Cubic resolvent of a quartic

For a depressed quartic $x^4+p{x}^2+qx+r$ with roots ${\alpha }_1,\dots ,{\alpha }_4$, the three pair-partition quantities ${\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4$ and its two analogues are the roots ofThe Galois action on these roots is the quotient of its action on the four quartic roots by the Klein four-group.

Let a separable polynomial $f$ of degree four be irreducible over a field $F$. If all three pair sums $r_1={\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4$, $r_2={\alpha }_1{\alpha }_3+{\alpha }_2{\alpha }_4$, and $r_3={\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3$ lie in $F$, its Galois group is the Klein four-group. Indeed $r_1-r_2=({\alpha }_1-{\alpha }_4)({\alpha }_2-{\alpha }_3)\ne 0$, and similarly the other pair sums differ. Every Galois automorphism therefore fixes each of the three partitions into pairs. The kernel of this action of $S_4$ on three partitions is $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. Irreducibility gives a transitive group action on the four roots, so the orbit-stabilizer theorem forces $4$ to divide the group order. Hence the Galois group equals $V_4$. The argument also works in characteristic two.

If an irreducible quartic over a field of characteristic zero has a cubic resolvent of a quartic whose Galois group of a polynomial is $S_3$, then the quartic has Galois group $S_4$. Irreducibility places the quartic group among the transitive subgroups of the symmetric group on four points, while the action on its three root pairings is surjective onto $S_3$. Of the possibilities $C_4,V_4,D_8,A_4,S_4$, only $S_4$ has this image.