C3.4

ALGEBRAIC

GEOM

ETRY

Math

ematicalIn

stitute,Oxford

.

Pro

f.AlexanderF.Ritter.

Comments

and

correctionsare

welcome:ritter@

math

s.ox.ac.uk

Contents

Preliminaries

1

AFFIN

EVARIE

TIE

S3

PROJECTIV

EVARIE

TIE

S11

CLASSIC

ALEMBEDDIN

GS

18

EQUIV

ALENCE

OFCATEGORIE

S23

PRODUCTSAND

FIB

RE

PRODUCTS

25

ALGEBRAIC

GROUPSAND

GROUP

ACTIO

NS

30

DIM

ENSIO

NTHEORY

34

DEGREE

THEORY

39

LOCALISATIO

NTHEORY

41

QUASI-PROJECTIV

EVARIE

TIE

S45

THE

FUNCTIO

NFIE

LD

AND

RATIO

NALMAPS

50

TANGENT

SPACES

53

BLOW

-UPS

57

SCHEMES

58

APPENDIX

1:Irreducible

decom

positionsan

dprimaryideals

69

APPENDIX

2:Differential

methodsin

algebraic

geom

etry

73

1.

Preliminaries

1.1.

COURSE

POLIC

Yand

BOOK

RECOM

MENDATIO

NS

C3.4

Coursepolicy:It

isessentialth

atyou

read

yournotesaftereach

lectu

re.

You

willnoticethat

formostPartC

courses,unlikepreviousyears,each

lecture

buildson

the

previous.

Ifyoudon

’tread

thenotes

then

within

alecture

ortw

oyoumay

feel

lost.For

PartC

courses,youshou

ldnot

expecteverydetailto

becovered

inlectures:

oftenit

isupto

you

tocheck

statem

ents

asexercises.

Thecourseassumesfamiliarity

with

algebra

(orth

atyou

are

willingto

read

up

on

it).

I’m

afraid

itwou

ldbeunrealisticto

expectcommutative

algebra

tobetaugh

tas

asubsetof

this

16-hou

rcourse.

Iwrite

“Fact”

ifyouarenot

required

toread

/know

theproof

(unless

weproveit),

andit

usually

refers

to:algebra

results,

ordiffi

cult

results,

orresultswedon

’thavetimeto

prove.

Algebraic

geom

etry

isadiffi

cult

andextrem

elybroad

subject,an

dIwilldomybestto

makeit

digestible.Butthis

willnot

hap

pen

byitself:it

requires

efforton

yourpart,

thinkingon

you

row

nab

outthenotes,theexam

ples,

theexercises.

Date:This

versionofthenoteswascreatedonFeb

ruary

20,2019.

1

2C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

RELEVANT

BOOKS

Basicalgebra

icgeometry

Reid,Undergraduate

algebraic

geometry.Start

from

Chp.II.3.

(Availab

leon

linefrom

theau

thor)

Fulton

,Algebraic

Curves.(A

vailable

onlinefrom

theau

thor)

Shafarevich,BasicAlgebraic

Geometry.

Harris,

Algebraic

Geometry,A

First

Course.

Gathman

n,Algebraic

geometry.(O

nlinenotes)

Back

gro

und

on

algebra

Atiyah

andMacDonald,Introductionto

commutative

algebra.

Reid,Commutative

algebra.

Beyond

this

course

Mumford,TheRed

Bookofvarietiesandschem

es.

Harshorne,

Algebraic

geometry.

Eisenbudan

dHarris,

Schem

es.

Griffithsan

dHarris,

PrinciplesofAlgebraic

Geometry.(T

his

iscomplexalg.geom.)

Matsumura,Commutative

ringtheory.

Eisenbud,Commutative

Algebra

withaview

toward

Algebraic

Geometry.

Vak

il,Foundationsofalgebraic

geometry.(O

nlinenotes)

RELATED

COURSES

PartC:C2.6In

troduction

toSch

emes,

andC3.7

EllipticCurv

es

Itmay

helpto

lookback

atnotesfrom

PartB:Algebra

icCurv

es,

Commuta

tivealgebra

.

1.2.

DIF

FERENTIA

LGEOM

ETRY

versusALGEBRAIC

GEOM

ETRY

You

may

haveencounteredsomedifferential

geom

etry

(DG)in

other

courses

(e.g.B3.2Geometry

ofSurfaces).

Herearethekeydifferenceswithalgebraic

geom

etry

(AG):

(1)In

DG

youallow

allsm

ooth

functions.

InAG

youon

lyallow

polynomials

(orra

tionalfunctions,

i.e.

fractionspoly/poly).

(2)

DG

isveryflexible,e.g.you

havebumpfunctions:

smooth

func-

tionswhichare

identicallyequalto

1onaneigh

bourhoodof

apoint,

andvanishou

tsideof

aslightlylarger

neigh

bourhood.

Moreovertw

osm

ooth

functionswhichareequalon

anop

ensetneednotequal

everywhere.

AG

isvery

rigid:ifapolynom

ialvanishes

onanon

-empty

open

setthen

itisthezero

poly-

nom

ial.

Inparticular,

twopolynom

ials

whichareequalon

anon

-empty

open

setare

equal

everywhere.

AG

ishow

ever

similar

tostudyingholom

orphic

functionsin

complexdifferen-

tial

geom

etry:non-zeroholom

orphic

functionsof

onevariable

haveisolated

zeros,

andmore

generally

holomorphic

functionswhichag

reeon

anon

-empty

open

setareequal.

(3)DG

studiesspaces

X⊂

Rnor

Cncutou

tbysm

ooth

equations.

AG

studiesX⊂

kncutou

tbypolynomialequationsover

anyfieldk.AG

canstudynumber

theory

problemsbyconsideringfieldsother

than

Ror

C,e.g.

Qor

finitefieldsF p

.

(4)

DG

cannotsatisfactorily

dealwithsingu

larities.

InAG,singu

larities

arisenaturally,

e.g.

x2+

y2−

z2=

0over

Rhas

asingu

larity

at0(see

picture).

AG

has

toolsto

studysingu

larities.

(5)DG

studiesmanifolds:

amanifoldis

atopologicalspacethatlocallylook

slikeRn,so

you

canthinkof

hav

ingacopyof

asm

allEuclideanballaroundeach

point.

Thisisan

especially

nicetopolog

y:Hau

sdorff,metrizable,etc.

AG

studiesvarieties.

They

are

topolog

ical

spaces,buttheirtopology

(Zarisk

ito

pology)

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

3

isnot

sonice.

Itis

highly

non

-Hau

sdorff:foran

yirreducible1variety,

anynon

-empty

open

setis

dense,an

dan

ytw

onon

-empty

open

sets

intersectin

anon

-empty

open

dense

set!

Avarietyis

locallymodeled

onkn.Thepoints

ofknarein

1:1correspon

den

cewithmax

imal

ideals

inR

=k[x

1,...,x

n].

Thecollection

ofallmax

imal

ideals

ofR

iscalled

Specm(R

),themaxim

alsp

ectrum.

Theirreducible

closed

subsets

ofknarein

1:1correspon

dence

withtheprimeideals

ofR.Thecollection

ofallprimeideals

ofR

iscalled

Spec(R

),the

spectrum.AG

canstudyvery

general

spaces,called

schemes:

simply

replace

Rbyan

ycommutative

ring,an

dstudyspaces

whicharelocallymodeled

onSpec(R

).In

AG

studying

varietiesreduceslocallyto

commutative

algebra.

2.

AFFIN

EVARIE

TIE

S

2.1.

VANIS

HIN

GSETS

k=

algebraically

closed

field,2e.g.

Cbutnot

Q,R,F p

.Fact.

kis

aninfiniteset.

kn=

{a=

(a1,...,a

n):aj∈k}is

avector

space/kof

dim

ension

n.

Wewillworkwiththefollow

ingk-algebra

3

R=

k[x

1,...,x

n]=

(polynom

ialring/

kin

nvariab

les).

Definition.X⊂

knis

anaffine(a

lgebra

ic)variety

ifX

=V(I)forsomeideal4

I⊂

R,where

V(I)=

{a∈kn:f(a)=

0forallf∈I}

Remark

.Moregenerally

wecandefineV(S

)foran

ysubsetS⊂

R.NoticeV(S

)=

V(I)forI=�S�

theidealgenerated

byS.

EXAM

PLES.

(1)V(0)=

kn.

(2)V(1)=∅=

V(R

).(3)V(x

1−

a1,...,x

n−

an)=

{thepoint(a

1,...,a

n)}⊂

kn.

(4)V(x

1)⊂

k2is

thesecondcoordinateax

is.

(5)V(f)⊂

kncalled

hypersurface.Specialcases:

n=

2:affine

plane

curv

e.E.g.ellipticcurv

esover

C:y2−

x(x−

1)(x−

λ)=

0for

λ�=

0,1,

isatoruswithapointremoved

(andit

isaRieman

nsurface).

n=

2,deg

f=

2:conic

section.E.g.thecircle

x2+

y2−

1=

0.

n=

2,deg

f=

3:cubic

curv

e.

E.g.

thecuspidal

cubic

y2−

x3=

0.Picturesare,

strictly

speaking,

meaninglesssince

wedraw

them

over

k=

R,

whichis

not

algebraically

closed

.Thinkof

thepicture

asbeingthereal

part5of

thepicture

fork=

C.

deg

f=

1:hyperp

lane:a·x

=a1x1+···+

anxn=

0has

normal

a�=

0∈kn.

a

1A

topologicalspace

Xis

irre

ducible

ifit

isnottheunionoftw

oproper

closedsets.

2Recall

this

meansk

containsall

theroots

ofany

non-constantpolynomialin

k[x].

Thustheonly

irreducible

polynomials

are

those

ofdegreeone,

andeverypoly

ink[x]factorizesinto

degree1polys.

Italsomeansthatforany

algeb

raic

fieldextensionk�→

Kthen

k=

K.Recallafieldextensionis

algebra

icifanyelem

entofK

satisfies

apoly

over

k,forexample

anyfinitefield

exte

nsion

(meaningdim

kK

<∞

)is

algeb

raic).

3A

k-algeb

rais

aringwhichis

alsoak-vectorspace,andtheoperations+,·,andrescalingsatisfyalltheobvious

axiomsyouwould

expect.

4Idealmeans:

0∈I,I+

I⊂

I,R

·I⊂

I.

5Youneedto

becarefulwiththis.Forexample,the“circle”x2+

y2=

1over

k=

Calsocontainsthehyperbola

x2−

y2=

1byreplacingybyiy.Also,disconnectedpictureslikexy=

1over

Rbecomeconnectedover

C(w

hy?).

4C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Fact.

kalgebraically

closed⇒

V(I)=∅⇔

1∈I(soiff

I=

R)

(see

Corollary

2.1)

This

failsforR:V(x

2+y2+1)=∅(realalgebraic

geom

etry

ishard!)

EXERCIS

ES.

(1)I⊂

J⇒

V(I)⊃

V(J

).(“Themoreequation

syouim

pose,thesm

aller

thesolutionset”.)

(2)V(I)∪V(J

)=

V(I

·J)=

V(I∩J).

(3)V(I)∩V(J

)=

V(I

+J).

(Note:�I∪J�=

I+J.)

(4)V(I),V(J

)aredisjointifan

don

lyifI,J

are

relativelyprime(i.e.I+J=�1�)

2.2.

HIL

BERT’S

BASIS

THEOREM

Fact.

Hilbert’s

BasisTheorem.R

=k[x

1,...,x

n]is

aNoetherianring.

Recallthefollow

ingareequivalentdefinitionsof

Noeth

erian

ring(intuitivelya“sm

allring”):

(1)Every

idealis

finitely

genera

ted(f.g.)

I=�f

1,...,f

N�=

Rf 1

+···+

Rf N

.

(2)ACC

(Ascen

dingChainCon

dition)on

ideals:

I 1⊂

I 2⊂

···ideals⇒

I N=

I N+1=

···eventually

allbecomeequal.

Note.(1)im

plies

that

affinevarietiesarecutoutbyfinitelyman

ypolynomialequations.

Soaffine

varietiesareintersection

sofhypersurfaces:

V(I)=

V(f

1,...,f

N)=

V(f

1)∩···∩

V(f

N).

(2)im

plies

that

everyidealiscontained

insomemaxim

alideal1

m(asotherwiseI�

I 2�

I 3�

···

wou

ldcontrad

ict(2)).2

Exercise.R

Noetherian⇒

R/I

Noetherian.

Coro

llary.Anyf.g.

k-algebra

Ais

Noetherian.

Proof.

Let

f:R

=k[x

1,...,x

n]→

A,sendingthexito

achoice

ofgenerators

forA.Then

R/I∼ =

AforI=

kerf(firstisom

orphism

theorem).

�

2.3.

HIL

BERT’S

WEAK

NULLSTELLENSATZ

Fact.

Hilbert’s

WeakNullstellensa

tz.(k

algebraically

closed

iscrucial)

Themax

imal

ideals

ofR

are

ma=

(x1−a1,...,x

n−an)

fora∈kn.

Warn

ing.Fails

over

R:

m=

(x2+1)⊂

R[x]

ismax

imal

since

R[x]/m∼ =

Cis

afield.It

isnot

max

imalover

C:

(x2+1)=

((x−i)(x

+i))⊂

(x−i).

Remark

.Theevaluationhomom

orphism

eva:R→

k,x

i�→

ai,

moregenerally

eva(f)=

f(a),

has

ker

eva=

ma,so

ma=

{f∈R

:f(a)=

0}.

Proof.

For

a=

0,k[x

1,...,x

n]→

k,xi�→

0(sof�→

theconstan

tterm

ofthepolynomialf)

obviouslyhas

kernel

(x1,...,x

n).

For

a�=

0dothelinearchan

geofcoordinatesxi�→

xi−ai.

�1m

�=R

isanidealandR/m

isafield.

2Foranyring(commutativewith

1),

anyproper

idealis

alway

scontained

insideamaxim

alideal.

How

ever,to

provethis

ingen

eralrequires

transfiniteinduction(Z

orn’s

lemma),

soin

practiceit

isnotclearhow

youwould

find

themaxim

alideal.

WhereasforNoetherianrings,

youknow

thatthealgorithm

whichkeepsfindinglarger

andlarger

ideals,I�

I 2�

I 3�

···,

willhaveto

stopin

finitetime.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

5

Upsh

ot.

1

{points

ofkn}↔

{max

imal

ideals

ofR}=

Specm(R

),themaxim

alsp

ectrum

a�→

(x1−a1,...,x

n−an)=

ma

�points

ofthevariety

X=

V(I)⊂

kn

�↔

�max

imal

ideals

m⊂

RwithI⊂

m

�=

Specm(R

/I).

Notice:

ifI�⊂

mathen

somef∈Isatisfies

f(a)�=

0,so

a/∈V(I).

Coro

llary

2.1.V(I)=∅⇔

1∈I⇔

I=

R.

Proof.

If1

/∈Ithen

Iis

aproper

ideal,

soit

lies

insidesomemax

imal

idealm.

BytheWeak

Nullstellensatz

m=

maforsomea∈An.ButI⊂

maim

plies

V(I)⊃

V(m

a)=

{a}.

�

Remark

.W

ithou

tassumingkalgebraically

closed

,amax

idealm⊃

Idefines

afieldextension

k�→

R/m∼ =

K

whereR/m∼ =

Ksendsxi�→

ai.

Thisdefines

apointa∈V(I)⊂

Kn,so

itisa“K

-point”

solvingou

rpolynom

ialequations,

butwedon

’t“see”this

pointover

kunless

a∈kn⊂

Kn.For

kalgebraically

closed

,k=

Kbecau

sek�→

Kisan

algebraicextension

bythefollow

ingFact,so

we“see”everything.

KeyFact.

Kf.g.k-algebra

+K

field⇒

Kf.g.as

ak-m

odule2⇒

k�→

Kfinite⇒

k�→

Kalgebraic.

(Because

theKey

Fact

implies

theWeakNullstellensatz

viatheRem

ark,theKey

Fact

issometim

esalsocalled

theWeakNullstellensatz).

Example.i∈V(x

2+1)⊂

Cbut∅=

V(x

2+1)⊂

R�→

C.

2.4.

ZARIS

KITOPOLOGY

TheZarisk

ito

pologyon

knis

defined

bydeclaring3that

theclosed

sets

aretheV(I).

Theop

ensets

arethe

UI

=kn\V

(I)

=kn\(

V(f

1)∩···∩

V(f

N))

=(k

n\V

(f1))∪···∪

(kn\V

(fN))

=D(f

1)∪···∪

D(f

N)

wheretheD(f

i)arecalled

thebasicopen

sets,where

D(f)=

Uf=

kn\V

(f)=

{a∈kn:f(a)�=

0}.

Exercise.Affinevarietiesarecompact:4an

yop

encover

ofan

affinevarietyX

has

afinitesubcover.

Definition.Affinesp

aceAn=

An kis

thesetAn=

knwiththeZariskitopology.

Example.A1 k=

khas

closed

sets∅,k,{finitepoints},

andop

ensets∅,k,an

d(thecomplementof

anyfinitesetof

points).

Itis

not

Hau

sdorffsince

anytw

onon

-empty

open

sets

intersect.

Theop

ensets

aredense

(astheon

lyclosed

setwithinfinitelyman

ypoints

isk,usingthat

kis

infinite).

Definition.TheZarisk

ito

pology

onanaffinevarietyX⊂

Anis

thesubspace

topology,so

the

closedsets

are

V(I

+J)=

X∩V(J

)foranyidealJ⊂

R(equivalently,

V(S

)forideals

I⊂

S⊂

R).

Anaffi

nesu

bvariety

Y⊂

Xis

aclosedsubset

ofX.

1Forthelast

equality,recall:

{ideals

J⊂

RwithI⊂

J}

↔{ideals

J⊂

R/I}

J�→

J=

{j=

j+

I∈R/I:j∈J}

J=

{j∈R

:j∈J}

←�

J.

2i.e.

ak-vectorspace.Clarification:in

analgeb

rayouare

allow

edto

multiply

gen

erators,in

amodule

youare

not.

3In

fact

itis

thesm

allesttopologysuch

thatpolynomials

are

continuousandanypointis

aclosedset.

4Historically

this

property

iscalled

quasi-compactn

ess

rather

than

compactness,

toremind

ourselvesthatthe

topology

isnotHausdorff

.

6C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

2.5.

VANIS

HIN

GID

EAL

For

anysetX⊂

An,let

I(X)=

{f∈R

:f(a)=

0foralla∈X}

EXAM

PLES.

(1)I(a)=

ma=

{f∈R

:f(a)=

0}.

(2)I(V(x

2))

=I(0)

=(x)⊂

k[x],so

I(V(I))�=

Iin

general.

Exercises.

(1)X⊂

Y⇒

I(X)⊃

I(Y).

(2)I⊂

I(V(I)).

Lemma2.2.V(I(V

(I)))=

V(I),

inparticularV(I(X

))=

XforanyaffinevarietyX.

Proof.

TakeV(·)

ofexercise

2ab

ove,

togetV(I(V

(I)))⊂

V(I).

Con

versely,

bycontradiction

,ifa∈

V(I)\V

(I(V

(I)))then

thereis

anf∈

I(V(I))

withf(a)�=

0.

Butsuch

anfvanishes

onV(I),an

da∈V(I).

�

Coro

llary.Foraffinevarieties,

X1=

X2⇔

I(X

1)=

I(X

2).

2.6.

IRREDUCIB

ILIT

YAND

PRIM

EID

EALS

Anaffi

nevarietyX

isreducible

ifX

=X

1∪X

2forproper

closedsubsets

Xi(soX

i�

X).

Otherwise,

callX

irreducible.1

Remark

.Som

ebook

srequirevarietiesto

beirreducible

bydefinition,andcallthegeneralV(I)

affinealgebra

icsets.Wedon

’t.

EXAM

PLES.

(1)V(x

1x2)=

V(x

1)∪V(x

2)is

reducible

(2)Exercise.X

irreducible⇔

anynon

-empty

open

subsetis

dense.

(3)Exercise.X

irreducible⇔

anytw

onon

-empty

open

subsets

intersect.

(4)In

aHau

sdorff

topological

space,

only

theem

pty

setan

don

epointsets

are

irreducible.

Theorem.X

=V(I)�=∅is

irreducible⇔

I(X)⊂

Ris

aprimeideal.2

Warn

ing.I⊂

Rneednotbeprime:

I=

(x2)is

not

primebutI(V(x

2))

=(x)is

prime.

Proof.

IfI(X)is

not

prime,

then

pickf 1,f

2satisfyingf 1

/∈I(X),f 2

/∈I(X),f 1f 2∈I(X).

Then

X⊂

V(f

1f 2)=

V(f

1)∪V(f

2)

sotakeX

i=

X∩V(f

i)�=

X(since

f i/∈I(X)).

Con

versely,

ifX

isnot

irreducible,X

=X

1∪X

2,X

i�=

X,so

(byLem

ma2.2)thereare

f i∈

I(X

i)\I

(X)butf 1f 2∈I(X),so

I(X)is

not

prime.

�

Notice,

abbreviatingI=

I(X),J=

I(Y),

{irred

ucible

varietiesX⊂

An}↔

{primeideals

I⊂

R}=

Spec(R

){irred

ucible

subvarietiesY

=V(J

)⊂

X=

V(I)⊂

An}↔

{primeideals

J⊃

IofR}

↔{p

rimeideals

JofR/I}=

Spec(R

/I).

Remark

.Spec(k)=

{0}

=just

apoint!3

So,

inseminars,

when

someonewritesSpec(k)�→

Spec(R

/I)they

arejust

saying“given

apointin

anaffi

nevariety...”.

1SoX

=X

1∪X

2forclosedX

iim

plies

Xi=

Xforsomei.

2I�=

Ris

anidealandR/Iis

anintegraldomain.

3Because

theonly

ideals

insideafieldkare

0,k

.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

7

2.7.

DECOM

POSIT

ION

INTO

IRREDUCIB

LE

COM

PONENTS

Theore

m.Anaffinevarietycanbe

decomposedinto

irre

ducible

components:thatis,

X=

X1∪X

2∪···∪

XN.

wheretheX

iare

irreducibleaffinevarieties,

andthedecompositionis

uniqueupto

reorderingifwe

ensure

Xi�⊂

Xjforalli�=

j.

Proof.

ProofofExisten

ce.Bycontrad

iction

,supposeitfailsforX.

SoX

=Y1∪Y

� 1forproper

subvars.

Soit

failsforY1or

Y� 1,W

LOG

Y1.

SoY1=

Y2∪Y

� 2forproper

subvars.

Soit

failsforY2or

Y� 2,W

LOG

Y2.

Con

tinueinductively.

Weob

tain

asequen

ceX⊃

Y1⊃

Y2⊃

···.

SoI(X)⊂

I(Y1)⊂

I(Y2)⊂

···.

SoI(YN)=

I(YN+1)=

···eventually

equal,since

Ris

Noetherian(H

ilbertBasis

Thm).

So,

byLem

ma2.2,

YN

=V(I(Y

N)))=

V(I(Y

N+1))

=YN+1whichis

not

proper.Con

trad

iction

.ProofofUniquen

ess.

SupposeX

1∪···∪

XN

=Y1∪···∪

YM,withX

i�⊂

Xjan

dYi�⊂

Yjfori�=

j.X

i=

(Xi∩Y1)∪···∪

(Xi∩YM)contrad

icts

Xiirreducible

unless

someX

i∩Y�=

Xi.

SoX

i⊂

Y�forsome�.

Sim

ilarly,Y�⊂

Xjforsomej.

SoX

i⊂

Y�⊂

Xj,contrad

ictingX

i�⊂

Xjunless

i=

j.Soi=

jan

dso

Xi=

Y�.

Given

i,the�isunique(dueto

Yi�⊂

Yjfori�=

j)an

dvice-versagiven

�thereisauniquesuch

i.�

Remark

.Thefact

that

Ris

aNoetherianringim

plies

that

affinevarietiesareNoeth

erian

topo-

logicalsp

aces,

i.e.

givenadescendingchain

X⊃

X1⊃

X2⊃

···

ofclosed

subsets

ofX,then

XN

=X

N+1=

···areeventually

allequal.

Proof.

Tak

eI(·)an

duse

theACC

onideals.SoI(X

N)=

I(X

N+1)=

···a

reeventually

equal.Then

takeV(·)

anduse

Lem

ma2.2.

�

2.8.

IRREDUCIB

LE

DECOM

POSIT

IONS

and

PRIM

ARY

IDEALS

This

Section

isnot

verycentral

tothecourse.

See

theAppendix,Section

16.

2.9.

I(V(·)

)AND

V(I(·)

)

Motivation.ByLem

ma2.2,

ifX

isavarietythen

V(I(X

))=

X.

Ofcourse,

theassumption

was

tobeexpected,since

V(·)

isalway

sclosed,so

forthisequalityto

hold

wecertainly

needX

tobeclosed

,i.e.

avariety.

Under

what

assumption

onan

idealIcanwegu

aran

tee

I(V(I))

? =I.

(2.1)

Thequestion

really

is,what

isspecialab

outtheidealswhichariseas

I(V(·)

)?Observethat

I(V(I))

isalway

sara

dicalideal:

ifit

containsapow

erfm

then

itmust

contain

f.Indeed,iffm(a)=

[f(a)]m

=0∈kthen

f(a)=

0.Weshow

nextthat

foran

yradical

idealI,(2.1)holds.

Definition.Thera

dical√IofanidealI⊂

Ris

defi

ned

by√I=

{f∈R

:fm∈Iforsomem}.

Iis

called

ara

dicalidealifI=√I.

8C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Example.V(x

3)=

{0}⊂

A1andI(V(x

3))

=�x�=

��x

3�.

So�x�i

sradical,but�x

3�isnot.

Exercise.Checkthat

V(I)=

V(√

I).

Exercise.I⊂

Ris

radical⇔

R/Ihasnonilpotent1

elem

ents,i.e.

R/I

isare

duced

ring.

Example.Anyprimeidealis

radical.

Motivation.Theproblem

isthat

V(·)

forgetssomeinform

ation.Oneshould

really

view

V(x

3)as

being0∈A1withamultiplicity

3of

vanishing.This

idea

isat

theheart

ofthetheory

ofschemes.

Loosely,

aschem

eshould

bea“variety”together

withachoice

ofaringoffunctions.

Theringof

functionsassociated

to(x

3)is

k[x]/x3,whichis

3-dim

ension

al,whereasfor(x)it

isk[x]/x,which

is1-dim

ension

al.The“a

dditionaldim

ension

s”canbethough

tof

asan

infinitesim

althickeningof

thevariety,

asit

keepstrack

ofad

ditionalderivatives.Rou

ghly:f=

a+

bx+

cx2∈

k[x]/x3has

∂xf(0)=

ban

d∂x∂xf=

2c,

whereask[x]/xonly

“sees”

f∼ =

a∈k[x]/x.

2.10.

HIL

BERT’S

NULLSTELLENSATZ

Theore

m2.3

(Hilbert’sNullstellensatz).

I(V(I))

=√I

Inparticular,

ifIis

radicalthen

I(V(I))

=I.

Proof.

Wewillprovethis

later.

�Coro

llary.Thereare

order-reversing2

bijections

{varieties}↔

{radicalideals}

{irreduciblevarieties}

↔{p

rimeideals}=

Spec(R

){points}↔

{maximalideals}=

Specm(R

)X

�→I(X)

V(I)←�

I.

Proof.

Thesearebijection

sbecause

V(I(V

(I)))=

V(I)byLem

ma2.2,

andI(V(I))

=Iforradical

ideals

IbyTheorem

2.3.

�TheNullstellensatz

(“Zeros

theorem”)

owes

itsnam

eto

theproofof

theexistence

ofcommon

zeros

foran

ysetof

polynomialequation

s(crucially,

ofcourse,

kis

algebraically

closed

):

Lemma2.4.ForanyproperidealI⊂

R,wehave

V(I)�=∅.

Proof.

Pickamax

imalidealI⊂

m⊂

R.

ByHilbert’sweakNullstellensatz,m

=m

a=

(x1−

a1,...,x

n−

an)forsomea∈kn.Hence

V(I)⊃

V(m

a)=

{a}⊃

V(R

)=∅.

�Pro

ofofth

eNullstellensa

tz.

Easydirection

:ab

oveweshow

edI(V(I))

isalwaysradical,weknow

I⊂

I(V(I)),so√I⊂

I(V(I)).

Rem

ainsto

show

I(V(I))⊂√I.

Given

g∈I(V(I)).

Trick:letI�=�I,yg−

1�⊂

k[x

1,...,x

n,y](theidea

being:

wegoto

anew

ringwhereg=

0is

impossible

inV(I

� )).

Observethat

V(I

� )=∅⊂

An+1.

ByLem

ma2.4,

I� =

k[x

1,...,x

n,y].

So1∈I� .

So1=

G0(x

1,...,x

n,y)·(yg−

1)+

�G

i(x1,...,x

n,y)·f

iforsomepolynomials

Gj,an

dwheref i

arethegeneratorsofI=�f

1,...,f

N�.

For

large�,

g�=

F0(x

1,...,x

n,gy)·(yg−

1)+

�Fi(x1,...,x

n,gy)·f

iforsomepolynom

ials

Fj

(notice3

thelast

variable

isnow

gyinsteadof

y).

1r∈R

isnilpotentifrm

=0forsomem

∈N.

2RecallX

⊂Y

⇒I(X)⊃

I(Y),

I⊂

J⇒

V(I)⊃

V(J

).3Example:ify3+y=

G(y)then

multiply

byg3to

get:g3y3+g3y=

(gy)3

+g2(gy)=

F(gy)whereF(z)=

z3+g2z.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

9

Since

yis

aform

alvariab

le,wemay

1replace

gyby1,

sog�=

�Fi(x1,...,x

n,1)·f

i∈I.

Sog∈√I.

�

2.11.

FUNCTIO

NS

Motivatingquestion

:what

map

sX→

A1dowewan

tto

allow?

Answ

er:an

ypolynom

ialin

thecoord

inate

functionsxi:a=

(a1,...,a

n)�→

ai.

Thefollow

ingaredefinitions(andnoticetheisom

orphismsarek-algebra

isos):

Hom

(An,A

1)

={p

olynom

ialmap

sAn→

A1,a�→

f(a),

somef∈R}

∼ =R.

Hom

(X,A

1)

={restriction

sto

Xof

such

map

s}∼ =

R/I(X

).

Noticethat

therestricted

map

sdonot

chan

geifwead

dg∈I(X)as

(f+

g)(a)=

f(a)fora∈X.

Wemay

putabar

fover

fas

areminder

that

wepassedto

thequotient,

sof+

g=

fifg∈I(X).

Remark

.Theab

oveareisom

orphismsbecau

sef 1

=f 2

asmap

sAn→

A1iff

f 1−f 2∈I(An)=

{0},

similarly

f 1=

f 2as

map

sX→

A1iff

f 1−

f 2∈I(X).

That

abstract

polynom

ials

canbeidentified

withtheirassociated

functionsrelies

onkbeinginfinite2

(whichholdsas

kisalgebraically

closed).

For

thefieldk=

Z/2therearefourfunctionsk→

kwhereask[x]containsinfinitelyman

ypolynom

ials.

2.12.

THE

COORDIN

ATE

RIN

G

Definition.Thecoord

inate

ringis

thek-algebra

generated

bythecoordinatefunctionsxi,

k[X

]=

R/I(X

).

EXAM

PLES.

1)k[A

n]=

k[x

1,...,x

n]=

R.

2)X

={(a,a

2,a

3)∈k3:a∈k}=

V(y−

x2,z−

x3),

then

3k[X

]=

k[x,y,z]/(y−

x2,z−

x3).

3)V

=(cuspidal

cubic)=

{(a2,a

3):a∈A1}=

V(x

3−

y2),

then

4k[V

]=

k[x,y]/(x

3−

y2).

Lemma2.5

(Thecoordinateringseparates

points).

Given

anaffinevarietyX,andpoints

a,b∈X,

iff(a)=

f(b)forallf∈k[X

]then

a=

b.

Proof.

Ifa�=

b∈X⊂

An,somecoordinateai�=

b i,so

f=

xi∈k[X

]has

f(a)=

ai�=

b i=

f(b).

�

2.13.

MORPHIS

MS

OF

AFFIN

EVARIE

TIE

S

F:An→

Am

isamorp

hism

(orpolynomialmap)ifit

isdefined

bypolynom

ials:

F(a)=

(f1(a),...,f m

(a))

forsomef 1,...,f

m∈R.

F:X→

Yisamorp

hism

ofaffinevarietiesifitistherestrictionof

amorphism

An→

Am

(here

X⊂

An,Y⊂

Am),

so

F(a)=

(f1(a),...,f m

(a))

forsomef 1,...,f

m∈k[X

].

1View

theequation

forg�asan

equation

inthevariable

(gy−

1)over

Rrather

than

ingy(this

isachangeof

variables),then

“puttinggy=

1”is

thesameassaying“compare

theorder

zero

term

ofthepolynomialover

Rin

the

variable

gy−1”.Algeb

raically,

thekey

is:k[x

1,...,x

n]�→

k[x

1,...,x

n,y]/(yg−1),

xi�→

xiis

aninjectivek-alg

hom.

2Hint.

Iff:A

n→

kvanishes,fixai∈k,then

f(λ,a

2,...,a

n)isapoly

inonevariable

λwithinfinitelymanyroots.

3Strictlysp

eaking,oneneedsto

checkthatI=

(y−

x2,z

−x3)is

aradicalideal,

since

k[X

]is

thequotientof

k[x,y,z]by√I=

I(X).

Notice

thatk[x,y,z]/(y

−x2,z

−x3)∼ =

k[t]via

x�→

t,y�→

t2,z�→

t3,withinverse

mapgiven

byt�→

x.Since

k[t]is

anintegraldomain,it

hasnonilpotents,so

Iis

radical(infact

wealsoproved

Iis

prime).We

remark

thatI(X)=

(y−

x2,z

−x3)now

follow

sbytheNullstellensatz:V(I)=

Xso

I(X)=

I(V(I))

=√I=

I.

4Again,weneedto

checkI(V)=

(x3−

y2).

Note

thatif(α

,β)∈

V(x

3−

y2)wecanpicka∈

kwitha2=

α(ask

isalg.closed).

Then

y2=

a6so

y=

±a3,andwecanget

+a3byreplacingaby−aifnecessary.SoV(x

3−

y2)⊂

V⊂

V(x

3−y2),

hen

ceequality.Wenow

show

(x3−y2)is

prime(hen

ceradical).Since

k[x,y,z]is

aUFD

(soirreducible

⇔prime),itisen

oughto

checkthatx3−y2isirreducible.Ifitwasreducible,then

x3−y2would

factorize

asapolynomial

inxover

theringk[y].

Sotherewould

bearootx=

p(y)forapolynomialp.This

isclearlyim

possible

(checkthis).

10

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

F:X→

Yis

anisomorp

hism

ifF

isamorphism

andthereis

aninversemorphism

(i.e.thereis

amorphism

G:Y→

Xsuch

that

F◦G

=id,G◦F

=id).

Example.(V

(xy−

1)⊂

A2)→

A1,(x,y)�→

xis

amorphism.Noticetheim

ageA1\{

0}is

not

asubvarietyof

A1.

Theorem.ForaffinevarietiesX⊂

An,Y⊂

Am

thereis

a1:1

corresponden

ce

Hom

(X,Y

)←→

Hom

k-alg(k[Y

],k[X

])=

{k-algebra

hom

sk[Y

]→

k[X

]}F

=ϕ∗:X→

Y←→

ϕ=

F∗:k[Y

]→

k[X

]

←→

ϕ=

F∗:Hom

(Y,A

1)→

Hom

(X,A

1),

g�→

F∗ g

=g◦F

wherek[X

]=

k[x

1,...,x

n]/I(X),

k[Y

]=

k[y

1,...,y

m]/I(Y)and

F∗ (y i)

=f i(x

1,...,x

n)=

ϕ(y

i)

ϕ∗ (a)

=(ϕ

(y1)(a),...,ϕ(y

m)(a))

=(f

1(a),...,f m

(a)).

Proof.

Thecorrespon

den

cemap

s,in

thetw

odirection

s,arewell-defi

ned.1

�(F

∗ )∗ (a)=

(F∗ (y 1)(a),...,F

∗ (y m

)(a))

=(f

1(a),...,f m

(a))

=F(a),

so(F

∗ )∗=

F.�

(ϕ∗ )

∗ (y i)=

ϕ(y

i),so

(ϕ∗ )

∗=

ϕ.�

�Remark

.Themap

sϕ∗ ,

F∗arecalled

pull-b

ack

s(orpull-backmap

s).

EXAM

PLES.

1)F

:A1→

V=

{(a,a

2,a

3)∈k3:a∈k},

F(a)=

(a,a

2,a

3)then

k[A

1]=

k[t]

F∗

←−

k[V

]=

k[x,y,z]/(y−x2,z−x3)

t←�

xt2

←�

yt3

←�

z

2)F

:A1→

V=

{(a2,a

3):a∈A1}=(cuspidal

cubic),

F(a)=

(a2,a

3)then

k[A

1]=

k[t]

F∗

←−

k[V

]=

k[x,y]/(x

3−y2)

t2←�

xt3

←�

y.

Exercise.

F:X→

Ymorph⇒

F−1(V

(J))

=V(F

∗ J)⊂

Xforan

yclosed

setV(J

)⊂

Y.

So

morphismsarecontinuou

sin

theZariskitopology.

EXERCIS

ES.

1)X

F →Y

G →Z⇒

(G◦F

)∗=

F∗◦G

∗:k[Z

]G

∗ →k[Y

]F

∗ →k[X

].

2)k[Z

]ψ →

k[Y

]ϕ →

k[X

]⇒

(ϕ◦ψ

)∗=

ψ∗◦ϕ

∗:X

ϕ∗ →Y

ψ∗ →Z.

Coro

llary.Foraffinevarieties,

X∼ =

Y⇔

k[X

]∼ =

k[Y

].

Proof.

IfX

F →Y

has

inverse

G,F◦G

=id

so(F◦G

)∗=

G∗◦F

∗=

id∗=

id.Sim

ilarly

forG◦F

.

Ifk[Y

]ϕ →

k[X

]has

inverseψ,ϕ◦ψ

=id

so(ϕ◦ψ

)∗=

ψ∗◦ϕ

∗=

id∗=

id.Sim

ilarly

forψ◦ϕ

.�

EXAM

PLES.

1)V

={(a,a

2,a

3)∈A3:a∈A1}∼ =

A1via

(a,a

2,a

3)↔

a,indeedk[V

]∼ =

k[t]∼ =

k[A

1]via

x↔

t.2)

Inthecuspidal

cubic

exam

ple

above,

Fisabijectivemorphism

butitcannot

bean

isom

orphism

becau

seF

∗is

not

anisom

orphism

(itdoes

not

hit

tin

theim

age).Theidea

isthat

Vhas

“few

erpolynom

ialfunctions”

than

A1dueto

thesingu

larity

at0.

Con

vince

yourselfthat

k[t],k[V

]arenot

isom

orphic

k-algebras,

sotherecannot

bean

yisom

orphism

A1→

V(stron

gerthan

just

Ffailing).

3)Exercise.IfF

:X→

Yisasurjectivemorphism

ofaffi

nevarieties,an

dX

isirreducible,then

Yis

irreducible.Show

that

itsufficesthat

Fis

dominant,

i.e.

has

dense

image.

Example.Y

={(t,t2,t

3):t∈k}is

irreducible

asitis

theim

ageof

A1→

Y,t�→

(t,t

3,t

3).

1In

particularϕ

∗ (X)⊂

Y⊂

Am,because

g(ϕ

∗ (a))

=ϕ(g)(a)=

0forallg∈I(Y)anda∈X,asg=

0∈k[Y

].

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

11

3.

PROJECTIV

EVARIE

TIE

S

3.1.

PROJECTIV

ESPACE

Notation:

k∗=

k\{

0}=

units,

i.e.

theinvertibles.

For

Van

yvector

space/k,definethepro

jectivisationby

P(V)=

(V\{

0})/(k

∗ -rescalingaction

v�→

λv,forallλ∈k∗ ).

Noticethis

alwayscomes

withaquotientmap

π:V

\{0}→

P(V),v�→

[v],where[v]=

[λv].

Bypickinga(linearalgebra)basis

forV,wecansupposeV

=kn+1.Wethen

obtain

Pn=

Pn k=

P(kn+1),called

pro

jectivesp

ace,defined

asfollow

s

Pn=

P(kn+1)

=(spaceof

straightlines

inkn+1through

0)

Write

[a0,a

1,...,a

n]or

[a0:a1:···:

an]fortheequivalence

classof

(a0,a

1,...,a

n)∈

kn+1\{

0},

whosecorrespon

dinglinein

kn+1isk·(a0,...,a

n)⊂

kn+1.Via

therescalingaction

,wethusidentify

[a0:...:an]=

[λa0:...:λan]

forallλ∈k∗ .

Asbefore,

wehaveaquotientmap

π:An+1\{

0}→

Pn,π(a)=

[a].

Thecoordinates

x0,...,x

nof

kn+1=

An+1arecalled

homogeneouscoord

inatesof

Pn,althou

ghnoticethey

arenot

well-defi

ned

functionson

Pn:xi(a)=

aibutxi(λa)=

λai.

EXAM

PLES.

1)For

k=

R(not

algebraically

closed

,butausefulexam

ple),

RPn

=Sn/(identify

antipodal

points

a∼−a)

becau

sethestraightlinein

Rn+1correspon

dingto

thegivenpointof

RPn

willintersecttheunit

sphereof

Rn+1in

twoan

tipodal

points.

a

−a

lineRa

2)For

k=

C,n=

1,

CP1

=P1 C

=C∪{infinity}∼ =

S2

thelast

isom

orphism

isthestereographic

projection,Above,

identify

[1:z]

withz∈

C,an

d[0

:1]

with∞.

Note[a

:b]

=[1

:z]if

a�=

0,taking

z=

b/a,usingrescalingbyλ=

a−1.For

a=

0,weget[0

:b]

=[0

:1],

rescalingbyλ=

b−1(note:

[0:0]

isnot

anallowed

point).

∞

S2

C

Wecanthinkof

Pnas

arisingfrom

“com

pactifying”

Anbyhyperplanes,planes,an

dpoints

atinfinity:

Pn=

{[1:a1:···:

an]}∪{[0:a1:···:

an]}

=An∪Pn

−1

=···

(byinduction)

=An∪An−1∪···∪

A1∪A0

whereA0is

thepoint[0

:0:···:

0:1].

3.2.

HOM

OGENEOUSID

EALS

Motivatingexample.Con

sider

f(x,y)=

x2+y3,an

d[a

:b]∈P1

.It

isnot

clearwhat

f[a

:b]=

0means,

since

[a:b]

=[3a

:3b]butf(a,b)=

a2+

b3=

0an

df(3a,3b)

=9a

2+

27b3

=0are

differentequations.

How

ever,forthehomogen

eouspolynom

ialF(x,y)=

x2y+

y3,theequations

F(a,b)=

a2b+b3

=0an

dF(3a,3b)

=27(a

2b+b3)=

0areequivalent,so

F[a

:b]=

0ismeaningful.

Nota

tion.R

=k[x

0,...,x

n]

(kalgebraically

closed)

Definition.F∈

Ris

ahomogeneouspolynomialofdegreedifallthemon

omials

xi 0 0···x

i n n

12

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

appearingin

Fhavedegreed=

i 0+···+

i n.Byconvention

,0∈R

ishom

ogeneousofeverydegree.

Noticean

ypolynomialf∈R

decom

poses

uniquelyinto

asum

ofhom

ogeneouspolynomials

f=

f 0+···+

f d,

wheref i

isthehom

ogeneouspartof

degreei,an

ddis

thehighestdegreethatarises.

