# ζ -and L -functions. ThepurposeofthisstudyistoproveDirichlet’stheorem.Webeginbydiscussingthetheoremforspecialcases.Thenweintroducecyclotomicpolynomials,proveGauss’sLemmaandusethislemmatoproveseveralpropertiesofcyclotomicpolynomials.Hereafter,usingcyclotom

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faculteit Wiskunde en Natuurwetenschappen

## Dirichlet’s Theorem

### Bachelor thesis Mathematics

July 2014

Student: D. R. Cirkel

First supervisor: Prof. Dr. J. Top

Second supervisor: Prof. Dr. H. L. Trentelman

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Abstract

The purpose of this study is to prove Dirichlet’s theorem. We begin by discussing the theorem for special cases. Then we introduce cyclotomic polynomials, prove Gauss’s Lemma and use this lemma to prove several properties of cyclotomic polynomials. Hereafter, using cyclotomic polyno- mials, we prove the theorem holds for a more general case. Finally, we will prove Dirichlet’s theorem with the use of complex numbers, Dirichlet series and ζ- and L-functions.

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### Contents

1 Introduction 3

2 Elementary proofs 4

2.1 Introduction . . . 4

2.2 Infinitely many prime numbers . . . 4

2.3 Infinitely many prime numbers 3 mod 4 . . . 4

2.4 Infinitely many prime numbers 1 mod 4 . . . 5

3 Cyclotomic polynomials 7 3.1 Introduction . . . 7

3.2 Gauss’s Lemma . . . 7

3.3 Defining cyclotomic polynomials . . . 9

3.3.1 Definitions . . . 9

3.3.2 Defining cyclotomic polynomials . . . 9

3.4 Properties of cyclotomic polynomials . . . 11

3.4.1 Definitions . . . 11

3.4.2 Properties of cyclotomic polynomials . . . 12

3.5 Special cases . . . 14

3.5.1 Infinitely many prime numbers 5 mod 8 . . . 14

3.5.2 Infinitely many prime numbers 7 mod 12 . . . 14

4 Dirichlet’s theorem 15 4.1 Introduction . . . 15

4.2 Characters . . . 15

4.2.1 Definitions . . . 15

4.2.2 Characters . . . 16

4.3 Ordinary Dirichlet series . . . 18

4.3.1 Definitions . . . 18

4.3.2 Partial sums . . . 19

4.3.3 Dirichlet series . . . 22

4.4 The zeta functions . . . 23

4.4.1 Definitions . . . 23

4.4.2 The zeta function . . . 24

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4.5 The L-functions . . . 27

4.5.1 Definitions . . . 27

4.5.2 The L-functions . . . 27

4.6 Characters modulo m . . . 28

4.6.1 Definitions . . . 28

4.6.2 Characters modulo m . . . 29

4.7 Density . . . 31

4.7.1 Definitions . . . 31

4.7.2 Density . . . 31

4.8 Dirichlet’s theorem . . . 32

4.8.1 Definitions . . . 32

4.8.2 Dirichlet’s theorem . . . 32

5 Conclusion 35

6 Acknowledgements 36

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### Introduction

Our goal is to prove the following theorem:

Dirichlet’s Theorem: Let a ≥ 1 and m ≥ 1 be relative prime integers.

There exist infinitely many primes p such that p ≡ a mod m.

The theorem was first suspected by Adrien-Marie Legendre and proved by Peter Dirichlet in 1837. 1

We will start with a few short basic proofs. These will show that for special cases the above theorem is quite easy to prove, without further knowledge other than some elementary arithmetic.

Hereafter, we will introduce cyclotomic polynomials and explain various properties of cyclotomic polynomials. Which can be used to show that for the case of a = 1 the theorem holds. That is, there is an infinite number of primes p such that p ≡ 1 mod m.

Finally, we wil look into the overall proof for Dirichlet’s theorem. In this chapter we will follow the proof written in A course in arithmetic from Jean- Pierre Serre  strictly. Even though the statement involves only integers, during the proof one will find out it requires a lot more knowledge.

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### 2.1 Introduction

In this chapter we will prove some basic results concerning special cases of Dirichlet’s theorem. We will start by proving that there are infinitely many primes in the first place, as this will be the base of other proofs.

### 2.2 Infinitely many prime numbers

Proposition 1. There are infinitely many primes.

Proof. There are primes p = 2, 3, 5, ... Let p1, . . . , pr be a sequence of r given primes. Define m as follows:

m := p1× p2× p3× . . . × pr+ 1

Then m > 1, hence m will have a prime divisor q ∈ Z. If we can prove that this q is a new prime, we have proven the proposition.

Assume q is not a new prime, in other words q = pj for some j ∈ [1, . . . , r].

Then

q | (p1× p2× p3× . . . × pr) but also q | (p1× p2× p3× . . . × pr+ 1), which obviously leads to a contradiction.

Therefore q 6= pj, which implies that at least r + 1 primes exist. Thus, we can conclude there are infinitely many primes.

### 2.3 Infinitely many prime numbers 3 mod 4

We will now show the same result for primes congruent to 3 mod 4.

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Proposition 2. There exists an infinite number of primes p such that p ≡ 3 mod 4.

Proof. There primes p ≡ 3 mod 4, for example 3 and 7 and 11. Consider:

m := 4 × p1× . . . × pr− 1 It is clear that m is congruent to 3 mod 4.

Since m > 1 and odd we can find at least one odd prime divisor q ∈ Z for this m. Hence, one gets q ≡ 1 mod 4 or q ≡ 3 mod 4. Suppose that all odd prime divisors are congruent to 1 mod 4. That is, qi ≡ 1 mod 4, qj ≡ 1 mod 4 for all 1 ≤ i ≤ j ≤ r. However, qiqj ≡ 12 ≡ 1 mod 4. This con- tradicts the fact that m is of the form 4k + 3. Thus, there is at least one primefactor q ≡ 3 mod 4.

Again, we will show by contradiction that q 6= pj. State q = pj for some j ∈ [1, . . . , r]. Then

q | (4 × p1× . . . × pr) and q | (4 × p1× . . . × pr− 1).

Which leads to a contradiction. Thus we have found a new prime congruent to 3 mod 4. Since this can be done for r tending to infinity, we can conclude that there are infinitely many primes p ≡ 3 mod 4.

### 2.4 Infinitely many prime numbers 1 mod 4

Proposition 3. There exists an infinite number of primes p such that p ≡ 1 mod 4.

Proof. Let p1, . . . , prbe a list of r primes congruent to 1 mod 4. For exam- ple p = 5, 13, 17, . . . Again we will make m as follows:

m := (2 × p1× . . . × pr)2+ 1 m > 2, odd and congruent to 1 mod 4.

