Dirichlet's Theorem and Applications

Transcription

1 University of Connecticut Honors Scholar Theses Honors Scholar Program Sring Dirichlet's Theorem and Alications Nicholas Stanford University of Connecticut - Storrs, mistersoccer@yahoo.com Follow this and additional works at: htt://digitalcommons.uconn.edu/srhonors_theses Part of the Mathematics Commons Recommended Citation Stanford, Nicholas, "Dirichlet's Theorem and Alications" (203). Honors Scholar Theses htt://digitalcommons.uconn.edu/srhonors_theses/286

2 Dirichlet s Theorem and Alications Nicholas Stanford B.S. Mathematics B.A. Music An Undergraduate Honors Thesis Submitted in Partial Fulfillment of the Requirements for the Degree of Bachelor of Science at the University of Connecticut May 203 i

3 Coyright by Nicholas Stanford May 203 ii

4 APPROVAL PAGE Bachelor of Science Honors Thesis Dirichlet s Theorem and Alications Presented by Nicholas Stanford, B.S. Math., B.A. Music Honors Major Advisor William Abikoff Honors Thesis Advisor Keith Conrad University of Connecticut May 203 iii

5 ACKNOWLEDGMENTS This thesis could not have been comleted without the hel of my thesis advisor, Professor Keith Conrad. Prof. Conrad offered to be my advisor during finals week of the fall semester of my senior year. I had not yet done any research toward my thesis and had taken just one class with Prof. Conrad. Yet he believed that I would be able to accomlish this and was willing to work tirelessly with me to make it haen. Prof. Conrad has been both relentless and atient while guiding me through this rocess. I can confidently say that very few of my eers had their theses scrutinized in their entirety as many times as mine was. His attention to detail, ensuring that every line of this thesis was as well written as ossible, was astounding. Furthermore, he ushed me to be exhaustive in my thinking. Secific direction was given when needed, but I was also allowed to consider the issues on my own. Perhas I mistook it as simly being sarcastic humor, but Prof. Conrad never seemed to grow uset with me or the rogress of my research (or lack thereof). Instead, he was understanding. He set feasible milestones for me to accomlish each week that ultimately led to the final roduct resented here. I sincerely thank Prof. Conrad for his hel in the comletion of this thesis, the comletion of my University of Connecticut Honors requirements, and my accetance into graduate school. I would like to acknowledge the entire Mathematics Deartment at UConn, including both the faculty and my eers. They have heled me grow as a student of math since my first day as an undergraduate. My rofessors rovided me with iv

6 v challenging material to learn, but also ensured that they were available to hel if necessary. Indeed, I have found that office hours are quite the valuable resource. The time sent collaborating with my eers, however, was just as beneficial. We discover new ways of thinking and strive to make our own work better when we work with others. The last four years sent learning with these individuals has reared me for the next stage of my education in math. Individuals outside of the Mathematics Deartment have also influenced me heavily. Music may not be my career, but it is my assion. The music faculty embraced me as a student and was encouraging of my studies. In articular, I would like to thank my iano teacher, Irma Vallecillo. She has been my most suortive teacher during my time at UConn. Irma not only taught me how to better exress myself when laying iano, but became a mentor and a close friend. To the rest of the University of Connecticut community, I thank you. I would be remiss if I did not recognize my friends and family here. While they may not have layed a direct role in the writing of this thesis, they have been invaluable nonetheless. My roommates ensured that I actually left my room occasionally for such trivial activities as dinner or a game of frisbee. They ket me hay and made my time at UConn worthwhile on a ersonal level. I will miss each of them deely when we go our searate ways in the fall. My family has loved and suorted me for as long as I can recall. Never has my mom questioned my desire to continue my education instead of finding a job. Instead, she has done all that is in her caacity in order to allow my sister and me to excel in our desired fields. She is an insiration to me. To my sister, thank you for utting u with me over the years; I really do areciate having you around. To my father, I still think of you often and am insired by you. Thank you once again to all of those who have heled me along my journey.

7 Dirichlet s Theorem and Alications Nicholas Stanford, B.S., B.A. University of Connecticut, May 203 ABSTRACT Dirichlet s theorem states that there exist an infinite number of rimes in an arithmetic rogression a + mk when a and m are relatively rime and k runs over the ositive integers. While a few secial cases of Dirichlet s theorem, such as the arithmetic rogression 2 + 3k, can be settled by elementary methods, the roof of the general case is much more involved. Analysis of the Riemann zeta-function and Dirichlet L-functions is used. The roof of Dirichlet s theorem suggests a method for defining a notion of density of a set of rimes, called its Dirichlet density, and the rimes of the form a+mk have a Dirichlet density /ϕ(m), which is indeendent of a. While the definition of Dirichlet density is not intuitive, it is easier to comute than a more natural concet of density, and the two notions of density turn out to be equal when they both exist. Dirichlet s theorem is often used to show a rime number exists satisfying a articular congruence condition while avoiding a finite set of bad rimes. For examle, it allows us to find the density of the set of rimes such that a given nonzero integer a is or is not a square mod. More generally, it lets us find the density of the set of rimes at which a finite set of integers have rescribed Legendre symbol values. vi

8 Contents Ch.. Introduction Ch. 2. Characters and Comlex Analysis 3 2. Characters Theorems from Comlex Analysis Ch. 3. The zeta-function and L-functions The Riemann zeta-function Dirichlet L-functions Theorems on Dirichlet Series Ch. 4. Dirichlet s Theorem Some Elementary Cases Proof of Dirichlet s Theorem Dirichlet Density Ch. 5. Alications of Dirichlet s Theorem The Legendre Symbol Statistics of One Legendre Symbol Statistics of Multile Legendre Symbols Bibliograhy 64 vii

9 Chater Introduction Dirichlet s theorem states that for two relatively rime integers a and m there exist infinitely many rimes a mod m. Table.0. below gives suorting numerical data for the case of rimes a mod 9 when (a, 9) =. We tabulate such N as N runs through owers of 0. N a : Table.0.: Number of rimes N satisfying a mod 9. Dirichlet s theorem says more: the roortion of rimes a mod m is /ϕ(m) when gcd(a, m) =. For examle, in Table.0.2 the roortions are all getting close to /ϕ(9) = /6 = The background that led to Dirichlet s theorem was Legendre s unsuccessful at-

10 2 N a : Table.0.2: Proortion of rimes N satisfying a mod 9. temt to rove quadratic recirocity. He broke u quadratic recirocity into eight cases, and to rove some of them he was led to conjecture that there are infinitely many rimes in any arithmetic rogression a + mk where a and m are relatively rime and k runs over the natural numbers [9,. 6 8]. The first roof of Legendre s conjecture was given by Dirichlet in 837, and his method was insired by Euler s analytic roof from 737 that there are infinitely many rimes: Euler showed the series / diverges, where runs over the rimes, and Dirichlet combined analysis with grou theory to show the series a mod m / diverges when gcd(a, m) =, so the set { a mod m} is infinite. Our treatment of Dirichlet s theorem will use comlex analysis, but Dirichlet s own roof did not. It receded Riemann s work on comlex analysis in number theory (for the zeta-function) by about twenty years. To rove Dirichlet s theorem, in Chater 2 we will introduce characters and discuss some essential results from comlex analysis. These results will then be used in Chater 3 in order to rove theorems about L-functions. These L-functions lay a critical role in roving Dirichlet s theorem in Chater 4. Lastly, in Chater 5 we will look at alications of Dirichlet s theorem to the behavior of Legendre symbols.

