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/ R. Hence Gal(Q(f (X))/Q) is isomorphic to a subgroup of S3 and so
∼ 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) ƒƒ Q( 2 ζ3 ) ii
ii
N N rr I
ƒƒƒ rr ii
ƒƒƒ
rr
ƒƒƒ ii
ƒƒƒ rr ii
ƒƒƒ rr
ƒƒƒ3 rr3
ƒƒƒ rr Q(ζ3 )
3
ƒƒƒ rr kkk Q
ƒƒƒ rr 2kkkkk
ƒƒƒ rr
ƒƒƒ k
ƒƒƒ rrr kkk
kkk
ƒƒƒr
ƒ 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
‰‰‰‰‰‰ ƒƒƒƒ
‰‰‰‰‰‰ yy
ƒƒƒƒ
‰‰‰‰‰‰ yy
‰‰‰‰‰‰ ƒƒƒƒƒƒ 2
yy
2 2‰‰‰ ƒƒƒ
‰‰‰‰‰‰  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,

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