Galois Theory Through Exercises by Juliusz Brzeziński

Author:Juliusz Brzeziński
Language: eng
Format: epub
Publisher: Springer International Publishing, Cham

9.3(b) We take in Exercise 9.​2(a)–(h) only the zeros of f(X) which are not in the ground field K.

(a) The zeros: numbered 1, 2, 3, 4, respectively. Then σ 0, σ 1, σ 2, σ 3 are (1), (1, 2), (3, 4), (1, 2)(3, 4), respectively.

(b) The zeros: numbered 1, 2, 3, 4, respectively. As in (a).

(c) The zeros: ε, ε 2, ε 3, ε 4 () numbered 1, 2, 3, 4, respectively. Then σ 0, σ 1, σ 2, σ 3 are (1), (1, 2, 4, 3), (1, 3, 4, 2), (1, 4)(2, 3), respectively.

(d) The zeros: numbered 1, 2, 3, 4, respectively. The four automorphisms defined by are {(1), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}.

(e) The zeros: numbered 1, 2, 3, 4, respectively. Then σ 0, σ 1, σ 2, σ 3 are (1), (1, 2)(3, 4), (1, 3, 2, 4), (1, 4, 2, 3), respectively.

(f) The zeros: numbered 1, 2, 3, respectively. Then σ 0, σ 1, σ 2, σ 3, σ 4, σ 5 are (1), (2, 3), (1, 2, 3), (1, 2), (1, 3, 2), (1, 3), respectively.

(g) In the notations of Exercise 9.​2(g), the zeros are α, −α, iβ, −iβ numbered 1, 2, 3, 4 in this order. Then the automorphisms σ 0, σ 1, σ 2, σ 3, σ 4, σ 5, σ 6, σ 7 are (1), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3), (3, 4), (1, 2), (1, 3, 2, 4), (1, 4, 2, 3), respectively.

(h) The zeros are ε, ε 2, where , numbered 1, 2 in the given order. The automorphisms σ 0, σ 1 are (1), (1, 2).

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