Let $K = \mathbb{Q}_3(i,\sqrt[4]{-3})$ and $L = K(\zeta_7)$ where $\zeta_7$ is a 7th root of unity.
Question What is $\min_K(\zeta_7)$?
Approach:
Over $\mathbb{F}_9$, the residue field of $K$, we obtain the factorization of $x^7-1$ in irreducible factors: $$ x^7-1 = (x+2)(x^3+(a+1)x^2+ax+2)(x^3+2ax^2+(2a+2)x+2). $$ Here, $a \in \mathbb{F}_9$ with $a^2+a+2=0$, i.e. $a$ is a primitive 8th root of unity in $\mathbb{F}_9$.
Let $\bar{f} = x^3+(a+1)x^2+ax+2 \in \mathbb{F}_9[x]$. If we take $b \in \mathbb{F}_{9^3}$ with $\min_{\mathbb{F}_9}(b) = \bar{f}$, then $b$ is a $7$-th root of unity in $\mathbb{F}_{9^3} = \mathbb{F}_9(b)$.
Suppose $\beta \in L$ is a lift of $b$. By Hensel's Lemma, there must be a lift $f \in K[x]$ of $\bar{f}$ with $f(\zeta_7)=0$. I expected $\beta = \zeta_7$. I thought it might be $$ f = \min_K(\zeta_7) = x^3+(\zeta_8+1)x^2+\zeta_8 x + 2 $$ where $\zeta_8$ is the lift of $a \in \mathbb{F}_9$ with $\min_{\mathbb{Q}_3}(\zeta_8) = x^2+x+2$ which is a primitive 8th root of unity. But with this minimal polynomial, I obtain $\beta^8 \neq \beta$, which would have to be true if $\beta = \zeta_7$.
Could someone please point out my mistake in my line of thought and help me finding the relation for $\zeta_7$ (and if necessary $\zeta_8$?
Thank you in advance!
What went wrong is that by the quadratic formula the zeros of $x^2+x+2=0$ are $(-1\pm \sqrt{-7})/2$, and these are not roots of unity in $K$. Because $-7\equiv-1\pmod3$ they are elements of $\Bbb{Q}_3(i)$ though. Those roots of unity are only congruent to these numbers modulo the maximal ideal of the ring of integers of $K$ (is that not what being "a lift" means?).
I would approach a problem like this using Galois theory of cyclotomic extensions of $\Bbb{Q}$. We know that $\Bbb{Q}(\zeta_7)/\Bbb{Q}$ is cyclic of degree six, and that the only quadratic intermediate field is $\Bbb{Q}(\sqrt{-7})$. This suggests strongly to me that having $\sqrt{-7}$ around gives your answer.
With $\zeta_7=e^{2\pi i/7}$ we have $$ (x-\zeta_7)(x-\zeta_7^2)(x-\zeta_7^4)=x^3+\frac{1-i\sqrt{7}}2x^2+\frac{-1-i\sqrt{7}}2x-1. $$ Replacing $(-1\pm \sqrt{-7})/2$ with appropriate zeros of $x^2+x+2=0$ in your field gives the factorization of the seventh cyclotomic polynomial over $\Bbb{Q}_3(i)=\Bbb{Q}_3(\sqrt{-7})$. It depends on the choice of $\sqrt{-7}$ which half of the seventh roots of unity have this cubic as their minimal polynomial; the other half will need the conjugate.