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

Statement

For every $n \geq 1$ the cyclotomic polynomial $Φ_{n}$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ) is irreducible in $Q [t]$ (Irreducible and prime elements of an integral domain).

Facts & Assumptions

Given: An integer $n \geq 1$ ; $Q$ is an ordered field (The rationals form a totally ordered field), so $m \cdot 1 > 0$ and in particular $m \cdot 1 \neq 0$ for $m \geq 1$ , whence $char Q = 0$ (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 $q$ ); a splitting field $E$ of $t^{n} - 1$ over $Q$ (Every nonzero polynomial over a field has a splitting field); a primitive $n$ -th root of unity $ζ \in E$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity); and $f \in Q [t]$ the minimal polynomial of $ζ$ over $Q$ .

[L1]

$t^{n} - 1$ is separable over $Q$ and $μ_{n} (E)$ is cyclic of order $n$ with exactly $φ (n)$ primitive $n$ -th roots of unity, which are its generators ( $t^{n} - 1$ 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 unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L2]

The roots of $Φ_{n}$ in $E$ are exactly the primitive $n$ -th roots of unity in $E$ ; $Φ_{n}$ is monic of degree $φ (n)$ (Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity, 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).

[L3]

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

[L4]

If $ξ$ is a primitive $n$ -th root of unity in $E$ , $g \in Q [t]$ is its minimal polynomial and $p$ is a prime with $p ∤ n$ , then $g (ξ^{p}) = 0$ (If $p$ is a prime not dividing $n$ , a rational minimal polynomial of a primitive $n$ -th root of unity also kills its $p$ -th power).

[L5]

Every integer $a \geq 1$ is a finite product of primes, the empty product being $1$ (The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ , Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L6]

In a cyclic group $⟨ g ⟩$ of finite order $m$ , $g^{a}$ generates the group if and only if $\gcd (a, m) = 1$ , that is when $a$ and $m$ are coprime (A cyclic group of order $n$ has exactly $φ (n)$ generators, Coprime integers: $\gcd (a, b) = 1$ ).

[L7]

A nonzero polynomial of degree $k$ over an integral domain has at most $k$ distinct roots in it (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

Proof

technique · direct

1.1

$ζ$ is a root of $Φ_{n}$ by [L2], so $f ∣ Φ_{n}$ in $Q [t]$ by [L3], and $f$ is monic and irreducible.

L2L3

1.2

Let $ξ \in E$ be any primitive $n$ -th root of unity. By [L1] it generates $μ_{n} (E) = ⟨ ζ ⟩$ , so $ξ = ζ^{a}$ for some integer $a$ , which may be taken with $a \geq 1$ after adding a multiple of $n$ ; and $\gcd (a, n) = 1$ by [L6]. Write $a = p_{1} p_{2} \dots p_{s}$ as a product of primes by [L5], with $s = 0$ when $a = 1$ . No $p_{i}$ divides $n$ , since $p_{i} ∣ a$ and $\gcd (a, n) = 1$ .

L1L5L6given

2.1

Put $ζ_{0} : = ζ$ and $ζ_{j} : = ζ_{j - 1}^{p_{j}}$ for $1 \leq j \leq s$ , so that $ζ_{s} = ζ^{a} = ξ$ . By induction on $j$ , each $ζ_{j}$ is a primitive $n$ -th root of unity and $f (ζ_{j}) = 0$ : at $j = 0$ this is the hypothesis on $ζ$ and step 1.1; and given it at $j - 1$ , the polynomial $f$ is monic irreducible and vanishes at the primitive $n$ -th root of unity $ζ_{j - 1}$ , so $f$ is the minimal polynomial of $ζ_{j - 1}$ by [L3], whence $f (ζ_{j}) = f (ζ_{j - 1}^{p_{j}}) = 0$ by [L4], while $ζ_{j}$ is primitive by [L6] because $\gcd (p_{j}, n) = 1$ .

step 1.1step 1.2L3L4L6

3.1

So $f$ vanishes at every primitive $n$ -th root of unity in $E$ , of which there are $φ (n)$ distinct ones by [L1]; hence $\deg f \geq φ (n)$ by [L7].

step 2.1L1L7

4.1

On the other hand $f ∣ Φ_{n}$ with $\deg Φ_{n} = φ (n)$ by [L2], so $\deg f \leq φ (n)$ ; therefore $\deg f = φ (n)$ , and $f$ and $Φ_{n}$ are monic with $f ∣ Φ_{n}$ , so $Φ_{n} = f$ is irreducible. At $n = 1$ this reads $Φ_{1} = t - 1$ , of degree $φ (1) = 1$ .

step 1.1step 3.1L2L3∎

Remarks

What the prime factorisation is doing. The lemma advances one prime at a time, and step 2.1 chains those advances along a factorisation of the exponent. The chain works only because each intermediate $ζ_{j}$ is again primitive and has the same minimal polynomial, which is what lets the lemma be reapplied rather than merely applied once.
Over other base fields the theorem is false. Irreducibility of $Φ_{n}$ depends on the base field, and over a finite field it usually fails (For $\gcd (n, q) = 1$ the reduction of $Φ_{n}$ in $F_{q} [t]$ is a product of distinct monic irreducibles, each of degree the order of $[q]$ modulo $n$ ). The criterion that separates the cases is $Φ_{n}$ is irreducible over $K$ exactly when $[K (ζ_{n}) : K] = φ (n)$ , exactly when the embedding into $(Z / n)^{\times}$ is onto.

Depends on

Used by

[ℚ(ζₙ):ℚ]=φ(n) and Gal(ℚ(μₙ)/ℚ)≅(ℤ/n)^× Corollary
FALSE: Φₙ is irreducible over every field False statement

Dependency tree · two levels

108 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, Theorem 5.10 (standard reference, not scraped)
P. L. Clark, Field Theory (course notes/monograph), Theorem 9.8 (standard reference, not scraped)
K. Conrad, Cyclotomic Extensions (expository blurb), Theorem 2.5 (standard reference, not scraped)

ΦnΦn​ is irreducible in Q[t]Q[t] for every n≥1n≥1