∼ S3 since the orders agree. We also have Q(f (X)) © R = Q(u1 ).

Gal(Q(f (X))/Q) =

The Galois group Gal(Q(f (X))/Q) contains an element of order 3 which corresponds to a

3-cycle when viewed as a permutation of the roots u1 , u2 , u3 ; we can assume that this is (1 2 3).

It also contains an element of order 2 obtained by restricting complex conjugation to Q(f (X));

this ¬xes u1 and interchanges u2 , u3 , so it corresponds to the transposition (2 3).

Remark 4.10. Such examples occur when the cubic polynomial f (X) has local maximum

and minimum at real values c+ and c’ with f (c+ ), f (c’ ) > 0 or f (c+ ), f (c’ ) < 0. This happens

for example with f (X) = X 3 ’ 3X + 3 which has local extrema at ±1 and f (1) = 1, f (’1) = 5.

Given a Galois extension E/K, we will next study subextensions L/K E/K and sub-

groups “ Gal(E/K), focusing on the relationship between objects of these types.

4.3. Subgroups of Galois groups and their ¬xed ¬elds

Let E/K a Galois extension and suppose that “ Gal(E/K). Consider the subset of

elements of E ¬xed by “,

E “ = {u ∈ E : ∀γ ∈ “, γ(u) = u}.

Lemma 4.11. E “ E is a sub¬eld of E containing K.

Proof. For u, v ∈ E “ and γ ∈ “,

γ(u + v) = γ(u) + γ(v) = u + v, γ(uv) = γ(u)γ(v) = uv.

Also, if u = 0,

γ(u’1 ) = γ(u)’1 = u’1 .

Finally, if t ∈ K then γ(t) = t, so K E “ .

50 4. GALOIS EXTENSIONS AND THE GALOIS CORRESPONDENCE

Definition 4.12. E “ E is the ¬xed sub¬eld of “.

By Proposition 3.73, the extensions E/E “ and E “ /K are separable. E/E “ is also normal,

so this is itself a Galois extension. We will identify its Galois group. Notice that

[E : E “ ] = (E : E “ ) = | Gal(E/E “ )|.

Now each element of Gal(E/E “ ) is also an element of Gal(E/K) and Gal(E/E “ ) Gal(E/K).

Gal(E/E “ ), so Lagrange™s Theorem implies that |“| divides

Notice that by de¬nition “

| Gal(E/E “ )|. In fact we have

Gal(E/K), we have Gal(E/E “ ) = “ and the equations

Proposition 4.13. For “

| Gal(E/K)|

[E : E “ ] = | Gal(E/E “ )| = |“|, [E “ : K] = .

|“|

Proof. We know that E/E “ is separable, so by the Primitive Element Theorem 3.75 it is

simple, say E = E “ (u). Now let the distinct elements of “ be γ1 = id, γ2 , . . . , γh , where h = |“|.

Consider the polynomial of degree h

f (X) = (X ’ u)(X ’ γ2 (u)) · · · (X ’ γh (u)) ∈ E[X].

Notice that f (X) is unchanged by applying any γk to its coe¬cients since the roots γj (u) are

permuted by γk . Hence, f (X) ∈ E “ [X]. This shows that

[E : E “ ] = [E “ (u) : E “ ] h = |“|.

Gal(E/E “ ), we also have

Since “

| Gal(E/E “ )| = [E : E “ ].

h = |“|

Combining these two inequalities we obtain

[E : E “ ] = | Gal(E/E “ )| = |“| = h

and therefore “ = Gal(E/E “ ).

4.4. Sub¬elds of Galois extensions and relative Galois groups

Let E/K a Galois extension and suppose that K L E. Then E/L is also a Galois

extension whose Galois group Gal(E/L) is sometimes called the relative Galois group of the

pair of extensions E/K and L/K. The following is immediate.

Lemma 4.14. The relative Galois group of the pair of extensions E/K and L/K is a subgroup

of Gal(E/K), i.e., Gal(E/L) Gal(E/K), and its order is | Gal(E/L)| = [E : L].

E. Then L = E Gal(E/L) .

Proposition 4.15. Let K L

Proof. Clearly L E Gal(E/L) . Suppose that u ∈ E ’ L. Then there is an automorphism

θ ∈ Gal(E/L) such that θ(u) = u, hence u ∈ E Gal(E/L) . This shows that E Gal(E/L)

/ L and

Gal(E/L) = L.

therefore E

We want to understand when Gal(E/L) Gal(E/K) is actually a normal subgroup. The

next result explains the connection between the two uses of the word normal which both derive

from the Galois theoretic usage.

