Version 1 lastupdate:11/27/149:47:29am Preliminar verison prone to errors and subjected to changes. The version number says all!

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1 7. Del Characters Versin 1 lastupdate:11/27/149:47:29am Preliminar verisn prne t errrs and subjected t changes. The versin number says all! Characters f finite abelian grups Let A be a finite abelian grup. Abelian grups can be bth additively and multiplicatively written, but in ur theretical deliberatins they will always be a multiplicative grups if the cntrary is nt explicitly stated. Recall the grup µ n C f n-th rts f unity; that is µ n = { z 2 C z n =1}. It is cyclic f rder n, ageneratrbeingexp 2 i/n. This is nt the nly generatr; indeed any pwer exp 2 im/n where (n, m) =1will generate. A generatr is called a primitive rt f unity. The grups Z/nZ and µ n are ismrphic. Chsing a generatr f µ n we can define an ismrphism Z/nZ! µ n by sending the residue class f a t the pwer a. This ismrphism depends n a chice f the primitive rt s Z/nZ and µ n are nt cannically ismrphism, and we shall distinguish between them. By a character f A we mean a grup hmmrphism : A! C.Aseveryelement in a in A is f finite rder, the character takes values in the subgrup f C f rts f unity. That is, if a n = e, then (a) n = (a n )= (e) =1,sthevaluesf belng t µ n if n = A. The set f characters f A is dented Â. Itisanabeliangrup;indeed,if 1 and 2 are tw characters, the prduct 1 2 is defined as usual by a 7! 1 (a) 2 (a), andne trivially sees that this is a grup hmmrphism (C is abelian). The neutral element in the character grup is the trivial character given as a 7! 1. It is usual written as 1 A r smetimes as 0. 1

2 The tw fllwing examples are fundamental: Example.. One has \ Z/nZ ' µ n.sinceeverycharacternz/nz takes values in µ n, there is the bvius map \ Z/nZ! µ n sending! (1). This is a grup hmmrphism by the definitin f the grup structure f the character grup Z/nZ, \ anditisbviusly injective since 1 generates Z/nZ. Tseeitissurjective,pickanyn-th rt f unity 2 µ n and take a lk at the hmmrphism Z! µ n sending a t a.itvanishesn nz, and thus furnishes us with a hmmrphism frm Z/nZ t µ n taking the value at 1. e Example.. One has ˆµ n ' Z/nZ. There is a natural map Z/nZ! ˆµ n sending a residue class [a] t the a-th pwer map 7! a. The pwer map is a grup hmmrphism µ n! µ n C that nly depends n the residue class f a mdul n (if a 0 = a + bn ne has a0 = a bn = a ). Hence the a map Z/nZ! ˆµ n is well defined, and it is easily checked t be a grup hmmrphism. It is injective since if a =1hlds fr all 2 µ n,ithldsinparticularfraprimitive n-th rt, and it fllws thta n a. Tseethatthemapissurjectivelet 0 be a primitive n-th rt. It generates µ n,s ( 0 )= 0 a fr sme integer a. Nwanyther 2 µ n is f the shape = 0 b with b 2 Z, andnehas ( ) = ( 0)= b ( 0 ) b = 0 ab = a. e Functriallity Assume that A 1 and A 2 are tw finite abelian grups and that : A 1! A 2 is agruphmmrphism.if is a character f A 2,thecmpsitin will be a character f A 1. This gives a map Â2! Â1, bviusly a grup hmmrphism, which is dented by ˆ, andbycmmnusageitiscalledthedual map. The fllwing lemma is usual expressed by saying that the hat cnstructin is functrial 1 : Lemma. One has id c A =idâ. Assumethat and are cmpsable grup hmmrphisms. Then ˆ ˆ = [. Prf: Obvius, but here are the details: ˆ( ˆ( )) = ˆ( )=( ) = ( ) = [ ( ). Example.. If B A and i dentes the inclusin map, then ˆ( )= i is nthing but the restrictin B f t B. e 1 Fr the cgnscenti: The map A 7!  is a cntravariant functr frm the categry f finite abelian grup t itself. It is a very special case f a general duality functr called Matlis duality. It can als be generalized in anther directin t what is callet Pntrjagin duality, 2

