quadratic extension

quadratic extension

Let

k

be a field and

K

be its

algebraic closure

. Suppose that

k

≠

K

. A

quadratic extension

E

over

k

is a field

k

<

E

≤

K

such that

E

=

k

⁢

(

α

)

for some

α

∈

K

-

k

, where

α

2

∈

k

.

If

a

=

α

2

, we often write

E

=

k

⁢

(

a

)

. Every element of

E

can be written as

r

+

s

⁢

a

, for some

r

,

s

∈

k

. This

representation

is unique and we see that

{

1

,

a

}

is a basis for the

vector space

E

over

k

. In fact, we have the following

Proposition

.

If the

characteristic

of

k

is not

2

, then

E

is a quadratic extension over

k

iff

dim

⁡

(

E

)

=

2

(as a vector space) over

k

.

Proof.

One

direction

is clear from the above discussion. So suppose

dim

⁡

(

E

)

=

2

over

k

and

{

1

,

β

}

is a basis for

E

over

k

. Then

β

2

=

r

+

s

⁢

β

for some

r

,

s

∈

k

. Set

α

=

β

-

s

2

. Then clearly

α

∈

E

-

k

and

{

1

,

α

}

is also a basis for

E

over

k

. Furthermore,

α

2

=

r

+

s

2

4

∈

k

. Thus,

k

⁢

(

α

)

is quadratic extension over

k

and

[

k

(

α

)

:

k

]

=

2

. But

k

⁢

(

α

)

is a

subfield

of

E

. Then

2

=

[

E

:

k

]

=

[

E

:

k

(

α

)

]

[

k

(

α

)

:

k

]

=

2

[

E

:

k

(

α

)

]

implies that

[

E

:

k

(

α

)

]

=

1

and

E

=

k

⁢

(

α

)

.
∎

In the proposition above, the

assumption

that

Char

⁡

(

k

)

≠

2

can not be dropped. If fact, quadratic extensions of

ℤ

2

do not exist, for if

α

2

∈

ℤ

2

, then

α

∈

ℤ

2

.

For the rest of the discussion, we assume that

Char

⁡

(

k

)

≠

2

.

Pick any

element

β

=

r

+

s

⁢

a

in

E

-

k

. Then

s

≠

0

and

(

β

-

r

)

2

=

s

2

⁢

a

∈

k

. So

β

is a root of the

irreducible polynomial

m

⁢

(

x

)

=

x

2

-

2

⁢

r

⁢

x

+

(

r

2

-

s

2

⁢

a

)

in

k

⁢

[

x

]

. If we define

β

¯

to be

r

-

s

⁢

a

, then

β

¯

is the other root of

m

⁢

(

x

)

, clearly also in

E

-

k

. This implies that the

minimal polynomial

of every element in

E

has degree at most 2, and splits into linear factors in

E

⁢

[

x

]

.

Since

Char

⁡

(

k

)

≠

2

,

β

≠

β

¯

are two distinct roots of

m

⁢

(

x

)

. This shows that

k

⁢

(

a

)

is

separable

over

k

.

Now, let

f

⁢

(

x

)

be any irreducible polynomial over

k

which has a root

β

in

E

. Then the minimal polynomial

m

⁢

(

x

)

of

β

in

k

⁢

[

x

]

must divide

f

. But because

f

is

irreducible

,

m

=

f

. This shows that

k

⁢

(

a

)

is normal over

k

. Since

k

⁢

(

a

)

is both separable and normal over

k

, it is a

Galois extension

over

k

.

Let

ϕ

be an

automorphism

of

E

=

k

⁢

(

a

)

fixing

k

. Then

ϕ

⁢

(

a

)

is easily seen to be a root of the minimal polynomial of

a

. As a result, either

ϕ

=

1

on

E

or

ϕ

is the

involution

that maps each

β

to

β

¯

. We have just proved

Theorem

.

Suppose

Char

⁡

(

k

)

≠

2

. Any quadratic extension of

k

is Galois over

k

, whose

Galois group

is isomorphic to

ℤ

/

2

⁢

ℤ

.

Remark

. A quadratic extension (of a field) is also known in the literature as a

2

-

extension

, a special case of a

p-extension

, when

p

=

2

.

Title

quadratic extension

Canonical

name

QuadraticExtension

Date of creation

2013-03-22 15:42:34

Last modified on

2013-03-22 15:42:34

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

20

Author

CWoo (3771)

Entry type

Definition

Classification

msc 12F05

Classification

msc 12F10

Synonym

2

-extension

Related topic

PExtension

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