11 Chater 2 Characters and Comlex Analysis 2. Characters Definition 2... For a finite abelian grou G, a character χ on G is a homomorhism from G to the unit circle S. Examle If G = Z/4, the functions χ(a mod 4) = i a and χ(a mod 4) = ( ) a are both characters of G. Multilication of two characters χ and ψ on G is defined by (χψ)(g) = χ(g)ψ(g) for g G. Theorem The set of all characters χ : G S forms a multilicative grou. Proof. Since the roduct of two elements in S is still in S, and S is abelian, the roduct of two characters still mas elements of G to S and is a character. The identity character is the character χ for which χ (g) = for all g G. We 3

12 4 call this the trivial character. The inverse of a character χ is χ, where we define χ(g) = χ(g) = χ(g) for all g G. We will denote the grou of characters on G by Ĝ. If G is nontrivial, the construction of a nontrivial character on G is based on extending characters from subgrous to the whole grou. Let us start with cyclic grous. Theorem Suose G is a finite cyclic grou with order n. There are n distinct characters of G, and each one is uniquely determined by its value on a generator g 0 by sending g 0 to each of the n th roots of unity in C. Moreover, Ĝ is cyclic. Proof. For any character χ of G and g G, χ(g) n = χ(g n ) = χ() =, so χ(g) is a root of unity with order dividing n. For any g G, we can write g = g 0 l for some l Z. Therefore, χ(g) = χ(g 0 l ) = χ(g 0 ) l, so the values that χ takes on G are comletely determined by the value that χ takes at g 0. We may assign the value of χ(g 0 ) to be any of the n th roots of in C: for ζ C with ζ n =, set χ(g 0 l ) = ζ l for all l Z. This is well-defined: if g 0 l = g 0 m then l m mod n, so ζ l = ζ m. It is easy to check that χ is a homomorhism, so χ is a character of G and χ(g 0 ) = ζ. Since there are n choices for the value of χ(g 0 ) and χ is comletely determined by this value, there are n distinct characters on G. Thus Ĝ = n. To show that Ĝ is cyclic, for 0 r < n, let ψ r be the character of G where ψ r (g 0 ) = e 2πir/n. Then ψ r (g 0 ) = ψ (g 0 ) r, so for all l Z ψ r (g l 0) = ψ r (g 0 ) l = e 2πirl/n = ψ (g l 0) r. This says ψ r (g) = ψ (g) r for all g G. Thus ψ r = ψ r, so Ĝ is generated by ψ.

13 5 Theorem For g, g 2 G where g g 2, there exists a character χ of G such that χ(g ) χ(g 2 ). Proof. To rove this, it will suffice to show that for every g G where g there exists a χ such that χ(g). Taking g = g g 2 then yields the desired result. Suose G is cyclic of order n with generator g 0. Every g G with g can be written as g = g 0 l with l < n. By the revious theorem, we can choose χ Ĝ such that χ(g 0 ) = e 2πi/n, ensuring that χ(g) = χ(g 0 l ) = e 2πil/n since l < n. For general G, we know by grou theory that G has a direct roduct decomosition G = C C r where C i is cyclic [3, ]. Let g G have the form (c,..., c r ) for c i G i. If g, then some c i. By the cyclic case, there exists a character ψ : C i S such that ψ(c i ). Let χ = ψ π i where π i : G C i is the rojection of G onto its ith factor. Then χ is a character on G and χ(g) = ψ(π i g) = ψ(c i ). Theorem Let G be a finite abelian grou and let Ĝ be its dual grou. Then G = Ĝ, and in fact Ĝ = G. Proof. If G is cyclic, then it is clear from Theorem 2..4 that Ĝ = G. To handle non-cyclic G, we first show for any finite abelian grous A and B that  B =  B. Let χ be a character on A B. Let χ A and χ B be the restriction of χ to A and B resectively, i.e., χ A (a) = χ(a, ) and χ B (b) = χ(, b). Then χ A and χ B are characters on A and B resectively, and χ(a, b) = χ((a, )(b, )) = χ A (a)χ B (b). In this manner, we obtain a maing φ:  B  B by χ (χ A, χ B ). We will now check that φ is an isomorhism. To show that φ is a homomorhism,

14 6 for χ and χ in  B we have φ(χχ ) = ((χχ ) A, (χχ ) B ) and φ(χ)φ(χ ) = (χ A, χ B )(χ A, χ B) = (χ A χ A, χ B χ B), so we need to show (χχ ) A = χ A χ A and (χχ ) B = χ B χ B. For a A, (χχ ) A (a) = (χχ )(a, ) = χ(a, )χ (a, ) = χ A (a)χ A(a) = (χ A χ A)(a), so (χχ ) A = χ A χ A. That (χχ ) B = χ B χ B follows in the same way. The homomorhism is injective since if χ A and χ B are trivial, then χ(a, b) = χ A (a)χ B (b) =, so χ is also trivial. To show φ is surjective, for any choice of (ψ, ψ ) in  B, define χ : A B S by χ(a, b) = ψ(a)ψ (b). This is a character on A B and χ A = ψ, χ B = ψ. Returning to the theorem, write G = C C m where each C i is cyclic. By induction on the number of terms,  B =  B generalizes to (A A m ) =  Âm for any finite abelian grous A,..., A m. Therefore, Ĝ = Ĉ Ĉm. From the cyclic case, Ĉi = C i, so Ĝ = Ĉ Ĉm = C C m = G. Theorem For any finite abelian grou G and g G, G if g =, χ(g) = χ Ĝ 0 if g. (2..)

15 7 where χ runs through all characters of G. More generally, for all g and h in G, G if g = h, χ(g)χ(h) = χ Ĝ 0 if g h. (2..2) Proof. Let S = χ χ(g). If g =, then χ(g) = for all χ Ĝ, so S is equal to the size of Ĝ, which is G by Theorem If g, choose a character ψ of G such that ψ(g). Such a ψ is guaranteed to exist by Theorem Multilying S by ψ(g) yields ψ(g)s = ψ(g) χ χ(g) = χ (ψχ)(g) = χ χ(g) = S. This imlies that either ψ(g) = or S = 0. But ψ(g) by assumtion, so S = 0. This roves (2..). To rove (2..2), write the left side as χ χ(g)χ(h ) = χ χ(gh ). Using gh for g in (2..), we get (2..2). Theorem For any finite abelian grou G and χ Ĝ, G if χ = χ, χ(g) = g G 0 if χ χ, (2..3) where g runs through all elements of G and χ is the trivial character of G. Proof. Set S = g χ(g). Suose χ = χ. Then it is clear that S = G.

