Number Theory: An Introduction to Mathematics

264 VI Hensel’sp-adic Numbers

Proposition 2Let F be a field with an absolute value||. Then the following properties
are equivalent:

(i)| 2 |≤ 1 ;
(ii)|n|≤ 1 for every positive integer n;
(iii)the absolute value||is non-archimedean.

Proof It is trivial that(iii)⇒(i). Suppose now that (i) holds. Then| 2 k|=| 2 |k≤ 1
for any positive integerk. An arbitrary positive integerncan be written to the base 2
in the form

n=a 0 +a 12 +···+ag 2 g,

whereai∈{ 0 , 1 }for alli<gandag=1. Then

|n|≤|a 0 |+|a 1 |+···+|ag|≤g+ 1.

Now consider the powersnk.Sincen< 2 g+^1 ,wehavenk< 2 k(g+^1 )and hence

nk=b 0 +b 12 +···+bh 2 h,

wherebj∈{ 0 , 1 }for allj<h,bh=1andh<k(g+ 1 ). Thus

|n|k=|nk|≤h+ 1 ≤k(g+ 1 ).

Takingk-th roots and lettingk→∞, we obtain|n|≤1, sincek^1 /k=e(logk)/k→ 1
and likewise(g+ 1 )^1 /k=e(log(g+^1 ))/k→1. Thus(i)⇒(ii).
Suppose next that (ii) holds. Then, since the binomial coefficients are positive
integers,

|x+y|n=|(x+y)n|=

∣

∣

∣

∣

∑n

k= 0

(

n

k

)

xkyn−k

∣

∣

∣

∣

≤

∑n

k= 0

|x|k|y|n−k

≤(n+ 1 )ρn,

whereρ=max(|x|,|y|).Takingn-th roots and lettingn→∞, we obtain|x+y|≤ρ.
Thus(ii)⇒(iii).

It follows from Proposition 2 that for an archimedean absolute value the sequence
(|n|)is unbounded, since| 2 k|→∞ask→∞. Consequently, for anya,b∈Fwith
a=0, there is a positive integernsuch that|na|>|b|. The name ‘archimedean’
is used because of the analogy with the archimedean axiom of geometry. It follows
also from Proposition 2 that any absolute value on a field of prime characteristic is
non-archimedean, since there are only finitely many distinct values of|n|.