Lemma3.1.Forf∈

R,if

fvanishes

atallpoints

ofthelinek·a⊂

An+1(correspondingto

the

point[a]∈Pn

)then

each

homogen

eouspart

offvanishes

at[a].

Proof.

0=

f(λa)=

f 0(a)+

f 1(a)λ+

···+

f d−1(a)λd−1+

f d(a)λdis

apolynomial/k

inλ

with

infinitely1man

yroots.

Soitis

thezero

polynom

ial,i.e.

thecoeffi

cients

vanish:f i(a)=

0,alli.

�

Exercise.F

ishom

ogeneousof

degreed⇔

F(λx)=

λdF(x)forallλ∈k∗ .

Definition.I⊂

Ris

ahomogeneousidealifit

isgeneratedbyhom

ogeneouspolynomials.

Exercise.I⊂

Ris

hom

ogeneous⇔

foran

yf∈I,allitshom

ogeneouspartsf i

alsoliein

I.

Example.For

R=

k[x,y],(x

2y+y3)=

k[x,y]·(x

2y+y3)is

hom

ogeneous.

Example.(x

2,y

3)=

R·x

2+R·y

3is

hom

ogeneous.

Non-example.(x

2+y3)is

not

hom

ogeneous:

itcontainsx2+y3butnotitshom.partsx2,y

3.

Exercise.2

Deduce

that

ahom

ogeneousidealis

generated

byfinitelyman

yhom

ogeneouspolys.

3.3.

PROJECTIV

EVARIE

TIE

Sand

ZARIS

KITOPOLOGY

Definition.X⊂

Pnis

apro

jectivevariety

if

X=

V(I)=

{a∈Pn

:F(a)=

0forallhomogeneousF∈I}

forsomehom

ogeneousidealI.

Definition.TheZarisk

ito

pology

onPn

has

closed

sets

precisely

theprojectivevarietiesV(I).

TheZariskitopology

onaprojectivevarietyX⊂

Pnis

thesubspacetopology,

sotheclosedsubsets

ofX

areX∩V(J

)=

V(I

+J)foran

yhom

ogeneousidealJ

(equivalently,

V(S

)forhomogeneous

ideals

I⊂

S⊂

R).

Apro

jectivesu

bvariety

Y⊂

Xis

aclosed

subsetofX.

EXAM

PLES.

1)Pro

jectivehyperp

lanes:

V(L

)⊂

PnwhereL=

a0x0+

···+

anxnis

homogeneousofdegree1

(alinearform

).In

particular,

thei-th

coord

inate

hyperp

laneis

Hi=

V(x

i)=

{[a0:...:ai−

1:0:ai+

1:...:an]:aj∈k}.

2)Pro

jectivehypersurface:V(F

)⊂

Pnforanon

-con

stan

thom

ogeneouspolynomialF∈

R.A

quadric(cubic,quartic,etc.)is

aprojectivehypersurfacedefined

byahomogeneouspolynomial

ofdegree2(respectively

3,4,

etc.).

For

exam

ple,theellipticcurv

esV(y

2z−x(x−z)(x−cz))⊂

P2(w

herec�=

0,1∈k)are

cubicsin

P2.

3)(P

rojective)

linearsu

bsp

aces:

theprojectivisationP(V)⊂

Pnof

anyk-vectorsubspace

V⊂

kn+1

isaprojectivevariety.

Itiscutoutbylinearhom

ogeneouspolynom

ials.Thecase

dim

kV

=1gives

apointin

Pn.Thecase

dim

kV

=2defines

the(projective)

linesin

Pn.Exam

ple:V

=spank(e

0,e

1)⊂

k3yieldstheline{[t 0

:t 1

:0]∈P2

:t 0,t

1∈k}=

{[1:t:0]:t∈k}∪

{[0:1:0]}∼ =

P1.

Exercise.

Usingbasic

linearalgebra

inkn+1,show

that

thereis

auniquelinethrough

anytw

odistinct

points

inPn

,an

dthat

anytw

odistinct

lines

inPn

meetin

exactlyonepoint.

3.4.

AFFIN

ECONE

For

aprojectivevarietyX⊂

Pn,theaffineconeX⊂

An+1istheunionofthestraightlines

inkn+1

correspon

dingto

thepoints

ofX.Thus,

usingthequotientmap

π:An+1\{

0}→

Pn,x�→

[x],

� X=

{0}∪

π−1(X

)⊂

An+1ifX�=∅,

and

� ∅=∅⊂

An+1.

1hereweuse

thatkis

aninfiniteset,

since

kis

algeb

raicallyclosed.

2RecalltheHilbertBasistheorem,i.e.

Ris

Noetherian.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

13

Exercise.If∅�=

X=

V(I)⊂

Pn,forsomehom

ogeneousidealI⊂

R,then

� Xis

theaffi

nevariety

associated

totheidealI⊂

R,

� X=

Vaffine(I)⊂

An+1.

Remark

.X=∅on

lyarises

ifI⊂

Rdoes

not

vanishon

anylinein

An+1.Byhom

ogeneity

ofI,this

forces

V(I)⊂

An+1to

beeither∅or

{0},

whichbyNullstellensatz

correspon

dsrespectivelyto

I=

R

orI=

(x0,...,x

n).

Wewan

t� X=∅so

I=

R.Theexercise

wou

ldfailfortheirre

levantideal

I irr=

(x0,...,x

n).

Noticethemax

imal

hom

ogeneousidealI irrdoes

not

correspon

dto

apointin

Pn([0]

isnot

allowed).

InSection

3.3wecould

havedefined

V(I)=

{a∈Pn

:f(α

)=

0forallf∈I,an

dallrepresentativesα∈An+1of

a},

sohereα∈π−1(a)is

anypointon

thelinek·α

defined

bya.

Exercise.CheckthisdefinitiongivesthesameV(I),byusingLem

ma3.1(sof(k

·α)=

0forces

all

hom

ogeneousparts

offto

vanishat

a∈Pn

).

Exercise.1

Show

that

V(I)=

π(V

affine(I)\0

).

3.5.

VANIS

HIN

GID

EAL

R=

k[x

0,...,x

n].

For

anysetX⊂

Pn,defineIh(X

)to

bethehom

ogeneousidealgenerated

bythehom

ogeneouspolys

vanishingon

X:

Ih(X

)=�F∈R

:F

hom

ogeneous,F(X

)=

0�.

Exercise.If

Iis

hom

ogeneous,

then

V(I

h(V

(I)))=

V(I)an

dI⊂

Ih(V

(I)).

Warn

ing.V(I

irr)=∅⊂

Pn,butIh(∅)=

R�=√I irr=

I irr.Sim

ilarly,if√I=

I irrthen

V(I)=

V(√

I)=∅an

dIh(V

(I))

=R.Thesearetheon

lycaseswheretheproj.Nullstellensatz

fails(Sec.3.6).

Lemma3.2.

Ih(X

)=

{f∈R

:f(α

)=

0foreveryα∈An+1representinganypointofX⊂

Pn}

=I(

� X).

Proof.

This

follow

sbyLem

ma3.1:

f∈Ih(X

)⇔

f(X

)=

0⇔

f(� X)=

0⇔

f∈I(

� X).

�

3.6.

PROJECTIV

ENULLSTELLENSATZ

Theore

m(P

rojectiveNullstellensatz).

Ih(V

(I))

=√I

foran

yhom

ogeneousidealIwith√I�=

I irr

Proof.

Vaffine(I)�=

{0}bytheaffi

neNullstellensatz,as√I�=

I irr.SoX

=V(I)=

π(V

affine(I)\0

)⊂

Pnisnon

-empty,so

itsaffi

neconeis

� X=

Vaffine(I).

UsingLem

ma3.2an

dtheaffi

neNullstellensatz

weob

tain:Ih(X

)=

I(� X)=

I(Vaffine(I))

=√I.

�

Remark

.From

Section

3.4,

ifX

=V(I)=∅,

then

I=

either

Ror

I irr,butIh(X

)=

R.

Theore

m.Thereare

1:1corresponden

ces

{proj.

vars.X⊂

Pn}↔

{homogen

eousradicalideals

I�=

I irr}

{irred.proj.

vars.X⊂

Pn}↔

{homogen

eousprimeideals

I�=

I irr}

{points

ofPn

}↔

{“maximal”

homogen

eousideals

I�=

I irr}

∅↔

{thehomogen

eousidealR}

1Hint.

Notice

thatV(I)=

X=

π(� X\0

)=

π(V

affi

ne(I)\0

).

14

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

wherethemapsare:X�→

Ih(X

)andV(I)←� I.

Thepointp=

[a0:···:

an]∈Pn

corresponds1

tothehomogen

eousideal

mp=�a

ixj−ajxi:alli,j�

={h

omogen

eouspolysvanishingata}

whichamongsthomogen

eousideals

differen

tfrom

I irris

maximalwithrespectto

inclusion.

Remark

.Themaxim

alideals

ofk[x

0,...,x

n]are�x

i−ai:alli�

inbijectionwithpoints

a∈An+1.

Theseideals

arenothom

ogeneousfora�=

0.In

fact,theonly

homogeneousmax

imalidealis

I irr

(thecase

a=

0).Thepoints

p∈Pn

correspon

dto

lines

inAn+1,so

they

are

primebutnotmax

imal

ideals.Thesearethehom

ogeneousideals

mp⊂

I irr⊂

k[x

0,...,x

n]show

nabove.

3.7.

OPEN

COVERS

Ui=

Pn\H

i=

{[x]∈Pn

:xi�=

0}is

called

thei-th

coord

inate

chart.

Exercise.

φi:Ui→

An

[x]=

[x0

xi:···:

xi−

1

xi

:1:xi+

1

xi

:···:

xn xi]→

(x0

xi,···

,xi−

1

xi,xi+

1

xi,···

,xn xi)

isabijection

,indeedahomeomorphism

intheZariskitopologies.2

Con

sequence:

X⊂

Pnprojectivevariety⇒

X=

�n i=

1(X∩Ui)

isanop

encoverof

Xbyaffinevarieties.

Example.X

=V(x

2+y2−z2)⊂

P2.

Uz=

{[x:y:1]

:x,y∈k}(thecomplementof

Hz=

{[x:y:0]:[x

:y]∈P1

}).

X∩Uz=

V(x

2+y2−1)⊂

A2is

a“circle”.

What

isX

outsideof

X∩Uz?

X∩H

z=

V(x

2+y2)gives

[1:i:0],[1

:−i:0]∈P2

(the“p

oints

atinfinity”of

X∩Uz).

Geometricexplanation:chan

gevariab

lesto

�y=

iythen

X∩Uz=

V(x

2−

�y2−

1)⊂

A2is

a“hyperbola”,withasymptotes

�y=

±x,so

y=

±ix

arethetw

olines

correspon

dingto

thetw

onew

points

[1:i:0],[1

:−i:0]

atinfinity.

3.8.

PROJECTIV

ECLOSURE

and

HOM

OGENIS

ATIO

N

Given

anaffi

nevarietyX⊂

An,wecanview

X⊂

Pnvia:

X⊂

An∼ =

U0⊂

Pn=

An∪Pn

−1.

Thepro

jectiveclosu

reX⊂

Pnof

Xis

theclosure

3of

thesetX⊂

Pn.

Remark

.X∼ =

X� �⇒

X∼ =

X� .

Example.V(y−x2),V(y−x3)in

A2are∼ =

A1,buttheirprojectiveclosuresare

notiso(see

Hwk).

Given

apolynomialf∈

k[x

1,...,x

n]of

degreed,write

f=

f 0+

f 1+

···+

f dwheref i

are

the

hom

ogeneousparts.Then

thehomogenisationof

fis

� f=

xd 0f 0

+xd−1

0f 1

+···+

x0f d

−1+f d.

EXAM

PLES.

1)x2+y2=

1in

A2becom

esx2+y2=

z2in

P2.

2)y2=

x(x−1)(x−c)

inA2becom

estheellipticcurvey2z=

x(x−z)(x−cz)in

P2.

Exercise.X

=V(� f)⊂

Pn⇒

X∩U0=

V(f)⊂

U0∼ =

An.

1Notice

thegen

erators

ofm

pare

the2×

2subdeterminants

ofthematrix

withrowsaandx,so

thevanishingof

thefunctionsin

mpsaythatxis

proportionalto

a.Another

way

tolookatthis,is

topickanaffinepatchU

i∼ =

An

containingp(soai�=

0).

Then

homogen

izethemaxim

alidealm

p,i=

�xj−

aj

ai:allj�=

i�thatyouget

forp∈A

n.

2Hints:to

show

itis

abijection,just

defi

neamapψ

iin

theother

directionsuch

thatψ

i◦φ

iandφi◦ψ

iare

iden

tity

maps.

Itremainsto

show

continuityofφi,ψ

i.Toshow

continuity,

youneedto

checkthatpreim

ages

ofclosedsets

are

closed.Soyouneedto

describetheideals

whose

vanishingsets

giveφ−1

i(V

(J))

andψ

−1

i(V

(I))

=φi(V

(I)).You

willfindthatin

onecase,youneedto

homogen

isepolynomials

withresp

ectto

thei-th

coordinate,so

f∈

J⊂

k[A

n]

becomes

� f=

xdegf

if(x0

xi,...,xn

xi)(butomitting

xi

xi),

andin

theother

case

youplugin

xi=

1andrelabel

variables.

3i.e.

thesm

allestZariskiclosedsetofPn

containingX.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

15

Exercise.For

anyf,g∈R,show

that

� fg=

� f·�g

.

Exercise.

You

canalso

dehomogen

iseahom

ogeneouspolynom

ialF∈

Rbysettingx0=

1,so

f=

F(1,x

1,...,x

n).

Checkthat

F=

x� 0� f,

some�≥

0.

Question:X

=V(� I)⊂

Pnforsomeideal

� I⊂

k[x

0,...,x

n].

Can

wefindan

ideal

� Ithat

works,

from

thegivenidealI⊂

k[x

1,...,x

n]whichdefines

X=

V(I)⊂

An?

Theore

m3.3.Wecan1take

� Ito

bethehomogenisationofI,

� I=

theidealgeneratedby

homogen

isationsofallelem

ents

ofI

=�� f

:f∈I�.

Remark

.In

general,itis

notsufficien

tto

homogen

izeonly

asetofgenerators

ofI(see

theHwk).

Proof.

Xaff

.var⊂

An≡

U0=

(x0�=

0)⊂

Pn.

Claim

.V(� I)=

X⊂

Pn.

Step

1.X⊂

V(� I).

Pf.

Itsufficesto

checkthat

thehom

ogeneousgeneratorsof

� Ivanishon

X.

Let

G∈

� Ibethehom

ogenisationof

someg∈I.

⇒G(1,a

1,...,a

n)=

g(a

1,...,a

n)=

0for(a

1,...,a

n)∈X

=V(I)

⇒G| U

0∩X

=G| X

=0

(viewingX⊂

U0,so

U0∩X

=X)

⇒X⊂

V(G

)⇒

X⊂

V(G

)(noteV(G

)is

alread

yclosed

)⇒

G| X

=0.�

Step

2.

�� I⊃

Ih(X

).(W

eknow

secretly

theseareequal,seetheCorollary

below

)

Itsufficesto

show

that

hom

ogeneousgeneratorsG∈Ih(X

)arein

�� I.

⇒G| X

=0.

⇒G| X

=0.

(Since

X⊂

X∩U0,indeedequalityholdsbytheab

oveexercise)

⇒f=

G(1,x

1,...,x

n)∈I(X).

⇒fm∈I,somem.(U

singtheNullstellensatz√I=

I(X))

⇒hom

ogenise:

� fm=

� fm∈

� I.⇒

Since

2G

=x� 0� f,

itfollow

sthat

Gm

=x�m 0

� fm∈

� I,so

G∈

�� I.

�Step

3.V(� I)⊂

X.

FollowsbyStep2:

V(� I)=

V(�

� I)⊂

V(I

h(X

))=

X.�

�Exercise.How

does

theab

oveproof

simplify,ifwestartwithI=

I(X)?

Lemma.Thehomogen

isation

� IofaradicalidealIis

alsoradical.

Proof.

First,theeasy

case:supposeG∈

�� Iis

hom

ogeneous.

ThusG

m∈

� Iforsomem,an

dweclaim

G∈

� I.G

m(1,x

1,...,x

n)=

(G(1,x

1,...,x

n))

m∈I

⇒f=

G(1,x

1,...,x

n)∈I,since

Iis

radical.

⇒hom

ogenising,

� f∈

� I.⇒

G=

x� 0� f∈

� I,some�(just

asin

Step2of

thepreviousproof).

�Secon

dly,thegeneral

case:g∈

�� I.

⇒g=

G0+

···+

Gd(decom

positioninto

hom

ogeneoussumman

ds).

⇒gm

=(G

0+

···+

Gd−1)m

+(termsinvo

lvingG

d)+

Gm d∈

� I,somem.

⇒G

m d∈

� I,since

� Iis

hom

ogeneous(G

m dis

thehom

ogeneoussumman

dof

gm

ofdegreedm).

1Theobviouschoiceis

totakeI=

I(X)and� I=homogen

isationofI(X).

How

ever,theTheorem

allow

syoualsoto

start

withanon-radicalI:just

homogen

iseandyouget

a(typicallynon-radical)

� Ithatworks,

soX

=V(� I)=

V(�

� I).

2Example:G

=x2 0(x

2 1−

x0x1),

f=

x2 1−

x1,� f=

x2 1−

x0x1haslost

thex2 0thatappearedin

G.

16

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

⇒G

d∈

� I,bytheeasy

case.

⇒(g−G

d)m

=(G

0+···+

Gd−1)m

=gm−(termsinvolvingG

d)−G

m d∈

� I.⇒

bythesameargu

ment,

Gm d−1∈

� Iso

Gd−1∈

� I.Con

tinueinductivelywithg−G

d−G

d−1,etc.

⇒G

0,...,G

d∈

� I,so

g∈

� I.�

�Coro

llary.In

Theorem

3.3,ifwetake

I=

I(X)then

� I=

(homogen

isationofI(X))

isradicaland

� I=

Ih(V

(� I))

byHilbert’s

Nullstellensatz.

3.9.

MORPHIS

MSOF

PROJECTIV

EVARIE

TIE

S

Motivation.Pn

isalread

ya“global”ob

ject,coveredbyaffi

nepieces.

Soit

isnot

reason

able

todefinemorphismsin

term

sof

Hom

(Pn,A

1).

Infact

Hom

(Pn,A

1)ou

ghtto

only

consist

ofconstan

tmap

s:Pn

=An∪Pn

−1,so

restrictingto

Anweou

ghtto

getHom

(An,A

1)∼ =

k[x

1,...,x

n],an

dthese

polynom

ials

(ifnon-con

stan

t)willblow-upat

thepoints

atinfinitywhichform

Pn−1⊂

Pn.

Definition.Forprojvars

X⊂

Pn,Y⊂

Pm,amorp

hism

F:X→

Ymeans:

foreveryp∈

Xthereis

anopenneighbourhoodp∈

U⊂

X,and

hom

ogeneouspolynomials

f 0,...,f

m∈

Rofthe

samedegree,

1with

F:U→

Y,F[a]=

[f0(a):···:

f m(a)].

Rmk1.Thefact

that

thedegrees

ofthef i

areequal

ensuresthat

themap

iswell-defined:F[λa]=

[f0(λa):···:

f m(λa)]=

[λdf 0(a):···:

λdf m

(a)]=

[f0(a):···:

f m(a)]=

F[a].

Rmk2.W

hen

constructingsuch

F,youmust

ensure

thef i

donot

vanishsimultan

eouslyat

anya

(andthat

Factually

landsin

Y⊂

Pm).

Rmk3.Anisomorp

hism

meansabijectivemorphism

whoseinverseis

also

amorphism.

EXAM

PLES.

1)TheVeron

eseem

beddingF

:P1→

V(xz−y2)⊂

P2,[s

:t]�→

[s2:st

:t2]is

amorphism.

Wewan

tto

buildan

inversemorphism.

Ifs�=

0then

[s:t]=

[s2:st].

Ift�=

0then

[s:t]=

[st:t2].

SodefineG

:V(xz−y2)→

P1by[x

:y:z]�→

[x:y]ifx�=

0,an

d[x

:y:z]�→

[y:z]ifz�=

0.This

isawell-defined

map

,since

ontheoverlapx�=

0,z�=

0wehave

[x:y]=

[xz:yz]=

[y2:yz]=

[y:z].

Itis

now

easy

tocheckthat

F◦G

=id,G◦F

=id.

2)Pro

jection

from

apoint.

Given

aprojvarX⊂

Pn,a

hyperplaneH

=V(L

)⊂

Pn,an

dapointp

/∈X

and

/∈H,

defineπp:X→

H∼ =

Pn−1by

πp(x)=

(thepoint∈H

wherethelinethrough

xan

dphitsH).

Example.p=

[1:0:···:

0],H

=V(x

0),then

πp[x

0:···:

xn]=

[0:x1:···:

xn].

H=

V(L

)⊂

Pn

X⊂

Pnx

πp(x)p

Exercise.

Show

that

byalinearchan

geof

coordinates

onAn+1thegeneral

case

reducesto

the

Exam

ple.(H

int.

Use

abasis

�p,h1,...,h

nwhere

�p∈An+1represents

p,andhjis

abasisforH.)

3)Pro

jective

equivalences.

An

isom

orphism

X∼ =

Yof

projectivevarietiesX,Y

⊂Pn

isa

projectiveequivalence

ifitis

therestrictionof

alinearisom

orphism

Pn→

Pn,[x]�→

[Ax]

i.e.

inducedbyalinearisom

orphism

An+1→

An+1,x�→

Ax,whereA∈GL(n

+1,k).

Since

[Ax]=

[λAx]weon

lycare

abou

tA

modulo

scalar

matricesλid,so

A∈PGL(n

+1,k)=

P(GL(n

+1,k)).

FACT.2

Thegrou

pAut(Pn

)of

isom

orphismsPn→

Pnis

precisely

PGL(n

+1,k).

1recall,byconven

tion,thatthezero

polynomialhaseverydegree.

2Hartshorne,

ChapterII,Example

7.1.1.This

requires

mach

inerybeyondthis

course.

Youmay

hav

eseen

thecase

ofholomorphic

isomorphismsP1

→P1

over

k=

C:youget

theMobiusmapsz�→

az+b

cz+dwhere(a

bc

d)∈PGL(2,C

).

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

17

Example.H

0∼ =

H1via

� 01

010

000Id

� .

Example.Puttingaprojectivelinearsubspaceinto

stan

dardform

:iff 1,...,f

marehom

ogeneous

linearpolyswhicharelinearlyindep

endent⇒

V(f

1,...,f

m)∼ =

V(x

1,...,x

m).

Non-example.P2⊃

H0∼ =

P1∼ =

V(xz−

y2)⊂

P2butthey

arenot

projectivelyequivalentsince

theirdegrees

aredifferent(w

ediscuss

degrees

inSec.9.1).

3.10.

GRADED

RIN

GS

and

HOM

OGENEOUS

IDEALS

RecallR

=k[x

0,...,x

n]=

�d≥0R

dwhereR

d=

hom

ogeneouspolysof

degreed,an

dR

0=

k,an

d

byconvention

0∈R

dforalld.In

particular,

theirrelevantidealis

I irr=

(x0,...,x

n)=

�d>0R

d.

Definition.Let

Abe

aring(commutative).

AnN-gra

din

gmeans

A=

� d≥0

Ad

asabeliangroups1

under

addition,andthegradingby

dis

compatiblewithmultiplication:

Ad·A

e⊂

Ad+e.

Theelem

ents

inA

darecalled

thehomogeneouselements

ofdegre

ed.

Noteeveryf∈A

isuniquelyafinitesum

�f d

ofhom

ogeneouselem

ents

f d∈A

d.

Anisomorp

hism

ofgra

dedringsA→

Bisan

isoof

rings

whichrespects

thegrad

ing(A

d→

Bd).

I⊂

Aideal,then

define

I d=

I∩A

d

whichis

asubgrou

pof

Adunder

addition.

Definition.I⊂

Ais

ahomogeneousidealif2

I=

� d≥0

I d.

EXERCIS

ES.

1)Ihom

ogeneous⇔

Igenerated

byhom

ogeneouselem

ents.

2)Ihom

ogeneous⇔

foreveryf∈I,also

allhom

ogeneousparts

f d∈I.

3)If

Ihom

ogeneous,

Iprimeideal⇔∀homogen

eousf,g∈A,f

g∈Iim

plies

f∈Ior

g∈I.

4)Sums,

products,

intersection

s,radicalsof

hom

ogeneousideals

arehom

ogeneous.

5)A

grad

ed,Ihom

ogeneous⇒

A/I

grad

ed,bydeclaring(A

/I) d

=A

d/I

d

(Soexplicitly:[�

f d]=

�[f

d]∈A/I

just

inheritsthegrad

ingfrom

A).

3.11.

HOM

OGENEOUS

COORDIN

ATE

RIN

G

R=

k[x

0,...,x

n]withgrad

ingdetermined

bytheusual

grad

ingof

R(sox0,...,x

nhavedegree1).

X⊂

Pnaprojectivevariety.

Thehomogeneouscoord

inate

ringS(X

)is

thegrad

edring3

S(X

)=

R/I

h(X

)=

R/I(� X)=

k[� X]

Example.S(P

n)=

R=

k[x

0,...,x

n].

Example.X

=V(yz−x2)⊂

P2(proj.closure

ofparab

olay=

x2)then

S(X

)=

k[x,y,z]/(yz−x2).

Remark

.f∈S(X

)defines

afunctionf:

� X→

k,butnot

X→

k(dueto

rescaling).

Lemma3.4.S(X

)∼ =

S(Y

)asgraded

k-algebras⇔

X∼ =

Yviaaprojectiveequivalence.

1so

Ad⊂

Ais

anadditivesubgroupandA

d∩A

e=

{0}ifd�=

e.2Recall�

meansthateach

f∈Icanbeuniquelywritten

asafinitesum

f=

f 0+

···+

f Nwithf d

∈I d,someN.

3here

� X⊂

An+1is

theaffineconeover

X,seeSection3.4.

18

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Proof.

(⇐)Let

ϕ:Pn→

PnbealinearisoinducingY∼ =

X.Soϕ∗ (xj)=

�A

jixiis

alinearpoly

inthehom

ogeneouscoordsxiof

Pn,whereA

isinvertible.Soϕ∗:S(X

) 1→

S(Y

) 1isavectorspace

iso(thexispan

thevector

spaces

S(X

) 1,S(Y

) 1).

Thisinducesaunique1

algebra

isoS(X

)→

S(Y

).(⇒

)Given

anisoψ

:S(X

)∼ =

S(Y

),it

restrictsto

alinearisoS(X

) 1→

S(Y

) 1,xj�→

�A

jixi.

Supposefirstthesimple

case

that

thexiarelinearlyindep

endentin

S(X

) 1,then

thexiare

linearly

indep

endentalso

inS(Y

) 1(indeed

S(X

) 1=

S(Y

) 1=

k[x

0,...,x

n] 1).

Then

Ais

awell-defi

ned

invertible

matrix.Thusϕ:Pn→

Pn,ϕ[a

0:...:an]=

[�A

0ia

i:...:�

Ania

i]is

alinearisoofPn

withϕ∗=

ψ,in

particu

larϕ∗ I(X

)⊂

I(Y)so

ϕ(Y

)⊂

X,an

dϕ:Y→

Xis

therequired

proj.equiv.

Now

theharder

case

when

xiarelinearlydep

endentin

S(X

) 1.Noticetheselineardep

endency

relation

sareprecisely

Ih(X

) 1.Supposed=

dim

kIh(X

) 1.Bypre-com

posingψbyalinearequivalence

ofPn

wemay

assumeIh(X

) 1=�x

n,x

n−1,...,x

n−d+1�.

Then

wecanview

X⊂

Pn−dsince

thelast

dcoordinates

vanishonX,an

dS(X

)willnot

havechan

gedupto

isom

orphism.Asdim

kS(X

) 1=

dim

kS(Y

) 1,wecandothesameforY

bypost-composingψbyan

other

projectiveequivalence.Now

wecanap

ply

thesimple

case

toX,Y

⊂Pn

−dto

obtain

aninvertible

matrixA∈GL(n−

d+

1,k).

Finally

use

� A0

0I

�forad×

didentity

matrix

Ito

obtain

therequired

projectiveequivalence

forthe

original

X,Y⊂

Pnupto

pre/p

ost-com

posingwithprojectiveequivalences.

�Non-E

xample.P2⊃

H2=

X∼ =

P1∼ =

Y=

ν 2(P

1)⊂

P2via

[x0:x1:0]�→

[x2 0:x0x1:x2 1],but

S(X

)=

k[x

0,x

1]an

dS(Y

)=

k[y

0,y

1,y

2]/(y

0y 2−

y2 1)arenot

isom

orphic

asgraded

algebras:

they

contain

adifferent2

number

oflinearlyindep

endentgenerators

ofdegree1.

Thusν 2(P

1)is(ofcourse)

not

projectivelyequivalentto

thehyperplaneH

2.

Warn

ing.X∼ =

Yproj.vars�⇒

� X∼ =

� Y,so

S(X

)is

not

anisom

orphism-invariantofX.3

Example.X

=P1∼ =

Y=

ν 2(P

1)⊂

P2via

[x0:x1]�→

[x2 0:x0x1:x2 1],butS(X

)=

k[� X]=

k[x

0,x

1]

andS(Y

)=

k[� Y

]=

k[y

0,y

1,y

2]/(y

0y 2−y2 1)are

notisom

orphic

k-algebrasbecause

thefirstisaUFD

butthesecondis

not(con

sider

thetw

ofactorisationsy 0y 2

=y2 1).

Alternatively,

onecan4show

that

theaffi

necones

� X=

A2,

� Y=

V(xz−y2)⊂

A3arenot

isom

orphic

usingmethodsfrom

Section13.

Hard

erexercise.An(ungraded

)k-algeb

raisom

orphism

S(X

)∼ =

S(Y

)im

plies

� X∼ =

� Y,butin

fact

italso

implies

that

X∼ =

Yvia

aprojectiveequivalence.5

4.

CLASSIC

AL

EM

BEDDIN

GS

4.1.

VERONESE

EM

BEDDIN

G

Example

4.1.TheVeroneseem

beddingP1

�→P2

is

ν 2:P1

�→P2

,[x

0:x1]�→

[x2 0:x0x1:x2 1].

Theim

age

ν 2(P

1)is

called

thera

tionalnorm

alcurv

eofdegree2,

ν 2(P

1)=

V(z

(2,0)z

(0,2)−z2 (1,1))⊂

P2

labellingthehomogen

eouscoordinatesonP2

by[z

(2,0):z (

1,1):z (

0,2)].

Example

4.2.Theim

age

ofν d

:P1

�→Pd

,[x0:x1]�→

[xd 0:xd−1

0x1:...:xd 1]is

called

rational

norm

alcurv

eofdegreed.

Motivation.Given

ahom

ogeneouspolynom

ialin

twovariab

les,

youcanview

itsvanishinglocus

astheintersection

ofν d(P

1)withahyperplane.

For

exam

ple,x2 0x1−

8x3 1=

0is

theintersectionof

1e.g.ϕ

∗ (x2 0x3+

7x3 5)=

ϕ∗ (x0)2ϕ

∗ (x3)+

7ϕ

∗ (x5)3.

2k[� X]has2,e.g.x0,x

1,andk[� Y

]has3,e.g.y0,y

1,y

2.Sodim

kS(X

) 1=

2anddim

kS(Y

) 1=

3.

3Meaning,X

∼ =Y

does

notim

ply

S(X

)∼ =

S(Y

),unlikethecase

ofaffinevarieties:

� X∼ =

� Y⇔

k[� X]∼ =

k[� Y

].4Proof:

� X=

A2is

non-singular,

but� Y

hasasingularity

at0since

thetangentspace

at(a,b,c)is

defi

ned

by

c(x−a)−2b(y−b)+a(z

−c)

=0,andat(a,b,c)=

0∈A

3thisequationisiden

ticallyzero.SoT0� Y=

A3�∼ =

A2∼ =

Tp� X.

5If

theambientdim

ensionsn,m

are

notthesame,

then

onegetsalinearinjection

An+1�→

Am

+1,butonecan

extendthatto

alinearisomorphism

Am

+1→

Am

+1byinsertingadditionalvariables.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

19

ν 3(P

1)⊂

P3withthehyperplanez (

2,1)−

8z(0,3)=

0usingcoordinates

[z(3,0):z (

2,1):z (

1,2):z (

0,3)]on

P3.TheVeron

esemap

,defined

below

,generalizes

this

toan

ynumber

ofvariab

les.

Definition

(Veron

eseem

bedding).TheVero

nese

mapis

ν d:Pn

�→P(

n+d

d)−

1,[x

0:...:xn]�→

[...

:xI:...]

runningoverall

monomials

xI=

xi 0 0xi 1 1···x

i n nofdegreed=

i 0+

···+

i n,whereyoupicksome

orderingoftheindices

I⊂

Nn+1whose

sum

ofallen

triesequals

d,e.g.

lexicographic

ordering.

Theim

age

ofν d

iscalled

aVero

nese

variety.

Remark

4.3

(Cou

ntingpolynom

ials).

Howmanymonomialsare

therein

n+1variables

x0,x

1,...,x

n

ofdegreed?Draw

n+

dpoints,e.g.

forn=

3,d=

4:

••

••

••

•Then

choosingdofthesepoints

determines

uniquelyamonomialofdegreed,e.g.

�•

��

•�

•meansx1 0x2 1x1 2x0 3(countupthestars

togetthepowers).Sothenumberofmonomials

is� n

+d

d

� .

Remark

4.4

(Veron

esesurface).Theim

age

of

ν 2:P2

�→P5

,[x

0:x1:x2]�→

[x2 0:x0x1:x0x2:x2 1:x1x2:x2 2]

iscalled

Vero

nese

surface.

Theore

m4.5.

Pn∼ =

Imag

e(ν d)

=V(z

Iz J−

z Kz L

:I+

J=

K+

L)

=�

I+J=K+L

(quadrics

V(z

Iz J−

z Kz L

))⊂

P(n+d

d)−

1

wherewerun

overall

multi-indices

I,J,K

,Loftype

(i0,...,i

n)∈

Nn+1withi 0

+···+

i n=

d.

Moreover,

theideal�z

Iz J−

z Kz L

:I+

J=

K+

L�i

sradical.1

Example.Forν 2

:P1→

P2,theequationz (

2,0)z

(0,2)−

z (1,1)z

(1,1)=

0is

thefamiliarxz−

y2=

0.

Proof.

That

imag

e(ν d)satisfies

theequationsz Iz J−z K

z L=

0isob

vioussince

z Iz J

=xIxJ=

xI+J.

Con

versely,

wefindan

explicitinversemorphism

forν d.Fix

J=

(i0,...,i

n)∈

Nn+1withd−

1=

i 0+

···+

i n,an

ddenoteJ�=

(j0,...,j

�+

1,...,j n)(sowead

don

ein

the�-th

slot

ofI,an

dthese

indices

now

addupto

d).

Define

ϕJ:∩(

thosequad

rics)��

�Pn

,[...:z I

:...]�→

[zJ0:z J

1:...:z J

n]

whichis

awell-defined

morphism

excepton

theclosed

setwhereallz J

�=

0.Example

toclarify.Forν 2(P

1),

J=

(0,1),

ϕJ:[z

(2,0):z (

1,1):z (

0,2)]�→

[z(1,1):z (

0,2)]corresponds

tothemap[x

2:xy:y2]�→

[xy:y2]=

[x:y]whichis

defi

ned

fory�=

0,andnotice

y=

(x,y)J.

TheϕJ,as

wevary

J,ag

reeon

overlaps.

Indeedforan

other

such

J� ,noticeJ�+J� ��=

J��+J� �(this

equalsJ+

J�plusad

d1in

thetw

oslots�,��),

hence

z J�z J

� ��=

z J��z

J� �,an

dthus2

ϕJ([z])=

ϕJ� ([z]).

Weclaim

ϕJisan

inverseof

ν dwherever

ϕJisdefined

.Thekeyob

servationis:xJ�=

xJ·x

�.Notice

ϕJ◦ν

d([x])=

[xJ0:...:xJn]=

[x0:...:xn](rescale

by1/

xJ).

1Non-examinableproof.

Trick

from

3.8:thehomogen

isationofaradicalidealisradical.

Soitsufficesto

checkitisa

radicalidealonanaffinepatch.Example

forν2:P1

→P2

:ontheaffinepatchz 1

�=0wecanputz 1

=1,so

z 3=

z2 2and

k[z

2,z

3]/(z

3−z2 2)∼ =

k[z

2]is

anintegraldomain,so

theidealis

radical.

Gen

eralcase:ontheaffinepatchz (

d,0,...,0)=

1,

bytheother

non-examinable

footnote

allz I

=xIare

determined

bythex�=

z J�forJ=

(d−1,0,...,0),

�=

0,...,n

,andthex�are

indep

enden

t.Sok[z

I:allI]/�z

Iz J

−z K

z L:I+J=

K+L�∼ =

k[x

0,...,x

n]whichisanintegraldomain.

2In

gen

eral,[x

0:...:xn]=

[y0:...:yn]∈Pn

⇔x,y

are

proportional⇔

all2×2minors

ofthematrix

(x|y)vanish.

20

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Now

consider

ν d◦ϕ

J([z I]).Abbreviate

xj=

z Jj,then

ϕJ([z I])=

[x0:...:xn]andν d◦ϕ

J([z I])=

[xI],

andon

ecan1checkthis

equals[z

I].

�

Theore

m.Y⊂

Pnproj.var.⇒

Pn⊃

Y∼ =

ν d(Y

)⊂

Pmis

aproj.subvar.

Proof.

This

isim

med

iate:ν d

:Pn

→Pm

isahom

eomorphism

onto

aclosed

set(hence

aclosed

embedding),

soit

sendsclosedsets

toclosed

sets.Wegive

below

another,explicit,

proof.

KeyTrick

:V(F

)=

V(x

0F,x

1F,...,x

nF)⊂

Pnsince

x0,...,x

ncannot

allvanishsimultan

eously.

So:

Y=

V(F

1,...,F

N)forsomeFihom

og.

ofvariou

sdegrees.

ByTrick:Y

=V(G

1,...,G

M)forsomeG

ihom

og.of

samedegree=

c·d

.So:

Gi=

Hi◦ν

dforsomeH

ihom

og.

ofsamedegreec.

So:

Pn⊃

Y=

V(G

1,...,G

m)

νd

−→V(H

1,...,H

M)⊂

Pm,

indeed:{a∈Pn

:G

i(a)≡

Hi(ν d(a))

=0∀i}−→

{b∈Pm

:H

i(b)

=0∀i}via

a�→

ν d(a)=

b.Soν d(Y

)=

ν d(P

n)∩V(H

1,...,H

M).

�

Example

4.6.Forν 2

:P2→

P5andY

=V(x

3 0+

x3 1)⊂

P2,

Y=

V(x

0(x

3 0+

x3 1),x1(x

3 0+

x3 1),x2(x

3 0+

x3 1))

=V(G

1,G

2,G

3)

forexample:G

1=

x0(x

3 0+

x3 1)=

(x2 0)2

+(x

0x1)x

2 1=

H1◦ν

2takingH

1=

z2 (2,0,0)+

z (1,1,0)z

(0,2,0).

Soν 2(Y

)=

ν 2(P

2)∩V(H

1,H

2,H

3)⊂

P5.

Example.Let

X⊂

Pnbeaprojectivevariety.

Con

sider

abasic

open

set

DF=

X\V

(F),

whereF

=�

aIxIis

ahomog

eneouspolynom

ialofdegreed.Abbreviate

N=

� n+d

d

� −1.Then

DF

canbeidentified

withan

affinevarietyin

AN

asfollow

s.Bythesameargu

mentas

intheMotivation

above,

ν d(V

(F))

lies

inthehyperplaneH

=V(�

aIz I)⊂

PN.Then,observethatwecanidentify

ν d(D

F)=

ν d(X

)\H

⊂PN

\H∼ =

AN

(you

canuse

alinearisom

orphism

tomap

Hto

thestandard

hyperplaneH

0,then

recallPN

\H0=

U0∼ =

AN

isahom

eomorphism).

Explicit

example.X

=V(x)=

[0:1]∈P1

,F

=x2+

y2.Then

ν 2(V

(F))⊂

V(X

+Z)⊂

P2since

ν 2([x:y])=

[X:Y

:Z]=

[x2:xy:y2]∈P2

.Also,

X=

V(xx,yx)(K

eyTrick

above),so

ν 2(D

F)=

V(X

Z−

Y2,X

,Y)\V

(X+

Z)⊂

P2.

Chan

gecoordinates:a=

X+Z,b=

Y,c=

Z.Soν 2(D

F)∼ =

V(a−c,b)\V

(a)⊂

U0=

(a�=

0)∼ =

A2

(usingcoordsb,cafterrescalingso

thata=

1)weobtain

theaffi

nevariety(a

point!)b=

0,c=

1.

4.2.

SEGRE

EM

BEDDIN

G

Below

,wehaven’t

actually

defined

whatPn

×Pm

meansas

aprojectivevariety(w

edonotuse

the

product

topolog

y,seeHwk).

Soitdoes

not

mak

esense

totalk

abou

t“m

orphism”yet.

Inreality,we

aredefi

ningthevarietyPn

×Pm

asbeingtheim

age

ofσn,m

inPl

argepower.See

Section

6.2.

1Non-examinable.

This

ismessy

tocheck.Wefirstneedto

checkthatz (

0,...,0,d

,0,...,0)cannotallvanishsimulta-

neously.

Suppose

bycontradictionthatthey

do.Weknow

somez I

isnon-zero(since

[zI]∈

projectivespace).

By

reorderingtheindices

(symmetry),

WLOG

i 0≥

i 1≥

···≥

i nwithi 0

+···+

i n=

d.Also,W

LOG,this

isthenon-zero

z Iwith

largestoccurringmaxim

alindex

i 0(soz K

=0if

Khasanyindices

kjlarger

than

i 0).

Weclaim

i 0=

d,

hen

ceI=

(d,0,...,0),

soz I

=z (

d,0,...)=

0,contradiction.Proof:

ifi 0

�=d,then

i 1≥

1andz Iz I

=z K

z K�where

K=

(i0+

1,i

1−

1,i

2,...),

K�=

(i0−

1,i

1+

1,i

2,...).

Butz K

=0since

i 0+

1>

i 0,forcingz I

=0,contradiction�

Now

,W

LOG

byreorderingindices

andthen

rescaling,z (

d,0,...)=

1.It

sufficesto

checkνd◦ϕ

J([z I])=

[xI]forasp

ecific

choiceofJ(since

thevariousϕ-m

apsagreeonoverlaps).WepickJ=

(d−1,0,...).

Sox0=

z (d,0,...),x1=

z (d−1,1,0,...),

x2=

z (d−1,0,1,0,...),etc.

Itisnow

astraightforw

ard

exercise

tocheckthat,usingthequadraticequations“z Iz J

=z K

z L”

oneobtainsxI=

zd−1

(d,0,...)z (

i 0d+i 1

(d−1)+

i 2(d

−1)+

...+

i n(d

−1)−

d(d

−1),i 1

,i2,...,in)=

z (i 0

,i1,...)=

z I.Asawarm

-up,trychecking

firstthatx1x2=

z (d,0,...)z (

d−2,1,1,0,...)=

z (d−2,1,1,0,...).

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

21

Definition

(Segre

embedding1).

