Is $[\mathbb Q(\sqrt[7]2):\mathbb Q]=7$ and $[\mathbb Q(w):\mathbb Q]=6$?

Question

Let $[\mathbb Q(\sqrt[7]2):\mathbb Q]=a$ and $[\mathbb Q(w=e^{2\pi i/7}):\mathbb Q]=b$, then
1) $a=b$
2) $a<b$
3) $a>b$

$[\mathbb Q(\sqrt[7]2):\mathbb Q]=7$ as $x^7 - 2$ is an irreducible polynomial over $\mathbb Q$ and for a primitive $n$th root of unity $\omega$, $[\mathbb Q(\omega):\mathbb Q]= \varphi(n)$, so $[\mathbb Q(w):\mathbb Q]=6$.

Is this correct?

Answer 1

You are correct. $x^7-2\in \mathbb{Q}[x]$ is irreducible of degree $7$. On the other hand, the "obvious" polynomial $x^7-1\in \mathbb{Q}[x]$ for the $7^{th}$ roots of unity already has a root in $\mathbb{Q}$, namely $1$. So, if we factor this out, we get $$ x^7-1=(x-1)\underbrace{(1+x+x^2+x^3+x^4+x^5+x^6)}_{\text{irreducible of degree}\:6}.$$ So, $[\mathbb{Q}[\omega]:\mathbb{Q}]=6$ as you propose.