Curso de Inverno — ICMC/USP São Carlos A.L.GORODENTSEV SOME GEOMETRIC FACES OF ALGEBRA AND ALGEBRAIC FACES OF GEOMETRY InAutumn2012SashaAnaninaskedmetogiveanintensiveIUM-style courseonbasicprojectivegeometryandaendantalgebraattheSão Carlos. I have made an aempt to combine a review of classical pro- jectivevarieties: quadrics,Segre,Veronese,andGrassmannianwitha self-contained introduction to linear, multilinear, and polynomial al- gebra and supplement this mixture with real affine convex geometry. Theimportantpartofthiscourseconsistsofexercisesandhometask problems. Thematerialofalmostallexercisesisintensivelyusedalong the course. Independent solution of problems followed by discussion ofproblemswithteachersallowstounderstandthethingsdeeper. The excitingtitlebelongstoCarlosHenriqueGrossiFerreira. Withouthis supporttheselectureswouldbeimpossible. SãoCarlos July2013 © A.L.Gorodentsev FacultyofMathematics&LaboratoryofAlgebraicGeometryattheHigherSchoolofEconomics, IndependentUniversityofMoscow, MathematicalPhysicsGroupattheInstituteofeoreticalandExperimentalPhysics, e-mail:[email protected], http://gorod.bogomolov-lab.ru Contents Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 §1 Numbers,functions,spaces,andfigures . . . . . . . . . . . . . . . . . . . . 4 1.1 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Vectorspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.3 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.4 Affinespaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.5 Projectivespaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.6 Projectivealgebraicvarieties . . . . . . . . . . . . . . . . . . . . . . . 14 1.7 Subspacesandprojections . . . . . . . . . . . . . . . . . . . . . . . . 17 1.8 Linearprojectivetransformations . . . . . . . . . . . . . . . . . . . . 18 1.9 Cross-ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Hometaskproblemsto§1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 §2 Projectiveadrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.1 Remindersfromlinearalgebra . . . . . . . . . . . . . . . . . . . . . . 27 2.2 Smoothnessandsingularities . . . . . . . . . . . . . . . . . . . . . . . 28 2.3 Conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.4 adraticsurfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 2.5 Linearsubspaceslyingonasmoothquadric . . . . . . . . . . . . . . 40 2.6 Digression: orthogonalgeometryoverarbitraryfield . . . . . . . . . 42 Hometaskproblemsto§2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 §3 TensorGuide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.1 TensorproductsandSegrevarieties . . . . . . . . . . . . . . . . . . . 51 3.2 Tensoralgebraandcontractions . . . . . . . . . . . . . . . . . . . . . 54 3.3 SymmetricandGrassmannianalgebras . . . . . . . . . . . . . . . . . 57 3.4 Symmetricandskew-symmetrictensors . . . . . . . . . . . . . . . . 62 3.5 Polarisationofcommutativepolynomials . . . . . . . . . . . . . . . . 64 3.6 Polarizationofgrassmannianpolynomials . . . . . . . . . . . . . . . 