The number of primitive polynomials of degree $m$ over a finite field $GF(p^m)$

Question

Why is it that over $GF(p^m)$ there are exactly $\phi(p^m − 1)/m$ primitive polynomials of degree $m$?

Answer 1

The roots of a primitive polynomial $f(X)\in\Bbb F_2[X]$ such that $\Bbb F_2[X]/(f)\cong GF(p^m)$ are generators of the group $GF(p^m)^\times$. As the latter is cyclic of order $p^m-1$, there are $\phi(p^m-1)$ such generators. As each polynomial has $m$ such roots and the sets of roots for distinct polynomials are disjoint (else the $\gcd$ would already produce $GF(p^m)$, we conclude that $m$ times the number of polynomials equals $\phi(p^m-1)$.

Answer 2

This is because the multiplicative group of a finite field is cyclic, isomorphic to the additive cyclic group $\mathbf Z/(p^m-1)\mathbf Z$.

A primitive polynomial is the minimal polynomial of a generator of this cyclic group, and the group $\mathbf Z/(p^m-1)\mathbf Z$ has exactly $\varphi(p^m-1)$ generators.