16 8 Suose χ χ, so there is g 0 G such that χ(g 0 ). Then χ(g 0 )S = g χ(g 0 )χ(g) = g χ(g 0 g) = g χ(g) = S. This imlies that χ(g 0 ) = or S = 0. But χ(g 0 ) by assumtion, so S = 0. In our study of Dirichlet s theorem, we will take G = (Z/m). A character of (Z/m) is called a Dirichlet character. In this case, the order of G is equal to ϕ(m). A Dirichlet character χ on (Z/m) can be defined on all of Z by setting χ(n mod m), if (n, m) =, χ(n) = 0, if (n, m). (2..4) Then χ(n n 2 ) = χ(n )χ(n 2 ) for all n, n 2 Z. We define the trivial Dirichlet character mod m, denoted m, as follows:, if (n, m) =, m (n) = 0, if (n, m). Note that χ(n) = χ(n) when n and m are relatively rime but not when n and m are not relatively rime because χ(n) = 0. However, χ(n) is still a valid exression for all n by defining χ(n) = χ(n). In articular, if (n, m), then χ(n) = χ(n) = 0 = 0. For Dirichlet characters, viewed as functions on Z, (2..2) becomes ϕ(m) if n a mod m, χ(n)χ(a) = χ 0 if n a mod m (2..5)

17 9 for all n Z, where the sum is over all Dirichlet characters mod m and (a, m) =. Formula (2..5) will be an imortant algebraic ingredient in the roof of Dirichlet s theorem. Our last general result about characters in this section concerns the existence of a character taking rescribed values at indeendent elements of the grou. Definition Let H be a finite abelian grou. We call a set of elements h i H multilicatively indeendent if i he i i = imlies that h e i i = for every i. Equivalently, no h j in the set can have a nontrivial ower written as a roduct of the owers of the h i for i j. Remark If the h i s all have order 2, then multilicative indeendence in H says no h j is a roduct of the h i for i j. Theorem 2... Let H be a finite abelian grou. For h,..., h r H, set ϕ : h h r h,..., h r by ϕ(h e,..., h er r ) = h e... h er r. Then {h,..., h r } is multilicatively indeendent in H if and only if ϕ is an isomorhism. Proof. To check that ϕ is a homomorhism, note from H being abelian that ϕ ( (h a,..., h ar r )(h b,..., h br r ) ) = ϕ ( h a +b ),..., h ar+br r = h a +b h ar+br r = h a h ar r h b h br r = ϕ (h a,..., h ar r ) ϕ ( h b,..., h br r ). Suose that ϕ(h a,..., h ar r ) =. Then i ha i i =. If the h i are multilicatively indeendent in H, we must have that h a i i = for all i. Hence ϕ is injective.

18 0 The function ϕ is clearly surjective since all elements in the image are of the form i ha i i. Therefore, multilicative indeendence of the h i in H imlies that ϕ is an isomorhism. Conversely, if ϕ is an isomorhism, then the h i must be multilicatively indeendent in H because the kernel of ϕ is trivial. Theorem In a finite abelian grou H, let {h,..., h r } be multilicatively indeendent. Let m i be the order of h i in H, and let ζ i S satisfy ζ m i i there exists a character χ Ĥ such that χ(h i) = ζ i for all i. =. Then To have such a χ, it is necessary that ζ m i i χ() =. = because ζ m i i = χ(h i ) m i = χ(h m i i ) = Proof. Define ψ : h,..., h r S by ψ(h a h ar r ) = ζ a ζ ar r. To show this function is well-defined, suose h a h ar r = h b h br r. Then = h a b h ar br r. Since the h i are multilicatively indeendent, this imlies that h a i b i i = for all i, so m i (a i b i ) and so a i b i mod m i. Because ζ m i i =, we conclude that ζ a i i = ζ b i i, so ζ a ζ ar r = ζ b ζ br r. One can also easily check that ψ is a homomorhism. By construction, ψ(h i ) = ζ i. To comlete the roof, we will extend ψ from a character on h,..., h r to a character on H. Let K = h,..., h r. If K = H, we are done. If K H, ick h H K and set K, h = {kh i : k K, i Z}. If K, h H, then K, h K, h, g for some g H such that g K, h. Since H is finite, we may reeat this rocess a finite number of times in order to make a tower of subgrous K K, h K, h, g H where we successively adjoin one new element each time. If we can extend the character ψ on K to a character ψ on K, h, that rocess can be

19 reeated to extend ψ to a character on H. It therefore suffices to show that ψ can be extended from K to K, h. Let m be the minimal ositive integer such that h m K. The value of ψ(h m ) is already known since h m K. Choose z C such that z m = ψ(h m ) so z S, and define ψ : K, h S by ψ (kh i ) = ψ(k)z i. To see that ψ is well-defined, suose k h i = k 2 h i 2. Then h i i 2 = k 2 k K, so m (i i 2 ) and i i 2 mod m. Let i 2 = i + mc for some integer c. Then k = k 2 h i 2 i = k 2 h mc = k 2 (h m ) c and h m K. This imlies that ψ(k ) = ψ(k 2 )ψ(h m ) c = ψ(k 2 )(z m ) c = ψ(k 2 )z mc = ψ(k 2 )z i 2 i ψ(k )z i = ψ(k 2 )z i 2, so ψ is a well-defined function on K, h. In articular, for k K, ψ (k) = ψ(k)z 0 = ψ(k), so ψ = ψ on K. Lastly, we must check that ψ is a homomorhism. Comutation shows that ψ ((k h i )(k 2 h i 2 )) = ψ(k k 2 )z i +i 2 = ψ(k )z i ψ(k 2 )z i 2 = ψ (k h i )ψ (k 2 h i 2 ). 2.2 Theorems from Comlex Analysis We will next discuss some theorems from comlex analysis that will be needed later. For an oen set U C, a function f : U C is called holomorhic (or analytic) if it is differentiable at each oint in U. Cauchy s integral formula says for holomorhic f that f(s) = 2πi γ f(ζ) ζ s dζ for each s U, where γ is any continuous ath in U that traces out a counterclockwise

