A Senior Seminar On Quadratic Reciprocity

78 Pages·2010·0.537 MB·English

by  Angel V. Kumchev

A Senior Seminar On Quadratic Reciprocity
AngelV.Kumchev

Contents
Chapter1. ElementaryNumberTheory: AReview 1
1.1. Divisibility 1
1.2. Thegreatestcommondivisoroftwointegers 2
1.3. Primenumbersandthefundamentaltheoremofarithmetic 5
1.4. Congruencemodulom 5
1.5. Completeandreducedsystemsmodulom 8
1.6. Exercises 9
Chapter2. TheLawOfQuadraticReciprocity 11
2.1. Polynomialcongruences 11
2.2. Linearcongruences 12
2.3. Quadraticcongruences. Quadraticresiduesandnonresidues 13
2.4. TheLegendresymbol 14
2.5. Quadraticreciprocity 16
2.6. Exercises 17
Chapter3. ComplexNumbers 19
3.1. Definitionandalgebraicproperties 19
3.2. Geometricinterpretation,moduliandconjugates 20
3.3. Exponentialform 22
3.4. Rootsofcomplexnumbers 23
3.5. Thecomplexexponential 23
3.6. Trigonometricandhyperbolicfunctionsofacomplexargument 24
3.7. Thecomplexlogarithmandpowerfunctions 26
3.8. Gausssums 28
3.9. Exercises 30
Chapter4. AlgebraOverTheComplexNumbers 33
4.1. Rootsofpolynomialswithcomplexcoefficients 33
4.2. Linearalgebraoverthecomplexnumbers 35
4.3. Reviewofdeterminants 40
4.4. Theresultantoftwopolynomials 44
4.5. Symmetricpolynomialsinseveralvariables 45
4.6. Exercises 46
Chapter5. FourierSeries 51
5.1. Definition 51
5.2. Calculusofcomplex-valuedfunctions 51
5.3. Anexample 53
5.4. RepresentingfunctionsbyFourierseries 54
i
ii CONTENTS
5.5. Poisson’ssummationformula 56
5.6. Exercises 57
Chapter6. AlgebraicNumbersAndAlgebraicIntegers 61
6.1. Definition 61
6.2. Theringofalgebraicintegers 61
6.3. Congruencesbetweenalgebraicintegers 63
6.4. AproofofTheorem2.17 64
6.5. Exercises 65
Chapter7. ProofsOfTheLawOfQuadraticReciprocity 67
7.1. S.Y.Kim’selementaryproof 67
7.2. Eisenstein’slemmaandcountinglatticepoints 68
7.3. AnotherproofofEisenstein’s 69
7.4. Analgebraicproof: Gausssumsarealgebraicintegers 69
7.5. AproofusingtheformulafortheGausssum 70
7.6. AnevaluationoftheGausssumusingFourierseries 71
7.7. AnevaluationoftheGausssumusingmatrices 72
7.8. Aproofusingpolynomialsandresultants 73
CHAPTER 1
Elementary Number Theory: A Review
As we said already in the Introduction, the focus of this course is a theorem from
number theory known as the law of quadratic reciprocity. Thus, it should not come as a
surprisethatweshallneedcertainfactsfromelementarynumbertheory.Inthischapter,we
reviewsomeofthebasicsofarithmetic: thedefinitionofdivisibility,thegreatestcommon
divisoroftwointegers,thedefinitionandbasicpropertiesofprimenumbers,andproperties
of congruences. It is very likely that you are familiar with much of the material in the
chapterfromearliercourses,soweaimmainlytorefreshyourknowledge,fillanypotential
gaps in it, and set up the notation and terminology for future reference. The pace of the
expositionisthusratherbrisk,andseveralproofsareleftasexercises.
1.1. Divisibility
Numbertheorystudiesthepropertiesoftheintegers,andthatstudyusuallystartswith
thenotionofdivisibilityofoneintegerbyanother.
Definition1.1. Ifa,b∈Zandb(cid:44)0,wesaythatbdividesaandwriteb|a,ifa=bqfor
someq∈Z. Ifbdoesnotdividea,wewriteb(cid:45)a. Whenb|a,wemayalsosaythatbisa
divisorofa,bisafactorofa,oraisamultipleofb.