69 Hometaskproblemsto§3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 §4 Grassmannians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 4.1 PlückerquadricandGr(2,4) . . . . . . . . . . . . . . . . . . . . . . . 74 4.2 LagrangiangrassmannianLGr(2,4) . . . . . . . . . . . . . . . . . . . 77 4.3 GrassmanniansGr(𝑘,𝑛) . . . . . . . . . . . . . . . . . . . . . . . . . . 78 4.4 Celldecomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 Hometaskproblemsto§4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 §5 Realaffineconvexgeometry . . . . . . . . . . . . . . . . . . . . . . . . . . 84 5.1 Linearaffinegeometry . . . . . . . . . . . . . . . . . . . . . . . . . . 84 5.2 Convexfigures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 5.3 Convexpolyhedrons . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 5.4 Convexpolyhedralcones . . . . . . . . . . . . . . . . . . . . . . . . . 92 5.5 Linearprogramming . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 2 Contents 3 Hometaskproblemsк§5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 Hintsandanswersforsomeexersices . . . . . . . . . . . . . . . . . . . . . . . . . 101 §1 Numbers,functions,spaces,andfigures 1.1 Fields.Aninteractionbetweenalgebraandgeometrystarswithfixationofthegroundfield denoted by 𝕜. is is a set of «constants» or «scalars» with two operations: addition (+) and multiplication(⋅)thatsharethefollowingpropertiesofrationalnumbers: 𝑎+𝑏 = 𝑏+𝑎 (commutativity) 𝑎⋅𝑏 = 𝑏⋅𝑎 (𝑎+𝑏)+𝑐 = 𝑎+(𝑏+𝑐) (associativity) (𝑎⋅𝑏)⋅𝑐 = 𝑎⋅(𝑏⋅𝑐) ∃0 ∶ ∀𝑎 0+𝑎 = 𝑎 (neutrals) ∃1 ∶ ∀𝑎 1⋅𝑎 = 𝑎 ∀𝑎 ∃−𝑎 ∶ 𝑎+(−𝑎) = 0 (opposites) ∀𝑎 ≠ 0 ∃𝑎−(cid:2869) ∶ 𝑎⋅𝑎−(cid:2869) = 1 distributivity: 𝑎⋅(𝑏+𝑐) = 𝑎⋅𝑏+𝑎⋅𝑐 non-triviality: 0 ≠ 1 us,theelementsof𝕜canbeadded,subtracted,multiplied,anddividedinthemannerofrational numbers. MostclosedinteractionbetweenAlgebraandGeometry¹takesplacewhen𝕜 = ℂisthefield of complex numbers or, more generally, when 𝕜 is algebraically closed. However, a significant pieceofclassicalgeometrymakessenseoveranyfield. Overafinitefield𝕜,allspacesandfiguresbecomefinitesetsandgeometricand/oralgebraic theorems obtain combinatorial flavour. To make these lectures self-contained, let us remember somedetailsconcerningfinitefields. Example1.1 Residue field 𝔽 = ℤ∕(7) consist of 7 residues [0], [1], [2], … , [6] modulo 7. ey ere added (cid:2875) andsubtractedlikethehoursroundedaboutclockfacedial: [0] [6] [1] [2]−[5] = −[3] = [−3] = [4]. emultiplicationislessvisualbutitisstilltruethat,say, [5] [2] [2]+[2]+[2]+[2] = [2]⋅[4] = [1]. [4] [3] Tablesofsquiresandinversesarelookingasfollows Fig.1⋄1. 𝑥 [0] [±1] [±2] [±3] 𝑥 [−3] [−2] [−1] [1] [2] [3] 𝑥(cid:2870) [0] [1] [4] [2] 1∕𝑥 [2] [3] [−1] [1] [−3] [−2] Exercise1.1. In𝔽 = ℤ∕(5)compute2013⋅[2]and[2](cid:2870)(cid:2868)(cid:2869)(cid:2871) (i.e. 2013-tiplesumandproductof (cid:2873) 2(mod 5)withitsel). 