I am a little bit confused on defining the automorphisms in Galois Groups. In particular, I was looking at the example in section 14.2 of Dummit and Foote where they find the galois group of the splitting field of $x^8 - 2$ over $\mathbb{Q}$. They say that the automorphisms are given:
$\theta \to \zeta^a\theta$, $a = 0,1,2,\ldots,7$
$i \to \pm i$
Where $\theta = 2^{1/8}$, $\zeta$ is a primitive 8th root of unity.
They conclude that there are 16 possible automorphisms. I understand that there should be 16 such maps, but do not understand why these are the ones. They note that it is necessary that it is necessary to provide justification that the maps are automorphisms and say that "this can be accomplished for example by using the extension theorems or by using degree considerations". However, I do not understand what this means.
Can anybody explain how we know that the above maps are automorphisms? More generally, how do we know that the maps we are defining are automorphisms?
Thank You
For each $a\in 0\ldots 7$ there is an isomorphism $\Bbb{Q}(2^{1/8} )\to \Bbb{Q}[x]/(x^8-2)\to \Bbb{Q}(\zeta_8^a 2^{1/8})$,
the minimal polynomial of $i$ is $x^2+1$ over $\Bbb{Q}(2^{1/8} )\subset \Bbb{R}$, since $x^2+1\in \Bbb{Q}[x]$ the minimal polynomial of $i$ stays $x^2+1$ over $\Bbb{Q}(\zeta_8^a 2^{1/8})$,
Thus the $16$ maps $(2^{1/8},i)\to (\zeta_8^a 2^{1/8},\pm i)$ are isomorphisms $\Bbb{Q}(2^{1/8},i )\to \Bbb{Q}(\zeta_8^a 2^{1/8},i)$,
and hence they are automorphisms of $\Bbb{Q}(2^{1/8},i )$ which is the splitting field of $x^8-2$.
There are no more because any automorphism must send $2^{1/8},i $ to one of the complex root of their $\Bbb{Q}$-minimal polynomials.