Finding the degree of a field extension over the rationals Let $\alpha = \sqrt{\sqrt{2}+\root 4 \of 2}$, $\beta =\sqrt{\sqrt{2}- \root 4 \of 2}$, $\gamma = \sqrt{-\sqrt{2}+i\root 4 \of 2}$ and $\delta = \overline{\gamma}=\sqrt{-\sqrt{2}-i\root 4 \of 2}$.
Let $L=\mathbb{Q}(\alpha , \beta , \gamma , \delta)$.
The goal is to find $[L:\mathbb{Q}\textbf{]}$.
A few facts I got so far:


*

*The minimal polynomial of $\alpha ,\beta ,\gamma$ and $\delta$ over $\mathbb{Q}$ is $f(t)=t^8-4t^4-8t^2+2$ and $[\mathbb{Q(\alpha )}:\mathbb{Q}\textbf{]}=[\mathbb{Q(\beta )}:\mathbb{Q}\textbf{]}=[\mathbb{Q(\gamma )}:\mathbb{Q}\textbf{]}=[\mathbb{Q(\delta )}:\mathbb{Q}\textbf{]}=8$.

*The extension $L:\mathbb{Q}$ is normal.

*We have $\mathbb{Q}(\alpha ^2)=\mathbb{Q}(\beta ^2)=\mathbb{Q}(\root 4 \of 2)$ and $\alpha \not \in \mathbb{Q}(\alpha ^2)$.

*The extension $\mathbb{Q}(\alpha):\mathbb{Q}(\alpha ^2)$ is normal and there exists $\chi \in \text{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q}(\alpha ^2))$ such that $\displaystyle \chi (\alpha)=-\alpha$. 

*We also have the following equalities: $\alpha \beta
    =\sqrt{2-\sqrt{2}}$, $\gamma \delta=\sqrt{2+\sqrt{2}}$ and $\alpha \beta \gamma \delta = \sqrt{2}$.

*The extension $\mathbb{Q}(\alpha \beta):\mathbb{Q}$ is normal and $[\mathbb{Q}(\alpha \beta):\mathbb{Q}\textbf{]}=4$. Furthermore $\mathbb{Q}(\alpha \beta)\neq \mathbb{Q}(\alpha ^2)$.

*It is true that $\beta \not \in \mathbb{Q}(\alpha)$, therefore $[\mathbb{Q}(\alpha ,\beta):\mathbb{Q}\textbf{]}=16$. Also $\gamma \not \in \mathbb{Q}(\alpha ,\beta)$ and $L=\mathbb{Q}(\alpha ,\beta ,\gamma)$.

*The extension $\mathbb{Q}(\alpha ,\beta):\mathbb{Q}(\sqrt{2})$ is normal and $[\mathbb{Q}(\alpha ,\beta):\mathbb{Q}(\sqrt{2})\textbf{]}=8$. There exist $\varphi \in \text{Gal}(\mathbb{Q}(\alpha ,\beta)/\mathbb{Q}(\beta))$ such that $\varphi (\alpha)=-\alpha$ and $\psi \in \text{Gal}(\mathbb{Q}(\alpha ,\beta)/\mathbb{Q}(\alpha))$ such that $\psi (\beta)=-\beta$. Also $\vert \varphi \vert$=$\vert \psi \vert=2$.

*There exists $\rho \in \text{Gal}(\mathbb{Q}(\alpha ,\beta)/\mathbb{Q}(\sqrt{2})$ such that $\rho (\alpha)=\beta$. Furthermore $\vert \rho \vert=4$.

*It is true that $\varphi , \psi \in \text{Gal}(\mathbb{Q}(\alpha ,\beta)/\mathbb{Q}(\sqrt{2}))$, $(\varphi \circ \psi)(\alpha)=\beta$, $(\varphi \circ \psi)(\beta)=\alpha$, $\vert \varphi \circ \rho \vert=2$ and $(\varphi \circ \rho)(\alpha \beta)=\alpha \beta \wedge (\varphi \circ \rho)(\alpha +\beta)=\alpha+\beta$.

*Finally $[L:\mathbb{Q}(\alpha ,\beta)\textbf{]}\in \{2,4\}$.


Can anyone find the exact value of $[L:\mathbb{Q}(\alpha ,\beta)\textbf{]}$?
If there's a way to find $[L:\mathbb{Q}\textbf{]}$ using a computer, I'd like to know the answer just for the sake of curiosity and peace of mind, but I'm still looking for a proof.
Appreciated.
 A: Using Sage I was able to calculate $[L: \mathbb Q]=64$. Basically you can create number fields in Sage and then create relative extensions of these number fields. I was able to factor the minimum polynomial of $\alpha$ over each extension and then adjoin a root until I reached a point at which the minimum polynomial split. I can try to throw together a sage notebook or something if you want to see this process. 
A: Let $K = \mathbb{Q}(\alpha^2, \beta^2, \gamma^2, \delta^2) = \mathbb{Q}(\sqrt[4]{2}, i)$ be the splitting field of $t^4 - 4 t^2 + 8t + 2$. Then $[K:\mathbb{Q}] = 8$ and $L/K$ is a Kummer extension formed by adjoining four square roots.
As per Kummer theory, we are thus interested in the subgroup of $K^*$ modulo squares that is generated by $\alpha^2, \beta^2, \gamma^2, \delta^2$. This is an abelian group of exponent 2 and order $1$, $2$, $4$, or $8$. (It can't be $16$ as we know their product is $2$, which is square)
We have $\alpha^2 = \sqrt[4]{2} (\sqrt[4]{2} + 1)$. The factor of $\sqrt[4]{2}$ shows that $\alpha^2$ is not a square, and so the group order cannot be $1$.
The group order cannot be 4 because of symmetry.
If the group order were $2$, it would mean $\alpha^2 \beta^2 = (\sqrt[4]{2})^2 (\sqrt{2} - 1)$ is a square. As it is in the real subfield, we can deduce that this in means $(\sqrt{2} - 1) = (\sqrt[4]{2}-1)(\sqrt[4]{2}+1)$ is a square in $\mathbb{Q}(\sqrt[4]{2})$.
Alas, these are units, which leaves me stuck. I want to believe $\sqrt[4]{2} \pm 1$ are the fundamental units and therefore this expression can't be square, but I don't know how to go about showing that. Mapping into a finite field might work, but the smallest prime that splits is $341$, I think, and that's beyond what I want to compute by hand at the moment.