Example1.2. Wehave3|15and6|324,but100(cid:45)2010.
Giventwointegersaandb (cid:44) 0, wesometimeswanttodivide abybeventhoughb
may not be a divisor of a: say, we may want to divide 37 candy bars among 4 children.
This problem leads to the concept of “division with a quotient and a remainder.” It is
summarized by the following theorem, known as the quotient-remainder theorem or the
divisionalgorithm.
Theorem1.3(Quotient-remaindertheorem). Ifa,b∈Zandb>0,thenthereexistunique
integersqandrsuchthat
a=bq+r, 0≤r<b.
Proof. “Existence.” Letqbethelargestintegermwithm ≤ a/bandletr = a−bq.
Then
q≤a/b =⇒ 0≤a−bq=r.
Furthermore,sinceqisthelargestintegernotexceedinga/b,wehave
a/b<q+1 =⇒ a<bq+b =⇒ r=a−bq<b.
“Uniqueness.” Supposethatq ,q ,r andr areintegerssuchthat
1 2 1 2
a=bq +r =bq +r , 0≤r ,r <b.
1 1 2 2 1 2
Then
bq +r =bq +r =⇒ r −r =b(q −q ).
1 1 2 2 1 2 2 1
1
2 1.ELEMENTARYNUMBERTHEORY:AREVIEW
Ontheotherhand,
0≤r ,r <b =⇒ −b<r −r <b.
1 2 1 2
Hence,
−b<b(q −q )<b =⇒ −1<q −q <1.
2 1 2 1
Sinceq −q isaninteger,weconcludethatq −q =0. Thus,q =q and
2 1 2 1 1 2
bq +r =bq +r =⇒ r =r . (cid:3)
1 1 2 2 1 2
1.2. Thegreatestcommondivisoroftwointegers
Definition1.4. Letaandbbeintegers,notboth0. Thegreatestcommondivisorofaand
b,denotedgcd(a,b)or(a,b),isthelargestnaturalnumberdsuchthatd |aandd |b.
Example1.5. Thepositivedivisorsof2are1and2,andthepositivedivisorsof6are1,
2,3and6. Thus,(2,6)=2.
Thepositivedivisorsof4are1,2and4,andthepositivedivisorsof6are1,2,3and
6. Thus,(4,6)=2.
Thepositivedivisorsof−4are1,2and4,andthepositivedivisorsof0areallnatural
numbers. Thus,(−4,0)=4.
Ifa(cid:44)0,thenanypositivedivisorofais≤|a|. Moreover,|a|(whichiseitheraor−a)
isadivisorofa,so|a|isthelargestpositivedivisorofa. Thus,(a,0)=|a|. (cid:3)
Intheaboveexample,wecomputed(a,0)foralla(cid:44)0. Wewanttofindamethodfor
computingthegcdofanytwonumbers. Thenextlemmareducesthatproblemtothecase
whenaandbarenaturalnumbers.
Lemma1.6. Ifaandbareintegersandb(cid:44)0,then(a,b)=(a,|b|).
Lemma1.7. Letb,q,andrbeintegers,andsupposethatbandrarenotboth0. Then
(bq+r,b)=(r,b).
Proof. Leta = bq+r, d = (a,b), andd = (r,b). Bythedefinitionof(r,b), d | r
1 2 2
andd |b,sob=d mandr=d nforsomem,n∈Z. Hence,
2 2 2
a=bq+r=(d m)q+d n=d (mq+n).
2 2 2
Since mq+n is an integer, it follows that d | a. Since d is a positive divisor of b, we
2 2
conclude that d is a natural number that divides both a and b. Then d does not exceed
2 2
thelargestnaturalnumberthatdividesbothaandb: thatis,d ≤d .
2 1
Next, bythedefinitionof(a,b), d | aandd | b, soa = d nandb = d mforsome
1 1 1 1
n,m∈Z. Hence,
r=(bq+r)−bq=a−bq=d n−(d m)q=d (n−mq).
1 1 1
Since n−mq is an integer, it follows that d | r. Since d is a positive divisor of b, we
1 1
concludethatd isanaturalnumberthatdividesbothrandb. Thend doesnotexceedthe
1 1
largestnaturalnumberthatdividesbothrandb: thatis,d ≤d .
