CorollaryStatement: AI-generatedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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For an odd prime $p$ , $Q (ζ_{p})$ has exactly one intermediate field of degree two over $Q$

Statement

Let $p$ be an odd prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) and let $ζ$ be a primitive $p$ -th root of unity (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity) in $Q (μ_{p})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ). Then there is exactly one intermediate field $F$ with

$Q \subseteq F \subseteq Q (ζ), [F : Q] = 2$

(The degree $[K : F] = \dim_{F} K$ of a finite field extension).

Intermediate, not proper. At $p = 3$ the Galois group has order two, the unique subgroup of index two is the trivial one, and the field it names is $Q (ζ_{3})$ itself. Reading the statement as "proper subfield" would make it false at the smallest case in scope.

Facts & Assumptions

Given: An odd prime $p$ and a primitive $p$ -th root of unity $ζ$ in the cyclotomic extension $Q (μ_{p}) = Q (ζ)$ ; write $G : = Gal (Q (μ_{p}) / Q)$ .

[L1]

$Q (μ_{p}) / Q$ is finite Galois ( $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ ) and the embedding $G \to (Z / p)^{\times}$ is an isomorphism ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ ).

[L2]

Every finite subgroup of the unit group of an integral domain is cyclic (Every finite subgroup of the unit group of an integral domain is cyclic); $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field, Field), hence an integral domain, and $(Z / p)^{\times}$ is its group of units (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L3]

$φ (p) = p - 1$ for a prime $p$ ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ ), and $φ (p) = ∣ (Z / p)^{\times} ∣$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L4]

In a cyclic group of finite order $m$ there is exactly one subgroup of each order dividing $m$ , and every subgroup has that form (A finite cyclic group has exactly one subgroup of each order dividing its own).

[L5]

For $K / F$ finite Galois with $G = Gal (K / F)$ , the maps $H \mapsto K^{H}$ and $E \mapsto Gal (K / E)$ are mutually inverse bijections between subgroups and intermediate fields, and $[K^{H} : F] = [G : H]$ (The fundamental theorem of finite Galois theory).

[L6]

For a finite group $G$ and $H \leq G$ , $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

By [L1] the group $G$ is isomorphic to $(Z / p)^{\times}$ , which is cyclic by [L2] and has order $p - 1$ by [L3]; so $G$ is cyclic of order $p - 1$ .

L1L2L3

2.1

Since $p$ is odd, $p - 1$ is even, so $2$ divides $p - 1$ and $(p - 1) / 2$ is a positive divisor of $p - 1$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ). By [L4] there is exactly one subgroup $H \leq G$ with $∣ H ∣ = (p - 1) / 2$ .

step 1.1L4given

2.2

By [L5] and [L6], an intermediate field $F$ of $Q (μ_{p}) / Q$ has $[F : Q] = [G : H] = ∣ G ∣ / ∣ H ∣$ for its corresponding subgroup $H$ , so $[F : Q] = 2$ holds exactly when $∣ H ∣ = (p - 1) / 2$ .

step 1.1L5L6

3.1

The correspondence of [L5] is a bijection, so the intermediate fields of degree two over $Q$ are in bijection with the subgroups of order $(p - 1) / 2$ , of which there is exactly one by step 2.1. Hence there is exactly one such field.

step 2.1step 2.2L5∎

Remarks

Which field it is, is a different question. The argument counts intermediate fields; it produces no generator of the one it counts, and no claim is made here about identifying it. Naming that field concretely requires a computation this proof does not carry out.
Oddness is needed. At $p = 2$ the field $Q (μ_{2})$ is $Q$ itself, of degree $φ (2) = 1$ , and it has no intermediate field of degree two at all; the step that fails is step 2.1, where $p - 1 = 1$ is odd.

Depends on

Used by

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Dependency tree · two levels

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Sources

K. Conrad, Cyclotomic Extensions (expository blurb), Section 4 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Chapter 5, cyclotomic extensions (standard reference, not scraped)

For an odd prime pp, Q(ζp)Q(ζp​) has exactly one intermediate field of degree two over QQ