20 2 loo once around s and can be shrunk to a oint in U. Also, f(s)ds = 0 for all γ closed aths γ in U that can be shrunk to a oint in U. Conversely, Morera s theorem [,. 208] says that if f : U C is continuous and f(s)ds = 0 for all triangles T T in U then f is holomorhic on U. Theorem If D is an oen disc in C and f : D C is holomorhic, then the ower series exansion of f at the center of D converges and equals f on all of D. Proof. Our roof is based on [4, ]. Let s 0 be the center of D. Pick a circle C centered at s 0 with radius less than the radius of D. For any s inside the oen disc at s 0 bounded by C, the Cauchy integral formula tells us that f(s) = 2πi C f(ζ) ζ s dζ. Write ζ s = ζ s 0 (s s 0 ) = ζ s 0 ( s s 0 ζ s 0 ). (2.2.) We wish to rewrite the last factor on the right of (2.2.) using its geometric series exansion. Since s s 0 < ζ s 0 for all ζ on C, define a distance function d : C [0, ) by d(ζ) = (s s 0 )/(ζ s 0 ). This is a continuous maing on a comact set, so there exists r [0, ) such that d(ζ) r < for all ζ C. (This is analogous to the extreme value theorem for continuous real-valued functions of a single real variable.) We may therefore write s s 0 ζ s 0 r <, so ( s s 0 ζ s 0 ) = ( ) n s s0, n=0 ζ s 0 where the series converges uniformly for all ζ C since the series is bounded termwise

21 3 by n=0 rn, which is convergent. This allows us to interchange the series with the integral when we combine the above equations, thereby obtaining f(s) = n=0 ( 2πi C ) f(ζ) dζ (s s (ζ s 0 ) n+ 0 ) n. (2.2.2) The coefficients of this series are the same for any s inside the disc bounded by C, and they are in fact indeendent of C, being f (n) (s 0 )/n!. This roves the ower series exansion of f converges and equals f on D. Theorem (Cauchy s estimate). Let f(s) be holomorhic on a closed disc D of radius r > 0 centered at a such that f(s) M for all s D. Then f (a) M/r. We had only defined holomorhic functions on oen sets in C. A function on a closed set in C is called holomorhic when it is the restriction to that set of a holomorhic function on a larger oen set. Proof. From Cauchy s integral formula or the coefficient in (2.2.2) at n =, f (a) = 2πi C f(ζ) (ζ a) 2 dζ, where C is the boundary of D, arameterized as ζ = a + re iθ for θ [0, 2π]. Thus f (a) 2π M 2π = M r. 2π 0 2π 0 f(a + re iθ ) (re iθ ) 2 dθ r rie iθ dθ

22 4 Remark By a similar argument for higher derivatives, Cauchy s estimate extends to f (n) (a) n!m/r n for all n 0. We will only need the case n =. Lemma For s C and r 0, let D(s, r) denote the closed disc centered at s of radius r. For any nonemty oen subset Ω of C and nonemty comact subset K of Ω, there is an r > 0 such that a K D(a, r) Ω, and this union is comact. Proof. Since Ω is oen, for each s Ω there exists a closed disc of ositive radius centered at s that is a subset of Ω. The union of the interiors of such discs, as s runs over all oints in K, is a covering of K. Since K is comact, the union of finitely many such oen discs covers K. Therefore the union of the closures of these finitely many discs also covers K and is a subset of Ω. Denote these oen discs by D,..., D n with centers s i and radii R i. For i n, define g i : K R by g i (a) = max{0, R i a s i }. Then g i is continuous and nonnegative on K. Define g ointwise by g(a) = max{g i (a)} for i n. Then g is continuous on K. If g(a) = 0 for some a K, then g i (a) = 0 for all i so R i a s i for all i. This means a D i for all i, a contradiction to the set of all D i covering K. Hence g(a) > 0 for all a K. Since K is comact, g attains a minimum value r on K where r > 0. We will show a K D(a, r) Ω for this r. By the definition of g, for each a K we have g(a) = g i (a) for some i. Since g(a) > 0, we must have that g i (a) = R i a s i when g(a) = g i (a). Hence r g(a) = R i a s i = a s i + r R i. (2.2.3) Then for all y D(a, r), we have from (2.2.3) and the triangle inequality y s i y a + a s i r + a s i R i,

23 5 so D(a, r) D(s i, R i ) = D i Ω. That roves a K D(a, r) Ω. To show K := a K D(a, r) is comact, we will show it is closed and bounded. By definition K = {s C : s a r for some a K}, (2.2.4) and K is bounded because it is comact, so (2.2.4) shows that K is also bounded. To show K is closed, we will show its comlement {s C: s a >r for all a K} is oen. For each s K, the inequality s a > r for all a K can be imroved to s a r + ε for some ε > 0 because K is comact: the function f s : K R given by f s (a) = s a r is continuous and ositive on K, so f s has a ositive lower bound. Let ε be such a lower bound. From s a r + ε, the oen ball D(s, ε) is in the comlement of K by the triangle inequality, since for z D(s, ε) and a K we have s a s z + z a < ε + z a, so z a > s a ε r. Theorem Let {f n } be a sequence of holomorhic functions on an oen subset Ω of C. If {f n } converges uniformly on every comact subset of Ω to a limit function f, then f is holomorhic on Ω and the derivatives f n converge uniformly on every comact subset of Ω to f. Proof. Our roof is based on [,. 24]. Since being holomorhic is a local roerty and f n f uniformly on any comact subset of any oen disc in Ω, to rove that f is holomorhic we can assume that Ω is an oen disc. Each f n is continuous and the sequence {f n } converges uniformly to f on every comact disc in Ω, so f is continuous on Ω since continuity is a local roerty and a uniform limit of continuous functions is continuous [5, ]. Let T be a triangle contained comletely within Ω. By triangle we mean the boundary, so it is a closed ath. Since f is continuous, it

24 6 is integrable. Then T f(s)ds T f n (s)ds T f(s) f n (s) ds f f n T length(t ), where f f n T is the suremum norm on T. Since T is comact, f f n T 0 as n. We can shrink T to a oint in Ω since we are taking Ω to be a disc, so by Cauchy s theorem f T n(s)ds = 0. Thus f(s)ds f f T n T length(t ) 0 as n, so f(s)ds = 0 for all triangles T in Ω. By Morera s theorem, we conclude T that f is holomorhic on Ω. To rove that f n f uniformly on each comact subset of Ω, let K be a comact subset of Ω. By Lemma there is an r > 0 such that K := a K D(a, r) is a subset of Ω, and K is comact. Let M n denote the maximum of f f n on K. For each a K we can aly Theorem to f f n on D(a, r) in order to obtain f (a) f n(a) M n /r. Since f n converges to f uniformly on each comact subset of Ω, such as K, M n 0 as n. Thus f n converges uniformly to f on K, and thus on its subset K, so f n f uniformly on each comact subset of Ω. Definition Let Ω be oen in C. f : Ω C is a holomorhic function L f A logarithm of a holomorhic function : Ω C such that e L f (s) = f(s) for all s Ω. Theorem If Ω is a connected and simly connected oen set in C then any nonvanishing holomorhic function f : Ω C admits a logarithm, and any two logarithms of f on Ω differ by an integral multile of 2πi. Proof. Our roof is based on [4,. 00-0]. Since Ω is connected and oen, it is

