Chapter 1 The Fundamental Theorem of Arithmetic

101 Pages·2001·0.4 MB·English

Chapter 1
The Fundamental Theorem of
Arithmetic
1.1 Prime numbers
Ifa,b ∈ Zwesaythatadividesb(orisadivisorofb)andwewritea | b,if
b = ac
forsomec ∈ Z.
Thus−2 | 0but0 (cid:45) 2.
Definition1.1 Thenumberp ∈ Nissaidtobeprimeifphasjust2divisorsinN,
namely1anditself.
Notethatourdefinitionexcludes0(whichhasaninfinityofdivisorsinN)and
1(whichhasjustone).
Writing out the prime numbers in increasing order, we obtain the sequence of
primes
2,3,5,7,11,13,17,19,...
which has fascinated mathematicians since the ancient Greeks, and which is the
mainobjectofourstudy.
Definition1.2 Wedenotethenthprimebyp .
n
Thusp = 11, p = 541.
5 100
Itisconvenienttointroduceakindofinversefunctiontop .
n
Definition1.3 Ifx ∈ Rwedenotebyπ(x)thenumberofprimes≤ x:
π(x) = (cid:107){p ≤ x : pprime}(cid:107).
1–1
374 1–2
Thus
π(1.3) = 0, π(3.7) = 2.
Evidently π(x) is monotone increasing, but discontinuous with jumps at each
primex = p.
Theorem1.1 (Euclid’sFirstTheorem)Thenumberofprimesisinfinite.
Proof (cid:73)Supposetherewereonlyafinitenumberofprimes,say
p ,p ,...,p .
1 2 n
Let
N = p p ···p +1.
1 2 n
Evidentlynoneoftheprimesp ,...,p dividesN.
1 n
Lemma1.1 Everynaturalnumbern > 1hasatleastoneprimedivisor.
Proof of Lemma (cid:66) The smallest divisor d > 1 of n must be prime. For otherwise
d would have a divisor e with 1 < e < d; and e would be a divisor of n smaller
thand. (cid:67)
Bythelemma,N hasaprimefactorp,whichdiffersfromp ,...,p . (cid:74)
1 n
Ourargumentnotonlyshowsthatthereareaninfinityofprimes;itshowsthat
p < 22n;
n
a very feeble bound, but our own. To see this, we argue by induction. Our proof
showsthat
p ≤ p p ···p +1.
n+1 1 2 n
Butnow,byourinductivehypothesis,
p < 221, p < 222, ..., p < 22n.
1 2 n
Itfollowsthat
p ≤ 221+22+···+2n
n+1
But
21 +22 +···+2n = 2n+1 −1 < 2n+1.
Hence
p < 22n+1.
n+1
Itfollowsbyinductionthat
p < 22n,
n
foralln ≥ 1,theresultbeingtrivialforn = 1.
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This is not a very strong result, as we said. It shows, for example, that the 5th
prime,infact11,is
< 225 = 232 = 4294967296.
In general, any bound for p gives a bound for π(x) in the opposite direction,
n
andviceversa;for
p ≤ x ⇐⇒ π(x) ≥ n.
n
Inthepresentcase,forexample,wededucethat
π(22y) ≥ [y] > y −1
andso,settingx = 22y,
π(x) ≥ log log x−1 > loglogx−1.
2 2
for x > 1. (We follow the usual convention that if no base is given then logx
denotesthelogarithmofxtobasee.)
ThePrimeNumberTheorem(whichweshallmakenoattempttoprove)asserts
that
p ∼ nlogn,
n
or,equivalently,
x
π(x) ∼ .
logx
This states, roughly speaking, that the probability of n being prime is about
1/logn. Note that this includes even numbers; the probability of an odd number
n being prime is about 2/logn. Thus roughly 1 in 6 odd numbers around 106 are
prime;whileroughly1in12around1012 areprime.
(The Prime Number Theorem is the central result of analytic number theory
since its proof involves complex function theory. Our concerns, by contrast, lie
withinalgebraicnumbertheory.)
