field extension with Galois group Q_8

field extension with Galois group

Q

8

Let

α

=

(

2

+

2

)

⁢

(

3

+

3

)

,

E

=

ℚ

⁢

(

α

)

. We will show that

E

is Galois over

ℚ

and that

G

=

G

⁢

a

⁢

l

⁢

(

E

/

ℚ

)

≅

Q

8

(the group of

quaternions

).

We begin by showing that

[

E

:

ℚ

]

=

8

. Let

F

=

ℚ

⁢

(

2

,

3

)

=

ℚ

⁢

(

2

+

3

)

. Claim that

F

⊊

E

. To show that they are not equal, we show that

α

∉

F

, i.e. that

(

2

+

2

)

⁢

(

3

+

3

)

is not a square in

F

. If it were, say

(

2

+

2

)

⁢

(

3

+

3

)

=

c

2

,

c

∈

F

, take

σ

∈

G

⁢

a

⁢

l

⁢

(

F

/

ℚ

)

to be the

element

σ

:

{

2

↦

2

3

↦

-

3

Then

(

2

+

2

)

⁢

(

3

+

3

)

⁢

σ

⁢

(

(

2

+

2

)

⁢

(

3

+

3

)

)

=

(

c

⁢

σ

⁢

(

c

)

)

2

, so

(

2

+

2

)

2

⁢

(

3

+

3

)

⁢

(

3

-

3

)

=

6

⁢

(

2

+

2

)

2

=

(

c

⁢

σ

⁢

(

c

)

)

2

But

c

⁢

σ

⁢

(

c

)

=

Tr

F

/

ℚ

⁢

(

2

)

⁡

(

c

)

∈

ℚ

⁢

(

2

)

, and thus

6

=

(

c

⁢

σ

⁢

(

c

)

2

+

2

)

2

, so

6

∈

ℚ

⁢

(

2

)

, a

contradiction

. Thus

F

≠

E

. We show that

F

⊂

E

by showing that

2

+

3

∈

E

.

(

2

+

2

)

⁢

(

3

+

3

)

=

6

+

3

⁢

2

+

2

⁢

3

+

6

∈

E

,

so

3

⁢

2

+

2

⁢

3

+

6

∈

E

,

so

(

3

⁢

2

+

2

⁢

3

+

6

)

2

=

36

+

12

⁢

(

2

+

3

+

6

)

∈

E

,

so

2

+

3

+

6

∈

E

,

so

3

⁢

2

+

2

⁢

3

+

6

-

2

-

3

-

6

=

3

+

2

⁢

2

∈

E

,

so

(

3

+

2

⁢

2

)

2

=

11

+

4

⁢

6

∈

E

,

so

6

∈

E

,

so

2

+

3

+

6

-

6

=

2

+

3

∈

E

So

F

⊊

E

and thus

[

E

:

F

]

=

2

. Then

[

E

:

ℚ

]

=

[

E

:

F

]

[

F

:

ℚ

]

=

8

.

Now, the

irreducible polynomial

f

⁢

(

x

)

for

(

2

+

2

)

⁢

(

3

+

3

)

over

ℚ

is the

product

of

x

-

τ

⁢

(

(

2

+

2

)

⁢

(

3

+

3

)

)

as

τ

ranges over

Gal

⁡

(

F

/

ℚ

)

:

f

⁢

(

x

)

=

(

x

-

(

2

+

2

)

⁢

(

3

+

3

)

)

⁢

(

x

-

(

2

-

2

)

⁢

(

3

+

3

)

)

⁢

(

x

-

(

2

+

2

)

⁢

(

3

-

3

)

)

⁢

(

x

-

(

2

-

2

)

⁢

(

3

-

3

)

)

so that

f

⁢

(

x

2

)

is a degree

8

polynomial

with

α

as a root. In fact,

f

⁢

(

x

2

)

=

x

8

-

24

⁢

x

6

+

48

⁢

x

4

-

288

⁢

x

2

+

144

This polynomial must be

irreducible

since

α

is of degree

8

, so

f

⁢

(

x

2

)

is the

minimal polynomial

for

α

over

ℚ

. The roots of

f

⁢

(

x

2

)

are obviously

±

(

2

±

2

)

⁢

(

3

±

3

)

Furthermore, it is easy to see that each of these roots lies in

E

, for

α

⁢

(

2

-

2

)

⁢

(

3

+

3

)

=

2

⁢

(

3

+

3

)

∈

F

α

⁢

(

2

+

2

)

⁢

(

3

-

3

)

=

6

⁢

(

2

+

2

)

∈

F

α

⁢

(

2

-

2

)

⁢

(

3

-

3

)

=

2

⁢

6

=

2

⁢

3

∈

F

so dividing through by

α

we see that

(

2

-

2

)

⁢

(

3

+

3

)

,

(

2

+

2

)

⁢

(

3

-

3

)

,

(

2

-

2

)

⁢

(

3

-

3

)

∈

E

Thus

E

is in fact Galois over

ℚ

, is the

splitting field

for

f

⁢

(

x

2

)

, and has

Galois group

G

=

Gal

⁡

(

E

/

ℚ

)

of order (

http://planetmath.org/OrderGroup

)

8

.

