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 $F⊊E$. 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

$$\sigma :\{\begin{array}{cc}\sqrt{2}\mapsto \sqrt{2}\hfill & \\ \sqrt{3}\mapsto -\sqrt{3}\hfill & \end{array}$$

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

$$(2+\sqrt{2})(3+\sqrt{3})=6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}\in E,\text{ so}$$

$$3\sqrt{2}+2\sqrt{3}+\sqrt{6}\in E,\text{ so}$$

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

$$\sqrt{2}+\sqrt{3}+\sqrt{6}\in E,\text{ so}$$

$$3\sqrt{2}+2\sqrt{3}+\sqrt{6}-\sqrt{2}-\sqrt{3}-\sqrt{6}=\sqrt{3}+2\sqrt{2}\in E,\text{ so}$$

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

$$\sqrt{6}\in E,\text{ so}$$

$$\sqrt{2}+\sqrt{3}+\sqrt{6}-\sqrt{6}=\sqrt{2}+\sqrt{3}\in E$$

So $F⊊E$ 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 $\mathrm{Gal}(F/\mathbb{Q})$:

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

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

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

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

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

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 $G=\mathrm{Gal}(E/\mathbb{Q})$ 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 ${\sigma }^2={\tau }^2\ne 1$.

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 $\sigma \tau =\tau {\sigma }^3$.

Putting this all together, we see that $G$ is generated by $\sigma ,\tau$, and that the generators satisfy the relations

$${\sigma }^4={\tau }^4=1,{\sigma }^2={\tau }^2\ne 1,\sigma \tau =\tau {\sigma }^3$$

Define $\phi :G\to Q_8$ by $\phi (\sigma )=i,\phi (\tau )=j$. This is easily seen to be a homorphism, and $\phi (\sigma \tau )=ij=k$, so $\phi$ is surjective and is thus an isomorphism since both groups have order (http://planetmath.org/OrderGroup) $8$. Thus $\mathrm{Gal}(E/\mathbb{Q})\cong Q_8$.

Title	field extension with Galois group $Q_8$
Canonical name	FieldExtensionWithGaloisGroupQ8
Date of creation	2013-03-22 17:44:28
Last modified on	2013-03-22 17:44:28
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	7
Author	rm50 (10146)
Entry type	Example
Classification	msc 12F10