Finding the degree of a field extension over the rationals

Question

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:

1. The minimal polynomial of $\alpha ,\beta ,\gamma$ and $\delta$ over $\mathbb{Q}$ is $f(t)=t^8-4t^4-8t^2+2$ and $[\mathbb{Q(\alpha )}:\mathbb{Q}\textbf{]}=[\mathbb{Q(\beta )}:\mathbb{Q}\textbf{]}=[\mathbb{Q(\gamma )}:\mathbb{Q}\textbf{]}=[\mathbb{Q(\delta )}:\mathbb{Q}\textbf{]}=8$.
2. The extension $L:\mathbb{Q}$ is normal.
3. We have $\mathbb{Q}(\alpha ^2)=\mathbb{Q}(\beta ^2)=\mathbb{Q}(\root 4 \of 2)$ and $\alpha \not \in \mathbb{Q}(\alpha ^2)$.
4. 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 $\displaystyle \chi (\alpha)=-\alpha$.
5. We also have the following equalities: $\alpha \beta =\sqrt{2-\sqrt{2}}$, $\gamma \delta=\sqrt{2+\sqrt{2}}$ and $\alpha \beta \gamma \delta = \sqrt{2}$.
6. The extension $\mathbb{Q}(\alpha \beta):\mathbb{Q}$ is normal and $[\mathbb{Q}(\alpha \beta):\mathbb{Q}\textbf{]}=4$. Furthermore $\mathbb{Q}(\alpha \beta)\neq \mathbb{Q}(\alpha ^2)$.
7. 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 $L=\mathbb{Q}(\alpha ,\beta ,\gamma)$.
8. 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 $\vert \varphi \vert$=$\vert \psi \vert=2$.
9. There exists $\rho \in \text{Gal}(\mathbb{Q}(\alpha ,\beta)/\mathbb{Q}(\sqrt{2})$ such that $\rho (\alpha)=\beta$. Furthermore $\vert \rho \vert=4$.
10. 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 $(\varphi \circ \rho)(\alpha \beta)=\alpha \beta \wedge (\varphi \circ \rho)(\alpha +\beta)=\alpha+\beta$.
11. 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.

Answer 1

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.

Answer 2

For convenience I will sometimes denote the number $\sqrt[4]{2}$ by $\theta$.

What we have here is a tower of square roots : To $\mathbb Q$ we can add successively $\sqrt{2},\sqrt[4]{2},i,\alpha,\beta,\gamma$. The main problem is that in this sequence, some square root might already be contained in the preceding field.

Remark Let $\mathbb A$ be a field and let $a$ be a nonsquare in ${\mathbb A}$ ; let $b\in {\mathbb A}(\sqrt{a})$. The extension ${\mathbb A}(\sqrt{a}) \subseteq {\mathbb A}(\sqrt{a},\sqrt{b})$ has degree $1$ or $2$.

To know which it is, there are two cases to consider :

Extension rule 1 Let ${\mathbb A},a,b$ be as above. If $b\in {\mathbb A}$, then the degree is $1$ iff $ab$ is a square in $\mathbb A$.

Extension rule 2 Let ${\mathbb A},a,b$ be as above. If $b\not\in {\mathbb A}$, so that $b=u+v\sqrt{a}$ with $v\neq 0$, then the degree is $1$ iff the equation $x^4-ux^2+\frac{av^2}{4}=0$ has a solution in $\mathbb A$.

(to see why rule 2 holds, not that for $x,y\in {\mathbb A}$ we have $(x+y\sqrt{a})^2=u+v\sqrt{a}$ iff $y=\frac{v}{2x}$ and $x^4-ux^2+\frac{av^2}{4}=0$).

Note that rule 2 unfortunately raises the degree to $4$ and complicates the recursive analysis, but in the present case we are lucky and will be able to succeed using only a small part of rule 2, namely

Extension rule 2' Let ${\mathbb A},a,b,u,v$ as in rule 2 above. If $\sqrt{u^2-av^2}\not\in {\mathbb A}$, then the degree of the extension is $2$.

(to see how this follows from rule $2$, not that $x^4-ux^2+\frac{av^2}{4}=0$ implies $(2x-u)^2=u^2-av^2$).

