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Let $P\in\mathbb{F}_q[X]$ irreducible of degree $d$ so that $\mathbb{F}_q[X]/(P)\simeq\mathbb{F}_{q^d}$, my question is when does $\overline{X}$ is a generator of the group $\mathbb{F}_{q^d}^*$ ? The first $q$ for which this fails is $16$ with $P=X^4+X^3+X²+X+1$, my guess is that such $P$ are very rare so that their density is $0$, so $$\lim\limits_{A\to+\infty}\frac{1}{A}\#\{P\in\mathbb{F}_q[X]\text{ monic irreducible with } \deg P\leqslant A \text{ and } \overline{X} \text{ generates }\mathbb{F}_q[X]/(P)\}=0. $$ Is this a known result, is there some elementary proof ?

Tuvasbien
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  • Do you mean $q^d=16$ rather than $q=16$? On a first reading I found it unclear what you meant by "such $P$": such $P$ that it is irreducible, or such $P$ that it fails? – Derek Holt Dec 19 '24 at 08:24
  • It is a bit unclear which density exactly you expect to go to zero? The number of irreducible monic polynomials of degree $d$ is known to be roughly $q^d/d$. The exact formula can be ggiven using the Möbius function. OTOH, the number of primitive monic polynomials of degree $d$ is $\Phi(q^d-1)/d$ as Schiepy explained. The interesting ratio could thus be roughly $\Phi(q^d-1)/q^d$. – Jyrki Lahtonen Dec 19 '24 at 11:58
  • See here for the formula. – Jyrki Lahtonen Dec 19 '24 at 11:59
  • But is your question really about the density of primitive versus non-primitive irreducible monic polynomial? That is $\Phi(q^n-1)$ vs. $M_n(q)\cdot n-\Phi(q^n-1)$ or? – Jyrki Lahtonen Dec 19 '24 at 12:02

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What you consider are called primitive polynomials in the finite fields context. They are a well-studied object and a lot of results are known. A very classical fact is that since $\mathbb{F}_{q^d}^{\ast}$ is cyclic and of order $q^d-1$ there are $\phi(q^d-1)$ generators of $\mathbb{F}_{q^d}^{\ast}$ and thus $\phi(q^d-1)/d$ (monic) primitive polynomials of degree $d$. Hence \begin{align*}\frac{1}{A}|\{P\in\mathbb{F}_q[x]&\mid P\text{ monic, primitive and }\deg(P)\le A \}|\\ &\ge \frac{1}{A}|\{P\in\mathbb{F}_q[x]\mid P\text{ monic, primitive and }\deg(P)= A \}|=\frac{\phi(q^A-1)}{A^2}\end{align*} and this should tend to $\infty$ if $A\to\infty$, so your guess is wrong.

schiepy
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