1.1.1 Residues: integernumbers.eringofresiduesℤ∕(𝑚)doesexistforanyinteger 𝑚 ⩾ 2. Itconsistsofequivalenceclasses[𝑘],𝑘 ∈ ℤ,suchthat[𝑘] = [𝑛]iff𝑘 = 𝑛+𝑚⋅𝑡forsome 𝑡 ∈ ℤ. eseclassesareexhaustedby[0], [1], … , [𝑚−1]. eoperationsarewelldefinedby therules[𝑘]+[𝑛] ≝ [𝑘+𝑛],[𝑘]⋅[𝑛] ≝ [𝑘𝑛]. Exercise1.2. Verifythattheresultsdependonlyontheclassesbutnotontheparticularchoices ofelementsinsidethem. ¹evenanequivalenceinsomeprecisesense 4 1.1.Fields 5 Proposition1.1 ℤ∕(𝑚)isafieldiff𝑚isprime¹. Proof. Given [𝑎] ≠ [0], to find [𝑥] = [𝑎]−(cid:2869), which satisfies [𝑎][𝑥] = [1], means to solve an equation𝑎𝑥+𝑚𝑦 = 1in𝑥,𝑦 ∈ ℤ. eminimalpositiveintegeroftheform𝑎𝑥+𝑚𝑦,𝑥,𝑦 ∈ ℤ,is g.c.d.(𝑎,𝑚). Hence,[𝑎]isinvertibleiffg.c.d.(𝑎,𝑚) = 1. isholdsforeach𝑎 = 1, 2, … , (𝑚−1) iff𝑚isprime. (cid:3) 1.1.2 Residues: polynomials.Let𝕜beanyfieldand𝕜[𝑥]denotetheringofpolynomials in𝑥 withcoefficientsfrom𝕜. For any non-constant polynomial 𝑓 ∈ 𝕜[𝑥] one can form the residue ring 𝕜[𝑥]∕(𝑓) in the samemannerasabove. eelementsof𝕜[𝑥]∕(𝑓)aretheequivalenceclasses[𝑔],𝑔 ∈ 𝕜[𝑥],such that [𝑔] = [ℎ] iff 𝑔 = ℎ + 𝑓 ⋅ 𝑞 for some 𝑞 ∈ 𝕜[𝑥]. ese classes are exhausted by [𝑔] with deg𝑔 < deg𝑓. Inotherwords,ifdeg𝑓 = 𝑛,thenalltheclasses [𝛼 +𝛼 𝑥+⋯+𝛼 𝑥(cid:3041)−(cid:2869)] = 𝛼 +𝛼 [𝑥]+⋯+𝛼 [𝑥](cid:3041)−(cid:2869) (1-1) (cid:2868) (cid:2869) (cid:3041)−(cid:2869) (cid:2868) (cid:2869) (cid:3041)−(cid:2869) are different for different choices of constants 𝛼 ,𝛼 ,…,𝛼 ∈ 𝕜 and exhaust the whole of (cid:2868) (cid:2869) (cid:3041)−(cid:2869) 𝕜[𝑥]∕(𝑓). Proposition1.2 𝕜[𝑥]∕(𝑓)isafieldiff𝑓 isirreducible². Proof. Given [𝑔] ≠ [0], to find [𝑟] = [𝑔]−(cid:2869), which satisfies [𝑔][𝑟] = [1], means to solve an equation𝑔𝑟+𝑓𝑠 = 1in𝑟,𝑠 ∈ 𝕜[𝑥]. emonic³polynomialofsmallestdegreerepresentableas 𝑔𝑟+𝑓𝑠,𝑟,𝑠 ∈ 𝕜[𝑥],isg.c.d.(𝑔,𝑓). Hence,[𝑔]isinvertibleiffg.c.d.(𝑔,𝑓) = 1. isholdsforeach 𝑔 ≠ 0withdeg𝑔 < deg𝑓 iff𝑓 isirreducible. (cid:3) 1.1.3 Primitive field extensions. Since 𝑓([𝑥]) = [𝑓(𝑥)] = 0 in 𝕜[𝑥]∕(𝑓), we can treat expressions(1-1)aspolynomialsofdegree< deg𝑓 in[𝑥]addedandmultipliedbytheusualdis- tributionrulesmodulotherelation𝑓([𝑥]) = 0,whichallowstoreducethedegreeofanexpression assoonitbecomes⩾ deg𝑓. Example1.2 Putℚ[√2] ≝ ℚ[𝑥]∕(𝑥(cid:2870)−2). Here[𝑥]satisfies[𝑥](cid:2870)−2 = [𝑥(cid:2870)−2] = [0],thatis[𝑥](cid:2870) = 2,andwe write√2for[𝑥]. efieldconsistsofall𝛼+𝛽√2with𝛼,𝛽 ∈ ℚ. Wehave 𝛼 +𝛽 √2 + 𝛼 +𝛽 √2 = (𝛼 +𝛼 )+(𝛽 +𝛽 )√2 (cid:3585) (cid:2869) (cid:2869) (cid:3586) (cid:3585) (cid:2870) (cid:2870) (cid:3586) (cid:2869) (cid:2870) (cid:2869) (cid:2870) 𝛼 +𝛽 √2 ⋅ 𝛼 +𝛽 √2 = (𝛼 𝛼 +2𝛽 𝛽 )+(𝛼 𝛽 +𝛼 𝛽 )√2 (cid:3585) (cid:2869) (cid:2869) (cid:3586) (cid:3585) (cid:2870) (cid:2870) (cid:3586) (cid:2869) (cid:2870) (cid:2869) (cid:2870) (cid:2869) (cid:2870) (cid:2870) (cid:2869) −(cid:2869) 𝛼 𝛽 𝛼+𝛽√2 = − √2. (cid:3585) (cid:3586) 𝛼(cid:2870)−2𝛽(cid:2870) 𝛼(cid:2870)−2𝛽(cid:2870) Exercise1.3. Showthat: a) 𝑥(cid:2870) −2isirreducibleinℚ[𝑥] b) 𝛼(cid:2870) −2𝛽(cid:2870) = 0onlyfor𝛼,𝛽 = 0 (assuming𝛼,𝛽 ∈ ℚ). ¹orirreducible,thatis𝑚=𝑟𝑠⇒𝑟or𝑠equals±1 ²i.e. 