A cyclotomic extension of a field F is a splitting field E for the polynomial

f(X) = X^{n} − 1

over F. The roots of f are called nth roots of unity. The n^{th} roots of unity form a multiplicative subgroup of the group E^{×} of non-zero elements of E, and thus must be cyclic. A primitive n^{th} root of unity is an n^{th} root of unity whose order in E^{×} is n. It is denoted ζ_{n}.

From now on, we will assume that we work in a characteristic char(F) such that char(F) does not divide n. (Otherwise, we have n = mchar(F) and 0 = ζ^{n}_{n} − 1 = (ζ^{m} − 1)^{char(F)} and the order of ζ^{n} is less than n.)

The n^{th} cyclotomic polynomial is defined by

Ψ_{n}(X) = ∏_{(i,n)} (X−ζ_{n}^{i})

where the product is taken over all primitive n^{th} roots of unity in C.

The degree of Ψ_{n}(X) is thus deg(Ψ_{n}) = φ(n)

We have

X^{n} −1 = ∏_{d|n} Ψ_{d}(X)

In particular, if n = p a prime, then d is either 1 or p and

X^{p} − 1 = Ψ_{1}(X)Ψ_{p}(X) = (X − 1)Ψ_{p}(X)

Ψ_{p}(X) = (X^{p} −1)/(X−1) = X^{p−1} +X^{p−2} +···+ X + 1

Proof:

Equality is proved by comparing the roots of both monic polynomials. If ζ is a nth root of unity, then by definition ζ^{n}_{n} = 0 and its order d divides n. Thus ζ is actually a primitive dth root of unity, and a root of Ψ_{d}(X). Conversely, if d|n, then any root of Ψ_{d}(X) is a d^{th} root hence a n^{th} root of unity.

The nth cyclotomic polynomial Ψn(X) satisfies

Ψ_{n}(X) ∈ Z[X].

Proof:

We proceed by induction on n. It is true for n = 1 since X − 1 ∈ Z[X]. Let us suppose it is true for Ψ_{k}(X) where k is up to n−1, and prove it is also true for n. Using the above proposition, we know that

X^{n} −1 = ∏_{d|n} Ψ_{d}(X)

_{n}(X) ∏

_{d|n,d<n}Ψ

_{d}(X).

_{n}(X) ∈ Z[X] :

_{n}(X) = (X

^{n}−1)/∏

_{d|n,d<n}Ψ

_{d}(X)

_{n}(X) has to be monic (by definition), and both X

^{n}− 1 and Ψ

_{d}(X) (by induction hypothesis) are in Z[X]. We can thus conclude invoking the division algorithm for polynomials in Z[X].