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Home , Root of unity

1 Theory and Graph Theory

Chapter 3

Arithmetic functions and roots of unity

By

A. Satyanarayana Reddy

Department of Shiv Nadar University Uttar Pradesh, India

E-mail: [email protected] 2

Module-4: nth roots of unity

Objectives

• Properties of nth roots of unity and primitive nth roots of unity.

• Properties of Cyclotomic .

Definition 1. Let n ∈ N. Then, a z is called

1. an of unity if it satisfies the xn = 1, i.e., zn = 1.

2. a primitive nth root of unity if n is the smallest positive for which zn = 1. That is, zn = 1 but zk 6= 1 for any k,1 ≤ k ≤ n − 1.

2πi ζn = exp( n ) is a primitive n-th root of unity.

k • Note that ζn , for 0 ≤ k ≤ n − 1, are the n distinct n-th roots of unity.

• The nth roots of unity are located on the circle of the , and in that plane they form the vertices of an n-sided regular polygon with one vertex at (1,0) and centered at the origin. The following points are collected from the article Cyclotomy and cyclotomic polynomials by B.Sury, Resonance, 1999.

1. Cyclotomy - literally circle-cutting - was a puzzle begun more than 2000 years ago by the Greek geometers. In their pastime, they used two implements - a ruler to draw straight lines and a compass to draw circles.

2. The problem of cyclotomy was to divide the circumference of a circle into n equal parts using only these two implements.

3. As these n points on the circle are also the corners of a regular n-gon, the problem of cyclotomy is equivalent to the problem of constructing the regular n-gon using only a ruler and a compass. 3

4. Euclid’s school constructed the equilateral triangle, the square, the regular pentagon and the regular hexagon.

5. For more than 2000 years mathematicians had been unanimous in their view that for no prime p bigger than 5 can the p-gon be constructed by ruler and compasses. The teenager proved a month before he was 19 that the regular 17-gon is constructible.

6. He did not stop there but went ahead to completely characterize all those n for which the regular n-gon is constructible! This achievement of Gauss is one of the most surprising discoveries in mathematics.

7. This feat was responsible for Gauss dedicating his life to the study of mathematics instead of philosophy in which he was equally proficient as well. He requested that the regular 17-gon be engraved on his tombstone! This wish was, however, not carried out.

8. If we view the plane as the complex plane, the has the equation z = eiθ ,0 ≤ θ < 2π. Since the arc length is proportional to the angle subtended at the center of the 2πik circle, the n complex exp( n ), for k = 0,1,...,n − 1 cut the circumference into n equal parts.

9. So the problem of cyclotomy is same as to find the nth roots of unity or to construct the 2π angle n only with the help of straight edge and compass.

Few properties of the nth roots of unity

n−1 k th 1. ∑ ζn = 0. That is, the sum of the n roots of unity is zero. k=0

n−1 n k 1−ζn 1−1 Proof. ∑ ζn = 1−ζ = 1−ζ = 0. k=0 n n

n− 1 k j 2. Let k ∈ N with k 6≡ 0 (mod n). Then, ∑ ζn = 0. j=0 4

k Proof. Since k 6≡ 0 (mod n), ζn 6= 1. Hence,

n−1 k n n k k j 1 − (ζn ) 1 − (ζn ) 1 − 1 ∑ ζn = k = k = k = 0. j=0 1 − ζn 1 − ζn 1 − ζn

k th Lemma 2. The n-th root ζn is a primitive n root of unity if and only if gcd(k,n) = 1. So, the number of primitive nth roots of unity is ϕ(n).

k th Proof. Part-1 Let gcd(k,n) = 1. Then, we claim that ζn is a primitive n root of unity. k m km Let if possible (ζn ) = 1, where 1 ≤ m < n. Then, (ζn) = 1. Hence, by definition km ≡ 0 (mod n). As gcd(k,n) = 1, by Theorem 2.7 of Module 4 of Chapter 2, we see that m ≡ 0 (mod n), a contradiction as 1 ≤ m < n.

k th Part-2 Conversely, let ζn be a primitive n root of unity. We need to show that gcd(k,n) = 1.

n Let it possible, gcd(k,n) = d ≥ 2 and m = d . Then, 1 < m < n and

k m km k ·(dm) n k (ζn ) = (ζn) = (ζn) d = (ζn ) d = 1,

k th a contradiction, as ζn is a primitive n root of unity. Hence, gcd(k,n) = 1.

Definition 3. (Cyclotomic ) Fix a positive integer n and let S = {k : 1 ≤ k ≤ n,gcd(k,n) = k 1}. Then, the polynomial Φn(x) = ∏ (x − ζn ) is called the n-th . k∈S

From Lemma 2, we see that deg(Φn(x)) = ϕ(n). In to see some more properties of the nth roots of unity and the corresponding cyclotomic polynomial, we need to review a few basic definitions from abstract algebra. For all the known results in algebra that have been used in this write-up but haven’t been defined, the readers are advised to see any standard text books of abstract algebra, (e.g., Contemporary Abstract Algebra by J A Gallian). 5

Let G be a non-empty set having a binary operation, say ‘·’ (called the product). Then, G is said to form a if the following three conditions hold:

1. a · (b · c) = (a · b) · c, for all a,b,c ∈ G, i.e, associativity holds in G.

2. There is an element e ∈ G such that a · e = a = e · a, for all a ∈ G. The element e is called the identity element of G.

3. For every a ∈ G, there exists an element b ∈ G such that a · b = e = b · a.

From now on, we will write ab, in place of a · b, for all a,b ∈ G.

• It can be easily verified that the identity element of a group G is unique. Moreover, for each a ∈ G, its inverse is also a unique element of G and is usually denoted by a−1.

