How to explicitly describe the generator of the Galois group of the extension defined by $x^4-3x^2+18$? Let $K=\mathbb{Q}_3$ and $L=K(\alpha)$ where the minimal polynomial of $\alpha$ over $\mathbb{Q}_3$ is $f = x^4-3x^2+18$. The LMFDB says that the Galois group is cyclic of degree $4$.
Suppose $\sigma$ is a generator of the Galois group. Since $L=K(\alpha)$, the generator $\sigma$ is uniquely determined by $\sigma(\alpha)$. Also, as $L$ is a $K$-vector space, the result of $\sigma(\alpha)$ can be written as a linear combination of powers of $\alpha$.
Question: How can I describe $\sigma(\alpha)$ explicitly as described above?
I know that there is a totally ramified subextension $L'=K(\alpha^2)$ which has minimal polynomial $x^2-3x+18$. Say that these roots in $L'$ are called $\beta$ and $\beta'$, and suppose $\alpha = \beta^2$. Then we may be able to describe $f$ in terms of these elements (maybe with $\beta+\beta' = 3$ or $\beta\beta' = 18$?) but I'm not sure if that really works.
 A: The Galois group of $x^4 - 3x^2 + 18$ over $\mathbf Q$ is dihedral of order 8, not cyclic of order $4$, so whatever argument is used has to involve something that is different between $\mathbf Q$ and $\mathbf Q_3$.
You can work out formulas for the roots of $f(x)$ by using the quadratic formula and square roots: $x^2 = 3(1 \pm \sqrt{-7})/2$, so
$$
x = \pm\sqrt{3\frac{1\pm\sqrt{-7}}{2}}.
$$
The number $-7$ is not a square in $\mathbf Q_3$ and $\beta := 3(1+\sqrt{-7})/2$ generates a quadratic extension $\mathbf Q_3(\beta) = \mathbf Q_3(\sqrt{-7})$, which as Lubin points out in a comment is unramified over $\mathbf Q_3$. Do you see why $\beta$ is a prime element of $\mathbf Q_3(\sqrt{-7})$? Letting $\alpha^2 = \beta$, the field $\mathbf Q_3(\alpha)$ has degree $4$ over $\mathbf Q_3$.
Let $\overline{\beta} = 3(1 - \sqrt{-7})/2$, so $\beta\overline{\beta} = 9(8)/4 = 9 \cdot 2$.  Since the ratio $2/(-7)$ is in $1 + 3\mathbf Z_3$, $2 = -7u^2$ for some $u \in \mathbf Q_3^\times$ (more precisely, $u \in 1 + 3\mathbf Z_3$).  In contrast, $2/(-7)$ is not a square in $\mathbf Q$ and that makes this a step that distinguishes studying the Galois closure of $x^3 - 3x^2 + 18$ over $\mathbf Q_3$ from studying it over $\mathbf Q$. Letting $\alpha'$ satisfy $\alpha'^2 = \overline{\beta}$, which only determines $\alpha'$ up to a sign, we have
$$
(\alpha\alpha')^2 = \beta\overline{\beta} = 9\cdot 2 = 9(-7u^2) = (3u)^2(-7), 
$$
so
$$
\alpha\alpha' = \pm 3u\sqrt{-7} = \pm (2\beta - 3)u = \pm (2\alpha^2 - 3)u.
$$
Therefore we can pin down the sign on $\alpha'$ by requiring $\alpha\alpha' = (2\alpha^2-3)u$, which gives us the following formula for $\alpha'$ in terms of $\alpha$.
$$
\alpha'  = \frac{(2\alpha^2-3)u}{\alpha} \in \mathbf Q_3(\alpha).
$$
(It is not true that $\alpha' \in \mathbf Q(\alpha)$, since $u = \sqrt{2/(-7)}$ is not rational.)
The four roots of $x^4 - 3x^2 + 18$ are $\pm \alpha$ and $\pm \alpha'$, which are all in $\mathbf Q_3(\alpha)$. At this point you should work out why this means the Galois group over $\mathbf Q_3$ is cyclic of order $4$ and not the other group of order $4$.
