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Degree Project 2012-06-01 Author: Zeynep Islek Subject: Mathematics Level: Bachelor Course code: 2MA11E Möbius Inversion Formula and Applications to Cyclotomic Polynomials
Degree Project
2012-06-01
Author: Zeynep Islek
Subject: Mathematics
Level: Bachelor
Course code: 2MA11E
Möbius Inversion Formula and Applications to
Cyclotomic Polynomials
Abstract
This report investigates some properties of arithmetic functions.We will prove Mobius inversion formula which is very important innumber theory. Also our report investigates roots of unity and cyclo-tomic polynomials over the complex numbers.
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Contents
1 Introduction 4
2 Arithmetic Functions 72.1 The Mobius Function . . . . . . . . . . . . . . . . . . . . . . 72.2 The Euler Function . . . . . . . . . . . . . . . . . . . . . . . . 102.3 Mobius Inversion Formula . . . . . . . . . . . . . . . . . . . . 13
3 Cyclotomic Polynomials and Roots of Unity 163.1 nth Root of Unity . . . . . . . . . . . . . . . . . . . . . . . . 163.2 Cyclotomic Polynomials . . . . . . . . . . . . . . . . . . . . . 18
4 Conclusion 23
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1 Introduction
The number theory is a very important part of mathematics. We can saythat it is a basis of mathematics. Carl Freidrich Gauss who is a very famousmathematician, was said that : ‘Mathematics is the queen of science andarithmetics is the queen of mathematics.’
The classic Mobius function is an important multiplicative function inthe number theory and combinatorics. In number theory there are two veryimportant multiplicative functions which are Mobius function and Euler’sfunction, denoted by µ and ϕ respectively. The Mobius µ introduced tosolve a problem which was related to Riemann zeta function. The Euler’sfunction ϕ introduced to generalize a congruence result of Fermat but inthis study we are not interested in congruence relating and Riemann zetafunction.
The classic Mobius function is defined by the August Ferdinant Mobius,in 1832. It is a function whose domain is the positive integers, and which isdefined as follows:
µ(n) =
1 if n = 10 if n is divisible by a square bigger than 1(−1)k if n is product of k distinct primes
Dedekind and Liouville reported the inversion theorem for sums, simul-taneously. In 1857 they gave some appliction to ϕ(m).
R.Dedekind redeploid the function in the reverse of series which is givenby Mobius. E.Laguerre described the function below, using the functionwhich is written by Dedekind. If
F (m) =∑
f(d)
where d ranges over the divisors of m, then
f(m) =∑d|m
µ(m
d)F (d). (1)
When these formulas and (1) were used, this formula was gotten:∑ϕ(d) = m.
F. Mertens noted that if n > 1,∑µ(d) = 0 where d ranges over the
divisors of n. A.F. Mobius recognized its arithmetical importance, in 1832.Mobius anaylsed the inverse of f which is an arbitrary function, using the
Dirichlet series. Liouville and Dedekind gave the finite form of the Mobiusinversion formula, in 1857, as follows
g(n) =∑d|n
f(d)⇐⇒ f(n) =∑d|n
µ(d)g(n
d).
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If you want to learn more details about history of these functions, pleaseread [1].
Now we will mention relationship between Mobius function and roots ofunity, but before we will give some definitions.
A polynomial, not identically zero, is said to be irreducible if it cannotbe written as a product of two or more non-trivial polynomials whose coeffi-cients are of specified type. If you want to learn more, please check [2, 3, 4].
Every non-zero polynomial over C can be factored as
p(x) = α(x− z1) . . . (x− zn)
where n is the degree, α is the leading coefficient and z1, . . . , zn the zeroesof p(x). If α ∈ C and rational numbers c1, . . . , cn exist satifying
αn + c1αn−1 + · · ·+ cn = 0
then α is called an algebraic numbers. If you want to read more details,please check [3, 5, 6, 7].
