TITLE: The Galois group for the following field extension
QUESTION [5 upvotes]: I am trying to find the Galois group of the extension $\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha)=K$ over $\mathbb{Q}$ where $\alpha$ is such that $\alpha^2=(9-5\sqrt{3})(2-\sqrt{2})$. 
Here is my attempt: First of all I try to calculate the degree of the extension. It is well known that $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=4$, and then when we add $\alpha$ we get a quadratic extension, so we have that $[K:\mathbb{Q}]=4\cdot 2=8$. 
If $\sigma$ is an element of $G$ (the Galois group), then we have that $\sigma(\sqrt{2})=\pm\sqrt{2}$ and $\sigma(\sqrt{3})=\pm\sqrt{3}$, but I cant really see what the action that $\sigma$ would have on $\alpha$. My gut feeling is that the answer will be $\sigma(\alpha)=\pm \alpha$ and then our group would be $C_2\times C_2\times C_2$, but I am not sure how to prove it. 
Thanks,

REPLY [4 votes]: Let $\sigma$ be the automorphism of $\mathbb Q(\sqrt 2, \sqrt 3)$ over $\mathbb Q$ sending $\sqrt 2$ to $-\sqrt 2$ and fixing $\sqrt 3$. Compute $\sigma(\alpha^2)/\alpha^2$. You will get $(2+\sqrt 2)/(2-\sqrt 2)$. This is $3+2\sqrt 2=(\sqrt 2 + 1)^2$. So $\sigma(\alpha^2)=\alpha^2(\sqrt 2 + 1)^2$. If $\alpha$ were in $\mathbb Q(\sqrt 2, \sqrt 3)$, then $(\sigma(\alpha))=\pm (\sqrt 2+1)\alpha$ and $\sigma(\sigma(\alpha))=\alpha(1+\sqrt 2)(1-\sqrt 2)=-\alpha$, a contradiction as $\sigma$ has order $2$. This proves $\alpha$ has degree $2$ over $\mathbb Q(\sqrt 2, \sqrt 3)$, as you indicate.
Extend $\sigma$ to an automorphism of the top field. This is possible by a standard theorem. We now see that $\sigma^4(\alpha)=\alpha$, so $\sigma$ generates a subgroup of order $4$ in the Galois group. 
Let $\tau$ be the automorphism of $\mathbb Q(\sqrt 2, \sqrt 3)$ over $\mathbb Q$ sending $\sqrt 3$ to $-\sqrt 3$ and fixing $\sqrt 2$. You can do a similar computation with $\tau$. You will find they are both of order $4$ and anti-commute. The only group of order $8$ with such elements is the quaternions, as the only non-commuative groups of order $8$ are $Q$ and $D_8$, and $D_8$ does not have two anti-commuting elements of order $4$.