Several proofs of the irreducibility of the cyclotomic polynomials [expository notes]

8 Pages·2012·0.073 MB·English

by  Steven H. Weintraub

SEVERAL PROOFS OF THE IRREDUCIBILITY OF THE CYCLOTOMIC
POLYNOMIALS
STEVENH.WEINTRAUB
ABSTRACT. We present a number of classical proofs of the irreducibility of the n-th
cyclotomicpolynomialF n(x). FornprimewepresentproofsduetoGauss(1801), in
boththeoriginalandasimplifiedform, Kronecker(1845), andScho¨nemann/Eisenstein
(1846/1850),andforgeneralnproofsduetoDedekind(1857),Landau(1929),andSchur
(1929).
LetF (x)denotethen-thcyclotomicpolynomial,definedby
n
F (x)=(cid:213) (x−z )
n
wherez rangesovertheprimitiven-throotsofunity. F (x)isalsogiveninductivelyby
n
xn−1
F (x)=
n (cid:213) F (x)
d
wheredrangesovertheproperdivisorsofn.Incasen=pisprime,F (x)=(xp−1)/(x−
p
1)=xp−1+xp−2+···+x+1.
ItisabasicresultinnumbertheorythatF (x)isirreducibleforeverypositiveinteger
n
n. Itisourobjectiveheretopresentanumberofclassicalproofsofthistheorem(certainly
notallofthem).
TheirreducibilityofF (x)for pprimewasfirstprovedbyGauss[4,article341],with
p
a simpler proof being given by Kronecker [5] and even simpler and more general proofs
beinggivenbyScho¨nemann[8]andEisenstein[3]. Wegivetheseproofshere. (Thelast
ofthesehasbecomethestandardproof.) Gauss’sproofisrathercomplicated, sowealso
giveasimplerproofalongthesamelines. TheirreducibilityofF (x)ingeneralwasfirst
n
proved by Kronecker [6], with simpler proofs being given by Dedekind [2], Landau [7],
andSchur[9]. Wegivethelastthreeoftheseproofshere. (AvariantofDedekind’sproof
hasbecomethestandardproof.)
WiththeexceptionofSchur’sproof,whichusessomeresultsaboutalgebraicintegers,
theseproofsalljustusebasicresultsaboutpolynomials. Wehaveorganizedthispaperto
collectbackgroundmaterialaboutpolynomialsinapreliminarysection,tohaveitavailable
whenwepresentthemainresults.
BACKGROUNDMATERIAL
ThefirstresultweneedaboutpolynomialsisGauss’sLemma[4,article42],whichwe
stateintheforminwhichwewilluseit.
Lemma. Let f(x)beamonicpolynomialwithintegercoefficients,andsupposethat f(x)=
g(x)h(x)whereg(x)andh(x)aremonicpolynomialswithrationalcoefficients. Theng(x)
andh(x)haveintegercoefficients.
2000MathematicsSubjectClassification. 12-03,Secondary12E05,01A55.
Keywordsandphrases. cyclotomicpolynomial,irreducibility.
1
2 STEVENH.WEINTRAUB
Proof. See[11,Corollary4.1.6].
Nowlet f(x)beanarbitrarymonicpolynomial, f(x)=xm+a xm−1+...+a . Let
m−1 0
f(x)haverootsr ,...,r ,sothat f(x)=(x−r )···(x−r ). Expandingthislastexpres-
1 m 1 m
sion we see that f(x)=xm+(cid:229) m (−1)is(r ,...,r )xm−i where s(r ,...,r ) is the i-th
i=1 i 1 m i 1 m
elementarysymmetricpolynomialintherootsr ,...,r , i.e., thesumofproductofthese
1 m
rootstakeniatatime. Thus
s (r ,...,r )=r +...+r ,
1 1 m 1 m
s (r ,...,r )=r r +...+r r ,
2 1 m 1 2 m−1 m
...
s (r ,...,r )=r ···r .
m 1 m 1 m
Comparing these twoexpressions we seethat, up tosign, the coefficients of f(x)are the
valuesofthesepolynomials;morepreciselya =(−1)is(r ,...,r ).
m−i i 1 m
Ingeneralapolynomialinmvariablesissymmetricifitisinvariantunderanypermuta-
tionofthevariables. Thesecondresultweneedisthefundamentaltheoremonsymmetric
polynomials,whichwasalreadywell-knownintheeighteenthcenturyandperhapsearlier.