Let’s take any algebraic number α. The minimal polynomial of α is theunique irreducible polynomial of the smallest degree p(x) with rational co-efficients such that p(α) = 0 and whose leading coefficients 1. If you wantto learn more details about minimal polynomials, please read [8].
In mathematics, a root of unity, is any complex number that equals 1when raised to some integer power n. Roots of unity is important in numbertheory.
An nth root of unity, is a complex number z satisfying the equationzn = 1 where n = 1, 2, 3, . . . is a positive interger.
An each element of this sum, which is shown e2πikn implies that nth root
and k for the kth power. Equivalentely, we can use (e2πin )k−n instead of
e2πikn , in the complex plane. An nth root of unity is primitive if it is not a
kth root of unity for some smaller k: zk 6= 1, k = 1, 2, . . . , n− 1.The zeroes of the polynomial
p(z) = zn − 1
are the nth roots of unity, each with multiplicity 1. There is a unique monicpolynomial Φn(x) having degree ϕ(n) whose root are the distinct primitiventh roots of unity, where ϕ is an Euler’s function. Φ is called a cyclotomicpolynomial.
Finally, we can say that about this study. In the first part of this the-sis, we emphasize the Mobius function and we prove the Mobius inversionformula. Using this Mobius inversion formula, we prove some theorems. In
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the second part we emphasize the roots of unity in the complex numbers,correspondingly we emphasize the cyclotomic polynomials in the complexnumbers, and we will see the connection between the Mobius inversion for-mula and the cyclotomic polynomials in the complex numbers.
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2 Arithmetic Functions
In this section we describe Mobius function and Mobius inversion formula,then we prove these functions. Also we prove some theorems which we needfor proving Mobius inversion formula. Now we start to give some definitionabout the number theory.
Definition 2.1. A real or complex valued function defined on the positiveintegers is called an arithmetic function. In set notation:
f : Z+ −→ R
orf : Z+ −→ C
If you want to see more details on arithmetic functions, check [9].
Definition 2.2. An arithmetic function f is called multiplicative if
f(mn) = f(m)f(n)
where m and n are relatively prime positive integers (i.e. (m,n) = 1).
Definition 2.3. An arithmetic function f is called completely multiplicativeif
f(mn) = f(m)f(n)
for every positive integers m and n.
Example 1. The function f(x) = 1x arithmetic function where f : Z+ −→
R, because since all x in Z+, the results are in R. Let’s take x = 2, thenf(x) = 1
x = 12 , and 1
2 is in R.
Now it is necessary to learn next definition for continue apprehensibly.
Definition 2.4. Let a, b ∈ Z and a 6= 0 such that b = ax if there existx ∈ Z, then we say that “a divides b” which can be denoted a|b, and a|b ifand only if b = ax for all x in Z.
2.1 The Mobius Function
The arithmetic function µ(n), defined for all natural numbers, is calledMobius function.
Definition 2.5. The Mobius function µ(n) is defined as follows
µ(n) =
1 if n = 10 if n is divisible by a square larger than 1(−1)k if n = p1 . . . pk where pi’s are relatively prime numbers
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Example 2. We have µ(1) = 1 it is clear to see from the definition. More-over
µ(2) = −1
because 2 is a prime number. So µ(2) = (−1)1 = −1. We have
µ(4) = 0
because 22|4 or we can say that 4 is divisible by a square. We have
µ(8) = 0
because 22|8. We haveµ(42) = −1
because 42 = 2 · 3 · 7, so 42 can be written as the product of three relativelyprime numbers. Thus µ(42) = (−1)3 = −1.
Theorem 2.1. The function µ(n) is multiplicative.
Proof. We will prove that µ(mn) = µ(m)µ(n) whenever m and n are rela-tively prime numbers. First, we consider m and n are square-free numbers.We assume that m = p1 . . . pk, where p1, . . . , pk are distinct primes, and n =q1 . . . qs, where q1, . . . , qs are distinct primes. From the definition of µ(n),we write that µ(m) = (−1)k and µ(n) = (−1)s, and mn = p1 . . . pkq1 . . . qs,again using the definition of µ(n), we write µ(mn) = (−1)k+s. Hence
µ(mn) = (−1)k+s = (−1)k(−1)s
= µ(m)µ(n).
