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
where g(z) is a polynomial with integer coefficients and leading coefficient 1 (by induction hypothesis). It follows by Gauss Lemma that Pn (z) has integer coefficients. We next establish that Pn (z) is irreducible over the field of rationals. The proof which follows is the one provided by Gauss and it uses modular arithmetic in very ingenious way. We will summarize the results needed as follows: 1. For a given prime p, the numbers 0, 1, 2, … , (p − 1) form a finite field under the operations of modulo addition and multiplication modulo p. 2. Since these numbers form a field, say Fp , we can talk about polynomials f(z) whose coefficients are in Fp . The set of all such polynomials, say Fp [z], has the unique factorization property i.e. any such polynomial can be factored as a product of irreducible polynomials in Fp [z] in a unique way apart from the order of the factors. The proof is same as that used for normal polynomials with rational coefficients.

3

Gauss and Regular Polygons: Cyclotomic Polynomials | Paramanand's Math Notes

3. If f(z) is a polynomial in Fp [z] then {f(z)}p = f(z p ). This is true for constant polynomials by Fermat's theorem which says that ap ≡ a mod(p). For higher degree polynomials this is achieved by induction by writing f(z) = az n + g(z) and using binomial theorem to raise both sides to power p. In so doing we only need to note that the binomial coefficients involved are divisible by p. The proof of irreducibility of Pn (z) is done in two stages: Stage 1: Let ζ be a primitive nth root of unity and let f(z) be its minimal polynomial i.e. f(z) is monic (leading coefficient 1), has rational coefficients and irreducible and f(ζ) = 0. Since ζ is also a root of z n − 1 = 0, it follows that f(z) divides (z n − 1) and by Gauss Lemma f(z) has integer coefficients as well. We now establish the following: If p is any prime which does not divide n then ζ p is a root of f(z) = 0. Proof: Since ζ is also a root of Pn (z) = 0 it follows that f(z) divides Pn (z). Thus we have

Pn (z) = f(z)g(z) where g(z) is also monic and has integer coefficients (by Gauss Lemma). Since p is coprime to n it follows that ζ p is also a primitive nth root. And therefore

Pn (ζ p ) = 0. Assuming that ζ p is not a root of f(z) = 0 (otherwise there is nothing to prove), we see that it must be a root of g(z) = 0. Therefore ζ is a root of g(z p ) = 0. Since f(z) is the minimal polynomial of ζ , it follows that f(z) divides g(z p ) so that g(z p ) = f(z)h(z) where h(z) is monic with integer coefficients. Also since Pn (z) is a factor of (z n − 1) so that we have

z n − 1 = Pn (z)d(z) where d(z) is again monic with integer coefficients. We thus have the following equations:

z n − 1 = f(z)g(z)d(z)

(3)

g(z p ) = f(z)h(z)

(4)

We now apply the modulo p operation to each of the equations above, i.e. we replace each coefficient in the polynomials involved with its remainder when it is divided by p. The resulting polynomials are all in Fp [z] and we will use the same letters to denote them. So the above equations are now to be interpreted as relations between some polynomials in Fp [z]. The second equation can now be equivalently written as

{g(z)}p = f(z)h(z)

(5)

Let k(z) in Fp [z] be an irreducible factor of f(z). Then from the above equation (5), k(z) divides {g(z)}p and so divides g(z). Thus from equation (3) the polynomial {k(z)}2 divides

(z n − 1). Thus (z n − 1) has repeated factors and therefore (z n − 1) and its derivative nz n−1 must have a common factor. Since n is coprime to p, therefore the derivative nz n−1 is non-zero polynomial and it clearly does not have any common factor with (z n − 1). We have reached a contradiction and therefore the initial assumption that ζ p is not the root of

4

Gauss and Regular Polygons: Cyclotomic Polynomials | Paramanand's Math Notes

f(z) = 0 is wrong. The result is now proved. Stage 2: We have thus established that if f(z) is the minimal polynomial for any primitive nth root of unity then for any prime p not dividing n, ζ p (which is again a primitive nth root) is also a root of f(z) = 0. And since f(z) is irreducible and monic, it will act as minimal polynomial for the primitive root ζ p .

The same logic can be applied repeatedly and we will get the result that f(z) is the minimal polynomial for ζ p1 p2 …pm where p1 , p2 , … , pm are any primes not dividing n. It follows that ζ k where k is coprime to n is also a root of f(z). Thus all the primitive nth roots of unity are roots of f(z) = 0. Hence Pn (z) divides f(z). Since f(z) is irreducible it follows that

f(z) = Pn (z) (both f(z) and Pn (z) are monic). We thus have established that Pn (z) is irreducible. Moreover since the roots of Pn (z) = 0 are not the roots of any z m − 1 = 0 for m < n, it follows that Pn (z) is the irreducible factor of (z n − 1) which is not the factor of any z m − 1 for m < n. And thus by our definition of cyclotomic polynomials we have Pn (z) = Φn (z). We have also justified the note about our definition of cyclotomic polynomials.

What we have achieved here can be summarized as: The roots of the polynomial equation Φn (z) = 0 are precisely the ϕ(n) primitive

nth roots of unity. And thus the degree of Φn (z) is ϕ(n). Also since Φn (z) is irreducible, we get:

Φn (z) is the minimal polynomial for each of the primitive nth roots of unity. The ground work in the theory of cyclotomic polynomials is now complete and we can then turn towards solving the cyclotomic equation Φn (z) = 0. The solution as provided by Gauss is beautiful, but somewhat involved and will therefore be divided into multiple posts.

By Paramanand Singh

Labels: Algebra

Saturday, December 26, 2009

Paramanand's Math Notes Shared under Creative Commons License

5