Let $f \in \mathbb{Q}[x]$ be an irreducible quartic, $L/\mathbb{Q}$ its splitting field. Label its roots $\alpha_1, \dots , \alpha_4$. Suppose that $$\mathbb{Q}(\alpha_i) \, \cap \, \mathbb{Q}(\alpha_j) = \mathbb{Q}$$ For all $i \neq j$. Determine $G:=\mathrm{Gal}(L/\mathbb{Q})$.
This is a practice problem given for my department's algebra exam. I'd like my work reviewed. EDIT: I made a mistake in my third step.
My solution: WLOG, take $f$ monic. Since $f$ has four distinct roots, $G$ is a subgroup of $S_4$. Furthermore, $f$ is irreducible, so $G$ acts transitively on its roots; this means $G$ must be one of $S_4$'s transitive subgroups:
$$S_4, \, A_4, \, V_4, \, D_{4\cdot 2}, \, \mathbb{Z}/4\mathbb{Z}.$$
(This fact ended up not being useful; all I needed to know was that $G$ is a subgroup of $S_4$.) We know that $|G| = [L:\mathbb{Q}]$, so if we calculate $[L:\mathbb{Q}]$ (or at least bound it sufficiently) we might be able to determine which subgroup $G$ is.
First step: we know $[\mathbb{Q}(\alpha_1) : \mathbb{Q}] = 4$ since $f$ is $\alpha_1$'s minimal polynomial over $\mathbb{Q}[x]$.
Second step: what is $[\mathbb{Q}(\alpha_1,\alpha_2):\mathbb{Q}(\alpha_1)]$? It equals the degree of $m:=$ the minimal polynomial of $\alpha_2$ over $\mathbb{Q}(\alpha_1)[x]$. This polynomial divides $f = (x-\alpha_1)\dots(x-\alpha_4)$. Since $\alpha_1$ is in the base field and is a root of $f$, we know we can divide $f$ by $x-\alpha_1$ and get an element of $\mathbb{Q}(\alpha_1)[x]$; hence $$g:=(x-\alpha_2)(x-\alpha_3)(x-\alpha_4) \in \mathbb{Q}(\alpha_1)[x].$$ So $m$ divides $g$. Now, since $g$ is cubic, it is irreducible iff it has no roots in the base field. And since $\alpha_2, \alpha_3, \alpha_4 \not \in \mathbb{Q}(\alpha_1)$ by hypothesis, it has no roots in the base field $\mathbb{Q}(\alpha_1)$. So $g$ is irreducible and $m = g$. Therefore, $[\mathbb{Q}(\alpha_1,\alpha_2):\mathbb{Q}(\alpha_1)] =3$.
Third step: I am not sure what to do here. Somehow I must determine $[\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3):\mathbb{Q}(\alpha_1,\alpha_2)].$
Fourth step: Whether $[\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3):\mathbb{Q}(\alpha_1,\alpha_2)]$ is $1$ or $2$ determines whether $G$ is $S_4$ or $A_4$.