1 2
Weprovedthatd ≤d andd ≤d . Therefore,d =d . (cid:3)
1 2 2 1 1 2
We can use Lemma 1.7 to calculate the gcd of any two positive integers. We illus-
tratethemainideabyanexample,andthenproveatheoremthatdescribesthemethodin
general.
1.2.THEGREATESTCOMMONDIVISOROFTWOINTEGERS 3
Example1.8. LetususeLemma1.7tocalculate(216,51). WeapplyLemma1.7repeat-
edlyasfollows:
(216,51)=(4·51+12,51)=(12,51)=(12,4·12+3)=(12,3)=3.
ItseemsplausiblethatweshouldbeabletoproceedsimilarlytoExample1.8tocom-
pute(a,b)forgeneralaandb. Thefollowingtheorem,knownastheEuclideanalgorithm,
establishesthatthisisindeedthecase.
Theorem1.9(Euclideanalgorithm). Letaandbbeintegers,with0<b<a.
i) Ifb|a,then(a,b)=b.
ii) Ifb(cid:45)a,thenthereexistintegersq1,r1,q2,r2,...,rm,qm+1suchthat
a=bq +r , 0<r <b;
1 1 1
b=r q +r , 0≤r <r ;
1 2 2 2 1
r =r q +r , 0≤r <r ;
1 2 3 3 3 2
.
.
.
r =r q +r , 0≤r <r ;
m−2 m−1 m m m m−1
rm−1 =rmqm+1;
and(a,b)=r .
m
Proof. i)Thispartisaneasyconsequenceofthedefinitionofgreatestcommondivi-
sor.
ii)Weuseinductiononatoprovethestatement:
Ifbisanintegersuchthat0<b<aandb(cid:45)a,thenthereexistintegers
q1,r1,q2,r2,...,rm,qm+1suchthat
a=bq +r , 0<r <b;
1 1 1
b=r q +r , 0≤r <r ;
1 2 2 2 1
r =r q +r , 0≤r <r ;
1 2 3 3 3 2
.
.
.
r =r q +r , 0≤r <r ;
m−2 m−1 m m m m−1
rm−1 =rmqm+1; rm =(a,b).
The base case of the induction is a = 3. When a = 3, the only admissible value of b is
b=2,andtheabovestatementholdswithm=1,q =r =1,q =2:
1 1 2
3=2·1+1, 0<1<2; 2=1·2; 1=(3,2).
Now, suppose that the above statement holds for all integers a with 2 < a < n, and
considerthecasea = n. Letbbeanintegerwith0 < b < nandb (cid:45) n. ByTheorem1.3,
thereexistintegersq,rsuchthat
n=bq+r, 0≤r<b.
Furthermore, since b (cid:45) n, the remainder r cannot be 0, so we can strengthen the above
inequalitytoget
n=bq+r, 0<r<b. (1.1)
Ifr|b,thenb=rq(cid:48)forsomeintegerq(cid:48),andLemma1.7andparti)give
(n,b)=(bq+r,b)=(r,b)=r.
4 1.ELEMENTARYNUMBERTHEORY:AREVIEW
Hence,theabovestatementforthepairn,bholdswithm=1,q =q,r =r,andq =q(cid:48).
1 1 2
Next, suppose that r (cid:45) b. Since 0 < r < b < n and r (cid:45) b, we can apply the inductive
hypothesistothepairb,r: thereexistintegersq1,r1,q2,r2,...,rm,qm+1suchthat
b=rq +r , 0<r <r;
1 1 1
r=r q +r , 0≤r <r ;
1 2 2 2 1
r =r q +r , 0≤r <r ;
1 2 3 3 3 2
.
.
.
r =r q +r , 0≤r <r ;
m−2 m−1 m m m m−1
rm−1 =rmqm+1; rm =(b,r).