σn,m

:Pn

×Pm

=P(kn+1)×

P(km+1)

�→P(kn+1⊗

km+1)∼ =

P(n+1)(m+1)−

1=

Pnm+n+m

([v],[w

])�→

[v⊗

w]

More

explicitly,

interm

softhestandard

bases,

(�xie

i,�

y jf j)�→

[�xiy

je i⊗

f j],thus:

([x0:···:

xn],[y

0:···:

y m])�→

[x0y 0

:x0y 1

:···:

x0y m

:x1y 0

:x1y 1

:···:

xny 1

:···:

xny m

]

usingthelexicographic

ordering.

TheSegre

variety

isΣn,m

=σn,m(P

n×

Pm)⊂

Pnm+n+m

Example.σ1,1:P1

×P1→

P3,([x:y],[a

:b])�→

[xa:xb:ya:yb],so

theim

ageis

defined

bythe

equationXW−

YZ

=0using[X

:Y

:Z

:W

]on

P3.

You

canthinkof

kn+1⊗

km+1∼ =

Mat

(n+1)×

(m+1)as

matrices(thecoeffi

cientof

e i⊗

f jbeingthe

(i,j)-entry),

then

σn,m([x],[y])

isthematrixproduct

ofthecolumnvector

xan

dtherow

vector

y,

givingthematrix[z

ij]=

[xiy

j].

Example.In

thepreviousexam

ple,forσ1,1,thematrixis

� xaxb

yayb

� =� X

YZ

W

� ∈P(Mat

2×2).

Theore

m4.7. Σn,m

=V(all2×

2minors

ofthematrix

(zij))⊂

P(Mat

(n+1)×

(m+1))

=V(z

ijz k

�−

z kjz i�:0≤

i<

k≤

n,0≤

j<

�≤

m)

Proof.

Exercise.Hint:

use

that

thecolumnsof

amatrixareproportion

aliff

all2×2minorsvanish.

Anexplicitinverseof

σn,m

is:

σn,m

:Σn,m

πcol×

πro

w−→

Pn×

Pm

whereπco

l:Σn,m→

Pnistheprojectionto

any(non

-zero)

column(theim

ages

arethesamesince

the

columnsareproportion

al).

Sim

ilarly,πrow:Σn,m→

Pmistheprojectionto

any(non

-zero)

row.

�

4.3.

GRASSM

ANNIA

NS

AND

FLAG

VARIE

TIE

S

Definition

(Grassman

nian).

TheGra

ssmannian(ofd-planes

inkn)is

Gr(d,n

)=

{alld-dim

ensionalvectorsubspacesV⊂

kn}

where1≤

d<

n.Forexample,Pn

=Gr(1,n+

1).

TheFlagvariety

Flag(d

1,...,d

s,n

)is

Flag(d

1,...,d

s,n

)=

{allflags

ofvectorsubspacesV1⊂

···⊂

Vs⊂

kn,dim

Vi=

di}.

Remark

4.8.Wecaniden

tify

Gr(d,n

)=

{d×

nmatrices

ofrankd}/

GLk(d)

byiden

tifyingthed-planeV∈

Gr(d,n

)withthematrix

whose

rowsare

anychoiceofbasisv i

for

V⊂

kn.

Twosuch

choices

ofbasesv i,�v

iare

relatedby

achange

ofbasismatrix

g∈

GLk(d):

�v i=

�g i

jv j

(soabove,GLk(d)actsby

left-m

ultiplication

on

d×

nmatrices).

More

abstractly:

Aut(V)∼ =

GLk(d)=

{d×

dinvertible

matrices

overk}.

1Recallthete

nso

rpro

ductoftw

ok-vectorspacesV

⊗W

isavectorspace

ofdim

ensiondim

V·d

imW

withbasis

v i⊗w

jwherev i,w

jare

basesforV,W

.SoR

n⊗R

m∼ =

Rnm.Youcanextendthesymbol⊗

toallvectors

bydeclaring

that(�

λiv i)⊗

(�µjw

j)=

�(λ

iµj)v

i⊗

wj.Notice

therefore

that0⊗

w=

0=

v⊗

0,so

donotconfuse

this

withthe

product

V×

Wwhichhasdim

ensiondim

V+

dim

W,e.g.R

n×

Rm

∼ =R

n+m.

Exerc

ise.ProveV

∗⊗

W∼ =

Hom(V

,W)forfinitedim

ensionalvectorspacesV,W

,whereV

∗is

thedualofV.

22

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

4.4.

PLUCKER

EM

BEDDIN

G

Definition

4.9

(Plucker

embedding).ThePlucker

mapis

defi

ned

by1

Gr(d,n

)�→

P(Λdkn)∼ =

P(n d)−

1

V�→

k·(v 1∧···∧

v d)

wherev i

isabasisforV.

Exercise4.10.Show

thatexplicitlythePlucker

mapis

Gr(d,n

)=

{d×

nmatrices

ofrankd}/

GLd(k)

�→P(

n d)−

1

[d×

nmatrix

A]�→

[alld×

dminors

Δi 1,...,idofA]

(Δi 1,...,idis

thedeterminantofthematrix

whose

columnsare

thei 1,...,i

d-thcolumnsofA).

Non-examinable

Fact.

Theim

age

ofthePlucker

map

isV(P

lucker

relations)⊂

P(Λdkn).

Wenow

describetherelation

s.2Let

z i1i 2...idbethehom

ogeneouscoordinates

onP(Λdkn),

i.e.

z i1i 2...idis

the

coeffi

cientof

thebasisvectore i

1∧···∧

e id∈Λdkn,wherei 1

<···<

i d.ThePlucker

relation

sare:

zi 1i 2...id·z

j 1j 2...jd=

�

1≤

�<d

�

r 1<r 2

<···<r �zi 1i 2...ir1−1j 1i r

1+1...ir2−1j 2i r

2+1...ir�−1j �i r

�+1...id·z

i r1i r

2...ir�j �

+1j �

+2...jd

Ontherigh

tweinterchan

ged

thepositionsof

j 1,...,j

�withthoseof

i r1,...,i

r �,in

thatorder.Notice

wedonot

allow

�=

d(thecase

r 1=

1,...,r d

=d).

Ontheright,

wetypicallymust

reorder

the

indices

onthez-variablesto

bestrictly

increasing:theconventionis

that

z ...i...j...=−z ...j...i...when

wesw

aptw

oindices

(this

equals

zero

iftw

oindices

areequal).

E.g.z 3

2=−z 2

3an

dz 2

2=

0.

Example

4.11.Gr(2,4):

thestandard

basisfork4is

e 1,e

2,e

3,e

4,so

abasisforΛ2k4is

e i∧e j

for

1≤

i<

j≤

4,explicitly:

e 1∧e 2,

e 1∧e 3,

e 1∧e 4,

e 2∧e 3,

e 2∧e 4,

e 3∧e 4.

Theircoeffi

cien

tsdefi

necoordinates[z

12:z 1

3:z 1

4:z 2

3:z 2

4:z 3

4]forP(Λ2k4)∼ =

P5,forexample

6e1∧e 4−

3e2∧e 4

hascoordinates[0

:0:6:0:−3:0].Then

weget

Gr(2,4)

=V(z

12z 3

4−

z 32z 1

4−

z 13z 2

4)=

V(z

12z 3

4−

z 13z 2

4+

z 23z 1

4)⊂

P5.

Inthenotationofthepreviousfootnote,in

thehomogen

eouscoordinate

ringS(P(Λ

2k4))

wehave

(e1∧e 2)·(e 3∧e 4)=

(e3∧e 2)·(e 1∧e 4)+

(e1∧e 3)·(e 2∧e 4).

1Recallthed-thexte

riorpro

ductΛ

dW

ofak-vectorspace

Wis

ak-vectorspace

ofdim

ension� d

imW

d

�gen

erated

bythesymbols

wi 1∧···∧

wi d

wherei 1

<···<

i d,wherew

iis

abasisforW

.Onecanextendthewed

ge-symbolto

all

vectors

bydeclaringit

tobealternating:w

i∧w

j=

−w

j∧w

i(inparticularw

i∧w

i=

0),

andmulti-linear:

(�λiw

i)∧(�

µjw

j)=

� i,j

λiµjw

i∧w

j=

� i<j

(λiµj−

µiλj)w

i∧w

j.

Exerc

ise.Given

anyvectors

v 1,...,v

d∈W

,letV

=span(v

1,...,v

d).

Then

foranyg∈Aut(V),

show

that

(gv 1)∧···∧

(gv d)=

(det

g)v 1

∧···∧

v d.

Ifyouthinkcarefully,

you’llnotice

this

isthedefi

nitionofdeterminant!

Sodefi

nition4.9

makessense:i.e.

thechoiceofbasisv i

forV

doesnotaffectthelinek·(v 1

∧···∧

v d)∈P(

Λdkn).

2Equivalently,

recallthehomogen

eouscoordinate

ringofP(

Λdkn)is

thepolynomialringin

thevariablesden

oted

bye i

1∧···∧

e id,withstrictly

increasingindices,wheree j

isthestandard

basisforkn.Then

thePlucker

relationsare

thequadraticpolynomialrelations,

given

by:

(v1∧·

··∧v d)·(

w1∧·

··∧w

d)=

�

i 1<···

<i �

(v1∧·

··∧v i

1−1∧w

1∧v

i 1+1∧·

··∧v d)·(

vi 1∧·

··∧v i

�∧w

�+1∧w

�+2∧·

··∧w

d)∈S(P

(Λdkn))

wherewesum

over

allchoices

except�=

d,andthesehold

forallv i

∈kn,w

j∈

kn(notice

thatifyouexpandthese

out,

usingthealternatingmulti-linearproperty

of∧,

then

they

becomequadraticpolynomialrelationsin

thevariables

e i1∧···∧

e id).

Foraminim

alsetofrelations,

youjust

needtheaboveforallv i,w

jpicked

amongst

thestandard

basis

vectors

e j(soexplicitly:v 1

=e j

1,...,v

d=

e jdwithj 1

<...<

j dandsimilarlyforthew’s).

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

23

Exercise4.12.Whatare

thePlucker

relationswritten

explicitlyin

term

softhed×dminors

Δi 1,...,id?

(e.g.checkthatin

theexample

Gr(2,4)

youjust

needonerelation:Δ

12Δ

34−Δ

13Δ

24+Δ

23Δ

14=

0.)

Sim

ilarly,usingthePlucker

map

s,forflag

varieties:

Flag(d

1,...,d

s,n

)�→

P(n d1)−

1×···×

P(n ds)−

1.

TheZariskitopologyon

Gran

dFlagisdefined

asthesubspacetopologyvia

thePlucker

embeddings.

Remark

4.13.Alltheem

beddings

above,overR

(respectivelyoverC),

are

infact

smooth

(respec-

tively

holomorphic)when

view

ingthespacesassm

ooth

(respectivelycomplex)

manifolds.

Lemma4.14.TheGrassmannianGr(d,n

)is

anirreduciblevariety.

Proof.

Let

W=

span

(e1,...,e

d)=

kd⊕0⊂

kn.Given

V=

span

(v1,...,v

d)∈Gr(d,n

)complete

this

toabasis

v 1,...,v

n,then

A∈GLn(k)withcolumnsv i

willmap

Wto

V.This

defines

asurjective

polynom

ialmap

GLn(k)→

Gr(d,n

),A�→

A(W

),wherewecanview

GLn(k)as

anaffi

nevariety

byidentifyingit

withV(z

·det−1)⊂

kn2+1via

A�→

(A,[det

A]−

1)(herezis

anew

variab

lethat

form

ally

invertsthedeterminan

t).Bythefinal

exam

ple

3in

Sec.2.13,

itremainsto

show

GLn(k)is

irreducible.This

iseasy

tocheck,1

since

GLn(k)is

dense

inkn2,an

dkn2is

irreducible.

�

Exercise.Show

that

Flag(d

1,...,d

s,n

)is

irreducible

byasimilar

argu

ment.

5.

EQUIV

ALENCE

OF

CATEGORIE

S

5.1.

REDUCED

ALGEBRAS

For

anyringA,f�=

0∈A

isnilpotentiffm

=0forsomem.

Ais

areduced

ringifithas

nonilpotents.

Lemma.A/I

isreduced⇔

Iis

radical.

Proof.

IfA/I

isreduced:

fm∈I

⇔fm

=0∈A/I

⇔f=

0∈A/I

⇔f∈I.

IfIisradical:fm

=0∈A/I

⇔fm

=0∈I

⇔f∈I

⇔f=

0∈A/I

.�

Upshot:2

{affinealgebraic

varieties}

→{f.g.reducedk-algeb

ras}

(X⊂

An)�→

k[X

]=

R/I(X

)?←�

A.

Af.g.⇒

onecanpickgeneratorsα1,...,α

n(som

en)

⇒determines

3ak-algebra

hom

f:R

=k[x

1,...,x

n]→

A,xi�→

αi

⇒I=

kerf⊂

Ris

radical

(since

Ais

reduced)

⇒A∼ =

R/I

,so

chooseX

=V(I).

Note.A

differentchoice

ofgeneratorscangiveacompletely

differentem

beddingX⊂

Am,somem.

Dueto

this

choice,thecorrectway

tophrase

theab

ove“correspon

dence”,

betweenvarietiesan

dalgebras,

isas

anequivalence

ofcategories,whichwenow

explain.

1In

gen

eral,

ifU

⊂X

isaden

seopen

setofan

irreducible

affinevariety

X,then

Uis

irreducible.

Indeed,if

U=

(C1∩U)∪(C

2∩U)forclosedC

1,C

2⊂

X,then

X=

U=

C1∪C

2,forcingC

i=

Xforsomei,

soU

=C

i∩U.

Finally,

notice

thatrelativelyclosedsubsets

V(I)∩GL

n(k)forGL

n(k)⊂

kn2

correspondprecisely

torelativelyclosed

sets

when

viewingGL

n(k)⊂

kn2+1.This

isbecause

given

anypoly

fforkn2+1,(det)N

fcu

tsoutthesamesubset

inGL

n(k)asfdoes,andit

cuts

outthesamesubsetifwealsoreplace

alloccurren

cesofz·d

etin

(det)N

fby1.So

WLOG

theequationsfusedto

defi

nearelativelyclosedsubsetofGL

n(k)⊂

kn2+1canbechosennotto

involvez.

2f.g.=

finitelygen

erated.

3recall,ak-algeb

rahom

istheiden

tity

maponk(since

itisk-linearand1�→

1),so

bylinearity

andmultiplicativity

itsufficesto

defi

nethehom

ongen

erators.

24

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

5.2.

WARM

-UP:EQUIV

ALENCE

OF

CATEGORIE

SIN

LIN

EAR

ALGEBRA

Weassumesomefamiliarity

withverybasic

category

theory

term

inology.

Category1:

CObjects:1kn

Morphisms:

Hom

(kn,k

m)=

Mat m

×n(k)(m

atrices).

Category2:

DObjects:

finitedim

ensional

vectorspacesover

k.

Morphisms:

Hom

(V,W

)=

{k-linearmapsV→

W}.

Linearalgebra

courses

secretly

provethat

thefunctor

F:C→

Dkn�→

kn

(matrix)�→

(linearmap

given

byleft

multiplication

bythatmatrix)

isan

equivalence

ofcategories.

Itis

notan

isomorphism

ofcategories

since

thereis

noinverse

functor

D→

C.Thereisanob

viousobject

toassociateto

V,namelyV�→

kdim

V,butat

thelevel

of

morphismsin

order

todefinealinearisomorphism

Hom(V

,V)→

Mat d

imV×dim

V(k)wewould

need

tochooseabasis

forV.

DefineG

:D→

Cas

follow

s:Pickabasis

v 1,...,v

nforeach

vector

space

V(heresy!)

For

knwestipulate

that

wechoosethestan

dard

basis

e 1,...,e

n.

Then

G:Hom

(V,W

)�→

Mat

m×n(k)(w

herem

=dim

W,n=

dim

V)is

defined

bysendingϕto

the

matrixforϕin

thechosen

basesforV,W

.G◦F

=id

Cbyconstruction,but

F◦G

:V→

kn

id →kn,Hom(V

,W)→

Mat m

×n

id →Mat m

×n

isnot

idD,so

Gis

not

aninverse

forF.Butforanequivalence

ofcategories,wejust

needthereto

beanaturalisom

orphism

F◦G

⇒id

D.

DefineF◦G

⇒id

Dbysending2

V�→

(morphism

FG(V

)=

kn→

id(V

)=

Vgiven

bye i�→

v i).

Ingeneral,to

find/d

efineG

isanuisance.Sooneusesthefollow

ingFACT:

Lemma5.1

(Criterion

forEquivalencesofCategories).

AfunctorF

:C→

Dis

anequivalence

ofcategories

ifitis

full,faithful,andessentiallysurjective.

Explanation:

FullmeansHom

(X,Y

)→

Hom

(FX,F

Y)is

surjective;

FaithfulmeansHom(X

,Y)→

Hom

(FX,F

Y)is

injective.

Sofullyfaithfulmeansyou

haveisomorphismsatthelevel

ofmorphisms.

Essentiallysu

rjectivemeans:

anyZ∈Ob(D

)is

isomorphic

toFX

forsomeX.

(intheab

oveexam

ple,anyvectorspaceV

isisomorphic

tosomekn,indeedtaken=

dim

V).

Exercise.Prove

theLem

ma.

1since

itis

just

asymbol,onecould

alsojust

label

theobjectsbyn∈N,andHom(n

,m)=

Mat m

×n(k).

2Thefact

thatit

isanaturaltransform

ationboilsdow

nto

thefollow

ingcommutativediagram

FG(V

)=

kn

��

FG(f

)=(m

atrix

forf)

��

id(V

)=

V

id(f

)=f∈Hom(V

,W)

��FG(W

)=

km

�� id(W

)=

W

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

25

5.3.

Equivalence

:AFFIN

EVARIE

TIE

SAND

F.G

.REDUCED

k-A

LGEBRAS

Theore

m.Thereis

anequivalence

ofcategories

1

{affinealgebraic

varietiesandmorphsofaff.vars.}

↔{f.g.reducedk-algsandhomsofk-algs}

op

XT �→

k[X

]

(XF →

Y)

T �→(F

∗:k[X

]←

k[Y

]).

Proof.

Tis

awell-defined

functor.�

Tis

faithful:

becau

se(F

∗ )∗=

F.�

Tis

full:given

ak-alg

hom

ϕ:k[X

]←

k[Y

],take

F=

ϕ∗then

F∗=

(ϕ∗ )

∗=

ϕ.�

Tis

essentially

surjective:

given

af.g.

reducedk-alg

A,choose

generatorsα1,...,α

nforA.Define

I A=

ker(k[x

1,...,x

n]→

A,xi�→

αi).

(5.1)

Then

A∼ =

k[x

1,...,x

n]/I A

=k[X

A]forX

A=V(I

A),usingI(X

A)=√I A

=I A

asA

isreduced.�

�Remark

.Theproof

ofLem

ma5.1,

inthis

particularexam

ple,wou

ldconstruct

afunctor

G:A�→

XA=

V(I

A)an

dG

:(ϕ

:A←

B)�→

(ϕ∗:X

A→

XB).

Then

mim

icSection

5.2.

Specm

nota

tion:ifA

isafinitelygenerated

reducedk-algebra,then

we’veshow

nthat

thereis

anaffi

nevarietyX

A(uniqueupto

isom

orphism)whosecoordinateringis

isom

orphic

toA.Write

Specm

A

forthisaffi

nevariety.

Section

15willdiscu

ssSpecm

properly.For

now

,recallthat

Specm(A

)as

aset

consistsof

themax

imal

ideals

ofA,whichindeedrepresentthegeom

etricpoints

ofX

A.How

ever,

torealisethis

asan

affinevariety(i.e.withachoice

ofem

beddingX

A⊂

Aninto

someAn)wehad

tomak

eachoice

ofgeneratorsforA.

5.4.

NO

EQUIV

ALENCE

FOR

PROJECTIV

EVARIE

TIE

S

Bycomposing(X

⊂Pn

)�→

(� X⊂

An+1)�→

(S(X

)=

k[� X])

weob

tain

amap

{proj.vars}→

�f.g.

reducedN-graded

algebrasA

generated

by

finitelyman

yelts

indegree1,

withA

0=

k

�

“Con

versely”,

given

such

analgebra

A,pickgeneratorsα0,...,α

nof

degree1,

this

determines

ahom

ϕ:R→

A,x

i�→

αi,then

X=

V(ker

ϕ)⊂

Pnsatisfies

S(X

)=

R/kerϕ∼ =

A(noticekerϕ

isahom

ogeneousideal).Thereis

noequivalence

ofcatego

ries

inthis

case:not

allalgebra

hom

omor-

phismsgiverise

toprojectivemorphismsof

theassociated

projectivevarieties(not

allmorphisms

� X→

� Ydescendto

X→

Y,becau

sethey

may

not

preservetherescalingk-action).

Ifwerequirethe

k-algebra

hom

sto

begrad

ing-preserving,

itbecom

estoorestrictive:

then

only

restrictionsof

linear

embeddings

Pn�→

Pmcanarise,

soforn=

mon

lyprojectiveequivalenceswou

ldbemorphs.

Asmention

edin

Section

3.11

,S(X

)is

not

anisom

orphism-invarian

t,so

therecannot

bean

equivalence

ofcatego

ries

ofprojectivevarietiesin

term

sof

thehom

ogeneouscoordinaterings

S(X

).

6.

PRODUCTS

AND

FIB

RE

PRODUCTS

6.0.

ALGEBRA

BACKGROUND:TENSOR

PRODUCTS

Thetenso

rpro

ductof

twok-vectorspaces

V⊗W

isavector

spaceof

dim

ension

dim

V·d

imW

withbasis

v i⊗

wjwherev i,w

jarebases

forV,W

.Example.Rn⊗

Rm∼ =

Rnm.

You

canextendthesymbol⊗

toallvectorsbydeclaringthat

(�λiv

i)⊗(�

µjwj)=

�(λ

iµj)v

i⊗wj.

Example.0⊗

w=

0=

v⊗

0,(e

1+

2e3)⊗

(7e 1

+e 2)=

7e1⊗

e 1+

14e 3⊗

e 1+

e 1⊗

e 2+

2e3⊗

e 2.

Exercise.V

∗⊗

W∼ =

Hom

(V,W

)forfinitedim

ension

alv.s.V,W

,whereV

∗is

thedual

ofV.

For

k-algebrasA

andB,thetensorproduct

A⊗B

(orA⊗

kB)is

thevector

spaceas

above,

and

1“op”is

theopposite

cate

gory

,so

arrow

s(m

orphs)

pointin

theopposite

directionthantheoriginalcategory.

26

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

multiplication

isdon

ecompon

entw

ise.

Thusageneral

elem

entisafinitesum

�ai⊗b i

withai∈A,

b i∈B,an

dtheproduct

is(�

ai⊗

b i)·(

�a� j⊗

b� j)=

�(a

ia� j)⊗

(bib

� j)summingover

allpairs

i,j.

Thetensorproduct

isdetermined

upto

uniquek-algeb

raisomorphism

byauniversalproperty.

Nam

ely,

A⊗

Bis

ak-algebra

together

withak-alg

hom

ϕ:A×

B→

A⊗

Bwhichis

abalanced

bihomomorp

hism.Bihomom

orphism

meansϕ(·,

b):A→

A⊗

Bis

ak-alg

hom

forallb∈B,and

similarly

forϕ(a,·)

.Balancedmeansϕ(λa,b)=

ϕ(a,λ

b)forallλ∈k,a∈A,b∈B.Theuniversal

property

isthat

anyk-alg

hom

ϕ� :

A×B→

Cwhichisabalan

cedbihom

omorphism

must

factorise

through

auniquek-alg

hom

ψ:A⊗

B→

C(soϕ� =

ψ◦ϕ

).Recallkis

analgebraically

closed

field(this

iscrucialforthenexttw

oresults).

Lemma

6.1.Let

Abe

afinitelygeneratedreducedk-algebra.If

a∈

Alies

inallmaximalideals

m⊂

A(equivalently:

a=

0∈A/m

),then

a=

0.

Proof.

Let

p∈X

=Specm(A

)beapoint.

Recallfrom

2.3that

pdefines

amax

imal

idealm

=m

p⊂

Aan

dan

evaluationisom

orphism:

ϕ:A/m

∼ = −→k.

Noticeϕ(a)=

a(p),

thusa�∈m

isequivalentto

thestatem

enta(p)�=

0.Finally,ifa∈k[X

]=

Ais

anon

-zerofunction(soa�∈I(X)),then

a(p)�=

0at

somep∈X.

�Theore

m6.2.Let

A,B

bek-algebras.

AssumeA

isfinitelygenerated.

(1)If

A,B

are

reduced,then

sois

A⊗

B.

(2)If

A,B

are

integraldomains,

then

sois

A⊗

B.

Proof.

(Non

-exam

inable.)

1)Say

c=

�ai⊗b i∈A⊗B

isnilpotent.

Bybilinearity,W

LOG

b iarelinearlyindep

endent/k.Any

max

idealm⊂

Ayieldsan

isoϕasin

Lem

ma6.1.

Con

sider

thek-algebra

hom

A⊗

B→

(A/m

)⊗

B∼ =

k⊗

B∼ =

B,

c=

�ai⊗

b i�→

�ai⊗

b i�→

�ϕ(a

i)⊗

b i�→

�ϕ(a

i)b i.

AsB

isreduced,thenilpotentelem

ent

�ϕ(a

i)b i

iszero,thusϕ(a

i)=

0byindep

enden

ce/k,so

ai=

0,thusai=

0byLem

ma6.1,

soc=

0.

2)Say

(�ai⊗b i)(

�a� j⊗b� j)=

0∈A⊗B,ag

ainW

LOG

b ilin.indep

./k,an

db� ilin.indep

./k.Applying

thehom

from

(1),(�

ϕ(a

i)b i)(

�ϕ(a

� i)b� i)

=0∈B.AsB

isan

I.D.,oneofthosetw

ofactorsiszero.

Bylinearindep

enden

ce,foreach

m,either

allϕ(a

i)=

0,or

allϕ(a

� j)=

0(orboth).

Thus,

either

allai∈m

oralla� j∈m

(butwedon

’tknow

ifthesamecase

amongthosetw

owillap

ply

forallm).

Geometricallythis

implies

X=

Specm(A

)=

V(a

i:alli)∪V(a

� j:allj).ButX

isirreducible

asA

isan

I.D.,so

WLOG

X=

V(a

i:alli),so

ai=

0∈A,thus

�ai⊗

b i=

0∈A⊗

B.

�

6.1.

PRODUCTS

OF

AFFIN

EVARIE

TIE

S

For

affinevarieties,

X=

V(f

1,...,f

N)⊂

An,

f j=

f j(x

1,...,x

n)∈k[x

1,...,x

n],

Y=

V(g

1,...,g

M)⊂

Am,

g i=

g i(y

1,...,y

m)∈k[y

1,...,y

m].

Theproduct

X×

Yis

theaffi

nevariety

X×

Y=

V(f

1,...,f

N,g

1,...,g

M)⊂

An+m

usingthecoordinateringk[A

n+m]=

k[x

1,...,x

n,y

1,...,y

m].

Abbreviate

I=

I(X),

J=

I(Y),

viewed

assubsets

ink[A

n+m]=

k[x

1,...,x

n,y

1,...,y

m].

Observethat:1

X×

Y=

V(I∪J)=

V(�I+

J�)⊂

An+m

where�I∪J�=

�I+

J�⊂

k[x

1,...,x

n,y

1,...,y

m].

HereI+J=

{f(x)+

g(y):f(x)∈I,g(y)∈J}a

swritten

isnot

yetan

idealin

k[x

1,...,x

n,y

1,...,y

m].

Itgenerates

theideal�I

+J�=

k[x

1,...,x

n,y

1,...,y

m]·(I

+J)=

k[y

1,...,y

m]·I+k[x

1,...,x

n]·J.

1Foraff./proj.

vars.,X

×Y

asasetis

theusual{(a,b):a∈

X,b

∈Y}.

It’s

theZariskitopologywhichis

subtle.

High-tech:allelem

ents

inSpecm

k[X

]⊗kk[Y

]hav

etheform

ma⊗m

b,butSpec

k[X

]⊗kk[Y

]alsohaselem

ents

whichare

notoftheform

℘1⊗

℘2:e.g.X

=Y

=A

1,thediagonalD

={(a,a

):a∈A

1}⊂

X×

Ycorrespondsto

℘=

�x1−

y1�.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

27

Atthecoordinateringlevel:1

k[X

×Y]

=k[x

1,...,x

n,y

1,...,y

m]/�I

+J�

∼ =k[x

1,...,x

n]/I⊗

kk[y

1,...,y

m]/J

=k[X

]⊗

kk[Y

]

byidentifyingxi∼ =

xi⊗

1an

dy j∼ =

1⊗

y j.Theisom

orphism

isexplicitlygiven

by

k[x

1,...,x

n,y

1,...,y

m]/�I

+J�→

k[x

1,...,x

n]/I⊗

kk[y

1,...,y

m]/J

�αiβ

i�→

�αi⊗

βi,

whereαi∈k[x

1,...,x

n],βi∈k[y

1,...,y

m].

Theinversemap

is�

αi⊗

βi�→

�αiβ

i.Exercise.Checkthat

thetw

omap

sarewell-defined.2

Lemma6.3.�I+J�=

k[y

1,...,y

m]·I

+k[x

1,...,x

n]·J

isaradicalidealin

k[x

1,...,x

n,y

1,...,y

m].

Proof.

ByTheorem

6.2.(1),since

I,J

areradical

wededuce

that

k[x

1,...,x

n]/I⊗

kk[y

1,...,y

m]/Jis

reduced.Bytheab

oveisom

orphism,it

follow

sthat

k[x

1,...,x

n,y

1,...,y

m]/�I

+J�isreduced.

�

Remark

.IfX,Y

areirreduciblethen

soisX×Y,byTheorem

6.2.(2)or

byageom

etricalargu

ment.3

6.2.

PRODUCTS

OF

PROJECTIV

EVARIE

TIE

S

For

proj.vars.X,Y

onecanuse

theab

oveaffi

neconstructionlocallyto

definetheZariskitopolog

yon

X×Y.Wenow

show

that

onecanequivalentlycarryou

taglob

alconstructionbyusingtheSegre

embeddingfrom

Section

4.2.

Recallfrom

that

Section

thenotation:σn,m

:Pn

×Pm

�→P(

n+1)(m+1)−

1,

theSegre

varietyΣn,m

=σn,m(P

n×

Pm)⊂

Pnm+n+m,an

dtheprojectionmap

sπco

l,πrow.

Definition(Zariskitopolog

yon

Products).TheZariskitopology

onPn

×Pm

isthesubspace

topology

onΣn,m⊂

Pnm+n+m

(i.e.wedeclare

thatσn,m

andπco

l×

πroware

isomorphisms).

TheZariskitopology

onX

×Y

isthesubspace

topology

onσn,m(X

×Y)⊂

Σn,m⊂

Pnm+n+m

(i.e.

wedeclare

thatσn,m

:X

×Y→

σn,m(X

×Y)is

ahomeomorphism).

Theore

m.X⊂Pn

,Y⊂Pm

proj.vars.⇒

X×Y

isaproj.var.isomorphic

toσn,m(X

×Y)⊂

Pnm+n+m.

Proof.

Itremainsto

show

that

σn,m(X

×Y)is

aprojectivevariety.

This

isan

exercise.

Hint:

SayX

=V(F

1,...,F

N),

Y=

V(G

1,...,G

M),

then

show

that

σn,m(X

×Y)=

Σn,m∩V(F

k(z

0j,...,z

nj),G

�(z i0,...,z

im):allk,�,i,j).

�

Ifweintersectwiththeop

ensets

U0,P

n=

(x0�=

0)=

{[1:x1:···:

xn]}

U0,P

m=

(y0�=

0)=

{[1:y 1

:···:

y m]}

then

σn,m((X×Y)∩

(U0,P

n×U0,P

m))

isdescribed

bythematrix[x

iyj]withfirstcolumn(1,x

1,...,x

n)

(since

x0=

y 0=

1)an

dfirstrow

(1,y

1,...,y

m).

SoDefinition

6.2ab

oveim

poses

precisely

the

vanishingof

f k=

Fk(1,x

1,...,x

n)an

dg �

=G

�(1,y 1,...,y

m)(theother

relation

sfrom

Σn,m

tell

usthat

theother

cols/row

shavenonew

inform

ation:they

arerescalings

ofthefirstcolumn/row

).Thustheglob

alconstructionwiththeSegre

embeddingag

rees

withthelocalaffi

neconstruction.

1Theisomorphism

isjustified

later.

Exerc

ise.Proveis

usingtheuniversalproperty

from

Sec.6.0.

2Example:iff i

∈I,then

f iβ∈�I+J�a

ndmapsto

fi⊗β=

0asfi=

0∈k[x

1,...,x

n]/I.Sim

ilarlyIβ→

I⊗β=

0.

3Hints.Bycontradiction,ifX

×Y

=C

1∪C

2forclosedsets

Ci,usingirreducibilityofY

show

thatX

=X

1∪X

2

whereX

i=

{x∈

X:x×

Y⊂

Ci}.

TheseX

iare

closed

(themap

X→

X×

Y,x

�→(x,y)is

continuousso

{x∈X

:(x,y)∈Z

i}is

closedforeach

y,now

intersecttheseover

ally∈Y).

Finallyuse

irreducibilityofX.

28

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

6.3.

CATEGORIC

AL

PRODUCTS

Category

Theory

:letC

beacatego

ry.

Examples.

Categoryof

Sets:

Objects=

sets,Morphisms=

allmapsbetweensets.

Categoryof

Vectorspaces:Obj=

vectorspaces,Morphs=

linearmaps.

Categoryof

Top

olog

icalspaces:Obj=

top.spaces,Morphs=

continuou

smap

s.Categoryof

Affinevarieties:

Obj=

aff.vars.,Morphs=

morphsof

affinevars.

Apro

ductof

X,Y

∈Ob(C

)(ifit

exists)is

anob

ject

X×

Y∈Ob(C

)withmorphismsπX,πY

toX,Y

s.t.

foran

yZ∈Ob(C

)withmorphsto

X,Y

wehave1

Z

∃unique ��

��

��

X×

YπY

��

πX

��

Y

X

Example.For

C=

Sets,

X×

Y=

{(x,y)∈X

×Y

:x∈X,y∈Y}is

theusual

product

ofsets.

Exercise.Show

X×

Yis

uniqueupto

canon

icalisom

orphism,ifit

exists.

Algebraically,weexpectthe“o

pposite”of

theproduct,so

thecopro

ductof

k[X

],k[Y

]:

k[Z

]��

∃unique

����

k[X

]⊗

kk[Y

]��π∗ Y

�� π∗ X

k[Y

]

k[X

]

whereπ∗ X(x

i)=

xi⊗

1,π∗ Y(y

j)=

1⊗

y j.Indeed,ifthegivenmap

sinto

k[Z

]wereϕ,ψ

,then

the

uniquemap

is�

αi⊗

βi�→

�ϕ(α

i)ψ(β

i).

This,together

withtheequivalence

ofcatego

ries

from

Sec.5.3,is

another

proof

oftheresult

from

Sec.6.1

that

k[X

×Y]∼ =

k[X

]⊗kk[Y

].Example.C

=Sets:

coproduct

X�Y

isthedisjointunion,withinclusion

sX→

X�Y

,Y→

X�Y

.Exercise.For

C=

VectorSpaces,thecoproduct

isthedirectsum

ofvector

spaces.

6.4.

FIB

RE

PRODUCTS

AND

PUSHOUTS

This

Sectionis

non-examinable.

Motivation.In

geometry,yo

ustudyfamiliesofgeometricob

jectslabeled

byaparam

eter

spaceB.

Sof:X→

Bwheref−1(b)is

thegeom

etricspace

inthefamilyassociated

totheparameter

b.

Example.

f:(V

(xy−

t)⊂

A2)→

A1,f(x,y,t)=

t,is

afamilyof

“hyperbolas”xy=

tin

A2

dep

endingon

aparameter

t∈k,whichat

t=

0degenerates

into

aunionoftw

olines

(thetw

oax

es).

Insettheory,thefibre

product

oftw

omap

sf:X→

B,g:Y→

B(overthe“b

ase”

B)is

X×

BY

={(x,y)∈X

×Y

:f(x)=

g(y)∈B}.

Example.Thefibre

f−1(b)is

thefibre

product

off:X→

Ban

dg=

inclusion:{b}→

B.

Example.Theintersection

X1∩X

2in

Xis

thefibre

product

oftheinclusion

sX

1→

X,X

2→

X.

Category

Theory

:letC

beacatego

ry.

Thefibre

pro

duct(orpullback

orCartesian

square

)of

f:X→

B,g:Y→

B(ifit

exists)is

1Convention:ifwewrite

adiagram,werequirethatit

commutes(unless

wesayotherwise).

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

29

anob

ject

X×

BY∈Ob(C

)withmorphismsπX,π

Yto

X,Y

s.t.

foran

yZ∈Ob(C

)withmorphsto

X,Y

(com

mutingwithf,g)wehave

Z

∃unique ��

��

��

X×

BY

πY

��

πX

��

Y

g ��X

f�� B

Exercise.X

×BY

isuniqueupto

canon

ical

isom

orphism,ifitexists.

Example.(Ifyouhaveseen

vector

bundles.)Given

avector

bundle

Y→

Bover

aman

ifold,an

da

map

f:X→

Bof

man

ifolds,

then

X×

BY

=� x

∈XYf(x

)is

thepullbackvector

bundle

f∗ Y→

X.

Algebraically,weexpectthe“opposite”,so

thepush

out1

k[Z

]��

∃unique

����

k[X

]⊗k[B

]k[Y

]��π∗ Y

�� π∗ X

k[Y

]�� g

∗

k[X

]��

f∗

k[B

]

whereπ∗ X(x

i)=

xi⊗1,

π∗ Y(y

j)=

1⊗y j,an

dwhere2

k[X

]⊗k[B

]k[Y

]=

k[X

]⊗kk[Y

]/�f

∗ (b)⊗1−1⊗g∗ (b)

:b∈k[B

]�.Example.For

C=

Sets,

thepushou

tof

theinclusion

sA∩B→

A,A∩B→

Bis

just

theunion

A∪B

(withob

viousinclusion

sfrom

A,B

).Thepushou

tof

general

map

sC→

A,C→

B,is

the

disjointunionA�B/∼

afteridentifyinga∼

bifa,b

areim

ages

ofsomecommon

c∈C.

Remark

.A

=k[X

]⊗

k[B

]k[Y

]may

havenilpotents

(asin

thenextExam

ple)in

whichcase

itdoes

not

correspon

dto

thecoordinateringof

anaffi

nevariety.

How

ever,wecanreducethealgebra:

Ared=

A/n

il(A

)wherethenilradical

nil(A

)is

thesubalgebra

ofnilpotentelem

ents.Then,as

we

wan

tan

affinevariety,

defineX

×BY

tobe“the”

affinevarietywithcoordinateringA

red.It

satisfies

thepushou

tdiagram

forallaffinevarietiesZ

(notenil(A

)→

{0}via

A→

k[Z

]as

k[Z

]is

reduced).

What

has

hap

pened

hereisthat

even

thou

ghk[X

]⊗k[B

]k[Y

]isthecorrectpushou

tin

thecategory

ofrings

(inparticular,also

inthecategory

ofk-algebras),itisnot

thecorrectpushou

tin

thecategory

off.g.

reducedk-algebras(equivalently,

thecategory

ofaffi

nevarieties),so

wehad

toreduce.

Example.Below

isthemostcomplicatedway

ofsolvingtheequationx2=

0(!)

Observethenextpicture.Wewan

tto

calculate

thefibre

product

over

0of

f:A1→

A1,a�→

a2.

A1×

A1{0}

��

��

{0} g=incl

��

k[x]⊗

k[b]k[b]/(b)∼ =

k[x]/(x

2)��

��k[b]/(b)

�� incl

∗

A1

f�� A

1k[x]��

f∗

k[b]

a✤

�� a2

x2��

✤ b

1In

theTopology&

Groupscourse,

youhaveseen

apushout:

intheVanKampen

theorem,when

youtakethefree

product

withamalgamationofthefirsthomotopygroups.

2we“identify”f∗ (b)

andg∗ (b),in

particular(f

∗ (b)x)⊗

y≡

x⊗

(g∗ (b)y),

butthereare

more

relationsaswetake

theidealgen

eratedbythose

identifications.

30

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

wherek[b]/(b)isthecoordinate

ringofthepointb=

0in

A1.Theabovediagram

proves

thatthefibre

f−1(0)is

Specm

(k[x]/(x))

wherewereduced(k[x]/(x

2))

red=

k[x]/(x),so

itis

V(x)=

{0}⊂

A1.

f

x2=

bx2=

0

b0

6.5.

GLUIN

GVARIE

TIE

S

This

Sectionis

non-examinable.

Therole

ofgeom

etry/algebra

above(pullback/p

ushout)

canalso

bereversed,asin

thecase

of

gluingvarieties.

TogluevarietiesX,Y

over

a“com

mon

”op

ensubsetU

�→X,U

�→Y,wepushout:

X×

UY��

��Y ��

X��

U

whichalgebraically

isthefibre

product

k[X

]×k[U

]k[Y

],nam

elythefunctionswhichagreeonU.As

usual,category

theory

helpsto

predictwhat

thean

swer

shou

ldbe,

butthereis

noguaranteethat

thepullback/p

ushou

texists

insidethecategory

weare

workingin.For

example,below

,wegluetw

oaffi

nevarietiesan

dweendupwithaprojectivevarietythat

isnot

affine.

Example.P1

=A1×

A1\{

0}A1is

thegluingof

twocopiesof

A1over

U=

A1\{

0}via

thegluing

map

sU→

A1,b�→

ban

dU→

A1,b�→

b−1.Algebraically:k[x]×

k[b,b

−1]k[y],determined

bythe

twohom

s(x,0)�→

b,(0,y)�→

b−1.This

correspon

dsto

pairs

ofpolynom

ialfunctionsf:A1→

k,

g:A1→

ksatisfyingf(b)=

g(b

−1),i.e.

agreeingon

theoverlapU

via

thegluingmaps.

Exercise.k[x]×

k[b,b

−1]k[y]∼ =

k.Indeedtheon

lyglob

alfunctionson

P1are

theconstan

tfunctions.

7.

ALGEBRAIC

GROUPS

AND

GROUP

ACTIO

NS

7.1.

ALGEBRAIC

GROUPS

Definition.G

isanalgebra

icgro

up1ifG

isanaffinevariety,

andithasagroupstructure

given

bymorphismsofaffinevarieties.

Explicitly:

multiplicationm

:G

×G→

Gandinversioni:G→

Gare

morphsofaff.vars.

Ahomomorp

hism

G→

Hofalg.groupsis

ahom

ofgroupswhichis

alsoamorphofaff.vars.

EXAM

PLES.

1)finitegrou

ps(viewed

asadiscretesetof

points).

2)SL(n,k)=

V(det−1)⊂

An2.

3)k∗=

k\{

0}∼ =

V(xy−

1)⊂

A2via

a↔

(a,a

−1),

withm

=multiplication.Recallthecoordinate

ringis

k[k

∗ ]=

k[x,y]/(xy−1)∼ =

k[x,x

−1].

4)k∼ =

A1withm

=addition.

5)GL(n,k)=(non-singu

larn×

nmatrices/k)∼ =

V(y

·det−1)⊂

An2+1,hence

anyZariskiclosed

subgrou

pwillalso

beanalgebraic

group.

Exam

plesof

such

subgrou

ps:

upper

triangularmatrices,2upper

unipotentmatrices,

3anddiagon

al

1Much

ofthetheory

isthealgeb

raic

analogueofthetheory

ofLie

groups(groupswhichare

alsomanifolds).

2M

ij=

0fori>

j.3M

upper

triangularandallM

ii=

1.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

31

matrices.