3 Prblem.. a) Let a 2 Z and let : Z/nZ! Z/nZ be the multiplicatin-by-a map. Shw that the dual map ˆ: µ n! µ n is the a-th-pwer map 7! a. b) Let a 2 Z and let : µ n! µ n be the map! a.shwthatˆ is multiplicatin by a. Prblem.. a) Shw that if : A 1! A 2 is a surjective map, then ˆ is injective. b) Let 0 be a primitive n-t rt f unity. Assume that d n. Shwthat d 0 is a primitive n/d-th rt f unity. Use this t shw that if i: µ d! µ n is the inclusin, then î: ˆµ n! ˆµ d is surjective. Prblem.. Let B A be tw finite abelian grups. Shw that any character n B extends t a character n A. Hint: Use inductin n the index [A : B], andthat C is divisible. Character grups and direct prducts Anaturalquestintaskishwthecharactergrupbehavesinrelatintdirect prducts. And, as we sn shall see, the behavir is immaculate: The character grup f a direct prduct is the direct prduct f the character grups f the factrs t be precise ne shuld say that it is cannically ismrphic t the direct prduct f the character grups. Cmbining this with the fundamental therem fr finite abelian grups that says a finite abelian grup is ismrphic t a direct prduct f cyclic grups and the examples. and. abve, we can cnclude that A and  are ismrphic. The ismrphism is nt cannical s care must be taken, but it tells us what the grup structure f  is. Fr example, the rder is the same as the rder f A. Let A be an abelian grup and let A i A be tw subgrups such that A is the direct prduct f A 1 and A 2. This amunts t the intersectin A 1 \ A 2 being trivial and A = A 1 A 2. Every element a 2 A may be written as a prduct a = a 1 a 2 with a i 2 A i in a unique way. The inclusins A i A induce maps Â! Âi sending t the restrictin Ai (which is just the same as ˆ i ( )= i if i : A i! A dente the inclusin maps). These are grup hmmrphisms, and tgether they define a map Â! Â1 Â2. Prpsitin. Given tw finite abelian grups A 1 and A 2.ThemapÂ! Â1 Â2 given by 7! ( A1, A2 ) is an ismrphism f grups. 3

4 Characters MAT4250 Høst 2014 Prf: First we check that it is injective. Assume that bth maps Ai are trivial. As any a 2 A may be written as a prduct a = a 1 a 2 with a i 2 A i,wefind (a) = (a 1 ) (a 2 )=1 1=1. We prceed by checking that the map is surjective. S let i 2 Âi fr i =1, 2 be tw given characters. Any a 2 A can be expressed as a prduct a = a 1 a 2 with unique elements a i 2 A i.hencewemayfrm (a) = (a 1 ) (a 2 ). One checks withut truble that in this way is well defined (because the factrizatin a = a 1 a 2 is unique), and that it is a character f A. Crllary. If {A i } i2i is a finite cllectin f finite abelian grups, then \Q i2i A i = Âi. Q i2i Prf: Inductin n the number f elements in I. Crllary. Let A be a finite abelian grup, then A and the character grup  have the same number f elements; that is,  = A. Prf: If A ' Q i2i Z/n iz, thenâ ' Q i2i µ n i by prpsitin. n page 3 and example. n 2; andfcurse, µ n = Z/nZ = n. The characters f the characters As already hinted at, the frmatin f characters is a kind f duality. In this it lies, amng ther things, that perfrming the hat peratin twice brings us back t the grup we started with. Fr every grup element a 2 A there is the character n the character grup  best described as the evaluatin at a : It is the map Â! C sending the character t the value (a) at a. InthiswaywearriveatamapA!  sending a 2 A t the evaluatin at a ; and f curse this is a grup hmmrphism (check it!). The cnstructin is natural, r functrial as ne says, in the sense that every grup hmmrphism : A 1! A 2 between finite abelian grups fits in the fllwing cmmutative diagram: A 1 /  1 A 2 /  2 ˆ Prblem.. Check that the diagram is cmmutative. Prpsitin. The map A!  is an ismrhism. 4