Now suppose at least one of m and n is divisible by a square of a prime,then mn is also divisible by the square of a prime. So µ(mn) = 0 and µ(m)or µ(n) is equal to zero. Now it is clear to see that the product of µ(m) andµ(n) is equal to zero. So µ(mn) = µ(m)µ(n).
On the other hand, from the definition of µ(n), we know that µ(4) = 0because 22|4 and µ(2) = −1. We can write that µ(4) = µ(2 · 2), butµ(4) = 0 6= µ(2)µ(2). Hence µ(n) is not completely multiplicative func-tion.
The Mobius function appears in many different places in number theory.One of its the most important properties is a formula for the divisor sum∑
d|n µ(d) , extended over the positive divisor of n. It leads to Mobius in-version formula.
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Theorem 2.2. For the Mobius function µ(n), the summatory function isdefined by ∑
d|n
µ(d) =
{1 if n = 10 if n > 1 .
We need the following theorem to prove Theorem 2.2 .
Theorem 2.3. If f is multiplicative function of n, and F is defined asfollows
F (n) =∑d|n
f(d)
then F is also multiplicative function.
Proof. We will show that F is multiplicative function. If F is multiplicativefunction, we write that when m and n are relatively numbers, then F (mn) =F (m)F (n). So, now let us choose (m,n) = 1. We have
F (mn) =∑d|mn
f(d).
Now, all divisiors of mn must be written as the product of relatively primenumbers. As we mentioned before, if F is multiplicative function, we writeF (mn) = F (m)F (n) when (m,n) = 1. So we write d = d1d2 as the productof relatively prime divisors d1 of m and d2 of n. Hence, we write
F (mn) =∑d1|md2|n
f(d1d2).
Since f is multiplicative and since (d1, d2) = 1, we can write that
F (mn) =∑d1|md2|n
f(d1)f(d2) =∑d1|m
f(d1)∑d2|n
f(d2)
= F (m)F (n).
Now we continue to prove the Theorem 2.2.
Proof. Consider, n = 1. It is clear to see that∑d|n
µ(d) =∑d|1
µ(d) = µ(1) = 1.
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Now we assume this formula for n > 1. Let us define an arithmeticfunction M as
M(n) =∑d|n
µ(d).
The Mobius function is multiplicative, then M(n) is multiplicative by The-orem 2.3. Let’s suppose that n which is the product of powers of r differentrelatively prime numbers such that
n =
r∏i=1
paii .
Then the results which are under the function of M are equal. So we write
M(n) =r∏i=1
M(paii ).
Now we are searching what is the result of M(n). If we find the resultof M(paii ), we find the result of M(n). Now we can write that M(paii ) =∑
d|paiiµ(d) using the Theorem 2.3, which is the way to find the result. We
have
M(paii ) =∑d|paii
µ(d)
= µ(1) + µ(pi) + µ(p2i ) + · · ·+ µ(paii )
= 1 + (−1) + 0 + · · ·+ 0
= 0.
For every integer bigger than 1, we proved that the sum function of Mobiusfunction is equal to zero.
The another example of an arithmetic function is Euler’s function. It isalso a multiplicative function.
2.2 The Euler Function
The function was introduced by Euler, in 1760, and is denoted by ϕ. Thisfunction is multiplicative which is one of the most important function innumber theory.
Definition 2.6. The Euler function ϕ(n) is the number of positive integersless than n which are relatively prime to n.
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Example 3. Here is some values of ϕ(n).
ϕ(1) = 1, ϕ(2) = 1, ϕ(3) = 2, ϕ(4) = 2, ϕ(5) = 4
ϕ(6) = 2, ϕ(7) = 6, ϕ(8) = 4, ϕ(9) = 6, ϕ(10) = 4
Now we find the values of the phi-function at primes powers.
Theorem 2.4. Let p be a prime number. Then ϕ(pα) = pα − pα−1.
