ANALYTIC NUMBER THEORY AND DIRICHLET S THEOREM

Transcription

1 ANALYTIC NUMBER THEORY AND DIRICHLET S THEOREM JOHN BINDER Abstract. In this aer, we rove Dirichlet s theorem that, given any air h, k with h, k) =, there are infinitely many rime numbers congruent to h mod k). Although this theorem lies strictly within the realm of number theory, its roof emloys a range of tools from other branches of mathematics, most notably characters from grou theory and holomorhic functions from comlex analysis. Contents. Arithmetic Functions 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 8.. Ste : The Convergence of Ls, χ) for Rs) > Ste 2: Ls, χ) is holomorhic for Rs) > Ste 3: Ls, χ ) has a simle ole at s = Ste 4: The function ζ k s) and its Dirichlet Series Ste 5: L, χ) = 0 imlies the convergence of ζ k for Rs) > Ste 6: The divergence of ζ k at s = φk) 5 References 6. Arithmetic Functions Analytic number theory is best described as the study of number theory through the use of functions, whose roerties can be examined using analytic techniques. The most basic tool of analytic number theory is the arithmetic function. Definition.. A function f : N C is called an arithmetic function. Examle.2. Any function f : C C defines an arithmetic function when its domain is restricted to the natural numbers. Examles.3. The following arithmetic functions are central to the study of number theory: Date: August 22, 2008.

2 2 JOHN BINDER ). 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 < uch that n, k) =. 4). Von Mangoldt s function Λ n), where log if n = m for some rime,.4) Λ n) = A subset of arithmetic functions is esecially useful to number theorists. These are the multilicative functions. Definition.5. An arithmetic function f is called a multilicative function if, for all m, n with m, n) =, we have fmn) = fm)fn). Moreover, f is called comletely multilicative if fmn) = fm)fn) for all integers m and n. Examle.6. The function fn) = n α is comletely multilicative for any α. Examles.7. The functions dn), σn), and φn) are all multilicative, but not comletely multilicative. 2. Dirichlet Products and Mobius Inversion Definition 2.. Let f and g be arithmetic functions. roduct of f and g by 2.2) f g) n) = n ) f d) g d d n We define the Dirichlet It should be noted that, though the Dirichlet roduct uses the same symbol that tyically signifies a convolution, the Dirchlet roduct is not a convolution in the strict sense since convolution requires a grou action, and the natural numbers do not form a grou under multilication. However, as seen in the following theorem, the Dirichlet roduct is itself a grou oeration on a subset of arithmetic functions, as seen below: Theorem 2.3. Let S be the set of arithmetic functions f such that f ) 0. Then S, ) forms an abelian grou. Proof. Since f g ) = f ) g ), then S is closed under Dirichlet multilication. Furthermore, simle algebraic maniulatiohows to be both commutative and associative. Let if n =, 2.4) e n) = 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 S, there exists an f S such that f f = e. Given f, we will construct f inductively. First, we need f f ) = e ), which occurs if and only if f ) f ) =. Since f )

