Dirichlet s Theorem. Calvin Lin Zhiwei. August 18, 2007

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1 Dirichlet s Theorem Calvin Lin Zhiwei August 8, 2007 Abstract This paper provides a proof of Dirichlet s theorem, which states that when (m, a) =, there are infinitely many primes uch that p a (mod m). The proof explained here mirrors Serre s proof in []. However, one distinction is that the Riemann zeta function is used as motivation to obtain results for the Dirichlet L-functions, which yields the above result. In fact, this paper can be read from the viewpoint of asking how do simple results of the zeta function depend on basic properties. For the rest of the paper, m is a fixed integer as given in the above theorem. Let P deonote the set of primes, P m,a denote the set of primes that are congruent to a modulo m and G m = (Z/mZ). We will use the Euler-phi function φ(m) which counts the numbers of positive integers less than m that are coprime to m. In particular, G m = φ(m). A sequence of complex numbers {a n } is called strictly multiplicative if a n a m = a nm, n, m N. The concept of analytic continuation will be useful in getting a deeper understanding of the proofs. However, prior knowledge is not necessary, and comments relating to analytic continuation can be ignored. Basic group theory is assumed. In particular, the result that abelian groups can be written as the product of cyclic groups is used without proof. Riemann-zeta Function In this section, we recall and derive certain basic properties of the Riemann zeta function. For each s C, Re(s) >, the Riemann zeta function is defined as

2 ζ(s) = n= Proposition. For Re(s) >, the zeta function converges. Proof. Using the integral test, n s t = s n= n+ n t s n= n = + s n=2 The result follows as we know that = s t s Proposition 2. ζ(s) = p P n + s p s n=2 l n within the domain. t = + s t s Proof. Let N be any natural number, A N be the set of natural numbers whose prime factors are not larger than N. Then we have n A N n s = p P,p N j=0 p js = The result follows by letting N tend to infinity. p P,p N p s Comment. Notice that the numerators are {,,,... } and form a strictly multiplicative sequence, which allows the function to be written in a product form. Proposition 3. ζ(s) = + ρ(s), where ρ(s) is holomorphic for Re(s) > 0. s Proof. Recall that Hence, s = t = n+ s n= n t s 2

3 ζ(s) = = n = s n= n= n+ n= = s + l n+ n n + ( s n= n+ n n s s n= n s t s n+ l t s ) ( ) Let ρ n (s) = l+, and ρ(s) = n= n s t s n= ρ n(s). It is clear that each ρ n (s) is well defined and holomorphic for Re(s) > 0. Hence, it remains to show that the series of functions converges uniformly on compact sets of Re(s) > 0. Observe that ρ n (s) sup n s t s, and n s t s = 0 when t = n. n t n+ By considering the derivative of n s t s which is st (s+), and letting x = Re(s), ρ n (s) sup n t n+ = s sup n t n+ n+ s n ts+ n+ n s sup n t n+ t s s x+ nx+ n 2 n+ n t s+ Thus, ρ n (s) converges uniformly for s k, Re(s) ɛ, ɛ > 0. Corollary. ζ(s) has a simple pole at s =. Comment. The equation ( ) gives the analytic continuation of ζ(s) in Re(s) > 0, s. Theorem. For Re(s) >, as s, p P log. s Proof. From proposition 3, since ρ(s) = ζ(s) = is holomorphic hence s bounded in a neighborhood of, thus log log ζ(s). Replacing ζ(s) by s its product form in Proposition 2, 3

4 log ζ(s) = log p = s p P p P where η(s) = p P,k 2 by the series, with x = Re(s) Hence,. η(s) p P,k 2 log p = s p P,k k p ks = p P + η(s). To show that η(s) is bounded, we majorize it k pks kp ks p P,k 2 p x (p x ) p P p P p ks p P,k 2 p(p ) log s log ζ(s) p P n=2 p kx n(n ) 2 Characters of finite abelian groups In this section, we obtain a basis for complex valued functions on G m = (Z/mZ) and study their properties. These arise naturally as an application of characters of G m. Let G be an abelian group with the operation written multiplicatively. Definition. A character of a group G is a homomorphism χ : G C. Example. The function that maps every element of G identically to is clearly a homomorphism. It is call the trivial character and denoted by χ. It is easy to check that the characters of G form a group Hom(G, C ) which is denoted by Ĝ, and called the dual of G. Proposition 4. Ĝ is a finite abelian group of the same order as G. Moreover, they are isomorphic as groups. 4