We can find a prime divisor q ∈ Z for this m, which will also be odd. Now, to conclude the proof, we will show q is of the form q ≡ 1 mod 4. Since q is a prime divisor of m the following holds:

m ≡ 0 mod q (2 × p1× . . . × pr)2+ 1 ≡ 0 mod q (2 × p1× . . . × pr)2≡ −1 mod q

(2 × p × . . . × p )q−1 ≡ (−1)q−12 mod q. (2.1)

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However, we know ab ≡ a mod b, which implies that aba−1 ≡ aa−1 mod b.

That is to say: ab−1= 1 mod b.

Now, by applying this to equation (2.1) one gets 1 mod q ≡ (−1)q−12 mod q.

Since q is odd, 1 6≡ −1 mod q. Therefore 1 = (−1)q−12 , which can only be true when 12(q − 1) is even. Thus,

q − 1

2 = 2n with n ∈ Z q = 4n + 1,

as desired.

Similarly as it has been shown in section 2.3 one can now prove q 6= pj for all j ∈ [1, . . . , r] and therefore the proposition holds.

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### 3.1 Introduction

With the use of cyclotomic polynomials, we will take a step towards the proof of Dirichlet’s theorem: we will prove there are infinite many primes of the form p ≡ 1 mod n. To prove this, we will first consider Gauss’s Lemma, as we will need it further on. From here on, we will define cyclotomic poly- nomials and prove several properties of cyclotomic polynomials. Ultimately, we will use cyclotomic polynomials to find other special cases.

### 3.2 Gauss’s Lemma

Gauss’s Lemma is the following:

Lemma 1. Let f ∈ Z[x] be a monic polynomial, and f = g · h for monic polynomials g, h ∈ Q[x]. Then all the coefficients of g and h are integers.

Example: Let f (x) = xn− 1 and let g ∈ Q[x] be an arbitrary monic divisor of f , then the coefficients of g are integers.

Proof. To prove Gauss’s Lemma we write f = fnxn+ fn−1xn−1+ . . . + f0, where the coefficients fn = 1, fn−1, . . . , f0 ∈ Z. Since fn = 1 it fol- lows that gcd(f0, f1, . . . , fn) = 1. From here and now on gcd(fj) denotes gcd(fn, fn−1, . . . , f0). Assume that the monic polynomials g and h satisfy f = g · h and g, h ∈ Q[x]. Write

g = xd+ gd−1xd−1+ . . . + g0 1, gd−1, gd−2, . . . , g0∈ Q, h = xn−d+ hn−d−1xn−d−1+ . . . + h0 1, hn−d−1, hn−d−2, . . . , h0∈ Q.

We have to prove that the coefficients of g and h do not contain any denom- inators. The next proposition helps to show this.

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Proposition 4. Let p, q be nonzero polynomials in Z[x]. The greatest com- mon divisor of the coefficients of the product p · q equals the product of the greatest common divisors of the coefficients of these polynomials.

Proof. Let, p, q be nonzero polynomials in Z[x]. Then p · q ∈ Z[x] and p · q 6= 0. Write

p =p0+ p1x + . . . q =q0+ q1x + . . . pq =a0+ a1x + . . . Let cp = gcd(pj) and cq= gcd(qj), then

p = cp· v q = cq· w

with v, w ∈ Z[x], gcd(vj) = 1, gcd(wj) = 1 and p · q = cpcqvw.

By definition, the coefficients of vw are (vw)i := cai

pcq ∈ Z, i ∈ [0, . . .). If we can show that gcd (vw)j = 1, we have proven the proposition.

Suppose gcd((vw)j) 6= 1, then gcd((vw)j) > 1. Therefore there exists a prime divisor p such that p | gcd((vw)j). If we take this prime and use it to calculate mod p we will get:

Z[x] → (Z/pZ)[x]

v 7→ v 6= 0

w 7→ w 6= 0

v · w 7→ v · w

Here, and from now on a = a mod p.

This map is multipicative. That is, rs = r · s. Since p | gcd((vw)j) we know that vw = 0, however v 6= 0, w 6= 0. This leads to a contradiction, since p is prime and therefore (Z/pZ)[x] has no zero divisors. Hence gcd((vw)j) = 1.

We now continue with the proof of Lemma 1. To prove that gj and hj do not contain any denominators we choose:

N > 0, N ∈ Z as small as possible, such that N · g ∈ Z[x], and M > 0, M ∈ Z as small as possible, such that M · h ∈ Z[x].

Since (N g)0= N and (M h)0 = M , we know gcd (N g)j | N and gcd (M h)j | M Moreover gcd(fj) = 1, hence gcd (N M f )0, . . . , (N M f )n = N M . By applying Proposition 4:

N M = gcd (N M f )j = gcd (N g · M h)j = gcd (N g)j · gcd (M h)j

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Therefore gcd (N g)j = N and gcd (M h)j = M . Since f ∈ Z[x], we can now state that all the coefficients of g and h are integers.

### 3.3 Defining cyclotomic polynomials

3.3.1 Definitions

Definition 1. We name the nth cyclotomic polynomial Φn(x). It is defined as the polynomial of the lowest degree, with rational coefficients and leading coefficient 1, such that e2πin is a zero.

Definition 2. We define the following polynomial:

fn(x) = xn− 1 3.3.2 Defining cyclotomic polynomials

Theorem 1. The roots of Φn are simple and exactly all e2πin m for m ∈ Z where gcd(n, m) = 1

To prove this theorem, we will divide it into three smaller propositions.

Proposition 5. The roots of Φn are simple.

Proof. Let ζ be the root of Φn, i. e. Φn(ζ) = 0,. We know that Φn divides fn. Indeed, let

gn:= gcd(Φn, fn) in Q[x].

Then by definition of gcd there exist polynomials a, b ∈ Q[x] such that a(x)Φn(x) + b(x)fn(x) = gn(x). Now ζ is a root of the left hand side, hence also of the right hand side. Moreover, the degree of gn is smaller than or equal to the degree of Φ > n. By definition of Φn, this implies that gn(x) = Φn(x). Hence Φn(x) | fn(x). In other words

xn− 1 = Φn(x)h(x), for some h ∈ Z[x].

Thus ζ is also a root of xn− 1. Assume ζ is not a simple root of Φn, then it is a multiple root of xn− 1 as well. Hence ζ is a root of the derivative of this polynomial. However, if one computes the derivative of fn(x), one gets fn0(x) = nxn−1, which has merely 0 as a root. Since ζ obviously is not equal to zero, ζ is a simple root.