3 ANALYTIC NUMBER THEORY AND DIRICHLET S THEOREM 3 0, f ) is uniquely determined. Now, assume that n > and f has been determined for all k < n. Then we have 2.5) f f n) = f d) f n ) f ) f n) = f d) f n ) d d d n d n, d> This uniquely determines f n). Thus, we may uniquely determine an f for all f S. One imortant alication of Dirichlet multilication involves recovering a function from a iecewise sum. For instance, given f, we wish to find a function g such that: 2.6) f n) = d n g d) We may solve this roblem using Dirichlet multilication. First, we write f = g, where n) = for all n. Then has an inverse function; call it µ. We then have g = g µ) = g ) µ = f µ Given the imortance of µ, we will construct it from the definition. We begin with a lemma. Lemma 2.7. Let f, g, h S, with h = f g. If both h and f are multilicative, theo is g. Proof. We must show that, for all m, n, with m, n) =, we have g mn) = g m) g n). We will show this by inducting uwards on the quantity mn. Since f is multilicative, we have f ) = f ) = f ) 2 f ) = 0 or. Since f has an inverse, we must have f ) =. Similarly, we have h ) =. Therefore, g ) =, so that g ) = g ) g). This roves the base case. Now for our inductive ste. Pick m, o that mn >. By definition, we have 2.8) h mn) = mn ) f d) g d d mn For all d, we may decomose d into d m d n, where d m m and d n n. Moreover, since m, n) =, this decomosition is unique. Therefore, we have 2.9) h mn) = d m m, d n n m f d m d n ) g d m ) n ) m Since m, n) =, then d m, d n ) = d m, n d n =. Since f is multilicative we have fd ) m d n ) ) = fd m )) fd n ), and by our inductive hyothesis we have g n m n d n = g d m g d n so long as d m d n < mn. Therefore, we may write. m d m 2.0) h mn) = d m m, d n n ) ) m n f d m ) f d n ) g g g m) g n) + g mn) d m d n d n

4 4 JOHN BINDER 2.) = d m m ) m f d m ) g d m ) n f d n ) g g m) g n) + g mn) dn n d n 2.2) = h m) h n) g m) g n) + g mn) Since h is multilicative, we must have g m) g n) = g mn). This comletes the inductive ste and therefore the roof. Since e is multilicative, we have the following corollary: Corollary 2.3. If f is multilicative, theo is f. Using lemma 2.7, we can construct µ, the Dirichlet inverse of. Theorem 2.4. Let 2.5) µ n) = Then µ = 0 if k > : k 2 n, ) m if n = 2... m Proof. Since is multilicative, theo too must be its inverse. Therefore, by determining µ m ) for each rime ower m, we determine µ. It is clear that we must have µ ) =. Consider µ ), for rime. Then 2.6) e) = 0 = µ ) = µ ) + µ ) µ ) = Next, consider 2. We have 2.7) 0 = µ 2) = µ ) + µ ) + µ 2) = + + µ 2) µ 2) = 0 Finally, assume µ k) = 0 for all 2 k m. Then we have 2.8) 0 = µ m ) = µ ) + µ ) + µ 2) µ m ) + µ m ) = µ m ) µ m ) = 0 This shows µ to be as stated in equation 2.5 for all rime owers. Because we know µ to be multilicative, then 2.5 correctly states µ for all natural numbers. This function µ is called the Mobius function. Remark 2.9. One should note that, even though is comletely multilicative, its inverse µ is not. In general, the inverse of a comletely multilicative function is not comletely multilicative. If we examine the roof of lemma 2.7, we see that we cannot uniquely decomose a divisor d of mn into a divisor of m and a divisor of n unless m and n are relatively rime. Therefore, roof of lemma 2.7 fails when m, n) at equation 2..

5 ANALYTIC NUMBER THEORY AND DIRICHLET S THEOREM 5 3. Dirichlet Characters Recall the definition of a character on an abelian grou. Definition 3.. Let G be a finite abelian grou. We call τ : G C a character on G if for all g G, τg) 0 and for all g, h G, we have τ gh) = τ g) τ h). Remark 3.2. For a finite abelian grou G, τ [G] T, since for each g G, there is an m > 0 such that g m =. Let τ be a character on the multilicative grou Z/kZ). We may extend the domain of this character to the entire set of natural numbers in the following manner: first, let h Z/kZ be the equivalence class modulo k containing h. We extend the domain of τ to the entire set of natural numbers in the most obvious way ossible: let τ n) if n, k) = χn) = Note that n Z/kZ) if and only if n, k) =. Therefore, this function is welldefined. We call χ a Dirichlet character modulo k. To formalize: Definition 3.3. A function χ : N C is called a Dirichlet character modulo k if there exists a character τ on the grou Z/kZ) such that τ n) if n, k) = χn) = It is easy to see that χ has two imortant roerties: first, it is k-eriodic that is, χa + k) = χa) for all a); and second, it is comletely multilicative that is, χm n) = χm)χn) for all m and n). The figures below show the Dirichlet characters modulo 5 and modulo 8. It is simle to check that no more exist, though we will rove this fact as a more general statement later. x mod 5) χ 0 χ 2 0 χ 3 0 i i χ 4 0 i i x mod 8) χ χ χ χ Remark 3.4. In both of the above tables, we have defined χ as the Dirichlet character modulo k that satisfies χ h) = if h, k) = and. This character is called the trivial or rincile) character modulo k. The trivial character is denoted by χ solely from convention, though this secific character will lay a secial role later.

