Some geometric faces of algebra and algebraic faces of geometry

105 Pages·2013·0.84 MB·English

by  Gorodentsev A.L

Curso de Inverno — ICMC/USP São Carlos
A.L.GORODENTSEV
SOME GEOMETRIC FACES
OF ALGEBRA
AND ALGEBRAIC FACES
OF GEOMETRY
InAutumn2012SashaAnaninaskedmetogiveanintensiveIUM-style
courseonbasicprojectivegeometryandaendantalgebraattheSão
Carlos. I have made an aempt 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,
MathematicalPhysicsGroupattheInstituteofeoreticalandExperimentalPhysics,
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 Projectiveadrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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'sliletheorem).
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 ⩾ 𝑛andaer
(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∗ wewilloenwrite⟨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),whichwewilldenotebythesameleer𝑓inspite
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