1Introduction. identicallywhenthesurfacehasbasepoints{thatis,parametervalues(s0;t0)forwhich

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1 OntheValidityofImplicitizationbyMovingQuadrics forrationalsurfaceswithnobasepoints DepartmentofMathematicsandComputerScience Amherst,MA,01002 AmherstCollege DavidCox DepartmentofComputerScience Houston,TX77025 RiceUniversity RonGoldman DepartmentofComputerScience Houston,TX77025 RiceUniversity MingZhang establishsucientpolynomialconditionsforthevalidityofimplicitizationbythe methodofmovingquadricsbothforrectangulartensorproductsurfacesofbi-degree (m;n)andfortriangularsurfacesoftotaldegreenintheabsenceofbasepoints. Techniquesfromalgebraicgeometryandcommutativealgebraareadoptedto Abstract

2 1Introduction istocomputethebivariateresultantofthethreepolynomials: equationofarationalsurface[sederberg&chen1995].theclassicalmethodfornding theimplicitequationofarationalparametricsurface Severalyearsago,TomSederbergintroducedanewtechniqueforndingtheimplicit Unfortunatelyformanyapplications,theresultantofthesethreepolynomialsvanishes x(s;t)?xw(s;t);y(s;t)?yw(s;t);z(s;t)?zw(s;t): x=x(s;t) w(s;t);y=y(s;t) w(s;t);z=z(s;t) identicallywhenthesurfacehasbasepoints{thatis,parametervalues(s0;t0)forwhich ExtensiveexperimentsrevealthatSederberg'stechniqueisgenerallyimpervioustobase points.thussederberg'smethod{whichhecalledthemethodofmovingquadrics(see complicatedperturbationtechniques[manocha&canny1992].sederberg'smethodof Section3){promisedtobeasubstantialimprovementoverclassicalmethodsbasedonresultants,whichcouldimplicitizerationalsurfaceswithbasepointsonlybyinvokingrather x(s0;t0)=0;y(s0;t0)=0;z(s0;t0)=0;w(s0;t0)=0: iscorrectintheory.noristhereanysystematicanalysisofpreciselyunderwhatconditions themethodmightfailorhowtorecovergracefullywhenitdoes. movingquadricsfrequentlyworksinpractice,thereisnorigorousproofthatthemethod {andasanaddedbonusrepresentstheimplicitequationasthedeterminantofamatrix movingquadricsusesonlyelementarylinearalgebra{solvingasystemoflinearequations one-fourththesizeoftheclassicalresultant. Sederbergtondtheimplicitequationofarationalcurve,andherethetheoreticalanalysis ismuchmorecomplete.infact,threepapershavebeenpublishedthatdemonstratethe Acomparabletechnique{calledthemethodofmovingconics{wasalsointroducedby Nevertheless,althoughthereissubstantialempiricalevidencethatthemethodof validityofthismethod.resultantsareusedin[sederbergetal1997]toshowthatthe methodofmovingconicsworkswheneverthereisnomovinglineoflowdegreethat followsthecurve(seesection3).sincetheexistenceofsuchamovinglineoflowdegree Bothoftheseproofsinvokeresultants.AttemptsbyChionhandtwooftheauthorsof a?basiswasintroducedin[coxetal1998].bytakingtheresultantofthe?basis,the foralmostallrationalcurves.toextendthemethodtoallrationalcurves,thenotionof isrepresentedbyapolynomialcondition,thisresultestablishesthatthemethodworks authorswereabletoprovethatthemethodofmovingconicsworksforallrationalcurves. thispaper(goldmanandzhang)toextendtheseresultantproofstosurfacesfailed.the 2

3 problemseemstobethatunlikeintheunivariatesetting,thereisnobivariateresultant oftherightsizeorthecorrectdegreeinthepolynomialcoecientstomodelthematrix athirdapproachforestablishingthevalidityofthemethodofmovingconicsbasedon Inthatpapertheauthorsexpressedthehopethatunliketheprevioustwoproofs,this generatedbythemethodofmovingquadrics.thereforethesethreeauthorsproposed newapproachcouldbeextendedtoananalysisofthemethodofmovingquadrics. factoringthedeterminantofthecoecientmatrixofthelinearsystem[zhangetal1999]. surfacehasnobasepoints,thenthemethodofmovingquadricswillsucceedprovided thatthereisnomovingplaneoflowdegreethatfollowsthesurface(seesection3). condition,thisresultestablishesthatthemethodofmovingquadricsworksforalmostall Sincetheexistenceofsuchamovingplaneoflowdegreeisrepresentedbyapolynomial rationalsurfaceswithoutbasepoints.aproofforsurfaceswherebasepointsarepresent isstillanopenproblem. Thispaperistherststepinsuchananalysis.Hereweshowthatiftherational relationshipsbetweenpolynomialfunctions.thealgebraicanalysisofsuchrelationships requiressophisticatedtoolsfromalgebraicgeometryandcommutativealgebra,includingcohen-macaulayrings,r-sequences,koszulcomplexes,andsheafcohomology.we Butforsurfacesweneedtostudysyzygiesofbivariatepolynomials{thatis,polynomial Forcurves,theproofrequiresonlysomestandardlinearalgebra,genericpropertiesofresultants,andafewsimplefactsaboutfactoringunivariatepolynomials[Zhangetal1999]. Evenintherelativelysimplesettingofnobasepoints,theproofisnotelementary. wanttoskipthedetails.onlythemeaningandvalidityofthetwomainpropositionsare presentanoverviewoftheseideasinsection2;themainresultsherearepropositions1 brieythemethodofmovingquadricsfortensorproductsurfaces.insection4.1we and2.thereadernotcomfortablewiththesetechniquesfromalgebraicgeometrymay requiredinordertounderstandtheremainderofthispaper. Proposition2ofSection2,butotherwisetheproofrequiresonlystraightforwardlinear weshowthatwhentherearenobasepointsthemovingquadriccoecientmatrixisnonsingularonlywhenthemovingplanecoecientmatrixisnon-singular.hereweinvoke Indeed,fromhereonthedetailsshouldbeeasytofollow.InSection3wereview introducethemovingplaneandmovingquadriccoecientmatrices,andinsection4.2 algebra.wethenusethisresulttoshowthatwhenthemovingplanecoecientmatrix fromtensorproductsurfacesofbi-degree(m;n)totriangularsurfacesoftotaldegreen. methodofmovingquadricsisguaranteedtosucceed.insection5weextendtheseresults FinallyweclosethispaperinSection6withafewopenquestionsforfutureresearch. isnon-singularthemethodofmovingquadricsisvalidfortensorproductsurfaces.we concludethatwhenthereisnomovingplaneoflowdegreethatfollowsthesurface,the 3

4 propositionsprovedherewillplayanimportantroleintheproofofourmaintheorems Thissectionwilldiscusssomeinterestingalgebraconnectedwithbivariateresultants.The laterinthepaper. 2ResultantsandSyzygies 2.1TriangularPolynomials whichhasthefollowingwellknowngeometricinterpretation:resultant(x;y;z)6=0ifand oftotaldegreen.inthissituation,wehavethemulti-polynomialresultantresultant(x;y;z), Webeginwiththreetriangularpolynomials onlyifitisimpossibletondapairofparameters(s0;t0)suchthat x(s;t)=nxi=0n?i Xj=0ai;jsitj;y(s;t)=nXi=0n?i Xj=0bi;jsitj;z(s;t)=nXi=0n?i Xj=0ci;jsitj(1) Moreprecisely,thismeansthatifwehomogenizex;y;zbyaddingathirdvariable,then thehomogenizedequationshavenocommonsolutionsinprojectivespace. whatisthealgebraicinterpretationofresultant(x;y;z)6=0?theanswerisgivenbythe followingproposition. Thisexplainsthegeometryoftheresultant,butwhatisthealgebra?Inotherwords, x(s0;t0)=y(s0;t0)=z(s0;t0)=0: (2) 0.Then,wheneverwehavepolynomialsA;B;C2C[s;t]satisfying Proposition1:Supposethatx;y;zaredenedasinEquation(1)andthatResultant(x;y;z)6= therearepolynomialsh1;h2;h32c[s;t]suchthat B=?h2x+h3z; C=?h1x?h3y: A=h1z+h2y; Ax+By+Cz=0; Furthermore,ifkisthemaximumdegreeofA;B;C,thenh1;h2;h3canbechosensothat theyhavedegreeatmostk?n. AnequationAx+By+Cz=0iscalledasyzygyonx;y;z.Ifwethinkofasyzygyasa Beforebeginningtheproof,let'smakesomecommentsonwhatthispropositionsays. 4