6 6 JOHN BINDER On the one hand, all Dirichlet characters are eriodic, comletely multilicative er the definition. However, interestingly, all nontrivial functions that have both of these roerties are Dirichlet characters: Theorem 3.5. If χ is an arithmetic function that is both eriodic and comletely multilicative, and it is not universally 0, then it is a Dirichlet character. Proof. Let k be the minimal eriod of χ. Since χ is k-eriodic, it is constant in each equivalence class modulo k. Moreover, since it is comletely multilicative, its values on Z/kZ) make it a character. Therefore, we need to show that χn) = 0 if and only if n, k). First, assume n, k) = ; then we have n φk) mod k). Since χ is comletely multilicative and k-eriodic, we have χ n) φk) = χ n φk)) = χ ). Thus, if χ n) = 0, then χ ) = 0. However, if χ ) = 0, then we have χ m) = χ ) χ m) = 0 for all m. Hence, if n, k) = and χ n) = 0, then χ is universally 0. On the other hand, assume there is an uch that n, k) > and χ n) 0. Then there is at least one rime such that k and χ ) 0. Consider this, and let m be any natural number. Because χ is k-eriodic and comletely multilicative, we have χ m) χ ) = χ m) = χ m + k) = χ ) χ m + k ) Since χ ) 0, we must have χ m) = χ m + k/) for all m. But this means that χ is k -eriodic. This violates the stiulation that k is the minimal eriod of χ. It should be remarked that the function n) = is technically a Dirichlet character modulo. Proosition 3.6. There are φ k) Dirichlet characters modulo k. Proof. We rove a more general roosition: let G be an abelian grou with G = N. Then there are N characters on G. First, since G is finite and abelian, we may decomose it into the direct roduct of grous C ai, where C ai are cyclic grous of order a i. Consider the cyclic grou C ai with generator x i. Then, since x ai i =, we have τx i ) ai =, this leaves a i ossibilities for τx i ), and each choice determines a character on C ai. Moreover, since x i is a generator, the choice of τx i ) uniquely determines τ. Consider the decomosition of G into cyclic grous; write G = C a... C am. That is, there are elements x i G such that x ai i = and each g G can be exressed uniquely as i xti i, with 0 t i < a i. Let τ i be a character on the subgrou generated by x i. Then it is clear that the function defined by τg) = i τ ix ti i ) is a character on G, and that i τ i = i τ i if and only if τ i = τ i for all i. We must now show that each character on G can be exressed as a roduct of τ i s. Let x i be as above.. Then, given g G, we may exress g uniquely as g = i xti i with 0 t i < a i. Thus, for all g G, we have τg) = i τx i) i. Moreover, the restriction of τ to the subgrou generated by x i is itself a character on that subgrou. Therefore, we may exress any character τ on G as a roduct of characters τ i, where τ i is a character on the subgrou generated by x i. But this subgrou is isomorhic to C ai, so there are a i characters on that subgrou. Hence, the number of characters on G is equal to i a i = G. Since each Dirichlet character modulo k is uniquely determined by a character on Z/kZ), the number of Dirichlet characters modulo k is Z/kZ) = φk).