Proof. Between 1 and pα there are pα integers. There are some numberswhich are not relatively prime to pα, they are p, 2p, . . . , pα−1. There areexactly pα−1 such integers. So there are pα−pα−1 integers less than pα thatare relatively prime to pα. Hence, ϕ(pα) = pα − pα−1.
Theorem 2.5. If p is a prime, then ϕ(p) = p− 1.
Proof. It is easy to see that from Theorem 2.4. Now we suppose α = 1 then
ϕ(pα) = pα − pα−1 = p1 − p1−1
= p− 1.
Example 4. We calculate the values of ϕ(n) for some prime numbers.
ϕ(53) = 53 − 52 = 100
ϕ(210) = 210 − 29 = 512
ϕ(112) = 112 − 111 = 110
The Euler’s ϕ function is multiplicative function. If you’re interested inproof of ϕ is multiplicative function, please read [3, 5, 7].
Example 5. Let’s calculate ϕ(756) using that the Euler’s ϕ function ismultiplicative.
This number can ben be written as 756 = 22 · 33 · 7. Hence
ϕ(756) = ϕ(22 · 33 · 7).
We know that ϕ function is multiplicative. So we write
ϕ(756) = ϕ(22) · ϕ(33) · ϕ(7).
Using the Theorem 2.4 and Theorem 2.5, we find
ϕ(22) = 22 − 2 = 2,
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ϕ(33) = 33 − 32 = 18,
ϕ(7) = 7− 1 = 6.
Soϕ(756) = 2 · 18 · 6 = 216.
Now we get the result.If you want to check more examples, read [5].
Theorem 2.6. For every positive integers d and n, we have∑
d|n ϕ(d) = n.
Proof. We will prove this by induction on the number of different primefactors. We consider the case n = pα, where p is a prime number. We have∑
d|n
ϕ(d) =∑d|pα
ϕ(d) = ϕ(1) + ϕ(p) + ϕ(p2) + · · ·+ ϕ(pα)
= 1 + (p− 1) + (p2 − p) + · · ·+ (pα − pα−1)= pα
= n.
Correspondingly, now suppose that the theorem holds for integers withk distinct prime factors. Let us take any integer N with k+1 distinct primefactors and pα be the highest power of p that divides N . Now we writeN = pαn where p and n are relatively prime numbers (i.e., (p,n)=1). Weknow when d ranges over the divisor of n, the set d, dp, dp2, . . . , dpα rangesover the divisors of N . Then
∑d|N
ϕ(d) =∑d|n
ϕ(d) +∑d|n
ϕ(dp) +∑d|n
ϕ(dp2) + · · ·+∑d|n
ϕ(dpα)
=∑d|n
ϕ(d) +∑d|n
ϕ(d)ϕ(p) +∑d|n
ϕ(d)ϕ(p2) + · · ·+∑d|n
ϕ(d)ϕ(pα)
=∑d|n
ϕ(d)[1 + ϕ(p) + ϕ(p2) + · · ·+ ϕ(pα)]
=∑d|n
ϕ(d)∑e|pα
ϕ(e)
= npα
= N.
We showed that∑
d|n ϕ(d) = n is true for every positive integers n, and dranges over n.
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Example 6. We give an example to understand the Theorem 2.6.∑d|12
ϕ(d) = ϕ(1) + ϕ(2) + ϕ(3) + ϕ(4) + ϕ(6) + ϕ(12)
= 1 + 1 + 2 + 2 + 2 + 4
= 12.
REMARK: As ϕ is an arithmetic function, we note that∑d|n
ϕ(d) =∑d|n
ϕ(n
d).
