Table Of ContentComputing Congruences of
Modular Forms and Galois Representations
Modulo Prime Powers
0
1 XavierTaixési Ventosa∗and GaborWiese†
0
2
21stJanuary 2010
n
a
J
1
2
Abstract
]
T ThisarticlestartsacomputationalstudyofcongruencesofmodularformsandmodularGalois
N representationsmoduloprimepowers. Algorithmsaredescribedthatcomputethemaximumin-
. tegermodulowhichtwo moniccoprimeintegralpolynomialshavea rootincommonina sense
h
t thatisdefined. Thesetechniquesare appliedto thestudyofcongruencesofmodularformsand
a
m modularGaloisrepresentationsmoduloprimepowers. Finally,somecomputationalresultswith
[ implicationsonthe(non-)liftabilityofmodularformsmoduloprimepowersandpossiblegener-
alisationsoflevelraisingarepresented.
2
v 2010MathematicsSubjectClassification: 11F33(primary);11F11,11F80,11Y40.
4
2
7
1 Introduction
2
.
9
0 Congruencesofmodularformsmoduloaprimeℓand–fromadifferentpointofview–modularforms
9
overF playanimportant role inmodern Arithmetic Geometry. Themostprominent recent example
0 ℓ
: isSerre’smodularityconjecture, whichhasjustbecomeatheoremofKhare,Wintenberger andKisin.
v
i
X We particularly mention the various techniques for Level Raising and Level Lowering modulo ℓ that
r werealready crucialforWiles’sproofofFermat’sLastTheorem.
a
Motivated by this, it is natural to study congruences modulo ℓn of modular forms and Galois
representations. However,asworkingovernon-factorialandnon-reducedringslikeZ/ℓnZintroduces
manyextradifficulties,oneisledtofirstapproachthissubjectfromanalgorithmicandcomputational
pointofview,whichisthetopicofthisarticle.
We introduce a definition of when two algebraic integers a, b are congruent modulo ℓn. Our
definition,whichmightappearnon-standardatfirst,wasforceduponusbythreerequirements: Firstly,
∗UniversitatPompeuFabra,Departamentd’EconomiaiEmpresa,RamonTriasFargas25-27,08005Barcelona
xavier.taixes@upf.edu
†UniversitätDuisburg-Essen,InstitutfürExperimentelleMathematik,Ellernstraße29,45326Essen,Germany
gabor@pratum.net,http://maths.pratum.net/
1
we want it to be independent of any choice of number field containing a, b. Secondly, in the special
case n = 1 a congruence modulo ℓ should come down to an equality in a finite field. Finally, if a, b
lieinsomenumberfieldK thatisunramifiedatℓ,thenacongruence ofaandbmoduloℓn shouldbe
acongruence moduloλn,whereλisaprimedividingℓinK.
Since algebraic integers are – up to Galois conjugacy – most conveniently represented by their
minimalpolynomials,weaddresstheproblemofdeterminingforwhichprimepowersℓntwocoprime
monic integral polynomials have zeros which are congruent modulo ℓn. We prove that a certain
number, called the reduced discriminant or – in our language – the congruence number of the two
polynomials, in all cases gives a good upper bound and in favourable cases completely solves this
problem. In the cases when the congruence number is insufficient, we use a method based on the
Newtonpolygonofthepolynomialwhoserootsarethedifferencesoftherootsofthepolynomialswe
startedwith.
With these tools at our disposal, we target the problem of computing congruences modulo ℓn
betweentwoHeckeeigenforms. Sinceourmotivation comesfromarithmetic, especially fromGalois
representations, our main interest is in Hecke eigenforms. It quickly turns out, however, that there
are several possible well justified notions of Hecke eigenforms modulo ℓn. We present two, which
we call strong and weak. The former can be thought of as reductions modulo ℓn of q-expansions of
holomorphic normalised Hecke eigenforms; the latter can be understood as linear combinations of
holomorphic modular forms, which are in general not eigenforms, but whose reduction modulo ℓn
becomesaneigenform (ourdefinitionisformulatedinadifferentway,butcanbeinterpreted tomean
this). We observe that Galois representations to GL (R), where R is an extension of Z/ℓnZ in the
2
sense of Section 2, can be attached to both weak and strong Hecke eigenforms (under the condition
ofresidual absolute irreducibility).
