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}.
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