Theorem. Let f(x)beamonicpolynomialwithinteger(respectively,rational)coefficients.
Letg(x)beanysymmetricpolynomialintherootsof f(x)withinteger(respectively,ratio-
nal) coefficients. Then g(x) can be written as a polynomial in the elementary symmetric
polynomialsoftherootsof f(x),andhenceasapolynomialinthecoefficientsof f(x),with
integer(respectively,rational)coefficients.
Proof. See[11,Lemma3.1.14andLemma3.1.15].
THERESULTSTHEMSELVES
Theorem1. Let pbeaprime. ThenthecyclotomicpolynomialF (x)isirreducible.
p
Proof (Gauss). This is trivial for p=2 so we suppose p is odd. We begin with some
generalconsiderations.
First, let f(x) be an arbitrary monic polynomial with rational coefficients, with roots
r ,...,r ,sothat f(x)=(x−r )···(x−r ),andletg(x)bethemonicpolynomialwhose
1 m 1 m
rootsarethek-thpowersoftherootsof f(x),forsomepositiveintegerk,sothatg(x)=(x−
rk)···(x−rk). Then the coefficients of g(x) are symmetric polynomials in {rk,...,rk},
1 m 1 m
and hence are symmetric polynomials in {r ,...,r }. Then by the fundamental theorem
1 m
onsymmetricpolynomialstheycanbeexpressedaspolynomialswithrationalcoefficients
inthecoefficientsof f(x). Henceg(x)hasrationalcoefficientsaswell.
Second, let j (x ,x ,...) be any polynomial with integer coefficients and let z be any
1 2
primitive p-th root of unity. Substituting xi =z ki for each i, we obtain a value for this
polynomialthatwemaywriteas
j (z k1,z k2,...)=A +A z +...+A z p−1
0 1 p−1
forsomeintegersA ,...,A ,andthenforanyt
0 p−1
j (z tk1,z tk2,...)=A +A z t+...+A z (p−1)t.
0 1 p−1
Then
j (1,1,...)=j (z pk1,z pk2,...)=A +A +...+A
0 1 p−1
SEVERALPROOFSOFTHEIRREDUCIBILITYOFTHECYCLOTOMICPOLYNOMIALS 3
and
j (z k1,z k2,...)+j (z 2k1,z 2k2,...)+...+j (z pk1,z pk2,...)=pA ,
0
andinparticularthissumisdivisibleby p.
NowsupposethatF (x)= f(x)g(x)fornonconstantmonicpolynomials f(x)andg(x),
p
with f(x) a polynomial of degree d. Then f(x) and g(x) each have integer coefficients.
(This is Gauss’s Lemma. Note that F (x) itself has integer coefficients.) Write f(x)=
p
xd+a xd−1+...+a x+a . Let W be the set of primitive p-th roots of unity. Let F
d−1 1 0
be the set of roots of f(x) and let G be the set of roots of g(x). Then F∪G=W and
F∩G={}. Let F′ be the set of reciprocals of the elements of F and let G′ be the set
of reciprocals of the elements of G. Then similarly F′∪G′ =W and F′∩G′ ={}. Let
f′(x) be the monic polynomial who roots are the elements of F′. Observe that f′(x)=
xd+(a /a )xd−1+...+(a /a )x+(1/a ). Wehavefourcases:
1 0 d−1 0 0
Case 1: F′ =F. Then f′(x)= f(x). In this case the roots of f(x) occur in conjugate
pairs, so f(x) is a product of d/2 factors each of the form (x−z )(x−z −1)=x2−(z +
z −1)x+1,andeachofthesefactorsispositiveforeveryrealnumberx. LetF betheset
k
ofk-thpowersoftheelementsofF,and f (x)themonicpolynomialwhoserootsarethe
k
elementsofF,foreachk=1,...,p−1. Thenthesamepropertyholdsforeach f (x). Let
k k
q = f (1) for k=1,...,p−1. Thus q ,...,q are all positive rational numbers. But
k k 1 p−1
infacteachpolynomial f (x)hasrationalcoefficients,bythefirstobservationabove,and
k
henceintegercoefficients(byGauss’sLemma),soq ,...,q areallpositiveintegers.