𝑓 =𝑔ℎ⇒𝑔orℎisaconstant ³weusethistermforpolynomialswhoseleadingcoefficientequals1 6 §1Numbers,functions,spaces,andfigures Example1.3 ℝ[√−1] ≝ ℝ[𝑥]∕(𝑥(cid:2870) +1) is the field of complex numbers ℂ. Indeed, 𝑓 = 𝑥(cid:2870) +1 is irreducible inℝ[𝑥],becauseithasnorealroots. Hence,ℝ[𝑥]∕(𝑓)isafield. Class[𝑥]satisfies[𝑥](cid:2870) = −1and canbedenotedby√−1. efieldconsistsof𝛼+𝛽√−1with𝛼,𝛽 ∈ ℝ. eoperationscoincide withthoseofℂ. Example1.4 Wecanrepeatthepreviousforafiniteresiduefield𝔽 = ℤ∕(3)intheroleofℝ. Namely,𝔽 consist (cid:2871) (cid:2871) of3elements: −1,0,1 (mod 3). Polynomial𝑓 = 𝑥(cid:2870)+1hasnorootsin𝔽 ,because𝑓(0) = 1and (cid:2871) 𝑓(±1) = −1. us,𝑓 isirreduciblein𝔽 [𝑥]and𝔽 [√−1] = 𝔽 [𝑥]∕(𝑥(cid:2870)+1)isafield. Itconsists (cid:2871) (cid:2871) (cid:2871) of 9 elements 𝛼 +𝛽√−1, where 𝛼,𝛽 = 0,1,−1, and is denoted by 𝔽 . e multiplication goes (cid:2877) (cid:2870) −(cid:2869) likeinℂbutmodulo3. Forexample: 1+√−1 = −√−1, 1+√−1 = −1+√−1. (cid:3585) (cid:3586) (cid:3585) (cid:3586) (cid:2870)(cid:2868)(cid:2869)(cid:2871) Exercise1.4. In𝔽 compute 2013⋅ 1+√−1 and 1+√−1 . (cid:2877) (cid:3585) (cid:3586) (cid:3585) (cid:3586) 1.1.4 Primesubfieldandcharacteristic.Givenafield𝔽,theintersectionofallsubfields 𝕜 ⊂ 𝔽iscalledtheprimesubfield of𝔽. Itisthesmallestsubfieldin𝔽w.r.t. inclusions. eprime subfieldcontainsallsums 1+1+ ⋯ +1 , 𝑝 ∈ ℕ. (1-2) ⎭⎪⎪⎬⎪⎪⎫ (cid:3043) Ifallthesesumsaredifferent,then𝔽 ⊃ ℤ. Hence,𝔽 ⊃ ℚandtheprimesubfieldof𝔽equalsℚ. In thiscasewesaythat𝔽haszero¹characteristic andwritechar𝕜 = 0. Ifsomeofsums(1-2)coincide,thenthecharacteristic char𝔽isdefinedasthesmallest𝑝 ∈ ℕ forwhichsum(1-2)vanishes. Inthiscasewesaythat𝔽hasfinitecharacteristic. echaracteristic hastobeaprimenumberbecauseoftheidentity 1+1+ ⋯ +1 = (1+1+ ⋯ +1)⋅(1+1+ ⋯ +1) ⎭⎪⎪⎬⎪⎪⎫ ⎭⎪⎪⎬⎪⎪⎫ ⎭⎪⎪⎬⎪⎪⎫ (cid:3038)(cid:3041) (cid:3038) (cid:3041) (vanishing of L.H.S. implies vanishing of some factor in R.H.S.). us, the prime subfield of a fieldofcharacteristic𝑝equals𝔽 = ℤ∕(𝑝). (cid:3043) 1.1.5 Frobeniushomomorphism.Letchar𝕜 = 𝑝befinite. eFrobeniusmap² (cid:3080)↦(cid:3080)(cid:3291) 𝐹 ∶ 𝕜 −−−−−−→ 𝕜 (1-3) (cid:3043) respectsmultiplication: 𝐹 (𝛼𝛽) = (𝛼𝛽)(cid:3043) = 𝛼(cid:3043)𝛽(cid:3043) = 𝐹 (𝛼)𝐹 (𝛽)aswellassummation: (cid:3043) (cid:3043) (cid:3043) 𝑝 𝑝 𝐹 (𝛼+𝛽) = (𝛼+𝛽)(cid:3043) = 𝛼(cid:3043) + 𝛼(cid:3043)−(cid:2869)𝛽+ ⋯ + 𝛼𝛽(cid:3043)−(cid:2869)+𝛽(cid:3043) = 𝛼(cid:3043) +𝛽(cid:3043) = 𝐹 (𝛼)+𝐹 (𝛽) (cid:3043) (cid:3585)1(cid:3586) (cid:3585)𝑝−1(cid:3586) (cid:3043) (cid:3043) (sinceeach(cid:3512)(cid:3043)(cid:3513) = 𝑝(𝑝−1)⋯(𝑝−𝑘+1)∕𝑘!,1 ⩽ 𝑘 ⩽ 𝑝−1,isdivisibleby𝑝). (cid:3038) ¹orinfinite(insomecontexts) ²orjustFrobeniusforshort 1.2.Vectorspaces 7 Frobeniuskeepsfixedeachelement[𝑛] = 𝑛⋅1,𝑛 ∈ ℕ,oftheprimesubfield𝔽 = ℤ∕(𝑝) ⊂ 𝕜, (cid:3043) because [𝑛](cid:3043) = ([1]+[1]+ ⋯ +[1])(cid:3043) = [1](cid:3043) +[1](cid:3043) + ⋯ +[1](cid:3043) = [1]+[1]+ ⋯ +[1] = [𝑛] ⎭⎪⎪⎪⎬⎪⎪⎪⎫ ⎭⎪⎪⎪⎪⎬⎪⎪⎪⎪⎫ ⎭⎪⎪⎪⎬⎪⎪⎪⎫ (cid:3041) (cid:3041) (cid:3041) (thisisknownasFermat'sliletheorem). Exercise 1.5. Look at the Frobenius action on the field 𝔽 from example 1.4. Does it coincide (cid:2877) with«conjugation»𝑎+𝑏√−1 ↦ 𝑎−𝑏√−1? 