Let G(E/F). Then Gal(E/F) if and only if (E) 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 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 E be such that E = F(). Further, let 1,...,n be the conjugates of over F. Then

(1) 1,...,n are distinct.

(2) For each i (1 < i < n), the exists an F-isomorphism of E into F such that i() = i. Moreover, i is unique and is given by i(b0 + b1 + ... + bn-1n-1) = b0 + b1i + ... + bn-1in-1,   bj F.

(3) There are exactly n F-isomorphisms of E into F and these are 1,...,n.

### 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) 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 G(E/F). Then Gal(E/F) if and only if (E) E.

Proof: Obvious.

It suffices to show that if (E) E, then (E) = E. Suppose the deg(E/F) = n, and let {1,...,n} be a basis of E over F. Since is an F-isomorphism, {(1),...,(n)} is a basis of (E) over F. Thus, deg((E)/F) = n. However, since (E) E,

deg(E/F) = deg((E)/F) · deg(E/(E)),

and thus deg(E/(E)) = 1, which implies that E = (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 E be such that E = F(). Further, let 1,...,n be the conjugates of over F and let i G(E/F) be such that i() = i. Then i Gal(E/F) if and only if i E. In particular, if E is a normal extension of F, then all i belong to E and Gal(E/F) contains n elements.

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

()(x) = ((x))

is an F-automorphism of E. Indeed, is a homomorphism of E into E such that ()(1) = 1. Therefore, is an isomorphism by Proposition 2 of the introduction to field theory. And since and are F-isomorphisms, if x F, we have ()(x) = ((x)) = (x) = x. Therefore, is an F-isomorphism of E into E, so that by Lemma 3, is an F-automorphism of E. Let us define the product of and to be the F-automorphism . 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 E. If Gal(E/F), let -1 denote the inverse function of , when is considered as a bijection from E onto E. Then -1 is an F-automorphism of E, and -1 = -1 = 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(). Then E is a normal extension of F and Gal(E/F) = {0, 1}, where

0() = ,    1() = -.

The mapping :Gal(E/F)Z2 defined by (a) = a (mod 2) is a surjective isomorphism. Therefore, Gal(E/F)Z2.

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

a(m) = ma.

Notice that

ab(m) = a(mb) = a(m)b = mab.

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

(1)
ab(m) = mqm+c = (mm)q mc = c(m)

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

(ab) = c(mod m) = ab(mod m) = (a)(b).

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

Gal(E/F)Zmx.

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

Gal(E/F) = {0},

where 0() = . 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.