Primitive polynomial over a finite field

Remarks

Equivalently, $f$ is a scalar multiple of the minimal polynomial over $\F_q$ of a primitive element of $\F_{q^m}$. Then all roots of $f$ are primitive elements, since they are the conjugates $\alpha^{q^i}$, which have the same order as $\alpha$. Another standard form: an irreducible $f$ of degree $m$ with $f(0) \ne 0$ is primitive if and only if the least $e \ge 1$ such that $f$ divides $x^e - 1$ is $q^m - 1$. There are $\varphi(q^m - 1)/m$ monic primitive polynomials of degree $m$ over $\F_q$, where $\varphi$ is Euler's totient function. Example: over $\F_2$, the polynomial $x^3 + x + 1$ is irreducible, and a root $\alpha$ has order dividing $7$ but not $1$, so its order is $7 = 2^3 - 1$, and the polynomial is primitive. Non-example: over $\F_2$, the polynomial $x^4 + x^3 + x^2 + x + 1$ is irreducible, but it divides $x^5 - 1$, so a root has order $5$, not $15$, and the polynomial is not primitive.