1.2 Vector spaces. A vector space over a field 𝕜 is an abelian group 𝑉 (with operation +) equippedwithmultiplicationbyelements𝜆 ∈ 𝕜insuchawaythat (𝜆+𝜇)𝑣 = 𝜆𝑣+𝜇𝑣 𝜆(𝑣+𝑤) = 𝜆𝑣+𝜆𝑤 𝜆(𝜇𝑣) = (𝜆𝜇)𝑣 1⋅𝑣 = 𝑣. Wesaythatvectors𝑤 ,𝑤 ,…,𝑤 span𝑉,ifeach𝑣 ∈ 𝑉canbeexpressedasalinearcombination (cid:2869) (cid:2870) (cid:3040) of𝑤 's,thatis (cid:3036) 𝑣 = 𝜆 𝑤 +𝜆 𝑤 +⋯+𝜆 𝑤 forsome𝜆 ,𝜆 ,…,𝜆 ∈ 𝕜. (cid:2869) (cid:2869) (cid:2870) (cid:2870) (cid:3040) (cid:3040) (cid:2869) (cid:2870) (cid:3040) Vectors𝑢 ,𝑢 ,…,𝑢 ∈ 𝑉 arecalledlinearlyindependent,if (cid:2869) (cid:2870) (cid:3041) 𝜆 𝑢 +𝜆 𝑢 +⋯+𝜆 𝑢 = 0 ⇒ each𝜆 = 0. (cid:2869) (cid:2869) (cid:2870) (cid:2870) (cid:3040) (cid:3040) (cid:3036) Lemma1.1(exchangelemma) Let𝑤 ,𝑤 ,…,𝑤 span𝑉 and𝑢 ,𝑢 ,…,𝑢 ∈ 𝑉 belinearlyindependent. en𝑚 ⩾ 𝑛andaer (cid:2869) (cid:2870) (cid:3040) (cid:2869) (cid:2870) (cid:3041) appropriaterenumberingof𝑤 'svectors𝑢 ,𝑢 ,…,𝑢 , 𝑤 , 𝑤 , … , 𝑤 dospan𝑉 aswell. (cid:3036) (cid:2869) (cid:2870) (cid:3041) (cid:3041)+(cid:2869) (cid:3041)+(cid:2870) (cid:3040) Proof. Assume inductively that for some 𝑘 < 𝑛 the vectors 𝑢 ,𝑢 ,…,𝑢 , 𝑤 , 𝑤 , … , 𝑤 (cid:2869) (cid:2870) (cid:3038) (cid:3038)+(cid:2869) (cid:3038)+(cid:2870) (cid:3040) span 𝑉 (the case 𝑘 = 0 corresponds to the initial situation). en 𝑢 is a linear combination (cid:3038)+(cid:2869) of these vectors. Since 𝑢 's are linearly independent, this linear combination contains some 𝑤 (cid:3036) (cid:3037) withnon-zerocoefficient. Renumbering𝑤 'sinordertohave𝑗 = 𝑘+1,weconcludethat𝑘 < 𝑚 (cid:3092) and 𝑤 is a linear combination of vectors 𝑢 ,𝑢 ,…,𝑢 , 𝑤 , 𝑤 , … , 𝑤 . Hence, they (cid:3038)+(cid:2869) (cid:2869) (cid:2870) (cid:3038)+(cid:2869) (cid:3038)+(cid:2870) (cid:3038)+(cid:2871) (cid:3040) span𝑉 andtheinductiveassumptionholdsfor𝑘+1aswell. (cid:3) 1.2.1 Basesanddimension.Linearlyindependentcollection𝑣 ,𝑣 ,…,𝑣 ∈ 𝑉thatspans (cid:2869) (cid:2870) (cid:3041) 𝑉iscalledabasisof𝑉. Anycollectionofvectorsspanning𝑉clearlydoescontainsomebasis. By lemma 1.1 each linearly independent collection of vectors can be completed to some basis and all the bases consist of the same number of vectors. is number is called dimension of 𝑉 and denoteddim𝑉. Exercise1.6. Finddimensionsof a) thespaceofsymmetric𝑛×𝑛-matrices b) thespaceof skew-symmetric 𝑛×𝑛-matrices c) the space of homogeneous polynomials of degree 𝑑 in𝑛variables. If dim𝑉 = 𝑛, then it follows from lemma 1.1 that any 𝑛 linearly independent vectors as well as any𝑛vectorsthatspan𝑉 formabasisfor𝑉. 1.2.2 Coordinates. Vectors 𝑒 ,𝑒 ,…,𝑒 ∈ 𝑉 form a basis iff each vector 𝑣 ∈ 𝑉 admits a (cid:2869) (cid:2870) (cid:3041) uniqueexpression𝑣 = 𝜆 𝑒 +𝜆 𝑒 +⋯+𝜆 𝑒 with𝑥 ∈ 𝕜. Inthiscasethemapping (cid:2869) (cid:2869) (cid:2870) (cid:2870) (cid:3041) (cid:3041) (cid:3036) 𝑉 ∋ 𝑣 = 𝜆 𝑒 +𝜆 𝑒 +⋯+𝜆 𝑒 ↦ (𝜆 ,𝜆 ,…,𝜆 ) ∈ 𝕜(cid:3041) (cid:2869) (cid:2869) (cid:2870) (cid:2870) (cid:3041) (cid:3041) (cid:2869) (cid:2870) (cid:3041) provides an isomorphism between 𝑉 and the coordinate space 𝕜(cid:3041). Numbers (𝜆 ,𝜆 ,…,𝜆 ) are (cid:2869) (cid:2870) (cid:3041) calledthecoordinatesof𝑣 w.r.t. thebasis{𝑒 }. (cid:3036) 8 §1Numbers,functions,spaces,andfigures