Finite Extensions and Bases The question reads: Find a basis of $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$ over $\mathbb{Q}$. Describe the elements of $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$.
My original thought on the approach was to find the minimum polynomial (which would have degree 6), and then just take create a basis where every element is $(\sqrt{2}+\sqrt[3]{4})^n$ for $n=0,\ldots,5$.
The hint in the back recommends adjoining $\sqrt[3]{4}$ first, but I'm not sure where to go once I've done that. Can anyone help guide me through the process the book wants me to use?
Thank you very much in advance.
 A: Dedekind's Product Theorem (which a friend of mine calls the "Royal Dutch Airlines Theorem") states that if $K\subseteq L\subseteq M$ are fields, then
$$[M:K] = [M:L][L:K].$$
The proof of the theorem is constructive: if $\{m_i\}_{i\in I}$ is a basis for $M$ as an $L$-vector space, and $\{\ell_j\}_{j\in J}$ is a basis for $L$ as a $K$-vector space, then one shows that $\{m_i\ell_j\}_{(i,j)\in I\times J}$ is a basis for $M$ as a $K$-vector space.
The Hint in the book is suggesting that you do this. Since $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4}) = \mathbb{Q}(\sqrt{2},\sqrt[3]{4})$, you can find a basis for $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$ by considering the tower of extensions
$$\mathbb{Q}\subseteq \mathbb{Q}(\sqrt[3]{4}) \subseteq \mathbb{Q}(\sqrt[3]{4})(\sqrt{2}).$$
It is easy to find a minimal polynomial for $\sqrt[3]{4}$ over $\mathbb{Q}$, from which you easily get a basis for the first extension; and it is easy to find a minimal polynomial for $\sqrt{2}$ over $\mathbb{Q}(\sqrt[3]{4})$, from which you get a basis for the second extension. Now you can combine them, using the argument of the Dedekind Product Theorem, to get a basis for $\mathbb{Q}(\sqrt{2},\sqrt[3]{4}) = \mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$ over $\mathbb{Q}$.
