What is the Galois group of x^8-1 over Q?

[chocok]

I was asked to find the Galois group of $$x^8-1$$ over Q,

I first find all the roots to it :
$$\pm i$$ , $$\pm \sqrt{i}$$ , $$\pm i \cdot \sqrt{i}$$, $$\pm 1$$.

Then since $$i \cdot \sqrt{i}$$ is just a multiple of i and sqrt(i)
so I had Q(i, sqrt(i)) being the splitting field for the equation over Q.
Next, [Q(i, sqrt(i)) :Q] = 4, so I conclude that the Galois group is a cyclic group of order 4.

is the above correct? If not, can someone please tell me what's wrong? Thanks!

 

[morphism]

It's not going to be the cyclic group of order 4.

In general I would avoid writing $\sqrt{i}$; which square root of i are you taking?

The roots of x^8 - 1 are going to be the 8th roots of unity, so the splitting field will be generated over Q by the primitive 8th root of unity $\zeta_8 = e^{2 \pi i / 8}$. That is, the splitting field is going to be $\mathbb{Q}(\zeta_8)$. Consequently $[\mathbb{Q}(\zeta_8) : \mathbb{Q}] = ?$ and $\text{Gal}(\mathbb{Q}(\zeta_8) / \mathbb{Q}) \cong ?$.

Try to take it from here.

 

[HallsofIvy]

The square roots of i are $(\sqrt{2}/2)(1\pm i)$.

 

[chocok]

thanks~ pls tell me this is right

for the 8 roots, eliminating the ones that can be expressed as multiples of others, i am left with $$e^{2\pi i/8}$$ and $$e^{\pi i/2}$$=i where their orders are 4 and 2 respectively so i get a splitting field over Q of order 8 =>Gal( Q( $$e^{2\pi i/8}$$ , $$e^{\pi i/2})$$ / Q) $$\cong$$ cyclic group of order 8 (D4)??

 

[morphism]

$$(e^{2 \pi i /8})^2 = e^{\pi i /2}$$

The splitting field is Q(e^(2pi*i/8)) which is of order 4 over Q. So the Galois group is either C_2 x C_2 or C_4. It's not going to be C_4 (why?).

[If you're comfortable with cyclotomic extensions, then this amounts to the fact that $\text{Gal}(\mathbb{Q}(\zeta_8) / \mathbb{Q}) \cong (\mathbb{Z}/8\mathbb{Z})^* \cong C_2 \times C_2$.]