proof of norm and trace of algebraic number

proof of norm and trace of algebraic number

Theorem 1

.

Let

K

be a

number field

and

α

∈

K

.  The norm

N

⁢

(

α

)

and the trace

T

⁢

(

α

)

of

α

in the

field extension

K

/

Q

both are

rational numbers

and especially

rational integers

in the case

α

is an

algebraic integer

.  If

β

is another element of

K

, then

N

⁢

(

α

⁢

β

)

=

N

⁢

(

α

)

⁢

N

⁢

(

β

)

and

T

⁢

(

α

+

β

)

=

T

⁢

(

α

)

+

T

⁢

(

β

)

.   If

[

K

:

Q

]

=

n

and

a

∈

Q

, then

N

⁢

(

a

)

=

a

n

and

T

⁢

(

a

)

=

n

⁢

a

.

Before proving this

theorem

, a lemma will be stated and proven.

Lemma

.

Let

K

be a number field with

[

K

:

Q

]

=

n

,

α

∈

K

such that

[

Q

(

α

)

:

Q

]

=

d

, and

N

*

⁢

(

α

)

and

T

*

⁢

(

α

)

denote the

absolute norm

(

http://planetmath.org/AbsoluteNorm

) and absolute trace (

http://planetmath.org/AbsoluteTrace

) of

α

, respectively.   Then

d

divides

n

,

N

⁢

(

α

)

=

(

N

*

⁢

(

α

)

)

n

d

,

and

T

⁢

(

α

)

=

n

d

⁢

T

*

⁢

(

α

)

.

Proof.

Note that

d

divides

n

because

n

=

[

K

:

ℚ

]

=

[

K

:

ℚ

(

α

)

]

[

ℚ

(

α

)

:

ℚ

]

=

[

K

:

ℚ

(

α

)

]

d

.

Note also that each of the

d

embeddings

of

ℚ

⁢

(

α

)

into

ℂ

extends to exactly

n

d

embeddings of

K

into

ℂ

. Thus,

N

⁢

(

α

)

=

∏

σ

⁢

emb. of

⁢

K

σ

⁢

(

α

)

=

∏

σ

⁢

emb. of

⁢

K

σ

|

ℚ

⁢

(

α

)

⁢

(

α

)

=

(

∏

τ

⁢

emb. of

⁢

ℚ

⁢

(

α

)

τ

⁢

(

α

)

)

n

d

=

(

N

*

⁢

(

α

)

)

n

d

and

T

⁢

(

α

)

=

∑

σ

⁢

emb. of

⁢

K

σ

⁢

(

α

)

=

∑

σ

⁢

emb. of

⁢

K

σ

|

ℚ

⁢

(

α

)

⁢

(

α

)

=

n

d

⁢

∑

τ

⁢

emb. of

⁢

ℚ

⁢

(

α

)

τ

⁢

(

α

)

=

n

d

⁢

T

*

⁢

(

α

)

.

∎

Now, the above theorem will be proven.

Proof of theorem 1.

Let

f

⁢

(

x

)

∈

ℚ

⁢

[

x

]

be the

minimal polynomial

for

α

over

ℚ

. Then

deg

⁡

f

=

d

, where

d

is as in the previous lemma. Note that

|

N

*

⁢

(

α

)

|

is equal to the

absolute value

of the

constant term

of

f

and that

T

*

⁢

(

α

)

is equal to the opposite of the

coefficient

of

x

d

-

1

of

f

. Thus,

N

*

⁢

(

α

)

,

T

*

⁢

(

α

)

∈

ℚ

. Therefore,

N

⁢

(

α

)

=

(

N

*

⁢

(

α

)

)

n

d

∈

ℚ

and

T

⁢

(

α

)

=

n

d

⁢

T

*

⁢

(

α

)

∈

ℚ

. Moreover, if

α

is an algebraic integer, then

f

⁢

(

x

)

∈

ℤ

⁢

[

x

]

,

N

*

⁢

(

α

)

,

T

*

⁢

(

α

)

∈

ℤ

,

N

⁢

(

α

)

=

(

N

*

⁢

(

α

)

)

n

d

∈

ℤ

, and

T

⁢

(

α

)

=

n

d

⁢

T

*

⁢

(

α

)

∈

ℤ

.