7 ANALYTIC NUMBER THEORY AND DIRICHLET S THEOREM 7 Proosition 3.7. The φ k) Dirichlet characters modulo k form a grou under multilication. Proof. We see that the trivial character is the identity element of the grou. If χ is a Dirichlet character, then its inverse is χ, the comlex conjugate of χ. Moreover, the set of characters modulo k is closed under multilication, since χ χ )gh) = χgh)χ gh) = χg)χh)χ g)χ h) = χ χ )g)χ χ )h), χ χ )n) = 0 if and only if n, k), and since the roduct of two k-eriodic functions is itself k- eriodic. 4. Orthogonality Relations of Characters In this section, we will rove the following identity, which will be useful in our roof of Dirichlet s theorem. Theorem 4.. Let χ,..., χ φk) be the Dirichlet characters modulo k. Then we have φk) φ k) if h mod k) and h, k) =, 4.2) χ i h)χ i ) = i= We will rove this theorem in general, for any Abelian grou. Since a Dirichlet character is simly an extension of a character on Z/kZ), the theorem will follow. Lemma 4.3. Let G be an abelian grou, where G = N, and let τ,..., τ N be the characters on G. Then we have τ g) = g G N if τg) = for all g, Proof. Obviously, the sum of the trivial character on this grou is the size of the grou, N. Otherwise, assume χ is not the trivial character, and ick b such that τ b). Moverover, the grou G is invariant under multilication by any of its elements. We therefore have 4.4) τ g) = τ b g) = τ b) τ g) = τ b) τ g) g G g g g Since τ b), we must have τ g) = 0. From here, we have another lemma: Lemma 4.5. Let A be the N N matrix [τ i g j )] i,j N and let A denote the conjugate transose of of A. Then we have AA = NI. Proof. Consider the i, j coordinate of AA. If v i denotes the vector of row i in A, then this quantity is equal to v i v j = g τ i g) τ j g). But τ i τ j is itself a character, and is the trivial character if and only if i = j. Therefore, we have AA ) ij = N if i = j and ; that is, AA = NI. From here, we rove the generalization of theorem 4..

8 8 JOHN BINDER Proosition 4.6. Given g i, g j G and τ,..., τ N the characters on G, we have N N if i = j τ l g i ) τ l g j ) = l= Proof. Since AA = NI, we have A A = NI; this imlies A A) ij = N if i = j and. Moreover, we can examine this quantity via matrix multilication. Let w i signify a column vector of A; then we have N A A) ij = w i w j = τ l g i ) τ l g j ). Given what we know about A A, this roves the roosition. Theorem 4. follows as an immediate corollary, since χ is an extension of a character on Z/kZ) and vanishes on those n not relatively rime to k. 5. An Analytic Proof of the Infinitude of Primes Almost two and a half millennia ago, Euclid gave the first, and most standard, roof of the infinitude of rimes. The argument is simle enough: assume there are finitely many; multily them all together and add one; the new quantity is not divisible by any rime in the list; contradiction. This argument, however, cannot be generalized to rove that there are infinitely many rimes congruent to h mod k), because numbers from one equivalence class may have divisors from a different class. Therefore, a different tye of argument is necessary to rove Dirichlet s theorem on rimes in arithmetic rogressions. We will summarize an analytic roof of the infinitude of rimes due to Euler; this roof will act as a starting oint for the roof of Dirichlet s theorem. l= Theorem 5.. There are infinitely many rimes. Proof. Define the function 5.2) ζ s) = This sum converges for all s >. However, as s +, we have ζ s). Consider the roduct + 2 s + 4 ) s s + 9 ) s s + ) 25 s where the roduct extends over all rimes. Because each natural number has a unique rime factorization, this roduct covers every natural number exactly once: i.e, we have 5.3) ζ s) = + s + ) 2s +... = ) s Taking the logarithm of both sides and alying the Taylor exansion for log + x) yields 5.4) log ζ s) = log ) s = n ns = s + n ns n=2

