Three Examples

In all our examples the ground field shall be Q and the extension field will be a subfield of the complex numbers C. We take as basic that the only nonzero rational numbers c for which √ c ∈ Q are those positive c for which each prime factor p appears an even number of times. In particular, √ 2, √ 3, 3/2 are all irrational. Example I: K = Q(√ 2, √ 3). As √ 2 has minimal polynomial x 2 − 2, Q(√ 2) has basis 1, √ 2 over Q. Now we need a simple result: Theorem 0.1 √ 3 ∈ Q(√ 2) Proof: If it were we would have √ 3 = a + b √ 2 with a, b ∈ Q. Squaring both sides 3 = a 2 + 2b 2 + 2ab √ 2 As 1, √ 2 is a basis the coefficient of √ 2 would need be zero. That is, 2ab = 0. So either a = 0 or b = 0. 1. b = 0: Then √ 3 = a ∈ Q, contradiction. 2. a = 0. Then √ 3 = b √ 2 so 3/2 = b ∈ Q, contradiction. From Theorem 0.1 and that √ 3 satisfies a quadratic (namely, x 2 − 3) over Q(√ 2), 1, √ 3 is a basis for Q(√ 2, √ 3) over Q(√ 2) and so 1, √ 2, √ 3, √ 6 is a basis for K over Q. That is, we may write K = Q(√ 2, √ 3) = {a + b √ 2 + c √ 3 + d √ 6 : a, b, c, d ∈ Q} (1) and every α ∈ K has a unique expression in this form. Now we turn to the Galois Group Γ(K : Q). Any σ ∈ K has σ(√ 2) = ± √ 2 and σ(√ 3) = ± √ 3 which gives four (Caution: this is 2 times 2) possibilities. The value of σ on √ 2, √ 3 determines the value on all of K. The four elements of the Galois Group are e, σ 1 , σ 2 , σ 3 where e(a + b √ 2 + c √ 3 + d √ 6) = a + b √ 2 + c √ 3 + d √ 6 σ 1 (a + b √ 2 + …