Compute in an extension field

Find an element \theta such that \mathbb{Q}(i,\sqrt[3]{2}) = \mathbb{Q}(\theta). Express \frac{4+i}{\sqrt[3]{2}} as a polynomial in \theta.


Note that i and \sqrt[3]{2} have minimal polynomials x^2+1 and x^3-2, with conjugates \alpha_i = \pm i and \beta_1 = \sqrt[3]{2}, \beta_2 = \sqrt[3]{2}(-1/2 + \sqrt{3}i/2), and \beta_3 = \sqrt[3]{2}(-1/2-\sqrt{3}i/2), respectively. Since \alpha_i+\beta_j = \alpha_1 + \beta_1 only if (i,j) = (1,1), we have \mathbb{Q}(i,\sqrt[3]{2}) = \mathbb{Q}(i+\sqrt[3]{2}). We let \theta = i + \sqrt[3]{2}. Evidently t(x) = x^6 + 3x^4 - 4x^3 + 3x^2 + 12x + 5 is satisfied by \theta.

Using some linear algebra, we see that \sqrt[3]{2} = \frac{1}{22}(91+100\theta -78\theta^2 + 40\theta^3 - 9\theta^4 + 12\theta^5) = a(\theta) and i = \frac{1}{22}(-91-78\theta + 78\theta^2 - 40\theta^3 + 9\theta^4 - 12\theta^5) = b(\theta).

Using the Euclidean algorithm, we see that \frac{1}{22}a(x)(-9x^5 + 4x^4 - 30x^3 + 53x^2 - 53x - 71) + \frac{1}{484}p(x)(108x^4 - 129x^3 + 432x^2 - 949x + 1389) = 1. Since p(\theta) = 0, (\sqrt[3]{2})^{-1} =  \frac{1}{22}(-9\theta^5 + 4\theta^4 - 30\theta^3 + 53\theta^2 - 53\theta - 71) = c(\theta); indeed we can verify that c(\theta)a(\theta) = 1. (See WolframAlpha)

Then \frac{4+i}{\sqrt[3]{2}} = (4 + b(\theta))*c(\theta) = \frac{1}{22}(57\theta^5 - 40\theta^4 + 190 \theta^3 - 354 \theta^2 + 453 \theta + 446).

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