DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS

Transcription

1 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS We la to rove the followig Theore (Dirichlet s Theore) Let (a, k) = The the arithetic rogressio cotais ifiitely ay ries a + k : = 0,, 2, } = : a od k} Euler gave a aalytic roof of the ifiitude of ries by showig that x I doig so he used roerties of the zeta fuctio ζ(s) = = ( ) s s = We will attet soethig siilar ad show that x a od k (for techical reasos we will i fact deostrate the divergece of a od k ) Note that the above su ca be writte as f a,k()/ where f a,k is the characteristic fuctio of the roerty a od k For exale, oe could take f a,k () = cos(π( a)/k) but this is suerficial ad o good to aybody It would be ice if our fuctio, or soe slight variat, ade sese as the coefficiets of a Dirichlet series ad allowed for a Euler roduct This would already ily ore deth tha our revious exale for f a,k Dirichlet characters Defiitio A Dirichlet Character odulo k is a arithetic fuctio χ : N C satisfyig () χ( + k) = χ() N (2) χ() = χ()χ(), N (3) χ() 0 (, k) = log

2 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS 2 Defiitio 2 The ricial character odulo k is the uique Dirichlet character χ such that χ () = (, k) = Firstly, ote that χ() = sice for (, k) = we have 0 χ() = χ( ) = χ()χ() Also, ote that χ s o-zero values are th-roots of uity sice for (, k) =, we have od k ad hece χ() = χ( ) = χ( + k) = χ() = Exales () k = : Oe character, χ() = (2) k = 2: Oe character, χ() = for odd, χ() = 0 for eve (3) k = 3: We have χ Suose χ χ The χ(2) ad χ(2) 2 = χ(4) = χ() = hece χ(2) = (4) k = 4: Siilar to k = 3 Note that by eriodicity we oly eed cosider oe reresetive fro a give residue class ˆ od k Also, we ay restrict attetio to those reresetatives with (, k) = sice χ() = 0 otherwise I other words, we ca restrict our attetio to the grou (Z/kZ) of uits odulo k By their ultilicative roerty, we see that ay give Dirichlet character χ : N C iduces a hooorhis χ : (Z/kZ) C = GL (C), ˆ χ() Coversely, give a hooorhis f : (Z/kZ) C we ca acquire a Dirichlet character by settig χ() = f(ˆ) whe (, k) = ad χ() = 0 whe (, k) > If we view the set of Dirichlet characters od k as hooorhiss fro (Z/kZ) C the they for a grou whe equied with the oitwise ultilicatio defied by (χ χ )() = χ()χ () The idetity eleet is give by χ ad the iverse of a eleet χ is give by χ sice (χ χ)() = χ() 2 = = χ () This eas that the Dirichlet characters odulo k ca be thought of as the eleets of soethig called the dual grou Ĝ of the grou G = (Z/kZ) A faous result of haroic aalysis (Potryagi Duality) ilies that G is isoorhic to Ĝ, ad hece we have the first art of the followig theore Theore 2 There are exactly Dirichlet characters odulo k Also, for ay give corie to k with od k there exists a χ such that χ() Theore 3 (Orthogoality) Let k be a ositive iteger ad let χ be a Dirichlet character odulo k The k ifχ = χ () χ() = 0 otherwise =

3 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS 3 Let be a ositive iteger The (2) χ od k χ() = if od k, 0 otherwise where the suatio is over all Dirichlet characters od k Let χ, χ be Dirichlet characters od k The k (3) χ()χ ifχ = χ () = 0 otherwise = For ositive itegers, 2 we have if 2 od k ad (, k) = (4) χ( )χ( 2 ) = 0 otherwise χ od k Proof For the first su the result is obvious if χ = χ If χ χ the ick a with (, k) = such that χ() The as rus through a reduced residue syste od k so does Therefore, k k k χ() χ() = χ() = χ() = = ad the result follows For the secod su ote that if (, k) > the all ters i the su are zero If od k the χ() = χ() = for all characters ad so the su sily equals the uber of characters od k, which is by Theore 2 So ow suose that (, k) = ad that od k Pick a χ such that χ () The as χ rus over the characters od k so does χ χ, sice they for a grou Therefore, χ () χ χ() = χ χ ()χ() = χ = (χ χ)() = χ χ() ad the su is therefore equal to zero i this case For the third su take χ = χχ i () Fially, for the fourth su ote that if either (, k) or ( 2, k) is > the the su is zero So suose (, k) = ( 2, k) = ad let 2 deote the iverse of 2 odulo k ie the uber satisfyig 2 2 od k We ow aly (2) with = 2 The su i questio is give by χ() = χ( )χ( 2 ) = χ( )χ( 2 ) χ χ χ This last equality follows o otig that χ( )χ( 2 ) = χ( 2 2 ) = χ() = ad hece χ( 2 ) = χ( 2 ) sice χ as oto the uit circle for itegers corie to k If 2

