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

f(X) = Xn − 1

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

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 = ζnn − 1 = (ζm − 1)char(F) and the order of ζn is less than n.)

The nth cyclotomic polynomial is defined by

Ψn(X) = ∏(i,n) (X−ζni)

where the product is taken over all primitive nth roots of unity in C.

The degree of Ψn(X) is thus deg(Ψn) = φ(n)

We have

Xn −1 = ∏d|n Ψd(X)

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

Xp − 1 = Ψ1(X)Ψp(X) = (X − 1)Ψp(X)

Ψp(X) = (Xp −1)/(X−1) = Xp−1 +Xp−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 ζnn = 0 and its order d divides n. Thus ζ is actually a primitive dth root of unity, and a root of Ψd(X). Conversely, if d|n, then any root of Ψd(X) is a dth root hence a nth root of unity.

The nth cyclotomic polynomial Ψn(X) satisfies

Ψn(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 Ψk(X) where k is up to n−1, and prove it is also true for n. Using the above proposition, we know that

Xn −1 = ∏d|n Ψd(X)

= Ψn(X) ∏d|n,d<n Ψd(X).
The aim is to prove that
Ψn(X) ∈ Z[X] :
Ψn(X) = (Xn −1)/∏d|n,d<n Ψd(X)
First note that Ψn(X) has to be monic (by definition), and both Xn − 1 and Ψd(X) (by induction hypothesis) are in Z[X]. We can thus conclude invoking the division algorithm for polynomials in Z[X].