DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS

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1 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS 1. Itroductio Questio: Let a, N be itegers with 0 a < N ad gcd(a, N) = 1. Does the arithmetic progressio cotai ifiitely may primes? {a, a + N, a + 2N, a + 3N,... } For example, if a = 4, N = 15, does the arithmetic progressio cotai ifiitely may primes? {4, 19, 34, 49,... } Aswer (Dirichlet, 1837): Yes. Ad furthermore, for fixed N the primes distribute evely amog the arithmetic progressios correspodig to differet values of a. For example, if N = 15, eight arithmetic progressios are cadidates to cotai primes: {1, , , ,... }, {2, , , ,... }, {4, , , ,... }, {7, , , ,... }, {8, , , ,... }, {11, , , ,... }, {13, , , ,... }, {14, , , ,... }. I fact, each of these progressios cotais ifiitely may primes, ad the primes distribute evely amog them. The phrase distribute evely will be defied more precisely later o. Recall some formulas: 2. Euler s proof of ifiitely may primes Geometric series: X ν = (1 X) 1, X C, X < 1, ν=0 Logarithm series: log(1 X) 1 = ν 1 X ν, X C, X < 1, ν=1 1

2 2 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS Telescopig series: ν=2 1 ν(ν 1) = 1. 1 (Proof: ν(ν 1) = 1 ν 1 1 ν.) First we establish Euler s idetity (i which P deotes the set of prime umbers): s = (1 p s ) 1, s > 1. Z + The Fudametal Theorem of Arithmetic asserts that ay Z + is uiquely expressible as = p e1 1 pe per r... with all e i N ad almost all e i = 0. Euler s idetity really just rephrases this fact: s = (2 s ) e = (1 2 s ) 1, =2 e e=0 s = (2 s ) e1 (3 s ) e2 = (1 2 s ) 1 (1 3 s ) 1, =2 e 1 3 e 2 =2 e 1 p er r. s = e 1=0 r i=1 e i=0 (p s e 2=0 i ) ei =. s = (1 p s ) 1. Z + r (1 p s i ) 1, With Euler s idetity i place, his proof that there are ifiitely may primes follows. Let ζ(s) = s = (1 p s ) 1, s > 1. Z + By the product expasio of ζ, log ζ(s) = log (1 p s ) 1 = log(1 p s ) 1 = ν 1 p νs. ν=1 That is, log ζ(s) = i=1 p s + ν 1 p νs. But the secod term i the previous display is small by a basic estimate, the the geometric sum formula, the compariso with the telescopig series, ν 1 p νs < p ν = 1 p 2 (1 p 1 ) = 1 p(p 1) < 1. ν=2 ν=2 Ad so ν=2 log ζ(s) 1 < p s < log ζ(s).

3 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS 3 By the sum expasio of ζ, lim s 1 + ζ(s) = because the harmoic series diverges. So lim s 1 + log ζ(s) =, ad thus lim p s =. s 1 + The oly way for the sum to diverge is if it is over a ifiite set of summads, so there must be ifiitely may primes. 3. Dirichlet characters Dirichlet augmeted Euler s idea by usig Fourier aalysis to pick off oly the primes p such that p a (mod N). Let G = (Z/NZ), a fiite abelia multiplicative group of order G = where φ is Euler s totiet fuctio. Defie G = {homomorphisms : G C }. The G forms a fiite abelia multiplicative group also. Specifically, for ay χ 1, χ 2 G, defie χ 1 χ 2 by the rule (χ 1 χ 2 )(g) = χ 1 (g)χ 2 (g), g G. The idetity elemet of G is the character χ such that χ(g) = 1 for all g G, ad we use the symbol 1 (or 1 N to emphasize N) to deote this character. The group G is called the dual group of G. It is fairly easy to show that G = G, but the isomorphism is ot caoical. Propositio 3.1 (Orthogoality Relatios). For each χ G, { G if χ = 1, χ(g) = 0 otherwise, Ad for each g G, g G χ(g) = { G if g = 1, 0 otherwise. For the secod orthogoality relatio, a argumet is eeded that if g 1 G the there is a character χ G such that χ(g) 1 C. For ay fuctio f : G C, the Fourier trasform of f is a correspodig fuctio o the dual group, ad the the Fourier series of f is f : G 1 C, f(χ) = f(x)χ(x 1 ), x G s f : G C, s f = f(χ)χ.