Proposition 4.16. Let E/K be a ¬nite Galois extension and L/K E/K a subextension.

(i) The relative Galois group Gal(E/L) of the pair of extensions E/K and L/K is normal

in Gal(E/K) if and only if L/K is normal.

(ii) If L/K is normal and hence a Galois extension, then there is a group isomorphism

∼

=

Gal(E/K)/ Gal(E/L) ’ Gal(L/K);

’ ± Gal(E/L) ’’ ±|L .

4.5. THE GALOIS CORRESPONDENCE 51

Proof. (i) Suppose that Gal(E/L) Gal(E/K). Then if ± ∈ Gal(E/L) and β ∈ Gal(E/K),

we have β±β ’1 ∈ Gal(E/L). Now if u ∈ L, then for any γ ∈ Gal(E/K) and ± ∈ Gal(E/L),

γ(u) ∈ E satis¬es

±γ(u) = γ(γ ’1 ±γ(u)) = γ(u),

since γ ’1 ±γ ∈ Gal(E/L). So L/K is normal.

Conversely, if L/K is normal, then for every • ∈ Gal(E/K) and v ∈ L, •(v) ∈ L, so for

every θ ∈ Gal(E/L), θ(•(v)) = •(v) and therefore

•’1 θ•(v) = v.

This shows that •’1 θ• ∈ Gal(E/L). Hence for every • ∈ Gal(E/K),

• Gal(E/L)•’1 = Gal(E/L),

which shows that Gal(E/L) Gal(E/K).

(ii) If ± ∈ Gal(E/K), then ±L = L since L/K is normal. Hence we can restrict ± to an

automorphism of L,

±|L : L ’’ L; ±|L (u) = ±(u).

Then ±|L is the identity function on L if and only if ± ∈ Gal(E/L). It is easy to see that the

function

Gal(E/K) ’’ Gal(L/K); ± ’’ ±|L

is a group homomorphism whose kernel is Gal(E/L). Thus we obtain an injective homomor-

phism

Gal(E/K)/ Gal(E/L) ’’ Gal(L/K)

for which

[E : K]

| Gal(E/K)/ Gal(E/L)| = = [L : K] = | Gal(L/K)|.

[E : L]

Hence this must be an isomorphism of groups.

4.5. The Galois Correspondence

We are now almost ready to state our central result which describes the Galois Correspon-

dence associated with a ¬nite Galois extension. We will use the following notation. For a ¬nite

Galois extension E/K, let

S(E/K) = the set of all subgroups of Gal(E/K);

F(E/K) = the set of all subextensions L/K of E/K.

These sets are ordered by inclusion. Notice that since every subgroup of a ¬nite group is

equivalent to its underlying set, S(E/K) is a ¬nite set. De¬ne two functions by

¦E/K : F(E/K) ’’ S(E/K); ¦E/K (L) = Gal(E/L),

˜E/K : S(E/K) ’’ F(E/K); ˜E/K (“) = E “ .

Theorem 4.17 (Main Theorem of Galois Theory). Let E/K be a ¬nite Galois extension.

Then the functions ¦E/K and ˜E/K are mutually inverse bijections which are order-reversing.

¦E/K

/

F(E/K) o S(E/K)

˜E/K

Under this correspondence, normal subextensions of E/K correspond to normal subgroups of

Gal(E/K) and vice versa.

Proof. We know from Proposition 4.15 that for an extension L/K in F(E/K),

˜E/K (¦E/K (L)) = ˜E/K (Gal(E/L)) = E Gal(E/L) = L.

Also, by Proposition 4.13 for H ∈ S(E/K) we have

¦E/K (˜E/K (“)) = ¦E/K (E “ ) = Gal(E/E “ ) = “.

52 4. GALOIS EXTENSIONS AND THE GALOIS CORRESPONDENCE

This shows that ¦E/K and ˜E/K are mutually inverse and so are inverse bijections.

Let L1 /K, L2 /K ∈ F(E/K) satisfy L1 /K L2 /K. Then Gal(E/L2 ) Gal(E/L1 ) since

L1 ⊆ L2 and so if ± ∈ Gal(E/L2 ) then ± ¬xes every element of L1 . Hence ¦E/K (L2 ) ¦E/K (L1 )

and so ¦E/K reverses order.