5 columnvector(a;b;c)t,thenwehavethreeobvioussyzygies: andthenaddingthemtogether,wegetthesyzygy Furthermore,multiplyingtherstofthesebyh1,thesecondbyh2,andthethirdbyh3, (y;?x;0)tcomingfromyx+(?x)y+0z=0; (z;0;?x)tcomingfromzx+0y+(?x)z=0; Thepropositiontellsusthatwhentheresultantdoesn'tvanish,allsyzygiesonx;y;zare h1(z;0;?x)t+h2(y;?x;0)t+h3(0;z;?y)t=(h1z+h2y;?h2x+h3z;?h1x?h3y)t: (0;z;?y)Tcomingfrom0x+zy+(?y)z=0: ProofofProposition1:Sincex;y;zhavenocommonzeros,theNullstellensatzimplies thattheygeneratetheunitidealinc[s;t].therstpartofthepropositionthenfollows generatedfromtheobviousonesinthisway. thematrices homogeneouspolynomialsofdegreeninthepolynomialringr=c[s;t;u].nowconsider fromtheargumentoflemma1ofsection2of[2].however,gettingthedegreebound willrequiremorework. Byhomogenizingwithrespecttoathirdvariableu,thepolynomialsx;y;zgenerate M0=?xyz NotethattheproductsM1M2andM0M1arezero.>FromM0;M1;M2,wegetmaps 0?xz 0?!RM2?!R3M1?!R3M0?!R (4) (3) calledthekoszulcomplexofx;y;z. providedweregardelementsofr3ascolumnvectorsofpolynomials.thissequenceis (A;B;C)T2R3,wehave thata;b;candh1;h2;h3arealsohomogeneous,whicheasilyimpliesthedesireddegree NotethatexactnessbetweenM1andM0impliesthatAx+By+Cz=0ifandonlyif A;B;Cisofthedesiredform.Furthermore,sincex;y;zarehomogeneous,wecanassume WesaythattheKoszulcomplexisexactbetweenM1andM0ifforeveryv= bound. M0v=0()v=M1wforsomew=(h1;h2;h2)T2R3: 5

6 algebraimplythatthekoszulcomplexisexact. zerosetofx;y;zhasco-dimension3inc3.fromhere,standardfactsincommutative ofx(s;t;u)=y(s;t;u)=z(s;t;u)=0inc3is(s;t;u)=(0;0;0).inotherwords,the exact,meaningitisexactbetweeneverytwomaps(includingbetween0andm2).with thisgoalinmind,werstnotethatresultant(x;y;z)6=0impliesthattheonlysolution Unfortunately,thedetailsrequireCohen-Macaulayrings,R-sequencesanddepth,all Itremainstoproveexactness.Theideaistoprovethattheentiresequence(4)is ofwhichrequiresubstantialexplanation.forreadersfamiliarwiththeseconcepts,the proofconsistsofthefollowingsteps: Sincex;y;zisanR-sequence,thesequence(4)isexact(Theorem16.5of[8]). Sincedepth(I)=3,x;y;zisanR-sequence(CorollarytoTheorem16.8of[8]). x;y;z2rgenerateanidealiwheredepth(i)=codim(i)(theorem18.7of[4]). codim(i)=3since,asnotedabove,thezerosetofx;y;zhasco-dimension3inc3. R=C[s;t;u]isaCohen-Macaulayring(Proposition18.9of[4]). Thiscompletestheproofoftheproposition. Wenowturnourattentiontotensorproductpolynomials.Considerthepolynomials 2.2TensorProductPolynomials theirzerosetincnhasco-dimensionk,thenthekoszulcomplexoff1;:::;fkisexact. Thisresultistruemoregenerally:iff1;:::;fkarepolynomialsint1;:::;tnsuchthat ofbi-degree(m;n).forthesepolynomials,thedixonresultantresultant(x;y;z)is nonzeroifandonlyifitisimpossibletond(s0;t0)suchthat Intheprojectiveplane,theseequationsalwayshavem2solutionsats=1andn2 x(s;t)=mxi=0nxj=0ai;jsitj;y(s;t)=mxi=0nxj=0bi;jsitj;z(s;t)=mxi=0nxj=0ci;jsitj(5) thenanysyzygy(a;b;c)tonx;y;zwouldhavetheform solutionsatt=1.sowhenwesaythat(6)hasnosolutions,wemeanthatthereareno additionalsolutions. Inthissituation,theanalogofProposition1wouldstatethatifResultant(x;y;z)6=0, x(s0;t0)=y(s0;t0)=z(s0;t0)=0: (A;B;C)T=h1(z;0;?x)T+h2(y;?x;0)T+h3(0;z;?y)T; 6

7 asinproposition1,andfurthermore,ifa;b;chadbi-degreeatmost(k;l),thenh1;h2;h3 couldbechosentohavebi-degreeatmost(k?m;l?n).unfortunately,thedegreebound polynomialsdonotvanishsimultaneously,sotheirdixonresultantisnon-vanishing.now considerthesyzygy canfailinthetensorproductcase.hereisasimpleexampletoshowwhatcangowrong. Example1:Letx=st,y=st+s+tandz=st+1.Oneeasilychecksthatthese Thepolynomialsx;y;zgeneratetheunitidealinC[s;t],sothatasabove,Lemma1of Section2of[2]implies forsomeh1;h2;h3.however,(?s2+s+1;?s;s2)thasbi-degree(2;0)andx;y;zhave (?s2+s+1;?s;s2)t=h1(z;0;?x)t+h2(y;?x;0)t+h3(0;z;?y)t (?s2+s+1)st+(?s)(st+s+t)+s2(st+1)=0: bi-degree(1;1),sothath1;h2;h3cannothavebi-degree(2;0)?(1;1)=(1;?1). However,thecrucialobservationisthatitdoesholdforcertainspecialbi-degrees,which isasyzygyonx;y;zofbi-degree(2m?1;2n?1).thentherearepolynomials(h1;h2;h3) suchthatthedixonresultantresultant(x;y;z)isnonzero.alsoassumethat(a;b;c)t ofbi-degreeatmost(m?1;n?1)suchthat areexactlytheoneswewilluselaterinthepaper.hereisthepreciseresultwewillneed. Proposition2:Supposethatx;y;zaretensorproductpolynomialsofbi-degree(m;n) ThisexampleshowsthatProposition1doesnotholdinthetensorproductcase. Proof:AsinProposition1,weneedtohomogenize,butherewehomogenizesandt separatelyusingnewvariablesuandv.thusx=x(s;t)becomes B=?h2x+h3z; C=?h1x?h3y: A=h1z+h2y; havebi-degree(0;1). ofbi-degree(m;n)since,asindicatedbythesemicolon,s;uhavebi-degree(1;0)andt;v andsimilarlyforyandz.hencex;y;zarebi-homogeneouspolynomialsins=c[s;u;t;v] IfP1istheprojectiveline,then(s;u;t;v)arehomogeneouscoordinatesforapointin x(s;u;t;v)=mxi=0nxj=0aijsium?itjvn?j; P1P1provided(s;u)6=(0;0)and(t;v)6=(0;0).Notealsothat (s;u;t;v)and(s;u;t;v) 7

