Exercise 11.2.5

Proof. Suppose that $\alpha \in F$ satisfies ${\alpha }^n=1$ for some $n>0$.

As the characteristic of $F$ is $p$, ${𝔽}_p$ is a subfield of $F$. Since $\alpha$ is a root of $x^n-1$, then $\alpha$ is algebraic over ${𝔽}_p$, and $\nu =[{𝔽}_p(\alpha ):{𝔽}_p]<\infty$, so $|{𝔽}_p(\alpha )|=p^{\nu }$ is finite, and ${𝔽}_p(\alpha )$ is a finite field with $q=p^{\nu }$ elements. Since $\alpha$ is in the group ${𝔽}_p{(\alpha )}^{\text{∗}}$, by Lagrange’s Theorem,

Let $d\ge 1$ the order of $\alpha$ in the group ${𝔽}_p{(\alpha )}^{\text{∗}}$. Then ${\alpha }^d=1$ and $d\mid p^{\nu }-1$, so $d$ is relatively prime to $p$. □