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Proposition 6. All roots of Φn are of the form e2πimn , with gcd(n, m) = 1 Proof. Let ζ be a root of Φn, then

fn(ζ) = ζn− 1

= Φn(ζ) · h(ζ)

= 0.

Therefore

ζn= 1 and ζ = e2πin ·m for m ∈ Z.

Suppose gcd(n, m) 6= 1, then gcd(n, m) = d, for some d > 1. It follows that n can be written as n = d · N and m as m = d · M . Since gcd(n, m) = d, we have that gcd(N, M ) = 1. Now the root can be rewritten as

ζ = e2πin m= e2πiN M.

This means that ζN = 1, which implies that Φn | fN. However, e2πin is a root of Φn, but not of xN − 1. This leads to a contradiction, since Φn| fN implies that every root of Φnshould be a root of fN. Therefore we can reject our assumption. Consequently, gcd(n, m) = 1.

Proposition 7. Φn has exactly all ζ = e2πin m, where gcd(n, m) = 1 as root.

Proof. To prove that all ζ with gcd(n, m) = 1 are exactly all roots of Φn, we will use a proof by contradiction.

Assume there exists a p with p - n, such that Φnp) 6= 0. Since Φn(x) is a divisor of fn(x) one finds that Φ(xp) is a divisor of fn(xp). Therefore

fn(xp) = Φn(xp)h(xp) for some h ∈ Z[x].

As ζ is a root of fn(xp) but not of Φn(xp), ζ must be a root of h(xp). This implies that Φn(x) is a divisor of h(xp). Thus h(xp) = Φn(x)g(x) for some g ∈ Z[x]. Therefore

fn(xp) = Φn(xpn(x)g(x).

To continue the proof, we need the use of the following lemma.

Lemma 2. Let p be a prime and 1 < j < p, then the binomial coeffecient

p

j is congruent to 0 mod p.

Proof. Assume pj

6≡ 0 mod p. Then pj

= m for some m ∈ Z, m 6≡ 0 mod p. By definition, pj = j!(p−j)!p! . This equation can be rewritten as:

m · j! = p!

(p − j)!.

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Since p is a prime, m · j! 6≡ 0 mod p. However, the right term of the equa- tion (p−j)!p! is equal to p(p − 1) . . . (p − j + 1), which obviously is congruent to 0 mod p. This is a contradiction.

Therefore, if p is a prime, pj ≡ 0 mod p, for 1 < j < p.

If one applies Lemma 2 and the fact that ap ≡ a mod p, one may write fn(xp) = xnp− 1

= (xn− 1)p. Moreover

Φn(xpn(x)g(x) = Φn(x)p+1g(x).

Thus

(xn− 1)p = Φn(x)p+1g(x) and Y

α

(x − α)p =Y

β

(x − β)p+1Y

γ

(x − γ)

This shows that every root of Φn(x) ∈ Z/pZ[x] is a root of xn− 1 ∈ Z/pZ[x]

with multiplicity ≥ p + 1. Therefore xn− 1 has roots with multiplicity ≥ 2.

If one computes the derivative of fn= (x − δ)2· h, one finds it is equal to nxn−1= 2(x − δ)h + (x − δ)2h0.

If we put x = δ, we have nδn−1 = 0. Since n 6= 0 =⇒ δ = 0. However, if we substitute this δ in xn− 1, we find that δn− 1 6= 0 and therefore the roots do not have multiplicity ≥ 2, which is a contradiction.

Theorem 1 is now clear.

### 3.4 Properties of cyclotomic polynomials

3.4.1 Definitions

Continuing on the outcomes of the previous section, one can redefine cyclo- tomic polynomials as:

Φn(x) = Y

gcd(n,m)=1



x − e2πin m , for m < n.

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3.4.2 Properties of cyclotomic polynomials Proposition 8. The coefficients of Φn are integers.

Proof. We again write

fn(x) = Φn(x)h(x).

By applying Lemma 1 one finds that the proposition holds.

Proposition 9. For each positive integer n, Φn is irreducible.

Proof. Suppose that Φn(x) = g(x)h(x) for polynomials g, h ∈ Q[x] of degree

≥ 1. By multiplying by a suitable nonzero constant, we may assume g and h are monic. Let ζ be a root of Φn. Then

Φn(ζ) = g(ζ)h(ζ) = 0.

This implies that ζ must be a root of either g or h. This contradicts with the fact that by definition Φn is the monic polynomial with ζ as root of lowest degree. Thus Φn is irreducible.

Proposition 10. Let n ≥ 2, then Φn has a constant term 1.

Proof. The degree of Φnis given by the number of elements m, with m ≤ n and gcd(n, m) = 1. This is equal to the number of elements in the set (Z/nZ) and can be calculated with use of the Euler φ-function. For n ≥ 3 the element −1 mod n has order 2 in the group (Z/nZ), hence 2 divides the order φ(n) of this group. This shows that φ(n) is even.

Now, by definition:

Φn(0) = Y

m<n gcd(n,m)=1

−e2πin m (3.1)

Since gcd(m, n) = 1 implies that gcd(n − m, n) = 1, the conjugate of ζ is equal to:

ζ = e2πin m= e2πin (n−m).

This and the fact that Φn has an even amount of factors, allows us to write (3.1) for n ≥ 3 as follows.

Φn(0) = Y

2m<n gcd(n,m)=1



e2πin m e2πin (n−m)

= Y

2m<n gcd(n,m)=1



e2πin (m+n−m)



= 1

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Thus the proposition holds for n ≥ 3. It is left to show that the proposition holds for n = 2, however in this case one simply gets

Φ2(0) = e2πi2 1 = 1.

Hence the desired result is clear.

Proposition 11. For every n ≥ 2 there are infinitely many primes congru- ent to 1 mod n.

To prove this, we will use the following lemma.

Lemma 3. Let n > 0 and k ∈ Z be integers. If p is a prime number such that p|Φn(nk), then p ≡ 1 mod n.

Proof. In case n = 1, this is trivially true because every number m ≡ 1 mod 1. So assume n > 1. Since Φn divides xn − 1 in Z[x], it follows that p|(nk)n− 1, or equivaltently, (nk)n ≡ 1 mod p. In particular this implies that p does not divide n. So the zeros in Fpof the polynomial xn− 1 ∈ Fp[x]

are all simple. If 0 < m < n and m|n, then Φndivides the polynomial xxmn−1−1. From the two observations above, it follows that every zero of Φnmod p has exact order n in Fp

. So this holds for the zero nk mod p as well. This shows that Fp contains an element of order n, and therefore n|(p − 1) or equivalently, p ≡ 1 mod n.

