This is a follow-up to my previous question; I wanted to give a different example to make sure I understand what is going on.
Compute $[ \mathbb Q(\zeta_8) : \mathbb Q(i)]$ and find a basis for the extension.
By the Tower theorem we have \begin{align*} [\mathbb Q(\zeta_8) : \mathbb Q] &= [\mathbb Q(\zeta_8): \mathbb Q(i)][\mathbb Q(i) :\mathbb Q]\\ 4 &= [\mathbb Q(\zeta_8): \mathbb Q(i)]\cdot 2 \end{align*}
because $[\mathbb Q(\zeta_8): \mathbb Q] = \varphi(8)=4$, and the minimal polynomial of $i$ over $\mathbb Q$ has degree $2$. Hence $[\mathbb Q(\zeta_8): \mathbb Q(i)] = 2$.
To find a basis, note that $\zeta_8 = e^{\pi i /4}= \cos(\pi/4) + i \sin(\pi/4) =\frac{\sqrt 2}{2} + i \frac{\sqrt 2}{2},$ so for any $a + b \zeta_8 \in \mathbb Q(\zeta_8)$ we can rewrite as $a + b/2(\sqrt 2) + i \cdot b/2 (\sqrt 2) $, which is a $\mathbb Q(i)$-linear combination of the vectors $\{1, \sqrt 2^\star\} \subset \mathbb Q(\zeta_8)$. Hence $\{1, \sqrt 2\}$ spans $\mathbb Q(\zeta_8)$, and they are clearly linearly independent, so this is a basis.
Is this correct?
${}^\star$ Some computation shows that $\sqrt 2 = \zeta_8 + \zeta_8^7$, so it indeed lies in $\mathbb Q(\zeta_8)$.