# field extension with Galois group $Q_{8}$

Let $\alpha=\sqrt{(2+\sqrt{2})(3+\sqrt{3})},E=\mathbb{Q}(\alpha)$. We will show that $E$ is Galois over $\mathbb{Q}$ and that $G=Gal(E/\mathbb{Q})\cong Q_{8}$ (the group of quaternions).

We begin by showing that $[E:\mathbb{Q}]=8$. Let $F=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$. Claim that $F\subsetneq E$. To show that they are not equal, we show that $\alpha\notin F$, i.e. that $(2+\sqrt{2})(3+\sqrt{3})$ is not a square in $F$. If it were, say $(2+\sqrt{2})(3+\sqrt{3})=c^{2},c\in F$, take $\sigma\in Gal(F/\mathbb{Q})$ to be the element

 $\sigma:\begin{cases}\sqrt{2}\mapsto\sqrt{2}\\ \sqrt{3}\mapsto-\sqrt{3}\end{cases}$

Then $(2+\sqrt{2})(3+\sqrt{3})\sigma((2+\sqrt{2})(3+\sqrt{3}))=(c\sigma(c))^{2}$, so

 $(2+\sqrt{2})^{2}(3+\sqrt{3})(3-\sqrt{3})=6(2+\sqrt{2})^{2}=(c\sigma(c))^{2}$

But $c\sigma(c)=\operatorname{Tr}_{F/\mathbb{Q}(\sqrt{2})}(c)\in\mathbb{Q}(\sqrt{2})$, and thus $6=\left(\frac{c\sigma(c)}{2+\sqrt{2}}\right)^{2}$, so $\sqrt{6}\in\mathbb{Q}(\sqrt{2})$, a contradiction   . Thus $F\neq E$. We show that $F\subset E$ by showing that $\sqrt{2}+\sqrt{3}\in E$.

 $\displaystyle(2+\sqrt{2})(3+\sqrt{3})=6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}\in E,% \text{ so}$ $\displaystyle 3\sqrt{2}+2\sqrt{3}+\sqrt{6}\in E,\text{ so}$ $\displaystyle(3\sqrt{2}+2\sqrt{3}+\sqrt{6})^{2}=36+12(\sqrt{2}+\sqrt{3}+\sqrt{% 6})\in E,\text{ so}$ $\displaystyle\sqrt{2}+\sqrt{3}+\sqrt{6}\in E,\text{ so}$ $\displaystyle 3\sqrt{2}+2\sqrt{3}+\sqrt{6}-\sqrt{2}-\sqrt{3}-\sqrt{6}=\sqrt{3}% +2\sqrt{2}\in E,\text{ so}$ $\displaystyle(\sqrt{3}+2\sqrt{2})^{2}=11+4\sqrt{6}\in E,\text{ so}$ $\displaystyle\sqrt{6}\in E,\text{ so}$ $\displaystyle\sqrt{2}+\sqrt{3}+\sqrt{6}-\sqrt{6}=\sqrt{2}+\sqrt{3}\in E$

So $F\subsetneq E$ and thus $[E:F]=2$. Then $[E:\mathbb{Q}]=[E:F][F:\mathbb{Q}]=8$.

Now, the irreducible polynomial  $f(x)$ for $(2+\sqrt{2})(3+\sqrt{3})$ over $\mathbb{Q}$ is the product    of $x-\tau((2+\sqrt{2})(3+\sqrt{3}))$ as $\tau$ ranges over $\operatorname{Gal}(F/\mathbb{Q})$:

 $f(x)=(x-(2+\sqrt{2})(3+\sqrt{3}))(x-(2-\sqrt{2})(3+\sqrt{3}))(x-(2+\sqrt{2})(3% -\sqrt{3}))(x-(2-\sqrt{2})(3-\sqrt{3}))$

so that $f(x^{2})$ is a degree $8$ polynomial    with $\alpha$ as a root. In fact,

 $f(x^{2})=x^{8}-24x^{6}+48x^{4}-288x^{2}+144$

This polynomial must be irreducible  since $\alpha$ is of degree $8$, so $f(x^{2})$ is the minimal polynomial for $\alpha$ over $\mathbb{Q}$. The roots of $f(x^{2})$ are obviously

 $\pm\sqrt{(2\pm\sqrt{2})(3\pm\sqrt{3})}$

Furthermore, it is easy to see that each of these roots lies in $E$, for

 $\displaystyle\alpha\sqrt{(2-\sqrt{2})(3+\sqrt{3})}$ $\displaystyle=\sqrt{2}(3+\sqrt{3})\in F$ $\displaystyle\alpha\sqrt{(2+\sqrt{2})(3-\sqrt{3})}$ $\displaystyle=\sqrt{6}(2+\sqrt{2})\in F$ $\displaystyle\alpha\sqrt{(2-\sqrt{2})(3-\sqrt{3})}$ $\displaystyle=\sqrt{2}\sqrt{6}=2\sqrt{3}\in F$

so dividing through by $\alpha$ we see that

 $\sqrt{(2-\sqrt{2})(3+\sqrt{3})},\sqrt{(2+\sqrt{2})(3-\sqrt{3})},\sqrt{(2-\sqrt% {2})(3-\sqrt{3})}\in E$

Thus $E$ is in fact Galois over $\mathbb{Q}$, is the splitting field  for $f(x^{2})$, and has Galois group  $G=\operatorname{Gal}(E/\mathbb{Q})$ of order (http://planetmath.org/OrderGroup) $8$.