Proof. (of Proposition 11) From Lemma 3 it follows immedialtely that primes p ≡ 1 mod n exist. Let p1, p2, . . . , pr be a list of such primes. Put

m := n × p1× p2× . . . × pr,

and consider Φn(mk) with k an integer big enough to ensure that φn(mk) >

1.

Since Φn(mk) > 1, it will be divisible by some prime q. Again, from Proposition 10 one gets q ≡ 1 mod n. From Proposition 10 we see that Φn(mk) ≡ 1 mod mk, hence in particular none of the pj can divide Φn(mk).

Therefore, q 6= pj for each j and the proposition holds.

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### 3.5 Special cases

3.5.1 Infinitely many prime numbers 5 mod 8

Let m be a positive integer, m ≤ 4. Then the fourth cyclotomic polynomial is given by:

Φ4(x) = Y

gcd(4,m)=1

(x − e2πi4 m)

= (x − eπi2 )(x − e3πi2 )

= x2+ 1

By applying Proposition 9 we now know that there are infinite many primes of the form p ≡ 1 mod 4.

Now, let α be an even positive integer, such that 4 - α. Then Φ4(α) = α2+ 1 ≡ 5 mod 8.

It follows that the prime divisors of Φ4(α) are all of the form 1 mod 4. That is, 1 mod 8 or 5 mod 8. Assume all prime divisors are congruent with 1 mod 8. Then pi ≡ 1 mod 8, pj ≡ 1 mod 8 and pipj ≡ 12 mod 8. This contradicts the fact that Φ4(α) ≡ 5 mod 8. Then at least one of the prime divisors must be congruent to 5 mod 8.

Since we know that there is at least one prime congruent to 5 mod 8, we can find at least one prime congruent to 5 mod 8 at every α, for 2 | α and 4 - α. Since α can increase infinitely, there are infinite many primes p ≡ 5 mod 8.

3.5.2 Infinitely many prime numbers 7 mod 12

Let n = 6, then Φ6(x) is of the form Φ6(x) = Y

gcd(6,m)=1

(x − e2πi6 m)

= x2− x + 1.

Let α be a positive integer. One has Φ6(α) =

 3 mod 4 if α ≡ 2 mod 4 1 mod 3 if 3 | α.

Now, let α be a positive integer such that α ≡ 2 mod 4 and 3 | α. Then Φ6(α) ≡ 7 mod 12, thus it has prime divisors congruent to 1 mod 12 and prime divisors congruent to 7 mod 12. One can prove in similar way as in 3.5.1 that Φ6(α) has at least one prime divisor congruent to 7 mod 12.

Hence there are infinite primes congruent to 7 mod 12.

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### 4.1 Introduction

Dirichlet’s theorem is as follows:

Theorem 2. Let a ≥ 1 and m ≥ 1 be relatively prime integers. Then there exist infinitely many primes p such that p ≡ a mod m.

As mentioned before, we will require many different fields of mathematics to conclude the proof. The fundamentals of the proof are in group theory.

However, we will also utilize complex numbers, Dirichlet series and ζ- and L-functions. This chapter will be devoted to proving several propositions, from which Dirichlet’s theorem will follow. Our exposition closely follows A course in arithmetic from Jean-Pierre Serre .

### 4.2 Characters

4.2.1 Definitions

Let G be a finite abelian group written multiplicatively.

Definition 3. A character χ is a homomorphism of G to the multiplicative group C. That is, χ : G → C.

Definition 4. bG denotes the group of all these characters χ and is called the dual of G. Let trivial character of bG be given by the character χ, which maps G to 1, χ(a) = 1, for all a ∈ G. We define the group operation on bG as follows: the product of characters χ1, χ2 is χ1χ2 defined by

χ1χ2 : x 7→ χ1(x)χ2(x).

Definition 5. Given any g ∈ G, let κ : bG → C be the function defined by κ : χ 7→ χ(g). κ is a character of bG since χ1χ2(g) = χ1(g)χ2(g) by definition. We obtain tha map  : G → bG .b

Definition 6. The group bG is called the bidual of G.

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4.2.2 Characters

Proposition 12. Let x ∈ G and χ ∈ bG. Then

X

χ∈ bG

χ(x) =

 n if x = 1 0 if x 6= 1,

where n denotes the number of characters χ in bG.

Proof. To prove the above equality, we will first prove the following formula and apply it onto the dual group. We have

X

x∈G

χ(x) =

 n if χ = 1

0 if χ 6= 1, (4.1)

where n denotes the number of elements in G and χ ∈ bG. The first equation is obvious.

For the second one, we choose y ∈ G such that χ(y) 6= 1. By definition χ is a homomorphism, thus the following holds

χ(y)X

x∈G

χ(x) =X

x∈G

χ(xy).

However, the map G → G : x 7→ xy is a bijection. Therefore, {x | x ∈ G} = {xy | x ∈ G}, thus

X

x∈G

χ(xy) = X

x∈G

χ(x).

Therefore

χ(y)X

x∈G

χ(x) =X

x∈G

χ(x) and

(χ(y) − 1)X

x∈G

χ(x) = 0.

This implies that either χ(y) = 1 or P χ(x) = 0 and since we have chosen y such that χ(y) 6= 1, it must be thatP χ(x) = 0.

If one applies formula (4.1) to the dual of group G, one gets

X

χ∈ bG

κ(χ) =

 m if κ = 1 0 if κ 6= 1.

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where m is the number of characters χ in bG and κ ∈ bG . Upon proving thatb m = n and that κ = 1 if and only if x = 1, the desired result follows. This can be accomplished using the following two lemmas.

Lemma 4. The order of G is equal to the order of its dual.

We will proof this by showing that G ∼= bG, then the lemma follows directly.

Proof. Since G is a finite abelian group, there is a unique list (d1, d2, . . . dt), with all d1 ∈ Z, 1 < d1 ≤ d2 ≤ . . . ≤ dt and di | di+1 for i = 1, . . . , t − 1, such that

G ∼= Z/d1Z × Z/d2Z × . . . × Z/dtZ.1 For convenience, we shall name Z/diZ as Gi.

Decompose G into two cyclic groups. That is, G = G1× G2. Define the map

ρ : bG → bG1× bG2 χ 7→ (χ| bG1, χ| bG2).

