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Saturday, October 26, 2013

Calculation of Galois Groups (14.2.10,12)

Dummit and Foote Abstract Algebra, section 14.2, exercises 10, 12:

MathJax TeX Test Page 10. Determine the Galois group of the splitting field over Q of x83.
12. Determine the Galois group of the splitting field over Q of x414x2+9.

Proof: (10) Letting θ=ζ8=22+22i, we have the 8 roots of this polynomial are θa83 for a=0,1,...,7. Therefore the splitting field for this polynomial is Q(83)(2)(i). We note x83 is irreducible over Q by Eisenstein, so the first extension is degree 8, and assuming x22 isn't irreducible over Q(83) leads to a solution(a0+a183+...+a7837)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+b183+...+b7837 for integers bi has for field of fractions Q(83), 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(83)(2) is of degree 16, and since this field is contained in R we see K=Q(83)(2)(i) is of degree 32 over Q.

We see there are 822=32 permutations of the roots, therefore these are all automorphisms, so Gal(K/Q) is a group generated by the automorphismsα : 83θ83β : 22γ : iiWe 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 3252721 mod 8 we observe Aut(Z8)=Z22, so letting φ be the isomorphism between these two groups, letting a generate Z8, and b,cZ22 be such that φ(b)(a)=a5 and φ(c)(a)=a3, we observe the same relations between these elements in Z22φZ8 also of order 32, so that finally we may sayGal(K/Q)=Z22φZ8 (12) Finding the roots of the polynomial in x2, we obtain the solutions α=7+210, β=7210, α, and β. We note αβ=72(210)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 x2α2 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 : αα, ψ : αβ=3/α, and φψ are all of order 2, soGal(Q(α)/Q)=Z22 

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