9 ANALYTIC NUMBER THEORY AND DIRICHLET S THEOREM 9 we may searate this sum as above since it is absolutely convergent for s > ). Since ζ s) as s +, so does its logarithm. Therefore, if we cahow that the second sum is bounded as s +, we will show that /s is unbounded as s aroaches. Fortunately, for all s we have 5.5) n=2 n ns < n = n=2 2 / < m=2 m m ) = Hence, we must have /s unbounded as s aroaches. This is only ossible if there are infinitely many rimes. Remark 5.6. In the above roof, the logarithm function was well-defined because our domain was restricted to the real numbers. However, the logarithm of a comlex number is not well-defined. Instead, we define the logarithm of a comlex number by its Taylor series; that is z n 5.7) log z) = n The identity log ab) = log a) + log b) still holds under this definition of the logarithm. 6. Dirichlet Series and L-functions Unfortunately, the zeta function by itself is insufficient to attack Dirichlet s theorem because it leaves no means for distinguishing between the residue classes modulo k. For this, we need a more general class of functions. Definition 6.. Let a n and λ n be two sequences of comlex numbers. s C, we define the function a n F s) = e λns Such a series is called a Dirichlet series. For all Examles 6.2. The zeta function is defined by a Dirichlet series, where a n = and λ n = log n). In fact, any series of the form a n / is a Dirichlet series where λ n = logn). The nice roerties of Dirichlet characters they are both eriodic and multilicative and interact nicely with one another) will be of great use in our attack on Dirichlet s theorem. Dirichlet series whose coefficients are given by Dirichlet characters are called L-functions. Definition 6.3. Let χ be a Dirichlet character modulo k, and let s be a comlex number. Then we define the function χ n) 6.4) L s, χ) = Such a function is called an L-function. Examle 6.5. Let χ be the nontrivial character modulo 4. Then we have L s, χ) = 3 s + 5 s 7 s Secifically, we have L, χ) = π 4. We now have all the concetual tools in lace to move on to the final stage of this aer: the roof itself.

10 0 JOHN BINDER 7. The Proof of Dirichlet s Theorem All necessary aaratus is now in lace to comlete the roof of Dirichlet s theorem. The thrust of the roof is similar to that of the roof of the infinitude of the rimes given iection 5. However, we will emloy L-functions in lace of the zeta function to isolate rimes in a residue class modulo k. The roof will consist of two arts: first, we will show that if L s, χ) is bounded and nonzero as s + for nontrivial χ, then Dirichlet s theorem holds. The second, and more challenging art, is to show that L, χ) is, in fact, bounded and nonzero for nontrvial χ. We will rove the easier ste first. This requires a small lemma. Lemma 7.. Let χ denote the trivial character modulo k. Then L s, χ ) is unbounded as s + Proof. We have L s, χ ) = n,k)= > qk) s = k s Since the sum on the right aroaches infinity as s aroaches, and the term /k s aroaches /k, this quantity must diverge. Theorem 7.2. If L s, χ) is bounded and nonzero as s for nontrivial χ, then we have lim s + s = This fact imlies Dirichlet s theorem. q= h mod k) Proof. We will examine L s, χ) as s +. Because χ is comletely multilicative, we may factor an L-function just as we did the zeta functioee equation 5.3). That is 7.3) L s, χ) = χ n) = q= q s + χ) ) s + χ2 ) 2s +... = χ ) ) s Recall from remark 5.6 that, although the logarithm function is not well-defined on the comlex lane, we may still define the logarithm by its Taylor series, which retains the roerty that log a)+log b) = log ab). Taking the logarithm of L s, χ) yields 7.4) log L s, χ)) = log χ ) ) s = χ ) n n ns Since this sum is absolutely convergent for s >, we may break it into two arts as shown: χ ) n ns = χ ) s + χ ) n ns n=2 We see that the second sum is bounded for s >, since χ ) n ns n < n Rs) n = 2 / < m m ) = n=2 n=2 n=2 m=2