8 givethesamepointinp1p1foranynonzero;2c. andtheintrinsicmeaningofresultant(x;y;z)6=0isthattheequationsx=y=z=0 havenosolutionsinp1p1. Proposition1wouldimplythattheKoszulcomplexofx;y;zwouldbeexact.Butsince inproposition1nolongerapply.wehavepolynomialsx;y;z2c[s;u;t;v].ifthezero setofx=y=z=0inc4hadco-dimension3,thentheremarkfollowingtheproofof Foranybi-homogeneousf2C[s;u;t;v],theequationf=0makessenseinP1P1, xisbi-homogeneousins;u;t;v,itvanisheswhenevers=u=0.thesameistruefory andz,sothattheplanes=u=0inc4liesinthezerosetofx;y;z.itfollowsthatthe zerosetcannothaveco-dimension3.henceweneedsomenewideas. Beforegivingtheproofoftheproposition,weshouldexplainwhythetechniquesused thewholering(aswedidinproposition1).letsk;ldenotethesetofallbi-homogeneous polynomialsofbi-degree(k;l).algebraicgeometryhastoolsforstudyingthesepieces, thoughitrequiresthelanguageofsheavesandsheafcohomology.standardreferencesare [5]and[6].WhatfollowsisabrieftutorialonhowsheaftheoryprovesProposition2. OnewaytoapproachtheproblemistoconcentrateonpiecesoftheringSratherthan alsou=t=1,s=v=1ands=t=1.thesecorrespondtofoursubsetsofp1p1, arefourwaystodehomogenizeit.thestandardwayistosetu=v=1,butthereare eachofwhichlookslikeacopyofc2,thoughwithdierentvariablesineachcase.we canwritethesesubsetsandthepolynomialfunctionsonthemasfollows: Webeginbynotingthatgivenabi-homogeneouspolynomialf2C[s;u;t;v],there inp1p1as(s;1;1;t?1).comparingthisto(s;1;1;v),weobtainv=t?1. ToseewhyweidentifyC[s;v]andC[s;t?1]inU2,notethat(s;1;t;1)givesthesamepoint U1=f(s;1;t;1)2P1P1g;andfunctionsonU1areC[s;t]; U2=f(s;1;1;v)2P1P1g;andfunctionsonU2areC[s;v]=C[s;t?1]; Inasense,thefourpolynomialringsandtherelationsbetweenthemgivenin(7) U3=f(1;u;t;1)2P1P1g;andfunctionsonU3areC[u;t]=C[s?1;t]; U4=f(1;u;1;v)2P1P1g;andfunctionsonU4areC[u;v]=C[s?1;t?1]:(7) tellusaboutthepolynomialfunctionsonp1p1.theremarkablefactisthatthereis onemathematicalobject,thestructuresheafo=op1p1,whichkeepstrackofallof thissimultaneously.intermsofthesheafo,(7)givesthesectionsofooverthesets U1;U2;U3;U4.OnealsohastheglobalsectionsofO,denotedby BesidesthesheafO,therearealsosheavesO(k;l)=OP1P1(k;l),whichhavetheproperty ThesearetheglobalpolynomialfunctionsonP1P1,whicharejusttheconstantsC. H0(P1P1;O): 8

9 thattheirglobalsectionsarethespacessk;lofbi-homogeneouspolynomialsofbi-degree (k;l).inotherwords, H0(P1P1;O(k;l))=Sk;l: (8) AcarefuldenitionofthesesheavescanbefoundinExercise5.6of[6,Ch.III](notethat whathartshornecallsqisthesameasp1p1byexercise2.15of[6,ch.i]). Forourpurposes,theimportantfactisthatthematricesM0;M1;M2from(3)give mapsofsheaves 0?!O(?1;?1)M2?!O(m?1;n?1)3M1?! O(2m?1;2n?1)3M0?!O(3m?1;3n?1)?!0: (9) Furthermore,thissequenceisexactbetweeneverypairofmaps,includingthezeromaps atthetwoends.toseewhythisisso,werstworkoveru1,wherethesequencebecomes 0?!C[s;t]M2?!C[s;t]3M1?!C[s;t]3M0?!C[s;t]?!0: ThissequenceistheKoszulcomplexofthedehomogenizedx;y;z.Thesepolynomials generatetheunitidealofc[s;t]sincetheydon'tvanishsimultaneously,andthekoszul complexofsuchpolynomialsisalwaysexact.anelementaryproofofthiscanbegiven alongthelinesoflemma1ofsection2of[2].ageneralproofcanbefoundin[4, Prop.17.14]. Similarly,oneseesthat(9)becomesexactoverU2,U3andU4sincethedehomogenized polynomialshavenocommonroots.sincep1p1istheunionofu1;u2;u3;u4,itfollows that(9)isexact. Wenextexplorehowtheexactnessof(9)relatestoglobalsections.Thisiswheresheaf cohomologycomesin.aintroductiontosheavesandsheafcohomologycanbefoundin [5,pp.34{41].TherststepistoemploythestrategyexplainedinExercise7of[1, Ch.6,x1]ofbreakingup(9)intotwoshorterexactsequences: 0?!O(?1;?1)M2?!O(m?1;n?1)3?!K?!0; 0?!K?!O(2m?1;2n?1)3M0?!O(3m?1;3n?1)?!0 suchthat=m1.eachoftheseshortexactsequencesgivesalongexactsequencein sheafcohomologywhich,using(8),canbewritten?!(sm?1;n?1)3?!h0(p1p1;k)?!h1(p1p1;o(?1;?1))?!;(10) 0?!H0(P1P1;K)?!(S2m?1;2n?1)3M0?!S3m?1;3n?1?!: (11) Below,wewilldiscussthemeaningofH1(P1P1;O(?1;?1)).Fornow,letusassume thatthissheafcohomologygroupvanishes.thishasthefollowingniceconsequence. 9

10 impliesthatv=(u)forsomeu2h0(p1p1;k).sinceh1(p1p1;o(?1;?1))=f0g, Supposethatv=(A;B;C)Tisasyzygyonx;y;z.ThenM0v=0,whichby(11) forsomew=(h1;h2;h3)t2(sm?1;n?1)3.then,usingm1=,weconcludethat ThisgivesthedesiredformulasforA;B;C,andsinceh1;h2;h3havebi-degree(m?1;n?1), itfollowsthat(u)=0in(10).sincethissequenceisalsoexact,wemusthaveu=(w) thepropositionfollows. onelearnshowtocomputethesheafcohomologyonpr,itisnothardtoshowthatthis particularsheafcohomologygroupvanishes.forexample,thisisdoneinpart(a)(2)of However,westillneedtoshowH1(P1P1;O(?1;?1))=f0g.Fortunately,once (A;B;C)T=v=(u)=((w))=M1w=M1(h1;h2;h3)T: Exercise5.6of[6,Ch.III].ThiscompletestheproofofProposition2. allsyzygiesonx;y;zofbi-degree(2m?1;2n?1).ifweletudenoteaparticularsyzygy, [5,pp.34{41]doesagoodjobofexplaininghowelementsinH1(X;F)canbeinterpreted thenwecandehomogenizeuinfourways,correspondingtothesetsu1;u2;u3;u4in(7). asobstructionstopatchingtogetherlocaldataintoaglobalobject. thoughneitherisparticularlyintuitiveatrstglance.thepreviouslymentionedreference Toseehowthisappliestooursituation,notethat(11)showsthatH0(P1P1;K)gives Sheavesandtheircohomologyareanimportantpartofmodernalgebraicgeometry, giveabi-homogeneous(h1;h2;h3)t?theobstructiontodoingsoistheelement inthefourringslistedin(7).butcanthesefourrepresentationsbepatchedtogetherto Since(9)isexact,wethengetfourrepresentationsofuintermsofpolynomialsh1;h2;h3 existenceoftherequiredbi-homogeneous(h1;h2;h3)t. groupvanishes,sodoeseveryobstruction(u),andthentheexactnessof(10)impliesthe sothatwecouldn'tndabi-homogeneous(h1;h2;h3)t.butsincetheentirecohomology whereisthemapin(10).if(u)werenonzero,theobstructionwouldbenon-vanishing, (u)2h1(p1p1;o(?1;?1)); thatis,rationalparametricsurfacesx(s;t) 3TheMethodofMovingSurfaces NowletusreviewbrieySederberg'smethodofmovingsurfaceswithparticularemphasis onmovingplanesandmovingquadrics.herewefocusonrationaltensorproductsurfaces w(s;t);y(s;t) 10w(s;t);z(s;t) w(s;t),where

