Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have recently been reading about non-Archimedean metrics on fields (in Koblitz: $p$-adic Numbers, $p$-adic Analysis, and Zeta-Functions), and came across the exercise:

Prove that a norm $\|.\|$ on a field $F$ is non-Archimedean if and only if $$\{x\in F : \|x\| < 1 \} \cap \{x\in F : \|x-1\| < 1 \} = \emptyset.$$

In one direction, the proof was trivial, and in the other direction somewhat harder, but my question is really about where this question comes from. If I were trying to think up exercises on this topic, I don't think I would have thought of this one in a million years.

I am (gradually) getting used to the "eccentricities" of non-Archimedean metrics, but if someone could give me some idea of the intuition that lies behind this particular property, I would be grateful.

share|improve this question
2  
$U^1 = \{x \in F : |x-1| < 1\}$ is an important group in number theory. I'm not sure what it's called in English, in German it's Einseinheiten. The norm residue symbol $(-,E/F)$ maps it surjectively onto the ramification group of $E/F$. It's a subgroup of the units of $\mathcal{O}_F$ such that $\mathcal{O}_F^{\times} / U^1 \cong \kappa^{\times}$ where $\kappa$ is the remainder field. Also, by local class field theory it's contained in the norm group $N_{E/F}E^{\times}$ if and only if $E/F$ is tamely ramified. –  Cocopuffs Jul 17 '12 at 11:28
    
Wow. It might take me a while to digest that! Thanks. –  Old John Jul 17 '12 at 11:31
    
So the intuition I think of this problem is to see that such one-units are really units in the normal since. –  Cocopuffs Jul 17 '12 at 11:35
    
Perhaps the down-voter would be good enough to give me a clue as to what is wrong with my question? –  Old John Dec 5 '13 at 18:48
add comment

1 Answer

up vote 5 down vote accepted

The defining property of ultrametrics is that in every triangle two longer sides are equal: more precisely, if ABC is a triangle (=triple of points) and $|AB|\ge |BC|\ge |AC|$ then $|AB|=|BC|$. Now, the exercise asks about the existence of a triangle in which one side has length 1 while the other two are strictly shorter: designed to be a contradiction to the definition.

share|improve this answer
    
Thanks, that makes it very clear (and I am annoyed that I didn't see it in that light myself!). So, would I be right in saying that the points $0$ and $1$ don't really have any special significance here, apart from being two elements that are bound to exist in a field? –  Old John Jul 17 '12 at 11:39
2  
@OldJohn It also matters that the distance between them is 1. –  user31373 Jul 17 '12 at 11:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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