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