It is clear that by definition of the group operation of bG, ρ is a homo- morphism. Furthermore it is to say that ρ is surjective. Indeed, for any given (χ1, χ2) ∈ bG1× bG2, we can find a character χ such that ρ(χ1, χ2) = χ1(g12(g2) = χ(g), for (g1, g2) = g. This can simply be done by defining χ as the restriction of χ1 to G1, and χ2 to G2. Ultimately, since G = G1× G2 it is to say that ρ is injective, thus isomorphic.

Now, by induction it follows that the following map is an isomorphism as well:

γ : bG → bG1× . . . × bGt

χ 7→ (χ| bG1, . . . , χ| bG)t.

If χ is a character of Gi, the element wi = χ(g) satisfies widi = 1, hence wi is a dith root of unity. The group of all dith roots of unity shall be denoted as µdi. Conversely, given any wi ∈ µdi defines a character of G by χ : ga7→ wai such that χ(g) = wi. Thus we see that the map

ζ : bG → µdi χ 7→ χ(g), is a bijection.

1Proposition 9.3.1, Chapter 9, .

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Furthermore, since (χ1χ2)(g) = χ1(g)χ2(g), ζ is a homomorphism. We now see that ζ is an isomorphism. Therefore bGi ∼= µdi and bGi is cyclic of order di. We conclude that Gi ∼= cGi. Since this can be done for each i, G ∼= bG holds.

Thus G is of the same order as its dual.

Lemma 5. The map  : G → bG is an isomorphism.b

Proof. To prove  is an isomorphism, we will start off by showing  is a homomorphism. In other words, that (xy) = (x)(y) holds for all x, y ∈ G.

 is defined as follows:

(x) = (χ 7→ χ(x)

| {z }

κ

)

Therefore (x)(y) maps χ to χ(x)χ(y), which equals χ(xy), since χ is a homomorphism. Thus it is to say that (x)(y) = (xy).

Now, in the same way one shows G and bG have the same order, one can also conclude that ord( bG) = ord( bG ) (= ord(G)). To conclude the proof, itb is left to show that  is injective. In other words, if x 6= 1 for x ∈ G, there exists a character of G such that χ(x) 6= 1.

Again write G ∼= Z/d1Z × Z/d2Z × . . . × Z/dtZ. As seen in Lemma 4, there is a wi6= 1 which suits the condition χ(g) = wi, for each i. Then by structure of G, x 6= 1 implies that there is a character χ, such that χ(x) 6= 1.

If we add to this that G and bG are of the same order, we can concludeb

 : G → bG is an isomorphism.b

Equation (4.1) is now clear. Indeed, since Lemma 4 shows that m = n and Lemma 5 proves  : G → bG is an isomorphism one can state that x 6= 1 ifb and only if κ(x) 6= 1.

Definition 7. χ is called a character modulo m, whenever χ is an element of the dual of (Z/mZ). χ(a) = 0, whenever a /∈ (Z/mZ).

For convenience χ can be viewed as a function defined as χ : (Z/mZ) → C in the following sections, such that χ(ab) = χ(a)χ(b).

### 4.3 Ordinary Dirichlet series

4.3.1 Definitions

Definition 8. An ordinary Dirichlet series is of the form X

n≥1

an

ns, with an∈ C and s ∈ C.

(21)

Definition 9. A function f : N → C is said to be multiplicative if f (mn) = f (m)f (n) for all m, n ∈ N, with gcd(m, n) = 1.

Definition 10. A function f : N → C is said to be strictly multiplicative if f (nn0) = f (n)f (n0) for all n, n0 ∈ N.

4.3.2 Partial sums

Proposition 13. If the series f (s) = Pan

ns converges for s = s0, it con- verges uniformly in every domain of the form Re(s − s0) ≥ 0, Arg(s − s0) ≤ α, with α < π2.

Proof. Let s0 = 0. We must prove that if the series P an

ns0 = P an con- verges, there is uniform convergence in every domain of the form Re(s) ≥ 0, Arg(s) ≤ α. Put Am,p = Pp

n=man and let bn = n−s = e−λns, where λn= log n. Consider the following equality.

Sm,m0 =

m0

X

n=m

an ns =

m0

X

n=m

ane−λns=

m0

X

n=m

anbn.

Replace anby Am,n− Am,n−1 and one has.

Sm,m0 =

m0

X

n=m

anbn=

m0

X

n=m

(Am,n− Am,n−1)bn

=

m0

X

n=m

Am,nbn

m0

X

n=m

Am,n−1bn

=

m0

X

n=m

Am,nbn

m0−1

X

n=m−1

Am,nbn+1

=

m0−1

X

n=m

Am,n(bn− bn+1) + Am,m0bm0+ Am,m−1bm

=

m0−1

X

n=m

Am,n(bn− bn+1) + Am,m0bm0 (4.2)

Note that Am,m−1=Pm−1

n=man, which is equal to zero since its upper index is smaller than the lower index.

Hence,

Sm,m0 =

m0−1

X

n=m

Am,n(e−λns− e−λn+1s) + Am,m0e−λm0s.

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Since the series P an converges, there is a N such that if m, m0 ≥ N , we have |Am,m0| ≤ .

|Sm,m0| ≤ 

m0−1

X

n=m

(e−λns− e−λn+1s) + e−λm0s

≤  |s|

x

m0−1

X

n=m

(e−λnx− e−λn+1x) + 1



Note that we have used that for α, β ∈ R, 0 < α < β,

|eαs− eβs| = s

Z β

α

e−tsdt

≤ |s|

Z β α

e−ts dt

= |s|

Z β

α

e−t(x+iy) dt

= |s|

Z β α

e−txe−tiy dt.

Let k = −ty and not that |eik| = | cos k + i sin k| = p

cos2k + sin2k = 1.

Hence

|eαs− eβs| ≤ |s|

Z β α

e−txdt

= |s|

x (e−αx− e−βx).

Moreover, |s|x is bounded by cos α1 = K, thus

|Sm,m0| ≤ (K(e−λnx− e−λn+1x) + 1)

≤ (K + 1).

Hence uniform convergence is clear.

Proposition 14. If f (s) converges uniformly for s = 0, it converges for Re(s) > 0 and the function is holomorphic.

Proof. This follow directly from the Weierstrass theorem. 2

2Theorem 1, Chapter 5, .

(23)

One can now redefine the ordinary Dirichlet series as follows:

Definition 11. The ordinary Dirichlet series is of the form f (s) =

X

n=1

an ns, with an∈ C and s ∈ C.

Proposition 15. If (an) is bounded, the ordinary Dirichlet series is abso- lutely convergent for Re(s) > 1.

First let us recall the definition of absolute convergence:

Definition 12. A sequenceP bn is absolutely convergent if P |bn| = K for some K ∈ R.