5 Proof. We will first prove this for cyclic groups G = n. Let x be a generator for the cyclic group G, then each χ Hom(G, C ) is uniquely determined by χ(x). Moreover, = χ() = χ(x n ) = χ(x) n shows that there are at most n possible characters of G. Conversely, given any n th root of unity ω, the map χ ω : G C given by χ ω (x a ) = ω a is a homomorphism. We will now prove the general statement for abelian groups G. By the classification of abelian groups, G H... H k, where each H i is a cyclic group of prime power h i. Let H i be generated by x i, then each χ Hom(G, C ) is uniquely determined by (χ(x ), χ(x 2 ),..., χ(x k )). Since χ(h i ) can take on at most H i values, Ĝ H i = G. Conversely, let π : G H i be the projection maending g to g (x g, x 2 g 2,..., x k k). Then, given any (n, n 2,..., n k ) N k, a i h i, consider χ(n, n 2,..., n k ) which takes g to e 2πi n i g h i i clearly defines a character. Thus, Ĝ = G, with the isomorphism given by the generators (hence not canonically) Proposition 5. Let G be a finite abelian group. Then G is canonically isomorphic to Ĝ. Proof. Given x G, the map φ x : χ χ(x) is a character of Ĝ. By the previous proposition, since G = Ĝ, it remains to show that the map x φ x is injective. With notation as in the previous proposition, consider the projection map G H i sending g to (x a,..., x a k k ). For g, a j 0. Without loss of generality, j =. Then, χ(, 0,..., 0)(g), which proves injectivity. Proposition 6. Let G = n, χ Ĝ. Then, { n χ = χ(x) = 0 χ x G Proof. The first formula follows from adding n s. For the second, choose y G such that χ(y). Then χ(y) x G χ(x) = x G χ(xy) = x G χ(x) 5

6 which implies that x G χ(x) = 0. Corollary. Let G = n, χ Ĝ. Then { n x = χ(x) = 0 x χ Ĝ Proof. This follows from the previous proposition applied to the dual group, and using the isomorphism given in proposition 5. Comment. Now consider functions from G to C. Since χ Ĝ form a linearly independent set of functions from G to C and G = Ĝ, we get a basis set of multiplicative functions which satisfy χ(xy) = χ(x)χ(y). For G m, we can extend the domain of these characters from G m to N, by defining { 0 (n, m) χ(n) = χ(a) n = a (mod m) Furthermore, it is clear that {χ(n)} is strictly multiplicative. These χ will be used to define the Dirichlet L-functions in Section Dirichlet series 3. Lemmas The following lemmas are useful for dealing with convergence of summations over complex numbers. Lemma 3.. Let {a n }, {b n } be two sequences of complex numbers and define A j,k = k n=j a n, and S j,j = j n=j a nb n. Then, S j,j = A j,j b j + j n=j A j,n(b n b n+ ). Proof. The statement follows by replacing a n with A j,n A j,n and regrouping terms. 6

7 Lemma 3.2. α, β R such that 0 < α < β. Let z = x + iy, x, y R and x > 0, then Proof. e αz e βz z x (e αx e βx ) e αz e βz = z β α e tz hence by taking absolute values, with x = Re(z), β e αz e βz z β e tz = z e tx = z α α x (e αx e βx ) The following is a standard lemma from complex analysis. Lemma 3.3. Let U be an open subset of C and f n a sequence of holomorphic functions on U which converge unifromly on every compact set to a function f. Then, f is holomorphic in U and the derivatives f n converge uniformly on all compact subsets to the derivative f of f. Proof. Let D be a closed disc contained in U and D be its boundary with usual orientation. By Cauchy s formula, z in the interior of D, f n (z 0 ) = f n (z) dz 2πi z z 0 Taking limits, uniform convergence allows us to exchange the integral at lmits, hence we get f(z 0 ) = lim f n (z 0 ) = lim f n (z) dz 2πi D z z 0 = lim f N(z) dz = f(z) dz 2πi D z z 0 2πi D z z 0 D which shows that f is holomorphic in the interior of D. The statement for derivatives is proved similarly using 7