25 7 ath connected. Fix s 0 Ω, and define L f : Ω C by L f (s) = γ f (w) f(w) dw + c 0, where γ is any ath in Ω connecting s 0 to s, and c 0 C satisfies e c 0 = f(s 0 ). To show L f (s) is indeendent of the choice of ath, let γ be another ath connecting s 0 to s in Ω. The ath ϕ obtained by traveling along γ from s 0 to s and then along γ backwards from s to s 0 is a loo that can be shrunk to a oint in Ω by simle connectedness. Since f is holomorhic on Ω, Cauchy s integral formula tells us 0 = ϕ f (w) f(w) dw = f (w) γ f(w) dw f (w) γ f(w) dw f (w) γ f(w) dw = f (w) γ f(w) dw, so L f (s) is indeendent of choice of ath from s 0 to s. To show that L f : Ω C is holomorhic, we will use the limit definition of the derivative: L L f (s + h) L f (s) f(s) = lim. h 0 h For small h 0, in L f (s + h) we can choose as the ath from s 0 to s + h the concatenation of the ath from s 0 to s used for L f (s) followed by the straight line ath from s to s + h. Then L f (s + h) L f (s) H = h [s,s+h] f (w) f(w) dw. Since the ath from s to s + h is a straight line we can arameterize w along it as

26 8 w = s + th for t [0, ], so f (w) h [s,s+h] f(w) dw = f (s + th) h 0 f(s + th) h dt = f (s + th) 0 f(s + th) dt. From continuity of f /f at s, f (s + th) lim h 0 0 f(s + th) dt = f (s) 0 f(s) dt = f (s) f(s), so L f (s) = f (s)/f(s). Thus L f is holomorhic. To show that e L f (s) = f(s), we calculate d ( ) f(s)e L f (s) = f (s)e L f (s) f(s)l ds f(s)e L f (s) = e L f (s) f(s) ( f (s)/f(s) L f(s) ) = 0, so f(s)e L f (s) is constant (since Ω is connected). Evaluating this exression at s 0 we get the value f(s 0 )e c 0 =, so f(s) = e L f (s) for all s Ω. Thus L f is a logarithm of f on Ω. Suose that f admits another logarithm L f on Ω, so f(s) = e L f (s) for all s Ω. Then for all s Ω, e L f (s) = e L f (s) e L f (s) L f (s) = L f (s) L f (s) 2πiZ. Since L f L f is continuous on Ω, its image is connected. The set 2πiZ is discrete, so the image of L f L f on Ω is a oint. Therefore there is an integer k such that L f (s) L f (s) = 2πik for all s Ω.

27 9 Examle On {s : s < }, a logarithm of /( s) is k sk /k. That is, ex( k sk /k) = /( s). To rove this, both sides are holomorhic on s < and equal for real s when 0 < s < by calculus, so they are equal for all s with s <.

28 Chater 3 The zeta-function and L-functions Proving Dirichlet s theorem requires the use of comlex-valued functions called the Riemann zeta-function and Dirichlet L-functions. In this chater, we will introduce these functions and rove various roerties of them that will be essential in the next chater. 3. The Riemann zeta-function The Riemann zeta-function is defined for Re(s) > by ζ(s) = n. s n= Following a common tradition in number theory, we will write the real and imaginary arts of s as s = σ + it with σ, t R. Theorem 3... The zeta-function is absolutely convergent for Re(s) >. 20

29 2 Proof. Note that n= n s n= n s = n= n = Re(s) n= n σ. Fix σ > 0. The function f σ (x) = /x σ is a monotonic decreasing ositive function for x > 0. By the integral comarison test, n= /nσ is convergent if and only if f σ (x)dx is convergent. Evaluating, f σ (x)dx = x σ σ = lim b b σ σ σ = σ lim b (σ )b σ. This is convergent if (and only if) σ >. Intuitively, taking the limit of ζ(s) as s + yields the (divergent) harmonic series, so we exect ζ(s) as s +. Let us rove this. Theorem For s R, the limit of ζ(s) as s + is infinity. Proof. For each s >, since x s is a monotonic decreasing function for x > 0, we have the inequality (n + ) s < n+ n x s dx < n s. (3..) Summing (3..) over n yields ζ(s) < x s dx < ζ(s), and x s dx = /(s ). Rearranging terms, we get < (s )ζ(s) < s, (3..2) so lim s +(s )ζ(s) =. For s >, ζ(s) > /(s ) by (3..2), so lim s + ζ(s) =.

30 22 By the next theorem, we will see that the zeta-function can be written as a roduct over the rimes. Theorem Let h : Z + C be totally multilicative such that h(n) < for all n and n= h(n) is absolutely convergent. Then n= h(n) = /( h()), where the roduct runs over the rimes and is also absolutely convergent. Proof. For x 2, consider the finite roduct P (x) = x {+h()+h(2 )+ } over all rimes x. Each series + h() + h( 2 ) + is + h() + h() 2 +, which is an absolutely convergent geometric series. Thus P (x) is a finite roduct of absolutely convergent series, so it may be exanded and rearranged into an absolutely convergent series whose general term is h( a )h( a 2 2 ) h( ar r ), which is h( a a 2 2 ar r ) since h is totally multilicative. By unique factorization in Z +, P (x) = n A h(n) where A is the set of ositive integers having all rime factors less than or equal to x. Therefore, n= h(n) P (x) = n B h(n), where B is the set of ositive integers having at least one rime factor greater than x, and n= h(n) P (x) n B h(n) n>x h(n). As x, we have n>x h(n) 0 since the series n h(n) converges. This imlies that lim x P (x) = n= h(n). Since h is totally multilicative, P (x) = x ( i=0 h()i ) = x /( h()). Taking the limit of this as x yields the desired result. The above theorem, using h(n) = /n s, shows that ζ(s) = n = s n= / s (3..3) for Re(s) >. The infinite roduct on the right is called the Euler roduct of ζ(s).

31 23 Euclid showed there are an infinitely many rimes by contradiction. Here is another roof of this, due to Euler in 737. It motivates the roof of Dirichlet s theorem. Theorem As s +, /s, where the sum is over all rimes. Proof. Our argument is based on [7,. 250]. Let λ N (s) = >N /( /s ) for Re(s) >. By (3..3), we have for Re(s) > that ζ(s) = / s = λ N(s) N / s where λ N(s) as N. (3..4) Taking the logarithm of both sides of (3..4) when s > and using the series exansion ln( x) = k= xk /k for x <, we get for s > that ln ζ(s) = ln λ N (s) + N k=. (3..5) kks Taking the limit as N of (3..5) yields ln ζ(s) = k = ks k= + s k=2. (3..6) kks The second double series on the right in (3..6) converges for each s > /2 since k < ks k=2 = ks k=2 2s ( / s ) /2 s < 2s / 2 ζ(2s). By Theorem 3..2 and that fact that lim x ln(x) =, we have ln ζ(s) as s +. For s >, k=2 k ks < ζ(2) / 2,

32 24 so by (3..6) we get /s as s +. Remark The roof used k 2 /kks only for s, not s > /2. We will need to use s > /2 (really, s > ε) more seriously in Corollary Corollary The series / diverges. If there were only finitely many rimes, then / would be convergent. Therefore, Corollary 3..6 shows that there are infinitely many rime numbers. Proof. For s >, / s < /, so /s /. Letting s +, Theorem 3..4 and the comarison test show that / =. Here is an alternate heuristic roof of Corollary 3..6, more in the sirit of Euler. Proof. Consider log( n /n). Using the Euler roduct at s = (which, strictly seaking, is not valid), log ( n ) ( ) = log = n / ( ) log = / ( log + ). Since e x = + x + x 2 /2! +, we have x > log(x + ) for x > 0. Therefore ( log + ) ( ) = log, n n so the divergence of n /n imlies that /( ) diverges. From, we 2( ) get / 2 /( ). Thus / diverges.