1 p−1
If j (x ,...,x )=(1−x )···(1−x ), then q =j (z k,...,z k), for k =1,...,p−1,
1 d 1 d k 1 d
whereF={z ,...,z },andj (z p,...,z p)=j (1,...,1)=0,sofromthesecondobserva-
1 d 1 d
tionaboveweseethatq +...+q isdivisiblebyp.Butalso f (x)···f (x)=F (x)d,
1 p−1 1 p−1 p
aseveryprimitive p-throotofunityisarootoftheleft-handsideofmultiplicityd. Hence,
lettingx=1,weobtainq ···q =pd.
1 p−1
Since pisaprimeandd< p−1,wemusthavegoftheintegersq ,...,q equalto
1 p−1
1 for some g>0, and then the rest of them are powers of p. But then q +...+q ≡
1 p−1
g(modp),andsothissumiscertainlynotdivisibleby p,acontradiction.
Case II: F 6=F′ but T =F∩F′ 6={}. Lett(x) be the monic polynomial whose roots
aretheelementsofT. Thent(x)isthegreatestcommondivisor(gcd)of f(x)and f′(x).
ThenbytheargumentofCaseIt(x)cannothaveallofitscoefficientsrational. But f(x)
and f′(x) are polynomials with rational coefficients, and hence their gcd is a polynomial
withrationalcoefficients,acontradiction.
CaseIII:G∩G′6={}. ApplyingtheargumentsofcasesIorIItog(x)yieldsthesame
contradiction.
CaseIV:G=F′andF =G′. Then
F (x)= f(x)f′(x)
p
=(xd+a xd−1+...+a x+a )(xd+(a /a )xd−1+...+(a /a )x+(1/a )),
d−1 1 0 1 0 d−1 0 0
andsettingx=1weobtain
a p=(1+a +...+a )2.
0 d−1 0
But(byGauss’sLemma) f′(x)hasintegercoefficients,soa =±1andweobtainthat±p
0
isaperfectsquare,acontradiction.
Proof(inthespiritofGauss). Wehavetheidentity
d d
(cid:213) (x−r)=(cid:229) (−1)is(r ,...,r )xd−i,
i i 1 d
i=1 i=0
4 STEVENH.WEINTRAUB
wherethes aretheelementarysymmetricfunctions.
i
Letj (r ,...,r )=(cid:213) d (1−r). Thenweseethat
1 d i=1 i
d
j (r ,...,r )=(cid:229) (−1)is(r ,...,r ).
1 d i 1 d
i=0
Thetheoremistrivialfor p=2sowemaysuppose pisanoddprime.
SupposethatF (x)isnotirreducibleandlet f (x)beanirreduciblefactorofF (x)of
p 1 p
degree d. Then f (x)=(x−z )···(x−z ) for some set of primitive p-th roots of unity
1 1 d
{z ,...,z }.Fork=1,...,p−1,let f (x)=(x−z k)···(x−z k).Thecoefficientsof f (x)
1 d k 1 d k
aresymmetricpolynomialsin{z k,...,z k},hencesymmetricpolynomialsin{z ,...,z },
1 d 1 d
hencepolynomialsinthecoefficientsof f (x),andso f (x)hasrationalcoefficients. Since
1 k
each f (x)dividesF (x),byGauss’sLemmainfacteach f (x)isapolynomialwithinteger
k p k
coefficients.
(Itiseasytoseethateach f (x)isirreducible,thatd mustdivide p−1,andthatthere
k
areexactly(p−1)/d distinctpolynomials f (x),butwedonotneedthesefacts.)
k
Since f (x)hasleadingcoefficient1andnorealroots, f (x)>0forallrealx. Also,
k k
p−1
F (x)d = (cid:213) f (x)
p k
k=1
sinceeveryprimitive p-throotof1isarootoftheright-handsideofmultiplicityd. Then
p−1
pd =F (1)d = (cid:213) f (1)
p k
k=1
andd< p−1,sowemusthave f (1)=1forsomeg>0valuesofk,and f (1)apower
k k
of pfortheremainingvaluesofk,andhence
p−1
(cid:229)
f (1)≡g6≡0(mod p).
k
k=1
But
j (z k,...,z k)= f (1)fork=1,...,p−1, andj (z p,...,z p)=j (1,...,1)=0.
