226 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

Lemma 9.4.31 G(K,L j) is a normal subgroup of G(K,F). Here K is a splitting field forsome polynomial having coefficients in F or more generally a normal extension of F.

Proof: Let η ∈ G(K,F) and let σ ∈ G(K,L j) . Is η−1ση ∈ G(K,L j)? First I needto verify it is a automorphism on K. After this, I need to show that it fixes L j. η−1ση

is obviously an automorphism on K because each in the product is. Does η−1ση fix L j?Let r ∈ L j with minimum polynomial f (x) having roots ri and coefficients in F. Then 0 =η f (r) = f (η (r)) and so η (r) is one of the roots of f (x) . It follows that η (r)∈L j becauseK is a normal extension and L j is an intermediate field so is also a normal extension. SeeProposition 9.4.19. Therefore, σ fixes η (r) and so η−1ση (r) = η−1η (r) = r. ■

Because of this lemma, it makes sense to consider the quotient group

G(K,F)/G(K,L j)

This leads to the following fundamental theorem of Galois theory.

Theorem 9.4.32 Let K be a splitting field of a separable polynomial p(x) having coeffi-cients in a field F. Let {Li}k

i=0 be the increasing sequence of intermediate fields between Fand K as shown above in 9.25. Then each of these is a normal extension of F (Proposition9.4.19) and the Galois group G(K,L j) is a normal subgroup of G(K,F) and

G(L j,F)≃ G(K,F)/G(K,L j)

where the symbol ≃ indicates the two groups are isomorphic.

Proof: All that remains is to check that the above isomorphism is valid. Let

θ : G(K,F)/G(K,L j)→ G(L j,F) , θ [σ ]≡ σ |L j

In other words, this is just the restriction of σ to L j. Thus the quotient group is well definedby Proposition 9.4.23. Is θ well defined? First of all, does it have values in G(L j,F)? Inother words, if σ ∈ G(K,F) , does its restriction to L j send L j to L j? If r ∈ L j it has aminimum polynomial q(x) with coefficients in F. σ (r) is one of the other roots of q(x)(Theorem 9.4.12) so, since K is a normal extension, being a splitting field of a separablepolynomial, σ (r) ∈K. But these subfields are all normal extensions so σ (r) ∈ L j.

Thus θ has values in G(L j,F) . Is θ well defined? If [σ1] = [σ2] , then by definition,σ−11 σ2 ∈ G(K,L j) so σ

−11 σ2 fixes everything in L1. Thus if r ∈ L1,σ

−11 σ2r = r and so

σ2r = σ1r. It follows that the restrictions of σ1 and σ2 to L j are equal. Therefore, θ iswell defined. It is obvious that θ is a homomorphism. Why is θ onto? This follows rightaway from Theorem 9.4.9. Note that K is the splitting field of p(x) over L j since L j ⊇ F.Also if σ ∈ G(L j,F) so it is an automorphism of L j, then, since it fixes F, p(x) = p̄(x) inthat theorem. Thus σ extends to ζ , an automorphism of K. Thus θζ = σ . Why is θ one toone? If θ [σ ] = θ [α] , this means σ = α on L j. Thus σα−1 is the identity on L j. Henceσα−1 ∈ G(K,L j) which is what it means for [σ ] = [α]. ■