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# An Introduction to Galois Theory by Andrew Baker

By Andrew Baker

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Example text

Hence the ℓi mj form a basis of M over K and so [M : K] = rs = [M : L] [L : K]. 28 We will often indicate subextensions in diagrammatic form where larger ﬁelds always go above smaller ones and the information on the lines indicates dimensions M✳ [M :L] ✮ ✩ L ✤ [L:K] ✏ ✕ ✚ [M :K]=[M :L] [L:K] K We often suppress ‘composite’ lines such as the dashed one. Such towers of extensions are our main objects of study. We can build up sequences of extensions and form towers of arbitrary length. Thus, if L1 /K, L2 /L1 , .

74, however a ‘try it and see’ approach will often be suﬃcient. 76. Example. Find a primitive element for the extension Q( 3, i)/Q. √ √ √ Q( 3, i) we ﬁnd Solution. Consider 3 + i. Then working over the subﬁeld Q( 3) √ that i ∈ / Q( 3) R and √ √ √ √ (X − ( 3 + i))(X − ( 3 − i)) = X 2 − 2 3X + 4 ∈ Q( 3)[X], hence Now taking √ X 2 − 2 3X + 4 = minpolyQ(√3),√3+i (X). √ √ (X 2 − 2 3X + 4)(X 2 + 2 3X + 4) = X 4 − 4X 2 + 16 ∈ Q[X], we see that minpolyQ,√3+i (X) | (X 4 − 4X 2 + 16) in Q[X]. Notice that √ √ √ √ 1 √ ( 3 − i) ( 3 − i) −1 √ ( 3 + i) = √ = = ( 3 − i) ∈ Q( 3 + i), 3+1 4 ( 3 + i)( 3 − i) √ √ since ( 3 + i)−1 ∈ Q( 3 + i).

Then if t ∈ F is a root of p(X), the evaluation homomorphism εt : K[X] −→ F factors through the quotient monomorphism εt : K[X]/(p(X)) −→ F whose image is K(t) F . Of course, there is one such monomorphism for each root of p(X) in F . If we ﬁx one such root t0 and identify K[X]/(p(X)) with K(t0 ) via εt0 , then each root of p(X) in F gives rise to a monomorphism φt = εt ◦ ε−1 t0 : K(t0 ) −→ F for which φt (t0 ) = t. φt =εt ◦ε−1 t 0 K(t0 ) o εt0 ∼ = K[X]/(p(X)) εt /* F Notice that if φ : K[X]/(p(X)) −→ F is any homomorphism extending the identity function on K, then the coset X + (p(X)) must be sent by φ to a root of p(X) in F , hence every such homomorphism arises this way.