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$Φ_{n}$ is irreducible over $K$ exactly when $[K (ζ_{n}) : K] = φ (n)$ , exactly when the embedding into $(Z / n)^{\times}$ is onto

Statement

Let $K$ be a field and $n \geq 1$ with $char K ∤ n$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise), let $E = K (μ_{n})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ), and let $ζ \in E$ 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). The following are equivalent.

The image of $Φ_{n}$ in $K [t]$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ) is irreducible (Irreducible and prime elements of an integral domain).
$[K (ζ) : K] = φ (n)$ (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 $n \geq 1$ ).
The embedding $Gal (E / K) \to (Z / n)^{\times}$ 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: A field $K$ , an integer $n \geq 1$ with $char K ∤ n$ , the extension $E = K (μ_{n})$ , a primitive $n$ -th root of unity $ζ \in E$ , and the minimal polynomial $m_{ζ} \in K [t]$ of $ζ$ over $K$ .

[L1]

The image of $Φ_{n}$ in $K [t]$ has as its roots in $E$ exactly the primitive $n$ -th roots of unity in $E$ (Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity).

[L2]

$Φ_{n}$ is monic of degree $φ (n)$ (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree); reduction into $K [t]$ preserves both.

[L3]

For $a$ algebraic over $K$ there is a unique monic irreducible $m_{a} \in K [t]$ with $f (a) = 0$ if and only if $m_{a} ∣ f$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[L4]

If $a$ is algebraic over $K$ with minimal polynomial of degree $m$ , then $[K (a) : K] = m$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[L5]

For this $n \geq 1$ , $E / K$ is finite Galois and $σ \mapsto [a_{σ}]_{n}$ is an injective homomorphism $Gal (E / K) \to (Z / n)^{\times}$ ( $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ ); moreover $E = K (ζ)$ for every primitive $n$ -th root $ζ$ ( $t^{n} - 1$ 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).

[L6]

$∣ (Z / n)^{\times} ∣ = φ (n)$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ), and $∣ Gal (E / K) ∣ = [E : K]$ for a finite Galois extension (Equivalent characterizations of a finite Galois extension).

[L7]

In an integral domain, an irreducible element is a nonzero nonunit every one of whose factorisations has a unit factor (Irreducible and prime elements of an integral domain).

Proof

technique · direct

1.1

$ζ$ is a root of the image of $Φ_{n}$ in $K [t]$ by [L1], so $m_{ζ}$ divides that image by [L3]; both are monic, and $m_{ζ}$ has positive degree because $ζ \neq 0$ is not a root of a nonzero constant. Write the image of $Φ_{n}$ as $m_{ζ} g$ with $g \in K [t]$ monic.

L1L2L3

1.2

For the equivalence of clauses 2 and 3: by [L5] one has $E = K (ζ)$ and the embedding is injective into a group of order $φ (n)$ by [L6], so it is surjective if and only if $∣ Gal (E / K) ∣ = φ (n)$ ; and $∣ Gal (E / K) ∣ = [E : K] = [K (ζ) : K]$ by [L6]. An injective homomorphism onto its target is an isomorphism.

L5L6

2.1

For the implication from clause 1 to clause 2: if the image of $Φ_{n}$ is irreducible, then in the factorisation of step 1.1 one factor is a unit by [L7], and $m_{ζ}$ is not, so $g$ is a nonzero constant; both $m_{ζ} g$ and $m_{ζ}$ being monic forces $g = 1$ and $m_{ζ} = Φ_{n}$ . Hence $[K (ζ) : K] = \deg m_{ζ} = φ (n)$ by [L2] and [L4].

step 1.1L2L4L7

2.2

For the implication from clause 2 to clause 1: if $[K (ζ) : K] = φ (n)$ then $\deg m_{ζ} = φ (n) = \deg Φ_{n}$ by [L2] and [L4], so $g$ in step 1.1 is monic of degree $0$ , that is $g = 1$ and the image of $Φ_{n}$ equals $m_{ζ}$ , which is irreducible by [L3].

step 1.1L2L3L4

3.1

Steps 2.1 and 2.2 give the equivalence of clauses 1 and 2, and step 1.2 the equivalence of clauses 2 and 3; so all three are equivalent.

step 2.1step 2.2step 1.2∎

Remarks

This is a criterion, not a theorem about $Q$ . Over $Q$ all three clauses hold for every $n$ ( $Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ ), but over $F_{q}$ they hold only when the class of $q$ generates $(Z / n)^{\times}$ (The reduction of $Φ_{n}$ is irreducible over $F_{q}$ exactly when $[q]$ generates $(Z / n)^{\times}$ ).

Depends on

Used by

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

Dependency tree · two levels

81 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

J. S. Milne, Fields and Galois Theory, v5.10, Lemma 5.9 (standard reference, not scraped)
K. Conrad, Cyclotomic Extensions (expository blurb), Section 2 (standard reference, not scraped)

ΦnΦn​ is irreducible over KK exactly when [K(ζn):K]=φ(n)[K(ζn​):K]=φ(n), exactly when the embedding into (Z/n)×(Z/n)× is onto