Modularformscanberepresentedbytheirq-expansions(e.g.inZ/ℓnZ),i.e.bypowerseries. For
computational purposes, such as uniquely identifying a modular form and comparing two modular
forms, itisessential that already afinitesegmentofacertain length oftheq-expansions suffices. We
noticethatasufficientlengthisprovidedbytheso-called Sturmbound, whichisthesamemoduloℓn
asincharacteristic 0.
The computational problem that we are mostly interested in is to determine congruences mod-
ulo ℓn between two newforms, i.e. equalities between strong Hecke eigenforms modulo ℓn. This
problem isperfectly suitedforapplying ourmethodsofdetermining congruences moduloℓn ofzeros
ofintegralpolynomials. ThereasonforthisisthattheFouriercoefficienta ofanormalisedHeckeei-
p
genformisazeroofthecharacteristicpolynomialoftheHeckeoperatorT actingonasuitableintegral
p
modular symbols space (see e.g.[S]or[W2]). Thus, inorder todetermine the primepowersmodulo
which twonewforms are congruent, wecompute the congruences between the roots of these charac-
teristic polynomials for a suitable number of p. One important point deserves to be mentioned here:
Ifthetwonewformsthatwewanttocomparedonothavethesamelevels(butthesameweights),one
cannot expect that they are congruent at all primes; a different behaviour is to be expected at primes
dividing the levels. Weaddress this problem by applying the usual degeneracy maps ‘modulo ℓn’ in
2
order to land in the same level. All these considerations lead to an algorithm, which we sketch. We
point out that this algorithm ismuch faster than the (naive) one which works withthe coefficients of
themodularformsasalgebraic integersina(necessarily big)numberfield.
Weimplemented thealgorithm andperformed manycomputations whichledtoobservations that
we consider very interesting. Some of the results are reported upon in Section 4. We are planning
to investigate questions like ‘Level Raising’ in more detail in a subsequent work. We remark that
the algorithm was already used in [DT] to determine some numerical examples satisfying the main
theorem ofthatarticle.
Acknowledgements
X.T.would like to thank Gerhard Frey for suggesting the subject of the article as PhD project. G.W.
wouldliketothankFrazerJarvis, LaraThomas,Christophe Ritzenthaler, IanKimingand,inparticu-
lar, Gebhard Böckle for enlightening discussions and e-mail exchanges relating tothe subject of this
article,aswellasKristinLauterforpointingoutthearticle[Pohst]. SpecialthanksareduetoMichael
Stollforsuggestingthebasicideaofonealgorithm,aswellastooneoftherefereeforalsosuggesting
it together with many other improvements in notation and presentation. Thanks are also due to the
secondrefereeforpointing outthatthereshouldbearelationtothepaper[ARS].
Both authors acknowledge partial support by the European Research Training Network Galois
Theory and Explicit Methods MRTN-CT-2006-035495. G. W. also acknowledges partial support by
theSonderforschungsbereich Transregio45oftheDeutscheForschungsgemeinschaft.
Notation
We introduce some standard notation to be used throughout. In the article ℓ and p always refer to
primenumbers. Byanℓ-adicfieldweshallunderstand afinitefieldextension ofQ . Wefixalgebraic
ℓ
closures Q of Q and Q of Q . By Z and Z we denote the integers of Q and Q , respectively. If K
ℓ ℓ ℓ ℓ
is either a number field or a local field, then O denotes its ring of integers. In the latter case, π
K K
denotes auniformiser, i.e.agenerator ofthemaximalidealofO ,andv isthevaluation satisfying
K K
v (π ) = 1. Moreover,v denotes thevaluation onK andonQ normalised suchthatv (ℓ)= 1.
K K ℓ ℓ ℓ
2 Congruences modulo ℓn
Inthissection wegiveourdefinition ofcongruences moduloℓn foralgebraic andℓ-adicintegers and
discusshowtocomputethem.
2.1 Definition
Sinceaquestion oncongruences isalocal question, weplace ourselves intheset-up ofℓ-adic fields.
