Finding primitive element of field extension. I wish to find the primitive element of $\mathbb{Q}(\sqrt[3]2,e^{\frac{2\pi i}{3}})$ over $\mathbb{Q}$. According to the proof of the theorem of primitive element. I suspect it is $\sqrt[3]2+e^{\frac{2\pi i}{3}}$ but I do not know how to show $\sqrt[3]2\in\mathbb{Q}(\sqrt[3]2+e^{\frac{2\pi i}{3}})$
Is there any hint to this?
 A: The field $K=\mathbf Q(\sqrt [3]2,j)$, where $j$ is a primitive 3rd root of $1$, is obviously the field of decomposition over $\mathbf Q$ of the polynomial $X^3 - 2$, so the extension $K/\mathbf Q$ is galois, say with Galois group $G$. As $\mathbf Q(\sqrt [3]2)/\mathbf Q$ is not normal, $G$ is non abelian. Since $K$ is the compositum of the linearly disjoint extensions $\mathbf Q(\sqrt [3]2)$ and $\mathbf Q(j)$, $G$ has order $6$, hence $G \cong S_3$, which is (up to isomorphism) the unique non abelian group of order $6$, consisting of Id,  the three transpositions fixing the three cubic subfields $\mathbf Q(\sqrt [3]2), \mathbf Q(j\sqrt [3]2), \mathbf Q(j^2\sqrt [3]2)$, and the two 3-cycles generating the cyclic subgroup of order $3$ which fixes the quadratic subfield $\mathbf Q(j)$. Summarizing, $K$ contains only five strict subextensions, and it is straightforward to check that $j+\sqrt [3]2$ does not belong to any of these. Consequently, $j+\sqrt [3]2$ is a primitive element of $K/\mathbf Q$.
