primitive element of biquadratic field

primitive element of biquadratic field

Theorem

.

Let

m

and

n

be distinct

squarefree

integers

, neither of which is equal to

1

. Then the

biquadratic field

Q

⁢

(

m

,

n

)

is equal to

Q

⁢

(

m

+

n

)

.

In other words,

m

+

n

is a

primitive element

(

http://planetmath.org/PrimitiveElement

) of

ℚ

⁢

(

m

,

n

)

.

Proof.

We clearly have

ℚ

⁢

(

m

+

n

)

⊆

ℚ

⁢

(

m

,

n

)

. For the reverse

inclusion

, it is

equivalent

(

http://planetmath.org/Equivalent3

) to show that

m

+

n

does not belong to any of the quadratic

subfields

of

ℚ

⁢

(

m

,

n

)

, which are

ℚ

⁢

(

m

)

,

ℚ

⁢

(

n

)

, and

ℚ

⁢

(

m

⁢

n

)

.

Suppose that

m

+

n

∈

ℚ

⁢

(

m

)

. Then

n

∈

ℚ

⁢

(

m

)

. Thus,

ℚ

⁢

(

n

)

=

ℚ

⁢

(

m

)

, which is proven to be false here (

http://planetmath.org/QuadraticFieldsThatAreNotIsomorphic

). By a ,

m

+

n

∉

ℚ

⁢

(

n

)

.

Suppose that

m

+

n

∈

ℚ

⁢

(

m

⁢

n

)

. Let

a

,

b

,

c

,

d

∈

ℤ

with

gcd

⁡

(

a

,

b

)

=

gcd

⁡

(

c

,

d

)

=

1

,

b

≠

0

, and

d

≠

0

such that

m

+

n

=

a

b

+

c

d

⁢

m

⁢

n

.

(1)

Now, we perform some basic

algebraic

manipulations.

b

⁢

d

⁢

m

+

b

⁢

d

⁢

n

=

a

⁢

d

+

b

⁢

c

⁢

m

⁢

n

b

⁢

d

⁢

m

+

b

⁢

d

⁢

n

-

a

⁢

d

=

b

⁢

c

⁢

m

⁢

n

(

b

⁢

d

⁢

m

+

b

⁢

d

⁢

n

-

a

⁢

d

)

2

=

(

b

⁢

c

⁢

m

⁢

n

)

2

b

2

⁢

d

2

⁢

m

+

2

⁢

b

2

⁢

d

2

⁢

m

⁢

n

-

2

⁢

a

⁢

b

⁢

d

2

⁢

m

+

b

2

⁢

d

2

⁢

n

-

2

⁢

a

⁢

b

⁢

d

2

⁢

n

+

a

2

⁢

d

2

=

b

2

⁢

c

2

⁢

m

⁢

n

b

2

⁢

d

2

⁢

m

+

b

2

⁢

d

2

⁢

n

+

a

2

⁢

d

2

-

b

2

⁢

c

2

⁢

m

⁢

n

=

2

⁢

a

⁢

b

⁢

d

2

⁢

(

m

+

n

)

-

2

⁢

b

2

⁢

d

2

⁢

m

⁢

n

Now, we use equation (

1

) to eliminate the

m

+

n

and obtain

b

2

⁢

d

2

⁢

m

+

b

2

⁢

d

2

⁢

n

+

a

2

⁢

d

2

-

b

2

⁢

c

2

⁢

m

⁢

n

=

2

⁢

a

⁢

b

⁢

d

2

⁢

(

a

b

+

c

d

⁢

m

⁢

n

)

-

2

⁢

b

2

⁢

d

2

⁢

m

⁢

n

.

