Cyclotomic Polynomial

The cyclotomic polynomial, denoted by Φ_n(x), is a unique monic polynomial whose roots are the primitive n^th roots of unity. These roots, represented as ζ, satisfy ζⁿ = 1 and ζ^k ≠ 1 for any positive integer k < n. 

Mathematically, the n^th cyclotomic polynomial is given by:

${\Phi_n(x) = \prod_{\substack{1 \leq k \leq n \\ \gcd(k, n) = 1}} \left( x – e^{\frac{2\pi i k}{n}} \right)}$

Here, 

• k and n are coprime integers
• ${e^{\dfrac{2\pi ik}{n}}}$ are the roots of Φ_n(x)

A few examples of cyclotomic polynomials are:

• Φ₁(x) = x – 1 (degree 1)
• Φ₂(x) = x + 1 (degree 1)
• Φ₃(x) = x² + x + 1 (degree 2)
• Φ₄(x) = x² + 1 (degree 2)
• Φ₅(x) = x⁴ + x³ + x² + x + 1 (degree 4)
• Φ₆(x) = x² – x + 1 (degree 2)

Note: For a prime p, the cyclotomic polynomial simplifies to:

Φ_p(x) = ${\dfrac{x^{p}-1}{x-1}}$ = x^{p – 1} + x^{p – 2} + … + x + 1

This is a geometric series summing all p^th roots of unity except 1

Cyclotomic polynomials are widely used in number theory, algebra, and in constructing field extensions, analyzing roots of unity, modular arithmetic, cryptography, coding theory, and determining the constructibility of regular polygons.

Properties

Recursive Factorization

Statement

Cyclotomic polynomials fit into the broader structure of polynomial factorization. They satisfy the recursive relation for any n: 

${x^n – 1 = \prod_{d \mid n} \Phi_d(x)}$, 

Here, 

• Φ_d(x) represents the d^th cyclotomic polynomial
• d runs over all divisors of n 

Proof 

Let us consider that r is a root of Φ_d(x) for all d ∣ n

Thus, r is the d^th root of unity.

Let q be the integer such that n = d · q, then

rⁿ = (r^d)^q = 1^q = 1

Thus, r is a root of xⁿ – 1

Since the roots of xⁿ – 1 are the n^th roots of unity, r is so.

Since d is the order of r, it will be a primitive d^th root of unity. 

Thus, r is a root of Φ_d(x). 

Since the n^th root of unity forms a group of n elements for xⁿ – 1 and d ∣ n, the roots of order d are the roots of Φ_d(x) for some d that divides n

Thus, roots of xⁿ – 1 include all n^th roots of unity, which can be expressed as the product of Φ_d(x) for all d ∣ n

Irreducibility

Statement

The cyclotomic polynomial Φ_n(x) is irreducible over the integers. This means that they cannot be factored into polynomials of lower degree with rational coefficients.

Proof

Let us assume that f(x) ∈ ℤ[x] is a monic, irreducible factor of Φ_n(x)

Since Φ_n(x) divides xⁿ – 1 in ℤ[x], there exists a polynomial g(x) ∈ ℤ[x] such that 

xⁿ – 1 = Φ_n(x)g(x) and Φ_n(x) = f(x)g(x) in ℤ[x]

Let r be a primitive n^th root of unity that is a root of f(x)

Let p be a prime such that p ∤ n, indicating gcd (p, n) = 1

By properties of roots of unity, r^p is also a primitive n^th root of unity.

Thus, r^p is a root of Φ_n(x), thus a root of either f(x) or g(x) 

If f(r^p) ≠ 0, g(r^p) = 0, which implies r is a root of g(x^p)

By definition, f(x) is monic and irreducible, and f(x) is the minimal polynomial of r over ℚ.

Thus, f(x) divides g(x^p) in ℚ[x].

Now, reducing f(x), g(x), and g(x^p) modulo p to obtain polynomials ${\overline{f}\left( x\right)}$, ${\overline{g}\left( x\right)}$, and ${\overline{g}\left( x^{p}\right)}$ in ℤ_p[x]

Using properties of reduction, we get

${\overline{g}\left( x^{p}\right)}$ = ${\overline{f}\left( x\right) \overline{h}\left( x\right)}$ in ℤ_p[x]

Since ℤ_p[x] is a unique factorization domain, ${\overline{f}\left( x\right)}$ and ${\overline{g}\left( x\right)}$ share a common irreducible factor, denoted as k(x)

Consequently, xⁿ – 1 in ℤ_p[x] has multiple roots, contradicting the fact that xⁿ – 1 and its derivative nx^{n – 1} are coprime (as p ∤ n)

The assumption f(r^p) ≠ 0 leads to a contradiction. Thus, f(r^p) = 0, meaning r^p is also a root of f(x)

Repeating this argument for all primitive n^th roots of unity ζ, we conclude that every root of Φ_n(x) is also a root of f(x)

Since f(x) is a monic, irreducible, and shares all roots with Φ_n(x), it follows that f(x) = Φ_n(x)

Thus, Φ_n(x) is irreducible of ℤ

Note: The cyclotomic polynomials Φ_n(x) are irreducible over ℚ.

By Gauss’s Lemma, if a polynomial is irreducible in ℤ[x], it is also irreducible in ℚ[x]. 

Thus, Φ_n(x) remains irreducible over ℚ.

Degree of Φ_n(x)

Statement

The degree of a cyclotomic polynomial Φ_n(x) is given by Euler’s totient function, φ(n), which counts the integers less than n and are coprime to n 

Proof 

Since ${\Phi_n(x) = \prod_{\substack{1 \leq k \leq n \\ \gcd(k, n) = 1}} \left( x – e^{\frac{2\pi i k}{n}} \right)}$ is a product of linear factors , every x terms in Φ_n(x) has a coefficient of 1.

As a result, when these linear factors are multiplied out, the x term with the largest exponent has a coefficient of 1.

Furthermore, the degree of a polynomial equals the number of its roots.

Since Φ_n(x) is monic and irreducible, the degree of Φ_n(x) will be the number of integers, k, such that 1 ≤ k ≤ n and gcd(k, n) = 1.

By definition, this is φ(n).

Integer Coefficients

Statement

For any positive integer n, the n^th cyclotomic polynomial Φ_n(x) has integer coefficients.

Proof

From Gauss’s Lemma, we conclude that if a polynomial with integer coefficients can be factored over ℚ, it must have integer coefficients. 

Since Φ_n(x) divides xⁿ – 1, which has integer coefficients, Φ_n(x) also has integer coefficients (i.e., Φ_n(x) ∈ ℤ[x]).

Relation to Galois Groups

Statement

The Galois group of Φ_n(x) over ℚ is isomorphic to (ℤ / nℤ)*, the group of units modulo n 

Proof  

The splitting field of Φ_n(x) is ℚ(ζ_n), where ζ_n is a primitive n^th root of unity. Automorphisms of the field correspond to multiplication by units modulo n, establishing the isomorphism.

Thus, the Galois group of the n^th cyclotomic field ℚ(ζ_n) is isomorphic to (ℤ / nℤ)*

Solved Example