Abstract Algebra Theory and Applications - Computer Science ...

HINTS AND SOLUTIONS 40518. Since α is algebraic over F of degree n, we can write any element β ∈ F (α)uniquely as β = a 0 +a 1 α+· · ·+a n−1 α n−1 with a i ∈ F . There are q n possiblen-tuples (a 0 , a 1 , . . . , a n−1 ).24. Factor x p−1 − 1 over Z p .Chapter 21. Galois Theory1. (a) Z 2 . (c) Z 2 × Z 2 × Z 2 .2. (a) Separable. (c) Not separable.3. [GF(729) : GF(9)] = [GF(729) : GF(3)]/[GF(9) : GF(3)] = 6/2 = 3 ⇒G(GF(729)/GF(9)) ∼ = Z 3 . A generator for G(GF(729)/GF(9)) is σ, whereσ 3 6(α) = α 36 = α 729 for α ∈ GF(729).4. (a) S 5 . (c) S 3 .5. (a) Q(i).7. Let E be the splitting field of a cubic polynomial in F [x]. Show that [E : F ]is less than or equal to 6 and is divisible by 3. Since G(E/F ) is a subgroup ofS 3 whose order is divisible by 3, conclude that this group must be isomorphicto Z 3 or S 3 .9. G is a subgroup of S n .16. True.20. (a) Clearly ω, ω 2 , . . . , ω p−1 are distinct since ω ≠ 1 or 0. To show that ω i isa zero of Φ p , calculate Φ p (ω i ).(b) The conjugates of ω are ω, ω 2 , . . . , ω p−1 . Define a map φ i : Q(ω) → Q(ω i )byφ i (a 0 + a 1 ω + · · · + a p−2 ω p−2 ) = a 0 + a 1 ω i + · · · + c p−2 (ω i ) p−2 ,where a i ∈ Q. Prove that φ i is an isomorphism of fields. Show that φ 2generates G(Q(ω)/Q).(c) Show that {ω, ω 2 , . . . , ω p−1 } is a basis for Q(ω) over Q, and considerwhich linear combinations of ω, ω 2 , . . . , ω p−1 are left fixed by all elements ofG(Q(ω)/Q).