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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 element of 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 subset of D subset of 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)
f(H) = {x element of E | sigma(x) = x for all sigma element of H}

Then we have

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

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

sigma(x) = x, sigma(y) = y    for all sigma element of H
impliessigma(x ± y) = sigma(x) ± sigma(y) = x ± y    for all sigma element of H
implies x ± y element of f(H).

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

sigma(x) = x,  sigma(y) = y   for all sigma element of H
implies(xy-1) = sigma(x)sigma(y-1) = sigma(x)sigma(y)-1
= xy-1    for all sigma element of H
impliesxy-1 element of f(H).

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

We have therefore shown how to go from a subgroup of H of G to an intermediate field f(H). One of the main results we will prove below states that every intermediate field is of the form f(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/f(H) is a Galois extension and

Gal(E/f(H)) = H.

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

(2)
H subset of Gal(E/f(H)).

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

(3)
s < t.

In order to show that Gal(E/f(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 alpha1,...,alphas+1 element of E which are linearly independent over f(H). The homogeneous system of equations

(4)
sum to s+1sigmaj(alphai)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 element of 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 alphai, we may assume that this solution has the form

(5)
(c1,c2,...,cr,0,0,...,0), cinot equal0   (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 f(H), then

sum to s + 1sigma1(alphai)ci = 0
implies sigma1(sum to s + 1alphaici) = 0 [since sigma1 is an f(H)-automorphism]
implies sum to s + 1alphaici = 0,

which is a contradiction to the assumed linear independence of alpha1,...,alphas+1 over f(H). Therefore, not all ci belong to f(H). Without loss of generality, we may assume that ci not an element of f(H). Then there exists sigma element of H such that sigma(c1)not equalc1. Apply sigma to the system of equations (4) to get

sum to s + 1sigmasigmaj(alphai)sigma(ci) = 0.

Therefore, since sigmasigmaj runs over H as sigmaj runs over H, we see that

(sigma(c1),sigma(c2),...,sigma(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

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

is a solution of (4). Since sigma(c1) - c1not equal0, 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

Hmapsf(H),

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

g(D) = {sigma element of G | sigma(x) = x for all x element of D}.

It is clear that

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

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

(7)
g(f(H)) = H.

Thus, if D is any intermediate field,

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

But it is clear that

f(g(D)) super set D,

so that by Equation (9),

f(g(D))/D) = 1 implies f(g(D)) = D.

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

(10)
f(g(D)) = D.

Not that this last equation implies that the intermediate field D is of the form f(H), where H = g(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 maps f(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 Hmapsf(H) is one to one follows from Equation (7), since f(H1) = f(H2) implies g(f(H1)) = g(f(H2)) implies H1 = H2. We have already noted above that every intermediate field is of the form f(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 subset of D subset of 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(square root of 2,square root of 3). Then Gal(E/F) is of order 4 and consists of the following elements:

psi 1   psi 2
psi 3   psi 4

The subgroups of Gal(E/F) are

H1 = {psi1}, H2 = {psi1, psi2}, H3 = {psi1, psi3},
H4 = {psi1, psi4}, H5 = {psi1,psi2, psi3, psi4}.

The corresponding fixed fields are

f(H1) = Q(square root of 2,square root of 3),
f(H2) = Q(square root of 3),
f(H3) = Q(square root of 2),
f(H4) = Q(square root of 6),
f(H5) = Q.

[The only computation which deserves comment is the one leading to the assertion f(H4) = Q(square root of 6). It is clear that square root of 6 = square root of 2 · square root of 3 element of f(H4), so that Q(square root of 6) subset of f(H4). But H4 has order 2, so that by the fundamental theorem and Proposition 4 of Galois groups, we see that deg(Q(square root of 2,square root of 3)/f(H4)) = 2. However, deg(Q(square root of 2,square root of 3)/Q(square root of 6)) = 2, so that deg(f(H4)/Q(square root of 6)) = 1 implies Q(square root of 6) = f(H4).]

As a consequence of the fundamental theorem, we see that Q(square root of 2,square root of 3) has only five subfields: Q, Q(square root of 6), Q(square root of 2), Q(square root of 3), and Q(square root of 2,square root of 3).

We shall refer to the correspondence H maps f(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 subset of H2 if and only if D1 superset of D2.

(2) The intermediate field corresponding to H1intersection ofH2 is D1D2 where by D1D2 we mean the smallest subfield of E containing both D1 and D2.

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

Proof: By the fundamental theorem

D1 = f(H1),    D2 = f(H2).

(1) It is clear that H1 subset of H2 implies that f(H1) superset of f(H2). Conversely, if f(H1) superset of f(H2), then g(f(H1)) subset of g(f(H2)), so that H1 subset of H2 by Equation (7).

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

(3) [H1 union H2] is the smallest subgroup of G containing both H1 and H2. Therefore, by (1) and the fundamental theorem, f([H1 union H2]) is the largest subfield contained in both f(H1) and f(H2). That is, f([H1 union H2]) = f(H1) intersection f(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 element of E. If sigma(x) = x for all sigma element of G, Then x element of F.

Proof: If sigma(x) = x for all sigma element of G, then x element of f(G). Now by the definition of G, g(F) = G and thus f(g(F)) = f(G). But by (10), f(g(F)) = F. Therefore, x element of 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 = g(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)isomorphic toG/H.

Proof: Let sigma element of G. Let us first show that

(11)
g(sigmaD) = sigmag(D)sigma -1.

For if eta element of g(D), then sigmaetasigma -1 is an F-automorphism of sigmaD and therefore g(sigmaD) superset of sigmag(D)sigma -1. If lambda element of g(sigmaD). then sigmalambdasigma -1 is an F-automorphism of D, so that sigma -1lambdasigma element of g(D) and lambda element of sigmag(D)sigma -1. Therefore, g(sigmaD) subset of sigmag(D)sigma -1. Thus (11) is proved.

(1)Note that

D/F is normal

equivalentsigmaD = D for all sigma element of Gal(E/F)

equivalent g(sigmaD) = g(D) for all sigma element of Gal(E/F) (by Theorem 6)

equivalent sigmaHsigma -1 = H for all sigma element of Gal(E/F) (by (11))

equivalent 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

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

which arises by mapping sigma element of 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 psi = {sigma element of Gal(E/F) | sigma = the identity on D}
= H.

Therefore, by the first isomorphism theorem,

G/Hisomorphic toGal(D/F).