Number of Distinct Elements in a Finite Field

Question

I'm working on this question:

For any element $a \in GF(p,(P(x))$, let $r(a)$ denote the number of distinct elements $b$ in $GF(p,P(x))$ such that $b^{p-1} = a.$ Show that if $a \neq 0$, then $r(a) = 0$ or $r(a)=p -1$.

I have a feeling that Fermat's Little Theorem would come into play here, but that is my only lead for this problem. How should one approach this problem?

Answer 1

By Fermat's little theorem in $\mathbb Z/p\mathbb Z$ you have $a^{p-1}=1$ for all $a\ne 0$. Now let $F$ be any field containing $\mathbb Z/p\mathbb Z$ and $a\in F^\times$. Let $b^{p-1}=a$ for some $b\ne 0$. Then $b^{p-1},(2b)^{p-1},\dots,((p-1)b)^{p-1}=a$ so you have at least $p-1$ roots, and $x^{p-1}-a=0$ can have at most $p-1$ roots, so $r(a)=p-1$.