Questioning a Basis for $\mathbb{Q}[\sqrt[3]{2}]$ over $\mathbb{Q}$ Let $\omega = e^{2 \pi i /3}$ and $\alpha = \sqrt[3]{2}$.  I'm seeing it claimed that $\mathcal{B} = \{\alpha, \alpha^2, \omega \alpha, \omega \alpha^2, \omega^2 \alpha, \omega^2 \alpha^2\}$ forms a basis for the vector space $\mathbb{Q}[\alpha, \omega]$ over $\mathbb{Q}$.
But the vector space $\mathbb{Q}[\alpha, \omega]$ over $\mathbb{Q}$ should have all of the elements of $\mathbb{Q}$ as members, and it's not clear to me how one could express some rational $q$ as a $\mathbb{Q}$ linear combination of the elements of $\mathcal{B}$.  This is because all of the members of $\mathcal{B}$ are either strictly irrational or strictly complex so that any linear combination of them will result again in a strictly irrational or strictly complex number (and hence not a member of $\mathbb{Q}$).
Am I missing something, or is $\mathcal{B}$ in fact not a basis for $\mathbb{Q}[\omega, \alpha]$?
EDIT: I miswrote the claimed basis $\mathcal{B}$.  I know that adding $1$ would make it a true basis.  But as now written, the supposed basis is $\mathcal{B} = \{\alpha, \alpha^2, \omega \alpha, \omega \alpha^2, \omega^2 \alpha, \omega^2 \alpha^2\}$.
 A: If $r$ is algebraic over $F$, with degree (of the minimum polynomial) $n$, then
$$
\{1,r,r^2,\dots,r^{n-1}\}
$$
is a basis for $F[r]$ over $F$. The dimension theorem says that, if you have a tower $F\subseteq K\subseteq L$ of finite field extension, $\{r_1,r_2,\dots,r_m\}$ and $\{s_1,s_2,\dots,s_n\}$ are bases of $K$ over $F$ and $L$ over $K$ respectively, then
$$
\{r_is_j:1\le i\le m, 1\le j\le n\}
$$
is a basis of $L$ over $F$. Since $\omega$ has minimum polynomial $X^2+X+1$, which is irreducible over $\mathbb{Q}[\alpha]$, having non real roots, you have that
$$
\{1\cdot1,\alpha\cdot1,\alpha^2\cdot1,1\cdot\omega,\alpha\omega,\alpha^2\omega\}
$$
is a basis of $\mathbb{Q}[\alpha,\omega]$ over $\mathbb{Q}$.
Of course there are other choices, but any basis must have six elements. Since $\{\omega,\omega^2\}$ is a basis for $\mathbb{Q}[\alpha,\omega]$ over $\mathbb{Q}[\alpha]$, a basis is also
$$
\{\omega,\alpha\omega,\alpha^2\omega,\omega^2,\alpha\omega^2,\alpha^2\omega^2\}.
$$
A: You need to add 1 to your basis. First of all, since $[\mathbb{Q}(\alpha,\omega):\mathbb{Q}]=[\mathbb{Q}(\alpha,\omega):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]$, then this dimension can't be 5 since each of the factors is non-trivial. Therefore your intuition is right and you have to to add 1. The dimension is actually 6.
EDIT Due to the edit I will say what the comments already say above: the new set is not linearly independent due to the fact that $\alpha+\alpha\omega+\alpha\omega^2=0$.