Example1.5(Lagrange'sinterpolation) Let𝑉 = {𝑓 ∈ 𝕜[𝑥]| deg𝑓 ⩽ 𝑛}. Given𝑛+1distinctpoints𝛼 ,𝛼 ,…,𝛼 ∈ 𝕜,thepolynomials (cid:2868) (cid:2869) (cid:3041) 𝐿 (𝑥) = ∏(𝑥−𝛼 )∕∏(𝛼 −𝛼 ), 𝑖 = 0, 1, … , 𝑛, (cid:3036) (cid:3092) (cid:3036) (cid:3091) (cid:3092)≠(cid:3036) (cid:3091)≠(cid:3036) satisfytherelations 1 for𝑖 = 𝑗 𝐿 (𝛼 ) = (cid:3036) (cid:3037) (cid:3696)0 for𝑖 ≠ 𝑗 whichforcethecoefficientsofanylinearcombination𝑔(𝑥) = ∑𝜆 𝐿 (𝑥)tobeequaltothevalues (cid:3036) (cid:3036) of𝑔: 𝜆 = 𝑔(𝛼 ). Weconcludethat𝐿 arelinearlyindependent,thus,formabasisof𝑉,thus,each (cid:3036) (cid:3036) (cid:3036) 𝑔 ∈ 𝑉 doeshave theexpansion𝑔(𝑥) = ∑𝑔(𝛼 )⋅𝐿 (𝑥). (cid:3036) (cid:3036) 1.2.3 Duality.Acovector onavectorspace𝑉 isalinearmap𝜑 ∶ 𝑉 → 𝕜,i.e. suchthat 𝜑(𝜆𝑣+𝜇𝑤) = 𝜆𝜑(𝑣)+𝜇𝜑(𝑤) ∀𝑣,𝑤 ∈ 𝑉, ∀𝜆,𝜇 ∈ 𝕜. Covectorsformavectorspacecalledthedualspace to𝑉 anddenotedby𝑉∗. Toemphasizethesymmetricrolesof𝑉and𝑉∗ wewilloenwrite⟨𝜑, 𝑣⟩forthevalue𝜑(𝑣) andcallitacontractionofacovector𝜑 ∈ 𝑉∗ andavector𝑣 ∈ 𝑉. If 𝑒 ,𝑒 ,…,𝑒 ∈ 𝑉 form a basis, then 𝑖-th coordinate mapping 𝑥 ∶ 𝑉 → 𝕜, which takes (cid:2869) (cid:2870) (cid:3041) (cid:3036) 𝑣 = 𝜆 𝑒 +𝜆 𝑒 +⋯+𝜆 𝑒 to 𝑥 (𝑣) ≝ 𝜆 , is a covector. is definition implies the following (cid:2869) (cid:2869) (cid:2870) (cid:2870) (cid:3041) (cid:3041) (cid:3036) (cid:3036) relationssimilartothosefromexample 1.5: 1 for𝑖 = 𝑗 (cid:3564)𝑥(cid:3036), 𝑒(cid:3037)(cid:3565) = (1-4) (cid:3696)0 for𝑖 ≠ 𝑗. Again,eachcovectoroftheform𝜓 = 𝜓 𝑥 +𝜓 𝑥 +⋯+𝜓 𝑥 ∈ 𝑉∗isforcedtotake𝜓(𝑒 ) = 𝜓 . (cid:2869) (cid:2869) (cid:2870) (cid:2870) (cid:3041) (cid:3041) (cid:3036) (cid:3036) Hence𝑥 arelinearlyindependent. Moreover,each𝜓 ∈ 𝑉∗ equals∑𝜓(𝑒 )⋅𝑥 ,becausetheboth (cid:3036) (cid:3036) (cid:3036) linearmaps𝑉 → 𝕜coincideonthebasis{𝑒 } ⊂ 𝑉. (cid:3036) Exercise 1.7. Verify that if two linear maps 𝑓,𝑔 ∶ 𝑈 → 𝑊 coincide on some collection of vectorsspanning𝑈,thentheycoincideeverywhereon𝑈. Definition1.1 Bases𝑥 ,𝑥 ,…,𝑥 ∈ 𝑉∗ and𝑒 ,𝑒 ,…,𝑒 ∈ 𝑉 satisfying(1-4)arecalleddual toeachother. (cid:2869) (cid:2870) (cid:3041) (cid:2869) (cid:2870) (cid:3041) Exercise1.8. Giventwocollectionsofvectors𝑥 ,𝑥 ,…,𝑥 ∈ 𝑉∗ and𝑒 ,𝑒 ,…,𝑒 ∈ 𝑉 satis- (cid:2869) (cid:2870) (cid:3040) (cid:2869) (cid:2870) (cid:3040) fying(1-4)showthattheyformdualbasesifoneofthefollowingconditionsholds: a) {𝑒 }span𝑉 b) {𝑥 }span𝑉∗ c) 𝑚 = dim𝑉 d) 𝑚 = dim𝑉∗. (cid:3036) (cid:3036) Proposition1.3 Assoonasdim𝑉 < ∞thereiscanonicalisomorphism𝑉 ⥲ 𝑉∗∗ sending𝑣 ∈ 𝑉 totheevaluation mapev ∶ 𝑉∗ → 𝕜,whichtakes𝜓 ↦ 𝜓(𝑣). (cid:3049) Proof. Itsendsanybasis{𝑒 }of𝑉 toabasisof𝑉∗∗ dualtothebasis{𝑥 }of𝑉∗ dualto{𝑒 }. (cid:3) (cid:3036) (cid:3036) (cid:3036) Exercise 1.9. Given a subspace 𝑈 in 𝑉 or in 𝑉∗ write Ann𝑈 for a subspace in the dual space (𝑉∗ or 𝑉 respectively) defined as Ann𝑈 = {𝜉| ∀𝑢 ∈ 𝑈 ⟨𝜉, 𝑢⟩ = 0}. Verify that the 1.3.Polynomials 9 correspondence 𝑈 ↦ Ann𝑈 establishes a self-inverse bijection between the subspaces of dualspaces𝑉and𝑉∗thatreversestheinclusions. Inotherwords,showthatAnnAnn𝑈 = 𝑈 and𝑈 ⊂ 𝑊⟺Ann𝑈 ⊃ Ann𝑊. Moreover,showthatittakesthelinearspansofcollections ofsubspacestotheintersectionsoftheirannihilatorsandviceversa. 