Now we show that for any multiplicative functions: If f is multiplicativefunction and not equal to zero, then
∑d|n f(d) or, equivalently,
∑d|n f(n|d)
denotes the sum of the values of a function f where d ranges over the positivedivisors of n. We write that∑
d|n
f(d) =∑d|n
f(n|d)
because since d ranges over n, nd ranges over n. For example;∑
d|18
f(d) = f(1) + f(2) + f(3) + f(6) + f(9) + f(18)
∑d|18
f(n
d) = f(
18
1) + f(
18
2) + f(
18
3) + f(
18
6) + f(
18
9) + f(
18
18)
= f(18) + f(9) + f(6) + f(3) + f(2) + f(1)
2.3 Mobius Inversion Formula
Theorem 2.7. If g is any arithmetic function and f is the sum function ofg, so that
f(n) =∑d|n
g(d)
theng(n) =
∑d|n
f(d)µ(n
d).
Equivalently, if d ranges over n, nd ranges over n. Hence, we can write∑
d|n
f(d)µ(n
d) =
∑d|n
f(n
d)µ(d).
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Proof. The equality ∑d|n
f(d)µ(n
d) =
∑d|n
f(n
d)µ(d)
is true from the remark.If d|n, we write n = ed from the definition (divisibility), where e is in Z.
Now let us take n = de, so e = nd . Then the previous sum can be written as∑de=n
f(d)µ(e)
and it is possible to write the last sum as,∑de=n
f(e)µ(d).
Now we must prove that the sum∑
d|n f(d)µ(nd ) is equal to g(n) or,equivalentely, the sum
∑d|n f(nd )µ(d) is equal to g(n). Using equality below
f(n
d) =
∑e|nd
g(e)
we write that ∑d|n
µ(d)f(n
d) =
∑d|n
(µ(d)∑e|nd
g(e)).
Since e divides nd , then e divides n. Inversely, each divisor of n is e which
divides nd if and only if d divides n
e . So d divides n. As have seen, thecoefficent of g(e) is
∑d|neµ(n) can be written as
∑d|ne
µ(n) =
{1 if n
e = 10 if n
e > 1
using the Theorem 2.2.That implies g(n) has only one coefficient g(e) which is not equal to zero.
So g(e) = 1. Then
g(n) =∑d|n
f(n
d)µ(d).
The Euler function is related to the Mobius function through the follow-ing formula.
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Theorem 2.8.
ϕ(n) = n∑d|n
µ(d)
d
where ϕ(n) is Euler’s ϕ function.
Proof. We know∑
d|n ϕ(d) = n from the Theorem 2.6. Take a function Fwhich is the sum of the Euler’s ϕ function is as
F (n) =∑d|n
ϕ(d) = n.
Use the Mobius inversion formula here
ϕ(n) =∑d|n
F (d)µ(n
d)
=∑d|n
F (n
d)µ(d)
=∑d|n
n
dµ(d)
= n∑d|n
µ(d)
d.
Example 7. Let’s calculate the value of ϕ(756), using Theorem 2.8.The divisors of this number are 1, 2, 3, 4, 7, 9, 12, 14, 18, 21, 27, 28, 36, 42,
54, 63, 84, 108, 126, 189, 252, 378, and 756. Now using the Theorem 2.8 wecan calculate ϕ(756).
ϕ(756) = n∑d|n
µ(d)
d
= 756 ·∑d|756
µ(d)
d
=∑d|n
n
dµ(d)
= 756 ·(µ(1)
1+µ(2)
2+ · · ·+ µ(378)
378+µ(756)
756
)And here is some values of µ(n).
µ(1) = 1, µ(2) = −1, µ(3) = −1, µ(6) = 1,
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µ(7) = −1, µ(14) = 1, µ(21) = 1, µ(42) = −1.
The value µ of the other numbers are equal to zero, because these numbersare divisible by a square larger than 1. Hence
756 ·(
1
1+−1
2+−1
3+
1
6+−1
7+
1
14+
1
21+−1
42
)= 756 · 2
7= 216
We solved same example using by ϕ function is a multiplicative function.That way is easy and short, because if you want to use the Theorem 2.8,you should know that the values of µ function.
