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$[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$

Statement

Let $n \geq 1$ and let $ζ$ be a primitive $n$ -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 $Q (μ_{n})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ). Then

$[Q (ζ) : Q] = φ (n)$

(The degree $[K : F] = \dim_{F} K$ of a finite field extension, The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ), and the embedding

$Gal (Q (μ_{n}) / Q) ⟶ (Z / n)^{\times}$

of $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ is an isomorphism.

Facts & Assumptions

Given: An integer $n \geq 1$ ; $Q$ is an ordered field (The rationals form a totally ordered field), so $char Q = 0$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise) and divides no $n \geq 1$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L1]

$Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ ( $Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ ).

[L2]

For a field $K$ with $char K ∤ n$ and a primitive $n$ -th root of unity $ζ$ in a splitting field, irreducibility of the image of $Φ_{n}$ in $K [t]$ , the equality $[K (ζ) : K] = φ (n)$ , and surjectivity of the embedding $Gal (K (μ_{n}) / K) \to (Z / n)^{\times}$ are equivalent ( $Φ_{n}$ is irreducible over $K$ exactly when $[K (ζ_{n}) : K] = φ (n)$ , exactly when the embedding into $(Z / n)^{\times}$ is onto).

Proof

technique · direct

1.1

Since $char Q = 0$ does not divide $n$ , [L2] applies with $K = Q$ .

L2given

2.1

The image of $Φ_{n}$ in $Q [t]$ is $Φ_{n}$ itself, irreducible by [L1]; so the first clause of [L2] holds, and therefore so do the other two: $[Q (ζ) : Q] = φ (n)$ , and the embedding is onto, hence an isomorphism, being injective.

step 1.1L1L2∎

Remarks

The isomorphism is canonical. It sends $σ$ to the class of the exponent $a_{σ}$ with $σ (ζ) = ζ^{a_{σ}}$ , and $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ shows that class does not depend on which primitive $n$ -th root of unity is chosen. So every subgroup of $(Z / n)^{\times}$ names an intermediate field of $Q (μ_{n}) / Q$ without any choice being made.

Depends on

Used by

Dependency tree · two levels

64 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), Corollary 9.9 (standard reference, not scraped)
K. Conrad, Cyclotomic Extensions (expository blurb), Theorem 2.5 (standard reference, not scraped)

[Q(ζn):Q]=φ(n)[Q(ζn​):Q]=φ(n) and Gal⁡(Q(μn)/Q)≅(Z/n)×Gal(Q(μn​)/Q)≅(Z/n)×