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

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Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity

Statement

Let $K$ be a field and $n \geq 1$ with $char K ∤ n$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise, Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ), and let $E$ be a splitting field of $t^{n} - 1$ over $K$ , that is $E = K (μ_{n})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ). Write $Φ_{n}$ also for the image of the integer polynomial $Φ_{n}$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ) in $K [t]$ . Then $Φ_{n}$ is separable over $K$ (Repeated roots in extension fields and separable polynomials), it splits over $E$ , and its roots in $E$ are exactly the $φ (n)$ primitive $n$ -th roots of unity in $E$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity, The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

Facts & Assumptions

Given: A field $K$ , an integer $n \geq 1$ with $char K ∤ n$ , a splitting field $E$ of $t^{n} - 1$ over $K$ , and the convention that for every finite subset $T \subseteq E$ the product $\prod_{ζ \in T} (t - ζ) \in E [t]$ means the finite product along any enumeration of $T$ ; because $E [t]$ is a commutative ring, the value is independent of the enumeration (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either, Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring). In particular, for each positive divisor $d$ of $n$ , let $S_{d} : = {ζ \in μ_{n} (E) : ord (ζ) = d}$ and $Ψ_{d} : = \prod_{ζ \in S_{d}} (t - ζ) \in E [t]$ .

[L1]

$t^{n} - 1$ is separable over $K$ when $char K ∤ n$ , and then $μ_{n} (E)$ is cyclic of order $n$ with exactly $φ (n)$ primitive $n$ -th roots of unity ( $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity, $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ ).

[L2]

For every $m \geq 1$ one has $\prod_{d ∣ m} Φ_{d} = t^{m} - 1$ in $Z [t]$ , each $Φ_{d}$ monic of degree $φ (d)$ (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ ); reduction into $K [t]$ preserves this identity.

[L3]

A monic $f$ of degree $m$ splits over $E$ when $f = \prod_{j = 1}^{m} (t - α_{j})$ with $α_{j} \in E$ , repetitions allowed, and a splitting field is generated over $K$ by the roots (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

[L4]

$f$ is separable over $K$ when no extension field of $K$ contains an $a$ with $(t - a)^{2}$ dividing the image of $f$ (Repeated roots in extension fields and separable polynomials).

[L5]

For an element $x$ of finite order $m$ , $x^{k} = e$ if and only if $m ∣ k$ (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 $j = k$ , 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).

[L6]

$R [x]$ is an integral domain when $R$ is (A polynomial ring over an integral domain is an integral domain); and $f (a) = 0$ if and only if $x - a$ divides $f$ (Factor theorem over a commutative ring).

Proof

technique · direct

1.1

By [L1] the polynomial $t^{n} - 1$ is separable over $K$ and $μ_{n} (E)$ is cyclic of order $n$ ; so $t^{n} - 1$ has $n$ distinct roots in $E$ , namely the elements of $μ_{n} (E)$ , and $t^{n} - 1 = \prod_{ζ \in μ_{n} (E)} (t - ζ)$ by [L3].

L1L3

1.2

Each $ζ \in μ_{n} (E)$ has order dividing $n$ by [L5], and for a positive divisor $m$ of $n$ the condition $ζ^{m} = 1$ says exactly $ord (ζ) ∣ m$ ; so $μ_{m} (E)$ is the disjoint union of the $S_{d}$ over positive divisors $d$ of $m$ .

L5given

2.1

For every positive divisor $m$ of $n$ the polynomial $t^{m} - 1$ divides $t^{n} - 1$ in $Z [t]$ , since $t^{n} - 1 = (t^{m} - 1) (t^{n - m} + t^{n - 2 m} + \dots + 1)$ ; hence it splits over $E$ with distinct roots, which are the elements of $μ_{m} (E)$ , and $t^{m} - 1 = \prod_{ζ \in μ_{m} (E)} (t - ζ)$ with $∣ μ_{m} (E) ∣ = m$ .

step 1.1L3L4algebra

3.1

Consequently $\prod_{d ∣ m} Ψ_{d} = \prod_{ζ \in μ_{m} (E)} (t - ζ) = t^{m} - 1$ for every positive divisor $m$ of $n$ .

step 2.1step 1.2

4.1

For every positive divisor $d$ of $n$ the image of $Φ_{d}$ in $E [t]$ is $Ψ_{d}$ , by induction on $d$ through the divisors of $n$ : at $d = 1$ both are $t - 1$ , since $S_{1} = {1}$ ; and if the claim holds for every positive divisor $e$ of $d$ with $e < d$ , then [L2] and step 3.1 give $(\prod_{e ∣ d, e < d} Ψ_{e}) Φ_{d} = t^{d} - 1 = (\prod_{e ∣ d, e < d} Ψ_{e}) Ψ_{d}$ , and cancelling the nonzero left factor in the integral domain $E [t]$ ([L6]) gives $Φ_{d} = Ψ_{d}$ .

step 3.1L2L6

5.1

Taking $d = n$ : the image of $Φ_{n}$ in $E [t]$ is $\prod_{ζ \in S_{n}} (t - ζ)$ , so $Φ_{n}$ splits over $E$ and its roots there are exactly the elements of $S_{n}$ , which are the elements of order $n$ in $μ_{n} (E)$ , that is the primitive $n$ -th roots of unity in $E$ ; there are $φ (n)$ of them by [L1], in agreement with $\deg Φ_{n} = φ (n)$ from [L2].

step 4.1L1L2

6.1

$Φ_{n}$ is separable over $K$ . The product identity [L2] shows that the image of $Φ_{n}$ divides $t^{n} - 1$ in $K [t]$ . If an extension field $L / K$ contained an $a$ for which $(t - a)^{2}$ divided the image of $Φ_{n}$ , then the same square would divide the image of $t^{n} - 1$ in $L [t]$ , making $a$ a repeated root of $t^{n} - 1$ . This contradicts the separability of $t^{n} - 1$ supplied by [L1]. Thus [L4] applies.

L1L2L4algebra∎

Remarks

Both hypotheses are used, and neither can be dropped. If the characteristic divides $n$ then $t^{n} - 1$ is not separable and the roots of $Φ_{n}$ do not separate into orders at all; and the statement is about the roots in a splitting field, not in $K$ itself, since $Φ_{n}$ may have no root in $K$ at all, as $Φ_{3}$ over $Q$ shows.
Irreducibility is a separate question. The theorem says what the roots of $Φ_{n}$ are; it says nothing about whether $Φ_{n}$ factors over $K$ . Over $Q$ it does not ( $Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ ), over a finite field it usually does (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 $n$ ), and in both cases the root description above is the same.

Depends on

Used by

Dependency tree · two levels

92 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

P. L. Clark, Field Theory (course notes/monograph), Theorem 9.7 (standard reference, not scraped)
K. Conrad, Cyclotomic Extensions (expository blurb), Section 5 (standard reference, not scraped)

Over a field whose characteristic does not divide nn, the roots of ΦnΦn​ are exactly the primitive roots of unity