Let $K=\Bbb Q (e^{\frac{2 \pi i}{7}})$ and $\alpha \in K - \Bbb Q$. Then show that, $\Bbb Q(\alpha)=K$ .
My attempt:
I have computed the minimal polynomial of$e^{\frac{2 \pi i}{7}}$ over $\Bbb Q$ to be $x^6+x^5+x^4+x^3+x^2+x+1 $ and thus, $[K:\Bbb Q]=6$
Since, $\alpha \in K - \Bbb Q$ , it follows that, $\Bbb Q(\alpha)$ is a subfield of $K$ containing $\Bbb Q$, moreover, $[\Bbb Q(\alpha):\Bbb Q] | [K:\Bbb Q] \implies [\Bbb Q(\alpha):\Bbb Q] =$1 or 2 or 3 or 6 .Since, $\alpha \notin \Bbb Q$, $[\Bbb Q(\alpha):\Bbb Q] \ne 1$ .
Since, for $\beta = e^{\frac{2 \pi i}{7}}$, $\{1,\beta,\beta^2,\beta^3 ,\beta^4,\beta^5\}$ is a basis of $K$ over $\Bbb Q$. then, $\exists a_j \in \Bbb Q , 0\le j \le 5$ with at least one of $a_j \ne 0 $ for $j \ge 1$ so that, $\alpha=a_0 +a_1 \beta+\dots+a_5 \beta^5$ .
How to proceed with then proof. Thanks in advance for help!
Note that this question is also present in here, here etc. but they use galois theory, but I only know Field theory and am also trying to solve the question using only that.
Let $\beta=e^{2\pi i/7}$, let $K={\bf Q}(\beta)$, let $\alpha=\beta+\beta^{-1}$. Then $$\alpha^3=\beta^3+3\beta+3\beta^{-1}+\beta^{-3}=\beta^3+3\beta+3\beta^6+\beta^4$$ and $$\alpha^2=\beta^2+2+\beta^{-2}=\beta^2+2+\beta^5$$ [thanks to Robert Lewis for drawing my attention to a typo here, which I have corrected] so $$\alpha^3+\alpha^2-2\alpha-1=1+\beta+\beta^2+\beta^3+\beta^4+\beta^5+\beta^6=0$$ It's easily seen that $x^3+x^2-2x-1$ is irreducible over the rationals, so it is the minimal polynomial for $\alpha$, so ${\bf Q}(\alpha)$ is a proper subfield of $K$, of degree 3.
Similarly, one can show that $\gamma=\beta+\beta^2+\beta^4$ is of degree two over the rationals.