There are several alternative proofs of Euclid’s Theorem. We shall give one
below. But first we must establish the Fundamental Theorem of Arithmetic (the
Unique Factorisation Theorem) which gives prime numbers their central roˆle in
numbertheory;andforthatweneedEuclid’sAlgorithm.
1.2 Euclid’s Algorithm
Proposition1.1 Suppose m,n ∈ N, m (cid:54)= 0. Then there exist unique q.r ∈ N
suchthat
n = qm+r, 0 ≤ r < m.
Proof (cid:73)Foruniqueness,suppose
(cid:48) (cid:48)
n = qm+r = q m+r ,
374 1–4
wherer < r(cid:48),say. Then
(q(cid:48) −q)m = r(cid:48) −r.
The number of the right is < m, while the number on the left has absolute value
≥ m,unlessq(cid:48) = q,andsoalsor(cid:48) = r.
We prove existence by induction on n. The result is trivial if n < m, with
q = 0, r = n. Supposen ≥ m. Byourinductivehypothesis,sincen−m < n,
n−m = q(cid:48)m+r,
where0 ≤ r < m. Butthen
n = qm+r,
withq = q(cid:48) +1. (cid:74)
Remark: One might ask why we feel the need to justify division with remainder
(as above), while accepting, for example, proof by induction. This is not an easy
questiontoanswer.
Kronecker said, “God gave the integers. The rest is Man’s.” Virtually all
number theorists agree with Kronecker in practice, even if they do not accept his
theology. In other words, they believe that the integers exist, and have certain
obviousproperties.
Certainly,ifpressed,onemightgobacktoPeano’sAxioms,whichareastan-
dard formalisation of the natural numbers. (These axioms include, incidentally,
proofbyinduction.) Certainlyanypropertiesoftheintegersthatweassumecould
easilybederivedfromPeano’sAxioms.
However, as I heard an eminent mathematician (Louis Mordell) once say, “If
youdeducedfromPeano’sAxiomsthat1+1 = 3,whichwouldyouconsidermost
likely, that Peano’s Axioms were wrong, or that you were mistaken in believing
that1+1 = 2?”
Proposition1.2 Suppose m,n ∈ N. Then there exists a unique number d ∈ N
suchthat
d | m, d | n,
andfurthermore,ife ∈ Nthen
e | m, e | n =⇒ e | d.
Definition1.4 We call this number d the greatest common divisor of m and n,
andwewrite
d = gcd(m,n).
Proof (cid:73) Euclid’s Algorithm is a simple technique for determining the greatest
common divisor gcd(m,n) of two natural numbers m,n ∈ N. It proves inci-
dentally — as the Proposition asserts — that any two numbers do indeed have a
greatestcommondivisor(orhighestcommonfactor).
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First we divide the larger, say n, by the smaller. Let the quotient be q and let
1
theremainder(allwearereallyinterestedin)ber :
1
n = mq +r .
1 1
Nowdividembyr (whichmustbelessthanm):
1
m = r q +r .
1 2 2
Wecontinueinthiswayuntiltheremainderbecomes0:
n = mq +r ,
1 1
m = r q +r ,
1 2 2
r = r q +r ,
1 2 3 3
...
r = r q +r ,
t−1 t−2 t−1 t
r = r q .
t t−1 t
The remainder must vanish after at most m steps, for each remainder is strictly
smallerthanthepreviousone:
m > r > r > ···
1 2
Now we claim that the last non-zero remainder, d = r say, has the required
t
property:
d = gcd(m,n) = r .
t
Inthefirstplace,workingupfromthebottom,
d = r | r ,
t t−1
d | r andd | r =⇒ d | r ,
t t−1 t−2
d | r andd | r =⇒ d | r ,
t−1 t−2 t−3
...
d | r andd | r =⇒ d | r ,
3 2 1
d | r andd | r =⇒ d | m,
2 1
d | r andd | m =⇒ d | n.
