Determine the Galois group of $x^4 -4x^2+2$

Question

Determine the Galois group of $x^4 -4x^2+2$

After discussing with some classmates , two approaches arised yielding different answers, so at least one should be wrong:

The roots are $\pm \alpha=\pm\sqrt{2+\sqrt 2} $ ; $\pm \beta=\pm\sqrt{2-\sqrt 2}$

Approach 1

Since $\sqrt 2 \in \Bbb Q (\sqrt{2+\sqrt 2})$ because $\sqrt{2+\sqrt 2}^2-2=2 $

Also $(\sqrt{2-\sqrt 2} \in \Bbb Q (\sqrt{2+\sqrt 2})$ because $\sqrt{2-\sqrt 2}= \frac{\sqrt2}{\sqrt{2+\sqrt 2} } $

Since there are four possibilities for an automorphism $\sigma :\sqrt{2+\sqrt 2} \mapsto: \sigma(\sqrt{2+\sqrt 2})$ which are each of the four roots and $(\sqrt{2-\sqrt 2}$ is already determined after this choice, the Galois group is isomorphic to a group with 4 elements

Approach 2

Using the tower rule

$[\Bbb Q(\alpha, \beta) : \Bbb Q]=[\Bbb Q(\alpha, \beta) : \Bbb Q(\alpha)][\Bbb Q(\alpha) : \Bbb Q]$.

$[\Bbb Q(\alpha) : \Bbb Q]=$4 , as the polinomial is irreducible over $\Bbb Q$

On another hand: $\beta=4-\alpha^2$ so the minimum polynomial of beta $m_{\beta}| (x^2+\alpha^2-4)$

$\alpha^2-4=\sqrt 2 -2 $ so if $(x^2+\alpha^2-4)$ splits , then it has zeros $\pm \sqrt{2-\sqrt 2} $ which are not in $\Bbb Q(\alpha)$ so $[\Bbb Q(\alpha, \beta) : \Bbb Q(\alpha)]=2$

and $[\Bbb Q(\alpha, \beta) : \Bbb Q]=8 $.Then the Galois group is isomorphic to $\langle \sigma, \tau \rangle$ defined as $\sigma(\alpha)=\beta$, $\sigma(\beta)=-\alpha$, $\tau(\alpha)=-\alpha, \tau (\beta)= \beta$ which has order 8

I think approach one is the correct one, but how do I determine what order-4 group is the galois group isomorphic to?

Edit:

Following with approach 1 and the suggestion in the comments:

For the case

1. $\sigma( \alpha)=\beta$

I have that $\sigma( \sqrt 2)=\sigma(\sqrt{2+\sqrt 2}^2)-\sigma(2)=(\sqrt{2-\sqrt 2}^2)-2=-\sqrt 2$

Then $\sigma( \beta)=\sigma(\sqrt{2-\sqrt 2})=\frac{\sigma(\sqrt 2)}{\sigma(\sqrt {2+\sqrt 2})}=\frac{-(\sqrt 2)}{\sqrt {2-\sqrt 2}}=- \sqrt {2+\sqrt 2}=-\alpha$

so $\sigma^2(\alpha)=\sigma(\beta)=-\alpha$

Similarly with the other cases:

2. $\sigma( \alpha)=-\beta$

I get $\sigma( \beta)=\alpha$ and $\sigma^2(\alpha)=\sigma(-\beta)=-\alpha$

3. $\sigma( \alpha)=\alpha$

I get $\sigma( \beta)=\beta$ and $\sigma^2(\alpha)=\sigma(\alpha)=\alpha$

4. $\sigma( \alpha)=-\alpha$

I get $\sigma( \beta)=-\beta$ and $\sigma^2(\alpha)=\sigma(-\alpha)=\alpha$

How does this help to get to the answer?

Answer 1

I wanted to leave this as a comment but don't have enough reputation, I wanted to suggest a useful resource: https://www.maths.ed.ac.uk/~tl/galois/lastyear/Galois.pdf on page 60, which proves the following theorem:

Let $f(X) = X^4+aX^2+b \in \mathbb{Q}[X]$ be irreducible.

(i) If $b$ is a square in $\mathbb{Q}$, then $\text{Gal} (\mathbb{Q}(f(X))/\mathbb{Q}) \cong \mathbb{Z} / 2 \times \mathbb{Z} /2$.

(ii) If $b(a^2-4b)$ is a square in $\mathbb{Q}$ then $\text{Gal} (\mathbb{Q}(f(X))/\mathbb{Q}) \cong \mathbb{Z} / 4$.

(iii) If neither $b$ nor $b(a^2-4b)$ is a square in $\mathbb{Q}$ then $\text{Gal} (\mathbb{Q}(f(X))/\mathbb{Q}) \cong D_8$.