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ANALYTIC NUMBER THEORY and DIRICHLET's THEOREM Contents 1. Arithmetic Functions 1 2. Dirichlet Products and Mobius Inversion 2

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ANALYTIC NUMBER THEORY and DIRICHLET's THEOREM Contents 1. Arithmetic Functions 1 2. Dirichlet Products and Mobius Inversion 2 ANALYTIC NUMBER THEORY AND DIRICHLET'S THEOREM JOHN BINDER Abstract. In this paper, we prove Dirichlet's theorem that, given any pair h, k with (h; k) = 1, there are infinitely many prime numbers congruent to h (mod k). Although this theorem lies strictly within the realm of number theory, its proof employs a range of tools from other branches of mathematics, most notably characters from group theory and holomorphic functions from complex analysis. Contents 1. Arithmetic Functions 1 2. Dirichlet Products and Mobius Inversion 2 3. Dirichlet Characters 4 4. Orthogonality Relations of Characters 6 5. An Analytic Proof of the Infinitude of Primes 7 6. Dirichlet Series and L-functions 9 7. The Proof of Dirichlet's Theorem 9 8. The Boundedness of log (L (s; χ)) for Nontrivial χ 11 8.1. Step 1: The Convergence of L(s; χ) for <(s) > 0 11 8.2. Step 2: L(s; χ) is holomorphic for <(s) > 0 12 8.3. Step 3: L(s; χ1) has a simple pole at s = 1 13 8.4. Step 4: The function ζk(s) and its Dirichlet Series 14 8.5. Step 5: L(1; χ) = 0 implies the convergence of ζk for <(s) > 0 14 −1 8.6. Step 6: The divergence of ζk at s = φ(k) 15 References 16 1. Arithmetic Functions Analytic number theory is best described as the study of number theory through the use of functions, whose properties can be examined using analytic techniques. The most basic tool of analytic number theory is the arithmetic function. Definition 1.1. A function f : N ! C is called an arithmetic function. Example 1.2. Any function f : C ! C defines an arithmetic function when its domain is restricted to the natural numbers. Examples 1.3. The following arithmetic functions are central to the study of number theory: Date: August 22, 2008. 1 2 JOHN BINDER 1). The divisor function d (n), which counts the number of divisors of a natural number n. 2). The divisor sum function σ (n), which takes the sum of the factors of a natural number n. 3). Euler's totient function φ (n), which counts the numbers k < n such that (n; k) = 1. 4). Von Mangoldt's function Λ (n), where ( log p if n = pm for some prime p; (1.4) Λ (n) = 0 otherwise A subset of arithmetic functions is especially useful to number theorists. These are the multiplicative functions. Definition 1.5. An arithmetic function f is called a multiplicative function if, for all m, n with (m; n) = 1, we have f(mn) = f(m)f(n). Moreover, f is called completely multiplicative if f(mn) = f(m)f(n) for all integers m and n. Example 1.6. The function f(n) = nα is completely multiplicative for any α. Examples 1.7. The functions d(n), σ(n), and φ(n) are all multiplicative, but not completely multiplicative. 2. Dirichlet Products and Mobius Inversion Definition 2.1. Let f and g be arithmetic functions. We define the Dirichlet product of f and g by X n (2.2) (f ? g)(n) = f (d) g d djn It should be noted that, though the Dirichlet product uses the same symbol that typically signifies a convolution, the Dirchlet product is not a convolution in the strict sense since `convolution' requires a group action, and the natural numbers do not form a group under multiplication. However, as seen in the following theorem, the Dirichlet product is itself a group operation on a subset of arithmetic functions, as seen below: Theorem 2.3. Let S be the set of arithmetic functions f such that f (1) 6= 0. Then (S; ?) forms an abelian group. Proof. Since f ? g (1) = f (1) g (1), then S is closed under Dirichlet multiplication. Furthermore, simple algebraic manipulation shows ? to be both commutative and associative. Let ( 1 if n = 1; (2.4) e (n) = 0 otherwise It is clear that e ? f = f for all arithmetic functions f, so that e is the identity element. Finally, we must show that, given f 2 S, there exists an f −1 2 S such that f ? f −1 = e. Given f, we will construct f −1 