8 f (z 0 ) = f(z) 2πi D (z z 0 ) dz Dirichlet series Let (λ n ) be a increasing sequence of positive real numbers tending to infinity. A Dirichlet series with exponents (λ n ) is a series of the form f(z) = a n e λnz, a n C, z C n= Example. The zeta function is a Dirichlet series with a n = and λ n = log n. The Dirichlet series with these exponents will be studied in Section 3.3 Proposition 7. If the Dirichlet series f(z) converges for z = z 0, then it converges uniformly in every domain of the form Re(z z 0 ) 0, Arg(z z 0 ) α, where α < π 2. Proof. Without loss of generality, z 0 = 0 and the hypothesis states that an is convergent. By Lemma 3.3, we have to show that there is uniform convergence in every domain of the form Re(z) 0, k. Given ɛ > 0, z Re(z) there exists N such that j, j N, A j,j ɛ. Applying Lemma 3. with b n = e λnz, we obtain j S j,j = A j,j e λ j z + A j,n (e λnz e λn+z ) n=j Applying Lemma 3.2, with x = Re(z), S j,j ɛ( + z j x (e λnx e λn+x )) ɛ( + k) n=j Thus, we have uniform convergence. Corollary. It follows that if f converges for z = z 0, then it converges for Re(z) > Re(z 0 ) by Lemma 3.3. Hence, the function is holomorphic in an open half plane, and its analytic continuation is given by the same formula. 8

9 Proposition 8. If the Dirichlet series f(z) has positive real coefficients {a j } and converges for R(z) > ρ, ρ R, and the function f can be extended analytically to a function holomorphic in a neighborhood of the point z = ρ, then there exists ɛ > 0 such that f converges for R(z) > ρ ɛ. Proof. Without loss of generality, assume ρ = 0, hence f is holomorphic in a disc z + ɛ and has a convergent Taylor series. The derivatives can be calculated using lemma 3.3, obtaining f (n) (z) = l a l ( λ l ) n e λ l z f (n) () = l a l ( ) n (λ l ) n e λ l The Taylor series expansion is Hence, f(z) = n=0 n! (z )n f (n) (), when z + ɛ f( ɛ) = n=0[ n! ( + ɛ)n ( ) n l a l ( ) n (λ l ) n e λ l ] is a double series with positive coefficient. By rearranging terms, f( ɛ) = l = l = l [a l e λ l n=0 a l e λ l e λ l(+ɛ) a l e λ lɛ n! ( + ɛ)n λ n l ] Hence, the Dirichlet series converges for z = ɛ and thus for Re(z) > ɛ from Proposition 7. Corollary. The set of convergence of a Dirichlet series is an open half plane bounded by a singularity. 9

10 3.3 Ordinary Dirichlet Series The ordinary Dirichlet series is obtained by setting λ n = log n, which gives us f(s) = n= a n n s, a n C, s C Corollary. Since n= converges for α >, if {a n α n } is bounded, then there is absolute convergence for Re(s) >. Proposition 9. If the absolute value of the partial sums A j,k = k n=j a n are bounded by K, then there is convergence for Re(s) > 0. Proof. This is similar to the proof given for ρ(s) in proposition 3. By applying Lemma 3., S j,j K ( j n=j ) n s (n + ) + s j s By Proposition 7, we may assume that s is real, hence the inequality simplifies to S j,j K j s, showing convergence. 3.4 Dirichlet L-function Recall that the strictly multiplicative function χ : N C was defined at the end of section 2. { 0 (n, m) χ(n) = χ(a) n = a (mod m) The corresponding Dirichlet L-function is defined by L(s, χ) = n= χ(n) n s Example. In the specific case where m =, ζ(s) = L(s, χ ). We now establish results for the L-functions that are analogous to Propositions, 2 and 3 given for the zeta function. 0

