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Let $g$ be an irreducible polynomial of degree $n$ over $\mathbb{Z}/p\mathbb{Z}$. $a$ is a generator of the group $(\mathbb{Z}/p\mathbb{Z}[x]/\langle g \rangle)^*$ if $(\mathbb{Z}/p\mathbb{Z}[x]/\langle g \rangle)^* = \{1, a, \ldots, a^{p^{\deg(g)} - 2}\}$. How many monic irreducible polynomials $h$ over $\mathbb{Z}/p\mathbb{Z}$ such that $h(a) = 0$ for some generator $a$ does exist?

My first observation is that all such polynomials must have degree $n$ since if there is an irreducible polynomial $h_0$ such that $\deg h_0 \neq \deg g$ and $h_0(a) = 0$ for a generator $a$ then $(\mathbb{Z}/p\mathbb{Z}[x]/\langle h_0 \rangle) \simeq \mathbb{Z}/p\mathbb{Z}(a) \simeq (\mathbb{Z}/p\mathbb{Z}[x]/\langle g \rangle)$. That's a contradiction since these two fields have different size. On the other hand, $x^{p^n}-x$ is the product of all monic irreducible polynomials with degree $d | n$. Thus it should contain all the polynomials we need as divisors. That's the point where I'm out of ideas.

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These are called primitive polynomials. The multiplicative group of the finite field $\mathbb{F}_{p^n}$ is cyclic of order $p^n - 1$, and so it has $\varphi(p^n - 1)$ generators where $\varphi(-)$ is the Euler totient function (exercise). A primitive polynomial has $n$ roots, each of which is a generator (exercise), and distinct primitive polynomials share no roots because they're irreducible, so there are

$$\frac{\varphi(p^n - 1)}{n}$$

primitive polynomials for $\mathbb{F}_{p^n}$. (Exercise: without knowing that this counts primitive polynomials, why is this expression an integer?)

(Final exercise: show that the product of all of the primitive polynomials for $\mathbb{F}_{p^n}$ is congruent to $\Phi_{p^n - 1}(x) \bmod p$ where $\Phi_n(-)$ denotes a cyclotomic polynomial. Note that $\Phi_n(x)$ always has degree $\varphi(n)$.)

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  • $\begingroup$ By the definition of the cyclotomic polynomial, each of the generators is a root of $\Phi_{p^n-1}$ and they're the only roots it has. Thus each of its irreducible divisors has only generator roots. On the other hand, as we know, each of these divisors is a primitive polynomial and no other primitive polynomials exist. Since each of these polynomials has degree $n$ there are $\phi(p^n-1)/n$ of them. Am I right that it proves that all roots of a primitive polynomial are generators? $\endgroup$ – Artur Riazanov Jan 9 '18 at 3:26
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    $\begingroup$ Proving that all of the roots of a primitive polynomial are generators is easier than that: they are all Galois conjugates of a generator (more specifically, they are all obtained by acting via the Frobenius automorphisms), and automorphisms preserve the property of being a generator. $\endgroup$ – Qiaochu Yuan Jan 9 '18 at 3:32

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