degree of algebraic number

degree of algebraic number

Theorem.

The degree (

http://planetmath.org/DegreeOfAnAlgebraicNumber

) of any

algebraic number

α

in the

number field

ℚ

⁢

(

ϑ

)

divides the degree of

ϑ

.  The zeroes of the

characteristic polynomial

g

⁢

(

x

)

of

α

consist of the

algebraic conjugates

of

α

, each of which having equal

multiplicity

as zero of

g

⁢

(

x

)

.

Proof.

Let the

minimal polynomial

of

α

be

a

⁢

(

x

)

:=

x

k

+

a

1

⁢

x

k

-

1

+

…

+

a

k

and all zeroes of this be

α

1

=

α

,

α

2

,

…

,

α

k

.  Denote the

canonical polynomial

of

α

with respect to the

primitive element

(

http://planetmath.org/SimpleFieldExtension

)

ϑ

by

r

⁢

(

x

)

; then

a

⁢

(

r

⁢

(

ϑ

)

)

=

a

⁢

(

α

)

=

0

.

If

a

⁢

(

r

⁢

(

x

)

)

:=

φ

⁢

(

x

)

,  then the

equation

φ

⁢

(

x

)

=

0

has

rational

coefficients

and is satisfied by

ϑ

.  Since the minimal polynomial

f

⁢

(

x

)

of

ϑ

is

irreducible

(

http://planetmath.org/IrreduciblePolynomial

), it must divide

φ

⁢

(

x

)

and all algebraic conjugates

ϑ

1

=

ϑ

,

ϑ

2

,

…

,

ϑ

n

of

ϑ

make

φ

⁢

(

x

)

zero.  Hence we have

a

⁢

(

α

(

i

)

)

=

a

⁢

(

r

⁢

(

ϑ

i

)

)

=

0

for

i

=

1

,

2

,

…

,

n

where the numbers

α

(

i

)

are the

ℚ

⁢

(

ϑ

)

-

conjugates

(

http://planetmath.org/CharacteristicPolynomialOfAlgebraicNumber

) of

α

.  Thus these

ℚ

⁢

(

ϑ

)

-conjugates are roots of the irreducible equation

a

⁢

(

x

)

=

0

, whence

a

⁢

(

x

)

must divide the characteristic polynomial

g

⁢

(

x

)

.  Let the power (

http://planetmath.org/GeneralAssociativity

)

[

a

⁢

(

x

)

]

m

exactly divide

g

⁢

(

x

)

, when

g

⁢

(

x

)

=

[

a

⁢

(

x

)

]

m

⁢

b

⁢

(

x

)

,

a

⁢

(

x

)

∤

b

⁢

(

x

)

.

Antithesis:

deg

⁢

(

b

⁢

(

x

)

)

≧

1

and

b

⁢

(

β

)

=

0

.

This implies that

g

⁢

(

β

)

=

0

,  i.e.

β

is one of the numbers

α

(

i

)

.  Therefore,

β

were a zero of

a

⁢

(

x

)

and thus

a

(

x

)

∣

b

(

x

)

, which is impossible.  Consequently,the antithesis is wrong, i.e.

b

⁢

(

x

)

is a constant, which must be 1 because

g

⁢

(

x

)

and

a

⁢

(

x

)

are

monic polynomials

.  So,

g

⁢

(

x

)

=

[

a

⁢

(

x

)

]

m

.  Since

a

⁢

(

x

)

=

(

x

-

α

1

)

⁢

(

x

-

α

2

)

⁢

⋯

⁢

(

x

-

α

k

)

,

it follows that

g

⁢

(

x

)

=

(

x

-

α

1

)

m

⁢

(

x

-

α

2

)

m

⁢

⋯

⁢

(

x

-

α

k

)

m

.

Hence

k

⁢

m

=

n

and

k

divides

n

, as asserted.  Moreover, each

α

j

is a zero of order

m

of

g

⁢

(

x

)

, i.e. appears among the roots

α

(

1

)

,

α

(

2

)

,

…

,

α

(

n

)

of the equation

g

⁢

(

x

)

=

0

m

times.

Title

degree of algebraic number

Canonical name

DegreeOfAlgebraicNumber

Date of creation

2013-03-22 19:08:51

Last modified on

2013-03-22 19:08:51

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11R04

Classification

msc 11C08

Classification

msc 12F05

Classification

msc 12E05

Related topic

KConjugates

Related topic

CharacteristicPolynomialOfAlgebraicNumber

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