Abstract Algebra Theory and Applications - Computer Science ...

374 CHAPTER 21 GALOIS THEORY{id, σ, τ, µ} ❅ ❅❅{id, σ} {id, τ} {id, µ}Q( √ 3, √ 5 ) ❅ ❅❅Q( √ 3 ) Q( √ 5 ) Q( √ 15 )❅❅❅{id} ❅❅❅Q Figure 21.1. G(Q( √ 3, √ 5 )/Q)1. The map K ↦→ G(E/K) is a bijection of subfields K of E containingF with the subgroups of G(E/F ).2. If F ⊂ K ⊂ E, then[E : K] = |G(E/K)| and [K : F ] = [G(E/F ) : G(E/K)].3. F ⊂ K ⊂ L ⊂ E if and only if {id} ⊂ G(E/L) ⊂ G(E/K) ⊂ G(E/F ).4. K is a normal extension of F if and only if G(E/K) is a normalsubgroup of G(E/F ). In this caseG(K/F ) ∼ = G(E/F )/G(E/K).Proof. (1) Suppose that G(E/K) = G(E/L) = G. Both K and L arefixed fields of G; hence, K = L and the map defined by K ↦→ G(E/K) isone-to-one. To show that the map is onto, let G be a subgroup of G(E/F )and K be the field fixed by G. Then F ⊂ K ⊂ E; consequently, E is anormal extension of K. Thus, G(E/K) = G and the map K ↦→ G(E/K) isa bijection.(2) By Proposition 21.5, |G(E/K)| = [E : K]; therefore,|G(E/F )| = [G(E/F ) : G(E/K)] · |G(E/K)| = [E : F ] = [E : K][K : F ].Thus, [K : F ] = [G(E/F ) : G(E/K)].(3) Statement (3) is illustrated in Figure 21.2. We leave the proof of thisproperty as an exercise.(4) This part takes a little more work. Let K be a normal extension ofF . If σ is in G(E/F ) and τ is in G(E/K), we need to show that σ −1 τσ