5 Prf: Since the grups n bth sides have the same number f elements, it suffices t shw that the map is injective. If a 2 A is nt the neutral element, we have t prvide acharacter nt vanishing 2 at a. We first examine the case A = Z/nZ. Leta 2 Z/nZ be a nn-zer element. If is aprimitiven-the rt, ne has a 6=1,andbyexample., there is a character with (1) =. Hence (a) = a 6=1. In the case f a general A, thereisfrsmen asurjectin : A! Z/nZ mapping a t a nn zer element, say a 0. This fllws frm the fundamental therem fr finite abelian grups. By what we just did, there is a character n Z/nZ nt vanishing n a 0,andhence is a character n A nt vanishing at a. The rthgnality relatins There are sme imprtant relatins between the different characters f an abelian grup A called the rthgnality relatins. These relatins cme in pairs and the frmulatins are dual t each ther; that is t say, the ne interpreted fr the dual grup  gives the ther fr the grup A. The rthgnality relatins are nt very mysterius, and finally they bil dwn t the equatin n 1 =0, satisfied by any nn-trivial n-th-rt f unity; ne sees this by factring the plynmial x n 1.Asanillustratin,assumethatA is cyclic f rder n with a generatr g. A character n A is given by the value (g) =. Nw (g i )= i,sif 6= 1the abve relatin becmes (g i )= (a) =0, 0applei<n a2a and f curse, if =1,weget a2a (a) = A. These tw relatins crrespnd t the nes belw with respectively 1 = and 2 =1. Recall that if 2 C and =1ne has 1 =. Prpsitin. (The first rthgnality relatin) Let A be a finite abelian grup. Then fr any pair f characters 1 and 2 n A the fllwing relatin hlds a2a 1(a) 2 (a) = ( 0 if 1 6= 2, A if 1 = 2. 2 In this setting vanishing means that (a) =1. That abelian grups can bth be additive and multiplicative smetimes creates inextricable linguistical knts. 5

6 Prf: If 1 = 2 then 2 (a) = 1 (a) 1 and the sum is bviusly equal t A. Assume that 1 6= 2. Then there is at least ne element b 2 A such that 1(b) 6= 2(b). LetF = P a2a 1(a) 2 (a). Nw, the prduct ab runs thrugh A when a des, and therefre ne has F = a2a 1(ab) 2 (ab) = 1 (b) 2 (b) a2a 1(a) 2 (a) = 1 (b) 2 (b)f, frm which ne infers that F =0,since 1(b) 2 (b) = 1 (b) 2 (b) 1 6=1. In a dual versin, that is refrmulated fr the character grup, these relatins becme: Prpsitin. (The secnd rthgnality relatins) ( 0 if a 1 6= a 2 (a 1 ) (a 2 )= A if a 1 = a 2 2Â Prf: By. A is the character grup f Â, anelementa 2 A crrespnding t the character 7! (a). The relatin in the prpsitin is then just the rthgnality relatin in prpsitin. translated t the dual setting. Dirichlet characters Jhann Peter Gustav Lejeune Dirichlet was a german mathematician living frm t. He was brn in the small twn Düren which t day has abut inhabitants. Düren lies in the western part f Germany nt far frm Aachen. At the time f Dirichlet s birth Düren was part f France, but after the Naplen wars it was ceded t Prussia. Dirichlet studied in Paris, held psitins in Breslau, Berlin and finally he became Gauss successr in Göttingen. When in Dirichlet prved his celebrated therem f primes in arithmetic prgressins, he intrduced what is nw called Dirichlet characters. They still play an irreplaceable rle in the prf, and in general they are priceless tls in number thery. Let m 2 N be a natural number greater than ne. The ring Z/mZ f residue classes mdul m has a unit grup Lejeune Dirichlet Z/mZ whse elements are the residue classes f integers relatively prime t m. By definitin f Euler s -functin the rder f Z/mZ is (m). Fr m = p aprimez/pz is the field with p elements which usually is dented by F p. The unit grup F p f nn-zer elements is a cyclic grup f rder p 1. Inthecase 6

