Galois group of a cubic polynomial

Galois group of a cubic polynomial

Definition 1

.

If

f

⁢

(

x

)

∈

k

⁢

[

x

]

, the

Galois group

of

f

⁢

(

x

)

is the Galois group

Gal

⁡

(

K

/

k

)

of a

splitting field

K

of

f

⁢

(

x

)

.

Theorem 1

.

For

f

⁢

(

x

)

∈

k

⁢

[

x

]

, the Galois group of

f

⁢

(

x

)

permutes the set of roots of

f

⁢

(

x

)

. Therefore, if the roots of

f

⁢

(

x

)

are

α

1

,

…

,

α

n

∈

K

, the Galois group of

f

⁢

(

x

)

is

isomorphic

to a

subgroup

of

S

n

.

Proof.

K

=

k

⁢

(

α

1

,

…

,

α

n

)

, so any

automorphism

σ

of

K

fixing

k

is determined by the image of each

α

i

. But

σ

must take each

α

i

to some

α

j

(where possibly

i

=

j

), since

σ

is a

homomorphism

of

K

and thus

f

⁢

(

σ

⁢

(

α

i

)

)

=

σ

⁢

(

f

⁢

(

α

i

)

)

=

0

. Thus

σ

permutes the roots of

f

⁢

(

x

)

and is determined by the resulting

permutation

.

We now restrict our attention to the case

k

=

ℚ

. If

f

⁢

(

x

)

∈

ℚ

⁢

[

x

]

is a cubic, its Galois group is a subgroup of

S

3

. We can then use the knowledge of the group

structure

of

S

3

to anticipate the possible Galois groups of a

cubic polynomial

. There are six subgroups of

S

3

, and the three subgroups of order

2

are

conjugate

. This leaves four essentially different subgroups of

S

3

: the

trivial group

, the group

⟨

(

1

,

2

)

⟩

that consists of a single

transposition

, the group

A

3

=

⟨

(

1

,

2

,

3

)

⟩

, and the full group

S

3

. All four of these groups can in fact appear as the Galois group of a cubic.

Let

K

be a splitting field of

f

⁢

(

x

)

over

ℚ

.

If

f

⁢

(

x

)

splits completely in

ℚ

, then

K

=

ℚ

and so the Galois group of

f

⁢

(

x

)

is

trivial

. So any cubic (in fact, a

polynomial

of any degree) that factors completely into linear factors in

ℚ

will have trivial Galois group.

If

f

⁢

(

x

)

factors into a linear and an

irreducible

quadratic term, then

K

=

ℚ

⁢

(

D

)

, where

D

is the

discriminant

of the quadratic. Hence

[

K

:

ℚ

]

=

2

and the order of

Gal

⁡

(

K

/

ℚ

)

is

2

, so

Gal

⁡

(

K

/

ℚ

)

≅

⟨

1

,

2

⟩

≅

ℤ

/

2

⁢

ℤ

; the

nontrivial element

of the Galois group takes each root of the quadratic to its

complex conjugate

(i.e. it maps

D

↦

-

D

). Thus any cubic that has exactly one

rational root

will have Galois group isomorphic to

⟨

1

,

2

⟩

≅

ℤ

/

2

⁢

ℤ

.

The Galois groups

A

3

and

S

3

arise when considering irreducible cubics. Let

f

⁢

(

x

)

be irreducible with roots

r

1

,

r

2

,

r

3

. Since

f

is irreducible, the roots are distinct. Thus

Gal

⁡

(

K

/

ℚ

)

has at least

3

elements

, since the image of

r

1

may be any of the three roots. Since

Gal

⁡

(

K

/

ℚ

)

⊂

S

3

, it follows that

Gal

⁡

(

K

/

ℚ

)

≅

A

3

or

Gal

⁡

(

K

/

ℚ

)

≅

S

3

and thus by the

fundamental theorem of Galois theory

that

[

K

:

ℚ

]

=

3

or

6

.

Now, the discriminant of

f

is

D

=

∏

i

<

j

(

r

i

-

r

j

)

2

This is a

symmetric polynomial

in the

r

i

. The

coefficients

of

f

⁢

(

x

)

are the

elementary symmetric polynomials

in the

r

i

: if

f

⁢

(

x

)

=

x

3

+

a

⁢

x

2

+

b

⁢

x

+

c

, then

c

=

r

1

⁢

r

2

⁢

r

3

b

=

-

(

r

1

⁢

r

2

+

r

1

⁢

r

3

+

r

2

⁢

r

+

3

)

a

=

r

1

+

r

2

+

r

3

Thus

D

can be written as a polynomial in the coefficients of

f

, so

D

∈

ℚ

.

D

≠

0

since

f

⁢

(

x

)

is irreducible and therefore has distinct roots; also clearly

D

∈

K

and thus

ℚ

⁢

(

r

1

,

D

)

⊂

K

. If

f

⁢

(

x

)

=

x

3

+

a

⁢

x

2

+

b

⁢

x

+

c

, then its discriminant

D

is

18

⁢

a

⁢

b

⁢

c

+

a

2

⁢

b

2

-

4

⁢

b

3

-

4

⁢

a

3

⁢

c

-

27

⁢

c

2

(see the article on the discriminant for a longer discussion).

If

D

∉

ℚ

, it follows that

D

has degree

2

over

ℚ

, so that

[

ℚ

(

r

1

,

D

)

:

ℚ

]

=

6

. Hence

K

=

ℚ

⁢

(

r

1

,

D

)

, so we can derive the splitting field for

f

by adjoining any root of

f

and the

square root

of the discriminant. This can happen for either

positive

or negative

D

, clearly.
Note in particular that if

D

<

0

, then

D

is

imaginary

and thus

K

is not a real field, so that

f

has one real and two imaginary roots. So any cubic with only one

real root

has Galois group

S

3

.