11 ANALYTIC NUMBER THEORY AND DIRICHLET S THEOREM Hence, we have log L s, χ)) = χ) + O ). s So far, we have worked only with a general Dirichlet character χ. We will now ick a k, and let χ, χ 2,... χ φk) be the Dirichlet characters modulo k. Pick h such that h, k) =, and consider the sum φk) χ i h) log L s, χ)) = φk) χ i h) φ k) φ k) i= i= χ i ) s + O ) Again, for s >, this sum is absolutely convergent. We may therefore rearrange the summations to yield φk) φ k) i= χ i h) log L s, χ)) = φ k) φk) s However, recall from theorem 4. that φk) φ k) if h mod k), χ i h) χ ) = Since h, k) =. Hence, we have i= h mod k) i= χ i h) χ ) + O ) s = φk) χ i h) log L s, χ i )) + O ) φ k) i= We know that L s, χ ) as s ; thus, so too does log L s, χ )). Consider the other characters modulo k: if L s, χ i ) is both bounded and nonzero as s, then its logarithm will be bounded. Moreover, if log L s, χ i )) is bounded for all nontrivial characters χ i, we must have φk) i= χ i h) log L s, χ i )), and therefore h mod k) /s, unbounded. Hence, if we cahow L s, χ i ) to be bounded and nonzero at s = for nontrivial χ, then Dirichlet s theorem will follow. 8. The Boundedness of log L s, χ)) for Nontrivial χ We now have the more difficult art of the roof: showing that log L s, χ)) is bounded as s aroaches for nontrivial Dirichlet characters χ. In order to do this, we must show that L, χ) is both convergent and nonzero. We will rove this iix stes. ). We will show that, for nontrivial χ, n χ n) /ns converges for all s in the half-lane R s) > 0. This not only shows L, χ) to be finite, but will also aid us ihowing that L, χ) 0. 2). We will use ste to show that L s, χ) is holomorhic in this domain. 3). We will show that L s, χ ) is holomorhic excet for a simle ole at s =. 4). Consider the function ζ k s) = χ L s, χ). From stes 2 and 3, we see that if there is a χ such that L, χ) = 0, then ζ k s) is holomorhic for all R s) > 0. We will show that, in the domain R s) >, we can exress ζ k as a convergent Dirichlet series with nonnegative real coefficients. 5). We will show that, if there is a χ such that L, χ) = 0, then the Dirichlet series defining ζ k is convergent for all s in the domain R s) > 0.

12 2 JOHN BINDER 6). Finally, we will use our findings ite 4 to show that the Dirichlet series defining ζ k is unbounded at s = φ k). This will contradict our assumtion that there exists a χ so that L, χ) = 0, comleting the roof. 8.. Ste : The Convergence of Ls, χ) for Rs) > 0. We will rove a generalized version of the needed result. Lemma 8.. Let a n be a sequence of comlex numbers so that the function S x) = n x a n is bounded say by J). Then the Dirichlet series a n converges for all s in the domain R s) > 0. Proof. We will show this sum to be Cauchy. By Abel s summation formula, we have M a n = S M) M s S m) M S t) m s + s dt m ts+ n=m+ Let σ denote the real art of s. Taking absolute values in the above equation yields M a n S M) S m) M S t) 8.2) M σ + m σ + s dt n=m+ m tσ+ ) 2 M J m σ + s dt 2 t σ J m σ + s σ M σ ) m σ 2J m σ + s ) σ m For a give where σ > 0, this quantity aroaches 0 as m aroaches. This shows the sum n a n/ to be Cauchy, and therefore convergent, for all s in the domain R s) > 0. The same logic can be used to rove a slightly more general version: Corollary 8.3. Consider a o that the sum n a n/0 is bounded. Then the Dirichlet series a n n=0 converges for all s in the domain R s) > R s 0 ). Geometrically, this corollary shows that a Dirichlet series converges in a halflane where we count both and C as half-lanes). Using lemma 8., we rove the convergence of L s, χ) in the desired domain for nontrivial χ. Proosition 8.4. If χ is a nontrivial character modulo k, then we have L s, χ) finite for all s with real art greater than 0. Secifically, L, χ) is finite. Proof. By lemma 8., it is sufficient to show that the sum S x) = n x χ n) is bounded.