Let α,β ∈ Z . In our definition of congruences modulo ℓn we are led by three requirements: (1) If
ℓ
3
n = 1, wewant that α ≡ β mod ℓ ifand only if thereductions ofαand β are equal inF . (2)If α
ℓ
and β are elements of some finite unramified extension K/Q , then we want α ≡ β mod ℓn if and
ℓ
onlyofα−β ∈ (πn). (3)WewantthedefinitiontobeindependentofanychoiceofK/Q containing
K ℓ
αandβ.
Weproposethefollowingdefinition.
Definition2.1 Letn ∈ N. Let α,β ∈ Z . Wesay that α is congruent to β modulo ℓn, for which we
ℓ
writeα≡ β mod ℓn,ifandonlyifv (α−β) > n−1.
ℓ
Notethatthisdefinitionsatisfiesourthreerequirements. Notealsothetrivialequivalence
α ≡ β mod ℓn ⇔ ⌈v (β −α)⌉ ≥ n. (2.1)
ℓ
Inthesequelofthisarticlewewilloftenspeakofcongruencesmoduloℓnof(global)algebraicintegers
byfixinganembeddingQ ֒→ Q . Thesamenotationwillbeusedalsointhissituationwithoutfurther
ℓ
comments.
2.2 Interpretation interms ofring extensions
Inthissectionweproposeaninterpretation oftheabovedefinitionofcongruencesmoduloℓninterms
of ring extension of Z/ℓnZ. This interpretation gives us a much better algebraic handle for working
withsuchcongruences becausewewillbeabletouseequalityinsteadofcongruence. Wewereledto
Definition2.1bythefollowingconsideration: LetK/Q beafiniteextensionandn ∈N. Whatisthe
ℓ
minimalmsuchthattheinclusionZ ֒→ O inducesaninjectionofZ/ℓnZintoO /(πm)? Inorder
ℓ K K K
toformulate theanswer, weintroduce afunction.
Definition2.2 Let L/K/Q be finite field extensions and let e denote the ramification index of
ℓ L/K
L/K. Forn ∈ N,letγ (n)= (n−1)e +1.
L/K L/K
Thisfunction satisfiesthefollowingsimpleproperties:
(i) Forn= 1,wehaveγ (1) = 1.
L/K
(ii) IfL/K isunramified, thenγ (n)= n.
L/K
(iii) Forextensions M/L/K,wehavemultiplicativity: γ (n) = γ (γ (n)).
M/K M/L L/K
(iv) For extensions L/K, the integer γ (n) is the minimal one such that the embedding O ֒→
L/K K
O induces aninjection O /(πn)֒→ O /(πγL/K(n)).
L K K L L
(v) Forα,β ∈ K/Q wehave:
ℓ
v (α−β) ≥ γ (n) ⇔ v (α−β) > n−1 ⇔ α ≡ β mod ℓn.
K K/Qℓ ℓ
4
Note that (i)–(iii) precisely correspond to the requirements (1)–(3) from Section 2.1. By (iv) we
haveproduced ringextensions
Z/ℓnZ ֒→ O /(πγK/Qℓ(n)) ֒→ O /(πγL/Qℓ(n)).
K K L L
Property(v)immediatelyyieldsareformulationofthecongruenceofαandβmoduloℓnasanequality
intheresidue ringO /(πγK/Qℓ(n)).
K K
In order to interpret congruences as equalities without always having to choose some finite ex-
tension of Q , we now make the following construction, which for n = 1 boils down to F . We
ℓ ℓ
define
Z/ℓnZ := lim O /(πγK/Qℓ(n)),
−→ K K
K
where K runs through all subextensions of Q of finite degree over Q and the inductive limit is
ℓ ℓ
taken with respect to themaps in(iv). Thenatural projections O ։ O /(πγK/Qℓ(n))give riseto a
K K K
surjective ringhomomorphism
π : Z ։ Z/ℓnZ.
n ℓ
Nowwecanmakeanother reformulation ofourdefinitionofcongruences moduloℓn: Letα,β ∈ Z .