(Allow

ingon

lynon

-singu

larmatrices)

6)If

G,H

alg.gp

s.then

theproduct

grou

pG

×H

isan

alg.gp

.Exam

ple:thealgebra

icto

rus1

Gm

=k∗×

···×

k∗is

analg.gp

.7)For

Galgebraic

grou

p,defineG

0=(the2

irreducible

compon

entcontaining1).Exercise:Show

that

G0is

analgebraic

grou

p.Show

that

theirreducible

compon

ents

ofG

arethecosets

ofG

0.

8)H⊂

Gasubgrou

pof

analgebraic

grou

p.Exercise:theclosure

His

analgebraic

subgrou

p.

9)ϕ:G→

Hamorphof

alg.gp

s.Exercise:kerϕ⊂

Gis

analgebraic

subgp

.Fact:

imϕ⊂

His

analgebraic

subgp

.10)Fact.

Every

alg.gp

.is

isom

orphic

toaclosed

subgp

ofsomeGL(n,k).

7.2.

GROUP

ACTIO

NSBY

ALGEBRAIC

GROUPSON

AFFIN

EVARIE

TIE

S

Definition.X

aff.var.,G

alg.gp.,then

anactionofG

onX

isamorphism

G×X→

X,(g,x

)�→

g·x

ofaff.vars.such

that1·x

=xandg 1

·(g 2

·x)=

(g1g 2)·x

.

Example.G

=k∗acts

onX

=A2byt·(a,b)=

(t−1a,tb).Theorbitsare:

O1=

{(0,0)}.

O2=

k∗·(1,0)

={(a,0):a∈k∗ }.

O3=

k∗·(0,1)

={(0,b)

:b∈k∗ }.

O(s)=

k∗·(1,s)

=V(xy−

s)=

{(t−

1,ts)

:t∈k∗ }

wheres∈k∗ .

Thepartition

byorbitsis

A2=

O1∪O

2∪O

3∪∪ s

∈k∗O(s).

Remark

.In

this

Exam

ple,afunctionf

:X→

kwhichis

G-invarian

twillbeconstan

ton

each

orbit.If

fis

continuou

s,then

ftakes

thesamevalueon

O1,O

2,O

3becau

seO

1⊂

O2,O

1⊂

O3.

ByLem

ma2.5,

thetopolog

ical

quotientA2/G

(thespaceof

orbits)

cannot

bean

affinevariety.

Our

goal

isto

defineabetternotionof

quotient,whichidentifies

theorbitsO

1,O

2,O

3so

that

this“g

ood

quotient”

isan

affinevariety.

7.3.

CATEGORIC

AL

QUOTIE

NT

and

REDUCTIV

EGROUPS

Definition.Theca

tegorica

lquotientY

(ifit

exists)is

an

affinevarietyY

withamorphism

F:X→

Ysuch

thatF

isconstantonorbits,

andF

is“universal”,meaning:

foranyother

such

data

Y� ,F

� :X→

Y�wehave

XF��

F���Y

∃uniquemorp

h��

Y�

Example.If

you

takeY

� =point,then

Y→

Y�map

severythingto

that

point.

Exercise.Show

that

Y,F

:X→

Yareuniqueupto

canon

ical

isom

orphism.

Remark

.Onedoes

not

requirethat

F:X→

Yis

surjective(categorically:an

epim

orphism).

Itis

not

diffi

cult

toshow

3that

foraffi

nevarietiesF

must

beadominantmorphism

(i.e.has

dense

imag

e).Attheendof

thesectionweconstruct

anon

-surjective

exam

ple.

TheG-actionon

Xalso

determines

aG-actionon

thecoordinateringk[X

]:g∈G

acts

by

k[X

]→

k[X

],f�→

fgwherefg(a)=

f(g

−1a).

1the“m”refers

tothefact

thatweuse

multiplication.

2Non-examinable:thereis

only

oneirreducible

componen

twhich

contains1.

Indeed,suppose

wehad

twosuch

componen

tsX,Y

.Weneedtw

ofacts:

(1)theim

ageofanyirreducible

varietyunder

acontinuousmapis

irreducible,

and(2)ifX,Y

are

irreducible

then

X×Y

isirreducible.Thustheim

ageunder

multiplicationm(X

×Y)is

irreducible

andcontainsboth

X,Y

(since

X=

m(X

×{1

}))hen

ceX

=Y

=m(X

×Y)byirreducibility.

3Given

acategoricalquotientY

⊂A

N,letY

�betheclosure

ofF(X

)⊂

AN,then

Y�alsosatisfies

theuniversal

property.Byexercise

sheet2,beingadominantmapis

equivalentto

hav

inginjectivepull-back

oncoordinate

rings,

sok[Y

� ]→

k[X

]isinjective.

Hen

cek[Y

]→

k[X

]isinjective,

since

bytheaboveuniversalproperty

itisthecompositionof

k[Y

]→

k[Y

� ]→

k[X

]wherethefirstmapisanisomorphism

bythepreviousexercise.SoF

isdominant(andY

=Y

� ).

32

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

This

isalinearaction,in

thesense

that

Gacts

linearlyon

thecoordinate

ring:1

(f1+

f 2)g(a)=

f 1(g

−1a)+f 2(g

−1a)=

fg 1(a)+

fg 2(a)and(λf)g(a)=

λf(g

−1a)=

λfg(a)forλ∈k,a∈X⊂

An.

Example.In

theaboveExam

ple,2k∗acts

onk[A

2]=

k[x,y]by3t·x

=tx

andt·y

=t−

1y.

TheG-invariantsu

balgebra

ofk[X

]consistsof

theinvarian

tfunctions

k[X

]G=

{f∈k[X

]:fg=

fforallg∈G}⊂

k[X

].

Example.In

theaboveExam

ple,k[x,y]G

=k[xy]∼ =

k[w

]∼ =

k[A

1]via

xy←� w

.

Lemma

7.1.If

amorphF

:X→

Yis

constantonorbitsthen

F∗:k[Y

]→

k[X

]Glandsin

the

invariantsubalg.

Proof.

(F∗ f

)g(x)=

(f◦F

)g(x)=

(f◦F

)(g−1x)=

f(F

(x))

=(F

∗ f)(x).

�Assume4forth

erest

ofth

isSection

7.3

thatth

ech

ara

cteristic

chark=

0.

Definition.G

isa(linearly)

reductivegroupifeveryrepresentation5ofG

iscompletely

reducible,

6

i.e.

isomorphic

toadirectsum

ofirreducibles.7

Examplesofreductivegro

ups.

(Whichwetreatas

facts)

1)Finitegrou

ps.

2)k∗ .

3)G

m=

k∗×···×

k∗ .

4)SL(n,k).

5)GL(n,k).

Non-example.

G=

k(w

ithad

dition)is

notreductive:

consider

theaction

8k�a�→

(1a

01)∈Aut(k2).

This

rephas

thesubrepk·� 1 0

�butwecannot

findacomplementary

subrep(exercise).

Theorem

(Nagata).Let

Gbe

areductivealg.gp.actingon

an

aff.var.

X.

Then

k[X

]Gis

af.g.

reducedk-alg,i.e.

k[X

]Gis

isomorphic

tothecoordinate

ringofanaff.var.

Remark

.k[X

]Gis

obviouslyreducedas

k[X

]is

reduced.It

ishardto

show

itis

finitelygenerated.

Specm

nota

tion:ifA

=k[X

]Gis

finitelygenerated,then

bySection5.3thereis

anaffi

nevariety

Specm

A(uniqueupto

isom

orphism)whosecoordinateringis

isom

orphic

toA.

Theorem.Let

Gbe

areductivealg.gp.actingonanaff.var.

X.Then

theinclusion

j:k[X

]G→

k[X

]

determines

acategoricalquotien

tgivenby

j∗:X→

X//G≡

Specmk[X

]G.

1Sok[X

]is

a(typicallyinfinitedim

ensional)

representationofG.

2Notice

thattheactionhas“dualized”onthecoordinate

ringlevel.

3Explicitly:x:A

2→

k,x(a,b)=

a,and(t·x

)(a,b)=

x(t

−1·(a,b))

=x(ta,t

−1b)

=ta

=(tx)(a,b).

4Thedefi

nitionsofreductiveandlinearlyreductiveare

differen

twhen

chark�=

0.Linearlyreductive(thedefi

nition

above)

implies

reductive,

buttheconverse

canfail.

5A

repre

senta

tionisa(finitedim

ensional)vectorspace

Vtogether

withahomomorphism

ρ:G

→Aut(V),where

Aut(V)are

thelinearisosV

→V

(bypickingabasisforV,youget

V∼ =

knandAut(V)∼ =

GL(n

,k),

soρallow

susto

“represent”

theactionofG

onV

via

asubgroupoftheinvertible

n×

nmatrices).

Weusuallyjust

say“the

representationV”,andwewrite

gvorg·v

insteadofρ(g)(v).

6Equivalently:(linearly)reductivemeanseveryG-stable

vectorsubspace

W⊂

VhassomeG-stable

vectorspace

complemen

tW

� ,i.e.

V=

W⊕

W�andtheactionofG

preserves

thesummands.

7Irreducible

meansnotreducible.A

repV

isre

ducible

ifthereis

asubrepresentation0�=

W�

V.A

subre

pre

-se

nta

tion

W⊂

Vis

aG-stable

vectorsubspace,meaningG

·W⊂

W(m

eaninggw

∈W

forallg∈G,w

∈W

).8(N

on-examinable)More

gen

erally,

“unipotentelem

ents

are

bad”.Thegen

eraldefi

nitionofre

ductiveexcludes

precisely

these.

Anelem

entrofaringis

unip

ote

ntifr−

1is

nilpotent.

Forexample,anyupper

triangularmatrix

with1in

each

diagonalen

try.

More

gen

erally,

amatrix

isunipotentifandonly

ifallofitseigen

values

are

1,since

after

conjugationit

canbeputinto

Jordannorm

alform

,yieldingsuch

anupper

triangularmatrix.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

33

Explicitly:

pickgenerators

f 1,...,f

Nfork[X

]G,then

theim

age

of

X→

AN,x�→

(f1(x),...,f N

(x))

isanaffinevarietywhichis

“the”

categoricalquotien

tofX

byG.

Remark

.Noticethat

j∗:X→

X//G

issurjectivebyconstruction,since

j∗(X

)=

V(ker

ϕ)=

X//G⊂

AN

whereϕ:k[x

1,...,x

N]→

k[X

]G,ϕ(x

i)=

f i.

Proof.

Step

1.j∗

isconstan

ton

orbits.

Proof.

Ifj∗(x)�=

j∗(gx),byLem

ma2.5thereissomef∈k[X

//G]=

k[X

]Gwithf(j

∗ x)�=

f(j

∗ (gx)).

⇒j(f)(x)=

(j∗∗f)(x)=

f(j

∗ x)�=

f(j

∗ (gx))

=(j

∗∗f)(gx)=

j(f)(gx).

⇒Con

trad

icts

that

j(f)∈k[X

]Gis

G-invarian

t.Step

2.j∗

isuniversal.

Xj∗��

F�

��X//G ∃uniquemorp

h?

��

k[X

]��

j

��

(F� )∗k[X

//G]=

k[X

]G�� ∃

uniquemorp

h?

Y�

k[Y

� ]

ByLem

ma7.1,

(F� )∗landsin

k[X

]G⊂

k[X

],an

dthediagram

ontherigh

tcommutesifthevertical

map

ontherigh

tis

(F� )∗:k[Y

� ]→

k[X

]G,an

dthis

istheuniquemap

that

works.

�EXAM

PLES.

1)In

theab

oveExam

ple(k

∗ -action

onA2)j:k[A

1]∼ =

k[xy]=

k[x,y]G→

k[x,y]=

k[A

2],j(xy)=

xy

determines

thecatego

ricalquotient

j∗:A2→

A1,j(a,b)=

ab.

Notice,

onorbits,

j∗map

sO(s)�→

s,whereasO

3,O

2,O

1allmap

to0∈A1.

Fact.

1Let

Xbean

affinevarietywithalinearlyreductivegrou

paction

byG.Given

anytw

odisjoint

G-invarian

tclosed

subsets

C0,C

1of

Xthereisafunctionf∈k[X

]Gwithf(C

0)=

0an

df(C

1)=

1.Exercise.Twoorbitsmap

tothesamepointin

thecatego

ricalquotient⇔

theirclosuresintersect.

Coro

llary

ofth

eexercise.

For

finitegrou

psG,thecatego

ricalquotientX//G

=X/G

canbe

identified

withtheorbit

space(since

points

areclosed).

2)G

=Z/

2actingon

A2by(−

1)·(a,b)=

(−a,−

b).

⇒G

acts

onk[A

2]=

k[x,y]by(−

1)·x

=−x,(−

1)·y

=−y.

⇒k[x,y]G

=k[x

2,x

y,y

2]∼ =

k[z

1,z

2,z

3]/(z

1z 3−

z2 2)=

k[Y

]whereY

=V(z

1z 3−

z2 2)⊂

A3.Sothe

catego

ricalquotientis

A2→

Y,(a,b)�→

(a2,ab,b2).

3)G

alg.gp

.,H⊂

Gan

yclosed

normal

subgp

.Fact.

2G/H

isan

algebraic

grou

pwithcoordinatering3

k[G

]H,so

G//H

=G/H

.

4)Thenon

-red

uctivegrou

pk,withad

dition,identified

withG

={(

1∗

01)},acts

onX

=SL(2,k)by

left

multiplication

ofmatrices.

Weclaim

that

C2is

acatego

ricalquotientX//G,withF

:X→

C2,

F(A

)=

(firstcolumnof

A).

NoticeF

isnot

surjectiveas

F(X

)=

C2\{

0}.Noticethat

k[X

]G⊂

k[X

]is

thek-algebra

k[x

11,x

21]⊂

k[x

ij]generated

bytheentriesof

thefirstcolumn.Then

theproof

oftheprevioustheorem

applies

tothis

case,since

k[X

]Gis

finitelygenerated.

1Algebraic

Urysohn’sLem

ma:ifC

0,C

1are

disjointclosedsets

inanyaff.var.

X,then

thereis

afunctionf∈k[X

]withf(C

0)=

0,f

(C1)=

1.Proof:

sayC

j=

V(I

j),

then

∅=

C0∩C

1=

V(I

0+

I 1)so

I 0+

I 1=

k[X

],so

forsome

f j∈

I jwehav

ef 0

+f 1

=1.Now

consider

f=

f 0.�

Inoursetup,wealsowantfto

beG-invariant.

Onedoes

this

byapplyingtheReynoldsoperatorR

:k[X

]→

k[X

]G,whichwehaven

’tconstructed

inthesenotes.

Forfinitegroups

G,it

iseasy

toconstruct:(R

f)(x)=

1 |G|�

g∈Gf(gx).

2It

isnotso

easy

toshow

thatSpecm

k[G

]H∼ =

G/H

are

homeomorphic.

3H

actsonk[G

]byfh=

f◦h

−1,so

k[G

]H⊂

k[G

]

34

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

8.

DIM

ENSIO

NTHEORY

8.1.

GEOM

ETRIC

DIM

ENSIO

N

Let

Xbeavariety(affineorprojective).A

chain

oflength

mmeansastrict

chainof

inclusions

∅�=

X0�

X1�

X2�

···�

Xm

(8.1)

whereeach

Xi⊂

Xis

anirreducible

subvariety.

OnecanstartwithX

0=

{p}apointof

X,an

difX

isirreducible

then

onecanendwithX

m=

X.

Definition.Theloca

ldim

ension

dim

pX

ofX

atapointp∈X

isthemaximum

overalllengths

ofchainsstartingwithX

0=

{p}.

Thedim

ensionofX

isthemaximum

ofthelengthsofallchains,

dim

X=

max m

(∃chain

X0�

X1�

X2�

···�

Xm)=

max

p∈X

dim

pX.

SayX

haspure

dim

ension

ifthedim

pX

are

equalforallp∈X.

Theco

dim

ension

ofanirreduciblesubvarietyY⊂

Xis1

codim

Y=

max m

(∃chain

Y�

X1�

X2�

···�

Xm−1�

Xm).

EXAM

PLES.

1.A0=

{0}=

V(x

1,...,x

n)⊂

A1=

V(x

2,...,x

n)⊂

···⊂

An−1=

V(x

n)⊂

Anso

dim

An≥

n.

2.X

=V(xy,x

z)=

(yz−

plane)∪(x−

axis).

Then

dim

pX

=2atallpoints

pin

theplane,

and

dim

pX

=1at

other

points.

3.X

=(pointp)�(line)⊂

A2(disjointunion).

Then

Y=

{p}⊂

Xhas

codim

=0.Noticethat

dim

X−

dim

Y=

1−

0=

1�=

codim

Y,whereasdim

pX−

dim

pY

=0−

0=

0=

codim

Y.

Exercise.If

X=

X1∪···∪

XN

isan

irreducible

decom

position

,then

dim

X=

max

dim

Xj.If

Xhas

pure

dim

ension

,then

dim

X=

dim

Xjforallj.

Exercise.Anaffi

nevarietywithdim

X=

0is

afinitecollection

ofpoints.

FACT.X

=V(I)⊂

Anis

afinitesetof

points⇔

k[X

]is

afinitedim

ension

alk-vectorspace.

Indeed,thenumber

ofpoints

isd=

dim

kk[X

],an

dk[X

]∼ =

kdas

k-algebras(exercise

2).

Sodonot

confuse

dim

k[X

]an

ddim

kk[X

].

Lemma8.1.If

X⊂

Ythen

dim

X≤

dim

Y.

IfX,Y

are

irreducibleandX

�Y

then

dim

X<

dim

Y.

(Soforirreducibles

X⊂

Y,ifdim

X=

dim

Ythen

X=

Y.)

Proof.

AnychainforX

isachain

forY.IfX�=

Yare

irredsthen

canextendfurther:X

m+1=

Y.

�

FACT.dim

Pn=

dim

An=

n.

8.2.

DIM

ENSIO

NIN

ALGEBRA

Let

Abearing(com

mutative

withunit).

Ach

ain

oflength

mmeansastrict

chain

ofinclusion

s

℘0�

℘1�

···�

℘m−1�

℘m

(8.2)

whereeach

℘i⊂

Ais

aprimeideal.

Onecanstartwithamax

ideal℘0=

m⊂

A.IfA

isan

integral

dom

ain

onecanendwith℘m

={0

}.FACT.For

ANoetherian,thedescendingchainconditionholdsforprimeideals,i.e.

(8.2)eventually

stop

s(how

ever,this

neednotholdforgeneral

ideals).

1W

hen

Xis

irreducible,onecantakeX

m=

X.Onecandefi

necodim

Yalsoforreducible

Yastheminim

um

ofall

codim

Y�forirreducible

subvarietiesY

�⊂

Y.Example:thedisjointunionY

=(point)

�(line)

⊂A

2hascodim

=1.

2Consider

theprimary

decomposition

ofI(X),

and

show

thattheminim

alprimes

I jare

pairwisecoprime,

then

use

theChineseremainder

theorem:foranyringA,if

I jare

coprimeideals

(meaningI i

+I j

=(1),

whichim

plies

I=

�I j

=∩I

j)then

A/I∼ =

�A/I j

via

theobviousmap.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

35

Definition.

Theheightht(℘)ofaprimeidealis

themaximallengthofachain

with℘0=

℘,

ht(℘)=

max m

(∃chain

℘�

℘1�

···�

℘m−1�

℘m).

TheKru

lldim

ension

isdim

A=

max

ht(m)

overmaxideals

m,i.e.

themaximallengthofchains.

ForanidealI⊂

Atheheightis

ht(I)=

minht(℘)overallprimeideals

℘containingI.

EXAM

PLES.

1.A

fieldhas

dim

ension

zero.

2.A

PID

has

dim

ension

1(unless

it’s

afield),

e.g.

dim

Z=

1.3.Minim

alprimeideals1areprecisely

thoseof

heigh

tzero.

4.(x

1,...,x

n)⊃

(x1,...,x

n−1)⊃

···⊃

(x1)⊃

{0}show

sdim

k[x

1,...,x

n]≥

n.

EXERCIS

ES.

1.2

Ifyouknow

abou

tlocalisation

(Sec.10),show

that

thecodim

ensioncodim

(℘)=

dim

A℘satisfies

codim

(℘)=

dim

A℘=

ht(℘).

2.3

Ifdim

A=

man

d(8.2)holds,

then

dim

A/℘

j=

m−

j.3.Deduce

that

dim

A≥

dim

(A/℘

)+

codim

(℘),

withequalityif℘=

℘jas

in(8.2)an

ddim

A=

m.

Wewillassumethefollow

ingtw

ofactsfrom

algebra,whichgeom

etricallysaythat

each

equation

weim

posecancutdow

nthedim

ension

byat

moston

e.Keepin

mind(see

Hom

ework2,

ex.1)that

itis

not

alwayspossible

tofindexactlyht(℘)generatorsfor℘.

Theore

m8.2

(Krull’s

principal

idealtheorem,Hau

ptidealsatz).

ForanyNoetherianringA,iff∈A

isneither

azero

divisornoraunit,then

ht((f))

=1.

Exercise.Byliftingachainfrom

A/(f)to

A,show

that

ht((f))

=1⇒

dim

A/(f)≤

dim

A−

1.

Example.WecheckKrull’s

theorem

inan

easy

case:forf∈

Airre

ducible

4an

dA

aUFD

(e.g.

k[x

1,...,x

n]).In

this

case,℘0=

(0)�

(f)is

achain,since

(f)is

prime.5Soht((f))≥

1.Wenow

show

0�

℘�

(f)is

impossible.Suppose0�=

g∈℘(w

ant:

f∈℘so

℘=

(f)).As℘⊂

(f),g=

fmh

forsomeh/∈(f).

Ash/∈(f)also

h/∈℘.As℘isprime,

fmh∈℘forces

fm∈℘an

dso

forces

f∈℘.

Theore

m(K

rull’s

heigh

ttheorem).

ForanyNoetherianringA,and�f

1,...,f

m��=

A,

ht(�f

1,...,f

m�)≤

m.

Sotheheightht(℘)is

atmost

thenumberofgenerators

of℘.Conversely,

if℘⊂

Ais

aprimeideal

ofheightm,then

℘is

aminim

alprimeidealoveranidealgeneratedby

melem

ents.6

Coro

llary.dim

k[x

1,...,x

n]=

n.

Proof.

Weknow

themax

imal

ideals

are�x

1−

a1,...,x

n−

an�,

sothey

haveheigh

tat

mostn

by

Krull’s

theorem,so

dim

k[x

1,...,x

n]≤

n.Theab

oveexam

ple

show

eddim

k[x

1,...,x

n]≥

n.

�1minim

alprimeidealmeansit

does

notcontain

anystrictly

smaller

primeideal.

2Hint:

recallthatprimeideals

inthelocalizationA

℘are

in1:1

corresponden

cewithprimeideals

ofA

inside℘.

3Hint:

recallthatprimeideals

ofA/Iare

in1:1

corresponden

cewithprimeideals

ofA

containingI.

4Recallanelem

entf∈A

ofaringisirre

ducible

ifitisnotzero

oraunit,anditisnottheproduct

oftw

onon-unit

elem

ents.Recallaunit

fis

aninvertible

elem

ent,

i.e.

fg=

1forsomeg∈A.

5Recall,in

anyintegraldomain,primeim

plies

irreducible,andin

aUniqueFactorizationDomain

theconverse

holds,

soprimes

andirreduciblescoincide.

Recallf∈A

isprimeiffis

notzero

andnotaunit,andf|ghim

plies

f|g

orf|h

(equivalently:A/(f

)is

anintegraldomain,i.e.

(f)⊂

Ais

anon-zeroprimeideal).

6Meaning℘correspondsto

aminim

alprimeidealofA/IwhereIis

anidealgen

eratedbym

elem

ents.

36

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Remark

.Moregenerally,

forA

Noetherian,dim

A[x]=

dim

A+

1.This

also

implies

theCorollary.

Thefollow

ingtw

ofactsfrom

algebra

ensure

thatfork-algebras,

dim

ension

theory

isnot

nasty:

Theore

m.Let

Abe

af.g.

k-algebra.1

Then

dim

A=

(maximalnumberofelem

ents

ofA

thatare

algebraicallyindepen

den

t/k).

If℘� ⊃

℘are

primeideals

inA,anytwosaturated2chainsfrom

℘�to

℘have

thesamelength.

Theore

m8.3.Let

Abe

af.g.

k-algebra

andanintegraldomain.3

Then

4

dim

A=

trdeg

kFrac(A).

Ifdim

A=

m,then

all

maximalideals

ofA

have

heightm,in

fact

everysaturatedchain

from

amaximalidealto

(0)haslengthm.Therefore

ht(℘)+

dim

(A/℘

)=

dim

A.

Thusthelengthofasaturatedchain

from

℘�to

℘is

ht(℘� )−

ht(℘)=

dim

A/℘−

dim

A/℘

� .

Asimple

applicationofthis

Theorem

is(comparetheExam

ple

afterTheorem

8.2):

Coro

llary

8.4.Forirreduciblef∈R

=k[x

1,...,x

n]thereis

amaximallengthchain

℘0�

···�

℘n−2�

℘n−1=

(f)�

℘n=

(0).

Notice

how

dim

R/(f)=

n−

1andht((f))

=1addupto

dim

R=

n.

Example.WeprovetheCorollaryusingtran

scendence

degrees.Asfcannot

beconstan

t,itinvolves

atleaston

evariab

le,say

xn.

Then

x1,...,x

n−1in

R/(f)are

algebraically

indep

endentover

k(w

hereasxnsatisfies

apolynom

ialrelation

over

k[x

1,...,x

n−1],so

k(x

1,...,x

n−1)�→

Frac(R/(f))

isan

algebraic

extension).

Sodim

R/(f)≥

n−

1,andbyKrulldim

R/(f)≤

n−

1.Hence

equality.

8.3.

GEOM

ETRIC

DIM

ENSIO

N=

ALGEBRAIC

DIM

ENSIO

N

Theore

m.If

X⊂

Anis

anaffinevarietythen

dim

X=

dim

k[X

]

ForaprojectivevarietyX⊂

Pn,dim

Xequals

themaximallengthofchains(8.2)ofhomogen

eous

primeidealswhichdonotcontain

theirrelevantideal(x

0,...,x

n),

inparticulardim

X=

dim

X−1.

Proof.

UsingHilbert’sNullstellensatz,thereis

abijection

between

chainsin

(8.1)an

dchainsin

(8.2):

℘j=

I(X

j)an

dX

j=

V(℘

j).

Theresult

foraprojectivevarietyfollow

sbytheprojective

Nullstellensatz

(so,

really,

bytheaffinecase

applied

totheaffi

neconeX).

�

Exercise.For

amax

imal

chain

asabove,

ht(℘j)=

codim

V(℘

j)=

n−

dim

V(℘

j).

Theore

m.ForanyirreducibleaffinevarietyX⊂

An,

dim

X=

n−

1⇔

X=

V(f)foranirreduciblef∈R

=k[x

1,...,x

n].

TheanalogousholdsforX⊂

Pnanirreducibleprojectivevarietyandfhomogen

eousin

k[x

0,...,x

n].

1Forexample,when

Ais

reduced,thecoordinate

ringofanaffinevariety.

2i.e.

achain

(8.2)thatcannotbemadelonger

byinsertingmore

primeideals.

3Thus,

thecoordinate

ringofanirreducibleaffinevariety.

4Foranintegraldomain,onecanconstruct

thefraction

field

Frac(A)(m

imickingtheconstructionofFrac(Z)

=Q).

Then

k�→

Frac(A)is

afield

extension.

Foranyfield

extension

k�→

Kthereexists

asubsetB

⊂K,called

transcendence

basis,

whose

elem

ents

are

algeb

raicallyindep

enden

tover

k(i.e.

they

donotsatisfyapolynomial

relationover

k)andsuch

thatk(B

)�→

Kis

analgeb

raic

extension.Herek(B

)den

otesthesm

allestsubfieldofK

containingk∪B.Thetranscendencedegre

etrdeg

kK

isthecardinality

ofB

(FACT:itisindep

enden

tofthechoice

oftranscen

den

cebasisB).

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

37

Proof.

(⇒):

dim

X=

n−1⇒

I(X)�=

(0)⇒∃f�=

0∈I(X).

Since

I(X)isprime,

itmust

contain

anirreducible

factor

ofthefactorizationof

f.SoW

LOG

fis

irreducible,hence

prime(R

isaUFD).

Then

X⊂

V(f)�

An,so

byLem

ma8.1,

dim

X≤

dim

V(f)<

dim

An=

nthusforcingX

=V(f)

since

dim

X=

n−

1.(⇐

):FollowsbyCorollary

8.4.

�Definition.ForanirreducibleaffinevarietyX,thefu

nction

field

is

k(X

)=

Frac(k[X

])

Thus,

byTheorem

8.3,

foran

yirreducible

affinevarietyX,

dim

X=

trdeg

kk(X

)

Rem

ark.

Elements

ofk(X

)areratios

ofpolynom

ials,so

they

definefunctionsX→

kwhichare

defined

onan

open

subsetof

X(thelocuswherethedenom

inator

does

not

vanish).1

Example.k(A

n)=

k(x

1,...,x

n)has

tran

scendence

basis

x1,...,x

nso

dim

An=

n.

Theore

m.ForX,Y

irreducibleaffinevarieties,

dim

(X×

Y)=

dim

X+

dim

Y.

Proof.

Exercise:2

comparethetrdeg

kfork[X

]=

k[x

1,...,x

n]/I(X),

k[Y

]=

k[y

1,...,y

m]/I(Y)an

dk[X

×Y]=

k[x

1,...,x

n,y

1,...,y

m]/�I(X

)+

I(Y)�∼ =

k[X

]⊗

kk[Y

].�

Remark

.Geometrically,

ht(I)is

thecodim

ension

ofthesubvarietyV(I)⊂

Spec(A

).For

anirred

affinesubvarY⊂

X,dim

X≥

dim

Y+

codim

X(Y

)(w

hichfollow

sfrom

k[Y

]∼ =

k[X

]/I(Y)).

Remark

.Aproj.var.X

iscalled

acomplete

intersectionifI(X)isgenerated

byexactlycodim

X=

htI(X)elem

ents.

Recallthetw

isted

cubic

X⊂

P3has

I(X)=�x

2−

wy,y

2−

xz,zw−

xy�⊂

k[x,y,z,w

],an

ditturnsou

tthat

I(X)cannot

3begenerated

by2=

htI(X)=

codim

Xelem

ents.

8.4.

NOETHER

NORM

ALIZ

ATIO

NLEM

MA

Theore

m8.5

(Algeb

raic

version).

Let

Abe

af.g.

k-algebra.Then

thereare

injectivek-alg

homs

k�→

k[y

1,...,y

d]�→

A(8.3)

wherey i

are

algebraicallyindepen

den

t/k,andA

isafinitemodule

overk[y

1,...,y

d].

Moreover,

ifA

isanintegraldomain,then d=

trdeg

kFrac(A).

Amorphof

affvars

f:X→

Yis

finiteiff∗:k[X

]←

k[Y

]is

anintegra

lextension(i.e.each

elem

entof

k[X

]satisfies

amon

icpolynom

ialwithcoeffi

cients

inf∗ k[Y

]).

Fact.

Iff:X→

Yis

afinitemorphof

irred.aff.vars.

then

1)fis

quasi-fi

nite,meaning:

each

fibre

f−1(p)is

afinitecollection

ofpoints;

1Thinkmeromorphic

functions.

2Non-examinableHints:Youwantto

show

thattheunionoftw

otranscen

den

cebases(f

i),(g

j)fork[X

],k[Y

]give

atranscen

den

cebasisfork[X

×Y],wheref i

∈k[x

1,...,x

n],g j

∈k[y

1,...,y

m].

Spanningis

easy

(hen

cedim

X×Y

≤dim

X+

dim

Y)butshow

ingalgeb

raic

indep

enden

ceis

harder.

Suppose

therewasadep

enden

cy,then

you

would

get

G1fI1+

···+

G�fI�∈

�I(X

)+

I(Y)�

⊂k[x

1,...,y

1,...]wheretheG’s

are

polynomials

intheg j,andthefIare

monomialsfi 1 1···f

i a ain

thegiven

f 1,...,f

a.Now

evaluate

they-variablesatanyp∈Y,to

ded

uce

G1(p),...,G

�(p)=

0byalgeb

raic

indep

enden

ceofthef i

ink[X

].Ded

uce

thatG

1,...,G

�∈I(Y),

andfrom

this

concludetheresult.

Another

approach,is

touse

Noether’s

Norm

alization

Lem

ma(Sec.8.4)to

get

finitesurjectivemorphismsX

→A

a,

Y→

Abandobtain

afinitesurjectivemorphism

ϕ:X

×Y

→A

a+b.Thelatter,im

plies

thatk[X

×Y]is

integralover

ϕ∗ (k[A

a+b])=

ϕ∗ (k[f

1,...,f

a,g

1,...,g

b]).TheGoingUp(andLyingOver)

Theorem

saysthatifaringB

isintegral

over

asubringA,then

anychain

ofprimeidealsin

Acanbelifted

toachain

ofprimeidealsin

B(such

thatintersecting

withA

gives

theoriginalchain).

Thusdim

k[X

×Y]≥

a+

b,asrequired

.Thatinequality

canalsobeobtained

more

gen

erallyfrom

thefactthatifϕ:X

→Y

isasurjectivemorphism

ofaffinevarieties,

then

dim

X≥

dim

Y.

Thatfact

isproved

usingresultsfrom

Sec.12.2

asfollow

s.First

replace

Ybyanirreducible

componen

tin

Yofmaxim

al

dim

ension.Then

replace

Xbyanirreducible

componen

tin

ϕ−1(Y

)whose

imageisden

sein

Y(checkitexists

byusing

surjectivityandirreducibilityofY).

Thus,

ϕis

now

adominantmorphism

betweenirreducible

affinevarieties.

This

inducesanextensiononthefunctionfieldsϕ

∗:k(Y

)�→

k(X

)whichbybasicfieldtheory

implies

trdeg

kY

≤trdeg

kX.

3X

=V(I)forI=

�yw−

x2,z

2w−

2xyz+

y3�,

butthis

idealis

notradical.

38

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

2)fis

aclosed

map

(f(closedset)

isclosed);

3)fis

surjective⇔

f∗is

injective.

Example.f:A1→

A1,f

(a)=

a2:seethepicture

inSec.6.4.Sof∗:k[b]→

k[x],f∗ (b)

=x2.Notice

xisintegral

over

k[b]:themonic

polyp(x)=

x2−bover

k[b]satisfies

p(x)=

x2−f∗ (b)

=0∈k[x].

Remark

.(N

on-examinable)

Quasi-finitedoes

not

imply

finite.

Let

f:V(xy−1)→

A1,f

(x,y)=

xbethevertical

projectionfrom

thehyperbola,

ithas

finitefibres.

Then

f∗:k[x]→

k[x,y]/(xy−

1)

istheinclusion

,butyis

not

integralover

k[x]as

xy−

1is

not

mon

ic.Thealgebra

isnothappy

abou

tthe“n

on-com

pactness”

phenom

enon

that

preim

ages

aredivergingnear0.

Notice

fis

nota

closed

map

.It

turnsou

tthatan

affinemorphism

f:X→

Yisfiniteifan

don

lyifitisuniversally

closed(m

eaning:

foreach

morphism

Z→

Ythefibre

product

X×

YZ→

Yis

aclosedmap).

Theorem

(Geometricversion).

Let

X⊂

Anbe

anirreducibleaffinevarietyofdim

ensionm.Then

thereis

afinitesurjective

morphism

f:X→

Am.

Sketchproof.

TakeA

=k[X

]in

Theorem

8.5,

andtake

Specm

of(8.3)to

obtain:X→

Ad→

point.

Therest

follow

sfrom

theab

oveFact.1

�

Soan

yirreducible

affinevarietyis

abra

nch

ed

coveringofaffi

nespace,

meaningamorphism

ofaffi

nevarietiesof

thesamedim

ensionwithdim

(“generic”fibersf−1(p))

=0andwhichresembles

thecoveringspaces

weknow

from

topologyover

thecomplementof

aclosed

subsetof“bad”points

pcalled

thebra

nch

locus.

Thera

mification

locusis

thepreim

agef−1(branch

locus).2

Oneway

tobuild

f:X→

Adis

by

linearprojection,takingy 1,...,y

dto

begeneric

linear

polynom

ials

inx1,...,x

n.

Theorem

(Algebraic

Version

2).When

Ais

af.g.

k-algebra

andanintegraldomain,onecanin

additionen

sure

thatfortheextensionsoffields

k�→

K=

k(y

1,...,y

d)�→

Frac(A)

thefirstis

apurely

transcen

den

talextension,thesecondis

aprimitive3

algebraic

extensionmeaning

FracA

=FracK[z]≡

K(z)

wherez∈A

isalgebraic

overK.Soonly

onepolynomialrelationis

needed:

G(y

1,...,y

d,z)=

0.

Theorem

(GeometricVersion2).ForX

anirreducibleaffvar,

k[y

1,...,y

d,z]�→

A=

k[X

]induces

amorphism

X→

Ad+1whichis

abirational

equivalence

4

X���V(G

)⊂

Ad+1.

Theconclusion

israther

striking:

everyirreducible

affinevarietyis

biration

alto

ahypersurface.

1Exerc

ise.

Show

directly

thatthefibresare

finiteby

using

thateach

xi∈

k[X

]satisfies

amonic

poly

over

k[y

1,...,y

d].

To

show

thefibre

f−1(p)is

non-empty,consider

f∗ �y1−

p1,...,y

d−

pd�⊂

k[X

].(Y

ou

may

need

Nakayama’s

lemma:foranyringsA

⊂B,ifB

isafiniteA-m

odule

then

aB

�=B

foranymaxim

alideala⊂

A).

2Compare

B3.2

Geometry

ofSurfaces:

non-constantholomorphic

mapsbetweenRiemannsurfacesare

locallyofthe

form

z�→

znwhichhasramificationlocu

s{0

}ifn>

1.Sonearmost

points

itis

alocalbiholomorphism.

3In

fact,oneproves

thatonecanchoose

y1,...,y

dso

thatk(y

1,...,y

d)�→

Frac(A)is

afiniteseparable

extension.

Then

theprimitiveelem

enttheorem

from

Galois

theory

applies.

4Wewillseetheselaterin

thecourse.

ArationalmapX

���Y

isamapdefi

ned

onanopen

subsetofX

defi

ned

usingrationalfunctionsin

k(X

)rather

thanpolynomialfunctionsin

k[X

].It

isbirationalifthereis

arationalmap

Y���X

such

thatthetw

ocompositesare

theiden

tity

wherethey

are

defi

ned

.Thinkofabirationalmapasbeing“an

isomorphism

betweenopen

den

sesubsets”.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

39

9.

DEGREE

THEORY

9.1.

DEGREE

Recall(Sec.3.3)alinearsu

bvariety

ofPn

isaprojectivisationL=

P(avectorsubspace

� L⊂

An+1).

X⊂

Pnproj.var.⇒

thedegre

eis

deg(X

)=

#intersectionpoints

ofX

withacomplemen

tary

linearsubvarietyin

generalposition

=generic#

L∩X

forlinearsubvarietiesL⊂

Pnwith

dim

L+dim

X=

n.

Wenow

explain

themeaningof

“general

position”an

d“g

eneric”.

TheGrassman

nianwhichparam

etrizesall

� L⊂

An+1ab

oveis

G=

Gr(n+

1−

dim

X,n

+1).

Fact.

Thereisanon

-empty

open

subsetU⊂

Gsuch

that

thenumber

ofintersection

points

#L∩X

forL∈U

isfinitean

dindep

endentof

L,an

dwecallthat

number

deg(X

).

Coro

llary

9.1.If

U� ⊂

Gis

anynon-empty

opensubset

such

that#L∩X

isfiniteandindepen

den

tofL∈U

� ,then

this

numberequals

deg(X

).

Proof.

Gisirreducible

byLem

ma4.14

,so

bySec.2.6

weknow

U∩U

� isnon

-empty

(anddense).

�

Thusthe“b

ad”L(yieldingadifferentfiniteor

infinitenumber)must

lieinsidesomeproper

closed

subsetV⊂

G,whichisthou

ghtof

as“small”

since

G\V

isop

enan

ddense.The“g

ood”

� L∈G\V

are

called

“ingeneral

position”,

andthat

finitenumber

deg(X

)is

oftencalled

the“g

eneric”number

orthe“expected”number

ofintersection

points.W

hen

Xisirreducible,deg(X

)isin

fact

themax

imal

possible

finitenumber

ofintersection

points

ofL∩X

forallL

(com

pareExam

ple

3below

).IfL� isagenericlinearsubspaceof

dim

ension

smallerthan

thecomplementary

dim

ension

n−dim

X,

then

L� ∩

X=∅.

Theidea

isas

follow

s.Con

sider

ageneric

linearsubspaceL

ofcomplementary

dim

ension

,then

L∩X

isafinitesetof

points.Onethen

checksthat

ageneric

proper

linearsubspace

L� ⊂

Lwillnot

contain

anyof

thosepoints,so

L� ∩

X=∅.

Examples.

1)X

=H

hyperplane⇒

deg

X=

1,forexam

ple

V(x

0)∩V(x

2,···

,xn)=

{[0:1:0:···:

0]}.

2)X

=Pn⊂

Pn,L=an

ypoint⇒

deg

Pn=

1.3)Thereducible

varietyX

=H

0∪{[1:0:1]}=

{[0:y:1]

:y∈k}∪

{[0:1:0],[1:0:1]}⊂

P2generically

intersects

alinein

onepoint,butL=

P(span

k(e

0,e

2))

={[x:0:1]

:x∈k}∪

{[1:0:0]}

intersects

Xtw

ice.

Ontheaffi

nepatch

z=

1,X

=(y-axis∪apointon

thex-axis),an

dL=

x-axis.

4)X

=V(xz−

y2)⊂

P2.

L=

V(ax+

by+

cz)

1:1

←→

(plane

� L⊂

A3)∈Gr(2,3)

1:1

←→

(normal

totheplane)

=[a

:b:c]∈P2

.Wenow

calculate

L∩X.Wewan

tto

goto

anaffi

nepatch

x�=

0,butmust

not

forget

intersection

points

outsideof

that.If

x=

0,then

y=

0,an

difc�=

0then

also

z=

0,but[0]is

not

allowed

inP2

.Thusassumec�=

0.Then

x�=

0,W

LOG

x=

1.Solving:

y=

−cz−a

bif

b�=

0an

d

z=

y2=

(−cz−a

b)2

gives

twosolution

szifthediscrim

inan

tof

thequad

raticequationis

non

-zero

(checkthediscrim

inan

tis

b2(b

2−

4ac)).

Thusdeg

X=

2,an

dthesetof

“bad

”L≡

[a:b:c]∈P2

form

sasubsetof

V(c)∪V(b)∪V(b

2(b

2−

4ac)),

hence

asubsetof

V(bc(b2−

4ac)).

Remark

.P1∼ =

V(xz−

y2)(V

eron

esemap

),yetdeg

P1=

1,deg

V(xz−

y2)=

2.Thusthedegree

dep

ends(unsurprisingly)on

theem

beddinginto

projectivespace.

Definition.X⊂

An≡

U0⊂

Pnaff

.var.⇒

deg

X=

deg

(X⊂

Pn).

Theore

m.F∈R

=k[x

0,...,x

n]homogen

eousofdegreedwithnorepeatedfactors⇒

deg

V(F

)=

d.

Proof.

L=an

yline,

X=

V(F

).⇒

X∩L=

V(F

| L)⊂

L∼ =

P1.

After

alinearchan

geof

coordinates,W

LOG

L=

V(x

2,...,x

n).

