The five-case resolvent classification of an irreducible quartic Galois group

Let $L$ be the splitting field of $f$ and $G=\mathrm{Gal}(L\mathrm{/}F)\le S_4$. Irreducibility makes $G$ transitive. The kernel of its action on the three pairing roots is $G\cap V_4$ by [L1]. The field generated by those roots is $M$, so the Galois correspondence identifies $M=L^{G\cap V_4}$.

If $R_f$ is irreducible, its pairing action is transitive, so [L1] leaves $S_4$ or $A_4$; by [L2] and the discriminant criterion, nonsquare $\mathrm{\Delta }$ gives $S_4$ and square $\mathrm{\Delta }$ gives $A_4$. If $R_f$ splits completely, the pairing action is trivial and transitivity on four roots forces $G=V_4$. If $R_f$ has exactly one root in $F$, its other two roots form one orbit, so the pairing image is $C_2$ and [L1] leaves $D_4$ or $C_4$. A cubic cannot have exactly two roots in $F$, and [L2] plus separability makes the resolvent separable, so these branches are exhaustive.

In the unique-root branch, $\mathrm{Gal}(L\mathrm{/}M)=G\cap V_4$. For $G=D_4$ this intersection is $V_4$, which acts transitively on the four roots, so $f$ remains irreducible over $M$. For $G=C_4$ the intersection has order two and two root orbits, so $f$ factors over $M$. The transitivity criterion proves both implications and completes the five-case classification.