Say we have the polynomial $x^6-1$, I want to find the Galois group of this over $\Bbb Q$.
$x^6-1=(x-1)(x+1)(x^2-x+1)(x^2+x+1)$
And using the quadratic formula on the two quadratic polynomials we get
$x=\tfrac{^+_-1^+_-i\sqrt{3}}{2}$
Which (together with $^+_-1$) are the 6th roots of unity
So a splitting field for $x^6-1$ over $\Bbb Q$ is $\Bbb Q(w)=\Bbb Q(\sqrt{3},i)$
where w is the 6th root of unity , so clearly the degree of the extension is 4.
Its also obvious that the extension is normal and separable so we have a Galois extension.
$\therefore |Gal(\Bbb Q(w)/\Bbb Q)|=|\Bbb Q(w): \Bbb Q|=4.$
Furthermore the minimum polynomial of $w$ ( $x^4+x^2+1$) has 4 roots
So $Gal(\Bbb Q(w)/\Bbb Q)\leq S_4$
So we know that the Galois group must be isomorphic to either the klein 4 group or the cyclic group $\Bbb Z_4$.
I know just intuitively it's $\Bbb Z_4$ but I'm a little stuck at showing that .
Of course we know that any $\sigma \in Gal(\Bbb Q(w)/ \Bbb Q)$ will map the roots of the min. poly of w to another root.
So say $\sigma:w \rightarrow w^2$
But the problem is this has order 3 not 4 ?
So why is this automorphism not working , how can I choose the correct automorphism ?
Edit: Because $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ we see that the min poly of $\tfrac{1^+_-i\sqrt{3}}{2}$ is $(x^2-x+1)$ and the min poly of {-1^+_-i\sqrt{3}}{2}$
Therefore any automorphism of the Galois group will map $\tfrac{1^+_-i\sqrt{3}}{2}\rightarrow \tfrac{1^+_-i\sqrt{3}}{2}$ and $\tfrac{-1^+_-i\sqrt{3}}{2}\rightarrow $\tfrac{-1^+_-i\sqrt{3}}{2}$$.
So the 4 automorphisms are :
$\sigma_1: \tfrac{1+i\sqrt{3}}{2}\rightarrow \tfrac{1+i\sqrt{3}}{2}, \tfrac{-1+i\sqrt{3}}{2} \rightarrow \tfrac{-1^+i\sqrt{3}}{2}$
$\sigma_2: \tfrac{1+i\sqrt{3}}{2}\rightarrow \tfrac{1-i\sqrt{3}}{2}, \tfrac{-1+i\sqrt{3}}{2} \rightarrow \tfrac{-1+i\sqrt{3}}{2}$
$\sigma_3: \tfrac{1+i\sqrt{3}}{2}\rightarrow \tfrac{1+i\sqrt{3}}{2}, \tfrac{-1+i\sqrt{3}}{2} \rightarrow \tfrac{-1-i\sqrt{3}}{2}$
$\sigma_4: \tfrac{1+i\sqrt{3}}{2}\rightarrow \tfrac{1-i\sqrt{3}}{2}, \tfrac{-1+i\sqrt{3}}{2} \rightarrow \tfrac{-1-i\sqrt{3}}{2}$
Testing all these elements we see that each has order two therefore the galois group is isomorphic to $\Bbb Z_2$
Best Answer
It is not true that $\Bbb{Q}(\omega)=\Bbb{Q}(\sqrt{3},i)$. In fact $\Bbb{Q}(\omega)=\Bbb{Q}(\sqrt{-3})$, which is a quadratic extension of $\Bbb{Q}$, so the Galois group is isomorphic to $\Bbb{Z}/2\Bbb{Z}$.
The map given by $\sigma:\ \omega\ \mapsto\ \omega^2$ is not even an automorphism. In stead, the only nontrivial automorphism is given by $\omega\ \mapsto\ \omega^5$.