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

1.1

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.

given

2.1

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

step 1.1

3.1

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

step 2.1algebra∎

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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)

Möbius inversion gives the closed formula for the number of monic irreducibles over FqFq​