proof of fundamental theorem of Galois theory

proof of fundamental theorem of Galois theory

The

theorem

is a

consequence

of the following lemmas, roughly
corresponding to the various assertions in the theorem. We assume

L

/

F

to be a

finite-dimensional

Galois extension

of fields with

Galois group

G

=

Gal

⁡

(

L

/

F

)

.

The first two lemmas establish the correspondence between

subgroups

of

G

and

extension fields

of

F

contained

in

L

.

Lemma 1

.

Let

K

be an extension field of

F

contained in

L

. Then

L

is
Galois over

K

, and

Gal

⁡

(

L

/

K

)

is a subgroup of

G

.

Proof.

Note that

L

/

F

is normal and

separable

because it is a Galois
extension; it remains to prove that

L

/

K

is also normal and
separable. Since

L

is normal and finite over

F

, it is the

splitting field

of a

polynomial

f

∈

F

⁢

[

X

]

over

F

. Now

L

is also
the splitting field of

f

over

K

(because

F

⊂

K

⊂

L

),
so

L

/

K

is normal.

To see that

L

/

K

is also separable, suppose

α

∈

L

, and let

f

F

α

∈

F

⁢

[

X

]

be its

minimal polynomial

over

F

. Then the
minimal polynomial

f

K

α

of

α

over

K

divides

f

F

α

, which has no double roots in its splitting field by the
separability of

L

/

F

. Therefore

f

K

α

has no double roots in
its splitting field for any

α

∈

L

, so

L

is separable over

K

.

The assertion that

Gal

⁡

(

L

/

K

)

is a subgroup of

G

is clear from the
fact that

K

⊃

F

.
∎

Lemma 2

.

The

function

ϕ

from the set of extension fields of

F

contained
in

L

to the set of subgroups of

G

defined by

ϕ

⁢

(

K

)

=

Gal

⁡

(

L

/

K

)

is an inclusion-reversing

bijection

. The

inverse

is given by

ϕ

-

1

⁢

(

H

)

=

L

H

,

where

L

H

is the

fixed field

of

H

in

L

.

Proof.

The definition of

ϕ

makes sense because of Lemma

1

.
The

ϕ

-

1

∘

ϕ

⁢

(

K

)

=

K

and

ϕ

∘

ϕ

-

1

⁢

(

H

)

=

H

for all subgroups

H

⊂

G

and all fields

K

with

F

⊂

K

⊂

L

follow from the properties of the Galois group. The fixed
field of

Gal

⁡

(

L

/

K

)

is precisely

K

; on the other hand, since

L

H

is the fixed field of

H

in

L

,

H

is the Galois group of

L

/

L

H

.

For

extensions

K

and

K

′

of

F

with

F

⊂

K

⊂

K

′

⊂

L

, we have

σ

∈

Gal

⁡

(

L

/

K

′

)

⇔

σ

∈

Gal

⁡

(

L

/

K

)

,

so

ϕ

⁢

(

K

)

⊃

ϕ

⁢

(

K

′

)

. This shows that

ϕ

is
inclusion-reversing.
∎

The following lemmas show that normal subextensions of

L

/

F

are
Galois extensions and that their Galois groups are

quotient groups

of

G

.

Lemma 3

.

Let

H

be a subgroup of

G

. Then the following are

equivalent

:

1.

L

H

is normal over

F

.

2.

σ

⁢

(

L

H

)

=

L

H

for all

σ

∈

G

.

3.

σ

⁢

H

⁢

σ

-

1

=

H

for all

σ

∈

G

.

In particular,

L

H

is normal over

F

if and only if

H

is a

normal
subgroup

of

G

.

Proof.

1

⇒

2

: Since for all

σ

∈

G

and

α

∈

L

H

,

σ

⁢

(

α

)

is a zero of the minimal polynomial of

α

over

F

, we have

σ

⁢

(

α

)

∈

L

H

by the
of

L

H

/

F

.

2

⇒

3

: For all

σ

∈

G

,

τ

∈

H

the equality

σ

⁢

τ

⁢

σ

-

1

⁢

(

x

)

=

σ

⁢

σ

-

1

⁢

(

x

)

=

x

holds for all

x

∈

L

H

(from the

assumption

it follows that

σ

-

1

⁢

(

x

)

∈

L

H

, which is fixed by

τ

). This implies that

σ

⁢

τ

⁢

σ

-

1

∈

Gal

⁡

(

L

/

L

H

)

=

H

for all

σ

∈

G

,

τ

∈

H

.

3

⇒

1

: Let

α

∈

L

H

, and let

f

be the minimal
polynomial of

α

over

F

. Since

L

/

F

is normal,

f

splits
into linear factors in

L

⁢

[

X

]

. Suppose

α

′

∈

L

is another zero
of

f

, and let

σ

∈

G

be such that

σ

⁢

(

α

′

)

=

α

(such a

σ

always exists). By assumption, for all

τ

∈

H

we have

τ

′

:=

σ

⁢

τ

⁢

σ

-

1

∈

H

, so that

τ

⁢

(

α

′

)

=

σ

-

1

⁢

τ

′

⁢

σ

⁢

(

α

′

)

=

σ

-

1

⁢

τ

′

⁢

(

α

)

=

σ

-

1

⁢

(

α

)

=

α

′

.

This shows that

α

′

lies in

L

H

as well, so

f

splits in

L

H

⁢

[

X

]

. We conclude that

L

H

is normal over

F

.
∎

Lemma 4

.

Let

H

be a normal subgroup of

G

. Then

L

H

is a Galois extension
of

F

, and the

homomorphism

r

:

G

→

Gal

⁡

(

L

H

/

F

)

σ

↦

σ

|

L

H

induces a natural identification

Gal

⁡

(

L

H

/

F

)

≅

G

/

H

.

Proof.

By Lemma

3

,

L

H

is normal over

F

, and because a
subextension of a

separable extension

is separable,

L

H

/

F

is a
Galois extension.

The map

r

is well-defined by the

implication

1

⇒

2

from
Lemma

3

. It is

surjective

since every

automorphism

of

L

H

that fixes

F

can be extended to an automorphism of

L

(if

L

≠

L

H

, for example, we can choose an

α

∈

L

∖

L

H

such that

L

=

L

H

⁢

(

α

)

using the

primitive element theorem

, and we
can extend

σ

∈

Gal

⁡

(

L

H

/

F

)

to

L

by putting

σ

⁢

(

α

)

=

α

). The kernel of

r

is clearly equal to

H

,
so the

first isomorphism theorem

gives the claimed identification.
∎

Title

proof of fundamental theorem of Galois theory

Canonical name

ProofOfFundamentalTheoremOfGaloisTheory

Date of creation

2013-03-22 14:26:38

Last modified on

2013-03-22 14:26:38

Owner

pbruin (1001)

Last modified by

pbruin (1001)

Numerical id

5

Author

pbruin (1001)

Entry type

Proof

Classification

msc 12F10

Classification

msc 11R32

Classification

msc 11S20

Classification

msc 13B05

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