1 d k 1 d
Thus
p−1 p−1
(cid:229) f (1)= (cid:229) j (z k,...,z k)
k 1 d
k=1 k=1
p
= (cid:229) j (z k,...,z k)
1 d
k=1
p d
= (cid:229) (cid:229) (−1)is(z k,...,z k)
i 1 d
k=1i=0
d p
=(cid:229) (−1)i(cid:229) s(z k,...,z k).
i 1 d
i=0 k=1
But s(r ,...,r ) is a sum of terms of the form r ···r , so each term in the inner sum
i 1 d j1 ji
aboveisasumofterms
p p
(cid:229) z k ···z k = (cid:229) (z ···z )k=0or p
j1 ji j1 ji
k=1 k=1
SEVERALPROOFSOFTHEIRREDUCIBILITYOFTHECYCLOTOMICPOLYNOMIALS 5
accordingasz ···z isaprimitive p-throotofunityorisequalto1. Thus
j1 ji
p−1
(cid:229)
f (1)≡ 0(mod p),
k
k=1
acontradiction.
Proof(Kronecker). Wefirstprovethefollowinglemma: Let f(x)beanarbitrarypolyno-
mialwithintegercoefficients. Letz beaprimitive p-throotof1. Then f(z )···f(z p−1)
and f(1)arebothintegersand
f(z )···f(z p−1)≡ f(1)p−1(modp).
To prove that the product f(z )···f(z p−1) is an integer, observe that it is a symmetric
polynomialin{z ,...,z p−1}so,bythefundamentaltheoremofsymmetricfunctions,isan
integerpolynomialinthecoefficientsofthepolynomialhaving{z ,...,z p−1}asroots.But
thispolynomialissimplyF (x)=xp−1+...+1. Toprovethatthecongruenceholds,let
p
g(x)= f(x)···f(xp−1)=(cid:229) A xn andconsider (cid:229) p−1g(z i). Thefirstexpressionforg(x)
n n i=0
givesthevalue f(1)p−1+(p−1)f(z )···f(z p−1)forthissumwhilethesecondexpression
forg(x)givesthevalue(cid:229) A p. Thus f(1)p−1+(p−1)f(z )···f(z p−1)≡
namultipleof p n
0(modp)andthelemmaimmediatelyfollows.
Now suppose F (x) is not irreducible and write F (x)= f(x)g(x), a product of non-
p p
constant polynomials. By Gauss’s lemma, f(x) and g(x) both have integer coefficients.
Then p=F (1)= f(1)g(1). One of these factors must be ±1, so suppose f(1)=±1.
p
Ontheonehand, f(z k)=0forsomek thatisnonzero(modp)(asthesearetherootsof
F (x)),so f(z )···f(z p−1)=0,butontheotherhand f(1)p−1≡1(modp),contradicting
p
theabovecongruence.
Proof(Scho¨nemann). Wehavethefollowingirreducibilitycriterion: Let f(x)beapolyno-
mialofdegreekwithintegercoefficients.Supposethat,forsomeprimep,andsomeinteger
a, f(x)=(x−a)k+pg(x)forsomepolynomialg(x)withintegercoefficientswithg(a)not
divisibleby p. (Aswemightphrasethisnowadays,supposethat f(x)≡(x−a)k (modp)
and f(a) is not divisible by p2.) Then f(x) is irreducible. Now, by the binomial theo-
rem,xp−1≡(x−1)p(modp),soF (x)= xp−1 ≡(x−1)p−1(modp),andalsoF (x)=
p x−1 p
xp−1+···+1soF (1)=p,andhenceF (x)satisfiesthehypothesesofthiscriterion.
