By Andrew Baker
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Extra info for An Introduction to Galois Theory, Edition: version 23 Jan 2013
Example. For n 1, show that each of the roots of f (X) = X n − 1 in C is simple. Solution. We have f ′ (X) = ∂(X n − 1) = nX n−1 , so for any root ζ of f (X), f ′ (ζ) = nζ n−1 ̸= 0. 59. Example. Show that 2i is a multiple root of f (X) = X 4 + 8X 2 + 16. Solution. We have f ′ (X) = 4X 3 +16X. Using Long Division and the Euclidean Algorithm we ﬁnd that gcd(f (X), f ′ (X)) = X 2 + 4, where 2i is also a root of X 2 + 4. Hence 2i is a multiple root of f (X). In fact, X 4 + 8X 2 + 16 = (X 2 + 4)2 , so this is obvious.
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 , .
We build up the list of monomorphisms in stages. √ √ First consider monomorphisms that ﬁx 3 2 and hence ﬁx the subﬁeld Q( 3 2). These form the subset √ √ 3 3 (Q( 2, ζ ), C) ⊆ Mono (Q( 2, ζ3 ), C). MonoQ( √ 3 3 Q 2) √ √ 3 3 We know that Q( 2, ζ3 ) = Q( 2)(ζ3 ) and that ζ3 is a root of the irreducible cyclotomic √ polynomial Φ3 (X) = X 2 + X + 1 ∈ Q( 3 2)[X]. So there are two monomorphisms id, α0 ﬁxing √ Q( 3 2), where α0 has the eﬀect (√ √ ) 3 2 −→ 3 2 α0 : . ζ3 −→ ζ32 √ √ Next we consider monomorphisms that send 3 2 to 3 2 ζ3 .