1
Thus
d | m,n;
sodiscertainlyadivisorofmandn.
Ontheotherhand,supposeeisadivisorofmandn:
e | m,n.
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Then,workingdownwards,wefindsuccessivelythat
e | mande | n =⇒ e | r ,
1
e | r ande | m =⇒ e | r ,
1 2
e | r ande | r =⇒ e | r ,
2 1 3
...
e | r ande | r =⇒ e | r .
t−2 t−1 t
Thus
e | r = d.
t
Weconcludethatourlastnon-zeroremainderr isnumberwearelookingfor:
t
gcd(m,n) = r .
t
(cid:74)
It is easy to overlook the power and subtlety of the Euclidean Algorithm. The
algorithmalsogivesusthefollowingresult.
Theorem1.2 Supposem,n ∈ N. Let
gcd(m,n) = d.
Thenthereexistintegersx,y ∈ Zsuchthat
mx+ny = d.
Proof (cid:73) The Proposition asserts that d can be expressed as a linear combination
(with integer coefficients) of m and n. We shall prove the result by working
backwards from the end of the algorithm, showing successively that d is a linear
combinationofr andr ,andso,sincer isalinearcombinationofr and
s s+1 s+1 s−1
r ,disalsoalinearcombinationofr andr .
s s−1 s
Tostartwith,
d = r .
t
FromthepreviouslineintheAlgorithm,
r = q r +r .
t−2 t t−1 t
Thus
d = r = r −q r .
t t−2 t t−1
Butnow,fromthepreviousline,
r = q r +r .
t−3 t−1 t−2 t−1
Thus
r = rt−3−q r .
t−1 t−1 t−2
374 1–7
Hence
d = r −q rt−1
t−2 t
= r −q (r −q r )
t−2 t t−3 t−1 t−2
= −q r +(1+q q )r .
t t−3 t t−1 t−2
Continuinginthisway,supposewehaveshownthat
d = a r +b r .
s s s s+1
Since
r = q r +r ,
s−1 s+1 s s+1
itfollowsthat
d = a r +b (r −q r )
s s s s−1 s+1 s
= b r +(a −b q )r .
s s−1 s s s+1 s
Thus
d = a r +b r ,
s−1 s−1 s−1 s
with
a = b , b = a −b q .
s−1 s s−1 s s s+1
Finally,atthetopofthealgorithm,
d = a r +b r
0 0 0 1
= a r +b (m−q r )
0 0 0 1 0
= b m+(a −b q )r
0 0 0 1 0
= b m+(a −b q )(n−q m)
0 0 0 1 0
= (b −a q +b q q )m+(a −b q )n,
0 0 0 0 0 1 0 0 0
whichisoftherequiredform. (cid:74)
Example: Supposem = 39, n = 99. FollowingEuclid’sAlgorithm,
99 = 2·39+21,
39 = 1·21+18,
21 = 1·18+3,
18 = 6·3.
Thus
gcd(39,99) = 3.
Also
3 = 21−18
= 21−(39−21)
= −39+2·21
= −39+2(99−2·39)
= 2·99−5·39.
374 1–8
ThustheDiophantineequation
99x+39y = 3
hasthesolution
x = 2, y = −5.
(By a Diophantine equation we simply mean a polynomial equation to which we
areseekingintegersolutions.)
This solution is not unique; we could, for example, add 39 to x and subtract
99 from y. We can find the general solution by subtracting the particular solution
wehavejustfoundtogiveahomogeneouslinearequation. Thusifx(cid:48),y(cid:48) ∈ Zalso
satisfies the equation then X = x(cid:48) − x, Y = y(cid:48) − y satisfies the homogeneous
equation
99X +39Y = 0,
ie
33X +13Y = 0,
thegeneralsolutiontowhichis
X = 13t, Y = −33t
fort ∈ Z. Thegeneralsolutiontothisdiophantineequationistherefore
x = 2+13t, y = −5−33t (t ∈ Z).
