# Having trouble computing some minimal polynomial of a complicated primitive element over $\mathbb{Q}_3(i,\sqrt[4]{-3})$

Let $$K = \mathbb{Q}_3(i,\sqrt[4]{-3})$$, $$L$$ and $$F$$ be the extensions of $$K$$ given by

• $$L = K(\sqrt[3]{2})$$
• $$F = K(\zeta_7)$$ where $$\zeta_7$$ is the primitive $$7$$-th root of unity given by $$\min_K x^3 + \frac{1-\sqrt{-7}}{2}x^2+\frac{-1-\sqrt{-7}}{2}x-1.$$

Let $$\alpha = \sqrt[3]{2}\left(\zeta_7^2 - \zeta_3 \zeta_7 + \frac{\zeta_3}{\zeta_3-1} \frac{1 - \sqrt{-7}}{2}\right)$$.

Question: How to determine $$\min_K(\alpha)$$ (quickly)?

Obviously, I could write down $$\alpha^0, \dots, \alpha^3$$ as linear combinations of $$\left(\sqrt[3]{2}^k \zeta_7^\ell \, | \, k,\ell=0,1,2 \right)$$ using the relations of the elements, and solve a linear system of equations. But practically, this is tedious and takes forever. Not to mention the mistakes I could make during the calculations...

I also tried to compute it with Sage but failed miserably. Looks like Sage is not good with extensions of local field in general. A source code is here:

Could someone help me with this problem? Any effort is highly appreciated!

If $$E/F$$ is a field extension of finite degree and $$\alpha\in E$$, the characteristic polynomial of left multiplication by $$\alpha$$ in $$E$$ (denoted by $$\ell_\alpha$$) is $$\chi_{\ell_\alpha}=\min_F(\alpha)^{[E:F(\alpha)]}.$$
Proof. Let $$m=[E:F(\alpha)]$$. If $$(e_1,\ldots,e_r)$$ is an $$F(\alpha)$$-basis of $$E$$, then $$\alpha^i e_j, i=0,\ldots,m-1, j=1,\ldots,r$$ is an $$F$$-basis of $$E$$. The matrix of $$\ell_\alpha$$ is this basis (choosing the order $$1,\ldots,\alpha^{m-1},e_1,e_1\alpha,\ldots,e_1\alpha^{m-1}...)$$ is a diagonal block matrix, where each block is the companion matrix of $$\min_F(\alpha)$$.
So you could compute the matrix of left multiplication by $$\alpha$$ is the basis $$i^k(\sqrt[4]{-3})^j,i=0,1,j=0,1,2,3$$. The characteristic polynomial will be your minimal polynomial, since your title suggests that you know that $$\alpha$$ is a primitive element.
(if it is not, then you will have to factor the characteristic polynomial over $$\mathbb{Q}_3$$ but it will be easier since you know it will be a power of a single polynomial).