unity. This is exactly where cyclotomic polynomials come into play. Cyclotomic polynomials are the irreducible factors o
Gauss and Regular Polygons: Cyclotomic Polynomials | Paramanand's Math Notes
Gauss and Regular Polygons: Cyclotomic Polynomials Introduction The word "Cyclotomy" literally means "cutting a circle". So the subtitle of the post suggests that the post is going to be about some polynomials which are related to cutting a circle. Cutting a circle actually refers to dividing a given circle into a number of arcs of same length. Supposing that we are able to divide a given circle into, say n, arcs of equal length by means of points
P0 , P1 , … , Pn−1 then joining the adjacent points we obtain a regular polygon P0 P1 … Pn−1 of n sides. Therefore cyclotomic polynomials are somehow related to the construction of regular polygons.
In the previous post about complex numbers, we were able to establish that construction of a regular polygon of n sides requires finding complex numbers z such that z n = 1. Thus we need to solve the equation
zn − 1 = 0
(1)
One trivial solution is z = 1, but we are rather interested in solutions other than z = 1. All the solutions of the above equation are said to be nth roots of unity. So how do we go about finding these nth roots of unity? Obviously we need to factorize the expression z n − 1 and then find any linear factors if possible. For n = 4 we can proceed as follows:
z4 − 1 = 0 ⇒ (z 2 − 1)(z 2 + 1) = 0 ⇒ (z − 1)(z + 1)(z 2 + 1) = 0 ⇒ z = 1, z = −1, z 2 + 1 = 0 ⇒ z = 1, z = −1, z 2 = −1 ⇒ z = 1, z = −1, z = i, z = −i Therefore we see that factorization of the polynomial z n − 1 is the key to finding nth roots of unity. This is exactly where cyclotomic polynomials come into play. Cyclotomic polynomials are the irreducible factors of the polynomial z n − 1. More precisely we have the following definition: For any given positive integer n, the nth cyclotomic polynomial, denoted by Φn (z), is that irreducible factor (over field of rationals) of polynomial z n − 1 which is not a factor of any polynomial z m − 1 where m < n.
(Note to the not so casual reader: It will be proved later in this post that there will be one and only one such irreducible factor which is not the factor of any z m − 1 with m < n.) For n = 1, it is clear that Φ1 (z) = z − 1. For n = 2, we have
z 2 − 1 = (z − 1)(z + 1)
1
Gauss and Regular Polygons: Cyclotomic Polynomials | Paramanand's Math Notes
and since (z − 1) is already used up in factorization for case n = 1, Φ2 (z) = z + 1. For n = 3, z 3 − 1 = (z − 1)(z 2 + z + 1) and the factor z 2 + z + 1 is irreducible so that
Φ3 (z) = z 2 + z + 1. Similarly Φ4 (z) = z 2 + 1. Primitive nth Roots of Unity The complex number ζ = cos(2π/n) + i sin(2π/n) is obviously an nth root of unity since
ζ n = cos (n ⋅
2π 2π ) + i sin (n ⋅ ) = cos(2π) + i sin(2π) = 1 n n
Now consider the powers of ζ . For each k, 0 ≤ k < n, ζ k is also an nth root of unity as
(ζ k )n = (ζ n )k = 1k = 1. Also each of these powers of ζ is distinct from another, i.e. ζ l ≠ ζ k , 0 ≤ k, l < n if l ≠ k. Thus the numbers 1, ζ, ζ 2 , … , ζ n−1 are all the "n" nth roots of unity. Consider any such nth root say ζ k , 0 < k < n. Obviously we have
ζ k = cos (
2kπ 2kπ ) + i sin ( ) n n
If k and n have a common factor we can reduce the fraction k/n into another fraction, say
l/m, where m < n, l < k. Then it follows that ζ k = cos (
2lπ 2lπ ) + i sin ( ) m m
is also an mth root of unity for some m < n. Clearly we are not interested in such repeated roots (i.e. they occur as smaller roots of unity), but would rather like to focus on those nth roots of unity which are not the mth roots of unity for any m < n. These are obtained as ζ k where k and n do not have any common factor, i.e. when k is coprime to n. Such nth roots of unity are called the primitive nth roots of unity. What we have achieved above can be stated as follows: The primitive nth roots of unity are ζ k where 1 ≤ k < n, k is coprime to n and
ζ = cos (
2π 2π ) + i sin ( ) n n
The number of positive integers k < n and coprime to n is denoted by ϕ(n) and is called the Euler's totient function. Also ϕ(1) is defined to be 1. Thus the number of primitive nth roots of unity is ϕ(n). From elementary number theory the following are easily deduced: 1. ϕ(mn) = ϕ(m)ϕ(n) if m is coprime to n. 2. ϕ(pk ) = pk − pk−1 where p is a prime number. 3. ϕ(n) = n (1 −
1 1 1 ) (1 − ) ⋯ (1 − ) where p1 , p2 , … , pk are all the distinct p1 p2 pk prime factors of n.
Primitive nth Roots and Cyclotomic Polynomials Let us consider the polynomial
2
Gauss and Regular Polygons: Cyclotomic Polynomials | Paramanand's Math Notes
Pn (z) =
∏
(z − ζ k )
1≤k≤n,(k,n)=1
where (k, n) refers to the GCD (greatest common divisor) of k and n. Note that in the above expression ζ can be taken as any of ϕ(n) primitive roots (this is easy to establish if one understands the fact that the positive integers less than n and coprime to n form a group under modulo n multiplication). Thus the polynomial Pn (z) has as its roots all the primitive
nth roots of unity. We will now identify it with the cyclotomic polynomial Φn (z) by proving that Pn (z) is an irreducible polynomial with integral coefficients. We first establish that Pn (z) has integer coefficients. To do so we will establish the following:
z n − 1 = ∏ Pd (z)
(2)
d∣n
Since any nth root of unity is of the form
cos (
2kπ 2kπ ) + i sin ( ) n n
and we can replace k/n with a fraction c/d in lowest terms so that d ∣ n, it follows that any
nth root of unity is a primitive d th root of unity for some divisor d of n and vice-versa. Also if d1 and d2 are two distinct divisors of n then none of primitive d1th roots of unity is a primitive d2th root of unity. Thus the above factorization of (z n − 1) as a product of polynomials Pd (z) is proved.
The above factorization (2) can be used to prove that Pn (z) has integer coefficients. Clearly
P1 (z) = z − 1 and we can apply induction by assuming that Pk (z) has integer coefficients for k < n. Then we have Pn (z) =
zn − 1 zn − 1 = ∏ Pd (z) g(z) d∣n, d
