PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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$μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$

Statement

Let $K$ be a field and $n \geq 1$ . Then $μ_{n} (K) = {x \in K : x^{n} = 1}$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity) is a finite cyclic subgroup of $K^{\times}$ whose order divides $n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ). It contains a primitive $n$ -th root of unity if and only if $∣ μ_{n} (K) ∣ = n$ , and in that case the primitive $n$ -th roots of unity in $K$ are exactly the generators of $μ_{n} (K)$ , of which there are $φ (n)$ (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 $n \geq 1$ , and the subgroup $μ_{n} (K)$ of $K^{\times}$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

[L1]

Let $D$ be an integral domain. Every finite subgroup $G \leq D^{\times}$ of the unit group of $D$ is cyclic (Every finite subgroup of the unit group of an integral domain is cyclic).

[L2]

A nonzero polynomial of degree $k$ over an integral domain has at most $k$ distinct roots in that domain (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

[L3]

For an element $x$ of finite order $m$ in a group: $x^{k} = e$ if and only if $m ∣ k$ ; and $∣ ⟨ x ⟩ ∣ = m = ord (x)$ (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).

[L4]

In a cyclic group $⟨ g ⟩$ of finite order $m$ , the element $g^{a}$ generates the group if and only if $\gcd (a, m) = 1$ , and the group has exactly $φ (m)$ generators (A cyclic group of order $n$ has exactly $φ (n)$ generators).

Proof

technique · direct

1.1L2given

$μ_{n} (K)$ is the set of roots in $K$ of the nonzero polynomial $t^{n} - 1$ , of degree $n$ ; a field is an integral domain, so $∣ μ_{n} (K) ∣ \leq n$ by [L2] and $μ_{n} (K)$ is finite.

2.1step 1.1L1L3

Being a finite subgroup of $K^{\times}$ , $μ_{n} (K)$ is cyclic by [L1]; write $m : = ∣ μ_{n} (K) ∣$ and fix a generator $ζ_{0}$ , so that $ord (ζ_{0}) = m$ by [L3].

3.1step 2.1L3

The order $m$ divides $n$ : $ζ_{0} \in μ_{n} (K)$ gives $ζ_{0}^{n} = 1$ , and [L3] turns this into $m ∣ n$ .

3.2step 1.1step 2.1L3

If $K$ contains a primitive $n$ -th root of unity $ζ$ , that is an element of order $n$ , then $ζ^{n} = 1$ puts $ζ$ in $μ_{n} (K)$ , and $⟨ ζ ⟩ \subseteq μ_{n} (K)$ has $n$ elements by [L3], so $n \leq m$ ; with step 1.1 this forces $m = n$ .

3.3step 2.1L3

Conversely, if $m = n$ then the generator $ζ_{0}$ of step 2.1 has order $n$ and so is a primitive $n$ -th root of unity in $K$ .

4.1step 2.1step 3.2step 3.3L3L4∎

Suppose $m = n$ . An element $x \in K$ of order $n$ lies in $μ_{n} (K) = ⟨ ζ_{0} ⟩$ and satisfies $∣ ⟨ x ⟩ ∣ = n = ∣ μ_{n} (K) ∣$ by [L3], so $⟨ x ⟩ = μ_{n} (K)$ and $x$ is a generator; conversely a generator has order $n$ by [L3]. So the primitive $n$ -th roots of unity in $K$ are exactly the generators of $μ_{n} (K)$ , and [L4] counts them as $φ (n)$ .

Remarks

Order dividing $n$ , not equal to $n$ . The two failures are different in kind. Over $Q$ the polynomial $t^{3} - 1$ simply does not split, so $μ_{3} (Q) = {1}$ ; over a field of characteristic three it cannot split into distinct factors at all, since $t^{3} - 1 = (t - 1)^{3}$ there (In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ ). Only the first failure is repaired by passing to a splitting field.

Depends on

Used by

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Sources

K. Conrad, Cyclotomic Extensions (expository blurb), Theorem 1.1 (standard reference, not scraped)
P. L. Clark, Field Theory (course notes/monograph), Lemmas 9.1 and 9.2 (standard reference, not scraped)

μn(K)μn​(K) is cyclic of order dividing nn, and has a primitive nn-th root of unity exactly when its order is nn