4 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS 4 od k the od k ad the above su equals by (2) If 2 the od k ad the reaiig case follows od k 2 Dirichlet L-fuctios Let χ be a Dirichlet character od k The the Dirichlet L-fuctio associated to χ is give by χ() (5) L(s, χ) =, s = σ + it s = Proositio 2 The series for L(s, χ) coverges absolutely for σ > ad for fixed δ > 0 it coverges uiforly for σ + δ It is therefore aalytic i the regio σ > Proof Postoed The derivative of L(s, χ) is give by the series (6) L χ() log (s, χ) = s = Proositio 22 If χ χ the L(s, χ) ad L (s, χ) coverge (coditioally) for σ > 0 I articular, their values at s = are defied Proof If χ χ the by artial suatio we have χ() (7) = x χ() + s χ() t s dt s x s x x By () the sus x χ() are bouded Therefore χ() x (8) x σ + s t σ dt s x This last exressio is O() for σ > 0 ad so the result follows o lettig x For the result ivolvig the derivative just use log t t ɛ i the above The followig is ot absolutely ecessary for our uroses but it does give ore eaig to soe of our results Theore 4 (Euler roduct) For σ > we have the absolutely coverget roduct (9) L(s, χ) = ( χ() ) s t

5 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS 5 Also, for σ > we have (0) L(s, χ) = = µ()χ() s ad the su is absolutely coverget i this regio Proof Note the factors i the roduct are the sus of geoetric series: ( χ() ) χ( ) = s s =0 Takig the roduct over ries less tha a give x gives χ( ) = χ( j j ) s ( j j ) = s x =0,, j x,, j 0 A(x) χ() s where A(x) = N : = x} Clearly, li x A(x) = N Now, L(s, χ) χ() χ() 0 s s >x A(x) as x sice the series = χ() s is coverget for σ > Sice a absolutely coverget roduct of o-zero ters is o-zero we see that L(s, χ) 0 for σ > Takig the recirocal gives L(s, χ) = ( χ() s Exadig this roduct we see that the resultat series has coefficiets µ()χ() ad so (0) follows If the rigour olice show u the just use the revious arguet Absolute covergece of the series follows o alyig the itegral test after usig µ()χ() ad s = σ ) 3 Dirichlet s Theore We are ow i a ositio to rove Dirichlet s Theore We shall i fact rove the followig Theore 5 Suose (a, k) = The log () x a od k = log x + O() The startig oit is to use (4) as a suitable characteristic fuctio

6 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS 6 Lea 3 Let (a, k) = The log (2) x a od k Proof By(4) we have (3) Now, (4) x a od k x log = log x + = x = x = χ () log log x χ χ(a) χ χ x χ(a)χ() χ() log log χ () + χ(a)χ() χ χ χ () log = x (,k)= log + = = log + O() x = log x + O() The above Lea ilies that if we ca show χ() log = O() x χ(a) χ χ x x x k log + O() χ() log for χ χ, the Dirichlet s Theore will follow This is ow our ai focus Iforal Discussio: As we have see before, icororatig rie owers ito a sus ivolvig log usually just adds a O() ter So we ca exect x χ() log = x χ()λ() + O() where Λ() = log if = ad Λ() = 0 otherwise Why are we always addig i rie ower ters if it does t really chage aythig? Well, this latter su adits a ore cocise descritio as the artial sus of the series for L (, χ)/l(, χ)

7 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS 7 Ideed, o logarithic differetiatio of the Euler roduct: χ() s ), we have L(s, χ) = ( (5) L (s, χ) L(s, χ) = log χ() s = (log )χ( ) s = =0 = χ()λ() s Fro this we see that (6) x χ() log L (, χ) L(, χ) ad this is bouded as log as L(, χ) 0 We ow attet to ake our discussio ito a ore rigorous arguet Our first ste is to deostrate a relatioshi siilar to (6), but i a ore quatitative for We the show that this for is bouded if L(, χ) 0 for χ χ Fially, we rove the latter Lea 32 We have (7) x χ() log = x χ()λ() + O() Proof We have (8) x χ()λ() =, x = x χ( ) log χ() log +, x, 2 χ( ) log Now, the secod su i the above is tha log = log ( ) < 2 log ( ) = O() Lea 33 For χ χ we have (9) x χ() log = L (, χ) x µ()χ() + O()

8 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS 8 Proof By the revious Lea ad the fact that Λ() = d µ(d) log(/d) we have (20) x χ() log + O() = x χ()λ() = x = χ() d χ(d)µ(d) d µ(d) log(/d) x/d χ() log after rearragig i ters of the divisors d ad usig the ultilicative roerties of χ Now, L χ() log (, χ) = = + χ() log, = Y >Y ad >Y χ() log = Y log Y χ() Y Y ( ) log t χ() dt log Y t Y Here we have used the fact that χ χ ad hece the su Y χ() is bouded Therefore, χ()λ() = L (, χ) ( ) χ(d)µ(d) log x/d + O d d x/d x (2) = L (, χ) χ(d)µ(d) + O () d t sice (log x log d) = x log x x log x + O(x) = O(x) Note that by equatio (0), the su x µ()χ() looks a lot like /L(, χ) ad so we have essetially established the relatio (6) Lea 34 Suose χ χ If L(, χ) 0 the µ()χ() (22) = O() x

