Find minimal polynomial of element

Question

Find minimal polynomial of element $3-2\sqrt[3]{2}-\sqrt[3]{4}$ $\in {\displaystyle \mathbb {Q}(\sqrt[3]{2}) } $ over the field $ {\displaystyle \mathbb {Q} } $.

Thank you for any help.

Answer 1

Here's a systematic way to find the minimal polynomial, which works in general setting. We know that $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ is a basis of $\mathbb{Q}(\sqrt[3]{2})$ as a vector space over $\mathbb{Q}$. Now consider the following relations:

$$(3-2\sqrt[3]{2}-\sqrt[3]{4})\cdot1 = 3-2\sqrt[3]{2}-\sqrt[3]{4}$$ $$(3-2\sqrt[3]{2}-\sqrt[3]{4})\cdot \sqrt[3]{2} = -2 + 3\sqrt[3]{2} - 2\sqrt[3]{4}$$ $$(3-2\sqrt[3]{2}-\sqrt[3]{4})\cdot \sqrt[3]{4} = -4 - 2\sqrt[3]{2} + 3 \sqrt[3]{4}$$

These three equations can be written in matrix notation in the following way:

$$ (3-2\sqrt[3]{2}-\sqrt[3]{4})\left[ \begin{array}{c} 1\\ \sqrt[3]{2}\\ \sqrt[3]{4} \end{array} \right] = \left[ \begin{array}{ccc} 3&-2&-1\\ -2&3&-2\\ -4&-2&3 \end{array} \right]\left[ \begin{array}{c} 1\\ \sqrt[3]{2}\\ \sqrt[3]{4} \end{array} \right] $$

This gives us that $(3-2\sqrt[3]{2}-\sqrt[3]{4})$ is an eigenvalue of the matrix on the right. Hence it's a zero of the polynomial $\det(xI-M)$. Obviously as the degree of it is $3$ and $(3-2\sqrt[3]{2}-\sqrt[3]{4}) \not \in \mathbb{Q}$ we get that it must be the minimal polynomial of $(3-2\sqrt[3]{2}-\sqrt[3]{4})$. Evaluating the determinant isn't a hard thing to do and you will get that the minimal polynomial is:

$$f(x) = x^3 - 9x^2 + 15x + 29$$