1.3 Polynomials.Let𝑥 ,𝑥 ,…,𝑥 ∈ 𝑉∗ formabasisof𝑉∗. Byapolynomial on𝑉 weunder- (cid:2869) (cid:2870) (cid:3041) standanelementofthepolynomialalgebra 𝑆𝑉∗ ≝ 𝕜[𝑥 ,𝑥 ,…,𝑥 ], (cid:2869) (cid:2870) (cid:3041) withcoefficientsin𝕜. Anotherchoiceofabasisin𝑉∗leadstoisomorphicalgebraobtainedfrom theinitialonebyaninvertiblelinearchangeofvariables. Wewrite𝑆(cid:3031)𝑉∗ ⊂ 𝑆𝑉∗ forthesubspaceofhomogeneouspolynomialsofdegree𝑑. Clearly,it isstableunderlinearchangesofvariablesandhasabasis𝑥(cid:3040) ≝ 𝑥(cid:3040)(cid:3117)𝑥(cid:3040)(cid:3118)…𝑥(cid:3040)(cid:3289) numberedbyall (cid:2869) (cid:2870) (cid:3041) collections𝑚 = (𝑚 ,𝑚 ,…,𝑚 ) ∈ ℤ(cid:3041) ofnon-negativeintegerswith∑𝑚 = 𝑑. (cid:2869) (cid:2870) (cid:3041) ⩾(cid:2868) (cid:3036) Exercise1.10. Showthatdim𝑆(cid:3031)𝑉∗ = (cid:3512)(cid:3041)+(cid:3031)−(cid:2869)(cid:3513)assoonasdim𝑉 = 𝑛. (cid:3031) In fact, the symmetric powers 𝑆(cid:3031)𝑉∗ and the whole symmetric algebra 𝑆𝑉∗ of the space 𝑉∗ admitanintrinsicfunctorialcoordinate-freedefinitionbutwepostponeituntiln∘3.3.1onp.58. Notethat 𝑆𝑉∗ = 𝑆(cid:3031)𝑉∗, where𝑆(cid:3038)𝑉∗⋅𝑆(cid:3040)𝑉∗ ⊂ 𝑆(cid:3038)+(cid:3040)𝑉∗. (cid:3575) (cid:3031)⩾(cid:2868) 1.3.1 Polynomial functions. Each polynomial 𝑓 = ∑𝑎 𝑥(cid:3040)(cid:3117)…𝑥(cid:3040)(cid:3289) ∈ 𝑆𝑉∗ produces a (cid:3040) (cid:2869) (cid:3041) (cid:3040) polynomialfunction𝑉 → 𝕜thattakes (cid:3040) (cid:3040) 𝑣 ↦ (cid:3540)𝑎(cid:3040)(cid:3564)𝑥(cid:2869), 𝑣(cid:3565) (cid:3117)…(cid:3564)𝑥(cid:3041), 𝑣(cid:3565) (cid:3289) (1-5) (cid:3040) (evaluationof𝑓 atthecoordinatesof𝑣). Wegetahomomorphism 𝑆𝑉∗ → {functions𝑉 → 𝕜}. (1-6) thattakesapolynomial𝑓tothefunction(1-5),whichwewilldenotebythesameleer𝑓inspite ofthenextclaimsayingthatthisnotationisnotcorrectforfinitefields. Proposition1.4 Homomorphism(1-6)isinjectiveifanonlyifthegroundfield𝕜isinfinite. Proof. If𝕜consistsof𝑞elements,thenthespaceofallfunctions𝑉 → 𝕜consistsof𝑞(cid:3044)(cid:3289) elements whereas the polynomial algebra 𝕜[𝑥 ,𝑥 ,…,𝑥 ] is definitely infinite. Hence, homomorphism (cid:2869) (cid:2870) (cid:3041) (1-6)cannotbeinjective. Let 𝕜 be infinite. For 𝑛 = 1 each non zero polynomial 𝑓 ∈ 𝕜[𝑥 ] vanishes in at most deg𝑓 (cid:2869) pints of 𝑉 ≃ 𝕜. Hence, the polynomial function 𝑓 ∶ 𝑉 → 𝕜 is not the zero function. For 𝑛 > 1 weproceedinductively. Writeapolynomial𝑓 ∈ 𝕜[𝑥 ,𝑥 ,…,𝑥 ]asapolynomialin𝑥 withthe (cid:2869) (cid:2870) (cid:3041) (cid:3041) coefficientsin𝕜[𝑥 ,𝑥 ,…,𝑥 ]: 𝑓 = 𝑓(𝑥 ,𝑥 ,…,𝑥 ; 𝑥 ) = ∑𝑓 (𝑥 ,𝑥 ,…,𝑥 )⋅𝑥(cid:3092). Let (cid:2869) (cid:2870) (cid:3041)−(cid:2869) (cid:2869) (cid:2870) (cid:3041)−(cid:2869) (cid:3041) (cid:3092) (cid:2869) (cid:2870) (cid:3041)−(cid:2869) (cid:3041) (cid:3092) the polynomial function 𝑓 ∶ 𝑉 → 𝕜 vanish identically on 𝕜(cid:3041). Evaluating the coefficients 𝑓 at (cid:3092) any 𝑤 ∈ 𝕜(cid:3041)−(cid:2869), we get polynomial 𝑓(𝑤;𝑥 ) ∈ 𝕜[𝑥 ] that produces identically zero function of (cid:3041) (cid:3041) 10 §1Numbers,functions,spaces,andfigures 𝑥 . Hence, 𝑓(𝑤;𝑥 ) = 0 in 𝕜[𝑥 ]. us, all coefficients 𝑓 (𝑤) are identically