• A subset S of G is said to generate G, denoted G = hSi, if for each g ∈ G, we can find

s1,s2,...,sk ∈ S such that g = s1s2 ···sk.

n • If there exists an element g ∈ G such that G = hgi = {g : n ∈ Z} then, G is called a and the element g is called a generator of the group.

• Let G0 be a non-empty subset of a group G. Then, G0 is called a subgroup of G if G0 is also a group with respect to the binary operation in G.

• If the number of elements in G is finite then, G is said to be a finite group, else G is an infinite group. Moreover, the number of elements in a finite group G is called the order of G and is denoted by |G|.

Definition 4. Let G be a group and let g ∈ G. Then, the order of g in G, denoted ◦(g), is defined to be the smallest positive integer n such that gn = e. If no such n exists then the order of g is said to be infinite. 6

n Let G be a finite group and let g ∈ G. Then, hgi = {g : n ∈ Z} is a subgroup of G and is called the cyclic subgroup generated by g. Thus, note that a finite group G is cyclic if and only if G contains an element whose order is equal to the order of G.

Theorem 5. (Lagrange’s Theorem) Let G be a finite group and H be a subgroup of G. Then order of H divides the order of G.

Corollary 6. Let G be a finite group and let g ∈ G. Then, ◦(g) divides |G|.

• A group G is said to be abelian if ab = ba, for all a,b ∈ G.

• If the binary operation of a group G is denoted by the symbol ‘+’ then, G is called an additive group and in this case we understand that G is an . For an additive group G, 0 denotes the identity element and −a denotes the of the element a ∈ G. Also, for a ∈ G, the element a + ··· + a is written as na. Similarly, (−n)a = −(na) and 0a = 0, | {z } n-times where the 0 on the left hand side is an integer and the 0 on the right hand side is the additive identity of G.

• For n ∈ N, let Zn denote the set of modulo n. Then, Zn is an abelian group with respect to the binary operation “addition modulo n”. Mathematically, the binary operation is written as

a ⊕n b = a + b (mod n).

The group Zn is known as the additive group of integers modulo n.

k 2 n−1 Remark 7. Let Rn = {ζn : 0 ≤ k ≤ n − 1} = hζni. Then, Rn = {1,ζn,ζn ,...,ζn } forms a cyclic k ` k+` (mod n) group with respect to multiplication as its binary operation given by ζn · ζn = ζn , for

0 ≤ k,` ≤ n − 1. As, |Rn| = n and Rn is a cyclic group, by Lemma 2, we see that Rn has ϕ(n) generators.

Now, note that using the sets Am, in the proof of Theorem 6 of Module 2 of Chapter 3, one obtains the following result. 7

th S If Mr denotes the set of all primitive r roots of unity then, Rn = Md. d|n

n n Lemma 8. Fix a positive integer n. Then, x −1 = ∏ Φd(x) is the of x −1 as product d|n of cyclotomic polynomials.

S Proof. Note that using Rn = Md, we have d|n

n−1 n k x − 1 = ∏(x − ζn ) = ∏ ∏ (x − ζ) = ∏Φd(x). k=0 d|n ζ∈Md d|n

n n Proof. Alternate: Let ζ ∈ C be a zero of x − 1. Then, ζ = 1 and hence there exists a unique d dividing n, such that ζ d = 1 and ζ ` 6= 1, for 1 ≤ ` ≤ d −1. Or equivalently, ζ is a primitive dth root

of unity, and hence Φd(ζ) = 0. d Conversely, let ζ ∈ C be a primitive dth root of unity for some d of n. Then, ζ = 1 and n n hence ζ = 1. Thus, we see that both the polynomials x − 1 and Πd|nΦd(x) have the same roots in C. 0 Now, let f (x) = xn − 1. Then, f (x) = nxn−1 is coprime to f (x). This implies that f (x) has no

repeated roots in C. Also, by definition Φd(x) has no repeated roots for all d. Moreover, Φd(x) and n Φe(x) have no common roots unless d = e. So, x − 1 = ∏d|nΦd(x) as both the polynomials have 1 as it’s leading coefficient.

As an immediate corollary, we get another proof of Theorem 6 of Module 2 of Chapter 3.

Corollary 9. Let n ∈ N.Then n = ∑ ϕ(d). d|n The following result relates sum of primitive n-th roots of unity to Mobius¨ function. 8

Theorem 10. Fix a positive integer n. Then, ∑ x = µ(n), where µ(n) is the Mobius¨ function x∈Mn  0, if n is not square free,   µ(n) = 1, if n is square free and has even number of prime factors,   −1, if n is square free and has odd number of prime factors.

Proof. The case n is not a square free integer is left an exercise. Now, suppose that n = p1 p2 ··· pk is the prime factorization of n into product of distinct primes. We use induction of k to get our result. th If k = 1 then, n is a . Hence, Rn = Mn ∪ {1}. As the sum of the distinct n roots of unity is zero, we have 0 = ∑ x = 1 + ∑ x and hence, ∑ x = −1 = (−1)1. x∈Rn x∈Mn x∈Mn Now, let us assume that the result is true for all positive integers n that are square free and whose

prime factorization has < k distinct primes. Then, for a positive integer n = p1 p2 ··· pk, we have

0 = ∑ x = ∑ ∑ x = ∑ x + ∑ x + ∑ x + ··· + ∑ x x∈Rn d|n x∈Md x∈M1 x∈Mpi x∈Mpi p j x∈Mn k k  k  = 1 + · (−1) + · (−1)2 + ··· + (−1)k−1 + x 1 2 k − 1 ∑ x∈Mn k = (1 + (−1))k − (−1)k + x. k ∑ x∈Mn

k k k Thus, ∑ x = k (−1) = (−1) . x∈Mn