ℓ
Thenwehave
α≡ β mod ℓn ⇔ π (α) = π (β).
n n
Inthesequel, wewillalwayschoose theπ inacompatible way,i.e.ifm < nwewantπ tobethe
n m
composition ofπ withthenatural mapZ/ℓnZ ։ Z/ℓmZ.
n
Remark2.3 We also point out a disadvantage of our choice of γ (n), namely that it is not ad-
K/Qℓ
ditive. This fact prevents us from defining avaluation on Z by saying that the valuation ofa ∈ Zis
ℓ
equal to the maximal n such that π (a) = 0. Defining γ (n) as n times the ramification index
n K/Qℓ
e would have avoided that problem. But then γ(1) = e 6= 1, in general, which is not in
K/Qℓ K/Qℓ
accordance withtheusualusageofmoduloℓ. Thisotherpossibility canbeunderstood asZ /ℓnZ .
ℓ ℓ
2.3 Computing congruences modulo ℓn
If one does not require one fixed embedding into the complex numbers, algebraic integers are most
easily represented by their minimal polynomials. Thus, it is natural to study congruences between
algebraic integersentirelythroughtheirminimalpolynomials. Thisisthepointofviewthatweadapt
anditleadsustoconsider thefollowingproblem.
Problem2.4 Wefix, onceandforall,foreveryncompatibly, ringhomomorphisms π : Z ֒→ Z ։
n ℓ
Z/ℓnZ. LetP,Q ∈ Z[X]betwocoprimemonicpolynomials andletn ∈ N.
Howcanwedecidethevalidity ofthefollowingassertion?
“Thereexistα,β ∈ Zsuchthat
5
(i) P(α) = Q(β) = 0and
(ii) π (α) =π (β)(i.e.α≡ β mod ℓn).”
n n
In this article, we will give two algorithms for treating this problem. The first one arose from
the idea that one could try to use greatest common divisors. This notion seems to be the right one
for n = 1, but it is not well behaved for n > 1 since the ring Z/ℓnZ[X] is not a principal ideal
domain. However, the algorithm for approximating greatest common divisors of two polynomials
over Z presented in Appendix A of [FPR] led us to consider the notion of congruence number or
ℓ
reduced resultant. Itcanbeusedtogivequiteafastalgorithm, which,however,doesnotalwaysgive
acompleteanswer.
Thesecond algorithm, whichwecalltheNewtonpolygonmethod, alwayssolvesProblem2.4but
tends to be slower (experimentally). Its basic idea was suggested to us by Michael Stoll after a talk
of the second author and was immediately put into practice. However, since the first version of this
article had already been finished, the algorithm wasnot included init, so that it wasagain suggested
tousbyoneofthereferees. Inthissectionwewillpresentbothalgorithmsindetail.
It should be pointed out explicitly that Problem 2.4 cannot be solved completely by considering
only the reductions of P and Q mod ℓn if n > 1. This is a major difference to the case n = 1. The
difference is due to the fact that in the problem we want α and β to be zeros of P and Q: if α and
β are elements in Z/ℓnZ such that inside that ring P(α) = Q(β) = 0, then it isnot clear if they are
reductions ofzerosofP andQ.
Congruencenumber
Thecongruence numberoftwointegralpolynomials provides anupperbound forcongruences inthe
senseofProblem2.4. Itisdefinedinsuchawaythatitcaneasilybecalculated onacomputer.
Definition2.5 LetR be any commutative ring. ByR[X] wedenote the R-module of polynomials
<n
of degree less than n. Let P,Q ∈ R[X] be two polynomials of degrees m and n, respectively. The
SylvestermapistheR-modulehomomorphism
R[X] ⊕R[X] → R[X] ,(r,s) 7→ rP +sQ.
<n <m <(m+n)
If R is a field, then the monic polynomial of smallest degree in the image of the Sylvester map
is the greatest common divisor of P and Q. In particular, with R a factorial integral domain and
P,Q primitive polynomials, the Sylvester map isinjective ifand only ifP and Q are coprime. Con-
sequently, ifP,Q ∈ Z[X]areprimitivecoprimepolynomials, thenanynon-zeropolynomialofsmal-
lestdegreeisaconstant polynomial.
Definition2.6 LetP,Q ∈ Z[X]becoprime polynomials. Wedefine thecongruence number c(P,Q)
ofP andQasthesmallestpositiveintegercsuchthattheconstantpolynomialcisintheimageofthe
Sylvester mapofP andQ.
6
We remark that for monic coprime polynomials P and Q via polynomial division the principal
ideal(c(P,Q))canbeseentobeequaltotheintersectionoftheidealofconstantintegralpolynomials
with the ideal in Z[X] generated by all polynomials rP + sQ when r,s run through all of Z[X].