p p
Proof(Eisenstein). Wehavethefollowingirreducibilitycriterion: Let f(x)=c xk+···+
k
c be a polynomial with integer coefficients. Suppose that, for some prime p, c is not
0 k
divisible by p, c ,...,c are divisible by p, and c is not divisible by p2. Then f(x)
k−1 0 0
is irreducible. Now F (x) is irreducible if and only if F (x+1) is irreducible. But,
p p
by the binomial theorem, F (x+1)= (x+1)p−1 =xp−1+a xp−2+···+a x+p, with
p x p−2 1
a ,...,a alldivisibleby p,whichsatisfiesthehypothesesofthiscriterion.
p−2 1
Remark. Scho¨nemann’sirreducibilitycriterionandhisproofoftheirreducibilityofF (x)
p
are little remembered now, with Eisenstein’s irreducibility criterion and his proof of the
irreducibility of F (x) being very well known. But in fact these two are equivalent. A
p
beautifuldiscussionofthispoint(boththemathematicsandthehistory)hasbeengivenby
Cox[1].
Theorem2. Letnbeanarbitrarypositiveinteger. ThenthecyclotomicpolynomialF (x)
n
isirreducible.
6 STEVENH.WEINTRAUB
Proof (Dedekind). Let f(x) be an irreducible factor of F (x). Since f(x) divides the
n
polynomialxn−1,itfollowsfromGauss’sLemmathat f(x)hasintegralcoefficients. Let
z beann-throotof1with f(z )=0. Itsufficestoprovethatif pisanyprimenotdividing
n,then f(z p)=0(asbyrepeatedapplicationsofthisresult,weobtainthat f(z j)=0for
any jrelativelyprimeton).
Suppose this is not the case. Let f(x) have roots z =z ,...,z , so that f(x)=(x−
1 k
z )···(x−z )=xk−c xk−1+...±c .Letg(x)=(x−z p)···(x−z p)=xk−d xk−1+
1 k k−1 0 1 k k−1
...±d . Thenc =s(z ,...,z )andd =s(z p,...,z p)foreachi,wheres isthei-thele-
0 i i 1 k i i 1 k i
mentarysymmetricfunction.Bythemultinomialtheorem,wehavethepolynomialidentity
s(xp,...,xp)≡s(x ,...,x )p(modp),i.e.,s(xp,...,xp)−s(x ,...,x )p=pt(x ,...,x )
i 1 k i 1 k i 1 k i 1 k i 1 k
for some polynomial t(x ,...,x ) with integral coefficients. Thus, for each i, d −cp =
i 1 k i i
s(z p,...,z p)−s(z ,...,z )p= pt(z ,...,z ). Butt(x ,...,x )isasymmetricpolyno-
i 1 k i 1 k i 1 k i 1 k
mial,sobythefundamentaltheoremonsymmetricpolynomialst(z ,...,z )isapolyno-
i 1 k
mialin{s (z ,...,z )=c }withintegercoefficients. Henced ≡cp≡c (modp)foreach
j 1 k j i i i
i,wherethelastcongruenceisFermat’sLittleTheorem,andsog(x)isalsoapolynomial
withintegercoefficientsandfurthermoreg(x)≡ f(x)(modp).
Thepolynomialg(x)isalsoirreducible(afactwhichfollowsfromtheobservationthat
z =(z p)q whereqisanintegerwith pq≡1(modn)),andg(x)6= f(x). Hence f(x)g(x)
divides m(x)=xn−1. Let e(x) denote the (modp) reduction of the polynomial e(x).
Then f(x)g(x)= f(x)2dividesm(x),whichisimpossibleasm(x)anditsformalderivative
m(x)′=nxn−1arerelativelyprime.
Remark. Thisproofcanbesimplifiedtoremovetheargumentaboutsymmetricfunctions.
ItfollowsimmediatelyfromthemultinomialtheoremandFermat’sLittleTheoremthatfor
anypolynomialk(x)withintegercoefficients,k(x)p≡k(xp)(modp)foranyprime p. Let
f(x) be as above and let g(x) be an irreducible polynomial having g(z p)=0. Clearly z
isarootofthepolynomialg(xp),so f(x)dividesg(xp). Reducing(modp), f(x)divides
g(xp)=g(x)p, so f(x) and g(x) have a common irreducible factor h(x). But then h(x)2
dividesm(x), whichisimpossibleasabove. Thissimplificationisalreadytobefoundin
vanderWaerden[10,article53].