Proof. Since (an) is bounded, |an| ≤ l for some l ∈ R. Also,

X

n=1

1 ns

=

X

n=1

1

|nx+iy| =

X

n=1

1

|nx||niy| =

X

n=1

1

|nx||eiy log n|. Let t = y log n and note that |eit| = | cos t + i sin t| = p

cos2t + sin2t = 1.

ThereforeP |n−s| =P n−x.

SinceP n−x converges for x > 1, it is to say thatP

n=1|n−s| tends to k for some k ∈ R. Hence the ordinary Dirichlet series is absolutely convergent for Re(s) > 1.

Proposition 16. If the partial sums Am,p = Pp

n=man are bounded, the Dirichlet series is convergent for Re(s) > 0.

Proof. Assume |Am,p| ≤ K and let bn= n−s. Now, consider

Sm,m0 =

m0

X

n=m

an

ns =

m0

X

n=m

anbn.

If one takes the absolute value of the above equation and applies equation (4.2) it can be written as:

|Sm,m0| =

m0

X

n=m

anbn

≤ K

m0−1

X

n=m

bn− bn+1

+

bm0



≤ K

m0−1

X

1

ns − 1 (n + 1)s

+

1 m0s



(4.3)

(24)

Whenever s is real, one may assume that 1

ns ≥ 1 (n + 1)s.

Hence, if one expands (4.3) for Re(s) > 0, one will find

|Sm,m0| ≤ K

 1

ms − 1

(m + 1)s + 1

(m + 1)s − 1

(m + 2)s + . . . + 1

(m0− 1)s − 1 m0s + 1

m0s



= K ms.

As a conclusion the convergence is clear.

4.3.3 Dirichlet series

Let f be a bounded multiplicative function. With Proposition 15 it is clear that the Dirichelet seriesPf (n)

ns is absolutely convergent for Re(s) > 1.

Proposition 17. The infinite product Y

p∈P

(1 + f (p)p−s+ . . . + f (pm)p−ms+ . . .)

is equal to the Dirichlet series and converges absolutely for Re(s) > 1.

Proof. Let S ⊆ P denote a set of prime numbers and let N(S) be de set of integers all of whose prime factors belong to S. One can now compose the next equality.

Y

p∈S

 X

m=0

f (pm) pms



= X

n∈N(S)

f (n) ns

When S increases, the right hand side tends more and more to the Dirichlet series.

Example: Let S = {2, 5, 11}. Now, one can evaluate N(S) = {2, 4, 5, 8, 10, 11, 16, 20, 22, 25, 33, . . .}. It is clear that with enlargement of S, N(S) tends to N and therefore

Y

p∈S



X

m=0

f (pm) pms



X

n=1

f (n) ns , whenever S increases.

Since the Dirichlet series converges absolutely for Re(s) > 1, it is to say that for S = P the infinite product is equal to the Dirichlet series and converges on the same domain. Therefore one can conclude that

X

n=1

f (n) ns = Y

p∈P

 X

m=0

f (pm) pms



for Re(s) > 1.

(25)

Furthermore if f is multipicative in strict sense, the following holds

X

n=1

f (n) ns = Y

p∈P

 X

m=0

f (p)m pms

 . Additionally, we know that for |x| < |y|

X

n=1

xn yn = 1

1 −xy. Therefore, for |f (p)| < |ps|,

X

n=1

f (n) ns = Y

p∈P

 X

m=0

f (p)m pms



= Y

p∈P

1 1 −f (p)ps

.

Thus, it suffices to show that |f (p)| < |ps|. However, in the next sections f (p) will be equal to either 1 or χ(p) and in both of these cases, |f (p)| < |ps| holds for Re(s) > 0.

### 4.4 The zeta functions

4.4.1 Definitions

Let U be an open subset of C and f : U → C .

Definition 13. A function f is said to be holomorphic in z0 if there exists

 > 0 such that the series f (z) =

X

n=1

f(n)(z0)

n! (z − z0)n converges for |z − z0| < .

Definition 14. The function f is said to be holomorphic on U if it is holomorphic in every z0∈ U .

Definition 15. The zeta function is defined as the ordinary Dirichlet series with f = 1. That is

ζ(s) =

X

n=1

1 ns = Y

p∈P

1 1 −p1s

. Let fn: U → C

Definition 16. A seriesP

n=0fn(x) converges normally if

X|| fn(x) ||:=

Xsup

S

|fn(x)| < ∞.

(26)

4.4.2 The zeta function

Proposition 18. Let f : U → C, f is holomorphic in z0 if and only if f is differentiable in an open environment of z0 and the derivative of f is a continuous function.

Proof. This follows simply from definition 13.

Proposition 19. The zeta function is a nonzero holomorphic function in the half plane Re(s) > 1.

Proof. It is clear that the zeta function is differentiable in an open envi- ronment of s, for every s. Also obviously, ζ(s) 6= 0 in the half plane of Re(s) > 1.

Proposition 20. One has:

ζ(s) = 1

s − 1+ φ(s), where φ(s) is holomorphic for Re(s) > 0.

Proof. We will start by proving ζ(s) is of the form (s − 1)−1+ φ(s).

By definition

ζ(s) =

X

n=1

1 ns

= 1

s − 1+

X

n=1

1 ns − 1

s − 1. Note that

1 s − 1 =

Z 1

t−sdt =

X

n=1

Z n+1 n

t−sdt. (4.4)

Here, the first equality can simply be checked by:

Z 1

t−sdt = t1−s 1 − s

t=1

= 1

s − 1, for Re(s) > 1.

By applying equation (4.4) to the zeta function one gets ζ(s) = 1

s − 1 +

X

n=1

 1 ns

Z n+1 n

t−sdt



= 1

s − 1 +

X

n=1

Z n+1 n

(n−s− t−sdt).

(27)

Hence, the zeta function is of the desired form. Put φn(s) =

Z n+1 n

(n−s− t−sdt) and φ(s) =

X

n=1

φn(s).

Consequently, the required result is found.

Next we will prove the second part of Proposition 20: φ(s) is holomorphic for Re(s) > 0. By Proposition 18 it is clear that φn(s) is holomorphic for all n ∈ [1, . . .). Now, all is left to show is that the series P φn(s) converges normally on all compact sets for Re(s) > 0.

One has:

||φn(s)|| = sup

n≤t≤n+1

Z n+1 n

(n−s− t−sdt)

≤ sup

n≤t≤n+1

Z n+1 n

|n−s− t−s|dt

≤ sup

n≤t≤n+1

|n−s− t−s|

We get that the derivative of the function n−s− t−s is equal to ts+1s . As a result

||φn(s)|| ≤ sup

n≤t≤n+1

s ts+1

. One applies the note from Proposition 15 to get:

||φn(s)|| ≤ |s|

nx+1 with x = Re(s).

