Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 18H (Galois Theory)

TakeIt is a nonzero polynomial, but every element of the finite field ${\mathbb{F}}_q$ satisfies ${\alpha }^q=\alpha$. Thus it vanishes on every point of ${\mathbb{F}}_q^n$. Formal nonzero polynomials and nonzero polynomial functions need not coincide over a finite field.

First make the irreducible polynomial monic; for a nonmonic polynomial restore its leading coefficient in front of the product below. If $\alpha$ is one root, $K={\mathbb{F}}_q(\alpha )$ has degree $n$ and cardinality $q^n$. Thus every element satisfies $x^{q^n}=x$. The Frobenius automorphism $x\mapsto x^q$ fixes ${\mathbb{F}}_q$ and permutes the roots. If the orbit of $\alpha$ had size $d<n$, its orbit product would be a degree-$d$ polynomial fixed by Frobenius, hence with coefficients in ${\mathbb{F}}_q$, contradicting irreducibility. ThereforeThe roots are distinct, since $X^{q^n}-X$ has derivative $-1$. They all lie in $K$ and include its generator $\alpha$, so the splitting field is ${\mathbb{F}}_{q^n}$. Frobenius has order $n$ there; a degree-$n$ extension has at most $n$ automorphisms, so

For $p\nmid n$, the reduction of the cyclotomic polynomial ${\Phi }_n$ has precisely the primitive $n$th roots as its roots. If $p\mid {\Phi }_n(x)$, the residue class of $x$ has multiplicative order $n$ in ${\mathbb{F}}_p^{\times }$, so Lagrange's theorem gives $n\mid p-1$. Conversely, if $n\mid p-1$, cyclicity of the multiplicative group supplies an element of order $n$, which is a root of ${\Phi }_n$; any integer representative supplies $x$. HenceThree explicit examples are ${\Phi }_3(X)=X^2+X+1$ with $x=2,p=7$; ${\Phi }_4(X)=X^2+1$ with $x=2,p=5$; and ${\Phi }_6(X)=X^2-X+1$ with $x=3,p=7$. Each prime is coprime to the relevant $n$ and has the stated congruence.