Let $N \geq 2$ be an integer.
I'm trying to decide when $\xi=e^{\frac{i \pi}{N}}$ belongs to the cyclotomic field $K=\mathbb{Q}(\mu_N)$. This is my reasoning so far: assume that $\xi$ belongs to $K$. Since $\xi$ is a primitive root of unity of order $2N$, this implies that $\mathbb{Q}(\mu_{2N}) \subseteq \mathbb{Q}(\mu_N)$. But, on the other hand, $\mathbb{Q}(\mu_{N})$ is always contained in $\mathbb{Q}(\mu_2N)$. Hence one needs $\mathbb{Q}(\mu_{2N})=\mathbb{Q}(\mu_N)$. Can one make the list of all $N$ for which this is true?
It works for $N=3$ since both fields are equal to $\mathbb{Q}(\sqrt{-3})$ in that case. And I have the impression it works for odd N, since then the degree of $\mathbb{Q}(\mu_{2N})$ is $$ \phi(2N)=\phi(2)\phi(N)=\phi(N), $$ which is also the degree of $\mathbb{Q}(\mu_{N})$. If $N$ is even, however, say of the form $2^k M$, then $$ \Phi(2N)=2^k \phi(N), $$ so one field is strictly smaller than the other. Is that correct?
Sure, it is correct. There does not seem to be any flaw in the proof. Oh, but to argue equality of fields through equality of degrees, you must emphasize that one extension is contained in the other (which you do in your first paragraph, but you need to do also in your extension degree argument).
For other readers: remember $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ both have degree $2$ but they are not equal.