a non-zero prime ideal in the Dedekind domain $A$ and $K$ its quotient field and I am supposed to show that this defines a discrete non-archimedean valuation on $K$, I need to show the properties in (2) but can at the same time use the properties of (1) for the exponent? (The examples speaks of $v_p$ only as a discrete valuation. However, then I fail to show that $|.|_{p,c}$ is a valuation if $v_p$ weren't an exponential valuation...)

Thanks for your help and explanations!

  • 0
    I'm a bit confused. If $v_p(x)$ is not an exponential valuation, there's no reason why $|\cdot|$ should be a valuation2017-02-07
  • 0
    @Mathmo123 Thank you! That's exactly what I was wondering. Maybe they stated this already somewhere else in the example and I missed it :) Do you maybe know if an exponential valuation is always denoted by $v_p$ and a valuation by $|.|$?2017-02-07
  • 3
    Yes that is the usual notation (although I'm used to calling $v_p$ a valuation, and $|\cdot|$ an *absolute value*)2017-02-07
  • 1
    I find the terminology “exponential valuation” strange, since it’s just the negative log of the absolute value.2017-02-08
  • 0
    @Mathmo123 Thank you - it's much clearer now.2017-02-08
  • 0
    @Lubin Thank you - I will look this up!2017-02-08

0 Answers 0