7 m is a cmpsite number, say m = ab with a and b relatively prime, the ring Z/mZ decmpses as the direct prduct Z/mZ = Z/aZ Z/bZ (this is the Chinese residue therem r the Sun-Tze therem that sme like t call it), and cnsequently the unit grup decmpses as well: Z/mZ = Z/aZ Z/bZ. The Dirichlet characters are intimately related t the characters f the unit grups Z/mZ, but there are subtle differences. We prefer t define a Dirichlet character mdul m as a functin : Z! C satisfying the fllwing three prperties: Peridicity: (n + m) = (n) fr all n 2 Z, Multiplicativity: (nn 0 )= (n) (n 0 ) fr all n, n 0 2 Z, Vanishing: (n) =0if and nly if (m, n) 6= 1. The last prperty specifies the value f n the integers having a cmmn factr with m, and this is cmpatible with the peridicity since (n + m, m) =(n, m). The last prperty implies that (1) 6= 0,andfrmthemultiplicativityweinferthat (1) = 1. There is a special character called the principal character mdul m. Itismstly dented by 0, but1 m wuld be a better ntatin since it depends n m. Ittakes the value 1 at n when (n, m) =1and, as impsed by the third cnditin, 1 m (n) =0 whenever (n, m) 6= 1; i.e., ne has ( 1 in case (n, m) =1, 0(n) =1 m (n) = 0 in case n and m have a cmmn factr. The first f the three requirements abve says that has m as a perid, hwever m is nt necessarily the smallest perid. The set f perids f is clsed under additin. It is thus a subgrup f the integers Z, and as such it has a unique psitive generatr. This is the smallest psitive perid f,anditiscalledthe cnductr 3 f. The Dirichlet character mdul m is said t be aprimitivecharactermdulm if the cnductr is equal t m; thatis,ifm is the smallest perid fr.ifthisisntcase,nesaysthat is imprimitive. Fraprimemdulusp, everycharacterisprimitive. One distinguishes between even and dd Dirichlet characters accrding t the value they take at 1. The character is even if ( 1) = 1 and dd if ( 1) = 1. By the first f the three prperties abve the value (n) depends nly n the residue class f n mdul m, hence induces a map 0 : Z/mZ! C. The map 0 is multiplicative since is, and hence it is a character f the multiplicative grup Z/mZ. Cnversely, given a character 0 n Z/mZ,nemaydefineaDirichletcharactermdul m by ( (n) = 0 if (n, m) 6= 1 ([n]) if (n, m) =1 3 Befre Wrld War II the german terminlgy fr the cnductr was der Führer. Frplitical reasns this was changed in

8 where, cnfrm t the cnventins, [n] stands fr the residue class f n mdul m. The nly thing t check is that behaves in multiplicative way als fr thse n having a cmmn factr with m. Butinthatcasenn 0 and m have a cmmn factr as well whatever the integer n 0 is, and hence bth (nn 0 ) and (n 0 ) (n) vanish. In this way ne establishes a ne-t-ne crrespndence between the characters f the unit grup Z/mZ and the Dirichlet characters mdul m. It als fllws that nn-zer values f a Dirichlet character are rts f unity whse rders divide (m). The Dirichlet characters mdul m frm a grup under multiplicatin; the prduct f tw is bviusly a Dirichlet character mdul m, andtheprincipalcharacteractsas a unit element. They are all invertible; the inverse f being the cmplex cnjugate.indeed,if(n, m) 6= 1,allcharactervanishatn, theprincipalneincluded,andif (n, m) =1,thevalue (n) is a rt f unity and therefre ne has (n) 1 = (n). T sum up what we have said s far, we have: Prpsitin. Let m>1 be a natural number. The Dirichlet characters mdul m frm a grup under multiplicatin f rder (m) with the principal character as the neutral element and the cmplex cnjugate as inversin. If : Z! Z/mZ is the natural 0 map, the assignment! 0 sets up a grup ismrphism between the character grup f Z/mZ and the grup f Dirichlet characters mdul m. Example.. It is high time t lk at a few examples with small m, andwestart with the simplest case m =2. The field F 2 has tw elements and the unit grup F 2 is reduced the trivial grup. The nly Dirichlet character is the principal ne. It vanishes n all even numbers and takes the value 1 at the dd nes. e Example.. Assume that m =3. The field F 3 has the tw units ±1, hencef 3 = µ 2 = {±1}, andtherearetwdirichletcharactersmdul3, theprincipalne 0 and anther ne given as (3n ± 1) = ±1. e Example.. Assume that m =4. The unit grup Z/4Z has tw elements, the residue classes f ±1. There are tw Dirichlet characters, the ne that is nt principal satisfies (4n +1) = 1 and (4n 1) = 1, andfcurseitvanishesntheeven numbers. Be aware f the subtle pint that the grups f units Z/3Z and Z/4Z are ismrphic and have the same characters, but the Dirichlet characters are certainly different. e Example.. Let us take a lk at the case m =5. The unit grup F 5 is f rder 4 cnsisting f ±1 and ±2; itiscyclicgeneratedbyeithertheresidueclassf2 r f 2. Therefre if 0 dentes the character n F 4 crrespnding t the Dirichlet character, ne sees that 0 (2) can be ne f ±i r ±1. Using this it is easy t fill in the fllwing table f the values the different characters take: 8

