Prove that this extension is Galois with galois group the quaternions Let $M=\mathbb Q(\sqrt{2},\sqrt{3})$ and let $E=M(\sqrt{(\sqrt{2}+2)(\sqrt{3}+3)})$. 
Prove that:


*

*M is Galois over $\mathbb{Q}$

*Show that $E$ is Galois over $\mathbb Q$ with Galois group the quaternion group.
I proved the first part, I don't know how could be useful to consider it for the second part. This is an exercise of Milne Field and Galois Theory Chapter 3.
 A: To check that $E$ is Galois over $\Bbb Q$ you need to check that the conjugates $x_2,x_3,x_4$ in $M$ of $x_1 = (\sqrt 2 + 2)(\sqrt 3 + 3)$ all have a square root in $E$.
Let $x_2 = (-\sqrt 2+ 2)(\sqrt 3+3), x_3 = (\sqrt 2+2)(-\sqrt 3+3)$ and $ x_4 = (-\sqrt 2+2)(-\sqrt 3+3)$
I don't think that any $x_i$ is a square in $M$ so we have to prove that $M(\sqrt{x_1}) = M(\sqrt {x_i})$, which in turn is equivalent to $x_1x_i$ being a square in $M$.
Now, since $(\sqrt 2 + 2)(-\sqrt 2 + 2) = 2 = (\sqrt 2)^2$ and $(\sqrt 3 + 3)(-\sqrt 3 + 3) = 6 = (\sqrt 2 \sqrt 3)^2$, we have that 
$x_1x_2 = (\sqrt 2 (3+\sqrt 3))^2, x_1x_3 = (\sqrt 6(2+\sqrt 2))^2, x_1x_4 = (2\sqrt 3)^2$ are all squares in $M$.
If you have a square root $y$ of $x_1$, then the square roots of $x_2$ are given by $\pm \sqrt{x_1 x_2} / \sqrt {x_1} = \pm (\sqrt 2(3 + \sqrt 3))/y = \pm \frac {\sqrt 2}{\sqrt 2+2}y$. 
You have similar formulas (defined over $M$ but not over $\Bbb Q$) to get the other $4$ conjugates of $y$, which gives you $8$ automorphisms of $E$ :
$$\sigma_1^\pm (\sqrt 2, \sqrt 3,y) = (+\sqrt 2,+\sqrt 3,\pm y) \\
\sigma_2^\pm (\sqrt 2, \sqrt 3,y) = (-\sqrt 2,+\sqrt 3,\pm \frac{\sqrt 2}{2+\sqrt 2}y) \\
\sigma_3^\pm (\sqrt 2, \sqrt 3,y) = (+\sqrt 2,-\sqrt 3,\pm \frac{\sqrt 6}{3+\sqrt 3}y) \\
\sigma_4^\pm (\sqrt 2, \sqrt 3,y) = (-\sqrt 2,-\sqrt 3,\pm \frac{2\sqrt 3}{(2+\sqrt 2)(3+\sqrt 3)}y) $$
$\sigma_2^+(\sigma_2^+(y)) = \sigma_2^+(\frac {\sqrt 2}{2+\sqrt 2} y) = \frac {-\sqrt 2}{2-\sqrt 2}\frac {\sqrt 2}{2+\sqrt 2}y = \frac {-2}{2}y = -y = \sigma_1^-(y)$. Similarly, we can check that $(\sigma_3^\pm)^2 = (\sigma_4^\pm)^2 = \sigma_1^-$.
$\sigma_3^+(\sigma_2^+(y)) = \sigma_3^+(\frac{\sqrt 2}{2+\sqrt 2}y) = \frac{\sqrt 2}{2+\sqrt 2}\frac{\sqrt 6}{3+\sqrt 3}y = \sigma_4^+(y)$, so $\sigma_4^+\sigma_3^+\sigma_2^+ = \sigma_1^-$, which shows that the Galois group is isomorphic to the quaternion group.