In [Pohst] the congruence number is called the reduced resultant. Note that in general the reduced
resultant is a proper divisor of the resultant. It makes sense to replace Z by Z everywhere and to
ℓ
define a congruence number as a constant polynomial in the image of the Sylvester map having the
lowestℓ-adicvaluation. Althoughthiselementisnotunique, itsvaluation is.
Thecongruence numbergivesanupperboundfortheninProblem2.4:
Proposition 2.7 LetP,Q ∈ Z[X]becoprimepolynomialsandletℓn betheexactpowerofℓdividing
c(P,Q). Thentherearenoα,β ∈ Zsuchthat
(i) P(α) = Q(β) = 0and
(ii) π (α) = π (β)(i.e.α ≡ β mod ℓn)foranym > n.
m m
Proof. By assumption there exist r,s ∈ Z[X] such that c = c(P,Q) = rP +sQ. Letα,β ∈ Z
bezerosofP andQ,respectively, suchthatπ (α) = π (β). Weobtain
m m
π (c) = π r(α)P(α)+s(α)Q(α) = π s(α) π Q(α) = π s(β) π Q(β) = 0.
m m m m m m
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
Thismeansthatℓm dividesc,whencem ≤ n. 2
Onthecomputationofthecongruencenumber
The idea for the computation of the congruence number is very simple: we use basic linear algebra
and the Sylvester matrix. The point is that the Sylvester map is described by the standard Sylvester
matrix S of P and Q (or rather its transpose if one works with column vectors) for the standard
basesofthepolynomialrings. Wedescribeinwordsthestraightforwardalgorithmforcomputingthe
congruencenumberc(P,Q)aswellasforfindingpolynomialsr,ssuchthatc(P,Q) = rP+sQwith
deg(r) < deg(Q)and deg(s) < deg(P). Thealgorithm consists ofbringing S into row echelon (or
Hermite) form, i.e. one computes an invertible integral matrix B such that BS has no entries below
thediagonal. Thecongruence numberc(P,Q)is(theabsolutevalueof)thebottomrightentryofBS
and the coefficients ofr and s arethe entries inthe bottom row ofB. Thisalgorithm worksover the
integersandoverℓ-adicringswithacertainprecision, i.e.Z/ℓnZ.
We note that by reducing BS modulo ℓ, one can read off the greatest common divisor of the
reductions ofP andQmodulo ℓ: itscoefficients (uptonormalization) aretheentries inthelastnon-
zerorowofthereductionofBSmoduloℓ. Thishasthefollowingtrivial,butnoteworthyconsequence.
Corollary 2.8 SupposethatP andQareprimitivecoprimepolyomialsinZ[X]. ThenP andQhave
anon-trivial commondivisor modulo ℓif and only ifthe congruence number ofP and Q isdivisible
byℓ. 2
7
Applicationsofthecongruencenumber
WenowexaminewhenthecongruencenumberisenoughtosolveProblem2.4forgivenP,Qandfor
alln. Incaseswhenitisnot,wewillgivealowerboundforthemaximumnforwhichtheassertions
oftheproblem aresatisfied.
Westartwiththeobservationthatthecongruencenumbersufficestosolveourproblemforn = 1.
Proposition 2.9 Let n = 1. Assume that P and Q are coprime monic polynomials in Z[X]. The
assertion inProblem2.4issatisfied ifandonlyifthecongruence numberc(P,Q)isdivisible byℓ.
Proof. Thecalculations oftheproofofProposition 2.7showthatiftheassertion issatisfied,then
ℓ divides c(P,Q). Conversely, if ℓ divides c(P,Q) then by Corollary 2.8 the reductions of P and Q
have anon-trivial common divisor and thus acommon zero in F . Allzeros in F lift to zeros in Z .
ℓ ℓ ℓ
2
Wefixan embedding Q ֒→ Q . Ourfurther treatment willbe based on the following simple ob-
ℓ
servation. LetM ⊂ Qbeanynumberfieldcontaining alltherootsofthemoniccoprimepolynomials
P,Q ∈ Z[X]andletc = c(P,Q) = rP +sQwithr,s ∈ Z[X],deg(r)< deg(Q),deg(s) < deg(P)
andfactorQ(X) = (X −β )inZ[X]. Thenforα ∈ZsuchthatP(α) = 0wehave
i i
Q
v (c) = v s(α) + v (α−β ). (2.2)
M M M i
i
(cid:0) (cid:1) X
Ouraimnowistofindalowerboundforthemaximumofv (α−β )depending onπ (c). Forthat
M i M
wediscussthetwosummandsintheequation separately.
