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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.

Rotdat
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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 $4$.

KCd
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    Also $K=\Bbb{Q}3(\sqrt3\zeta{16})$ and the Galois group is generated by the restriction of the Frobenius of $\Bbb{Q}3(\zeta{16},\sqrt3)/\Bbb{Q}3(\sqrt3)$ which sends $\sqrt{3}\zeta{16} \to \sqrt{3}\zeta_{16}^3$. – reuns Nov 25 '20 at 01:17
  • @KCd: I know this question is coming late but do you mind explaining why $-2/7 \in 1+\mathbb{Z}_3$ implies $2 = -7u^2$ for some $u \in 1+\mathbb{Z}_3$? I feel like it just follows from some properties of $3$-adic integers but I have trouble seeing that. – Rotdat Dec 09 '20 at 02:40
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    For each odd prime $p$, every number $a$ in $1 + p{\mathbf Z}_p$ is a perfect square in $1 + p{\mathbf Z}_p$: solve $x^2 - a = 0$ by Hensel's lemma by starting with approximate root $1$. Apply that with $a = -2/7$. This is like saying that since $2/7 > 0$ in $\mathbf R$, we can write $2/7 = u^2$ for some $u > 0$ in $\mathbf R$. To be a guaranteed $2$-adic perfect square you need to be a bit closer to $1$: it holds for all $a$ in $1 + 8\mathbf Z_2$ but not necessarily all $a$ in $1 + 2\mathbf Z_2$ (try $a = 3$). – KCd Dec 09 '20 at 02:57
  • @KCd: Thanks, that helps! I have another question (I realized that I did not understand everything just as I thought): In the end, I showed that the Galois group is cyclic by showing that the cubic resolvent has no roots over $\mathbb{Q}_3$. But then I realized that this has nothing to do with the answer. Could you explain to me why it could not be the Klein four group? I tried to derive a contradiction but also failed in that... – Rotdat Dec 09 '20 at 03:53
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    Forget resolvents: show the $\mathbf Q_3$-autormorphism of $\mathbf Q_3(\alpha)$ that sends $\alpha$ to $\alpha' = (2\alpha^2-3)u/\alpha$ has order 4. – KCd Dec 09 '20 at 03:58
  • @KCd: Oh, that's way easier than I thought, how embarrassing. Thanks for elucidating me! – Rotdat Dec 09 '20 at 04:52