Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 18I / Solution

Let $G$ be the Galois group of a polynomial of the irreducible quartic. Irreducibility makes $G$ a transitive subgroup of $S_4$. The three roots of the given cubic resolvent of a quartic correspond to the three partitions of the four quartic roots into two unordered pairs, and the kernel of the $S_4$-action on these partitions is the Klein four-group $V_4$. If the resolvent has Galois group $S_3$, the image of $G$ in this action is all of $S_3$. Among the transitive subgroups of the symmetric group on four points, only $S_4$ has this image. Hence

Now put $f_p(t)=t^4+pt+p$. The Eisenstein criterion at $p$ proves that $f_p$ is an irreducible polynomial, and its cubic resolvent isIts discriminant of a depressed cubic isFor odd $p$, the $p$-adic valuation of ${\Delta }_p$ is three because $p\nmid 256$; for $p=2$, ${\Delta }_2=2^4\cdot 101$. Thus ${\Delta }_p$ is never a square in $\mathbb{Q}$.

By the rational root theorem, a reducible $g_p$ has an integer root among $\pm 1,\pm p,\pm p^2$. Reduction modulo $p$ shows that the root is divisible by $p$; writing it as $pk$, with $k\in \{\pm 1,\pm p\}$, givesThe only possibilities are $k=-1,p=3$ and $k=1,p=5$, corresponding to roots $-3$ and $5$. For every other prime, $g_p$ is irreducible with nonsquare discriminant, so its Galois group is $S_3$ and the criterion above gives the quartic group $S_4$. For $p=3$ or $5$, the reducible resolvent makes the quartic group a proper subgroup. Therefore