DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS

Transcription

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 p s = log ζ(s) + ε, ε < 1.

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. Oe ca show that G = G by usig the elemetary divisor structure of fiite abelia groups (or by usig the Su Ze theorem ad the structure of the groups (Z/p e Z) ), 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 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, f : G 1 C, f(χ) = f(x)χ(x 1 ), x G ad the the Fourier series of f is s f : G C, s f = χ G 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 χ G y G f(y) y G χ 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. χ G 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 Yet more o Dirichlet characters Let G be a fiite abelia group, writte additively, ad let H be a subgroup. Suppose that χ : H C is a homomorphism. We show that χ exteds homomorphically to G. Ideed, cosider ay elemet g of G that does ot lie i H. Some multiple dg does lie i H, ad we cosider the smallest such positive d. Cosider the direct sum H Zg, which eed ot be a subgroup of G. Cosider also the subgroup Z( dg dg) of the direct sum. The quotiet (H Zg)/Z( dg dg) is isomorphic to the supergroup H + Zg (odirect sum) of H i G. Exted χ from H to the direct sum H Zg by defiig χ(h 0) = χ(h) for all h H ad defiig χ(0 g) to be ay complex umber whose d-th power is χ(dg); there are d such extesios of χ. This exteded χ is trivial o Z( dg dg), ad so it desceds to the quotiet. That is, χ is defied o the supergroup H + Zg of H i G. Repeat the process to exted the character util it is defied o all of G. Now retur to the settig of this writeup, with G = (Z/NZ) for some N. This sectio has show that ay Dirichlet character of ay subgroup H of G exteds to a Dirichlet character of G, ad i fact there are G / H such extesios. 6. 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 (χ), ε o (χ) < 1.

6 6 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS 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. 7. 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 7.1. 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 log ζ(s) p s, meaig that lim s 1 + log ζ(s) = 1, p s is the substace of Euler s proof.

7 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS 7 8. The secod idea of Dirichlet s proof Recall that for s > 1, p s = 1 χ(a 1 ) log L(χ, s) + ε, ε < 1. p=a(n) χ 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) = χ G 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 or ζ N (s) is meromorphic o {s : Re(s) > 0} with a simple pole at s = 1 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. 9. Meromorphy of ζ N (s) at s = 1 Lemma 9.1. 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, χ G (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, ad this is the character that is evaluated at p. Whe p N, the process described i the previous paragraph merely reproduces χ(p + NZ), ow referrig to the origial χ. More geerally, 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 1 χ(p)t o the left side of the lemma s equality is otrivial if ad oly if the origial χ factors through (Z/N p Z). To repeat: oly the characters i G that factor through (Z/N p Z) cotribute somethig other

8 8 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS tha 1 to the left side of the lemma s equality. Furthermore, ay character i G that does factor, χ = χ Np π N,Np, is determied by χ Np. Thus, to prove the lemma we may cosider oly characters modulo N p. 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 ; here p + N p Z is the multiplicative group geerated by p modulo 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 a character χ modulo N p that takes p to ρ j where ρ is a primitive complex f-th root of uity. This is clear for characters of the subgroup p + N p Z of (Z/N p Z), ad we have discussed the fact that such characters exted to (Z/N p Z). Puttig together the ideas of this paragraph, for each such j there exist g characters χ modulo N p that take p to ρ j, idepedetly of k. Now the proof of the lemma is immediate. 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 χ G 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 9.2. 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 χ G χ G = (1 χ(p)p s ) 1 = (1 p fs ) g. χ G The product coverges absolutely for Re(s) > 1, justifyig the rearragemets. Theorem 9.3. 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, ν=0 ν=0

9 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS 9 ad so for s > 1/, 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. 10. 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 + The correspodig ideas i Dirichlet s proof were L(χ, s) = χ() s = (1 χ(p)p s ) 1, Z + Cosequetly, p a(n) I other words, p a(n) p s 1 p s 1 lim ζ N (s) = s 1 χ G χ(a) 1 log L(χ, s), where ζ N (s) = χ(a) 1 log L(χ, s) 1 χ G p a(n) p s χ G L(χ, s). log ζ(s) 1 p s. lim = 1 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. 11. 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). For positive real s it is the egative sum of small areas above the y = t s curve but below the =1

10 10 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS circumscribig box of the curve over each uit iterval, ad hece it is bouded absolutely by 1. More geerally, for complex s with positive real part we ca quatify the smalless of the sum as follows. Sice for all t [, + 1] we have it follows that t s s = s t x s 1 dx s +1 t (t s s ) dt s, Re(s)+1 x Re(s) 1 dx s Re(s) 1, 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. Propositio 11.1 (Summatio by Parts). Let {a } 1 ad {b } 1 be complex sequeces. Defie A = a k for 0 (icludig A 0 = 0), so that Also defie k=1 a = A A 1 for 1. 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 k=1

11 DIRICHLET S THEOREM ON ARITHMETIC PROGRESSIONS 11 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, {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) = 1 s q 1(q) s = (1 q 1 s )ζ(s), Re(s) > 1. 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, because if s = σ + it the q 1 s = e (1 σ) l q e it 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, the prime q is arbitrary, ad 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.