4 4 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS The secod orthogoality relatio shows that the Fourier series reproduces the origial fuctio, s f (x) = 1 f(y)χ(xy 1 ) = 1 y G f(y) y G χ(xy 1 ) = f(x). Because the group G is fiite, o qualificatios o the fuctio f, ad o covergece issues of ay sort, are ivolved here. Returig to the Dirichlet proof, specalize the fuctio f to pick off a (mod N), { 1 if x = a, f(x) = 0 otherwise. The for ay χ G, the χth Fourier coefficiet of f is simply f(χ) = χ(a 1 )/, ad the relatio s f (x) = f(x) is ievitably just the secod orthogoality relatio, { 1 χ(xa 1 1 if x = a, ) = 0 otherwise. The Dirichlet proof is cocered with the sum p=a(n) p s. The idicator fuctio f lets us take the sum over all primes istead ad the replace f by its Fourier series s f = (1/) χ χ(a 1 )χ to get p=a(n) p s = We will retur to this formula soo. f(p)p s = 1 χ(a 1 ) χ(p)p s. 4. More o Dirichlet Characters Associate to ay character χ G a correspodig fuctio from Z to C, also called χ, as follows. First, there exists a least positive divisor M of N such that χ factors as χ = χ o π M : (Z/NZ) χ π M (Z/MZ) χo C. The iteger M is the coductor of χ, ad the character χ o is primitive. Note that χ o ( + MZ) = χ( + NZ) if gcd(, N) = 1, but if gcd(, M) = 1 while gcd(, N) > 1 the χ o ( + MZ) is defied ad ozero eve though χ( + NZ) is udefied. Secod, redefie the origial symbol χ to deote the primitive character χ o exteded to a multiplicative fuctio o the positive itegers, { χ : Z + χ o ( + MZ) if gcd(, M) = 1, C, χ() = 0 if gcd(, M) > 1. The followig relatio, with the ew χ o the left ad the origial χ o the right, χ() = χ( + NZ) if gcd(, N) = 1,

5 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS 5 justifies the multiple use of the symbol χ. (For example, the orthogoality relatios are udisturbed if we apply the ew χ to coset represetatives rather tha applyig the origial χ to cosets.) For gcd(, N) > 1, χ() is defied ad possibly ozero, while χ( + NZ) is udefied. By default, we pass all Dirichlet characters through the process described here, suppressig further referece to χ o from the otatio. I particular, if N > 1 the the trivial character 1 N G does ot exted directly to the costat fuctio 1 o the positive itegers. However, 1 N has coductor M = 1, ad the primitive trivial character 1 modulo 1 is idetically 1 o (Z/1Z) = {0}. The primitive trivial character lifts to the costat fuctio 1() = 1 for all Z. 5. L-fuctios ad the first idea of Dirichlet s proof Recall that G = (Z/NZ), a G, ad the goal is to show that the set {p P : p a (mod N)} is ifiite. For each χ G (with its correspodig χ : Z C) defie L(χ, s) = χ() s = (1 χ(p)p s ) 1, s > 1. Z + (Equality of the sum ad product follow from a straightforward aalogue to the proof of Euler s idetity, sice characters are homomorphisms.) The log L(χ, s) = ν Z + ν 1 χ(p ν )p νs = χ(p)p s + ν 2 ν 1 χ(p ν )p νs, ad the secod term has absolute value at most 1 by the argumet i Euler s proof. Equivaletly, χ(p)p s = log L(χ, s) + ε o, ε o < 1. Recall the formula that came from the Fourier series of the idicator fuctio of a (mod N), p s = 1 χ(a 1 ) χ(p)p s. p=a(n) χ The previous two displays show that the desired sum is close to the liear combiatio of {log L(χ, s)} whose coefficiets are the Fourier coefficiets of the idicator fuctio, p=a(n) p s = 1 χ(a 1 ) log L(χ, s) + ε, ε < 1. χ Now the goal is to show that the right side goes to + as s 1 +. Already we kow that the summad for the trivial character does so. The crux of the matter will be that the fiite value L(χ, 1) for otrivial χ is ozero.