3 Cyclotomic Polynomials and Roots of Unity
3.1 nth Root of Unity
Assume that a be an nth root of a number b, this means that bn = a. Inparticular the square root of 1 is 1 because 1 · 1 = 1, but (−1)(−1) = 1,so −1 also a square root of 1. There are two square roots of 1. If we takethe cube root of 1, then −1 is not a solution because (−1)(−1)(−1) = −1.So 1 has only one solution if we study on real numbers, but in the complexnumbers, there are three roots of 1. All cube roots of 1 can also be definedas powers of the negative interval. We see them below
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REMARK: Let n be an integer and x be a complex number (and, inparticular, a real number), the Euler’s function states that
ei(nx) = cos(nx) + isin(nx).
C be the field of complex numbers, there are exactly n different nth rootsof 1. If you divide the unit circle into n equal parts, using n points, it iseasy to find them.
Definition 3.1. A complex number z is called an nth root of unity for apositive integer n, if zn = 1.
We will show that zn = 1. Let’s take any complex number z. If we writethis complex number on the polar coordinates, we get
z = cosθ + isinθ.
If we take the nth power of z,
zn = (cosθ + isinθ)n = cos(n · θ) + isin(n · θ).
We can write this equality, because this is de Moivre’s formula. So
zn = 1
cos(n · θ) + isin(n · θ) = cos(0) + isin(0).
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Then nθ = 0 + 2kπ, where k = 0, 1, . . . , n− 1. Thus θ = 2kπn . The roots of
unity is then ine2πi/n
for k = 0, 1, . . . , n− 1.There are n different solutions for zn = 1, namely, e2πi/n, e2πi2/n, . . . ,
e2πin/n. We usually assume that ζn = e2πi/n so that ζn, ζ2n, . . . , ζ
nn are the
nth roots of zn = 1.
Definition 3.2. An nth root of unity is primitive if it is of the form ζkn withk and n relatively prime numbers, i.e., (k, n) = 1. If ζn is a primitive nthroot of unity and (ζkn)m = 1 then n|m.
If you want to read more details, check [6].
3.2 Cyclotomic Polynomials
Definition 3.3. Let n be a positive integer and let ζkn be the primitiventh root of unity ( ζn is the complex number e2πi/n). The nth cyclotomicpolynomial Φn(x) is
Φn(x) =∏
1≤k≤ngcd(n,k)=1
(x− ζkn)
whose roots are the primitive nth roots of unity.
Theorem 3.1. Let n be a positive integer. Then
xn − 1 =∏d|n
Φd(x)
where d ranges over the divisor of n.
Proof. The roots of xn−1 are exactly nth roots of unity. On the other hand,if ζ is an nth root of unity and the order of ζ is d, then ζ is a primitive dthroot of unity. So ζ is a root of Φd(x). But d|n, so ζ is a root of the righthand side. It follows that the polynomials on the left and right hand sidehave the same roots. Thus they are equal.
Another way to find the nth cyclotomic polynomial:If n > 1, then
Φn(x) =xn − 1∏d Φd(x)
(2)
where d ranges over, except n, the divisor of n.
If you want to see more detail about cyclotomic polynomials, check [4,10].
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Example 8. Here is some value of cyclotomic polynomials.
Φ1(x) = x− 1
Φ2(x) =x2 − 1
Φ1(x)=x2 − 1
x− 1= x+ 1
Φ3(x) =x3 − 1
Φ1(x)=x3 − 1
x− 1= x2 + x+ 1
Φ4(x) =x4 − 1
Φ1(x)Φ2(x)= x2 + 1
Φ5(x) =x5 − 1
Φ1(x)= x4 + x3 + x2 + x+ 1
Φ6(x) =x6 − 1
Φ1(x)Φ2(x)Φ3(x)= x2 − x+ 1
Now we know that xn − 1 =∏d|n Φd(x), and conversely, by using the
Mobius function, we can write the following theorem.
Theorem 3.2. Let n be a positive integer and µ(n) denotes the Mobiusfunction. Then
Φn(x) =∏d|n
(xd − 1)µ(nd).
Proof. To prove this formula, first we use the equality xn− 1 =∏d|n Φd(x),
then we take complex logarithm of this equality, and finally use the Mobiusinversion formula. We have
xn − 1 =∏d|n
Φd(x).
