# Showing that $\mathbb{Q}(\sqrt{2},\sqrt{3}, \sqrt{(9 - 5\sqrt{3})(2-\sqrt{2})})$ is normal over $\mathbb{Q}$, and finding its Galois group

If $K=\mathbb{Q}(\sqrt{2},\sqrt{3}, u)$, where $u^2 = (9 - 5\sqrt{3})(2-\sqrt{2})$, show that $K/\mathbb{Q}$ is normal, and find $\operatorname{Gal}(K/\mathbb{Q})$.

I found that the minimal polynomial of $u^2$ over $\mathbb{Q}$ is $f = t^4 - 72t^3 + 720t^2 - 864t + 144$, so that $u$ satisfies $g = t^8 - 72t^6 + 720t^4 - 864t^2 + 144$, but I'm not sure how to show that $K$ is a splitting field of $g$. Any help would be greatly appreciated!

• In your steps each time when you squared an equation, consider changing the sign on one side to arrive to another root of $f$ or $g$. – Berci Jun 10 '15 at 22:05

Let $M = \Bbb Q(\sqrt 2, \sqrt 3)$. Clearly, $\Bbb Q \subset M$ is normal of degree $4$ and $M \subset M(u)$ is normal of degree $2$, so you have to show that every $\Bbb Q$-conjugate of $u^2$ also has its square roots in $K$.

So for each automorphism $\sigma \in Gal(M/\Bbb Q)$, you have to show that $\sigma(u^2)u^2$ is a square in $M$. In particular it's enough to show that $(9-5\sqrt 3)(9+5\sqrt 3)$ and $(2-\sqrt 2)(2+\sqrt 2)$ are squares in $M$.

$(9-5\sqrt 3)(9+5\sqrt 3) = 81-75 = 6$, and $\sqrt 6 \in M$.
$(2-\sqrt 2)(2+\sqrt 2) = 2$, and $\sqrt 2 \in M$.

This shows that the conjugates of $u$ are $\pm u, \pm \frac {\sqrt 2}{2-\sqrt 2}u, \pm \frac{\sqrt 6}{9-5\sqrt 3}u$, and $\pm 2\sqrt 3 /u$, and they are all in $K$.

You get the action of those transformations on $\sqrt 2$ and $\sqrt 3$ by looking at what they do to $u^2$.

Then you can compute the multiplication table of the Galois group and see it is isomorphic to the quaternion group.

• You say that $M(u)/M$ is clearly of degree $2$. It is clearly of degree at most $2$, but I'd say some argument is needed to show $u$ is not in $M$, i.e., $(9-5\sqrt{3})(2-\sqrt{2})$ is not a square in $M$. – KCd Jun 11 '15 at 12:27