12. Determine the Galois group of the splitting field over Q of x4−14x2+9.
Proof: (10) Letting θ=ζ8=√22+√22i, we have the 8 roots of this polynomial are θa8√3 for a=0,1,...,7. Therefore the splitting field for this polynomial is Q(8√3)(√2)(i). We note x8−3 is irreducible over Q by Eisenstein, so the first extension is degree 8, and assuming x2−2 isn't irreducible over Q(8√3) leads to a solution(a0+a18√3+...+a78√37)2=2However, we notice the coefficient of the basis element 1 here isa20+6a1a7+6a2a6+6a3a5+3a24=2We notice that the integral domain of elements of the form b0+b18√3+...+b78√37 for integers bi has for field of fractions Q(8√3), because the latter contains the former and the former contains the latter by writing fractional coefficients under a common denominator. Thus by Gauss's lemma we may assume the ai are integers, and reducing modulo 3 the equality is impossible. Therefore Q(8√3)(√2) is of degree 16, and since this field is contained in R we see K=Q(8√3)(√2)(i) is of degree 32 over Q.
We see there are 8∗2∗2=32 permutations of the roots, therefore these are all automorphisms, so Gal(K/Q) is a group generated by the automorphismsα : 8√3↦θ8√3β : √2↦−√2γ : i↦−iWe see these elements satisfy α8=β2=γ2=1, βγ=γβ, βα=α5β, and γα=α3γ, and also these relations on a free group of three generators is sufficient to write any element in the form αaβbγc, of which there are 32 combinations, so this is precisely the set of relations.Gal(K/Q)=<α,β,γ | α8=β2=γ2=1, βγ=γβ, βα=α5β, γα=α3γ>Now, since 32≡52≡72≡1 mod 8 we observe Aut(Z8)=Z22, so letting φ be the isomorphism between these two groups, letting a generate Z8, and b,c∈Z22 be such that φ(b)(a)=a5 and φ(c)(a)=a3, we observe the same relations between these elements in Z22⋊φZ8 also of order 32, so that finally we may sayGal(K/Q)=Z22⋊φZ8 (12) Finding the roots of the polynomial in x2, we obtain the solutions α=√7+2√10, β=√7−2√10, −α, and −β. We note αβ=√72−(2√10)2=3, so that β=3/α and the splitting field is merely Q(α). We see that Q(α) is of degree 4 since the polynomial points to ≤4 and Q(√10)⊂Q(α) (for the proper inclusion, consider a solution to x2−α2 over Z(√10) by Gauss's lemma). The automorphisms of this extension must be the permutations of α about the roots of its minimal polynomial, so we observe φ : α↦−α, ψ : α↦β=3/α, and φψ are all of order 2, soGal(Q(α)/Q)=Z22 ◻
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