# Finding the Galois group over $\Bbb{Q}$.

If K is the splitting field of $X^8-2$ over $\Bbb{Q}$, I want to find the galois group.

We know that $K=\Bbb{Q}(2^{1/8}, \zeta_8)$.

So first I want to look at $Gal(\Bbb{Q}(\zeta_8)/\Bbb{Q})$ and then look at $Gal(\Bbb{Q}(2^{1/8})/\Bbb{Q})$, since there is a homomorphism $\rho: Gal(\Bbb{Q}(2^{1/8},\zeta_8)/\Bbb{Q}) \rightarrow Gal(\Bbb{Q}(\zeta_8)/\Bbb{Q})$.

Since $Gal(\Bbb{Q}(\zeta_8)/\Bbb{Q}) \cong Aut(<\zeta_8>) \cong Z^{\times}_8 = \{1, 3, 5, 7\}$, we know that we have four subgroups in $Gal(Q(\zeta_8)/\Bbb{Q})$ (Let $\zeta_8 = \zeta$):

$\sigma_1(\zeta) = \zeta$

$\sigma_3(\zeta) = \zeta^3$

$\sigma_5(\zeta) = \zeta^5$

$\sigma_7(\zeta) = \zeta^7$

And now I have to relate this to the homomorphism $\rho$ in order to find the rest of the permutations, right? But I'm a bit confused. I've spent hours trying to do it and I'm seriously stuck...could anybody help me with this?

• @Artus How specifically do you want it to be done? You're going to end up getting the same group either way (it's a semi-direct product of $\mathbb Z_{n_1}$ and $\mathbb Z_{n_2}$ for some $n_1, n_2$.) May 8, 2013 at 21:08
• @Artus Could you fix the $\LaTeX$ in that comment? May 8, 2013 at 21:21
• If we adjoin the $\sqrt[8]{2}$ first, we get a degree $8$ extension. Then factoring the $8$th cyclotomic polynomial over $\mathbb R$, we see the minimal polynomial of $\zeta_8$ over $\mathbb Q(\sqrt[8]{2})$ is $x^2+\sqrt 2x +1$ (and this can't factor because $\zeta_8$ is complex). So the whole extension is degree $8$. May 8, 2013 at 21:36
• So you need to find $2$ extensions for each $\sigma_i$ you found, and you will know that is all of them. (More forthcoming) May 8, 2013 at 21:36
• @Potato $\mathbb{Q}(\sqrt[8]{2})$ and $\mathbb{Q}(\sqrt[8]{3})$ are pretty different. Notice that, if $\zeta$ is a primitive $8$-th root of unity, then $\zeta+\zeta^{-1} = \pm \sqrt{2}$. May 8, 2013 at 23:52

$\zeta:=\zeta_8$ is of degree $2$ over $\mathbb{Q}(\sqrt[8]{2})$, hence $K:=\mathbb{Q}(\sqrt[8]{2},\zeta)$ is of degree $16$ over $\mathbb{Q}$ and of degree $4$ over $\mathbb{Q}(\zeta)$. The minimal polynomial of $\sqrt[8]{2}$ over $\mathbb{Q}(\zeta)$ is $X^4-\sqrt{2} = X^4 - (\zeta + \zeta^{-1})$.

Hence $Gal(K/\mathbb{Q})$ is an extension of $(\mathbb Z_8)^\times$ by $\mathbb Z_4$.

If $\sigma \in Gal(K/\mathbb{Q})$, then $\sigma$ satisfies: $\sigma(\zeta) = \zeta^a$ and $\sigma(\sqrt[8]{2})=\zeta^b \sqrt[8]{2}$ for some $a \in \mathbb Z_8^\times$, $b \in \mathbb Z_8$ such that $\zeta^{4b} \sqrt{2}= \zeta^a + \zeta^{-a}$ which means $b = \tfrac{a-1}{2} \pmod 2$.

EDIT: Note that $\zeta_8 = \exp(2i\pi/8) = \exp(2i\pi/8)$. We know that take any value in $\{ 1,3,5,7 \} =\mathbb Z_8^\times$, and we have :

• $\exp(1 \times i \pi/4) + \exp(-1.i \pi/4) =\sqrt{2}$.
• $\exp(7 \times i \pi/4) + \exp(-7.i \pi/4) =\sqrt{2}$.
• $\exp(3 \times i \pi/4) + \exp(-3.i \pi/4) =-\sqrt{2}$.
• $\exp(5 \times i \pi/4) + \exp(-5.i \pi/4) =-\sqrt{2}$.

Depending on $a$, the value $(\zeta^b)^4$ must be $+1$ or $-1$. In any case they are exactly $4$ values of $b$ allowed.

• Is the very end supposed to be mod 2 or mod 8?
– user58289
May 10, 2013 at 21:10
• I solved it with mod 8 (I wrote the solution above), but that didn't work either because I ended up with fractions, and they didn't make sense.
– user58289
May 10, 2013 at 21:37
• @Artus: The condition $\sqrt{2}=\zeta^{a-4b}+\zeta^{-a-4b}$ is NOT equivalent to $a-4b=1 \mod 8$ (neither implies the other). To solve the equation, you have to notice that $\zeta^a+\zeta^{-a} = (-1)^{(a-1)/2} \sqrt{2}$ and $\zeta^{4b} = (-1)^b$. May 10, 2013 at 22:05
• The formula is not important. All you have to do is to take each $a$ (there are only 4 of them) and check what sign should have $(-1)^b$. May 10, 2013 at 22:41
• You are wrong for $b$. For example if you take $a=3$, then you have $\zeta^3+\zeta^{-3} = exp(3i\pi/4)+exp(-3i\pi/4) = -\sqrt{2}$, hence $b$ must satisfy $(-1)^b=-1$ i.e. $b \in \{1,3,5,7\}$. The formula is missleading, I should have written : $b \in \{0,2,4,6\}$ if $a \in \{1,7\}$ and $b \in \{1,3,5,7 \}$ if $a \in \{3,5\}$. May 10, 2013 at 23:00