Say we that we have a $pth$ root of unity where $p>2$ is prime. Let $E=\Bbb Q(w)$ and let $\alpha=w+w^{-1}$.
I want to figure this question out asked on a past exam:
1) Compute | E : $\Bbb Q(α)$ |, and |$\Bbb Q(α)$ : $\Bbb Q$|.
If $w$ is a primitive pth root of unity then it's Galois grouo over $\Bbb Q$ is $\Bbb Z/p \Bbb Z \cong C_{p-1}\cong C_i\times……\times C_2.$ By the Chinese remainder theorem. We see that $\alpha$ must be fixed by a transposition of the elements $w^1,w^{p-1}$. So we choose the automorphism $\phi:w\rightarrow w^{p-1}$ which is order two as $\phi(w^{p-1})=w^{p^2-2p+1}=w$. The subgroup of automorphism fixing $\alpha$ then is $\langle \phi^{p-1}\rangle \cong C_2.$ And so by Galois Correspondence $| E : \Bbb Q(α) |=2$ while |$\Bbb Q(α)$ : $\Bbb Q|=p/2$.
2) It says find the Galois group $G = Gal(E/\Bbb Q(α))$? But I did that in the first part infact I don't see another way to determine part (1) then by determining the group first, Is there ?
3)What are the isomorphisms in $G=Gal(\Bbb Q(α)/ \Bbb Q)?$ i.e. How do they act on $α?$
By the Fundamental theorem of Galois $Gal(\Bbb Q(α)/ \Bbb Q)= \cong\tfrac{Gal(E/F)}{Gal(E/Q(\alpha)}$ so $Gal(\Bbb Q(α)/ \Bbb Q)\cong C_{p-1}/C_2$.
This part confuses me, specifically because if we consider factors $C_{p-1}/C_2$ over different $p$ then we get different groups So I don't understand how this can be considered generally . Unless the question is asking you to deduce it in some other manner . In which case I Know we could say that isomorphisms in $G$ are those which fix $\Bbb Q $ but not $\Bbb Q(\alpha)$. But this a trivial.
Any advice anyone ?
Here's how to find the degree of $E$ over ${\bf Q}(\alpha)$ without finding the group.
$\omega$ satisfies a quadratic with coefficients in ${\bf Q}(\alpha)$, namely, $\omega^2-\alpha\omega+1=0$, so the degree we're looking for is either $1$ or $2$. But $\omega$ is not real, while $\alpha$ is (and, hence, all elements of ${\bf Q}(\alpha)$ are), so the degree can't be $1$, so it must be $2$.