Let $K = \mathbb{Q}_3$ and $L=K(\alpha)$ where $\min_K(\alpha) = x^4 - 3x^2 + 18$. Furthermore, let $F/K$ be the unique unramified extension of degree $4$. It can be shown that $F/K$ is generated by a primitive $5$-th root of unity $\zeta_5$.
Now consider the automorphism $\varphi \in Gal(FL/K)$ given by $\zeta_5 \mapsto \zeta_5^3$ and $\alpha \mapsto \frac{(2 \alpha^2 - 3)u}{\alpha}$ where $u = \sqrt{-2/7}$ (this is another root of $x^4-3x^2+18$, cf. this post). The order of $\varphi$ is $4$.
Let $L' = (LF)^{\langle \varphi \rangle}$. By construction, the extension $LF/L'$ is cyclic of degree $4$.
Question: What is a generator $\gamma$ of $L'$?
This $\gamma$ must be an eigenvector of $\varphi$ with eigenvalue $1$ but it must be $\gamma \not\in K$.
I suppose that a basis of $LF$ as a $K$-vector space is given by $(1,\zeta_5, \zeta_5^2, \zeta_5^3, \alpha, \alpha \zeta_5, \alpha \zeta_5^2, \alpha \zeta_5^3)$. If I consider the representation matrix of $\varphi$ wrt. this basis, we will receive a $8 \times 8$-matrix with two quadratic block matrices of size $4$ on the top left and the bottom right. I think that the bottom right matrix will have eigenvalue $1$ which will also give us the desired $\gamma$. But I am stuck with my computations here (the numbers seem to be really messy).
It would have been a bit easier if we would be in a Kummer-theoretic setting (i.e. the base field has a root of unity of degree $4$) with which I am a bit more familiar. But $\mathbb{Q}_3$ does not have a $4$-th root of unity.
Could you please guide me through my computation (or at least say if my approach is correct)?
