Calculation of Galois Groups (14.2.10,12)

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

10. Determine the Galois group of the splitting field over $\mathbb{Q}$ of $x^8-3$.
12. Determine the Galois group of the splitting field over $\mathbb{Q}$ of $x^4-14x^2+9$.

Proof: (10) Letting $θ=\zeta_8=\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}i$, we have the $8$ roots of this polynomial are $θ^a\sqrt[8]{3}$ for $a=0,1,...,7$. Therefore the splitting field for this polynomial is $\mathbb{Q}(\sqrt[8]{3})(\sqrt{2})(i)$. We note $x^8-3$ is irreducible over $\mathbb{Q}$ by Eisenstein, so the first extension is degree $8$, and assuming $x^2-2$ isn't irreducible over $\mathbb{Q}(\sqrt[8]{3})$ leads to a solution$$(a_0+a_1\sqrt[8]{3}+...+a_7\sqrt[8]{3}^7)^2=2$$However, we notice the coefficient of the basis element $1$ here is$$a_0^2+6a_1a_7+6a_2a_6+6a_3a_5+3a_4^2=2$$We notice that the integral domain of elements of the form $b_0+b_1\sqrt[8]{3}+...+b_7\sqrt[8]{3}^7$ for integers $b_i$ has for field of fractions $\mathbb{Q}(\sqrt[8]{3})$, 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 $a_i$ are integers, and reducing modulo $3$ the equality is impossible. Therefore $\mathbb{Q}(\sqrt[8]{3})(\sqrt{2})$ is of degree $16$, and since this field is contained in $\mathbb{R}$ we see $K=\mathbb{Q}(\sqrt[8]{3})(\sqrt{2})(i)$ is of degree $32$ over $\mathbb{Q}$.

We see there are $8*2*2=32$ permutations of the roots, therefore these are all automorphisms, so $\text{Gal}(K/\mathbb{Q})$ is a group generated by the automorphisms$$α~:~\sqrt[8]{3}↦θ\sqrt[8]{3}$$$$β~:~\sqrt{2}↦-\sqrt{2}$$$$γ~:~i↦-i$$We see these elements satisfy $α^8=β^2=γ^2=1$, $βγ=γβ$, $βα=α^5β$, and $γα=α^3γ$, and also these relations on a free group of three generators is sufficient to write any element in the form $α^aβ^bγ^c$, of which there are $32$ combinations, so this is precisely the set of relations.$$\text{Gal}(K/\mathbb{Q})=<α,β,γ~|~α^8=β^2=γ^2=1,~βγ=γβ,~βα=α^5β,~γα=α^3γ>$$Now, since $3^2≡5^2≡7^2≡1~\text{mod }8$ we observe $\text{Aut}(Z_8)=Z_2^2$, so letting $φ$ be the isomorphism between these two groups, letting $a$ generate $Z_8$, and $b,c∈Z_2^2$ be such that $φ(b)(a)=a^5$ and $φ(c)(a)=a^3$, we observe the same relations between these elements in $Z_2^2 \rtimes_φ Z_8$ also of order $32$, so that finally we may say$$\text{Gal}(K/\mathbb{Q})=Z_2^2 \rtimes_φ Z_8$$ (12) Finding the roots of the polynomial in $x^2$, we obtain the solutions $α=\sqrt{7+2\sqrt{10}}$, $β=\sqrt{7-2\sqrt{10}}$, $-α$, and $-β$. We note $αβ=\sqrt{7^2-(2\sqrt{10})^2}=3$, so that $β=3/α$ and the splitting field is merely $\mathbb{Q}(α)$. We see that $\mathbb{Q}(α)$ is of degree $4$ since the polynomial points to $≤4$ and $\mathbb{Q}(\sqrt{10}) ⊂ \mathbb{Q}(α)$ (for the proper inclusion, consider a solution to $x^2-α^2$ over $\mathbb{Z}(\sqrt{10})$ 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 $2$, so$$\text{Gal}(\mathbb{Q}(α)/\mathbb{Q})=Z_2^2~\square$$