11 Triangularsurfaces thatis,surfacesoftotaldegreen willbediscussedinsection5. x(s;t)=mxi=0nxj=0ai;jsitj; z(s;t)=mxi=0nxj=0ci;jsitj; y(s;t)=mxi=0nxj=0bi;jsitj; Wewillonlyconsidersurfaceswithoutbasepoints.Abasepointforsurface(12)isa w(s;t)=mxi=0nxj=0di;jsitj: (12) pairofparameters(s0;t0)suchthat Itiswellknownthatinprojectivespacethebi-degreepatch(12)alwayshasm2base pointsats=1andn2basepointsatt=1.sowhenwesaysurface(12)hasnobase points,wemeanthatthesurfacedoesnothaveanyadditionalbasepoints. Amovingplaneofbi-degree(1;2)isanimplicitequationoftheform x(s0;t0)=y(s0;t0)=z(s0;t0)=w(s0;t0)=0: Xi=02 Similarly,amovingquadricofbi-degree(1;2)isanimplicitequationoftheform Foreachxedvalueofsandt,Equation(13)istheimplicitequationofaplaneinC3. Xj=0(Ai;jx2+Bi;jy2+Ci;jz2+Di;jxy+Ei;jxz+Fi;jyz+Gi;jxw+Hi;jyw+Ii;jzw+Ji;jw2)sitj=0: Xi=02 Xj=0(Ai;jx+Bi;jy+Ci;jz+Di;jw)sitj=0: Again,whens;tarexed,Equation(14)istheimplicitequationofaquadricinC3. Themovingplane(13)ormovingquadric(14)issaidtofollowsurface(12)if Xi=02 1Xj=0(Ai;jx(s;t)+Bi;jy(s;t)+Ci;jz(s;t)+Di;jw(s;t))sitj0; (15) (14) ByequatingthecoecientsofallthemonomialssitjinEquations(15)or(16)tozero, or weobtainasystemoflinearequationsintheindeterminatesfai;j;bi;j;ci;j;di;jgor Xi=02 1Xj=0(Ai;jx2(s;t)+Bi;jy2(s;t)++Ji;jw2(s;t))sitj0: 11 (16)

12 fai;j;bi;j;ci;j;di;j;ei;j;fi;j;gi;j;hi;j;ii;j;ji;jg,0i1,0j2.solvingthis systemgivesusacollectionofmovingplanesormovingquadricsthatfollowsurface(12). Forcertainspecialvaluesof1;2,itturnsoutthatwecannd(1+1)(2+1)linearly independentmovingplanesormovingquadrics.themethodofmovingplanes(ormoving quadrics)thenconstructstheimplicitequationoftheparametricsurface(12)bytaking thedeterminantofthecoecientmatrixofthe(1+1)(2+1)linearlyindependent movingplanes(ormovingquadrics)thatfollowtheparametricsurface(12). Specically,formovingplanes,wecanchoose1=2m?1,2=n?1.Then fromequation(15)weobtainahomogeneouslinearsystemof6mnequationswith8mn unknowns.thissystemhasatleast2mnlinearlyindependentsolutions: L12m?1 Xi=0n?1 Xj=0(A1i;jx+B1i;jy+C1i;jz+D1i;jw)sitj=0;. L2mn2m?1 Xi=0n?1 Xj=0(A2mn i;jx+b2mn i;jy+c2mn i;jz+d2mn i;jw)sitj=0: Eachofthesesolutionsisamovingplanethatfollowssurface(12).Thedeterminantof thecoecientsofsitjofthese2mnmovingplanes,i.e. A10;0x+B10;0y+C10;0z+D10;0wA12m?1;n?1x++D12m?1;n?1w A2mn 0;0x+B2mn 0;0y+C2mn 0;0z+D2mn 0;0wA2mn 2m?1;n?1x++D2mn 2m?1;n?1w vanisheswhenever(x;y;z;w)liesonthesurface.henceifthisdeterminantdoesnot vanishidentically,thenthisdeterminantisamultipleoftheimplicitequationofsurface (12).Moreover,thedegreeinx;y;z;wofthisdeterminantis2mn,whichisthegeneric degreeofsurface(12).therefore,thisdeterminantisagoodcandidatefortheimplicit equationofsurface(12). Formovingquadrics,wecanchoose1=m?1,2=n?1.Thenweobtaina homogeneouslinearsystemof9mnequationswith10mnunknownsfromequation(16). Thissystemhasatleastmnlinearlyindependentsolutions Q1m?1 Xi=0n?1 Xj=0(A1i;jx2+B1i;jy2++J1i;jw2)sitj=0;. Qmnm?1 Xi=0n?1 Xj=0(Amn i;jx2+bmn i;jy2++jmn i;jw2)sitj=0: 12

13 Eachsolutionisamovingquadricfollowingsurface(12).Thedeterminant theimplicitequationofsurface(12)becausethisdeterminantvanisheswhenthepoint (x;y;z;w)liesonthesurfaceandthedegreeofthisdeterminantis2mninx;y;z;w. ofthecoecientsofsitjofthesemnmovingquadricsisagainagoodcandidatefor A10;0x2+B10;0y2++J10;0w2A1m?1;n?1x2++J1m?1;n?1w2 Example2(MovingPlanes):x(s;t)=st+1; Amn 0;0x2+Bmn 0;0y2++Jmn. 0;0w2Amn. m?1;n?1x2++jmn. m?1;n?1w2 Itiseasytondtwomovingplanesfollowing(17): w(s;t)=s+t+1: z(s;t)= y(s;t)= (w?x?2y?z)+s(w?y)=0: (w?x?y?z)+sz=0; t; (17) Thedeterminant Example3(MovingQuadrics):x(s;t)=s2t2+1; givestheimplicitequationforsurface(17). w?x?y?zz w?x?2y?zw?y=w2?wx?2wy+xy+y2?2wz+xz+3yz+z2 Choose1=2=1.UsingMathematicawendthattherearefourmovingquadricsof bi-degree(1;1)thatfollowsurface(18): Q1??3wy+3y2?wz+3xz+2yz?4z2+s?wz?4yz?2z2 w(s;t)=s2+t2+1: z(s;t)=t+1; y(s;t)=s2+s+1; (18) Q2??3w2+6wx+20wy?3xy?19y2?9wz?12xz?4yz+24z2 +t??3wy+3y2+4z2+st?2z2=0; +s?3wx?2wy?3xy+2y2?18wz+3xz+34yz+6z2 13

