Let F be a field. A normal extension of F is an extension of F obtained by adjoining all zeros of a finite set of polynomials {f1,...,fm}, fj F[X]. If E is a normal extension of F, then E/F is called a Galois extension. An F-automorphism of E is an F-isomorphism of E onto itself. The set of all F-automorphisms of E will be denoted Gal(E/F). Gal(E/F) is called the Galois group of the extension E over F

### The Fundamental Theorem of Galois Theory

Let us now determine what information about an extension E of F can be gleaned from the Galois group. Usually, very little can be said. For example, Example 3 of the section on Galois groups, shows that E can be of degree > 1, while the Galois group is of order 1. In this case one would not expect the Galois group to contain very much information about the extension. In some sense, the Galois group is "too small." However, we have shown that if E is a normal extension of F, then the order of Gal(E/F) equals deg(E/F). This situation looks more hopeful.

Definition 1: If E is a normal extension of F, then we say that E/F is a Galois extension.

Let us fix a Galois extension E/F of degree n and let G = Gal(E/F). By an intermediate field D we will mean a field such that

F D E.

The fundamental theorem of Galois theory allows a complete description of all intermediate fields D in terms of the Galois group G.

One way of manufacturing an intermediate field is as follows: Let H be a subgroup of G, and set

(1)
(H) = {x E | (x) = x for all H}

Then we have

Proposition 2: (H) is an intermediate field called the fixed field of H.

Proof: It is clear that (H) E. Moreover, since H implies that is an F-automorphism of E, we see that F (H). Therefore, it suffices to show that (H) is a field. If x,y (H), then

(x) = x, (y) = y    for all H
(x ± y) = (x) ± (y) = x ± y    for all H
x ± y (H).

Thus, (H) is an additive subgroup of E, If x,y (H) - {0}, then

(x) = x,  (y) = y   for all H
(xy-1) = (x)(y-1) = (x)(y)-1
= xy-1    for all H
xy-1 (H).

Therefore, the nonzero elements of (H) form a group under multiplication and (H) is a field.

We have therefore shown how to go from a subgroup of H of G to an intermediate field (H). One of the main results we will prove below states that every intermediate field is of the form (H) for some subgroup H of G. Thus, the subgroups of G "parametrize" the intermediate fields for the extension E/F.

Proposition 3: Let E/F be a Galois extension, H a subgroup of Gal(E/F). Then E/(H) is a Galois extension and

Gal(E/(H)) = H.

Proof: Since E/F is finite and normal, the same is true of E/(H). Therefore, E/(H) is a Galois extension. Moreover, Gal(E/(H)) consists of all (H)-automorphisms of E. However, if H, then by the definition of (H), is an (H)-automorphism of E. Therefore, Gal(E/(H)) and

(2)
H Gal(E/(H)).

Suppose that H has s elements and Gal(E/(H)) has t elements. Then, by (2),

(3)
s < t.

In order to show that Gal(E/(H)) = H, it therefore suffices to show that s = t. Let us assume that s < t and reason by way of contradiction. Since s + 1 < t, we can find 1,...,s+1 E which are linearly independent over (H). The homogeneous system of equations

(4)
j(i)Xi = 0   (1 < j < s).

has s equations in s+1 unknowns. By Theorem 6 of the section on linear transformations, this system has a solution (c1,...,cs+1),ci E, ci not all zero. Among all nonzero solutions of (4), let us assume that (c1,...,cs+1 is one with as few nonzero entries as possible. By possibly renumbering the i, we may assume that this solution has the form

(5)
(c1,c2,...,cr,0,0,...,0), ci0   (1 < i < r)

Since the system (4) is homogeneous, we may normalize cr to be 1 by replacing (5) by the solution (c1cr-1,c2cr-1,...,1,0,0,...,0). If all ci belong to (H), then

1(i)ci = 0
1(ici) = 0 [since 1 is an (H)-automorphism]
ici = 0,

which is a contradiction to the assumed linear independence of 1,...,s+1 over (H). Therefore, not all ci belong to (H). Without loss of generality, we may assume that ci (H). Then there exists H such that (c1)c1. Apply to the system of equations (4) to get

j(i)(ci) = 0.

Therefore, since j runs over H as j runs over H, we see that

((c1),(c2),...,(cr-1),1,0,0,...,0)

is a solution of the system (4). But the difference of the two solutions of (4) is also a solution of (4), so that we see that

((c1) - c1, (c2) - c2,...,(cr-1) - cr-1,0,0,...,0)

is a solution of (4). Since (c1) - c10, we see that this solution is nonzero having at most r - 1 nonzero entries. But this is a contradiction to our original choice of the solution (c1,...cr,0,0,...0).

In our discussion above, we constructed the correspondence

H(H),

which associates to a subgroup H of Gal(E/F), the intermediate field (H). It is also possible to construct a correspondence which associates to an intermediate field D a subgroup (D) of Gal(E/F) - let us set

(D) = { G | (x) = x for all x D}.

It is clear that

(6)
(D) = Gal(E/D).

Therefore, by Proposition 3, if H is any subgroup of Gal(E/F), then

(7)
((H)) = H.

Thus, if D is any intermediate field,

(8)
(((D))) = (D)    [Equation (7) with H = (D)]
deg(E/((D))) = |(((D)))|
[Equation (6) with D replaced by ((D)), and Theorem 4 of Galois groups.]
(9)
= |(D)| [Equation (8)]
= deg(E/D)
[Equation (6) and Theorem 4 of Galois groups]

But it is clear that

((D)) D,

so that by Equation (9),

((D))/D) = 1 ((D)) = D.

