ultrametric triangle inequality

ultrametric triangle inequality

Theorem 1

.

Let

K

be a field and

G

an

ordered group equipped with zero

.  Suppose that the function

|

⋅

|

:

K

→

G

satisfies the postulates 1 and 2 of

Krull valuation

.  Then the

non-archimedean

or

ultrametric triangle inequality

3.

|

x

+

y

|

≦

max

⁡

{

|

x

|

,

|

y

|

}

in the field is with the condition

(*)

|

x

|

≦

1

⇒

|

x

+

1

|

≦

1

.

Proof.

The value

y

=

1

in the ultrametric triangle inequality gives the (*) as result.  Secondly, let’s assume the condition (*).  Let

x

and

y

be non-zero elements of the field

K

(if

x

⁢

y

=

0

then 3 is at once verified), and let e.g.

|

x

|

≦

|

y

|

.  Then we get

|

x

y

|

=

|

x

|

⋅

|

y

|

-

1

≦

1

,  and thus according to (*),

|

x

+

y

|

⋅

|

y

|

-

1

=

|

x

+

y

y

|

=

|

x

y

+

1

|

≦

1

.

So we see that

|

x

+

y

|

≦

|

y

|

=

max

⁡

{

|

x

|

,

|

y

|

}

.

Theorem 2

.

The Krull valuation (and any

non-archimedean valuation

(

http://planetmath.org/

Valuation

))

|

⋅

|

of the field

K

satisfies the sharpening

|

x

+

y

|

=

max

⁡

{

|

x

|

,

|

y

|

}

for

⁢

|

x

|

≠

|

y

|

of the ultrametric triangle inequality.

Proof.

Let e.g.

|

x

|

>

|

y

|

.  Surely

|

x

+

y

|

≦

|

x

|

,  but also

|

x

|

=

|

(

x

+

y

)

-

y

|

≦

max

⁡

{

|

x

+

y

|

,

|

y

|

}

;  this maximum is

|

x

+

y

|

since otherwise one would have

|

x

|

≦

|

y

|

.  Thus the result is:

|

x

+

y

|

=

|

x

|

.

Note.

The metric defined by a non-archimedean valuation of the field

K

is the

ultrametric

of

K

.  Theorem 2 implies, that every triangle of

K

with vertices

A

,

B

,

C

(

∈

K

) is isosceles:  if

|

B

-

C

|

≠

|

C

-

A

|

,  then

|

A

-

B

|

=

max

⁡

{

|

B

-

C

|

,

|

C

-

A

|

}

.

Theorem 3

.

The valuation (

http://planetmath.org/Valuation

)

|

⋅

|

:

K

→

ℝ

of the field

K

is

archimedean

if and only if the set

{

|

1

|

,

|

1

+

1

|

,

|

1

+

1

+

1

|

,

…

}

of the “values” of the

multiples

of the unity is not

bounded

.

Proof.

If

|

⋅

|

is non-archimedean, then

|

n

⋅

1

|

=

|

1

+

…

+

1

|

≦

max

⁡

{

|

1

|

}

=

1

,  and the multiples are bounded.  Conversely, let

|

n

⋅

1

|

<

M

⁢

∀

n

∈

ℤ

+

.  Now one obtains, when

|

x

|

≦

1

:

|

x

+

1

|

n

≦

∑

j

=

0

n

|

(

n

j

)

|

⋅

|

x

|

j

<

(

n

+

1

)

⁢

M

,

or

|

x

+

1

|

<

(

n

+

1

)

⁢

M

n

for all

n

.  As

n

tends to

infinity

, this

n

th

root has the limit 1.  Therefore one gets the limit

inequality

|

x

+

1

|

≦

1

,  i.e. the valuation is non-archimedean.

References

1

Emil Artin

:

Theory of

Algebraic Numbers

.  Lecture notes.  Mathematisches Institut, Göttingen (1959).

Title

ultrametric triangle inequality

Canonical name

UltrametricTriangleInequality

Date of creation

2013-03-22 14:54:15

Last modified on

2013-03-22 14:54:15

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

25

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13F30

Classification

msc 13A18

Classification

msc 12J20

Classification

msc 11R99

Related topic

MaximalNumber

Related topic

PAdicCanonicalForm

Related topic

UltrametricSpace

Related topic

MinimalAndMaximalNumber

Related topic

ExponentValuation2

Defines

non-archimedean triangle inequality

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