ItisclearthattheEuclideanAlgorithmgivesacompletesolutiontothegeneral
lineardiophantineequation
ax+by = c.
Thisequationhasnosolutionunless
gcd(a,b) | c,
in which case it has an infinity of solutions. For if (x,y) is a solution to the
equation
ax+by = d,
andc = dc(cid:48) then(c(cid:48)x,c(cid:48)y)satisfies
ax+by = c,
andwecanfindthegeneralsolutionasbefore.
374 1–9
Corollary1.1 Supposem,n ∈ Z. Thentheequation
mx+ny = 1
hasasolutionx,y ∈ Zifandonlyifgcd(m,n) = 1.
ItisworthnotingthatwecanimprovetheefficiencyofEuclid’sAlgorithmby
allowing negative remainders. For then we can divide with remainder ≤ m/2 in
absolutevalue,ie
n = qm+r,
with−m/2 ≤ r < m/2. TheAlgorithmproceedsasbefore;butnowwehave
m ≥ |r /2| ≥ |r /22| ≥ ...,
0 1
sotheAlgorithmconcludesafteratmostlog msteps.
2
This shows that the algorithm is in class P, ie it can be completed in polyno-
mial (in fact linear) time in terms of the lengths of the input numbers m,n — the
lengthofn,iethenumberofbitsrequiredtoexpressninbinaryform,being
[log n]+1.
2
Algorithms in class P (or polynomial time algorithms) are considered easy or
tractable, while problems which cannot be solved in polynomial time are consid-
eredhardorintractable. RSAencryption—thestandardtechniqheforencrypting
confidentialinformation—restsonthebelief—anditshouldbeemphasizedthat
thisisabeliefandnotaproof—thatfactorisationofalargenumberisintractable.
Example: Takingm = 39, n = 99,asbefore,theAlgorithmnowgoes
99 = 3·39−18,
39 = 2·18+3,
18 = 6·3,
giving(ofcourse)
gcd(39,99) = 3,
asbefore.
1.3 Ideals
We used the Euclidean Algorithm above to show that if gcd(a,b) = 1 then there
wecanfindu,v ∈ Zsuchthat
au+bv = 1.
Thereisamuchquickerwayofprovingthatsuchu,v exist,withoutexplicitly
computingthem.
Recall that an ideal in a commutative ring A is a non-empty subset a ⊂ A
suchthat
374 1–10
1. a,b ∈ a =⇒ a+b ∈ a;
2. a ∈ a, c ∈ A =⇒ ac ∈ a.
Asanexample,themultiplesofanelementa ∈ Aformanideal
(cid:104)a(cid:105) = {ac : c ∈ A}.
Suchanidealissaidtobeprincipal.
Proposition1.3 Everyideala ⊂ Zisprincipal.
Proof (cid:73)Ifa = 0(byconventionwedenotetheideal{0}by0)theresultistrivial:
a = (cid:104)0(cid:105). Wemaysupposethereforthata (cid:54)= 0.
Then a must contain integers n > 0 (since −n ∈ a =⇒ n ∈ a). Let d be the
leastsuchinteger. Then
a = (cid:104)d(cid:105).
Forsupposea ∈ a. Dividingabyd,
a = qd+r,
where
0 ≤ r < d.
But
r = a+(−q)d ∈ a.
Hencer = 0;forotherwiser wouldcontradicttheminimalityofd. Thus
a = qd,
ieeveryelementa ∈ aisamultipleofd. (cid:74)
Nowsupposea,b ∈ Z. Considerthesetofintegers
I = {au+bv : u,v ∈ Z}.
ItisreadilyverifiedthatI isanideal.
AccordingtothePropositionabove,thisidealisprincipal,say
I = (cid:104)d(cid:105).
Butnow
a ∈ I =⇒ d | a, b ∈ I =⇒ d | b.
Ontheotherhand,
e | a, e | b =⇒ e | au+bv
=⇒ e | d.