9 Class md n +2 i -i n n 2 -i i n n The character grup is cyclic generated by either f the tw dd characters 1 r 2. The even nn-principal character 3 generates a subgrup f rder 2. e Prblem.. This exercise is abut the Dirichlet characters mdul 8. Shwthat Z/8Z is ismrphic t µ 2 µ 2,andcnsistsftheresidueclassesf±1 and ±3. One has (±3) 2 =1and 3 and 3 generate Z/8Z.Andhencethatanycharacterisgiven as (8n +3)= 1 and (8n 3) = 2 where 1 and 2 are elements frm {±1}, andall cmbinatins the tw signs can ccure. Shw that (8n 1) = 1 2,andthat (2n) =0. Verify the table Class md n n n n Shw that is an imprimitive nn-trivial character which in fact cincides with the nn-trivial character mdul 4. Shwthatthetwthernn-trivialcharactersare primitive. Prblem.. Cnstuct the table like in example. fr the case m =7. Prblem.. Shw that if p is an dd prime, then there is nly ne real nn-trivial character mdul p. Can yu indentify it? Example.. The cncept f a primitive character is slightly subtle, s hpefully this example, treating the case m =15, will be clarifying. One has Z/15Z = Z/3Z Z/5Z, and the unit grup Z/15Z is the grup Z/3Z Z/5Z.Itisfrder8 and is ismrphic t µ 2 µ 4. There are thus altgether fur nn-trivial imprimitive Dirichlet characters mdul 15. Three induced frm the three nn-trivial characters 1, 2 and 3 f Z/5Z and ne frm the nly nn-trivial character n Z/3Z. Additinally, there are three nn-trivial primitive characters. They are the three prducts i with i =1, 2 and 3, andfcurse,thereistheprincipalne. 9

10 We take a clser lk at the three characters i. These all have perid 5, andthey vanish n the set 5Z [ 3Z f integers having 3 r 5 as a factr. Nw, the same three characters n Z/5Z induce Dirichlet characters mdul 5 as well, ur friends frm example.. These all have perid 5, butcntrarytthepreviuscase,theydnt vanish n multiples f 3 unless they als are divisible by 5. Sthepintwewantt illustrate, is that the same characters n the same unit grup Z/5Z induce different Dirichlet characters mdul 5 and mdul 15! e Orthgnality relatins In view f the crrespndence between Dirichlet characters mdul m and the characters n grup Z/mZ,therthgnalityrelatinsweprvedfrthecharacters f an abelian grup in prpsitins. and. n page 5 migrate immediately t crrespnding rthgnality relatins fr the Dirichlet characters mdul m: Prpsitin. Let m>1 be a natural number and let a and b be tw integers. Then (a) (b) = ( (m) if a b md m and (a, m) =(b, m) =1 0 therwise, where the sum is taken ver all Dirichlet characters mdul m. Prf: If either a r b has a cmmn factr with m, alltermsfthesumntheleft vanish. If (a, m) =(b, m) =1,thefrmulasarejusttherthgnalityrelatinsfrthe characters n the unit grup Z/mZ, i.e., prpsitin. n page 5. When prving Dirichlet therem abut primes in arithmetic prgressins, we shall apply the previus prpsitin in the fllwing frm Prpsitin. Let m>1 be an natural number and let a 2 Z. Then (a) = ( (m) if a 1 md m 0 therwise, where the sum is taken ver all Dirichlet characters f mdulus m. Prf: Just take b =1in the previus prpsitin. Prblem.. Translate the secnd rthgnality relatin in prpsitin. n page 6 t a statement abut Dirichlet characters. Prblem.. Describe all Dirichlet characters mdul 12 and mdul