13 ANALYTIC NUMBER THEORY AND DIRICHLET S THEOREM 3 If χ is nontrivial, we have S l k) = 0 for all l =, 2,.... Moreover, we have if n, k) = χ n) = We therefore have S n) φ k) Ste 2: Ls, χ) is holomorhic for Rs) > 0. We wish to show that L s, χ) is holomorhic in the domain R s) > 0. To do this, we make use of a well-known lemma from comlex analysis. Lemma 8.5. Let f n be a sequence of functions, holomorhic iome subset A of C, that converge uniformly on A to a function f. Then f is holomorhic on A, and d dz f = lim n d dz f n. Furthermore, it is sufficient to show that f n converges to f uniformly in each comact subset of A. Proosition 8.6. Let K be a comact subset of s C R s) > 0}, and let the sum n a n be bounded again by J). Then the Dirichlet series F s) = n χ n) /ns converges uniformly on K for all nontrivial characters χ. It follows that F is holomorhic in the given half-lane. Proof. We will use equation 8.2, which states M a n 2J n=m+ m σ + s σ where σ denotes the real art of s, for all nontrivial χ. In articular, we will use this to show that the Dirichlet series is uniformly Cauchy on K. Let α = inf z K R z). Since K is comact, we must have α > 0. Furthermore, let β = su z K z. Again, since K is comact, β is well-defined and finite. Therefore, for all s K, we have M a n 2J m α + β ) α n=m+ This quantity aroaches 0 as m. Moreover, the constants α and β deend only on K and are indeendent of the individual elements of K. Therefore, the Dirichlet series defining L is uniformly Cauchy, and therefore uniformly convergent, on K. Thus, F s) is holomorhic in its domain of convergence. Again, the same logic shows that if n a n/0 is bounded, then the function F s) defined by the Dirichlet series n a n/ is holomorhic in the domain R s) > R s 0 ). Secifically, a Dirichlet series is holomorhic in its half-lane of convergence. Of course, since the artial sums of nontrivial Dirichlet characters are bounded, we have the following corollary: Corollary 8.7. For nontrivial χ, we have L s, χ) holomorhic in the domain R s) > 0. Ite 5, we will make use of the derivative of a function defined by a Dirichlet series. Fortunately, lemma 8.5 tells us that we may differentiate a Dirichlet series termwise in its domain of convergence. )

14 4 JOHN BINDER Corollary 8.8. Let F s) = n f n) /ns be a Dirichlet series that converges for R s) > 0. Then F k) s) = ) k f n) log n) k Proof. Termwise differentiation yields d ds F s) = n f n) log n) /ns. The corollary follows inductively Ste 3: Ls, χ ) has a simle ole at s =. Proosition 8.9. The function L s, χ ) is holomorhic in the entire domain R s) > 0 excet for a simle ole at s =. Proof. We have L s, χ ) = n,k)= /ns. Factoring this sum yields L s, χ ) = ) s = ζ s) ) s k k The zeta function is holomorhic for R s) excet for a simle ole at s =. Moreover, for a given, we have / s holomorhic everywhere and nonzero at s =. Since there are only finitely many to consider, we must have Ls, χ) holomorhic excet for a simle ole at s = Ste 4: The function ζ k s) and its Dirichlet Series. We consider the function ζ k s) = L s, χ) χ where the roduct is taken over all Dirichlet characters modulo k. Since we know L s, χ ) has a simle ole at s = and that L s, χ) is holomorhic for all s with R s) > 0, then this function will be holomorhic in this domain if L s, χ) = 0 for some χ. Therefore, examination of this function is sufficient to rove Dirichlet s theorem. Fortunately, this function has some nice roerties. Proosition 8.0. For R s) >, ζ k s) is defined by a convergent Dirichlet series with ositive integer coefficients. Proof. We have 8.) ζ k s) = χ L s, χ) = χ k χ ) ) s = k χ χ ) ) s Consider χ χ ) s ). Let f ) be the order of modulo k; then the values χ ) are the f ) th roots of unity. Moreover, if we consider the character grou, the set of characters χ such that χ ) = w where w is a root of unity for f )) is a coset of the subgrou χ χ ) = }. Hence, there are exactly g ) = φ k) /f ) characters that attain each w. We therefore have 8.2) χ χ ) ) s = w f) = w s ) ) g) = ) ) g) sf)