33 Dirichlet L-functions For any Dirichlet character χ, viewed as a function on Z, set L(s, χ) = n= χ(n) n s (3.2.) for Re(s) >. This is called the Dirichlet L-series, or Dirichlet L-function, of χ. Since χ(n) = (or 0), we may follow the roof of Theorem 3.. to see that a Dirichlet L-series is absolutely convergent for Re(s) >. While Dirichlet only used real s in his work, here we allow s to be a comlex number. Examle When χ 4 is the nontrivial character mod 4, L(s, χ 4 ) = /3 s + /5 s /7 s +..., which converges for all real s > 0, and L(, χ 4 ) = π/4 by Leibniz s formula for /3 + /5 / By Theorem 3..3, there is an Euler roduct reresentation L(s, χ) = χ()/ s (3.2.2) for Re(s) >. Examle If χ = m, the trivial character mod m, then L(s, χ) = not dividing m / s = m ( s ) ζ(s). We will see later (Corollary 3.3.6) that the L-series of every nontrivial Dirichlet character converges for Re(s) > 0, although the Euler roduct in (3.2.2) has no obvious convergence when 0 < Re(s). The behavior of L(s, χ) near s = is

34 26 central to the roof of Dirichlet s theorem. 3.3 Theorems on Dirichlet Series We collect here several basic analytic facts about the tyes of series like ζ(s) and L(s, χ). Definition A Dirichlet series is a function of the form f(s) = n a n/n s where the a n are comlex constants and s is a comlex variable. Theorem Suose that a n /n σ converges for at least one real σ and diverges for at least one other real σ. Then there exists a real number σ 0 such that a n /n s converges absolutely for Re(s) > σ 0 but does not converge absolutely for Re(s) < σ 0. Proof. When s C with real art σ, a n /n s is absolutely convergent if and only if an /n σ is absolutely convergent since a n /n s = a n /n σ. Define D to be the set of real numbers σ such that a n /n σ diverges. By assumtion, D and the comlement of D are both nonemty. By the comarison test, if σ D then σ D for all σ < σ, and if σ D then σ D for all σ > σ. Therefore, since the comlement of D is nonemty, D is bounded above. Denote the suremum of D by σ 0. The series must diverge for all σ < σ 0 by the comarison test: if σ < σ 0 then there is a σ D such that σ < σ < σ 0, so σ D. The series converges for all σ > σ 0 because σ 0 is the suremum of D and hence an uer bound on all real numbers such that the series diverges. Theorem If the sequence {a n } is bounded, then n a n/n s converges absolutely for Re(s) >.

35 27 Proof. Since {a n } is bounded, there exists a number B R such that a n B for all n. Hence n a n/n s B n /nσ, which converges for σ > by Theorem 3... Theorem If {a n } is a sequence such that the artial sums a + + a n are bounded for all n, then n a n/n s converges for Re(s) > 0. Proof. We will use summation by arts: N u n (v n v n ) = u N v N u v 0 n= for any sequences u,..., u N and v 0, v,..., v N. N n= v n (u n+ u n ) Set b n = a + + a n and let b n b for all n. Then a n = b n b n where b 0 = 0. For N 2, we have N n= a n n s = N n= b n b n n s N = n (b s n b n ) n= = b N N N ( b s n (n + ) ) by summation by arts s n s n= = b N N N + n+ s b s n dx xs+ n= n= n = b N N N + s n+ dx b s n. (3.3.) xs+ If σ = Re(s) > 0 then b N /N s b/n σ 0 as N. Also, b n n+ n dx x s+ n n+ b n dx x σ+ < b n σ+,

36 28 so the series n b n n+ n dx/x s+ is absolutely convergent when σ > 0 by comarison with n b/nσ+ = bζ(σ + ). Therefore the series in (3.3.) converges as N, so n a n/n s converges when Re(s) > 0. Remark The convergence of n a n/n s in Theorem need not be absolute, e.g., the series /2 s + /3 s /4 s +... with alternating coefficients ± converges for Re(s) > 0 but not absolutely for 0 < Re(s). Corollary For any nontrivial Dirichlet character χ, the Dirichlet L-series L(s, χ) = n χ(n)/ns converges for Re(s) > 0. Proof. Let χ be a nontrivial Dirichlet character mod m. Since χ is nontrivial, by Theorem 2..8 we know that m n= χ(n) = n (Z/m) χ(n) = 0. Furthermore, m+i n=i χ(n) = 0 for all i. The artial sums of the coefficients in L(s, χ) are bounded due to their cyclic vanishing, so L(s, χ) is convergent for nontrivial χ and Re(s) > 0 by Theorem We will need not just convergence of L(s, χ) in a half-lane, but analyticity too. This is justified by the next theorem. Theorem If the Dirichlet series n a n/n s converges at s = s 0, then it converges in the oen half-lane {s : Re(s) > Re(s 0 )}, and in fact its artial sums converge uniformly on comact subsets of this half-lane. Proof. See [,. 235]. Corollary If the Dirichlet series n a n/n s converges in the oen half-lane {s : Re(s) > σ 0 }, then it is holomorhic here and its derivative can be comuted termwise as n ( a n log n)/n s.

37 29 Proof. For ε > 0, Theorem with s 0 = σ 0 +ε imlies the artial sums of n a n/n s converge uniformly on comact subsets of {s : Re(s) > σ 0 + ε}, so Theorem imlies n a n/n s is holomorhic on {s : Re(s) > σ 0 + ε}, and its derivative is comutable termwise since the artial sums N n= a n/n s are holomorhic with derivative N n= a n(log n)/n s. Since {s : Re(s) > σ 0 } is the union of {s : Re(s) > σ 0 + ε}, we are done. Corollary The function ζ(s) = n /ns is holomorhic on Re(s) >, and the function L(s, χ) = n χ(n)/ns is holomorhic on Re(s) > 0 for nontrivial χ. Proof. Combine Theorem 3.. and Corollary to see that ζ(s) is holomorhic on Re(s) >. Combine Corollary and Corollary to see that L(s, χ) is holomorhic on Re(s) > 0 for nontrivial χ. Theorem For nontrivial χ, a logarithm of L(s, χ) for Re(s) > is χ(k ) k ks k = (χ()/ s ) k. (3.3.2) k k Proof. Since χ( k )/k ks / kσ and k / kσ is absolutely convergent for σ >, we conclude by the comarison test that the series in (3.3.2) is absolutely convergent for σ >. Exonentiating (3.3.2), we find that ( ) (χ()/ s ) k ex = k k ( ) (χ()/ s ) k ex = k k = L(s, χ) χ()/s by using Examle 2.2.8, the fact that χ()/ s < since Re(s) >, and the Euler roduct (3.2.2). We can extend ζ(s) analytically to Re(s) > 0, like L(s, χ), excet there is a ole

