Let $\alpha \in \mathbb{C}$ a zero of the polynomial $x^5+x+1$. I want to prove that $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^2)$ but I am out of ideas since the polynomial is not irreducible, but I want to prove this without computing the factors of the polynomial.
Furthermore, what is the degree of the extension $\mathbb{Q} \subset \mathbb{Q}(\alpha \sqrt{2})$? And why is $x^7-6$ irreducible over $\mathbb{Q}(\alpha \sqrt{2})$?
I am lost on this exercise, so any hint will be useful.
Hint for the first part: Since $\alpha^5+\alpha+1=0,$ we have $\alpha^6+\alpha^2+\alpha=0.$