Wefirsttreatv s(α) . ByF wedenotethereduction moduloℓofanintegralpolynomial F.
M
(cid:0) (cid:1)
Proposition 2.10 Suppose thatℓdividesc(P,Q).
(a) IfsandQarecoprime, thenv s(α) = 0forallα ∈ Zwithπ (Q(α)) = 0.
M 1
(cid:0) (cid:1)
(b) If one of P or Q does not have any multiple factors, then there is α ∈ Z such that P(α) = 0,
π (Q(α)) = 0 and v (s(α)) = 0, or there is β ∈ Z such that Q(β) = 0, π (P(β)) = 0 and
1 M 1
v (r(β)) = 0.
M
(c) If P is an irreducible polynomial in F [X] and Q is irreducible in Z [X], then s and Q are
ℓ ℓ
coprimeandv s(α) = 0forallα ∈ Zwithπ (Q(α)) = 0.
M 1
(cid:0) (cid:1)
Proof. (a)SincesandQarecoprime, thereduction ofαcannotbearootofbothofthem.
(b) We prove that there exists y ∈ F which is a common zero of P and Q, but not a common
ℓ
zero of r and s at the same time. Assume the contrary, i.e. that r(y) = s(y) = 0 for all y ∈ F
ℓ
withP(y) = Q(y) = 0. LetG ∈ F [X]bethemonic polynomial ofsmallest degree annihilating all
ℓ
y ∈ F withtheproperty P(y) = Q(y) = 0. ThenGdivides P,Qaswellasbyassumption r ands.
ℓ
Hence,wehave
2
0 = rP +sQ = G r P +s Q
1 1 1 1
(cid:0) (cid:1)
8
withcertainpolynomials r ,P ,s ,Q ∈ F [X]. Weobtaintheequation
1 1 1 1 ℓ
0 =r P +s Q (2.3)
1 1 1 1
and wealso have deg(r ) < deg(Q )and deg(s ) < deg(P ). As either P or Q does not have any
1 1 1 1
multiplefactor, itfollowsthatP andQ arecoprime. Thiscontradicts Equation2.3.
1 1
Hence, wehave y ∈ F withP(y) = Q(y) = 0 and r(y) 6= 0 or s(y) 6= 0. If r(y) 6= 0then we
ℓ
lifty toazeroβ ofQ. Intheothercasewelifty toazeroαofP.
a
(c)Theassumptions implythatQ = P forsomea. Asthedegreeofsissmallerthanthedegree
of P, it follows that s and P are coprime. Thus also, s and Q are coprime and we conclude by (a).
2
Wenowtreattheterm v (α−β ).
i M i
P
Proposition 2.11 Supposethatℓdividesc(P,Q)andthatαisarootofP whichiscongruenttosome
root of Q modulo ℓ (which exists by Proposition 2.9). Assume without loss of generality that β is a
1
rootofQwhichisclosesttoα,i.e.suchthatv (α−β ) ≥ v (α−β )foralli.
M 1 M i
(a) Suppose that Q has no multiple factors (i.e. the discriminant of Q is not divisible by ℓ, or, equi-
valently, thecongruence numberofQandQ′ isnotdivisible byℓ).
Then v (α−β ) =v (α−β ).
i M i M 1
(b) IngenPeralwehavev (α−β ) ≥ ⌈ 1 v (α−β ) ⌉.
M 1 deg(Q) i M i
(cid:0)P (cid:1)
Proof. (a) If Q does not have any multiple factors, then v (β −β ) = 0 for all i 6= 1. Con-
M 1 i
sequently, v (α−β )= v (α−β +β −β )= 0fori 6= 1.
M i M 1 1 i
(b)istrivial. 2
We summarise of the preceding discussion in the following corollary, solving Problem 2.4 if P
andQdonothaveanymultiplefactors, andgivingapartialanswerintheothercases.
