quartic polynomial with Galois group D_8

quartic polynomial with Galois group

D

8

The

polynomial

f

⁢

(

x

)

=

x

4

-

2

⁢

x

2

-

2

is Eisenstein at

2

and thus

irreducible

over

ℚ

. Solving

f

⁢

(

x

)

as a quadratic in

x

2

, we see that the roots of

f

⁢

(

x

)

are

α

1

=

1

+

3

α

3

=

-

1

+

3

α

2

=

1

-

3

α

4

=

-

1

-

3

Note that the

discriminant

of

f

⁢

(

x

)

is

-

4608

=

-

2

9

⋅

3

2

, and that its

resolvent cubic

is

x

3

+

4

⁢

x

2

+

12

⁢

x

=

x

⁢

(

x

2

+

4

⁢

x

+

12

)

=

0

which factors over

ℚ

into a linear and an irreducible quadratic. Additionally,

f

⁢

(

x

)

remains irreducible over

ℚ

⁢

(

-

4608

)

=

ℚ

⁢

(

-

2

)

, since none of the roots of

f

⁢

(

x

)

lie in this field and the discriminant of

f

⁢

(

x

)

, regarded as a quadratic in

x

2

, does not lie in this field either, so

f

⁢

(

x

)

cannot factor as a

product

of two quadratics. So according to the article on the

Galois group

of a

quartic polynomial

,

f

⁢

(

x

)

should indeed have Galois group

isomorphic

to

D

8

. We show that this is the case by explicitly examining the

structure

of its

splitting field

.

Let

K

be the splitting field of

f

⁢

(

x

)

over

ℚ

, and let

G

=

Gal

⁡

(

K

/

ℚ

)

.

Let

K

1

=

ℚ

⁢

(

α

1

)

=

ℚ

⁢

(

α

3

)

and

K

2

=

ℚ

⁢

(

α

2

)

=

ℚ

⁢

(

α

4

)

. Clearly

K

contains both

K

1

and

K

2

and thus contains

K

1

⁢

K

2

=

ℚ

⁢

(

α

1

,

α

2

)

. But obviously

f

⁢

(

x

)

splits in

K

1

⁢

K

2

, so that

K

=

K

1

⁢

K

2

. We next determine the degree of

K

over

ℚ

.

Note that

K

1

≠

K

2

since

K

1

is a real field while

K

2

is not. Thus

K

1

∩

K

2

⊊

K

1

,

K

2

. Clearly

[

K

1

:

ℚ

]

=

[

K

2

:

ℚ

]

=

4

, so

[

K

1

∩

K

2

:

ℚ

]

≤

2

. But

3

=

(

1

+

3

)

2

-

1

=

-

(

1

-

3

)

2

+

1

so

3

∈

K

1

∩

K

2

. Hence

K

1

∩

K

2

=

ℚ

⁢

(

3

)

; call this field

F

.

Since

K

1

≠

K

2

, we also have

K

=

K

1

⁢

K

2

≠

K

1

and

K

=

K

1

⁢

K

2

≠

K

2

; thus

K

is a

quadratic extension

of each and

[

K

:

F

]

=

4

.

Putting these results together, we see that

[

K

:

ℚ

]

=

[

K

:

F

]

[

F

:

ℚ

]

=

8

so that

G

has order

8

.

Now, neither

K

1

nor

K

2

is Galois over

ℚ

(since the

Galois closure

of either one is

K

), so that the

subgroup

of

G

fixing (say)

K

1

is a nonnormal subgroup of

G

. Thus

G

must be

nonabelian

, so must be isomorphic to either

D

8

or

Q

8

(the

quaternions

). But the subgroups of

G

corresponding to

K

1

and

K

2

are distinct subgroups of order

2

in

G

, and

Q

8

has only one subgroup of order

2

. Thus

G

≅

D

8

. (Alternatively, note that all subgroups of

Q

8

are normal, so

G

≅

D

8

since it has a nonnormal subgroup).

Title

quartic polynomial with Galois group

D

8

Canonical name

QuarticPolynomialWithGaloisGroupD8

Date of creation

2013-03-22 17:44:09

Last modified on

2013-03-22 17:44:09

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

6

Author

rm50 (10146)

Entry type

Example

Classification

msc 12D10

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