38 30 at s =. Theorem The zeta-function has a meromorhic continuation from Re(s) > to Re(s) > 0 that is holomorhic everywhere excet for a simle ole at s = with residue. Proof. Consider the two series ζ 2 (s) = 2 s + 3 s 4 s + = n ( ) n n s and ζ 3 (s) = + 2 s 2 3 s + 4 s + 5 s 2 6 s +, where the coefficients in ζ 3 (s) are eriodically,, 2. By Theorem 3.3.4, these series converge for Re(s) > 0, so they are holomorhic there by Corollary Note that ζ(s) ζ 2 (s) = 2 2 s s + = 2 2 s ( + 2 s + 3 s + ) = 2 2 s ζ(s), so ζ(s) = ζ 2 (s). (3.3.3) /2s The right side of (3.3.3) is meromorhic for Re(s) > 0 excet erhas for simle oles at s = + 2kπi/ log 2, k Z, which is where /2 s =. Similarly, ζ(s) ζ 3 (s) = 3 ζ ζ(s), so ζ(s) = 3s 3 (s). (3.3.4) /3s The right side of (3.3.4) is meromorhic for Re(s) > 0 excet ossibly for simle oles at s = + 2lπi/ log 3 where l Z. Since ζ(s) has two meromorhic continuations to Re(s) > 0, the uniqueness of meromorhic continuation imlies that ζ(s) can only have a ole for Re(s) > 0 when

39 3 s = + 2kπi/ log 2 = + 2lπi/ log 3 for some integers k and l. These formulas imly k log 3 = l log 2, so 2 l = 3 k. This imlies k = l = 0, so ζ(s) is holomorhic on Re(s) > 0 away from s =. At s =, there is a simle ole since ζ(s) is meromorhic at and lim s +(s )ζ(s) = by (3..2), which tells us that the residue of ζ(s) at s = is. Lemma (Landau). Suose that a Dirichlet series f(s) = n= a n/n s with nonnegative coefficients converges for Re(s) > σ 0. If f(s) extends analytically to some disc centered at σ 0, then the Dirichlet series n= a n/n s converges on the half-lane Re(s) > σ 0 ε for some ε > 0. Proof. Our roof is based on [,. 237]. Using f(s + σ 0 ) = n= a nn σ 0 /n s in lace of f(s), we can assume σ 0 = 0 and f(s) has an analytic continuation to a neighborhood of 0, so to some disc B(0, δ). There is a small ε < δ such that the disc {s : s < + ε} is entirely inside B(0, δ) {s : Re(s) > 0}. By Theorem 2.2., the ower series of f(s) at s = converges to f(s) for any s such that s < + ε. For σ < + ε with σ R, we have the ower series f(σ) = k=0 f (k) () (σ ) k. k! By reeated use of Corollary 3.3.8, we can comute f (k) () by differentiating f(s) = n= a n/n s termwise to get f (k) () = a n ( log n) k = ( ) k s= n= n s a n (log n) k. n n=

40 32 Therefore f(σ) = ( ) k k=0 n= a n (log n) k (σ ) k = k!n a n (log n) k ( σ) k. (3.3.5) k!n k=0 n= When ε < σ <, all terms on the right side of (3.3.5) are nonnegative, so this series can be rearranged to f(σ) = n= a n n ( ) (log n) k ( σ) k = k! k=0 n= a n n e(log n)( σ) = n= a n n n σ = n=0 a n n σ. con- a n Therefore f(σ) = n when ε < σ <, so the Dirichlet series a n σ n s n= n= verges for all s C such that Re(s) > ε by Theorem

41 Chater 4 Dirichlet s Theorem In this chater we will rove Dirichlet s theorem: for two relatively rime integers a, m, there exist infinitely many rimes a mod m. The roof uses the Riemann zeta-function, Dirichlet characters, and Dirichlet L-series. These toics were covered in Chaters 2 and 3. In this chater, we will first give roofs of elementary cases of Dirichlet s theorem and then discuss a hard theorem about L-series in detail, which is needed to rove Dirichlet s theorem in general. 4. Some Elementary Cases To better areciate the roof of Dirichlet s theorem, we will work out here some secial cases by elementary algebraic methods that do not extend to the general case. Our roofs are based on [5, ]. Theorem 4... There are infinitely many rimes mod 4. 33

42 34 Proof. One such rime is 5. If we have finitely many mod 4, say,..., r, we will construct another one. Let N = (2 r ) 2 +. Clearly neither 2 nor any i divides N. Since N >, it has a rime factor. Let be some rime dividing N, so is odd. By Fermat s little theorem, using a = 2 r, a 2 mod ( ) ( )/2 a mod ( ) ( )/2 = mod 4. Hence N has a rime factor congruent to mod 4 that is not any i. Theorem There are infinitely many rimes 3 mod 4. Proof. One such rime is 3. If we have finitely many 3 mod 4, say,..., r, we will construct another one. Let N = 4 r, so N 3 mod 4. Therefore N is odd and, since N >, at least one of the rime factors of N is congruent to 3 mod 4. (If this were not the case, N would be congruent to mod 4.) Since N mod i, no i divides N. Therefore N has a rime factor congruent to 3 mod 4 that is not any i. Iteration of this rocess roduces infinitely many rimes 3 mod 4. The set of rimes congruent to 3 mod 4 created using the above algorithm is robably not all of them. For instance, starting with the rime = 3, the next few rimes generated (using the least rime factor of N that is congruent to 3 mod 4 at each ste) are 2 =, 3 = 3, and 4 = 7, 29. Each of these numbers is equal to N in the above rocess, but that need not be the case (in fact, at the 6 th iteration N is comosite); the rocess guarantees that we will find a rime 3 mod 4 dividing N, not that N itself is rime. Theorem There are infinitely many rimes mod 3.