40

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

⇒F| L

=degreedhomog

.poly1in

x0,x

1(ifdeg

F| L

<dthen

Lis

not

generic

enou

gh).

⇒#(zeros

ofpoly)≤

d,an

dgenerically2it

hasdzeros.

�

Fact.

(WeakBezout’sTheore

m)3

Let

X,Y⊂

Pnbeproj.vars.ofpure

dim

ensionwith4dim

X∩Y

=dim

X+

dim

Y−

n,then

deg

X∩Y≤

deg

X·d

egY.

2·3

=6

intersections

cubic

quadric

9.2.

HIL

BERT

POLYNOM

IAL

X⊂

Pnproj.var.

Wenow

relate

thedegreeto

Sections3.10

and3.11

.

S(X

)=

k[� X]=⊕

m≥0S(X

) m,whereS(X

) mis

thevectorspacek[x

0,...,x

n] m

/I(X

) m.Define

hX

:N→

N,

hX(m

)=

dim

kS(X

) m=

� m+n

m

� −dim

kI(X) m

EXAM

PLES.

1)hPn(m

)=

� m+n

m

� =(m

+n)!

m!n!

=1 n!(m

+n)···(m

+1)=

1 n!m

n+

lower

order.

2)X

=V(F

)⊂

P2,forF

irred.hom

og.

ofdeg

d.Then

I(X) m

={α

F:deg

α=

m−

d}.

Thus

hX(m

)=

� m+2

m

� −� m

−d+2

m−d

� =(m

+2)(m+1)

2−

(m−d+2)(m−d+1)

2

=1 2(m

2+

3m

+2−

(m2−

2md+3m

)−

(d−

2)(d−

1))=

dm−

(d−1)(d−2)

2+

1.

Fact.

(Degre

e-genusform

ula

foralgebra

iccurv

es).g=

genus(X)=

(d−1)(d−2)

2.

ThushX(m

)=

dm−

g+

1.

FACT.

X⊂

Pnproj.var.

⇒thereexists

pX∈k[x]an

dthereexists

m0such

that

forall5m≥

m0,

hX(m

)=

pX(m

).

pX

iscalled

theHilbert

polynomialof

X⊂

Pn.Moreover,

thelead

ingterm

ofpX

is

deg

X

(dim

X)!·m

dim

X

Remark

.pX

dep

endsontheem

beddingX⊂

Pn.

Remark

.Other

coeffi

cients

ofpX

arealso“d

iscreteinvariants”of

X.Soweon

ly“care”to

compare

varietieswithequal

Hilbertpolynom

ial.

Remark

.X,Y⊂

Pn,ifX≡

Yarelinearlyequivalent6

then

pX

=pY.

1Putt=

x1/x0to

get

a(non-homogen

eous)

poly

inonevariable,andyoufindallroots

(explicitly,

ift=

ais

aroot

then

theoriginalhomog.poly

hadarootfor[x

0:x1]=

[1:a],andit

remainsto

checkwhether

[0:1]wasaroot).

2Thereis

agen

eralnotionofdiscrim

inant(essen

tiallythere

sultantpolynomialorthesquare

oftheVander-

mondepolynomial),andgen

ericityis

ensuredifthediscrim

inantis

non-zero.

3Remark

.FornprojectivehypersurfacesX

1,...,X

n⊂

Pnofdegrees

d1,...,d

nthen

#(X

1∩·

··∩X

n)=

d1d2···d

n

gen

erically(itisalsod1d2···d

nifitisnotinfinite,

provided

thatonecounts

intersectionswithmultiplicities).

Thekey

trickis:Y

⊂Pn

proj.var.,dim

X=

δ,deg

X=

d1,H

⊂Pn

hypersurf

ofdeg

H=

d2notcontainingirredcomponen

tsofX,then

X∩H

hasdim

=δ−

1anddeg

=d1d2.

4This

dim

ensionconditionis

whatyouwould

get

forvectorsubspacesX,Y

⊂knwithX

+Y

=kn.

5Think:“forlargem,hX

reallyis

apolynomial”.

6i.e.

X∼ =

Yis

inducedbya(linear)

isomorphism

Pn∼ =

Pn.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

41

9.3.

FLAT

FAM

ILIE

S

Aflatfamilyof

varietiesis1aproj.var.

X⊂

Pntogether

withasurjectivemorphism

π:X→

B

whereB

isan

irred.proj.var.

(orquasi-proj.var.)an

dthefibersX

b=

π−1(b)havethesameHilb.poly.

Example.

φ:P1→

P1,[x]�→

[f0(x):f 1(x)]

wheref 0,f

1arehom

ogeneousof

thesamedegree.

Assumef 0,f

1arelinearlyindep

endent/k

(soaf 0−

bf1�=

0forall(a,b)∈

k2\{(0,0)}).

Then

φ−1[a

:b]

=V(bf 0−

af 1)⊂

P1is

ahypersurf

ofdegreed,hence

(byHom

ework3,

ex.2)they

have

thesameHilbertpolynom

ialforalla,b

(infact

theHilb.polyis

theconstan

td).

Non-example.Theblow-u

pof

A2at

theorigin

is

B0A2=

{anylinethrough

0in

A2together

withan

ychoice

ofpointon

theline}

together

withthemap

π:B

0A2→

A2whichprojectsto

thechosen

pointon

theline.

Explicitly:

P1×

A2⊃

V(xw−

yz)=

B0A2→

A2,([x:y],(z,w

))�→

(z,w

).

If(z,w

)�=

0,π−1(z,w

)=

([z:w],(z,w

))=on

epoint2

(soB

0A2is

thesameas

A2exceptover

the

point0).W

hereasover

30:

π−1(0,0)=

{([x

:y],(0,0))}∼ =

P1.Noticethedim

ension

ofthefibers

jumpsat

0.Com

pactifyingtheab

ove4

weob

tain

theblow-upπ:B

pP2→

P2of

P2at

p=

[0:0:1],

whichis

not

aflat

family(thedegreeof

theHilbertpolyof

thefibersjumpsat

p).

10.

LOCALIS

ATIO

NTHEORY

10.1.

LOCALIS

ATIO

NIN

ALGEBRA

Let

Abearing(com

mutative

with1).

Definition

10.1.S⊂

Ais

amultiplica

tivese

tif5

1∈S

and

S·S⊂

S.

EXAM

PLES.

1).S=

A\{

0}foran

yintegral

dom

ainA.

2).S=

A\℘

foran

yprimeideal℘⊂

A.

3).S=

{1,f,f

2,...}foran

yf∈A.

Thedefinitionof

localisation

ofA

atS

mim

icstheconstructionof

thefraction

fieldFrac(A)foran

integral

dom

ainA,so

mim

ickingFrac(Z)

=Q.RecallFrac(A)consistsof

fraction

sr s,whichform

ally

arethou

ghtof

aspairs

(r,s)∈A×

(A\{

0}),

subject

toidentifyingfraction

sr s∼

r�

s�ifrs

� =r�s.

Definition

10.2.Theloca

lisa

tion

ofA

atS

is

S−1A

=(A

×S)/∼

whereweabbreviate

thepairs(r,s)by

r s,andtheequivalence

relationis:

r s∼

r� s�⇐⇒

t(rs

� −r�s)

=0forsomet∈S.

(10.1)

Weshou

ldexplain

whytap

pears

in(10.1).Algebraically

tensuresthat∼

isan

equivalence

relation

.Exercise.Checkthat∼

isatran

sitiverelation

(noticeyo

uneedto

use

aclever

t).

Inman

yexam

ples,

tis

not

necessary:if

Ais

anintegral

dom

ain

and

0/∈

S,then

(10.1)

forces

rs� −

r�s=

0(since

therearenozero

divisorst�=

0in

S).

Geometric

Motivation.

Thetplaysacrucial

role

inensuringthat

localisation

identifies

the

1This

defi

nitionis

equivalentto

theusualdefi

nitionofflatfamily(see

HartshorneIII.9).

2Given

anon-zeropointin

A2,thereis

auniquelinethroughthepointand0.

3Thinkofπ−1(0)∼ =

P1asparametrizingthetangentialdirectionsalongwhichlines

inA

2approach

theorigin.

4π:B

pP2

→P2

isanisomorphism

onthecomplemen

tofπ−1(p).

5S·S

⊂S

meansst

∈S

foralls,t∈S.Somebooksrequirethat0/∈S,butwedonot.

42

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

functionsthat

oughtto

bethough

tof

asequal.

Consider

X=

V(xy)=

(x-axis)∪(y-axis)⊂

A2an

dA

=k[X

]=

k[x,y]/(xy).

What

arethe“localfunctions”

nearthepointp=

(1,0)?

Wewan

tto

form

ally

invertallthosefunctionsf∈A

whichdonotvanishatp:

S=

{f∈A

:f(p)�=

0}

=A\I

(p)

={f∈A

:f/∈�x−

1,y�}.

For

exam

ple,x∈

Ssince

itdoes

not

vanishatp=

(1,0).

Consider

theglobalfunctions0andy:

thesearedifferentin

A.How

ever,on

cewelocalise

nearp,byrestricting0andyto

aneighbourhood

ofpsuch

as(x-axis)\0

=X

\V(x),

then

thelocalfunctions0an

dybecom

eequal.

Sowewant

y=

y 1=

0 1=

0in

S−1A.Indeed,t·(y·1−

0·1

)=

0∈

Ausingt=

x∈

S.W

ithouttin

(10.1)

this

wou

ldhavefailed

.Moreover,wewantthelocalfunctionsofX

nearpto

agree

withthelocal

functionsof

theirreducible

componentV(y)=(x-axis)nearp,so

weexpect(andweprovelater)

that

S−1A

isisom

orphic

tothek-algebra

k[x]afterinvertingallh∈k[x]\I

(p):

k[x] I(p)=

k[x][1 h:h(p)�=

0]⊂

Frac(k[x])=

k(x).

Exercise.1

S−1A

=0⇔

0∈S.

Exercise.Show

that r s

=0∈S−1A⇔

(tr=

0forsomet∈S)⇔

r∈

� t∈S

Ann(t).

Inparticular,

foran

integral

domain

A,r s=

0⇔

r=

0(assuming0/∈S).

EXAM

PLES.

1).A

f=

S−1A

isthelocalisation

ofA

atS=

{1,f,f

2,...}.

So

Af=

{r fm

:r∈A,m

≥0}

/∼

whereforexam

ple

r fm

=rf

fm

+1,an

dmore

generally

r fm

=r�

fn⇔

fN(rfn−r�fm)=

0forsomeN≥

0.

•iffis

nilpotent,so

fN

=0∈S

forsomeN,so

Af=

{0}.

Indeed:A

f=

0⇔

fis

nilpotent.

•ifA

isan

integral

dom

ain,

Af=

A[1 f]⊂

Frac(A).

2).A

=k[x,y]/(xy),

S=

{1,x

,x2,...}then

y=

y 1is

zero

since

yis

annihilatedbyx∈S.Thus

S−1A∼ =

k[x] x

=k[x,x

−1]⊂

Frac(k[x])=

k(x).

Exercise.In

general,A

f∼ =

A[z]/(zf−

1)(w

ehaveseen

this

trickbefore).

S−1A

isaringin

anaturalway: r s+

r� s�=

rs� +

r�s

ss�

r s·r

� s�=

rr�

ss�

withzero

0=

0 1an

didentity

1=

1 1,an

dit

comes

withacanonicalringhomom

orphism

π:A→

S−1A,

a�→

a 1

whichhas

kernel

kerπ=

{a∈A

:ta

=0forsomet∈S}=

� t∈S

Ann(t).

IfA

isan

integra

ldomain

then

π:A

�→S−1A

isinjective(assuming0/∈S).

Exercise.Checktheab

ovestatem

ents

(inparticular,

that

theop

erationsare

well-defined

).

1Hint.

Consider

1.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

43

EXAM

PLES.

1).S=

A\℘

,then

thelocalisa

tion

ofA

atth

eprimeideal℘is1

A℘=

{r s:r∈A,s

/∈℘}/∼

.

2).For

anintegral

dom

ainA,letS=

A\{

0},then

thelocalisation

at℘=

(0)is:

S−1A

=A

(0)=

Frac(A).

Definition

10.3.A

isaloca

lringifithasauniquemaximalidealm⊂

A.

ThefieldA/m

iscalled

residuefield.

Exercise.2

Ais

local⇔

thereexists

anidealm

�A

such

that

allelem

ents

inA\m

areunits.

Lemma10.4.A

℘is

alocalringwithmaximalideal℘A

℘=

{r s:r∈℘,s

/∈℘}/∼.

Proof.

Notice℘·A

℘is

anideal.

Suppose

r s/∈℘A

℘.Then

r/∈℘.So

r sis

aunit

since

s r∈A

℘.

�

KeyExercise.For

Aan

integral

dom

ain,

A=

�

maxm⊂

A

Am

=�

prime℘⊂

A

A℘⊂

Frac(A).

Exercise.3

Let

ϕ:A→

Bbearinghom

,an

d℘⊂

Baprimeideal.

Abbreviate

ϕ∗ ℘

=ϕ−1(℘

).Show

thereis

anaturallocalringhom

4

Aϕ∗ ℘→

B℘.

(10.2)

Example.LocalisingZat

aprime(p):

Z (p)=

{a b:p�b}h

asmax

idealm

p=

pZ (

p)=

{a b:p|a,p�b}.

Exercise.Theresiduefieldis

Z (p)/m

p∼ =

Z/(p),

a b�→

ab−

1.

Asan

exercise

inalgebra,tryprovingthefollow

ing:

FACT.Thereis

a1:1correspon

dence

{primeideals

I⊂

AwithI∩S=∅}

↔{p

rimeideals

J⊂

S−1A}

I�→

J=

I·S

−1A

={i s

:i∈I,s∈S}

I=

π−1(J

)=

{i∈A

:i 1∈J}←�

J.

Inparticular,

foraprimeideal℘⊂

A,

{primeideals

I⊂

℘⊂

A}↔

{primeideals

J⊂

A℘}

I=

π−1(J

)↔

J=

IA

℘.

Exercise.If

Ais

Noetherian,then

S−1A

isNoetherian.

Exercise.S−1(A

/I)∼ =

(S−1A)/(IS−1A),

inparticu

lar

(A/I

) ℘∼ =

A℘/I

A℘.

Example.

Con

sider

againA

=k[x,y]/(xy)=

k[X

],so

X=

X1∪X

2whereX

1=

V(y)=

(x-

axis)an

dX

2=

V(x)=

(y-axis).

Con

sider

p=

(1,0)∈

X1\X

2an

dm

p=

I(p).

Recallan

yf∈(y)⊂

k[X

2]=

k[y]becom

eszero

inA

mpbecau

sexf=

0∈A,wherex∈S=

k[X

]\m

p.Solet

I=

yA⊂

A,then

IA

mp=

0⊂

Am

p.Thus,

since

A/I∼ =

k[x]=

k[X

1]:

k[X

] mp=

Am

p∼ =

Am

p/I

Am

p∼ =

(A/I

) mp∼ =

k[X

1] m

p=

k[x][1 h:h(p)�=

0]⊂

k(x)

aspromised.In

general,ifyou

localize

atapointpwhichon

lybelon

gsto

oneirreducible

compon

ent,

then

thelocalringat

pag

rees

withthelocalringof

theirreducible

compon

entat

p.

Exercise.S−1√I=√S−1I,in

particu

larlocalisingradical

ideals

gives

radical

ideals.

1Don’t

get

confusedwithA

f.ForA

fweinvertf.ForA

℘weinverteverythingexceptwhat’sin

℘!

2Hints.Toshow

mis

maxim

al:

m�

I⊂

Aim

plies

Icontainsaunit,so

I=

A.Conversely,ifu∈A\m

werenota

unit,then

thereis

amaxim

alidealcontainingtheideal�u�,

andthis

cannotequalm.

3Hint.

IfS⊂

Aismultiplicativesuch

thatϕ(S

)⊂

Bconsistsofunits,there’sanobvioushom

S−1A

→B,

r s�→

ϕ(r

)ϕ(s

).

4ahom

oflocalringsR

1→

R2sendingthemaxidealm

1to

(asubsetof)

themaxidealm

2.

44

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

10.2.

LOCALIS

ATIO

NFOR

AFFIN

EVARIE

TIE

S:regularfunctionsandstalks

Motivation.Wenow

wantto

consider

thek-algebra

offunctionsthatare

naturallydefi

ned

near

apointp,andweexpect

thatanyfunctionwhichdoesn’tvanishatpshould

beinvertible

nearp.

For

any

topolog

icalspace

X,agerm

ofa

function

nearapointp∈

Xmeansafunction

f:U→

kdefined

onaneigh

bourhoodU⊂

Xof

p,whereweidentify

twosuch

functionsU→

k,

U� →

kifthey

agreeon

asm

allerneigh

bou

rhoodofp.Soagerm

isanequivalence

class[(U,f

)].

Let

Xbean

affinevariety,

andp∈X.A

functionf:U→

kdefined

onaneigh

bou

rhoodof

pis

called

regularatpifon

someop

enp∈W⊂

U,thefollow

ingfunctionsW→

kareequal,

f=

g hsomeg,h∈k[X

]an

dh(w

)�=

0forallw∈W.

Wewrite

OX(U

)forthek-algebra

offunctionsf:U→

kregu

laratallpoints

inan

open

U⊂

X.

Thestalk

OX,p

isthek-algebra

ofgermsofregu

larfunctionsat

p,so

equivalence

classesof

pairs

(U,f

)withp∈U⊂

Xop

enan

df:U→

karegu

larfunction,whereweidentify

(U,f

)∼

(V,g)if

f| W

=g| W

onan

open

p∈W⊂

U∩V.Exercise.Checkthis

isak-algeb

rain

theob

vioussense.

EXAM

PLES.

1)For

anyf∈k[X

],f:X→

kisregularateach

point(consider

U=

Xan

df=

f 1).

Wewillshow

inTheorem

11.2

thatfunctionsregularat

each

pointof

Xalwaysarisein

this

way.So

OX(X

)∼ =

k[X

].

2)For

X=

A1,m∈N,f:U

=D

x=

A1\{

0}→

k,f(x)=

1 xm

isregu

larat

anyp∈U,so

f∈O(U

).3)

Moregenerally,foran

yf∈k[X

],recallD

f=

X\V

(f).

Corollary

11.3

willshow

that

OX(D

f)∼ =

k[X

] f.

4)Let

X=

V(xy)⊂

A2(theunionof

thetw

oax

es).

Let

U=

X\V

(y)=

(x-axis)\{

0}.Then

f:U→

k,f(x,y)=

y x∈O(U

),but(U

,f)∼

(U,0)as

y=

0:U→

k,so

[(U,f

)]=

0.

Lemma10.5.Atp∈X,thestalk

ofthestructure

sheafO

Xis:

OX,p∼ =

k[X

] mp

wherem

p=

I(p)=

{f∈k[X

]:f(p)=

0}is

themaximalidealcorrespondingto

p.

Proof.

Theisom

orphism

isdefined

by

(U,f

)�→

g hwheref| U

=g hforg,h∈k[X

],h(p)�=

0.Themapiswell-defined:h(p)�=

0⇒

h/∈m

p⇒

g h∈k[X

] mp.

Moreover,

if(U

,f)∼

(U� ,f� ),so

g h=

g�

h�on

abasic

open

p∈

Ds⊂

U∩U

� ,wheres∈

k[X

],then

gh� −

g� h

=0on

Ds.Since

s(p)�=

0,wehaves/∈m

p.Thuss·(gh� −

g� h)=

0everywhereonX,so

s·(gh� −

g� h)=

0in

k[X

].Thus

g h=

g�

h�in

k[X

] mp.

Webuildtheinverse

map

:forh

/∈m

p,letU

=D

h,then

send

g h�→

(U,g h).

Moreover,

ifg h=

g�

h�in

k[X

] mp,then

s·(gh� −

g� h)=

0forsomes∈k[X

]\m

p.Then

s(p)�=

0so

p∈D

s,an

dgh� −

g� h

=0

onD

s.Thus

g h=

g�

h�asfunctionsD

s→

k,as

required.

Byconstruction,thetw

omap

sare

inverseto

each

other,so

wehavean

isom

orphism.

�

Example.For

anirreducible

varietyX,wegetan

integral

dom

ainA,so

Lem

ma10

.5becom

es:

OX,p=

k[X

] mp=

k[X

][1 h:h(p)�=

0]⊂

Frac(k[X

])=

k(X

)

andtheKey

Exercise,from

Section

10.1,im

plies

1

k[X

]=

� p∈X

OX,p⊂

OX,p⊂

k(X

).

1Recallthefirstequality

implies

thetheorem

“regularatallpoints

implies

polynomial”

foranirred.aff.var.

(Theorem

11.2).

Thatkis

algeb

raicallyclosedcomes

into

play:allmaxideals

arise

asm

p=

I(p)forp∈X.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

45

TheFACT,from

Section

10.1,tran

slates

into

geom

etry

asthe1:1correspon

dence:

{irred

ucible

subvarietiesY⊂

Xpassingthrough

p}↔

{primeideals

inO

X,p}

Y=

V(J

)↔

J·O

X,p=

{f∈O

X,p:f(Y

)=

0}.

whereJ=

I(Y).

Inparticular,thepointY

={p

}correspon

dsto

themax

imal

idealm

pO

X,p⊂

OX,p.

ByLem

ma10.4,m

pO

X,p⊂

OX,pis

theuniquemax

imal

ideal.

Thequotientrecovers

ourfieldk:

K(p)=

OX,p/m

pO

X,p∼ =

k,

g h�→

g(p)

h(p).

(10.3)

Warn

ing.Not

allfunctionspaces

ariseas

alocalisation

ofk[X

].For

exam

ple

f=

x y=

z w∈k(X

)

whereX

=V(xw−

yz)⊂

A4defines

aregu

larfunction

f∈

OX(D

y∪D

w).

Butit

turnsou

tthat

onecannot

write

f=

g hon

allof

Dy∪D

wforg,h∈

k[X

](this

iscausedbythefact

that

k[X

]=

k[x,y,z,w

]/(xw−

yz)is

not

aUFD).

SoO

X(D

y∪D

w)is

not

alocalisation

ofk[X

],unlike

OX(D

y)=

k[X

] y,O

X(D

w)=

k[X

] w,O

X(D

y∩D

w)=

OX(D

yw)=

k[X

] ywwhicharealllocalisation

s.

10.3.

HOM

OGENEOUSLOCALIS

ATIO

N:pro

jectivevarieties

Let

A=⊕

m≥0A

mbean

N-graded

ring.

Let

S⊂

Abeamultiplicativesetconsistingon

lyof

hom

ogeneouselem

ents.

Then

S−1A

=⊕

m∈Z

(S−1A) m

has

aZ-grad

ing:

ifr∈

A,s∈

Sare

hom

ogeneouselem

ents

then

m=

deg

r s=

deg(r)−deg(s)∈Z.

Exercise.Show

that

(S−1A) 0⊂

S−1A

isasubring.

Example.For

A=

k[x

0,...,x

n],(S

−1A) 0

isim

portant:

they

aretheration

alfunctions

F(x

0,...,x

n)

G(x

0,...,x

n)

forF,G

hom

ogeneouspolysof

equal

degree,

soF(p)

G(p)∈kis

well-defined

1forp∈Pn

withG(p)�=

0.

Definition

10.6.Thehomogeneo

usloca

lisa

tionis

thesubring(S

−1A) 0

ofS−1A.Abbreviate

byA

(f)=

(Af) 0

theh.localisationat{1,f,f

2,...}forahomogen

eouselem

entf∈A;andA

(℘)=

(A℘) 0

fortheh.localisationatallhomogen

eouselem

ents

inA\℘

forahomogen

eousprimeideal℘⊂

A.

Let

X⊂

An≡

U0⊂

Pnbean

affinevariety.

Wenow

comparetheaffi

nelocalisation

k[X

] mpwith

thehom

ogeneouslocalisation

S(X

) (m

p)at

apointp∈

X,whereX⊂

Pnis

theprojectiveclosure,

mp=

{f∈k[X

]:f(p)=

0},an

dm

p=

{F∈S(X

):F(p)=

0}.

Lemma10.7.

k[X

] mp∼ =

S(X

) (m

p)

Proof.

Themutually

inversemorphismsaregiven

byhom

ogenisingan

ddehom

ogen

ising.

Explicitly,

whered=

max

(deg(f),deg(g)),

f(x

1,...,x

n)

g(x

1,...,x

n)�→

xd 0f(x

1x0,...,xn

x0)

xd 0g(x

1x0,...,xn

x0)

F(x

0,x

1,...,x

n)

G(x

0,x

1,...,x

n)�→

F(1,x

1,...,x

n)

G(1,x

1,...,x

n).

�

Exercise.[See

Hwksheet1,

ex.5.]

Show

that

theprojectivisationX⊂

P2of

X=

V(y−

x3)⊂

A2

isnot

isoto

P1bycomputingthelocalringO

X,pat

p=

[0:1:0]

(com

parewithlocalrings

ofP1

).

Show

V(y−x3)∼ =

V(y−x2)as

affinevarietiesin

A2,buttheirprojectivisationsin

P2arenot

iso.

11.

QUASI-PROJECTIV

EVARIE

TIE

S

11.1.

QUASI-PROJECTIV

EVARIE

TY

Aim

:Defi

nealargeclass

ofvarietieswhichcontainsboth

affinevars,projectivevars,andopensets

e.g.

k∗⊂

k,such

thatanyopensubset

ofavarietyin

this

class

isalsoin

this

class.

1i.e.

unchanged

under

thek∗ -rescalingactionwhichdefi

nes

Pn.

46

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Definition.A

quasi-p

rojectivevariety

X⊂

Pnis

anyopensubset

ofaprojectivevariety,

so

X=

UJ∩V(I)

whereUJ=

Pn\V

(J),

soX

isanintersection1ofanopenandaclosedsubset

ofPn

.Notice

Xis

alsothedifferen

ceoftwoclosedsets:X

=V(I)\V

(I+

J).

Aquasi-p

rojectivesu

bvariety

X�of

Xis

asubset

ofX

whichis

alsoaquasi-projectivevariety,

soX

� =UJ�∩V(I

� )forI⊂

I� ,J� ⊂

J.

EXAM

PLES.

1)AffineX⊂

An:then

X=

An∩X

(exercise

2).

2)ProjectiveX⊂

Pn:then

X=

Pn∩X.

3)A2\{

0}=

(U0∩(U

1∪U2))∩P2

(viewing3A2≡

U0⊂

P2).

4)Anyop

ensubsetof

aq.p.var.is

also

aq.p.var.,since

UJ�∩(U

J∩V(I))

=(U

J�∩UJ)∩V(I).

Definition.A

morp

hism

ofq.p.vars.X→

Yis

defined

just

asforproj.vars.,so

locally

p�→

[F0(p):···:

Fm(p)]

forhom

ogeneouspolysF0,...,F

mof

thesamedegree(w

hereX⊂

Pn,Y⊂

Pm).

Remark

.For

X,Y

affine,

this

agrees

withthedefinitionof

morphof

aff.vars.:

[x0:···:

xn]

[F0(x):···:

Fm(x)]

[1:y 1

:···:

y n]✤

�� [1:f 1(y):···:

f m(y)]=

[xd 0:xd 0f 1(y):···:

xd 0f m

(y)]

wherey i

=xi/x0(x

0�=

0),d=

max

deg

f i,an

dF0(x)=

xd 0,Fi(x)=

xd 0f i(y)(noticedeg

Fi=

d).

Coro

llary.X⊂

An,Y⊂

Am

q.p.vars.If

thereare

mutuallyinversepolynomialmapsX→

Yand

Y→

X,then

X∼ =

Yasq.p.vars.

Warn

ing.Theconverseis

false:

A2⊃

V(xy−1)∼ =

A1\0

q.p.vars,

butnot

via

apolynom

ialmap

:

(x,y)

x

P2�[x

:y:1]

=[x

:x−1:1]

✤�� [x:1]∈P1

[x2:1:x]��

✤ [x:1]

[x2:y2:xy]��

✤ [x:y]

P2⊃

V(xy−z2)�[x

:y:z]✤

�� [x:y]

Definition.A

q.p.var.X

isaffi

neifitis

isomorphic

(asq.p.vars)to

anaff.var.

Y=

V(I)⊂

An.

Wewilloften

write

k[X

]when

wemeank[Y

]=

k[x

1,...,x

n]/I(Y).

Example.k∗⊂

kis

affine.

11.2.

QUASI-PROJECTIV

EVARIE

TIE

SARE

LOCALLY

AFFIN

E

Lemma11.1.X

aff.var.,f∈k[X

].Then

Df=

X\V

(f)is

anaffineq.p.var.

with4

k[D

f]∼ =

k[X

] f.

1such

sets

are

called

locally

close

dsu

bse

ts.

2RecallX

⊂Pn

istheprojectiveclosure

ofX

⊂A

n≡

U0⊂

Pn,andrecallTheorem

3.3.

3{[1:∗:∗]}=

{[x0:x1:x2]:x0�=

0}andweexcludethecase

x1=

x2=

0bytakingU

1∪U

2=

P2\V

(x1,x

2).

4ForX

notirreducible,wemay

worryaboutthedefi

nitionoflocalisation:

g fa=

h fb∈

k[X

] f⇔

f�(f

bg−

fah)=

0

forsome�≥

0.Butevaluatingatp∈D

f(thusf(p)�=

0)im

plies

f(p)bg(p)−

f(p)ah(p)=

0∈k,so

g(p

)f(p

)a=

h(p

)

f(p

)b.So

alsothefunctions

g fa=

h fb:D

f→

kagree.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

47

Remark

.k[D

f]=

{g fm

:D

f→

kwhereg∈

k[X

],m≥

0}∼ =

k[X

][1 f]∼ =

k[X

] f.For

Xirreducible,

onecanview

k[X

][1 f]as

thesubalgebra

{g fm

:g∈k[X

],m≥

0}⊂

k(X

)=

Frack[X

].Butin

general,

wedefinek[X

][1 f]≡

k[X

][xn+1]/(fxn+1−

1),so

weintroducedaform

alinverse“x

n+1=

1 f”.

The

identification

withthelocalisation

k[X

] fis:xn+1�→

1 f(andg∈k[X

]to

g 1)withinverse

g fa�→

gxa n+1.

Proof.

Define

� I=�I(X

),xn+1f−

1�⊂

k[x

1,...,x

n,x

n+1]

⇒V(� I)⊂

An+1is

affinewithanew

coordinatefunctionxn+1whichis

reciprocalto

f,

k[V(� I)]=

k[X

][xn+1]/(fxn+1−

1)≡

k[X

][1 f].

Subclaim

.ϕ:D

f→

V(� I)is

anisoof

q.p.vars,

via

a=

(a1,...,a

n)�→

(a1,...,a

n,

1f(a))

withinverse

(b1,...,b

n)←� (b 1,...,b

n,b

n+1).

PfofSubclaim

.View

Df⊂

An≡

U0⊂

Pnvia

(a1,...,a

n)↔

[1:a1:···:

an]an

dV(� I)⊂

An+1≡

U0⊂

Pn+1via

(a1,...,a

n+1)↔

[1:a1:···:

an+1].

Then

ϕis

therestrictionof

F:Pn→

Pn+1,

[1:a1

a0:···:

an

a0]✤

“ϕ”

�� [1:a1

a0:···:

an

a0:

1f(a1

a0,···,

an

a0)]

[a0:a1:···:

an]✤

F�� [a0� f(a):a1� f(a):···:

an−1� f(a):adeg

f+1

0]

wherewehom

ogenised:

� f(a)=

� f(a0,...,a

n)=

adeg

f0

f(a

1a0,···

,an

a0),

and

inthesecond

vertical

identification

werescaled

bya0� f(a).

Thelocalinverseis

[a0:···:

an]←� [a

0:···:

an+1]∈U0(the

composites

givetheidentity,usingthat

� f(a)�=

0on

Df,so

wemay

rescaleby

1 � f(a)).

�

Theore

m.Every

q.p.var.

hasafiniteopencoverby

affineq.p.subvars.In

particular,

affineopen

subsetsform

abasisforthetopology.

Proof.

Pn⊃

X=

UJ∩

V(I)=

V(F

1,...,F

N)\V(G

1,...,G

M)(w

herewepick

generatorsfor

J,I).

WLOG

1it

sufficesto

checktheclaim

ontheop

enU0∩X.Then

U0∩X

isV(f

1,...,f

N)\

V(g

1,...,g

M)=∪ j

V(f

1,...,f

N)\V

(gj)=∪ j

DgjwhereD

gjis

thebasic

open

subset(g

j�=

0)⊂

V(f

1,...,f

N),

andwheref 1

=F1| x 0

=1∈

k[x

1,...,x

n]so

f 1(a)=

F1(1,a)etc.

Now

apply

Lem

ma

11.1.

�

11.3.

REGULAR

FUNCTIO

NS

Motivation.A1\{

0}∼ =

V(xy−

1)⊂

A2.Wewan

tto

allow

thefunction

1 xm

=ym.

Definition.X

aff.var.,U⊂

Xopen.

OX(U

)=

{regularfunctionsf:U→

k}

={f

:U→

k:fis

regularateach

p∈U}

Recall,fregularatpmeans:

onsomeopenp∈W⊂

U,thefollowingfunctionsW→

kare

equal,

f=

g hsomeg,h∈k[X

]andh(w

)�=

0forallw∈W.

Example

1.U

=D

x=

A2\V

(x)⊂

A2,f:D

x→

k,f(x,y)=

y x∈O

X(D

x).

2.For

anyg,h∈k[X

],withh�=

0,wehave

g h∈O

X(D

h).

REM

ARKS.

1)Som

ebook

sjust

sayh(p)�=

0,an

dthis

isenou

gh2since

wecanalwaysreplace

WbyW∩D

h.

2)Wearenotsayingthat

f=

g hholdson

allof

U,on

lylocally.

1because

Pnhasanopen

cover

byU

i.

2AlthoughIfindthemeaningoftheequality

f=

g hunclear,

onthelarger

W.

48

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Wearenotsayingthat

g,h

areunique(e.g.in

Q,2 3=

4 6).

3)Noticeab

ovewerequired

g,h

tobeglob

alfunctionson

X.Wearenotlosingouton

anything,

since

ifweinsteadrequired

g� ,h�∈

k[D

β]forabasicop

ensubsetp∈

Dβ⊂

X,then

g�=

g/βa,

h� =

h/β

b,forsomeg,h∈k[X

],so

g� /h� =

g/(hβa−b)or

(gβb−

a)/h(dep

endingonwhether

a≥

bor

a<

b)show

swecanwrite

g� /h�asaquotientofglob

ally

defined

functions.

4)Later

wewillprove

thatifU∼ =

Y⊂

Anisaffi

ne,

then

OX(U

)isisom

orphic

totheclassicalk[Y

].Bymak

ingW

smaller,wecanalwaysassumeW

isabasic

open

setD

βforsomepolynom

ialfunction

β:X→

k(andβ(p)�=

0).AsD

βisaffi

ne,

OX(D

β)=

k[D

β]=

k[Y

] β,therefore

f=

α βN

asfunctions

Dβ→

k,forsomeα,β∈k[X

],N∈N.ByreplacingβbyβN,wecanassumef=

α β(soN

=1).

5)Som

ebook

salwaysab

breviate

k[U

]=

OX(U

),butwewilltryto

avoidthisto

preventconfusion

.

Definition.X

q.p.var.,U⊂

Xopen.

OX(U

)=

{F:U→

k:F

isregularateach

p∈U}.

Fregularatpmeans:

onsomeaffi

neopenp∈W⊂

U,F| W

isregularatpaspreviouslydefi

ned.

REM

ARKS.

1)Recalltheaffi

neopen

coveringUi=

(xi�=

0)⊂

Pn.Supposep∈

X∩Ui.

Note

thatX∩Ui

isan

open

setin

Ui∼ =

An.Then

nearp,F

isequal

toaratioof

twopolynom

ials

inthevariab

les

x0,...,x

i−1,x

i+1,...,x

nwhosedenom

inator

does

not

vanishat

p.FollowingRem

ark

4above,

we

canalso

pickan

affineop

enD

β⊂

Ui∼ =

Anso

that

F=

α βas

afunctionD

β→

korequivalentlyas

anelem

entof

thelocalisation

k[D

β]∼ =

k[U

i]β=

k[x

0,...,x

i−1,x

i+1,...,x

n] β.If

youwan

tto

view

Fas

afunctionPn→

kdefined

near1

p,youneedto

hom

ogenisebyreplacingeach

xjbyxj/x

i.Clearingdenom

inatorswillgivearatioof

hom

ogeneouspolynomials

ofthesamedegree.

Solocally

nearp∈X,F

isrepresentedbyan

elem

entof

thehom

ogeneouslocalisationS(X

) mp(see

Sec.10.3).

2)Gluing

regularfunctions.

Given

open

sets

U1,U

2in

aq.p.var.

X,and

regularfunctions

f 1∈O

X(U

1)an

df 2∈O

X(U

2),ob

servethat

thenecessary

andsufficientconditionto

beable

tofind

aglued

regu

larfunctionf∈O

X(U

1∪U

2)(m

eaning,

itrestrictsto

f ion

Ui)isthatf 1| U

1∩U

2=

f 2| U

1∩U

2.

Indeed,definef=

f ion

Ui,

then

f:U1∪U2→

kis

well-defined,an

dregularity

follow

sbecau

seregu

larity

isalocalconditionan

dwealread

yknow

itis

satisfied

byf 1,f

2onU1,U

2.

Exercise.(N

on-examinab

le)UsingRem

ark2an

dSec.15.5,show

OX

isasheaf(ofk-algs)

onX.

3)Let

ϕ:X∼ =

Ybeisom

orphic

q.p.vars,andU⊂

Xan

open

set,so

V=

ϕ(U

)⊂

Yisanopen

set.

Then

wehavean

isoϕ∗:O

Y(V

)∼ =

OX(U

),F�→

F◦ϕ

.(H

int:

firstread

Sec.11.4).

Warn

ing.For

f∈O

X(U

),it

may

not

bepossible

tofindafraction

f=

g hthatworksonallofU.

Example.For

theaffi

nevarietyX

=V(xw−

yz)⊂

A4,f=

x y=

z w∈

k(X

)=

Frack[X

]defines

aration

alfunction

f∈

OX(D

y∪D

w)on

theq.p.var.

U=

Dy∪D

wsince

x y∈

OX(D

y)and

z w∈O

X(D

w),buton

ecannot

2findaglobal

expressionf=

g hdefined

onallofU.

Theorem

11.2.X

affinevariety⇒

OX(X

)=

k[X

].

Proof.

3Claim

1.k[X

]⊂

OX(X

).Proof.

f∈k[X

]⇒

f=

f 1on

X,so

itis

regulareverywhere.

�Claim

2.O

X(X

)⊂

k[X

].Proof.∀p∈X,∃op

enp∈Up⊂

X:

OX(X

)�f=

gp

hpas

map

sUp→

k,

1Thefunctionisnotdefi

ned

onallofPn

astheden

ominatormay

vanish(recallglobalmorphsPn

→kare

constant).

2this

isessentiallycausedbythefact

thatk[X

]is

notaUFD.

3Theproofis

easier

when

Xis

irreducible:insteadofusingtheidealJ

andthecover

Di∩D

j,oneargues

that

g ihj=

hig j

onD

i∩D

jforces

X=

Di∩D

j⊂

V(g

ihj−hig j)since

Di∩D

jis

anopen

den

sesetforirreducible

X,and

thusg ihj=

hig j

holdsonallofX.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

49

whereg p,h

p∈k[X

],an

dhp�=

0at

allpoints

ofUp.Since

basic

open

sets

areabasis

fortheZariski

topolog

y,wemay

assumeUp=

D� p

forsome� p∈k[X

](possibly

mak

ingUpsm

aller).Wenow

need:1

Trick

.gp

hp=

gp� p

hp� p

onD

� p.Replacingg p,h

pby

g p� p,h

p� p,wemayassumeg p

=hp=

0onV(�

p).

As

hp�=

0atpoints

ofUp=

D� p,wededuce

Dhp=

D� p.Sof=

gp

hponUp=

Dhp,andg p

=0onV(h

p).

Now

consider

theidealJ=�h

p:p∈X�⊂

k[X

].

Then

V(J

)=∅since

hp(p)�=

0.ByHilbert’sNullstellensatz,J=

k[X

]=�1�s

o1=

�αih

pi∈k[X

]forsomefinitecollection

ofpi∈X,an

dαi∈k[X

].Abbreviate

hi=

hpi,g i

=g p

i,D

i=

Upi=

Dhpi.

Notethat

1=

�αih

iim

plies

2that

theD

iarean

open

coverof

X.OntheoverlapD

i∩D

j,weknow

gj

hj=

f=

gi

hi,so

hig

j=

hjg i

onD

i∩D

j.Bytheab

oveTrick,hig

j=

hjg i

also

holdson

V(h

i)=

X\D

i

since

g i=

hi=

0there,

andalso

onV(h

j)=

X\D

jsince

g j=

hj=

0there.

Thushig

j=

hjg i

holds

everywhereon

Xas

X=

(Di∩D

j)∪V(h

i)∪V(h

j).

Thus,

onX,wededuce

f=

g j hj=

1·g

j

hj=

� i

αih

i·g

j

hj=

� i

αihig

j

hj

=� i

αihjg i

hj

=� i

αig

i∈k[X

].�

Coro

llary

11.3.D

h⊂

Xforanaff.var.

X⊂

An,then

OX(D

h)=

{g hm

:D

h→

k,wherem≥

0,g∈k[X

]}∼ =

k[X

][1 h]∼ =

k[X

] h.

Proof.

Followsfrom

Lem

ma11

.1an

dTheorem

11.2.Onecanalso

proveit

directly,

bymim

icking

thepreviousproof:f=

gp

hpon

Dh∩Up,then

V(�hp�)⊂

V(h),

sobyNullstellensatz

hm∈�h

p�,

and

argu

ingas

aboveon

ededuceshm

=�

αih

i,then

hmf=

�αig

ian

dfinally

f=

�αigi

hm

∈k[D

h].

�

Example.3

Let

X=

A2\{

0}.Then

OX(X

)=

k[x,y](w

hichim

plies

that

Xisnot

affine4).

Indeed,

A2\{

0}=

Dx∪D

y,so

f∈k[X

]defines

regu

larfunctionsf 1

=f| D

x∈k[D

x],f 2

=f| D

y∈k[D

y]which

agreeon

theoverlap:f 1| D

x∩D

y=

f| D

x∩D

y=

f 2| D

x∩D

y∈

k[D

x∩D

y](con

verselysuch

compatible

regu

larf 1,f

2determineauniqueglued

f∈k[D

x∪D

y]).Com

parek[D

x],k[D

y]insideFrack[A

2]=

k(x,y),

sok[X

]=

k[D

x]∩

k[D

y]⊂

k(x,y),

and5k[D

x]∩

k[D

y]=

k[x,y] x∩k[x,y] y

=k[x,y].

Exercise.O

P1(P

1)=

k,i.e.

theconstan

tfunctions.

11.4.

REGULAR

MAPS

ARE

MORPHIS

MS

OF

Q.P.V

ARIE

TIE

S

Definition.X,Y

q.p.vars.,

F:X→

Yis

aregularmapif∀p∈

X,∃openaffines

p∈

U⊂

X,

F(p)∈V⊂

Y(inparticularU∼ =

ZU⊂

AnandV∼ =

ZV⊂

Am

are

affine)

such

that

F(U

)⊂

Vand

ZU∼ =

UF| U −→

V∼ =

ZV⊂

Am

isdefi

ned

bym

regularfunctions6.

Lemma.F

isaregularmap⇔

Fis

amorphofq.p.vars.

Proof.