Now take the complex logarithm both side of equality. It doesn’t effect tothe equality.
log(xn − 1) = log
∏d|n
Φd(x)
Let’s assume d = {d1, d2, . . . , ds}, where di’s are divisor of n and are notequal to n.
log
∏d|n
Φd(x)
= log (Φd1(x) · Φd2(x) · · ·Φds(x))
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Using the property of logarithm, we write
log (Φd1(x) · Φd2(x) · · ·Φds(x)) = log(Φd1(x)) + log(Φd2(x)) + · · ·+ log(Φds(x))
=∑d|n
log(Φd(x)).
Using Mobius inversion formula, we can write
log(Φn(x)) =∑d|n
µ(n
d) ·(
log(xd − 1))
=∑d|n
(log(xd − 1)µ(
nd))
= log((xd1 − 1)µ( nd1
)) + log((xd2 − 1)
µ( nd2
)) + · · ·+ log((xds − 1)µ(
nds
))
= log(
(xd1 − 1)µ( nd1
) · (xd2 − 1)µ( nd2
) · · · (xds − 1)µ(nds
))
= log
∏d|n
(xd − 1)µ(nd)
If we cancel out the logarithm, we get the result
Φn(x) =∏d|n
(xd − 1)µ(nd).
Now we can see the connection between Mobius inversion formula andcyclotomic polynomials. When
xn − 1 =∏d|n
Φd(x),
it is possible to write that
Φn(x) =∏d|n
(xd − 1)µ(nd).
Example 9. We are going to find cyclotomic polynomials using the formula
Φn(x) =∏d|n
(xd − 1)µ(nd)
for n = 1, 2, . . . , 20.Φ1(x) = x− 1Φ2(x) = x+ 1Φ3(x) = x2 + x+ 1
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Φ4(x) = x2 + 1Φ5(x) = x4 + x3 + x2 + x+ 1Φ6(x) = x2 − x+ 1Φ7(x) = x6 + x5 + x4 + x3 + x2 + x+ 1Φ8(x) = x4 + 1Φ9(x) = x6 + x3 + 1Φ10(x) = x4 − x3 + x2 − x+ 1Φ11(x) = x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x+ 1Φ12(x) = x4 − x2 + 1Φ13(x) = x12 + x11 + x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x+ 1Φ14(x) = x6 − x5 + x4 − x3 + x2 − x+ 1Φ15(x) = x8 − x7 + x5 − x4 + x3 − x+ 1Φ16(x) = x8 − 1Φ17(x) = x16 + x15 + x14 + x13 + x12 + x11 + x10 + x9 + x8 + x7 + x6 + x5 +x4 + x3 + x2 + x+ 1Φ18(x) = x6 − x3 + 1Φ19(x) = x18 +x17 +x16 +x15 +x14 +x13 +x12 +x11 +x10 +x9 +x8 +x7 +x6 + x5 + x4 + x3 + x2 + x+ 1Φ20(x) = x8 − x6 + x4 − x2 + 1
We have already found some cyclotomic polynomials in example 8. Inthis example you don’t need to know values of Mobius function. If you wantto find cyclotomic polynomials using the Theorem 3.2, you should knowthe values of Mobius function. As have seen, the coefficients of cyclotomicpolynomial are often −1, 0 and 1, but for n ≥ 105 some coefficients aredifferent from this set. For example n = 105, then
Φ105(x) =x48 + x47 + x46 − x43 − x42 − 2x41 − x40 − x39 + x36 + x35 + x34
+ x33 + x32 + x31 − x28 − x26 − x24 − x22 − x20 + x17 + x16 + x15
+ x14 + x13 + x12 − x9 − x8 − 2x7 − x6 − x5 + x2 + x+ 1.
We see that there is a coefficient −2 which is not included in {−1, 0, 1}.
Theorem 3.3. The coefficients of Φn(x) are integers.
