## Express some elements in a given field extension

Show that $\sqrt{2}, \sqrt[3]{2} \in \mathbb{Q}(\sqrt{2} - \sqrt[3]{2})$ by explicitly writing them as polynomials in $\theta = \sqrt{2} - \sqrt[3]{2}$.

First, we will find a polynomial satisfied by $\theta$. Note that the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $a(x) = x^2 - 2$, and that the minimal polynomial of $-\sqrt[3]{2}$ is $b(x) = x^3+2$. So these elements have conjugates $\alpha_1 = \sqrt{2}$ and $\alpha_2 = -\sqrt{2}$ and $\beta_1 = -\sqrt[3]{2}$, $\beta_2 = \sqrt[3]{2}(\frac{1}{2} + \frac{\sqrt{3}}{2}i)$, and $\beta_3 = \sqrt[3]{2}(\frac{1}{2} - \frac{\sqrt{3}}{2}i)$. Then $\theta$ is a root of $\prod_i \prod_j (x - \alpha_i - \beta_j) = t(x)$, where $t(x) = x^6 - 6x^4 + 4x^3 + 12x^2 + 24 x - 4$. (I recommend using a computer to carry out this multiplication; for instance, WolframAlpha can do it.) We don’t claim that $t(x)$ is the minimal polynomial of $\theta$ (it is); all we need to know at the moment is that we only need to consider powers of $\theta$ up to the fifth.

Let $\eta = \sqrt[6]{2}$, so that $\theta = \eta^3 - \eta^2$. Then $\theta^2 = 2 + \eta^4 - 2 \eta^5$, $\theta^3 = -2 + 6\eta - 6\eta^2 + 2\eta^3$, $\theta^4 = 4 + 2\eta^2 - 8\eta^3 + 12\eta^4 - 8\eta^5$, and $\theta^5 = -40 + 40\eta - 20\eta^2 + 4\eta^3 - 2\eta^4 + 10\eta^5$. Now setting $a + b\theta + c\theta^2 + d\theta^3 + e\theta^4 + f\theta^5 = z_1 + z_2\eta + z_3\eta^2 + z_4 \eta^3 + z_5\eta^4 + z_6 \eta^5$ and comparing coefficients, we have the following system of linear equations.

$\left[ \begin{array}{cccccc} 1 & 0 & 2 & -2 & 4 & -40 \\ 0 & 0 & 0 & 6 & 0 & 40 \\ 0 & -1 & 0 & -6 & 2 & -20 \\ 0 & 1 & 0 & 2 & -8 & 4 \\ 0 & 0 & 1 & 0 & 12 & -2 \\ 0 & 0 & -2 & 0 & -8 & 10 \end{array} \right] \left[ \begin{array}{c} a \\ b \\ c \\ d \\ e \\ f \end{array} \right] = \left[ \begin{array}{c} z_1 \\ z_2 \\ z_3 \\ z_4 \\ z_5 \\ z_6 \end{array} \right]$

Solving this system for appropriate $z_i$, we have

$\frac{1}{310}(364 + 152\theta + 156\theta^2 - 160\theta^3 - 9\theta^4 + 24\theta^5) = \sqrt[3]{2}$

and

$\frac{1}{310}(364 + 4620\theta + 156\theta^2 - 160\theta^3 - 9\theta^4 + 24\theta^5) = \sqrt{2}$.