Exercise.7

�1Wecannotuse

Rem

ark

4above,

otherwisewehav

eacircularargument.

Also,weneedthetrick,because

otherwise

laterin

theproofg jhi=

g ihjwillonly

hold

onD

i∩D

j,so

f| D

j=

gj

hj=

�iαihigj

hj=

�iαig i

willonly

hold

on∩ i

Di.

2∅=

V(�hi�)

=∩ i

V(h

i)so

X=

X\∩

iV(h

i)=

∪ iX

\V(h

i)=

∪ iD

i.Equivalently,

ifx∈X

\∪D

ithen

hi(x)=

0for

alli,contradictingtheequation�

αihi=

1.

3Notice,this

says:

ifyouare

regularonA

2\{

0}then

youmust

beregularalsoat0.Theanalogousstatementholds

forholomorphic

functionsof2(ormore)variables(H

artogs’

extensiontheorem),unlikethe1-dim

ensionalcase

A1\{

0}

wherepolesandessentialsingularities

canarise.

4If

Xwereaffine,

itwould

beisomorphic

toA

2,asit

hasthesamecoordinate

ring.Atthecoordinate

ringlevel,

weobtain

someisomorphism

ϕ:k[A

2]=

OA2(A

2)→

OX(X

).Thepreim

ageoftheprimeidealI=

�x,y�⊂

OX(X

)yieldsaprimeidealJ=

ϕ−1(I)⊂

k[A

2].

ButV(I)=

∅⊂

X,so

V(J

)=

ϕ∗ (V(I))

=∅⊂

A2,so

J=

k[x,y]bytheaffine

Nullstellensatz.Butϕ

isanisomorphism,so

I=

ϕ(J

)=

k[x,y],contradiction.

5f/xk=

g/y�⇔

y�f=

xkg∈k[x,y]⇔

xk|f,y

�|g,so

f/xk∈k[x,y].

6In

other

words,

ZU→

ZV

isdefi

ned

bypolynomials

usingtheA

n,A

mcoordinates.

7Hint.

foranaffineopen

U⊂

X,thereisanaff.var.

Zsuch

thatU

∼ =Z

⊂A

n.CheckthatO

X(U

)∼ =

OZ(Z

)∼ =

k[Z

],usingTheorem

11.2

forthelast

iso.Therefore

amapdefi

ned

byregularfunctionsis

locallyapolynomialmap.

50

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

11.5.

THE

STALK

OF

GERM

SOF

REGULAR

FUNCTIO

NS

Definition.Theringofgerm

sofregularfu

nctionsatp(orthestalk

ofO

Xatp)is

OX,p=

{pairs(f,U

):anyopenp∈U⊂

X,anyfunctionf:U→

kregularatp}/∼

where(f,U

)∼

(f� ,U

� )⇔

f| W

=f� | W

someopenp∈W⊂

U∩U

� .

For

aqpvX⊂

Pnandp∈

X,pickanaffi

neop

enp∈

W⊂

X,then

wecanview

thestalkin

several

equivalentways:

OX,p∼ =

S(X

) (m

p)∼ =

k[W

] mp∼ =

OW,pbyLem

ma10

.7.

For

F:X→

Yamorphof

q.p.vars.

wegetaringhom

onstalks,

F∗ p:O

Y,F

(p)→

OX,p,F

∗ p(U

,g)=

(F−1(U

),F

∗ g)

whereF

∗:O

Y(U

)→

OX(F

−1(U

)),F

∗ g=

g◦F

.

Lemma.“KnowingF

∗ pforallp∈X

determines

F”.

More

precisely:ifF,G

:X→

Ysatisfies

F∗ p=

G∗ p∀p∈X

then

F=

G.

Proof.

Exercise

(com

pareHomew

ork

3,ex.4).

�Remark

.Alltheab

ovearestepstowardstheproofthat

OX

isasheafonX,called

stru

ctu

resh

eaf,an

d(X

,OX)is

alocallyringedspace,

indeedaschem

e(since

itis

locallyaffi

ne),seeSec.15.

12.

THE

FUNCTIO

NFIE

LD

AND

RATIO

NAL

MAPS

12.1.

FUNCTIO

NFIE

LD

For

anirred.aff.var.X,k[X

]is

anintegral

domain,so

wecan1definethefunction

field

k(X

)=

Frack[X

]=

{f=

g h:g,h∈k[X

]}/(g h=

�g � h⇔

g� h=

�gh)

Example.

g h∈k(X

)⇒

g h∈O

X(D

h)is

aregularfunctionon

theop

enD

h=

X\V

(h)⊂

X.

Example.Let

X=

V(xw−

yz)⊂

A4.Then

f=

x y=

z w∈k(X

).Noticef∈O

X(D

y∪D

w).

Lemma12.1.U,U

� �=∅affineopensin

anirredaffvarX⇒∀basicopen∅�=

Dh⊂

U∩U

� ,

k(U

)∼ =

k(D

h)∼ =

k(U

� ).

Proof.

U∼ =

Z=

V(I)⊂

An,so

2k[D

h]∼ =

k[Z

] h,so

k(D

h)∼ =

Frac(k[Z

] h)∼ =

Frac(k[Z

])∼ =

k(Z

)=

k(U

).�

Remark

.Thereisan

obviousrestrictionmap

ϕ:k(U

)→

k(D

h),

f g�→

π(f

)π(g)usingthecanon

ical

map

π:k[U

]=

k[Z

]�→

k[Z

] h=

k[D

h].Theaboveproves

ϕisbijective.

Theserestrictionsarecompatible:

thecompositek(U

)→

k(D

h)→

k(D

hh� )equalsk(U

)→

k(D

h� )→

k(D

hh� )(note:

Dhh�=

Dh∩D

h� ).

Exercise.For

irreducible

affineX,wecancompare

variou

srings

insidethefunctionfield:3

k[U

]=

OX(U

)=

�

Dh⊂U

OX(D

h)=

� p∈U

OX,p

⊂O

X,p=

k[X

] mp

⊂Frac(k[X

])=

k(X

)

Definition

12.2.ForX

anirredq.p.v.and∅�=

U⊂

Xanaffineopen,defi

nek(X

)=

k(U

).

Exercise.Show

that

thisfieldisindep

endent(upto

iso)on

thechoice

ofU.(H

int.

aboveLem

ma.)

1Remark

.ForX

reducible

(k[X

]notanintegraldomain)theanalogueofFrack[X

]is

thetotalringoffractions:

localize

k[X

]atS=

{allf∈k[X

]whichare

notzero

divisors}.

Fork[X

](oranyNoetherianreducedring),S

−1k[X

]∼ =

�Frac(k[X

]/℘i)where℘iare

theminim

alprimeideals

(geometrically,

theirredcomponen

tsX

iofX).

This

isnota

field:itisaproduct

offieldsk(X

i).

Anelem

entin

S−1k[X

]isonerationalfunctiononeach

Xicompatibly

onX

i∩X

j.

2To

clarify:

h:X

→k

isa

polynomialmap,defi

ning

Dh

=(h

�=0)⊂

X.

Since

Dh

⊂U,wealso

have

Dh=

(h| U

�=0)⊂

Ufortherestricted

function

h| U

:U

→k.

Alsoh

defi

nes

apolynomialfunction

h�on

Zvia

Z∼ =

U⊂

X→

k(aboveweabusivelycalled

h�again

h)defi

ningD

h�=

(h��=

0)⊂

Z.Now

Dh,D

h�are

isomorphic,so

theircoordinate

ringsare

alsoiso.Explicitly:k[D

h]∼ =

k[X

] h∼ =

OX(D

h)∼ =

OU(D

h| U)∼ =

OZ(D

h� )

∼ =k[Z

] h� .

3Sec.10defi

nes

localisation,andLem

ma10.5

show

sO

X,p

∼ =k[X

] mp⊂

k(X

)consistsoffractions

f gwithg∈k[X

]\m

p.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

51

12.2.

RATIO

NAL

MAPS

AND

RATIO

NAL

FUNCTIO

NS

Motivation.Let

X,Y

beirredaff

vars.Recallk-alg

hom

sk[X

]→

k[Y

]arein

1:1correspon

den

cewithpolynom

ialmap

sX←

Y.Dok-alg

hom

sk(X

)→

k(Y

)correspon

dto

map

sgeom

etrically?

Example.k(t)→

k(t),t�→

1 tcorrespon

dsto

A1←

A1given

bya�→

1 a,defined

ontheop

enA1\{

0}.

Definition12.3.ForX

anirredq.p.v.,ara

tionalmapf:X

���Y

isaregularmapdefi

ned

ona

non-empty

opensubset

ofX,andweiden

tify

rationalmapswhichagree

onanon-empty

opensubset.

Remark

.Soaration

almap

isan

equivalence

class[(U,F

)]where∅�=

U⊂

Xisop

en,F

:U→

Yis

amorphof

q.p.v.’s.

Weidentify

(U,F

)∼

(U� ,F

� )ifF| U

∩U�=

F� | U

∩U� .Bydefinitionof

regu

larmap

,wecanalwaysassumethat

F:U→

V⊂

Yisapolynom

ialmap

betweenaffi

neop

ensU⊂

X,V⊂

Y.

Remark

.Since

Xis

irred,U⊂

Xis

dense,so

fis

“defined

almosteverywhere”.X

irreducible

ensuresthat

intersection

sof

finitelyman

ynon

-empty

open

subsets

arenon

-empty,op

enan

ddense.

EXAM

PLES.

1).Pn

���Pn

−1,[x

0:···:

xn]→

[x0:···:

xn−1]is

defined

onU

=Pn

\{[0

:···:

0:1]}.

2).f:X

���Andetermines

regu

larf 1,...,f

n:U→

A1in

OX(U

)on

someop

en∅�=

U⊂

X.

3).f i∈O

X(U

i)forop

ens∅�=

Ui⊂

Xyield

f=

(f1,...,f

n):X

���An,defined

onU

=∩U

i.

4).Anexam

ple

of(2)/(3):

g h∈k(X

)determines

X���A1,a�→

g(a)

h(a),defined

onU

=D

h⊂

X.

Definition

12.4.A

rationalfu

nction1is

arationalmapf:X

���A1.

Lemma12.5.ForX

anirredq.p.v.,

k(X

)∼ =

{rational

functionsf:X

���A1},

g h�→

[(D

h,g h)].

Remark

.Analogou

sto:forX

affvar,k[X

]∼ =

{polynom

ialfunctionsf:X→

A1},

g�→

(Xg →A1).

Proof.

WLOG

(byrestrictingto

anon

-empty

open

affinein

X)wemay

assumeX

isan

irreducible

affinevariety.

Bydefinition,aration

alfunctionisdetermined

byarepresentative

onan

ynon

-empty

open

subset,so

wecanpickan

(arbitrarily

small)basic

open

subsetD

h⊂

Xwith2f=

[(D

h,g h)]for

some

g h∈k(D

h).

ByLem

ma12.1,this

correspon

dsto

auniqueelem

entin

k(D

h)∼ =

k(X

),an

dthe

elem

entconstructed

isindep

endentof

thechoice

ofD

hbytheRem

arkunder

Lem

ma12.1.

�

Warn

ing.Rational

map

smay

not

compose:

A1���A1,a�→

0an

dA1���A1,a�→

1 a.

f=

[(U,F

)]:X

���Y,g=

[(V,G

)]:Y

���Z

haveawell-defined

compositeg◦f

:X

���Z

ifF(U

)∩V�=∅:

then

g◦f

isdefined

ontheop

enF

−1(F

(U)∩V).

Toensure

composites

withfare

alwaysdefined

,indep

endentlyof

g,wewan

tF(U

)to

hit

everyop

enin

Y,i.e.

F(U

)⊂

Yis

dense.

Definition.f=

[(U,F

)]:X

���Y

isdomin

antiftheim

age

F(U

)⊂

Yis

den

se.

Exercise.Thedefinitionis

indep

endentof

thechoice

ofrepresentative

(U,F

).Exercise.Let

f:X

���Y

bedom

inan

t,an

dg:Y

���X

aration

almap

satisfyingg◦f

=id

X(an

equalityof

ration

almap

s,i.e.

g◦f

=id

Xon

somenon

-empty

open

set).Show

gis

dom

inan

t.

1Cultura

lRemark

.Chow’s

theore

m:everycompact

complexmanifold

X⊂

Pn(holomorphicallyem

bed

ded

)is

asm

ooth

projvar;

everymeromorphic

functionis

arationalfunction;holo

mapsbetweensuch

mfdsare

regularmaps.

Example

(CoursesB3.2/B3.3):

forX

acompact

connected

Riemann

surface,k(X

)=

{meromorphic

functions

X��

�A

1 C=

C}=

{holomorphic

mapsX

→P1

}\{constantfunction∞

}.Thefollow

ingcategories

are

equivalent:

(1)non-singularirredpro

jectivealgebra

iccurv

es(i.e.dim

=1)over

Cwithmorphsthenon-constantregularmaps,

(2)compact

connectedRiemannsurfaceswithmorphsthenon-constantholomorphic

maps,

(3)theopposite

ofthecategory

ofalgebra

icfu

nction

fieldsin

onevariable/C

(meaning:af.g.

field

extension

C�→

Kwithtrdeg

CK

=1,so

afinitefieldextensionC(t)�→

K)withmorphsthefieldhomsfixingC.

Soanytw

omeromorphic

functionsare

algeb

raicallydep

enden

t/k,andcompact

connectedRiemannsurfacesare

isoiff

theirfunctionfieldsare

iso(this

may

failforsingularcu

rves,andcompactnessis

crucialto

ensure

Xis

algeb

raic).

The

“non-constant”

conditionen

suresthemapsare

dominant.

2If

f=

ghN

wecanalw

aysreplace

hbyhN

toassumeN

=1.

52

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Definition.A

birationaleq

uivalence

f:X

���Y

isadominantrationalmapbetweenirreducible

q.p.v.’swhichhasarationalinverse,

i.e.

thereexists

arationalmapg:Y

���X

withf◦g

=id

Y

andg◦f

=id

X(equalities

ofrationalmaps).WesayX�

Yare

birational.

EXAM

PLES.

1).An�

Pnarebirational

via

theinclusion

An∼ =

U0⊂

Pn,whichhas

rationalinversePn→

An,

[x0:···:

xn]→

(x1

x0,···

,xn

x0)defined

onU0.

2).For

anirredq.p.v.X⊂

Pn,X�

Xvia

theinclusionX

�→X.

3).For

anirredq.p.v.X⊂

Pn,X∩Ui�

Xvia

theinclusion

,assumingX∩Ui�=∅(i.e.X�⊂

V(x

i)).

4).

TheCremona

transform

ation

P2���

P2,[x

:y

:z]�→

[yz

:xz

:xy],

defined

on

the

open

whereat

leasttw

ocoordsarenon

-zero.

Dividingbyxyz,this

rational

mapis

equivalentto

[x:y:z]�→

[1 x:

1 y:

1 z],defined

ontheop

enwhereallcoordsarenon-zero.This

mapis

itsow

n

inverse,

sobirational.

Lemma12.6.ForX,Y

irredaffvars,f:X

���Y

determines

ak-alg

hom

f∗:k[Y

]→

k(X

)via

(y:Y→

A1)�→

(f∗ y

=y◦f

:X

���A1).

Moreover

f∗injective⇔

fdominantin

whichcase

wegetak-alg

hom

f∗:k(Y

)→

k(X

),g h�→

f∗ g

f∗ h.

Proof.

f=

[(U,F

)]defines

f∗ y

=[(U,F

∗ y)]=

[(U,y◦F

)].Thelack

ofinjectivityofthelinearmap

F∗dep

endson

itskernel.For

y�=

0,

F∗ y

=0⇔

y(F

(u))

=0∀u∈U⇔

F(u)∈V(y)∀u∈U⇔

F(U

)⊂

V(y)⊂

Y.

F(U

)not

dense⇔

F(U

)⊂(som

eproper

closedsubsetsayV(J

)�=

X)⊂

V(y),anyy�=

0∈J.

For

thefinal

claim:f∗ h�=

0ifh�=

0(since

f∗inj).

�

12.3.

Equivalence

:IR

REDUCIB

LEQ.P.V

ARS.AND

F.G

.FIE

LD

EXTENSIO

NS

Theorem

12.7.Thereis

anequivalence

ofcategories

1

{irred

q.p.v.X

,withrationaldominantmaps}

→{f.g.fieldextensionsk�→

K,withk-alg

homs}

op

X�→

k(X

)(f

=ϕ∗:X

���Y)�→

(ϕ=

f∗:k(X

)←

k(Y

))

Inparticular,

thefollowingpropertieshold:

(1)f∗∗

=fandϕ∗∗

=ϕ;

(2)X

f ���Y

g ���Z⇒

(g◦f

)∗=

f∗◦g

∗:k(X

)f∗

←−

k(Y

)g∗

←−

k(Z

);

(3)k(X

)ϕ ←−

k(Y

)ψ ←−

k(Z

)⇒

(ϕ◦ψ

)∗=

ψ∗◦ϕ

∗:X

ϕ∗

���Y

ψ∗

���Z.

(4)X�

Ybirationalirreducibleq.p.v.’s⇔

k(X

)∼ =

k(Y

)isok-algs.

Remark

.Recalltheequiv

{affinevars,aff

morphs}→

{f.g.reducedk-algs,k-alg

homs},X�→

k[X

].This

was

not

anisoof

cats:to

buildX

from

thek-alg

A,on

echooses

generators

g 1,...,g

n∈A

togetϕ:k[x

1,...,x

n]�

A,xi�→

g i,so

ϕ:k[x

1,...,x

n]/kerϕ∼ =

A.Then

X=V(ker

ϕ)⊂

An.

Proof.

Claim

1.finducesϕ=

f∗ .

Pf.

WLOG

X,Y

areaffi

ne(since

fis

representedbyan

affinemap

F:U→

Vonopen

affines

and

k(U

)=

k(X

),k(V

)=

k(Y

)bydefinition).

ByLem

ma12.6,f:X

���Y

determines

k(Y

)→

k(X

),g h�→

f∗ g

f∗ h

.

Claim

2.For

fieldextensionsk�→

A,k�→

B,anyk-alg

hom

A→

Bis

afieldextension(i.e.inj).

Pf.

Let

J=

ker(A→

B).

AsJis

anidealin

afieldA,it

iseither

0(don

e)orA

(false:

1�→

1).

�1A

field

extension

k�→

Kis

finitely

genera

ted

ifthereare

elem

ents

α1,...,α

nsuch

thatthehomomorphism

k(x

1,...,x

n)=

Frack[x

1,...,x

n]→

K,xi�→

αiis

surjective.

Notice

weallow

fractions,

unlikefinitelygen

erated

k-algeb

raswhereyouonly

allow

polynomials

inthegen

erators.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

53

Claim

3.For

X,Y

irredaffi

ne,

ak-alg

hom

ϕ:k(Y

)→

k(X

)determines

abirational

f:X

���Y.

Pf.

ByClaim

2,ϕ

isinjective(inparticularan

injectionk[Y

]�→

k(Y

)→

k(X

)).Let

y 1,...,y

nbe

generatorsof

k[Y

](ifY⊂

Anthen

k[Y

]is

generated

bythecoordinatefunctionsy j).

Then

ϕ(y

j)=

g j hj∈k(X

).

Let

U=∩D

hj,then

ϕ(y

j)∈

OX(U

).Since

k[Y

]is

generated

bythey j,also

ϕ(k[Y

])⊂

OX(U

).

WLOG

Uis

affine(rep

lace

Ubyasm

allerbasic

open).

Then

1O

X(U

)=

OU(U

)=

k[U

].The

inclusion

ϕ:k[Y

]�→

k[U

]correspon

dsto

amorphϕ∗:U→

Yof

affvars

(see

aboveRem

ark),

and

ϕ∗is

dom

inan

tsince

ϕis

injective(L

emma12

.6),

soit

represents

adom

inan

tϕ∗:X

���Y.

Remark

.Explicitly,

foru∈U⊂

X,

u�→

(ϕ(y

1)(u),...,ϕ(y

n)(u))

=(g1(u

)h1(u

),...,gn(u

)hn(u

))∈Y⊂

An.

Claim

4.For

X,Y

q.p.v.’s,

ak-alg

hom

ϕ:k(Y

)→

k(X

)determines

abirational

f:X

���Y.

Pf.

k(X

)=

k(U

),k(Y

)=

k(V

)foraffi

neop

ensU,V

.ByClaim

3,k(V

)=

k(Y

)→

k(X

)=

k(U

)defines

U��

�V,whichrepresents

X��

�Y.�

Claim

5.For

anyf.g.

k�→

K,thereis

anirredq.p.v.X

withK∼ =

k(X

).Pf.

Pick

generatorsk1,...,k

nof

K,letR

=k[x

1,...,x

n],

defineϕ

:R→

K,xj�→

kj.

Let

J=

kerϕ,then

R/J

�→K,so

Jisaprimeidealas

Kisan

integral

dom

ain.Let

X=

V(J

)⊂

Anbe

theirreducible

affinevarietycorrespon

dingto

R/J

.Then

k(X

)∼ =

Ksince

k(X

)=

FracR/J

�→K

containsthegeneratorskjin

theim

age.

�Exercise.Prove

properties

(1)-(4)(thesefollow

from

analog

ousknow

nclaimsforaffi

nemorphs).

Claim

6.Thefunctor

intheclaim

isan

equivalence

ofcatego

ries.

Pf.

It’s

fullyfaithfulbyf∗∗

=f,ϕ∗∗

=ϕ(P

roperty

(1)).

It’s

essentially

surjectivebyClaim

5.�

�

Coro

llary

12.8.Anyirreducibleaffinevarietyis

birationalto

ahypersurface

insomeaffinespace.

Proof.

WLOG

Xis

affine(restrictto

anaffi

neop

en).

ByNoether

normalisation(Section

8.4),for

anirred.aff.var.X,

k�→

k(y

1,...,y

d)�→

k(X

)∼ =

k(y

1,...,y

d,z)=

Frack[y

1,...,y

d,z]/(G

)

wherey 1,...,y

darealgebraically

indep

endent/k,d=

dim

X=

trdeg

kk[X

],an

dz∈

k[X

]satisfies

anirreducible

polyG(y

1,...,y

d,z)=

0.Since

V(G

)⊂

An+1has

k[V(G

)]=

k[y

1,...,y

d,z]/(G

),the

aboveisok(X

)∼ =

k(V

(G))

implies

via

Theorem

12.7

that

X��

�V(G

)arebirational.

�

Definition

12.9.A

q.p.v.X

isra

tionalifitis

birationalto

Anforsomen.

Remark

.BytheThm,X

rational⇔

k(X

)∼ =k(x

1,...,x

n)is

apurely

transcen

den

talextensionofk.

13.

TANGENT

SPACES

13.1.

TANGENT

SPACE

OF

AN

AFFIN

EVARIE

TY

Foramore

detailed

discussionofthetangentspace,wereferto

theAppendix

Section17.

F∈k[x

1,...,x

n].

p=

(p1,...,p

n)∈An.

Thelinearpolynom

ialdpF∈k[x

1,...,x

n]is

defined

by

dpF

=dF| x=

p·(x−

p)=

�∂F

∂xj(p)·(xj−

pj)

Example.p=

0,F(x)=

F(0)+

a0x0+

···+

anxn+

quad

ratic+

higher.Thelinearpartof

this

Taylorexpan

sion

isd0F

=�

ajxj.

1RecalltheTheorem:X

anaffinevariety⇒

OX(X

)=

k[X

].

54

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Definition.Thetangentsp

ace

toanaff.var.

X⊂

An,withI(X)=�F

1,...,F

N�,

is

TpX

=V(d

pF1,...,d

pFN)=∩kerdFi⊂

An

REM

ARKS.

1)TpX

isan

intersection

ofhyperplanes

V(d

pFi),so

itis

alinearsubvariety.

2)TpX

istheplanewhich“b

est”

approxim

atesX

nearp.Notice

p∈TpX.

3)Bytran

slating,−p+TpX,weobtain

thevectorspace

which“b

est”

approxim

atesX

nearp(w

ith

0“correspon

ding”

top).

This

isalso

oftencalled

thetangentspace.

Silly

example.X

=An,I(An)=

{0}so

TpAn=

An.

Example.Thecusp

X=

V(y

2−

x3)=

{(t2,t

3):t∈

k}is

determined

by

F=

y2−x3.Atp=

(t2,t

3),

dF

=−3x

2dx+2y

dy=

� −3x2

2y

�

dpF

=−3t

4(x−t2)+2t

3(y−t3).

For

t�=

0,TpV

=kerdpF

isthe(1-dim

ension

al)straightlineperpendicular

to(−

3t4,2t3).

Butat

t=

0,dpF

=0so

TpX

=V(0)=

k2is

2-dim

ension

al.

p

TpX

dpF

X

Exercise.

Recallalinethroughphas

theform

�(t)

=p+

tvforsomev∈

kn.A

lineis

called

tangentto

Xat

pifFi(�(t))has

arepeated1root

att=

0.Show

that

TpX

=∪(

lines

tangentto

Xat

p).

Definition.p∈X

isasm

ooth

poin

tif

2

dim

kTpX

=dim

pX.

p∈X

isasingularpoin

t3ifdim

kTpX

>dim

pX.Abbreviate

Sing(X

)=

{allsingularpoints}⊂

X.

Theorem.Let

Xbe

anirreducibleaff.var.

ofdim

ensiondwithI(X)=�F

1,...,F

N�.

⇒Sing(X

)⊂

Xis

aclosedsubvarietygivenby

thevanishingin

Xofall(n−

d)×

(n−

d)minors

oftheJacobianmatrix

Jac

=

�∂Fi

∂xj

�.

Proof.

TpX

isthezero

setof ϕ

p:

�x1 . . . xn

��→

∂F1

∂x1

� � � p···

∂F1

∂xn

� � � p

. . .∂FN

∂x1

� � � p···

∂FN

∂xn

� � � p

·

x1−p1

. . .xn−pn

Hence

p∈SingX⇔

dim

ϕ−1

p(0)>

d⇔

dim

kerJac

p>

d⇔

all(n−d)×(n−d)minors

vanish.4

�

Example.For

thecusp:F

=y2−

x3,Jac

=(−

3x2

2y),

n=

2,d=

1.So1×

1minors

allvanish

precisely

when

(x,y)=

(0,0).

1Weknow

t=

0is

aroot,

since

theFivanishatp∈X.

2Recall:dim

pX

=(thedim

ensionoftheirreducible

componen

tofX

containingp),

Section8.1.

3(N

on-examinable)Fact:dim

TpX

≥dim

pX

alway

sholds.

Intuitively:ifthedpFiare

linearlyindep

enden

tthen

the

Fiare

also“indep

endentnearp”,so

each

equationFi=

0cu

tsdow

nbyonethedim

ensionofX

atp.Over

complex

numbers,

this

isaconsequen

ceoftheim

plicitfunction

theorem.

More

gen

erally,

oneway

toprovethis

isvia

the

Noether

Norm

alizationLem

ma(G

eometricVersion2)from

Sec.8.4

andapplyingthefollow

ingfact

totheprojection

from

thetangent“bundle”TX

={(p,v)∈

X×

An:v∈

TpX}→

X,(p,v)�→

p.Fact.

Given

anyregularsurjective

mapf:X

→Y

ofirreducible

q.p.vars,then

dim

F≥

dim

X−

dim

Yforanycomponen

tF

off−1(y),

andanyy∈Y.

Moreov

er,dim

f−1(y)=

dim

X−

dim

Yholdsonanon-empty

open

(hen

ceden

se)subsetofy∈Y.

4Otherwisewewould

findn−

dlinearlyindep

enden

tcolumns(thecolumnsinvolved

inthatminor),andhen

cethe

rankwould

beatleast

dim

=n−

d,so

thekernel

would

beatmost

dim

=d.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

55

13.2.

INTRIN

SIC

DEFIN

ITIO

NOF

THE

TANGENT

SPACE

OF

AVARIE

TY

Theore

m.X

aff.var.,p∈X,andrecallm

p=

{f g∈O

X,p:f(p)=

0}⊂

OX,p.Then

,canonically,

TpX∼ =

� mp/m

2 p

� ∗

(thevectorspace

mp/m

2 p,before

dualization,is

called

theco

tangentsp

ace

).

Proof.

WLOG

(after

alinearisoof

coords)

assumep=

0∈An.

Nota

tion.Toavoidconfusion

,wefirstlist

below

themax

imal

ideals

that

willarisein

theproof:

k[A

n]⊃

m=

{f:An→

k:f(0)=

0}=�x

1,...,x

n�

k[X

]⊃

m=

{f:X→

k:f(0)=

0}=

m·k

[X]=

m+

I(X)

OX,0

⊃m

0=

{f g:f,g∈k[X

],g(0)�=

0,f(0)=

0}=

m·O

X,0.

Step

1.Weprove

itforX

=An.

d0F

=�

∂F

∂xi

� � � 0·x

iis

alinearfunctional

An≡

T0An→

k,so

d0F∈(T

0An)∗.Thus

d0:k[x

1,...,x

n]→

(T0An)∗,F�→

d0F

andd0is

linear.1Restrictingto

themax

imal

idealm

=(x

1,...,x

n)of

thoseF

withF(0)=

0,

d0| m

:m→

(T0An)∗.

d0| m

islinearan

dsurjective.2

Subclaim

.kerd0| m

=m

2,hence

d0| m

:m/m

2→

(T0An)∗

isan

iso.

Proof.

d0F

=0⇔

∂F

∂xi(0)=

0∀i⇔

(Fon

lyhas

mon

omials

ofdegrees≥

2)⇔

F∈m

2.�

Step

2.Weprove

itforgeneral

X.

Theinclusion

j:T0X

�→T0Anis

injective,

sothedual

map

3is

surjective,

j∗:m/m

2∼ =

(T0An)∗→

(T0X)∗

⇒j∗◦d

0:m→

(T0X)∗

surjective.

Subclaim

.4kerj∗◦d

0=

m2+

I(X)=

m2⊂

k[X

],hence

m/(m

2+

I(X))∼ =

(T0X)∗.

Proof.

F∈ker(j∗◦d

0)⇔

j∗d0F

=d0F| T 0

X=

0⇔

d0F∈I(T0X)

⇔d0F∈�d

0F1,...,d

0FN�w

hereI(X)=�F

1,...,F

N�.

⇔d0F

=�

aid0Fiwhereai∈k[x

1,...,x

n].

⇔d0(F−

�aiF

i)=−

�(d

0ai)·F

i(0)

=0(since

0=

p∈V(F

1,...,F

N)).

⇔F−

�aiF

i∈kerd0| m

=m

2.

⇔F∈I(X)+

m2.�

Finally

5

(T0X)∗∼ =

m/(m

2+

I(X))∼ =

m/m

2

wherethelast

isois

oneof

the“isomorphism

theorems”.6

Now

localise:

Claim

.ϕ:m/m

2∼ =

m0/m

2 0,f�→

f 1(thetheorem

then

follow

s).

Proof.

Subclaim

1.ϕis

surjective.

Proof.

For

f g∈m

0,letc=

g(0)�=

0.

⇒ϕ(fc)−

f g=

f c−

f g=

f 1·(

1 c−

1 g)∈m

2 0(since

f 1∈m

0an

d(1c−

1 g)∈m

0).

⇒ϕ(fc)=

f gmodulo

m2 0.�

Subclaim

2.ϕis

injective.

1d0(λF

+µG)=

λd0F

+µd0G.

2d0xi=

xiare

abasisfor(T

0A

n)∗.

3explicitly,

j∗is

just

therestrictionmap:j∗F

=F

◦j=

F| T 0

X:T0X

j →T0A

nF →

k.

4m

den

otestheim

ageofm

inthequotien

tk[X

]=

R/I(X).

5usingthatI(X)⊂

m,since

f| X

=0⇒

f| p

=0.

6wequotientnumeratorandden

ominatorbyacommonsubmodule,I(X).

Explicitly:m

→m/m

2is

surjectiveand

thekernel

iseasily

seen

tobem

2+

I(X).

56

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Proof.

Need

toshow

kerϕ

=0.

Suppose

f 1∈

m2 0.

Thus1

f 1=

�gi

hi·g

� ih� iwhereg i,g

� i∈

mand

hi,h� i∈k[X

]\m.Take

common

denominators

(andredefi

neg i)to

get

f 1=

�gig� i

hforsomeh∈k[X

]\m.

Then

s·(fh−

�g ig� i)=

0∈k[X

]forsomes∈k[X

]\m.Thussf

h∈m

2=

m2+I(X).

Since

f∈m,

also

2(sh−

s(0)h(0))f∈

m2.

Thuss(0)h(0)f∈

m2,forcing3

f∈

m2.

So

f 1=

0∈

m/m

2as

required.

�

Remark

.That

also

provedTpX∼ =

(Ip/I

2 p)∗

whereI p

=�x

1−p1,...,x

n−pn�⊂

k[X

].

Coro

llary.TpX

only

4depen

dsonanopenneighbourhoodofp∈X.

Proof.

BytheTheorem,iton

lydep

endsonthelocalringO

X,p(anditsuniquemaxim

alidealm

p).

�

Definition.ForX

aq.p.var.

wedefi

nethetangentspace

atp∈X

byTpX

=� m

p/m

2 p

� ∗.

Remark

.In

practice,

youpickan

affineneigh

bou

rhoodof

p∈X,then

calculate

theaffinetangent

spaceusingtheJacobian.

13.3.

DERIV

ATIV

EM

AP

Lemma.ForF

:X→

Yamorphofq.p.vars.,onstalksF

∗:O

Y,F

(p)→

OX,p

isalocal5

ringhom

mF(p)→

mp,g�→

F∗ g

=g◦F

.

Proof.

g(F

(p))

=0im

plies

(F∗ g)(p)=

0.�

F:X→

Ymorphofq.p.vars.

Wewantto

construct

thederivativemap

DpF

:TpX→

TF(p)Y.

BytheLem

ma,

F∗ (m

F(p))⊂

mp,so

F∗ (m

2 F(p))⊂

m2 p,an

dthus6

F∗:m

F(p)/m

2 F(p)→

mp/m

2 p.

Itsdual

defines

thederivativemap:

DpF

=(F

∗ )∗:(m

p/m

2 p)∗→

(mF(p)/m

2 F(p))

∗ .

Exercise.Show

that

locally,onaffi

neop

ensarou

ndp,F

(p),youcanidentify

DpF

withtheJacobian

matrixof

F.Moreprecisely:locallyF

:An→

Am,p=

0an

dF(p)=

0,andJacF

=(∂Fi

∂xj)acts

by

left

multiplication

An≡

T0An→

Am≡

T0Am.

Example.F

:A1→

V(y−x2)⊂

A2,F(t)=

(t,t

2),F(0)=

(0,0).

For

A1:m

0=

t·k

[t] (t)⊂

k[t] (t)(w

einvert

anythingwhichis

not

amultiple

oft).

For

A2:m

F(0)=

(x,y)·(k[x,y]/(y−x2))

(x,y)⊂

(k[x,y]/(y−x2))

(x,y).

F∗:ax+by

+higher∈m

F(0)/m

2 F(0)�→

at+bt

2=

at∈m

0/m

2 0.

⇒D

0F

=(F

∗ )∗:t∗�→

x∗ ,

wheret∗(at)

=aan

dx∗ (ax+by)=

a.

⇒D

0F

=� 1 0

�in

thebasis

x∗ ,y∗on

thetarget

(andbasis

t∗on

thesource).

This

agrees

withtheJacob

ianmatrixof

partialderivatives:

D0F

=� ∂

tF1| 0

∂tF2| 0� =

� 1 2t�� � t=

0=

� 1 0

� .

1Bydefi

nitionm

2 0is

gen

eratedbyproductsofanytw

oelem

ents

from

m0,so

itinvolves

asum

andnotjust

one

gg�

hh�.

2since

sh−

s(0)h(0)andfboth

vanishat0.

3since

s,hdonotvanishat0.

4Soitisindep

enden

tofthechoiceofFiwithI(X)=

�F1,...,F

N�,

anditisindep

enden

tofthechoiceofem

bed

ding

X⊂

An,i.e.

itis

anisomorphism

invariant.

5meaning:maxideal→

maxideal.

6This

F∗is

called

thepullback

maponcotangen

tspaces.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

57

14.

BLOW

-UPS

14.1.

BLOW

-UPS

Theblow-upof

Anat

theorigin

isthesetof

lines

inAnwithagiven

choice

ofpoint:

B0An=

{(x,�):An×

Pn−1:x∈�}

=V(x

iyj−

xjy i)⊂

An×

Pn−1

usingcoords(x

1,...,x

n)on

An,[y

1:···:

y n]on

Pn−1.That

x∈�means(x

1,...,x

n)an

d(y

1,···

,yn)

areproportion

al,equivalentlythematrixwiththoserowshas

rank1so

2×

2minorsvanish.

Exercise.Via

thelinearisox�→

x−

p,describetheblow-upB

pAnat

p.

Themorphism

π:B

0An→

An,π(x,[y])=

x,

isbirational

withinverse1

An→

B0An,x�→

(x,[x])

defined

onx�=

0.Thefibre

π−1(x)is

apoint

withtheexception

oftheexceptionaldivisor2

E0=

π−1(0)=

{0}×

Pn−1.

Thusπ:B

0An\E

0→

An\0

isan

iso,

andπcollap

sesE

0to

thepoint0.

Infact

E0∼ =

Pn−1∼ =

P(T0An)istheprojectivisationof

thetangentspace:3theclosure

ofthepreim

age

{(vt,[vt]):t�=

0}of

thepuncturedlinet�→

tv,t�=

0,containsthenew

point(0,[v])

(usingthat

[vt]=

[v]∈Pn

−1byrescaling).

Definition.For

X⊂

Anan

aff.var.w

ith0∈X,thepro

pertransform

(orblow-u

pof

Xat

0)is

B0X

=closu

re(π−1(X

\{0}

))⊂

B0An.

Aga

inπ:B

0X→

Xis

birational,an

d

E=

π−1(0)∩B

0X

istheexception

aldivisor.B

0X

only

keepstrackof

direction

sE⊂

E0at

whichX

approaches

0,unlike

theto

taltransform

π−1(X

)=

B0X∪E

0.

Example.X

=V(xy)=

(x-axis)∪(y-axis)⊂

A2.Then

π−1(X

\0)=

{((x,y),[a

:b])∈A2×

P1:xb−

ya=

0,xy=

0,(x,y)�=

(0,0)}.

Solving:

((x,0),[1

:0])forx�=

0,((0,y),[0

:1])fory�=

0.Then

B0X

istheclosure:(A

1×0,[1

:0])�(0

×A1,[0:1]),adisjointunionof

lines!Theexception

aldivisor

Econsistsof

twopoints:((0,0),[1:0]),

((0,0),[0:1]),

the2direction

sof

thelines

inX.

14.2.

RESOLUTIO

NOF

SIN

GULARIT

IES

Blow-upsareim

portantbecau

sethey

provideaway

todesingu

larise

avarietyX,i.e.

findinga

smoothvarietyX

�whichis

birational

totheoriginal

varietyX.Ofcourse,

X�is

not

unique.

Example.Thecuspidal

curveX

=V(y

2−

x3)⊂

A2is

singu

larat

0.Use

coords((x,y),[a

:b])on

B0A2,xb−

ya=

0.NoticeD

a=

X∩(a�=

0)canbeviewed

asasubsetof

A2usingcoords(x,b),

since

WLOG

a=

1,then

y=

xb(for

a=

0,werescaleb=

1,butthen

bothx=

0an

dy=

0).

Substitute

into

ourequation:0=

y2−x3=

x2b2−x3.Theproper

tran

sform

isob

tained

bydropping

thex2factor:b2−

x=

0(checkthis).

ThusB

0X

={(b2,b

3),[1

:b]):b∈

k}is

asm

oothcurve,

isom

orphic

totheparab

olax=

b2in

A2,an

dit

isbirational

toX.

1anon-zeroxdetermines

thelineuniquely:�=

[x1:···:

xn].

2Divisorherejust

meanscodim

ension1subvariety,

althoughmore

gen

erallydivisorrefers

toform

alZ-

linearcom-

binationsofsuch

(theseare

called

Weildivisors).

3more

accurately,ofthenorm

alspace

to{0

}=

T00⊂

T0A

n:wekeeptrack

ofhow

xconverges

norm

allyinto

0.

58

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Hironaka’s

Theorem

(Hard

!).Assumechar

k=

0.Foranyp.v./q.p.v.

X,thereis

asm

ooth

p.v./q.p.v.X

�andamorphπ:X

� →X

whichis

birational,such

that

π:X

� \π−1(Sing(X

))→

Xsm

ooth

=X

\Sing(X

)

isaniso.If

Xis

affine,

then

X� =

BI(X

)canbe

constructed

astheblow-upofX

alonga(possibly

non-radical)

idealI⊂

k[X

](see

Section14.3),

with

V(I)=

Sing(X

).

14.3.

BLOW

-UPS

ALONG

SUBVARIE

TIE

SAND

ALONG

IDEALS

Definition.For

affineX,an

dI=�f

1,...,f

N�⊂

k[X

],defineB

I(X

)to

bethegra

phoff:X

���

PN−1,f(x)=

[f1(x):···:

f N(x)],meaning:

BIX

=closu

re({(x,f

(x))

:x∈X

\V(I)})⊂

X×PN

−1.

Themorph

π:B

I(X

)→

X,π(x,[v])=

x

isbirational

withinverse

x�→

(x,f

(x))

(defined

onX

\V(I)).Theexceptionaldivisoris

E=

π−1(V

(I)).

Definition.Theblow-u

palongasu

bvariety

Yis

BYX

=B

I(Y)X

.

Exercise.For

Y=

{0}(soI=

I(0)=

(x1,...,x

n)),show

BYX

istheproper

transform

B0X.

Remark

.B

IX

isindep

endentof

thechoiceof

generatorsf j,butitdep

ends1

onIandnotjust

V(I).

Definition.For

q.p.v.X⊂

Pn,an

dI⊂

S(X

)hom

og.,pickhom

og.gensf 1,...,f

Nofthesame

degree2.Thusf:X

���PN

−1determines

BIX⊂

X×PN

−1asbefore,an

ddefine

BIX

=B

IX∩(X

×PN

−1).

15.

SCHEM

ES

Section15is

anintroductionto

modernalgebraic

geometry.It

isconceptuallycentralto

thesubject.

However,forthepurposesofexams,

alm

ost

allofsection15is

non-examinable.

Theonly

topicsyou

needto

know

are:(1)thedefi

nitionofSpec,Specm

in15.1;(2)theZariskitopology

onspectrain

15.2;(3)morphismsbetweenspectrain

15.3.

15.1.

SpecOF

ARIN

Gand

THE

“VALUE”OF

FUNCTIO

NS

ON

Spec

A=

anyring(com

mutativewith1).

Theaffinescheme3forA

isthespectrum

Spec

A,where

Spec

A=

{primeideals

℘⊂

A}⊃

{max

ideals

m⊂

A}=

Specm

A

HereA

playstherole

ofthecoordinatering

A=

O(Spec

A)=

“ringof

glob

alregu

larfunctions”

whereO

iscalled

thestru

ctu

resh

eaf(m

oreon

this

later).

Remark

.NoticeO(Spec

k[x]/x2)=

k[x]/x2remem

bersthat0is

adou

ble

rootofx2,whereasthe

affinecoordinateringk[V(x

2)]=

k[x]/xdoes

not.

Question

:In

what

sense

areelem

ents

ofA

“functions”

onSpec

A?

f∈A⇒

“function”

Spec

A→

??℘�→

f(℘

)∈K(℘

)=

Frac(A/℘)

1e.g.B

(x2,y

)A

2is

singularbutB

(x,y

)A

2is

smooth,althoughV(x

2,y)=

V(x,y).