If

a

∈

ℚ

, then

d

=

1

,

N

⁢

(

a

)

=

(

N

*

⁢

(

a

)

)

n

=

a

n

, and

T

⁢

(

a

)

=

n

⁢

T

*

⁢

(

a

)

=

n

⁢

a

.

Finally, if

α

,

β

∈

K

, then

N

⁢

(

α

⁢

β

)

=

∏

σ

⁢

emb. of

⁢

K

σ

⁢

(

α

⁢

β

)

=

∏

σ

⁢

emb. of

⁢

K

σ

⁢

(

α

)

⁢

σ

⁢

(

β

)

=

(

∏

σ

⁢

emb. of

⁢

K

σ

⁢

(

α

)

)

⁢

(

∏

σ

⁢

emb. of

⁢

K

σ

⁢

(

β

)

)

=

N

⁢

(

α

)

⁢

N

⁢

(

β

)

and

T

⁢

(

α

+

β

)

=

∑

σ

⁢

emb. of

⁢

K

σ

⁢

(

α

+

β

)

=

∑

σ

⁢

emb. of

⁢

K

σ

⁢

(

α

)

+

σ

⁢

(

β

)

=

∏

σ

⁢

emb. of

⁢

K

σ

⁢

(

α

)

+

∑

σ

⁢

emb. of

⁢

K

σ

⁢

(

β

)

=

T

⁢

(

α

)

+

T

⁢

(

β

)

.

∎

Theorem 2

.

An algebraic integer

ε

is a unit if and only if its

N

*

⁢

(

ε

)

=

±

1

,

.   Thus, in the minimal polynomial of an algebraic unit is always

±

1

.

Proof.

Let

K

=

ℚ

⁢

(

ε

)

. Since

ε

is an algebraic integer,

d

=

[

K

:

ℚ

]

is finite. Let

𝒪

K

denote the

ring of integers

of

K

.

If

N

*

⁢

(

ε

)

=

±

1

, then let

f

⁢

(

x

)

∈

ℤ

⁢

[

x

]

be the minimal polynomial of

ε

over

ℚ

. Let

a

1

,

⋯

,

a

d

-

1

∈

ℤ

such that

f

⁢

(

x

)

=

x

d

+

∑

j

=

1

d

-

1

a

j

⁢

x

j

±

1

. Then

0

=

f

⁢

(

ε

)

=

ε

d

+

∑

j

=

1

d

-

1

a

j

⁢

ε

j

±

1

. Thus,

ε

⁢

(

ε

d

-

1

+

∑

j

=

1

d

-

1

a

j

⁢

ε

j

-

1

)

=

±

1

. Since

ε

d

-

1

+

∑

j

=

1

d

-

1

a

j

⁢

ε

j

-

1

∈

𝒪

K

, it follows that

ε

is a unit in

𝒪

K

.

Conversely

, let

ε

be a unit in

𝒪

K

. Let

υ

∈

𝒪

K

with

ε

⁢

υ

=

1

. Since

N

*

⁢

(

ε

)

⁢

N

*

⁢

(

υ

)

=

N

*

⁢

(

ε

⁢

υ

)

=

N

*

⁢

(

1

)

=

1

and

N

*

⁢

(

ε

)

,

N

*

⁢

(

υ

)

∈

ℤ

, it follows that

N

*

⁢

(

ε

)

=

±

1

. ∎

References

1

Marcus, Daniel A.

Number Fields

. New York: Springer-Verlag, 1977.

Title

proof of norm and trace of algebraic number

Canonical name

ProofOfNormAndTraceOfAlgebraicNumber

Date of creation

2013-03-22 15:58:53

Last modified on

2013-03-22 15:58:53

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

27

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 11R04

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