14 Q3??2wy+3y2+xz?2z2+s?wy?y2+wz?xz?4yz Q4??5w2+9wx+34wy?3xy?33y2?11wz?21xz?12yz+42z2 +t?18wy?18y2+3wz?6xz?8yz?12z2 +st(?3xz+2yz)=0; +t??2wy+3y2+2z2+st?wy?y2=0; +s??w2+6wx?2wy?6xy+3y2?31wz+6xz+60yz+12z2 3xz+2yz?4z2?3wy+3y2?wz+ stofthesefourmovingquadrics: Theimplicitequationisobtainedbytakingthedeterminantofthecoecientsof1,s,t, +t?30wy?30y2+5wz?9xz?12yz?24z2+st(wz?6xz)=0:?3w2+6wx+20wy??2wy+3y2+xz?2z2wy?y2+wz?xz?4yz?2wy+3y2+2z2 3xy?19y2?9wz? 12xz?4yz+24z2 wz?4yz?2z2 3wx?2wy?3xy+2y2? 18wz+3xz+34yz+6z218wy?18y2+3wz??3wy+3y2+4z2 6xz?8yz?12z2 2z2?5w2+9wx+34wy? 3xy?33y2?11wz? wy?y2?3xz+2yz 21xz?12yz+42z2 (Resultant(x(s;t)?xw(s;t);y(s;t)?yw(s;t);z(s;t)?zw(s;t)))aremovingplanes metricsurfacehasnobasepoints.infact,therowsofthedixonresultantmatrices [Dixon1908].Themethodofmovingquadricshasbeenshowntoworkempirically,correctlyimplicitizingawidevarietyofsurfacesinmanyexperiments[Sederberg&ChenantswithsizeonlyaquarterofthestandardDixonresultant.Forinstance,toimplicitize 1995].Themovingquadricmethodproducestheimplicitequationintermsofdetermi- atensorproductsurface(12)ofbi-degree(m;n),thedixonresultantcomputesadeterminantofsize2mn2mn,whilethemovingquadricmethodcomputesadeterminant ofsizemnmnorevensmaller.themovingquadricmethodalsooftenworksinthe presenceofbasepoints[sederberg&chen1995]wherethedixonresultantfails. purposeofthispaperistoprovideasucientconditionwhichguaranteesthatthemethod ofmovingquadricssucceeds. Exactlywhenthemethodofmovingquadricsworksisstillanunsolvedproblem.The Themethodofmovingplanesalwaysgeneratestheimplicitequationifthepara- 6xy+3y2?31wz+?w2+6wx?2wy? 6xz+60yz+12z2 9xz?12yz?24z2 30wy?30y2+5wz? wz?6xz 14

15 4.1MovingPlaneandMovingQuadricCoecientMatrices 4ImplicitizingTensorProductSurfacesbythe Considertherationaltensorproductsurface(12).LetMPbethecoecientmatrixof thepolynomialssitjx;sitjy;sitjz;sitjw,0im?1,0jn?1.thatis, MethodofMovingQuadrics thelinearsystemgeneratedbythemovingplanesofbi-degree(m?1;n?1)thatfollow ThenMPisasquarematrixoforder4mn.NotethatMPisthecoecientmatrixof therationalsurface(12). =[1t2n?1s2m?1s2m?1t2n?1]MP xyzwsm?1tn?1xsm?1tn?1ysm?1tn?1zsm?1tn?1w 0im?1,0jn?1.Thatis, ThenMQisamatrixofsize9mn10mn.Moreover,MQisthecoecientmatrix Similarly,letMQbethecoecientmatrixofthepolynomialssitjx2;sitjy2;;sitjw2, ofthelinearsystemgeneratedbythemovingquadricsofbi-degree(m?1;n?1)that followsurface(12).letmqwbethesubmatrixofmqwiththecoecientsofsitjw2 =[1t3n?1s3m?1s3m?1t3n?1]MQ: x2y2z2xyxzyzxwywzww2sm?1tn?1(x2w2) BelowweshallshowthattheconverseisalsotruewhenResultant(x;y;z)6=0. deleted;mqwisthenasquarematrixoforder9mn.whenmqwisnon-singular,the linearsystemofmovingquadricsofbi-degree(m?1;n?1)hasexactlymnlinearly pendenciesonthecolumnsofmpgeneratelineardependenciesonthecolumnsofmqw. independentsolutions. 4.2TheValidityoftheMethodofMovingQuadricsforTensor ItiseasytoseethatwhenjMPjvanishes,jMQwjalsovanishes,becauselinearde- Ifsurface(12)hasnobasepoints,thenforthepurposeofimplicitization,wecanassume thatresultant(x;y;z)6=0.thereasonisasfollows:ifresultant(x;y;z)=0,then x;y;zhaveacommonroot(eitheraneoratinnity);sincewdoesnotvanishatthis orinnite)forsomeconstantc.thusresultant(x+cw;y;z)6=0.theeectofsuch commonroot,thepolynomialsx+cw;y;zdonothaveacommonroot(eithernite ProductSurfaces atransformationisjustasimpletranslationofsurface(12).itiseasytoseethatnding 15

16 theimplicitequationoftheoriginalsurfaceisequivalenttondingtheimplicitequation non-zeroleadingterm,i.e.thecoecientofsmtnisnotzero.suppose,withoutlossof oftheshiftedsurface. generality,thatx(s;t)hasanon-zeroleadingcoecient;then allhavenon-zeroleadingcoecients.again,thesetransformationsonlyinducesimple translationsoftheoriginalsurface. y(s;t)+constantx(s;t);z(s;t)+constantx(s;t);w(s;t)+constantx(s;t) Moreover,ifResultant(x;y;z)6=0,thenatleastoneofx(s;t);y(s;t);z(s;t)hasa Proof:IfjMQwj=0,thenthecolumnsofMQwarelinearlydependent.Thusthereexist impliesjmpj=0. Theorem1:Forthetensorproductpatch(12),ifResultant(x;y;z)6=0,thenjMQwj=0 9mnscalarssuchthatthelinearcombinationofthe9mncolumnsofMQwwiththese andthatx(s;t);y(s;t);z(s;t);w(s;t)allhavenon-zeroleadingcoecients. Thereforebelow,wewillassume,withoutlossofgenerality,thatResultant(x;y;z)6=0 9mnscalarsisidenticallyzero.Wecanwritethislinearcombinationas ofthepolynomials: Eachpolynomialpi(s;t)isofbi-degree(m?1;n?1),sinceMQwconsistsofthecoecients whereeachofthepolynomialspi(s;t)consistsofthecoecientsofoneofx2;y2;;zw. x2;y2;z2;xy;xz;yz;xw;yw;zw;;sm?1tn?1(x2;y2;z2;xy;xz;yz;xw;yw;zw): p1x2+p2y2+p3z2+p4xy+p5xz+p6yz+p7xw+p8yw+p9zw0; (19) Ifp3(s;t)60orp9(s;t)60,wewanttoprovethat RewriteEquation(19)as thesepolynomialsash1andh3willsoonbecomeclear).toprovethis,notethatinthe forsomebi-degree(m?1;n?1)polynomialsh1(s;t);h3(s;t)(thereasonforlabeling (p1x+p4y+p5z+p7w)x+(p2y+p6z+p8w)y+(p3z+p9w)z0: p3z+p9w+h1x+h3y0; (20) terminologyofsection2,equation(20)isthesyzygyonx;y;zgivenby Furthermore,sinceeachpi(s;t)hasbi-degree(m?1;n?1),thissyzygyhasbi-degree therearepolynomialsh1;h3ofbi-degree(m?1;n?1)suchthat (2m?1;2n?1).SinceResultant(x;y;z)6=0,Proposition2ofSection2impliesthat (A;B;C)T=(p1x+p4y+p5z+p7w;p2y+p6z+p8w;p3z+p9w)T: C=?h1x?h3y: 16

