$F_{p^n}=\mathbb{Z}_p(\alpha)$, determine whether $\alpha$ is the generator of multiplicative group $F_{p^n}^*=F_{p^n}\backslash \{0\}$.

Question

$F_{p^n}=\mathbb{Z}_p(\alpha)$, determine whether $\alpha$ is the generator of multiplicative group $F_{p^n}^*=F_{p^n}\backslash \{0\}$.

My try: Since $\alpha$ is the root of a irreducible polynomial of degree $n$, the order of the multiplicative group is $p^n-1$. Also the multiplicative froup is cyclic. Hence if we want to prove $\alpha $ is a generator, we need to prove its order is $ord(\alpha) =p^n-1 $. I am stucked at this step.

Answer 1

A generator of the field need not be a generator of the multiplicative group. Consider $n=3$. Then there are $p^3-p$ choices for $\alpha$, but there are fewer generators of the cyclic group of order $p^3-1$. In particular, $m^{th}$ roots of unity provide counterexamples, where $m=p^2+p+1$.