Size of field extension is at most $p^d$

Question

I have a field extension $\mathbb{F}_p(\alpha)$ where $\alpha$ is a root of the irreducible polynomial $f \in \mathbb{F}_p[t]$ and I know that $\alpha ^{p^d} = \alpha$, where $p$ is some prime.

I'm trying to show that then $|\mathbb{F}_p(\alpha)|$ is at most $p^d$. I'm struggling on how to get started with this and would appreciate some help. The only fact I can think of is that $\mathbb{F}_p(\alpha)$ has size $p^{deg(f)}$. I feel like I may be missing something obvious.

Thanks

Answer 1

Show that

either

• the elements $\alpha, \alpha^p, \alpha^{p^2}, \cdots, \alpha^{p^{d-1}}$ are all distinct,

or

• $\alpha^{p^{c}} = \alpha$ where $c$ is some proper divisor of $d$, so that the sequence $(\alpha, \alpha^p, \alpha^{p^2}, \cdots, \alpha^{p^{d-1}})$ of length $d$ is $d/c$ repetitions of the sequence $(\alpha, \alpha^p, \alpha^{p^2}, \cdots, \alpha^{p^{c-1}})$.

Thus, $f(t)$, the minimal polynomial of $\alpha$ in $\mathbb F_p[t]$, is of degree $d$ or of degree $c$, a divisor of $d$. In either case, $\mathbb F_p(\alpha)$ has at most $p^d$ elements.