Solution
= Solution
By the allowed cyclicity theorem, $K^\times$ has a generator $z$. Every nonzero element of $K$ is a <power function of a statistical test>[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.