Corollary 2.12 LetP,QbecoprimemonicpolynomialsinZ[X](orZ [X])andletℓn bethehighest
ℓ
power of ℓ dividing the congruence number c := c(P,Q) and let r,s ∈ Z[X] (or Z [X]) be polyno-
ℓ
mialssuchthatc= rP +sQwithdeg(r) < deg(Q)anddeg(s) < deg(P).
(a) Ifn = 0,thennorootofP iscongruent moduloℓtoarootofQ.
(b) Ifn = 1,thenthereareα,β inZ(inZ ,respectively) withP(α) = Q(β) = 0suchthattheyare
ℓ
congruent modulo ℓ, and there are no α , β in Z (in Z , respectively) with P(α) = Q(β) = 0
1 1 ℓ
suchthattheyarecongruent moduloℓ2.
(c) Suppose nowthatn≥ 1andthatoneofthefollowingproperties holds:
(i) P does not have any multiple factors and Q does not have any multiple factors (i.e. ℓ ∤
c(P,P′)andℓ ∤c(Q,Q′)).
9
(ii) Qdoesnothaveanymultiplefactors andsandQarecoprime.
(iii) P doesnothaveanymultiplefactors andrandP arecoprime.
Thenthereareα,βinZ(inZ ,respectively)withP(α) = Q(β) = 0suchthattheyarecongruent
ℓ
modulo ℓn andthere arenoα ,β inZ(inZ ,respectively) withP(α ) = Q(β ) = 0such that
1 1 ℓ 1 1
theyarecongruent moduloℓn+1.
(d) Suppose thatn ≥ 1.
(i) IfsandQarecoprime, letm = ⌈ n ⌉.
deg(Q)
(ii) IfrandP arecoprime, letm = ⌈ n ⌉.
deg(P)
(iii) If(i)and(ii)donothold,letm = 1
Thenthereareα,βinZ(inZ ,respectively)withP(α) = Q(β) = 0suchthattheyarecongruent
ℓ
moduloℓm andtherearenoα ,β inZ(inZ ,respectively) withP(α ) = Q(β ) = 0suchthat
1 1 ℓ 1 1
theyarecongruent moduloℓn+1.
Proof. In the proof we use the notation introduced above. The upper bounds in (b)-(d) were
provedinProposition 2.7.
(a)followsfromProposition 2.9.
(b)Theexistence ofacongruence followsfromCorollary 2.8.
(c)Incase(i),byProposition2.10(b)wecanchooseα,β ∈ ZcongruentmoduloℓwithP(α) = 0
and β ∈ Z with Q(β) = 0 such that v (s(α)) = 0 or v (r(β)) = 0. Without loss of generality
M M
(afterpossiblyexchangingtherolesof(P,r)and(Q,s))wemayassumetheformercase. Incase(ii),
by Proposition 2.10 (a) anyα ∈ Zwith P(α) = 0and π (Q(α)) = 0 willsatisfy v (s(α)) = 0. In
1 m
bothcases, fromProposition 2.11andEquation2.2weobtaintheequality
v (c) = v (ℓn) = v (α−β ),
M M M 1
where β comes from Proposition 2.11. This gives the desired result. Case (iii) is just case (ii) with
1
therolesof(P,r)and(Q,s)interchanged.
(d)alsofollowsfromPropositions 2.10and2.11andEquation2.2. Moreprecisely, incase(i)we
havetheinequality
v (c) en n n
M
v (α−β ) ≥ ⌈ ⌉ = ⌈ ⌉ ≥ ⌈ ⌉−1 e+1 = γ (⌈ ⌉),
M 1 deg(Q) deg(Q) deg(Q) M/Qℓ deg(Q)
(cid:0) (cid:1)
whereeistheramificationindexofM/Q . Hence,π (α−β ) = 0withm = ⌈ n ⌉. Case(ii)is
ℓ m 1 deg(Q)
case(i)withtherolesof(P,r)and(Q,s)interchanged. 2
Remark2.13 It is straightforward to turn Corollary 2.12 into an algorithm. Say, P,Q ∈ Z[X] are
coprimemonicpolynomials. Firstwecomputethecongruencenumbersc(P,P′)andc(Q,Q′). Ifany
of these iszero, then wefactor P (respectively, Q)in Z[X]into irreducible polynomials P = P
i i
Q
10