“2 , then E “2 E “1 since if w ∈ E “2 then it is

Similarly, if “1 , “2 ∈ S(E/K) with “1

¬xed by every element of “1 which is a subset of “2 . Hence ˜E/K reverses order.

There is an immediate consequence of the Main Theorem 4.17 which is closely related to

Proposition 3.20.

Corollary 4.18. Let E/K be a ¬nite Galois extension. Then there are only ¬nitely many

subextensions L/K E/K.

Proof. Since the set S(E/K) is ¬nite, so is F(E/K).

When dealing with a ¬nite Galois extension E/K, we indicate the subextensions in a diagram

with a line going upwards indicating an inclusion. We can also do this with the subgroups of

the Galois group Gal(E/K) with labels indicating the index of the subgroups. In e¬ect, the

Galois Correspondence inverts these diagrams.

Example 4.19. Figure 4.1 displays the Galois Correspondence for the extension of Exam-

ple 3.30.

√

E = Q( 3 2, ζ3 )

ee kk H ii

eeeeee k ii

eeeeee kkkkkk

e ii

eeeee2 2

k

eeeeee ii

kkk 2

√ eeeeeeee k ii

√2

√ ii

3

3

3 3

Q( 2 ζ3 )

Q( 2) Q( 2 ζ3 ) ii

ii

N N rr I

rr ii

rr

ii

rr ii

rr

3 rr3

rr Q(ζ3 )

3

rr kkk Q

rr 2kkkkk

rr

k

rrr kkk

kkk

r

kkk

QS

¦E/K

Gal(E/Q)

‚‚‚

kkk v ‚‚‚ 2

kkk vvvv ‚‚‚

kk v ‚‚‚

kkk v ‚‚

kkk vv

k

3 kkkk 3 vv

v

k Gal(E/Q(ζ3 ))

kkk vv 3

kkk vv yy

kk v

kkk yy

vv

k yy

kkk vv

√ kkk √v yy

√

v y

2 )) 3 yy

3

3 3

Gal(E/Q( 2 ζ3 yy

Gal(E/Q( 2)) Gal(E/Q( 2 ζ3 ))

yy

yy

yy

2

yy

2 2

y

{id}

√

Figure 4.1. The Galois Correspondence for E = Q( 3 2, ζ3 )/Q

√

As noted at the end of Example 3.38, the Galois group here is Gal(Q( 3 2, ζ3 )/Q) ∼ S3 . It

=

is useful to make this isomorphism explicit. First take the √ roots of the polynomial X 3 ’ 2 for

3

√√ 2

which E is the splitting ¬eld over Q; these are 2, 2 ζ3 , 3 2 ζ3 which we number in the order

3 3

4.6. GALOIS EXTENSIONS INSIDE THE COMPLEX NUMBERS AND COMPLEX CONJUGATION 53

they are listed. Then the monomorphisms id, ±0 , ±1 , ±1 , ±2 , ±2 extend to automorphisms of E,

each of which permutes these 3 roots in the following ways given by cycle notation:

±0 = (2 3), ±1 = (1 2 3), ±1 = (1 2), ±2 = (1 3 2), ±2 = (1 3).

We ¬nd that

√

Gal(E/Q(ζ3 )) = {id, ±1 , ±2 } ∼{id, (1 2 3), (1 3 2)}, Gal(E/Q( 2)) = {id, ±0 } ∼{id, (2 3)},

3

= =

√ √2

Gal(E/Q( 2 ζ3 )) = {id, ±2 } ∼{id, (1 3)}, Gal(E/Q( 2 ζ3 )) = {id, ±1 } ∼{id, (1 2)}.

3 3

= =

Notice that {id, (1 2 3), (1 3 2)} S3 and so Q(ζ3 )/Q is a normal extension. Of course Q(ζ3 )

is the splitting ¬eld of X 3 ’ 1 over Q.

4.6. Galois extensions inside the complex numbers and complex conjugation

When working with Galois extensions contained in the complex numbers it is often useful

to make use of complex conjugation as an element of a Galois group. Let E/Q be a ¬nite Galois

extension with E/Q C/Q. Setting ER = R © E, we have Q ER E.

Proposition 4.20. Complex conjugation ( ) : C ’’ C restricts to an automorphism of E

over Q, ( )E/Q : E ’’ E.

(i) ( )E/Q which agrees with the identity function if and only if ER = E.

(ii) If ER = E then

( )E/Q = {id, ( )E/Q } ∼ Z/2,