How can you write $\mathbb{Q}[\sqrt[3]{2},\omega_3]$ using a single algebraic element $\mathbb{Q}[\alpha]$?

Question

Looking at the basis of $\mathbb{Q}[\sqrt[3]{2},\omega_3]$ gives me no idea on how to generate it using $\{1, \alpha, \alpha^2,\alpha^3,\alpha^4,\alpha^5\}$ for some $\alpha$ algebraic over $\mathbb{Q}$.

Answer 1

If $K$ is an infinite field and $\alpha,\beta$ are separable algebraic over $K$, then there is some $u \in K$ such that the $\alpha_i + u \beta_j$ are pairwise distinct, where $\alpha_i,\beta_j$ are the conjugates of $\alpha,\beta$. This is exactly what happens in the proof of the primitive element theorem. One then proves that $\alpha + u \beta$ is a primitive element of $K(\alpha,\beta)$. Actually in many cases $u=1$ will do the job.

Let $\alpha=\sqrt[3]{2}$ and $\beta=\omega_3$. The conjugates of $\alpha$ are $\alpha \beta^i$ with $i=0,1,2$. The conjugates of $\beta$ are $\beta^j$ with $j=1,2$. One computes (with a computer algebra software for instance) that the elements $\alpha \beta^i + \beta^j$ are pairwise distinct. Hence, $\sqrt[3]{2}+\omega_3$ is a primitive element of $\mathbb{Q}(\sqrt[3]{2},\omega_3)$.