Thus

X

n=1

||φn(s)|| < ∞ for Re(s) > 0,

and therefore φn(s) is normally convergent, thus uniformly convergent on all compact sets for Re(s) > 0. Concluding, by Weierstrass theorem φ(s) is holomorphic for Re(s) > 0.3

Note: It is clear the zeta function has a simple pole for s = 1.

Proposition 21. When s → 1, one has X

p∈P

1

ps ∼ log 1 s − 1.

3Theorem 1, Chapter 5, .

(28)

Proof. Let f, g be two functions. By notation, f ∼ g for s → 1 signifies lims→1 f

g = 1.

We will prove this proposition by showing that P

p∈P 1

ps ∼ log ζ(s) as it follows from the note in Proposition 20 that log ζ(s) and log1−s1 both have a pole for s = 1, and therefore log ζ(s) ∼ log1−s1 .

Due to the fact that log is not technically speaking a function, it is necessary to explain what we mean by it. Even for Re(s) > 0, the zeta-function could still be complex valued. Therefore we choose the branch of log ζ(s) in which becomes zero when s → ∞ on the real axes.

Definition 17. For Re(s) > 1, each factor of ζ(s) is of the form 1−α1 , with α < 1. We define log1−α1 as the series P

n=1 αn ns. By definition:

log ζ(s) = logY

p∈P

1 1 −p1s

=X

p∈P

log(1 − 1 ps)−1

=X

p∈P

X

n≥1

1 npns

=X

p∈P

1 ps +X

p∈P

X

n≥2

1 npns.

Note that we have used that for complex numbers − log(1 − z) =P n=1zn

n

for Re(z) < 1.

We write log ζ(s) =P 1

ps + ψ(s), where ψ(s) =P P np−ns. Now it suffices to show ψ is bounded. To show this, take s to be real and write

ψ(s) =X

p∈P

X

n=2

1 npns

≤X

p∈P

X

n=2

1 pns. Furthermore, for |k| > 1 we know that

X

n=2

1

kn = k−2 1 1 − k−1.

(29)

Thus the following holds

ψ(s) ≤X

p∈P

p−2s 1 1 − p−s

=X

p∈P

1 ps(ps− 1)

≤X

p∈P

1 p(p − 1)

X

n=1

1 n(n − 1)

= 1.

Concluding, ψ(s) is bounded and the proposition holds.

### 4.5 The L-functions

4.5.1 Definitions

Let m ≥ 1 and let χ be a character modulo m.

Definition 18. The L-function is defined by the Dirichlet series L(s, χ) =

X

n=1

χ(n) ns . 4.5.2 The L-functions

Proposition 22. For χ 6= 1 the series L(s, χ) converges in the half plane Re(s) > 0.

Proof. To prove this proposition, we will focus on the upper term of the fraction. That is, P

n=1χ(n).

Let

Au,v =

v

X

n=u

χ(n), for u ≤ v.

For n = u, . . . , u + m − 1 Propostition 12 shows that An,m−1=

u+m−1

X

n=u

χ(n) = 0.

It is to say that χ(a+m) = χ(a), thus χ can be viewed as a periodic function, hence |Au,v| is bounded by the number of elements of a period. |Au,v| ≤

(30)

Proposition 23. For χ 6= 1 one has L(s, χ) = Y

p∈P



1 −χ(p) ps

−1

and the series L(s, χ) converges absolutely in the half plane Re(s) > 1.

Proof. This follows directly from Proposition 17.

Proposition 24. For χ = 1, one has

L(s, 1) = F (s)ζ(s) with F (s) =Y

p|m

(1 − 1 ps)

Proof. From Proposition 23, we know that L(s, χ) = Q

p∈P



1 − χ(p)ps

−1

. Now, for χ = 1, we have

L(s, 1) =Y

p-m

 1 − 1

ps

−1

= Y

p∈P

 1 − 1

ps

−1

Y

p|m

 1 − 1

ps



= ζ(s)Y

p|m

 1 − 1

ps

 . Hence, the desired result is clear.

Proposition 25. For χ = 1 the L-function L(s, 1) extends analytically for Re(s) > 0 and has a simple pole for s = 1.

Proof. Since these properties apply to the ζ-function and Proposition 24 holds, it is obvious that these properties also apply to the L-function.

### 4.6 Characters modulo m

4.6.1 Definitions

Let us recall the definition of a character modulo m.

Definition 19. Let m, n ≥ 1, for m, n ∈ Z. χ is called a character modulo m, whenever χ is an element of the dual of (Z/mZ). That is, χ : (Z/mZ)→ C. χ(a) = 0, whenever a /∈ (Z/mZ).

Definition 20. We define ζm(s) as the product of all L(s, χ) over all char- acters modulo m. That is

ζm(s) =Y

χ

L(s, χ).

(31)

4.6.2 Characters modulo m

Proposition 26. For n ∈ Z it is to say that n mod m ∈ (Z/mZ) if and only if gcd(n, m) = 1.

Proof. Let n ∈ Z. If n mod m ∈ (Z/mZ), then by definition one can find b ∈ Z such that nb mod m = (n mod m)(b mod m) = 1 mod m. This implies that nb − 1 = mq, for some q ∈ Z. However, stating nb − mq = 1 is equivalent with stating gcd(n, m) = 1.

Note that for all n with gcd(n, m) 6= 1, by definition χ(n) = 0.

Let n be a positive integer such that gcd(m, n) = 1. We define a new function g(n), which denotes the order of the quotient of (Z/mZ) by the subgroup (n) generated by n.

g(n) := φ(m) ord(n).

Here φ(m) is the Euler φ-function [and ord(n) the smallest integer such that nord(n)≡ 1 mod m].

Proposition 27. The function ζm(s) can be written as ζm(s) =Y

p-m

1



1 −pord(p)s1

g(p).

Proof. By definition ,

ζm(s) =Y

χ

L(s, χ)

=Y

χ

Y

p∈P

1 1 −χ(p)ps

.

Now, let µw be the set of ord(p)th roots of unity. Then, for p - m one has the identity

Y

w∈µw

(1 − wT ) = Y

w∈µw

word(p) Y

w∈µw

(1 w − T )

= eiπ(n−1) Y

w∈µw

(w − T )

= (−1)n−1(−1)n Y

w∈µw

(T − w)

= 1 − Tord(p).