6 6 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS 6. Aalytic Properties of L(χ, s) We eed to study the behavior of L(χ, s) as s 1 +. Eve though s is real, L(χ, s) still takes complex values. Brig complex aalysis to bear o the matter by viewig s as a complex variable. Begi by extedig the defiitio of L(χ) to L(χ, s) = χ() s = (1 χ(p)p s ) 1, s C, Re(s) > 1. Z + Here s = e s l for Z +. The sum expressio for L(χ, s) coverges absolutely o the half plae {s : Re(s) > 1}, ad the covergece is uiform o compacta. Its summads, hece its partial sums, are aalytic. So L(χ, s) is aalytic o the half plae. Propositio 6.1. The fuctio L(χ, s) has a meromorphic cotiuatio to the right half plae {Re(s) > 0}. If χ = 1 the the exteded fuctio ζ(s) has a simple pole at s = 1 with residue 1. If χ 1 the the exteded fuctio L(χ, s) is aalytic. Elemetary argumets to be give at the ed of this writeup establish the propositio, but such argumets do ot scale up beyod the situatio at had. I a separate writeup, results that subsume the propositio are proved by methods that have scope. We reiterate here that the idetity meaig that is the substace of Euler s proof. log ζ(s) p s, lim s 1 + log ζ(s) = 1, p s 7. The secod idea of Dirichlet s proof Recall that for s > 1, 1 χ(a) 1 log L(χ, s) 1 < p a(n) p s 1 χ(a) 1 log L(χ, s). Also, L(1, s) as s 1 +. We will show that for χ 1, L(χ, 1) 0 ad thus log L(χ, 1) is fiite. Sice χ(a) 1 = 1 for all χ G, it follows that lim χ(a) 1 log L(χ, s) s 1 + χ G = + ad Dirichlet s proof is complete. So we eed to study the fuctio ζ N (s) = L(χ, s). Sice L(1, s) is meromorphic o {s : Re(s) > 0} with a simple pole at s = 1 ad all other L(χ, s) are aalytic o {s : Re(s) > 0}, there are two possibilities. Either ζ N (s) is meromorphic o {s : Re(s) > 0} with a simple pole at s = 1

7 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS 7 or ζ N (s) is aalytic o {s : Re(s) > 0}. We must rule out the secod possibility to complete the proof. The fuctio ζ N (s) has aother defiitio as the cyclotomic Dedekid zeta fuctio. A separate writeup describes ζ N (s) as the cyclotomic Dedekid zeta fuctio, but i doig so it must ivoke some laguage ad some results from algebraic umber theory. 8. Meromorphy of ζ N (s) at s = 1 Lemma 8.1. Let p be prime. Let N = p d N p with p N p. Let f be the order of p i (Z/N p Z), i.e., the smallest positive iteger such that p f 1 (mod N p ). Let g = φ(n p )/f. The for ay idetermiate T, (1 χ(p)t ) = (1 T f ) g. (See the commet immediately below for a careful parsig of the product i the previous display.) O the left side of the equality asserted by the lemma, the expressio χ(p) cootes that the character χ G has bee reduced to the primitive character χ o modulo M where M N is the coductor of χ, the exteded M-periodically to χ : Z C, which is evaluated at p. Whe p N the whole process merely reproduces χ(p + N Z), ow referrig to the origial χ. The process produces a ozero value χ(p) if ad oly if p does ot divide the coductor M of the origial χ. That is, the multiplicad o the left side of the lemma s equality is otrivial if ad oly if the origial χ desceds to a character ot ecessarily primitive modulo N p. Ay character i G that does ot desced to a character modulo N p is irrelevat to the lemma s equality. A character χ modulo N p takes p to 1 if ad oly if it factors through the quotiet group (Z/N p Z) / p + N p Z. This quotiet group has order φ(n p )/f = g, ad so it has g characters. That is, g characters χ modulo N p take p to 1. Also, for each j {0, 1,, f 1} there exist characters χ modulo N p that take p to ρ j where ρ is a primitive complex fth root of uity. Thus for each such j there exist g characters χ modulo N p that take p to ρ j, idepedetly of k. Ad as already metioed, every character modulo N that is ot defied modulo N p takes p to 0. Proof. Let ρ be a primitive fth root of uity i C. The f 1 1 T f = (1 ρ j T ), ad cosequetly, because g characters χ G take p to ρ j for each j, j=0 f 1 (1 T f ) g = (1 ρ j T ) g = (1 χ(p)t ). j=0

8 8 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS I the lemma we could have let H = (Z/N p Z), which equals G for all p N, ad the stated the lemma s formula as a product over χ H rather tha fussig about it holdig for G. Our reaso for isistig o G is maifest i the proof of the ext propositio. Propositio 8.2. ζ N (s) = (1 p fs ) g for Re(s) > 1. Proof. Compute, usig the lemma at the last step, ζ N (s) = L(χ, s) = (1 χ(p)p s ) 1 = (1 χ(p)p s ) 1 = (1 p fs ) g. The product coverges absolutely for Re(s) > 1, justifyig the rearragemets. Theorem 8.3. ζ N (s) has a simple pole at s = 1. Therefore L(χ, 1) 0 for each otrivial character χ modulo N. Proof. Otherwise ζ N (s) is aalytic o {s : Re(s) > 0} so that its product expressio coverges there. But for s R +, ( ) g (1 p fs ) g = p νfs p νs = (1 p s ) 1, ad so for s > 1/, ν=0 ν=0 ζ N (s) (1 p s ) 1 = ζ(s). Now lettig s approach 1/ from the right shows that the product expressio of ζ N diverges there. This gives a cotradictio. We ote that the complex aalysis is beig treated somewhat loosely here. 9. Review of the proofs Let the otatio f(s) g(s) mea lim s 1 + f(s)/g(s) = 1. The three ideas i Euler s proof were ζ(s) = s = (1 p s ) 1, Z + p s log ζ(s), lim ζ(s) =. s 1 +

