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:

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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).


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