Show field extension is Galois via constructing separable polynomial Looking to show that $\mathbb{Q}(\alpha,\beta)$ where $\alpha = \sqrt{1+\sqrt{3}}, \beta = \sqrt{1 - \sqrt{3}}$ is Galois via showing that $\mathbb{Q}(\alpha,\beta)$ is the splitting field for a separable polynomial (or some easier method of showing that the field is Galois). So the minimial polynomial in $\mathbb{Q}$ for $\alpha$ is $x^{4}-2x^{2}-2$ and the minimal polynomial in $\mathbb{Q}$ for $\beta$ is the same i.e.  $x^{4}-2x^{2}-2 \in \mathbb{Q}$, hence we need to extend by $i$ to $\mathbb{Q}(\alpha,\beta)$ to include all roots. I know that $[\mathbb{Q}(\alpha,\beta): \mathbb{Q}]=8$ which is the degree of the field extension and the corresponding Galois group, but am wondering:
(1) how to construct a separable polynomial which is (i) of degree 8 and (ii) splits linearly (thus is separable) over $\mathbb{Q}(\alpha,\beta)$;
(2) whether the elements of the Galois group $G(\mathbb{Q}(\alpha,\beta))$ are just those automorphisms that (in addition to the identity automorphism) send $\alpha$ to $-\alpha, \pm\beta$ and $\beta$ to $-\beta, \pm\alpha$ or whether (as this doesn't give me 8), I need to include $\alpha\beta$ and $\alpha \pm \beta$ here too.
 A: $\beta$ is not real, so $\beta \notin \Bbb Q(\alpha)$.
As you noticed, $\beta$ and $\alpha$ share the same minimal polynomial, whose roots are $\alpha,-\alpha,\beta,-\beta$.
Therefore, $(X-\beta)(X+\beta) = (X^4 -2X^2-2)/(X-\alpha)(X+\alpha) \in \Bbb Q(\alpha)[X]$ is the minimal polynomial of $\beta$ over $\Bbb Q(\alpha)$, so $\Bbb Q(\alpha,\beta)$ has degree $8$ and is the splitting field of the polynomial $X^4-2X^2-2$.
Now about the Galois group. Since the extension is Galois it has order $8$. One automorphism is determined by the image of $\alpha$ (among the $4$ candidates), which determines the image of $- \alpha$, and then the image of $\beta$ among the two remaining roots.
We can write them down explicitly and find its group structure :
Let $\phi$ be defined by $\phi(\alpha,\beta) = (\beta,- \alpha)$.
Then $\phi^2(\alpha,\beta) = \phi(\beta, -\alpha) = (-\alpha,-\beta)$
$\phi^3(\alpha,\beta) = \phi(-\alpha,-\beta) = (-\beta,\alpha)$
$\phi^4(\alpha,\beta) = \phi(-\beta,\alpha) = (\alpha,\beta)$, so $\phi$ has order $4$.
Let $\psi$ be defined by $\psi(\alpha,\beta) = (-\alpha,\beta)$.
$\psi$ has order two, and the remaining automorphisms are
$\psi\phi(\alpha,\beta) = \psi(\beta,-\alpha) = (\beta,\alpha)$
$\psi\phi^2(\alpha,\beta) = (\alpha,-\beta)$
$\psi\phi^3(\alpha,\beta) = (-\beta,-\alpha)$  
$\phi\psi(\alpha,\beta) = \phi(-\alpha,\beta) = (-\beta,-\alpha) = \psi\phi^3(\alpha,\beta)$ so $\phi\psi = \psi\phi^3$ : the group is not commutative.
All of this shows that the group is isomorphic to the diedral group of order $8$,  $D_8$.
$D_8$ has three subgroups of order $4$ and $5$ subgroups of order $2$, corresponding to $\Bbb Q(\sqrt {-2}), \Bbb Q(\sqrt 3),\Bbb Q(\sqrt {-6}),\Bbb Q(\sqrt {-2},\sqrt 3)
,\Bbb Q(\alpha), \Bbb Q(\beta), \Bbb Q(\alpha+ \beta), \Bbb Q(\alpha-\beta)$.
See if you can find a simplified expression of $\alpha+\beta$ and $\alpha-\beta$
In order to get an irreducible polynomial of order $8$ that splits completely, you need the minimal polynomial of an element of $\Bbb Q(\alpha,\beta)$ that is not fixed by any element of the Galois group.
For example you can pick the minimal polynomial of $\alpha+2\beta$
