Lectures on the Algebraic Theory of Fields - Tata Institute of ...

7. Galois extensions 41σ −1 τσω=ωorτ(σω)=σω. This being true for allτ, it follows thatσω∈L for allω in L ′ . ThusσL ′ ⊂ L ′ . We can similarly prove thatσ −1 L⊂L ′ whichproves our statement.In particular let L/k be a normal extension of k which is containedin K. ThenσL=L for allσ∈G(K/k). This means that G(K/L) is anormal subgroup of G(K/k). On the other hand if L is any subfield suchthat G(K/L) is a normal subgroup of G(K/k) then by aboveσL=L forallσ∈G(K/k) which proves that L/k is normal. Thus5) Let k⊂L⊂K. Then L/k is normal⇐⇒ G(K/L) is a normalsubgroup of G(K/k)6) If L/k is normal, then G(K/k)≃G(L/k)/G(K/L).Letσ∈G(K/k) and ¯σ the restriction ofσto L. Then ¯σ is an automorphismof L/k so that ¯σ∈G(L/k). Nowσ→ ¯σ is a homomorphismfo G(K/k) into G(L/k). Forστω=στω=σ(τω)= ¯σ¯τω for allω∈L. Thus ¯σ¯τ=στThe homomorphism is onto since every automorphism of L/k can be 49extended into an automorphism of K/k. Now ¯σ is identity if and only if¯σω=ωfor allω∈L. Thusσ∈G(K/L). Also everyσ∈G(K/L) has thisproperty so that the kernel of the homomorphism is G(K/L).Let K/k be a finite galois extension. Every isomorphism of K/k inΩ is an an automorphism. Also K/k being separable, K/k has exactly(K : k) district isomorphisms. This shows that(K : k)=order of G.We shall now prove the converse7) If G is a finite group of automorphisms of K/k having k as thefixed field then (K : k)=order of G.