Now, we perform some more basic algebraic manipulations.

b

2

⁢

d

2

⁢

m

+

b

2

⁢

d

2

⁢

n

+

a

2

⁢

d

2

-

b

2

⁢

c

2

⁢

m

⁢

n

=

2

⁢

a

2

⁢

d

2

+

2

⁢

a

⁢

b

⁢

c

⁢

d

⁢

m

⁢

n

-

2

⁢

b

2

⁢

d

2

⁢

m

⁢

n

b

2

⁢

d

2

⁢

m

+

b

2

⁢

d

2

⁢

n

-

a

2

⁢

d

2

-

b

2

⁢

c

2

⁢

m

⁢

n

=

2

⁢

b

⁢

d

⁢

(

a

⁢

c

-

b

⁢

d

)

⁢

m

⁢

n

Since

m

⁢

n

∉

ℚ

,

b

≠

0

, and

d

≠

0

, we must have

a

⁢

c

-

b

⁢

d

=

0

. Thus,

c

d

=

b

a

. (Note that we have

a

≠

0

since

a

⁢

c

=

b

⁢

d

≠

0

.) Using this in equation (

1

), we obtain

m

+

n

=

a

b

+

b

a

⁢

m

⁢

n

.

Now we perform calculations as before.

a

⁢

b

⁢

m

+

a

⁢

b

⁢

n

=

a

2

+

b

2

⁢

m

⁢

n

a

⁢

b

⁢

m

+

a

⁢

b

⁢

n

-

a

2

=

b

2

⁢

m

⁢

n

(

a

⁢

b

⁢

m

+

a

⁢

b

⁢

n

-

a

2

)

2

=

(

b

2

⁢

m

⁢

n

)

2

a

2

⁢

b

2

⁢

m

+

2

⁢

a

2

⁢

b

2

⁢

m

⁢

n

-

2

⁢

a

3

⁢

b

⁢

m

+

a

2

⁢

b

2

⁢

n

-

2

⁢

a

3

⁢

b

⁢

n

+

a

4

=

b

4

⁢

m

⁢

n

a

2

⁢

b

2

⁢

m

+

a

2

⁢

b

2

⁢

n

+

a

4

-

b

4

⁢

m

⁢

n

=

2

⁢

a

3

⁢

b

⁢

(

m

+

n

)

-

2

⁢

a

2

⁢

b

2

⁢

m

⁢

n

a

2

⁢

b

2

⁢

m

+

a

2

⁢

b

2

⁢

n

+

a

4

-

b

4

⁢

m

⁢

n

=

2

⁢

a

3

⁢

b

⁢

(

a

b

+

b

a

⁢

m

⁢

n

)

-

2

⁢

a

2

⁢

b

2

⁢

m

⁢

n

a

2

⁢

b

2

⁢

m

+

a

2

⁢

b

2

⁢

n

+

a

4

-

b

4

⁢

m

⁢

n

=

2

⁢

a

4

+

2

⁢

a

2

⁢

b

2

⁢

m

⁢

n

-

2

⁢

a

2

⁢

b

2

⁢

m

⁢

n

a

2

⁢

b

2

⁢

m

+

a

2

⁢

b

2

⁢

n

-

a

4

-

b

4

⁢

m

⁢

n

=

0

Since

b

2

divides

a

4

and

gcd

⁡

(

a

,

b

)

=

1

, we must have

b

2

=

1

. Plugging into the equation above yields

a

2

⁢

m

+

a

2

⁢

n

-

a

4

-

m

⁢

n

=

0

.

Now for yet some more algebraic manipulations.

a

2

⁢

m

-

a

4

-

m

⁢

n

+

a

2

⁢

n

=

0

a

2

⁢

(

m

-

a

2

)

-

n

⁢

(

m

-

a

2

)

=

0

(

m

-

a

2

)

⁢

(

a

2

-

n

)

=

0

Thus,

m

=

a

2

or

n

=

a

2

, a

contradiction

. It follows that

ℚ

⁢

(

m

+

n

)

=

ℚ

⁢

(

m

,

n

)

.
∎

Title

primitive element of biquadratic field

Canonical name

PrimitiveElementOfBiquadraticField

Date of creation

2013-03-22 17:54:17

Last modified on

2013-03-22 17:54:17

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

9

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 11R16

Related topic

PrimitiveElementTheorem

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