CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Möbius inversion gives the closed formula for the number of monic irreducibles over $F_{q}$

Statement

Let $F_{q}$ be a finite field of order $q$ , and let $N_{q} (n)$ denote the number of monic irreducible polynomials of degree $n$ in $F_{q} [t]$ . Then for every $n \geq 1$ ,

$N_{q} (n) = \frac{1}{n} \sum_{\begin{array}{c} d ∣ n \\ d > 0 \end{array}} μ (d) q^{n / d} .$

Facts & Assumptions

Given: A finite field $F_{q}$ of order $q$ and an integer $n \geq 1$ .

Proof

technique · direct

1.1given

By $\sum_{d ∣ n} d N_{q} (d) = q^{n}$ for the counts $N_{q} (d)$ of monic irreducibles of degree $d$ over $F_{q}$ , one has $\sum_{d ∣ n} d N_{q} (d) = q^{n}$ . Regard $f (d) : = d N_{q} (d)$ and $g (n) : = q^{n}$ as arithmetic functions of the degree variable.

2.1step 1.1

Applying Classical Möbius inversion over positive divisors to those functions gives $n N_{q} (n) = \sum_{d ∣ n} μ (d) q^{n / d}$ .

3.1step 2.1algebra∎

Since $n \geq 1$ , division by $n$ yields the claimed closed formula for $N_{q} (n)$ .

Depends on

Used by

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Sources

Keith Conrad, Finite Fields, Section 6 (standard reference, not scraped)
Kiran S. Kedlaya, An Introduction to Analytic Number Theory, Section 3.2 (standard reference, not scraped)

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