Download PDF by Andrew Baker: An Introduction to Galois Theory

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79, f (X) splits in E. Also, E is generated by some of the roots of f (X). Hence E is a splitting field for f (X) over K. Now suppose that E is a splitting field for g(X) ∈ K[X], so that E = K(v1 , . . , vk ), where v1 , . . , vk are the distinct roots of g(X) in E. Now any monomorphism θ ∈ MonoK (E, K) must map these roots to θ(v1 ), . . 34). Since θ permutes the roots vj , we have θE = θK(v1 , . . , vk ) = K(θ(v1 ), . . , θ(vk )) = K(v1 , . . , vk ) = E. 81. Corollary. Let E/L and L/K be finite extensions.

We have the following tower of extensions. √ √ Q( 2, 3) √ √ √ [Q( 2, 3):Q( 2)] √ Q( 2) √ √ √ √ √ [Q( 2, 3):Q]=2[Q( 2, 3):Q( 2)] 2 Q √ √ √ We will show that [Q( 2, 3) : Q( 2)] = 2. 30 √ √ √ √ √ √ Notice that if u ∈ Q( 2, 3) = Q( 2)( 3) then u = a + b 3 for some a, b ∈ Q( 2), √ √ √ √ √ √ so 1, 3 span Q( 2, 3) over Q( 2). But if these are linearly dependent then 3 ∈ Q( 2). Writing √ √ 3=v+w 2 with v, w ∈ Q, we find that √ v 2 + 2w2 + 2vw 2 = 3 ∈ Q, √ and hence 2vw 2 ∈ Q. The possibilities v = 0 or w = 0 are easily ruled out, while v, w ̸= 0 √ √ √ would implies that 2 ∈ Q which is false.

Definition. Let L/K be a finite extension. The separable degree of L over K is (L : K) = | MonoK (L, K)|. 69. Lemma. For a finite simple extension K(u)/K, (K(u) : K) = | Roots(minpolyK,u , K)|. If K(u)/K is separable, then [K(u) : K] = (K(u) : K). 50 Proof. 34 applied to the case L = K. 1) K(u1 )/K, K(u1 , u2 )/K(u1 ), · · · , L = K(u1 , . . , uk )/K(u1 , . . , uk−1 ). So we can use the following to compute (L : K) = (K(u1 , . . , uk ) : K). 70. Proposition. Let L/K and M/L be finite extensions.

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An Introduction to Galois Theory by Andrew Baker


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