Quaternion Galois Group (14.2.27)

Dummit and Foote Abstract Algebra, section 14.2, exercise 27:

Let $α=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ and consider the extension $E=\mathbb{Q}(α)$.
(a) Show that $a=(2+\sqrt{2})(3+\sqrt{3})$ is not a square in $F=\mathbb{Q}(\sqrt{2},\sqrt{3})$.
(b) Conclude from (a) that $[E~:~\mathbb{Q}]=8$. Prove that the roots of the minimal polynomial over $\mathbb{Q}$ for $α$ are the 8 elements $\pm \sqrt{(2 \pm \sqrt{2})(3 \pm \sqrt{3})}$
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(f) Conclude $\text{Gal}(E/\mathbb{Q})≅Q_8$.

Proof: (a) We observe that if $a=c^2$ then $aφa=c^2φc^2=(cφc)^2$ where $φ$ is the automorphism of $F$ fixing $\sqrt{2}$ and negating $\sqrt{3}$. Since $cφc$ is actually the image of $c$ under $N_{F/\mathbb{Q}(\sqrt{2})}$ we have $cφc=\sqrt{aφa}=\sqrt{6(3+\sqrt{3})^2}=\pm (3\sqrt{2}+3\sqrt{6})∈\mathbb{Q}(\sqrt{2})$ and now $\sqrt{6}∈\mathbb{Q}(\sqrt{2})$, a contradiction.

(b-f) (We take a slightly different path from the authors, especially for parts c and d) We have shown $[\mathbb{Q}(\sqrt{2},\sqrt{3},α)~:~\mathbb{Q}]=8$. Since $\dfrac{α^2}{2+\sqrt{2}}-3=\sqrt{3}$, we have $\mathbb{Q}(\sqrt{2},\sqrt{3},α)=\mathbb{Q}(\sqrt{2},α)=E(\sqrt{2})$. Assume $[E(\sqrt{2})~:~E]=2$; then $E(\sqrt{2})$ is Galois of degree $2$ over $E$ (splitting $x^2-2$), and the map $φ:\sqrt{2}↦-\sqrt{2}$ is an automorphism of $E(\sqrt{2})$ fixing $E$. But $φ$ is in particular an isomorphism of $F$ allowing us to observe $φ(α^2)=φ((2+\sqrt{2})(3+\sqrt{3}))=(2-\sqrt{2})(3+\sqrt{3}) \neq α^2$, a contradiction. So $E=E(\sqrt{2})$ and $[E~:~\mathbb{Q}]=8$.

Now as we saw above, $E$ is the splitting field for $x^2-α^2$ over $F$. Therefore every automorphism of $F$ extends to an automorphism of $E$. This is an order $4$ subgroup of $\text{Aut}(E/\mathbb{Q})$. Since $E/F$ is Galois we also have the automorphism $ψ:α↦-α$ fixing $F$. Letting $n=|\text{Aut}(E/\mathbb{Q})|$ by Lagrange we see $4~|~n$, by Galois we see $n≤8$, and by counting we see $n > 4$, so that $n=8$ and $E/\mathbb{Q}$ is Galois.

We see that the 8 elements mentioned above are distinct by observing squares, and since $φ(x^2)=φ(x)^2$ for general automorphisms we have $φ(x)=\pm \sqrt{φ(x^2)}$. Letting $H=\text{Aut}(E/F)$ we see $1,ψ$ form a set of right coset representatives for $H$ in $\text{Aut}(E/\mathbb{Q})$. By letting $λ∈H$ fix $\sqrt{3}$ and negate $\sqrt{2}$, for example, we see $λψ(α^2)=(2-\sqrt{2})(3+\sqrt{3})$ and thus $λψ(α)=\pm \sqrt{(2-\sqrt{2})(3+\sqrt{3})}$. Similarly, we can see any automorphism maps $α$ to one of the 8 forms above, and thus must map to all of them, and these are the 8 distinct roots of the minimal polynomial for $α$ over $\mathbb{Q}$.

Let $σ$ map $α$ to $β=\sqrt{(2-\sqrt{2})(3+\sqrt{3})}$. We see $σ(α^2)=β^2$ so that $σ(\sqrt{2})=-\sqrt{2}$ and $σ(\sqrt{3})=\sqrt{3}$, and together with $αβ=\sqrt{2}(3+\sqrt{3})$ we have $σ(αβ)=-αβ$ and thus $σ(β)=-α$. Now $σ$ can be seen to be of order $4$ and together with $\tau$ mapping $α$ to $γ=\sqrt{(2+\sqrt{2})(3-\sqrt{3})}$ we similarly find the relations $σ^4=\tau^4=1$, $σ^2=\tau^2$, and $σ\tau = \tau σ^3$ (keeping in mind $β=\dfrac{\sqrt{2}(3+\sqrt{3})}{α}$), so that $\text{Gal}(E/\mathbb{Q})≅Q_8$.$~\square$