Thus, we see that if D is any intermediate field, then

(10)
((D)) = D.

Not that this last equation implies that the intermediate field D is of the form (H), where H = (D).

Finally we may state

Theorem 4 (Fundamental Theorem of Galois Theory): Let E/F be a Galois extension with Galois group G. Then the correspondence

H (H)

is a one-to-one mapping of the set of subgroups G onto the set of intermediate fields for the extension E/F. Moreover, if H corresponds to D under this correspondence, then E/D is a Galois extension and Gal(E/D) = H.

Proof: The face that the correspondence H(H) is one to one follows from Equation (7), since (H1) = (H2) ((H1)) = ((H2)) H1 = H2. We have already noted above that every intermediate field is of the form (H), and thus the correspondence is onto. This last assertion follows from Proposition 3.

Corollary 5: Let E/F be a Galois extension. There are only finitely many intermediate fields D such that F D E.

This result is somewhat surprising.

Proof: Gal(E/F) is a finite group and therefore has only finitely many subgroups. Thus, by the fundamental theorem, there are only finitely many intermediate fields.

Example 1: Let F = Q, E = Q(,). Then Gal(E/F) is of order 4 and consists of the following elements:

The subgroups of Gal(E/F) are

H1 = {1}, H2 = {1, 2}, H3 = {1, 3},
H4 = {1, 4}, H5 = {1,2, 3, 4}.

The corresponding fixed fields are

(H1) = Q(,),
(H2) = Q(),
(H3) = Q(),
(H4) = Q(),
(H5) = Q.

[The only computation which deserves comment is the one leading to the assertion (H4) = Q(). It is clear that = · (H4), so that Q() (H4). But H4 has order 2, so that by the fundamental theorem and Proposition 4 of Galois groups, we see that deg(Q(,)/(H4)) = 2. However, deg(Q(,)/Q()) = 2, so that deg((H4)/Q()) = 1 Q() = (H4).]

As a consequence of the fundamental theorem, we see that Q(,) has only five subfields: Q, Q(), Q(), Q(), and Q(,).

We shall refer to the correspondence H (H) as the Galois correspondence. Let us conclude this section by proving two very fundamental facts concerning the Galois correspondence.

Theorem 6: Let E/F be a Galois extension with Galois group G. Let H1 and H2 be subgroups of G, and let D1 and D2, respectively, correspond to H1 and H2 under Galois correspondence. Then

(1) H1 H2 if and only if D1 D2.

(2) The intermediate field corresponding to H1H2 is D1D2 where by D1D2 we mean the smallest subfield of E containing both D1 and D2.

(3) The intermediate field corresponding to [H1 H2] is D1 D2.

Proof: By the fundamental theorem

D1 = (H1),    D2 = (H2).

(1) It is clear that H1 H2 implies that (H1) (H2). Conversely, if (H1) (H2), then ((H1)) ((H2)), so that H1 H2 by Equation (7).

(2) H1 H2 is the largest subgroup of G contained in both H12. Therefore, by (1) and the fundamental theorem, (H1 H2) is the smallest subfield of E containing (H1) and (H2).

(3) [H1 H2] is the smallest subgroup of G containing both H1 and H2. Therefore, by (1) and the fundamental theorem, ([H1 H2]) is the largest subfield contained in both (H1) and (H2). That is, ([H1 H2]) = (H1) (H2).

Perhaps the most useful consequence of the fundamental theorem is the following result, which will be used many times throughout this chapter.

Corollary 7: Let E/F be a Galois extension with Galois group G, and let x E. If (x) = x for all G, Then x F.

Proof: If (x) = x for all G, then x (G). Now by the definition of G, (F) = G and thus ((F)) = (G). But by (10), ((F)) = F. Therefore, x F.

Theorem 8: Let E/F be a Galois extension and let G = Gal(E/F). Further, let D be an intermediate field and let H = (D). Then

(1) D/F is a normal extension if and only if H is a normal subgroup of G.

(2) If H is a normal subgroup of G, then D/F is a Galois extension and Gal(D/F)G/H.

Proof: Let G. Let us first show that

(11)
(D) = (D) -1.

For if (D), then -1 is an F-automorphism of D and therefore (D) (D) -1. If (D). then -1 is an F-automorphism of D, so that -1 (D) and (D) -1. Therefore, (D) (D) -1. Thus (11) is proved.

(1)Note that

D/F is normal

D = D for all Gal(E/F)

(D) = (D) for all Gal(E/F) (by Theorem 6)

H -1 = H for all Gal(E/F) (by (11))

H is a normal subgroup of G.

(2) By part(1), if H is a normal subgroup of G, then D/F is normal. Moreover, it is clear that D/F is finite. Thus D/F is a Galois extension. Let us consider the mapping

:Gal(E/F) Gal(D/F)

which arises by mapping Gal(E/F) into its restriction to D. It is clear that this mapping is a group homomorphism. Moreover, since every F-automorphism of D can be extended to an F-automorphism of E, this homomorphism is surjective. Finally,

ker = { Gal(E/F) | = the identity on D}
= H.

Therefore, by the first isomorphism theorem,

G/HGal(D/F).