$G$ acts transitively on the roots of $f(x^{2})$, and $E=\mathbb{Q}(\alpha)$, so an element of $G$ is determined by the image of $\alpha$. Thus the elements of $G$ are the automorphisms      of $E$ that map $\alpha$ to any of the eight roots of $f(x^{2})$. Let

 $\alpha=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ $\beta=\sqrt{(2-\sqrt{2})(3+\sqrt{3})}$ $\gamma=\sqrt{(2+\sqrt{2})(3-\sqrt{3})}$ $\delta=\sqrt{(2-\sqrt{2})(3-\sqrt{3})}$

and let $\sigma:\alpha\mapsto\beta,\tau:\alpha\mapsto\gamma$ be elements of $G$.

$\sigma(\alpha^{2})=\beta^{2}$, so $\sigma(2+\sqrt{2})\sigma(3+\sqrt{3})=(2-\sqrt{2})(3+\sqrt{3})$. This is an equation in $F$, so regarding $\sigma$ as an automorphism of $F/\mathbb{Q}$, it must be the automorphism $\sqrt{2}\mapsto-\sqrt{2},\sqrt{3}\mapsto\sqrt{3}$. Since $\alpha\beta=\sqrt{2}(3+\sqrt{3})$, we have $\sigma(\alpha\beta)=-\alpha\beta$ and thus that $\sigma(\beta)=-\alpha$. It follows that $\sigma$ is an element of order (http://planetmath.org/OrderGroup) $4$ in $G$.

Similarly, $\tau(\alpha^{2})=\gamma^{2}$, so $\tau(2+\sqrt{2})\tau(3+\sqrt{3})=(2+\sqrt{2})(3-\sqrt{3})$, so that $\tau$, regarded as an automorphism of $F/\mathbb{Q}$, must be $\sqrt{2}\mapsto\sqrt{2},\sqrt{3}\mapsto-\sqrt{3}$. Since $\alpha\gamma=\sqrt{6}(2+\sqrt{2})$, we have $\tau(\alpha\gamma)=-\alpha\gamma$, so that $\tau(\gamma)=-\alpha$, and $\tau$ is also an element of order (http://planetmath.org/OrderGroup) $4$ in $G$. Note also that $\sigma^{2}(\alpha)=-\alpha=\tau^{2}(\alpha)$, so that $\sigma^{2}=\tau^{2}\neq 1$.

Looking at $\sigma\tau$,

 $\sigma\tau(\alpha)=\sigma(\gamma)=\sigma\left(\frac{\alpha\gamma}{\alpha}% \right)=\frac{\sigma(\sqrt{6}(2+\sqrt{2}))}{\sigma(\alpha)}=\frac{-\sqrt{6}(2-% \sqrt{2})}{\beta}=-\frac{\beta\delta}{\beta}=-\delta$

while

 $\tau\sigma(\alpha)=\tau(\beta)=\tau\left(\frac{\alpha\beta}{\alpha}\right)=% \frac{\tau(\sqrt{2}(3+\sqrt{3}))}{\tau(\alpha)}=\frac{\sqrt{2}(3-\sqrt{3})}{% \gamma}=\frac{\gamma\delta}{\gamma}=\delta$

and thus $\tau\sigma^{3}(\alpha)=\tau\sigma\sigma^{2}(\alpha)=-\tau\sigma(\alpha)=-% \delta=\sigma\tau(\alpha)$. So $\sigma\tau=\tau\sigma^{3}$.

 $\sigma^{4}=\tau^{4}=1,\qquad\sigma^{2}=\tau^{2}\neq 1,\qquad\sigma\tau=\tau% \sigma^{3}$

Define $\varphi:G\to Q_{8}$ by $\varphi(\sigma)=i,\varphi(\tau)=j$. This is easily seen to be a homorphism, and $\varphi(\sigma\tau)=ij=k$, so $\varphi$ is surjective  and is thus an isomorphism      since both groups have order (http://planetmath.org/OrderGroup) $8$. Thus $\operatorname{Gal}(E/\mathbb{Q})\cong Q_{8}$.

Title field extension with Galois group $Q_{8}$ FieldExtensionWithGaloisGroupQ8 2013-03-22 17:44:28 2013-03-22 17:44:28 rm50 (10146) rm50 (10146) 7 rm50 (10146) Example msc 12F10