Absolute Values
Let K be a field. An absolute value v on K is a real-valued function x ↦ |x| v on K satisfying the following three properties: AV 1 We have |x| v ≧ 0 for all x ∊ K, and |x| v = 0 if and only if x = 0. AV 2 For all x, y ∊ K, we have |xy| v = |x| v |y| v .
- PDF / 3,099,026 Bytes
- 35 Pages / 439 x 666 pts Page_size
- 11 Downloads / 257 Views
XII
Absolute Values
§1.
DEFINITIONS, DEPENDENCE, AND INDEPENDENCE
Let K be a field. An absolute value v on K is a real-valued function x f---+ Ix Iv on K satisfying the following three properties : AV 1.
We have Ixl v ~ 0 for all x E K , and [x], = 0 if and only if x =
AV 2.
For all x, yE K , we have
AV 3.
For all x, y E K , we have [x
o.
Ixylv = Ixlv lylv '
+ ylv ~
[x] ,
+ IYlv'
If instead of AV 3 the absolute value satisfies the stronger condition AV 4.
[x
+ ylv ~
max(lxl v, Iylv)
then we shall say that it is a valuation, or that it is non-archimedean. The absolute value which is such that Ix], = 1 for all x =1= 0 is called trivial. We shall write Ix Iinstead of Ix Ivif we deal with just one fixed absolute value . We also refer to v as the absolute value. An absolute value of K defines a metric. The distance between two elements x, y of K in this metric is Ix - y I. Thus an absolute value defines a topology on K . Two absolute value s are called dependent if they define the same topology. If they do not , they are called ind ependent. We observe that III = 11 21 = I( -1) 2 1= 111 2 whence
111=1-11=1. Also, I - x I = Ix I for all x E K, and IX- I I = Ix 1- 1 for x
=1=
o. 465
S. Lang, Algebra © Springer Science+Business Media LLC 2002
466
ABSOLUTE VALUES
XII, §1
Proposition 1.1. Let I II and I 12 be non-trivial absolute values on afield K. They are dependent if and only if the relation
implies Ix 12 < 1. If they are dependent, then there exists a number A > 0 such that [x ], = I xl~ for a ll x E K. Proof If the two absolute values are dependent, then our condition is satisfied , because the set of x E K such that Ix I, < I is the same as the set such that lim x" = 0 for n -+ co. Conversely, assume the condition satisfied. Then [x I, > I implies [x Iz > I since Ix-'I, < 1. By hypothesis, there exists an element xoEK such that IXo/1> 1. Let a = IX o/ I and b = IX oI2 . Let
A = log b. log a Let x E K, x # o. Then IxI, = IXoI~ for some number IX. If m,n are integers such that min> IX and n > 0, we have
Ixll > IXoIT/n whence
and thus n
Ix lxO'I2 < 1. This implies that Ix 12
Data Loading...
