Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 18H (Galois Theory)

Over a finite field ${\mathbb{F}}_q$, the splitting field is a finite extension ${\mathbb{F}}_{q^m}$, where $m$ is the least common multiple of the degrees of the distinct irreducible factors. Its Galois group is cyclic of order $m$, generated by $x\mapsto x^q$.

Over ${\mathbb{F}}_2$, the polynomial has no linear factor: its values at zero and one are both one. The only irreducible quadratic is $x^2+x+1$; reduction modulo it gives $x$, not zero. A reducible quartic without linear factors would split into quadratics, so $f$ is irreducible. Thus its splitting field is ${\mathbb{F}}_{16}$ and

Direct division gives $f=(x-1)(x^3+2x^2+2x+2)$ over ${\mathbb{F}}_3$. The cubic takes values $2,1,1$ at $0,1,2$, so is irreducible. Finite-field extensions are cyclic Galois extensions, and the splitting field degree is the least common multiple of these factor degrees, namely three. Hence

The splitting field over ${\mathbb{F}}_2$ is ${\mathbb{F}}_{16}$, and it contains the unique subfield ${\mathbb{F}}_4$. Over ${\mathbb{F}}_4$, Frobenius is the square of the order-four ${\mathbb{F}}_2$ Frobenius, so it has two orbits of length two on the four roots. Thus $f$ factors into two irreducible quadratics, andgenerated by $x\mapsto x^4$ on ${\mathbb{F}}_{16}$.

Irreducibility modulo two implies irreducibility over $\mathbb{Q}$ by Gauss lemma for polynomials, so the Galois group embeds transitively in $S_4$. The discriminant is $229$, hence primes two and three are unramified. The factor-degree theorem says that square-free reduction with factor degrees $d_1,\dots ,d_r$ provides an element of the Galois group with cycle lengths $d_1,\dots ,d_r$. The reductions already found therefore give a 4-cycle and a 3-cycle.

The group order divides 24 and is divisible by four and three, so is 12 or 24. An order-12 subgroup of $S_4$ has index two and must be $A_4$: in its quotient every transposition has the same nontrivial image, since transpositions are conjugate and generate $S_4$, making the quotient map the sign. But a 4-cycle is odd. Therefore the order is 24 andThe discriminant can also be obtained from $f^{\prime }=x^2(4x+3)$ and the resultant, giving $256-27=229$.