zero functions of (cid:3041) (cid:3041) (cid:3041) (cid:3092) 𝑤 ∈ 𝕜(cid:3041)−(cid:2869). Byinduction,theyarezeropolynomials. (cid:3) Exercise1.11. Giveanexplicitexampleofnon-zeropolynomial𝑓 ∈ 𝔽 [𝑥]thatproducesiden- (cid:3043) ticallyzerofunction𝐹 ∶ 𝔽 → 𝔽 . (cid:3033) (cid:3043) (cid:3043) 1.3.2 Digression: list of finite fields. Each finite field 𝕜 of characteristic 𝑝 is a finite dimensional vector space over the prime subfield 𝔽 ⊂ 𝕜. Let dim 𝕜 = 𝑛. en 𝕜 consists of (cid:3043) 𝔽 (cid:3291) 𝑝(cid:3041) elements. us,cardinalitiesoffinitefieldsareexhaustedbythepowersofprimes. Foreachpower𝑞 = 𝑝(cid:3041) afield𝔽 ofcardinality𝑞isconstructedasfollows. Considerpolyno- (cid:3044) mial𝑓(𝑥) = 𝑥(cid:3044) −𝑥 ∈ 𝔽 [𝑥]andusetheconstructionn∘1.1.3onp.5tobuildafield𝕜 ⊃ 𝔽 such (cid:3043) (cid:3043) thatin𝕜[𝑥]𝑓 becomesaproductoflinearfactors. Exercise1.12. Explainpreciselyhowtobuiltsuchafield𝕜. Since𝑓(cid:3807)(𝑥) = −1,polynomial𝑓hasprecisely𝑞distinctrootsin𝕜. eyformafield,becausefor anytworoots𝛼 = 𝛼(cid:3044) and𝛽 = 𝛽(cid:3044) wehave 𝛼+𝛽 = 𝛼(cid:3043)(cid:3289) +𝛽(cid:3043)(cid:3289) = 𝐹(cid:3041)(𝛼)+𝐹(cid:3041)(𝛽) = 𝐹(cid:3041)(𝛼+𝛽) = (𝛼+𝛽)(cid:3044), (cid:3043) (cid:3043) (cid:3043) 𝛼𝛽 = 𝛼(cid:3044)𝛽(cid:3044) = (𝛼𝛽)(cid:3044), (−𝛼) = (−𝛼)(cid:3044), 1∕𝛼 = (1∕𝛼)(cid:3044). Usingelementarygroup-theoreticalarguments,wecansaymore: Proposition1.5 Anyfield𝔽ofcardinality𝑞 = 𝑝(cid:3041) isisomorphictothefield𝔽 constructedabove. (cid:3044) Proof. Sincethemultiplicativegroup𝔽∗ ≝ 𝔽∖{0}hasorder𝑞−1,eachnon-zeroelement𝑎 ∈ 𝔽∗ satisfiesanequation𝑎(cid:3044)−(cid:2869) = 1. us,𝔽consistsof𝑞 distinctrootsofpolynomial𝑥(cid:3044) −𝑥. Exercise 1.13 (group exponent). Let 𝐴 be an abelian group and 𝑎,𝑏 ∈ 𝐴 have finite orders 𝛼 and𝛽 respectively. Constructanelement𝑐 ∈ 𝐴oforderl.c.m.(𝛼,𝛽). Deducefromthisthat ifthereexisttheleastcommonmultiple¹𝜇 fortheordersofallelements𝑎 ∈ 𝐴 (e.g. if𝐴 is finite)thenthereisanelement𝑚 ∈ 𝐴oforder𝜇. Write𝑑 fortheexponent²of𝔽∗ andfixanelement𝜁 ∈ 𝔽∗ oforder𝑑. Weclaimthat𝑑 = 𝑞−1: otherwise𝑞elementsof𝔽wouldbetherootsofpolynomial𝑥(cid:3031)+(cid:2869)−𝑥ofdegree𝑑+1 < 𝑞. us, 𝔽 = {0, 1, 𝜁, 𝜁(cid:2870), … , 𝜁(cid:3044)−(cid:2870)}. Let 𝑔 ∈ 𝔽 [𝑥] be the minimal polynomial³ of 𝜁 over 𝔽 . en 𝑔 is (cid:3043) (cid:3043) anirreduciblefactorof 𝑓 in 𝔽 [𝑥]. eevaluationmapev ∶ 𝔽 [𝑥]∕(𝑔) → 𝔽, [ℎ(𝑥)] ↦ ℎ(𝜁), is (cid:3043) (cid:3085) (cid:3043) welldefined,because𝑔(𝜁) = 0,andsurjective,because𝑥(cid:3038) ↦ 𝜁(cid:3038). us,𝔽 ≃ 𝔽 [𝑥]∕(𝑔). (cid:3043) Exercise1.14. Verifythateachnon-zerohomomorphism⁴offieldsisinjective. On the other hand, since 𝑓 has 𝑞 roots in 𝔽 , substituting them into factorization 𝑓 = 𝑔𝑟, we (cid:3044) conclude that 𝑔 also has a root 𝜉 in 𝔽 . en the evaluation map ev ∶ 𝔽 [𝑥]∕(𝑔) → 𝔽 , (cid:3044) (cid:3093) (cid:3043) (cid:3044) [ℎ(𝑥)] ↦ ℎ(𝜉),iswelldefinedinjection. Bythecardinalityreasons𝔽 [𝑥]∕(𝑔) ≃ 𝔽 . (cid:3) (cid:3043) (cid:3044) ¹itiscalledanexponent ofgroup𝐴 ²seeexrs.1.13above ³i.e. monicpolynomialofminimalpossibledegreesuchthat𝑔(𝜁)=0 ⁴i.e. takingatleastonenon-zerovalue