Conjugate elements in a field extension

Question

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?

Answer 1

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 $\Bbb Q(\sqrt{2})$.