CorollaryStatement: 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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The reduction of $Φ_{n}$ is irreducible over $F_{q}$ exactly when $[q]$ generates $(Z / n)^{\times}$

Statement

Let $F_{q}$ be a finite field of order $q$ and $n \geq 1$ with $\gcd (n, q) = 1$ (Coprime integers: $\gcd (a, b) = 1$ ). The image of $Φ_{n}$ in $F_{q} [t]$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ) is irreducible (Irreducible and prime elements of an integral domain) if and only if $[q]_{n}$ generates $(Z / n)^{\times}$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ). In particular this can happen only when $(Z / n)^{\times}$ is cyclic.

Facts & Assumptions

Given: A finite field $F_{q}$ of order $q$ and $n \geq 1$ with $\gcd (n, q) = 1$ ; write $d : = ord ([q]_{n})$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity) and ${\overset{ˉ}{Φ}}_{n}$ for the image of $Φ_{n}$ in $F_{q} [t]$ .

[L1]

${\overset{ˉ}{Φ}}_{n}$ is a product of pairwise distinct monic irreducible polynomials, each of degree $d$ , and there are $φ (n) / d$ of them (For $\gcd (n, q) = 1$ the reduction of $Φ_{n}$ in $F_{q} [t]$ is a product of distinct monic irreducibles, each of degree the order of $[q]$ modulo $n$ ).

[L2]

$Φ_{n}$ is monic of degree $φ (n)$ (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L3]

$∣ (Z / n)^{\times} ∣ = φ (n)$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ), and $∣ ⟨ x ⟩ ∣ = ord (x)$ for an element of finite order (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Proof

technique · direct

1.1

By [L1] the number of monic irreducible factors of ${\overset{ˉ}{Φ}}_{n}$ , counted without repetition and with none repeated, is $r = φ (n) / d$ .

2.1

If $r = 1$ then ${\overset{ˉ}{Φ}}_{n}$ is itself one of those monic irreducible polynomials, hence irreducible; if $r \geq 2$ then ${\overset{ˉ}{Φ}}_{n}$ is a product of $r$ polynomials each of degree $d \geq 1$ , none of them a unit, so it is not irreducible. Hence ${\overset{ˉ}{Φ}}_{n}$ is irreducible exactly when $r = 1$ , that is exactly when $d = φ (n)$ .

step 1.1L1L2

3.1

By [L3] the subgroup $⟨ [q]_{n} ⟩$ has order $d$ and $(Z / n)^{\times}$ has order $φ (n)$ , so $d = φ (n)$ holds exactly when $⟨ [q]_{n} ⟩ = (Z / n)^{\times}$ , that is exactly when $[q]_{n}$ generates the unit group. With step 2.1 this proves the equivalence, and a group with a generator is cyclic.

step 2.1L3∎

Remarks

When the criterion cannot be met at all. If $(Z / n)^{\times}$ is not cyclic then no class generates it, so the reduction of $Φ_{n}$ is reducible over every finite field of order coprime to $n$ ; the smallest such $n$ is $8$ , where $(Z / 8)^{\times}$ has three elements of order two and no element of order four.

Depends on

Used by

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Sources

K. Conrad, Cyclotomic Extensions (expository blurb), Corollary 5.7 (standard reference, not scraped)
K. Conrad, Finite Fields (expository blurb), Section 5 (standard reference, not scraped)

The reduction of ΦnΦn​ is irreducible over FqFq​ exactly when [q][q] generates (Z/n)×(Z/n)×