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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

1.1

$μ_{n} (K)$ 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.

L2given

2.1

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].

step 1.1L1L3

3.1

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

step 2.1L3

3.2

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 $m = n$ .

step 1.1step 2.1L3

3.3

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 $K$ .

step 2.1L3

4.1

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 $φ (n)$ .

step 2.1step 3.2step 3.3L3L4∎

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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55 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, 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