If

D

∈

ℚ

, then any element of

Gal

⁡

(

K

/

ℚ

)

must fix

D

. But a transposition of two roots does not fix

D

- for example, the map

r

1

↦

r

2

,

r

2

↦

r

1

,

r

3

↦

r

3

takes

D

=

(

r

1

-

r

2

)

⁢

(

r

1

-

r

3

)

⁢

(

r

2

-

r

3

)

↦

(

r

2

-

r

1

)

⁢

(

r

2

-

r

3

)

⁢

(

r

1

-

r

3

)

=

-

D

Then

Gal

⁡

(

K

/

ℚ

)

does not include transpositions and so it must in this case be isomorphic to

A

3

. Thus

[

K

:

ℚ

]

=

3

, so

K

=

ℚ

⁢

(

r

1

)

=

ℚ

⁢

(

r

1

,

D

)

since

D

∈

ℚ

. This proves:

Theorem 2

.

Let

f

⁢

(

x

)

∈

Q

⁢

[

x

]

be an irreducible cubic and

K

its splitting field. Then if

α

is any root of

f

,

K

=

ℚ

⁢

(

α

,

D

)

where

D

is the discriminant of

f

. Thus if

D

is

rational

,

[

K

:

Q

]

=

3

and the Galois group is isomorphic to

A

3

, otherwise

[

K

:

Q

]

=

6

and the Galois group is isomorphic to

S

3

.

Thus if

f

is an irreducible cubic with Galois group

A

3

, it must have three real roots. However, the

converse

does not hold (see Example 4 below).

of looking at the above

analysis

is that for a “general” polynomial of degree

n

, the Galois group is

S

n

. If the Galois group of some polynomial is not

S

n

, there must be

algebraic

relations

among the roots that restrict the available set of permutations. In the case of a cubic whose discriminant is a rational square, this relation is that

D

, which is a polynomial in the roots, must be preserved.

Example 1

f

⁢

(

x

)

=

x

3

-

6

⁢

x

2

+

11

⁢

x

-

6

. By the rational root test, this polynomial has the three rational roots

1

,

2

,

3

, so it factors as

f

⁢

(

x

)

=

(

x

-

1

)

⁢

(

x

-

2

)

⁢

(

x

-

3

)

over

ℚ

. Its Galois group is therefore trivial.

Example 2

f

⁢

(

x

)

=

x

3

-

x

2

+

x

-

1

. Again by the rational root test, this polynomial factors as

(

x

-

1

)

⁢

(

x

2

+

1

)

, so its Galois group has two elements, and a splitting field

K

for

f

is derived by adjoining the square root of the discriminant of the quadratic:

K

=

ℚ

⁢

(

-

1

)

. The nontrivial element of the Galois group maps

-

1

↔

-

-

1

.

Example 3

f

⁢

(

x

)

=

x

3

-

2

. This polynomial has discriminant

-

108

=

-

3

⋅

6

2

. This is not a rational square, so the Galois group of

f

over

ℚ

is

S

3

, and the splitting field for

f

is

ℚ

⁢

(

2

3

,

-

108

)

=

ℚ

⁢

(

2

3

,

-

3

)

. This is in agreement with what we already know, namely that the

cube roots

of

2

are

2

3

,

ω

⁢

2

3

,

ω

2

⁢

2

3

where

ω

=

-

1

+

-

3

2

is a

primitive

cube root of unity.

Example 4

f

⁢

(

x

)

=

x

3

-

4

⁢

x

+

2

. This is irreducible since it is Eisenstein at

2

(or by the rational root test), and its discriminant is

202

, which is not a rational square. Thus the Galois group for this polynomial is also

S

3

; note, however, that

f

has three real roots (since

f

⁢

(

0

)

>

0

but

f

⁢

(

1

)

<

0

).

Example 5

f

⁢

(

x

)

=

x

3

-

3

⁢

x

+

1

. This is also irreducible by the rational root test. Its discriminant is

81

, which is a rational square, so the Galois group for this polynomial is

A

3

. Explicitly, the roots of

f

⁢

(

x

)

are

r

1

=

2

⁢

cos

⁡

(

2

⁢

π

/

9

)

,

r

2

=

2

⁢

cos

⁡

(

8

⁢

π

/

9

)

,

r

3

=

2

⁢

cos

⁡

(

14

⁢

π

/

9

)

and we see that

cos

⁡

(

14

⁢

π

/

9

)

=

cos

⁡

(

4

⁢

π

/

9

)

=

2

⁢

cos

2

⁡

(

2

⁢

π

/

9

)

-

1

cos

⁡

(

8

⁢

π

/

9

)

=

2

⁢

cos

2

⁡

(

4

⁢

π

/

9

)

-

1

Let’s consider an automorphism of

K

sending

r

1

to

r

3

, i.e. sending

cos

⁡

(

2

⁢

π

/

9

)

↦

cos

⁡

(

14

⁢

π

/

9

)

. Given the relations above, it is clear that this

mapping

uniquely determines the image of

r

3

as well, since

r

3

=

2

⁢

cos

2

⁡

(

2

⁢

π

/

9

)

-

1

↦

2

⁢

cos

2

⁡

(

4

⁢

π

/

9

)

-

1

=

r

2

and thus we see how the relation imposed by the discriminant actually manifests itself in terms of

restrictions

on the

permutation group

.

Title

Galois group of a cubic polynomial

Canonical name

GaloisGroupOfACubicPolynomial

Date of creation

2013-03-22 17:40:33

Last modified on

2013-03-22 17:40:33

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

10

Author

rm50 (10146)

Entry type

Topic

Classification

msc 12F10

Classification

msc 12D10

Related topic

CardanosFormulae

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