Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 18I (Galois Theory)

An irreducible degree-$n$ polynomial over ${\mathbb{F}}_q$ is separable and its roots lie in ${\mathbb{F}}_{q^n}$, with any one root generating that field. Hence its splitting field is ${\mathbb{F}}_{q^n}$ andgenerated by the Frobenius automorphism $x\mapsto x^q$.

If $n_i=\mathrm{deg}{h}_i$ and $N=\mathrm{lcm}(n_1,\dots ,n_r)$, the splitting field is the compositum of the fields ${\mathbb{F}}_{q^{n_i}}$, namely ${\mathbb{F}}_{q^N}$. Thereforeagain generated by Frobenius.

The multiplicative group of $L={\mathbb{F}}_{q^N}$ is cyclic of order $m=q^N-1$. For a primitive element $\zeta$, $L={\mathbb{F}}_q(\zeta )$, so this is a cyclotomic extension. The cyclotomic character isIts image is the cyclic subgroup $⟨q⟩$, of order $N$.

Irreducibility makes the quartic Galois group a transitive subgroup of $S_4$. Its square discriminant puts it inside $A_4$, leaving $V_4$ or $A_4$. The cubic resolvent isThe rational root theorem shows it is irreducible over $\mathbb{Q}$. Thus the Galois group acts transitively on the three pairings of four roots, excluding $V_4$. Consequently