I'm trying to create the Galois group lattice and the corresponding field lattice for the polynomial $f(x) = x^3-3x+4$ over $\mathbb{Q}$, but I'm having a small issue, first I'll explain what I've done so far.
I noticed that $f$ is irreducible in $\mathbb{F}_5[x]$, and thus $f$ is indeed irreducible over $\mathbb{Q}$. Furthermore, the splitting field $K_{/\mathbb{Q}}$ contains a root of $f$, and since $f' \neq 0$ in $K$, I know that $f$ contains no multiple roots, implying that $f$ is separable.
Finally, $f$ has discriminant $D(f) = -324$ which isn't a square in $\mathbb{Q}$, implying from a Galois theory theorem that $\text{Gal}(K_{/\mathbb{Q}}) \cong D_3 = \langle a,b \: | \: a^3=b^2=e, \: ba=a^2b \rangle$. Hence I know that $f$ has a single real root $\alpha$ and a pair of complex conjugate roots $\beta,\overline{\beta}$, implying that $K = \mathbb{Q}(\alpha,\beta)$.
Then I noticed that $i \in K$, since $\text{Re}(\beta), \text{Im}(\beta) \in K$, since $\frac{\beta+\overline{\beta}}{2} \in K$ and $\frac{\beta-\overline{\beta}}{2} \in K$, implying that: $$(\beta-\text{Re}(\beta))\text{Im}(\beta)^{-1} = i \in K \implies \mathbb{Q}(i) \subset K.$$
Now, I think I have enough to start making the lattice. I know that:
- $\langle e \rangle \sim K$
- $D_3 \sim \mathbb{Q}$
- I have to match up the groups $\langle a \rangle$, $\langle b \rangle$, $\langle ab \rangle$, and $\langle a^2b \rangle$.
Hence, I just need to find automorphisms $\sigma_a,\sigma_b$ corresponding to the generators $a,b$ of $D_3$, and I should be done. The obvious automorphism I thought of for $\sigma_b$ was: $$\sigma_b:\begin{cases} \alpha \mapsto \alpha \\ \beta \mapsto \overline{\beta} \\\end{cases}$$
I chose this one since then $|\sigma_b|=2$, and hence has index $3$, corresponding to the field $\mathbb{Q}(\alpha)$. Thus, we have that $\langle b \rangle \sim \mathbb{Q}(\alpha)$. This leaves $\mathbb{Q}(\beta) = \mathbb{Q}(i)$ (since the real and imaginary parts of $\beta$ are in here by above). This is a degree $2$ extension over $\mathbb{Q}$ and so I figured I would want $\sigma_a$ to be the automorphism whos fixed field is $\mathbb{Q}(i)$.
$\textbf{(Problem starts here})$, so I need $\sigma_a$ to fix $i$, but I don't know what $\sigma_a$ does to $\text{Re}(\beta), \text{Im}(\beta)$. I know it $\textbf{cant}$ fix these elements (im pretty sure atleast) since if $\sigma_a$ fixed $\beta$, then it fixes $\overline{\beta}$, and hence it must fix $\alpha$ too, but this is just the identity map.
- $\textbf{First Issue}$: I don't know what $\sigma_a$ should be, as I dont know what $\alpha,\beta$ actually are.
- $\textbf{Second Issue}$: I still need to match $\langle ab \rangle$ and $\langle a^2b \rangle$ to intermediate fields of $K$, which I don't know yet, but I beleive this is doable once I know that $\sigma_a$ is, as then I can find out what $\sigma_a \sigma_b$ and $\sigma_a^2 \sigma_b$ both fix, and go from there.
I'd really appreciate a respond on how to proceed from here. Thanks!