11 Corollary. From proposition 2, L(s, χ ) = F (s)ζ(s) where F (s) = p m ( r s ). Thus L(s, χ ) extends analytically for Re(s) > 0 and has a simple pole at s =. Proposition 0. For χ χ, L(s, χ) converges for Re(s) > 0 and converges absolutely for Re(s) >. Moreover, L(s, χ) = for Re(s) > p P χ(p) Proof. Since the absolute value of the coefficients χ(n) are bounded by, it follows that the series is absolutely convergent for Re(s) >. Let N be any natural number, let A N be the set of natural numbers whose prime factors are not larger than N. Then we have χ(n) = n s n A N p P,p N l=0 χ(p) l p ls = p P,p N χ(p) As N tends to infinity, we establish the product formula of L(s, χ) as above. From Proposition 6, j+m 2 i=j χ(i) = 0, hence A j,j is bounded by φ(m). By Proposition 9, L(, χ) converges in Re(s) > 0. Thus, we have convergence in Re(s) > 0 from Proposition Product of Dirichlet L-functions for fixed m For any prime p m, let p be the image of p in (Z/mZ), f(p) be the order of p, and g(p) = φ(m)/f(p). Proposition. Let ζ m (s) = χ L(s, χ). Then ζ(s) is a ordinary Dirichlet series, with positive integral coefficients, converging in the half plane Re(s) >, and ζ m (s) = ( ) g(p) p m p f(p)s Proof. For a fixed p, let ω p denote the f(p)-th roots of unity. From the construction of characters in Proposition 4, it follows that the multi-set {χ(p)} is g(p) copies of ω p.

12 Since ω ω p ( ωt ) = ( T f(p) ), we have χ ( χ(p)t ) = ( T f(p) ) g(p). Convergence in Re(s) > is guaranteed as we are multiplying finitely many convergent functions. Replace each L-function by its product form and set T =, ζ m (s) = χ = p m = p m p m χ(p) χ χ(p) ( ) g(p) p f(p)s Expansion of this product form shows that ζ m (s) is a ordinary Dirichlet series with has positive coefficients. Theorem 2. χ, L(, χ) 0. Proof. Proof by contradiction. Suppose that for some χ, L(, χ) = 0, then ζ m would be holomorphic at s =, and hence in Re(s) > 0 by Proposition 0. The product form given in Proposition would thus be the analytic continuation of ζ m (s). By Proposition 8, since the Dirichlet series has positive coefficients, the series would converge for all Re(s) > 0. However, ( ) g(p) is dominated by p, φ(m)s p f(p)s it follows that the coefficients of ζ m are greater than p m sequence diverges for s = as φ(m) p m p = p m ( p ) n n p φ(m)s This 2

13 Corollary. ζ m (s) has a simple pole at s = Comment. Observe that ζ (s) = ζ(s), so we have a proof of the corollary to Proposition 3, which is independent of Proposition 3. 4 Proof of Dirichlet s Theorem The result stated in Theorem allows us to define the analytic density of a set of primes A P. We say that A has an analytic density k if the ratio p / s p A p P tends to k as s + on the reals. It is clear that if A is finite, then it has an analytic density of 0. We can now formulate a stronger version of Dirichlet s Theorem. Theorem 3. The set P m,a has an analytic density of φ(m). Comment. In the spirit of this paper, this result is analogous to Theorem. To imitate the proof, we need to understand log ζ m (s). The principal determination of the logarithm is log = α n as given by the Taylor α expansion. For L(s, χ), define log L(s, χ) = log = log p P χ(p) p P = χ(p) p P,l N χ(p) l lp ls This series is clearly convergent in Re(s) >. Equivalently, we take the branch of log L(s, χ) in Re(s) > which becomes 0 as s tends to infinity on the real axis. For χ ˆ G m, define f χ (s) = p m,p P This function is convergent for s >. χ(p) Proof. We require the following 2 lemmas. = p P χ(p) 3

14 Lemma 4.. For χ =, f χ (s) log s. Proof. This follows from Theorem as f χ (s) differs from by the finitely many terms p m. Lemma 4.2. For χ, f χ (s) is bounded as s. Proof. To prove this, we consider the logarithm of L(s, χ). F χ (s) l 2 so is f χ (s). log L(s, χ) = f χ (s) + F χ (s) lp ls where F χ (s) = p P l 2 χ(p l ) lp ls, and by Theorem 2, log L(s, χ) is bounded, hence Now, for the proof of Dirichlet s theorem, let g(s) = p P m,a. Then, φ(m) χ Ĝ χ(a) f χ (s) = φ(m) = φ(m) = φ(m) = g(s) χ Ĝ p P p P χ Ĝ p P,a p χ(a) χ(p) χ(a p) φ(m) where the third equality follows from Proposition 6. Thus, g(s) = φ(m) χ Ĝ χ(a) f χ (s) φ(m) (a) f (s) φ(m) log s References [] Jean-Pierre Serre, A course in Arithmetic,