determining the continuations of exponent

determining the continuations of exponent

Task.

Let

ν

0

be the 3-adic (triadic) (

http://planetmath.org/PAdicValuation

)

exponent valuation

of the field

ℚ

of the

rational numbers

and let

𝔬

be the ring of the

exponent

.  Determine the

integral closure

𝔒

of

𝔬

in the

extension field

ℚ

⁢

(

-

5

)

and the

continuations

of

ν

0

to this field.

The triadic exponent (

http://planetmath.org/ExponentValuation

) of

ℚ

at any non-zero rational number

3

n

⁢

u

v

, where

u

and

v

are

integers

not

divisible

by 3, is defined as

ν

0

⁢

(

3

n

⁢

u

v

)

:=

n

.

Any number of the

quadratic field

ℚ

⁢

(

-

5

)

is of the form

r

+

s

⁢

-

5

with

r

and

s

rational numbers.  When

α

=

r

+

s

⁢

-

5

belongs to

𝔒

, the rational

coefficients

of the

quadratic equation

x

2

-

2

⁢

r

⁢

x

+

(

r

2

+

5

⁢

s

2

)

=

0

,

satisfied by

α

, belong to the ring

𝔬

, whence one has

ν

0

⁢

(

-

2

⁢

r

)

≧

0

,

ν

0

⁢

(

r

2

+

5

⁢

c

2

)

≧

0

.

The first of these

inequalities

implies that

ν

0

⁢

(

r

)

≧

0

since

-

2

is a unit of

𝔬

.  As for

s

, if one had

ν

0

⁢

(

s

)

<

0

,  then

ν

0

⁢

(

5

⁢

s

2

)

=

2

⁢

ν

0

⁢

(

s

)

<

0

,  and therefore one had

ν

0

⁢

(

r

2

+

5

⁢

s

2

)

=

min

⁡

{

ν

0

⁢

(

r

2

)

,

ν

0

⁢

(

5

⁢

s

2

)

}

<

0

.

Thus we have to have

ν

0

⁢

(

s

)

≧

0

,  too.  So we have seen that for

r

+

s

⁢

-

5

∈

𝔒

,  it’s necessary that

r

,

s

∈

𝔬

.  The last condition is, apparently, also

sufficient

.  Accordingly, we have obtained the result

𝔒

=

{

r

+

s

-

5

⋮

r

,

s

∈

𝔬

}

.

Since the degree (

http://planetmath.org/Degree

) of the field extension

ℚ

⁢

(

-

5

)

/

ℚ

is 2, the exponent

ν

0

has, by the

theorem

in the parent entry (

http://planetmath.org/TheoremsOnContinuation

), at most two continuations to

ℚ

⁢

(

-

5

)

.  Moreover, the same entry (

http://planetmath.org/TheoremsOnContinuation

) implies that the

intersection

of the rings of those continuations coincides with

𝔒

, whose non-associated (

http://planetmath.org/

Associate

)

prime elements

determine the continuations in question.

We will show that there are exactly two of those continuations and that one may choose e.g. the

conjugate

numbers

π

1

:=

1

+

-

5

,

π

2

:=

1

-

-

5

for such prime elements.

Suppose that

π

1

splits in

𝔒

into factors (

http://planetmath.org/DivisibilityInRings

) as

π

1

=

α

⁢

β

where

α

=

a

0

+

a

1

⁢

-

1

,

β

=

b

0

+

b

1

⁢

-

5

(

a

i

,

b

i

∈

𝔬

).  Then also

π

2

=

α

′

⁢

β

′

where

α

′

=

a

0

-

a

1

⁢

-

1

,

β

′

=

b

0

-

b

1

⁢

-

5

.  We perceive that

π

1

⁢

π

2

=

6

=

α

⁢

α

′

⋅

β

⁢

β

′

=

(

a

0

2

+

5

⁢

a

1

2

)

⁢

(

b

0

2

+

5

⁢

b

1

2

)

,

but according to the entry

ring of exponent

, the only

prime numbers

of

𝔬

are the associates of 3.  Now we have factorised the prime number

6

of

𝔬

into a

product

of two factors (

http://planetmath.org/Product

)

α

⁢

α

′

and

β

⁢

β

′

, and consequently, e.g.

α

⁢

α

′

is a unit of

𝔬

and hence of

𝔒

, too.  Thus

α

and

α

′

are units of

𝔒

, which means that

π

1

and

π

2

have only

trivial

factors.  The numbers

π

1

and

π

2

themselves are not units, because

1

1

±

-

5

=

1

6

∓

1

6

⁢

-

5

∉

𝔒

;

π

1

and

π

2

are not associates of each other, since

π

1

π

2

=

1

+

1

3

⁢

-

5

∉

𝔒

.  So

π

1

and

π

2

are non-associated prime elements of

𝔒

.  This ring has no other prime elements non-associated with both

π

1

and

π

2

, because otherwise

ν

0

would have more than two continuations.

According to the entry ring of exponent (

http://planetmath.org/RingOfExponent

), any non-zero

element of

the field

ℚ

⁢

(

-

5

)

is uniquely in the form

ξ

=

ε

⁢

π

1

m

⁢

π

2

n

,

with

ε

a unit of

𝔒

and

m

,

n

integers.  The both continuations

ν

1

and

ν

2

of the triadic exponent

ν

0

are then determined as follows:

ν

1

⁢

(

ξ

)

=

m

,

ν

2

⁢

(

ξ

)

=

n

.

Title

determining the continuations of exponent

Canonical name

DeterminingTheContinuationsOfExponent

Date of creation

2013-03-22 18:00:16

Last modified on

2013-03-22 18:00:16

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Example

Classification

msc 11R99

Classification

msc 13A18

Classification

msc 12J20

Classification

msc 13F30

Related topic

ExampleOfRingWhichIsNotAUFD

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