Abstract Algebra Theory and Applications - Computer Science ...

370 CHAPTER 21 GALOIS THEORYin K. All of these zeros have multiplicity 1, since E is separable over F .Since F is infinite, we can find an a in F such thata ≠ α i − αβ − β jfor all i and j with j ≠ 1. Therefore, a(β − β j ) ≠ α i − α. Let γ = α + aβ.Thenγ = α + aβ ≠ α i + aβ j ;hence, γ − aβ j ≠ α i for all i, j with j ≠ 1. Define h(x) ∈ F (γ)[x] byh(x) = f(γ − ax). Then h(β) = f(α) = 0. However, h(β j ) ≠ 0 for j ≠ 1.Hence, h(x) and g(x) have a single common factor in F (γ)[x]; that is, theirreducible polynomial of β over F (γ) must be linear, since β is the onlyzero common to both g(x) and h(x). So β ∈ F (γ) and α = γ − aβ is inF (γ). Hence, F (α, β) = F (γ).□21.2 The Fundamental TheoremThe goal of this section is to prove the Fundamental Theorem of GaloisTheory. This theorem explains the connection between the subgroups ofG(E/F ) and the intermediate fields between E and F .Proposition 21.9 Let {σ i : i ∈ I} be a collection of automorphisms of afield F . ThenF {σi } = {a ∈ F : σ i (a) = a for all σ i }is a subfield of F .Proof. Let σ i (a) = a and σ i (b) = b. Thenandσ i (a ± b) = σ i (a) ± σ i (b) = a ± bσ i (ab) = σ i (a)σ i (b) = ab.If a ≠ 0, then σ i (a −1 ) = [σ i (a)] −1 = a −1 . Finally, σ i (0) = 0 and σ i (1) = 1since σ i is an automorphism.□Corollary 21.10 Let F be a field and let G be a subgroup of Aut(F ). Thenis a subfield of F .F G = {α ∈ F : σ(α) = α for all σ ∈ G}