43 35 Proof. One such rime is 7. If we have finitely many such rimes, say,..., r, set N = (2 r ) Then N mod 2 and N mod 3, so 2, 3, and i do not divide N. Obviously N >. Let be a rime factor of N, imlying that 3 is a square mod. We will show that 3 mod imlies mod 3. Write 3 b 2 mod. In C, a cube root of unity is ( + 3)/2. Analogously, in Z/ the number ( + b)/2 cubes to : ( ) 2 + b = 2 2b + b2 4 = b 2 so ( ) 3 + b = + b b = b2 4 =. And ( +b)/2 mod since if ( +b)/2 mod then b 3 mod, which after squaring would imly 3 9 mod so 2, a contradiction. Therefore, ( + b)/2 has order 3 in Z/. Since (Z/) has size, this imlies that 3 ( ) and hence mod 3. Infinitely many rimes that are mod 3 can be constructed in this manner. Theorem There are infinitely many rimes 2 mod 3. Proof. One such rime is 2. If we have finitely many such rimes, say,..., r, set N = 3 r. Clearly no i divides N. Since N 2 mod 3 and N >, at least one rime factor of N must be congruent to 2 mod 3. Since {,..., r }, infinitely many 2 mod 3 can be added to the list by reeating this rocess. The roofs of these theorems involved creating secific formulas for N such that some rime dividing N satisfies certain conditions. These conditions varied in each case and do not aear to generalize in a uniform way. This suggests that the general method of roving Dirichlet s theorem must be of a very different form.

44 Proof of Dirichlet s Theorem The following technical result is the hardest ste in the roof of Dirichlet s theorem. Theorem For any nontrivial Dirichlet character χ, the Dirichlet L-function L(s, χ) is nonzero at s =. Proof. We will show that L(, χ) 0 by contradiction. Our argument is adated from [2]. Set H(s) = ζ(s) 2 L(s, χ)l(s, χ) for Re(s) > 0. Since ζ(s) is analytic for Re(s) > 0 excet for a simle ole at s = by Theorem 3.3., and L(s, χ) and L(s, χ) are analytic for Re(s) > 0 by Corollary 3.3.9, H(s) is analytic for Re(s) > 0 excet ossibly at s =, where it has at worst a double ole. We have for Re(s) > 0 that L(s, χ) = n χ(n) n s = ( ) χ(n) = n s n n χ(n) n s = L(s, χ). Setting s =, we get L(, χ) = L(, χ), so L(, χ) = 0 = L(, χ) = 0. The double ole at s = in ζ(s) 2 is therefore canceled by zeros at s = of the two L-functions in H(s). Thus H(s) is analytic at s =, and hence for all s with Re(s) > 0. For Re(s) >, the functions ζ(s), L(s, χ), and L(s, χ) can be reresented by Euler

45 37 roducts: ζ(s) = / s, L(s, χ) = χ()/ s, L(s, χ) = χ()/ s. (4.2.) Because /( z) = ex( k zk /k) for z < (Examle 2.2.8), we have by (4.2.) that for Re(s) > ζ(s) 2 L(s, χ)l(s, χ) = ( ) 2 ( ) ( ) / s χ()/ s χ()/ s = e k 2/kks e k χ()k /k ks ( ) 2 + χ() k + χ() k = ex k k ks e k χ()k /k ks ( ) 2 + χ() k + χ() k = ex, (4.2.2),k where k runs over the natural numbers and over the rimes. All series are absolutely convergent on Re(s) >, which justifies rearrangements of series above and in what follows. k ks Look at the Dirichlet series in the exonential of (4.2.2). If divides the modulus of χ, then the coefficient of / ks is 2/k. If does not divide the modulus of χ, we can write χ() = e iθ. In that case, the coefficient of / ks is 2 + χ() k + χ() k k = 2 + eikθ + e ikθ k = cos(kθ ) k = 2( + cos(kθ )) k 0. In both cases, the coefficient of / ks is nonnegative, so H(s) is the exonential of a Dirichlet series with nonnegative coefficients that converges (absolutely) on Re(s) >.

46 38 Since the ower series of e s and,k (2+χ()k +χ() k )/k ks have nonnegative coefficients, we may exand and rearrange (4.2.2) to get a Dirichlet series reresentation of H(s) with nonnegative coefficients for Re(s) >. We will now show that this series in fact converges on the larger half-lane Re(s) > 0. Let C be the set of σ > 0 at which n a n/n σ converges, so (, ) C. Let σ 0 be the infimum of C, so 0 σ 0. We will show σ 0 = 0. First we will show n a n/n s converges when Re(s) > σ 0 : for each such s there is a σ C such that σ 0 < σ < Re(s), so by the comarison test Re(s) C, and by absolute convergence the series n a n/n s converges. If σ 0 > 0, then Landau s lemma (Lemma 3.3.2) tells us that n a n/n σ converges slightly to the left of σ 0 since H(s) is analytic for Re(s) > 0. This contradicts σ 0 being the infimum of C, so σ 0 = 0. Therefore n a n/n s converges and equals H(s) on the half-lane Re(s) > 0. We will next calculate the coefficient of / 2s in the Dirichlet series reresentation of H(s). Writing H(s) = n a n/n s, for a fixed rime the subseries of H(s) over owers of is ( ) a r 2 + χ() k + χ() k = ex rs k ks r 0 k = ( ) 2 + χ() k + χ() k ex k s k ks ( ) 2 + χ() + χ() = ex ex ( 2 + χ() 2 + χ() 2 2 2s ) ex = (2 + χ() + χ()) j (2 + χ() 2 + χ() 2 ) j j! js 2 j (j!) 2js j 0 j 0 j 0 ( ) 2 + χ() + χ() (2 + χ() + χ())2 = s 2 ( 2s χ()2 + χ() s ( 2 + χ() 3 + χ() 3 3 3s ) ( χ()3 + χ() 3 3 3s + ) (2 + χ() 3 + χ() 3 ) j 3 j (j!) 3js ).

47 39 Exanding roducts and rearranging terms, the coefficient of / 2s is (2 + χ() + χ()) χ()2 + χ() 2 2 = 3 + χ() 2 + χ() 2 + 2χ() + 2χ() + χ()χ() = (χ() + χ() + ) 2 χ()χ() + 2. Since χ()+χ() = 2Re(χ()) R and χ()χ() = χ() 2 is 0 (if χ() = 0) or (if χ() 0), we have χ()χ() + 2 = 2 or, so the coefficient of / 2s is real and. Since a 2 and each coefficient in the Dirichlet series for H(s) is nonnegative, for real s > 0 we have H(s) a 2/2s /2s. Plugging in s = /2 gives us H(/2) /, but this series diverges by Corollary This is a contradiction to the Dirichlet series for H(s) being convergent for Re(s) > 0. We therefore conclude that L(, χ) 0 for every nontrivial Dirichlet character χ. Corollary For any nontrivial χ, χ()/s converges as s +. Proof. By Theorem 3.3.0, for Re(s) > a logarithm of L(s, χ) is l(s, χ) :=,k χ() k k ks = χ() s + k 2 χ() k. (4.2.3) kks The second term on the right in (4.2.3) converges absolutely for σ > /2 by the roof of Theorem 3..4, and it is holomorhic on the half-lane σ > /2 by Corollary From (4.2.3), for Re(s) > we may write χ() s = l(s, χ) k 2 χ() k. (4.2.4) kks Since L(s, χ) 0 at s = by Theorem 4.2., L(s, χ) 0 near s = because