G

acts transitively on the roots of

f

⁢

(

x

2

)

, and

E

=

ℚ

⁢

(

α

)

, so an element of

G

is determined by the image of

α

. Thus the elements of

G

are the

automorphisms

of

E

that map

α

to any of the eight roots of

f

⁢

(

x

2

)

. Let

α

=

(

2

+

2

)

⁢

(

3

+

3

)

β

=

(

2

-

2

)

⁢

(

3

+

3

)

γ

=

(

2

+

2

)

⁢

(

3

-

3

)

δ

=

(

2

-

2

)

⁢

(

3

-

3

)

and let

σ

:

α

↦

β

,

τ

:

α

↦

γ

be elements of

G

.

σ

⁢

(

α

2

)

=

β

2

, so

σ

⁢

(

2

+

2

)

⁢

σ

⁢

(

3

+

3

)

=

(

2

-

2

)

⁢

(

3

+

3

)

. This is an

equation

in

F

, so regarding

σ

as an automorphism of

F

/

ℚ

, it must be the automorphism

2

↦

-

2

,

3

↦

3

. Since

α

⁢

β

=

2

⁢

(

3

+

3

)

, we have

σ

⁢

(

α

⁢

β

)

=

-

α

⁢

β

and thus that

σ

⁢

(

β

)

=

-

α

. It follows that

σ

is an element of order (

http://planetmath.org/OrderGroup

)

4

in

G

.

Similarly,

τ

⁢

(

α

2

)

=

γ

2

, so

τ

⁢

(

2

+

2

)

⁢

τ

⁢

(

3

+

3

)

=

(

2

+

2

)

⁢

(

3

-

3

)

, so that

τ

, regarded as an automorphism of

F

/

ℚ

, must be

2

↦

2

,

3

↦

-

3

. Since

α

⁢

γ

=

6

⁢

(

2

+

2

)

, we have

τ

⁢

(

α

⁢

γ

)

=

-

α

⁢

γ

, so that

τ

⁢

(

γ

)

=

-

α

, and

τ

is also an element of order (

http://planetmath.org/OrderGroup

)

4

in

G

. Note also that

σ

2

⁢

(

α

)

=

-

α

=

τ

2

⁢

(

α

)

, so that

σ

2

=

τ

2

≠

1

.

Looking at

σ

⁢

τ

,

σ

⁢

τ

⁢

(

α

)

=

σ

⁢

(

γ

)

=

σ

⁢

(

α

⁢

γ

α

)

=

σ

⁢

(

6

⁢

(

2

+

2

)

)

σ

⁢

(

α

)

=

-

6

⁢

(

2

-

2

)

β

=

-

β

⁢

δ

β

=

-

δ

while

τ

⁢

σ

⁢

(

α

)

=

τ

⁢

(

β

)

=

τ

⁢

(

α

⁢

β

α

)

=

τ

⁢

(

2

⁢

(

3

+

3

)

)

τ

⁢

(

α

)

=

2

⁢

(

3

-

3

)

γ

=

γ

⁢

δ

γ

=

δ

and thus

τ

⁢

σ

3

⁢

(

α

)

=

τ

⁢

σ

⁢

σ

2

⁢

(

α

)

=

-

τ

⁢

σ

⁢

(

α

)

=

-

δ

=

σ

⁢

τ

⁢

(

α

)

. So

σ

⁢

τ

=

τ

⁢

σ

3

.

Putting this all together, we see that

G

is generated by

σ

,

τ

, and that the

generators

satisfy

the

relations

σ

4

=

τ

4

=

1

,

σ

2

=

τ

2

≠

1

,

σ

⁢

τ

=

τ

⁢

σ

3

Define

φ

:

G

→

Q

8

by

φ

⁢

(

σ

)

=

i

,

φ

⁢

(

τ

)

=

j

. This is easily seen to be a homorphism, and

φ

⁢

(

σ

⁢

τ

)

=

i

⁢

j

=

k

, so

φ

is

surjective

and is thus an

isomorphism

since both groups have order (

http://planetmath.org/OrderGroup

)

8

. Thus

Gal

⁡

(

E

/

ℚ

)

≅

Q

8

.

Title

field extension

with Galois group

Q

8

Canonical name

FieldExtensionWithGaloisGroupQ8

Date of creation

2013-03-22 17:44:28

Last modified on

2013-03-22 17:44:28

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

7

Author

rm50 (10146)

Entry type

Example

Classification

msc 12F10

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