Fact 1 The four numbers $g_1=2-\sqrt{2},\sqrt{2}g_1=-2+2\sqrt{2},g_2=2+\sqrt{2}$ and $\sqrt{2}g_2=-2-2\sqrt{2}$ are all nonsquares in ${\mathbb Q}(\sqrt{2})$.

Proof of fact 1 Use rule 2' with ${\mathbb A}={\mathbb Q}, a=2$, and $(u,v)=(2,1)$ or $(-2,2)$ (yielding $u^2-av^2=2$ or $-4$, both nonsquares in $\mathbb Q$).

Fact 2 The seven numbers $g_1=2-\sqrt{2}$, $g_2=6+4(\theta+\theta^2+\theta^3)$, $g_3=6-4(\theta-\theta^2+\theta^3)$, $g_4=4-2\sqrt{2}$, $g_5=\sqrt{2}+\sqrt[4]{2}$, $g_6=\sqrt{2}-\sqrt[4]{2}$, and and $g_7=2+\sqrt{2}$ are all nonsquares in ${\mathbb Q}(\sqrt[4]{2})$.

Proof of fact 2 For the numbers $g_1$ and $g_7$, use fact 1 above and rule 1 with ${\mathbb A}={\mathbb Q}(\sqrt{2}), a=\sqrt{2}$. For the numbers $g_5$ and $g_6$, use fact 1 above and rule 2' with ${\mathbb A}={\mathbb Q}(\sqrt{2}), a=\sqrt{2}$. For the other numbers, note that $g_2=g_1(2+\theta+\theta^2+\theta^3)^2, g_3=g_1(2-\theta+\theta^2-\theta^3)^2$ and $g_4=g_1((\sqrt{2})^2)$.

Fact 3 The seven numbers $h_1=-\sqrt{2}+i\sqrt[4]{2}$, $h_2=-2-\sqrt[4]{8}+i(\sqrt{2}+\sqrt[4]{8})$, $h_3=-2+\sqrt[4]{8}+i(-\sqrt{2}+\sqrt[4]{8})$, $h_4=2-2\sqrt{2}+i(2\sqrt{2}-\sqrt[4]{8})$, $h_5=\sqrt{2}+\sqrt[4]{2}$, $h_6=\sqrt{2}-\sqrt[4]{2}$, and $h_7=2-\sqrt{2}$ are all nonsquares in ${\mathbb Q}(\sqrt[4]{2},i)$.

Proof of fact 3 Use fact 2 above and rule 2' with ${\mathbb A}={\mathbb Q}(\sqrt[4]{2}), a=-1$ : for any $k\in \lbrace 1,2,3,4 \rbrace$, we can write $h_k=u_k+v_ki$ where $u_k$ and $v_k$ are both in ${\mathbb Q}(\sqrt[4]{2})$, and $u_k^2-av_k^2=g_k$. Thus $h_k$ is a nonsquare in ${\mathbb Q}(\sqrt[4]{2},i)$ because $g_k$ is a nonsquare in ${\mathbb Q}(\sqrt[4]{2})$, when $k\in \lbrace 1,2,3,4 \rbrace$. For numbers $h_5$ to $h_7$ use rule 1 with ${\mathbb A}={\mathbb Q}(\sqrt[4]{2})$ and $a=-1$.

Fact 4 The seven numbers $\gamma^2$, $(\alpha\gamma)^2$, $(\beta\gamma)^2$, $(\alpha\beta\gamma)^2$, $\alpha^2$, $\beta^2$ and $(\alpha\beta)^2$ are all nonsquares in ${\mathbb Q}(\sqrt[4]{2},i)$.

Proof of fact 4 This is because of fact 3 above and $\gamma^2=h_1$ ,$(\alpha\gamma)^2=h_2$, $(\beta\gamma)^2=h_3$, $(\alpha\beta\gamma)^2=h_4$, $\alpha^2=h_5$, $\beta^2=h_6$, $(\alpha\beta)^2=h_7$.

Starting from fact 4 and iterating rule 1, we see successively that $\alpha$ has degree $2$ on ${\mathbb Q}(\sqrt[4]{2},i)$, then that $\beta$ has degree $2$ on ${\mathbb Q}(\sqrt[4]{2},i,\alpha)$, then that $\gamma$ has degree $2$ on ${\mathbb Q}(\sqrt[4]{2},i,\alpha,\beta)$. The total degree is therefore $64$.

Answer 3

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.