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Let sigma element of G(E/F). Then sigma element of Gal(E/F) if and only if sigma(E) subset of E. 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 one element of a residue class modulo n is relatively prime to n, every element of the residue class is relatively prime to n. When this occurs we say that the residue class is reduced. Let Zxn denote the set of all reduced residue classes modulo n. The set of rational numbers. The set residue classes mod n. Let E be an extension of F of degree n, and let beta element of E be such that E = F(beta). Further, let beta1,...,betan be the conjugates of beta over F. Then

(1) beta1,...,betan are distinct.

(2) For each i (1 < i < n), the exists an F-isomorphism sigma of E into F such that sigmai(beta) = betai. Moreover, sigmai is unique and is given by sigmai(b0 + b1beta + ... + bn-1betan-1) = b0 + b1betai + ... + bn-1betain-1,   bj element of F.

(3) There are exactly n F-isomorphisms of E into F and these are sigma1,...,sigman.

The Galois Group of an Extension

Let F be a field and let E be a finite extension of F. Our objective at this point is to associate a finite group to the extension E/F, called the Galois group. We will show how it is possible to utilize the Galois group to describe the arithmetic of the extension. In this section we will limit ourselves to defining the Galois group and computing some examples.

Definition 1: 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).

Since an F-automorphism of E is an F-isomorphism of E into F, we see that Gal(E/F) subset of G(E/F). Therefore, Theorem 4 of the section on conjugates implies the following result.

Lemma 2: Suppose that deg(E/F) = n. Then Gal(E/F) contains at most n elements.

The following lemma helps to pick out the elements of Gal(E/F) from G(E/F).

Lemma 3: Let sigma element of G(E/F). Then sigma element of Gal(E/F) if and only if sigma(E) subset of E.

Proof: implies Obvious.

backwards implication It suffices to show that if sigma(E) subset of E, then sigma(E) = E. Suppose the deg(E/F) = n, and let {alpha1,...,alphan} be a basis of E over F. Since sigma is an F-isomorphism, {sigma(alpha1),...,sigma(alphan)} is a basis of sigma(E) over F. Thus, deg(sigma(E)/F) = n. However, since sigma(E) subset of E,

deg(E/F) = deg(sigma(E)/F) · deg(E/sigma(E)),

and thus deg(E/sigma(E)) = 1, which implies that E = sigma(E).

By Lemma 3 and Theorem 4 of the section on conjugates, we can deduce the following description of Gal(E/F):

Theorem 4: Let E be an extension of F of degree n, and let beta element of E be such that E = F(beta). Further, let beta1,...,betan be the conjugates of beta over F and let sigmai element of G(E/F) be such that sigmai(beta) = betai. Then sigmai element of Gal(E/F) if and only if betai element of E. In particular, if E is a normal extension of F, then all betai belong to E and Gal(E/F) contains n elements.

Let us define a group structure on Gal(E/F). If sigma,eta element of Gal(E/F). the the composite function etasigma, defined by

(etasigma)(x) = eta(sigma(x))

is an F-automorphism of E. Indeed, etasigma is a homomorphism of E into E such that (etasigma)(1) = 1. Therefore, etasigma is an isomorphism by Proposition 2 of the introduction to field theory. And since eta and sigma are F-isomorphisms, if x element of F, we have (etasigma)(x) = eta(sigma(x)) = eta(x) = x. Therefore, etasigma is an F-isomorphism of E into E, so that by Lemma 3, etasigma is an F-automorphism of E. Let us define the product of eta and sigma to be the F-automorphism etasigma. With respect to this law of multiplication, Gal(E/F) is a group. Indeed, the identity automorphism i defined i(x) = x for all x element of E. If eta element of Gal(E/F), let eta-1 denote the inverse function of eta, when eta is considered as a bijection from E onto E. Then eta-1 is an F-automorphism of E, and etaeta-1 = eta-1eta = i.

Definition 5: Gal(E/F) is called the Galois group of the extension E over F.

Let us compute Gal(E/F) in a few examples.

Example 1: F = Q, E = Q(square root of 2). Then E is a normal extension of F and Gal(E/F) = {sigma0, sigma1}, where

sigma0(square root of 2) = square root of 2,    sigma1(square root of 2) = -square root of 2.

The mapping psi:Gal(E/F)mapsZ2 defined by psi(sigmaa) = a (mod 2) is a surjective isomorphism. Therefore, Gal(E/F)isomorphic withZ2.

Example 2: F = Q, E = Q(zetam), where zetam is a primitive mth root of 1. Then E is a normal extension of F of degree phi(n). The conjugates of zetam are zetama (0 < a < m - 1, (a,m) = 1). Therefore, Gal(E/F) has order phi(n) and consists of the elements sigmaa (0 < a < m -1, (a,m) = 1) defined by

sigmaa(zetam) = zetama.

Notice that

sigmaasigmab(zetam) = sigmaa(zetamb) = sigmaa(zetam)b = zetamab.

Therefore, if q and c are integers such that ab = qm + c, 0 < c < m - 1, we have

(1)
sigmaasigmab(zetam) = zetamqm+c = (zetamm)q zetamc = sigmac(zetam)

since zetamm = 1. Let psi:Gal(E/F)mapsZmx be the mapping defined by psi(sigmaa) = a (mod m). Then formula (1) implies that

psi(sigmaasigmab) = c(mod m) = ab(mod m) = psi(sigmaa)psi(sigmab).

Therefore, psi is a homomorphism. It is clear that psi is a bijection, and thus

Gal(E/F)isomorphic withZmx.

Example 3: F = Q, E = Q(cube root of 2), where cube root of 2 denotes the real cube root of 2. The conjugates of cube root of 2 over Q are cube root of 2, cube root of 2omega, cube root of 2omega2, where omega = (-1 + square root of 3i)/2. The only conjugate of cube root of 2 which lies in E is cube root of 2, so that

Gal(E/F) = {sigma0},

where sigma0(cube root of 2) = cube root of 2. Note that deg(E/F) = 3, but the order of Gal(E/F) is 1. This is not a contradiction to our above discussion since E is not a normal extension of F.