Galois Theory at Work: Concrete Examples

1. Examples Example 1.1. The field extension Q(√ 2, √ 3)/Q is Galois of degree 4, so its Galois group has order 4. The elements of the Galois group are determined by their values on √ 2 and √ 3. The Q-conjugates of √ 2 and √ 3 are ± √ 2 and ± √ 3, so we get at most four possible automorphisms in the Galois group. See Table 1. Since the Galois group has order 4, these 4 possible assignments of values to σ(√ 2) and σ(√ 3) all really exist. σ(√ 2) σ(√ 3) √ 2 √ 3 √ 2 − √ 3 − √ 2 √ 3 − √ 2 − √ 3 Table 1 Each nonidentity automorphism in Table 1 has order 2. Since Gal(Q(√ 2, √ 3)/Q) contains 3 elements of order 2, Q(√ 2, √ 3) has 3 subfields K i such that [Q(√ 2, √ 3) : K i ] = 2, or equivalently [K i : Q] = 4/2 = 2. Two such fields are Q(√ 2) and Q(√ 3). A third is Q(√ 6) and that completes the list. Here is a diagram of all the subfields. Q(√ 3) Q(√ 6) q q q q q q q q q q q q Q In Table 1, the subgroup fixing Q(√ 2) is the first and second row, the subgroup fixing Q(√ 3) is the first and third row, and the subgroup fixing Q(√ 6) is the first and fourth row (since (− √ 2)(− √ 3) = √ 2 √ 3).