I want to describe the Galois group of $x^4-6x^2+7$ over $\mathbb{Q}$ (by computing its degree and seeing if it is abelian or not).
I know that its roots are
$$\sqrt{3+\sqrt{2}}\qquad-\sqrt{3+\sqrt{2}}\qquad\sqrt{3-\sqrt{2}}\qquad-\sqrt{3-\sqrt{2}}$$ so I can do the extension:
$$\mathbb{Q}\subset\mathbb{Q}\left(\sqrt{3+\sqrt{2}}\right)$$
Question: Does $\mathbb{Q}\left(\sqrt{3+\sqrt{2}}\right)$ contain the root $\sqrt{3-\sqrt{2}}$?
If so, then that extension is actually the splitting field and all I have to do is to describe the Galois group to check whether it is abelian or not.
If not, I must also adjoint $\sqrt{3-\sqrt{2}}$ in order to have the splitting field.
How to check if the (somehow) conjugate belongs to an extension? What kind of computations do you recommend?
Checking memberships in quadratic extensions allows for some simplifications.
Assuming you have a field $F$ and a quadratic extension $F(\sqrt{\alpha})$. Then for $x\in F(\sqrt{\alpha})$ to satisfy $x^2 \in F$, then either $x\in F$ or $x^2$ must be of the form $y^2\alpha$ for some $y\in F$. This is done by writing $x$ in the form $a+b\sqrt{\alpha}$, then $x^2 = a^2+b^2\alpha + 2ab\sqrt{\alpha}$. Now $a=0$ or $b=0$ follows from the fact that $1,\sqrt{\alpha}$ form a basis for $F(\sqrt{\alpha})$ over $F$.
In this case ($F = \Bbb Q(\sqrt{2}), \alpha = 3+\sqrt{2}$) you must first check whether $3-\sqrt{2}$ is a square in $\Bbb Q(\sqrt{2})$, and then you check to see whether $\frac{3+\sqrt{2}}{3-\sqrt{2}} = \frac{(3+\sqrt{2})^2}{7}$ is a square in $\Bbb Q(\sqrt{2})$.