9 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS 9 Proof Note that (23) S := x µ()χ() x/ χ() =, x χ()µ() after rearragig We ow grou together the ters, writig = l say, ad rearrage to give (24) S = χ(l) µ() = l l x after usig O the other had, (25) S = x = x µ() = l l if l =, 0 otherwise µ()χ() L(, χ) µ()χ() =L(, χ) x =L(, χ) x >x/ L(, χ) + O(/x) µ()χ() µ()χ() + O ( x + O () χ() ) χ()µ() x Sice L(, χ) 0, we ay divide through ad the result follows 4 The o-vaishig of L(, χ) for o-ricial characters As the sectio headig suggests, we la to rove the followig Theore 6 If χ χ the L(, χ) 0 The roof is slit ito two cases: oe where χ is real, ad the other where χ is colex We will start with the real case Lea 4 Suose χ is a real Dirichlet character od k Let (26) A() = d χ(d)

10 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS 0 The A() 0 for all ad A() if is a square Proof First ote A() is ultilicative sice it is the covolutio of two ultilicative fuctios, aely () := ad χ() Therefore, it is deteried by its values at rie owers We have χ() =0 = A( α ) =, χ() = = A( α ) = d( α ) = α +, χ() = = A( α if α is eve, ) = 0 if α is odd Writig = α we see A() = A(α ) ad hece A() 0 for all If is square the all the owers α i the rie decoositio of are eve ad so A() Proositio 42 Suose χ is a real o-ricial Dirichlet character od k Let (27) B(x) = x A() /2 The () B(x) as x (2) B(x) = 2x /2 L(, χ) + O() ad hece L(, χ) 0 Proof For the first art ote that by the Lea B(x) x = 2 = χ(d) d /2 A() /2 x = 2 For the secod art we have B(x) = χ(d) = /2 x d 2 (28) /2 = log x 2 x /2 χ(d) d /2, after rearragig /2 x/d ( x /2 + O(), usig Euler suatio d) χ(d) d =2x /2 ( =2x /2 L(, χ) d>x =2x /2 L(, χ) + O() ( ) + O χ(d)d /2 χ(d) d ) + O()

11 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS where we have used d>x χ(d) d = χ(d) + x x χ(d) dt = O(x ) t 2 d t We ca ow tur to the case where χ is colex cotradictio We will roceed towards a Lea 43 If χ χ ad L(, χ) = 0 the (29) L (, χ) x µ()χ() = log x + O() Proof We have (30) S := x =, x µ()χ() x/ χ() log(x/) µ()χ() log(x/) Oce agai, we grou together the ters, writig = l say, ad rearrage to give (3) S = l x χ(l) log(x/l) l µ() = log x l after usig µ() = l if l =, 0 otherwise

12 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS 2 O the other had, (32) S = x = x = x µ()χ() µ()χ() µ()χ() x/ x/ χ() ( log + log(x/)) χ() log ( L (, χ) + + log(x/) x/ ) χ() log >x/ ( + log(x/) L(, χ) >x/ χ() ) χ() = ( ) µ()χ() log(x/) L (, χ) + O x/ x ( log(x/) + log(x/)l(, χ) + O x/ Sice we re assuig L(, χ) = 0 this equals (33) L (, χ) ( µ()χ() ) + O log(x/) x x The result ow follows o otig that the su i big O ter is x, as has bee see reviously Let (34) S(k) = ad let x } χ od k : χ χ, L(, χ) = 0 (35) N(k) = S(k) Now, if L(, χ) = 0 the L(, χ) = 0 ad so if χ S(k) the χ S(k) (ote these characters are distict sice χ χ for colex characters) Therefore, N(k) is eve ) Proositio 44 We have (36) x (k) log = N(k) log x + O()

13 DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS 3 This ilies that N(k) = 0 sice otherwise N(k) 2 ad hece the right had side of the above would be egative for large eough x, cotrary to the su o the left beig ade u solely of ositive ters Proof By Leas 3 ad 33 we have log = log x + (37) x (k) = log x χ χ x χ() log L (, χ) χ χ x + O() µ()χ() + O() Now, by Lea 34 the su over x is O() if L(, χ) 0 If L(, χ) = 0 the the su ties L (, χ) is log x + O() by Lea 43 Therefore, after writig P for the characteristic fuctio of a give roerty P, we have (38) x (k) log = log x χ χ L(,χ) 0 O() + L(,χ)=0 ( log x + O() ) + O() = N(k) log x log x + O()