I have been working on these two minimal polynomial questions and am particularly concerned about (b)
Find the minimal polynomial for $\sqrt[3]{4}+\sqrt[3]{2}$
(a) over $\mathbb{Q}$
By setting $\alpha=\sqrt[3]{4}+\sqrt[3]{2}$, I found the minimal polynomial to be $\alpha^3-6\alpha-6=0$ (edited). This method involved just squaring out terms and was fairly lengthy – is this the standard procedure?
(b) over $\mathbb{Q}(\sqrt{2})$
What does it mean to be the minimal polynomial over $\mathbb{Q}(\sqrt{2})$? Over $\mathbb{Q}$, I see the minimal polynomial as the polynomial of lowest degree such that $\alpha$ is a root but I cannot see what is going on here.
From (a), we know that $[\mathbb{Q}(\alpha) : \mathbb{Q}]=3$. So:
$$[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}]=[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}(\alpha)][\mathbb{Q}(\alpha) : \mathbb{Q}]=3[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}(\alpha)]$$
So since the degree is a multiple of $3$, the degree of the minimal polynomial of $\alpha$ over $\sqrt{2}$ is also a multiple of $3$. Can we automatically conclude it is $3$? I do not believe so since we are considering a larger field ($\mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}$
Best Answer
Another way to do this is to interpret $\alpha$ as a $\mathbb{Q}$-linear map from $\mathbb{Q}(\sqrt[3]{2})$ to itself. With respect to the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$, this is represented by the matrix $$\begin{pmatrix} 0 & 2 & 2 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix}.$$ The characteristic and minimal polynomial is $$t^3 - (0+0+0) t^2 + 3 \cdot \mathrm{det} \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix} t - \mathrm{det} \begin{pmatrix} 0 & 2 & 2 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix}$$ $$= t^3 - 6t - 6.$$