How to prove whether $i\in\mathbb{Q}(\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5})$ or not

Question

I've been solving problems from my Galois Theory course, and I need help with some detail in this one. It says:

Calculate how many subfields has the splitting field of $P=X^7+4X^5-X^2-4$ over $\mathbb Q$.

What I first noticed is that $P$ can be expressed as product of irreducibles as: $$P=(X-1)(X^4+X^3+X^2+X+1)(X^2+4),$$ so the roots of $P$ are $\{1,\omega,\omega^2,\omega^3,\omega^4,2i,-2i\}$, where $\omega=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}$. Hence, the splitting field of $P$ over $\mathbb{Q}$ is $$\mathbb{Q}(1,\omega,\omega^2,\omega^3,\omega^4,2i,-2i)=\mathbb{Q}(i,\omega).$$ Now, I want to find $\operatorname{Gal}(\mathbb{Q}(i,\omega)/\mathbb Q)$, but to correctly find the automorphisms I first need to know the degree of the extension $\mathbb Q(i,\omega)/\mathbb Q$ (then I would know that degree equals the number of automorphisms, since it's a Galois extension because $\mathbb{Q}(i,\omega)$ is splitting field over $\mathbb Q$ of a separable polynomial). My problem is finding this degree. I know $$[\mathbb Q(i):\mathbb Q]=2, \quad [\mathbb Q(\omega): \mathbb Q]=4,$$ but I'm not sure if the degree of $\mathbb Q(i,\omega)/\mathbb Q$ is $4$ or $8$. My intuition tells me the easiest path will be checking if $i\in\mathbb Q(\omega)$ (implying degree $4$) or if $i\notin\mathbb{Q}(\omega)$ (implying degree $8$). However, I don't figure out an easy way to prove this.

How can I find is $i$ is or not inside $\mathbb Q(\omega)$? Is there an easier way to find this degree? Any help or hint will be appreciated, thanks in advance.

Answer 1

Note that $\Bbb Q(i)\cap \Bbb Q(\zeta_5)=\Bbb Q$, see here:

Clarification on proof that $\mathbb Q (\zeta _ m) \cap \mathbb Q(\zeta_n) = \mathbb Q$ when $\gcd(m,n) = 1$

It follows that $i\not\in \Bbb Q(\zeta_5)=\Bbb Q(\omega)$ and you are done.

Answer 2

You can also get this using values of $\sin(2\pi/5),\cos(2\pi/5)$.

First of all let us observe that $$2\cos (2\pi/5)=\omega+\omega^{-1}$$ and hence if $i\in\mathbb {Q} (\omega) $ then $\sin(2\pi/5)\in\mathbb {Q} (\omega) $.

But the minimal polynomial of $$a=4\sin(2\pi/5)=\sqrt{10+2\sqrt{5}}$$ is $$x^4-20x^2+80$$ which is irreducible by Eisenstein criterion with $p=5$.

Hence $\mathbb{Q} (a) $ is of degree $4$ over $\mathbb {Q} $. By our assumption we have $\mathbb{Q} \subseteq \mathbb {Q} (a) \subseteq \mathbb {Q} (\omega) $ and then $\mathbb {Q} (a) =\mathbb{Q} (\omega) $. This is totally absurd as $\mathbb {Q} (a)\subseteq \mathbb {R} $ and hence we must have $i\notin\mathbb {Q} (\omega) $.