11 The quadratic character One f the famus functins in number thery is the Legendre symbl appearing in the frmulatin f quadratic reciprcity. Fr an dd prime p and n 2 Z the Legendre symbl is defined as 8 n p = >< 1 if n is a square mdul p 1 if n is nt a square mdul p >: 0 if (n, p) 6= 1 The Legendre symbl is as we shall see, a primitive Dirichlet character. It is called the quadratic character mdul p. We state withut prf, the famus law f quadratic reciprcity discvered by Euler and prven by Gauss: Therem. Let p and q be tw dd primes. Then p q q p =( 1)(p 1)/2 (q 1)/2. The therem says that if either p r q is f the frm 4k +1then p asquaremdq if and nly if q is a square md p, andifbtharefthefrm4k +3,thennefthem is a square mdul the ther, while the ther is nt a square mdul the first. Prblem.. Shw that 17 is nt a square mdul 107. Prblem.. Let p be an dd prime. Recall that the set A = {±k 0 apple k apple (p 1)/2 } is called the set f least representatives fr the residues classes mdul p. In this way the least representatives are divided int a negative part and a psitive part. Amng the p 1 numbers n, 2n,...,(p 1)n acertainnumber,sayµ has n aleastrepresentativeinthenegativepartfa. ShwGauss lemma: =( 1) µ. p Hint: Shw that the residue classes f n, 2n,...,(p 1)n is a full set f representatives fr the nn-zer residue classes md p. UsesWilsn stherem:(p 1)! 1 md p. Prblem.. Shw that 2 p =( 1) (p2 1)/8. The grup f units F p in the field F p with p elements is a cyclic grup f rder p 1. Inside this grup is sitting a cpy f µ 2 ;indeednehas{±1} F p since p is dd. Frm Fermats little therem saying that a p 1 =1when a 2 F p we infer that a (p 1)/2 2 µ 2. Hence there is the character : F p! µ 2 given by a 7! a (p 1)/2. which we dente by fr shrt. Interpreted as a functin n the integers it is described as ( 0 if (n, p) 6= 1 (n) n (p 1)/2 if (n, p) =1 11,

12 Appealing t the little Fermat therem nce mre, ne has (a 2 ) (p 1)/2 =1,and whith this, it is easily verified that wed have the fllwing exact sequence: 1 / µ 2 / F p s / F p / µ 2 / 1, where s dentes the squaring-map s(a) =a 2.Anelementa 2 F p is therefre a square if and nly if a (p 1)/2 =1. We have shwn that the Dirichlet character assciated t cincides with the Legendre symbl. The definitin f the Legendre symbl is restricted t p being a prime, but it can be gerealised t any dd dd number, and it is then called the Jacbi symbl. Assume m = p e 1 Q 1 p er r is an dd cmpsite number. There is a surjectin Z/mZ ' i Z/pe i i Z! Q i Z/p iz,andfreachfthefactrsz/p i Z we have the quadratic character i : Z/p i Z! µ 2.Hencetheirprduct 1 r is a character n Z/mZ with values in µ 2,aswellasisthecmbinatin e r er. This is called the Jacbi symbl, andisthequadraticcharactermdulm. Itisdentedby n,andnehas m n m = n p 1 e1 n p r e r Prblem.. Shw that n m is multiplicative in bth n and m. n Prblem.. It is n lnger true that =1implies that n is a sqaure mdul m n m. Giveanexamplethatmanifeststhis.Hwever,if = 1, thenn cannt be a m square mdul m. Shwthis. Addendum Fr the benefit f thse in the audience wh are nt cmpletely cmfrtable with Euler s -functin, we give a quick an dirty expsitin f its main prperties. And in view f their prminent rle in the thery f Dirichlet characters we use the pprtunity t describe the grups f units in the finite rings Z/nZ. The Euler -functin There are tw definitins f easily seen t be equivalent. On the ne hand (m) is the number residue classes n such that n and m are relatively prime. This is equivalent t m being invertible in the ring Z/mZ; indeed(n, m) =1is equivalent t there being arelatin1=an + bm with a, b 2 Z, andthisinitsturn,isequivalenttn being invertible md m (then inverse is the class f a). Hence (m) is the rder f the unit grup Z/mZ, i.e., (m) = Z/mZ. The cmputatin f (m) hinges n the tw fllwing prpsitins. 12

