Explicit basis for normal closure of $\Bbb Q(\alpha)$ over $\Bbb Q$ Let $\alpha$ denote the real $6^{th}$ root of $2$, and let $L$ denote the normal closure of $\Bbb Q(\alpha)$ over $\Bbb Q$.
I want to give an explicit basis for $L$ as a $\Bbb Q$-vectorspace.

In the answer below, I have attempted to rigorously find this explicit basis.
 A: We have that $\alpha$ is the real $6^{th}$ root of $2$. The normal closure of $\Bbb Q(\alpha)$ is the smallest normal extension over $\Bbb Q(\alpha)$.
Since $f(x)=x^6-2\in \Bbb Q(\alpha)$ has root $\alpha$, the normal closure of $\Bbb Q(\alpha)$ must be the splitting field of $f$. Now let $\zeta$ be a primitive $6^{th}$ root of $1$, then $\alpha\zeta^i$ where $i\in \{0,\cdots,5\}$ are the roots of $f(x)$.
Now $\zeta$ has minimal polynomial $x^2-x+1$, and clearly this minimal polynomial is irreducible(rational roots) so $\zeta\not\in \Bbb Q$ so $[\Bbb Q(\zeta):\Bbb Q]=2$
Now then $[\Bbb Q(\zeta,\alpha):\Bbb Q]\leq 12$ and $x^6-2$ splits over $\Bbb R$ while $x^2-x+1$ has complex roots. I.e. $\Bbb Q(\alpha)\subseteq \Bbb R$ and $[\Bbb Q(\alpha):\Bbb Q]=6$, and $\Bbb Q(\zeta)\not\subseteq \Bbb R$, so $[\Bbb Q(\alpha,\zeta):\Bbb Q]=12$.
And since $\Bbb Q(\alpha)$ has basis $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4,\alpha^5\}$ as a $6$-dimensonal $\Bbb Q$-vectorspace, and $\Bbb Q(\zeta)$ has basis $\{1,\zeta\}$ as a $2$-dimensional $\Bbb Q$-vectorspace. 
We know that the basis of the vectorspace basis formed by adjoining both is meant to be $u_iv_j$(the product of each basis element)
So we obtain basis:
$$\{1,\alpha,\alpha^2,\alpha^3,\alpha^4,\alpha^5,\zeta,\alpha\zeta,\alpha^2\zeta,\alpha^3\zeta,\alpha^4\zeta,\alpha^5\zeta\}$$
