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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$K (μ_{m}) K (μ_{n}) = K (μ_{lcm (m, n)})$

Statement

Let $K$ be a field and let $m, n \geq 1$ be integers such that $char K$ divides neither $m$ nor $n$ (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 $q$ ). Put $ℓ : = lcm (m, n)$ (Common multiple, and the least common multiple $lcm (a, b)$ , taken to be $0$ when $a = 0$ or $b = 0$ ) and let $Ω$ be a splitting field of $t^{ℓ} - 1$ over $K$ (Every nonzero polynomial over a field has a splitting field). Then $char K ∤ ℓ$ ; the subfields $K (μ_{m} (Ω))$ and $K (μ_{n} (Ω))$ of $Ω$ are cyclotomic extensions of $K$ of orders $m$ and $n$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ); and their compositum inside $Ω$ is

$K (μ_{m}) K (μ_{n}) = K (μ_{ℓ}) = Ω .$

Facts & Assumptions

Given: A field $K$ , integers $m, n \geq 1$ with $char K$ dividing neither, $ℓ = lcm (m, n)$ , and a splitting field $Ω$ of $t^{ℓ} - 1$ over $K$ ; the characteristic of a field is $0$ or a prime (The characteristic of a field is zero or a prime number), and $0$ divides no positive integer.

[L1]

$lcm (a, b)$ is a common multiple of $a$ and $b$ (Common multiple, and the least common multiple $lcm (a, b)$ , taken to be $0$ when $a = 0$ or $b = 0$ ); every common multiple of $a$ and $b$ is a multiple of $lcm (a, b)$ , and $\gcd (a, b) lcm (a, b) = ∣ a b ∣$ (Every common multiple of $a$ and $b$ is a multiple of $lcm (a, b)$ , and $\gcd (a, b) \cdot lcm (a, b) = ∣ a b ∣$ ).

[L2]

If a prime $p$ divides $a b$ then $p ∣ a$ or $p ∣ b$ (Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ , Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L3]

$t^{k} - 1$ is separable over $K$ exactly when $char K ∤ k$ , and then a splitting field $E$ has $μ_{k} (E)$ cyclic of order $k$ and $E = K (ζ)$ for any primitive $k$ -th root 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 $n$ , The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

[L4]

A polynomial is separable over $K$ when no extension field contains a repeated root (Repeated roots in extension fields and separable polynomials); a splitting field is generated over $K$ by the roots (Polynomials that split and splitting fields of a polynomial or a family of polynomials, Finitely generated field extensions $F (a_{1}, \dots, a_{r})$ ).

[L5]

For a finite group $G$ and $H \leq G$ , $∣ H ∣$ divides $∣ G ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ); $⟨ S ⟩$ is the smallest subgroup containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups); 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 $j = k$ , 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).

Proof

technique · direct

1.1

$char K ∤ ℓ$ . If $char K = 0$ this is immediate; if $char K = p$ is a prime dividing $ℓ$ , then $ℓ$ divides $m n$ because $\gcd (m, n) ℓ = m n$ by [L1] and $\gcd (m, n) \geq 1$ , so $p ∣ m n$ and [L2] gives $p ∣ m$ or $p ∣ n$ , contrary to hypothesis.

L1L2given

2.1

By [L3] and step 1.1 the polynomial $t^{ℓ} - 1$ is separable over $K$ , the group $μ_{ℓ} (Ω)$ is cyclic of order $ℓ$ , and $Ω = K (μ_{ℓ} (Ω))$ .

step 1.1L3L4

3.1

For every positive divisor $k$ of $ℓ$ one has $∣ μ_{k} (Ω) ∣ = k$ and $K (μ_{k} (Ω))$ is a splitting field of $t^{k} - 1$ over $K$ , hence a cyclotomic extension of $K$ of order $k$ : indeed $t^{k} - 1$ divides $t^{ℓ} - 1$ , since $t^{ℓ} - 1 = (t^{k} - 1) (t^{ℓ - k} + t^{ℓ - 2 k} + \dots + 1)$ , so it splits over $Ω$ , and a repeated root of it would be a repeated root of $t^{ℓ} - 1$ , excluded by step 2.1 through [L4]; so its $k$ roots are distinct and they are the elements of $μ_{k} (Ω)$ , which generate $K (μ_{k} (Ω))$ over $K$ .

step 2.1L3L4algebra

4.1

$m$ and $n$ are positive divisors of $ℓ$ by [L1], so step 3.1 applies to both: $K (μ_{m} (Ω))$ and $K (μ_{n} (Ω))$ are cyclotomic extensions of $K$ of orders $m$ and $n$ , and each contains a primitive root of unity of its order by [L3].

step 3.1L1L3

5.1

Both are contained in $Ω = K (μ_{ℓ} (Ω))$ , since $μ_{m} (Ω)$ and $μ_{n} (Ω)$ are subsets of $μ_{ℓ} (Ω)$ by $m ∣ ℓ$ and $n ∣ ℓ$ ; hence their compositum inside $Ω$ is contained in $K (μ_{ℓ})$ .

step 2.1step 4.1L1

5.2

For the reverse inclusion, fix a primitive $m$ -th root of unity $ζ_{m} \in μ_{m} (Ω)$ and a primitive $n$ -th root of unity $ζ_{n} \in μ_{n} (Ω)$ , and let $H : = ⟨ ζ_{m}, ζ_{n} ⟩ \leq μ_{ℓ} (Ω)$ . By [L5] the orders $m = ∣ ⟨ ζ_{m} ⟩ ∣$ and $n = ∣ ⟨ ζ_{n} ⟩ ∣$ both divide $∣ H ∣$ , so $∣ H ∣$ is a common multiple of $m$ and $n$ and therefore a multiple of $ℓ$ by [L1]; and $∣ H ∣$ divides $ℓ$ by [L5]. Hence $∣ H ∣ = ℓ$ and $H = μ_{ℓ} (Ω)$ .

step 2.1step 4.1L1L5

6.1

Both $ζ_{m}$ and $ζ_{n}$ lie in the compositum $K (μ_{m}) K (μ_{n})$ , which is a field, so $H \subseteq K (μ_{m}) K (μ_{n})$ and therefore $μ_{ℓ} (Ω) \subseteq K (μ_{m}) K (μ_{n})$ by step 5.2; hence $Ω = K (μ_{ℓ} (Ω)) \subseteq K (μ_{m}) K (μ_{n})$ . With step 5.1 this gives $K (μ_{m}) K (μ_{n}) = K (μ_{ℓ}) = Ω$ .

step 2.1step 5.1step 5.2∎

Remarks

The intersection is not the mirror image of this. The compositum identity holds over every base field of admissible characteristic, but the corresponding identity $K (μ_{m}) \cap K (μ_{n}) = K (μ_{\gcd (m, n)})$ does not: it is proved here only over $Q$ ( $Q (μ_{m}) \cap Q (μ_{n}) = Q (μ_{\gcd (m, n)})$ ), and the companion page gives a finite-base-field failure in $F_{3} (μ_{5}) \cap F_{3} (μ_{7})$ is larger than $F_{3}$ although five and seven are coprime ↗.

Depends on

Used by

ℚ(μₘ)∩ℚ(μₙ)=ℚ(μ_gcd(m,n)) Theorem

Dependency tree · two levels

91 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), Section 3 (standard reference, not scraped)
P. L. Clark, Field Theory (course notes/monograph), Chapter 9, Section 1 (standard reference, not scraped)

K(μm)K(μn)=K(μlcm⁡(m,n))K(μm​)K(μn​)=K(μlcm(m,n)​)