15 ANALYTIC NUMBER THEORY AND DIRICHLET S THEOREM 5 Therefore, when we take the roduct over all rimes, we have ζ k s) = ) g) sf) k For Rs) >, each term in the roduct is equal to + / s f) + / 2s f) +...) g). Thus, in this domain, we may exress ζ k as a Dirichlet series with ositive integer coefficients Ste 5: L, χ) = 0 imlies the convergence of ζ k for Rs) > 0. We have shown that ζ k is defined by a Dirichlet series with nonnegative real coefficients in the domain R s) >. We now wish to show that if it is holomorhic in the half lane R s) > 0, then the series defining it is convergent in the same domain. We will rove a lemma; the roosition will immediately follow. Lemma 8.3. Let F s) be a function defined in the half-lane R s) > c by the convergent Dirichlet series f n) F s) = where f n) is real and nonnegative. Moreover, let F be holomorhic in the domain Rs) > c, with c < c. Then the Dirichlet series defining F converges for all R s) > c. Proof. Consider some oint a > c. Then F is holomorhic, and therefore analytic, at a, and therefore we may define F by its ower series F k) a) F s) = s a) k k! k=0 Moreover, from ste 3, we know that F k) a) = ) k f n) log n) k We may therefore write the Taylor series in this form: a s) k f n) log n) k F s) = k! k=0 Pick b > c. Then there is a neighborhood O of a that both contains b and lies entirely in the set z Rz) > c }. Therefore, F is holomorhic in the entirety of O, so that the ower series reresentation for F is valid for b. Moreover, since each term in this double-sum is real and nonnegative, we may rearrange the order of summation, giving F b) = f n) n a k=0 a b) log n)) k = k! = f n) n = a a b) f n) n b f n) logn) n a ea b) Thus, the Dirichlet series converges at b for any b > c, and therefore in the entire half-lane Rs) > c.

16 6 JOHN BINDER Corollary 8.4. If ζ k is holomorhic at s =, the Dirichlet series defining it is convergent in the entire domain where R s) > 0. Proof. Recall from ste 4 that if ζ k is holomorhic at s =, it is holomorhic in the entire half-lane where R s) > 0. Also recall that the Dirichlet series defining it is convergent for Rs) >. Therefore, by lemma 8.3, this Dirichlet series must converge for Rs) > Ste 6: The divergence of ζ k at s = φk). Proosition 8.5. The Dirichlet series defining ζ k s) diverges at s = φk). Proof. Recall from ste 4 that we have ζ k s) = sf) k ) g) and consider the factor corresonding to. We have ) g) = + sf) + 2sf) +... sf) Since f ) φ k) and g ), this sum dominates the sum + sφk) + 2sφk) +... ) g) Therefore, the series defining ζ k s) dominates the series n,k)= n sφk). However, this sum diverges at s = /φ k). We have shown that, if there is a χ so that L, χ) = 0, then the function ζ k s) is holomorhic in the entire half-lane R s) > 0. This would imly that the Dirichlet series defining ζ k converges in this domain. However, we have just found at least one s for which this Dirichlet series does not converge. Therefore, there cannot exist a χ so that L, χ) = 0, which in turn imlies the boundedness of log L, χ)) = 0 for nontrivial χ. Theorem 7.2 shows that this fact imlies Dirichlet s theorem. Hence, the roof is comlete. References [] J. P. Serre. Course in Arithmetic. Sringer-Verlag, 973. [2] Tom M. Aostol. Introduction to Analytic Number Theory. Sringer-Verlag, 976.