Find the splitting field $\mathbb K$, the isomorphism type of the Galois group Gal$(\mathbb K / \mathbb Q)$, and the action of its generators on $\mathbb K$ for $x^4 + x^2 -6$ (from $\mathbb Q [x]$)
$f=x^4 +x^2 -6 = (x-i\sqrt3)(x-i\sqrt3)(x-\sqrt2)(x+\sqrt2)$
So for the splitting field, we have $\mathbb K =\mathbb Q (i\sqrt3 , -i\sqrt3, \sqrt2, -\sqrt2 )= \mathbb Q (i\sqrt3, \sqrt2 )$.
I found that $\{1, i\sqrt3, \sqrt2, i\sqrt6 \}$ is a basis of $\mathbb K$ so the dimension of it is $4$. So the isomorphism type is either $C_2 \times C_2$ or $C_4$.
Corollary: Let $K$ be a field, $f ∈ K[x]$ and let $L$ be the splitting field for $f$ over $K$. Write $f$ as $ag_1^{s_1} g_2 ^{s_2} ···g_k^{s_k}$ , where $a ∈ K$ and all $g_i ∈ K[x]$ are monic irreducible and pairwise different. (Since all $g_i$ are monic, $a$ is the leading coefficient of $f$). Let $R$ be the set of roots of $f$ in $L$, and $R_i$ be the set of roots of $g_i, i = 1,...,k$. Then
$\{R_i|1 ≤ i ≤ k\}$ is a partition of $R$; and
the orbits of Gal$(L/K)$ on $R$ are precisely the sets $R_1,...,R_k$.
$G=\text{Gal}(\mathbb K / |\mathbb Q)$ acts faithfully on the set of roots of $f$, which is $R=\{i\sqrt3,-i\sqrt3,\sqrt2,-\sqrt2 \}$ (do we need to prove this statement???).
Since $f=(x^2+3)(x^2-2)$
Both $(x^2+3), (x^2-2)$ are irreducible (wont include proof of that here) and monic. So by this corollary, $G$ has two orbits on $R$ which are $R_1= \{i\sqrt3,-i\sqrt3 \}$ and $R_2=\{\sqrt2, -\sqrt2 \}$. So the order of $G$ is two or four.
Action on generators:
$$ \begin{matrix} \hline ~ & i\sqrt3 & \sqrt2 \\ \hline g_1 & i\sqrt3 & -\sqrt2 \\ g_2 & -i\sqrt3 & \sqrt2 \\ g_3 & -i\sqrt3 & -\sqrt2 \\ g_4 & i\sqrt3 & \sqrt2 \\ \hline \end{matrix} $$
Then appartently all of these elements have order $2$ so the isomorphism type is $C_2 \times C_2$.
Have i done this correct? I feel i either done unnecessary stuff (like find basis) or i haven't included important stuff.