I am trying to prove that $i \notin \mathbb{Q}(\alpha)$ where alpha is the 5th root of the Unit, that is, $\alpha = e^{\frac{2\pi i}{5}}$ and I have tried it in two different ways.
First I tried to do this by comparing coeficients. But the calculations were very big and I really could not conclude by there.
Second option, I supposed that $i \in \mathbb{Q}(\alpha)$, then $[\mathbb{Q}(\alpha,i),\mathbb{Q}]=1$ and since $[\mathbb{Q}(\alpha),\mathbb{Q}]=4$ and $[\mathbb{Q}(i),\mathbb{Q}]=2$ we conclude that $[\mathbb{Q}(\alpha,i),\mathbb{Q}(i)]=2$ by Tower Law. Then, there exist a quadratic polynomial in $\mathbb{Q}(i)[X]$ such that $\alpha$ is a root. The first automatic idea for the next step was to use that $\overline{\alpha}$ is also a root of this polynomial and then get a contradiction since $\alpha+\overline{\alpha}$ is not rational. But then I realized that I cannot do this because I dont know if $\overline{\alpha}$ is in fact a root.
Can someone give me a hint about how to do the next step?? Or give me another way to do this (just hints, not the answer) if you see that this approach is not going anywhere!
Thank you in advance!
To complete the question with an answer, I give some details of the solution in the comments. Since we have $\mathbb{Q}(\zeta_n,\zeta_m)=\mathbb{Q}(\zeta_{[m,n]})$, see here, Theorem $3.4$ we have $$ [\mathbb{Q}(i,\zeta_5):\mathbb{Q}]=[\mathbb{Q}(\zeta_{[4,5]}):\mathbb{Q}]=\phi (4\cdot 5)=8, $$ whereas $[\mathbb{Q}(\zeta_5):\mathbb{Q}]=4$. Hence $i=\zeta_4\not\in \mathbb{Q}(\zeta_5)$.