Galois group of a quartic polynomial

Galois group of a quartic polynomial

Consider a general (monic)

quartic polynomial

over

ℚ

f

⁢

(

x

)

=

x

4

+

a

⁢

x

3

+

b

⁢

x

2

+

c

⁢

x

+

d

and denote the

Galois group

of

f

⁢

(

x

)

by

G

.

The Galois group

G

is

isomorphic

to a

subgroup

of

S

4

(see the article on the Galois group of a

cubic polynomial

for a discussion of this question).

If the quartic splits into a linear factor and an

irreducible

cubic, then its Galois group is simply the Galois group of the cubic portion and thus is isomorphic to a subgroup of

S

3

(embedded in

S

4

) - again, see the article on the Galois group of a cubic polynomial.

If it factors as two irreducible quadratics, then the

splitting field

of

f

⁢

(

x

)

is the

compositum

of

ℚ

⁢

(

D

1

)

and

ℚ

⁢

(

D

2

)

, where

D

1

and

D

2

are the

discriminants

of the two quadratics. This is either a biquadratic extension and thus has Galois group isomorphic to

V

4

, or else

D

1

⁢

D

2

is a square, and

ℚ

⁢

(

D

1

,

D

2

)

=

ℚ

⁢

(

D

1

)

and the Galois group is isomorphic to

ℤ

/

2

⁢

ℤ

.

This leaves us with the most interesting case, where

f

⁢

(

x

)

is irreducible. In this case, the Galois group acts transitively on the roots of

f

⁢

(

x

)

, so it must be isomorphic to a

transitive

(

http://planetmath.org/GroupAction

) subgroup of

S

4

. The transitive subgroups of

S

4

are

S

4

A

4

D

8

≅

{

e

,

(

1234

)

,

(

13

)

⁢

(

24

)

,

(

1432

)

,

(

12

)

⁢

(

34

)

,

(

14

)

⁢

(

23

)

,

(

13

)

,

(

24

)

}

⁢

and its conjugates

V

4

≅

{

e

,

(

12

)

⁢

(

34

)

,

(

13

)

⁢

(

24

)

,

(

14

)

⁢

(

23

)

}

ℤ

/

4

⁢

ℤ

≅

{

e

,

(

1234

)

,

(

13

)

⁢

(

24

)

,

(

1432

)

}

⁢

and its conjugates

We will see that each of these transitive subgroups actually appears as the Galois group of some class of irreducible quartics.

The

resolvent cubic

of

f

⁢

(

x

)

is

C

⁢

(

x

)

=

x

3

-

2

⁢

b

⁢

x

2

+

(

b

2

+

a

⁢

c

-

4

⁢

d

)

⁢

x

+

(

c

2

+

a

2

⁢

d

-

a

⁢

b

⁢

c

)

and has roots

r

1

=

(

α

1

+

α

2

)

⁢

(

α

3

+

α

4

)

r

2

=

(

α

1

+

α

3

)

⁢

(

α

2

+

α

4

)

r

3

=

(

α

1

+

α

4

)

⁢

(

α

2

+

α

3

)

But then a short computation shows that the discriminant

D

of

C

⁢

(

x

)

is the same as the discriminant of

f

⁢

(

x

)

. Also, since

r

1

,

r

2

,

r

3

∈

ℚ

⁢

(

α

1

,

α

2

,

α

3

,

α

4

)

, it follows that the splitting field of

C

⁢

(

x

)

is a

subfield

of the splitting field of

f

⁢

(

x

)

and thus that the Galois group of

C

⁢

(

x

)

is a

quotient

of the Galois group of

f

⁢

(

x

)

. There are four cases:

•

If

C

⁢

(

x

)

is irreducible, and

D

is not a

rational

square, then

G

does not fix

D

and thus is not

contained

in

A

4

. But in this case, where

D

is not a square, the Galois group of

C

⁢

(

x

)

is

S

3

, which has order

6

. The only subgroup of

S

4

not contained in

A

4

with order a

multiple

of

6

(and thus capable of having a subgroup of index

6

) is

S

4

itself, so in this case

G

≅

S

4

.

•

If

C

⁢

(

x

)

is irreducible but

D

is a rational square, then

G

fixes

D

, so

G

≤

A

4

. In

addition

, the Galois group of

C

⁢

(

x

)

is

A

3

, so

3

divides the order of a transitive subgroup of

A

4

, which means that

G

≅

A

4

itself.

•

If

C

⁢

(

x

)

is

reducible

, suppose first that it splits completely in

ℚ

. Then each of

r

1

,

r

2

,

r

3

∈

ℚ

and thus each element of

G

fixes each

r

i

. Thus

G

≅

V

4

.

•

Finally, if

C

⁢

(

x

)

splits into a linear factor and an irreducible quadratic, then one of the

r

i

, say

r

2

, is in

ℚ

. Then

G

fixes

r

2

=

(

α

1

+

α

3

)

⁢

(

α

2

+

α

4

)

but not

r

1

or

r

3

. The only possibilities from among the

transitive groups

are then that

G

≅

D

8

or

G

≅

ℤ

/

4

⁢

ℤ

. In this case, the discriminant of the quadratic is not a rational square, but it is a rational square times

D

.

Now,

G

∩

A

4

fixes

ℚ

⁢

(

D

)

, since

G

fixes

D

up to sign and

A

4

restricts our attention to

even permutations

. But

|

G

:

G

∩

A

4

|

=

2

, so the

fixed field

of

G

∩

A

4

has

dimension

2

over

ℚ

and thus is exactly

ℚ

⁢

(

D

)

. If

G

≅

D

8

, then

G

∩

A

4

≅

V

4

, while if

G

≅

ℤ

/

4

⁢

ℤ

, then

G

∩

A

4

≅

ℤ

/

2

⁢

ℤ

; in the first case only,

G

∩

A

4

acts transitively on the roots of

f

⁢

(

x

)

. Thus

G

∩

A

4

≅

V

4

if and only if

f

⁢

(

x

)

is irreducible over

ℚ

⁢

(

D

)

.

So, in summary, for

f

⁢

(

x

)

irreducible, we have the following:

Condition

Galois group

C

⁢

(

x

)

irreducible,

D

not a rational square

S

4

C

⁢

(

x

)

irreducible,

D

a rational square

A

4

C

⁢

(

x

)

splits completely

V

4

C

⁢

(

x

)

factors as linear times irreducible quadratic,

f

⁢

(

x

)

irreducible over

ℚ

⁢

(

D

)

D

8

C

⁢

(

x

)

factors as linear times irreducible quadratic,

f

⁢

(

x

)

reducible over

ℚ

⁢

(

D

)

ℤ

/

4

⁢

ℤ

References

1

D.S. Dummit, R.M. Foote,

Abstract Algebra

, Wiley and Sons, 2004.

Title

Galois group of a quartic polynomial

Canonical

name

GaloisGroupOfAQuarticPolynomial

Date of creation

2013-03-22 17:41:35

Last modified on

2013-03-22 17:41:35

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

7

Author

rm50 (10146)

Entry type

Topic

Classification

msc 12D10

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