Proof. We prove it using inductive method. Clearly, Φ1(x) = x− 1 ∈ Z[x].Now we suppose that, for k < n, the coefficients of Φn(x) are integers.
Let f(x) =∏
d|nd<n
Φd(x). Then we say that f(x) ∈ Z[x], from the inductive
method. Using the formula (2), we write if
Φn(x) =xn − 1
f(x)
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thenxn − 1 = Φn(x)f(x)
On the other hand, xn − 1 ∈ Z[x]. Also using the division algorithm,we can write xn − 1 = f(x)g(x) + r(x) for some g(x), r(x) ∈ Z[x]. By theuniqueness, we take r(x) = 0. So xn − 1 = f(x)g(x). It is easy to seeg(x) = Φn(x). Since g(x) ∈ Z[x], Φn(x) ∈ Z[x]. Finally we say that thecoefficients of Φn(x) are integers.
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4 Conclusion
In this study, the Mobius inversion formula has been introduced and proved.The roots of unity and cyclotomic polynomials have been introduced.
In the first cheapter, some informations about Mobius function, Euler’sϕ function, Mobius inversion formula, roots of unity and cyclotomic poly-nomials have been introduced. And also some definitions which will be usedin the next cheapters have been mentioned.
In the second cheapter, some arithmetic functions and properties of thesefunctions have been studied. The Mobius function which is defined on arith-metic functions, has been studied. The Euler’s ϕ function and its applica-tions have been mentioned. The Mobius inversion formula has been intro-duced. A theorem which is related with Euler’s ϕ function and µ function,using Mobius inversion formula, has been proved, that is
ϕ(n) = n∑d|n
µ(d)
d.
In the third cheapter, the roots of unity in the complex numbers havebeen introduced. Accordingly, the primitive roots of unity have been intro-duced. Later, the cyclotomic polynomials whose roots are the primitive nthroots of unity, have been investigated. Accordingly, the connection betweenMobius inversion formula and cyclotomic polynomials has been shown. Theconnection between Mobius inversion formula and cyclotomic polynomialsin the complex numbers has been shown. Since this is the formula below,
xn − 1 =∏d|n
Φd(x).
and with the Mobius inversion formula and using the complex logarithm
Φn(x) =∏d|n
(xd − 1)µ(nd)
has been proved.
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References
[1] L. E. Dickson, History of the theory of numbers, Vol. I, Divisibility andPrimality, Chelsea Publishing Company, New York (1992)
[2] Gareth A. Jones and J. Mary Jones, Elementary number theory,Springer-Verlag London Limited (1998)
[3] H. E. Rose, A course in number theory, Oxford University Press, NewYork, second edition (1994)
[4] I. N. Herstein, Abstract algebra, Prentice-Hall, Inc., third edition (1996)
[5] J. H. Silverman, A friendly introduction to number theory, Prentice-Hall, Inc., second edition (2001)
[6] B. L. van der Waerden, Algebra, Vol. I, Springer-Verlag New york, Inc.,(1991)
[7] K. H. Rosen, Elementary number theory and its applications, Assison-Wesley Publishing Company, third edition (1993)
[8] I. Niven, H. S. Zuckerman, An introduction to the theory of numbers,John Wiley & Sons, Inc., second edition (1967)
[9] G. H. Hardy, E. M. Wright, An introduction to the theory of number ,Oxford University Press, Oxford, second edition (2008)
[10] T. Nagell, Introduction to number theory, Almqvist-Wiksell, Sweden(1951)
[11] K. Ireland, M. Rosen, A classic introduction to modern number theory,Springer-Verlag, New York (1982)
[12] N. Lauritzen, Concrete abstract algebra: from numbers to Grobnerbases, Cambridge University Press, New York, USA (2003)
[13] I. S. Luthar, I. B. S. Passi, Algebra, Vol. 4, Field Theory, Alpha ScienceInternational Ltd., Harrow, U.K. (2004)
[14] W. J. LeVeque, Topics in number theory, Vol. I, Addison-Wesley Pub-lishing Company (1956)
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