17 Hence whichprovesthatthecolumnsofmparelinearlydependent.thereforejmpj=0. SinceResultant(x;y;z)6=0,x;ycannothaveanycommonfactor.Otherwise,x;y;z Ifp3(s;t)p9(s;t)0,thenEquation(20)becomes 0C+h1x+h3y=p3z+p9w+h1x+h3y; wouldhavecommonrootsotherthanthem2+n2rootsats=1andt=1andhence mustbeafactorofp2y+p6z+p8w.weexaminethefollowingtwocases: Resultant(x;y;z)=0contrarytoassumption.Therefore,ifp2y+p6z+p8w60,thenx Ifp260,orp660,orp860,thenEquation(21)impliesthat (p1x+p4y+p5z+p7w)x+(p2y+p6z+p8w)y0: (21) Otherwise,p2p6p80.ThenbyEquation(21), (m?1;n?1)ins;t. forsomepolynomialh(s;t).butbyassumption,xhasnon-zeroleadingcoecient, columnsofmparelinearlydependent,sincep2;p6;p8;hareallofbi-degree soitiseasytoseethath(s;t)isofbi-degree(m?1;n?1).itfollowsthatthe p2y+p6z+p8w=h(s;t)x; (m?1;n?1)followingsurface(12),wehavethefollowing SincejMPj6=0isequivalenttothefactthattherearenomovingplanesofbi-degree ofmparelinearlyindependent,sojmpj=0. wherep1;p4;p5;p7arenotallzeropolynomials.thisagainprovesthatthecolumns p1x+p4y+p5z+p7w=0; quadricscomputestheimplicitequationofsurface(12). planesofbi-degree(m?1;n?1)followingsurface(12).thenthemethodofmoving Theorem2:Supposethatsurface(12)hasnobasepoints,andthattherearenomoving Proof:AssumewithoutoflossofgeneralitythatResultant(x;y;z)6=0.Abi-degree (m?1;n?1)movingquadrichastheform Themovingquadric(22)followssurface(12)ifandonlyif m?1 Xi=0n?1 Xj=0?Ai;jx2(s;t)+Bi;jy2(s;t)+Ci;jz2(s;t)++Ji;jw2(s;t)sitj0:(23) Xi=0n?1 Xj=0?Ai;jx2+Bi;jy2+Ci;jz2+Di;jxy++Ji;jw2sitj=0: (22) 17

18 Thus(A0;0;;J0;0;;Am?1;n?1;;Jm?1;n?1)Tisasolutiontothelinearsystem RewriteEquation(24)as MQw0 MQ(A0;0;;J0;0;;Am?1;n?1;;Jm?1;n?1)T=0: (24) Am?1;n?1 A I0;0. Im?1;n?1 1 CA=?Coe(w2;;sm?1tn?1w2)0@J0;0 0im?1,0jn?1.Thatis,movealltermsinvolvingJi;jonthelefthandside wherecoe(w2;;sm?1tn?1w2)consistsofthecoecientsofthepolynomialssitjw2, Jm?1;n?11A;. (25) (J0;0;;Jm?1;n?1)tothestandardunitvectorsek=(0;;0;1;0;;0),0k fromequation(25).thuswewillgetmnlinearlyindependentsolutionswhenweset isalsonon-singular.therefore,wecansolvefor(a0;0;;i0;0;;am?1;n?1;;im?1;n?1) mn?1. followingsurface(12).sinceresultant(x;y;z)6=0,weknowfromtheorem1thatmqw inequation(24)totherighthandsideofequation(25). Byassumption,jMPj6=0becausetherearenomovingplanesofbi-degree(m?1;n?1) whereq;correspondstosetting Q;=m?1 Nowmnmovingquadricscanbeconstructedfromthesemnsolutions: ThusQ;=w2st+termswithoutw2;0m?1;0n?1: Xi=0n?1 Xi=0A; i;jx2++j; (J0;0;;Jm?1;n?1)=en+: i;jw2sitj=0;0m?1;0n?1; ThereforethecoecientsofthemonomialsstfromthemovingquadricsQ;forman mnmnmatrixm=264w2+w2+...w ;

19 termw2.sincemcontainsw2onlyinthediagonalentries,jmjcontainsthetermw2mn. Sothisdeterminantisnotidenticallyzero.SinceeachentryinMisquadraticinx;y;z;w, Q;.Notethattheo-diagonalentriesarequadraticinx;y;z;wbutdonothavethe whereeachrowconsistsofthecoecientsofthemonomialssitjfromamovingquadric 0m?1,0n?1,representsamovingquadricthatfollowssurface(12),so thetotaldegreeofjmjis2mninx;y;z;w.moreover,byconstruction,eachrowq;, equationoftherationalsurface. forpointsonsurface(12).ontheotherhand,thedegreeofthe(irreducible)implicit 5ImplicitizingTriangularSurfacesbytheMethodof forpointsonthesurface,thecolumnsofmarelinearlydependent;hencejmjvanishes equationofsurface(12)is2mn[coxetal1998].therefore,jmjmustbetheimplicit surfacesoftotaldegreenwithnobasepoints. Inthissection,weestablishthevalidityofthemethodofmovingquadricsforimplicitizing Arationalsurfacex(s;t) MovingQuadrics x(s;t)=x w(s;t);y(s;t) i+jnai;jsitj; w(s;t);z(s;t) w(s;t)isoftotaldegreenif y(s;t)=x Justasinthetensorproductcase,weconsidermovingplanesandmovingquadricsthat z(s;t)=x i+jnci;jsitj; w(s;t)=x i+jnbi;jsitj; quadricsoftotaldegreen?1. followthistriangularsurface.hereweshallbeinterestedinmovingplanesandmoving i+jndi;jsitj: (26) Amovingplaneofdegreen?1 followssurface(26)ifitvanishesidenticallywhenthepolynomialsx(s;t),y(s;t),z(s;t), 5.1MovingPlaneandMovingQuadricCoecientMatrices w(s;t)aresubstitutedforx;y;z;winequation(27). i+jn?1(ai;jx+bi;jy+ci;jz+di;jw)sitj=0 X 19 (27)

20 i+jn?1.thatis, Notethatthenumberofmonomialssitj,wherei+jk,is?k+2 LetMPbethe(movingplane)coecientmatrixofthepolynomialssitj(x;y;z;w), =[1tst2n?1s2n?1]MP: [xyzwtn?1(xyzw)sn?1(xyzw)] and #columnsofmp=4n+1 #rowsofmp=2n+1 2=2n2+n: 2=2n2+2n 2.Therefore, parametricsurfaceoftotaldegreen. Thustherealwaysexistatleastnmovingplanesofdegreen?1thatfollowarational indiceswithjij=n.denempitobethecoecientmatrixofthepolynomials ToobtainasquaresubmatrixofMP,letIf(k;l)j0k+ln?1gbeasetof sitj(x;y;z);0i+jn?1; columnsandthesamenumberofrows.thereforempiisasquaresubmatrixofmpof has Thatis,toobtainMPI,weremovethencolumnssitjw,(i;j)2IfromMP.ThusMPI sitjw;(i;j)62i: order2n2+n. i+jn?1?ai;jx2+bi;jy2+ci;jz2+di;jxy+ei;jxz+fi;jyz+gi;jxw+hi;jyw+ii;jzw+ji;jw2sitj=0 XSimilarly,amovingquadricofdegreen?1 4n+1 2?n=2n2+n xy;xz;yz;xw;yw;zw;w2),i+jn?1.thatis, w(s;t)aresubstitutedforx;y;z;winequation(28). followssurface(26)ifitvanishesidenticallywhenthepolynomialsx(s;t),y(s;t),z(s;t), LetMQbethe(movingquadric)coecientmatrixofthepolynomialssitj(x2;y2;z2, x2y2w2tn?1(x2y2w2)sn?1(x2y2w2) (28) Then =1tst3n?1s3n?1MQ #columnsofmq=10n+1 202=10n2+10n 2