inductively. First, we need f ? f −1 (1) = e (1), which occurs if and only if f (1) f −1 (1) = 1. Since f (1) 6= ANALYTIC NUMBER THEORY AND DIRICHLET'S THEOREM 3 0, f −1 (1) is uniquely determined. Now, assume that n > 1 and f −1 has been determined for all k < n. Then we have X n X n (2.5) f ? f −1 (n) = f (d) f −1 ) −f (1) f −1 (n) = f (d) f −1 d d djn djn, d>1 This uniquely determines f −1 (n). Thus, we may uniquely determine an f −1 for all f 2 S. One important application of Dirichlet multiplication involves recovering a func- tion from a piecewise sum. For instance, given f, we wish to find a function g such that: X (2.6) f (n) = g (d) djn We may solve this problem using Dirichlet multiplication. First, we write f = g ? 1, where 1 (n) = 1 for all n. Then 1 has an inverse function; call it µ. We then have g = g ? (1 ? µ) = (g ? 1) ? µ = f ? µ Given the importance of µ, we will construct it from the definition. We begin with a lemma. Lemma 2.7. Let f, g, h 2 S, with h = f ? g. If both h and f are multiplicative, then so is g. Proof. We must show that, for all m, n, with (m; n) = 1, we have g (mn) = g (m) g (n). We will show this by inducting upwards on the quantity mn. Since f is multiplicative, we have f (1) = f (1 · 1) = f (1)2 ) f (1) = 0 or 1. Since f has an inverse, we must have f (1) = 1. Similarly, we have h (1) = 1. Therefore, g (1) = 1, so that g (1 · 1) = g (1) · g(1). This proves the base case. Now for our inductive step. Pick m, n so that mn > 1. By definition, we have X mn (2.8) h (mn) = f (d) g d djmn For all d, we may decompose d into dm · dn, where dmjm and dnjn. Moreover, since (m; n) = 1, this decomposition is unique. Therefore, we have X m n (2.9) h (mn) = f (dm · dn) g · dm dn dmjm, dnjn m n Since (m; n) = 1, then (dm; dn) = ; = 1. Since f is multiplicative dm dn we have f(dm · dn) = f(dm) · f(dn), and by our inductive hypothesis we have m n m n g · = g g so long as dm · dn < mn. Therefore, we may write. dm dn dm dn (2.10) 0 1 X m n h (mn) = @ f (dm) f (dn) g g A − g (m) g (n) + g (mn) dm dn dmjm, dnjn 4 JOHN BINDER 0 1 0 1 X m X n (2.11) = @ f (dm) g A @ f (dn) g A − g (m) g (n) + g (mn) dm dn dmjm dnjn (2.12) = h (m) h (n) − g (m) g (n) + g (mn) Since h is multiplicative, we must have g (m) g (n) = g (mn). This completes the inductive step and therefore the proof. Since e is multiplicative, we have the following corollary: Corollary 2.13. If f is multiplicative, then so is f −1. Using lemma 2.7, we can construct µ, the Dirichlet inverse of 1. Theorem 2.14. Let ( 0 if 9k > 1 : k2jn; (2.15) µ (n) = m (−1) if n = p1 · p2 · ::: · pm Then µ = 1−1 Proof. Since 1 is multiplicative, then so too must be its inverse. Therefore, by determining µ (pm) for each prime power pm, we determine µ. It is clear that we must have µ (1) = 1. Consider µ (p), for p prime. Then (2.16) e(p) = 0 = 1 ? µ (p) = µ (1) + µ (p) ) µ (p) = −1 Next, consider p2. We have (2.17) 0 = 1 ? µ p2 = µ (1) + µ (p) + µ p2 = 1 + −1 + µ p2 ) µ p2 = 0 Finally, assume µ pk = 0 for all 2 ≤ k ≤ m − 1. Then we have (2.18) 0 = 1 ? µ (pm) = µ (1) + µ (p) + µ p2 + ::: + µ pm−1 + µ (pm) = 1 + −1 + 0 + ::: + 0 + µ (pm) ) µ (pm) = 0 This shows µ to be as stated in equation 2.15 for all prime powers. Because we know µ to be multiplicative, then 2.15 correctly states µ for all natural numbers. This function µ is called the Mobius function. Remark 2.19. One should note that, even though 1 is completely multiplicative, its inverse µ is not. In general, the inverse of a completely multiplicative function is not completely multiplicative. If we examine the proof of lemma 2.7, we see that we cannot uniquely decompose a divisor d of mn into a divisor of m and a divisor of n unless m and n are relatively prime. Therefore, proof of lemma 2.7 fails when (m; n) 6= 1 at equation 2.11. ANALYTIC NUMBER THEORY AND DIRICHLET'S THEOREM 5 3. Dirichlet Characters Recall the definition of a character on an abelian group. Definition 3.1. Let G be a finite abelian group. We call τ : G ! C a character on G if for all g 2 G, τ(g) 6= 0 and for all g, h 2 G, we have τ (gh) = τ (g) τ (h).
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