Combiningtheseconditionswith(1.1),wefindthattheintegersq,r,q1,r1,q2,...,rm,qm+1
haveallthedesiredpropertiesrelativetothepairn,b,withthepossibleexceptionof(n,b)=
r . Toverifythelatterproperty,weuseLemma1.7andtheidentity(b,r)=r above:
m m
(n,b)=(bq+r,b)=(r,b)=r . (cid:3)
m
The Euclidean algorithm is very effective for computational purposes. However, it
doesnotrelate(a,b)directlytoaandb.Thenexttheoremprovidesexactlysucharelation.
Theorem1.10. Letaandbbeintegers,notboth0,andletd = (a,b). Thendistheleast
positiveintegerintheset
S=(cid:8)ax+by| x,y∈Z(cid:9).
Inparticular,thereexistintegersx,ysuchthatd =ax+by.
TherearetwostandardwaystoproveTheorem1.10: onebasedonthewell-ordering
axiom of the natural numbers, the other based on the Euclidean algorithm. Both proofs
aresketchedintheexercises(cf. Exercises1.4and1.5). Wenowstateseveralimportant
consequencesofTheorem1.10,whichwewillusenumeroustimesthroughoutthecourse.
Theorem1.11. Ifa,b,careintegerssuchthat(a,b)=(a,c)=1,then(a,bc)=1.
Proof. Suppose that a,b and c are integers such that (a,b) = (a,c) = 1. Then by
Theorem1.10,thereexistintegersx,y,u,vsuchthat
ax+by=(a,b)=1, au+cv=(a,c)=1.
Hence,
1=ax+by=ax+by(1)=ax+by(au+cv)=a(x+byu)+(bc)(yv).
Sincex+byuandyvareintegers,thisshowsthat1isapositiveintegerfromtheset
S=(cid:8)am+(bc)n|m,n∈Z(cid:9).
Sincethereisnopositiveintegerlessthan1, itfollowsthat1istheleastpositiveinteger
in S. By Theorem 1.10, the least positive integer in S is (a,bc), so we conclude that
(a,bc)=1. (cid:3)
Corollary1.12. Ifk≥2anda,b ,...,b areintegerssuchthat
1 k
(a,b )=(a,b )=···=(a,b )=1,
1 2 k
then(a,b b ···b )=1.
1 2 k
Theorem1.13. Ifa,b,careintegerssuchthata|bcand(a,b)=1,thena|c.
Theorem1.14. Ifa,b,careintegerssuchthata|c,b|cand(a,b)=1,thenab|c.
1.4.CONGRUENCEMODULOm 5
1.3. Primenumbersandthefundamentaltheoremofarithmetic
Wenowrecallthedefinitionofprimeandcompositenumbers.
Definition1.15. Aninteger p>1iscalledaprimenumber,orsimplyaprime,ifitsonly
positive divisors are 1 and p. An integer n > 1 which is not prime is called a composite
number. Thenumbers0and1areneitherprime,norcomposite.
Example1.16. Theintegers2,3,13and89areprime;4,6,51and837arecomposite.
The next theorem provides a necessary and sufficient condition for primality that is
oftenmoreusefulinnumber-theoreticproofsthanthedefinitionofaprime.
Theorem1.17. Apositiveinteger p>1isprimeifandonlyif phastheproperty:
(∀a,b∈Z)(p|ab =⇒ p|a or p|b). (1.2)
Proof. “⇒”. Suppose that p is a prime number and a and b are integers such that
p | ab. If p | a, statement (1.2) is true. Now, suppose that p (cid:45) a. By the definition of
a prime, the only positive divisors of p are 1 and p. Hence, (p,a) is 1 or p. However,
(p,a) (cid:44) p,because p (cid:45) a. Hence,(p,a) = 1. Since p | aband(p,a) = 1,itfollowsfrom
Theorem1.13that p|b. Therefore,(1.2)istrueagain.
“⇐”. Supposethat p > 1hasproperty(1.2). Wemustshowthat pisprime. Suppose
that p is composite. Then p has a positive divisor other than 1 and p: p = ab, where
a,b∈Nand1<a< p. Inparticular,sincea< p,weconcludethat p(cid:45)a. Notealsothat
p=ab, a>1 =⇒ b< p =⇒ p(cid:45)b.
However,sincep| p=ab,itfollowsfromproperty(1.2)thatp|aorp|b;acontradiction.