(32)

When we apply this to the product of all characters χ of (Z/mZ) we use the fact that for all w ∈ µw there exists exactly g(p) characters χ such that χ(p) = w. Thus

Y

χ

(1 − χ(p)T ) = (1 − Tord(p))g(p).

Knowing this, one can apply the formula to the ζm-function by putting T = p−s and get the following

ζm(s) =Y

p-m

1



1 −pord(p)s1

g(p).

Proposition 28. L(1, χ) 6= 0 for all χ 6= 1.

Proof. One can prove this by use of contradiction. Assume L(1, χ) = 0, for some χ 6= 1. In Proposition 25 we have seen L(s, 1) has a pole at s = 1.

Then in de product L(s, χ)L(s, 1) the pole of L(s, 1) and the root of L(s, χ) would cancel each other out. Since Proposition 23 demonstrates that all other L(s, χ) are holomorphic it follows that ζm would be holomorphic at s = 1, thus also for all s such that Re(s) > 0. Since it is a Dirichlet series with positive coefficients, it is to say that ζm would converge for all s in the same domain.4 Suppose this is true.

Observe the pth factor of ζm, which is equal to 

1 − 1

pord(p)s

−g(p)

. In the same way as shown in Proposition 17 the following holds.

1



1 −pord(p)s1

g(p) =

 X

k=0

p−k ord(p)s

g(p)

X

k=0

p−k ord(p)g(p)s

=

X

k=0

 1

pφ(m)s

k

.

4Proposition 7, Chapter VI, .

(33)

Then in general, one has

ζm(s) =Y

p-m

1



1 −pord(p)s1

g(p)

≥Y

p-m

X

k=0

 1

pφ(m)s

k!

= X

gcd(n,m)=1

1

nφ(m)s (4.5)

It follows that ζm has all coefficients greater than those of (4.5), which diverges for s = φ(m)−1. This contradicts with our assumption and therefore L(1, χ) 6= 0 for all χ 6= 1.

Remark. Since L(1, χ) 6= 0 for al χ 6= 1 and Proposition 24 holds, we know that L(s, 1) had a simple pole at s = 1. Now we can conclude that ζm also has a simple pole at s = 1.

### 4.7 Density

4.7.1 Definitions

Definition 21. Let B ⊆ A be two sets of positive numbers. The Dirichlet density δ of B is defined as

δ(B) = lim

s→1

P

x∈Ax−s P

x∈Bx−s,

It is to say that the density of a set B is equal to the ratio of elements B in A. One may observe that δ ∈ [0, 1], with δ(B) = 1 in the case of B = A.

4.7.2 Density

Let P again denote the set of all prime numbers and let D ⊂ P . The density δ of D is as follows:

δ(D) = P

p∈D 1 ps

P

p∈P 1 ps

, (4.6)

when s tends to 1.

As we have seen in Proposition 21

X 1

ps ∼ log 1 s − 1,

(34)

when s tends to 1. Therefore equation 4.6 can be rewritten as

δ(D) = P

p∈D 1 ps

logs−11 , when s tends to 1.

Proposition 29. The density of a finite set is 0.

Proof. Let A be a infinite set and let B ⊂ A be a finite set. By definition the density of B is as follows

δ(B) = lim

s→1

fB(s) fA(s).

However, since we can assume that the denominator tends to infinity for s → 1, while the numerator does not, δ(B) = 0 for s → 1.

### 4.8 Dirichlet’s theorem

4.8.1 Definitions

Definition 22. Let m ≥ 1 and let a be such that gcd(a, m) = 1. Pa denotes the set of primes such that p ≡ a mod m.

4.8.2 Dirichlet’s theorem

Theorem 3. The density of Pa is equal to φ(m)1 . The density of Pa is given by the following ratio

P

p∈Pa

1 ps

logs−11 ,

which tends to a certain k for s → 1. Theorem 3 states that k = φ(m)1 . For this to be true, one must prove that P

p∈Pa

1

psφ(m)1 logs−11 .

By definition every element p of Pa is congruent to a mod m. In other words a−1p ≡ 1 mod m for all p ∈ Pa. By Proposition 12 we know that

X

χ

χ(a−1p) =

 n if p ∈ Pa 0 otherwise.

Since n denotes the number of elements in (Z/mZ), one writes X

χ

χ(a−1p) = φ(m) for all p ∈ Pa.

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Distance between sister chromatids (μm) was measured and plotted against the number of Rif1/PICH-positive UFBs. Indicated numbers of anaphases from one representative experiment

Staff outcomes from the caring for aged dementia care REsident study (CADRES): A cluster randomised trial. International Journal of Nursing Studies. Barbosa A, Lord K, Blighe

• On each sheet of paper you hand in write your name and student number!. • Do not provide just

Given that in the first ten minutes fifteen passengers have been submitted to the superficial inspection, what is the probability that in the same period exactly 4 passengers have

• Het gebruik van een computer, rekenmachine, dictaat of boeken is niet

In each case state (with proof) whether the rela- tion is an equivalence relation or not. Problem E) For each of the following statements decide if it is true

Binding of apoE-rich high density lipoprotein particles by saturable sites on human blood platelets inhibits agonist- induced platelet aggregation.. The effect of blood

The main policy implication of this study is that higher education institutions possess the necessary resources to support social innovation initiatives acting as agents of

On my orders the United States military has begun strikes against al Qaeda terrorist training camps and military installations of the Taliban regime in Afghanistan.. §2 These

One can also relate the ideal class group to the Galois group of abelian extension of the field K. But to do so, we must first relate the ideals of the order O to ideals of the

Keywords Path-following gradient method · Dual fast gradient algorithm · Separable convex optimization · Smoothing technique · Self-concordant barrier · Parallel implementation..

Belgian customers consider Agfa to provide product-related services and besides these product-related services a range of additional service-products where the customer can choose

Or, you can use your newly created AMS-TEX or AMSPPT format file to typeset the file amsguide.tex; even if you print out this guide from another source, we recommend using the

In turn, these bounds are conjectured to be optimal for the finite-field setting, which would imply that over finite fields, Szemer´ edi’s theorem with random differences cannot

Since our replacement theorem is easily applicable to all versions of Thompson subgroups and there is a gap in the literature whether the ZJ-theorem holds for other versions of

Door middel van het project 'Expertisecentrum Dierenwelzijn' wordt getracht dierenwelzijn in het hogere groene onderwijs meer in te bedden2. Hiermee wordt een impuls gegeven

I want to thank Erik Brave, Gijs Scheenstra, Ivo Becker, Walter Bruins, and in particular Mark Bouts for many memorable evenings, talking about the meaningful (and sometimes the

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