2Recallthetrick:V(f

)=

V(z

0f,z

1f,...,z

nf).

Sowecanget

f jofequaldegree.

3Thesewillbethelocalmodelsforgen

eralschem

es.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

59

whereweneedto

explain

1what

f(℘

)is,insidethefraction

fieldof

theintegral

dom

ainA/℘

:

A→

A/℘

�→K(℘

)

f�→

f�→

f(℘

)=

f 1∈K(℘

)

Remark

.It

isnot

actually

afunction:thetarget

K(℘

)is

afieldwhichdep

endson

thegiven

℘!

Example.A

=Z.

Spec

A=

{(0)}∪

{(p):pprime}.

K(0)=

Frac(Z

/0)=

Q,K(p)=

Frac(Z

/p)=

Z/p.

Con

sider

f=

4.f((0))=

4∈Q.

f((3))=

(4mod3)

=1∈Z/

3.f((2))=

0∈Z/

2,since

4∈(2).

Exercise.

f(℘

)=

0⇔

f∈℘

When

℘=

mis

amax

imal

ideal,A/m

isalread

yafield,so

K(m

)=

A/m

,thus:

f(m

)=

(fmodulo

m)∈A/m

.

Example.A

=k[x]correspon

dsto

theaffi

nevarietySpecm

A=

A1.Con

sider

apolynom

ialf(x)∈

A,an

dtheidealm

=(x−

2).Then

f(m

)=

(fmodx−

2)∈k[x]/(x−

2)correspon

dsto

thevalue

f(2)∈kvia

theidentification

K(m

)=

k[x]/(x−

2)∼ =

k,x�→

2.

Remark

.For

anaffi

nevarietyX⊂

An,so

takingA

=k[X

],themax

imal

ideals

ma=�x

1−

a1,...,x

n−

an�correspon

dto

thepoints

a∈

X⊂

An,an

dthe“function”f

atm

ajust

means

reducingfmodulo

ma.Butk[X

]/m

a∼ =

kvia

theevaluationmap

g(x)�→

g(a),

sowegetan

actual

functionon

themax

imal

ideals: f:Specm

A→

k,m

a�→

f(m

a)=

f(a)

inother

words,thisisthepolynom

ialfunctionV(I)→

kdefined

bythepolynom

ialf∈k[x

1,...,x

n]/I,

sothevaluef(a)is

obtained

byplugg

ingin

thevalues

xi=

aiin

f.

Example.A

=k[X

]=

R/I

foran

affinevarietyX⊂

An,whereR

=k[x

1,...,x

n].

For

f∈A,weob

tain

f:X

=Specm

A→

kas

remarked

above,

andthis

isthepolynom

ialfunction

obtained

via

k[X

]∼ =

Hom

(X,k).

Exam

ple:xi∈A

defines

thei-th

coordinatefunctionxi:X→

k.

For

℘⊂

Aaprimeideal,weob

tain

asubvarietyY

=V(℘

)⊂

X,an

dyo

ushou

ldthinkof

f(℘

)as

the

restrictionto

Yof

thepolynom

ialfunctionf:X→

k,so

f(℘

):Y→

k.Indeed,letA

=k[Y

]=

A/℘

andf=

(fmod℘)∈

A.Then

therestrictionf| Y

:Y→

kequalsthefunctionf:Specm

A→

kwhichcorrespon

dsto

the“function”f(℘

)argu

ingas

before.2

Remark

.Thevalues

f∈K(℘

)“d

etermine”

theim

ageof

fin

anyfieldFunder

anyhom

omorphism

ϕ:A→

F.Indeed(assumingϕisnot

thezero

map

),℘=

kerϕisaprimeidealsince

A/℘

�→Fisan

integral

dom

ain,so

ϕfactorises

asA→

A/℘

�→K(℘

)�→

Fsince

K(℘

)isthesm

allest

fieldcontaining

A/℘

,so

ϕ(f)is

determined

byf∈K(℘

)an

dthefieldextension

K(℘

)�→

F.

15.2.

THE

ZARIS

KITOPOLOGY

ON

Spec

Motivation.Wewan

tthefollow

ingto

beabasic

closed

setin

Spec

A,foreach

f∈A:

V(f)=

{℘∈Spec

A:f(℘

)=

0}=

{℘∈Spec

A:℘�f}.

Thus,

wedefinetheZarisk

ito

pologyon

Spec

Aan

dSpecm

Abydeclaringas

closed

sets:

V(I)=

{℘∈Spec

A:℘⊃

I}⊂

Spec

AV(I)=

{m∈Specm

A:m⊃

I}⊂

Specm

A

1herewewrite

fto

meanfmodulo

℘,so

thecosetf+

℘∈A/℘.

2iden

tifyingK(m

)∼ =

kvia

evaluation,foranymaxidealm

⊂A,i.e.

amaxidealm

⊂A

whichcontains℘.

60

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

foran

yidealI⊂

A.Noticeallf∈Iwillvanishin

A/℘

for℘∈V(I),equivalentlyf(℘

)=

0∈K(℘

).Moregenerally,forasubsetS⊂

A,wewrite

V(S

)to

meanV(�S�).

Aga

inwehavebasicop

ensets

Df=

{℘:f(℘

)�=

0}=

{℘:f/∈℘}⊂

Spec

AD

f=

{m:f(m

)�=

0}=

{m:f/∈m}⊂

Specm

A

foreach

f∈A,whichdefineabasis

forthetopolog

y.Exercise.Spec

A\V

(℘)=

{primeideals

notcontaining℘}=∪ f

∈℘D

f.

Theelem

ents

ofSpecm

Aarecalled

theclosed

points

1of

Spec

A.A

pointofatopolog

ical

spaceis

called

genericifit

isdense.2

Soagenericpoint℘∈Spec

Ais

apointsatisfyingV(℘

)=

Spec

A.

Examples.

1.For

A=

R=

k[x

1,...,x

n],then

Specm

A≡

knvia

ma=�x

1−

a1,...,x

n−

an�

1:1

←→

a

V(I)=

{ma:m

a⊃

I}⊂

Specm

A1:1

←→

{a∈kn:{a

}=

Vclassical(m

a)⊂

Vclassical(I)}

=Vclassical(I)⊂

An.

SoSpecm

R∼ =

Anarehom

eomorphic,an

dO(A

m)=

R.

Spec

Rcontainsallirreducible

subvarietiesY

=V(℘

)⊂

An:

Spec

A1:1

←→

Specm

A∪{p

rimeideals

℘⊂

Rwhicharenot

maxim

al}

1:1

←→

An∪{irred

subvars

Y⊂

Anwhicharenot

points}

1:1

←→

{allirredsubvars

Y⊂

An}

This

isunlike

theEuclideantopology

(for

k=

Ror

C)wheretheon

lynon

-empty

irreducible

sets

aresinglepoints,so

wedon’t

noticeinteresting“p

oints”ap

artfrom

An.

2.For

X⊂

Anaff

.var.,letI=

I(X),

sok[X

]=

R/I

whereR

=k[x

1,...,x

n].

X∼ =

Specm

(R/I

)are

homeomorphic,an

dO(X

)=

k[X

]=

R/I

a=

V(m

a)=

{f∈k[X

]:f(a)=

0}wherem

a=

ma+

I⊂

R/I

=k[X

].

3.For

A=

Z,Spec

Z=

{theclosed

points

{p}forpprime}∪{thegeneric

point(0)}

Note:

(p)is

max

imal,V(p)=

{(p)},and(0)is

generic

since

V((0))=

Spec

Zas(0)⊂

(p)forallp.

4.For

A=

k[x],

Spec

k[x]=

{(x−

a):a∈k}∪

{(0)}↔

An∪(generic

point).

Note:

0is

generic

asV((0))=

Spec

k[x]as

(0)⊂�x−

a�.

5.For

A=

k[x]/x2,

Specm

A=

Spec

A=

{(x)}

=on

epoint

O(Spec

A)=

A=

k[x]/x2

A�f=

a+

bx:Spec

A→

k,(x)�→

a=

(fmod(x)∈K((x))∼ =

A/x∼ =

k).

Sowehaveatw

o-dim

ensional

space

offunctions(twoparameters:

a,b∈k),

even

thoughwhen

we

consider

thevalues

ofthefunctionsweon

lyseeoneparam

eter

worth

offunctions.

Sotheringof

functionsO(Spec

A)also

remem

berstangential

inform

ation:3

thetangentvector

∂ ∂x

� � x=0,namelythe

operator

actingon

functionsasfollow

s,∂ ∂x

� � x=0f=

b.

1“closed”because

V(m

)=

{m}.

2i.e.

itsclosure

iseverything.

3More

gen

erally:aclosedpointm

∈Spec

Acorrespondsto

ak-alg

hom

A→

k(w

ithkernelm),whichcorrespondsto

amap{p

oint}

=Spec

k→

Spec

A,andthesameholdsifwereplace

k∼ =

k[x]/x.W

hereasamapSpec

k[x]/x2→

Spec

Acorrespondsto

ak-alg

hom

A→

k[x]/x2whichdefi

nes

aclosedpointtogether

witha“tangen

tvector”.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

61

Whyis

this

areason

able

definition?

The“ringedspace”

Spec

Ais

not

thesameas

Spec

k[x]/x:it

remem

bersthat

itaroseas

adeformationof

Spec

B=

{twopoints

α,β∈A1}as

α,β→

0where

B=

k[x]/(x−

α)(x−

β)∼ =

k⊕

k

whereα,β∈karenon

-zerodistinct

deformationparam

eters,

andthesecondisom

orphism

1is

eval-

uationat

α,β

respectively.Sof=

a+

bx�→

(a+

bα)⊕

(a+

bβ),

sowecanindep

endentlypickthe

twovalues

offat

thetw

opoints

{α,β

}=

Specm

B,givingatw

o-dim

ension

alfamilyof

functions.

Thederivative∂xf| x=

0=

b=

lim

f(α

)−f(β

)α−β

asweletα,β

convergeto

0.

Exercise.

Anaffi

nevarietyX⊂

Anis

irreducible

ifan

don

lyif

Spec

k[X

]has

ageneric

point.

Exercise.Know

ingthevalueof

f∈A

atageneric

pointdetermines

thevalueof

fat

allpoints.

Example.f∈Z,

then

f((0))=

f 1∈K((0))=

Qdetermines

f((p))∈K((p))

=Z/

p(red

uce

modp).

15.3.

MORPHIS

MS

BETW

EEN

Specs

Apartfrom

themotivationcomingfrom

deformationtheory,an

other

convincingreason

forprefer-

ringSpec

Aover

Specm

A,is

that

wegetacatego

ryof

affineschem

esbecau

sewehavemorphisms:

Definition.Themorphisms2

Hom

(Spec

A,Spec

B)=

{ϕ∗:Spec

A→

Spec

Binducedby

ringhomsϕ:B→

A}

whereϕ∗ (℘)=

ϕ−1(℘

)⊂

A,foranyprimeideal℘⊂

B.

Exercise.Thepreim

ageof

aprimeidealunder

aringhom

isalwaysprime.

Warn

ing.Thisexercise

failsformax

imal

ideals.Example.For

theinclusion

ϕ:Z→

Q,ϕ−1(0)=

(0)⊂

Zis

not

max

imal

even

thou

gh(0)⊂

Qis

max

imal.Sim

ilarly,fortheinclusion

ϕ:k[x]→

k(x)=

Frack[x],ϕ−1(0)=

(0)is

not

max

imal

since

(0)⊂

(x).

Remark

.Wedid

not

noticethis

issuewhen

dealingwithaffi

nevarieties,

whichwas

thestudyof

Specm

off.g.

reducedk-algs,

becau

sein

that

case

morphismsexistbetweentheSpecm.

Exercise.3

Moregenerally:foran

yf.g.

k-algebrasA,B

,an

dϕ

:A→

Bak-alg

hom

,prove

that

Specm

A←

Specm

B:ϕ∗is

well-defined

,nam

elyϕ∗ (m)=

ϕ−1(m

)is

alwaysmax

imal.

15.4.

LOCALIS

ATIO

N:RESTRIC

TIN

GTO

OPEN

SETS

Rem

ark.Wealread

yencounteredlocalisation

inSection

10,so

wewillbebrief.

Question:Whatare

thefunctionsonabasicopenset?

RecallD

f=

{℘:f(℘

)�=

0}⊂

Spec

A,so

weshou

ldallow

thefunction

1 fon

Df.Thuswe“d

efine”

O(D

f)=

Af=

localisation

ofA

atf

whichwillensure

that

Spec

Af∼ =

Df.W

hen

Ais

anintegral

dom

ain,

Af=

{a fm∈FracA

:a∈A,m

∈N}.

Example.

A1\0=

Dx⊂

A1,an

dweview

A1\0∼ =

V(xy−

1)⊂

A2as

anaffi

nevarietyvia

t↔

(t,t

−1).

Bydefinition,k[A

1\0

]=

k[x,y]/(xy−

1)∼ =

k[x,x

−1]∼ =

k[x] x

isthelocalisation

atx.

Question:Whatare

thefunctionsonageneralopensetU⊂

Spec

A?

1This

istheChineseRem

ainder

Theorem.Explicitly:1=

x−β

α−β+

α−x

α−β,so

k[x]/(x

−α)(x−

β)∼ =

k[x]/(x

−α)⊕

k[x]/(x

−β)via

g�→

x−β

α−βg⊕

α−x

α−βg.Finally,

k[x]/(x

−γ)∼ =

kvia

f�→

f(γ).

2Categorically:Spec

isafunctorRings→

Topopfrom

thecategory

ofrings(commutative)

totheopposite

ofthe

category

oftopologicalspacesandcontinuousmaps.

3Hints.k⊂

A/ϕ

−1(m

)⊂

B/m

∼ =somefield.W

hen

kisalgeb

raicallyclosed,weknow

B/m

∼ =k,so

weare

done.

For

gen

eralk,wealreadyknow

ϕ−1(m

)isprimeso

A/ϕ

−1(m

)isadomain.Finallyuse:(1)f.g.k-alg

+field⇒

algeb

raic/k

⇒finitefieldextension/k;anduse

(2)domain

+algeb

raic/k⇒

fieldextensionofk.

62

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Weknow

U=∪D

fisaunionof

basic

open

sets.Loosely,1the“functions”

inO(U

),called

sections

s U,aredefined

asthefamilyof

functionss f∈O(D

f)=

Afwhichagreeontheoverlaps

s f| D

g=

s g| D

f∈O(D

f∩D

g)=

O(D

fg)=

Afg.

Remark

.Not

allop

ensets

arebasic

open

sets.For

X=

V(xw−yz)⊂

A4,theunionD

y∪D

w⊂

Xisnot

basican

dO(D

y∪D

w)does

not

arise

asalocalisation

ofk[X

].Indeedf=

x y=

z w∈O(D

y∪D

w)

cannot

bewritten

asafraction

whichis

simultan

eouslydefined

onbothD

y,D

w.

Question:Whatare

thegerm

soffunctions?

Recallthegerm

ofafunctionnearapointa∈X

ofatopologicalspace,

meansafunctionU→

kdefined

onaneigh

bou

rhoodU⊂

Xof

a,an

dweidentify

twosuch

functionsU→

k,U

� →kifthey

agreeon

asm

allerneigh

bourhoodof

a(sothegerm

isan

equivalence

classoffunctions).Write

O℘

forthegerm

sof

functionsat

℘∈Spec

A,this

iscalled

thestalk

ofO

at℘.It

turnsoutthat2

O℘=

A℘=

localisation

ofA

atA\℘

i.e.

welocalise

atallf

/∈℘,byallowing

1 fto

beafunctionwhenever

fdoes

notvanishat℘.We

explained

this

ingreater

detailin

Sec.15.10.

When

Ais

anintegral

dom

ain,3

A℘=

{a b∈FracA

:b/∈℘(i.e.b(℘)�=

0)}=

� f/∈℘

Af⊂

FracA.

Example.Let

A=

k[x,y]/(xy).

Theaffi

nevarietyX

=Specm(A

)∼ =

V(xy)⊂

A2consistsofthe

x-axis

andy-axis.Thex-axis

isthevanishinglocusof

theprimeideal℘=

(y).

Thefunctionf=

xdoes

not

vanishat

℘,since

x�=

0∈

(k[x,y]/(xy))/℘

∼ =k[x],

so1 x∈

A℘is

agerm

ofafunction

onSpec(A

)defined

near℘.This

shou

ldnot

beconfusedwithgermsof

functionsdefined

nearthe

closure

V(℘

),i.e.

germ

sof

functionsdefined

nearthex-axis.Indeed,thegermsoffunctions0and

yaredifferenton

anyneigh

bourhood4of

V(℘

).How

ever,in

thelocalisationA

℘thefunctions0and

yareidentified

,becau

sexy=

0forces

0=

1 x·x

y=

y.Also,

1 xis

not

awell-defined

functiononall

ofV(℘

),as

itis

not

defined

atx=

0,it

ison

lydefined

ontheop

ensubsetV(℘

)∩D

fofV(℘

).So

functionsin

A℘aredefined

nearthegeneric

point℘

ofV(℘

)butneednot

extendto

afunctionon

allof

V(℘

).

TheringO

℘=

A℘is

alocalring,meaningithas

precisely

onemax

imalideal,namely

m℘=

A℘·℘⊂

A℘.

SoSpecmA

℘=

onepoint,

nam

ely

m℘,which

you

shou

ldthink

ofas“representing℘”because

Specm

A℘→

Spec

Amap

sthepointto

℘.

Exercise.Show

that,indeed,at

thealgebra

level

A℘←

Amap

sm

℘←� ℘.

Thevalueof

f∈A

at℘lives

intheresiduefield

5ofthatlocalring

f(℘

)∈O

℘/m

℘=

A℘/m

℘∼ =

K(℘

).

Exercise.Prove

thatA

℘/m

℘∼ =

FracA/℘

=K(℘

).Example.Con

sider

A=

Z.Either

℘=

(p)forprimep,or

℘=

(0):

1Form

ally:O(U

)=

lim ←−

O(D

f)is

theinverse

limit

forD

f⊂

U,taken

over

therestrictionmapsO(D

f� )

←O(D

f)

forD

� f⊂

Df⊂

U(thesemapsare

thelocalisationmapsA

� f←

Af).

This

meansprecisely

thatforeach

basicopen

set

insideU

wehav

eafunction,andthesefunctionsare

compatible

witheach

other

under

restrictionsto

overlaps.

2Form

ally:O

℘=

lim −→

O(U

)is

thedirectlimit

foropen

subsets

Ucontaining℘,taken

over

therestrictionmaps

O(U

)→

O(U

� )forU

⊃U

��

℘.

So

wehavesectionss U

∈O(U

)and

weiden

tify

sectionss U

∼s V

when

ever

s U| W

=s V

| Wforsomeopen

℘∈W

⊂U

∩V.

3This

requires

care:

a b=

c d⇔

ad=

bc(thedefi

nitionofFrac),so

theremay

bemanyexpressionsforthesame

elem

ent.

InA

℘wewantsomeexpressionto

haveaden

ominatorwhichdoes

notvanishat℘.Example:℘=

(2)⊂

A=

Z,then

2 3∈A

(2)⊂

FracZ=

Qsince

3/∈(2),

whereas

4 6failsthecondition6/∈(2)even

thoughit

equals

2 3.

4such

neighbourhoodscontain

allbutfinitelymanypoints

ofthey-axis,so

0�=

yasfunctions.

5thefieldobtained

byquotientingalocalringbyitsuniquemaxim

alideal.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

63

Op=

Z (p)=

{a b∈Q

:p�|b

},m

p=

p·Z

(p)=

{a b∈Q

:p|a,p�|b

},an

d1K(p)=

Op/m

p∼ =

F p=

Z/p.

O0=

Z (0)=

FracZ=

Q,

m0=

(0)⊂

Q,

and

K(0)=

O0/m

0∼ =

Q.

15.5.

SHEAVES

Given

atopolog

ical

spaceX,ash

eafS

ofringson

Xmeansan

association2

(open

subsetU⊂

X)�→

(ringS(

U)).

Elements

ofS(

U)arecalled

sectionsoverU.

Werequirethat

forallop

enU⊃

Vthereis

arestriction,nam

elyaringhom

omorphism S(U)→

S(V),s�→

s| Vsatisfyingtw

oob

viousrequirem

ents:S(

U)→

S(U)istheidentity

map

,an

d“restrictingtw

iceisthe

sameas

restrictingon

ce”.

3Wealso

requiretw

olocal-to

-globalconditions:

(1).

“Twosectionsequal

ifthey

equal

locally”.

4

(2).

“You

canbuildglob

alsectionsbydefininglocalsectionswhichag

reeon

overlaps”.5

Withou

tthelocal-to-global

conditions,

itwou

ldbecalled

apre

sheaf.

Given

asheaf(orpresheaf)

Son

X,thestalk

S pat

p∈X

istheringof

germ

sof

sectionsat

p.6

EXAM

PLES.

1.X

=Spec

A,an

dS(

U)=

O(U

)as

inSection

15.4.For

exam

ple,O(D

f)=

Af,an

dD

f⊃

Dfg

determines

therestrictionwhich“localisesfurther”,

Af→

Afg,

a fm�→

agm

(fg)m

.

2.Sheafof

continuou

sfunctions:

S(U)=

C(U

,k)=(con

tinuou

sfunctionsU→

k).

3.Sheafof

sectionsof

amap

7π:E→

B:takeS(

U)=

sections8s:U→

π−1(U

)⊂

E.

4.Skyscraper

sheafat

p∈X

fortheringA:S(

U)=

Aifp∈U,an

dS(

U)=

0ifp/∈U.Exercise:

show

thestalksareS p

=A

andS q

=0forq�=

p.

Non-example.

Thepresheafof

constan

tfunctions(orconstan

tpresheaf):S(

U)=

Aforop

enU�=∅,

andS(∅)

=0,

isnot

asheafforA

=Z/

2an

dX

={p

,q}withthediscretetopolog

y.Indeed,

takes| {

p}(p)=

0,s| {

q}(q)

=1:

theselocalsectionsdonot

glob

aliseto

aglob

alconstan

tfunction

s:X→

Acontrad

icting(2).

15.6.

SHEAFIF

ICATIO

N

Onecanalwayssh

eafify

apresheafP

toob

tain

asheafS

byartificially

imposinglocal-to-global:

S(U)=

{s=

(sp)∈

� p∈U

P p:∀p∈Uthereis

anop

enp∈V⊂

Uan

ds V∈P(V

)withs V

| p=

s p}.

Noticehow

weim

posethat

locallyallgerm

sarisefrom

restrictingalocalsection.Wenow

explain

this

inmoredetail.

For

anysheafS

onatopolog

ical

spaceX,thereis

anob

viousrestrictionS(

U)→

S x,f�→

f xto

stalks,

foreach

x∈U.Beingasheafensuresthelocal-to-global

property:

Iff x

=g x

atallx∈U,then

f=

g∈S(

U)

1via

a b↔

ab−

1modp.

2Categorically:apre

sheafisafunctorOpen

op

X→

RingswheretheobjectsofOpen

Xare

theopen

sets

andtheonly

morphismsallow

edare

inclusionmaps;

andamorp

hism

ofpre

sheavesis

anaturaltransform

ationofsuch

functors.

Forsh

eavesweim

pose

theabovelocal-to-globalconditionsforsections,

butnoextraconditiononmorphs.

3ForU

⊃V

⊃W

,S(

U)→

S(V)→

S(W

)agrees

withS(

U)→

S(W

).4Forf,g

∈S(

U),

U=

∪Ui,f| U

i=

g| U

iforalli⇒

f=

g.

5U

=∪U

i,s i

∈S(

Ui),

s i| U

j=

s j| U

i∈S(

Ui∩U

j)⇒

thereis

somes∈S(

U)withs| U

i=

s i(andsis

uniqueby(1)).

6A

germ

atpis

anequivalence

class

ofsections.

Itis

determined

bysomesections U

∈S(

U),

foranopen

p∈

U.

Weiden

tify

twosectionss U

∼s V

ifs U

| W=

s V| W

,foranopen

p∈W

⊂U

∩V.

7forexample,avectorbundle

Eover

amanifold

B.

8here“section”meansit

iscompatible

withtheprojectionπ,so

π(s(u))

=u.Soateach

uin

thebase,thesection

spicksanelem

entin

thefibre

s−1(u)over

u.

64

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

becau

sef x

=g x

meansthatf,g

equal

onasm

allneigh

bou

rhoodof

x.Sofiscompletely

determined

bythedata(f

x) x

∈U.Not

alldata

(fx) x

∈Uarisesin

thisway,thedata

has

tobecompatible:locally,

onsomeop

enV

arou

ndan

ygiven

point,

thef x

arisefrom

restrictingsomeF∈

S(V).

SoS(

U)

consistsof

compatible

families(f

x) x

∈Uan

dtherestrictionmapforop

enV⊂

Uextractssubfamilies:

S(U)→

S(V),

(fx) x

∈U�→

(fx) x

∈V.

Sothesheafification

ofapre-sheafP

is

S(U)=

{com

patible

familiesof

germ

s{s

x} x

∈Uwheres x∈P x

}.

This

isavery

usefultrick,wewilluse

itin

Section

s15

.8and15

.12.

Exercise.Show

thatthesheafificationof

thepre-sheafof

constantk-valued

functionson

atopolog-

ical

spaceX

isthesheafof

locallyconstantfunctions(i.e.constan

toneach

connectedcompon

ent).

Example.For

Xanaffinevariety,

letP(U

)=

{functionsf:U→

k:f=

g hsomeg,h∈k[X

],with

h(u)�=

0forallp∈U}.

This

isapresheaf,whose

sheafification

defines

O(U

),seeSec.10.2.

15.7.

MORPHIS

MS

OF

SHEAVES

Amorphism

ψ:S 1→

S 2of

sheavesover

Xmeansan

association

(open

subsetU⊂

X)�→

(ringhom

ψU:S 1

(U)→

S 2(U

))

whichis

compatible

withrestrictionmap

s.1

Exercise.Show

that

this

inducesaringhom

onstalks:

ψp:S 1

,p→

S 2,p.

Exercise.ψ:S 1→

S 2is

anisom

orphism⇔

itis

anisomorphism

onstalks(allψpare

isos).

Exercise.If

ψ:X→

Yis

acontinuou

smap

oftopolog

icalspaces,an

dS

isasheafonX,then

ψinducesasheafon

Ycalled

directim

agesh

eafψ∗S

,defined

by

(ψ∗S

)(U)=

S(ψ−1(U

)).

15.8.

RIN

GED

SPACES

Aringed

space(X

,S)is

atopolog

icalspaceX

together

withasheafof

rings,S.

Example.Theaffi

neschem

e(Spec

A,O

)is

aringedspace.

Amorp

hism

ofringed

spaces(X

1,S

1)→

(X2,S

2)meansacontinuou

smap

f:X

1→

X2

together

withamorphism

ofsheaves

over

X2,

f∗:f ∗S 1←

S 2so

explicitlyf∗ (U)map

sS 1

(f−1(U

))←

S 2(U

)forU⊂

X2,an

don

stalksf∗ p:(S

1) p←

(S2) f

(p).

Example.ϕ:A→

Baringhom⇒

f=

ϕ∗:Spec

A←

Spec

Ban

d

ψ=

f∗:O

A→

f ∗O

B

soψU

:O

A(U

)→

OB((ϕ∗ )

−1(U

)).NoticeψSpec

A:A→

Bis

just

ϕ,on

basic

open

sets

ψis

the

relevantlocalisationofϕ,an

don

stalksweget

thelocalisedmap

2ψϕ∗ ℘

:A

ϕ∗ ℘→

B℘for℘∈Spec

B.

Alocallyringed

spacemeanswead

ditionally

requirethestalksS p

tobelocalrings,so

they

have

auniquemax

imal

idealm

p⊂

S p.A

morphism

oflocallyringedspaces

isad

ditionally

required

topreservemax

imal

ideals,i.e.

f∗:m

p←

mf(p)(butthis

neednot

bebijective).

Example.3

Show

thatSpec

A←

Spec

Bis

amorphof

locallyringedspaces.

1ForV

⊂U

⊂X,acommutativediagram

relatesψ

U,ψ

VwiththerestrictionmapsresU V

,so:resU V

◦ψU=

ψV◦r

esU V.

2Explicitly:

a a��→

ϕ(a

)ϕ(a

� )wherea�∈A\ϕ

−1(℘

)(soϕ(a

� )∈B

\℘).

3Youneedto

checkthatϕ

∗ ℘·A

ϕ∗℘mapsinto

℘·B

℘via

f ϕ∗℘.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

65

15.9.

SCHEM

ES

Anaffineschemeis

alocallyringedspaceisom

orphic

to(Spec

A,O

)forsomeringA.

Ascheme(X

,S)is

alocallyringedspacewhichis

locallyan

affineschem

e.1

Wenow

describetheaffi

neschem

eX

=Spec(A

)as

alocallyringedspace(X

,OX)(L

emma10

.4willprovethat

thestalksO

X,℘

ofthestructure

sheafarelocalrings).

Bydefinition,

{ringhom

sϕ:A→

B}

1:1

←→

{morphismsϕ∗:Spec(A

)←

Spec(B

)}whereϕ∗ ℘

=ϕ−1(℘

).Onecancheckthat

aringhom

A→

Binducesalocalringhom

onstalks

OA,ϕ

∗ ℘→

OB,℘

(Equation(10.2)).

Wesketched

onedefinitionof

thestructure

sheafO

=O

Xon

X=

Spec(A

)in

Section

15.4.We

now

explain

anequivalentdefinitionusingsheafification

(Sec.15.6).For

U⊂

Xan

open

subset,O(U

)consistsof

compatible

familiesof

elem

ents

{f℘∈O

℘} ℘

∈U.RecallO

℘∼ =

A℘isthelocalisation

ofA

attheprimeideal℘,so

weform

ally

invert

allelem

ents

inA\℘

.Soequivalently,

thesearefunctions

f:U→

� ℘∈U

A℘,

℘�→

f ℘.

Com

patible

means:

foran

yq∈

U,thereis

abasic

open

setq∈

Dg⊂

U(sog

/∈q)

andsome

F∈

Ag=

O(D

g)such

that

thef ℘

aretherestrictionsof

F(m

eaning,

Ag→

A℘,F�→

f ℘forall

℘∈D

g).

Therestrictionhom

sforop

enV⊂

U,aresimply

defined

bytakingsubfamilies:

O(U

)→

O(V

),(f

℘) ℘

∈U�→

(f℘) ℘

∈V.

The“value”

f(℘

)∈K(℘

)of

f(Sec.15.1)

istheim

ageof

f ℘via

thenaturalmap

O℘→

O℘/m

℘∼ =

K(℘

).Exercise.After

read

ingSection

11,checkthat

theab

oveis

consistentwiththeexplicitdefinition

ofO

X,O

X,pforaquasi-projectivevarietyX,carriedou

tin

Section

s11

.3an

d11

.5.

15.10.

LOCALIS

ATIO

NREVIS

ITED:affinevarieties

For

Xan

affinevarietyan

d℘⊂

k[X

]aprimeideal,thestalkO

X,℘

means“g

ermsof

functionson

Spec

k[X

]defined

near℘”,

whichwenow

explain.It

sufficesto

consider

basic

neigh

bou

rhoodsD

f,

forf∈

k[X

]withf�=

0∈

k[X

]/℘.Then

OX,℘

consistsof

pairs

(Df,F

)withf�=

0∈

k[X

]/℘,U

open,F

:U→

kregu

lar,

andidentifying(D

f,F

)∼

(Dg,G

)⇔

F| D

h=

G| D

hon

anop

enD

hwith

Dh⊂

Df∩D

gan

dh�=

0∈k[X

]/℘.Algebraically

this

isthedirectlimit

OX,℘

=lim −→ ℘∈D

f

OX(D

f)=

lim −→ f/∈℘

k[X

] f

over

allbasic

open

neigh

bou

rhoodsD

fof

℘.It

iseasy

toverify

algebraically

that

lim −→ f/∈℘

k[X

] f∼ =

k[X

] ℘,

indeedweareform

ally

invertingallelem

ents

that

donot

belon

gto

℘.Thisisthean

alog

ueof

Lem

ma

10.5,whichshow

edO

X,m

p∼ =

k[X

] mp,nam

elythecase

when

℘is

amax

imal

ideal(correspon

dingto

ageom

etricpointin

X).

Recallthat

analog

ouslyto

(10.3),wegetafieldextension

ofk:

K(℘

)=

Frac(A/℘

).

Wethinkof

theuniqueprimeideal(0)of

thisfieldas

correspon

dingto

thepoint℘∈Spec(A

)=

X:

theringhom

ϕ:A→

A/℘

�→K(℘

)correspon

dsto

thepoint-inclusion

ϕ∗:Spec(K

(℘))

�→Spec(A

),(0)�→

℘.In

Section

15.1

weusedK(℘

)to

definethe“value”

of“functions”

f∈A,bysayingthat

f(℘

)=

f∈A/℘

�→K(℘

).

Exercise.f(℘

)�=

0∈K(℘

)⇔

f/∈℘⇔

℘∈D

f.

Example.A

=Z,

p∈Z

prime,

K(p)=

Z/p=

F p.For

f∈A,f(p)=

(fmodp)∈F p

.

Example.Con

sider

X=

(x-axis)∪(y-axis),

k[X

]=

k[x,y]/(xy)an

d℘=

(y),

soV(℘

)=

(x-axis).

Then

k[X

] ℘∼ =

k(x),

indeedweinvert

everythingou

tsideof

(y),

wealread

ysaw

that

invertingx

1X

=∪U

i,U

i∼ =

Spec

AisomeringsA

i,S|

Ui∼ =

OA

i(thestructure

sheafforA

i).

66

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

gives

k[X

] x∼ =

k[x,x

−1],butnow

wealso

invertan

ypolynom

ialin

xso

weget

k(x)=

Frac(k[x]).

Oneshou

ldnot

interpret“germsnear℘”as

meaning“germsnearV(℘

)”,since

thefunctionsyand

0arenot

equal

onan

yneigh

bou

rhoodof

V(℘

)=

(x-axis).

Inparticular,

1 xis

notwell-defined

onall

ofV(℘

).Thecorrectinterpretationof

k[X

] ℘is:ration

alfunctionsdefined

onanon-empty

(dense)

open

subsetof

V(℘

).

Exercise.For

Xan

irreducible

affinevariety,

i.e.

A=

k[X

]an

integral

domain,show

that

OX(U

)=

�

Df⊂U

O(D

f)=

�

Df⊂U

k[X

] f⊂

Frac(k[X

])=

k(X

),

usingthat

theD

fareabasis

forthetopology,

andthat

afunctionis

regu

lariff

itis

locallyregular.

When

Xisnot

irreducible,then

wecannot

definethefraction

fieldofA

=k[X

]in

whichto

take

the

aboveintersection

(k[X

] fandk[X

] gdon

’tlivein

alarger

common

ringwherewecanintersect).So

instead,algebraically,on

ehas

totaketheinverselimit:

OX(U

)=

lim ←−

Df⊂U

OX(D

f)=

lim ←−

Df⊂U

k[X

] f

taken

over

allrestrictionmap

sk[X

] f→

k[X

] gwhereD

g⊂

Df⊂

U.Explicitly,

theseare

familiesof

functionsFf∈k[X

] fwhicharecompatible

inthesense

that

Ff| D

g=

Fg(w

hereFf| D

gis

theim

age

ofFfvia

thenaturalmap

k[X

] f→

k[X

] g).

This

definitionmakessense

also

foran

yq.p.v.X.

Finally,theFACT

from

Section

10.1,im

plies

a1:1correspon

den

ce

{irred

ucible

subvarietiesY⊂

XcontainingV(℘

)}1:1

←→

{primeideals

ofk[X

] ℘}.

15.11.

WORKED

EXAM

PLE:THE

SCHEM

ESpec

Z[x]

Som

ebasic

algebra

implies

that

Spec

Z[x]

={(0)}∪

{(p):p∈Zprime}∪

∪{(f):f∈Z[x]non

-con

stan

tirreducible}∪

∪{(p,f

):p∈Zprime,f∈Z[x]irreducible

modp}

Con

sider

theprojectionπ:Spec

Z[x]→

Spec

Zinducedbytheinclusion

Z→

Z[x].

Exercise.Explicitlyπ(℘

)=(allconstan

tpolynom

ials

in℘).

Below

isan

imaginativegeom

etricpicture

1ofπ.

ThebaseSpec

Zhas

primeideals

(p)an

d(0).

Since

(0)is

ageneric

pointit

isdrawnbyasquiggly

symbol

toremindourselvesthat(0)is

dense

inSpec

Z.Thefibre

over

(p)is

π−1((p))

=V((p)),i.e.

primeidealsin

Z[x]whichcontain

p,an

dπ−1((0))consistsof

allother

primeideals,i.e.

those

which

donot

contain

anon-zeroconstan

tpolynom

ial.

Thefibre

π−1((p))

containsthegeneric

point(p),

andwedraw

itbyasquiggly

symbol

becau

seit

isdense

inV((p)).Thepoint(0)∈

Spec

Z[x]is

generic,becau

seeveryidealin

Z[x]contains0,

soweuse

alargesquigglysymbol.

When

look

ingforgenerators

ofan

idealin

π−1(p)(apartfrom

p),

wemay

reduce

thepolynomial

coeffi

cients

modp.Exam

ple:for(5,x

+j)∈π−1(5)weon

lyneedto

consider

thecasesj=

0,1,...,4.

1anadaptationofafamouspicture

byDav

idMumford,TheRed

BookofVarietiesandSchem

es.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

67

��

��

��

��

��

��

��

��

��

��

��

��

��������������������������

��������������������������

��

�� �� ��

�� ����

��

��

��

����

����

����

��

��

��

�� ������

��

��

��

��

��

��

��

��

��

����

����

��

��

��

�� ����

��

��

��

��

��

��

��

(3,x

)(7,x

)

(5)

(3)

(7)

(2)

(3,x

+1)

(2,x

+1)

(2,x

)

V((3))

V((5))

V((7))

(3,x

+2)

(5,x

)

(5,x

+1)

(5,x

+2)

(5,x

+3)

(5,x

+4)

V((2))

Spec

Z[x]

Spec

Z(2)

(7)

π

(0)

(x2+1)

(x)

(3)

(5)

V((x2+1))

(0)

(2,x

2+x+1)

···

···

···

···

···

Exercise.π−1(p)=

V((p))∼ =

Spec

F p[x]arehom

eomorphic,whereF p

=Z/

p.

Bydefinition,(x

2+

1)is

dense

(hen

ceageneric

point)

inV((x2+

1)),

sowedraw

itbyasquiggly

symbol

lyingon

the“curve”

V((x2+1)).

This“curve”

containsthepoints

(2,x

+1),(5,x+2),(5,x+3),

etc.,that

is:weclaim

(x2+

1)is

contained

inthoseideals.

Example.Z[x]/(5,x

+2)∼ =

F 5[x]/(x

+2)

byfirstquotientingby(5).

This

isois

given

by“reduce

mod

5”.

Now

x2+

1is

divisible

by(x

+2)

mod

5,becau

se−2is

aroot

ofx2+

1mod

5.So

x2+1=

0∈F 5

[x]/(x

+2)∼ =

Z[x]/(5,x

+2),so

(x2+1)⊂

(5,x

+2).Therootsof

x2+1mod5are

precisely

2,3,

whichexplainsthepoints

(5,x

+2),(5,x

+3)

onthe“curve”

V((x2+

1)).

Remark

.Noticethepoints

onV((x2+1))encodethesquarerootsof−1over

F p.A

classicalresult

innumber

theory

saysthat

solution

sexist⇔

p≡

1mod4or

p=

2.

Wewan

tto

prove

theab

ovedescription

ofSpec

Z[x],usingthefibre

product

machinery.1

InSection

6.4,

workingwithaffi

nevarietiesover

analgebraically

closed

fieldk,weexplained

that

thefibre

ofX→

Yover

a∈Y

isSpecm

of

k[X

]⊗

k[Y

]k

wherek∼ =

k[Y

]/m

a=

Frack[Y

]/m

a=

K(a),

wherem

ais

themax

imal

idealcorrespon

dingto

a.

When

workingwithrings,an

dthemap

Spec

A→

Spec

Binducedbysomeringhom

A←

B,the

schem

e-theoreticfibre

over

℘∈Spec

Bis

theSpec

ofthefollow

ingring:

A⊗

BK(℘

)

wherethere

siduefield

K(℘

)at

℘is K(℘

)=

Frac(B

/p)∼ =

B℘/m

℘

Remark

.(L

ater

inthecourse.)Primeideals

inthelocalisation

B℘arein

1:1correspon

dence

with

primeideals

ofB

contained

in℘,an

d℘correspon

dsto

theuniquemax

idealm

℘⊂

B℘.

Exercise.After

read

ingab

outlocalisation

inSection

10,prove

Frac(B

/℘)∼ =

B℘/m

℘.

1Ofcourse,

Spec

Z[x]is

theunionofthefibresofπ,explicitly:℘∈Spec

Z[x]lies

inπ−1(π

(℘)).

68

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Thediagram

forthefibre

product

is Spec

(A⊗

BK(℘

))��

��

Spec

A

��Spec

K(℘

)�� Spec

B

(onepoint(0))

✤�� (℘)

Inou

rcase,π:Spec

Z[x]→

Spec

Z,so

A=

Z[x]

B=

Z.For

℘=

(p):

K((p))

=Frac(B

/p)∼ =

F pA⊗

BK((p))

=Z[x]⊗

ZF p∼ =

F p[x].

For

℘=

(0):

K((0))=

Frac(B

/0)∼ =

QA⊗

BK((0))=

Z[x]⊗

ZQ∼ =

Q[x].

Sothetw

odiagram

sforthefibre

product

are:

Spec

F p[x]

��

��

Spec

Z[x]

π ��

Spec

Q[x]

��

��

Spec

Z[x]

π ��Spec

F p�� Spec

ZSpec

Q�� Spec

Z(0)✤

�� (p)

(0)✤

�� (0)

RecallF p

[x]is

aprincipal

idealdom

ain,so

Spec

F p[x]=

{(0)}∪

{(f):f∈F p

[x]irred}.

RecallQ[x]is

aprincipalidealdomain,so

Spec

Q[x]=

{(0)}∪

{(f):f∈Q[x]irred}.

Finally,recallGauss’s

lemma:anon

-con

stan

tpolynom

ialf∈Z[x]is

irreducible

ifandonly

ifit

isirreducible

inQ[x]an

ditis

primitive1

inZ[X].

Com

biningthesetw

ocalculation

s,wededuce

theab

ovedescription

ofSpec

Z[x].

15.12.

PROJ:th

eanalogueofSpecforpro

jectivevarieties

Recallweassociated

anaffi

neschem

eto

aring,

which

fork[x

1,...,x

n]recoversAn.

Can

we

associateaschem

eto

anN-graded

ring,

whichfork[x

0,...,x

n]withgrad

ingbydegreerecoversPn

?

Let

A=⊕

m≥0A

mbeagrad

edring.

Theirrelevantidealis

A+

=⊕

m>0A

m(inanalogywith

(x0,...,x

n)⊂

k[x

0,...,x

n]in

Section

3.6).Then

define

Proj(A)=

{hom

ogeneousprimeideals

inA

not

containingtheirrelevantidealA

+}

withtheZariskitopology

:closed

sets

areV(I)=

{℘∈Proj(A):℘⊃

I}forallhomogeneousideals

I⊂

A.Thebasic

open

sets

areD

f=

{℘∈Proj(A):f/∈℘}forhom

ogeneousf∈A.

Example.For

A=

k[x

0,...,x

n],themax

imal

ideals

inProjA

correspondto

the(closed)points

{[a]}

=Vclassical(m

a)of

Pn,wherem

a=�a

ixj−

ajxi:alli,j�(

Sec.3.6).