21 Thustherealwaysexistatleast(n2+7n)=2movingquadricsofdegreen?1thatfollow and #rowsofmq=3n+1 2=9n2+3n matrixoftheremainingpolynomials aparametricsurfaceoftotaldegreen. aswellasthe3ncolumnssitj(xw;yw;zw),(i;j)2i,anddenemqitobethecoecient TogetasquaresubmatrixofMQ,weremoveallthesitjw2columns,0i+jn?1, 2: columnsandthesamenumberofrows.therefore,mqiisasquaresubmatrixofmqof ThusMQIhas sitj(xw;yw;zw);(i;j)62i: sitj(x2;y2;z2;xy;xz;yz);0i+jn?1; order(9n2+3n)=2. 5.2TheMethodofMovingQuadricsforTriangularSurfaces 9n+1 2?3n=9n2+3n 2 movingquadrics.notice,however,thatthedegreeinx;y;z;wofthedeterminantofsuch matrixoforder(n2+n)=2,whoseentriesconsistofthecoecients(ofsitj)of(n2+n)=2 Thereare(n2+n)=2monomialssitj,0i+jn?1,intheexpressionforamoving insection4.2forrationaltensorproductsurfaces,wewouldneedtoconstructasquare triangularsurfacebythemethodofmovingquadricsinamannersimilartoourapproach an(n2+n)=2(n2+n)=2matrixisn2+nsinceeachentryisquadraticinx;y;z;w, quadricofdegreen?1.therefore,tocomputetheimplicitequationofadegreenrational movingplanesofdegreen?1thatfollowsurface(26),sincempisofsize(2n2+n) whereasthedegreeinx;y;z;woftheimplicitequationofsurface(26)isn2.therefore, (2n2+2n).Nowthis(n2+n)=2(n2+n)=2matrixconsistsofthecoecients(ofsitj) theentriesofthis(n2+n)=2(n2+n)=2matrixcannotallbethecoecientsofmoving movingplanes.fromsection5.1,weknowthattherealwaysexistnlinearlyindependent of(n2?n)=2movingquadricsandnmovingplanes.therefore,thedeterminantofthis squarematrixisofdegreen2=2(n2?n)=2+n1inx;y;z;w,whichisexactlywhat wedesire. Tolowerthedegreeofthisdeterminant,weshallreplacenmovingquadricsbyn planestomakesurethatthedeterminantisnotidenticallyzero.forexample,ifoneof Nevertheless,wemustbecarefulhowwechoosethesemovingquadricsandmoving 21

22 themovingplanesisp(x;y;z;w)=0andoneofthemovingquadricsisq(x;y;z;w) (26).Tillnowthesemovingplanesandmovingquadricshavebeenchoseninanadhoc xp(x;y;z;w)=0,thenthedeterminantwillvanishidentically.themethodofmoving quadricsassertsthat,ingeneral,itispossibletochoosethe(n2?n)=2movingquadrics manner.inthenextsection,wepresentasystematicwaytochoosetherightmoving 5.3TheValidityoftheMethodofMovingQuadricsforTriangularSurfaces quadricsandmovingplanes. andnmovingplanessothatthisdeterminantactuallyistheimplicitequationofsurface showhowtoexploitthisrelationshiptochoosetherightmovingquadricsandmoving Proceedinginamannersimilartoourapproachtotensorproductsurfaces,belowwerst explorearelationshipbetweenthetwocoecientmatricesmpiandmqi.thenweshall planestoimplicitizeatriangularsurface. ItiseasytoseethatifjMPIjvanishes,thenjMQIjalsovanishes,becauselinear impliesjmpij=0. Proof:SupposejMQIj=0.ThenasintheproofofTheorem1,thereexistpolynomials dependenciesonthecolumnsofmpigeneratelineardependenciesonthecolumnsof p1(s;t);;p9(s;t)suchthat Theorem3:Forthetriangularsurface(26),ifResultant(x;y;z)6=0,thenjMQIj=0 MQI.BelowweshallseethattheconverseisalsotruewhenResultant(x;y;z)6=0. wherep1;p2;p3;p4;p5;p6;p7;p8;p9areoftotaldegreen?1ins;t,buttheexponents ofthemonomialsinp7;p8;p9arenotintheindexseti. RewriteEquation(29)as p1x2+p2y2+p3z2+p4xy+p5xz+p6yz+p7xw+p8yw+p9zw=0; (p1x+p4y+p5z+p7w)x+(p2y+p6z+p8w)y+(p3z+p9w)z=0: (29) MPIarelinearlydependent;hencejMPIj=0. Sincetheexponentsofthemonomialsinp9arenotinI,itfollowsthatthecolumnsof Section2,thereexistpolynomialsh1;h3oftotaldegreen?1suchthat Ifp360orp960,thenEquation(30)isasyzygyonx;y;z.HencebyProposition1of Ifp3p90,butp260,orp660,orp860,thenasintheproofofTheorem1, p2y+p6z+p8w=h(s;t)x p3z+p9w=?h1x?h3y: 22

23 forsomepolynomialh(s;t).withoutlossofgenerality,wecanassumethatthetotal degreeofxisn;thereforeh(s;t)isoftotaldegreeatmostn?1.sincetheexponentsof themonomialsinp8arenotini,thisagainprovesthatthecolumnsofmpiarelinearly dependent,sojmpij=0. i.e.jmpij=0. whilep1;p4;p5;p7arenotallzeropolynomials.sincetheexponentsofthemonomialsin p7arenotini,theaboveequationshowsthatthecolumnsofmpiarelinearlydependent, Ifp3p90,andp2p6p80,thenbyEquation(30), IforwhichMPIisnon-singular. NextwearegoingtoshowthatifMPhasmaximalrank,thenthereisanindexset p1x+p4y+p5z+p7w=0; ofmparelinearlyindependent. Lemma1:Forthetriangularsurface(26),ifResultant(x;y;z)6=0,thenthecolumns polynomialsp1(s;t);p2(s;t);p3(s;t)oftotaldegreen?1ins;tsuchthat Proof:Supposethecolumnssitj(x;y;z)arelinearlydependent.Thenthereexistthree sitjx;sitjy;sitjz;0i+jn?1; SinceResultant(x;y;z)6=0,byProposition1ofSection2,theabovesyzygyimpliesthat p1x+p2y+p3z=0: forthreepolynomialsh1;h2;h3ofdegree?1.thush1=h2=h30.hencep1=p2= p30.therefore,thecolumnsp1=h1z+h2y; p2=?h2x+h3z; arelinearlyindependent. sitjx;sitjy;sitjz;0i+jn?1; p3=?h1x?h3z; Lemma2:IfResultant(x;y;z)6=0,andMPhasmaximalrank,thenthereexistsan indexseti,jij=n,suchthatmpiisnon-singular. Proof:ByLemma1,weknowthatsitj(x;y;z),0i+jn?1,arelinearlyindependent. Considerthefollowingalgorithm: 23

24 =fsitj(x;y;z)j0i+jn?1g; I=(emptyset);?=fsitjwj0i+jn?1g; IfMPhasmaximalrank2n2+n,then2n2+noutofthe2n2+2ncolumnsinMPare while?6= linearlyindependent.thatis,theabovealgorithmterminateswithjj=2n2+nand Otherwise,add(i;j)toI. Ifsitjwislinearlyindependentfromthecolumnsin,thenadd Selectacolumnsitjwfrom?,andremoveitfrom?; jij=n.nowbydenitionmpiconsistsofthecolumnsin,andbyconstructionthese columnsarelinearlyindependent.therefore,mpiisnon-singular. sitjwto; themethodofmovingquadricssuccessfullyimplicitizestriangularsurfaces. MQIisnon-singular.Thenexttheoremshowsthatthesetwoconditionsguaranteethat linearlyindependentmovingplanesofdegreen?1thatfollowsurface(26).thenthe Theorem4Supposethatsurface(26)hasnobasepointsandthatthereareexactlyn methodofmovingquadricscomputestheimplicitequationofsurface(26). FromTheorem3,itfollowsthatwhenMPIisnon-singularandResultant(x;y;z)6=0, jmpij6=0forsomeindexsetjij=n.hencebytheorem3,mqiisnon-singular. (26)hasexactlynlinearlyindependentmovingplanesofdegreen?1isequivalentto Proof:Since,byassumption,oursurfacehasnobasepoints,wecanassume,without lossofgenerality,thatresultant(x;y;z)6=0.bylemma2,theconditionthatsurface SincejMQIj6=0,wecansolveforAi;j,Bi;j,Ci;j,Di;j,Ei;j,Fi;j,i+jn?1andGi;j, Hi;j,Ii;j,(i;j)2I. Hi;j,Ii;j,(i;j)62I,intermsoftheundeterminedcoecients:Ji;j,i+jn?1;Gi;j, Thelinearsystemgeneratedfromthedegreen?1movingquadrics[Equation(28)]is (31)canbewrittenas Inparticular,setGi;j=Hi;j=Ii;j=Ji;j=0for(i;j)2I.Thenthelinearsystem MQ[Ai;jBi;jIi;jJi;j]T=0: (31) MQI264. Ai;j Ii;j.375=?Coe?sitjw2ji+jn?1;(i;j)62I ; (32)