9 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS 9 The correspodig ideas i Dirichlet s proof were L(χ, s) = χ() s = (1 χ(p)p s ) 1, Z + p s 1 χ(a) 1 log L(χ, s), Cosequetly, p a(n) I other words, p a(n) p s 1 lim ζ N (s) = s 1 where ζ N (s) = χ(a) 1 log L(χ, s) 1 lim s 1 + p a(n) p s p s = 1. L(χ, s). log ζ(s) 1 p s. That is, ot oly is the set {p P : p a (mod N)} ifiite, but furthermore i some limitig sese it cotais 1/ of all the primes. This is the sese i which the primes distribute evely amog the cadidate arithmetic progressios a + N Z. 10. Place-holder Cotiuatio Argumets Oe way to cotiue the Euler Riema zeta fuctio from {Re(s) > 1} to {Re(s) > 0} is as follows. Compute that for Re(s) > 1, s 1 = t s dt = t s dt = ζ(s) + (t s s ) dt. 1 =1 This last sum is a ifiite sum of aalytic fuctios; call it ψ(s). all t [, + 1] we have it follows that t s s = s t x s 1 dx s +1 t =1 (t s s ) dt s, Re(s)+1 x Re(s) 1 dx s Re(s) 1, Sice for ad so the sum ψ(s) coverges o {s : Re(s) > 0}, uiformly o compact subsets, makig ψ(s) aalytic there. Thus ζ(s) = ψ(s) + 1, Re(s) > 1. s 1 But the right side is meromorphic for Re(s) > 0, its oly sigularity for such s beig a simple pole at s = 1 with residue 1. The previous display exteds ζ ad gives it the same properties. Oe way to exted L(χ, s) to Re(s) > 0 for χ 1 uses the discrete aalogue of itegratio by parts.

10 10 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS Propositio 10.1 (Summatio by Parts). Let {a } 1 ad {b } 1 be complex sequeces. Defie A = a k for 0 (icludig A 0 = 0), k=1 so that a = A A 1 for 1. Also defie b = b +1 b for 1. The for ay 1 m, the summatio by parts formula is 1 k=m a k b k = A 1 b A m 1 b m Proof. The formula is easy to verify i cosequece of 1 k=m A k b k. a k b k + A k b k = A k b k+1 A k 1 b k, k 1. Returig to L(χ, s) = Z + χ() s where χ is otrivial, the first orthogoality relatio gives 0+N = 0 χ() = 0 for ay 0 Z +. Let {a } = {χ()} ad {b } = { s }, ad ote that {A } is bouded while b Re(s) 1. Summatio by parts gives 1 1 L(χ, s) = lim a k b k = lim A k b k, k=1 ad the right side coverges o {s : Re(s) > 0}, uiformly o compacta. Thus L(χ, s) is aalytic o {s : Re(s) > 0}. Summatio by parts gives a secod argumet for the cotiuatio of the zeta fuctio as well. For ay prime q, itroduce the sequece of coefficiets {a } cosistig of q 1 times 1, the a sigle 1 q, the q 1 more times 1, the aother 1 q, ad so o, k=1 {a } = {1, 1,, 1, 1 q, 1, 1,, 1, 1 q, 1, 1,, 1, 1 q, }. ad cosider the Dirichlet series f q (s) = 1 a s. The sequece of partial sums of the coefficiets is (startig at idex 0 here) {A } = {0, 1, 2,, q 1, 0, 1, 2,, q 1, 0, 1, 2,, q 1, 0, }. Ad so summatio by parts shows that the Dirichlet series f q (s) is aalytic o Re(s) > 0. Compute that for Re(s) > 1 (where we have absolute covergece ad therefore may rearrage terms freely), f q (s) = s q 1 1(q) s = (1 q 1 s )ζ(s), Re(s) > 1.

11 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS 11 Sice f q (s) is aalytic o {Re(s) > 0} ad agrees with (1 q 1 s )ζ(s) o {Re(s) > 1}, it follows that (1 q 1 s )ζ(s) cotiues aalytically to {Re(s) > 0}. Therefore ζ(s) cotiues meromorphically to {Re(s) > 0} with poles possible oly where q 1 s = 1. The coditio q 1 s = 1 readily works out to s 1 + 2πiZ/ l q. Thus the oly possible poles of ζ(s) i {Re(s) > 0} are distributed evely alog the lie Re(s) = 1 with spacig 2π/ l q. However, q is arbitrary, ad a small exercise shows that the sets 2πZ/ l q ad 2πZ/ l q meet oly at 0 for distict primes q ad q. Thus the oly possible pole of the exteded ζ(s) is at s = 1. This completes the proof.