PropositionStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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$\sum_{d ∣ n} d N_{q} (d) = q^{n}$ for the counts $N_{q} (d)$ of monic irreducibles of degree $d$ over $F_{q}$

Statement

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

$\sum_{d ∣ n} d N_{q} (d) = q^{n},$

the sum being over the positive divisors $d$ of $n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ , The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form). At $n = 1$ the identity reads $N_{q} (1) = q$ .

Facts & Assumptions

Given: A finite field $F_{q}$ of order $q \geq 2$ and an integer $n \geq 1$ ; monic polynomials are as in Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree.

[L1]

In $F_{q} [t]$ one has $t^{q^{n}} - t = \prod P (t)$ , the product being over the monic irreducible $P \in F_{q} [t]$ whose degree divides $n$ , each such $P$ occurring once (Over $F_{q}$ , $x^{q^{n}} - x$ is the product of all monic irreducibles whose degrees divide $n$ ).

[L2]

If $R$ is an integral domain and $f, g \in R [x]$ are nonzero, then $f g \neq 0$ and $\deg (f g) = \deg f + \deg g$ (Over an integral domain, degrees add under multiplication of nonzero polynomials).

[L3]

If $R$ is an integral domain, then $R [x]$ is an integral domain (A polynomial ring over an integral domain is an integral domain).

Proof

technique · direct

1.1

For each $d \geq 1$ the monic polynomials of degree $d$ in $F_{q} [t]$ are the $t^{d} + a_{d - 1} t^{d - 1} + \dots + a_{0}$ with $a_{0}, \dots, a_{d - 1} \in F_{q}$ , so there are exactly $q^{d}$ of them and $N_{q} (d) \leq q^{d}$ is finite.

givenalgebra

2.1

Consequently the family of monic irreducible $P \in F_{q} [t]$ with $\deg P ∣ n$ is finite, having at most $\sum_{d ∣ n} q^{d}$ members, so the product in [L1] is a finite product of nonzero polynomials in the integral domain $F_{q} [t]$ ([L3]).

step 1.1L1L3

3.1

Taking degrees in [L1] and applying [L2] repeatedly to that finite product gives $\deg (t^{q^{n}} - t) = \sum_{\deg P ∣ n} \deg P$ , where the sum runs over the same finite family; and $\deg (t^{q^{n}} - t) = q^{n}$ because $q^{n} > 1$ .

step 2.1L1L2algebra

4.1

Splitting that sum according to the degree of $P$ : the possible degrees are exactly the positive divisors $d$ of $n$ , there are $N_{q} (d)$ monic irreducibles of degree $d$ , and each contributes $d$ ; hence $\sum_{d ∣ n} d N_{q} (d) = q^{n}$ .

step 1.1step 3.1algebra

5.1

At $n = 1$ the only positive divisor is $d = 1$ , so the identity reads $N_{q} (1) = q$ , in agreement with the fact that the monic polynomials of degree one are the $t - a$ for $a \in F_{q}$ and each is irreducible.

step 1.1step 4.1algebra∎

Remarks

What the identity does not give. It determines $N_{q} (n)$ only once every $N_{q} (d)$ for proper divisors $d$ of $n$ is known, so it is a recursion rather than a formula. Inverting it into a closed form is a separate matter and is not carried out here.

Depends on

Used by

The divisor-sum identity at q=2, n=3 finds exactly two monic irreducible cubics Example

Dependency tree · two levels

32 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Conrad, Roots and Irreducibles (expository blurb), formula (6.2) and Example 6.2 (standard reference, not scraped)
K. Conrad, Finite Fields (expository blurb), Section 6 (standard reference, not scraped)

∑d∣nd Nq(d)=qn∑d∣n​dNq​(d)=qn for the counts Nq(d)Nq​(d) of monic irreducibles of degree dd over FqFq​