25 wherecoe(sitjw2ji+jn?1;(i;j)62i)isthecoecientmatrixofthepolynomialssitjw2,(i;j)62i.setting(;ji;j;),(i;j)62i,tothestandardunitvectors (0;;0;1;0;;0)inC(n2?n)=2,weobtain(n2?n)=2linearlyindependentsolutions. Thesesolutionsgenerate(n2?n)=2movingquadrics: manner,nmovingplanes Ontheotherhand,sinceMPIisnon-singular,wecanalsond,inananalogous Pi;j=wsitj+termsnotinvolvingwsktlwith(k;l)2I;(i;j)2I: Qi;j=w2sitj+termswithoutw2; (i;j)62i: thecoecientsofthesemovingquadricsandmovingplanes,weobtainthematrix Altogetherwenowhave(n2?n)=2movingquadricsandnmovingplanes.Collecting 264w2+...w2+w+...w+ movingplanespi;j,(i;j)2i.notethatherethecolumnsareindexedby wheretherst(n2?n)=2rowsconsistofthecoecients(ofsitj)ofthe(n2?n)=2moving quadricsqi;j,(i;j)62i,andthelastnrowsconsistofthecoecients(ofsitj)ofthen 3 75; Thedeterminantofthismatrixhasthetermwn2,whichshowsthatthisdeterminantis notidenticallyzero.moreover,itiseasytoseethatthisdeterminantvanisheswhenever thepoint(x;y;z;w)liesontheoriginalsurface(26),becauseeachrowrepresentsamoving planeoramovingquadricthatfollowssurface(26).sincethisdeterminantisoftotal 1;t;s;t2;st;s2;;sitj;tn?1;;sn?1 {z };sitj (i;j)2i: {z} 6OpenQuestions Someimportantquestionsregardingthemethodofmovingquadricsstillremainopen. degreen2,whichisthetotaldegreeoftheimplicitequationofsurface(26)[coxetal Belowwedescribethethreemostimportantunresolvedissues. 1998],thisdeterminantisindeedtheimplicitequationofsurface(26). 25

26 Throughoutthispaperwehaveassumedthatourrationalsurfaceshavenobase pointsinordertoprovethetheoremsinsections4and5.whathappensifthe presenceofbasepoints,somemovingquadricscanbereplacedbymovingplanes surfacesdohavebasepoints?eachbasepointwill,ingeneral,lowerthedegreeof tocorrectlycomputetheimplicitequation.itisstillnotclearhowtomakethese theimplicitequationbyone.experiments[sederbergetal1995]showthatinthe replacementssystematicallywhensurfaceshavebasepoints.moreover,ifthereare Forrationalcurves,thedeterminantofthemovingconiccoecientmatrix(MCw) movingquadricsormovingplanes.whatdegreesshouldweuseandhowcanwe maketheadjustmentautomatically? oftheimplicitequationislessthanmn.thuswecannotusebi-degree(m?1;n?1) example,whenabi-degree(m;n)surfacehasmorethanmnbasepoints,thedegree toomanybasepoints,weneedtomodifythemethodofmovingquadrics.for canbefactoredintermsofthedeterminantofthemovinglinecoecientmatrix (ML)andtheresultantoftwocoordinatepolynomials[Zhangetal1999].Indeed, jmpij)andtheresultantofthreecoordinatepolynomials?inparticular,isittrue CanwefactorthedeterminantjMQwj(orjMQIj)similarlyintermsofjMPj(or thatjmqwj=jmpj3resultant(x;y;z)(fortensorproductsurfaces),(33) jmcwj=jmlj2resultant(x;y): boththelefthandsideandtherighthandsideofequations(33)and(34)have Webelievethatthesefactorizationsarecorrectforthefollowingtworeasons:First, thusjmljisadoublefactorofjmcwj.forrationalsurfaces,eachlinearrelation univariatesetting,eachlinearrelationamongthecolumnsofmlgeneratestwo linearrelationsamongthecolumnsofmcw(multiplyingtherelationbyx(t);y(t)); thesamedegreesinthecoecientsofx(s;t);y(s;t);z(s;t);w(s;t).second,inthe jmqij=jmpij3resultant(x;y;z)(fortriangularsurfaces)?(34) betweenthecolumnsofmp(ormpi)generatesthreelinearrelationsbetween TheproofsofTheorems2and4relyonTheorems1and3,which,inturn,depend haveyettosucceedinestablishingsucharesult. isanirreduciblepolynomialinthecoecientsofx(s;t);y(s;t);z(s;t);w(s;t).we MQI).However,arigorousproofwouldrequireustoshowthatjMPj(orjMPIj) Therefore,weconjecturethatjMPj(orjMPIj)isatriplefactorofjMQwj(or thecolumnsofmqw(ormqi)(multiplyingtherelationbyx(s;t);y(s;t);z(s;t)). uponthetwopropositionsinsection2,andtheproofsofthesetwopropositions 26

27 requireadvancedknowledgeinalgebraicgeometryandcommutativealgebra.is thereanelementaryprooffortheorems2and4?willthefactorizationdiscussed Acknowledgments Wehopethatweshallbeabletoanswerthesequestionsinfuturepapers. inthepreviousparagraphleadtosuchastraightforwardproof? References RonGoldmanandMingZhangarepartiallysupportedbyNSFgrantCCR formanystimulatingdiscussionsduringajanuary1999visittobrighamyounguniversity. DavidCoxandRonGoldmanwouldliketothankTomSederbergandZhengJianmin [1]D.Cox,J.Little,D.O'Shea.UsingAlgebraicGeometry.Springer-VerlagNewYork, [2]D.Cox,T.W.Sederberg,F.Chen.TheMovingLineIdealBasisofPlanarRational [3]A.L.Dixon.TheEliminantofThreeQuanticsinTwoIndependentVariables.Proc. Inc.,1998. [4]D.Eisenbud.CommutativeAlgebra.Springer-VerlagNewYork,Inc.,1995. Curves.ComputerAidedGeometricDesign,15:803{827,1998. [6]R.Hartshorne.AlgebraicGeometry.Springer-VerlagNewYork,Inc.,1977. [5]P.Griths,J.Harris.PrinciplesofAlgebraicGeometry.JohnWiley&Sons,1978. LondonMathematicsSociety,6:49{69,473{492,1908. [7]D.ManochaandJ.F.Canny.AlgorithmsforImplicitizingRationalParametricSurfaces.ComputerAidedGeometricDesign,9:25{50,1992ceedingsofSIGGRAPH, ,1995. MethodofMovingAlgebraicCurves,JournalofSymbolicComputation,23:153{175, [10]T.W.Sederberg,R.N.Goldman,H.Du.ImplicitizingRationalCurvesbythe [8]H.Matsumura,CommutativeRingTheory.CambridgeUniversityPress,1986. [9]T.W.Sederberg,F.Chen.ImplicitizationUsingMovingCurvesandSurfaces,Pro- 27

28 [11]M.Zhang,E.W.Chionh,R.N.Goldman.OnaRelationshipbetweentheMoving LineandMovingConicCoecientMatrices,toappearinComputerAidedGeometric Design,