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 Fautomorphism of E is an Fisomorphism of E onto itself. The set of all Fautomorphisms of E will be denoted Gal(E/F).
Since an Fautomorphism of E is an Fisomorphism 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 Fisomorphism, {(_{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
is an Fautomorphism 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 Fisomorphisms, if x F, we have ()(x) = ((x)) = (x) = x. Therefore, is an Fisomorphism of E into E, so that by Lemma 3, is an Fautomorphism of E. Let us define the product of and to be the Fautomorphism . 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 Fautomorphism 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
The mapping :Gal(E/F)Z_{2} defined by (_{a}) = a (mod 2) is a surjective isomorphism. Therefore, Gal(E/F)Z_{2}.
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 _{m}^{a} (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
Notice that
_{a}_{b}( _{m}) = _{a}( _{m}^{b}) = _{a}( _{m}) ^{b} = _{m}^{ab}.
Therefore, if q and c are integers such that ab = qm + c, 0 < c < m  1, we have
(1)
_{a}_{b}( _{m}) = _{m}^{qm+c} = ( _{m}^{m}) ^{q }_{m}^{c} = _{c}( _{m})
since _{m}^{m} = 1. Let :Gal(E/F)Z_{m}^{x} be the mapping defined by (_{a}) = a (mod m). Then formula (1) implies that
( _{a}_{b}) = c(mod m) = ab(mod m) = ( _{a}) ( _{b}).
Therefore, is a homomorphism. It is clear that is a bijection, and thus
Gal(E/F) Z_{m}^{x}.
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.
