# 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:

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.

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What if you start from $[\mathbb{Q}(\alpha^2, \beta^2, \gamma^2, \delta^2) : \mathbb{Q}]= 8$? Then L is obtained by adjoining four square roots, and the challenge is to see that three are independent. –  Hurkyl Mar 10 '13 at 23:10
You have some $a^2$ in step 3. It should be probably $\alpha^2$. –  Cortizol Sep 9 '13 at 11:50
@Cortizol Yes, thank you. –  Git Gud Sep 9 '13 at 12:17

## 3 Answers

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 $64$.

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Thank you for your answer. –  Git Gud Mar 16 '13 at 16:53

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.

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Thank you for your trouble. I've never tried Sage. I think I might install it to try to see how it works. –  Git Gud Feb 3 '13 at 8:14

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.

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Can't you use Dirichlet's unit theorem to tell you the structure of $K^\times$? –  user38268 Mar 11 '13 at 0:36
@Hurkyl Thank you for your effort. To be honest I had never heard of Kummer Theory before, but if you happen to get a complete solution I'll read about it and try to understand it. –  Git Gud Mar 11 '13 at 7:05
I'm not too well versed on it; I mainly view it as stating the "obvious", but I've only ever used it for simple things -- in this case, it gives a precise statement of what it means for the square roots to be independent/dependent, and a theorem that says the extension formed by adjoining them behaves exactly like it ought to behave. –  Hurkyl Mar 11 '13 at 9:01
@Hurkyl : could you elaborate on why “the group order cannot be 4 because of symmetry” for the benefit of slow-minded people like me ? –  Ewan Delanoy Mar 11 '13 at 12:02
@Ewan: There are lots of ways to see it. The way I saw it is that none of $\alpha^2,\beta^2,\gamma^2,\delta^2$ can be the identity (because then they all would have to be). But no pair of them can be equal (because then all pairs would have to be). Thus contradiction. The "right" way is probably looking at automorphism groups or something. –  Hurkyl Mar 11 '13 at 15:05