ThefullProjA

corresponds

geom

etricallyto

thecollection

ofalltheirreducible

projectivesubvarietiesVclassical(℘)⊂

PnofPn

.Example.Wewilldescribeblow-upsin

term

sofProjin

Section

15.13.

Wedefinethestru

ctu

resh

eafO

=O

Xon

X=

Proj(A):

forU⊂

Xanopen

subset,

O(U

)consistsof

compatible

families{f

℘∈O

℘} ℘

∈U,equivalentlyfunctions

f:U→

� ℘∈U

A(℘

),℘�→

f ℘,

whereO

℘∼ =

A(℘

)is

thehom

ogeneouslocalisation

whichwedefined

inSection

10.3.

RecallA

(℘)

consistsof

allfraction

sF G

ofhom

ogeneouselem

ents

ofA

ofthesamedegree,

whosedenominatorG

isnot

in℘,equivalentlyG(℘

)�=

0∈K(℘

)=

Frac(A/℘

).Com

patibilityis

defined

asbefore:locally,

onabasic

neigh

bou

rhoodD

G,thereis

acommon

function

F G∈

A(G

)=

O(D

G)whose

restriction

gives

theelem

ents

f ℘,for℘∈D

G(hereA

(G)is

thehom

ogeneouslocalisationatthemultiplicative

setgenerated

byahom

ogeneouselem

entG

ofA,so

weform

ally

invert

G).

1A

polynomialis

primitiveiftheg.c.d.ofthecoeffi

cien

tsis

aunit.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

69

15.13.

THE

BLOW

-UP

AS

APROJ

Themodern

definition

ofblow-upsis

via

theProjconstruction.

Let

R=

k[x

1,...,x

n].

For

anaff

.var.Y⊂

An=

Specm

R,withdefiningidealI=

Ih(Y

),theblow-upof

Analon

gY

(i.e.alongthe

idealI)is

BYAn=

Proj

∞ � d=0

Id=

Proj(R⊕

I⊕

I2⊕

···)

whereI0=

R,so

thehom

ogeneouscoordinateringis

S=⊕

d≥0Id.Theexception

aldivisor

is

E=

Proj

∞ � d=0

Id/I

d+1=

Proj(R/I⊕

I/I

2⊕

I2/I

3⊕

···)

whichcanbeinterpretedas

follow

s:I/I

2canbethou

ght1

ofas

thevector

spacewhichis

“normal”

toY,an

dwewan

tto

take

theprojectivisationof

this

vectorspace.

Com

parePn

=P(An+1):

we

taketheirrelevantidealJ

=�x

0,x

1,...,x

n�⊂

k[x

0,...,x

n],then

thek-vectorspaceJ/J

2canbe

identified

withAn+1,an

dto

projectivisewetakeProj⊕

d≥0Jd/J

d+1.Equivalently,

this

istheProj

ofthesymmetricalgebra

Sym

RJ/J

2≡

k[x

0,...,x

n].

Example.For

Y=

{0},

I=�x

1,...,x

n�,

wehaveasurjectivehom

ϕ:R[y

1,...,y

n]→

S=⊕Id/I

d+1,

y i�→

xi.

Then

J=

kerϕ=�x

iyj−xjy i�d

efines

anaff

.var.V(J

)⊂

An×Anwhichishow

weoriginally

defined

theblow-upB

0An(after

projectivisingthesecondAnfactor,i.e.

V(J

)⊂

An×

Anis

theconeof

B0An⊂

An×

Pn−1).

16.

APPENDIX

1:Irreducible

deco

mpositionsandprimary

ideals

This

Appendix

isnon-examinable.

Recall,ifX

isan

affinevariety,

then

ithas

adecom

positioninto

irreducible

affinevarieties

X=

X1∪X

2∪···∪

XN

(16.1)

whichis

uniqueupto

reordering,

provided

2weim

poseX

i�⊂

Xjforalli�=

j.This

implies

I(X)=

I(X

1)∩I(X

2)∩···∩

I(X

N)

(16.2)

wherePj=

I(X

j)⊂

R=

k[x

1,...,x

n]aredistinct

primeideals

(inparticular,

radical).

Question.Canwerecover(16.2)

byalgebra

methods?

(then

recover(16.1)

bytakingV(·)

).

Thean

swer

isyes,an

dtheaim

ofthis

discussionis

toexplain

thefollow

ing:

1A

tangentvectorv∈TpA

nnorm

alto

TpY

actsonfunctionsbytakingthedirectionalderivativeoffatpin

the

directionv.In

thenorm

alspace

(thequotientofvectorspacesTpX/TpY),

weview

vaszero

ifv∈

TpY.Byonly

allow

ingfunctionsf∈I(i.e.vanishingalongY)ween

sure

thatvactsaszero

ifv∈TpY,since

fdoes

notvary

inthe

TpY

directions.

Since

differen

tiationonly

caresaboutfirstorder

term

s,weonly

care

aboutthequotientclass

f∈I/I2

(because

d(I

2)�

d(�

aib i)=

�aidb i

+�

b idai=

0alongY

astheai,b

i∈

IvanishonY).

Sothenorm

alspace

isthedualvectorspace

(I/I2)∗

=(linearfunctionals

v:I/I2→

k).

Example:Y

={p

}(point)

then

I=

mp,andthe

norm

alspace

equals

TpX

=(m

p/m

2 p)∗.

2e.g.sillyway

sto

makeit

non-uniqueare:takeX

N+1=

∅orX

N+1=

{p}forsomep∈X

N.

70

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

FACT.(L

ask

er1-N

oeth

erTheore

m)ForanyNoetherianringA,andanyidealI⊂

A,

I=

I 1∩···∩

I N(16.3)

whereI j

are

primary

ideals

(Defi

nition16.1).

Thedecompositionis

called

reducedifthePj=

�I j

are

alldistinct

andtheI j

are

irredundant2.

Areduceddecompositionalways

exists,andthePjare

uniqueupto

reordering.

Theprimeideals

Pj

are

called

theassociated

primesofI,den

oted3

Ass(I)=

{P1,...,P

N}.

Moreover,

view

ingM

=A/I

asanA-m

odule,

Ass(I)=

{allannihilators

AnnM(m

)⊂

Awhichare

primeideals

ofA}.

RecallAnnM(m

)=

{a∈A

:am

=0∈M

},so

forsomenon-uniqueaj∈A,

Pj=

AnnM(a

j)=

{r∈A

:r·a

j=

0∈M

}=

{r∈A

:r·a

j∈I}.

Definition16.1

(Primary

ideals).

I�

Aisaprimary

idea

lifallzero

divisors

ofA/I

are

nilpotent.

Such

anIis

P-p

rimaryif√I=

P.Thedecomposition(16.3)

isaprimary

decompositionofI.

Remark

s.Beingprimary

isweakerthan

beingprime(inwhichcase

zero

divisors

ofA/Iare

zero).

Exercise.4

Iprimary⇒

P=√Iis

prime,

infact

thesm

allestprimeidealcontainingI.

Examplesofprimary

ideals.

1).

Theprimaryideals

ofZ

are

(0)an

d(p

m)forpprime,

anym≥

1.The(p

m)are

(p)-primary.

2).

Ink[x,y],I=

(x,y

2)is

(x,y)-primary.

Indeedthezero

divisorsof

k[x,y]/I∼ =

k[y]/(y

2)liein

(y)an

darenilpotentsince

y2=

0.Notice(x,y)2

�(x,y

2)�

(x,y),

soprimaryideals

neednot

be

apow

erof

aprimeideal.

(Conversely,apow

erofaprimeidealneednot

beprimary,

althou

ghit

istrueforpow

ersof

maxim

alideals).

Exercise.Show

thefollow

ingare

equivalentdefinitionsforIto

beprimary:

•zero

divisorsofA/I

are

nilpotent

•∀f

,g∈A,iffg∈A

then

f∈Iorg∈Ior

bothf,g∈√I.

•∀f

,g∈A,iffg∈A

then

f∈Iorgm∈Iforsomem∈N.

•∀f

,g∈A,iffg∈A

then

fm∈Iorg∈Iforsomem∈N.

Exercise.I,Jboth

P-primary⇒

I∩J

isP-primary.

IfI=∩I

jis

aprimary

decom

positionwithPi=√I i

=�

I j=

Pj,then

wecanreplace

I i,I

jwith

I i∩I

jsince

that

isaga

inPi-primary(bythelast

exercise).

Thisway,onecanalwaysad

just

aprimary

decom

positionso

that

itbecom

esreduced(see

thestatem

entofLasker-N

oether).

Examplesofprimary

decompositions.

1This

isin

fact

alsothefamouschessplayer,Emanuel

Lasker,worldchesschampionfor27years.

2meaningnosm

aller

subcollectionoftheI j

gives

I=

∩Ij.

3This

unfortunate

notationseem

sto

bestandard.Alleged

ly,theBourbakigroupwasthinkingof“assassins”.

4LEM

MA.ForanyNoetherianringA,

nilra

dicalofA

=nil(A

)def

={a

llnilpotentelem

ents

ofA}

=intersectionoftheprimeideals

ofA

radicalofI

=√I

def

={f

∈A

:fm

∈Iforsomem}

=intersectionoftheprimeideals

containingI

=preim

ageofnil(A

/I)via

thequotienthom

A→

A/I

Pro

of.

Forthefirstclaim

,suppose

f∈A

isnotnilpotent.

Let

Pbeanidealthatis

maxim

al(forinclusion)amongst

ideals

satisfyingfn

/∈P

forall

n≥

1(usingA

Noetherian).

Then

Pis

primebecause:if

xy∈

Pwith

x,y

/∈P,

then

(x)+

Pand(y)+

Pare

larger

thanP,hen

cesomefn∈

(x)+

P,fm

∈(y)+

P,hen

cefnm

∈(xy)+

P⊂

P,

contradiction.Sonil(A

)⊂

∩(primeideals),

andtheconverse

iseasy.Thesecondclaim

follow

sbythecorresponden

cetheorem:primeideals

inA/Icorrespondprecisely

totheprimeideals

inA

containingI.

�

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

71

1).

A=

Z,I=

(n),

sayn

=pa1

1···p

aN

Nis

thefactorizationinto

distinct

primes

pj.

Then

I=

(pa1

1)∩···∩

(paN

N)is

theprimarydecom

position.SoI j

=(p

aj

j)an

dPj=

(pj)=

AnnZ/

(n)(

n pj).

2).

I=

(y2,x

y)⊂

k[x,y],hereareseveral

possible

primarydecom

positions

I=

(y)∩(x,y)2

=(y)∩(x,y

2)=

(y)∩(x

+y,y

2).

Ineach

case,P1=

�(y)=

(y)=

Ann(x)an

dP2=√I 2

=(x,y)=

Ann(y).

3).

A=

Z[√−5]

isan

integral

dom

ainbutnot

aUFD:uniquefactorizationinto

irreduciblesfails:

6=

2·3

=(1

+√−5)(1−√−5)

whereyo

ucancheckthat

2,3,1±√−5areallirreducibles(butnot

primes.1)Noticethat

(1+√−5)

isnot

primary:2·3

=0∈A/(1+√−5)

butthezero

divisor

2is

not

nilpotent.2W

hereas(2),

(3)

areprimary.3In

this

case,I=

(6)=

I 1∩I 2

forI 1

=(2),

I 2=

(3),

and4

P1=

�(2)=

(2,1−√−5)

=AnnA/(6)(3+

3√−5)

P2=

�(3)=

(3,1−√−5)

=AnnA/(6)(2+

2√−5).

Theoriginal

goal

oftheLasker-Noether

theorem

was

torecovera“u

niquefactorization”theorem

insuch

situations.

Note:

itis

auniquefactorizationtheorem

forideals,rather

than

elem

ents.

Exercise.5

ANoetherian⇒

primary

decompositionsalways

exist.

Theminim

al6elem

ents

ofAss(I)arecalled

minim

alprimeideals

orisolated

primeideals

inI,theothersarecalled

embedded

primeideals

inI.TheV(P

i)⊂

V(I)arecalled

associated

reduced

components

ofV(I),

andit

iscalled

anembeddedcompon

entifV(P

i)�=

V(I).

Geometrically,

forX

=V(I)an

dI⊂

R=

k[x

1,...,x

n],

theminim

alPiaretheirreducible

compon

ents

Xi=

V(P

i)=

V(I

i),an

dtheem

bedded

Piareirreducible

subvarietiescontained

inside

theirreducible

compon

ents

(ifP1⊂

P2then

V(P

1)⊃

V(P

2)).

Example.I=

(y2,x

y)⊂

k[x,y]then

I=

(y)∩(x,y)2

soAss(I)=

{(y),(x,y)}.SoP1=

(y)is

minim

al,an

dP2=

(x,y)is

embedded.Geometrically,

V(I)=

X1=

{(a,0):a∈k}∼ =

A1is

alread

yirreducible,V(y)=

V(I)is

anassociated

compon

ent,

theorigin

V(x,y)=

{(0,0)}�

V(I)is

anem

bedded

compon

ent.

NoticeX

2=

{(0,0)}does

not

arisein

theirreducible

decom

position(16.1)

since

X2⊂

X1,an

din

(16.2)

wegetI(X)=

(y)=

P1becau

sewedecom

posed

I(X)=√Inot

I.

GEOM

ETRIC

MOTIV

ATIO

N.

Asyou

canseefrom

thelast

exam

ple,primarydecom

positionis

not

very

interestingin

classical

algebraic

geom

etry

(i.e.reducedk-algebras).It

becom

esim

portantin

modernalgebraic

geom

etry,

when

you

consider

theringof

“functions”

O(Spec(A

))=

A(Section

15.1).

Examples.

1).

I=

k[x

2,y]an

dA

=k[x,y]/I.Then

IisP-primary,

whereP

=(x,y)correspon

dsto

theorigin

(0,0)∈A2.W

hat

dothefunctionsA

onSpec(A

)meangeom

etrically?

Write

f=

a0+

a10x+

a01y+

a20x2+

a11xy+

a02y2+

higher∈k[x,y].

Reducingmodulo

Igives

f=

a0+

a10x∈A.

1e.g.1±

√−5are

zero

divisors

inA/(2).

2brute

force:

2m

=(a

+b√

−5)(1+

√−5)=

(a−

5b)

+(a

+b)√−5forces

b=

−aand2m

=6a,im

possible.

3e.g.A/(2)hasazero

divisor1+

√−5,butit

isnilpotent(1

+√−5)2

=−4+

2√−5=

0∈A/(2).

4byLasker-N

oether,wejust

need

toverifythatthose

annihilators

are

prime.

This

holdsasboth

quotients

are

integraldomains:

Z/3∼ =

A/(2,1

−√−5)via

2�→

√−5,andZ/

3∼ =

A/(3,1

−√−5)via

2�→

2.

5Hints:firstshow

thateveryidealis

anintersectionofindecomposa

ble

ideals

(I⊂

Ais

indecomposa

ble

ifI=

J∩K

implies

I=

JorI=

K).

Dothis

byconsideringamaxim

alelem

entamongst

indecomposable

ideals

(thatamaxim

alelem

entexists

usesthatA

isNoetherian).

Then

show

thatforNoetherianA,indecomposable

implies

primary.Forthis

notice

thatI⊂

Ais

indecomposable/primary

iff0⊂

A/Iis

indecomposable/primary,so

youreduce

tostudyingthecase:fg=

0andAnn(g)⊂

Ann(g

2)⊂

···⊂

Ann(g

m)⊂

···(again

now

use

thatA

isNoetherian).

6minim

alwithresp

ectto

inclusion.Onecanshow

thattheseare

infact

minim

alamongst

allprimeidealscontaining

I,andallsuch

minim

alprimeideals

arise

intheAss(I).

72

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

The“values”of

fatprimeideals

℘∈

Spec(A

)only

1“see”

a0.Buttheabstract

functionf∈

Aalso

remem

bersthepartial

derivativea10=

∂xf| (0

,0).

SoSpec(A

)should

bethough

tofasapoint

(0,0)∈A2together

withthetangentvector

∂xin

thehorizontalx-direction

.

2).

For

I=

(x,y)2

=(x

2,x

y,y

2),f∈A

=k[x,y]/Iremem

bers∂xfan

d∂yfat

zero

(namelya10,a

01)

andthusbylinearity

itremem

bersallfirstorder

directional

derivatives.ThusSpec(A

)should

be

thou

ghtof

astheorigin(0,0)∈A2together

withafirstorder

infinitesim

alneighbourhoodof

0.

(Sim

ilarly,Spec(A

)forI=

(x,y)n

isan(n−

1)-thorder

infinitesim

alneighbourhoodof

zero:the

ringof

functionsremem

berstheTaylorexpan

sion

offupto

order

n−

1).

3).

I=

(x2)⊂

k[x,y]correspon

dsto

they-axis

inA2together

with

afirstorder

infinitesim

al

neigh

bou

rhoodof

they-axis.It

remem

bersallcoeffi

cients

a0m,a

1m

off,allm≥

0,so

itremem

bers

allvalues

offan

d∂xfat

anypointon

they-axis.

4).

Theprimarydecom

position

I=

(x2,x

y)=

(x)∩(x,y)2

correspondsto

they-axisin

A2together

withafirst-order

neigh

bou

rhoodof

theorigin.Thefact

thatI=

(x)∩(x

2,y)is

another

primary

decom

positionreflects

thegeometricfact

thatifa“function”f∈

A=

k[x,y]/Iremem

bersallthe

values

onthey-axis,then

itau

tomaticallyremem

bersallthevalues

of∂yfalongthey-axis,so

the

only

additional

inform

ationcomingfrom

thefirst-order

neighbourhoodoftheorigin

isthehorizontal

derivative∂xf| (0

,0)(com

parethediscu

ssionof(x

2,y)in

1)above).

There

mainderofth

isSection

isless

important(a

nd

non-examinable).

Weexplain

below

thelast

piece

oftheproofoftheLasker-N

oether

theorem:whyAss(A

/I)are

the

primean

nihilatorsof

theA-m

odule

M=

A/I.

Lemma16.2.If

Jis

aP-primary

idealforI,then

P=√J=√AnnM(a)foranya∈A\J

.

Proof.

Ifra∈Jthen

,since

Jisprimary,either

rm∈J(sor∈√J=

P)or

a∈J(false,

a∈A\J

).ThusAnn(a)⊂

P.Con

versely,

ifr∈P

then

somerm∈J,so

rm∈Ann(a),

sor∈√Ann(a).

�

Exercise.If

a∈A\P

then

AnnM(a)=

J.If

a∈J

then

AnnM(a)=

A.

Exercise.Show

that√IJ=√I∩J=√I∩√J.Hen

ceitfollowsfrom

(16.3)that:

√I=

P1∩P2∩···∩

PN.

Now

,forI=∩I

j,notice

that:AnnM(a)=

�AnnA/∩I

j(a)=

�AnnA/I j(a)so

bythetw

oexercises,

√AnnM(a)=

� j

√AnnA/I j(a)=

� a/∈I

j

Pj.

Exercise.Let

Abe

aring,

I j⊂

Aideals,P⊂

Aaprimeideal.

Then

:If

P=∩J

jthen

P=

Jjforsomej.

IfP⊃∩J

jthen

P⊃

Jjforsomej.

Bytheexercise,itfollow

sthat

if√AnnM(a)isprime,

then

itequalssomePj.Thisistheconverse

ofLem

ma16

.2.It

also

follow

sbythelast

twoexercisesthatan

yprimeidealofA

containingImust

contain

aminim

alprimeideal:

P⊃

I=∩I

jthen

P=√P⊃∩�

I j=∩P

jso

P⊃

Pj.

Lemma16.3.A

maximal2elem

entofthecollection{A

nnM(a):a�=

0∈M

}is

aprimeidealin

A.

Proof.

Noticethat

a�=

0ensuresthat

1/∈AnnM(a)⊂

Aare

proper

ideals.Suppose

P=

Ann(a)is

max

imal

amon

gstan

nihilators.

Ifxy∈P

andy/∈P,then

xya=

0∈M

,ya�=

0.SoP⊂

Ann(ya)

must

bean

equality,

bymax

imality.

Butx∈Ann(ya),

sox∈P.

�

1explicitly:f(℘

)=

(fmod℘)=

a0∈K(℘

)=

Frac(A/℘)since

x2∈I⊂

℘im

plies

x∈℘,because

℘is

prime.

2under

inclusion.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

73

For

ANoetherian,theLem

maim

plies

1that

�

Pj∈A

ss(I)

Pj=

{allzero

divisorsof

A/I

}.

Lemma16.4.FortheA-m

odule

M=

A/I

,

(P=

AnnM(m

)is

prime,

forsomem∈M

)⇐⇒

(Mcontainsasubm

odule

Nisomorphic

toA/P

)

forexample

N=

Am⊂

M.Moreover,

P=

AnnM(n)foranyn∈N.

Proof.

TheA-m

odule

hom

A→

Am,1�→

mbydefinitionhas

kernel

P,so

A/P

∼ =Am

asA-m

ods.

AsP

isprime,

A/P

has

nozero

divisorsso

an=

0∈Am

forces

a∈P,so

AnnM(n)=

P.Con

versely

anisoA/P

∼ =N⊂

Mis

asurjectivehom

ϕ:A→

N,1�→

mwithP

=kerϕ=

AnnM(m

).�

Lemma16.5.

1).

Iis

P-primary⇔

Ass(I)=

{P}.

2).

IfA

isNoetherian,andIis

P-primary,then

P=

AnnA/I(β

)forsomeβ∈A/I

.

Proof.

(1)follow

sbydefinition:I=

Iis

aprimarydecom

position.Lem

ma16.3

implies

(2).

�Lemma.ForA

Noetherian,letM

=A/I

,

Ass(I)=

{allannhiliators

AnnM(a)whichare

primeideals

inA}

Remark

.Noticewedon

’tneedto

taketheradicalsof

thean

nihilators.

Proof.

Con

sider

areducedprimarydecom

positionI=∩I

j,so

Pj=

�I j

aretheelem

ents

inAss(I).

Con

sider

theinjectivehom

2

ϕ:M

=A/I

�→�

A/I

j.

ByLem

ma16.4

applied

toI i,A/P

∼ =N⊂

A/I

i.Noticethat

ϕ(M

)∩N�=∅b

ecau

sebyirredundan

cythereis

somem∈∩ j

�=iI

j\I

i,so

ϕ(m

)is

only

non

-van

ishingin

theA/I

isumman

d.Pickan

ysuch

m∈

ϕ−1(N

\{0}),then

ϕdefines

anisoof

A-m

odsA/I

⊃Am∼ =

Aϕ(m

)=

N⊂

A/I

i(by

Lem

ma16.4,N

=Aϕ(m

)).

SoA/I

also

containsan

A-submodisoto

A/P

,so

byLem

ma16.4

P=

AnnM(m

).�

17.

APPENDIX

2:Differentialmeth

odsin

algebra

icgeometry

This

Appendix

isnon-examinable.

THE

TANGENT

SPACE

INDIF

FERENTIA

LGEOM

ETRY

Inphysics,wethinkof

atangentvector

toasm

oothman

ifoldM

(e.g.asm

oothsurface)

atapoint

p∈

Mas

thevelocity

vector

γ� (0)

ofasm

oothcurveγ:(−

ε,ε)→

Mpassingthrough

γ(0)=

p.

Mathem

atically,wedefinethetangentspaceTpM

asthecollection

ofallequivalence

classes[γ]of

smoothcurves

through

γ(0)=

p,identifyingtw

ocurves

ifin

localcoordinates

they

havethesame

velocity

γ� (0).TheTaylorexpan

sion

3of

γat

t=

0in

localcoordinates

is

γ(t)=

p+tv

+(t

2-termsan

dhigher)

(17.1)

soγ(0)=

p,γ� (0)

=v,an

dv∈Rnis

thetangentvector

inlocalcoordinates.

Notice:

reducingmodulo

t2wegetγ(t)=

p+tv∈R[t]/t2,an

dthis

determines

thepair(p,v).

Thecurveγalso

defines

adifferential

operator:forasm

oothfunctionf:M→

R,γ“operates”

onfbytellingustherate

ofchan

geof

falon

gγat

p:

f�→

∂ ∂t

� � t=0f(γ(t))

=D

pf·γ

� (0)

=D

pf·v∈R.

1If

ra=

0∈

A/I,then

themaxim

alannihilatorcontainingAnn(a)willbeanassociatedprimeidealcontainingr.

Conversely,ifr∈∪P

j,then

rm∈I j

forsomej,

m,so

picka∈∩ i

�=jI i

\Ij(usingirredundancy)then

rma=

0∈A/I

show

sthatris

azero

divisorofA/I.

2ThequotientmapA

→⊕A/I j

issurjectiveandhaskernel

∩Ij=

I.

3Notallsm

ooth

functionsare

equalto

theirTay

lorseries

(e.g.e−

1/x2

haszero

Tay

lorseries

atx=

0).

This

will

notbeanissueforussince

weonly

care

aboutthebestlinearapproxim

ation.

74

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

Sowecanalso

defineTpM

asthevectorspaceof

derivationsat

p,meaningR-linearmap

sL

:C

∞(M

)→

Ractingon

smooth

functionsan

dsatisfyingtheLeibniz

rule:

L(fg)=

L(f)·g

(p)+

f(p)·L

(g).

(17.2)

Theγin

(17.1)

correspon

dsto

theoperator

L(f)=

Dpf·v

=�∂f(p),v�=

�∂xif(p)·v

i

sotheinner

product

betweenv=

γ� (0)

andthevector

(∂x1f,...,∂

xnf)| x

=pof

partialderivatives.

Example.

For

M=

Rn,γ(t)=

(t,0,...,0)correspondsto

thestandard

basis

vector

v=

e 1=

(1,0,...,0)an

ditop

erates

byf�→

Dpf·e

1=

∂f

∂x1,so

wethinkof

γas

theop

erator∂x1.

Con

sider

theidealof

smoothfunctionsvanishingat

p:

mp=

I(p)=

{f∈C

∞(M

):f(p)=

0}.

Then

consider

theab

ovelinearmap

L:m

p→

Rrestricted

tom

p.Notice

that

L(m

2 p)=

0byLeibniz

(17.2),since

f,g

vanishatp.ThuswegetanR-linearmap:

L:m

p/m

2 p→

R.

(17.3)

Con

versely,

given

such

alinearmap

Lcanwerecover

thederivation

L?For

f∈C

∞(M

),write

f=

f(p)+(f−

f(p))∈R⊕

mp.

(17.4)

AderivationL

alwaysvanishes

onconstan

tfunctions:

L(1)=

L(1

·1)=

L(1)·1

+1·L

(1)=

2L(1),

soL(1)=

0,so

bylinearity

L(R

)=

0.Sogiven

(17.3),wedefineL

via

L(f)=

L(f−

f(p)).Is

La

derivation?Abbreviatingf(p)=

f p,g(p)=

g p,an

dusingthat

Lvanishes

on(f−f p)·(g−g p)∈m

2 p,

L(fg)

=L(fg−

f pg p)

=L((f−

f p)·g

p+

f p·(g−

g p)+

(f−

f p)·(g−

g p))

=L(f−

f p)·g

p+

f p·L

(g−

g p)

=L(f)·g

p+

f p·L

(g).

(17.5)

Example.For

M=

Rn,p=

0,then

mp/m

2 p∼ =

Rx1+

···+

Rxn(asavectorspace),andknow

ing

what

Ldoes

oneach

xidetermines

L.IndeedL=

�v i∂xicorrespon

dsto

L(x

i)=

v i.

SowecandefineTpM

asthevectorspace

oflinearfunctionals(17.3):

TpM∼ =

(mp/m

2 p)∗.

SupposeweTay

lorexpandf∈C

∞(M

)at

pin

localcoordinates,

f=

f(p)+

�ai(xi−

pi)+

�aij(x

i−

pi)(x

j−

pj)+

(higher

order

(xi−

pi)).

whereai=

∂xif| x=

p∈R.Com

posingwith(17.1)an

ddroppingt2

term

s:

f◦γ

(t)=

f(p)+

�aiv

it∈R[t]/t2.

(17.6)

Werecoverp,v

bytakingf=

xi:

f◦γ

=pi+

v it∈

R[t]/t2.Soeach

γdefines

anR-algebra

hom

C∞(M

)→

R[t]/t2,f�→

f◦γ

andsuch

ahom

ϕdetermines

p,v

via

ϕ(x

i)=

pi+

v it.

Thus1

ϕ(f)=

ϕ[f(p)+

�ai(xi−

pi)+

···]

=f(p)+

�aiL(x

i−

pi)t.

(17.7)

SoL,L

,ϕcompletely

determined

each

other.Sovia

v it=

ϕ(x

i−

pi)

weget:

TpM

={ϕ∈Hom

R-alg(C

∞(M

),R[t]/t2):(ϕ

(xi)

modt)

=pi∈R[t]/t}.

Supposenow

that

themanifoldisalreadyem

bedded

inEuclideanspace,

soM⊂

Rn(e.g.theunit

sphereS2⊂

R3),

then

wecanthinkof

TpM

assittinginsideRnasfollow

s.

SupposeP

:Rm

�→M⊂

Rnis

alocalparam

etrization

ofM

,withP(p

0)=

p.

Example.Spherical

coordinates

(θ,ϕ

)∈R2giveP(θ,ϕ

)=

(sin

θcosϕ,sin

θsinϕ,cos

θ)∈S2⊂

R3.

1R-algeb

rahomssend1to

1,so

C∞

⊃R·1

→R·1

⊂R[t]/t2

istheiden

tity

map.

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

75

Alocalcurveγ(t)=

p0+

v 0t+

···∈

Rm

then

givesrise

toacurveP◦γ

(t)=

p+

vt+

···∈

Rn.

Bythechainrule,v=

∂t| t

=0P◦γ

=D

p0P·v

0.Solocaltangentvectors

v 0∈Rm

=Tp0Rm

correspon

dto

vectors

Dp0P

·v∈RnsittinginsideRn.So

TpM

=Im

age(D

p0P)=

Dp0P

·Rm⊂

Rn.

This

isavector

subspaceof

Rn.Finally,ifM

islocallydefined

bythevanishingof

functions

M=

V(F

1,...,F

N)locallynearp

(e.g.S2⊂

R3is

defined

byF

=X

2+

Y2+

Z2−

1=

0),then

foran

ycurveγ⊂

M⊂

Rn,all

Fj(γ(t))

=0.

Differentiatingvia

thechainrule:allD

pFj·γ

� (0)

=0.

Equivalently:

γ� (0)

=v∈kerD

pF1∩···∩

kerD

pFN.

(17.8)

Con

versely,

aγsatisfying(17.8)

isacurveγ(t)on

whicheach

Fjvanishes

tosecondorder

orhigher.

SoTpM

canbeidentified

withthevectorsubspacekerD

pF1∩···∩

kerD

pFN⊂

Rn.Theaffi

neplane

p+

TpM⊂

Rnis

theplanewhichbestap

proxim

ates

M⊂

Rnat

pan

dit

istheplanewhichwe

usually

visualisein

picturesas

thetangentspace.

Since

γan

d�(t)

=p+

tvareequal

modulo

t2,i.e.

equivalentcurves

inRn,

p+

TpM

=�

{lines

�:�(t)

=p+

tv∈Rn,each

Fj◦�

vanishes

toorder≥

2at

t=

0}⊂

Rn.

These�arenot

curves

inM

usually,they

arecurves

inRn.SowearedescribingTpM

asavector

subspaceof

TpRnbydecidingwhichtangentvectorsof

Rnarealso

tangentto

M.Theab

ovedescribes

p+TpM

astheunionof

straightlines

which“tou

ch”M

atp(m

eaning,

toorder

atleasttw

o,indeed

tangentlines

ariseas

limitsof

secantlines

whichintersectM

atleasttw

icenearp).

Onesometim

esab

breviatesbydpfthelinearpartof

theTaylorexpan

sion

offat

p,so

dpf=

�∂xif(p)·(xi−

pi).

(17.9)

Inthis

notation,theaffi

neplanep+

TpM⊂

Rncanbedescribed

succinctly

as:

p+

TpM

=V(d

pF1,...,d

pFN)⊂

Rn.

THE

TANGENT

SPACE

INALGEBRAIC

GEOM

ETRY

For

Xan

affinevariety,

recallthestalkO

X,p=

k[X

] I(p)consistsof

germ

sof

regu

larfunctionsat

p,

andthis

isalocalringwhoseuniquemax

imal

idealis:

mp=

I(p)·O

X,p=

{g h:g,h∈k[X

],g(p)=

0,h(p)�=

0}.

Ak-algebra

Ais

ak-vectorspacewhich

isalso

aring(com

mutative

with

1),such

that

the

operationsarecompatible

intheob

viousway.Soin

particular,

Acontainsacopyof

k=

k·1

.A

k-algebra

homomorp

hism

ϕ:A→

Bmeans:

ϕisk-linearan

dϕisaringhom

(inparticular,

this

requires

ϕ(1)=

1).Soin

particularϕis

theidentity

map

onk·1→

k·1

.A

k-d

erivation

L∈

Der

k(A

,M)from

ak-algebra

Ato

anA-m

odule

Mmeansak-linearmap

A→

MsatisfyingtheLeibniz

rule

L(ab)

=L(a)b

+aL(b).

Theore

m17.1.Let

X=

V(F

1,...,F

N)⊂

An.Thefollowingdefi

nitionsare

equivalent:1

(1)Writing� v(t)=

p+

tvforthestraightlinein

Anthrough

pwithvelocity

v,

p+

TpX

=�

{�v:allFj(�

v(t))

vanishto

order≥

2att=

0}⊂

An

(2)Recallthenotationdpf=

�∂xif(p)·(

xi−

pi).Then

p+TpX

isanintersectionofhyperplanes:

p+

TpX

=V(d

pF1,...,d

pFN)⊂

An

(3)RecallthenotationD

pf·v

=�

∂xif(p)·v

i.Then

TpX

isthevectorspace

TpX

=kerD

pF1∩···∩

kerD

pFN⊂

kn

1Clarification.W

hatwecalled

TpX

inSection13.1

correspondsto

p+

TpX

inthis

Section(w

enow

wantTpX

toden

ote

thevectorspace

notthetranslatedaffineplane).

76

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

(4)Let

Jac(F

)=(∂Fi

∂xj)be

theJacobianmatrix

ofF=(F

1,...,F

N):A

n→

AN,so

X=

F−1(0).

TpX

=kerJac(F)

(5)ViewingkasanO

X,p-m

odule

viaK(p)=

OX,p/m

p∼ =

k,

g h�→

g(p)

h(p),

TpX

=Der

k(O

X,p,k)

(6)Theco

tangentsp

ace

atpis

thek-vectorspace

mp/m

2 p.Itsdualis

TpX

=(m

p/m

2 p)∗

(7)

TpX

=Hom

k−alg(O

X,p,k[t]/t2)

Remark

.(6)is

theofficialdefi

nition.In

schem

etheory

onereplaceskby

K(℘

)=

Frac(O

X,℘/℘

).

Proof.

Weshow

(1)⇔

(2).

NoteFj(�(0))

=Fj(p)=

0asp∈

X.So(F

j(�(t))

=order

t2)⇔

(the

derivativeat

0vanishes)⇔

(thelinearpartdpFjin

theTay

lorseries

vanishes

atx=

�(t)

=p+tv).

Weshow

(1)⇔

(3):

∂t| t

=0Fj(�(t))

=0⇔

DpFj·�

� (0)

=0⇔

�∂xiFj(p)·v

i=

0⇔

v∈

�kerD

pFj.

(alternatively(2)⇔

(3)since

dpFj(�(t))

=dpFj(p

+tv)=

�∂xiFj(p)·tv i).

That

(3)⇔

(4)is

clear:

therowsofthematrixJac(F

)arethelinearfunctionalsD

pFi.

Now

(5)⇔

(6):

derivation

sL:O

X,p→

kvanishon

k·1

andm

2 pbyLeibniz

(17.2).

Just

as(17.4),

OX,p∼ =

k⊕

mp

ask-vectorspaces,an

dm

p∼ =

(mp/m

2 p)⊕m

2 p.So,

argu

ingasin

(17.5),L

isdetermined

byak-linear

L:m

p/m

2 p→

k.

Now

(6)⇔

(7).

Let

ϕ:O

X,p→

k[t]/t2

beak-alg

hom

ϕ:O

X,p→

k[t]/t2.

Claim

.ϕ(m

p)⊂

(t).

Sub-p

roof.

Com

poseϕ

withthequotientmap

k[t]/t2→

k[t]/t∼ =

kto

get

ϕ:O

X,p→

k.Since

ϕ(1)=

1,ϕ

issurjective,

soO

X,p/kerϕ∼ =

k.Sokerϕ⊂

OX,p

isamax

imalidealso

itmust

equal

theuniquemax

imalidealm

p.Finallyϕ(m

p)=

0im

plies

ϕ(m

p)⊂

(t)

�Soϕ(f−

f(p))∈(t).

Werecover

Lvia

ϕ(f−

f(p))

=L(f−

f(p))t.

So:

ϕ(f)=

ϕ[f(p)+

(f−

f(p))]=

f(p)+

L(f−

f(p))t∈k[t]/t2.

Now

(3)⇔

(7):

theanalog

ueof

(17.6),forf∈k[X

],is

that

f(�(t))

=f(p

+tv)=

f(p)+

�∂xif(p)·v

it=

ϕ(f)∈k[t]/t2

defines

ak-alg

hom

ϕ:k[X

]→

k[t]/t2.Indeed,

ϕ(fg)=

f(p)g(p)+

�(∂

xif(p)·g

(p)+

f(p)·∂

xig(p))

·vit=

ϕ(f)·ϕ

(g)

modulo

t2.

Con

versely,

given

ϕ,definev i

via

ϕ(x

i−

pi)

=v it.

Then

since

Fj=

0∈

k[X

](by

definition

k[X

]=

k[x

1,...,x

n]/√�F

1,...,F

N�),wehaveϕ(F

j)=

0.So,usingFj(p)=

0andt2

=0,

weget

0=

ϕ(F

j)=

ϕ[F

j(p)+

�∂xiFj(p)·(xi−

pi)+

(termsin

I(p)2)]=

�∂xiFj(p)·v

it.

�

Lemma17.2.ForX

=V(J

)⊂

An,letI p

=I(p)·k

[X]⊂

k[X

]then

mp/m

2 p∼ =

I p/I2 p∼ =

I(p)/(I(p)2

+J)

Proof.

Apply

thethirdisomorphism

theorem

1usingthat

J⊂

I(p)since

p∈X.

�

Theore

m17.3.ThedisjointunionTX

ofalltangentspacesTpX,aswevary

p∈X,is:

TX

=Hom

k-alg(k[X

],k[t]/t2)

(i.e.morphismsSpec(k[t]/t2)→

X)

1ForR-m

odulesS⊂

M⊂

B(“sm

all,m

edium,big”),

B/M

∼ =(B

/S)/(M

/S).

Apply

this

toJ⊂

I(p)2

+J⊂

I(p).

C3.4

ALGEBRAIC

GEOMETRY,

PROF.ALEXANDER

F.RIT

TER

77

Proof.

Given

ak-algeb

rahom

ϕ:k[X

]→

k[t]/t2,composewiththequotientk[t]/t2→

k[t]/t∼ =

kto

getak-alg

hom

ϕ:k[X

]→

k.This

issurjective(since

1�→

1)so

thekernel

isamax

imal

ideal

ofk[X

](ask[X

]/ker∼ =

k).

Butthemax

imal

ideals

ofk[X

]areprecisely

theI(p)forp∈X.Thus

ϕ(I(p))

=0,

soϕ(I(p))⊂

(t).

Localisingϕat

I(p),

gives

ϕ:O

X,p→

k[t]/t2.

�Exercise.For

ak-alg

A,themodule

ofKahlerdifferentials

istheA-m

odΩA/kgenerated

over

Abythesymbolsdf

forallf∈A,modulo

therelation

smak

ing

d:A→

ΩA/k,f�→

df

ak-derivation.1

For

anyk-m

odM

,show

there’sanaturaliso

Der

k(A

,M)∼ =

Hom

A(Ω

A/k,M

),L�→

(ΩA/k→

M,df�→

L(f)).

IfA

isalso

alocalring,

with

max

idealm

and

residuefield

A/m

∼ =k,show

2that

thereis

anisom

orphism

m/m

2∼ =

ΩA/k⊗

Ak,f�→

df.

DenoteΩX,p=

ΩO

X,p/kforaffi

neX.Show

that

3

mp/m

2 p∼ =

ΩX,p⊗

OX

,pk,f�→

df

Der

k(O

X,p,k)∼ =

Hom

OX

,p(Ω

X,p,k),

∂∂xj| x=

p�→

(dxj)∗

(17.10)

wherek∼ =

OX,p/m

p=

K(p)as

OX,p-m

od,an

d(dxj)∗

isdefined

by(dxj)∗(dxi)=

dxi(

∂∂xj)=

δ ij.

Remark

.Globally,TX

andΩXaresheaves(thetangentsheafan

dthecotangentsheaf),an

d(17.10)

saysthey

aredual

inthesense

that:

TX

=Der(O

X)∼ =

Hom

OX(Ω

X,O

X).

Thenon

-singu

larpoints

ofX

arein

fact

thosewhereΩX,pisafree

OX,p-m

odule,i.e.

whereΩX

isa

vector

bundle.

Example.WedescribeTpAn=

An.

Using(1):

I(An)=

{0}an

d(0◦�

)(t)

vanishes

toinfiniteorder

for�(p)=

p+tv,an

yv∈An.

Using(2),

(3)or

(4):

I(An)=

{0}so

ker

Dp0=

ker0=

An.

Using(5):

OAn,p=

{f=

g h:h(p)�=

0}⊂

k(x

1,...,x

n),so

Der

k(O

An,p,k)∼ =

kL1⊕···⊕

kLnwhere

Lj=

∂∂xj| x=

p.

Using(6):

mp=

(x1−

p1,...,x

n−

pn)·O

X,p

={g h

:g(p)=

0,h(p)�=

0}⊂

k(x

1,...,x

n).

Thus

mp/m

2 p∼ =

ke 1⊕···⊕

ke n∼ =

knas

vectorspaces

wherethebasis

ise i

=xi−pi.

Thus

(mp/m

2 p)∗∼ =

kL1⊕···⊕

kLn∼ =

kn

usingthedual

basis

Lj=

∂∂xj| x=

p:m

p/m

2 p→

k.

Using(7):

Hom

k-alg(O

X,p,k[t]/t2)∼ =

kϕ1⊕···⊕

kϕnwhereϕj(f)=

p+Lj(f)t.

Using(17.10):

ΩX,p⊗

OX

,pk∼ =

kdx1⊕···⊕

kdxn.

Exercise.DescribeTpX

forthecuspidal

cubic

X=

V(y

2−x3)at

p=

0.Show

that

bytheLem

ma,

mp/m

2 p∼ =

(x,y)/(x

2,x

y,y

2,y

2−x3)∼ =

kx⊕ky,an

dΩX,p⊗

OX

,pk=

kdx⊕kdy.

1so

dis

k-linearandd(f

g)=

f(dg)+

(df)g.

2Toshow

injectivityit

may

beeasier

toshow

surjectivityofthedualmapHom

k(Ω

A/k,k

)→

Hom

k(m

/m

2,k

).If

a∈A

equals

c+

m∈k⊕

m,consider

L(a)=

L(m

)forL∈(m

/m

2)∗.

3Forf:X

→kthinkofdf

asthelinearfunctionalD

pf:TpX

→Tf(p

)k∼ =

k.Such

Dpfsatisfyrelations,

e.g.in

V(y

2−

x3),

Dp(y

2−

x3)=

0im

plies

2p2dy−

3p2 1dx=

0.The·⊗

OX

,pkjust

meansevaluate

coeffi

cien

tfunctionsatp.