Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 18H / ii / Solution

The three resolvent cubic roots correspond to the partitions $12|34$, $13|24$, $14|23$. Permuting the four roots of the quartic permutes these three quantities and defines a homomorphism $G\to S_3$. They are distinct: because the quartic has zero cubic coefficient, their differences includesince characteristic zero and irreducibility imply separability. Thus the kernel of the action is exactly $V_4$. The splitting field of the resolvent is the subfield generated by its roots; restriction from the quartic splitting field is surjective onto its Galois group, since automorphisms of a normal separable intermediate extension extend to the splitting field. HenceIn the four cases these groups are respectively the trivial group, $C_2$, $C_3$ and $S_3$. Computing the symmetric functions also gives $Q(X)=X^3+2b{X}^2+(b^2-4d)X-c^2$ with the source's root convention.