Put the $K$ is a finite extension field over the field $F$.
If so, $[K;F]$ = (The number of the basis of the K over F )
e.x.) $[\mathbb{Q}(\sqrt[3]{2}) ; \mathbb{Q} ] = 3 $ (Since the basis are $\sqrt[3]{2}, \sqrt[3]{4} ,\sqrt[3]{8}$ )
As a vector space, $\mathbb{Q}(\sqrt[3]{2}) = \langle \sqrt[3]{2}, \sqrt[3]{4} ,\sqrt[3]{8}\rangle$
So Here is the my question begins.
Say $f(x) = x^n -1$ and $g(x) = x^{2n}-1$(or $x^n +1)$ with each roots $\alpha = e^{{2\pi i} \over {n}}$ and $\beta = e^{{2\pi i} \over 2n }$ respectively.
Let the $K_1=\mathbb{Q}(\alpha)$ and $K_2=\mathbb{Q}(\beta)$ be a splitting field of $f$ and $g$ over the $\mathbb{Q}$ respectively.
Then $[K_1 ; \mathbb{Q} ] = \phi (n)$ and $[K_2 ; \mathbb{Q} ] = \phi (2n)$
And it is clear that $K_1 = \langle 1, \alpha, \alpha^{2}, ... \alpha^{n-1}\rangle$ and $K_2 = \langle 1, \beta, \beta^{2}, ... \beta^{n-1}\rangle$
(When the $\beta$ case, because of the $\beta^n = -1, \beta^{n+1} = -\beta, ..., \beta^{2n} = 1$)
First question, Does two equations (1) and (2) hold?
(1) $K_1 = \langle \alpha ^{i} \vert (i,n)=1$ or $i =n\rangle $ = $\langle 1, \alpha, \alpha^{2}, ... \alpha^{n-1}\rangle$
(2) $K_2 = \langle \beta^{i} \vert (i,2n) =1$ or $i =2n \rangle$ = $\langle 1, \beta, \beta^{2}, ... \beta^{n-1}\rangle$
In my thought, considering the case $n=8$, $\alpha^2 \not \in \ \langle \alpha ^{i} \vert (i,8)=1$ or $i =8\rangle$ for $\alpha^2(\in \mathbb{Q}(\alpha))$
Hence, $\{\alpha ^{i} \vert (i,n)=1$ or $i =n\}$ is not a basis(generator) of the $Q(\alpha)$.
I'm very confused 'cause it looks like a basis on the fact that $[K_1 ; \mathbb{Q}] = \phi(n)$
Sencond question, Then What is generator(or basis) of the $Q(\alpha)$ and $Q(\beta)$?
It is clear that generator of the $K_1$ and $K_2$ are $\{1, \alpha, \alpha^{2}, ... \alpha^{n-1} \}$ and $\{1, \beta, \beta^{2}, ... \beta^{n-1} \}$ respectively.
If the equations in first question don't hold, What is the basis $Q(\alpha)$ and $Q(\beta)$??
Your bases should be $K_1 = \langle 1, \alpha, \alpha^2, \dots, \alpha^{\phi(n) -1} \rangle$ and $K_2 = \langle 1, \beta, \beta^2, \dots, \beta^{\phi(2n) - 1} \rangle$.
As for why $[\mathbf{Q}(e^{\frac{2 \pi i}{n}}) : \mathbf{Q}] = \phi(n)$, even though $\alpha = e^{\frac{2 \pi i}{n}}$ is a root of $x^n - 1$, the latter isn't irreducible over $\mathbf{Q}$. Instead, the minimal polynomial of $\alpha$ is $$p(x) = \prod_{1 \leq i < n\\(i,n) = 1} \big ( x - \alpha^i \big ),$$
which clearly has degree $\phi(n)$.
You should be able to find a reference for this in pretty much any algebra book.