Therefore, pisprime. (cid:3)
Theimportanceofprimenumberscomesfromthenexttheorem,whichsaysroughly
that one can use multiplication to build any integer n > 1 from primes, and that there is
essentially a unique way to do so. In other words, one can view the primes as the basic
buildingblocksoftheintegersundermultiplication.
Theorem1.18(Fundamentaltheoremofarithmetic). Letn > 1beaninteger. Thennhas
auniquefactorizationoftheform
n= p p ···p , p ≤ p ≤···≤ p ,
1 2 k 1 2 k
where p ,p ,...,p areprimenumbers.
1 2 k
1.4. Congruencemodulom
Definition1.19. Letm ∈ N,withm > 1,andleta,b ∈ Z. Wesaythataiscongruenttob
modulom,andwritea≡b (mod m),ifm|(a−b).
Theorem1.20. Leta,b,m∈Zandm>1. Thena≡b (mod m)ifandonlyifthereexists
anintegerksuchthata=b+mk.
Proof. Bythedefinitionsofcongruenceanddivisibility,
a≡b (mod m) ⇐⇒ m|(a−b)
⇐⇒ a−b=mk
⇐⇒ a=b+mk,
wherek∈Z. (cid:3)
6 1.ELEMENTARYNUMBERTHEORY:AREVIEW
Thenexttheoremsummarizessomebasicpropertiesofcongruences.Propertiesi)–iii)
demonstratethatcongruencemodulomisanequivalencerelationontheintegers. Proper-
tiesiv)–vi)introducethebasicarithmeticoperationswithcongruences.
Theorem1.21. Letm∈Nanda,b,c,d ∈Z. Then:
i) a≡a (mod m);
ii) ifa≡b (mod m),thenb≡a (mod m);
iii) ifa≡b (mod m)andb≡c (mod m),thena≡c (mod m);
iv) ifa≡b (mod m)andc≡d (mod m),thena+c≡b+d (mod m);
v) ifa≡b (mod m)andc≡d (mod m),thenac≡bd (mod m);
vi) ifac≡bc (mod m)and(c,m)=d,thena≡b (mod m/d).
Proof. i)Sincem|0=(a−a),wehavea≡a (mod m).
ii)ByTheorem1.20,a=b+kmforsomeintegerk,whence
b=a−km=a+(−k)m.
Since−kisalsoaninteger,itfollowsfromTheorem1.20thatb≡a (mod m).
iii)ByTheorem1.20,
a≡b (mod m) =⇒ a=b+km, b≡c (mod m) =⇒ b=c+lm,
wherek,l∈Z. Hence,
a=b+km=(c+lm)+km=c+(l+k)m=c+nm,
wheren=l+kisalsoaninteger. Thus,a≡c (mod m),byTheorem1.20.
iv)ByTheorem1.20,
a≡b (mod m) =⇒ a=b+km, c≡d (mod m) =⇒ c=d+lm,
wherek,l∈Z. Hence,
a+c=(b+km)+(d+lm)=(b+d)+(k+l)m=(b+d)+nm,
wheren=k+lisalsoaninteger. Thus,a+c≡b+d (mod m),againbyTheorem1.20.
v)ByTheorem1.20,
a≡b (mod m) =⇒ a=b+km, c≡d (mod m) =⇒ c=d+lm,
wherek,l∈Z. Hence,
ac=(b+km)(d+lm)=(bd)+(kd+lb+klm)m=(bd)+nm,
wheren=kd+lb+klmisalsoaninteger. Thus,byTheorem1.20,ac≡bd (mod m).
vi)ByTheorem1.20,
ac≡bc (mod m) =⇒ ac=bc+km,
where k ∈ Z. Moreover, because (c,m) = d, there exist u,v ∈ Z such that cu+mv = d.
Combiningthesetwoidentities,wefindthat
ac=bc+km =⇒ acu=bcu+kmu
=⇒ a(d−mv)=b(d−mv)+kmu
=⇒ ad =bd−bmv+kmu+amv=bd+nm
=⇒ a=b+n(m/d),
wheren=−bv+ku+avisalsoaninteger. Thus,yetanotherappealtoTheorem1.20gives
a≡b (mod m/d). (cid:3)