Proof(Landau). Let f(x)beanirreduciblepolynomialwithintegercoefficientsofdegree
dwith f(z )=0forsomen-throotofunityz .Bythedivisionalgorithm,forany jthereare
uniquepolynomialsq (x)andr (x)with f(xj)= f(x)q (x)+r (x)andr (x)ofdegreeless
j j j j j
thand (perhapsr (x)=0). Sincethevalueof f(z j)onlydependson j(modn),wehave
j
afiniteset{r (x),...,r (x)}ofpolynomialssuchthat, foranyinteger j, f(z j)=r(z )
0 n−1
forr(x)somepolynomialinthisset. Furthermore,ifs(x)isanypolynomialofdegreeless
thand with f(z j)=s(z ),thenwemusthaves(x)=r(x)(asotherwisez wouldbearoot
ofthenonzeropolynomials(x)−r(x)ofdegreelessthand,whichisimpossible).
Inparticular,foranyprimep, f(z p)= f(z p)−f(z )p=r(z )forsomesuchpolynomial
r(x). But f(xp)≡ f(x)p (modp), so f(xp)− f(x)p = pg(x) for some polynomial g(x).
But,againbythedivisionalgorithm,thereisauniquepolynomialh(x)ofdegreelessthan
dwithh(z )=g(z ). Thusr(z )=ph(z )withr(x)and ph(x)bothofdegreelessthand,so
r(x)= ph(x). Inparticular,allthecoefficientsofr(x)aredivisibleby p. NowifAisthe
largestabsolutevalueofthecoefficients ofallofthepolynomials{r (x)}, wemusthave
j
f(z p)=r(z )=0forp>A,andso f(z m)=0foranyintegermnotdivisiblebyanyprime
p≤A.
Nowletkbeanyintegerrelativelyprimeton,andconsiderm=k+n(cid:213) q,whereqruns
overalltheprimes≤Athatdonotdividek. Let pbeanyprime≤A. If pdividesk,then,
SEVERALPROOFSOFTHEIRREDUCIBILITYOFTHECYCLOTOMICPOLYNOMIALS 7
sincekandnarerelativelyprime, pdoesnotdividen(cid:213) qandhencedoesnotdividem. If
pdoesnotdividek,then pdividesn(cid:213) qandhencedoesnotdividem. Thusweseethatm
issuchaninteger,andm≡k(modn),so f(z k)= f(z m)=0. ThusF (x)hasz k asaroot
n
foreverykrelativelyprimeton,andsoF (x)isirreducible.
n
Remark. Thisproof,writteninLandau’susualtelegraphicstyle,takes8linesintheorigi-
nal.
Proof (Schur). Let g(x)=xn−1, and let D be the discriminant of g(x), i.e., the product
ofthesquaresofthedifferencesofthedistinctroots. ThenD =±nn, asweseefromthe
followingcomputation:
D =(cid:213) (z i−z j)2
i<j
=±(cid:213) (z i−z j)
i6=j
=±(cid:213) z i(1−z j−i)
i6=j
=±(cid:213) z i¡(cid:213) (1−z k)¢
i k6=0
=±(cid:213) z i(n)=±nn.
i
Inthiscomputation,theequality(cid:213) (1−z k)=ncomesfromthefactthattheleft-hand
k6=0
sideisthevalueh(1)forh(x)thepolynomialh(x)=(cid:213) (x−z k)=(xn−1)/(x−1)=
k6=0
xn−1+...+1.
Now suppose that f(x) is a factor of xn−1. Let z be a root of f(x) and let p be
any prime not dividing n. We claim that z p is also a root of f(x). Suppose not. Then
f(x)=(x−z )···(x−z ) for some n-th roots of unity z =z ,z ,...,z , not including
1 k 1 2 k
z p. Thus 06= f(z p) is a product of differences of n-th roots of unity, so is an algebraic
integer dividing nn. But f(xp)≡ f(x)p(modp), so f(z p)≡ f(z )=0 (modp), i.e., p
divides f(z p). Butthatimplies pdividesnn,acontradiction. (Arationalintegeradivides
arationalintegerbasrationalintegersifandonlyifadividesbasalgebraicintegers.)
Remark. Schur observes that Landau’s proof and his proof are both simplifications of
proofsduetoMertens.
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