13 Prpsitin. The Euler -functin is multiplicative, i.e., if n and n 0 are tw relatively prime natural numbers, ne has: (nn 0 )= (n) (n 0 ) Prf: Since n and n 0 are relatively prime, the Chinese remainder therem gives a ring ismrphism Z/nn 0 Z ' Z/nZ Z/n 0 Z which induces an ismrphism between the unit grups: Z/nn 0 Z ' Z/nZ Z/n 0 Z. The prpsitin fllws. p In case m = p is a prime, the ring Z/pZ is a field, and f curse Z/pZ is f rder 1. This prves the first case =1f the fllwing prpsitin. Prpsitin. Assume that p is a prime and anaturalnumber.then (p )= p 1 (p 1). We wuld like t use the fllwing very general lemma: Lemma. Let A and B be cmmutative rings with 1 an let : A! B be a surjective ring hmmrphism with kernel I. AssumethatI 2 =0.Thenthereisanexactsequence f unit grups: 0 / 1+I / A / B / 1 where dentes the restrictin f t the units. Prf: Frm I 2 =0it fllws that 1+a is a unit fr all a 2 I; indeed,1 a is an inverse t 1+a: (1 + a)(1 a) =1 a 2 =1.Inthesamevain1+I is clsed under multiplicatin since (1+a)(1+b) =1+a+b+ab =1+a+b, andclearly1+i =Ker. Observe that the multiplicative grup 1+I is ismrphic t the additive grup I. What is left is t see that is surjective. S take any unit b 2 B.Liftb t sme a in A and b 1 t sme a 0.Onehasaa 0 =1+ with 2 I, butthisgivesaa 0 (1 ) =1 and a is invertible. Prf f Prpsitin.: The prf ges by inductin n the expnent. If =1, we are thrugh as already remarked just befre the prpsitin. If >1, there is the exact sequence 0 / p Z/p +1 Z / Z/p +1 Z / Z/p Z / 0, where is the natural reductin md p hmmrphism. The kernel p Z/p +1 Z is ismrphic t Z/pZ (an ismrphism sends the class f a md p t the class f p a md p +1 ), and by the general lemma abve, there is an exact sequence Cunting rders, we get 1 / 1+p Z/p +1 Z / Z/p +1 Z / Z/p Z / 1. Z/p +1 Z = Z/p Z p Z/pZ = p 1 (p 1)p = p (p 1), and the prpsitin fllws. 13

14 We sum up the prperties f the -functin in the fllwing prpsitin: Prpsitin. The fllwing hld fr the Euler -functin: (nn 0 )= (n) (n 0 ) when (n, n 0 )=1, (p )=p 1 (p 1) when p is a prime, (n)/n = Q p n (1 p 1 ). Prf: The tw first frmulas are already shwn. Fr the last, write n = p 1 1 p r with the p i s different primes. Using the tw first prperties we find the expressin (n) = Y i p i 1 i (p i 1). Divide thrughut by n t arrive at the third frmula in the prpsitin. The behavir f the Euler -functin is rather erratic. Fr example, A. Schinzel has shwn that the fractins f tw cnsecutive values f (n) frm a dense subset f the set f all psitive real numbers; that is, the fractin (n +1)/ (n) can be as clse as yu want t any number in R +.Hweverthequtient (n)/n behaves smehw mre regularly. Obviusly (n)/n < 1, andifp is a prime ne has (p)/p =1 1/p, which is clse t ne. There being infinity many prime we infer that lim sup n!1 (n)/n =1. This is illustrated in the figure belw, where the values (n) is pltted against n fr n up t 800. The clud f pltted pints is clearly bunded abve by the line y = x. On the ther hand, if n has a lt f prime factrs the qutient (n)/n tends t be small. In lemma. n page 5 in chapter 6 we established the inequality Y (1 p 1 ) < 1/ lg x, papplex and taking n = Q papplen p, a number with an awful lt f prime factrs, we btain a number with (n)/n < 1/ lg x. This shws that lim inf n!1 (n)/n =0. This behavir is nt apparent in the figure, the reasn being that n has t be a very large number fr (n)/n t be very small, certainly much larger than 800. Ofthe numbers f the frm Q papplen p cnsidered abve, nly is less than 800! 14

15 (n) against n fr n apple 800 Prblem.. On the figure the line y =8/35 x is printed in red. Can yu explain why (the line is printed, nt why it is red)? There are three blue dts n it; explain that. Can yu predict the crdinates f thse pints? And what is the next blue dt n the line? Prblem.. There seems t be a lt f blue dts n the line y =1/2 x. Why? The grups f units Z/nZ Let n = Q p n p p be the prime factrizatin f the natural number n. Byurfriendsthe chineses the ring Z/nZ decmpses as the direct prduct f rings Z/nZ ' Q p n Z/p p Z, and hence there is a crrespnding decmpsitin f the grup f units Z/nZ = Q p n Z/p p Z. Thus, knwing the grup structure f each f the factr grups we knw the structure f the grup Z/nZ. In what fllws we determine the grup structure f Z/p Z fr p aprime. The cases p dd and p =2are slightly different, althugh the underlying structure is the same. In bth cases there is an bvius reductin map Z/p Z t a cyclic grup, respectively Z/pZ and Z/4Z.Incasep =2the grup Z/4Z is the simplest factr grup that gives smething, Z/2Z being trivial.we shall see that these sequences are split, and in bth cases the kernel will be cyclic. In the dd case the result is simply that Z/p Z is cyclic, while if p =2,theunit grups are ismrphic t direct prducts Z/2 2 Z Z/2Z if 3, andz/4z = µ 2 and Z/2Z is trivial. That is, ne has: 15

