Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / A / A2

Regard $K$ as a vector space over $F$. Since $K$ is finite, a maximal linearly independent list has finite length $d$ and is a basis. Every element is uniquely a linear combination of its $d$ basis vectors, with $q$ choices for each coefficient. Hence $\boxed{|K|=q^d}$, with $d=[K:F]$. This includes $d=1$ when the extension is trivial.

Write $q=p^r$, where $p$ is the characteristic. In a splitting field of $X^{q^m}-X$ over $F$, let $K$ be its set of roots. The derivative of this polynomial is $-1$, so all $q^m$ roots are distinct. The Frobenius power is additive and multiplicative: $(x+y{)}^{q^m}=x^{q^m}+y^{q^m}$ and $(xy{)}^{q^m}=x^{q^m}{y}^{q^m}$. Thus roots are closed under addition, negatives and multiplication; for a nonzero root, $(x^{-1}{)}^{q^m}=x^{-1}$. They include $0,1$ and form a field.

Every $a\in F$ satisfies $a^q=a$, by Lagrange's theorem on $F^{\times }$ and separately at zero. Iterating gives $a^{q^m}=a$, so $F\subseteq K$. The constructed field therefore contains $F$ and has exactly $\boxed{q^m\text{ elements}}$.

By the allowed cyclicity theorem, $K^{\times }$ has a generator $z$. Every nonzero element of $K$ is a power of $z$, and zero also belongs to $F(z)$. Thus $K\subseteq F(z)\subseteq K$, giving $\boxed{K=F(z)}$. This is a primitive element argument for finite fields using the multiplicative group, not a claim that the additive group is cyclic.

For $\alpha \ne 0$, Lagrange's theorem in the multiplicative group of order $q^m-1$ gives ${\alpha }^{q^m-1}=1$; at zero the desired identity is immediate. Thus all elements satisfy ${\alpha }^{q^m}-\alpha =0$.

For an irreducible degree-$m$ polynomial $f\in F[X]$, the quotient $L=F[X]/(f)$ is a field of dimension $m$, hence has $q^m$ elements by part 1. Its element $\alpha =[X]$ therefore satisfies ${\alpha }^{q^m}-\alpha =0$. The evaluation homomorphism $F[X]\to L$ has kernel $(f)$, so