# $\mathbb Q_8$ as Galois group

Let $\alpha =\sqrt {(2+\sqrt2)(3+\sqrt3)}$ and consider the extension $\mathbb Q(\alpha)/\mathbb Q$. Prove that $\mathbb Q(\alpha)/\mathbb Q$ is a Galois extension with Gal$(\mathbb Q(\alpha)/\mathbb Q) \simeq \mathbb Q_8$.

I did the basic calculation to find a polynomial which is satisfied by $\alpha$ and that is of the form: $x^8 -24 x^6+144x^4-288x^2+144 =0$. I am unable to show this is irreducible. Thanks for kind help.

Ok. From this link now it is clear that why this extension is galois. But what about the Galois group?

Let $L$ be your degree 8 field, $G = \text{Gal}(L/\mathbb{Q})$, and let $K_2 = \mathbb{Q}(\sqrt 2)$, $K_3 = \mathbb{Q}(\sqrt 3)$, $K_6 = \mathbb{Q}(\sqrt 6)$ be its three quadratic subfields. Up to isomorphism, there are exactly two non abelian groups of order $8$ : the dihedral $D_8$ and the quaternionic $H_8$, characterized by the number of their cyclic subgroups of order 4: $D_8$ admits exactly 1 such subgroup; for $H_8$ , all its 3 subgroups of order 4 are cyclic. To decide whether G is $D_8$ or $H_8$, you just have to compute the Galois groups of $\text{Gal}(L/K_2)$, etc. If Lucky, you don't even need to do the complete calculations, because there is a criterion which says that $L/K_i$ is embeddable into a cyclic extension of degree 4 iff $-1$ is a norm in $L/K_i$ . My "guess" : $G$ is $H_8$ .