16 Prpsitin. Let p be a prime number. If p 6= 2,thenthegrupfunitsZ/p Z is cyclic f rder p 1 (p 1). If p =2and 3, thegrupfunitsz/2 Z is ismrphic t the direct prduct Z/2 2 Z Z/2Z. Z/4Z is cyclic f rder 2, andz/2z is trivial. The prf will be a series f lemmas, the first being: Lemma. Fr any natural number a ne has (1 + p) pa 1+p a+1 md p a+2. Prf: Inductin n a, thestarta =1is clear. S assume that (1 + p) pa 1+p a+1 md p a+2. p Nw, since the binmial cefficients fr k 6= 1,p all are divisible by p, ithldsthat k if x y md p s,thenx p y p md p s+1. This gives (1 + p) pa+1 (1 + p a+1 ) p =1+p p a+1 + p p k(a+1) 1+p a+2 md p a+3 k k 2 since k(a +1) a +3as k 2. We start with treating the case f dd p, and start with the simplest case Z/pZ. This is a field, and there is the general esult Prpsitin. If k is a field and G k afinitesubgrup,theng is cyclic. Prf: Let d be the expnent f G; i.e., the least cmmn multiple f the rders f all the elements in G. AfinitegrupG is cyclic if and nly if d = G ; indeed,ifg is cyclic this is clear, and fr the implicatin the ther way use that there is always an element f rder d in G. The elements in G satisfy the equatin x d =1, which has at mst d slutins. It fllws that d = G, andg is cyclic. Lemma. Assume that p is an dd prime. The the grup f units Z/p Z is cyclic f rder p 1 (p 1). Prf: We determined the rder in previus paragraph. The reductin mdul p map induces an exact sequence 1 / K / Z/p Z / Z/pZ / 1 16

17 where the kernel K is f rder p 1.LiftageneratrfZ/pZ t an element g 2 Z/p Z, then g p 1 = a 2 K, andsincep 1 is relatively prime t the rder p 1 f K, thereis an element b 2 K with b p 1 = a. Therefre (ga) p 1 =1,andthesequencesplits 4 ;that is Z/p Z ' K Z/pZ.SinceZ/pZ is cyclic f rder p 1, itsufficestshwthat K icyclic(frderp 1 ). This we shall d by shwing that 1+p is a generatr. By lemma. abve ne has in Z/p Z (that is mdul p ) (1 + p) p 2 =1+p 1 6=1 since p 1 is nn-zer in Z/p Z,andwecncludethattherderf1+p is equal t p 1. Lemma. The grup f units (Z/4Z) is a cyclic grup f rder 2, andz/2z is trivial. If v>2, ithldsthatz/2 v Z is ismrphic t a prduct f tw cyclic grups respectively f rders 2 and 2 2. Prf: The residue classes in Z/4Z are 0, ±1 and 2 and ±1 are the nly units, hence Z/4Z is ismrphic with µ 2.Inthegeneral,casetheexactsequence 1 / K / Z/2 Z / Z/4Z / 1 crrespnding t the reductin mdul 4 map splits, indeed {±1} Z/2 Z gives a splitting. Just as in the case with p>2, weshwthatthekernelk is cyclic. The first element yu think f in kernel, is 5, anditturnsuttbeageneratr: 5 2v 2 =(1+4) 2v 2 = =1, and 5 is f rder a pwer f 2. Furthermrenhas =(1+4) 2v 3 = = =1. This shws that the rder f 5 is 2 2, which is the same as the rder f the kernel K. Hence the kernel is cyclic. 4 This is a general fact. If in an exact sequence f abelian grups, the tw extreme grups are f relatively prime rder, the sequence splits. (It is even true, but much deeper, fr nn abelian finite grups.) 17