# 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.

• I don't understand what you mean by "we need to extend by $i$ to $\Bbb Q(\alpha,\beta)$ to include all roots." What does that mean exactly?
– anon
Commented May 27, 2014 at 3:13

$\beta$ 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 $\alpha+2\beta$

• Thanks mercio. I've worked out the $\textbf{D}_{8}$ correspondence and am now working out the corresponding Galois sub-fields for each subgroup. Should I just choose $\gamma=\alpha\beta$ as the normal basis generator (that seaturtle mentions above) to which the automorphisms in the subgroups should be applied? The way we learnt in class is a bit cumbersome as it involves applying the automorphisms to the general form of polynomials in $L$, then comparing with the general form of such polynomials to see which factors remain. Commented May 27, 2014 at 11:14
• Yes I was having some trouble inserting the polynomial into the comments part so I posted in the main section...and I just noticed I didn't even insert it in the end! Commented May 27, 2014 at 11:22
• You don't necessarily need a normal basis, you only need to find for every subgroup $H$ some element $x \in L$ whose invariant subgroup is $H$. First it's better to pick an element at random (like $\alpha\beta$ or $\alpha$), then compute its invariant subgroup then poof you have a pair subfield <-> subgroup. To get a normal basis you need an element $\gamma$ that is changed by every nontrivial automorphism. This is not the case with $\alpha\beta$. Moreover, this is not sufficient. Picking $\alpha+2\beta$ will not give you a free family, for example. Commented May 27, 2014 at 11:41
• Ok thanks for the feedback Commented May 27, 2014 at 12:04